TITLE: Reference Request: Vector bundles in rigid analytic geometry
QUESTION [7 upvotes]: In algebraic geometry it is well-known (see Hartshorne Exercise II.5.16 for example) that there is a 1-1 correspondence between rank $n$ (geometric) vector bundles $\pi\colon Y\to X$ on a scheme $X$ and locally free sheaves of $\mathcal{O}_X$-modules of rank $n$.
As I would imagine is well-known to experts in the subject, there is an analogous result in rigid analytic geometry (in the sense of Tate). Here the analogue of a trivial geometric vector bundle of rank $n$ on a rigid $K$-analytic space $X$ is the fibre product $X\times_K \mathbb{A}^{n,an}$ together with its natural projection onto $X$. Here $\mathbb{A}^{n,an}$ denotes the rigid analytic space obtained by gluing together polydiscs of larger and larger radius as in Example 9.3.4.1 of Non-Archimedean analysis by Bosch, Güntzer and Remmert. One can verify that the sections of the projection map $X\times_K \mathbb{A}^{n,an}\to X$ are naturally a free $\mathcal{O}_X(X)$-module of rank $n$. Given this it is not difficult to make a definition of a general geometric vector bundle on rank $n$ on a rigid $K$-analytic space in such a way that the sections of such a bundle form a locally free sheaf of $\mathcal{O}_X$-modules (that is a vector bundle on $X$ in the only sense I can find in the literature). Moreover, as in the algebraic setting this actually defines a 1-1 correspondence between the two notions of vector bundle. 
My question is whether anyone has written this up in a form that can be easily cited.

REPLY [2 votes]: My guess is that in chapter 4, specially section 4.7, of 
Rigid Analytic Geometry and Its Applications, by Jean Fresnel and Marius Van Der Put, Progress in Mathematics, vol. 218, Birkhäuser, Boston, 2004.
you could find the result you are looking for.<|endoftext|>
TITLE: Hilbert's Theorem on $L_2$ norm of polynomials in $\mathbb{Z}[X]$ - Explicit construction and a converse?
QUESTION [23 upvotes]: Consider the set of polynomials with real coefficients as a vector space with the following inner-product: $\langle f, g \rangle = \int_{a}^{b} f(x)g(x) dx$.
Hilbert showed, in a paper from 1894, that the norm (with respect to this inner-product) of a non-zero polynomial in $\mathbb{Z}[X]$ can get arbitrarily small when $b-a < 4$. In other words, $\min_{0 \neq p \in \mathbb{Z}[X]} \int_{a}^{b} p^2(x) dx = 0$.
My questions are: 

What happens when $b-a \ge 4$? Can the norm get arbitrarily small in this case? Or is there some (positive) lower bound for $\min_{0 \neq p \in \mathbb{Z}[X]} \int_{a}^{b} p^2(x) dx$?
When $b-a < 4$, is there an explicit construction of a sequence $p_n \in \mathbb{Z}[X],n\ge 1$, with norm tending to 0?

I posted similar questions in MSE and got no responses.
(Hilbert's proof was as follows: Minimizing the norm for (non-zero) polynomials of degree less than $n$ is equivalent to minimizing a certain positive-definite quadratic form. The corresponding matrix $A_n$ has entries $a_{i,j} = \langle x^i, x^j \rangle = \frac{b^{i+j+1} - a^{i+j+1}}{i+j+1}, 0 \le i,j \le n-1$. A calculation using an orthonormal basis for our inner-product space shows that $\det A_n = (\frac{b-a}{4})^{n^2}n^{-1/4} (2 \pi)^n c_n$ where $c_n$ converges to a positive constant. A result by Minkowski shows that in general, the minimal value of a positive quadratic form $\langle v, Av \rangle$ in $n$ variables is at most $n (\det A)^{1/n}$. Since   $\lim_{n} n (\det A_n)^{1/n} = 0$ for $b-a<4$, the result follows.)

REPLY [18 votes]: For Question 1: Once $b-a \geq 4$, the norm cannot get arbitrarily small.
Suppose $p(x)$ has leading term $c_n x^n$.
The minimum of $\int_a^b p^2(x) \, dx$
over all polynomials $p(x) = c_n x^n + O(x^{n-1})$ with real coefficients
is a multiple of $P_n(l(x))$, where $P_n$ is the $n$-th
Legendre
polynomial and $l$ is the affine-linear transformation
$l(x) = (2x-(a+b))/(b-a)$ that takes $(a,b)$ to $(-1,1)$.
Now $P(x/2)$ has leading coefficient
$$
\lambda_n := \frac12 \frac34 \frac56 \cdots \frac{2n-1}{2n}
 \sim \frac1{\sqrt{\pi n}}
$$
and norm $\int_{-2}^2 P(x/2)^2 \, dx = 4/(2n+1)$.  Thus in general
$$
\int_a^b p^2(x) \, dx
  \geq \frac4{(2n+1)\lambda_n^2} \left(\frac{b-a}{4}\right)^{\!2n+1} c_n^2,
$$
with the factor $r_n := 4/(2n+1)\lambda_n^2$ approaching $2\pi$
as $n \rightarrow \infty$.
Since $|c_n| \geq 1$, we deduce a lower bound on $\int_a^b p^2(x) \, dx$
that approaches $2\pi$ when $b-a=4$,
and increases exponentially with $n$ when $b-a > 4$.
[Added later, in connection with the recent
MO Question 188807:
since $\{ r_n \}_{n \geq 0}$ is an increasing sequence and the
$n=0$ bound $b-a$ is attained by the integer polynomial $p(x) = 1$
(or $p(x) = -1$), it also follows that $b-a$ is the minimum over
all nonzero $p \in {\bf Z}[X]$ once $b-a \geq 4$.]
It follows also that if $b-a>4$ then for any $M$ there are
only finitely many $p \in {\bf Z}[X]$ for which $\int_a^b p^2(x) \, dx < M$.
When $b-a = 4$ this is not true when $a,b$ are integers;
I do not know what happens otherwise.
[For the integer case: it is enough to prove it for $(a,b) = (-2,+2)$.
let $T_n$ be the $n$-th
Chebyshev
polynomial; then $p(x) = 2T_n(x/2)$ has integer coefficients and
$$
\int_{-2}^2 p(x)^2 \, dx \leq \int_{-2}^2 2^2 \, dx = 16.
$$
(In fact for large $n$ the integral approaches $8$.)]<|endoftext|>
TITLE: Using schemes to prove things about rings
QUESTION [18 upvotes]: I apologize for asking a big list question, I've tried to avoid doing so for a while. I'll give my justification in a moment.
The question is as follows:

What are examples of strict applications of the language of schemes/stacks/algebraic geometry to commutative rings?

Here a "strict" application means that the statement of the problem can be formulated without using any algebro-geometric language (stick to rings and modules and complexes, etc.) but a solution either requires or is very naturally obtained by using algebro-geometric language.
I don't know examples of this phenomenon off the top of my head, but here are two examples from algebraic topology:

Work on exotic spheres via homotopy theory (an example where this is the only known method to produce the results.)
(one of) Quillen's proof(s) of the Atiyah-Swan conjecture. While there is a purely algebraic, group-cohomological proof, it turns out to be very natural to prove this theorem using spaces with an action, as opposed to specializing to when the space is a point.

Motivation
Thanks to work of (insert all the usual suspects here), we now have a very strong theory of spectral algebraic geometry, i.e. algebraic geometry done with commutative ring spectra as opposed to commutative rings. While I don't know of any (hence this question), I am positive there exist strict applications of algebraic geometry to ring theory. It would be very neat if we could transplant these into strict applications of spectral algebraic geometry to the theory of ring spectra. Obviously I don't expect this to be straightforward, or literally possible, but I maintain that answers to this question would provide a useful insight in how to think about the relationship between non-affine and affine phenomena.

REPLY [2 votes]: We prove by elementary scheme-theoretic methods the following classical result:
CLAIM: Let $A$ be an artinian ring with maximal ideals $\mathsf{Spec} \, A=\lbrace \mathfrak m_1,\ldots,\mathfrak m_N\rbrace$. Then there is an isomorphism $$A\simeq \prod_{i=1}^NA_{\mathfrak m_i}.$$
PROOF: Let $X=\mathsf{Spec} \, A$. denote by $\mathcal{O}_X$ its structure sheaf. Since $X$ is discrete, by the gluing axiom, we have $$A\simeq \mathcal{O}_X(X)=\prod_{i=1}^N\mathcal{O}_X(\mathfrak m_i),$$ but $\mathcal{O}_X(\mathfrak{m}_i)\simeq A_{\mathfrak m_i}$.<|endoftext|>
TITLE: Does there exist a surface bundle over a surface of genus at least 2 that fibers in three distinct ways?
QUESTION [17 upvotes]: Let 
$$
\Sigma_g \to E \to \Sigma_h
$$
be a surface bundle over a surface. Unless otherwise stated, I'll assume $g, h \ge 2$. The theory of Thurston norm shows that surface bundles over $S^1$ often fiber in infinitely many ways (e.g. with fibers of infinitely many genera). For surface bundles over surfaces, however, the Euler characteristic (which is the product of the characteristics of the base and the fiber) provides an arithmetic constraint on the possible genera of the base and the fiber. When the genus of the base and the fiber are both at least two, this shows that there are only finitely many possible fiber genera, and an analysis of the fundamental group shows that there are only finitely many possibilities for the subgroup of $\pi_1E$ corresponding to the fiber (this follows, for instance, by work of F.E.A. Johnson). Moreover, Johnson's work also shows that when the monodromy representation $\rho: \pi_1 \Sigma_h \to \operatorname{Mod}(\Sigma_g)$ is non-injective, there are at most two fiberings, so that any bundle fibering in at least three ways must have injective monodromy.
I am aware of the example of Atiyah-Kodaira, which shows that it is possible for bundles with injective monodromy to fiber in at least two ways (of course, product bundles are a trivial example of multiply-fibered total spaces as well). However, I haven't seen any example of a bundle with three distinct fiberings when the base genus is at least two. When the base genus is one, there are trivial constructions one can do: if $M$ is any three-manifold fibering over $S^1$ in infinitely many ways, then $M\times S^1$ will fiber over $S^1\times S^1$ in infinitely many ways. I don't know of an example of a torus bundle over a surface that fibers in infinitely many ways, but I would be interested to see this, too. 

With all this said, does anyone know of an example of a surface bundle over a surface of genus $h \ge 2$ that fibers in at least three ways (i.e these are pairwise non-fiberwise diffeomorphic)? What about a surface bundle over a closed $2k$-manifold of nonzero Euler characteristic that fibers in at least $k+2$ ways? (Note that a product of $k+1$ surfaces gives an example where there are $k+1$ fiberings).

REPLY [12 votes]: Yes. I recently found methods of constructing surface bundles over surfaces with at least $n$ fiberings for any $n$. The idea is to perform a fiberwise connect-sum of trivial bundles in such a way that the two different fiberings for each component can be compatibly assembled into a bundle with many fiberings. See this preprint.<|endoftext|>
TITLE: Geometric fibers of schemes.
QUESTION [10 upvotes]: I would like to get an understanding of the notion of geometric fibers of scheme morphisms: If $f: X \rightarrow Y$ is a morphism of schemes, then its geometric fiber is defined to be $X \times_{Y} \overline{k(p)}$ for the quotient field $k(p)$ at $p \in Y$. I would like to know, why this is a good choice for the notion of "fiber". Why does one pick such an abstract notion and why does exactly this the right job?
Every help will be appreciated.

REPLY [3 votes]: Jacob has already given a good answer. I just want to add something that convinces me that the definition of geometric fibre is indeed a good one.
Let's fix a field $k$, and consider the concept of smoothness of a variety $X$ over $k$ (or more generally, schemes locally of finite type over $k$).
It's quite natural to call some (closed) point $x \in X$ a smooth point if   $\Omega_{X/k,x}$ if free of rank equal the local dimension of $X$ at $x$. Geometrically, this simply states there are enough independent differentials locally. 
We would like to express this condition in terms of the local ring $O_{X,x}$. In the case $k$ is algebraically closed, this condition is (as is well known) that $O_{X,x}$ is  a regular local ring.
However, in general, the equivalent condition is $O_{X,x}\otimes_k \bar{k}$ is regular. (This means each localization is regular, as in general this ring is only semi-local). Here geometric fibre arises naturally. 
The reason for this is somehow technical, there is a result characterizing regular local rings in terms of Kaehler differentials, see for example Hartshorne Thm II.8.8 (notice that there the assumption that $k$ being perfect is superfluous).
Theorem: 
Let $B$ be a local ring containing a field $k$ isomorphic to its residue field. Assume furthermore that $B$ is a localisation of a finitely generated $k$-algebra. Then $\Omega_{B/k}$ is a free $B$-module of rank equal to dim $B$ if and only if $B$ is a regular local ring.
The first assumption is clearly satisfied for varieties over an algebraically closed field, but fails in general. So a base extension to the algebraic closure arises quite naturally.<|endoftext|>
TITLE: Most understandable notes on Jacquet-Langlands?
QUESTION [13 upvotes]: I am particularly interested in the comparison of the trace formula part of Jacquet-Langlands. But I found the original text hard to read.

REPLY [6 votes]: I think that you may like the notes by Ioan Badulescu: see 
http://www-math.univ-poitiers.fr/~badulesc/pub/JLtata1.pdf
They are a 20 pages modern essentially complete exposition of the proof of Jacquet-Langlands not only for $Gl_2$ but for higher higher rank linear groups and their inner form as well.<|endoftext|>
TITLE: Classification of Tori of GL2, up to conjugation
QUESTION [8 upvotes]: Over an algebraically closed field $k$, every one-dimensional torus embedded (as a closed algebraic subgroup) into GL2 is diagonalisable, and the embedding is $t\mapsto (t^m,t^n)$ for some integers $m,n$ with $gcd(m,n)=1$.
What happens when the field $k$ is not algebraically closed? Can I find somewhere the classification of all subtori of GL2, up to conjugation?
By a torus I of course mean a closed algebraic subgroup of GL2, which is isomorphic to $K^{*}$ over the algebraical closure $K$ of $k$.
One example is, when $k=\mathbb{R}$, the group of rotations. This group is algebraically given by matrices of the form $[a,b,c,d]$ (sorry it seems that matrices do not appear correctly, but you should see what I mean) with $a=d$, $b=-c$ and $a^2+b^2=1$. The same example works over any field, and it is not diagonalisable if $-1$ is not a square in the field.
I would like to have a complete classification.

REPLY [2 votes]: A finite dimensional commutative $k$-algebra is étale if it is isomorphic to a product of
separable field extensions of $k$. For any $n \ge 1$, the conjugacy classes of maximal
$k$-tori in $\operatorname{GL}_n = \operatorname{GL}(V)$ are in 1-1 correspondence with isomorphism classes of étale $k$-algebras $E$ of dimension $n$ over $k$.
Here is how the correspondence goes: given a maximal torus $T$ of $\operatorname{GL}(V)$, the Lie algebra $\operatorname{Lie}(T)$ is an $n$-dimensional etale subalgebra of $\operatorname{End}_k(V)$.
Conversely, given $E$ an étale algebra of dimension $n$ over $k$, view $E$ as a left $E$-module.
The module structure determines a $k$-algebra embedding $E \to \operatorname{End}_k(E)$ and hence an injective homomorphism
of algebraic groups $E^\times \to \operatorname{GL}(E) = \operatorname{GL}_n$, 
where $E^\times$ is the "unit group of $E$ viewed as an algebraic group". The image of this homomorphism is the desired maximal torus of $\operatorname{GL}_n$.
It remains to see that étale subalgebras $E$ and $F$ of $\operatorname{End}_k(V)$ are conjugate by an element of $\operatorname{GL}_n(k)$ if and only if they are isomorphic $k$-algebras. This follows from the observation that
if $E$ is an étale subalgebra of $\operatorname{End}_k(V)$ with $\dim_k E = \dim_k V$, then viewed as $E$-module, $V$ is isomorphic to the regular representation of $E$.
From the point of view of Galois cohomology, here is what is going on. Write $G = \operatorname{GL}_n$, let $T_0$ be a split maximal $k$-torus, and let $N = N_G(T_0)$. The $k$-variety $\mathcal{T}$ of all maximal tori of $G$ is $\mathcal{T} = G/N$, and there is an exact sequence of pointed sets
$$1 \to N(k) \to G(k) \to \mathcal{T}(k) \to H^1(k,N) \to 1$$
since $H^1(k,G)$ is trivial (Hilbert 90).
Now, $N$ is the semidirect product of the split torus $T_0$ and the symmetric group $S_n$.
Since $H^1(k,T_0)$ is trivial (Hilbert 90 again), and since the quotient homomorphism $N \to S_n$ has a section, $H^1(k,N)$ identifies with $H^1(k,S_n)$, which in turn identifies with the set $\operatorname{Et}_n(k)$ of iso. classes of étale $k$-algebras of dimension $n$. 
Thus the mapping $\mathcal{T}(k) \to H^1(k,N) = \operatorname{Et}_n(k)$ is onto,
and by "twisting" one identifies the fibers of this mapping with the $G(k)$-orbits on $\mathcal{T}(k)$; i.e. with the conjugacy classes of maximal $k$-tori in $\operatorname{GL}_n$.<|endoftext|>
TITLE: relations between  log schemes and toric varieties
QUESTION [6 upvotes]: The theory of regular log scheme, roughly specking is that scheme theory with singularities like those of toric varieties.
Dose anyone could explain this idea in detail or give a reference about the relations between  log schemes and toric varieties?

REPLY [9 votes]: The source text
Kato, Kazuya
Logarithmic structures of Fontaine-Illusie. Algebraic analysis, geometry, and number theory (Baltimore, MD, 1988), 191–224, Johns Hopkins Univ. Press, Baltimore, MD, 1989. 
available here 
http://www.math.brown.edu/~abrmovic/LOGGEOM/Kato-log.pdf.
Example (3.7) (1)
the bottom line is
"... essentially, the theory of toroidal embeddings is nothing but the theory of schemes with smooth fine log. str.'s over a field (with respect to the trivial log. str. on the base field)."<|endoftext|>
TITLE: Left/right exact functor "in nature" which is not a right/left adjoint
QUESTION [18 upvotes]: Let $\mathcal{F}$ and $\mathcal{G}$ be abelian categories.  It is well-known that if a functor $\phi : \mathcal{F} \rightarrow \mathcal{G}$ has a right-adjoint (so $\phi$ is itself a left-adjoint to some other functor), then $\phi$ is right exact.  Similarly, if $\phi$ has a left-adjoint (so $\phi$ is itself a right-adjoint to some other functor), then $\phi$ is left-exact.
I was going through the list of left- and right-exact functors I know, and they all were covered by the above condition.  Can someone give me examples that appear "in nature" (so aren't too artificial) of left- or right-exact functors that are not adjoints?
I'm aware of the adjoint functor theorem, but its conditions are much stronger than just being left- or right-exact.

REPLY [2 votes]: Inverse image of sheaves is exact, but generally only has a right adjoint (direct image). 
Also, for locally compact Hausdorff spaces, direct image with proper supports is left exact, but does not have any adjoints in general unless we pass to the derived category.<|endoftext|>
TITLE: Applications for intersection (co)homology and for the Decomposition Theorem for students?
QUESTION [9 upvotes]: Which applications of intersection (co)homology and of the (Topological) Decomposition Theorem  have most chances to be understood by students?

REPLY [13 votes]: I gave a course on the decomposition theorem (and its failure with positive char coefficients) in Bochum. Notes and exercises are here:
http://people.mpim-bonn.mpg.de/geordie/bochum/
I found the example of Weierstraß family of elliptic curves quite instructive. One considers the family
$Y^2 = (X-\lambda Z)(X-Z)X$ of elliptic curves over $\mathbb{C}$ (parametrised by $\lambda$).
In a series of exercises (available at the above link) the topology of this example can be pursued in several steps:
1) one tries to understand the local system of $H^1$ away from the singular points 0 and 1. Using ramified covers of $\mathbb{P}^1$ students can calculate the monodromy around the singular points.
(Here one sees the first remarkable fact predicted by the decomposition theorem: even though the monodromy is unipotent around each singularity, the global representation of the free group on two letters is simple. So the decomposition theorem fails in the complex analytic category.)
2) one can calculate the the cohomology of the singular points and sees that it agrees with the invariants of the monodromy.
(Here one sees why the definition of the IC sheaf is "correct" for local systems on $\mathbb{C}$ minus points. This is also an example of the invariant cycle theorem, as Donu points out.)
(More advanced: This also gives an example of a non pointwise pure IC sheaf.)
3) The local systems of $H^0$ and $H^2$ are constant. One can use the hard Lefschetz theorem along the fibres of the map to deduce the splitting of the direct image.
(Here one sees the relative hard Lefschetz theorem in play.)
Other examples that I find really instructive:
1) resolutions of Kleinian surface singularities. (These are also discussed in the notes above.) Here one can connect the decomposition theorem to the non-degeneracy of intersection forms, as in the beautiful work of de Cataldo and Migliorini.
There are notes from a beautiful course Migliorini gave here:
http://people.mpim-bonn.mpg.de/geordie/Migliorini.pdf
2) I find the proof of Deligne's theorem on the decomposition theorem for smooth maps quite instructive. Here one really sees "relative Hodge theory" in action: the hard Lefschetz theorem implies the degeneration of the spectral sequence, and to get the semi-simplicity of the local systems of cohomology one needs to use the fact that they are all polarised by an ample line bundle on the fibres, hence admit invariant forms, hence are semi-simple.
3) Another example where the situation is much simpler is the Hilbert scheme of points on a curve, viewed as a resolution of the symmetric power (the Hilbert-Chow morphism).<|endoftext|>
TITLE: Unramified Galois representations not from smooth and proper stacks
QUESTION [11 upvotes]: Are there any irreducible representations of $\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)$ over $\mathbb Q_l$  that are unramified away from $l$ and crystalline at $l$ but are not known to arise from the etale cohomology of smooth and proper stacks over $\operatorname{Spec}\mathbb Z$?

Thee simplest examples of unramified crystalline Galois representations known, other than powers of the cyclotimc character, are the two-dimensional Galois representations arising from level $1$, weight $n+1$ modular forms. These live inside the etale cohomology of the stack $\overline{M}_{1,n}$, which is indeed smooth and proper. More examples come from other $\overline{M}_{g,n}$, but these are the only examples I am aware of!
The situation for schemes is very different where, as far as I can tell, no Galois representation other than powers of the cyclotomic character is known to arise from smooth and proper schemes over $\mathbb Z$.

REPLY [7 votes]: Here's a class of examples which isn't in your list as far as I know. Take an even unimodular lattice, e.g. the $E_8$ lattice. The corresponding orthogonal group is a reductive group over $\mathbf{Q}$ which is split at every finite place and compact at $\infty$ (see Gross, "Groups over $\mathbf{Z}$", Inventiones 124 (1996)). There will be lots of algebraic automorphic representations of this group $G$ if you take the weight large enough, and these will give you automorphic representations of $GL(n)$ which are self-dual and unramified at all finite places. It's known that these have Galois representations attached, which will be crystalline at $\ell$ and unramified everywhere else; but I don't think these are known to come from smooth proper stacks over $\mathbf{Z}$.
There is much more on level 1 automorphic representations in the recent preprint of Chenevier and Renard, http://www.math.polytechnique.fr/~chenevier/articles/dimform.pdf.<|endoftext|>
TITLE: Subgroups of $\mathbb{Z}^n$
QUESTION [17 upvotes]: I hope that the following problem isn't actually elementary (at least, for the sake of the fact that I'm posting it here), and I apologize if it is.  I did try hard to solve it first.
Let $V$ be a $\mathbb{Q}$-vector subspace of $\mathbb{Q}^n$, and let $G = V \cap \mathbb{Z}^n$.
Does there exist a linearly independent generating set for $G$ (i.e. a subset of $G$ such that every element of $G$ can be expressed uniquely as as $\mathbb{Z}$-linear combination of elements of this subset)?
Is there an algorithm to find it (given a basis for $V$)?

REPLY [14 votes]: EDIT: I typed this up before everyone piled on to say that this is too elementary. Moving to CW rather than deleting, as I distinctly remember not seeing this during my education on groups and rings (but then maybe I missed class that day). And don't even go down the line of "grad school courses"...
Feel free to downvote if you feel like it, of course.

Interesting question! The following is not a full answer, but seemed worth a bit more than a comment.
If I understand correctly, the first part asks if G is free abelian, in which case the answer is apparently yes, see this MO question from someone reputable. (I confess this is not something I knew, although I did have my suspicions.)
My guess is that some modification of Gauss-Jordan elimination should provide an algorithm for extracting a "Z-basis" from a Z-generating set, but it's not clear to me right now how one would get a Z-generating set from a given Q-basis of your original V.<|endoftext|>
TITLE: Second order difference implies differentiability
QUESTION [8 upvotes]: Suppose that a function $f$ on the line satisfies $|f(x+2h)-2f(x+h)+f(x)|\le |h|^{3/2}$ for all $x,h$ real. Is it true that $f$ is differentiable and its derivative satisfies
$|f'(x+h)-f'(x)|\le c |h|^{1/2}$ for all $x,h$?

REPLY [3 votes]: Some remarks on the issue of weakening the continuity assumptions on $f$ to locally integrable, and about generalizing to other exponents of $|h|$.
Keeping track of the big O term in the first part of Terry Tao's proof, we may state it as an a priori bound on the Hölder norm of $f'$: Let $f\in C^1(\mathrm{R})$: 

If $f$ satisfies, for some $0 < \alpha \le 1$ and for all $x$ and $h$
  $$|f(x+2h)-2f(x+h)+f(x)|\le C|h|^{1+\alpha} \qquad \qquad(1)$$   then
  its derivative is $\alpha$-Hölder, and in fact $$|f'(y)-f'(x)|\le\frac{C}{2^\alpha -1 } |h|^{\alpha}\, .\,\qquad \qquad(2) $$

The same conclusion holds if we only assume $f\in L^1_{loc}(\mathbb{R})$. Indeed, we may consider the standard approximation of $f$ by convolution, $f_\epsilon:=f*\phi_\epsilon$ 
with
$\phi_\epsilon(x):=(1/\epsilon)\phi(x/\epsilon)$, 
for  $\phi\in C^\infty_c(\mathbb{R}) _ + $ with $\int_\mathbb{R}\phi\, dx=1\, .$  Then the $f_\epsilon$ are in $C^{1,\alpha}(\mathbb{R})$ and satisfy the above hypothesis (1) with the same $C$, so the $f _\epsilon'$ are equicontinuous. By the Ascoli-Arzelà theorem, since $f _\epsilon \to f$ locally uniformly, this is sufficient to conclude that $f$ is also in $C^{1,\alpha}(\mathbb{R})$, with the same bound on the Hölder norm.
Rmk: the condition (1) for $\alpha > 1$ became trivial: a (locally integrable) $f$ satisfying it is then linear.<|endoftext|>
TITLE: subalgebra of a simple forcing
QUESTION [7 upvotes]: Let $\alpha > 0$ be any ordinal.  Consider the forcing $Add(\omega,\alpha) * Add(\omega_1,1)$.  Let $B$ be its boolean completion.  Let $\dot{X}$ be the canonical $B$-name for the generic subset of $\omega_1$ added by the second iterand.  By forcing folklore, there is a complete subalgebra $A \subseteq B$ that adds $X$, completely generated by the boolean values $|| \check{\beta} \in \dot{X} ||$  (see Jech p. 247).  One can check that for each $\beta < \omega_1$, $|| \check{\beta} \in \dot{X} || = (1,\check{ \lbrace (\beta,1) \rbrace } ) $.
Questions:
(1) What is the nature of $A$ in relation to $B$?  Although $A$ adds all the reals of $B$, I think I can show that it is always a proper subset of $B$ (even for $\alpha = 1$).  This leads to the second question.
(2) If $G$ is generic for $A$, what is the nature of the quotient algebra $B/G$?  One can show that it is $(\omega,\infty)$-distributive.  In the case where $\alpha$ is countable, it is atomic.  Is it nonatomic for uncountable $\alpha$?  Is it strategically closed?
(3) If $G$ is generic for $A$, does $V[G]$ always contain some $Add(\omega,\alpha)$ generic, like in the case $\alpha$ is countable?
Edit: I thank J.D. Hamkins for his answer, but I already knew about the case $\alpha < \omega_1$, as stated in the original post.  I would really like some insight on the case $\alpha \geq \omega_1$.

NEW RELATED QUESTION
Consider instead $Col(\mu,<\kappa)*Add(\kappa,1)$, where $\mu< \kappa$ are regular.  Let $B$ be the completion, and let $A$ be the complete subalgebra generated by the canonical name for the Cohen subset of $\kappa$ added by the second part.  Let $H \subseteq A$ be generic.
Does $B/H$ have a $\mu$-closed dense subset in $V[H]$?  Any proof or refutation would be great, or a discussion of other related structural aspects of $A,B$, and $B/H$.  The properties of this forcing in relation to large cardinals are important to my research, so all nontrivial information is welcome.

REPLY [2 votes]: I have answers the original questions, by way of counterexample, which I will sketch.  The "new related question" remains unsolved.  I am grateful to Mohammad Golshani for his answer to this question, which I realized helps to answer the present question.
Assume CH holds in the ground model $V$.  Let $B_0$ be the completion of $Col(\omega,<\omega_2) * Add(\omega_2^V,1)$.  Let $\mathbb{P}$ be the countable support product of $Col(\omega,\alpha)$ for $\alpha < \omega_2$, and let $B_1$ be the completion of $\mathbb{P} * Add(\omega_2^V,1)$.
Let $A_0 \subseteq B_0$ be the complete subalgebra generated by the name for the Cohen subset of $\omega_2^V$ from the second iterand, and let $A_1 \subseteq B_1$ be the same with respect to $B_1$.  The following is a special case of a more general lemma that is part of my work in progress, to be submitted soon.  (CH is used)

Claim: $A_0 \cong A_1$, and $X \subseteq \omega_2^V$ is generic for $A_0$ iff it is generic for $A_1$.

Now, Mohammad Golshani gives examples of two $\Sigma_1$ properties, $\varphi_0, \varphi_1$, in parameters from $V$, such that whenever $G_0 \subseteq Col(\omega,<\omega_2)$ is generic, and $G_1 \subseteq \mathbb{P}$ is generic, $V[G_0] \models \varphi_0 \wedge \neg \varphi_1$, and $V[G_1] \models \varphi_1 \wedge \neg \varphi_0$.  By the Claim, if $X \subseteq \omega_2^V$ is generic for $A_0$ (and $A_1$), then we necessarily have $V[X] \models \neg \varphi_0 \wedge \neg \varphi_1$.
Now we note that $Col(\omega,<\omega_2)$ is equivalent to $Col(\omega,\omega_1) \times Add(\omega,\omega_2^V)$.  Let $g: \omega \to \omega_1^V$ be generic for the left side.  Then $B_0 / g$ is equivalent to $Add(\omega,\omega_1^{V[g]}) * Add(\omega_1^{V[g]},1)$, like in the original question.  For any $G_0 * X$ generic for $B_0$ extending $g$, we have $g \in V[X]$, and $V[X]$ contains no filter over $Add(\omega,\omega_2^V)$ which is generic over $V[g]$, using Mohammad's lemma and the above factoring.  This answers question (3).
It also answers question (2).  The quotient in this case is indeed atomless, and it is not $<\omega_1$-strategically closed.  If it were, then we could construct an $Add(\omega,\omega_1^{V[g]})$-generic (over $V[g]$) filter, by recursively deciding initial segments.  This also sheds some light on question (1), although that is a broad and vague question.<|endoftext|>
TITLE: Higher-dimensional Fáry's theorem?
QUESTION [9 upvotes]: Fáry's theorem says that every finite simple planar graph admits a planar embedding with straight line edges.
For which $(k,d)$ is it true that every finite $k$-dimensional simplicial complex embeddable in $\mathbb{R}^d$ has an embedding which is linear on every face?
It is true when $d \ge 2k+1$ by putting things in general position.
I am especially interested to know if anything is known about the case $(k,d)=(2,3)$.
I vaguely remember an old conjecture of Branko Grünbaum that every triangulation of the torus admitting a "straight" embedding in $\mathbb{R}^3$ but I don't know a reference (or whether this is still open).

REPLY [6 votes]: The answer is negative for all pairs $(k,d)$ with $k+1\leq d\leq 2k$, as long as $k \ge 2$.
Brehm [1] constructed a triangulation of the Möbius strip that does not admit a geometric (simplexwise linear) embedding into $\mathbb{R}^3$.
More generally, for every pair $(k,d)$ with $k+1\leq d\leq 2k$, Brehm and Sarkaria [2]  constructed an example of a $k$-dimensional simplicial complex that admits a piecewise 
linear embedding into $\mathbb{R}^d$, but no geometric embedding.
Moreover, for any given integr $r \geq 0$, there is such a $K$ such that even the $r$-fold barycentric subdivision of $K$ is not geometrically embeddable into $R^d$. 
Furthermore,
for certain values of the parameters, e.g., for $k=d-1$ and $d\geq 5$, it is known that 
there is no recursive bound on the complexity of the subdivision needed to embed a finite
$k$-dimensional simplicial complex piecewise linearly into $\mathbb{R}^d$ (see [3, Corollary 1.2]).
[1] U. Brehm. A nonpolyhedral triangulated Möbius strip. Proc. Amer. Math. Soc., 89(3), 519–522, 1983. 
[2] U. Brehm and K. Sarkaria. Linear vs. piecewise-linear embeddability of simplicial complexes. Tech. Report 92/52, Max-Planck-Institut für Mathematik, Bonn, 1992. Available for
download from Karanbir Sarkaria's website http://kssarkaria.org/List.htm
[3] J. Matousek, M. Tancer, and U. Wagner. Hardness of embedding simplicial complexes in $\mathbb{R}^d$. J. Eur. Math. Soc. 13, 259–295, 2011.<|endoftext|>
TITLE: Unitalization internal to monoidal categories
QUESTION [5 upvotes]: Let $C$ be a monoidal category, not assumed to be symmetric. Assume that the underlying category of $C$ is nice enough, for example cocomplete, perhaps even presentable. A semigroup object in $C$ is a pair $(X,\mu)$ consisting of an object $X \in C$ and a morphism $\mu : X \otimes X \to X$ satisfying the associativity law $\mu \circ (X \otimes \mu) = \mu \circ (\mu \otimes X)$. Does the forgetful functor from monoid objects in $C$ to semigroup objects in $C$ have a left adjoint? In other words, is there an unitalization internal to $C$?
The cases $C=\mathsf{Set}$ and $C=\mathrm{Mod}(R)$ are well-known. More generally, the answer is yes when $\otimes$ preserves coproducts in each variable. Then the unitalization of $(X,\mu)$ is $(1 \oplus X,\mu',\eta)$ with the obvious morphism $\mu' : (1 \oplus X) \otimes (1 \oplus X) = 1 \oplus X \oplus X \oplus X \otimes X \to 1 \oplus X$ and $\eta : 1 \to 1 \oplus X$.
Actually I am interested in the case that $C=(\mathrm{End}(D),\circ,\mathrm{id})$ for a (nice) category $D$, thus I would like to know if every semi-monad can be made into a monad. Here $\otimes$ preserves colimits in the left variable, but not in the right variable. Actually $D$ is even a presentable symmetric monoidal category and $\mathrm{End}(D)$ refers to enriched endofunctors, i.e. I am interested in strong (semi) monads.

REPLY [3 votes]: If $C$ is locally presentable and $S$ is a semi-monad whose underlying functor is accessible, then there exists a unitalization of $S$. Here is a proof modeled after an idea discussed at the nLab at the page free monad.
Define an algebra of a semi-monad $S: C \to C$ in the expected way, as an object $X$ of $C$ equipped with a morphism (an "action") $SX \to X$ satisfying the usual associativity law for an action. Morphisms between algebras are also defined in the expected way, so that there is a full embedding $S$-$\mathrm{Alg}_\mathrm{smd} \hookrightarrow S \downarrow C$ into the comma category. (I use the subscripts "smd" and "mnd" to indicate algebras qua semi-monads and monads.)
The main thing to check is that the forgetful functor $S$-$\mathrm{Alg}_\mathrm{smd} \to C$ is monadic in the "evil" sense, so that there is an isomorphism $F$-$\mathrm{Alg}_\mathrm{mnd} \simeq S$-$\mathrm{Alg}_\mathrm{smd}$ in $Cat/C$ for some monad $F$. The claim is that then $F$ is the free monad on the semi-monad $S$. For in that case, given a monad $M$ on $C$ we have natural bijections between

Semi-monad morphisms $S \to M$,

$S$-algebra structures $S U_M \to U_M$ where $U_M:$ $M$-$\mathrm{Alg}_\mathrm{mnd} \to C$ is the forgetful functor,

Morphisms $M$-$\mathrm{Alg}_\mathrm{mnd} \to S$-$\mathrm{Alg}_\mathrm{smd}$ in $Cat/C$,

Morphisms $M$-$\mathrm{Alg}_\mathrm{mnd} \to F$-$\mathrm{Alg}_\mathrm{mnd}$ in $Cat/C$,

$F$-algebra structures (qua algebras over a monad) $F U_M \to U_M$,

Monad morphisms $F \to M$


so that $F$ is evidently the free monad on the semi-monad $S$.
So now we check monadicity, using the precise monadicity theorem. It is straightforward that the forgetful functor $U: S$-$\mathrm{Alg}_{\mathrm{smd}} \to C$ creates (not just reflects!) $U$-split coequalizers, so we just have to check that $U$ has a left adjoint. However, since the 2-category of locally presentable categories and accessible functors inherits lax limits from $Cat$, and since $S$-$\mathrm{Alg}_\mathrm{smd}$ is a lax limit in $Cat$ (for essentially the same reason that Eilenberg-Moore categories for monads are lax limits), we see that $U: S$-$\mathrm{Alg}_\mathrm{smd} \to C$ is an accessible functor between locally presentable categories. In this situation, existence of a left adjoint to $U$ is equivalent to preservation of limits by $U$. But limit-preservation is clear. So the conditions of the precise monadicity theorem are satisfied.<|endoftext|>
TITLE: totally ordered chain in the powerset with big cardinality
QUESTION [13 upvotes]: Let $B$ be some set. The problem is to find a set $A\subset\mathcal{P}(B)$ of subsets of $B$ which is totally ordered by inclusion and such that there exists a bijection $A\leftrightarrow \mathcal{P}(B)$.
This is an easy exercise if $B$ is countable where one can explicitly construct such a set (one identifies $B$ with $\mathbb{Q}$ and takes $A$ to be the set of sets of the form $(-\infty, r)$ for all $r\in \mathbb{R}$)
My question now is the following:
For which sets $B$ can the existence of such an $A$ be shown using any combination of (reasonable) additional hypotheses (e.g. AC, CH, GCH, ...)?
Under which circumstances can one find $B$'s that don't admit such an $A$?

REPLY [16 votes]: Let's think about the countable case like this: think of the
binary tree $2^{\lt\omega}$, which has size $\omega$, but has
$2^\omega$ many branches. Each branch describes a cut in the
natural lexical order on the nodes, and so we have a countable
linear order with $2^\omega$ many cuts.
So consider a cardinal $\kappa$ and the tree $2^{\lt\kappa}$,
which admits a similar lexical order. If this tree has size
$\kappa$, then we get a linear order of size $\kappa$ with
$2^\kappa$ many cuts. And such a family of cuts turns into a chain
just as you point out in the question.
Thus, if $B$ has size $2^{\lt\kappa}$, then we can find $A$ of size
$2^\kappa$. In particular, whenever $2^{\lt\kappa}=\kappa$, then $P(\kappa)$
has a chain of size $2^\kappa$, as you desire.
In particular, under the GCH, the phenomenon will occur for every
infinite cardinal, since GCH implies $2^{\lt\kappa}=\kappa$ for
all infinite cardinals $\kappa$.
Lastly, let me point out that this argument method can be turned
into a characterization. Namely, the sets $B$ for which there is
an $A$ as you desire are exactly the sets $B$ of size $\kappa$ for
which there is a linear order on $\kappa$ with $2^\kappa$ many
cuts. The one direction we've established, and for the converse,
when there is such a chain $A$ in $P(B)$ of that size, then we may
place a pre-order on $B$ according to the order in which points
are added to sets in the chain (and extend this to a linear
order). Each element of $A$ gives a cut in this order, and so we
have a linear order on $\kappa$ with $2^\kappa$ many cuts.
So the phenomenon occurs for exactly those sets $B$ of size
$\kappa$ for which there is a linear order on $\kappa$ with
$2^\kappa$ many cuts.<|endoftext|>
TITLE: Local model of virtual fundamental cycle
QUESTION [8 upvotes]: The following baby version of virtual fundamental cycle is well known:
Let $M\subset V$ be the zero locus of a section $s$ of a vector bandle $E \to V$, in general $s$ is not transversal to the zero section and $M$ does not have the expected dimension, then one uses excess intersection theory to define the virtual fundamental class $[M]^{vir}
$ to be $0_E^{!}[C_{M/V}]\in A_{vdim}(M)$, where $C_{M/V}\subset E|_M$ is the normal cone of $M$ in $V$. 
If $V$ is smooth, one can reformulate these into Behrend-Fantachi's language by setting $E^{-1}=E^{\vee}|_M, E^0=\Omega_V|_M$, and get a perfect obstruction theory $E^* \to L_{M}$ associated to the model $(V,E,s)$.
My question is whether the reformulation can be reversed:
Given a perfect obstruction theory $E^* \to L_{M}$, can we return to the baby version locally? I mean whether we can find open covering (in some topology) of $M$, such that for each open subset $U$ in the covering, we can find $U\subset V$,a vector bundle $F \to V$ and $U$ is realized as zero locus of a section $s$. And $[E^* \to L_{M}]|_U$ coincides with the perfect obstruction theory associated to the local model $(V,F,s)$. I will appreciate if you could provide the details.
Thank you!

REPLY [6 votes]: The answer is yes.  Let $[E^{-1} \rightarrow E^0]$ be a perfect obstruction theory on $M$.  After localizing in $M$ we can assume that the map $E^0 \rightarrow \Omega_M$ is induced as $\mathcal{O}_M \otimes \Omega_{\mathbf{A}^n} \rightarrow \Omega_M$ for some map $M \rightarrow \mathbf{A}^n$.(*) As the map on differentials is surjective the relative differentials vanish and the map is unramified.  By EGA IV.18.4.7, we can (at least after localizing further in $M$) factor the map $M \rightarrow \mathbf{A}^n$ as a closed embedding $M \rightarrow V$ followed by an étale map $V \rightarrow \mathbf{A}^n$.
Let $I$ be the ideal of $M$ inside $V$.  Then we have an exact sequence
$I/I^2 \rightarrow \mathcal{O}_M \otimes \Omega_V \rightarrow \Omega_M \rightarrow 0$
where the first two terms are $\tau_{\geq -1} L_M$.  By the definition of an obstruction theory, we are given a map $E^{-1} \rightarrow I/I^2$ inducing a surjection $H^{-1}(E) \rightarrow H^{-1} L_M$.  By the 5-lemma, $E^{-1} \rightarrow I/I^2$ is surjective.
Localizing further in $V$ and $M$, we can assume that the map $E^{-1} \rightarrow I/I^2$ is induced from a map $F \rightarrow I$ of coherent sheaves on $V$ with $F$ locally free.  By Nakayama's lemma we can assume, possibly after further localization, that $F \rightarrow I$ is surjective.  Then consider the section of the vector bundle $F^\vee$ dual to the composition $F \rightarrow I \subset \mathcal{O}_V$.  The vanishing locus of this section is $M$ and the induced complex $F \rightarrow \mathcal{O}_M \otimes \Omega_V$ is the obstruction theory we started with.
(*)  We can represent $\tau_{\geq -1} L_M$ as $[ J / J^2 \rightarrow \Omega_W ]$ for some closed embedding $M \subset W$ and we are given a map of complexes $E^\bullet \rightarrow \tau_{\geq -1} L_M$.  Replacing $E^\bullet$ with a quasi-isomorphic complex, we can assume that the map $E^0 \rightarrow \Omega_W$ is surjective.  But $E^0$ and $\Omega_W$ are vector bundles, so after localizing in $M$ we can assume that $\Omega_W$ is a direct summand of $E^0$.  Choosing a basis for the complementary summand, we can identify $E^0 \cong \Omega_{W \times \mathbf{A}^n}$.<|endoftext|>
TITLE: An element $g$ in a group such that neither $g=1$ nor $g\ne 1$ can be proved.
QUESTION [5 upvotes]: Edited (this question contains two versions of a similar question)
Is there some finitely presented group $G$ generated by $g_1,...,g_n$ such that 
there is an element $g\in G$ expressed as a finite word in $g_i$'s so that it is impossible to prove neither $g=1$ nor $g\ne 1$?
Is such a group $G$ exists, what would be a relatively simple example? 
Adjusted question. Is there $G,g$ so that $g\ne 1$ in $G$ but is it impossible to prove this in finite time?

REPLY [9 votes]: The following is an abridged translation of a philosophical digression from my
lectures on Group Theory (in Russian).

Metatheorem 3.5.1.
For any finite (group) presentation $G$ with non-solvable
word problem, there exists a word $w$
(called bad) such that
it is possible to prove neither that
$w=1$ in $G$ nor that
$w\ne 1$ in $G$.
Proof.
Note that any reasonable definition
of the
notion of
proof has the following property:
Any proof can be written in some
fixed formal language; and there exists an algorithm
VERIFICATION, which takes as input any text in this language and
any assertion (also written in this language) and
says whether or not the given
text is a proof of the given assertion.
Suppose now that there are no bad words for a given presentation. Then
it is easy to construct an algorithm solving the word problem:
for any input word $v$, we simply search all texts and feed them to the
algorithm VERIFICATION until we find a text which is
either a proof that $v=1$ in $G$ or a proof
that $v\ne1$ in $G$.
Remark. Any word equal to a bad word in the
group
$G$ is bad itself (because for any two words
equal in $G$, their equality
can
always be proven).
So, we can speak about bad elements of a group
(not only bad words). Moreover, it is easy to show that the badness of
an element $g\in G$ does not depend on the choice of finite presentation
of $G$.
As a corollary, we obtain the following strange-looking fact.
Metatheorem 3.5.2.
Any bad word is, actually, not equal to 1.
(Because 1 is not a bad element, obviously.)
Although bad words exist, no particular
example can be constructed.
Metatheorem 3.5.3.
For any bad element, it is impossible to prove that this element is bad.
Proof.
Indeed,
a proof of the badness of an element $g$ would prove, in particular,
that $g\ne1$ (by Metatheorem 3.5.2) that contradicts the definition
of a bad element.

A similar argument and the Adyan$-$Rabin theorem show that:

There exists a bad group, i.e. a finitely presented group such that
it is possible to prove neither that this group is trivial nor that this
group is non-trivial.
Each bad group is non-trivial.
For any bad group, it is impossible to prove that this group is bad.<|endoftext|>
TITLE: Axiom of Choice and continuous functions
QUESTION [8 upvotes]: Do you know if the following statement is an equivalent form of the axiom of choice or not?

If $X$ is a compact metric space, then every continuous function $f: X \longrightarrow \mathbb{R}$ is uniformly continuous.

If you know any references, please let me know.

REPLY [2 votes]: Look Ma, no axiom of choice!
THEOREM 0   Let   $X$   be a compact space. Let   $\Phi$   be a non-empty family of closed subsets of   $X$,   $F := \bigcap \Phi$,   and   $G\supseteq F$   an open subset of   $X$.   Then there exists a finite   $\Phi_0\subseteq \Phi$   such that   $\bigcap\Phi_0\subseteq G$.
PROOF   Family   $\Gamma\ :=\ G\cup\{X\setminus A : A \in \Phi\}$   is an open covering of   $X$. Etc.   END of PROOF

As an instant corollary we get:
THEOREM 1   Let   $X$   be a compact space. Let   $\Phi$   be a non-empty family of closed subsets of   $X$,   $F := \bigcap \Phi$,   and   $G\supseteq F$   an open subset of   $X$.   Assume also that   $\Phi$   is linearly ordered by   $\subseteq$.   Then there exists   $A\in \Phi$   such that   $A \subseteq G$.
THEOREM 2   Let   $(X\ d)$   be a compact metric space. Let   $W$   be an open subset of   $X^2$,   such that   $\Delta_X\subset W$,   where   $\Delta_X := \{(x\ x):x\in X\}$.   Then there exists   $\delta > 0$   such that   $V_X(\delta)\subseteq W$,   where   $V_X(\delta) := \{(x\ y)\in X^2 : d(x\ y)\le\delta\}$.
PROOF   Apply Theorem 1 to   $X^2$ as a replacement of   $X$   of Theorem 1; etc.   END of PROOF
THEOREM 3   Let   $f : X\rightarrow Y$   be an arbitrary continuous function of a metric compact space   $(X\ d_X)$   into an arbitrary metric space   $(Y\ d_Y)$.   Then function   $f$   is uniformly continuous.
PROOF   Let   $\epsilon > 0$.   Let   $W\subseteq X^2$   be the inverse image of   $V_Y(\epsilon)$   under function   $f\times f$.   There exists, by Theorem 2,   $\delta > 0$   such that   $V_X(\delta)\subseteq W$.   Then   $d_Y(f(x')\ f(x''))\le\epsilon$   for every   $x'\ x''\in X$   such that   $d_X(x'\ x'')\le\delta$.   END of PROOF<|endoftext|>
TITLE: Motivic cohomology and cohomology of Milnor K-theory sheaf
QUESTION [14 upvotes]: Let $X$ be a smooth variety over a field $k$. (Assume $k$ has characteristic 0 if it helps; in fact I'd be happy to assume that $k$ is a finite extension of either $\mathbf{Q}$ or $\mathbf{Q}_p$). 
Then there is a sheaf $\mathscr{K}_m^M$ on $X$ (in the Zariski topology), for each $m \ge 1$, which comes from sheafifying the Milnor $K$-theory of function rings of affine opens of $X$. So we can take sheaf cohomology groups $H^*(X, \mathscr{K}_m^M)$. We can also consider Voevodsky's motivic cohomology groups $H^*(X, \mathbf{Z}(i))$, where $\mathbf{Z}(i)$ is Voevodsky's motivic complex.

Is it true that there are isomorphisms 
  $$H^r(X, \mathscr{K}_s^M) = H^{r+s}(X, \mathbf{Z}(s))$$
  for all integers $r, s \ge 0$?

(Note that the right-hand side is known to be coincide with Bloch's higher Chow group $CH^s(X, s-r)$.) 
Here is why I think this. Both sides are zero if $r > s$ or $r > \dim X$. For $r \le s$, one has a candidate for the isomorphism using Kerz's Gersten complex for Milnor K-theory and a construction due to Landsburg; and for $r = s$ or $r = s-1$ this map is indeed known to be an isomorphism. On the other hand, for $r = 0$ and $X = \operatorname{Spec} k$, this is just the isomorphism of Nesterenko--Suslin--Totaro, $$H^s(\operatorname{Spec} k, \mathbf{Z}(s)) = K_s^M(k).$$ 
(This is related to my earlier question When do the $\gamma$-filtration and codimension filtration of K-theory agree?, where I asked essentially the same thing for the Quillen $K$-theory sheaf instead of the Milnor one.  It was pointed out there that for $r = 0$ and $X = \operatorname{Spec} k$ the motivic cohomology is just the Milnor K-theory, which leads me to wonder if one does get an isomorphism using the Milnor $K$-theory sheaf; for $r = s$ or $r = s-1$ the cohomology of the Milnor and Quillen $K$-theory sheaves agrees.)

REPLY [9 votes]: The previous answer contained a major error/misconception, and I apologize for the dealy in correcting it. The answer to the question is "yes" locally in the Zariski topology but "no" globally. 
Comparison of Milnor K-cohomology and motivic cohomology via edge maps:
Motivic cohomology has Zariski descent and hence there is a 
descent/hypercohomology/Brown-Gersten-Quillen/coniveau spectral sequence computing the motivic cohomology of a smooth scheme $X$. On the $E_1$-page, the Gersten resolution of the sheaves $\mathcal{H}^q(\mathbb{Z}(n))$ (Zariski sheafification of motivic cohomology) appears. The entry $E_1^{p,q}$ is $\bigoplus_{z\in X^{(p)}}{\rm H}^{q-p}(\kappa(z),\mathbb{Z}(n-p))$ and the differential $d_1:E_1^{p,q}\to E_1^{p+1,q}$ is given by the appropriate residue maps. 
The $E_2$-spectral sequence has the form
$$
E^{p,q}_2={\rm H}^p(X,\mathcal{H}^q(\mathbb{Z}(n)))\Rightarrow {\rm H}^{p+q}(X,\mathbb{Z}(n))
$$
and the differentials go $d_s^{p,q}:E^{p,q}_s\to E^{p+s,q-s+1}_s$. 
This spectral sequence can then be used to relate Milnor K-cohomology to motivic cohomology. The relevant thing to note is the identification 
$$
\tau_{\geq s}\mathbb{Z}(s)_X\to\mathbf{K}^{\rm M}_{s,X}[-s]
$$
which can be found e.g. in Section 3.1, Corollary 3.2 of

K. Rülling and S. Saito. Higher Chow groups with modulus and relative Milnor K-theory. arXiv:1504.02669v2.

This statement basically says that $\mathcal{H}^q(\mathbb{Z}(n))=0$ for $q>n$ and $\mathcal{H}^n(\mathbb{Z}(n))=\mathbf{K}^{\rm M}_n$, where $\mathcal{H}^q$ denotes the motivic cohomology sheaves. (It also says that the answer is yes locally in the Zariski topology.)
From this vanishing statement, we get edge maps ${\rm  H}^p(X,\mathbb{Z}(n))\to {\rm H}^{p-n}(X,\mathcal{H}^n(\mathbb{Z}(n)))$ because in the line $q=n$ there are no incoming differentials and so $E^{p,n}_\infty$ is a subgroup of ${\rm H}^{p-n}(X,\mathcal{H}^n(\mathbb{Z}(n)))$. Since $E^{p,n}_\infty$ is the last subquotient of the filtration of ${\rm H}^p(X,\mathbb{Z}(n))$ this provides the above edge maps. These would be the natural comparison maps (and they are also mentioned in Thi's answer.) 
Comparison isomorphisms for $r=s$ and $r=s-1$:
To get the Bloch formula and the identification of ${\rm H}^{2n-1}(X,\mathbb{Z}(n))\cong {\rm H}^{n-1}(X,\mathbf{K}^{\rm M}_n)$, we need to show that there can be no outgoing differentials either. In the $E_2$-page, the relevant differentials for Bloch's formula are $d_s^{p,q}:{\rm H}^n(X,\mathcal{H}^n(\mathbb{Z}(n)))\to {\rm H}^{n+s}(X,\mathcal{H}^{n-s+1}(\mathbb{Z}(n)))$. The latter group is trivial, since on the $E_1$-page, the coefficients would be $\mathcal{H}^{1-2s}(\mathbb{Z}(-s))$ and the sheaves $\mathbb{Z}(-s)$ are trivial for $s>0$. The other relevant differentials are $d_s^{p,q}:{\rm H}^{n-1}(X,\mathcal{H}^n(\mathbb{Z}(n)))\to {\rm H}^{n+s-1}(X,\mathcal{H}^{n-s+1}(\mathbb{Z}(n)))$ with the coefficients in the $E_1$-page given by ${\rm H}^{2-2s}(\mathbb{Z}(1-s))$. For $s>1$, this is again trivial. In particular, all the outgoing differentials are trivial, so that $E^{n,n}_\infty={\rm H}^n(X,\mathbf{K}^{\rm M}_n)$ and $E^{n-1,n}_\infty={\rm H}^{n-1}(X,\mathbf{K}^{\rm M}_n)$. This also implies that the only $E^{n+s,n-s}_\infty$ contributing to ${\rm H}^{2n}(X,\mathbb{Z}(n)))$ is ${\rm H}^n(X,\mathbf{K}^{\rm M}_n)$, yielding Bloch's formula. To get ${\rm H}^{n-1}(X,\mathbf{K}^{\rm M}_n)\cong {\rm H}^{2n-1}(X,\mathbb{Z}(n))$ we need to show that ${\rm H}^{n+s-1}(X,\mathcal{H}^{n-s}(\mathbb{Z}(n)))=0$. The relevant coefficients in the $E_1$-page are $\mathcal{H}^{1-2s}(\mathbb{Z}(1-s))$. For $s=1$, we get $\mathcal{H}^{-1}(\mathbb{Z}(0))$ which is trivial, all the other terms are again trivial because $\mathbb{Z}(1-s)=0$ for $s>1$.
Failure of global comparison:
The natural comparison isomorphism fails to be an isomorphism in general. The simplest counterexample (pointed out by Jens Hornbostel) is actually $\mathbb{P}^1$. In this case, we have 
$$
{\rm H}^r(\mathbb{P}^1,\mathbf{K}^{\rm M}_s)\cong \left\{\begin{array}{ll} {\rm K}^{\rm M}_s(F) & r=0\\
{\rm K}^{\rm M}_{s-1}(F) & r=1\end{array}\right.
$$
where $F$ denotes the base field. On the other hand, the projective bundle formula for motivic cohomology implies 
$$
{\rm H}^{r+s}(\mathbb{P}^1,\mathbb{Z}(s))\cong {\rm H}^{r+s}(F,\mathbb{Z}(s))\oplus {\rm H}^{r+s-2}(F,\mathbb{Z}(s-1)).
$$
For $r=0$, we have ${\rm H}^0(\mathbb{P}^1,\mathbf{K}^{\rm M}_s)\cong {\rm K}^{\rm M}_s(F)$ and ${\rm H}^s(\mathbb{P}^1,\mathbb{Z}(s))\cong {\rm K}^{\rm M}_s(F)\oplus {\rm H}^{s-2}(F,\mathbb{Z}(s-1))$, so Milnor K-cohomology and motivic cohomology differ whenever ${\rm H}^{s-2}(F,\mathbb{Z}(s-1))\neq 0$. A particular instance where that happens is $s=3$ in which case ${\rm H}^1(F,\mathbb{Z}(2))\cong {\rm K}^{\rm ind}_3(F)$. 
In the formulation of the descent spectral sequence, the spectral sequence is contained in the two columns for $p=0,1$ and degenerates at the $E_2$-term. The failure of the global comparison follows since $\mathcal{H}^2(\mathbb{Z}(3))$ has nontrivial first cohomology over $\mathbb{P}^1$, given by ${\rm H}^1(F,\mathbb{Z}(2))$.<|endoftext|>
TITLE: Goormaghtigh equation
QUESTION [5 upvotes]: For fixed prime $p$ and $q$ is it true that the following equation has at most one pair $(m,n)$ of solution?
$$\frac{p^n-1}{p-1}=\frac{q^m-1}{q-1}$$ where $m,n>1$.

REPLY [7 votes]: Yes for fixed $p,q$ it is know there is at most one solution.  Even without the condition $p,q$ prime. 
This was proved by He and Togbé, On the number of solutions of Goormaghtigh equation for given x and y (Indagationes Math, 2008)<|endoftext|>
TITLE: Masas in second duals of Banach algebras
QUESTION [8 upvotes]: Lel $B$ be a Banach algebra and give $B^{**}$ one of the Arens products in order to make it a Banach algebra. Then the canonical embedding $\kappa\colon B\to B^{\ast\ast}$ is a homomorphic embedding wrt to either Arens product. 
Suppose $M$ is a maximal abelian subalgebra of $B$. Can infer from this that $\overline{\kappa(M)}^{w^*}$ is a maximal abelian subalgbra of $B^{\ast\ast}$? I suspect it need not be the case... How about when $B$ is Arens regular, that is, when the Arens products in $B^{\ast\ast}$ coincide?

REPLY [8 votes]: It need not be so. The simplest example would be the Toesplitz algebra $\cal T$. It is the ${\rm C}^\ast$-algebra generated by the unilateral shift $S$ on $\ell_2{\bf N}$, $Se_n=e_{n+1}$. The algebra ${\cal T}$ contains the ideal $\cal K$ of the compact operators, and the quotient is $\pi\colon{\cal T} \to {\cal T}/{\cal K} \cong C({\bf T})$. The diagonal subalgebra $D:=\ell_\infty{\bf N} \cap {\cal T}$ is a masa of $\cal T$, which coincides with $c_0{\bf N}+{\bf C}1$. This follows from the facts that $(c_0{\bf N})'\cap{\bf B}(\ell_2{\bf N})=\ell_\infty{\bf N}$ and that $\langle xe_n,e_n\rangle$ converges for every $x\in{\cal T}$ (to $\int_{\bf T} \pi(x)\ dm$). One has $$\bigl( D^{\ast\ast}\subset {\cal T}^{\ast\ast} \bigr) \cong \bigl( (c_0{\bf N})^{\ast\ast}\oplus {\bf C}1\subset {\cal K}^{\ast\ast} \oplus C({\bf T})^{\ast\ast} \bigr)$$
and $D^{\ast\ast}$ is not a masa.
In fact, I think the weak*-closure of a masa is never a masa in the second dual $B^{\ast\ast}$ unless the situation is very degenerate, e.g., $c_0 \subset {\cal K}$ or the masa has finite index. This is because every quasi-equivalence class of a *-representation $\pi$ of $B$ has a corresponding central projection $c([\pi])$ in $B^{\ast\ast}$, which satisfies $\pi(B)'' \cong B^{\ast\ast}c([\pi])$. That $\overline{M}^{w\ast}$ contains them all means that the quasi-equivalence class of a *-representation $\pi$ of $B$ is uniquely determined by its restriction $\pi|_M$ to $M$. This is rather rare.<|endoftext|>
TITLE: Equivalence of two well-known forms of (RH): reference-request
QUESTION [5 upvotes]: This is a reference-request about a very simple statement.
The Riemann hypothesis is well-known to be equivalent to
$$(1)\ \ \  \pi(x) = \mathrm{Li}(x)+O(x^{1/2} \log x)$$
and to
$$(2)\ \   \theta(x)=x+O(x^{1/2} \log^2 x).$$
Here as usual, $\pi(x) = \sum_{p \leq x} 1$ and $\theta(x) = \sum_{p \leq x} \log p$.

I am looking for a reference giving the proof of the equivalence between (1) and (2).

All the analytic number theory textbooks I have looked at gives at best a proof that
$\theta(x) \sim \pi(x) \log x$, which is clearly not enough.

REPLY [6 votes]: To complement Greg Martin's response, here is a proof that (1) implies (2).
Write $\pi(t)=\mathrm{Li}(t)+f(t)$, so that $f(t)=O(t^{1/2}\log t)$ by assumption. Then
$$ \theta(x) = \int_{2-}^x\log t\ d\pi(t)=\int_{2-}^x\log t\ d\mathrm{Li}(t)+\int_{2-}^x\log t\ d f(t). $$
Here the first integral is
$$ \int_{2-}^x\log t\ d\mathrm{Li}(t) = \int_2^x\log t\frac{dt}{\log t} = x-2 $$
and the second integral is
$$ \int_{2-}^x\log t\ df(t) = f(x)\log x - \int_2^x \frac{f(t)}{t}dt = 
O(x^{1/2}\log^2 x) +  \int_2^x O(t^{-1/2}\log t)\ dt. $$
The last integral is clearly
$$ O(\log x)\int_2^x t^{-1/2}\ dt = O(x^{1/2}\log x),$$
hence altogether
$$ \theta(x) = x-2 + O(x^{1/2}\log^2 x)+O(x^{1/2}\log x) = x+O(x^{1/2}\log^2 x). $$<|endoftext|>
TITLE: On Tamarkin's proof of Etingof-Kazhdan quantization of Lie bialgebra
QUESTION [19 upvotes]: This question is motivated by the desire to understand better the interplay between Drinfeld associators, algebraic structures up to homotopy and related results.
Recall that a Lie bialgebra is a Lie algebra $\mathfrak g$ with a Lie algebra structure on $\mathfrak g^*$ (cobracket) plus some compatibility conditions. These are infinitesimal versions of compatible Poisson structures on Lie group. A quantization of $\mathfrak g$ is a $\mathbb C[[\hbar]]$ deformation of the envelopping algebra of $\mathfrak g$ the coproduct of which gives back the cobracket of $\mathfrak g$ at the quasi-classical limit. Etingof-Kazhdan theorem states that every Lie bialgebra can be quantized, in a functorial way. Tamarkin's proof of this results relies on the formality of the little disc operad, and in fact so does the original proof in a slighty different language.
The original proof use the statement "the choice of an associator leads to a quasi-triangular quasi-Hopf algebra (aka braided monidal category) from any metrizable Lie algebra (i.e. equipped with an invariant symmetric bilinear form)". Indeed, any metrizable Lie algebra leads to a representation of a certain operad in Lie algebra $\mathfrak t$ and an associator is nothing but an isomorphism from some completion of the group algebra of pure braid operad to $U(\mathfrak t)$.
Then the proof goes a follow:

To any Lie bialgebra $\mathfrak g$ is asociated its double $\mathfrak d$ which quasi-triangular and in particular metrizable. $\mathfrak d=\mathfrak g\oplus \mathfrak g^*$ as vector space and the pairing is just the canonical one.
Use the above statement to turns $U(\mathfrak d)[[\hbar]]$ into a quasi-triangular quasi-Hopf algebra
Solve the so-called "twist equation" which say that the quasi-Hopf algebra $U(\mathfrak d)[[\hbar]]$ can be turned into an honest Hopf algebra $U_{\hbar}(\mathfrak d)$. An important fact is that the product is not modified, i.e. $U_{\hbar}(\mathfrak d)$ is isomorphic to $U(\mathfrak d)[[\hbar]]$ as an algebra.
That we have a quasi-triangular quantization means roughly that we get a quantum version of the pairing (given by the quantum R-matrix) which can be used to identify a sub-Hopf algebra $U_{\hbar}(\mathfrak g)$ of $U_{\hbar}(\mathfrak d)$ which is the quantization we were looking for. One can also show that $U_{\hbar}(\mathfrak d)$ is the quantum Drinfeld double of $U_{\hbar}(\mathfrak g)$.

Note that thinking of $U_{\hbar}(\mathfrak g)$ as a subalgebra of $U_{\hbar}(\mathfrak d)$ is a bit misleading, and that it's somehow better to see $U(\mathfrak d)$ as an auxiliary space acting on $S(\mathfrak g)$ by differential operators. In this sense the "twist" really leads to a star product and a "star coproduct" on $S(\mathfrak g)$ whose coefficients are given by the action of some differential operators.
Now Tamarkin's proof goes as follow:

Any Lie bialgebra $\mathfrak g$ leads to a Gerstenhaber structure on the free graded commutative algebra $H=S(\mathfrak g_{\hbar}[-1])$ where in $\mathfrak g_{\hbar}$ the cobracket is multiplied by $\hbar$.
The operad of Gerstenhaber algebras is quasi-isomorphic to $BU(\mathfrak t)$ (where $B$ is the bar construction).
The little disc operad is formal, which is roughly a refinement of the above statement on the relation between braids and $\mathfrak t$, applied at the level of chain.
The operad of chain of the little disc operad is quasi-isomorphic to the operad of brace algebras (this is a version of the so-called Deligne conjecture).
Therefore we get a brace structure on $H$, which induces a differential graded Hopf algebra structure on the cofree coalgebra $C(H[1])$. 
Finally, so far I understand, one can show that the 0th cohomolgoy group of the dg Hopf algebra obtained this way is a Hopf algebra quantizing $\mathfrak g$.

Admittedly I'm less confortable with this proof, though I can roughly understand how it goes, except maybe the last step. Anyway, I'm interested in understanding how thse proofs relate. Clearly the key fact is quite similar, though different: in the first case the (trivial deformation of) the category of modules over $\mathfrak d$ is an algebra over the operad $\mathfrak t$, while in the second case we construct a space which is an algebra over some operad of complex attached to $\mathfrak t$ through some combinatorial quasi-isomorphism. Then the same isomorphism coming from an associator is applied but in different worlds. So are related Etingof-Kahdan 2 and Tamarkin 3.
Now I'd like to understand the relations between the other steps. The relation between the pairing on $\mathfrak d$ and the complex associated to $\mathfrak g$ is probably well known, but I didn't find a reference. The second part is probably more tricky, and I would really be happy to learn that there is indeed some relations between the Deligne conjecture and the statement that "a quasi Hopf algebra obtained from a Drinfeld associator can be turned into an Hopf algebra". Since the role of the Drinfeld double is to somehow merge the (Lie) algebra and the (Lie) coalgebra structure, it's temptating to think that there is a sort of correspondance between coalgebra deformations of $U(\mathfrak d)$ and bialgebra deformations of $U(\mathfrak g)$, and Etingof-Kahdan's proof show that it is indeed the case, in a highly non obvious way. Can this fact also be related to the Deligne conjecture ?
Edit I just came across these very interesting notes By Pavel Safronov which explain Tamarkin's proof along the same lines as Kevin's answer below.

REPLY [15 votes]: I'm not sure I can answer everything Adrien asked, but perhaps I can explain a little about part 6 in Tamarkin's proof, and about how the Drinfeld double appears in Tamarkin's story.
First, I'd like to explain a slightly different way to phrase Tamarkin's story. It's relies on Koszul duality, and throughout, there are some subtleties about completness, finiteness, etc. which always appear in Koszul duality. These can be dealt with, but I'll surpress the details.
The key thing we need to know is that any augmented $E_2$ algebra produces a Hopf algebra. (This is part 6 in Tamarkin's proof, as explained in the question). We can see this using Tanakian theory and Koszul duality.  If $A$ is an augmented associative algebra, with a map $A \rightarrow k$ to the ground field, we can form a Koszul dual associative algebra
$$
A^! = \mathrm{Hom}_k(A,A).
$$
You can find (in good situations) an equivalence
$$
A^!\mbox{-}\mathrm{mod} = A\mbox{-}\mathrm{mod}.
$$
Actually, a precise version of this statement is pretty subtle; see e.g. Positselski's work. The coalgebra version of this statement (where we use comodules over $k \otimes_A k$) works better.
Let's assume for simplicity that $A^!$ has cohomology in degree $0$, so we don't have to worry about homotopical analogs of Hopf algebra structures.
Now, if $A$ is an $E_2$ algebra then left modules over $A$ form a monoidal category $A\mbox{-}\mathrm{mod}$ (this is explained in Lurie's higher algebra, for instance).   Thus $A^!\mbox{-}\mathrm{mod}$ is a monoidal category, and you can check that the forgetul functor
$$ 
A^!\mbox{-}\mathrm{mod} \rightarrow \mathrm{dgVect}
$$
is monoidal.  Thus by Tannakian theory $A^!$ has a Hopf algebra structure.
There's a similar story for $E_3$ algebras. Suppose $A$ is an $E_3$ algebra which is augmented as an $E_2$ algebra.  Then $A\mbox{-}\mathrm{mod}$ has an $E_2$ (i.e. braided monoidal) structure, so that $A^!\mbox{-}\mathrm{mod}$ is a braided monoidal category such that the forgetful functor to vector spaces is monoidal.  This implies, I believe, that $A^!$ has the structure of a quasi-triangular Hopf algebra.
Let $P_2$ be the homology operad of $E_2$. Formality of the $E_2$ operad gives an isomorphism between $E_2$ and $P_2$, and allows us to turn any $P_2$ algebra into an $E_2$ algebra. Lie bialgebras give rise to augmented $P_2$ algebras, as follows. If $\mathfrak{g}$ is a Lie bialgebra, then $C^\ast(\mathfrak{g})$ is a commutative algebra, with a Poisson bracket defined on generators $\mathfrak{g}^\vee$ by the Lie coalgebra structure on $\mathfrak{g}$.
This story (due to Tamarkin, of course) produces a Hopf algebra from any Lie bialgebra (subject to various caveats about completeness, etc. etc. which is why we find a formal quantization instead of an actual quantization)
The question asked about the Drinfeld double.  I would guess that this works as follows. The Drinfeld double of a Hopf algebra is a quasi-triangular Hopf algebra. Quasi-triangular Hopf algebras are Koszul dual to $E_3$ algebras, and under Koszul duality, Drinfeld double should corresponds to $E_2$ Hochschild cohomology. By the higher analog of Deligne's conjecture, the $E_2$ Hochschild cohomology of an $E_2$ algebra is an $E_3$ algebra.
Some aspects of this story about $E_2$ algebras and Hopf are explained in detail in a paper I plan to put on the arxiv soon (which is closely related to Johnson-Freyd and Gwilliam's work).  Of course, the picture is not due to me, but to Tarmarkin, but I don't think he showed that there's an equivalence of monoidal categories in this story.<|endoftext|>
TITLE: Is $\kappa$-distributive the same as $\kappa$-strategically closed?
QUESTION [7 upvotes]: References for the definitions are Jech's Set Theory Definition 15.5, and Cummings paper in the Handbook of set theory Definition 5.15.

REPLY [8 votes]: Forcing with a Souslin tree is $\omega$-distributive, but it is not countably strategically closed. No $\omega_1$-tree can be countably strategically closed, since if it were, we could construct a subtree of order type $2^{\lt\omega}$, which had all limits. This would give rise to an uncountable level in the tree, contrary to the assumption that it was an $\omega_1$-tree. 
Another example (in ZFC rather than merely relatively consistent) would be the forcing to shoot a club through a stationary co-stationary subset $S\subset \omega_1$, with conditions consisting of closed bounded subsets of $S$.  This is $\omega$-distributive when $S$ is stationary, by a reflection argument reflecting $\omega_1$ to a point in $S$, but it is not countably strategically closed when $S$ is co-stationary, since there would have to be ordinals not in $S$ which were in a sense closed under the strategy (go to a countable elementary substructure), and one could find a play according to the strategy which would force this ordinal to appear in the club, even though it is not in $S$.<|endoftext|>
TITLE: Upper bound for the order of the group of automorphisms of Riemann surfaces of genus 2
QUESTION [7 upvotes]: By the Hurwitz's automorphisms theorem there is an upper bound $|\text{Aut}(C)|\leq 84(g-1)$ for all Riemann surfaces $C$ with $g(C)\geq 2$, but it is not sharp if $g=2$. What is the sharp upper bound for genus $2$ Riemann surfaces? Also, I'm asking for the simple proof of the fact that there is no genus $2$ Riemann surfaces with $|Aut(C)|=84$.

REPLY [7 votes]: It follows immediately from Hurwitz' theorem on the genus that a curve of genus 2 cannot have an automorphism of order 7, by considering the quotient map by this action.  Hurwitz himself proved that if 84(g-1) does not occur, the next largest possible bound is 48(g-1).  I.e. his method gives a list of possible bounds, not just one maximal bound.
Mathematische Annalen
1892, Volume 41, Issue 3, pp 403-442
This is offered as an argument for reading the original work.
see also:
Hurwitz's automorphisms theorem with deformations<|endoftext|>
TITLE: problems from the scottish book
QUESTION [10 upvotes]: Which of the problems from the Scottish Book (pdf of English version) by Stefan Banach are still open? I know that one of the problems was solved by Per Enflo for which he got a live goose from Stanislaw Mazur.

REPLY [3 votes]: The second edition of the Mauldin book has information up-to-date as of 2015.<|endoftext|>
TITLE: what prevents a manifold to be symplectic?
QUESTION [11 upvotes]: Are there any obstructions known which prevent an even dimensional orientable manifold from
being symplectic? I am a novice in this area so I unfortunately I cannot make the question more precise. What I have in mind is raw and is as follows:from the point of handle body theory you can put a symplectic structure on each handle as you are building the manifold, so there must something go wrong on the boundary of the handle which is a three manifold, for example in the case of spheres of dimension >2.  I just want to know if there is something related to this question in the literature.

REPLY [16 votes]: For a closed manifold of arbitrary dimension, there are two algebraic conditions which must be satisfied. Firstly as Russell states there must be a reduction of the structure group from $GL(2n)$ to $Sp(2n)$, or equivalently from $SO(2n)$ to $U(n)$. This is also necessary for open symplectic manifolds, and is equivalent to the existence of an almost complex structure.
Secondly, for a closed manifold, there must be a homology class $[\omega] \in H^2_{dR}$ so that the $n$-fold wedge product of $[\omega]$ generates $H^{2n}_{dR}$. This only holds for closed manifolds since there are exact symplectic structures on open manifolds.
For 4-manifolds there are further known obstruction due to Taubes, which tell us that the Gromov invariant must be equal to the Seiberg-Witten invariant, in particular the Seiberg-Witten invariant must be non-zero. This shows that the connected sum of three copies of $\mathbb{CP}^2$ admits no symplectic form, even though it satisfies the previous two conditions. As far as I know this is the only known geometric obstruction. For higher dimensional manifolds, it's my understanding that it's an open question whether the two algebraic conditions suffice.<|endoftext|>
TITLE: Metric on tangent bundle
QUESTION [8 upvotes]: Let $(M,g)$ be a compact Riemannian manifold, no boundary. 
I would like to know when is it possible to endow the tangent bundle $TM$ with an euclidean Riemannian metric. 
Do you know of a good reference?

REPLY [2 votes]: One of the ways of defining naturally a metric on the tangent bundle $TM$ is indeed (as Peter says above) a Sasaki metric defined in 1968. You can prove that $(TM, g_{\mathrm{Sas}})$ is flat if and only if $(M,g)$ is flat. The reference is for example the paper by Sigmundur Gudmundsson and Elias Kappos On the Geometry of Tangent Bundles. They give explicit calculations for Levi-Civita connection for this metric as well as curvature tensor.<|endoftext|>
TITLE: Math Annotate Platform?
QUESTION [72 upvotes]: Suppose most mathematical research papers were freely accessible online. 
Suppose a well-organized platform existed where responsible users could write comments on any paper (linking to its doi, Arxiv number, or other electronic identifier from which it could be retrieved freely), or even ``mark it up'' (pointing to similar arguments elsewhere, catch and correct mistakes, e.g.), and where you could see others' comments and mark-ups. 
Would this be, or evolve into, a useful tool for mathematical research? What features would be necessary, useful, or to-be-avoided-at-all-costs? 
This is not a rhetorical question: a committee of the National Research Council is looking into what could be built on top of a World Digital Math Library, to make it even more useful to the mathematical community than having all the materials available. This study is being funded by the Sloan Foundation. 
Input from the mathematical community would be very useful.

REPLY [3 votes]: Having read and followed numerous discussion on the subject,
it seems that possible downsides arise from having comments or discussions unwanted by the authors of the article. 
However, enabling the authors to moderate all comments themselves would seem
to address those concerns. That is, when the authors would receive the comments first and then decide whether to make them public. Of course, that would include  the possible decision by the authors not to receive any comments at all, a decision that they should also be able to reverse at any time in the future.
Basically, since the authors are the ones who wrote the paper, they should own all the corresponding rights, including any comments and moderation.<|endoftext|>
TITLE: why is it called the diamagnetic inequality?
QUESTION [5 upvotes]: How is the Diamagnetic inequality born? Why is it call this name?
Diamagnetic inequality： $\big|\nabla|u|(x)\big|\leq \big|(\nabla+iA)u(x)\big|$.

REPLY [9 votes]: Why is it called this name?

Diamagnetism is a quantum effect in which material under an applied external magnetic field generate their own "opposite" magnetic field. The most prominent example happens in superconductors. 
A consequence of this is that matter under an applied external magnetic field "gain" energy (or rather, removing magnetic fields decreases kinetic energy). See, for example, this paper. Now, the quantum mechanical description of the momentum operator is just $i\nabla$, and the associated (free) Hamiltonian is $-\triangle = (i\nabla)^2$. The mathematical description of a quantum mechanical system under an external magnetic field uses the gauge potential $A$ to capture the magnetic field, and postulates that the Hamiltonian is $H = (i\nabla + A)^2$. And hence results which bound the lower energy $i\nabla$ by the higher energy $i\nabla + A$ became known as "diamagnetic inequalities". See also this mathscinet review where this is briefly discussed. 
Note that originally inequalities of the type are (among many) called Kato's inequalities. Now they are called diamagnetic inequalities because it conveniently reminds us of the physics. 
For how to derive the inequality and some more historical and practical notes, see sections 7.19 - 7.22 in Lieb and Loss, Analysis.<|endoftext|>
TITLE: Formal power series over a henselian ring
QUESTION [5 upvotes]: Let $R$ be a henselian ring. Is $R[[x]]$ also a henselian ring?

REPLY [11 votes]: Yes, it is. 
More generally, the following result holds.

Proposition. If $R$ is henselian at the maximal ideal $\mathfrak{m}$, then $R[[x_1, \ldots, x_n]]$ is henselian at the maximal ideal lying over $\mathfrak{m}$.

A reference is the paper by N. Sankharan A Theorem on Henselian Rings, Canad. Math. Bull. 11 (1968), 275-277. See in particular Corollary 2.
Remark. The Proposition above is no longer valid if one takes the polynomial ring instead of the power series ring. For instance, if $K$ is a field than $K$ is henselian but $K[x]$ is not.<|endoftext|>
TITLE: A description of cellular boundary maps in terms of a Morse function
QUESTION [8 upvotes]: I'm writing a paper on classical Morse Theory and I'm interested in applying Morse functions to the computation of homology groups of a compact manifold $M$. The standard way in which this is done is by using Theorem 3.5 in Milnor's textbook, which proves that a Morse function $f: M \to \mathbb{R}$ induces a CW-structure on $M$, up to homotopy equivalence; one can then try to compute homology groups using cellular homology. What I have noticed though is that none of the textbooks I have used provide details regarding the attaching maps used in the construction of this CW-complex, only the number of cells of each dimension is readily available (i.e. the number of critical points with index equal to the dimension). This information is sufficient in the computation of homology groups of manifolds where the Morse function has no critical points with consecutive indices (like $\mathbb{C}P^n$), since in this case all cellular boundary maps (i.e. boundary maps in the cellular chain complex) are necessarily trivial. So here's the question:

Is there a simple description of cellular boundary maps in terms of the function $f$, which could be used to compute the (cellular, integral) homology of a compact manifold using a Morse theory? Or are there reasons to believe such a description would be difficult to find in general?

My guess is that if one does exists, quite a bit of work would be needed to obtain it, since, at least following Milnor's approach, theorem 3.5 is the result of cellular approximation and a couple of lemmas involving CW-complexes, in addition to local Morse-theoretic results regarding sublevel sets. Matsumoto proves the result (4.18) in a more direct fashion, but I still can't see an easy way to obtain what I'm looking for.
Something I noticed when trying to compute the homology of $\mathbb{R}P^n$ using Morse Theory was that, even though a there exists a Morse function $f \colon \mathbb{R}P^n \to \mathbb{R}$ which has exactly one critical point with index $k$, for $k = 0,\ldots, n$ and that critical values are in the same (ascending) order as indices, it isn't evident what the cellular boundary maps might be. My initial reason for thinking otherwise was that cells are attached to sublevel sets in increasing dimension (hence no need for cellular approximation) and by embeddings $S^{k-1} \to M_{c_k - \varepsilon}$ (the sublevel set, with $c_k$ critical value) for $k >0$. However in dimension $0$ the attaching map isn't injective, and this invalidates possible inductive attempts to describe subsequent attaching maps to the $(k-1)$-skeleton of the CW-complex which is homotopy equivalent to $M$. Incidentally, these maps are the $2$-sheeted covering projections $S^k \to \mathbb{R}P^k$, but I know this for a different reason (see Hatcher p.144). So even in this "well behaved" case I wasn't able to come up with anything significant.
I also know there is a more modern, homology-oriented approach to Morse Theory, which uses gradient flow lines to compute homology directly. However I am not familiar with any details and I'm looking mainly for ways to compute the homology of $M$ using Milnor's theorem 3.5.
Thank you for any insight!

REPLY [7 votes]: Under certain conditions, (Morse-Smale being one, but not sufficient) the  stratification by unstable manifolds of a Morse flow on a compact manifolds gives a cellular decomposition; see the paper On the Space of Trajectories of a Generic Vector Field by Burghelea & co, and the paper On Moduli Spaces and CW Structures Arising from Morse Theory On Hilbert Manifolds of Lizhen Qin.
Having a cellular decomposition, does not make reading the boundary map any easier.    However, the boundary map has a dynamical description, and the complex     with this description is usually referred to as the Floer complex of a Morse-Smale gradient flow.
If you are interested in $\mathbb{Z}$-coefficients things can be tricky due to  various sign issues. Over $\mathbb{Z}/2$ these sign issues are no longer    relevant and the computations are much easier.
For (real) Grassmanians, it is relatively easy to compute their $\mathbb{Z}/2$-Betti numbers using Morse theory.  A  particularly nice paper is Integrable Gradient Flows and Morse Theory by Dynnikov and Veselov.
To see how  a combination of Morse theory and the technques of Harvey-Lawson work,  check my paper Schubert calculus on the Grassmannian of hermitian lagrangian spaces (doi:10.1016/j.aim.2010.02.003) where I  use  Morse theoretic techniques to produce a (real) Schubert calculus on $U(n)$.   In particular, this yields a geometric description of the integral  cohomology ring of $U(n)$   Arguments similar to the ones in this paper can be used to compute the  integral homology of any grassmanian, real or complex, by Morse theoretic means. These include the projective space $\mathbb{RP}^n$.
For more on this subject I would suggest     you have a look at  Sec. 2.5 and Chaper 4 of the 2nd Edition of my book, An Invitation to Morse Theory where I go through these issues in great detail.<|endoftext|>
TITLE: Diff(M) and connectedness
QUESTION [9 upvotes]: Can anything be said about connectedness of a smooth manifold M from some property of Diff(M) in an analogous way Like C(X) has no idempotents iff X is connected.

REPLY [9 votes]: $M$ is connected if and only if the connected component of $Diff(M)$ (equivalently, of $Diff_c(M)$) acts transitively on $M$.
Edit: I just remembered, that the Lie algebra of compactly supported vector fields determines the base manifold up to diffeomorphism, see:
MR0064764 (16,331a) Reviewed 
Shanks, M. E.; Pursell, Lyle E.
The Lie algebra of a smooth manifold. 
Proc. Amer. Math. Soc. 5, (1954). 468–472. 
This is also true for larger Lie algebras, and for complex Stein manifolds, see:
MR0516602 (80g:57036) Reviewed 
Grabowski, J.
Isomorphisms and ideals of the Lie algebras of vector fields. 
Invent. Math. 50 (1978/79), no. 1, 13–33. 
Moreover, the group of compactly supported diffeomorphisms determines the base manifold completely, but I cannot find the relevant paper now.<|endoftext|>
TITLE: Lie algebra embeddings and the center of their enveloping algrabras
QUESTION [5 upvotes]: Let $\mathfrak{g}_1\subset \mathfrak{g}_2$ be a Lie algebra embedding. Assume both are semisimple. For instance take the standard diagonal embedding  $\mathfrak{sl}(2, \mathbb{C})\subset \mathfrak{sl}(3, \mathbb{C})$. This lifts to an embedding $U(\mathfrak{sl}(2, \mathbb{C}))\subset U(\mathfrak{sl}(3, \mathbb{C}))$ of the corresponding enveloping algebras. Consider their centers $Z(\mathfrak{sl}(2, \mathbb{C})),Z(\mathfrak{sl}(3, \mathbb{C}))$. It is relatively easy to see that under the previous embedding we have that $Z(\mathfrak{sl}(2, \mathbb{C}))\cap Z(\mathfrak{sl}(3, \mathbb{C})) = \mathbb{C}$. Is there any reasonable way to "connect" these centers? Let me be a bit more clear. For instance, we can pass the same question to the symmetric algebras of $\mathfrak{g}_1,\mathfrak{g}_2$. Here we have $S(\mathfrak{sl}(2, \mathbb{C})^*)^{SL(2,\mathbb{C})}$ the algebra of $SL(2,\mathbb{C})$-invariant functions on $\mathfrak{sl}(2, \mathbb{C})$ and $S(\mathfrak{sl}(3, \mathbb{C})^{\*} )^{SL(3,\mathbb{C})}$. There is a natural embedding of $S(\mathfrak{sl}(3, \mathbb{C})^{\*} )^{SL(3,\mathbb{C})}$ into $S(\mathfrak{sl}(3, \mathbb{C})^{\*} )^{SL(2,\mathbb{C})}$ and a projection of $S(\mathfrak{sl}(3, \mathbb{C})^{\*} )^{SL(2,\mathbb{C})}$ onto $S(\mathfrak{sl}(2, \mathbb{C})^{\*} )^{SL(2,\mathbb{C})}$ given by restriction. Are there any results that could describe how these centers behave under restrictions?

REPLY [4 votes]: Take a look at "Shifted Schur Functions"
Andrei Okounkov, Grigori Olshanski
http://arxiv.org/abs/q-alg/9605042
Section 10: "Coherence property of quantum
immanants and shifted Schur polynomials"
In particular formulas 10.4, 10.5 - they discuss "averaging operators"
Z(U(gl(n)) -> ZU(gl(N)) , n < N
and later prove certain "good" (coherence) property of special generators of the centers
Z(U(gl(k)) which has been studied by the authors and M. Nazarov.
Hope this helps...
What is very interesting for me personally - is try to generalize such things
to the case of loop algebras Z(U(\hat gl)).
Here certain "good" elements of the centers has been constructed by Talalaev's formula,
it is natural to expect that Okounkov-Olshanski-... story can be generalized to loop algebra case

REPLY [2 votes]: I'm doubtful about getting nice relationships between centers of universal enveloping algebras, if you look at embeddings for arbitrary pairs of semisimple Lie algebras; maybe there are subtle connections in special cases, however.   For one thing, centers and algebras of invariant polynomial functions for a given semisimple Lie algebra are related only   indirectly in the work of Chevalley and Harish-Chandra.   Moreover, embeddings into special linear algebras of unbunded rank occur when you start with a fixed semisimple Lie algebra and consider all of its faithful irreducible representations.   But the respective centers of the smaller and larger enveloping algebras are isomorphic to polynomial algebras in the number of indeterminates given by the respective ranks.   
If there is a useful way to relate centers, it should be visible in diagonal embeddings for Lie algebras both of small rank such as you describe.   But it's not immediately visible to me. 
P.S. A tag lie-algebras would be useful.<|endoftext|>
TITLE: Relating curvature and torsion of a connection to those of a curve
QUESTION [9 upvotes]: I'm currently trying to relate two descriptions of the curvature and torsion of a connection and am running into some confusion.
I know that an affine connection $A$ on an $n$-dimensional manifold $M$ can be split into two parts $A = \omega + e$, where $\omega$ takes values in the Lie algebra of rotations $\mathfrak{so}(n)$ while $e$ takes values in the Lie algebra of translations $\mathfrak{t}(n)$.
The curvature form $\Omega = d\omega + \omega \wedge \omega$ can then be obtained from the $\mathfrak{so}(n)$ part, and the torsion form $\theta = de + \omega \wedge e$ from the $\mathfrak{t}(n)$ part.
However, I am also aware that Cartan also related the curvature and torsion of an affine connection on $M$ to the older idea of curvature and torsion of curves in $M$ (I assume this is the origin of the word 'torsion' for this quantity?). If you take a curve from $M$ and develop it in a flat Euclidean space, then the curvature and torsion of the connection on $M$ both induce 'extra' curvature and torsion in the developed curve. I think that the modern version of this development would be a horizontal lift of the curve in $M$ into the principal bundle over $M$.
I'm struggling to see how these two ideas fit together. In particular, a curve in $\mathbb{R}^n$ has $n-1$ of these Frenet-Serret invariants, not just curvature and torsion.

What I would like to understand is why only the first two invariants of the curve appear in the affine connection, seeing as there are $n-1$ of these for a curve in $\mathbb{R}^n$. And what does the torsion of a curve have to do with the translation group? 

I would really appreciate any help on understanding this, or any reference suggestions.

REPLY [13 votes]: This is not really an answer, but I think it might help you look at things a little bit differently.  It is not at all clear that Cartan chose the word 'torsion' to describe the 'translation' component of the curvature because it was related to the torsion of a curve in flat space or had anything to do with developing maps associated to what are now called "Cartan connections".  In fact, I rather suspect that this is a red herring.  I think he chose the term because the geometric picture that he had in mind was connected with something physically 'twisting'.
The first article that he wrote that uses the word 'torsion' in this sense seems to be Sur une généralization de la notion de courbure de Riemann et les espaces à torsion  (C. R. Acad. Sci. 174 (1922), 593–595).  In that article he tries to give the feature of holonomy associated to his generalized 'connection' (a word he does not use in this article) that corresponds to translation a physical meaning, at least in $3$ dimensions.  He explicitly compares 'torsion' (twisting(?)) to 'tension ou pression' (tension or pressure).  For example, the first use of the word 'torsion' in the body of the article is in the sentence On a ansi une image géométrique d'un milieu matèrial continu en équilibre sous la seule action de ses forces élastiques, mais dans le cas où ces forces se manifesteraient sur chaque élément de surface, non seulement par une forçe unique (tension ou pression), mais par un couple (torsion). 
In his later articles, he generalized this notion of torsion to apply to spaces with affine, conformal, or projective connections; and he kept the word 'torsion' to describe similar features in all of them, but I don't think that there is any place in Cartan (at least, I'm not aware of one) in which he tries to relate this notion of 'torsion' specifically to the notion of 'torsion' that one encounters in the standard theory of curves in Euclidean $3$-space.
Later, when I have time, I'll try to say something more about Cartan's notion of 'generalized spaces with torsion', which seems to be somewhat poorly understood these days.  That may clarify why there isn't any apparent connection with the more classical notion of torsion of space curves.<|endoftext|>
TITLE: Is it true that every f.g. infinite simple group has exponential growth?
QUESTION [16 upvotes]: Is it true that every finitely generated infinite simple group has
exponential (word-)growth?
Remark: As Mark Sapir has pointed out, the question whether
every finitely generated group of subexponential growth is even residually finite
has been answered in the negative in
Anna Erschler. Not residually finite groups of intermediate growth,
commensurability and non-geometricity, J. Algebra 272 (2004), no. 1, 154–172,
http://www.sciencedirect.com/science/article/pii/S0021869303006410.

REPLY [10 votes]: No, there exists a finitely generated infinite simple group of intermediate 
growth. This has meanwhile been found out by Volodymyr Nekrashevych, cf.
Palindromic subshifts and simple periodic groups of intermediate growth,
arXiv, 2016.<|endoftext|>
TITLE: A nice generating set for the symmetric power of an algebra
QUESTION [5 upvotes]: I'm looking for a reference for the following fact.
Suppose $A$ is a finitely generated associative commutative unital algebra over an algebraically closed field of characteristic zero.  Let $S^n(A)$ be the $n$-th symmetric power of $A$.  That is, $S^n(A)$ is the subalgebra of $A^{\otimes n}$ fixed by the natural action of $S_n$ on this tensor product by permuting the factors.  I would like to know a reference for the fact that $S^n(A)$ is generated by elements of the form
$(a \otimes 1 \otimes \cdots \otimes 1) + (1 \otimes a \otimes 1 \otimes \cdots \otimes 1) + \cdots + (1 \otimes \cdots \otimes 1 \otimes a)$
for $a$ in $A$.
Note that if $A = \mathbb{C}[t]$, then $S^n(A)$ is the ring of symmetric polynomials in $n$ variables and then the result follows from the fact that the power sum polynomials generate this ring.
I call the above a "fact" since I've seen it used in papers (without proof -- and I've contacted the authors and they don't know of a reference) and a colleague of mine says he could write out a proof.  However, I can't help but think that there must be a reference for this in some text.  But I'm unable to find one...
Alternatively, if someone knows of a very short proof, that would be the next best thing to a reference.  I want to use this fact in a paper I'm writing and I'd like to avoid having to digress from the main thread of the exposition by including a long proof of this fact.

REPLY [4 votes]: Since your algebra is finitely generated, you only really need this result in the case when $A=\mathbb{C}[t_1,\ldots,t_n]$. 
It seems that this result first appeared in 
F. Junker, Über symmetrische Funktionen von mehreren Reihen von Veränderlichen, Math. Ann. 43 (1893), 225-270 
(see here for a scanned version of it). 
A much more modern reference (from which I know about Junker's paper) is 
J. Dalbec, Multisymmetric functions, Beiträge zur Algebra und Geometrie, Vol. 40 (1999), No. 1, pp. 27-51. 
(see here for an electronic version).
Edit: as Abdelmalek Abdesselam suggests, this result goes back to 
L. Schläfli, Über die Resultante eines Systemes mehrerer algebraischer Gleichungen Ein Beitrag zur Theorie der Elimination. (A. d. IV Bd. der Denkschriften der math.-naturw. Classe der kais. Acad. d. Wiss. bes. abgedruckt), Wien 1852
(see here for a scanned version).<|endoftext|>
TITLE: Which limits does group cohomology commute with?
QUESTION [6 upvotes]: For a discrete group G, if $M$ is a direct/inverse limit of $M_i$, is $H^i(G, M)$ the direct/inverse limit of the $H^i(G, M_i)$? Of course, cohomology commutes with finite direct sums, but how about general limits/colimits over finite categories? Thank you!

REPLY [7 votes]: Ken Brown shows in

Homological criteria for finiteness, Comment. Math. Helv. 50 (1975), 129–135, doi:10.1007/BF02565740, (free author version)

that group cohomology for a group G commutes with direct limits iff G is of type $FP_\infty$. That is the trivial module $\mathbb Z$ has a projective $\mathbb ZG$ resolution which is finitely generated in each degree.<|endoftext|>
TITLE: Maximal ideals of the algebra of measurable functions
QUESTION [17 upvotes]: It is a well-known fact that for any compact topological space $\rm S$, the set its points is in bijection with the set of maximal ideals of $\mathcal C(\rm S)$, the algebra of continuous functions $\rm S \to \mathbb R$.
Now consider the circle $\rm S^1$ with its usual $\sigma$-algebra. Note $\mathcal M$ the algebra of measurable functions $\rm S^1 \to \mathbb R$ and $\mathcal N$ the ideal of functions vanishing almost everywhere.
By Zorn's lemma, the algebra $\mathcal M / \mathcal N$ has at least one maximal ideal. However in this example, it is not possible to get a maximal ideal by considering functions vanishing at one point. 

What is an example of a maximal ideal of this algebra?

REPLY [2 votes]: If, instead of all measurable functions, you use the bounded measurable functions, then you probably want to consult the literature known as "lifting" ... for example:
A & C Ionescu Tulcea, Topics in the Theory of Lifting (Ergebnisse der Mathematik und ihrer Grenzgebiete. 2. Folge)<|endoftext|>
TITLE: Malcev Lie algebra and associated graded Lie algebra
QUESTION [7 upvotes]: Suppose $L$ is a nilpotent finite-dimensional Lie algebra over $\mathbb{Q}$ of class $c$. We can define an associated graded Lie algebra to $L$ that, as a vector space, is:
$$\bigoplus_{i=1}^c \gamma_i(L)/\gamma_{i+1}(L)$$
where $\gamma_i$ denotes the $i^{th}$ member of the lower central series, and where the Lie bracket is obtained from the Lie bracket of $L$ in a natural fashion (as follows: the Lie bracket of $L$ induces a map $\gamma_i(L) \times \gamma_j(L) \to \gamma_{i+j}(L)$ which descends to a bilinear map $\gamma_i(L)/\gamma_{i+1}(L) \times \gamma_j(L)/\gamma_{j+1}(L) \to \gamma_{i+j}(L)/\gamma_{i+j+1}(L)$. The Lie bracket in the associated graded is obtained from these constituent maps being summed up.
I believe that, in general, $L$ and its associated graded Lie algebra do not necessarily have to be isomorphic. However, I'm having a little trouble coming up with an explicit example where they're not. In particular: (i) if $L$ has class two, then it is isomorphic to its graded Lie algebra, (ii) if $L$ is a "free nilpotent" Lie algebra on a certain number of generators, then again it is isomorphic to its graded Lie algebra.
This has some relation with the Malcev Lie correspondence, but I preferred not to introduce the complication of nilpotent groups into the question. If viewed that way, the question is: for a rationally powered finitely generated nilpotent group, is the Malcev Lie ring the same as its associated graded Lie ring? Some of the discussion at Relationship between the cohomology of a group and the cohomology of its associated Lie algebra. may also be relevant.

REPLY [7 votes]: The example Misha refers to is 7-dimensional.
The smallest examples are 5-dimensional, there are exactly two of them. The first $\mathfrak{g}$ is the Lie algebra with basis $(x_1,x_2,x_3,x_4,x_5)$ with nonzero brackets $[x_1,x_3]=x_4$, $[x_1,x_4]=[x_2,x_3]=x_5$. The associated Carnot graded Lie algebra $\mathfrak{g}'$ (often called "associated graded" but this is misleading) has the same basis with nonzero brackets $[x_1,x_3]=x_4$, $[x_1,x_4]=x_5$. They are not isomorphic (because $\mathfrak{g}$ has 1-dimensional center and $\mathfrak{g}'$ has 2-dimensional center (actually it splits as a direct product product of the subalgebras with basis $(x_1,x_3,x_4,x_5)$ and $(x_2)$).
The second example $\mathfrak{h}$ has basis $(y_1,y_2,y_3,y_4,y_5)$ with nonzero brackets $[y_1,y_i]=y_{i+1}$ for $i=2,3,4$ and $[y_2,y_3]=y_5$. The associated Carnot graded Lie algebra $\mathfrak{h}'$ has the same basis with nonzero brackets $[y_1,y_i]=y_{i+1}$ for $i=2,3,4$; they are not isomorphic, for instance because $\mathfrak{h}'$ has an abelian 1-codimensional ideal unlike $\mathfrak{h}$.
In the first case, although the Betti numbers are the same, the groups are not quasi-isometric because the real cohomology algebras are not isomorphic as graded real algebras, and we can apply Sauer invariance theorem (which improves Shalom's).
In the second case, it can be checked that the real cohomology algebras are isomorphic as graded algebras, and it is not known whether the nilpotent groups are quasi-isometric, it's indeed the smallest open case.
In dimension 6 and beyond, there are plenty of examples.<|endoftext|>
TITLE: Cantor's diagonal argument and ZF
QUESTION [12 upvotes]: Cantor's diagonalization construction, on a certain view, furnishes functions 
$$d_X:{\rm Injections}(X,P(X))\rightarrow P(X)$$ 
that satisfy $\forall X\forall i\ \ d_X(i)\not\in i(X)$ 
In ZF, can one prove the existence of such functions with the added requirement
that $d_X(i)$ actually depends only on the image $i(X)$?

REPLY [6 votes]: Joel David Hamkins, Asaf Karagila and I have made some progress characterizing which sets have such a function. There is still one open case left, but Joel's conjecture holds so far.
Let $[Y]^{X}$ denote the set of all $A \subseteq Y$ such that $A \approx X$. The question is to characterize the sets $X$ for which there is a function $d:[\mathcal{P}(X)]^{X}\to\mathcal{P}(X)$ such that $d(A) \notin A$ for all $A \in [\mathcal{P}(X)]^X$. We will instead try to answer the more general question when there is such a function $d:[Y]^X\to Y$ for arbitrary sets $X, Y$ with $X \preceq Y$. (If $X \npreceq Y$ then $[Y]^X = \varnothing$ and the question is not interesting.) An obviously necessary condition is that $\newcommand{\napprox}{\not\approx}X \napprox Y$ but this is not sufficient as Ricky illustrated. Joel conjectured that such a function exists if and only if $\aleph(X) \preceq Y$, where $\aleph(X)$ is the Hartog number of $X$, the smallest ordinal that does not inject into $X$. We will show that this conjecture is true for all Dedekind infinite sets $X$ (i.e. when $\aleph_0 \preceq X$). Since the statement is obviously true when $X \prec \aleph_0$, the only remaining case is when $X$ is infinite but Dedekind finite (i.e. when $n \prec X$ for every $n \prec \aleph_0$ but $\aleph_0 \npreceq X$).
If there is an injection $f:\aleph(X)\to Y$, then there is such a $d:[Y]^X\to Y$ can be defined as $d(A) = f(\alpha_0)$ where $\alpha_0 = \min\lbrace \alpha \lt \aleph(X) : f(\alpha) \notin A\rbrace$. This last set is always nonempty otherwise composing $f:\aleph(X)\to A$ with a bijection from $A$ onto $X$ contradicts the fact that $\aleph(X) \npreceq X$.
For the converse, the hope is to define an injection $f:\aleph(X)\to Y$ by transfinite recursion where at each stage $\alpha\lt\aleph(X)$, we choose some $f(\alpha) \notin \lbrace f(\beta) : \beta \lt \alpha \rbrace$. To make these choices we would need a function $\hat{d}:[Y]^{\prec\aleph(X)}\to Y$ such that $\hat{d}(A) \notin A$ for every $$A \in [Y]^{\prec\aleph(X)} \colon= \lbrace Z \subseteq Y : Z \prec \aleph(X)\rbrace.$$ What we are given is a function $d:[Y]^X\to Y$ with $d(A) \notin A$ for every $A \in [Y]^X$. A simple idea is to fix some $A_0 \in [Y]^X$ and define $$\hat{d}(A) = d(A \cup A_0)$$ for all $A \in [Y]^{\prec\aleph(X)}$ but this only makes sense when $A \cup A_0 \approx X$. In general, we only know that $$X \preceq A \cup A_0 \preceq A + X,$$ so this strategy will work provided that $\alpha + X \approx X$ for every $\alpha \lt \aleph(X)$. This last statement holds precisely when $X$ is empty or Dedekind infinite. Indeed, $X \approx 1+X$ already implies that $X$ is Dedekind infinite and then $X \approx \alpha + X$ follows from the fact that $\alpha + \alpha \preceq \max(\aleph_0,|\alpha|)$ for every ordinal $\alpha \lt \aleph(X)$.

Asaf Karagila and I have made a little more progress on the case where $X$ and $Y$ are both Dedekind finite. In that case $Y \approx X + Z$ and the complement in $Y$ of any element of $[Y]^X$ has size exactly $Z$ and vice versa. Therefore, the existence of a $d:[Y]^X\to Y$ such that $d(A) \notin A$ for each $A \in [Y]^X$ is precisely equivalent to the existence of a choice function $c:[Y]^Z \to X$. In particular, if $Y \approx X+1$ then there is such a $d:[Y]^X \to Y$ since there clearly is a choice function $c:[Y]^1\to Y$. This does contradict the extension of Joel's conjecture to arbitrary $Y$ but not Joel's original conjecture where $Y = \mathcal{P}(X)$. Unfortunately, we still do not know what happens when $X$ and $Y = \mathcal{P}(X)$ are both infinite but Dedekind finite.
The existence of choice functions $[Y]^Z\to Y$ is a very intricate problem. For example, it is known that the existence of a choice function $[Y]^2\to Y$ is equivalent to the existence of a choice function $[Y]^4\to Y$ but that this does not imply the existence of a choice function $[Y]^3\to Y$! These intricate implications have been examined by John Conway in the article Effective implications between the "finite" choice axioms [Cambridge Summer School in Mathematical Logic, Lecture Notes in Mathematics 337 (1973), 439–458. MR0360275, doi:10.1007/BFb0066784].<|endoftext|>
TITLE: Nontrivial trivial integrals
QUESTION [9 upvotes]: I posted this question to stackexchange and after 24 hours it's got five votes and no answers, so let's see if mathoverflow can say more than that.
Consider two propositions in geometry:

Circumscribe a right circular cylinder about a sphere.  The surface area of the cylinder between any two planes orthogonal to the cylinder's axis equals the surface area of the sphere between those two planes.  (Archimedes showed this.)
Let a curve running from the south pole to the north pole on a perfectly spherical earth meet every meridian of longitude at the same angle $\gamma$.  Then the length of the curve is proportional to $\sec\gamma$.  (In particular, it's infinite of $\gamma$ is a right angle, and one could even go beyond a right angle and consider it an oriented length, so that the oriented length is negative if the curve goes from the north pole to the south pole.  In that case, the infinite length would be the $\infty$ that's at both ends of the real line rather than $+\infty$ or $-\infty$, so the length depends continuously on $\gamma$.)

Both propositions admit the same kind of proof: For the first, consider what happens when the distance between the two planes is infinitesimal; then show that if it works then, then it also works for larger-than-infinitesimal distances.  For the second, show that the length of an infinitesimal increment of the curve is $\sec\gamma$ times the latitude-component of the distance, just by drawing an infinitesimal right triangle.
You might think that in both cases one is finding an integral, but it's an integral of a constant: just the constant itself times the length of the interval over which one integrates.  So it's as if you don't need to know about integrals and you certainly don't need to find any antiderivatives, but the method of proof doesn't work for larger-than-infinitesimal quantities.
One would like to say things like "energy is force times distance; it is only when the force is not constant that one must use integrals and multiply the force at an instant by the infinitesimal distance and then integrate".  But if proving the force is constant itself requires an infinitesimal viewpoint, then despite the triviality of computing the integral by multiplying two numbers, it seems necessary to the logic of the argument to view it as an integral.
So:

What other instances besides these two are known?  (I suspect there are many in geometry.)  
What else of interest can be said about this phenomenon?

REPLY [2 votes]: In a static universe with homogeneous density of stars, the solid angle covered by the stars of a shell with radius $r$ and thickness $dr$ is $C r^{-2} r^2 \, dr $. The solid angle covered by all stars of the sphere with radius $r$ is $Cr$. (Olber's paradox)<|endoftext|>
TITLE: When can we make a digraph acyclic by fliping groups of arcs?
QUESTION [5 upvotes]: We have a digraph D=(V,A) and its arc set A is partitioned into classes. We can flip the classes, which means changing the direction of all the arcs in the class.
Is there any result on when can we make the digraph acyclic with this operation?
Even such special case would be interesting for me such as:

the underlying graph is planar, and
each classes form a path of same length in the underlying graph 
and each of these undirected paths (the classes) has alternating orientation.

Thanks for the answers.

REPLY [3 votes]: A teacher of mine (Zoltán Király) proved it to be NP-complete by reducing the 3-SAT problem to this problem: for each clause there should be an indirected circle with 4 edges. The classes and the starting orientation are:

one edge from each circle as a referential edge directed clockwise.
the other 3 edges of each circle represent the variables in the clause, the edges representing the same literals form one class. The orientation of the edges in a circuit representing literals of the related clause are clockwise (the same as the referential edge) if the literal is the variable, and counterclockwise if it's the negated variable.

It's easy to see that finding an acyclic orientation by fliping the classes representing the literals (we can suppose that the oriention of the referential edges is fixed) is the same as to satisfy the 3-SAT instance.<|endoftext|>
TITLE: Distribution of distances in permutations
QUESTION [8 upvotes]: Let's consider all possible permutations of N numbers. Suppose for each permutation we calculate the sum of absolute differences between consecutive elements. Thus, for (1,2,3) one would have abs(1-2)+abs(2-3)=2. Is it possible to obtain a distribution of such sums for given N? For instance, for N=3 one would have 3!=6 permutations and possible sums as 2 and 3. The number of sums of 2 is 2 and for 3 it's equal 4.

REPLY [4 votes]: Andrew King pointed out that $E[X]$ is $(n^2-1)/3$. 
So, to calculate the variance $E[X^2] - E[X]^2$ we need to find $E[X^2]$.
Let $\delta(i) = |\pi(i+1)-\pi(i)|$, so $X = \sum_{i=1}^{N-1} \delta(i)$.
$$\begin{eqnarray}X^2 &=& \bigg(\sum_{i=1}^{N-1} \delta(i)\bigg)^2 \newline &=& \sum_{i=1}^{N-1} \delta(i)^2 + 2 \sum_{i=1}^{N-2} \delta(i)\delta(i+1) + \sum_{|i-j|\gt 1} \delta(i)\delta(j) \end{eqnarray}$$
We can compute the expected values of each of these terms to find $E[X^2]$. By symmetry, we only need to find $(N-1)E[\delta(1)^2]$, $2(N-2)E[\delta(1)\delta(2)]$, and $((N-2)(N-3))E[\delta(1)\delta(3)].$
$$\begin{eqnarray}E[\delta(1)^2] &=& \frac{1}{N(N-1)} ~\sum_{a,b} (b-a)^2  \newline &=& \frac{1}{N(N-1)} \frac{N^4-N^2}{6} \newline &=&\frac{n^2+n}{6}\end{eqnarray}$$
I used Mathematica for the sums.
$$\begin{eqnarray} E[\delta(1)\delta(2)] &=& \frac{1}{N(N-1)(N-2)} \sum_{a,b,c~ \text{distinct}} |b-a||c-b| \newline &=& \frac{1}{N(N-1)(N-2)}\bigg(\sum_{a,b,c} |b-a||c-b| - \sum_{a,b,c | a=c} (b-a)^2 \bigg) \newline &=& \frac{1}{N(N-1)(N-2)} \bigg( \frac{7N^5 - 15N^3 - 8N}{60} - \frac{N^4-N^2}{6} \bigg)\newline &=& \frac{7N^2+11N+4}{60}\end{eqnarray}$$
The diagonal terms evaluate to $0$ or resemble the first sum.
$$\begin{eqnarray} E[\delta(1)\delta(3)] &=& \frac{1}{N(N-1)(N-2)(N-3)} \sum_{a,b,c,d ~ \text{distinct}} |b-a||d-c|\newline &=& \frac{1}{N(N-1)(N-2)(N-3)} \bigg( \sum_{a,b,c,d} |b-a||d-c| \newline & & - 4 \sum_{a,b,c,d | b=d} |b-a||c-b|  + 2 \sum_{a,b,c,d|a=c,b=d} (b-a)^2 \bigg) \newline &=& \frac{1}{N(N-1)(N-2)(N-3)} \bigg( \frac{N^6-2N^4 + N^2}{9} \newline & &  - 4\frac{7N^5-10N^4-15N^3+10N^2+8N}{60}+ 2 \frac{N^4-N^2}{6} \bigg) \newline &=& \frac{5N^5 - 21 N^4 + 35N^3+45N^2-40N -24}{45(N-1)(N-2)(N-3)} \newline &=& \frac{5N^4-16N^3+19N^2+64N+24}{45(N-2)(N-3)}\end{eqnarray}$$
If I calculated correctly, then 
$$\begin{eqnarray}E[X^2] &= &\frac{N^3-N}{6} + \frac{7N^3-3N^2-18N-8}{30} \newline & & + \frac{5N^4-16N^3+19N^2+64N+24}{45} \newline &=& \frac{10N^4+4N^3+29N^2+59N+24}{90}\end{eqnarray}$$
$$\begin{eqnarray}\text{Var}[X] &=& E[X^2] -E[X]^2 \newline &=& \frac{4N^3+49N^2+59N+14}{90} \end{eqnarray}$$
For $N=100$, the standard deviation is $\sqrt\frac{249773}{5}=223.505$.
I believe a similar (but simpler) technique can be used to approximate higher moments of $X$. The dominant term is where no differences are adjacent, and the moments should be approximately the same as for random functions to $\lbrace 1, ..., n \rbrace$ instead of permutations. I think the method of moments would allow one to prove that the distribution is asymptotically normal.<|endoftext|>
TITLE: many expected streaks imply high probability for a streak
QUESTION [5 upvotes]: In CLRS, "Intro to Algorithms" section 5.4.3 the following is shown.  If a fair coin is flipped n times, the expected number of streaks of consecutive heads of length (1/2)log(n) is $ \Theta (\sqrt n)$.  So, for large n, "we expect there are a large number of streaks of length (1/2)log(n)."  
Then the book continues, "Therefore, one streak of such a length is likely to occur."
Is that a correct conclusion?  To conclude that there is high probability for such a streak isn't it neccessary to bound the variance besides for showing a large expectation?
Thanks.

REPLY [2 votes]: The broken logic in the book can bite in many places.  Here is a simple example. If we make a random graph with $n$ vertices and edge probability $3/n$, the expected number of hamiltonian cycles goes to infinity exponentially fast, yet the probability of having any hamiltonian cycle at all goes to 0 exponentially fast.<|endoftext|>
TITLE: A linear order obtained by forcing with P(omega)/fin
QUESTION [9 upvotes]: In their paper "A barren extension", Henle, Mathias, and Woodin considered the forcing whose conditions are the sets of natural numbers ordered by inclusion up to finite error.
If we force over $L(\mathbb{R})$ and assume that this is a model of determinacy, they show that this will add a free ultrafilter U on the natural numbers, and moreover in the 
extension there are no new (ordinal-indexed) sequences of elements of the ground model. 
Let us order the functions $f\colon\omega\rightarrow\omega$ in $L(\mathbb{R})[U]$ by domination mod $U$. We obtain a linear order $(L,\leq)$. 
(1) Is there a strictly increasing $\omega_1$-sequence in $(L,\leq)$? 
(2) Is there a strictly increasing $\omega_1$-sequence of gaps in $(L,\leq)$?

REPLY [8 votes]: The answer to the first question is no, at least if one is willing to assume large cardinals in $V$. I'd have to research what the optimal hypothesis is. It is shown in the two Neeman-Zapletal papers that if there are infinitely many Woodin cardinals below a measurable cardinal and above some cardinal $\kappa$, then forcing with a c.c.c. partial order of cardinality less than $\kappa$ cannot change the theory of $L(\mathbb{R})$ with ordinal parameters. Since the cardinal invariant $\mathfrak{t}$ (the tower number) can be made arbitrarily large by c.c.c. forcing, this shows from the same hypothesis that the intersection of any wellordered collection of dense open subsets of $\mathcal{P}(\omega)/\mathrm{Fin}$ in $L(\mathbb{R})$ is dense open. For this specific problem, we need to intersect only $\omega_{1}$ many dense sets. Since c.c.c. forcings don't change $\omega_{1}$, we can do with the ostensibly weaker hypothesis that c.c.c. forcing doesn't change the theory of $L(\mathbb{R})$ (not allowing for ordinal parameters), which gives that intersections of $\omega_{1}$ many dense open subsets in $\mathcal{P}(\omega)/\mathrm{Fin}$ are dense open. I don't know if this can be proved just assuming $\mathrm{AD}$ + $V = L(\mathbb{R})$, or even whether one can deduce it from the Henle-Mathias-Woodin result. 
Suppose now that $\langle \tau_{\alpha} : \alpha < \omega_{1} \rangle$ is a sequence of $\mathcal{P}(\omega)/\mathrm{Fin}$ names for nonempty sets of reals. By the assumptions of the previous paragraph, there is an infinite $a \subseteq \omega$ such that for each $\alpha < \omega_{1}$, the set $B_{\alpha}$, consisting of those reals forced by $a$ to be in $\tau_{\alpha}$, is nonempty. The sequence $\langle B_{\alpha} : \alpha < \omega_{1} \rangle$ is in $L(\mathbb{R})$. If $a$ forces each $\tau_{\alpha}$ to be a mod-$U$ equivalence class in the Baire space, then the members of each $B_{\alpha}$ must agree mod-finite on $a$. So, restricting to $a$, we get an $\omega_{1}$-sequence of disjount countable subsets of the Baire space, which contradicts the conjunction of the Perfect Set Property and the Baire Property. So there are no $\omega_{1}$-sequences of distinct mod-$U$ equivalence classes in $L(\mathbb{R})[U]$, regardless of the relationship between the classes. 
A similar argument gives the Perfect Set Property in $L(\mathbb{R})[U]$, which was originally proved by Di Prisco and Todorcevic. It seems that the same approach should give a negative answer to the second question, but I don't see it.<|endoftext|>
TITLE: Question about unusual highest weight modules for $U_q(sl(2))$
QUESTION [6 upvotes]: Background
Let $U_q(sl(2))$ be the quantum group associated with $sl(2)$ i.e. the associative algebra with 1 over $Q(q)$ generated by $x^+,x^-,K,K^{-1}$ with relations 
$$KK^{-1}=K^{-1}K=1$$
$$Kx^+K^{-1}=q^2x^+,Kx^-K^{-1}=q^{-2}x^-$$
$$x^+x^{-}-x^{-}x^+=\frac{K-K^{-1}}{q-q^{-1}}$$
Here $q$ is indefinite, in particular not a root of unity.  A $U_q(sl(2))$-module $V$ has highest weight $\omega \in Q(q)$ if there exists a vector $v \in V$ such that
$$ U_q(sl(2))\cdot v=V $$
$$ x^+\cdot v=0 $$
$$ K\cdot v=\omega v$$
Most standard texts (e.g. Chari Pressley, etc.) focus mainly on the case where $\omega$ equals $q$ to some power (i.e. $K$ acts as $q^m$ for some integer $m$ or $-q^m$ though this case is equivalent to the first by tensoring with a 1-dimensional module).  If $\omega$ is not of this form then the module in question is necessarily infinite dimensional.
Question
Has anyone considered what happens when $K$ acts by something other than a power of $q$, say $2q-1$?  I am especially interested in what happens in the $q=1$ specialization of these modules.  Any references would be appreciated.

REPLY [4 votes]: When talking about modules for such a quantized enveloping algebra, you have to be explicit about whether the module is finite dimensional or not.   (For affine or other infinite dimensional Lie algebras of interest, the parallel question is whether modules are integrable or not.)    In Jantzen's introductory AMS text Lectures on Quantum Groups, for instance, he starts out with analogues of the traditional finite dimensional representations.   Here the associated highest weights are (up to sign) given by powers of $q$ in the rank one case.   The infinite dimensional modules with dominant integral highest weights then have reasonable properties as well.   
It's possible in this special case to look at infinite dimensional modules with arbitrary highest weights, including simple modules, but typically these are far less understood.  There is a lot of literature by now on quantum groups, but I'll mention one earlier paper by Lusztig:   Quantum deformations of certain simple modules over enveloping algebras, Adv. in Math. 70 (1988).   In Section 2 he defines very general highest weight modules, but then specializes to the integrable ones where more can be said.   
I'm not at all sure where the general case leads.   Keep in mind that even in the classical situation, infinite dimensional simple modules with arbitrary highest weights pose difficult problems.<|endoftext|>
TITLE: Results from Differential Cohomology
QUESTION [17 upvotes]: I've been working through some notes on differential cohomology for the past few months. I feel like I have a pretty decent grasp on the concepts and its construction, at least for differential extensions of ordinary cohomology, like Deligne cohomology. 
My friend recently asked me about some of the uses of differential cohomology and I was unable to provide a good answer. What are some results of differential cohomology? Or some theorems that have been proven using these new tools from differential cohomology? I'd also be happy with some computations computed using differential cohomology. This is a general question not specific to Deligne cohomology, so answers pertaining to any differential cohomology theory (like differential K-theory) are very much encouraged.
Thanks!

REPLY [7 votes]: The paper
Chern, Shiing Shen; Simons, James
Characteristic forms and geometric invariants. 
Ann. of Math. (2) 99 (1974), 48–69
seems to be seminal in the subject, and gives a flavour of how differential cohomology came about. It suggests that Chern and Simons discovered the theory "by accident", when looking for a combinatorial formula for the first Pontrjagin number of $4$-manifolds.
They soon realised that their discovery allowed certain topological invariants to be promoted to geometric invariants. 
A very concrete application given in Section 5 is the use of differential Pontrjagin  classes to show that $SO(3)\approx\mathbb{R}P^3$ (with the usual metric of constant curvature) does not admit a conformal immersion in $\mathbb{R}^4$.<|endoftext|>
TITLE: Is there an Enriques–Kodaira-like classification of Fano threefolds?
QUESTION [7 upvotes]: I am mainly interested in varieties over an algebraic closed field $k$ (or $\mathbb{C}$). The classification of complex surface is established in the last century and known as Enriques–Kodaira classification. In higher dimension, people realized that mild singularities must be taken into account to build a reasonable theory and Mori and others studied Minimal model program, which is the birational classification of algebraic varieties. However, I am more interested in a concrete description of classification, like the Enriques–Kodaira classification (maybe except for general ones). If I understand correctly, Fano threefolds are pretty well-studied in a concrete manner by Mukai and others. Here is my question:

Is there an Enriques–Kodaira-like classification of Fano threefolds? Maybe over $\mathbb{C}$?

REPLY [8 votes]: Yes, Fano threefolds have been completely classified and one has explicit projective models of them. See V.A. Iskovskih, Yu. G. Prokhorov: Algebraic Geometry V: Fano varieties. Encyclopaedia of Math. Sciences 47, Springer-Verlag, Berlin 1999. See also Andreas Ott's thesis on Fano threefolds of picard number $\rho\ge 2$.
In some sense the most complicated Fanos are the ones with Picard number one, i.e., $Pic(X)\simeq \mathbb Z$. Here there are 18 families of such threefolds and they have been classified by Iskovskih. The most basic invariant here is the $index$ of $X$, i.e., the maximal integer $r$ such that $K_X$ is divisible by $r$ in $Pic(X)$. Fano threefolds of higher Picard number was classified by Mori and Mukai. This was one of the early triumphs of Mori theory.<|endoftext|>
TITLE: Box removing operators on partitions
QUESTION [11 upvotes]: Consider the box removing operators $d_i$ for $i\geq 1$ defined as follows.
Let $\lambda$ be a partitions. Then $d_i(\lambda)$ is the partition obtained by removing a box from a part of length $i$ in $\lambda$ so as to obtain another partition. (Here removing a box is an operation performed on the Ferrers diagram of $\lambda$.)
If there is no part of length $i$ in $\lambda$, then $d_i(\lambda)=0$.
(As pointed out by Darij in the comments below, $0$ is not to be confused with the empty partition $\varnothing$.)
It is easy to see that the following equalities are satisfied.
$$
d_id_j=d_jd_i \text{ if } \lvert i-j \rvert\geq 2$$
$$d_id_{i+1}d_i=d_id_id_{i+1}$$
$$d_id_{i+1}d_{i+1}=d_{i+1}d_id_{i+1}$$ 
Now consider words in the $d_i$. Two words $w_1$ and $w_2$ are called equivalent if $w_1(\lambda)=w_2(\lambda)$ for all partitions $\lambda$.
Is it true that if $w_1, w_2$ are equivalent then one can be obtained from the other by using the 3 relations mentioned above?

REPLY [4 votes]: These are the Coxeter relations, and I believe equivalence comes from a general theorem: Tit's Theorem. For example, if you are willing to accept the Edelemann-Greene correspondence between Young Tableau and reduced words of permutations, then it would be equivalent to show that one can move freely between two reduced words using the above relations (now correspondingly, $d_i$ act on permutations via adjacent transpositions: $d_i(i,i+1)=(i+1,i)$. You can find a proof of reduced word equivalence here, page 4, Theorem 1.1.2. Essentially, the proof for tableaux should be of the same flavor: show that you can pass from any word to a carefully chosen canonical word.<|endoftext|>
TITLE: Classifying Globally Generated Holomorphic Line Bundles over a Flag Manifold
QUESTION [6 upvotes]: I was recently looking back at an old question of mine, where I asked about the classification of the line bundles over a general complex flag manifold. Pavel Etingof gave the following excellent answer:

The partial flag manifold X which you mentioned is the set of flags $0 ⊂ V_1 ⊂ \ldots ⊂ V_m=ℂ^n$, such that $\dim(V_j/V_{j−1})=k_j$. So we have vector bundles $V_j$ on $X$ and line bundles $L_j$ on $X$ which are the top exterior powers of the bundles $V_j$, $j = 1,...,m−1$. Any line bundle on $X$ is a tensor product of powers of these line bundles, and the powers are uniquely determined. This is why line bundles on $M$ are labeled by a set of $m−1$ integers.

I then began to wonder how this relates to a more recent question of mine regarding the definition of a globally generated holomorphic vector bundle. So my question is: With respect to the above line bundle classification, and the obvious choice   of holomorphic structure for these bundles (I am assuming that there is just one obvious choice here), which of these bundles are globally generated?
For the simplest case of complex projective space, we have of course that the line bundles are classified by the integers, and the positive ones are globally generated (or the negative, depending on convention). Now for the Grassmannians  the line bundles are again classified by the integers, and I would again guess that the globally generated ones are exactly those of positive charge. So how does this extend?

REPLY [7 votes]: If I am not mistaken, the criterion is: such a line bundle is globally generated if and only if the space of its holomorphic sections is non-zero. To wit, both the flag variety and these bundles are homogeneous w.r.t. the natural action of $GL_n(\mathbb C)$. Thus if a section of such a bundle $L$ does not vanish at a point $x$, then the section $g^*s$ does not vanish at the point $y=g^{-1}x$, where $g\in GL_n(\mathbb C)$; the point $y$ may be made arbitrary.
The criterion for the existence of a non-zero section may be found in Chapter 1 of the book http://www.amazon.com/Complex-Geometry-Grundlehren-mathematischen-Wissenschaften/dp/3540613781<|endoftext|>
TITLE: Reference for my monads?
QUESTION [10 upvotes]: I'm looking for a reference for a certain pair of monads on $Cat$. One problem is that I don't know the modern way of thinking about some basic things, so excuse me if my presentation is naive.
First some notation. Let $C$ and $D$ be small categories and let $F\colon C\to Cat\;$ and $G\colon D\to Cat\;$ be functors. Let $(F\Uparrow G)$ denote the following category. $$Ob(F\Uparrow G)=\{(c,d,\ell)\;|\;c\in Ob(C), d\in Ob(D), \ell\colon F(c)\to G(d)\}$$ and whose morphisms are ``natural transformation squares", i.e. $$Hom_{(F\Uparrow G)}((c,d,\ell),(c',d',\ell')=\{(f,g,\alpha)\;|\;f\colon c\to c', g\colon d\to d', \alpha\colon G(g)\circ\ell\to\ell'\circ F(f)\}.$$
With the same setup, let $(F\Downarrow G)\;$ denote the category with the same objects but slightly different morphisms, the only difference being that the natural transformation point the other way:
$$Ob(F\Downarrow G)=\{(c,d,\ell)\;|\;c\in Ob(C), d\in Ob(D), \ell\colon F(c)\to G(d)\}$$
$$Hom_{(F\Downarrow G)}((c,d,\ell),(c',d',\ell')=\{(f,g,\beta)\;|\;f\colon c\to c', g\colon d\to d', \beta\colon\ell'\circ F(f)\to G(g)\circ\ell\}.$$
Let $D=\{\star\}$, and denote a functor $G\colon D\to Cat\;$ by $\{G\}$. Now let $C=FCat$, some skeleton of the category of finite categories. Then we have a functors $$(FCat\Uparrow\{-\})\colon Cat\to Cat \;\;\;\;\text{ and }\;\;\;\;(FCat\Downarrow\{-\})\colon Cat\to Cat.$$
I think that each is the functor part of some kind of 2-monad on $Cat$. The unit is "constant" and the multiplication is "Grothendieck construction". 
Proving that this is associative, etc, looks laborious, and I don't want to reinvent notation, etc. Is there a good reference for these monads, if they really are monads?
Thanks.

REPLY [9 votes]: What you are describing is an example of Max Kelly's notion of club, closely connected with the concept of operad. The original references date back to the 70's; one reference is 

G.M.Kelly. On clubs and doctrines. In Category Seminar, Sydney 1972/1973. Springer LNM 420, pp. 181-257 (1974). 

Actually, a club is defined to be a monoid in a monoidal category $M$ whose objects are pairs $(C, F: C \to Cat)$ where $C$ is a small category, and where a morphism $(C, F) \to (D, G)$ consists of a functor $H: C \to D$ and a natural transformation $G H \to F$; the monoidal product is a kind of wreath product. There is an "actegory" structure (an action of the monoidal category $M$) on $Cat$, 
$$\wr: M \times Cat \to Cat,$$ 
so that each club = monoid in $M$ induces a monad on $Cat$. You can also find a succinct description of this material here. 

As a reality check, here is a more direct description of the (underlying functor of the) monad on $Cat$, associated with the club structure on the inclusion $i: \mathrm{FCat} \hookrightarrow Cat$, which can be extracted by applying the construction given in Borisov's paper, pp. 3-4. The monad takes a category $D$ to a category which I will denote $\mathrm{FCat} \wr D$ (Borisov uses a semi-direct product symbol). The objects of $\mathrm{FCat} \wr C$ are pairs $(C, l: i(C) \to D)$ where $C \in \mathrm{FCat}$ is a finite category and $l: i(C) \to D$ is a functor. Morphisms $(C, l) \to (C', l')$ of $\mathrm{FCat} \wr D$ are pairs $(F: C \to C', \phi: l \to l' \circ i(F))$ where $\phi$ is a natural transformation. 
Unless I have misunderstood something, this gives one of the monads described in the OP. I think the other monad is obtained by a process of dualization (apply $(-)^{op}$, then apply the first monad, then apply $(-)^{op}$ again), so we really only need to worry about the first. As I said, all the laborious technical details were covered long ago in the "Australian school". 
Other familiar examples of clubs are where we take the inclusion $i: Set \to Cat$, mapping each set $S$ to the discrete category on $S$; the corresponding monad is the free coproduct completion monad. Another is the composite inclusion $\mathbb{P} \hookrightarrow Set \hookrightarrow Cat$ of the groupoid of finite permutations; here the corresponding monad is the free symmetric (strict) monoidal category construction.<|endoftext|>
TITLE: Pullbacks and pushouts in the category of (small) dg-categories
QUESTION [6 upvotes]: It is a well-known result (by G. Tabuada, as I recall) that the category $\mathbf{dgcat}_k$ of (small) dg-categories over a fixed commutative ring $k$ admits a model category structure such that the weak equivalences are the quasi-equivalences. Then, necessarily, $\mathbf{dgcat}_k$ is small complete and cocomplete, in particular it admits pullbacks and pushouts. However, I could not find any description of them. Actually, I don't even know what should be the product (or coproduct) of two dg-categories, nor could I find any kind of reference.
Do you have any clue?

REPLY [3 votes]: Small dg-categories are categories enriched over $\mathcal{V} = \textbf{Ch}(k)$, the category of chain complexes over a commutative ring $k$.  
It is a general result that if $\mathcal{V}$ is (co)complete, then so is the category $\mathcal{V}\textbf{-Cat}$ of small $\mathcal{V}$-enriched categories. 
See the text around Corollaries 7.3.6 and 7.3.7 of Christina Vasilakopoulou's thesis for a sketch of the proof and other references of this fact.
This paper by Harvey Wolff proves that $\mathcal{V}\textbf{-Cat}$ is cocomplete by first showing that $\mathcal{V}\textbf{-Graph}$ is cocomplete ($\mathcal{V}$-graphs are just $\mathcal{V}$-categories without identities or composition). The construction for coproducts is easy enough, but things get quite involved when constructing coequalizers even for just $\mathcal{V}$-graphs.
In the remainder of this answer, I'll just show how to construct coproducts of $\mathcal{V}$-categories.
Let $\mathcal{C}_i, i \in I$ be a collection of $\mathcal{V}$-categories for some indexing set $I$. Form a $\mathcal{V}$-category $\mathcal{C}$ in the following manner:

$\text{Ob}(\mathcal{C}) := \coprod_{i \in I} \text{Ob}(\mathcal{C}_i)$
For $x,y \in \text{Ob}(\mathcal{C})$ where $x \in \text{Ob}(\mathcal{C}_i)$ and $y \in \text{Ob}(\mathcal{C}_j)$, define $\mathcal{C}(x,y)$ to be  $\mathcal{C}_i(x,y)$ if $i = j$ and $0$ otherwise, where $0$ is the initial object of $\mathcal{V}$.
For $x,y,z \in \text{Ob}(\mathcal{C})$, we need to produce composition maps $$\mu_{x,y,z}\colon \mathcal{C}(x,y) \otimes \mathcal{C}(y,z) \to \mathcal{C}(x,z).$$ If $x,y,z$ are all from the same $\mathcal{C}_i$, then just use the composition maps from  $\mathcal{C}_i$. Otherwise, at least one of $\mathcal{C}(x,y)$ and $\mathcal{y,z}$ is $0$. Since $0 \otimes X = X \otimes 0 = 0$ for any $X \in \mathcal{V}$, the domain of $\mu_{x,y,z}$ is $0$, so take $\mu_{x,y,z}$ to be the unique map $0 \to \mathcal{C}(x,z)$.

It's easy to check that composition as defined above is associative and unital.<|endoftext|>
TITLE: Compact surface with genus$\geq 2$ with Killing field
QUESTION [6 upvotes]: Let M be a compact Riemannian surface of genus$\geq 2$.
Can M have a globally defined Killing field ?
Can M have a Killing field defined on M-(finite set of points)?

REPLY [5 votes]: Of course, there's always the Killing field $X\equiv0$. :)
Seriously, here's a different proof and an argument that addresses the 'suppose one leaves out a finite number of points' question:
Taking the orientation double cover if necessary, we can assume that $M$ is connected and orientable with $g\ge 2$.  The hypotheses imply that the Euler characteristic of $M$ is negative.  However, if $X$ were a nonzero Killing field on $M$, then it would have isolated zeros of index $+1$, and, since the sum of the indices of the zeros of $X$ equals the Euler characteristic, we have a contradiction.
As for omitting points, any Killing vector field $X$ defined on a punctured disk must extend smoothly across the puncture (and remain Killing). To see this, remember that $X$ is also a conformal vector field and hence is the real part of a holomorphic vector field $Z$ on the punctured disk (once one fixes the orientation and hence the underlying complex structure).  However, $X = \mathrm{Re}(Z)$ must remain bounded in size near the puncture (because the Killing equation is an overdetermined linear system of finite type with coefficients that are smooth across the puncture), so the usual removable singularities theorem in complex analysis tells us that $Z$, and hence $X$, extends smoothly across the puncture.  Since $X$ is Killing everywhere except at the puncture, it must be Killing there, too.<|endoftext|>
TITLE: Vector chromatic number and Lovasz theta
QUESTION [6 upvotes]: For $\alpha \ge 2$, an $\alpha$-vector coloring of a graph $X$ is an assignment of unit vectors to the vertices of $X$ such that vectors assigned to adjacent vertices have inner product less than or equal to $\frac{1}{1-\alpha}$. A strict $\alpha$-vector coloring requires equality for the inner product.
The vector chromatic number of a graph $X$, denoted $\chi_{vec}(X)$, is defined to be the minimum $\alpha$ for which $X$ has an $\alpha$-vector coloring. Strict vector chromatic number is defined analogously. It is known that the strict vector chromatic number of $X$ is equal to the Lovasz theta function of the complement of $X$.
Are there any finite graphs known for which the vector and strict vector chromatic numbers are different?
For reference, vector and strict vector colorings and the relation to Lovasz theta come from "Approximate Graph Coloring by Semidefinite Programming" by Karger, Motwani, and Sudan.

REPLY [4 votes]: Apparently, Schrijver defined a function $\vartheta'$ in the paper A Comparison of the Delsart and Lovasz Bounds which (I am pretty sure) is equivalent to vector chromatic number of the complement. Furthermore, he gives an example of a vertex transitive graph $X$ which has $\vartheta'(X) < \vartheta(X)$. The graph $X$ has the $01$-strings of length 6 as its vertices, and two such strings are adjacent if they are at Hamming distance at most 3. Taking complements obviously provides an example answering my question in the affirmative.
Considering the graph $Y = \overline{X}$, two strings are adjacent if they are at Hamming distance 4 or more. We can then replace 01-strings with $\pm 1$ vectors in the obvious way, and then normalize these vectors. Now, adjacency will correspond to two vectors having inner product less than or equal to $-1/3$. This then implies that $\chi_{vec}(Y) \le 4$, which happens to be the clique number of $Y$, and thus $\chi_{vec}(Y) = 4$ (this value is given in Schrijver's paper, but it is not discussed how he comes to this value). According to Schrijver, $\bar{\vartheta}(Y) = 16/3 > 4$. Obviously, the way $Y$ was defined gave rise to a "natural" vector coloring, and I would guess that other similarly defined graphs might be examples of graphs with vector chromatic number less than strict vector chromatic number.
Interestingly, Schrijver's paper was written in 1979, about 15 years before Karger, Motwani, and Sudan defined vector chromatic number.<|endoftext|>
TITLE: Distribution of maximum of random walk conditioned to stay positive
QUESTION [17 upvotes]: I have an $n$ step random walk which starts at zero $X_0 = 0 = S_0$ where the steps $X_i$ are independent uniform random variates in $[-1,1]$, but the walk is conditioned on the hypothesis that it stays in the right half-line $[0,\infty)$ (that is, $S_k = X_0 + X_1 + \dots + X_k \geq 0$ for $k \in 0, \dots, n$). 
I'm interested in the maximum of the walk, or the random variable $M_n = \max_{0 \leq i \leq n} S_i$.  

What is the distribution of $M_n$ (as a function of $n$)?
Can one derive (upper or lower) bounds on the probability $P[M_n < a]$ that the walk is actually contained in the interval $[0,a]$ instead of $[0,\infty)$? Both types of bounds would be interesting, for different reasons, in a polymer physics question.

I've looked at the classical references (eg. Feller, or Spitzer, Chapter 4), but Spitzer at least seems to deal only with the integer valued case. However, it seems that I'm either missing it, or I can't quite find the right reference. 
Are these standard results with a standard reference? It's clearly related to A random walk with uniformly distributed steps, which deals with the overall volume of the space of walks with uniform random variates as steps conditioned to stay positive, but not the maximum.

REPLY [2 votes]: 1) I believe it is possible to obtain $P(M_n<a\ | \tau>n)$, where $\tau:=\inf\{n\ge 1: S_n\le 0\}$ in more or less explicit way only for the case of the simple random walk. These computations can be found in Billingsley, Convergence of Probability measures, Chapter 2.11. For large $n$ these computations become universal as you can apply FCLT, see below.
2) Generally, one can use Brownian motion approximations as was suggested by fedja. Mathematic justification is given in the references below.  Before going into this I will say a few words about conditioning of random walks to stay positive. 
There are two ways to condition random walk to be positive. First way is to condition on $\{\tau>n\}$, where $\tau:=\inf\{n\ge 1: S_n\le 0\}$. Second way, is to ''condition'' on $\{\tau=\infty\}$. Here you cannot use direct conditioning as $\mathbf P(\tau=\infty)=0$ (for the zero-mean finite variance random walk). So, the way to proceed is to use Doob's $h$ transform. As a result one obtains a Markov chain (Lamperti Markov chain, in fact) which is a discrete-time analogue of Bessel process.  
Now I can move on to approximations as $n\to\infty$. 
a) When $a$ is of order $\sqrt n$, you need to use corresponding Functional Central Limit Theorem (FCLT). You need functional central limit theorem as the maximum depends on the whole trajectory of the random walk. 
The way to use functional limit theorems is described in the above Billinsley's book, you can see an example for unconditioned random walk in Chapter 2.11. 
For the first type of conditioning FCLT  is proved in Iglehart, 1974, Functional Central Limit Theorems for Random Walks Conditioned to Stay Positive (doi:10.1214/aop/1176996607), see also Bolthausen, 1976, On a Functional Central Limit Theorem for Random Walks Conditioned to Stay Positive. Conditioning of the second type was considered in Bertoin and Doney (1994), On Conditioning a Random Walk to Stay Nonnegative (doi:10.1214/aop/1176988497). FCLT for the second type of conditioning can be found in Caravenna, Chaumont (2008), Invariance principles for random walks conditioned to stay positive (doi:10.1214/07-AIHP119). Finally, in higher dimensions you can use FCLT from http://arxiv.org/abs/1110.1254. 
b) When $a$ is much larger then $\sqrt n,$ then $\mathbf P(M_n<a\ |\ \tau>n)\to 1$. 
c) When $a<<\sqrt n$ you are dealing with small deviations probability. 
If you write $\mathbf P(M_n<a | \tau>n)=\mathbf P(M_n<a , \tau>n)/\mathbf P(\tau>n)$, then approximation for $\mathbf P(\tau>n)\sim Cn^{-1/2}, n\to\infty$, and 
$$
\mathbf P(\tau>n, M_n<a)=\mathbf P(\mbox{random walk $S_k$ stays in (0,n) for all } 0<k\le n)
$$
Now, if $a\sim An^{1/2-\delta}$ for $\delta\in(0,1/2)$ you can split this trajectory in $n/a^2$ parts of length $a^2$ and use independence to obtain estimate 
 $$
\mathbf P(\tau>n, M_n<a)\le \left(\mathbf P(\mbox{random walk $S_k$ stays in (0,2a) for all } 0<k\le a^2)\right)^{n/a^2}
$$
Now, the probability inside the parenthesis is on the right scale and you can apply the FCLT to get 
$$
\mathbf P(\mbox{random walk $S_k$ stays in (0,2a) for all } 0<k\le a^2)\to C(A)\in(0,1).
$$
This results in the following upper bound
$$
\mathbf P(\tau>n, M_n<a) \le C(A)^{n/a^2}\approx C(A)^{n^{2\delta}}.
$$
As $C(A)$ is in $(0,1)$ this probability is quite small. .<|endoftext|>
TITLE: Non-tame 3-manifolds covered by the Euclidean space
QUESTION [8 upvotes]: An open 3-manifold is tame if it is homeomorphic to the interior of a compact manifold. Is there a (known) example of an open 3-manifold that is not tame, has finitely generated fundamental group and universal cover homeomorphic to $\mathbb R^3$?

REPLY [6 votes]: I believe the original such examples (with fundamental group $\mathbb{Z}$) were due to Scott and Tucker.<|endoftext|>
TITLE: Quotients with unconditional bases
QUESTION [5 upvotes]: Gowers' dichotomy theorem asserts that every Banach space either contains an HI subspace or a subspace having an unconditional basis. There are examples of HI spaces without quotients having unconditional bases (was Argyros the first who proved that?). This strange phenomenon tempts me to ask 
whether every reflexive non-HI space has a quotient with an unconditional basis? 
My apologies if this is well-known.

REPLY [6 votes]: The space of Gowers and Maurey is reflexive and H.I. , and as proved by Ferenczi every of its quotients and its dual is also  H.I. (as  mentioned in the answers of Bill Johnson (giving the reference) and Kevin Beanland). The results of Argyros and Felouzis are answering to the question.
Moreover the remark  of Bill Johnson is to the point. S. Argyros and myself  provided a reflexive and unconditionally saturated Banach space $X$  with an H.I. dual $X^*$. Thus every quotient of   $X$ is indecomposable. Additionally we show that every quotient of $X$ has a  further quotient which is H.I.   
S. Argyros,  A. Tolias,  Indecomposability and unconditionality in duality. Geom. Funct. Anal. 14 (2004), no. 2, 247–282.<|endoftext|>
TITLE: Stabilizers for nilpotent adjoint orbits of semisimple groups
QUESTION [5 upvotes]: Let $G$ be a connected, simply-connected, complex, semisimple Lie group with Lie algebra $\frak{g}$. Suppose that $X\in\frak{g}$ is a nilpotent element (i.e. that $ad_X:\frak{g}\rightarrow\frak{g}$ is a nilpotent endomorphism), and let $C_G(X)\subseteq G$ denote its stabilizer with respect to the adjoint representation of $G$. What is known about the topology of $C_G(X)$ (ex. topological invariants)? I would appreciate any and all references/suggestions, particularly those concerning the stabilizers of irregular nilpotent elements of $\frak{g}$.
Thanks!

REPLY [3 votes]: To supplement what Francois Ziegler says, I'd point out that the structure of semisimple complex Lie groups has been developed piecemeal over a century or so.  The basic results on nilpotent elements in the Lie algebra are by now fairly old, but refinements continue to be made.   The "topology" involved in the group centralizers seems to be limited mainly to the study of the component groups: groups of connected components in the centralizer group of a given nilpotent element.   These groups have a topological interpretation as fundamental groups, summarized in Chapter 6 of the cited book by Collingwood–McGovern.   (Unfortunately, that slim 1993 book has become unaffordably expensive and doesn't exist in all math libraries.)
It should be kept in mind that what Collingwood–McGovern do is mostly not original, so there are older sources in the literature.   Moreover, most of the ideas here generalize nicely to semisimple algebraic groups over an algebraically closed field of arbitrary "good" characteristic (sometimes avoiding 2, 3, 5).   Here the basic results on centralizers and component groups, though not so explicitly topological, are much the same.   Detailed accounts and tables are found in the 1985 book Finite groups of Lie type on finite simple groups by Roger Carter, while a recent AMS monograph by Liebeck and Seitz, Unipotent and nilpotent classes in simple algebraic groups and Lie algebras, goes into considerable detail about the structure of centralizers in all characteristics.
The older ideas involving nilpotent orbits or unipotent classes have continued to be explored in research papers, often in connection with representation theory and algebraic geometry.  See for example the rethinking of component groups for unipotent elements (in good characteristic) by McNinch and Sommers in Component groups of unipotent centralizers in good characteristic (Journal of Algebra 260 (2003) 323-337, doi:10.1016/S0021-8693(02)00661-0).
However you approach the nilpotent elements in the Lie algebra, case-by-case work is inevitable: the centralizers aren't as a rule reductive, but on the other hand the number of orbits is always finite.   Subregular orbits have been especially well studied, in connection with Dynkin curves and the like in algebraic geometry.<|endoftext|>
TITLE: Topologically embedding curves in Jacobian
QUESTION [11 upvotes]: Let $X$ be an irreducible smooth projective algebraic curve over $\mathbb{C}$. Then, the Abel-Jacobi map gives embedding $X \hookrightarrow Jac(C)$ of $C$ into it's Jacobian. This map induces an isomorphism on $H_1$.
Question: Can it be made obvious, using purely topological reasoning, that a connected, oriented compact 2-dimensional topological manifold of genus $g$ embeds into a torus $\phi: C \rightarrow (S^1)^{2g}$, so that $\phi$ induces an isomorphism on $H_1$?
I don't know how to see this without using the Abel-Jacobi theorem.

REPLY [17 votes]: Yes. We will come up with $g$ maps from the Riemann surface to the two-dimensional torus. We can then take the product of these maps. As long as one of them is nonconstant at each point, it will be an embedding.
View the Riemann surface in the standard way as a handlebody that looks like a bunch of tori glued together. 

(Presumably the classification of Riemann surfaces, which we use to put it in this form, counts as purely topological reasoning!) There are $g$ tori. Pick one, and contract all but that one to a point. If that is one of the tori at the ends, then all the rest of the surface gets contracted to a single point. If it's in the middle, the left half gets contracted to a point on the left side of the torus and the right side gets contracted to a point on the right side of the torus.
This set of $g$ maps gives us the associated kind of map. Clearly, this gives an isomorphism on homology. To make sure this is an embedding, we have to make sure that no point on the torus gets contracted in every single map. We will do this by defining the maps a little generously, such that the stuff that is contracted for the leftmost torus ends a little bit to the right of where the stuff contracted for the second-leftmost torus begins.
Hopefully this explanation makes sense! My picture-drawing skills are not up to the task of realizing this.

REPLY [8 votes]: I'll do things a little more abstractly than Will.
Of course, things are trivial in genus $1$.  To simplify things a bit, I'll restrict myself to a surface $\Sigma_g$ of genus $g \geq 3$ (the case $g=2$ can be handled with a bit more care).  Both the surface $\Sigma_g$ and the torus $(S^1)^{2g}$ are Eilenberg-MacLane spaces.  Recall that (modulo basepoint issues) homotopy classes of maps between Eilenberg-MacLane spaces are in bijection with homomorphisms between their fundamental groups.  Fixing an identification of $\pi_1((S^1)^{2g})$ with $H_1(\Sigma_g;\mathbb{Z})$, there thus exists a unique homotopy class of continuous maps $\phi : \Sigma_g \rightarrow (S^1)^{2g}$ inducing the surjection $\pi_1(\Sigma_g) \rightarrow H_1(\Sigma_g;\mathbb{Z})$; since the target is abelian, there is no need to worry about basepoints.  Since the dimension of $(S^1)^{2g}$ is greater than $5=2 \cdot 2 + 1$, the usual Whitney argument shows that we can homotope $\phi$ a little bit to assure that $\phi$ is an embedding.<|endoftext|>
TITLE: What is quantum Brownian motion?
QUESTION [22 upvotes]: It seems that the current state of quantum Brownian motion is ill-defined. The best survey I can find is this one by László Erdös, but the closest the quantum Brownian motion comes to appearing is in this conjecture (p. 30):

[Quantum Brownian Motion Conjecture]: For small [disorder] $\lambda$ and [dimension] $d \ge 3$, the location of the electron is governed by a heat equation in a vague sense: $$\partial_t \big|\psi_t(x)\big|^2 \sim \Delta_x \big|\psi_t(x)\big|^2 \quad \Rightarrow \quad \langle \, x^2 \, \rangle_t \sim t, \quad t \gg 1.$$ 
  The precise formulation of the first statement requires a scaling limit. The second
  statement about the diffusive mean square displacement is mathematically precise, but
  what really stands behind it is a diffusive equation that on large scales mimics the
  Schrödinger evolution. Moreover, the dynamics of the quantum particle converges to
  the Brownian motion as a process as well; this means that the joint distribution of
  the quantum densities $\big|\psi_t(x)\big|^2$ at different times $t_1 < t_2 < \dots < t_n$  converges to the corresponding finite dimensional marginals of the Wiener process.

This is the Anderson model in $\mathbb R^d$ with disordered Hamiltonian $H = -\Delta + \lambda V$. The potential $V$ is disordered, and is generated by i.i.d. random fields; the parameter $\lambda$ controls the scale of the disorder.

Classical Brownian motion admits many characterizations and generalizations. For example, Wiener measure leads to the construction of an abstract Wiener space, which is the appropriate setting for the powerful Mallivin calculus. The structure theorem of Gaussian measures says that all Gaussian measures are abstract Wiener measures in this way. I would love to know what all this theory looks like in the language of non-commutative probability theory.
The QBM Conjecture states roughly that a quantum particle in a weakly disordered environment should behave like a quantum Brownian motion. This is an important open problem, but it doesn't quite capture what a QBM is, nor what different types of QBM may exist. Thus my question:

What kind of precise mathematical object is a quantum Brownian motion?

REPLY [3 votes]: (In words explained below:) Quantum Brownian motion (QBM) is a class of possible dynamics for an open, quantum, continuous degree of freedom in which the reduced dynamics are specified by a quadratic Hamiltonian and linear Lindblad operators in the phase-space variables $x$ and $p$.
Consider the arbitrary time-evolution of a system's density matrix when it is in contact with an environment:
\begin{align}
\rho = \rho_{\mathcal{S}}(t) = \mathrm{Tr}_{\mathcal{E}} [U_t \sigma^0_{\mathcal{SE}} U_t^\dagger],
\end{align}
where $\sigma^0_{\mathcal{SE}}$ is the initial global state (both $\mathcal{S}$ and $\mathcal{E}$) and $U_t$ is the unitary governing the global evolution.  Then the system is said to evolve according to a special case of quantum Brownian motion — a QBM quantum dynamical semigroup — when the evolution of its density matrix obeys a Lindblad master equation
\begin{align}
\partial_t \rho = -i [\hat{H},\rho] + \sum_i \left(V_i \rho V_i^\dagger - \frac{1}{2} \{V_i^\dagger V_i, \rho\} \right),
\end{align}
generated by a time-independent Hamiltonian that is a quadratic polynomial in $x$ and $p$
\begin{align}
\hat{H} = \frac{1}{2m}\hat{p}^2 + \frac{\mu}{2} \{\hat{x},\hat{p}\} + \frac{m\omega^2}{2} \hat{x}^2,
\end{align}
with $\mu$, $m$, and $\omega^2$ real, and by time-independent Lindblad operators that are linear polynomials in the same
\begin{align}
V_i = a_i \hat{p} + b_i \hat{x},  \qquad (i=1,2)
\end{align}
with $a_i$ and $b_i$ complex. The master equation can be re-written as
\begin{align}
\partial_t \rho = -i &[\hat{H},\rho] + i (\lambda/2) [\hat{p},\{\hat{x},\rho\}] - i (\lambda/2) [\hat{x},\{\hat{p},\rho\}] \\
&- D_{pp}[\hat{x},[\hat{x},\rho]] - D_{xx}[\hat{p},[\hat{p},\rho]] + D_{xp}[\hat{p},[\hat{x},\rho]] + D_{px}[\hat{x},[\hat{p},\rho]]
\end{align}
with coefficients
\begin{align}
D_{xx} &= \frac{\vert a_1 \vert^2 + \vert a_2 \vert^2}{2} \quad , \quad &  D_{pp} &= \frac{\vert b_1 \vert^2 + \vert b_2 \vert^2}{2},\\
D_{xp} &= D_{px} = -\mathrm{Re} \frac{a_1^* b_1 + a_2^* b_2}{2} \quad , \quad &  \lambda &= \mathrm{Im} (a_1^* b_1 + a_2^* b_2),
\end{align}
More generally, we say a system undergoes quantum Brownian motion when it evolves according to the above master equation, regardless of whether it forms a quantum dynamical semigroup.  If it obeys the master equation with time-independent coefficients then the QBM is time-homogeneous (in the sense of a Markov process); otherwise it is time-inhomogeneous.  The class of all possible instantaneous QBM dynamics is parameterized by $\mu$, $m$, $\omega^2$, $a_i$, and $b_i$.
The resulting dynamics take a particularly beautiful form in the Wigner representation.  The above master equation for $\rho$ is equivalent to the following dynamical equation for the Wigner function $W(x,p)$:
\begin{align}
\partial_t W = -\frac{p}{m}\partial_x W + m\omega^2 & x \partial_p W + (\lambda - \mu)\partial_x (x W) + (\lambda + \mu)\partial_p (p W)\\
&+D_{pp} \partial^2_x W + D_{xx} \partial^2_p + (D_{xp}+D_{px}) \partial_x \partial_p W.
\end{align}
More compactly:
\begin{align}
\partial_t W (\alpha) &= \left[ F_{ab} \partial_a \alpha_b + D_{ab} \partial_a \partial_b \right] W(\alpha)
\end{align}
where 
\begin{align}
F_{ab} = \left( \begin{array}{cc} \lambda - \mu & -1/m \\ m \omega^2 & \lambda+\mu \end{array} \right) \quad, \quad D_{ab} = \left( \begin{array}{cc} D_{xx} & D_{xp} \\ D_{px} & D_{pp} \end{array} \right) 
\end{align}
are matrices with real elements. Above, the phase-space indices $a,b$ take the values $x,p$, with Einstein summation assumed, so that $\alpha_a$ is a vector in phase space.  (The directional derivative $\partial_a$ is just shorthand for $\partial_{\alpha_a}$.)
This is identical in form to a Klein-Kramers equation (more generally a Fokker-Planck-type equation) for the phase-space probability distribution of a classical point particle undergoing Brownian motion. 
This is remarkable because such equations were originally derived for true probability distribution, but they also apply to the Wigner function.  As a bonus, this gives us an immediate and simple physical interpretation for each of the terms in the QBM master equation
The best comprehensive modern statement of the above definition is probably

Isar et al. "Open Quantum Systems"

which includes comparisions to important special cases discussed by other authors.  Here are some more references I found useful in compiling the above: 

Alicki and Lendi, arXiv:quant-ph/0205188.
Alicki, Quantum Dynamical Semigroups and Applications.
Lindblad, G. (1976). "On the generators of quantum dynamical semigroups". Commun. Math. Phys. 48 (2) 119.
Breuer and Petruccione, The Theory Open Quantum Systems.
C. Caves, Completely positive maps, positive maps, and the Lindblad form.
G. Lindblad, "Brownian motion of a quantum harmonic oscillator".
H. Dekker, "Quantization of the linearly damped harmonic oscillator"
A. Sandulescu and H. Scutaru, "Open quantum systems and the damping of collective modes in deep inelastic collisions".
A.O. Caldeira and A.J. Leggett, "Path integral approach to quantum Brownian motion".
B. Hu, J. Paz, and Y. Zhang, "Quantum Brownian motion in a general environment: Exact master equation with nonlocal dissipation and colored noise".
J. Halliwell and T. Yu, "Alternative derivation of the Hu-Paz-Zhang master equation of quantum Brownian motion".
W Zurek, "Decoherence, einselection, and the quantum origins of the classical".<|endoftext|>
TITLE: Two questions on complex geometry
QUESTION [5 upvotes]: I have two questions on complex geometry.
First one is that why the existence of almost complex structure on tangent bundle on real 2n-dimensional manifold is a topological question?
Wikipedia describes it as a topological question. I think that mean there is some homology or cohomology group associated to topological property whose vanishing or nonvanishing pertains to the existence.
Would you explain why it should be a topological question intuitively and may I suggest exact topological invariant which captures the existence property?
Second, in Huybrechts's book 'Complex Geometry',
when it comes to Euler exact sequence on $P^n$, he mentioned that there is natural inclusion map such that $O(-1)\rightarrow\oplus_{j=0}^{n}O$.
I can hardly come up with any idea on what this inclusion map is.
If possible, would you let me know the exact expression for this map?
(here, $O(-1)$ is the tautological line bundle sheaf on $P^n$ and $O$ is the holomorphic sheaf of the trivial line bundle) 
Hoping to get some shedding light in your reply.

REPLY [3 votes]: If $(x_0:x_1:\dots:x_n)$ are the homogeneous coordinates on $P^n$ then the map $O(-1) \to O^{n+1}$ is given by $s \mapsto (sx_0,sx_1,\dots,sx_n)$.<|endoftext|>
TITLE: Are constructible derived categories invariant up to weak homotopy equivalence?
QUESTION [6 upvotes]: Let $X$ and $Y$ be two topological spaces and $R$ be commutative ring.  Let $D_c^b(X, R)$ and $D_c^b(Y,R)$ be their respective bounded derived categories of constructible sheaves of $R$-modules.  I have three questions:

If $f:X\rightarrow Y$ is a weak homotopy equivalence does it induce (under any of the standard cohomological operations) an equivalence between $D_c^b(X, R)$ and $D_c^b(Y,R)$?
If 1. is not true in general, is there some interesting subcategory of topological spaces where it does become true? Perhaps CW-complexes?
If 1. is not true does it become so if we restrict to the full subcategories of $D_c^b(X, R)$ and $D_c^b(Y,R)$ whose objects have local system cohomology?

I suppose what I'm really asking is whether it is possible to naturally associate a bounded derived categories of constructible sheaves of $R$-modules to a homotopy type?

REPLY [6 votes]: I don't know much about these things but I think 1. fails even if $X$ is a point and $Y$ is a line. Let's say also $R$ is a field. Then the category of constructible sheaves on $X$ is just $R$-Vect which is in particular semisimple, but the category of constructible sheaves on $Y$ is not semisimple: if we choose a sheaf on a point $p$ and on the complement $U$ of that point, then there are in general many non-isomorphic choices of a sheaf on $Y$ with given restrictions to $p$ and $U$ which fit in a short exact sequence. 
So their derived categories are not equivalent either.

Addendum: Maybe I can say something about how the usual homotopy invariance of cohomology is visible in the 'six functor' formalism. For any space $X$ let $\pi_X$ be the projection to a point, then the cohomology of $X$ (with any coefficients) just corresponds to functor $R \pi_{X\ast} \circ \pi_X^\ast$. If $f \colon X \to Y$ is an arbitrary map, then note that $\pi_Y \circ f = \pi_X$ which implies 
$$ R\pi_{Y\ast} \circ Rf_\ast \circ f^\ast \circ \pi_Y^\ast = R \pi_{X\ast} \circ \pi_X^\ast;$$
and now the morphism $\mathbf 1 \to Rf_\ast \circ f^\ast$ coming from the adjunction gives a map $f^\#  \colon R \pi_{Y\ast} \circ \pi_Y^\ast \to R \pi_{X\ast} \circ \pi_X^\ast$. Of course evaluating this equation on a choice of coefficients we get the usual map $H^\bullet(Y) \to H^\bullet(X)$. The correct way to express homotopy invariance is now that if $f$ and $g$ are homotopic maps $X \to Y$, then $f^\#$ and $g^\#$ are equal. This can be generalized to the relative situation, when $X$ and $Y$ are spaces over some base space $S$ and we consider the derived pushforward to $S$ instead of to a point, and we take homotopies over $S$.<|endoftext|>
TITLE: What is the intuition behind the inertia orbifold (or stack)?
QUESTION [11 upvotes]: I am studying orbifolds with view towards Chen-Ruan cohomology. I have been struggling with inertia orbifolds but have no intuition about them at this point. I would appreciate your motivating me by answering the following questions:

What is the intuition behind inertia orbifolds? How should one think of them? 
Why are inertia orbifolds important? What are the useful for? 

Of course, given an orbifold, inertia orbifold comes for free. So it is indeed a natural mathematical object to think about. I hope to know some down-to-earth answers.
Thank you for sharing your ideas with me.

REPLY [22 votes]: I am not sure what kind of answer you are looking for. But if you have a stack $X$, then the inertia stack $IX$ is basically the gadget parametrizing pairs $(x,\sigma)$ where $x$ is a point of $X$ and $\sigma$ is in the isotropy group at $x$ (an automorphism of $x$). Informally, the locus where you have automorphism group $G$ becomes "doubled" $|G|$ times. Example: for the stack $[\mathbb A^1/\mu_2]$ the inertia stack is $[\mathbb A^1/\mu_2] \sqcup B\mu_2$, the extra $B\mu_2$ corresponding to the origin being doubled because it has an extra automorphism. 
One way to think about it is as a kind of "infinitesimal loop space", where instead of taking maps to $X$ from a circle we take maps from the homotopically equivalent object $B\mathbb Z$. This is pleasant because the inertia stack is the fibered product $X \times_{X\times X} X$, and for a topological space $X$ the homotopy fibered product $X \times_{X\times X}^h X$ is the space of free loops on $X$. 
You can motivate the inertia stack through Gromov--Witten theory. If $X$ is a variety, then there is an evaluation map from the stack of stable $n$-pointed maps to $X$ to $X^n$. If $X$ is a stack, then the correct notion is that of a twisted stable map, and in this case the evaluation maps do not land on $X$ but in its inertia stack $IX$! (In fact it lands on the rigidified inertia stack, where some automorphisms have been removed from the picture, but nevermind this). So quantum cohomology of a stack is not extra structure on the cohomology ring of $X$ itself, but on the cohomology ring of $IX$.  

Since it was discussed in some now deleted comments, let me say a few words about the example $X = BG$. Then $X$ has a single point (point = $\mathbb C$-point), corresponding to the trivial torsor over a point. What are the automorphisms of the trivial torsor? For every $g \in G$ we have an automorphism given by the action of $G$. But these will not correspond to distinct points of the inertia stack $IX$ in general, because these automorphisms may be isomorphic. So we should figure out what is a morphism between automorphisms of the trivial torsor. By thinking a bit one sees that morphisms are given by conjugation in the group, i.e. a morphism between the automorphisms "acting by $g$" and "acting by $g'$" is an element $h$ such that $g = hg'h^{-1}$. In other words, $IX$ is given by the stack quotient $[G/G]$, where the first $G$ is the underlying set of $G$, and the group $G$ acts on itself by conjugation. Equivalently, $IX = \coprod_{[g]} BC_G(g)$ where the disjoint union is taken over conjugacy classes in $G$ and $C_G(g)$ is the centralizer of $g$. Using unnecessarily fancy words, $IX$ is the classifying stack of the "loop groupoid" of the finite group $G$. This illustrates the informal statement earlier, that the locus with automorphism group $G$ becomes "doubled" $|G|$ times: the stack $[G/G]$ is of course $|G|$ times larger than $[pt/G]$, in a natural sense.<|endoftext|>
TITLE: A simple (?) combinatorial question
QUESTION [6 upvotes]: Does anybody know a proof (or a reference to proof) of the following combinatorial question: Let $a,b,c,\ldots=1,\ldots,n$ enumerate elements of a set of size $n$, e.g. vertices of an $n-1$ symplex. Let $C_{ab}$ be a cocycle, i.e. an object satisfying $C_{ab}=-C_{ba}$ and $$C_{ab}+C_{bc}+C_{ca}=0.$$ 
This can be interpreted as the sum over edges of a triangle $abc$, where the weight that each edge is assigned is the cocycle. Then the following identity holds: $$\sum_a (-1)^{a+1} \prod_{b<c,\, b,c\not=a} C_{bc}=0.$$
This can be interpreted as a sum over $n-1$ simplices of an $n$ simplex, with what is summed over being the product of the cocycles for all edges of the corresponding $n-1$ symplex (with signs). Note that for $n=3$ the above identity is just the cocycle condition itself. 
This identity was checked (with Mathematica) for some low $n$'s, so I am reasonably confident it holds for a general $n$. It is also so simple that it must be known (if true). This is why I am hoping that somebody will point out a reference. Yet another question is: Is there any cohomological interpretation of this identity?

REPLY [13 votes]: Choose an arbitrary number $D_n$ and define  $D_i = C_{in}+D_n$ for $i \neq n$. So $C_{ij} = D_i - D_j$.
The identity we want to prove is
$$\sum_{a=1}^n (-1)^{a+1} \prod_{\begin{matrix} 1 \leq b < c \leq n \\ b,c \neq a\end{matrix}} (D_b-D_c) = 0.$$
The product is the Vandermonde determinant $\det (D_j^k)_{ 1 \leq j \leq n,\ j \neq a}^{0 \leq k \leq n-2}$. So the sum is the row expansion of 
$$
\det \begin{pmatrix}
1 & 1 & 1 & \cdots & 1 \\
1 & 1 & 1 & \cdots & 1 \\
D_1 & D_2 & D_3 & \cdots & D_n \\
D_1^2 & D_2^2 & D_3^2 & \cdots & D_n^2 \\
\vdots \\
D_1^{n-2} & D_2^{n-2} & D_3^{n-2} & \cdots & D_n^{n-2} \\
\end{pmatrix}.$$
Since this matrix has a duplicate row, its determinant is $0$.<|endoftext|>
TITLE: What manifolds are boundaries of euclidian spaces ?
QUESTION [20 upvotes]: I would like to know if there are compact (n-1)-manifolds $N$ that are not spheres but such that there is a manifold with boundary $M$ which satisfies the following two properties:

$\partial M\cong N$
$M-\partial M\cong \mathbb{R}^n$ 

I am primarily interested in this question in the category of smooth manifolds but I would be interested to know the answer in the topological case as well.

REPLY [15 votes]: The answer is in fact given in the question Misha links to in his comment. For completeness, I wanted to give the details in a comment, but that became too long, so I turned it into this answer. The idea is the same as in Ryan Budney's answer, but the arguments are a bit more detailed and direct. I do apologize for the repetition. Please upvote Misha's comment, the answer he links to, and Ryan Budney's answer.$\newcommand{\RR}{\mathbb{R}}
\newcommand{\int}{\operatorname{int}}$
Edit: I have slightly refined the argument below so that it goes through when the interior of the smooth manifold $M$ is only homeomorphic to $\RR^{n+1}$, not necessarily diffeomorphic. By the way, this weaker assumption makes no difference when $n\neq 3$, as the uniqueness (up to diffeomorphism) of differentiable structures on $\RR^{n+1}$ implies that the interior of $M$ is then actually diffeomorphic to $\RR^{n+1}$. So the easier, more elementary argument below assuming the interior of $M$ is diffeomorphic to $\RR^{n+1}$ is enough when $n\neq 3$.
The final answer is: if a compact smooth manifold $M$ has interior homeomorphic to $\RR^{n+1}$, then its boundary $\partial M$ is diffeomorphic to $S^n$, as long as $n>4$. In dimension $n=4$, we get only a homeomorphism between the boundary and $S^n$ by using Freedman's h-cobordism theorem instead of the smooth h-cobordism theorem in the argument below. We still get a diffeomorphism in dimension $n=3$, by the recently proved original Poincaré conjecture.
The argument goes as follows. It is simpler when the interior is actually diffeomorphic to $\RR^{n+1}$, so let us assume that at first. Let $M$ be a compact smooth manifold with interior diffeomorphic to $\RR^{n+1}$. Then removing the interior of the closed unit disc $D$ from $\RR^{n+1}$ (which we identify with the interior of $M$), we get a smooth cobordism $N=M\setminus(\int D)$ between $\partial M$ and $\partial D=S^n$ (the boundary of the removed disc).
On the one hand, the inclusion of $\partial D=S^n$ into $N$ is a homotopy equivalence, as $N$ is actually homotopy equivalent to $N\setminus \partial M=\RR^{n+1}\setminus(\int D)$ (since $\partial M$ has a collaring in $M$ and thus in $N$).
On the other hand, the inclusion of $\partial M$ into $N$ is also a homotopy equivalence. Again, note that $\partial M$ has a collar in $N$, and the complement of this collar is compact in the interior of $M$, i.e. in $\RR^{n+1}$. So by pushing off to infinity radially within $\RR^{n+1}\setminus(\int D)$ (via a suitable homotopy with bounded support starting at the identity of $\RR^{n+1}\setminus(\int D)$), we obtain a deformation of $N$ into the collar of $\partial M$ which fixes $\partial M$ pointwise. Then the collar of $\partial M$ deformation retracts onto $\partial M$, and these two deformations together give a deformation retraction of $N$ onto $\partial M$.
To modify the above argument when the interior of $M$ is only homeomorphic to $\RR^{n+1}$ via a homeomorphism $\varphi:\int M \to \RR^{n+1}$, make the following changes:

$N=M\setminus(\int D)$ is now defined to be $M$ minus the interior of a closed disc $D$ smoothly embedded in the interior of $M$. So $N$ is still a smooth cobordism between $\partial M$ and $\partial D$.
Importantly, note that the image $\varphi(\partial D)$ need not be a standard sphere in $\RR^{n+1}$. However, $\varphi(\partial D)$ is locally flat in $\RR^{n+1}$ (since $\partial D$ is locally flat in $\int M$), so the Shoenflies theorem implies that it is taken to a standard sphere in $\RR^{n+1}$ via some homeomorphism of $\RR^{n+1}$. By modifying $\varphi$, we can then assume that $\varphi(\partial D)$ is the standard sphere $S^n\subset \RR^{n+1}$.
Since $\varphi(\partial D)=S^n$ is now the standard sphere in $\RR^{n+1}$, we can apply, essentially unchanged, the above two arguments showing that the inclusions of $\partial M$ and $\partial D$ into $N$ are deformation retracts.

In conclusion, $N$ is a smooth h-cobordism between $\partial M$ and $\partial D$, where $\partial D$ is diffeomorphic to the standard sphere $S^n$. Since $S^n$ is simply connected, the h-cobordism theorem implies that $\partial M$ is diffeomorphic to $S^n$ when $n>4$.
As stated earlier, in dimension $n=4$, we only get a homeomorphism between $\partial M$ and $S^n$ by Freedman's h-cobordism theorem. In dimension $n=3$ we can use the Poincaré conjecture in dimension 3 (recently proved) to conclude that $\partial M$ is diffeomorphic to $S^3$, since those manifolds are homotopy equivalent to $N$ and thus to each other. In dimensions $n<3$, the classification of closed $2$-manifolds and $1$-manifolds shows we get a diffeomorphism again.
Finally, note that the elementary argument above (before the modification requiring Schoenflies theorem) showing that $N$ is a h-cobordism between $\partial M$ and $\partial D$ holds equally well for a compact topological manifold $M$ whose interior is homeomorphic to $\RR^{n+1}$. Then $\partial M\simeq N\simeq \partial D\cong S^n$. Since the Poincaré conjecture holds topologically in all dimensions, we conclude that the boundary of $M$ is homeomorphic to $S^n$.<|endoftext|>
TITLE: Are all models of ZF + DC + "All set of reals are lebesgue measurable" also models of CH?
QUESTION [5 upvotes]: Possible Duplicate:
Lebesgue Measurability and Weak CH 

I have studied a little set theory and I found that Solovay constructed a model of ZF+DC+"All set of reals are Lebesgue measurable" and I found that the same model satisfies the "perfect set property" which implies that every uncountable subset of $\mathbb{R}$ is bijectable with $\mathbb{R}$. In other words, the Continuum Hypothesis (CH) is true in that model. I also found that the Axiom of Determinacy (AD) implies that every set of reals is Lebesgue measurable and it also implies the perfect set property.
Therefore, I have a question: Does ZF+DC+"All set of reals are Lebesgue measurable" always imply CH?

REPLY [5 votes]: The question seems to be an open research problem; it was posed in 2011 on MO (and has remained unanswered), see: 
Lebesgue Measurability and Weak CH<|endoftext|>
TITLE: Is there a connection between the closed forms of these two infinite products?
QUESTION [6 upvotes]: Take the following two infinite products that have closed forms. 
Assume: $\gamma_n > 0 \in \mathbb{R};s,a,x \in \mathbb{C}; x \ne 0,a \pm ix\gamma_n \ne 0$
The first product:
$$\displaystyle H_{int}(s,a,x) := \prod_{n=1}^\infty \left(1- \frac{s}{a + i x \gamma_n} \right) \left(1- \frac{s}{{a - i x \gamma_n}} \right) = \frac{\xi_{int}(0 -\frac{a}{x}+\frac{s}{x})}{\xi_{int}(0-\frac{a}{x})}$$
has $\gamma_n = n$, so runs through the integers with: $\xi_{int}(s) = \frac{\sinh(\pi s)}{s}$.  
The second product, for which the closed form can be derived assuming RH is true,
$$\displaystyle H_{rie}(s,a,x) := \prod_{n=1}^\infty \left(1- \frac{s}{a + i x \gamma_n} \right) \left(1- \frac{s}{{a - i x \gamma_n}} \right) = \frac{\xi_{rie}(\frac12 - \frac{a}{x}+\frac{s}{x})}{\xi_{rie}(\frac12 - \frac{a}{x})}$$
has $\gamma_n = \Im(\rho_n)$, so runs through the non-trivial zeros $\rho_n$ with: $\xi_{rie}(s)= \frac12 s(s-1) \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2}) \zeta(s)$.
The products clearly have a comparable structure and share the following characteristics:
1) They are both reflexive: $H(s,a,x) = H(2a-s,a,x)$ (since this is fully independent of $\gamma_n$).
2) Their $\xi(s)$-functions are reflexive as well:
$$\xi_{int}(s)=\xi_{int}(2(0)-s)$$
$$\xi_{rie}(s)=\xi_{rie}(2(\frac12)-s)$$
3) They both have a similar 'base' Hadamard product (for $a=0$ and $a=\frac12$ respectively):
$$\displaystyle \xi_{int}(s) = \xi_{int}(0) \prod_{n=1}^\infty \left(1- \frac{s}{0+i n} \right) \left(1- \frac{s}{{0- i n}} \right) \rightarrow \xi_{int}(0)=\pi$$
$$\displaystyle \xi_{rie}(s) = \xi_{rie}(0) \prod_{n=1}^\infty \left(1- \frac{s}{\frac12 + i \Im(\rho_n)} \right) \left(1- \frac{s}{{\frac12 - i \Im(\rho_n)}} \right) \rightarrow \xi_{rie}(0)=\frac12$$
4) They are both entire functions and the "undesired" zeros/poles from their meromorphic components $\sinh(s)$ and $\zeta(s)$ are annihilated via $s$ and $\frac12 s(s-1), \Gamma(\frac{s}{2})$ respectively. Both meromorphic elements can be expressed as an infinite product as well as an infinite sum (over a certain domain):
$$\sinh(s) = s \prod_{k=1}^\infty \left(1+ \frac{s^2}{k^2\pi^2)} \right)=\sum_{n=1}^\infty \frac{s^{2n-1}}{\Gamma(2n)}$$ 
$$\zeta(s) = \prod_{p} \left(\frac{1}{1-p^{-s}} \right)=\sum_{n=1}^\infty \frac{1}{n^s}$$ 
5) And both of the above can be analytically continued throughout the entire complex plane via:
$$\sinh(0-s) = -\sinh(s)$$
$$\zeta(1-s) = \chi(s)\zeta(s)$$
My questions:
1) Is there any other possible (or known) set of values for $\gamma_n$ that could yield yet another closed form? Or is this all there is, i.e. are the ("via parameter $x$ linearly scalable") integers $\gamma_n=n$ at $a=0$ and $\gamma_n =\Im(\rho_n)$ at $a=\frac12$, the only possible choices? 
2) Since the $\rho_n$'s contain information about the (distribution of) primes and the primes in turn can induce the integers via unique multiplication, could there be a connection made between (the closed forms of) these two products?
Thanks.

REPLY [3 votes]: In my quest for similar closed forms as the above, I did find:
$$\displaystyle H_{int^2}(s,a,x) := \prod_{n=1}^\infty \left(1- \frac{s}{a + i x \gamma_n} \right) \left(1- \frac{s}{{a - i x \gamma_n}} \right) = \frac{\xi_{int^2}(0 -\frac{a}{x}+\frac{s}{x})}{\xi_{int^2}(0-\frac{a}{x})}$$
with $\gamma_n = n^2$, running through the squared integers and $\xi_{int^2} = {\frac {\sin \left(  \left(\frac12-\frac12i \right)  \sqrt{2s} \ \pi \right) \sin \left(  \left(\frac12+\frac12i \right)  \sqrt{2s} \ \pi \right)}{s}}$
I do believe more closed forms exist, however only for $\gamma_n = n^{2k}$ with ($k= 1,2,3...)$ and would like to conjecture that for the 'base' Hadamard product it is true that:
$$\displaystyle \xi_{int^{2k}}(s) = \xi_{int^{2k}}(0) \prod_{n=1}^\infty \left(1- \frac{s}{0+i n^{2k}} \right) \left(1- \frac{s}{{0- i n^{2k}}} \right) \rightarrow \xi_{int^{2k}}(0)=\pi^{2k}$$
However, my (maybe too big) dream was to find a closed form for $\gamma_n = p_n$ with $p_n$ being the n-th prime. Via brute force calculations, I did find a few (all explainable) 'leads' for:
$$\displaystyle H_{prime}(s,a,x) := \prod_{n=1}^\infty \left(1- \frac{s}{a + i x p_n} \right) \left(1- \frac{s}{{a - i x p_n}} \right)$$
$$H_{prime}(1,1,1) = \frac{\pi^2}{15} = \frac{\zeta(4)}{\zeta(2)}$$
$$H_{prime}(1,\frac12,1) = 1$$
$$H_{prime}(i,i,1) = \frac{\pi^2}{6} = \zeta(2)$$
and the general rules, that also apply to the above:
$$H_{prime}(s,a,x) = H_{prime}(2a-s,a,x)$$
$$H_{prime}(s,a,x) = \frac{1}{H_{prime}(s,1-a,x)}$$
I can also prove that a closed (entire) form must have the shape: $\dfrac{\xi_{prime}(\beta -\frac{a}{x}+\frac{s}{x})}{\xi_{prime}(\beta-\frac{a}{x})}$, however the key snag is that the meromorphic component of $\xi_{prime}(s)$ that generates the zeros (similar to $\zeta(s)$), must do so only at the primes, whilst all extra "undesired and maybe more trivial" zeros need to annihilated by other components of the entire function $\xi_{prime}(s)$. 
This brought me immediately to Wilson's formula, where I used the $\Gamma$-function instead of the factorial and replaced the $mod$ by the $cos$ to produce a function that generates zeros only at integers that are prime:
$$\cos\left(\dfrac{\pi}{2}\dfrac{\Gamma(s)+1}{s}\right)$$
however it unfortunately also generates a tremendous amount of 'undesired, but hopefully more trivial' (non-integer) zeros that I need to annihilate. Is there any known function (other than using the floor function) that could accomplish this? Any other ideas\steers?
Thanks.<|endoftext|>
TITLE: integer surgeries on knots
QUESTION [7 upvotes]: I have constructed a list of surgery coefficients which yield spherical space forms. For instance, there are two knots with different Alexander polynomials on which 29-surgery will give a small Seifert fibered space of $S^2(2,3,5)$-type. Now I am just wondering does anyone know any knot on which 58 or 83 surgery will give a spherical space form. I would appreciate any help very much.

REPLY [9 votes]: Lens spaces of orders 58 and 83 may be obtained by integral surgeries on torus knot; see Moser's "Elementary surgeries on torus knots" which gives the classification up to homeomorphism. (Mind Moser's conventions.) 
More specifically:
58/1 surgery on the (3,19)-torus knot yields the lens space L(58,-9), and
83/1 surgery on the (2,41)-torus knot yields the lens space L(83,-4).
Here I'm using the convention that -p/q surgery on the unknot yields the lens space L(p,q) so that +5/1 surgery on (2,3)-torus knot yields L(5,-4).
Edit: Of course any non-zero surgery on unknot is spherical, but I reckon you were looking for something less trivial.<|endoftext|>
TITLE: What can the degrees of constructibility be?
QUESTION [11 upvotes]: If $r, s\in\mathbb{R}$, we say $r$ is constructible relative to $s$ - and write $r\le_cs$ - if $r\in L[s]$. Modding out by the induced equivalence relation $\equiv_c$, we get a partial order, the degrees of constructibility $\mathcal{C}$. This poset is extremely dependent on the ambient set theory. If $V=L$, the poset is trivial, whereas the existence of a Cohen real automatically leads to a very complicated structure (see, e.g., the paper "The degrees of constructibility of Cohen reals" by Abraham and Shore). 
My question is basically, what is known about the possible posets which $\mathcal{C}$ could be? I am especially interested in properties of $\mathcal{C}$ which follow from large cardinals. For example, if $0^\sharp$ exists, then we can deduce that $\mathcal{C}$ has size $2^{\aleph_0}$, which is the maximum possible; presumably this actually gives us a lot more.
As a sub-question, what is a good source for learning about techniques for building a model of $ZFC$ in which $\mathcal{C}$ has some prescribed properties?
ADDED: I'm most interested in what can be said when the size of $\mathcal{C}$ is assumed to be $2^{\aleph_0}$. As a particular example, the paper "Hinges and automorphisms of the degrees of non-constructibility" by P. Farrington (http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.96.3272) has, as Lemma 2.5, the following (CAVEAT: Farrington's paper actually gives the wrong definition of "hinge," which was corrected in Lubarsky's review in the JSL in 1989): 

Suppose that $\omega_2^L<\omega_1$. Then there is no nontrivial automorphism of $\mathcal{C}$.

(Now $\omega_2^L<\omega_1$ (and much more!) is a consequence of the existence of $0^\sharp$, so this is a very weak hypothesis.) Results like this are exactly what I'm looking for: properties $\mathcal{C}$ which follow from large cardinals, and in particular can hold under the assumption $\vert\mathcal{C}\vert=2^{\aleph_0}$.

REPLY [9 votes]: In addition to the work on what can occur in the constructibility degrees, there's a theorem of Bob Lubarsky about what cannot occur.  If the lattice of constructibility degrees is countable and has a top element, then it must be a complete lattice.  [Lattices of c-degrees, Ann. Pure Appl. Logic 36 (1987) 115-118]

REPLY [7 votes]: The article Initial segments of the degrees of constructibility by Marcia Groszek and Richard Shore (Israel Journal of Mathematics
June 1988, Volume 63, Issue 2, pp 149-177) shows that 


Any constructible, constructibly countable, (dual) algebraic lattice is isomorphic to the degrees of constructibility of reals in some generic extension of L.


And there is a lot of further work on this topic by Groszek, Slaman and others.<|endoftext|>
TITLE: Rigid monoidal abelian category without an exact tensor functor to Vect
QUESTION [9 upvotes]: A Tannakian category is defined to be a rigid monoidal abelian category over a field with an exact tensor functor to the category of vector spaces. (It is always equal to the category of representations of some pro-algebraic group.)
Is there such a thing as a rigid monoidal abelian category over a field without an exact tensor functor to the category of vector spaces?
What could such a category look like?

REPLY [2 votes]: Representations of quantum groups at roots of unity provide a particularly helpful example (the $x^2=1+x$ example above corresponds to quantized sl(2) at some low index root of unity).  These are rigid monoidal abelian categories (not symmetric) whose objects in general have noninteger dimensions, so they clearly cannot have a functor to Vect.  However, the objects are representations of a Hopf algebra, so there is a functor to Vect of most of the structure.  The tensor product is the ordinary tensor product of representations (again a representation for a Hopf algebra), but then you quotient by a subrepresentation (which can be thought of very loosely as the nonsemisimple part).<|endoftext|>
TITLE: When is spectral norm of AB equal to that of BA?
QUESTION [8 upvotes]: I have $A^{1/2} B A^{1/2} \preceq I$ for two PSD matrices $A$ and $B$, and I'd like to know if that implies $\|AB\|_2 \leq 1.$ 
The argument I was using to show this is that for any two square matrices $A$ and $B,$ it is always the case that $\|AB\|_2 = \|BA\|_2.$ I thought I read that this equality does hold in a reputable source, but I don't have access to it right now and I was unsuccessful in reproducing a proof. 
I know the eigenvalues of $AB$ and $BA$ are the same modulo possibly having different numbers of zeros, so I'm worried that I might have ``remembered'' something that isn't true!
Any references/counterexamples for either question?

REPLY [3 votes]: Going back to the original motivation, the following observations might still be useful. 

Denis already mentioned this: $A^{1/2}BA^{1/2} \sim A^{1/2}A^{1/2}BA^{1/2}A^{-1/2}$ for strictly positive $A$. This implies that $\sigma(A^{1/2}BA^{1/2})=\sigma(AB)$.
If $A$ and $B$ are arbitrary (square) matrices such that their product $AB$ is normal, then one can show that $\sigma(AB) \prec_w \sigma(BA)$ (where $\prec_w$ denotes weak majorization); equivalently, for any unitarily invariant norm, we have $\|AB\| \le \|BA\|$.
If $\|AB\| \le 1$ for positive definite $A$ and $B$, then $\|A^{1/2}B^{1/2}\| \le 1$.<|endoftext|>
TITLE: Non-$\Sigma$ $E_n$ algebras?
QUESTION [11 upvotes]: Any symmetric operad can be considered as an non-$\Sigma$ operad by throwing away permutations. Does anyone know what sort of structure one gets for algebras over $C_n$ the little n-cubes operad, or some equivalent variant, considered in this way? One thing which is strange is that there is a non-$\Sigma$ operad splitting of the map $Ass \to Com$ (where $Ass$ and $Com$ are the symmetric associative and commutative operads, respectively), and so any non-$\Sigma$ $C_n$ algebra should be a loop space, but I do not see what, if any, extra structure is present.

REPLY [9 votes]: Justin, that is a cute question, and I've never thought about it.  One starting thought free guess (acting on connected spaces) is that a non-$\Sigma$ $C_n$-space has $n$ possibly inequivalent but definitely related loop space structures.  Do you see examples, or is this just curiosity? As to Ass and Com, I see no interest in building in the permutations to define Ass and then throwing them away.  Of course, a non-$\Sigma$ `Com'-algebra is the same thing as an Ass-algebra, but the operadic modules over these kinds of algebras are different (bimodules and left modules).  David, I think Justin is asking a concrete question, not the sort that model category structures shed light on.<|endoftext|>
TITLE: The Paley-Wiener theorem and exponential decay.
QUESTION [9 upvotes]: Consider a function whose Fourier transform is supported on a half-ray:
$$
A(t)=\int_0^\infty \omega(E) e^{-iEt}d E,
$$
where I can suppose $\omega(E)\geq 0$ and any suitable regularity conditions on $\omega(E)$ in the limit $E\rightarrow\infty$. I am interested in results constraining the rate of decay of $A(t)$ in the limit $t\rightarrow\infty$. Specifically, I would like to rule out the asymptote $\mathbf{|A(t)|\sim e^{-\Gamma t}}$.
The reference I have [1] uses the original 1934 Paley-Wiener theorem [2], which states that under these (or similar) assumptions for $A$ the integral
$$
\int_{-\infty}^\infty \frac{\left|\ln|A(t)|\right|}{1+t^2}dt<\infty
$$
must converge. This is strong enough to rule out the asymptote.
However, I have looked up the proof in Paley and Wiener and I find it far too technical and non-self-contained for me to follow with any ease; it also has an air of old mathematics that has probably been replaced with cleaner arguments by now. I do get some of the intuition behind the appearance of the $1/(1+t^2)$ factor. (Namely, a unitary transform of the upper half-plane $z$ space into the unit circle in $\zeta=i\frac{z+1}{z-1}$, where the measure transforms as $\frac{d\zeta}{\zeta}\approx\frac{dz}{1+z^2}$.) I still don't find, however, any intuition into how the can't-be-too-fast decay of $A$ correlates with the support of $\omega$, or at least no intuition that can be turned into a rigorous argument.
I am looking for references or arguments that prove in a clearer fashion that exponential decay of $A$ is impossible with such a Fourier domain, and particularly for ones that have clear intuition behind them that can be turned into a solid argument, even if the rigorous details are fiddly.

L. Fonda, G. C. Ghirardi and A. Rimini. Decay theory of unstable quantum systems. Rep. Prog. Phys. 41, pp. 587-631 (1978). Page 592.
R. Paley and N. Wiener. Fourier Transforms in the Complex Domain (Providence, Rhode Island: American Mathematical Society, 1934). Theorem XII, p16.

REPLY [17 votes]: The idea behind this theorem is actually simple. 
For the Fourier transform to make sense (with usual understanding of the integral) the function
$\omega$ must be summable, that is in $L^1(0,+\infty)$. This immediately implies that the Fourier integral converges not only on the real line but also in the lower half-plane, so $f$
is an analytic function in the lower half-plane which is bounded. Indeed,
$$|f(t)|\leq\int_0^\infty|\omega(t)|e^{\Im t}dt\leq\int_0^\infty|\omega(t)|dt,$$
because $\Im t\leq 0$.
Now, there is a general principle that a bounded analytic function cannot be too small.
This is easier to see in the unit disc. $\log|f(z)|$ is subharmonic which means that
$$\log|f(0)|\leq\frac{1}{2\pi}\int_0^{2\pi}\log|f(e^{i\theta})|d\theta.$$
The integrand is bounded from above, so the integral of the NEGATIVE part of $\log|f(e^{i\theta})|$
must converge. (If $f(0)=0$ this is still true: just divide $f$ by an appropriate power of $z$;
this will not change the right hand side). This is called Jensen's inequality.
Now, this fact that the negative part of $\log|f |$ is integrable, when transfered from the 
disc to the half-plane via conformal mapping, means exactly that the logarithmic integral
$$\int_0^\infty\frac{\log|f(t)|}{1+t^2}dt$$
cannot diverge to $-\infty$, that is the Wiener-Paley theorem.
Of course, one can argue directly in the half-plane without the reference to the unit disc.
Boundedness of $\log|f|$ from above implies that the Poisson integral of its boundary value
on the real line must converge, which is exactly the same condition.
Now Beurling-Malliavin theorem says essentially that the condition of convergence of
the log integral is best possible (subject to some technical regularity assumptions). Even if you 
consider $\omega$ of arbitrarily small compact support.
A good modern reference for all these things is the books of P. Koosis.<|endoftext|>
TITLE: realization of maps between classifying spaces of categories
QUESTION [7 upvotes]: The classifying space $B\mathcal{C}$ of a small category $\mathcal{C}$ is by definition the geometric realization of the nerve of $\mathcal{C}$. Now let $\mathcal{C}_1$ and $\mathcal{C}_2$ be two small categories. Any functor $F : \mathcal{C}_1 \rightarrow \mathcal{C}_2 $ induces a continuous map $BF: B\mathcal{C}_1 \rightarrow B\mathcal{C}_2$.
My question is the following: when is a continuous map $f: B\mathcal{C}_1 \rightarrow B\mathcal{C}_2$ homotopic to a map of the form $BF$, for some functor $F:\mathcal{C}_1 \rightarrow \mathcal{C}_2 $
Thanks.

REPLY [3 votes]: Here's one instance where the answer is affirmative: when the target category is a groupoid (a little more general than my remark on John Klein's answer).
Let $\mathcal{C}$ be any category and $\mathcal{G}$ be a groupoid. Then $B\mathcal{G}$ is a 1-type (i.e., all $\pi_n$ with $n\ge 2$ vanish, with any base points), and therefore, for any space $X$, $\mathrm{maps}(X, B\mathcal{G})$ is a 1-type too, and in fact is just weakly homotopy equivalent to $B \mathrm{Fun}(\pi_{\le 1}(X), \mathcal{G})$ where $\pi_{\le1}(X)$ is the fundamental groupoid of $X$ (notice that since $\mathcal{G}$ is a groupoid, so is any functor category $\mathrm{Fun}(\mathcal{A}, \mathcal{G})$). Applying this to $X=B\mathcal{C}$, we get a weak homotopy equivalence $\mathrm{maps}(B \mathcal{C}, B \mathcal{G}) \simeq B \mathrm{Fun}(\pi_{\le 1}(B\mathcal{C}), B\mathcal{G})$. Now, $\pi_{\le 1}B \mathcal{C}$ is just $\mathcal{C}[\mathcal{C}^{-1}]$, the localization of $\mathcal{C}$ obtained by adding formal inverses for all morphisms of $\mathcal{C}$; and since $\mathcal{G}$ is a groupoid, any functor $\mathcal{C} \to \mathcal{G}$ automatically factors through $\mathcal{C}[\mathcal{C}^{-1}]$, so composition with the canonical functor $\mathcal{C} \to \mathcal{C}[\mathcal{C}^{-1}]$ induces an equivalence of categories $\mathrm{Fun}(\mathcal{C}, \mathcal{G}) \cong \mathrm{Fun}(\mathcal{C}[\mathcal{C}^{-1}], \mathcal{G})$. Equivalences of categories induce homotopy equivalences of classifying spaces, so we get a weak equivalence $\mathrm{maps}(B \mathcal{C}, B \mathcal{G}) \simeq B\mathrm{Fun}(\mathcal{C}, \mathcal{G})$, which is a little better than just saying that every function between the classifying spaces is homotopic to one coming from a functor.<|endoftext|>
TITLE: Injective morphism from curves to $\mathbb CP^2$
QUESTION [14 upvotes]: Is there a smooth compact complex curve that does not admit an injective holomorphic map to $\mathbb CP^2$ ? Let me stress, that the image of the curve in $\mathbb CP^2$ can have singularities.
I should probably add, that I see this question as a baby version of the following question of Fancesco Polizzi: Surfaces in $\mathbb{P}^3$ with isolated singularities
Remark. Any curve $x^n+y^n+z^n=0$ admits injective morphisms to $\mathbb CP^2$ of arbitrary high degree (the proof is given by Jeremy Blanc here: Injective morphism from an elliptic curve to $\mathbb CP^2$. )

REPLY [8 votes]: Update. 
Estimates coming from the Bogomolov-Miyaoka-Yau inequality yield:
Corollary. (to Theorem A in this paper of Sakai)  There is no $M$ such that every curve admits an injective morphism to a curve in $\mathbb{P}^2$ whose points all have multiplicity at most $M$.
Proof. The geometric genus $g$ curves of degree $d$ move in a family of dimension $3d + g - 1$, so if these are to fill the moduli space we must have $2g \le 3d + 2$.  On the other hand, Theorem A of this paper of Sakai gives after rearranging that for any genus $g$, degree $d$, unibranch plane curve,
$$ 2g \ge \frac{d^2}{2m+1} + O(md) $$ 
where $m$ is the maximal multiplicity of a point on the curve.  For any bounded $m$, these inequalities cannot both be satisfied for large $g$. $\square$
Remark: the argument also shows that the same statement holds restricted to hyperelliptic curves.  This is not incompatible with Jeremy's answer below, which requires a cusp of multiplicity $2g$ to inject a hyperelliptic curve of genus $g$.

I leave the older argument below because it is a different and possibly still useful approach. 
Theorem. There is no finite set $\Sigma$ of topological types of unibranch singularities such that every smooth curve admits an injective map to a curve in $\mathbb{P}^2$ with only singularities from $\Sigma$. 
Proof. I will argue that curves with unibranch singularities of total cogenus $\delta$ occur in the linear system $|\mathcal{O}(d)|$ with codimension at least $\mathrm{min}(\kappa, 3d - 1 + \delta)$, where $\kappa$ is the codimension of the equiclassical locus in the versal deformation of the singularities; it is $2\delta$ in the case of curves with only cusps, and for any finite set of singularities is bounded below by $(1 + \epsilon)\delta$ for $0 < \epsilon \le 1$ some constant depending on the set of singularities.  Let us first see why this implies the result. 
Saying a geometric genus $g$ curve appears in degree $d$ in codimension $3d - 1 + \delta = 3d + (d-1)(d-2)/2 - g - 1$ means it comes in a $g + 1 + (d+1)(d+2)/2 - (d-1)(d-2)/2 -3d = g + 3$ dimensional family. On the other hand $\kappa$ is given by $\delta + r$, where $r$ is the total ramification divisor on the normalization of the curve (for the normalization map), i.e. the sum over the branches of the singularities of (multiplicity - 1).  So in this case the curves come in a family of dimension $g + 4 + 3d - r$.  I don't know what to do with this, but under the further assumption that the set of singularities from which we choose must be finite and so we get $\kappa = (1 + \epsilon) \delta$ as above (e.g., for only cusps, $\epsilon = 1$), then the argument in Will's answer above shows this can fill the moduli space for only finitely many $g$. 
To get the advertised codimension estimate, let $C$ be such a curve of degree $d$.  Let $\Lambda$ be the tangent space to the versal deformation of the singularities of $C$.  Then there exists a linear subspace $L \subset \Lambda$ such that

equigeneric ideal $\ge$ L $\ge$ equiclassical ideal
aside from the above restriction, L may be taken so that the dimension of the equigeneric ideal mod L is up to $3d - 1$.
the image of $T_C|\mathcal{O}(d)| \to \Lambda$ is transverse to $L$. 

This is essentially classical, for the sort of argument to prove it you can see e.g. the statement of Lemma 8 and the proof of Lemma 10 in my paper with Steve. $\square$
It's not yet clear to me how to improve this to answer the OP, maybe the better sort of bounds of Greuel, Lossen, and Shustin will help?  (Or maybe there's just some easy argument not involving deformation theory...)<|endoftext|>
TITLE: How should a professor feel peace of mind when a student leaves academia?
QUESTION [7 upvotes]: I apologize if this is the wrong forum for this question -- if so, could somebody please point me to the correct one?
I'm a professor who recently started advising graduate students, and I'm trying to find a way to feel good about the time I spend on my students when they leave academia.  I devote loads of time and energy into helping my students reach their full potential, and giving them every benefit of my experience, in part because this will contribute to the development of mathematics.  But I'm having trouble coming up with a justification for putting so much effort into mentoring students, if the students will wind up leaving academia.  Probably many people on this forum have thought through this and come up with some kind of answer, and I would love to hear about these.
Let me clarify that I understand that advising grad students is part of my job.  But there are only so many hours in the day, so I need to find a balance between time spent mentoring students and time spent doing research, writing papers, etc.  So far I've made my students my top priority, but then when they leave academia, I have a hard time justifying to myself that I should have sacrificed effort on all other fronts in order to help the student develop.

REPLY [7 votes]: How you feel about this will depend on who you are, maybe also on who the student is, and also other circumstances. What are the student's reasons for leaving academia?
If the student leaves academia because his thesis does not seem to be good enough to get him the kind of academic job that he wants, I could imagine feeling a couple of other things than "I shouldn't have put in so much effort". I might even feel that I should have put in even more effort; I might feel that the student has let me down and should have put in more effort; I might feel that in hindsight I shouldn't have agreed to work with that particular student and resolve to be choosier in the future but give it my all when I have chosen; I might feel relieved that the student is not embarking on a very hard road with little chance of satisfaction. 
If a student who had done good work then surprised me by seeming to reject the academic life and its values in favor of making big bucks on Wall Street, I might feel a bit resentful and taken advantage of, but I think I would not really feel that I had wasted my time: the student has put in more time and effort than I have, and really who am I to tell him where to go from here?
How would you feel if a student told you right up front that she wanted to study for a PhD under your guidance and then go into business or industry? Would you try to change her mind? Would you decline to take her on as a student?
EDIT Maybe you should consider cutting down on how much of your time you allot to this part of your job anyway -- not because sometimes your students leave academia, but partly because you have other kinds of work to do, too, and partly because the student ought to have primary responsibility for the success of his or her graduate studies.<|endoftext|>
TITLE: A question about Universal sets.
QUESTION [5 upvotes]: In several set theories, among which Quine's NF is one of the best known and most extensively
investigated, the existence of a Universal set V can be proved. V is the set of all sets-indeed
all "proper classes", including V itself, are elements of V. Although it might be considered
"meaningless" to ask what the "cardinal number" of V is, there are some questions of this type
which one could ask. As far as I know, the consistency of NF relative to ZF is still an open
question. But have any of the well known "large cardinal axioms" (assuming that they can be 
expressed in NF) been proved to be inconsistent with NF? If so, this might throw some light
on the question of how "large" V really is.

REPLY [6 votes]: Concerning the consistency of NF, Peter Smith made the following post last November on Logic Matters: 


Randall Holmes has now announced “I believe that I am in possession of a fairly accurate outline of a proof of the consistency of New Foundations.”
He goes on to say “NF has the same consistency strength as TST + Infinity, has the same kinds of extensions as NFU in the same ways, has no interesting consequences for the combinatorics of small sets, etc. No surprises, this is a rather boring outcome in my opinion …” Well, I don’t know about boring! If Randall has indeed cracked this long-standing problem, it’s a major achievement. He is still editing the document, so nothing is released yet. However, Thomas Forster is organizing a conference here in Cambridge in the spring and Randall says he will “certainly be discussing this.”
And Thomas confirms  ”I am indeed organising an NF meeting in Cambridge in the spring, current intention is last week of March and first week of April. The idea is that before Randall arrives there will be a warm-up act wherein the background and some preparatory material is set out for people who are not already familiar with it. Thus when Randall arrives we will all be primed and ready to go. … I am not at this stage soliciting other offers of talks, tho’ that may change. If you have something you think I may find irresistible by all means try to twist my arm. And – of course – contact me if you want to come.”


I'm not sure whether there has been any further news about this since then.<|endoftext|>
TITLE: Fundamental groups of symplectic leaves
QUESTION [7 upvotes]: Let $X = \operatorname{Spec}(R)$ be a conical Poisson variety over $\mathbb{C}$.  This means
that $R$ is a non-negatively graded Poisson algebra over $\mathbb{C}$ 
with $R_0 = \mathbb{C}$ and that the Poisson
bracket is homogeneous of negative degree.  Assume that $X$ admits a conical symplectic resolution.
Examples to keep in mind are:

the nilpotent cone in a simple Lie algebra
a Slodowy slice to a nilpotent orbit inside the nilpotent cone
the symmetric scheme of $n$ points on a Kleinian singularity
a hypertoric variety
a quiver variety.

Let $S$ be a symplectic leaf of $X$.
Q:  Is it possible for the fundamental group of $S$ to be infinite?
In the first four examples listed above, the fundamental group of every symplectic leaf is finite.  I don't know whether or not this property holds for quiver varieties, or for other examples.

REPLY [4 votes]: I do not know the answer. I believe that it is so. See http://arxiv.org/abs/1301.1008 for algebraic fundamental groups.
If you consider quiver varieties of affine types, they are moduli spaces of instantons on ALE spaces. If you replace the base space by $\mathbb R^4$ and assume rank is $2$, then the finiteness follows from Atiyah-Jones conjecture for rank $2$. The Atiyah-Jones conjecture actually says much stronger: there is a homotopy equivalence between moduli spaces and the infinite dimensional space of all connections up to dimension explicitly given by the instanton number. The conjecture was proved by Boyer et al.
There are lots of generalizations of the Atiyah-Jones conjecture. So it probably applies to quiver varieties of affine types.<|endoftext|>
TITLE: Betti numbers of Proper nonprojective varieties
QUESTION [9 upvotes]: This is a question about pathologies.
Let $X/\mathbb{C}$ be an irreducible projective variety smooth over $\mathbb{C}$. Then, the singular cohomology groups $H^i(X, \mathbb{C})$ have a hodge decompositon, and hodge theory tells us that the betti numbers (and hodge numbers) are not completely random. Eg, the odd betti numbers $b_{2i+1}$ are even integers, Hard Lefschetz theorem, even betti numbers $b_{2i}$ are nonzero, ... 
We can also see that the even Betti numbers of a variety $X$ as above are nonzero in "another" way: $X$ has a finite surjective map to $\mathbb{P}^n$ where $n= \dim X$. Then, because $X$ is Kahler, such a map induces an injective map on singular cohomology. Hence, the result follows from the corresponding result for projective space.
$\textbf{Question:}$
1.) Are there any nontrivial restrictions on the Betti numbers of smooth, irreducible proper (but non-projective) varieties?
For example, can I have such a 4-fold with Betti numbers $b_0 = b_8 = 1$, but all other $b_i = 0$?
I find the situation a little disconcerting: for example, suppose I have a connected compact topological (or complex) manifold. If I know it's betti numbers are as directly above, then I would like to immediately say that it doesn't have the structure of an algebraic variety. However, the only thing I can say now is that a smooth complex variety with these Betti numbers can't be projective. If such a variety existed, then (by Ehresmann's thm) we can't put it in a proper smooth family over a curve with any projective variety. I'm no expert but that sounds pretty bad.

REPLY [10 votes]: I think, that many basic restrictions, that you have for complex projective varieties still hold for proper smooth complex varieties. Let me show that $b_2>0$, and $b_{2n-2}>0$.
Suppose that $X^{2n}$ is a proper smooth complex variety. Then there is a birational morphism $\phi: Y^{2n}\to X^{2n}$ from a projective variety $Y$ to $X$. Let $E_1,...,E_n$ be all the exceptional divisors of $\phi$. Since $Y$ is projective there is a divisor $D\subset Y$ that has a point $x$ in $Y^{2n}\setminus (E_1,...E_n)$ and there is a curve $C$ in $Y$ passing through $x$ and not contained in $D$. Consider finally  the divisor $\phi(D)$ and the curve $\phi(C)$ in $X^{2n}$. Note that they intersect positively (since they have common point $\phi(x)$ and $\phi(C)$ does not belong to $ \phi(D)$).
So they represent non-zero cycles in $H_2(X^{2n})$ and $H_{2n-2}(X^{2n})$.
Also, it is very easy to see that $b_1$ should be even. Indeed the fundamental group of a smooth complex variety is a birational invariant, and $b_1$ depends only on $\pi_1$.
I have an idea how using weak factorization of birational maps one can prove that all other $2k+1$ Betty numbers are even. But this requires some work, so I'll write this if I manage to work out details (I believe this should be a very well known fact).<|endoftext|>
TITLE: What do generic Torelli theorems claim? 
QUESTION [5 upvotes]: I have a question on "Torelli theorems" in algebraic geometry. "Torelli theorems" say about how the period map of a given family of varieties behave. If I understand correctly, global Torelli theorems are the most strong ones and basically classify the varieties you have, that is, the period map is isomorphism. Local Torelli theorems state that the differential of the period map is injective. What do generic Torelli theorem claim then?

REPLY [4 votes]: I would formulate it as follows: global Torelli says that the period map is injective/an immersion (but not necessarily an isomorphism!!), local Torelli says that the period map has injective differential, and generic Torelli says that the period map is generically injective (injective on a Zariski open dense set).<|endoftext|>
TITLE: Pullback measures
QUESTION [35 upvotes]: Why do all measure theory textbooks present the concept of push-forward measure, but never the concept of pull-back measure? Doesn't the latter exist?
It's true that the naive treatment of such a concept would sometimes lead to contradictions. For instance, let $p:\mathbb{R}^2\rightarrow\mathbb{R}$ be given by $p(x,y)=x$, $\lambda$ be the Lebesgue measure on $\mathbb{R}$, and $A=[0,1]\times[0,1]$. If one decomposes $A=A_1 \cup A_2$ where $A_1=[0,1]\times[0,\frac{1}{2}]$ and $A_2=[0,1]\times(\frac{1}{2},1]$, then $(p^* \lambda)(A)=1$ while $(p^* \lambda)(A_1)+(p^* \lambda)(A_2)=2$, showing that the naive definition of the pull-back does not lead to a measure.
Similarily, if $i:\mathbb{R}\rightarrow\mathbb{R}^2$ is given by $i(x)=(x,0)$, then the pull-back of the Lebesgue measure on $\mathbb{R}^2$ would be 0.
Given the two situations presented above, can one define a fruitful concept of pull-back measure?

REPLY [10 votes]: A measure $\mu$ on a set $X$ is a linear functional  from a  vector space of functions $F(X)$ on that set $X$ $\newcommand{\bR}{\mathbb{R}}$
$$ \mu : F(X)\to\bR,\;\; f\mapsto \langle \mu, f\rangle:=\int_X f(x) \mu(dx). $$
We can view the space of measures $M(X)$ on $X$  as a dual to $F(X)$. The natural  operation  on functions is that of pullback.  The natural operation on measures would be the dual operation, pushforward. Thus, if $\Phi: X\to Y$ is a map then we get two linear maps
$$ \Phi^*: F(Y)\to F(X),\;\;\Phi_\ast: M(X)\to M(Y) $$
related by the equality
$$ \langle \mu , \Phi^* f\rangle= \langle \Phi_* \mu, f\rangle,\;\;\forall f\in F(Y),\;\;\mu\in M(X). $$<|endoftext|>
TITLE: Which topological spaces are (topological) groups?
QUESTION [16 upvotes]: General literature does not seem to offer a characterisation of topological groups among all topological spaces. Of course, being completely regular (uniform) is necessary, but separation properties, or indeed any sort of "niceness" like pseudo-metrisability are not sufficient, since topological groups, for instance, cannot have fixed point property.

REPLY [3 votes]: A necessary condition for a Hausdorff compact space to admit the structure of a topological group is the Suslin condition (I hope I am using proper terminology)
every family of pair-wise disjoint open sets is countable.

This is so because Hausdorff compact topological groups admit Haar measure.<|endoftext|>
TITLE: "Arithmetic genus" of a plane curve singularity.
QUESTION [12 upvotes]: I believe that the following questions are very basic, but I don't know how to get a reference. 
Consider a curve in the plane $C\in \mathbb C^2$ with a singularity at $0$ and suppose it is 
unibranch at zero (i.e. analytically irreducible). Then I guess one should be able to define "arithmetic genus defect" of the curve at $0$. Namely if one smooths analytically $C$, its geometric genus will grow by a positive number (in case of the cusp $x^2=y^3$ it will grow by one), and let us call this number the defect. 
Question 1. Is this defect well defined (independent of a smoothing)? How is it called and how one should calculate it (say it terms of the local ring of $C$ at $0$)?
Question 2. Suppose we have an explicit local parametrisation of $C$ at $0$, say by two holomorphic functions $f(t), g(t)$ (polynomials if you wish). Is it possible to find this "defect" as a certain invariant of this pair of functions at $t=0$? 
Question 1 is settled in the answer of unknown and Question 2 in comments to it by Roy and Vivek

REPLY [8 votes]: Another way to compute the delta invariant in pratice: let $m_1$ be the multiplicity of the curve at the point $p_1$ you are looking for. You blow-up $p_1$, and look for singular points of the strict transform of the curve which lie on the exceptional curve obtained. Denote by $m_2,\dots,m_k$ the multiplicities obtained (which satisfy $m_2+\dots+m_k\le m_1$). Blow-up all these points. Repeat the process until the curve has no singular point infinitely near to $p_1$. You get a set of multiplcities $m_1,\dots,m_l$ (with $l\ge k$). The delta you want is exactly $\sum_{i=1}^l m_i(m_i-1)/2$. This is a direct consequence of adjunction formula.
It depends of the situation, but sometimes it is easier to compute than the formula of Milnor. The multiplicities are easy to get from either the equation of the curve or a parametrisation.<|endoftext|>
TITLE: abelian group objects category
QUESTION [5 upvotes]: Suppose that $\mathcal{C}$ is a cartesian closed category. When is the category of abelian group objects $\mathcal{Ab}(\mathcal{C})$ a symmetric monoidal closed with respect to something substituting tensor product in the classical case $\mathcal{C}=\mathcal{Set}$?
I need reference or sketch of an argument in the case when $\mathcal{C}$ is a topos.

REPLY [5 votes]: I previously indicated how to construct the tensor product for $\textbf{Ab}(\mathcal{V})$ when $\mathcal{V}$ is a sufficiently nice cartesian closed category. (The comments are probably more helpful than the post itself.) A Grothendieck topos is certainly nice enough, but so is an elementary topos with a natural numbers object. (See Theorem D5.3.5 in Sketches of an elephant.)
Basically, we mimic the standard construction of tensor products by generators and relations; the only stumbling block is the construction of free internal abelian groups. Once this is done, $\textbf{Ab}(\mathcal{V}) \to \mathcal{V}$ will be monadic, say with left adjoint $F : \mathcal{V} \to \textbf{Ab}(\mathcal{V})$, and $\textbf{Ab}(\mathcal{V})$ will have coequalisers for reflexive pairs. (See Lemma D5.3.2 in Sketches of an elephant.) One can then define a certain object $I$ in $\mathcal{V}$ for each pair of internal abelian groups $A$ and $B$ and a pair of morphisms $I \rightrightarrows F (A \times B)$ in $\mathcal{V}$ such that
$$F I \rightrightarrows F (A \times B) \to A \otimes B$$
presents $A \otimes B$ as a coequaliser of a reflexive pair in $\textbf{Ab}(V)$. (We could take
$$I = (A \times A \times B) \amalg (A \times B \times B) \amalg (F1 \times A \times B) \amalg (A \times F1 \times B)$$
so that we can code the four equations
\begin{align}
(a + a') \otimes b & = a \otimes b + a' \otimes b \newline
a \otimes (b + b') & = a \otimes b + a \otimes b' \newline
(r a) \otimes b & = r (a \otimes b) \newline
a \otimes (r b) & = r (a \otimes b)
\end{align}
but one can probably get away with less.)
It turns out to be much easier to construct internal homs for $\textbf{Ab}(\mathcal{V})$: you just need $\mathcal{V}$ to be cartesian closed with equalisers. Of course, once you have both the tensor product and internal homs, then they make $\textbf{Ab}(\mathcal{V})$ into a symmetric monoidal closed category. 
One can also carry out this construction in much greater generality for any commutative monad on any sufficiently nice symmetric monoidal closed category $\mathcal{V}$. This, I think, is an old result of Kock [1971, 1972] generalising Linton [1966].<|endoftext|>
TITLE: Why is it important that partial derivatives commute?
QUESTION [22 upvotes]: I am asking this in the context of differential geometry (specifically Riemannian).
When the Levi-Civita Connection is defined, we require that the torsion tensor is 0, which in local coordinates translates to the requirement that $\Gamma_{ij}^{k} = \Gamma_{ji}^{k}$; which is the covariant derivative version of saying partial derivatives commute: $\nabla_{\partial_i}(\partial_j)=\nabla_{\partial_j}(\partial_i)$.
This is obviously true in the Euclidian settings, and I understand all the details of the proofs. But why is this such an essential property? Why does this capture our intuitive sense of derivatives?

REPLY [2 votes]: The Levi-Civita connection is just a very special one - torsion free. It is interesting that the same geometry may be described by switching to another, non-torsion-free connection. E.g. there is such a version of general relativity which is called teleparallel formulation. While the curvature tensor (based on the new connection) vanishes, all the deviation from flatness has been shifted to the torsion tensor (better: vector-valued 2-form). Einstein exchanged his ideas with Cartan in the 1920 about that.
Torison has also an equivalent in physics as dislocation density (disclocations are defects in crystals). The theory has been developed in the 1950's by Kondo, Bilby and Kröner. See also the book Ricci calculus by J.A. Schouten.
To summarize, it is not that important that the connection is symmetric, it is merely a matter of choice. The metric, for instance, is independent of the connection.<|endoftext|>
TITLE: what is a spinor structure?
QUESTION [18 upvotes]: There are of course lots of definitions and references for this, but in the same way that, on a manifold $M$,

a Riemannian metric is a section of positive definite symmetric bilinear forms on $TM$
or an almost complex structure is a section $J$ of $\textrm{End}(TM)$ which is everywhere an anti-involution (i.e. $J_x^2 = - \mathrm{Id}_{T_x M} $)
or an orientation is a non-vanishing section of $\Lambda^m TM$ 

is 
a spin structure, a section of quadratic forms $Q$ on $TM$ (of type $(s,t)$) and a vector bundle $S$ so that $\textrm{End}(S) \simeq \textrm{C}\ell(TM,Q)$?
...or something of the like? any references where it may be stated in this fashion?

REPLY [15 votes]: A spin structure on a real vector space V equipped with a real quadratic form μ
is an invertible bimodule (i.e., a Morita equivalence)
from Cl(V,μ) to Cl(Rdim(V),ν).
Here ν is the direct sum of dim(V) copies of the canonical
quadratic form on R.
A spinc structure on a complex vector space V equipped with a complex quadratic form μ
is an invertible bimodule
from Cl(V,μ) to Cl(Cdim(V),ν).
Here ν is the direct sum of dim(V) copies of the canonical
quadratic form on C.
Note that spin and spinc structures form
a category instead of a mere set, just as we would expect.
Of course, these definitions immediately extend
to vector bundles, with the obvious requirement
that invertible bimodules form bundles themselves.
A spin or spinc structure on a smooth manifold
is a spin or spinc structure on its real
or complex tangent bundle.<|endoftext|>
TITLE: Mod-p cohomology of $GL(n,p^d)$ 
QUESTION [6 upvotes]: In the classic paper On the Cohomology and K-Theory of the General Linear Groups Over a Finite Field, Quillen proved (Theorem 6): 

$H^i(GL(n,p^d),\mathbb{F}_p)=0$ for $0 < i < d(p-1)$ and all $n$. 

On the other hand, the cohomology of a finite group doesn't completely vanish (this is nicely discussed on MO: Non-vanishing of group cohomology in sufficiently high degree). So there is a minimal integer $i_0=i_0(n,d) \ge d(p-1)$ satisfying $$H^{i_0}(GL(n,p^d),\mathbb{F}_p) \neq 0$$  Is Quillen's lower bound $d(p-1)$ sharp ? (I couldn't find any information about sharpness in Quillen's paper). If not, is the precise value of $i_0$ known ?

REPLY [6 votes]: There are some results known in this direction; see the paper On the vanishing ranges for the cohomology of finite groups of Lie type, by Christopher Bendel, Daniel Nakano, and Cornelius Pillen (Int. Math. Res. Not. (2012), 2817-2866). In particular, they show that if $p \geq n+2$, then the first degree in which nonzero cohomology appears is $d(2p-3)$.
Bendel, Nakano, and Pillen also have a sequel paper posted on the arXiv that treats a number of cases for other Lie types that aren't already handled in the reference I gave above.<|endoftext|>
TITLE: Linearization instability and singular points of algebraic varieties
QUESTION [9 upvotes]: In a well known 1973 paper, Fischer and Marsden pointed out (with similar, contemporary remarks made in the physics literature by Brill and Deser) that the space of solutions of some non-linear partial differential equations, like the Einstein or Yang-Mills equations, on a manifold with compact Cauchy slices cannot be manifold (of an appropriate infinite dimensional kind). This phenomenon was given then name linearization instability; I'll explain it in more detail below.
As far as I can tell, this terminology is only used in the literature that followed this observation. However, the actual mathematical phenomenon is very similar to the fact that algebraic varieties are also not manifolds, since they possess singular points. My question is this: what terminology is used in algebraic geometry for the various aspects of this phenomenon? I'll be more specific below.
Given a non-linear PDE, $E[u] = 0$, where $u$ is the unknown function, defined on some manifold $M$. Let its linearization about a background solution $E[U]=0$ be $E_U[u] = 0$, that is, $$E[U+\varepsilon u]=\varepsilon E_U[u] + O(\varepsilon^2).$$ Let $X$ denote the set of solutions of the full equation (exact solutions) and let $\ker E_U$ denote the set of all solutions of the linearized equation (linearized solutions). The background solution $U$ is said to be linearization stable if each linearized solution $u\in \ker E_U$ can be extended to a 1-parameter family $v_\varepsilon = U + \varepsilon u + O(\varepsilon^2)$ of exact solutions, $E[v_\varepsilon] = 0$. If $U$ is linearization stable, then one can show that the space $X$ of exact solutions that are close to $U$ is an (infinite dimensional) manifold, modeled on the space $\ker E_U$.
On the other hand, it is possible that there exists a function $Q[u]$ (typically involving some integral of a local expression over a submanifold of $M$) such that a linearized solution $u\in \ker E_U$ cannot be extended to a 1-parameter family $v_\varepsilon = U + \varepsilon u + O(\varepsilon^2)$ of exact solutions unless $Q(u)=0$. If such a non-trivial function $Q[u]$ exists, the background $U$ is said to be linearization unstable. In the examples of Einstein or Yang-Mills equations, $U$ happens to be linearization unstable if it possess some symmetries and the corresponding $Q(u)$ is quadratic. One can then show that the space $X$ of exact solutions of $E[v]=0$ around $v=U$ is not a manifold, since it has neighborhoods that are not homeomorphic to any vector space (in particular not $\ker E_U$) but rather to a conical subset of $\ker E_U$ defined by $Q[u]=0$.
Sometimes it is possible to identify a set of functions $S_i[u]$ such that the intersection of the zero set $S_i[u]=0$, for all $i$, with the space $X$ of exact solutions consists only of linearization unstable backgrounds. In the example of Einstein's equations, the $S_i[u]=0$ could consist of the Killing equations, which identify metrics with non-trivial symmetries.
As I've already mentioned, a linearization unstable point in $X$ is very similar to the singular point of an algebraic variety. Namely, consider the following analogy. Instead of a function on a manifold, $u$ is a finite dimensional vector. Instead of a PDE, $E[u]=0$ is a polynomial equation, which defines an algebraic variety $X$. A linearization unstable point $U\in X$ is, I believe, called a singular point of $X$. For instance, we could have $u=(x,y,z)$, $E[u]=x^2+y^2-z^2$, $U=(0,0,0)$, $E_U[u] = 0$, $Q[u] = x^2+y^2-z^2$, $S[u] = z$.
Could someone provide a translation dictionary into the appropriate terminology used in algebraic geometry (preserving grammatical structure if possible)?

linearization sability/instability --- ?
linearization stable/unstable point $U\in X$ --- ?
linearized solutions $\ker E_U$ --- ?
functions like $Q[u]$ --- ?
the zero set $Q[u]=0$ in $\ker E_U$ --- ?
functions like $S_i[u]$ --- ?
the zero set $S_i[u]=0$ --- ?

Also, do functions like $S_i[u]$ play a significant role in blowing up singular points?

REPLY [6 votes]: Maybe the best answer I can give is ''learn about deformation theory''.  But I will take a shot at writing the dictionary you request, in the case where $E(u)$ is a polynomial function

linearization stability <--> smoothness / instability <--> singularity
linearization stable/unstable point <--> singular / smooth point
linearized solutions <--> Zariski tangent space
functions like $Q[u]$ <--> obstructions
the zero set $Q[u]=0$ <--> (Edited.) the tangent cone
functions like $S_i[u]$ <--> the functions $\partial E/\partial u$, or maybe it is better to say, a certain Fitting ideal of the sheaf of Kahler differentials on $X$
the zero set $S_i[u] = 0$ <--> on $X$, you would I guess call it the singular locus; in the whole ambient space I don't know.

Regarding ''learn about deformation theory'', the point is that you started with some very explicit algebraic variety $X$ and then decided to think about its structure near a point.  But deformation theory lets you think about the following thing: say you want to think about the space of all algebraic ''solutions'' of some sort near a given ''solution'', but to a less explicit problem like ''describe all holomorphic maps from a genus 5 curve into projective space'', which of course can also be written as a PDE.  It may be difficult or worse to construct such a space globally, but still you can start to think about the local infinitesimal structure near a given solution using infinitesimal methods; and then there are very powerful algebraic tools (the Artin approximation theorem) which let you under good conditions construct an algebraic neighborhood of your given solution.<|endoftext|>
TITLE: When does $A^A=2^A$ without the axiom of choice?
QUESTION [48 upvotes]: Assuming the axiom of choice the following argument is simple, for infinite $A$ it holds: $$2\lt A\leq2^A\implies 2^A\leq A^A\leq 2^{A\times A}=2^A.$$
However without the axiom of choice this doesn't have to be true anymore. For example if $A$ is an amorphous set (infinite set that cannot be written as a disjoint union of two infinite sets), then it is actually true that $2^A<3^A<4^A<\ldots< A^A$. The reason these inequalities hold is that $A^A$ is actually Dedekind-finite, so whenever we remove elements we strictly decrease in cardinality.
Of course there are still sets that obey the equation $A^A=2^A$, even if $A$ cannot be well-ordered. For example given any set $A$ it is not hard to verify that $A^\omega$ has the property $A^\omega\times A^\omega=A^\omega$. From this follows:
$$2\lt A^\omega\leq 2^{A^\omega}\implies 2^{A^\omega}\leq\left(A^\omega\right)^{A^\omega}\leq \left(2^{A^\omega}\right)^{A^\omega}=2^{A^\omega}$$
(In fact we can replace $\omega$ by any set $\tau$ such that $\tau+\tau=\tau$)
But I have a hard time to believe that these two things are equivalent: $$A^A=2^A\iff A\times A=A.$$

Question I. Is there anything known on the properties of sets for which $A^A=2^A$?
Question II. If $2^A=A^A$ does not characterize the sets for which $A\times A=A$, does the axiom "For every infinite $A$, $2^A=A^A$", imply the axiom of choice?

If the answer is unknown, does this question (or variants, or closely related questions) appeared in the literature?
It seems like a plausible question by Tarski or Sierpiński. I found several other questions I have asked before to be questions that have been asked in one a paper or another.

REPLY [28 votes]: David Pincus in "A note on the cardinal factorial" (Fundamenta Mathematicae vol.98(1), pages 21-24(1978)) proves that $A^A=2^A$ does not imply the axiom of choice, therefore it does not characterise the sets for which $A=A\times A$. The counterexample is the model from his paper "Cardinal representatives", Israel Journal of Mathematics, vol.18, pages 321-344 (1974). 
As Pincus writes on the last lines of [Pincus78]:

c. Our arguments [ in the proof of "4.The Main Theorem; If $x$ is infinite then $2^x=x!$" ] have made little use of the particular definition of $x!$. Indeed let $\mathcal{F}$ be any set valued operation which satisfies: 
(1) The predicate $y\in\mathcal{F}$ is absolute from $M$ to $V$.
(2) $\mathsf{ZF}$ proves $|y|\leq x \Rightarrow |\mathcal{F}(y)|\leq |\mathcal{F}(y)|$ and $|2x|=|x|\Rightarrow 2^x\leq|\mathcal{F}(x)|$ for infinite $x$.
(3) $\mathsf{ZFC}$ proves $2^x=|\mathcal{F}(x)|$ for infinite $x$. 
The statement "For every infinite $x$, $2^x=|\mathcal{F}(x)|.$", holds
  in $M$ (and therefore is not an equivalent to the axiom of choice).
  Examples of $\mathcal{F}$, apart from $x!$, are $x^x$ and $x^x-x!$.

Therefore, in this Pincus model, $\mathsf{ZF}$ + $\lnot\mathsf{AC}$ + "for all infinite x, $2^x=x!=x^x=x^x-x!=|\mathcal{F}(x)|$" holds for any set valued operator $\mathcal{F}$ as above.
I should say that, after failing to come up with an answer myself, I found out about this by searching into my good old friend the "Consequences of the axiom of choice" by Howard and Rubin, Form 200, and Note 64. I try to reference this book whenever I can :)<|endoftext|>
TITLE: spectacular applications of functional analysis in resolutions of apparently unrelated problems
QUESTION [7 upvotes]: What are some of the spectacular applications of functional analysis to apparently unrelated problems.One that 
comes to my mind is Per Enflo's resolution of Hilbert's 5th problem.There are also reformulations of RH in Hilbert Space.I would be happly to hear about it's nice applications in geometry,number theory,topology,etc. specially in context to solving conjectures.

REPLY [5 votes]: One should probably mention Gelfand's proof of Wiener's theorem that, if a nowhere zero periodic $f$ has absolutely convergent Fourier series, then so does $1/f$.
There is also Kantorovich's famous note "On a problem of Monge":

Monge, in his memoirs of 1781, considering the problem of the most rational ways of transporting earth from an embankment to an excavation, proposed the following problem: divide two equal volumes into infinitesimal particles and associate them one to another so that the sum of the path lengths multiplied by the volumes of the particles be minimum possible.
In connection with this problem, Monge created the geometrical theory of congruences. As to the problem itself, he conjectured, but did not proved rigorously, that the paths of the mass translocation form a family of normals to a certain family of surfaces.
The same problem was studied later by Dupin, but a rigorous proof of the Monge theorem was given only a century later, in 1884, in a 200-page memoirs by Appell. (...)
Meanwhile, this assertion follows immediately from the abstract theorem mentioned above. (...)

REPLY [5 votes]: Here is one application, which may not seem spectacular to the modern  mathematician, but it has  many profound applications.
Suppose that $X$ is a reflexive  Banach space $E: X\to (-\infty, \infty]$ a convex function such that
$$ \lim_{\Vert x\Vert\to\infty}  E(X)=\infty, $$
and
$$ E(x)\leq \liminf_{y\to x} E(x). \;\;\forall x\in  X. $$
Then there exists $x_0\in X$ such that
$$ E(x_0)\leq E(x),\;\;\forall x\in X. $$
For example, one can use this to settle the so called Dirichlet principle which  generated many debates in the 19th century.

REPLY [3 votes]: For every smooth function $g$ the linear partial differential equation with constant coefficients $P(D)f=g$ is solvable in convex sets. Although the statement has nothing to do with functional analysis Malgrange's proof heavily
relied on Frechet space theory (and, of course, Fourier transformation).
The same holds if $g$ is a distribution (but one may object that distribution theory is part of functional analysis).<|endoftext|>
TITLE: Untangling entwined rigid chains in 3-space
QUESTION [6 upvotes]: I am interested in exploring the degree of "tangledness"
of two rigid chains in space.
A polygonal chain is a simple (non-self-intersecting) path
of segments in
$\mathbb{R}^3$, viewed as a rigid body.  Think of it as a bent
wire in space.
Define a move of a chain as either a translation of the whole chain
in a fixed direction, or a rotation of the chain about a fixed point axis.
I.e., a translation moves each point $p$ of the chain along the path
$p(t) = p + t v$ for a fixed direction $v$ as $t$ varies from $0$ to $1$.
Say that two chains are separable if there is a sequence of moves
during which the chains never touch one another, and eventually
end up on opposite sides of a plane.

Q1.
  If each chain consists of $n$ segments, is there an upper bound as
  a function of $n$ (and independent of the geometry of the chains)
  on the number of moves needed to separate a separable pair?

Here is an example for which I believe $O(n)$ moves suffices to separate:

          


A more focussed question seeks difficult-to-untangle chains:

Q2.
  Is there a pair of chains that requires $\Omega(n^2)$ moves to separate?

My explorations have not led to an example that clearly requires
many moves to separate.  For example, I think this pair of spirals also only
needs $O(n)$ moves.

          
          


A simplification would also be of interest:

Q3.
  The same questions but with "moves" restricted to translations.

Thanks for ideas or pointers to relevant literature!

REPLY [2 votes]: Hopefully, someone with better graphics-making ability than me will be able to post a picture of what I'm thinking of (or show that it doesn't work), but here's an attempt to show a pair that require $\Omega(n^2)$ moves.
For now, I'm restricting to translations only.
The first chain is the standard square wave form in the first figure above. For the second chain, take a square spiral on a plane perpendicular to the square wave. The spiral should have largest side length equal to the vertical height of the square wave (so the only way to move a vertical segment through is by pushing it through that one slot. Now, to force the wave to travel the spiral over and over, we add another piece to the second chain. It should be on a plane parallel to the plane of the spiral, and should be a modified square that is just too small for the wave to fit through. We modify the square on the pair of horizontal sides, adding a small detour to each edge, pushing out a small section and adding little extra space, so that the square wave can fit through only if it is centered on the square. Then, line the two up so the center of the square is lined up with the center of the spiral.
Then, the square wave starts going through the center of both parts of the second chain. In order to get each vertical segment through the spiral, we need to go around the spiral to the outside slot, but then to get that vertical segment through the square, we need to spiral back around to the middle.
Does that make sense?
If it does, it seems plausible to me that we'd be able to make 'rotation guards' to add to the second chain, that force us to do those motions even if rotations are allowed, but I'm not sure.<|endoftext|>
TITLE: How many "elementary" characterizations of twisted SU(2) representation varieties are known?
QUESTION [8 upvotes]: If $\Sigma_g$ is a genus-$g$ surface, $g \geq 2$, then let $\mathcal{M}(\Sigma_g)$ be its twisted SU(2) representation variety, i.e. $$\mathcal{M}(\Sigma_g) := \{ (A_1, B_1, \ldots, A_g, B_g) \in SU(2)^{2g} \:|\: [A_1,B_1]\cdots[A_g,B_g] = -I \}/SO(3),$$ where $SO(3) = SU(2)/\{\pm 1\}$ acts on these tuples by conjugating each component.  Then $\mathcal{M}(\Sigma_g)$ is a smooth, compact $(6g-6)$-dimensional symplectic manifold, and it's symplectomorphic to the moduli space of stable bundles over a genus-$g$ Riemann surface with rank $2$ and determinant equal to some fixed line bundle of odd degree (this moduli space is a Kaehler manifold).
In his 1968 paper "Stable bundles of rank 2 and odd degree over a curve of genus 2", Peter Newstead showed that $\mathcal{M}(\Sigma_2)$ is isomorphic to the intersection of two generic quadric hypersurfaces in $\mathbb{P}^5$.  In a recent paper, Ivan Smith used this characterization to get a lot of information about the Fukaya category of $\mathcal{M}(\Sigma_2)$ in terms of the Fukaya category of $\Sigma_2$.  (My impression is that people care a lot about the Lagrangian intersection theory of $\mathcal{M}(\Sigma_g)$ because if you understand that, you understand the instanton Floer homology of 3-manifolds.)  My question is:

are there any similarly elementary characterizations of $\mathcal{M}(\Sigma_g)$ for $g \geq 3$?

I know that bits and pieces about $\mathcal{M}(\Sigma_g)$ are known, e.g. its symplectic volume and characteristic classes (maybe even its cohomology?); I would really like to know about more elementary things that are known to be isomorphic to it as symplectic/Kaehler manifolds.

REPLY [3 votes]: I think Nate might know this by now, but in case anyone else is curious, there is a generalization of the intersection-of-quadrics-in-$\mathbb{P}^5$ picture to general genus, proved first in this paper, I believe

Classification of Vector Bundles of
  Rank 2 on Hyperelliptic Curves, U.V.
  Desale and S. Ramanan, Inventiones

The story is that $\mathcal{M}_g$ is homeomorphic (in fact, symplectomorphic), to the space of $g-2$ dimensional linear subspaces in the intersection of two quadrics in $\mathbb{P}^{2g+1}$.<|endoftext|>
TITLE: Decomposition of induced representations / Refinement of Mackey's criterion
QUESTION [14 upvotes]: There are already some questions with almost the same title, but they are more restrictive.
Let $G$ be a finite group, $H$ a subgroup, $V$ an irreducible representation of $H$,
and $W=Ind_H^G V$ the induced representation. 

Is there a simple way to compute the number of irreducible components of $W = Ind_H^G V$?

Let me precise my question by saying what kind of answers I would hope for. 
There is a well-known description of $Res_G^H Ind_H^G V = Res_G^H W$ as follows. Let $S$ be a set of representatives of the double classes of $H \backslash G / H$. For $s \in S$, let $H^s$
be the group $sHs^{-1} \cap H$ and $V^s$ the representation of $H^s$ on the space $V$ where $x \in H^s$ acts by $s^{-1}xs$ on $V$. Then
$$Res_G^H W = \oplus_{s \in S} Ind_{H_s}^H V^s.$$
Using this description, one gets using Frobenius reciprocity twice that $$Hom_G (W,W) = Hom_H(V,Res_G^H W) = \oplus_{s \in S} Hom_H(V,  Ind_{H^s}^H V^s) = \oplus_{s \in S} Hom_{H^s} (V,V^s)$$
which implies the well-known criterion of Mackey: $W$ is irreducible if and only if for all $s \in S$, $V$ and $V^s$ have no $H^s$-irreducible subrepresentations in common.
So assuming we can compute $S$, the groups $H^s$, the representations $V^s$, and how $V$ and $V^s$ decomposes as representations of $H^s$ (this may be hard in practice, but let's assume we can do that), then we know when $W$ is irreducible, and in general we know how
to determine the dimension of $Hom_G(W,W)$. (Of course all of this is standard cf. Serre Linear Representations for instance). 
Yet this falls short of answering the question,
because if for example you know that the dimension of $Hom_G(W,W)$ is $4$, that doesn't tell you if $W$ is the sum of 4 non-isomorphic irreducible representations, or of 2 copies of the same irreducible
representation. What I'd like would be a way to tell us, by looking at the $H^s$ and the $V^s$ etc, of determine which one it is. Or more generally, how to "read" the decomposition of $W$ into irreducible reps. in terms of the $V^s$, $H^s$ etc. If it is not possible, I would also 
like an explanation why.

REPLY [2 votes]: The proof of Mackey's theorem on intertwiners actually tells you how to construct the endomorphism algebra of an induced representation, not just its dimension. So, if you work a little harder, you may be able to get the additional information that you want.
To see how this algebra can be found, you may refer to my notes
http://www.imsc.res.in/~amri/html_notes/notesch1.html#x4-70001.4
A convolution product can be defined on $\Delta$'s in the notes, which will correspond to multiplication in the endomorphism algebra.
Added:
An interesting special case is decomposing parabolically induced representations of general linear groups over finite fields, which is beautifully explained in the notes of Howe and Moy Harish-Chandra homomorphisms for $p$-adic Groups.
Such algebras are often called Hecke algebras, and there is a vast literature on them.<|endoftext|>
TITLE: Is strong multiplicity one (obviously) stronger than multiplicity one?
QUESTION [5 upvotes]: In the theory of automorphic representations one says that G satisfies a "multiplicity one" property if every cuspidal representation occurs with multiplicity one in $L^2(G(F)\backslash G(A))$.
One also says that G satisfies "strong multiplicity one" if a cuspidal representation is uniquely determined up to isomorphism by its behaviour at cofinitely many places of F. 
My question is: does the latter always imply the former (as the name would suggest), and is there a decent reason?
This question was prompted while thinking about Jacquet-Langlands, whose proof (via the trace formula) seems to give multiplicity one as a by-product and whose statement also directly implies strong multiplicity one (for inner forms of GL2 using the result for GL2 itself). However, I didn't see any way to deduce multiplicity one directly from the usual statement.
I also can't see how strong multiplicity one for classical modular forms should directly imply the q-expansion principle (which I guess is roughly the same as multiplicity one). Apologies if I'm missing something extremely obvious on this dozy Sunday evening.
Thanks,
Tom.

REPLY [4 votes]: I'm not sufficiently privileged to leave comments yet, but in support of the semantics I'll mentioned the following paper:
http://www2.math.ou.edu/~rschmidt/papers/sqfree.pdf
In this paper, Schmidt refers to his Corollaries 3.16 and 3.17 as "strong multiplicity one theorems".  This would agree with Tom's definition of "strong multiplicity one".  In the case of $GSp(4)$ "multiplicity one" is not known (representations do not necessarily have Whittaker models, which complicates things), so even though Schmidt can say isomorphic at cofintely many place implies isomorphic, he is not able to say that the global representations are literally equal because the latter may (but are not expected to) occur with multiplicity greater than one in $L^2(GSp(4, \mathbb{Q}) \backslash GSp(4, \mathbb{A})$.
So, from an appropriate point of view, Bump's statement of Theorem 3.3.6 is a "multiplicity one" + "strong multiplicity one" (the former of which is known to be true for $GL_2$ of course).<|endoftext|>
TITLE: For which metric measure spaces is the Hardy-Littlewood maximal operator not of weak type (1,1)?
QUESTION [6 upvotes]: Let $(X,d,\mu)$ be a metric measure space, i.e. $\mu$ is a Borel measure on the metric space $(X,d)$. I'll denote the Hardy-Littlewood maximal operator - either centred or uncentred, I don't mind which - by $M$. More precisely, for a function $f$ on $X$, define
$$ Mf(x) := \sup_{B \ni x} \frac{1}{\mu(B)} \int_B f(y) \; dy,$$
with the supremum taken over either all balls centred at $x$ or over all balls containing $x$.
There are plenty of theorems of the form 'suppose $(X,d,\mu)$ has property A. Then $M$ is of weak-type (1,1)', (for example, property $A$ could be '$(X,d,\mu)$ is doubling') thus implying that $M$ is bounded on $L^p(X)$ for $1 < p \leq \infty$ (since $L^\infty$ boundedness of $M$ is always obvious).
I'm interested in the converse. What is an example of a space $(X,d,\mu)$ is not of weak-type (1,1)? And how would one go about proving such a result?
Further, if there are any spaces with $M$ not of weak-type (1,1), do we still have $L^p$ boundedness for any $p > 1$?

REPLY [3 votes]: Such counterexamples are known. For instance, P. Sjögren has shown that the associated uncentered maximal to the $2$-dimensional Gaussian measure on $\mathbb{R}^2$ isn't weak type $(1,1)$. In a later paper, Forzani, Scotto, Sjögren and Urbina showed that this operator is strong type $(p,p)$ for $p>1$.
A well-known problem in this area is to determine if the ball (with Lebesgue measure) in $\mathbb{R}^n$ satisfies a weak type $(1,1)$ bound with a constant independent of the dimension. Related to this, there are a number of constructions of infinite families of metric spaces with nice properties, but for which the weak type $(1,1)$ constant can't be taken independent of the space in the family.  See, for instance, the section ``Lower bounds" (page 8) of this paper of  Naor and Tao for some such constructions.<|endoftext|>
TITLE: What is the order of the lower tail of a Chi-Squared distribution?
QUESTION [8 upvotes]: Let X be a random variable with having a chi-squared distribution with n degrees of freedom and let y be some real number at most n. Is it known how P (X < y) behaves at least in some reasonable range of y? For instance, could one determine exactly the order of the latter probability when y=cn (for some $0\lt c \lt 1$) when n becomes large?
In the end, I am interested in the upper bound for the latter probability, but I am not sure if the usual Chernoff bound gives the correct magnitude.

REPLY [13 votes]: You can get the correct exponential order of the decay of the probability from a large deviation principle. If $X_n$ has chi-squared distribution with $n$ degrees of freedom then $$X_n = Z_1^2 + Z_2^2 + \cdots Z_n^2,$$ where the $Z_i$ are independent standard normal random variables. Then, Cramer's theorem gives the exponential order of decay for both the left and right tails. That is, 
$$
 \lim_{n\rightarrow\infty} \frac{1}{n} \log P( X_n < c n ) = - I(c), \quad 0 < c < 1,
$$
and
$$
 \lim_{n\rightarrow\infty} \frac{1}{n} \log P( X_n > c n ) = - I(c), \quad 1 < c < \infty,
$$
where 
$$
 I(c) = \frac{c-1-\ln(c)}{2}.
$$
Since you said that you're looking for an upper bound, it should also be noted that an examination of the proof of Cramer's theorem shows that you can actually get uniform exponential upper bounds (and not just asymptotics as stated above). In fact, 
$$
 P(X_n < cn) \leq e^{-n I(c)}, \quad \forall n\geq 1, \text{ and } 0 < c < 1.
$$<|endoftext|>
TITLE: de rham model for relative cohomology
QUESTION [14 upvotes]: In GTM82, I read a model for the relative cohomology of (M,N) with N a submanifold of M. 
And in the page: 
Relative De Rham cohomologies, 
I got to know that there is another model for relative cohomology by using differential forms of M, like those forms which restrict to 0 on N. 
Taladris called this Godbillon model. And Johannes Ebert said this model is used by jost in his book "Riemannian geometry and geometric analysis". But I can't find this model although  I have checked that book(the 6th edtion) twice.
I wanna know in which paper,book, or article I can find the definition of the second model.

REPLY [13 votes]: Suppose that $C$ is a closed subset of $M$. Denote by $\newcommand{\eO}{\mathscr{O}}$ $\eO$ its complement.  
The DeRham cohomology of $M$ is in fact the cohomology  associated to a particular soft resolution  of the constant sheaf $\newcommand{\ur}{\underline{\mathbb{R}}}$ $\ur$ on $M$.
To any   sheaf $\newcommand{\eS}{\mathscr{S}}$ $\eS$ on $M$ we can associate a sheaf $\eS_{\eO}$, also on $M$  whose stalk $\eS_{\eO}(x)$ at $x\in M$ is $\eS(x)$ if $x\in \eO$, and $0$ if $x\in M\setminus \eO$.  For any open subset $U\subset M$  the space $\Gamma(U, \eS_{\eO})$ of sections of $\eS_{\eO}$ over $U$ consists of the section $s\in \Gamma(\eO\cap U,\eS)$ such that the support of $s$ is closed in $U$.  The operation $\eS\mapsto \eS_{\eO}$ is an exact functor. Moreover if $\eS$ is soft, so is $\eS_{\eO}$. Thus the DeRham resolution
$$ 0\to\ur\to\Omega^0\to\Omega^1\to\cdots, $$
$\Omega^k=$ the sheaf of smooth $k$-forms  on $M$,  produces a soft resolution of $\ur_{\eO}$
$$ 0\to\ur_{\eO}\to\Omega_{\eO}^0\to\Omega_{\eO}^1\to\cdots . $$
Hence, the cohomology of the sheaf $\ur_{\eO}$ is computed by the the cohomology of the complex
$$ \Gamma(M, \Omega_{\eO}^0)\stackrel{d}{\to}\Gamma(M, \Omega^1_{\eO})\stackrel{d}{\to}\cdots, \tag{1} $$
where, as explained above $\Gamma(M, \Omega^k_{\eO})$   consists of smooth $k$-forms on $\eO$  whose support is a closed  subset in $M$. Equivalently $\Gamma(M, \Omega^k_{\eO})$ consists of forms on $M$ whose supports do not intersect $C$. If $M$ is compact, then   $\Gamma(M, \Omega^k_{\eO})$ consists of form  with compact support contained in $\eO$.
On the other hand, the cohomology of $\ur_{\eO}$ can be identified with the relative cohomology $H^\bullet(M,C;\mathbb{R})$.  The complex (1) is  the a DeRham model for this  cohomology. For more details I recommend the comprehensive book of   Kashiwara and Schapira Sheaves on Manifolds  or my notes which are less comprehensive, but may guide you through the literature.  In particular, see  Remark 2.17 of my notes.<|endoftext|>
TITLE: Polynomial with all zeros on a circle and many real coefficients
QUESTION [8 upvotes]: On a circle (or a line) $\Omega$ in the complex plane that is not symmetric w.r.t. the real axis, choose $n\ge5$ distinct points $z_1,...,z_n$ and consider the polynomial $p(z)=\prod_j(z-z_j)=z^n+a_{n-1}z^{n-1}+\cdots+a_0$. Question: How many of the $a_k$'s can be reals at most ?
Intuitively, imagining something like a Java applet, there are "more or less" $n-1$ degrees of freedom to move the points around on $\Omega$, so we might reasonably expect that it is possible to choose them (say for an appropriate $\Omega$) such that all but one $a_{k_0}$ are reals.

How could that be rigorously proven? 
If it is true, can it be done for any $k_0\in\lbrace0,...,n-1\rbrace$? 
How to find a concrete solution? 
Can such a polynomial even be rational, i.e. such that $a_{k_0}\in\mathbb Q[i]$ and $a_k\in\mathbb Q$ for $k\ne k_0$?

REPLY [3 votes]: Suppose it is a line, passing through the origin under the angle $\phi$.
Then your polynomial must be
$$z^n+a_1z^{n-1}+...+a_0=\prod_{j=1}^n(z-t_je^{i\phi}),$$
where $t_j$ are real. Vjeta's formulas give
$$a_k=\pm\sum t_{i_1}...t_{i_k} e^{ik\phi},$$
Now how can $a_k$ be real?
First way: $e^{ik\phi}$ is real. For how many $k=1...n$ this can happen, is easy to find out.
Second way: $$b_k:=\sum t_{i_1}...t_{i_k}=0.$$
Of course this can happen for all $k$ if all $t_k=0$. 
If you want to exclude $t_j=0$ than the question is reduced to
"how many zero coefficients can have a polynomial with all roots real and non-zero ?". I mean the real
polynomial $\prod(z-t_k)$, whose coefficients are $\pm b_k$.
For this real polynomial, you can use the following theorem of Descartes:
The number of positive zeros of a real polynomial is at most the number of sign changes
in the sequence of coefficients (which is at most the number of non-zero coefficients minus 1).
Same applies to the number of negative zeros if you make the change of the variable $x\to-x$,
which changes the sign switches but does not change the number of non-zero coefficients.
If you want all roots to be distinct, at most one of them is zero.
I leave the details to you.<|endoftext|>
TITLE: Looking for a copy of Leo Harrington's unpublished notes on the first nonprojectible ordinal
QUESTION [18 upvotes]: Sometime around 1975, Leo Harrington wrote a set of notes, apparently 13 pages long, entitled Kolmogorov's $R$-operator and the first nonprojectible ordinal.  I do not know how widely they were circulated (or if they were ever available from the UCB library or anywhere).
From what I understand, these notes relate recursion on the first stable ordinal in the first nonprojectible to recursion in a certain explicitly defined type-3 functional.  A more precise statement is given in Thomas John's paper “Recursion in Kolmogorov's $R$-operator and the ordinal $\sigma_3$” (J. Symbol. Logic 51 (1986) 1–11), but the proof is not reproduced there.  A related but slightly different result of Harrington's, with a different type-3 operator, is quoted (as example 4.10) by Stephen Simpson's “Short course on Admissible Recursion Theory” (355–390 in: Fenstad &al. eds., Generalized Recursion Theory II (Oslo 1977), North-Holland 1978).
I'd very much like to see a copy of these notes, or a proof of any closely related result (e.g., the one quoted by Simpson's paper mentioned above).
The author has been kind enough to see if he can find them, but he isn't too optimistic.  I've also written to a number of people who worked in the subject around that time (Sacks, Shore, Simpson and Soare), but without success.  So I now turn to MO in the hope that someone has heard of these notes or knows where a copy might be found.
[Xref: link to meta thread]
PS: While I'm aware that offering prizes other than reputation points is frowned upon on MO, if someone should go through the (real-world!) trouble of copying, scanning or mailing these notes for me, I think it would be appropriate that I should respond with a small (real-world!) token of thanks, like an Amazon gift card or something. :-)

REPLY [23 votes]: I asked Alekos Kechris.  He had a copy and made a scan of it. Here is a link to it.
http://dl.dropbox.com/u/2566697/Harrington.pdf
Regards,
Ted<|endoftext|>
TITLE: Small-index subgroups of SL(3,Z)
QUESTION [20 upvotes]: I would like to know the smallest-index subgroups of ${\rm SL}(3,\mathbb{Z})$.
The smallest I could find has even entries $a_{3,1}$ and $a_{3,2}$,
along the bottom row.  I could not figure out whether there are
subgroups of index $2$ or $3$.
A search found lots of information about ${\rm SL}(2,\mathbb{Z})$,
but not about ${\rm SL}(3,\mathbb{Z})$.

REPLY [9 votes]: You don't need the congruence subgroup property to see that this group has no proper subgroups of index less than $7$, in other words that it has no nontrivial homomorphism to $S_6$. 
Nor do you need a presentation of the group; all you need is that it is generated by six elements satisfying certain relations, and a little patience.
Let $x_{i,j}$ be the matrix with $1$ on the diagonal, $1$ in the $(i,j)$, spot, and $0$ otherwise. It is not hard to see, by row reduction, that $SL(3,\mathbb Z)$ is generated by these elements. When $i$, $j$, and $k$ are distinct then $x_{i,j}$ is the commutator of $x_{i,k}$ and $x_{k,j}$, and also $x_{i,j}$ commutes with $x_{i,k}$ and with $x_{k,j}$.
Now suppose for contradiction that $S_6$ has elements $x_{i,j}$ satisfying these same relations and not all equal to the identity. No $x_{i,j}$ can be the identity because then by taking commutators they all would be the identity. Being a commutator, $x_{i,j}$ must belong to $A_6$. It is the commutator of $x_{i,k}$ and $x_{k,j}$, two elements of $A_6$ that commute with it. Looking at all cases, you see that the only nontrivial elements of $A_6$ whose centralizers are nonabelian are those of the form $(ab)(cd)$, product of two disjoint $2$-cycles. If $x_{i,j}=(ab)(cd)$ then $x_{i,k}$ and $x_{k,j}$ must be elements of its centralizer and again of the same type. But the only such elements are $(ab)(cd)$, $(ac)(bd)$, $(ad)(bc)$, $(ab)(ef)$, and $(cd)(ef)$. There are only a few possibilities for pairs of these whose commutator is $(ab)(cd)$. You'll find that none of them will lead to a solution, a choice of $x_{i,j}$ for all $i$ and $j$.<|endoftext|>
TITLE: Is the Binomial Expectation of Convex Function Convex in p?
QUESTION [6 upvotes]: Suppose $X$ has a binomial distribution with success probability $p$ and $n$ trials and let $h(\cdot)$ be a positive convex real-valued function.
Is the function $g(p)=\mathbb{E}[h(X)\ |\ p]$ convex in $p$?
Related questions:
Binomial Expectation of Convex Function and expected values over binomial distributions
Note: Based on trial-and-error analysis, it appears to be convex. If anyone is interested, I can send them a spreadsheet with some numbers.

REPLY [3 votes]: This is known, for more general exponential families in place of the binomial one,
from papers of Shaked (1980) and Schweder (1982), see http://www.jstor.org/stable/10.2307/2984960 and http://www.jstor.org/stable/10.2307/4615873.<|endoftext|>
TITLE: Elliptic curves over QQ with isomorphic n-torsion
QUESTION [16 upvotes]: Hey all,
I just ran into a question of Mazur from his 1978 paper 'Rational isogenies of Prime Degree' and I wonder what is the status of his question:
"Are there examples of elliptic curves $E/\mathbf{Q}$, $E'/\mathbf{Q}$ which are not isogenous over $\mathbf{Q}$ such that $E[N] \cong E'[N]$ (as $\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ modules) for some $N \geq 7$?"
Also the case $N=6$ is interesting, as it is the only genus $1$ case. For $N\leq 5$ the genus of $X(N)$ is zero, hence for all twists it will have infinitely many rational points.
Thanks for your answers.

REPLY [12 votes]: Tom Fisher has a paper where he shows that there are infinitely many examples for $N = 9$ and $N = 11$. See this link.<|endoftext|>
TITLE: bornological vector spaces over a non-archimedean field
QUESTION [11 upvotes]: Let $k$ be a complete non-archimedean field. In definitions I have seen of bornological vector spaces over $k$ there are usually some extra assumptions on the non-archimedean field. For instance in 'Espaces analytiques relatives et theorem de finitude' by Houzel it is assumed that the valuation is non-discrete and that $k$ is maximally complete. On page 43 of 'Seminaire Banach' published as Springer Lecture notes in Mathematics volume 277, it is assumed that the valuation is non-discrete. What is the main reason behind these restrictions? I am interested in bornological vector spaces over a field with trivial valuation. Banach spaces over such a field make sense and therefore one gets a metric space and therefore a bornological set where the bornology is compatible with the linear structure. This should still be a (complete) bornological vector space hopefully. For what parts of the theory are these extra restrictions needed or useful? Are there some pathologies about the category of bornological vector spaces over a general complete non-archimedean field that are not present when you add these extra assumptions?

REPLY [7 votes]: I try to reply to your questions:
1) In the first part of "Espaces analytiques relatives et theorem de finitude" Houzel says that the field under consideration is supposed to be maximally compact. I don't see any point where he use this hypothesis in his article neither in the result that he recall from the "Seminarie Banach". Moreover, in the second part, where he expose the sheaf-theoretic (global) version of the result obtained in the first part he recall the notation and remove this hypothesis (see on page 29 the first two lines of the paragraph "Faisceaux bornologiques"). Hence my idea with respect to this hypothesis is that it is simply a misprint.
2) For sure you can develop a theory of bornological or topological vector spaces over any valued field (also over any field with more exotic structures). The problem is that already in the case of trivially valued fields you find pathologies. First, the notion of convexity is quite strange: since $k = k^\circ$ then you find that the natural generalization of the absolute convex hull of a subset $X \subset E$ (where $E$ is a $k$-vector space) is given by $\Gamma(X) = \left \{ \sum \lambda_i x_i | \lambda_i \in k^\circ = k, x_i \in X \right \}$. Hence $\Gamma(X)$ is the linear span of $X$. 
The main problem in this situation is that there is no analog of the duality between bornological and topological vector spaces, which is the main topic of the seminaire Banach. Houzel construct two adjoint functors $t: Born \to Top$ and $b: Top \to Born$ where $Born$ is the category of bornological vector spaces of convex type and $Top$ is the category of locally convex topological vector spaces. For the $t$ functor you consider on a bornological vector space of convex type the vector space topology given by bornivorous sets (i.e. sets which absorbs all bounded sets). And for the $b$ functor you consider on locally convex topological vector space the Von Neumann bornology. 
So if you perform this construction on a seminormed vector space $E$ over a non-trivially valued field you get that $E \cong b(t(E))$ and $E \cong t(b(E))$ (but this is also true for more general $k$-vector space topologies and bornologies). This is false for the trivial valued field even thinking of $k$ as a $1$-dimensional vector space over itself. Cause if $k$ is trivially valued, then all subsets of $k$ are bounded. In particular $k$ itself is bounded, so the only bornivorus set for this bornology is $k$. Therefore, in this case, $t(k)$ has the indiscrete topology but the trivial valuation gives to $k$ the discrete topology. Moreover, if you try to describe the Von Neumann bornology for the discrete topology, you don't find a bornology cause $0$ is a neighborhood of itself and $0$ doesn't absorb nothing else than itself. This mean the $0$ would be the only bounded set for the Von Neaumann bornology but this doesn't make sense.
This lack of duality is (i think) the main issue for which Houzel exclude the trivial valuation in his work.<|endoftext|>
TITLE: Real forms of Drinfeld-Jimbo quantum groups
QUESTION [7 upvotes]: A real form of a Hopf algebra $H$ over $\mathbb{C}$ is defined to be a $\ast$-structure on $H$ which is compatible with the coproduct.  Compatibility of the $\ast$-structure with the counit and antipode then follows.
The real forms of the Drinfeld-Jimbo quantum groups $U_q(\mathfrak{g})$ have been classified.  This result is stated as Theorem 20 in Chapter 6 of Quantum Groups and their Representations, by Klimyk and Schmudgen, and also Proposition 9.4.2 of A Guide to Quantum Groups, by Chari and Pressley.
For a given $\mathfrak{g}$, there are $\ast$-structures when $q \in \mathbb{R}$ or when $|q|=1$.  These $\ast$-structures depend on a diagram automorphism of the Dynkin diagram of $\mathfrak{g}$, plus some extra parameters, and it is understood which sets of parameters give equivalent $\ast$-structures.
There is also an exceptional case, but the two sources I have cited differ on what the exceptional case is.  Klimyk and Schmudgen say that the exceptional case is when $\mathfrak{g} = \mathfrak{sp}_{2n}$ and $q \in i \mathbb{R}$, while Chari and Pressley say that the exceptional case is $\mathfrak{g} = \mathfrak{so}_{2n+1}$ and $q \in i \mathbb{R}$.
Neither book contains a proof, nor cites a source, although Chari-Pressley gives a sketch of the idea of the proof.  So I would be interested in knowing the following things:

Which is correct?
What is the original reference for the classification of the real forms of $U_q(\mathfrak{g})$?


To set the record straight
It appears that the paper Real forms of $U_q(\mathfrak{g})$, by Eric Twietmeyer, is the original reference.  According to that paper, it is $\mathfrak{sp}_{2n}$ that has real forms for $q \in i \mathbb{R}$.
Many thanks to Uwe Franz for digging up that reference!

REPLY [3 votes]: Two references I recall are
E. Twietmeyer, Real forms of Uq (g), Lett. Math. Phys. 24, 49-58, 1992.
V. Lyubashenko, Real and imaginary forms of quantum groups, Lecture Notes in Math. 1510, 1992, pp 67-78.<|endoftext|>
TITLE: Central Limit Theorem (and Berry-Esseen theorem) for non-independent variables
QUESTION [8 upvotes]: Consider the triangular array $X_{n,k}$ such that, for each $n>0$, the variables $(X_{n,1},\cdots,X_{n,n})$ have the following properties:

For any given $1 \le L \le n$, all
subsets of
$(X_{n,1},\cdots,X_{n,n})$ of size
$L$ have the same joint distribution
(even after applying an arbitrary permutation). 
Each $X_{n,k}$ has
zero mean, variance
$0<\sigma^2<\infty$, and third
absolute moment $0<\rho<\infty$
Each $(X_{n,k_1},X_{n,k_2})$ pair
($k_1 \ne k_2$) has covariance
$\frac{-C}{n}$ for some $0 < C < \infty$ (hence the variables are all negatively correlated, and the correlation tends to zero)

Let $S_n = \sum_{k=1}^{n} X_{n,k}$.  Is $\frac{S_n}{\sqrt{\mathrm{Var}[S_n]}}$ asymptotically distributed as $N(0,1)$?  If so, is a convergence rate of $O(\frac{1}{\sqrt{n}})$ achieved (cf. Berry-Esseen theorem)?  What about in the multivariate setting where each $X_{n,k}$ is a random vector (and the relevant quantities above are replaced by vectors/matrices)?
If it helps, we can also assume that the $X_{n,k}$ are uniformly bounded in $n$ and $k$ with probability one.  Answers with further assumptions than the ones listed will also be appreciated.

REPLY [3 votes]: No.  You can even suppose the variables are all identically distributed, bounded and the covariances are all zero, and you still won't have convergence to a normal distribution.  To see this, take my main example $(X_n)$ from here (for any fixed $N>1$), arrange it into a triangular array by letting $Y_{n,k} = X_k$, and then let $X_{n,k} = Y_{n,\pi_n(k)}$ where $\pi_n$ is a random permutation (with all permutations equally likely) of $1,...,n$.  Since $Y_{n,1},...,Y_{n,n}$ is $N$-tuplewise independent, the variables in the randomly permuted row will still have zero covariance.  And since the $(X_n)$ don't satisfy the CLT, neither will the $(X_{n,k})$.<|endoftext|>
TITLE: group actions on blow-ups 
QUESTION [5 upvotes]: Hi friends,
Let $X$ be a projective variety over a field $k$ of characteristic zero. Assume that $X$ comes with the action of a finite group $G$. Now let $Z$ be a closed subvariety stable under the action of $G$. Let $\pi: \tilde{X} \to X$ be the blow-up of $X$ along $Z$. 
How can one extend the action of $G$ from $X$ to $\tilde{X}$? 
Any help will be appreciated.

REPLY [15 votes]: Yes you can extend the action. One can prove this using the universal property of blow-ups (see Hartshorne Corollary II.7.15). 
As $Z$ is invariant under the action of $G$, the inverse image of $Z$ with respect to the morphism $G \times X \to X$ is $G \times Z$. Therefore on applying the universal property of blow-ups to this morphism we obtain a morphism $G \times \widetilde{X} \to \widetilde{X}$. Now, by assumption this morphism satisifies the identities $$(gh)x = g(hx), \quad ex = x$$ for all $x$ in $\widetilde{X}\setminus  E$, where $E$ denotes the exceptional divisor of the blow-up. However since any two morphisms which are equal on an open dense subset must be equal on the whole space, we see that these identities hold for all $x$ in $\widetilde{X}$, i.e. the morphism $G \times \widetilde{X} \to \widetilde{X}$ gives an action of $G$ on $X$.
Note that in this argument we did not use the fact that $G$ was finite, it works for any algebraic group.<|endoftext|>
TITLE: On the stable splitting of loops on a suspension
QUESTION [12 upvotes]: Let $X$ be a connected, based CW complex. Then the James splitting
of $\Sigma\Omega\Sigma X$ gives, in particular, a weak equivalence of spectra
$$
\Sigma^{\infty} \Omega\Sigma X_+  \quad \simeq \quad  \Sigma^{\infty} (S^0 \vee X \vee X^{[2]} \vee X^{[3]} \vee \cdots ) ,
$$
where $X^{[n]}$ denotes the $n$-fold smash product of $X$ with itself. Each side  of this equivalence has the structure of $A_\infty$-ring spectrum (the structure on the right side is induced by
concatenation $X^{[n]} \wedge X^{[m]} \cong X^{[m+n]}$, and the right side can be seen as the tensor algebra over the sphere spectrum on the points of $X$). 
Now, my understanding$^\dagger$ is that the Cartan formula for Hopf invariants
implies that the above splitting is multiplicative up to homotopy.
Question:  Can the above splitting be 
enriched to an equivalence of $A_\infty$-rings?
If so, can anyone provide me with a reference?
$^\dagger\tiny \text{From being once upon a time in Bill Richter's orbit.}$

REPLY [12 votes]: Hi John, this is not a complete proof, but it should give the idea of one.  There is a clever proof of the splitting of suspension spectra of $\Omega^n\Sigma^n X$ due to Ralph Cohen in his
paper "Stable proofs of stable splittings".  His proof is also given in section VII.5 of
LMS "Equivariant stable homotopy theory".  It applies more simply to give the splitting that you are interested in.  Modernizing using EKMM Section II.4 (or symmetric or orthogonal spectra), the right side of your equivalence is equivalent to the free $A_\infty$ ring spectrum generated by $\Sigma^{\infty} X$. With minor finagling (about base points in particular), the freeness will give a map of $A_{\infty}$ ring spectra from right to left. It should be an equivalence by inspection of the cited proof.<|endoftext|>
TITLE: On a Hirzebruch surface
QUESTION [16 upvotes]: I am trying to solve exercise in Huybrechts's book 'Complex geometry'
While solving problems, one problem kept me from going forward.
That is,
The surface $\Sigma_n=\mathbb{P}$ $(\mathcal{O}_ {\mathbb{P}^1}\oplus \mathcal{O}_{\mathbb{P}^1} (n))$ is n-th Hirzebruch surface.
Show that $\Sigma_n$ is isomorphic to the hypersurface
{${([x_0 , x_1 ],[y_0,y_1,y_2]):x_0^n y_1 - x_1^n y_2 =0}$}$\subset \mathbb{P}^1 \times \mathbb{P}^2$.
How to attack this?
I hope someone shed me a light!

REPLY [2 votes]: The holomorphic section of $\mathcal O_\mathbb{P^1}(n)$ is canonically isomorphic to $\mathbb C[z_0,z_1]_n$ of all homogeneous polynomial of degree n, which is an n+1 dimension vector space. Let $s_0$, $s_1$ be two holomorphic section corresponds to the polynomial $z_{0}^n$ and $z_{1}^n$ respectively. 
Then we have a map $\mathcal{O}_{\mathbb{P}^1}\oplus\mathcal{O}_{\mathbb{P}^1}\longrightarrow\mathcal{O}_{\mathbb{P}^1}(n) $ by sending $(f,g)$ to $fs_0+gs_1.$ This map will induces an isomorphism from $\Sigma_n$ to $\mathbb{P}(\mathcal{O}_ {\mathbb{P}^1}\oplus \mathcal{O}_{\mathbb{P}^1} (n))$.
If you want to write the isomorphism explicitly. Here we define the embbeding $\phi:\mathbb{P}(\mathcal{O}_ {\mathbb{P}^1}\oplus \mathcal{O}_{\mathbb{P}^1} (n))\longrightarrow \mathbb P^1\times \mathbb P^2$
On $\mathbb{P}(\mathcal{O}_ {\mathbb{P}^1}\oplus \mathcal{O}_{\mathbb{P}^1} (n))\mid_{U_0}$, this map is give by
$U_{0}\times \mathbb P^{1}\longrightarrow \mathbb P^1\times \mathbb P^2\\
[x_0,x_1]\times[a,b]\mapsto[x_0,x_1]\times[a,b\frac{{x_1}^n}{{x_0}^n},b]$
On $\mathbb{P}(\mathcal{O}_ {\mathbb{P}^1}\oplus \mathcal{O}_{\mathbb{P}^1} (n))\mid_{U_1}$, this map is give by
$U_{1}\times \mathbb P^{1}\longrightarrow \mathbb P^1\times \mathbb P^2\\
[x_0,x_1]\times[a,b]\mapsto[x_0,x_1]\times[a,b,b\frac{{x_0}^n}{{x_1}^n}]$
With the definition above $\phi$ is well defined and is the embedding we are looking for.<|endoftext|>
TITLE: Repertory of the different sorts of operads
QUESTION [5 upvotes]: Many different types of operads have emerged in recent years (symmetric, shuffle, cyclic, anticyclic, coloured, etc.).
I would like, for any of these, list the following data:

Description of the free object (combinatorial objects involved, expression for the composition maps, expression for auxiliary map (as e.g., the action of the symmetric group or the cyclic action);
Given a collection $G$ of generators and its generating series $G(t)$, give an expression for the Hilbert series $H_G(t)$ of the operad generated by $G$;
Give some examples of worthwile and well-studied operads that fit into this kind of operads.

References are naturally welcome but I would like to create here a big list (with one kind of operad by answer if possible).

REPLY [2 votes]: One operad that's popped up in my work is the kind of operad that has the same formal properties as the endomorphism operad of a space $X$, provided $X$ is equipped with an action of a topological group $G$.  Level $n$ of the endomorphism operad is just the space of maps $Map(X^n, X)$, but we think of this space as one with an action of the group
$$ G \times (\Sigma_n \ltimes G^n) \equiv G \times (\Sigma_n \wr G). $$
I sometimes denote this group $\Sigma^*_n \wr G$, and the family of groups $\Sigma^* \wr G$. I've been calling these operads $\Sigma^* \wr G$-operads as I thought the name was kind of harmless and more or less descriptive. 
The free objects over this operad look like a disjoint union of rooted trees, where the vertices of the trees are decorated by points in the generating $\Sigma^*_k \wr G$-spaces.  There is an equivalence relation on these decorated trees, generated by one relation for each edge of the trees -- corresponding to the endomorphism operad's equivariance. 
These types of operads come up (and are useful) with $G$ being various orthogonal groups, or diffeomorphism groups of balls (or other automorphism groups of balls) for certain embedding spaces, like spaces of knots.  For classical knots, like knots in the 3-sphere, there is an almost free operad of this type that describes the homotopy-type of the space of knots completely, up to the computation of certain (finite) symmetry groups of some hyperbolic links in $S^3$.  The operads I'm talking about I call splicing operads.  These are subspaces of $Map(X^n,X)$ but in this case, $X$ is a ball, and the maps are smooth maps with various restrictions on them (to get knots of various types).<|endoftext|>
TITLE: Heat Kernel Asymptotics on Manifold with Boundary
QUESTION [15 upvotes]: This is crosspost from math.stackexchange https://math.stackexchange.com/questions/311213/heat-kernel-asymptotics-on-manifold-with-boundary where the question did not yield any answer
On a closed Riemannian manifold $M$, the heat kernel $k_t(x, y)$ of the Laplace-Beltrami operator (or more general of any generalized symmetric Laplace-type operator acting on sections of a vector bundle) admits an asymptotic expansion of the form
$$  k_t(x, y) \sim \exp\left( -\frac{1}{4t}d(x, y)^2\right) \sum_{j=0}^\infty t^j \Phi_j(x, y) $$
where $d(x, y)$ denotes the Riemannian distance and $\Phi_j$ are appropriate smooth functions, not depending on $t$. This is meant in the sense that for each $N \in \mathbb{N}$, there exists a constant $C>0$ such that for all $x, y \in M$, we have
$$ \left| k_t(x, y) -  \chi(x, y)\exp\left( -\frac{1}{4t}d(x, y)^2\right) \sum_{j=0}^Nt^j \Phi_j(x, y) \right| < C t^{N+1}$$
where $\chi(x, y)$ is an appropriate cutoff function that is $\equiv 1$ near the diagonal.
In the case that $M$ is still compact but has a boundary, in many books there can be found an asymptotic expansion of the trace, but I could not find an asymptotic expansion of the kernel itself, uniform on $M \times M$. Is there such an expansion?
Remark 1: When $M$ has a totally geodesic boundary, such an expansion is easy to get with help of the "Riemannian double". But of course, this case is quite unlikely (it is not even fulfilled for domains in Euclidean space).
Remark 2: I did not specify any boundary conditions, but one can assume that we are in the simplest case, i.e. the Laplace-Beltrami operator acting on functions with either Dirichlet or Neumann boundary conditions, whichever you like.

Edit: Clearly, the same asymptotics as above cannot hold on a manifold with boundary. One could however expect results like
$$\Bigl|k_t(x, y) - (2\pi t)^{-n/2}e^{-\frac{d(x, y)^2}{4t}} \pm (2\pi t)^{-n/2}e^{-\frac{\sigma(x, y)^2}{4t}} \Bigr| \leq C t^{-n/2+\varepsilon},$$
where $\sigma(x, y)$ is the distance "over the boundary" (i.e. the length of the shortest path from $x$ to $y$ that touches the boundary somewhere) and the sign $\pm$ depends on the boundary conditions (+ for Dirichlet, - for Neumann). (Addendum: The whole statement is: For every compact set $K$ in the $M\times M$ minus cutpoints (which may include boundary points), there exists a constant $C$ such that the above statement is true for. $x,y \in K$).
Results like this come from physical arguments (like reflecting paths etc.) and are used with handwavy arguments in physics literature, but I do not know if something like this is ever rigourously proved (probably this is even wrong). 
Edit: This result does hold in the case that the boundary is totally geodesic, but will most definitely be wrong otherwise.

REPLY [10 votes]: Uniformity on compact sets interior to the domain
The boundary condition has no real  impact on the small-time asymptotics of the heat kernel for interior points. More precisely, consider a Riemannian manifold $\mathbb{M}$ and $\Omega$ a relatively compact smooth open domain in $\mathbb{M}$. Denote by $p_\Omega$ the heat kernel associated with a self-adjoint extension of the Laplace-Beltrami operator on $\Omega$ and $p_\mathbb{M}$ the usual heat kernel on $\mathbb{M}$. If $K$ is a compact subset of $\Omega$, then we have for $ t \le T$, $x,y \in K$
$
| p_\mathbb{M}(t,x,y) - p_\Omega(t,x,y) | \le C_1 e^{-C_2/t} \quad  (\star)  
$
where $C_1$ depends on $T$ and $C_2$ on $K$. As a consequence the small time asymptotics of  $p_\mathbb{M}$ leads to a small-time asymptotics of $ p_\Omega$ and we obtain the following theorem:

Let $K$ be any compact subset of $\Omega$.  For every $N \ge 0$, there exist $\delta> 0$, smooth functions $\Phi_0,  \Phi_1, \cdots, \Phi_N$ and a constant $C$ such that for every $x,y \in K$ with $d(x,y) \le \delta$, $t \in [0,1]$,
  $
\left| p_\Omega(t,x,y) -\frac{e^{-\frac{d^2(x,y)}{4t}} }{t^{n/2}} \left( \sum_{k=0}^N  \Phi_k (x,y) t^k \right) \right| \le C \frac{e^{-\frac{d^2(x,y)}{4t}} }{t^{n/2-(N+1)}}
$

This is how we can prove the inequality $ (\star)  $.  Fix $x \in \Omega$, and consider a smooth domain $\tilde{\Omega}$ that contains $x,y$. Denote by $p^D_{\tilde{\Omega}}$ the Dirichlet heat kernel on $\tilde{\Omega}$. We have
$
p_\Omega(t,x,y)=p^D_{\tilde{\Omega}}(t,x,y)+\mathbb{E}_x \left( 1_{ t \ge T_{\tilde{\Omega}}} p_\Omega(t-T_{\tilde{\Omega}},X_{T_{\tilde{\Omega}}},y) \right)
$
where $X$ is the diffusion process associated with $p_\Omega$ and $T_{\tilde{\Omega}}$ the hitting time of the boundary of $\tilde{\Omega}$.
We easily deduce from that
$
0\le p_\Omega (t,x,y) - p^D_{\tilde{\Omega}}(t,x,y) \le C_1 \mathbb{P}_x ( t \ge T_{\tilde{\Omega}} ) \le C_1 e^{-C_2/t}$
Similarly we obtain
$
0 \le p_\mathbb{M}(t,x,y) -p^D_{\tilde{\Omega}}(t,x,y)   \le C'_1 e^{-C'_2/t}
$
and we get the inequality  $ (\star)  $ by combining the two above inequalities.
Uniformity on compact sets intersecting the boundary
In general, we can not hope for any estimate  of the type
$
\left|  p_\Omega (t,x,y) -\frac{e^{-\frac{d^2(x,y)}{4t}} }{t^{n/2}} \Phi (x,y)\right| \le \frac{C}{t^{n/2-\varepsilon}}
$
which would be uniform on a compact $K$ intersecting the boundary of $\Omega$. Indeed, take for $p_\Omega$ the Dirichlet heat kernel. If the above estimate is satisfied we would have
$
\left| t^{n/2} p_\Omega (t,x,x) - \Phi (x,x)\right| \le C t^{\varepsilon}
$
If $x \in \partial \Omega \cap K$, then $ p_\Omega (t,x,x) =0$, so we deduce $\Phi (x,x)=0$. If $x \in \Omega \cap K$, $\lim_{t \to 0} t^{n/2} p_\Omega (t,x,x) =\frac{1}{(4\pi)^{n/2}}$, so $\Phi(x,x)=\frac{1}{(4\pi)^{n/2}}$. 
So $\Phi(x,x)$ is not continuous, which is a contradiction because $t^{n/2} p_\Omega (t,x,x) $ is continuous and supposed to converge uniformly to $\Phi(x,x)$.
However as OP observed, in some cases like when the boundary is totally geodesic and the boundary condition is Neumann or Dirichlet, there exists a modified asymptotics 
$
\left|  p_\Omega (t,x,y) -\left( \frac{e^{-\frac{d^2(x,y)}{4t}} }{t^{n/2}} \pm \frac{e^{-\frac{\sigma^2(x,y)}{4t}} }{t^{n/2}}  \right) \Phi (x,y)\right| \le \frac{C}{t^{n/2-\varepsilon}}
$
which holds uniformly on a compact intersecting the boundary. 
We can not   hope for such "reflection principle" in  great generality for the Neumann kernel  because under some convexity assumption  of the boundary we have for $x,y \in \partial{\Omega}$
$
p_\Omega (t,x,y)\simeq \frac{\Phi_1 (x,y)}{t^{n/2+1/6}}  e^{-\frac{d^2(x,y)}{4t}-\frac{\Phi_2( d(x,y))}{t^{1/3}}} $
where $\Phi_1$ and $\Phi_2$ are directly related to the normal curvature of the boundary.
The result was proved by Ikeda and Hsu, using probabilistic methods
SHORT-TIME ASYMPTOTICS OF THE HEAT KERNEL ON A CONCAVE BOUNDARY<|endoftext|>
TITLE: A question of Allan on infinite divisibility
QUESTION [5 upvotes]: Loosely speaking the question is: If an element of a commutative ring is infinitely divisible by $x$, is it the product of $x$ with an infinitely divisible element?
More preceisely:
For a fixed element $x$ of a commutative unital ring $R$ let $I(x)=\bigcap \lbrace x^nR :n\in \mathbb N \rbrace$ be the ideal of elements which are divisible by all powers of $x$.

Is $I(x)= x  I(x)$?

This question is due to Graham R. Allan in this article and it appeared in his investigations of embedding the algebra $\mathbb C[[X]]$ of formal power series into Banach algebras (surprisingly, this is indeed possible). Originally, Allan asked this for so-called stable elements, which means that the infinite system $a_n - x a_{n+1} =b_n$ is solvable for all sequences $b_n$ in $R$. For stable elements of Banach or Frechet algebras the answer is positive (even more, it is positive for stable elements of algebras having a sub-multiplicative norm - the reason is that for stable elements Baire's theorem for $R^{\mathbb N}$ endowed with the product of the discrete topologies implies something to work with for the given norm).

REPLY [4 votes]: What's to stop you from just freely building a counterexample?  That is, let the ring be generated by elements $x,a,b_1,b_2,\dots,b_n,\dots$ subject to (only) the relations $a=x^nb_n$ for all positive integers $n$.  Then $a$ is in $I(x)$.  Unless I'm making a stupid mistake, the only solutions $q$ of $a=xq$ are finite linear combinations (with integer coefficients adding to 1) of the elements $x^{n-1}b_n$, and none of those are in $I(x)$.<|endoftext|>
TITLE: Reference request on symmetric polynomials
QUESTION [5 upvotes]: A version of this question on stackexchange got a few comments from one person and no answers.
Let $e_k$ be the $k$th-degree elementary symmetric polynomial in variables $x_1,\ldots,x_n$ (so in particular $e_k=0$ if $k>n$).  I don't know much about these polynomials.
Consider the rational function
$$
\frac{e_1+e_3+e_5+\cdots}{e_0+e_2+e_4+\cdots}.\tag{1}
$$
The simplest case is with just two variables: $$x_1\circ x_2=\dfrac{x_1+x_2}{1+x_1 x_2}.\tag{2}$$

Is it known, in the sense of being in the refereed literature or at least widespread folklore, that the binary operation defined by $(2)$ is associative?
Is it known that $x_1\circ \cdots\circ x_n$ equals the expression in $(1)$ (the proof is trivial, but is the fact mentioned in the literature)?
Is it known that $\pm1$ are absorbing elements for this operation, so that if just on of the $x$s in $(1)$ equals $1$ then the whole expression in $(1)$, depending on $x_1,\ldots,x_n$, equals $1$, and similarly $-1$?  (If $x_1=1$ and $x_2=-1$, then you get $0/0$ and if I'm not mistaken, the singularity is not removable.)

REPLY [2 votes]: The fact that the binary operation defined by (2) is associative can be found in the Wikipedia article on formal groups (section 2) http://en.wikipedia.org/wiki/Formal_group. Presumably one might find more in the literature of formal groups.<|endoftext|>
TITLE: Faithful representation of the projective unitary group with the lowest dimension?
QUESTION [7 upvotes]: What is the lowest dimension of a faithful ordinary representation (as compared with projective representation) of the projective unitary group $\rm{PU}(d)$? Is it $d^2-1$?

REPLY [10 votes]: If $G\subset GL_n({\mathbb C})$ is a compact Lie group, and $G({\mathbb C})$ is the Zariski closure of $G$ in $GL_n$, then every continuous representation of $G$ extends uniquely to an algebraic representation of $G({\mathbb C})$. Consequently, if we take $G=PU(d)$, then $G({\mathbb C})=PGL(d)=SL(d) \quad modulo \quad centre$, and the question of minimality of irreps for $G$ becomes one for $PGL(d)$. By the Borel -Weil Theorem,  every irreducible representation of $SL(d)$ is of the form $V_{\chi}=Ind_B ^G (\chi)$ where $B$ is the Borel subgroup of upper triangular matrices, and $\chi$ is an anti-dominant character. In order that this representation descend to $PGL(d)$ it is necessary and sufficient that $\chi$ be trivial on the centre of $SL(d)$. 
By Using the Weyl dimension formula, one can then see that the smallest $V_{\chi}$  is  the adjoint representation. To see this, let $\lambda$ be the highest weight of the representation $V_{\chi}$. In the usual notation, 
$$\lambda =(m_1,\cdots, m_{d-1})$$ where $m_i$ are non-negative integers which are decreasing. Set $a_i=m_i+d-i$. The dimension of this representation $V_{\chi}$ is the product 
$$\prod _{1\leq i< j\leq d} \frac{a_i-a_j}{i-j}.$$  Since we have that $\lambda $ is trivial on the centre of $SU(d)$, it follows that $m_1+\cdots+m_{d-1}$ is divisible by $d$. Hence $m_1\geq 2$. This is easily seen to imply that the dimension of $V_{\chi}$ is at least $d^2-1$ with equality if and only if $m_1=2$ and $m_2=\cdots =m_{n-1}=1$. That is, $\lambda$ is the highest weight of the adjoint representation.<|endoftext|>
TITLE: Is there a compact space with no countably generated dense subspace?
QUESTION [10 upvotes]: This is a reformulation of this MO question which recieved little or no attention due to the fact that the OP gave no motivation whatsoever. I found the question quite interesting and decided to give it another try (if this is not OK please let me know and I´ll delete this post).
It is known that if $G$ is an abelian compact topological group then it contains a dense subgroup $H$ which is countably tight (in fact Frechet-Urysohn). However the following is open (at least it was a few years ago):

If $G$ is a compact group, must $G$ contain a dense subspace of countable tightness?

This is problem 4.1.1 in "Topological Groups and Related Structures" by A.Arhangelskii and M.Thachenko. Problem 4.1.7 (also open) in the same book is:

Is it true that every homogeneous compact space contains a dense subspace of countable tightness?

My guess is that there should be known examples of (non-homogeneous) compact spaces such that any dense subspace has uncountable tightness, but I could not find any. So I have two questions:
1) Is there such a compact space?
2) For a cardinal $\kappa > 2^{\aleph_0}$ what is a dense subspace of $[0,1]^\kappa$ that has countable tightness?
Perhaps the answer to 2) is that there is none, and $[0,1]^\kappa$ is indeed a counterexample for the second quoted question, but I wouldn´t expect that. Note that if $\kappa \leq 2^{\aleph_0}$ then $[0,1]^\kappa$ is separable and any countable dense subspace would do the trick.
Edit: As Santi suggests in a comment to his answer of 2), the space $X=\beta\mathbb{N} \setminus \mathbb{N}$ is a good candidate for 1), but I still don´t know for sure. Some facts about $X$ that might be relevant:
a) $X$ is compact,
b) any dense subset of $X$ has size at least $2^{\aleph_0}$,
c) the tightness of $X$ is $2^{\aleph_0}$,
d) there are $R$-points in $X$, i.e. there exist an open $U \subseteq X$ and a point $x \in \overline{U}$ such that $x \notin \overline{A}$ for any $A \subseteq U$ with $|A|<2^{\aleph_0}$.

REPLY [3 votes]: An $\eta_1$-set is a linearly ordered set $(Q, \leq)$ with the property that if A and B are countable subsets of $Q$ satisfying $a < b$ for all $a \in A, b \in B$, then there is an $x \in Q$ satisfying $a < x < b$ for all $a \in A, b \in B$. (This definition and the notation are from Gillman and Jerison.) Let $(X, \leq)$ be the Dedekind compactification of an $\eta_1$-set, that is, $X$ is constructed from the $\eta_1$-set $Q$ in analogy to the construction of the reals from the rationals using Dedekind cuts, and then a first and last element is added. Then $X$ is a connected compact space. I can't recall a reference, but $X$ is the disjoint union of three dense sets: the set of P-points (which includes the endpoints), the set of points that are limits of strictly increasing sequences, and the set of points that are limits of strictly decreasing sequences. Furthermore, in $X$ a point is a limit of a non-trivial sequence if and only if it is an accumulation point of a countable set. If $D$ is a dense subset containing a P-point, it obviously does not have countable tightness.  If $D$ is a dense subset containing a limit $x$ of a strictly increasing sequence, then $x$ is not a limit of a strictly decreasing sequence, so the set $S$ of elements of $D$ larger than $x$ contains $x$ in its closure, but no countable subset of $S$ has $x$ in its closure. Similarly, a dense subset containing a limit of a strictly decreasing set does not have countable tightness.<|endoftext|>
TITLE: Essential geometric morphisms on the étale site. 
QUESTION [5 upvotes]: Hi!
There is a risk that this question might be utterly trivial, and if it is, my sincerest apologies, but I haven't been able to find anything in the literature. 
Let $f:X \rightarrow Y$ be an étale morphism of schemes. This gives rise to a geometric morphism of topoi $f^\ast:Sh(Y) \rightleftarrows Sh(X) : f_\ast$, and my question is now:
In what cases do $f^\ast$ have a left adjoint? Are there any explicit descriptions for this for say affine schemes?

REPLY [3 votes]: As Jonathan said, the answer is always, and the left adjoint $f_!$ has a simple description as the "extension by $\emptyset$" functor. Think of an open immersion $j:U\rightarrow X$ of topological spaces, then we have the usual extension by $\emptyset$ functor by sheafififying the presheaf
$j_!\mathcal{F}(W)=\mathcal{F}(W)$ if $W\subset U$
$j_!\mathcal{F}(W)=\emptyset$ otherwise. 
It's an easy calculation to show that this is left adjoint to $j^*$.
Translating this to the étale topology, if $f:X\rightarrow Y$ is an étale morphism of schemes, we have the extension by $\emptyset$ functor by sheafififying the presheaf
$f_!\mathcal{F}(W\overset{f}{\rightarrow} Y)=\coprod_{g\in\mathrm{Hom}_Y(W,X)}\mathcal{F}(W\overset{g}{\rightarrow} X)$
and again it's fairly easy to verify that $f_!$ does what we want it to.
It's worth noting that to get an adjoint for abelian sheaves, we need to replace this coproduct by the coproduct in the category of abelian sheaves, i.e. direct sum. Thus $f_!\mathcal{F}$ depends on whether we are considering $\mathcal{F}$ as a sheaf of sets or of abelian groups.
It's possibly also worth noting that we don't get a geometric morphism $f^*:\mathrm{Sh}(Y)\leftrightarrows \mathrm{Sh}(X):f_!$ because $f_!$ doesn't preserve limits. Despite this, the version of $f_!$ for abelian sheaves is actually exact!<|endoftext|>
TITLE: Simply connected simple algebraic groups
QUESTION [12 upvotes]: Before asking the question I should say that I don't know much about algebraic groups and I'm not sure if the question has the right level for MO. If not, please let me know and I will delete the question that emerged when I was trying to understand some theorems in a paper on the vanishing range of the cohomology of finite groups of Lie type. 
Call an algebraic group (over an algebraically closed field) simple if it is simple as an abstract group and almost simple if it has finite center. Every almost simple algebraic group is of one of the following types: 
$$A_n, B_n, C_n, D_n, E_6, E_7, E_8, F_4, G_2$$
Question: What are the simply connected simple algebraic groups (over an alg. closed field with no assumption on the characteristic) and of which type are they ? 
Counter-examples: $SL_{n+1}$ is simply conntected almost simple of type $A_n$, but not simple while $PSL_{n+1}$ is simple but not simply connected of the same type. 
Added: The mathematical problem is solved by Jay's answer. But I'm still confused about the terminology: In contrast to finite groups, where each simple group has a definite name, simple algebraic groups doesn't seem to have a name on their own (apart from the classical types). For example, by comments of Yves Cornulier and Jim Humphreys, for each type there is a unique simply connected simple alg. group. In my opinion it would be reasonable to give them a special name, for example $E_6^{sc}$. However the typical wording in the literature is like "let $G$ be a simply connected, connected complex algebraic group of type $E_6$". 
Is there a particular reason for this convention ?

REPLY [15 votes]: If $G$ is an adjoint algebraic group then it is always simple as an abstract group, (EDIT: This is because any proper normal subgroup of a simple algebraic group must be finite and lie in the centre). In general assume $G$ is connected reductive algebraic group and let $\pi : G \to G_{ad}$ be an adjoint quotient of $G$. Then $G$ is simple as an abstract group if the kernel of $\pi$ is trivial because $\pi$ is then an isomorphism of abstract groups. If $\pi$ does not have a trivial kernel then it is clear that $G$ is not simple as an abstract group.
For an example: in characteristic 2 the symplectic group $Sp_{2n}$ is simple as an abstract group as it is isomorphic (as an abstract group) to the corresponding adjoint group. However these two groups are not isomorphic as algebraic group (only isogenous).
EDIT: I suppose this didn't really explicitly answer your question. Explicitly we have a simple simply connected group is simple as an abstract group if and only if it is on the following list:

$G_2$, $F_4$ or $E_8$ any characteristic.
$B_n$, $C_n$, $D_n$ (n>2) or $E_7$ in characteristic 2.
$E_6$ in characteristic 3.
$A_n$ if $n+1$ is a power of the characteristic.

REPLY [7 votes]: The almost simple algebraic groups $G$ are classified by their root datum (defined here): in short, this is a quadruple $(X,X^{\ast},R,R^{\ast})$ with $X,X^{\ast}$ dual finitely generated free abelian groups and $R\subset X, R^{\ast}\subset X^{\ast}$ finite subsets (satisfying some conditions). Think of $X$ as the character group of some maximal torus in $G$ and $X$ a collection of roots (in particular, $R$ will define a root system once we tensor $X$ with $\mathbb{R}$). Then, $R^{\ast}\subset X^{\ast}$ are the corresponding co-objects (cocharacters, coroots). This datum characterises almost simple algebraic  groups in the following sense: the type of $G$ is determined by the root system determined by $R\subset X$ (type ABCDEFG, as you've mentioned) and the extra information supplied by knowing the co-objects determines the weight lattice of this root system. The simply connected groups are those groups for which the weight lattice of the root system of $G$ is equal to $X$; this is the same as those groups of each type with the `largest' (finite) centre. The simple groups are those groups for which $\mathbb{Z}R = X$, that is, when the root lattice equals the character lattice. Hence, the simple simply connected groups are those $G$ whose weight lattice is equal to its root lattice.
Some nice, introductory notes are available here. Also, see Chevalley's Collected Works here.
I suppose this does not completely answer your question (I haven't given a list of the simply connected simple groups) but hopefully this will give you an idea of the general framework.
EDIT: I appear to have been too slow! As I was off looking for the required list @Jay Taylor has given a complete answer.<|endoftext|>
TITLE: Can sine be made into a homomorphism?
QUESTION [6 upvotes]: Consider the usual sine function $\mathbb{R}\rightarrow \mathbb{R}$. Is there some (single) group structure we can put on $\mathbb{R}$ with respect to which sine becomes a homomorphism?
I suspect the answer is either no for a trivial reason, or yes by a simple set-theoretic argument (probably providing a great many such group structures of no interest). 
This latter seems plausible if I can "replace" the reals (and the sine function) by some arbitrary equal-sized set (and sufficiently similar function). Indeed, if I only asked that we have a pair of group structures $*_1$ and $*_2$ so that sine is a homomorphism from one to the other, then such an argument does go through (if I'm not mistaken). 
So I'm putting this question forward mostly in case there's a pleasant surprise. Depending on the answer, one could of course ask for further restrictions (abelian, continuous (probably impossible), torsion, torsion-free etc), but for now I'll leave as is.

REPLY [6 votes]: If you require the group structure to be continuous, this is impossible.  Indeed, in that case, the image $[-1,1]$ would have to be a group as well.  But $[-1,1]$ is not homogeneous, so it cannot be a topological group.
Without requiring continuity, it is possible.  Let me first give a construction where you restrict from $\mathbb{R}$ to $X=[-\pi/2,\pi/2]$; I will write $f:X\to X$ for the sine function.  In that case, the (set-theoretic) dynamical system given by $f$ has a particularly simple structure: $f$ is injective with a unique fixed point 0, $X_n=f^n(X)\setminus f^{n+1}(X)$ has cardinality $2^{\aleph_0}$ for all $n$, and $\bigcap f^n(X)=\{0\}$.  Such a dynamical system is completely determined by giving the sequence of sets $X_n$ for $n\geq 0$ with bijections $X_n\to X_{n+1}$.  Let $Y$ be the free $\mathbb{Q}$-vector space on $X$; then it is easy to see that the induced homomorphism $\tilde{f}:Y\to Y$ will have the same properties.  Choosing compatible bijections $X_n\to Y_n$, we get a group structure on $X$ for which $f$ is a homomorphism.
Extending this to all of $\mathbb{R}$ is now easy: identify $\mathbb{R}$ with $X\times \mathbb{Z}$ in such a way that the projection $X\times\mathbb{Z}\to X$ sends each $x\in \mathbb{R}$ to the unique $y\in [-\pi/2,\pi/2]$ such that $\sin x=\sin y$.  Then the sine map $\mathbb{R}\to \mathbb{R}$ can be identified with the projection to $X$ followed by $f:X\to X$.
This construction is quite flexible if you want the group structure to have various properties.  For instance, you could replace $Y$ with the free group on $X$, or the free abelian group on $X$, or the free $\mathbb{F}_p$-vector space on $X$, or many other constructions; you could also replace $X\times\mathbb{Z}$ with any semidirect product of $X$ with a countable group.<|endoftext|>
TITLE: k3 surface as ramified double cover of $\mathbb{P}^2$
QUESTION [9 upvotes]: I read that one example of k3 surface is a double cover of $\mathbb{P}^2\mathbb{C}$ ramified over a sextic. My question is why a sextic? i believe that the sextic is isomorphic to the ramification divisor, but why is this? also how can i see that k3 surfaces realized this way are deformation equivalent? 
thank you

REPLY [22 votes]: Q1: Hurwitz formula + canonical divisor of $\mathbb P^2$.
Q2: Move the curve in $\mathbb P^2$.<|endoftext|>
TITLE: Kunen's inconsistency result
QUESTION [11 upvotes]: A well-known result of Kunen says that there is no non-trivial elementary embedding $j: V \rightarrow V.$ There are several proofs of this theorem (see Kanamori, The higher infinite). I wonder to know if the following argument works:
Assume on the contrary that such an embedding exists. Let $\Phi(\alpha)$ be the following statement:
'' $\alpha=crit(j)$ for some non-trivial $j: V \rightarrow V.$''
Let $\kappa$ be the least such that $\Phi(\kappa)$ and let $j: V \rightarrow V$ witness this. Using $j$ we can conclude that $j(\kappa)$ is also the least such that $\Phi(j(\kappa))$ and this is impossible.

REPLY [13 votes]: Although your argument doesn't prove the full Kunen inconsistency,
for the reasons already explained by Ali Enayat, it does prove the following form of the Kunen inconsistency, ruling out definable nontrivial elementary embeddings. What is more, it is often only this form of the result that one finds expressed by set theorists. For example, the version of the Kunen inconsistency stated in Kanamori's text is only this weaker form (see footnote 1 in the paper linked below). 
Theorem. There is no definable class $j$ that is a nontrivial elementary embedding $j:V\to V$.
Proof: This is a theorem scheme of ZFC, with one theorem for each possible defining formula of $j$. Suppose first that $j:V\to V$ is a nontrivial elementary embedding and that $j(x)=y$ is defined by a formula
$\psi(x,y)$. We may then define the critical point of $j$, just as in your argument, and deduce by elementarity that $j(\kappa)$ would also have to be
the critical point, which is a contradiction. Thus, there is no
definable nontrivial elementary embedding $j:V\to V$. 
One can adapt the argument to show that there is no nontrivial
elementary embedding $j:V\to V$ that is definable from parameters.
To do this, if $j(x)=y$ is defined by $\psi(x,y,z)$, with
parameter $z$, then you may assume $z$ is chosen so that this
embedding gives rise to the smallest possible critical point among
any parameter for which the definition does define an elementary
embedding (and this is itself definable, by asserting that the relation is $\Sigma_1$-elementary and cofinal). This defines $\kappa$
without parameters, and so again one attains a contradiction. QED
In particular, when one chooses to formalize the Kunen inconsistency as a result in ZFC, as opposed to the second order theories GBC or KM, then it admits of this very soft proof. Furthermore, this argument does not use the axiom of choice, and so actually it is provable in ZF. This argument was published by Suzuki in 1998. You will also find an
account of it and further generalizations in my paper
Generalizations of the Kunen Inconsistency, joint with Norman Perlmutter and Greg Kirmayer.
Our view, which we explain in the paper, is that this ZFC formulation of the Kunen inconsistency does not capture the full power of his argument. Kunen himself proved his theorem in Kelly-Morse KM set theory, although his argument and the other combinatorial proofs can be given in Gödel-Bernays GBC set theory, which gives a stronger result. Our paper contains several further generalizations of this argument, to treat nontrivial embeddings $j:V[G]\to V$ or $j:V\to V[G]$ that might be definable in a forcing extension $V[G]$ and many other cases.<|endoftext|>
TITLE: First mention of the fundamental bigroupoid of a space?
QUESTION [8 upvotes]: The fundamental bigroupoid $\Pi_2(X)$ of a space $X$ was independently described by Hardie, Kamps and Kieboom (paywall) and Stevenson (arXiv) around the year 2000. HKK cite Baez-Dolan's seminal HDA0, but the idea there is just very roughly sketched, and cited Grothendieck's Pursuing Stacks for inspiration. The idea goes back to a letter in February 1975 from Grothendieck to Breen describing the ideas later expanded in Pursuing Stacks:

However, I would be surprised if the notion wasn't some kind of folklore. What other sources apart from those I've cited mention $\Pi_2(X)$?
Bénabou doesn't mention bigroupoids in his paper introducing bicategories, so a subsidiary question is: when were bigroupoids were first mentioned?

REPLY [7 votes]: I feel, pacé Grothendieck, that the use of fundamental groupoid or $n$-groupoid for a strict structure in dimension $1$ and a weak structure in higher dimensions is confusing! But I seem to be out on a limb on this. 
In any case, the higher dimensional strict structures have advantages for calculating homotopical invariants,  as shown in my papers  with Higgins and with Loday. These higher homotopy groupoid structures are known to be well defined  for certain structured spaces, in fact for filtered spaces, and for $n$-cubes of spaces. This relates them well to classical homotopical structures, i.e. to relative homotopy groups, and to $n$-ad homotopy groups, and yield new calculations and understanding of of these. 
I hope it will be useful to explain the route to this conclusion. 
Philip and I defined a strict fundamental double groupoid $\rho(Y,X,x)$ of a pair of pointed spaces in 1974, but Frank Adams' opposition  delayed the publication till 1978. I had been looking for a homotopy double groupoid of a space since 1965, but could not do it. In 1974 Philip and I agreed: 

Whitehead's subtle theorem (1941-1949), proved using transversality and knot theory,  that the crossed module $ \delta: \pi_2(X \cup _\lambda e^2 _\lambda,X,x) \to \pi_1(X,x) $ is free on the $2$-cells was an example  of a universal property in $2$-dimensional homotopy theory;
If our conjectured $2$-dimensional van Kampen was true in some form, then it should imply Whitehead's theorem. 
Whitehead's theorem involves relative homotopy groups. 
We should therefore try to define a homotopy double groupoid in a relative situation. 
Given we had little time remaining in Philip's stay at Bangor, we should try the simplest idea, and that seemed to be to consider homotopy classes of maps of a square $I^2$ into $Y$ which mapped the edges into the subspace $X$ and the vertices to the base point. Note that this a symmetric definition, and requires no choice of which edges to be mapped to the base point, as in the usual (and so unaesthetic?)  definition of relative homotopy groups. 
This was a sensible definition; the work that had already been done by then with Chris Spencer on double groupoids with connections, commutative cubes, and the precise relation of these double groupoids  with crossed modules, allowed a proof of the van Kampen theorem and,  and given work by Philip on induced crossed modules, a deduction of Whitehead's theorem, and indeed a generalisation to a theorem on $\delta:\pi_2(X\cup CA,X,x)\to \pi_1(X,x)$, with Whitehead's theorem being the case $A$ is a wedge of circles. 

This suggests the advantage of a proper strategic analysis, even if delayed! 
Note that one of the advantages of strict structures is that the calculation of colimits of them is reasonably clear; in particular, the calculation of colimits of crossed modules is an interesting extension of the calculation of colimits of groups. In algebraic topology, one use of homotopical invariants is to show that objects are not equivalent, and this needs precise answers. 
It was then not hard to see the putative extension to filtered spaces; the work was in fact quite hard technically and conceptually, but was completed in 1977, with two CRAS notes published in that year, and the full papers in 1981.  
Actually the interest of Whitehead's theorem seems not so much appreciated. It allows one,  for the usual representation of the Klein bottle as an identification of a square $\sigma$,  to write the more precise 
$$\delta \sigma = a+b-a +b $$
instead of the usual $\partial \sigma = 2b$. See this presentation.  
In 1982 Loday published a paper giving a definition of a fundamental cat$^n$-group of a pointed $n$-cube of spaces, and showed that such structures modelled all weak, pointed homotopy $n$-types. In a visit of mine to Strasbourg towards the end of 1981, I talked about my work with Philip Higgins, and together Loday and I  conjectured a van Kampen type theorem for his cat$^n$-groups. This was proved by 1984 and published in 1987. It allows some quite new calculations of homotopical invariants, e.g. the homotopy $3$-type of $SK(G,1)$, the suspension of an Eileneberg-Mac Lane space. 
These theorems have strong limitations, as is only to be expected, but they do allow new determinations and open out new prospects. 
So one possible moral is that the concentration on bare topological spaces, without any further structure, is a more restricted endeavour. This fits with Section 5 of Grothendieck's "Esquisse d'un programme", where he argues that the needs of geometry require more than just a topology, and he argues for kinds of stratifications. One way of looking at this is to say  that the specification of a space requires some kind of data, and that data will have some kind of structure; so, conjecturally, one should look for invariants of spaces with that kind of structure. 
It would be  interesting to compare these higher homotopy groupoid structures, and the weak structures in common current parlance,  with the vision of the algebraic topologists of the early 20th century  for higher dimensional versions of the nonabelian fundamental group. Since the finding in 1932 that Cech's definition of higher homotopy groups, which he submitted to the ICM at Zurich,  led to abelian groups, this has seemed to be a mirage.<|endoftext|>
TITLE: Cohomology of the complement of an elliptic arrangement
QUESTION [5 upvotes]: In the paper "Cohomology of the complement of an elliptic arrangement", the authors (Levin and Varchenko) consider the complement to an arrangement of (elliptic) hyperplanes in a cartesian
power of an elliptic curve and describe its cohomology with coefficients in a non-trivial rank
one local system (Actually, their result concerns generic rank one local systems).
What about the case of the trivial rank one local system $\mathbb C$?
I'm mainly interested in the case of the square of an elliptic curve.
Maybe this is already known. In this case, a good reference would be welcome
Thanks for your answers.

REPLY [2 votes]: The following preprint of Christin Bibby seems to solve your problem completely: http://arxiv.org/abs/1310.4866<|endoftext|>
TITLE: Residually nilpotent vs residually p
QUESTION [15 upvotes]: A well-known theorem of Gruenberg implies that a finitely generated residually torsion-free nilpotent group is residually $p$-finite for all primes $p$. What about the converse?
Question:
Are there any examples of a finitely generated group $G$ with the properties that $G$ is residually $p$-finite for all primes $p$, but $G$ is not residually torsion-free nilpotent?

REPLY [13 votes]: Browsing through the archive of solved problems of Kourovka Notebook, I accidentally saw that the same question was asked  by Yu.V. Kuz'min in 1999 (see question 14.52). Apparently the required examples were produced a long time ago by B. Hartley (Hartley, B.,
`On residually finite p-groups.' Symposia Mathematica, Vol. XVII (Convegno sui Gruppi Infiniti, INDAM, Roma, 1973), pp. 225–234. Academic Press, London, 1976. MR0419611 (54 #7629)). 
Hartley showed that for any set of primes $\Pi$ there exists a $3$-generated center-by-metabelian group $G$ which is not residually torsion-free nilpotent and which is residually $p$-finite iff $p \in \Pi$.<|endoftext|>
TITLE: Results and conjectures on bounds on degrees of isogenies
QUESTION [7 upvotes]: given an isogeny between two abelian varieties $\varphi: A\rightarrow B$ (everything definied over a number field $K$), we can factor $\varphi$ through a multiplication-by-$n$-endomorphism on $A$ and a `remaining' isogeny $\psi:A\rightarrow B$.

What do we know or expect about an upper bound $N$ on the degree of $\psi$?

For elliptic curves over $\mathbf{Q}$ it is known that $N=163$ is such a bound. It is even known that precisely the 26 cases $N= 1, \ldots , 19, 21, 25, 27, 37, 43, 67, 163$ do occur. This is the classification of cyclic isogenies of elliptic curves over $\mathbf{Q}$ and is due to Mazur and Kenku (1978-1982).

I am wondering if for bounded dimension $d$ of the abelian varieties and bounded degree $n$ of the number fields one expects that there is such a bound $N(d,n)$ for the possible degrees of $\psi$?

Is such a bound expected? (Is it clear that such a bound exists in case one fixes the number field and only varies the abelian varieties up to some bounded dimension?)
A subquestion would be the existence of rational $N$-torsion. Here it is known that for elliptic curves over a number field $K$ the order of the $K$-rational torsion is bounded by a constant that only depends on $[K:\mathbf{Q}]$. This is due to Merel (1996). What do we expect or know about the torsion in higher dimensions?
Thanks a lot.

REPLY [7 votes]: Mein lieber Stefan,
I think that these sorts of questions are wide-open. As far as I know, even if you fix the number field, and vary over AVs up to some fixed dimension, then it is not clear that there exists a bound on the possible degrees of your $\psi$ (which I'm guessing is a cyclic isogeny). You in any case have to be careful of complex multiplications; for example, if you have an elliptic curve over $\mathbb{Q}(i)$ with CM by the maximal order, then you can create $\mathbb{Q}(i)$-rational prime-degree isogenies for every prime that splits in $\mathbb{Z}[i]$. But if you exclude these CM isogenies, then I think it is believed that such a bound on cyclic isogeny degrees for AVs exists. 
But even pushing Mazur's isogeny theorem to number fields is fairly recent stuff; we can now say that, if $K$ is a number field not containing the Hilbert Class Field of an imaginary quadratic field, and GRH is true, then there are only finitely many prime-degree isogenies for elliptic curves over $K$; this goes back to Momose, but with some refinements by Agnès David; see also the work of Larson and Vaintrob. These bounds are, however, quite big. I once computed the bound for $\mathbb{Q}(\sqrt{5})$ by computing all the constants in David's paper "Caractère d'isogénie..." and got something like $10^{120}$!
Serre's open image theorem holds for abelian varieties $A/K$ of dimension 2,6 and odd, where $End(A) = \mathbb{Z}$; so there is a bound $C(A,K)$ such that, for all primes $l > C(A,K)$, the mod-$l$ representation is surjective. It is believed that this constant can be made independent of $A$. So, if you fix your base $K$, and vary over all AVs $A/K$ of fixed dimension $d$ (=2,6,or odd)  which have no extra endomorphisms, then the prime-degree isogenies are bounded by a constant depending only on $K$. But as you well know, this uniformity conjecture is still open even for elliptic curves over $\mathbb{Q}$!
The paper "Expander Graphs, gonality, and variation of Galois Representations" by Ellenberg, Hall, and Kowalski has some interesting results about families of AVs; see theorems 4 and 7, for example. In particular, if you have a family of AVs over a number field $K$, then there is an absolute constant $C$ such that, if $l > C$, then almost all members of the family have "large" mod-$l$ image. And (morally) "large" image means you don't get isogenies or torsion. See their paper for more details. 
Finally, regarding your subquestion about torsion: François gave a link to a paper of Clark and Xarles where they address the question of bounding torsion primes for certain classes of AVs. In particular, they prove that a "generalised Szpiro conjecture" implies uniform boundedness of torsion for all "Hilbert-Blumenthal" abelian varieties. But it seems that doing this for all abelian varieties of a fixed dimension is very hard.<|endoftext|>
TITLE: "Mathai-Quillen-type" form on $M\times M$?
QUESTION [9 upvotes]: Let $(M,g)$ be a compact, oriented, $(2n)$-dimensional Riemannian manifold.  I'm wondering whether there is a "canonical" construction of a $(2n)$-form $\eta_g$ on $M\times M$, such that

$\eta_g$ is a de Rham representative
of the Poincare dual of the
diagonal;
The pullback under the diagonal
inclusion $\iota:M\to M\times M$
is proportional to the curvature Pfaffian:
$\iota^*\eta_g=\tfrac{1}{(2\pi)^n}\text{Pf}(\text{Rm})$.

There are plenty of "non-canonical" such forms, I think.  For instance, pick a neighbourhood of the diagonal which is diffeomorphic to $TM$, and transfer the Mathai-Quillen Thom form on $TM$ to this neighbourhood using a suitable diffeomorphism.  (By the way, the Mathai-Quillen Thom form on the total space of a bundle-with-metric-and-connection is the model I have in mind here for what "canonical" should mean -- a form whose value at each point depends only on local invariants.)
Motivation:  Such a form would yield a proof of the Chern-Gauss-Bonnet theorem which is both quick and natural.  Namely, 
$\chi(M)=PD(\Delta)\cup PD(\Delta)=\int_\Delta \eta_g =\tfrac{1}{(2\pi)^n}\int_M \text{Pf}(\text{Rm})$.

REPLY [5 votes]: I was styding the paper of Jean Michael Bismut and Christophe SOULÉ, Complex Immersions and Arakelov Geometry 
The following answer may help
In the case when $M$ is a vector space or co-adjoint orbit via KKS symplectic form we know the answer
Take $\pi : T^*M\to M$ be a cotangent bundle and $s:M\to T^*M$ be as section then from the work of Bismut and Soule it is known that
$$s^*\mu_0(T^*M)=(2\pi)^{-m/2}\text{Pf}(D^2)$$
Here the Thom class $\mu_t(T^*M)=(-1)^{m(m-1)/2}(2\pi)^{-m/2}B(e^{-\omega_t})$ considered when $t\to 0$ and  $B\{·\}$ is the Berezin integral and $\omega_t=\frac{1}{2}t^2|x|^2+tDx+D^2$
Now in the special case when $M$ is a vector space then $T^*M\cong M\times M$ (and hence $s$ is just $\iota$). So when $M$ is the vector space this gives an answer to your question see this paper and  
also
If we change your question a little bit and if we Take $M=G$ as lie gorup and consider $T^*G\cong G\times T_eG\to G$ then we can find a relation between pull-back of Kirillov–Kostant—Souriau symplectic form on coadjoint orbit $\mathcal O_\mu$ and Mathai-Quillen form. In page 11 of my Master expose in Marseille, 2014 you can find the canonical construction of KKS form and Pfaffian and by pervious construction you can relative it to Mathai-Quillen form
and also the book of Heat Kernels and Dirac Operators by Berline, Nicole, Getzler, Ezra, Vergne, Michèle
http://www.springer.com/br/book/9783540200628<|endoftext|>
TITLE: Injective morphism from an elliptic curve to $\mathbb CP^2$. 
QUESTION [9 upvotes]: Let $E$ be the elliptic curve $x^3+y^3+z^3=0$. 
Question. Are there injective morphisms $E\to \mathbb CP^2$ of arbitrary high degree?
Comments. 1) There are injective morphisms $E\to \mathbb CP^2$ of degree $3$ (the obvious one) and of degree $6$ - whose image is the curve dual to the cubic.  2) Clearly for $\mathbb CP^1$ there are injective morphisms to $\mathbb CP^2$ of arbitrary high degree. 3) I don't know any (smooth projective) curve for which on can prove that it does not admit an injective morphism to $\mathbb CP^2$ of arbitrary high degree. 
This question is related to Injective morphism from curves to $\mathbb CP^2$

REPLY [11 votes]: The answer is yes. And more generally, one has the following result:
Lemma: for any elliptic curve $X$, the degree of injective morphisms $X\to \mathbb{P}^2_{\mathbb{C}}$ (or $\mathbb{C}P^2$ if you want) is not bounded. 
Proof: The curve $X$ is isomorphic to a plane cubic of equation $X^3+Y^3+Z^3=\lambda XYZ$ for some $\lambda$ (Hessian form). Then, the curve has nine inflexion points. These are points $[0:1:\omega]$, $[1:0:\omega]$, $[1:\omega:0]$ with $\omega^3=1$. Choose three of them, $p_1,p_2,p_3$ such that the three corresponding tangent lines $L_1,L_2,L_3$ intersect at three points $L_1\cap L_2$, $L_1\cap L_3$, $L_2\cap L_3$, different from $p_1,p_2,p_3$.
We change the coordinates and assume that $L_1,L_2,L_3$ are the lines $x=0$, $y=0$ and $z=0$. The points $p_1,p_2,p_3$ become points of the form $[0:1:a_1]$, $[a_2:0:1]$, $[1:a_3:0]$ with $a_1a_2a_3\not=0$.
Take now the birational map of $\mathbb{P}^2$ given locally by $(x,y)-->(x,x^ny)$, and globally by $[x:y:z]-->[xz^n:x^ny:z^{n+1}]$, for $n\ge 2$. Its base-points are $[0:1:0]$ and $[1:0:0]$, which do not belong to the curve. The curves contracted are $z=0$ and $x=0$, and $y=0$ is fixed. Outside of the triangle $xyz=0$, the map is an automorphism. In consequence, the image of the curve has degree $3n$ and only cuspidal singularities. This yields an injective map from $X$ to $\mathbb{P}^2$.<|endoftext|>
TITLE: Topology in Arithmetic
QUESTION [5 upvotes]: Conjectured by Mordell and later proved by Faltings, a non-singular algebraic curve of genus $g$ over $\mathbb{Q}$ has finitely many rational points if $g > 1$. Since the genus of the Fermat curve $x^{n} + y^{n} = 1$ is $\frac{n(n-1)}{2}$ by the degree formula, Faltings' Theorem implies that it can only have finitely many rational solutions for $n > 2$, which proves a weak form of FLT.
Are there other topological invariants of curves or surfaces which bound the genus or (more directly) the number of rational points that can exist on them, e.g., arithmetic/geometric genus, (Zariski) multiplicity, Milnor number, Hirzebruch signature, Casson Invariant, Rohklin Invariant, etc.?
For example, if $f = \sum_{i=0}^{n} z_{i}^{a_{i}}$ with positive integers $a_0, \dots, a_n$, then one defines a Brieskorn-Pham manifold $\Sigma(a_0, \dots, a_n)$ as the intersection of the corresponding hypersurface of $f$ with a sufficiently small sphere, namely, $f^{-1}(0) \cap S^{2n+1}_{\epsilon}$. Brieskorn-Pham manifolds give examples of exotic spheres in certain odd dimensions for certain exponents. Is there a connection between certain topological attributes of $\Sigma(a_0, \dots, a_n)$, including topological invariants, and the number of rational points on the curve $f = 1$?
NB: This question was first asked on Math.SE, but went unanswered.

REPLY [6 votes]: Dear user02138, you may be interested in the Bombieri-Lang conjecture. It has a weak form and a strong form. The latter is:

If $X$ s a smooth variety over a number field $K$ which is of general type over a number field, then there are only finitely many curves of (geometric) genus $\leq 1$ contained in $X$, and outside those curves $X$ has only finitely many
  $K$-rational points. 

The key word here is "general type" which is a geometric condition (defined for a proper smooth variety as the canonical divisor $K$ being big, that is the order magnitude of the dimensions of the space of global section of the bundle $L(nK)$ is $n^d$ as $n$ goes to infinity), which for proper smooth curves is equivalent to the Mordell's condition $g \geq 2$. 
Under this condition (which some experts in algebraic geometry may discuss in more detail that I can -- for example is it a condition that can be read on $X(\mathbb C)$ as topological space
like for curves?), you get strong results on the number of rational points. Of course this conjecture is wide-open, except in the cases of curves or more generally subvarieties of abelian varieties, proved by Faltings.<|endoftext|>
TITLE: When polynomial f(x^2) can be factored as g(x)·g(-x) ?
QUESTION [10 upvotes]: In relation to my question Expression for the sum of square roots of zeros of a polynomial
How to characterize polynomials $f(x)$ with rational coefficients such that $f(x^2)=g(x)\cdot g(-x)$ where $g(x)$ is also a polynomial with rational coefficients? 
Is there a computationally efficient way to identify if given polynomial $f(x)$ is such without factoring $f(x^2)$ ?

REPLY [2 votes]: (Caveat:  nothing in this response has been checked or even really thought through.)
If it's true with integral coefficients, it's true mod p for all p.  Now the question of whether f(x^2) splits as g(x)g(-x) for f in F_p[x] is the question of whether f is a norm in the quadratic extension obtained by adjoining a square root of x, which should just be a question about whether each irreducible factor p(x) appearing an odd number of times in f splits in that quadratic extension.  By quadratic reciprocity (I think) this comes down to whether each of these irreducible factors p(x) has p(0) a quadratic residue.  This is easy enough to check for lots of p.
Of course, to have any hope, you need f(0) to be a square (as an integer) so I think in practice what I'd do would be to take a long list of primes p, reduce f mod p for each p, and if f factors into p_1(x) ... p_k(x) mod p, check that each p_i(0) is a residue.  And if this keeps on happening you should gain confidence that your f(x^2) actually factors in this way.<|endoftext|>
TITLE: Self-dual finite-dimensional complex normed spaces
QUESTION [8 upvotes]: Suppose $X$ is a complex normed space of dimension 2 or 3 and $X$ is isometrically isomorphic to its dual.  Is $X$ a Hilbert space?
Remarks: There are easy counterexamples in the real case, and in higher dimensions one can construct counterexamples from sums of 2-dimensional spaces which are not isometric to their duals.  Similarly a 3-dimensional counterexample can be constructed from a 2-dimensional counterexample.

REPLY [5 votes]: A self-dual norm on $\mathbf{C}^2$ need not be Hilbertian. Here is an example: 
$$ \|(z_1,z_2)\| = \begin{cases} |z_1|+|z_2| & if \ \ \ |z_1| \leq |z_2| \\ 2|z_1| & if \ \ \ |z_1| \geq |z_2|. \end{cases} $$
The dual norm $\|\cdot\|^*$ satisfies $\|(z_1,z_2)\|^*=\frac{1}{2}\|(z_2,z_1)\|$.
Here a picture for $z_1$, $z_2$ real with $\|\cdot\|=1$ in blue and $\|\cdot\|^*=1$ in red.<|endoftext|>
TITLE: Do isogenies with rational kernels tend to be surjective?
QUESTION [18 upvotes]: Dear MO Community, 
this is a pretty vague title, so let me tell you the precise observation I have made.
Consider the family of elliptic curves over $\mathbf{Q}$ having a rational $5$-torsion point $P$. They are given by
$$E_d: Y^2 + (d+1)XY +dY=X^3+dX^2,$$
for $d \in \mathbf{Q}^*$ and $P=(0,0)$. Let $\eta: E_d \rightarrow E_d'$ the isogeny whose kernel consists exactly of the five rational $5$-torsion points. 
Now assume that $E_d$ has rank $1$. After modding out torsion the Mordell-Weil group is isomorphic to $\mathbf{Z}$, and hence, $\eta$ induces an injective group homomorphism $\mathbf{Z} \rightarrow \mathbf{Z}$ which is either an isomorphism or has cokernel of size $5$.

It seems to me that this map tends to be an isomorphism.

To be precise, among all $d$, such that the numerator and denominator is bounded by $100$, there are $3,038$ elliptic curves of analytic rank equal to $1$ (out of $6,087$ total curves), and among those the above map $\mathbf{Z} \rightarrow \mathbf{Z}$ is an isomorphism in $91.2\%$ of the cases. 

So I wonder, what you should expect on average? $50\%$? $100\%$?

Maybe, this was just a coincidence in a small database. Maybe someone has seen a similar behaviour somewhere else. Maybe this is nothing new and I just haven't heard about it. I am curious to read what you think about it.
Many thanks.

REPLY [2 votes]: I tried the same as Chris, but I don't think it gives you an argument for the surjectivity. As it is too long for a comment, I post an answer.
Assume we are given an elliptic curve $E$ over $\mathbf{Q}$ with rational $5$-torsion and denote by $\eta:E \rightarrow E'$ the isogeny modding out the rational $5$-torsion. As I mentioned in my question this curve is given by a rational non-zero number $d$.
Write $d=u/v$, with $u$ and $v$ coprime non-zero integers. Now we apply the equation of Cassels and Tate which encodes the invariance of BSD:
$$ \frac{\# sha(E/\mathbf{Q})}{\# sha(E'/\mathbf{Q})} = \frac{R_{E'}}{R_E} \cdot \frac{ \# E(\mathbf{Q})^2_{tor} }{\# E'(\mathbf{Q})^2_{tor} } \cdot \frac{P_{E'}}{P_E} \cdot \prod_{p \leq \infty}\frac{c_{E',p}}{c_{E,p}}.$$
Now assume, that the elliptic curve has rank 1.
As Chris already stated, the regulator quotient equals $5^a$, for $a \in \{\pm 1 \}$, where $a=1$ if and only if the induced map of $\eta$ on the free part of the Mordell-Weil group is surjective. Hence this is the case we are interested in.
For the torsion quotient we have, that it equals $5^b$, for $b \in \{0,2 \}$, where $b=2$ if and only if $d$ is not a fifth power. This is true for $100\%$ of the cases.
For the periods and Tamagawa numbers we have, that they are equal to $5^{c-1}$, for
$$c=\# \{p \equiv 1(5),\ p \mid u^2+11uv-v^2\} + \# \{p=5,\ p^3 \mid u^2+11uv-v^2 \}$$
$$-\# \{p,\ p\mid uv \},$$
where $p$ ranges over the finite primes. (The results on the torsion quotient and on the period and Tamagawa quotient (= local quotient) can be found here http://arxiv.org/abs/1206.1822/)
(The case $5^3\mid u^2+11uv-v^2$ happens if and only if $u \equiv 7v (25)$.)
Hence, in 100% of the cases we have 
$$ \frac{\# sha(E/\mathbf{Q})}{\# sha(E'/\mathbf{Q})} = 5 \cdot 5^{\{\pm 1 \}} \cdot 5^c.$$
I don't see any reason why a negative value of $c$, which you should expect quite often, should force $a$ to be $1$, and not be (completely) subsumed in the Sha quotient.
So, instead of having a pro argument for $a=1$, this is just a pro argument, that the isogeny $\eta$ can alter the $5$-primary part of Sha in an arbitrary way.
(In case you wonder whether $c$ can be positive, look at $u=1$, $v=76971487$. Then $uv$ is prime and $u^2+11uv-v^2$ factors as $-1 \cdot 11 \cdot 31 \cdot 41 \cdot 61 \cdot 131 \cdot 151 \cdot 331 \cdot 1061$, hence $c=7$.)
(In my database of 3,038 curves, there is the following situation: $c=0$ for 247 curves. For all of them $\eta$ is NOT surjective. $c=-2$ for 2258 curves. For 20 of them $\eta$ is NOT surjective. $c=-4$ for 533 curves. For all of them $\eta$ is surjective.)<|endoftext|>
TITLE: Modern developments in finite-dimensional linear algebra
QUESTION [21 upvotes]: Are there any major fundamental results in finite-dimensional linear algebra discovered after early XX century? Fundamental in the sense of non-numerical (numerical results, of course, are still interesting and important); and major in the sense of something on the scale of SVD or Jordan normal form.   
(EDIT) As several commenters observed, using Jordan normal form as a benchmark sets the bar way too high. Let's try lowering it to Weyl's inequality.

REPLY [6 votes]: Since you lowered the level to Weyl's Inequalities (1912), it is worth mentionning the improvements of these inequalities made by Ky Fan, Lidskii and others. They culminated in a much involved conjecture by A. Horn (1961), eventually proved by Knutson & Tao on the turn of the century.<|endoftext|>
TITLE: Vassilliev invariants of knots and their cables 
QUESTION [7 upvotes]: The following is perhaps a standard question, but i could not find a plain enough answer 
by simply searching online.
Q: Given a knot $K$ and its $(p,q)$-cable $K_{p,q}$ what is a relation
between the Vassiliev invariants of $K$ and $K_{p,q}$?
In particular, I would be happy with a formula for the 2nd coefficient
of the Conway polynomial. (One may attempt to
solve this via the A-polynomial since it has a simple relationship for
satellites, however relating the resulting A-polynomial back
to the Conway p. seems nontrivial, ... at least for me. Also it is
seems likely that somebody could have already worked this out.)

REPLY [9 votes]: As I mentioned in a comment, for the degree $2$ invariant $v_2$ which is the coefficient of $z^2$ in the Conway Polynomial, we have that $v_2(K_{p,q})=av_2(K)+b$. If $K$ is the unknot, this implies that $b=v_2(T_{p,q})$, where $T_{p,q}$ is the $(p,q)$-torus knot (assuming here $p,q$ are relatively prime.) Alvarez and Labastida wrote down formulas for Vassiliev invariants of torus knots, and in particular they showed $$v_2(T_{p,q})=\frac{1}{24}(p^2-1)(q^2-1).$$ So that gives you your constant term. Using Ryan's formula for the Alexander polynomial, one should be able to show that $a=p$. This is because when you make the substitution $t\mapsto t^p$, in the conversion to the the Conway polynomial we have $z^2=t+t^{-1}-2$, and so $z^2\mapsto t^p+t^{-p}-2$. It's a lemma that $t^p+t^{-p}=2+pz^2+\cdots$, so the coefficient of $z^2$ will get multiplied by $p$. So the answer will be
$$v_2(K_{p,q})=pv_2(K)+\frac{1}{24}(p^2-1)(q^2-1).$$<|endoftext|>
TITLE: Exceptional Schur-Weyl Duality
QUESTION [7 upvotes]: Are there any known analogs of Schur-Weyl duality for the exceptional groups?
In particular, I am looking to decompose the tensor powers of the action of $E_6$ on its 27-dimensional module.  Any advice or references to this end would be greatly appreciated.

REPLY [3 votes]: There is an article by Jing-Song Huang and Chen-Bo Zhu called Weyl's construction and tensor power decomposition for G2 that builds on an invariant theory for $G_2$ (see the linked review for details). There's quite recent article Algèbres de Jordan et théorie des invariants which deals with these invariants also for other exceptional cases and I would guess there'll be some others.<|endoftext|>
TITLE: Access to a preprint by D. N. Verma
QUESTION [11 upvotes]: Some work I am doing is connected with a sequence 1, 3, 40, 1225, 67956, $\dots$ which agrees with http://oeis.org/A012250 for all eight terms. The only useful information in OEIS on this sequence is the reference D. N. Verma, Towards Classifying Finite Point-Set Configurations, preprint, 1997. Does anyone know how to obtain this preprint? Verma himself passed away last year.

REPLY [16 votes]: I don't have Verma's preprint, but there are more modern references on the subject, which don't seem to be on OEIS. Basically one looks at the GIT quotient $(\mathbb{P^1})^n//\operatorname{SL}_2$, and your sequence corresponds to its degree under a certain embedding in projective space for values of $n=4,6,8,...$. 
See section 2.11 in the paper "The moduli space of n points on the line is cut out by simple quadrics when n is not six", by B. Howard, J. Millson, A. Snowden and R. Vakil. An even more recent reference that also contains a sort of formula for these degrees is "The ring of evenly weighted points on the line", by M. Hering, B. Howard. Hope this helps.<|endoftext|>
TITLE: Research level applications of "row rank = column rank"?
QUESTION [17 upvotes]: No less an authority than Gilbert Strang frames "row rank equals column rank" (and a couple of other facts) as "The Fundamental Theorem of Linear Algebra."
I'd simply like to assemble (for teaching purposes) a list of research level applications of this basic fact.
Applications can be theoretical or practical, and I would particularly appreciate learning what value this fact has in the minds of physicists.
(It goes without say) please do not start a debate concerning the centrality of this fact (that's not what MO is for).  My question merely seek insights or pointers to the literature that would support making a positive case for centrality.  So if you use linear algebra all the time but never this fact, no need to chime in.
"Row rank equals column rank" has the consequence for square matrices that $A$ singular makes $A^T$ singular; I'm sure this case comes up everywhere.  Here I'm specifically looking for applications of the one-the-nose numerical equality of ranks.
Feel free to offer a philosophical take on linear algebra that would support the centrality of "row rank equals column rank" even if that philosophy isn't grounded in specific results.  (For example, does this simple statement offer a hidden paradigm for whole sophisticated theories.)

REPLY [4 votes]: In some sense you can view the singular value decomposition as a sharpening of this theorem (for real and complex matrices, anyway).  This, in turn, is useful all over the place.<|endoftext|>
TITLE: Can the SL_2 character variety of a three-manifold be nonreduced?
QUESTION [15 upvotes]: Let $M^3$ be a three-manifold and consider the representation variety and the character variety of $M$:
$$Y=\operatorname{Hom}(\pi_1(M^3),\operatorname{SL}(2,\mathbb C))$$
$$X=\operatorname{Hom}(\pi_1(M^3),\operatorname{SL}(2,\mathbb C))//\operatorname{SL}(2,\mathbb C)$$
I would like to think of these as schemes.  They are usually singular; in fact the trivial representation is almost always a singular point.
Are there any known examples where $Y$ (or $X$) are nonreduced as schemes?
I really want to know the answer for $\pi_1(M^3)$; examples with just any finitely presented group would be less interesting.

REPLY [15 votes]: There is actually an old (ca 1986) example of nonreduced $SL(2, {\mathbb C})$-representation scheme of a 3-manifold group. Take an oriented Seifert manifold $M$ which fibers over the $S^2(3,3,3)$ orbifold (sphere with three cone points of order 3). The fundamental group of the base-orbifold is von Dyck group with presentation
$$
\Gamma=\langle a, b, c | a^3, b^3, c^3, abc\rangle. 
$$ 
It is an old observation of Lubotzky and Magid (in their book "Representation varieties of finitely-generated groups") that the $SL(2, {\mathbb C})$-representation scheme of $\Gamma$ is nonreduced at a representation $\rho_0$ whose image is isomorphic to ${\mathbb Z}_3$ ($\rho_0$ sends all generators to an order $3$ element). Namely, $H^1(\Gamma, Ad \rho_0)$ is $1$-dimensional, while the representation $\rho_0$ s locally rigid. There is a nice geometric explanation of this phenomenon: Take a spherical equilateral triangle in $S^3=SU(2)$ contained in a great circle. Then this triangle is locally rigid but admits a nontrivial 1st order deformation in $S^3$. Now, $\rho_0$ lifts to an $SL(2, {\mathbb C})$-representation $\tilde\rho_0$ of the central extension $\pi=\pi_1(M)$ of the group $\Gamma$ (killing the center of $\pi$). This is your example. It is locally rigid but has 1-st order nontrivial infinitesimal deformations.  A drawback of this example is that the image has large centralizer. 
Below is a more difficult "universality" result:
Theorem 1. Let $X$ be an affine scheme over ${\mathbb Q}$ and $x\in X$ be a rational point. Then there exists an open subscheme $X'\subset X$ containing $x$, a natural number $N$, a closed 3-dimensional manifold $M$ with fundamental group $\pi$, a unitary representation $\rho: \pi\to SU(2)\subset SL(2, {\mathbb C})$ (whose image is dense in $SU(2)$) and an open subscheme 
$$
R'\subset Hom(\pi, SL(2, {\mathbb C}))
$$
containing $\rho$, so that $R'$ admits a regular etale covering over $X'\times SL(2, {\mathbb C})^N$ (with abelian group of deck transformations) and the covering sends $\rho$ to $x$. In particular, the centralizer of $\rho(\pi)$ is $\{\pm 1\}$. 
For instance, the analytic germ of the character scheme 
$$
Hom(\pi, SL(2, {\mathbb C}))//SL(2, {\mathbb C})$$ 
at $[\rho]$ could be isomorphic to the germ at $0$ of the scheme 
$$
\{x^{100}=0\} \times {\mathbb C}^{k} 
$$
(for some $k$). In particular, the character scheme of 3-manifold groups could be nonreduced at points of Zariski density. 
Proof of Theorem 1 could be now found here. 
The proof is a combination of my old work with Millson (see here) with the recent theorem of Panov and Petrunin, which deserves to be better known:
Theorem 2. For every finitely presented group $\Gamma$ there exists a  closed 3-dimensional orbifold $O$ so that the fundamental group of the underlying space of $O$ is isomorphic to $\Gamma$.  
It is a difficult open problem if Theorem 1 holds for 3-manifolds which are homology spheres.<|endoftext|>
TITLE: Decomposition of $\mathrm{End}(V)$ as $S_n\times S_n$-module
QUESTION [8 upvotes]: Let $V$ be a finite-dimensional, complex vector space and set $\newcommand{\Gl}{\mathrm{Gl}}G:=\Gl(V)\times\Gl(V)$. Let $E:=\mathrm{End}(V)$ and consider its coordinate ring $\mathbb C[E]$, the space of all polynomial functions on $E$. It is well-known (see 9.7 in Claudio Procesi's book on Lie Groups) that as a $G$-module, $\mathbb C[E]$ decomposes as 
$$\begin{align*}
\mathbb C[E]_d &= \bigoplus_{\lambda\mathrel\vdash d} \mathbb S_\lambda(V^\ast)\otimes\mathbb S_\lambda(V)
\end{align*}$$
where $\mathbb S_\lambda$ denotes the Schur functor. Now, simply by choosing a basis and restricting to permutation matrices, we have an action of $S:=S_n\times S_n$ on $E$ and therefore, also on $\mathbb C[E]$. Hence, there must be some decomposition
$$\begin{align*}
\mathbb C[E]_d &= \bigoplus_{\lambda,\mu \mathrel\vdash n} (\mathbb V_\lambda\otimes \mathbb V_\mu)^{\oplus N_d(\lambda,\mu)}
\end{align*}$$
where $\mathbb V_\lambda$ is the Specht module and $N_d(\lambda,\mu)\in\mathbb N$ are certain multiplicities. 
My question is: What, if anything, is known about the $N_d(\lambda,\mu)$?

REPLY [2 votes]: You can give a formula for these numbers in terms of plethysm and internal product and inner product.
The starting point is Exercise 7.74 of Enumerative Combinatorics II by Richard Stanley which is a formula for the Schur functors of the defining representation of the symmetric group in terms of plethysm and inner product.<|endoftext|>
TITLE: Is there a statistical interpretation of Green's theorem, Stokes' theorem, or the divergence theorem?
QUESTION [18 upvotes]: This is cross-posted from math.stackexchange and stats.stackexchange.  Probably there is no great answer to this question, but I thought I'd give it a shot here.
I'm teaching a class on integration of functions of several variables and vector calculus this semester. The class is made up most of economics majors and engineering majors, with a smattering of math and physics folks as well. I taught this class last semester, and I found that a lot of the economics majors were rather bored during the second half. I was able to motivate multiple integrals by doing some calculations with jointly distributed random variables, but for the vector analysis part of the course the only motivation I could think of was based on physics.
So I'm wondering if anybody knows a statistical/probabilistic interpretation of any of the main theorems of vector calculus. This might require having such an interpretation of div, grad, and curl, and it's not so obvious what it might be. Anyone have any ideas?

REPLY [2 votes]: This doesn't quite answer the question but multivariable calculus is baby differential topology and there are a few topological theorems that economists quote frequently.
The most common one is probably Brouwer fixed point theorem, which is required for the existence of Nash equilibria of a pretty wide class of games. Game theory is the main ingredient of microeconomics. (Actually the relevant general theorem is Kakutani's fixed point theorem for compact- and convex-valued correspondences but that is a stretch for most econ folks.)
Also, the Hairy Ball theorem has an economic interpretation. In a market, for each corresponding price there is an excessive demand. If there are $n$ goods, then price and demand are vectors in $\mathbb{R}^n$. When excessive demand is zero, economics says the market clears and is in equilibrium. So Hair Ball theorem says that the market clears for some price.
If you cover Lagrange multipliers, economics students taking intermediate microeconomics and macroeconomics see that everyday. The econ view Lagrange multiplier is the "shadow value of money", meaning that if the budget constraint is relaxed by $\epsilon$ at the optimal bundle (in the direction of the gradient), consumer utility increases by $\epsilon \cdot \lambda$. The equation
$$\nabla u = \lambda \cdot \nabla g$$
describes the consumer substituting between goods in his basket as he compares marginal utilities (entries in $\nabla u$) and marginal cost (entries in $\nabla g$).
As for the gradient and Maximal Likelihood estimation procedure, at the very basic level it's just a calculation. Suppose you have observations $\{ x_i \}$ drawn independently from the probability space $(\mathbb{R}, f(x, \theta)dx)$, where $\theta$ lies in some compact parameter space $\subset \mathbb{R}^n$. To estimate $\theta$, you maximize the likelihood function
$$
L(\theta) = \Pi_i f(x_i, \theta).
$$
The gradient of $\log L$ is called the score function. On a deeper level, although I am sure how much of it you can mention to your undergrads, MLE is the beginning point of information geometry, where statistics and differential geometry interact.<|endoftext|>
TITLE: Classification of generously transitive groups
QUESTION [7 upvotes]: A permutation group $G \lt S_n$ is called generously transitive, if for each $i,j$ there exists a permutation that interchanges them. Is there a reasonable classification of such (finite) groups?

REPLY [6 votes]: Clearly every generously transitive permutation group $G < {\rm S}_n$ acts transitively on $\{1,\dots,n\}$. Therefore we can find all generously
transitive permutation groups of degree $n$ by searching through the
transitive groups of degree $n$. The latter are available in the
Transitive Permutation Groups Library of GAP, for all $n \leq 30$.
The following GAP function tests whether a given group is generously
transitive:
IsGenerouslyTransitive := function ( G )

  local  n, i, j;

  n := LargestMovedPoint(G);
  if not IsTransitive(G,[1..n]) then return false; fi;
  if Transitivity(G,[1..n]) >= 2 then return true; fi;
  for i in [1..n] do
    for j in [i+1..n] do
      if   RepresentativeAction(G,[i,j],[j,i],OnTuples) = fail
      then return false; fi;
    od;
  od;
  return true;
end;

Now let's check how many of the transitive groups of degree $\leq 20$
are even generously transitive:
gap> List([2..20],NrTransitiveGroups);                                         
[ 1, 2, 5, 5, 16, 7, 50, 34, 45, 8, 301, 9, 63, 104, 1954, 10, 983, 8, 1117 ]
gap> Sum(last);
4722
gap> List([2..20],n->Number(AllTransitiveGroups(DegreeAction,n),
>                           IsGenerouslyTransitive));
[ 1, 1, 4, 4, 11, 5, 37, 22, 39, 6, 191, 7, 47, 63, 1239, 9, 605, 5, 870 ]
gap> Sum(last);
3166

Hence about two thirds of the transitive groups of degree $\leq 20$
are even generously transitive.
-- Therefore, perhaps rather than asking for a classification of generously
transitive groups, one might ask for a classification of those transitive
permutation groups which are not generously transitive.<|endoftext|>
TITLE: A malnormal embedding theorem?
QUESTION [9 upvotes]: Let $Q$ be a recursively presented group. Is it possible to embed $Q$ into a finitely presented group $G$ such that the image of $Q$ is malnormal in $G$?
Note that a subgroup $H$ of $G$ is malnormal if $H^g\cap H\neq 1$ implies that $g\in H$. That is, $H$ intersects each of its proper conjugates trivially.
This seems far to strong (far too good!) to be true. However, I cannot find anything in the literature that even hints at the existence or non-existence of such an embedding.
(More generally, I am wondering if there is some kinds of embedding which preserves the centraliser of a given element of $Q$. That is, if we fix $x\in Q$ then can $Q$ be embedded in a finitely presented group $G$ such that $C_Q(x)=C_G(x)$ (or even $C_Q(x)\cong C_G(x)$)? Malnormality guarantees this, and is a more interesting question, but I thought I should mention this too...)

REPLY [6 votes]: As I wrote in Remark 5.23 cited by Ian Agol, the embedding from that paper should answer the first question. It certainly preserves all centralizers, that is basically proved in the paper. That is if $G$ is a finitely generated recursively presented group, $Q>G$ is the group constructed in that paper, then for every $x\ne 1$ in $G$, $C_Q(x)=C_G(x)$. In fact the version of Higman embedding in Olʹshanskii, A. Yu.; Sapir, M. V. The conjugacy problem and Higman embeddings. Mem. Amer. Math. Soc. 170 (2004), no. 804 satisfies this property too. If I had more time, I would list lemmas from that paper which imply that. I am not sure if that embedding answers your first question. It might.<|endoftext|>
TITLE: Isometric but differently shaped closed surfaces in $\mathbb{R}^3$
QUESTION [5 upvotes]: Starting from the following inclusions for surfaces $M_1,M_2$ in $\mathbb{R}^3$:

     $M_1,M_2$ have the same shape, i.e. are related by an ambient isometry
→ $M_1,M_2$ have the same metric
→ $M_1,M_2$ have the same Gaussian curvature

the only examples of isometric but differently shaped surfaces in $\mathbb{R}^3$ I have seen so far do have boundaries:

cone and cylinder
catenoid and helicoid 
other associate families of isometric minimal surfaces. 

I wonder if there are no (examples of) isometric but differently shaped closed surfaces, and why that could be. (I am particularly interested in smooth surfaces.)
And I am still looking for (closed) surfaces with the same Gaussian curvature but different metrics.
A picture gallery would be highly welcome, because I really would like to see two such (non-)isometric surfaces.

REPLY [7 votes]: Here is the standard explict example of 2-spheres $S_1, S_2$ embedded in $R^3$ with identical curvature functions (under suitable parameterization), so that $S_1$ is not isometric (as a Riemannian manifold) to $S_2$. Both surfaces will be surfaces of revolution. Take two cylinders of revolution $C_i$, $i=1, 2$ with unit radius and different heights. Now, attach isometric rotationally-symmetric caps $C_i^\pm, i=1,2$ at the top and the bottom of the cylinders $C_i$. One can easily write explicit functions for the cups (similarly to writing equations for the bump-functions) so that the resulting surfaces $S_i$ are smooth. Now, it is clear that, under suitable parameterization, surfaces $S_1, S_2$ have the same curvature functions, but will not be isometric to each other (since their regions of zero curvature are not isometric as they have different heights). I think, you can easily draw pictures of such surfaces yourself.<|endoftext|>
TITLE: Extension of measures from the ball sigma-algebra to the borel sigma-algebra
QUESTION [8 upvotes]: Let $X$ be a metric space, $\Sigma_{1}$ the borel sigma algebra and
$\Sigma_{2}$ the sigma algebra generated by balls (open and closed). 
If $\mu$ is a probability measure on $\Sigma_{2}$ can it be extended to a
measure on $\Sigma_{1}?$

REPLY [11 votes]: Take a set $X$ of power $\aleph_1$, with the discrete metric where two distinct points have distance $1$.  The balls are singletons and the whole space.  The ball sigma-algebra is the countable and co-countable sets.  Let countable sets have measure zero, co-countable sets have measure 1.  
Now all subsets are open, so the Borel sigma-algebra is the power set.  There is no extension of this measure!<|endoftext|>
TITLE: Rational and log canonical singularity that is not log terminal
QUESTION [9 upvotes]: It is well-known that log terminal singularities are both rational and log
canonical. It is also well-known that log canonical singularities are not necessarily
log terminal and also not necessarily rational. However, it seems that the usual
example(s) given to show that show this at the same time. What I mean is that the
most common example of a log canonical, but not necessarily log terminal singularity
is a simple elliptic singularity (e.g., a cone over an elliptic curve), which is also
not rational.
I was trying to find an example that's log canonical and rational, but not log
terminal. (Obviously one cannot find one that's log terminal, but not rational).  Is
there a relatively simple example of a rational log canonical singularity which is
not log terminal?

REPLY [14 votes]: This is a good question and it is not entirely trivial, because there is no such example using a simple cone construction. More generally, if the resolution graph of the (normal) singularity is a chain of rational curves, then if the singularity is log canonical, then it is necessarily log terminal. (I won't include the calculation of this, but it is very similar to the calculation below, relatively straightforward using adjunction.)
So, to get an example like that one needs a singularity with a slightly more complicated  resolution graph. The simplest graph that is not a chain is a triple fork. So, let's suppose the exceptional curves are $E_0,E_1,E_2,E_3$ such that $E_0\cdot E_i=1$ for all $i\neq 0$ and $E_i\cdot E_j=0$ for all $i\neq 0, j\neq 0$. Furthermore let $E_i^2=-n_i$ with $n_i\in\mathbb N$ for all $i$. By Artin's criterion this is a rational singularity for any choices where $n_i\geq 2$. 
Writing down adjunction for each curve $E_i$ gives the equations:
$$-n_0(a_0+1) +a_1 +a_2+a_3 =-2$$
and 
$$ -n_j(a_j+1) + a_0 =-2 $$
for all $j\neq 0$.
These imply that if $a_0\geq -1$, then 
$$ a_j + 1 = \frac{a_0+2}{n_j} \geq \frac{-1\ \ \ }{n_j} > 0,$$
so
$$ a_j> -1 $$
and then solving for $a_0$ gives
$$ a_0 = -2 + \frac{n_0-1}{n_0-(\frac 1{n_1}+\frac 1{n_2}+\frac 1{n_3})}$$
so this singularity is log canonical, i.e., $a_0\geq -1$ iff 
$$
\frac 1{n_1}+\frac 1{n_2}+\frac 1{n_3}\geq 1
$$
and it is log terminal iff there is strict inequality here.
In particular it is log canonical, but not log terminal if there is equality which happens for example if $n_1=2, n_2=3, n_3=6$ and $n_0\geq 2$ arbitrary.
Finally, you might ask whether such a singularity exists, but that's easy. By blowing up appropriate points you can certainly create a configuration as above. Then you check that the intersection matrix is negative definite and use Artin's criterion to contract this configuration. The resulting singularity will be your example.<|endoftext|>
TITLE: Kissing Number of Spheres in Non-Euclidean Geometry
QUESTION [7 upvotes]: There has been much work done on the kissing number problem (of determining the greatest number of congruent spheres which can touch a single sphere in a packing) in Euclidean space for dimensions $1$ to $24$ and even some asymptotic work. My question is whether or not the kissing number problem has been studied in Non-Euclidean geometries such as the other seven Thurston Geometries $\mathbb{S}^2 \times \mathbb{R}, \mathbb{H}^2 \times \mathbb{R}, \mathbb{S}^3, \mathbb{H}^3, \widetilde{SL_{2}(\mathbb{R})}$, Nilgeometry, or Solvgeometry.
Any references or ideas about this topic would be very interesting!

REPLY [5 votes]: In one sense, which you are free to see as unimportant, the kissing number in the standard hyperbolic plane is unbounded, as it increases with increasing radii of the disks. One of the Laws of Cosines says
$$ \cos \alpha = 1 - \frac{1}{1 + \cosh a},  $$
where $a$ is the side length of an equilateral triangle and $\alpha$ is the vertex angle. As $a$ increases without bound, $\alpha$ decreases without nonzero lower bound. So, with discs of large enough radius, we can place as many disjoint-interior disks as we like around a given central one. The equilateral triangle has vertices at the centers of three mutually tangent circles. 
I'm just saying.<|endoftext|>
TITLE: Status of Grothendieck's conjecture on homomorphisms of abelian schemes
QUESTION [15 upvotes]: In [1] Grothendieck posits the following:
Conjecture. Let $S$ be a reduced connected scheme, locally of finite type over Spec($\mathbf{Z}$) or a field $k$, $A$ and $B$ two abelian schemes over $S$, $l$ a prime number, $u_l: T_l(A) \rightarrow T_l(B)$ a homomorphism, and suppose that there exists a point $s\in S$ such that $u_{ls}$ comes from a homormophism $u_s: A_s \rightarrow B_s$ of abelian schemes over $k(s)$. Then there exists an integer $n>0$ and a homomorphism $v: A \rightarrow B$ such that $T_l(v)=n u_l$.
Next Grothendieck remarks that the conjecture follows from Tate's conjectures on algebraic cycles. He then proceeds to a give a proof, using the Serre-Tate theorem, for the case of a characteristic zero base field for which the statement holds with $n=1$.

Question: What is the current status of this conjecture? If not resolved, are there any partial results for the positive characteristic case?

Here is what I've spotted in the literature so far, all characteristic zero:
Deligne in the second of three papers on Hodge theory, proves results about homomorphisms of abelian schemes over schemes $S$ of finite type over $\mathbf{C}$. For instance:
Proposition (Deligne 4.4.12 [2]) : Let $f: X \rightarrow S$ be an abelian scheme over a smooth scheme $S$. The following conditions are equivalent:
$(i)$ For every abelian scheme $g: Y \rightarrow S$, we have
$$ \mathrm{Hom}_S(X,Y) \cong \mathrm{Hom}_S(R_1 f_* \mathbf{Z}, R_1 g_* \mathbf{Z} ) $$
$(ii)$ The condition $(i)$ is verified for $X=Y$, and the centre $Z$ of $\mathrm{End}_S(X)\otimes \mathbf{Q}$ does not admit a complex place $\rho: Z \rightarrow \mathbf{C}$ such that the direct factor $R_1 f_*\mathbf{Q} \otimes_{Z,\rho} \mathbf{C}$ of $R_1 f_* \mathbf{C}$ is of pure Hodge type (-1,0).

Along similar lines, S.G. Tankeev in a 1976 paper [3] proves if $S$ is a connected smooth curve over $\mathbf{C}$, and $\pi_i: X_i \rightarrow S, i=1,2$ are two abelian schemes, then under some natural conditions, the canonical homomorphism
$$ \mathrm{Hom}_S(X_1,X_2) \rightarrow \mathrm{Hom}(R_1 \pi_{1*} \mathbf{Z}, R_1 \pi_{2*} \mathbf{Z})$$
is an isomorphism. In a paper the following year [4] Tankeev proves a Tate module variant of the above, of the form
$$ \mathrm{Hom}_k(I_1,I_2)\otimes_\mathbf{Z} \mathbf{Z}_l \cong \mathrm{Hom}_G(T_l(I_1), T_l(I_2)),$$
where $G=\mathrm{Gal}(\overline{k}/k)$, and $I_1, I_2$ are abelian varieties over $k(t)$ whose Néron models admit compactifications with certain properties.
Are there other well-known results along the same lines? I would especially like to know about the characteristic $p$ case, as well any results that actually spread a homomorphism of abelian varities over a point of S to maps of abelian schemes over S, as in the conjecture. 

[1] A. Grothendieck Un théorème sur les homomorphismes de schémas abéliens Invent. Math. 2 59-78 (1966) 
[2] P. Deligne Théorie de Hodge: II Publications mathématiques de l'I.H.E.S, tome 40 (1971) p.5-57
[3] S.G. Tankeev On homomorphisms of abelian schemes Math. USSR Izvestija Vol. 10 (1976), No. 4
[4] S.G. Tankeev On homomorphisms of abelian schemes II Math. USSR Izvestija Vol. 11 (1977), No. 6

REPLY [11 votes]: This is an expansion of my comments. Grothendieck's conjecture is now a theorem, except perhaps when $S$ is in characteristic $p$ and $\ell=p$. The explanation that follows is presumably what Grothendieck had in mind in his comment regarding the Tate conjecture.
First, we make some reductions: We can certainly assume that $S$ is integral. In fact, we can assume that $S$ is normal. This follows from Proposition 1.2 of Grothendieck's article.
Finally, we can assume that the generic point $\eta$ of $S$ is the spectrum of a finitely generated field. This can be shown as in 2.2 of Grothendieck's article.
In this scenario, by Proposition 2.7 in Chai-Faltings, 'Degeneration of Abelian varieties', the natural restriction map $Hom(A,B)\to Hom(A_{\eta},B_{\eta})$ is a bijection. Suppose now that $\eta$ is the spectrum of a finitely generated field. Then, by the Tate conjecture for homomorphisms of abelian varieties over finitely generated fields (proved by Zarhin in finite characteristic, and Faltings in characteristic $0$), the natural map of $\mathbb{Z}_{\ell}$-modules:
$$Hom(A,B)\otimes\mathbb{Z}_{\ell}\to Hom_G(T_{\ell}(A_{\eta}),T_{\ell}(B_{\eta}))$$
is an isomorphism. Here, on the right hand side, $G$ is the absolute Galois group of the spectrum of $\eta$. In particular, $u_{\ell}$ gives rise to an element of $Hom(A,B)\otimes\mathbb{Z}_{\ell}$.
Now, suppose that the specialization of $u_{\ell}:T_{\ell}(A)\to T_{\ell}(B)$ over a point $s\in S$ arises from an honest homomorphism $u_s:A_s\to B_s$. Then we find that, for some $n\in\mathbb{Z}_{\geq 1}$, $nu_{\ell}$ arises from a homomorphism $u':A\to B$. Essentially, the map
$$Hom(A,B)_{\mathbb{Q}}\to Hom(A_s,B_s)_{\mathbb{Q}}$$
identifies the left hand side with a vector sub-space of the right hand side. In fact, if $S$ is of characteristic $p$, we can take $n$ to be a power of $p$, since the specialization map will have saturated image after inverting $p$.<|endoftext|>
TITLE: Is the isomorphism problem for amenable groups decidable?
QUESTION [16 upvotes]: Is it algorithmically decidable if two finitely presented amenable groups are isomorphic?
Or slightly different:
Does there exist a family of amenable groups (indexed by natural numbers) for which one cannot algorithmically decide if two elements of the family are isomorphic?

REPLY [7 votes]: This answer is to point out that the particular way in which you
have posed the second question is not actually the question you
meant to ask, for it admits a trivial answer, requiring almost no knowledge about amenable groups. Namely, you ask,


Does there exist a family of amenable groups (indexed by natural numbers) for which one cannot algorithmically decide if two elements of the family are isomorphic?


The answer is yes. Fix any non-computable set
$A\subset\mathbb{N}$, and then enumerate the groups you are
interested in $G_0,G_1,G_2,\ldots$ (assuming there are at least
two non-isomorphic such groups) in such a way so that the indices
$n$ of one of the isomorphism classes is exactly $A$, and use the
rest of the indices for the rest of the groups (or just some of
them) in an arbitrary manner. That is, we make all $G_n$ for $n\in
A$ isomorphic, and not isomorphic to any other $G_m$ for $m\notin
A$. With such an enumeration, the isomorphism problem is not
decidable, simply because for a fixed $k\in A$, we cannot tell if
$G_n\cong G_k$, because this would provide a decision procedure
for $n\in A$, which is undecidable.
For example, a similar argument shows that there is an enumeration $G_0,G_1,G_2,\ldots$ of copies of the two groups $\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}$, such that the isomorphism problem is not decidable. Simply make $G_n\cong\mathbb{Z}$ if $n\in A$ and otherwise $G_n\cong \mathbb{Z}/2\mathbb{Z}$, for a fixed undecidable set $A$. 
Of course this isn't the answer you or anyone is interested in,
even though it does answer the question you actually
asked. So the point is that you shouldn't consider such arbitrary
enumerations when asking decidability questions about finitely
presented groups, but rather you want to ask about decidability
questions for the more natural enumerations of the presentations
that arise from a natural indexing of the presentations, by means
of a coding of the syntax of the presentation.<|endoftext|>
TITLE: Are exponentials in categorical models of linear logic harmful?
QUESTION [6 upvotes]: Categorical models for linear logic with $\otimes$, $1$, $\&$, $\top$, $\oplus$, $0$, and $\multimap$ are typically symmetric monoidal closed categories (for modeling $\otimes$, $1$, and $\multimap$) with products (for modeling $\&$ and $\top$) and coproducts (for modeling $\oplus$ and $0$). Is it harmful to additionally require that such a category has exponentials? Exponentials would model a binary operator $\Rightarrow$ for which proofs of $A \vdash B \Rightarrow C$ correspond to proofs of $A \mathbin\& B \vdash C$. Are there sensible categorical models for linear logic that have exponentials, or does the introduction of exponentials make the structure collapse?

REPLY [8 votes]: Michal already answered the question where we consider intuitionistic multiplicative linear logic, without an involutive linear negation; there are many such models where we can have a symmetric monoidal closed structure concurrent with a cartesian closed structure on a category. Another easy example (in fact a special case of one of his constructions) is the topos of $G$-sets where $G$ is an abelian group, made into a symmetric monoidal closed category by using the Day convolution. 
I'd like to supplement his answer by mentioning that if you do demand an involutive negation as well (i.e., work with classical multiplicative linear logic), which we categorically model using $\ast$-autonomous categories, then we do get a collapse: every such model is posetal. More is true: a cartesian closed category $C$ that is self-dual must be a poset; notice that a classical linear negation gives a self-duality, i.e., an equivalence 
$$\neg: C^{op} \stackrel{\cong}{\to} C.$$ 
This is because the initial object in a cartesian closed category is strict, i.e., for any object $X$, if there is a morphism $X \to 0$, then $X \cong 0$. Given that fact, consider morphisms $1 \to (A \Rightarrow B)$. Then, letting $(-)^\ast$ be the self-duality functor, these are in bijective correspondence with morphisms 
$$(A \Rightarrow B)^\ast \to 0$$ 
and if such a morphism exists, then $(A \Rightarrow B)^\ast \cong 0$ by strictness. In particular, there is at most one morphism $(A \Rightarrow B)^\ast \to 0$, and therefore at most one morphism $1 \to A \Rightarrow B$, and hence at most one morphism $A \to B$ for any two objects $A$, $B$.<|endoftext|>
TITLE: Weighted Jaccard Similarity
QUESTION [9 upvotes]: I'd like to calculate the similarity between two sets using Jaccard but temper the results using the relative frequency of each item within a corpus.
Jaccard is defined as the magnitude of the intersection of the two sets divided by the magnitude of the union of them both.
$jaccard(A, B) = \frac{|A \bigcap B|}{ |A\bigcup B|}$
If I use inverse document frequency (the log of the number of documents divided by the frequency of the item) ...
$idf(i) = log\frac{| D |}{f(i)+1}$
$|D|$ is the number of documents
$|f(i)|$ is the frequency of the item in the documents.
... can I define my weighted Jaccard similarity function as the sum of the IDFs of the items in the intersections divided by the sum of the IDFs of the union? (Sorry describing this in LaTeX reaches the limits of my knowledge of notation.)  Will this scale the similarity appropriately? Are a collection of weights better suited to a cosine similarity?

REPLY [2 votes]: Yes! There's a paper that does exactly that using exponentially distributed hashes for idf weighting.
https://www.robots.ox.ac.uk/~vgg/publications/2008/Chum08a/chum08a.pdf
If your hash function maps to (0,1], then apply the transformation $\frac{-\log(h(x))}{w_i}$ where $w_i$ is the weight of item i.
Then apply the original minhash algorithm using these biased hashes instead of the uniform ones, and it has the properties you desire.
Check out the section on "probability distributions" in this article for a derivation of how it works. https://moultano.wordpress.com/2018/11/08/minhashing-3kbzhsxyg4467-6/<|endoftext|>
TITLE: When is this braiding not a symmetry?
QUESTION [6 upvotes]: Given a topological space $X$ instead of forming the fundamental groupoid $\pi(X)$ which is the category whose objects are the points and morphisms the homotopy classes of paths one can also form the fundamental 2-groupoid which is the bicategory with objects = points, 1-morphisms = paths and 2-morphisms = homotopy classes of homotopies between paths. Composition of 1-morphisms is the "standard" composition $a\otimes b$ of paths given by $(a\otimes b)(t) = a(2t)$ for $0 \leq t \leq \frac{1}{2}$ and $(a\otimes b)(t) = b(2t-1)$ for $\frac{1}{2} \leq t \leq 1$.
By restricting to loops around a fixed point $e \in X$ one therefore gets a bicategory with one object, i.e., a monoidal category.
When moreover $X$ has the structure of a topological monoid it is easy to see that the structure map $\mu: X \times X \rightarrow X$ induces a braiding $\gamma_{a,b}: a\otimes b \stackrel{\sim}{\rightarrow} b\otimes a$ on this monoidal category. 
My question is: Is there an (elementary) example where this braiding is not a symmetry? One obvious necessary condition is that $X$ has to be noncommutative in order to give such an example.

REPLY [5 votes]: If $X=\Omega Y$, that braided monoidal category (indeed groupoid) classifies the homotopy type of $P_3Y$, the $3$-type of $Y$. Such $3$-type is completely determined by the map $\eta^*\colon \pi_2(Y)\rightarrow \pi_3(Y)$ defined by precomposition with the Hopf map $\eta\colon S^3\rightarrow S^2$. This map is quadratic, i.e. $\eta^* (x)=\eta^* (-x)$ and the map defined by 
$\eta^* (x|y)=\eta^* (x+y)-\eta^* (x)-\eta^* (y)$ is bilinear. One can recover this map from the monoidal category, essentially $\eta^*(x)$ corresponds to the braiding $\gamma_{x,x}\colon x\otimes x\cong x\otimes x$. The category is symmetric if and only if $\eta^* $ is a homomorphism. This does not always happen, since any quadratic map between abelian groups $A\rightarrow B$ can be realized by some appropriate $Y$. For instance, if you take $Y=S^2$ the quadratic map is $\eta^* \colon \mathbb{Z}\rightarrow \mathbb{Z}$ is $\eta^* (n)=n^2$, hence you get an example.<|endoftext|>
TITLE: Do finitely many plurigenera determine the Kodaira dimension?
QUESTION [21 upvotes]: Let $X$ be a smooth projective variety over a field of characteristic $0$ and let $K_X$ be the canonical bundle.
Recall that the Kodaira dimension $\kappa(X)$ is defined as the number $\kappa$ such that $$\alpha m^{\kappa}\le h^0(X,mK_X) \le \beta m^{\kappa}$$ for $\alpha,\beta>0$ and $m$ sufficiently large and divisible (or $\kappa=-\infty$ if $h^0(X,mK_X)=0$ for all $m$). It is well-known that $\kappa(X)$ is a birational invariant. A natural question is how large $m$ we need to take to determine $\kappa$. More precisely:

Is there an integer $M>0$, depending
  only $\dim X$, such that the values
  $h^0(X,K_X),h^0(X,2K_X),\ldots,h^0(X,MK_X)$
  determine $\kappa(X)$?

In particular, is there an $M>0$, depending on $\dim X$, such that $h^0(K_X)=\ldots=h^0(MK_X)=0$ implies that $\kappa=-\infty$?

REPLY [2 votes]: I guess in general the answer is NO. 
For instance if you take $X$ to be an $n$ dimensional variety which is $Y\times E$, where $E$ is an elliptic curve and $Y$ is an $n-1$ dimensional variety of general type with fast growing pluricanonical general, say $Y$ is a hypersurface of large degree $d$, then for any fixed $M$,  if you choose $d$ sufficiently large, $h^0(X,iK_X)$ can beat any sequence of numbers $a_i$ $(0\le i\le M)$ which you wrote for a fixed $n$-dimensional general type variety.<|endoftext|>
TITLE: discrete subgroups of Lie groups and actions on homogeneous spaces
QUESTION [14 upvotes]: Let $\Gamma$ be a discrete subgroup of a connected finite dimensional Lie group $G$. Let $K$ be a maximal compact subgroup of $G$ and denote $X=G/K$. It is well-known that $\Gamma$ acts properly on $X$ via multiplication on the left.
Now suppose that a discrete group $\tilde{\Gamma}$ contains $\Gamma$ as a finite index subgroup (I'm not assuming that $\tilde{\Gamma}$ is a subgroup of $G$.)
When can the action of $\Gamma$ on $X$ be extended to a proper action of $\tilde{\Gamma}$ on $X$?
Can you give an example where this is not the case?
Edit: Let my emphasize that I only require the action of $\tilde{\Gamma}$ on  $X$ to be proper (i.e. any continuous action with discrete orbits and finite point stablizers). I am not asking whether $\tilde{\Gamma}$ is a subgroup of $G$. The action of $\tilde{\Gamma}$ on $X$ is allowed to have a finite kernel.
Thanks.

REPLY [10 votes]: Take $\tilde{\Gamma}$ to be a higher genus Baumslag-Solitar group, see section 9 of 
M. Kapovich, B. Kleiner, Coarse Alexander duality and duality groups, Journal of Differential Geometry,  Vol. 69, (2005)  Number 2, p. 279-352. Group $\tilde\Gamma$ cannot act properly discontinuously on any contractible 3-manifold, but it contains a finite index subgroup $\Gamma$ which is is Gromov-hyperbolic and isomorphic to the fundamental group of a compact 3-manifold with boundary. Hence, by a corollary of Thurston's geometrization theorem, $\Gamma$ acts isometrically and discretely on hyperbolic 3-space, i.e., is a discrete subgroup of 
$G=PO(3,1)$. 
Edit: Here is an easier example. Note first that every finite group admitting an effective topological action on $R^2$ is either cyclic or dihedral. (B. von Kérékjártó, Über die endlichen topologischen Gruppen der Kugelfläche, Nederl. Akad. Wetensch. Proc. Ser. A vol. 22 (1919) pp. 568-569: There is probably a more accessible reference for this result, but I do not have one.) Now, take, say, a simple noncyclic  finite group $F$ and set $\tilde\Gamma= F *  F$. Then $\tilde\Gamma$ is virtually free, so it contains a free finite index subgroup $\Gamma$, which embeds in $PSL(2, R)$. On the other hand, every topological $\tilde\Gamma$-action on the plane has to be trivial since each free factor has to act trivially. 
A side note: If you are willing to replace the symmetric space $X$ with the $n$-fold product $X^n=X\times ... \times X$, then every isometric action of $\Gamma$ extends to an isometric action of $\tilde\Gamma$ on $X^n$ (where $|\tilde\Gamma:\Gamma|=n$): This is the induced representation construction. Incidentally, the following is an open problem:
Let $\Gamma$ be a discrete isometry group of hyperbolic $n$-space $H^n$ and $\Gamma \subset \tilde\Gamma$ is a finite-index extension. Is it true that $\tilde\Gamma$ admits a properly discontinuous isometric action on $H^m$ for some $m$?<|endoftext|>
TITLE: Contracting a curve of negative self-intersection on a surface
QUESTION [8 upvotes]: It is easy to show using birational factorization that the only curves on a surface which can be contracted to get an algebraic, smooth surface are smooth $(-1)$-curves. Furthermore, I know of examples of smooth curves of some other negative self-intersection which can be contracted in the algebraic category to result in a singular surface (where the order of the singularity is equal to the negative self-intersection?)

Question 1: Given a curve of negative self-intersection on a complex surface, what is the construction (in the analytic category) of its contraction?
Question 2: Given the curve, what conditions on it that determine whether the contraction is algebraic or not?

Good, clear, elementary references would be fine, as an alternative to an answer!

REPLY [11 votes]: The best numerical criterion of contrability for curves on surfaces is perhaps the following result, due to Michael Artin.

Proposition. Let $V$ be a surface and $X=\bigcup X_i \subset X$ be a connected curve. Then the following are equivalent:
$\boldsymbol{(i)}$ $X$ is contractible and if $\pi \colon V \to \bar{V}$ is the contraction map, then $\chi(\mathcal{O}_V) = \chi(\mathcal{O}_{\bar{V}})$;
$\boldsymbol{(ii)}$ the intersection matrix $|(X_i \cdot X_j)|$ is negative definite and for any cycle $Z$ supported in $X$ one has $p_a(Z) \leq 0$.
Moreover, under these conditions, if $V$ is a normal projective surface then $\bar{V}$ is also projective.

In other words, if $V$ is normal, projective and $X$ is "sufficiently rational and negative", then $X$ is contractible and the contraction is algebraic.
For further detais, see 
M. Artin: Some Numerical Criteria for Contractability of Curves on Algebraic Surfaces,American Journal of Mathematics
Vol. 84, No. 3 (Jul., 1962), pp. 485-496,
in particular Theorem 2.3 p. 491.<|endoftext|>
TITLE: finite subsets of torsion-free groups
QUESTION [5 upvotes]: Does there exist a torsion-free group $(G,.)$ with the following property?
There exists finite, non-empty subsets $S_1,S_2\subset G$ such that for all $s \in S_1.S_2$ there are at least two solutions to the equation $s=s_1.s_2$ where $s_1\in S_1$ and $s_2\in S_2$.
-
I imagine that such a group $(G,.)$ does exist. If you've found one, please could you explain how you constructed it also?
-
The motivation for this question is to prove that the group ring (for example over $\mathbb{Z}$) of any torsion-free group has no zero divisors, as conjectured by Kaplansky. If such $S_1$ and $S_2$ never exist then Kaplansky's conjecture would be true for lots of rings (any ring with no zero divisors in fact).
Thank you in advance.

REPLY [9 votes]: These are the "unique product groups". Not every torsion-free group satisfies this property. See Promislow, S. David, A simple example of a torsion-free, nonunique product group. 
Bull. London Math. Soc. 20 (1988), no. 4, 302–304.<|endoftext|>
TITLE: The critical exponent in the multiplicative order of 2 modulo primes
QUESTION [8 upvotes]: This is a sequel to this MO question: 
The multiplicative order of 2 modulo primes
As shown in Charles Matthews' paper linked to there, it is not hard to show that for each $\delta > 0$ there is a $c = c(\delta) > 0$ such that the set of primes $p$ for which the multiplicative order $n_p := \mathrm{ord}_p(2)$ of $2$ modulo $p$ satisfies $n_p < c\sqrt{p}$, has density $< \delta$. In particular, the set of primes with $n_p < p^{\frac{1}{2} - \epsilon}$ have zero density.
My question is, is $1/2$ really the critical exponent? For $\epsilon > 0$, is there a positive density of primes $p$ with $n_p < p^{\frac{1}{2} + \epsilon}$. Moreover, is there a $C < \infty$ for which the set of $p$ with $n_p < C\sqrt{p}$ has positive density?
I would also like to ask about the analogous question for elliptic curves: is $1/3$ really the critical exponent there? Given a point $P$ of infinite order, is there a $C < \infty$ for which the set of $p$ such that the order of $P \mod{p}$ is $< C \sqrt[3]{p}$ has positive density?

REPLY [5 votes]: The Generalized Riemann Hypothesis (for Dedekind zeta functions of certain Kummerian fields) would show that the "critical exponent" for the multiplicative order problem is $1$. In fact, GRH implies that if $g$ is any function tending to infinity, then (asymptotically) 100% of primes $p$ satisfy $n_p > p/g(p)$. This follows from the methods of Hooley in his conditional solution of Artin's conjecture; an explicit reference is Theorem 23 in this paper of Kurlberg and Pomerance:
http://www.math.dartmouth.edu/~carlp/PDF/par13.pdf
I suspect something similar can be said about your 2nd problem assuming GRH for Dedekind zeta functions associated to division fields. There's lots of relevant work in this area by Cojocaru, David, and others. But I'm not so knowledgeable about this, so I'll bow out now.<|endoftext|>
TITLE: Can we have many 1-dimensional rep, and very few high dimensional reps in a finite group?
QUESTION [7 upvotes]: Let $G$ be a finite group. Is it possible that there are "many" one-dimensional representations, and "very few" high-dimensional irreducible representations?
Originally, I thought it's impossible for the following reason. If $\chi_1, \chi_2$ are two one-dimensional representations such that $\chi_2 \chi_1^{-1} \not= 1$, and let $\rho$ be a high-dimensional irreducible representation. Then $\chi_1 \rho$ and $\chi_2 \rho$ are two nonequivalent irreps. When I tried to prove this formally, I feel it's not the case. 
Also, I find a finite group $G$ which has $2^{k-1}$ one-dimensional irreps, and two $2^{k-1}$-dimensoinal irreps, which is defined as follows. Let $G$ be generated by $g_1, g_2, \ldots, g_{2k-1}, -1$ such that $g_i^2 = -1$ and $g_i g_j = -g_j g_i$.
It seems that I already see the answer. But I hope to see some better explanation about it.

REPLY [6 votes]: There is another way to produce such examples. Take $V$ to be an $n$ dimensional ${\mathbb F}_2$ vector space, let $V^*$ be its dual and consider the Heisenberg group $H$ of $(n+1)\times (n+1)$ matrices with ${\mathbb F}_2$ entries, which consist of unipotent upper triangular  matrices, all of whose upper off diagonal entries are zero except those in the first row or the $n+1)$-th column. Thus we have an exact sequence 
$$1 \rightarrow {\mathbb F}_2 \rightarrow H \rightarrow V\oplus V^* \rightarrow 1. $$    It is easily seen that there is one irreducible representation $\rho$ of the Heisenberg group $H$,  of dimension $2^n$ where the centre ${\mathbb F}_2$ acts non-trivially. All the other irreducible representations  are one dimensional (and are characters of the quotient group $V\oplus V^*$.<|endoftext|>
TITLE: Example of wall-crossing formulae?
QUESTION [9 upvotes]: In a nutshell my question is "Are there any easy, educational wall-crossing formulae?".
Recently I often hear the word "(Kontsevich-Soibelman's etc) wall-crossing formula" in algebraic geometry talks. I wonder what they are like but am not really ready to read those relevant papers with high technology. I understand that this kind of formula expresses how your invariants of moduli spaces varies when you change the stability condition to construct the moduli spaces. Could someone provide me with a toy example? I am looking for a "wall-crossing formula" which can be understood by those with basic AG and maybe some GIT. 
Thank you very much.

REPLY [23 votes]: I can tell you  the gist of a  "wall crossing formula".    Typically you have a  space of parameters $\newcommand{\eS}{\mathscr{S}}$ $\Lambda$,  a configuration space   $\newcommand{\eC}{\mathscr{C}}$ $\eC$, and a parametrized moduli space $\newcommand{\eM}{\mathscr{M}}$ $\eM$ which is a  subvariety $\eM\subset \eC\times \Lambda$.
We have a natural projection
$$\pi:\eM\to \Lambda. $$
The fiber of this map over $\lambda\in\Lambda$, denoted by $\eM_\lambda$,  is called the moduli space corresponding to the  parameter $\lambda$.
One could associate to $\eM_\lambda$ various invariants. For example, there might exists a sheaf $\eS\to\eM$ which restricts to a  sheaf $\eS_\lambda\to\eM_\lambda$. We denote   by  $e(\lambda)$ the Euler characteristic  of $H^\bullet(\eM_\lambda,\eS_\lambda)$.
It could happend that  $e(\lambda)$ depends on $\lambda$, but it could depend in a rather  specal way. Namely, there could exists  real codimension one  subvarieties  $W_i\subset \Lambda$, $i\in I$, called walls,  so that, if
$$ W:=\bigcup_{i\in I} W_i, $$
then $\lambda\to e(\lambda)$ is continuous on $\Lambda^\ast:=\Lambda\setminus W$.  In particular, the  function $\lambda\to e(\lambda)$ is constant on the connected components   of $\Lambda^*$, which are called chambers.
If    two chambers $C_0, C_1$ are adjacent, i.e.,  they sit on opposite sides of a wall  $W_i$ ,   then  $e(\lambda)$  has constant values $e(C_0)$ and $e(C_1)$ in these two chambers and a wall crossing formula  will tell you what the difference  $e(C_1)-e(C_0)$ is.
Here is a  trivial example.    Let $\Lambda=\mathbb{R}^2$.  We let $(b,c)$ denote the coordinates  of a point in $\Lambda$.   The configuration space $\eC$ is $\mathbb{R}$ and the parametrized moduli space is
$$\eM= \bigl\lbrace (t,b, c)\in\eC \times \Lambda;\;\;t^2+bt+c=0\,\bigr\rbrace. $$
The moduli space $\eM_{b,c}$ can be identified with the set of real roots of the quadratic polynomial $t^2+bt+c$. We denote by $e(b,c)$ the number of such roots. In other words, $e(b,c)$ is the (topological) Euler characteristic of the space $\eM_{b,c}$.
In this case we have a single  wall
$$ W=\bigl\lbrace (b,c)\in \Lambda;\;\; b^2- 4c=0\bigr\rbrace, $$
with two chambers,
$$ C^\pm=\bigl\lbrace \;\pm(b^2-4c)>0\;\bigr\rbrace. $$
In this case the Wall crossing formula is
$$ e(C^+)-e(C^-)= 2. $$<|endoftext|>
TITLE: Series defined by a fixed-point functional equation
QUESTION [12 upvotes]: Description
I my work, I met fixed-point functional equations of a very specific form. Since I am not expert in the domain of functional equations, I have not pushed the study of these very far. Here is a description of the objects.
Let $X := \lbrace x_1, \dots, x_k \rbrace$ be an alphabet of mutually commuting indeterminates and $F(x_1, \dots, x_k)$ be the formal power series
\begin{equation}
    F(x_1, \dots, x_k) := 
    \sum_{\alpha_1, \dots, \alpha_k} 
    \lambda_{\alpha_1 \dots \alpha_k} \: 
    x_1^{\alpha_1} \dots x_k^{\alpha_k},
\end{equation}
where all coefficients $\lambda_{\alpha_1 \dots \alpha_k}$ are nonnegative integers.
The series $F(x_1, \dots, x_k)$ is defined by the fixed-point functional equation
\begin{equation}
    F(x_1, \dots, x_k) = x_1 + F(P_1(x_1, \dots, x_k), \dots, P_k(x_1, \dots, x_k)),
\end{equation}
where for any $i \in [k]$, $P_i(x_1, \dots, x_k)$ is a polynomial with nonnegative integer coefficients. Of course, there are some necessarily conditions to ensure that $F(x_1, \dots, x_k)$ is well-defined.
Here are two celebrated examples of fixed-point equations of this kind.

The series $S(x)$ defined by
\begin{equation}
    S(x) = x + S(x^2 + x^3).
\end{equation}
This functional equation has been studied by A. M. Odlyzko (see [Odl82]). One can compute first coefficients of $S(x)$ by iteration:
\begin{equation}
    S_1(x) = x,
\end{equation}
\begin{equation}
    S_2(x) = x + x^2 + x^3,
\end{equation}
\begin{equation}
    S_3(x) = x + x^2 + x^3 + x^4 + 2x^5 + 3x^7 + 3x^8 + x^9,
\end{equation}
and so on. This generating series is of the form
\begin{equation}
    S(x) = x + x^2 + x^3 + x^4 + 2x^5 + 2x^6 + 3x^7 + 4x^8 + 5x^9 + 
    8x^{10} + 14x^{11} + 23x^{12} + \cdots
\end{equation}
and its coefficients have a combinatorial interpretation: the coefficient of $x^n$ is the number of $2,3$-balanced trees (see [Odl82] and also A014535) with $n$ leaves. 
The series $B(x, y)$ defined by
\begin{equation}
    B(x, y) = x + B(x^2 + 2xy, x).
\end{equation}
By iteration, one finds
\begin{equation}
    B_1(x, y) = x,
\end{equation}
\begin{equation}
    B_2(x, y) = x + 2xy + x^2,
\end{equation}
\begin{equation}
    B_3(x, y) = x + 2xy + x^2 + 4x^2y + 2x^3 + 4x^2y^2 + 4x^3y + x^4.
\end{equation}
The coefficient of $x^n$ in the specialization $B(x) := B(x, 0)$ is the number of balanced binary trees (see [BLL94] and also A006265) with $n$ leaves.

In my work, I encountered fixed-point functional equations rather more complicated:
\begin{equation}
    F(x, y, z) = x + F(x^2 + xy + yz, x, xy),
\end{equation}
\begin{equation}
    G(x, y, z) = x + G(x^2 + 2xy + yz, x, x^2 + xy),
\end{equation}
\begin{equation}
    H(x, y, z, t) = x + H(x^2 + 2yt + yz, x, x^2 + xy, yt + yz).
\end{equation}
Coefficients of these count some tree-like structures.
It seems that the theory is fairly well-known when the generating series are univariate (for instance, A. M. Odlyzko [Odl82] gives asymptotic formulas and recurrence relations for the coefficients of $S(x)$). In the multivariate case, it seems that only few things are known and up to my knowledge, there are no other way to express the series $B(x)$ counting balanced binary trees.
Questions

(1) Have you encountered other fixed-point functional equations of the described form?

Given a fixed-point functional equation for a series $F$ of the given form:

(2) What about an other method to extract coefficients of $F$ otherwise than iteration?
(3) What about recurrence relations between the coefficients of $F$?
(4) What about asymptotic approximations for the coefficients of $F$?
(5) What about rationality, algebraicness, d-finiteness, or other of $F$?

References
[Odl82] A. M. Odlyzko. Periodic Oscillations of Coefficients of Power Series That Satisfy Functional Equations. Advances in Mathematics, 44:180–205, 1982
[BLL94] F. Bergeron, G. Labelle, and P. Leroux. Combinatorial Species and Tree-like Structures. Cambridge University Press, 1994.

REPLY [2 votes]: This maybe off-topic answer, but since OP asked 
(1) Have you encountered other fixed-point functional equations of the described form?
I will put an example that I have seen. 
$$
S(x)=x-S(x^2)
$$
This gives the power series
$$
S(x)=\sum_{k=0}^{\infty} (-1)^k x^{2^k}.
$$
This is the famous example given by Hardy. 
Using Tauberian Theory, one can prove that this is not Abel summable. i.e.
$$
\lim_{x\rightarrow 1-} S(x) \textrm{ does not exist.}
$$
cf) G. H. Hardy, “On certain oscillating series”, Quart. J. Math. 38 (1907), 269–288.<|endoftext|>
TITLE: Coherent Sheaves and Holomorphic Vector Bundles
QUESTION [7 upvotes]: For a complex manifold $M$, I'm trying to understand (from a differential geometry point of view) what its category of coherent sheaves is. If I understand correctly, then the sheaf of holomorphic sections of a complex vector bundle is a coherent sheaf for $M$. Is there some case when the category of coherent sheaves is equivalent to the category of holomorphic vector bundles? In other words can I just forget about the algebraic geometry and understand the definition in terms of complex geometry? Or alternatively, can one somehow "generate" the category of coherent sheaves from the category of holomorphic vector bundles

REPLY [3 votes]: There are two approaches to this question, one by Toledo and Tong (using Cech covers or hypercovers) and one by Block (using the Dolbeaut algebra). Block's paper is at http://arxiv.org/abs/math/0509284<|endoftext|>
TITLE: Is the operadic butterfly symmetric?
QUESTION [8 upvotes]: The operadic butterfly is a diagram in the category of operads in vector spaces. It extends the short exact sequence relating commutative, associative and Lie operads.
$$\begin{array}{ccccc}
  & Dend & & & & Dias & \newline
 \nearrow & &\searrow & &\nearrow & &\searrow\newline
Zinb &  &  &Ass & & & \quad Leib \newline
 \searrow & &\nearrow & &\searrow & &\nearrow\newline
  & Comm & & & & Lie & \newline 
\end{array}$$
Here is a paper by Loday in which some discussion can be found. As the reference explains, the Koszul duality functor, $\mathcal{O}\mapsto \mathcal{O}^!$, gives the above diagram symmetry about the vertical axis.
Is the butterfly also symmetric about the horizontal axis? 
In other words, does there exist a functor $F : Operad(C) \to Operad(C)$, where $C$ is vector spaces or chain complexes, such that $F\circ F \simeq 1$, $F$ fixes the operads: $Zinb$, $Ass$ and $Leib$, and exchanges the pairs: $(Dend,Dias) \leftrightarrow (Comm,Lie)$.

REPLY [6 votes]: I have never seen such a functor and I doubt that it would exist. For instance, the operad Com is symmetric and one-dimensional in every arity. Whereas the operad Dend is regular (coming from a non-symmetric one) and is generated by two (non-symmetric) generators. I also now no non-trivial functor which preserves Zinb or Leib. 
Can I ask why you are looking for such a functor (despite the symmetric shape of the operadic butterfly)?<|endoftext|>
TITLE: Is every projective $\mathbf{Z}[x]$-module free?
QUESTION [14 upvotes]: Is every finitely generated projective $\mathbf{Z}[x]$-module free?

REPLY [12 votes]: When $R$ is a PID, then every finitely generated projective $R[x]$-module is free. As Steven already said, this is due to Seshadri. Here is the reference:

Seshadri, C.S., Triviality of vector bundles over the affine space $K^2$, Proc. Nat. Acad. Sci. USA 44 (1958), 456-458.<|endoftext|>
TITLE: When adding a constant makes a multivariate polynomial reducible?
QUESTION [11 upvotes]: Given a multivariate polynomial $f(x_1,\dots,x_n)$ with integer coefficients, how to find an integer $m$ (if it exists) such that $f(x_1,\dots,x_n) + m$ factors into polynomials of smaller degrees? 
Are there any simple criteria to identify cases when such $m$ does not exist?
Is it possible that more than one suitable values of $m$ exist?

REPLY [5 votes]: Suppose that $f$ cannot be written as $f(x_1,x_2,\dots,x_n)=h(g(x_1,x_2,\dots,x_n))$ with $h\in\mathbb C[x]$ of degree $\ge2$ and $g\in\mathbb C[x_1,x_2,\dots,x_n]$. For $m\in\mathbb C$ let $a_m$ be the number of irreducible factors of $f(x_1,x_2,\dots,x_n)+m$. Then Ewa Cygan proved (see here for a review and reference) that $\sum_{m\in\mathbb C}(a_m-1)\le\deg f-1$. (The case $n=2$ was shown previously by Josef Stein, see here.)
So, as hinted in the comments, there can be infinitely many $m$ with $f(x_1,x_2,\dots,x_n)+m$ reducible. But in that case $f$ has a rather specific shape.
The question also asks to decide if there is at least one $m$ making $f+m$ reducible. Here is a sketch on how to find the $m$'s in case that there are only finitely many: Suppose that $f+m=uv$ with non-constant polynomials. Pick $\bar x\in\mathbb C^n$ with $u(\bar x)=v(\bar x)=0$. (There probably is no such $\bar x$, so one should better work in the projective completion.) Then the partial derivatives $\frac{\partial f}{\partial x_i}=\frac{\partial u}{\partial x_i}v+u\frac{\partial v}{\partial x_i}$ vanish in $\bar x$.
So the recipe is as follows: Determine all (usually finitely many) common roots $\bar x$ of the partial derivatives $\frac{\partial f}{\partial x_i}$, set $m=-f(\bar x)$, and check if $f(x_1,\dots,x_n)+m$ is reducible.
As said above, one should better work projectively in order to avoid missing solutions $m$ coming from singularities at infinity.<|endoftext|>
TITLE: Are there non-compact, non-smoothable manifolds?
QUESTION [12 upvotes]: There do exist manifolds which do not admit any smooth structure at all. But the only examples I've heard of are all compact.

Are there any non-compact, non-smoothable manifolds?

REPLY [30 votes]: The Cairns-Hirsch theorem says that a PL manifold $M$ is smoothable if and only if $M\times \mathbb{R}$ is smoothable, so you can take $M$ to be any one of the known compact, PL examples such as Kervaire's manifold and then $M\times\mathbb{R}^n$ is non-smoothable for $n \geq 1$.<|endoftext|>
TITLE: Is there a constructive proof of Cantor–Bernstein–Schroeder theorem ?
QUESTION [36 upvotes]: An important feature of the Cantor-Schroeder-Bernstein theorem is that it does not rely on the axiom of choice. However, its various proofs are non-constructive, as they depend on the law of excluded middle.
There are several well-known proof strategies. For example, there is a simple proof which uses Tarski's fixed point theorem.
My question pertains to the well known proof attributed to Julius Konig, and given here: Proof by Konig. 
What I have not been able to grasp is,  how this is not a constructive proof? Given functions $f$ and $g$, isn't it possible to actually arrive at the bijection by forming biinfinite sequences as given in the proof?

REPLY [4 votes]: It is well-known that Grothendieck toposes and realizability toposes (for different reasons) are models of Intuitionistic Zermelo–Fraenkel set theory. Therefore both Andrej and Todd showed in their answers (in essentially different way) that Cantor–Bernstein–Schroeder cannot be proved in IZF.
Of course, this does not mean that Cantor–Bernstein–Schroeder property is incompatible with constructive mathematics --- it just shows that IZF is too weak to prove CBS. Therefore, a complementary question could be: what are the implications of IZF+CBS; does IZF+CBS make the logic collapse to the boolean case?
If I am not mistaken, the answer is no, and the counterexample is constructed below. However, I shall start with a negative observation.
Let $\Omega$ be a Heyting algebra with countable unions. An element $v \in \Omega$ is complementable if there exists an element $w \in \Omega$ such that $v \vee w = 1$ and $v \wedge w = 0$.
We say that $\Omega$ is a boolean algebra if its every element is a finite union of complementable elements (equivalently, if every element is complementable).
We say that $\Omega$ is a pro-boolean algebra if its every element is a countable union of complementable elements.
The claim is that every elementary topos with countable colimits satisfying Cantor–Bernstein–Schroeder propery is pro-boolean (i.e. its subobject classifier is a pro-boolean algebra).
Let $v \colon V \rightarrow 1$ be a "truth" value (a monomorphism into terminal object). We shall construct two objects $X = \coprod_{\mathbb{N}} 1$ and $Y = V \sqcup X$. There are obvious monomorphisms ${\iota_X}\colon{X}\rightarrow{Y}$ given by the coproduct injection, and ${v \sqcup \mathit{id}}\colon{Y}\rightarrow{X}$. Therefore, by Cantor–Bernstein–Schroeder there is an isomorphism ${b}\colon{Y}\rightarrow{X}$. Since coproducts in a topos are extensive, we may divide each component ${\iota_0}\colon V \rightarrow{Y}$, ${\iota_k}\colon 1 \rightarrow{Y}$ of the coproduct $Y$ along $b$ through coproduct's injections ${\iota_l} \colon 1 \rightarrow{X}$ obtaining elements $\alpha_{k,l}$ such that $Y \approx \coprod_k \coprod_l \alpha_{k,l} \approx \coprod_l \coprod_k \alpha_{k,l}$ and $b = \coprod_l b_l$ with each ${b_l}\colon{\coprod_l \alpha_{l,k}}\rightarrow{1}$ being an isomorphism (using again extensivity of coproducts and the fact that pullback of an iso is iso). 
Because unions in a topos are effective (and coproducts are disjoint) $1 \approx \coprod_l \alpha_{l,k} = \bigcup_l \alpha_{l,k}$ and so each $\alpha_{l,k}$ is complemented by $\bigcup_{x \neq l} \alpha_{x,k}$. Since every subterminal value can sit in exactly one way in the terminal object, $V = \bigcup_l \alpha_{l,0}$ is a countable disjoint union of complementable elements.
(We cannot get more from this construction; for example in the category of sheaves over rational numbers with the usual topology, $\coprod_{\mathbb{N}} 1 \approx V \sqcup \coprod_{\mathbb{N}} 1$ for (non-complementable) truth value $V$ corresponding to the open ball $(-1, 1)$; unfortunately there are other objects in this category that can serve as counterexamples for CBS. 
Let me also point that the standard procedure of constructing an isomorphism from two monomorphisms would not work in this case. However, by the above argument it is clear that such an isomorphism may be constructed for any pro-boolean topos. The solution is to not shift uniformly the whole $V$ (or its pseudocomplement), but to move each of (countably many) complementable parts of $V$ separately.)

[I am terribly sorry, now I see that there is an error in the following argument; I will try to fix it (on condition it is possible --- I am not sure now). I should have checked all relevant detail before posted this as an answer.]
For a positive result, consider set:
$$\mathcal{D} = \lbrace 0, 1, \frac12, \frac13, \frac14, \dotsc\rbrace$$
with topology inherited from $\mathbb{R}$ and construct the category $\mathit{Sh}(\mathcal{D})$ of sheaves over $\mathcal{D}$. Every open set in $\mathcal{D}$ can be build from singletons $\lbrace\frac1n\rbrace$ and a set of the form $[0, \frac1k]$.
Let $F, G \colon \mathcal{D}^{op} \rightarrow \mathbf{Set}$ be any sheaves and assume there are monomorphisms $m \colon F \rightarrow G$ and $n \colon G \rightarrow F$. A monomorphism between sheaves is an injection on its each component, therefore by CBS theorem for sets $F(U) \approx G(U)$ for every open set $U$. Let $\phi_U \colon F(U) \approx G(U)$ be a collection of such isomorphisms. We shall construct an isomorphism $\alpha \colon F \rightarrow G$ between sheaves inductively:

$\lambda_{[0, 1]} = \phi_{[0, 1]}$ 
for every nonempty $F(\lbrace\frac1k\rbrace)$ choose an element $1_{\frac1k} \in F(\lbrace\frac1k\rbrace)$; if $F(\lbrace\frac1k\rbrace)$ is empty then $\lambda_{[0, \frac1{k+1}]} = \phi_{[0, \frac1{k+1}]}$, otherwise $\lambda_{[0, \frac1{k+1}]} = G([0, \frac1{k+1}] \subset [0, \frac1{k}]) \circ h_{\frac1k}$, where $h_{\frac1k} \colon F([0, \frac1{k+1}]) \rightarrow F([0, \frac1{k}])$ is the unique morphism to the product  $F([0, \frac1{k}]) = F([0, \frac1{k+1}]) \times F(\lbrace\frac1k\rbrace)$ induced by $F([0, \frac1{k+1}]) \overset{!}\rightarrow 1 \overset{1_{\frac1k}}\rightarrow F(\lbrace\frac1k\rbrace)$ and the identity on $F([0, \frac1{k+1}])$
similarly, for every nonempty $F([0, \frac1{k+1}])$ choose an element $1_{[0, \frac1{k+1}]} \in F([0, \frac1{k+1}])$; if $F([0, \frac1{k+1}])$ is empty then $\lambda_{\lbrace\frac1{k}\rbrace} = \phi_{\lbrace\frac1{k}\rbrace}$, otherwise $\lambda_{\lbrace\frac1{k}\rbrace} = G(\lbrace\frac1{k}\rbrace \subset [0, \frac1{k}]) \circ h_{{[0, \frac1{k+1}]}}$
if $U$ is a disjoint union of the form $[0, \frac1k] \sqcup \bigcup_i\lbrace\frac{1}{n_i}\rbrace$, where $[0, \frac1k]$ is the largest interval contained in $U$, then $\lambda_{U} = \lambda_{[0, \frac1k]} \times \prod_i \lambda_{\lbrace\frac{1}{n_i}\rbrace}$, where the products are determined by structures of the sheaves.

In the second and third step we have chosen the components of $\lambda$ to be upward compatible, and in the fourth step the naturality condition follows from the universal property of products. Thus $F \approx G$.
[EDIT: Let me argue that $\phi_U$ may be chosen in such a way that each $\lambda_U$ is really an isomorphism. Assume, that all $F(\lbrace \frac1k\rbrace)$ are nonempty. Define $\mathit{colim}F([0, \frac1k])$ to be the colimit of the diagram:
$$F([0, 1]) \rightarrow F([0, \frac12]) \rightarrow \cdots \rightarrow F([0, \frac1k]) \rightarrow \cdots $$
We have:
$$(\mathit{colim}_kF([0, \frac1k])) \times (\prod_i F(\lbrace\frac1i\rbrace))  \approx F([0, 1]) \times \mathit{colim}\_k \prod\_{i > k} F(\lbrace\frac1i\rbrace) \approx F([0,1])$$
and similarly for $G$. Since in a locally presentable category monomorphisms are stable under directed colimits, both:
$$\mathit{colim}F([0, \frac1k]) \overset{\mathit{colim}\left(m_{[0, \frac1k]}\right)}\rightarrow \mathit{colim}G([0, \frac1k])$$
and:
$$\mathit{colim}F([0, \frac1k]) \overset{\mathit{colim}\left(n_{[0, \frac1k]}\right)}\leftarrow \mathit{colim}G([0, \frac1k])$$
are monomorphisms, thus by CBS for sets $\mathit{colim}F([0, \frac1k]) \overset{\phi_0}\approx \mathit{colim}G([0, \frac1k])$.
Therefore, $\phi_{[0, 1]}$ may be assumed to be of the form $\phi_0 \times \prod \phi_{\lbrace\frac1k\rbrace}$. Likewise every $\phi_{[0, \frac1k]}$.
 ]
(BTW, I think we are not really that far from the inverse of the above theorem, but  that is for another story...)<|endoftext|>
TITLE: Higher dimensional Bezout via Hilbert polynomials: a reference
QUESTION [7 upvotes]: For the purposes of teaching my elementary course in algebraic geometry I am looking for a reference (or notes) that contains a complete proof of a higher-dimensional weak Bezout theorem. I only want to learn about a proof that is based on the approach when dimensions and degrees of projective varieties are introduced via Hilbert polynomials.
Weak Bezout. Let $F_1,...,F_n$ be homogeneous polynomials of degrees $d_1,...,d_n$
such that hypersurfaces $F_1=0$,..., $F_n=0$ have only finite number of intersections in
$\mathbb P^n_K$ (with $K$ algebraically closed). Then the number of intersections is 
at most $d_1\cdot...\cdot d_n$.
Justification of the question. Of course there is a large amount of proofs of this statement in many books on algebraic geometry
(Shafarevich, Harris, Hasset, ect.) but these proofs usually come after 200 pages of text, while 
I want an honest proof that is contained in a complete set of notes of 40 pages (this will be the maximal 
length of my notes and I don't want to spend  more than 10 pages on higher dimensional Bezout). Also these proofs often develop dimension theory basing on transcendence
degree of the field of rational functions and I don't want to use this approach, (since it will require 
2-3 additional lectures which I don't have time to give). 
I know one place where the approach that I want to use (namely everything is based on Hilbert polynomials)
is taken, these are the notes of Manin (end of 60ties). Unfortunately it seems to me that there is 
a problem with the proof he proposes. Basically everything works if $F_1,...,F_n$
form a regular sequence, but to understand why in the condition of the theorem $F_1,...,F_n$
do form a regular sequence is left in the notes as something "not hard to do"...
In the book of Hasset, it seems to me there is a similar problem (i.e. it is not explained 
why $F_1,...,F_n$ form a regular sequence if the number of intersections of hypersurfaces is finite). I hope there is a nice complete proof somewhere... 
In other words, what will completely satisfy me is a short proof of the following statement:
If the number of intersections of $F_i=0$ is finite, then $F_i$ form a regular sequence.
Is there a short proof of this statement?
PS. I think that the answer of Sandor shows that probably there is no "magic easy" solution to the question (if one sticks to Hilbert polynomial approach instead of using higher-dimensional resultants), so I decided to accept it.

REPLY [2 votes]: I'm writing out what is essentially Looijenga's proof for my own reference. Let $Z_r = \{ F_1 = F_2 = \cdots = F_r = 0 \}$. From basic dimension theory (eg Shavarevich Theorem I.6.2.5, much easier than Macaulay's result), all components of $Z_r$ have dimension $\geq n-r$. If any of them had dimension $>n-r$, then $Z_n$ would have a component of dimension $>0$, a contradiction. So all components of $Z_r$ have dimension $n-r$.
Let $A_r = k[x_0, x_1, \ldots, x_n]/\langle f_1, f_2, \ldots, f_r \rangle$. Look at the exact sequence of graded modules:
$$0 \to K_r \to A_r \overset{\cdot f_{r+1}}{\longrightarrow} A_{r}[d_r] \to A_{r+1}[d_r] \to 0$$
where $K_r$ is defined to be the kernel of the map it precedes. 
If we knew unmixedness, we would know that $K_r=0$. Although we don't know that, we do know that the support of $K_r$ does not contain any irreducible component of $Z_r$. That is because, if $W \subset Z_r$ were an irreducible component of $Z_r$ on which $f_{r+1}$ was $0$, then $W$ would be an $n-r$ dimensional component of $Z_{r+1}$, contradiction. 
So the Hilbert polynomial of $K_r$ has degree $\leq n-r-1$. (Of course, in reality, it is zero.) We now inductively prove that the Hilbert polynomial of $A_r$ has leading term $d_1 d_2 \cdots d_r \frac{x^{n-r}}{(n-r)!}$, as desired.<|endoftext|>
TITLE: What is a Gaussian measure?
QUESTION [23 upvotes]: Let $X$ be a topological affine space. A Gaussian measure on $X$ is characterized by the property that its finite-dimensional projections are multivariate Gaussian distributions.
Is there a direct characterization of a Gaussian measure which does not rely on finite-dimensional projections? This definition is analogous to describing a duck as the animal whose shadows look like $2$-dimensional ducks. The definition is sufficient for doing analysis, but to me it misses the essence of what a Gaussian measure is as a mathematical object in and of itself. 

Here is the precise definition of a Gaussian measure that I usually work with, which relies on the fact that Gaussians are entirely described by their covariance structure.
For $X$ a topological affine space as above, let $X^*$ denote its dual space of affine functionals. The dual space is a linear space, since there there is a natural zero functional $0 \in X^*$.
Let $K : X^* \to X$ be a continuous affine operator which is symmetric and non-negative-definite. i.e., $f'(Kf) = f(Kf')$ and $f(Kf) \ge 0$ for all $f, f' \in X^*$. Let $m_K := K(0)$ denote the image of the zero functional.
There is a unique Gaussian measure $P_K$ on $X$ with mean point $m_K \in X$ and covariance operator $K : X^* \to X$. That is, if $\pi : X \to \mathbb R^n$ denotes a finite-dimensional projection, then the push-forward measure $\pi_* P_K := P_K \circ \pi^{-1}$ is an $n$-dimensiona Gaussian distribution with mean vector $\pi(m_K) \in \mathbb R^n$ and covariance matrix $\pi K \pi^*$, where $\pi^* : (\mathbb R^n)^* \to K^*$ denotes the formal adjoint operator.
Furthermore, the structure theorem for Gaussian measures states that all Gaussian measures arise in this way. Consequently, we may parametrize the space of Gaussian measures by the space $\mathcal K(X)$ of symmetric, non-negative operators from $X^*$ to $X$.
This provides a weak answer to the question stated at the top of this post: yes, Gaussian measures can be directly characterized by their covariance structure. Consequently, here is the stronger form of my question:

Is there a geometric description of the space $\mathcal K(X)$ of Gaussian covariance operators?

For example, is the space $\mathcal K(X)$ an infinite-dimensional manifold? What is its symmetry group?

Edit: My above post implicitly defines the covariance form incorrectly. In the affine setting, the covariance form is defined by $\langle f', f \rangle_K := f'(Kf) - f'(0)$, and the conditions of symmetry and non-negative-definiteness are $\langle f', f \rangle_K = \langle f, f' \rangle_K$ and $\langle f, f \rangle_K \ge 0$, respectively. It is an easy exercise to verify that this defines a bilinear form on the dual space $X^*$ of affine functionals.

REPLY [12 votes]: You could alternatively try defining Gaussian measures as $2$-stable distributions. This does remove any reliance on finite dimensional projections, and even removes reference to topology. Let $V$ be a measurable vector space (by which, I mean a real vector space $V$ with sigma-algebra $\mathcal{F}$ with respect to which addition and multiplication are measurable).
A probability measure $\mu$ on $V$ is then a centered Gaussian iff, for any independent pair $X,Y$ of $V$-valued random variables each with measure $\mu$, then $aX+bY$ also has measure $\mu$ for all $a,b\in\mathbb{R}$ with $a^2+b^2=1$.
If $A$ is a (measurable) affine space with underlying vector space $V$, then we could similarly say that $\mu$ is Gaussian iff there exists an $m\in A$ such that $X-m$ is a centered Gaussian on $V$ for a random variable $X$ with measure $\mu$.
Several facts should then follow quickly from this:

Affine maps take Gaussians to Gaussians, and linear maps take centered Gaussians to centered Gaussians.
Linear combinations of independent (centered) Gaussians are again (centered) Gaussians.
On separable Banach spaces, the definition is equivalent to the standard one as measures whose one-dimensional projections are Gaussian. More generally, this holds for any locally convex space on which addition is jointly Borel measurable (e.g., separable Frechet spaces).
The definition even makes sense for, e.g., separable F-spaces which can have trivial dual. (Whether it is actually useful to consider Gaussians in such spaces is another question).

This seems to give an answer first paragraph of the question, and does not depend on projections. I'm not sure if it is going in the direction that the question was asking for though, as it says nothing about the stronger form of the question further down and didn't mention covariance operators at all.<|endoftext|>
TITLE: Largest number of k-arithmetic progressions without a (k+1)-arithmetic progression
QUESTION [9 upvotes]: Suppose $A \subseteq \{1,\dots,n\}$ does not contain any arithmetic progressions of length $k+1$. What is the largest number of $k$-term arithmetic progressions that $A$ can have? (one may also wish to put some lower or upper on the size of $A$) We can work over $\mathbb{Z}_p$ if it makes the answer any easier. The "degenerate" case $k=2$ asks for the largest size of the set without arithmetic progressions and it is known that there exist $A$'s with this property of almost linear size.

REPLY [2 votes]: I assume that $k$ as fixed. The answer to this problem is closely related to the maximum density $p=p_{k+1}(n)=r_{k+1}(n)/n$ of a subset of $\{1,\ldots,n\}$ without a $(k+1)$-term arithmetic progression. 
Indeed, let $S \subset \{1,\ldots,n\}$ be a set of cardinality $r_{k+1}(n)$ without a $(k+1)$-term arithmetic progression. Now construct a random subset $A \subset [4kn]$ as follows. The set $A$ consists of the union of $k$ random translations $S_i$ of $S$, where $S_i=S+d_i$ and $d_i$ for $1 \leq i \leq k$ is picked uniformly at random from the interval $\{4(i-1)n,4(i-1)n+1,\ldots,4(i-1)n+2n-1\}$ of $2n$ integers. It is easy to check that $A$ cannot contain a $(k+1)$-term arithmetic progression. It is also not difficult to check that the expected number of $k$-term arithmetic progressions in $A$ is at least $(cp)^k n^2$ where $c>0$ is an absolute constant. Hence, letting $N=4kn$ and changing $c$ slightly, there is a subset of $[N]$ with no $(k+1)$-term arithmetic progression but the number of $k$-term arithmetic progressions is at least $(cp)^k N^2$. 
In the other direction, any subset $B \subset \{1,\ldots,N\}$ with no $(k+1)$-term arithmetic progression has density at most $2p$. A $k$-term arithmetic progression is determined by its first two terms. The number of possible first two terms in $B$ is at most ${2pN \choose 2} < 2p^2N^2$. Hence, there are at most $2p^2N^2$ total $k$-term arithmetic progressions in $B$. 
In summary, the number of $k$-term arithmetic progression in $\{1,\ldots,N\}$ which can be in a set with no $(k+1)$-term arithmetic progression is between $(cp)^k N^2$ and $2p^2N^2$. However, there is still a large gap between Rankin's lower bound and Gowers' upper bound on $p=r_{k+1}(n)/n$. Hence, the bound on this problem is closely tied to quantitative bounds for Szemerédi's theorem.<|endoftext|>
TITLE: Probability that a random distance function is metric
QUESTION [10 upvotes]: Take a random $n \times n$ nonnegative symmetric matrix $D$ with zero diagonal. What is the probability that it is an abstract distance matrix, i.e. satisfies $D_{xy}+D_{yz} \geq D_{xz}$ for all index triples $x,y,z$?
In other words: what is the probability that a nonnegative function on $V \times V$, where $V$ is some finite set, defines a finite metric space?

REPLY [10 votes]: Since you haven't given a distribution, let me make an observation giving the right form of the answer in the case where the $D_{xy}$ are independent uniform $[0,1]$ random variables. 
I want to claim that the probability, $p$, that the matrix defines a metric satisfies
$\alpha^{n^2}\le p\le \beta^{n^2}$ for constants $\alpha$ and $\beta$.
First, notice that if all the $\binom n2$ edge lengths are in $[\frac12,1]$, then the triangle inequality is satisfied, so that $p\ge 2^{-\binom n2}$.
Conversely, consider the triples $(b,b+a,b+2a)$ where $a$ is an odd number in the range $1$ to $\frac n4$  and $1\le b\le a$. Notice that no two of these triples contain two elements in common. There are $\Theta(n^2)$ such triples. For such a triple, consider the event $E_{a,b}$ that $D_{b,b+2a}\le D_{b,b+a}+D_{b+a,b+2a}$. These events all have the same probability that is strictly less than 1. They are also independent, since no edge occurs in two events. Hence we obtain the upper bound.<|endoftext|>
TITLE: Which level structures on elliptic curves are twist-invariant?
QUESTION [9 upvotes]: Let $N \geq 5$ be a prime, and $H$ a subgroup of $GL_2(\mathbb{F}_N)$. As shown in Chapter IV of [DeRap], there is a curve $X_H(N)$, defined over $K_N := \mathbb{Q}(\zeta_N)^{\det H}$, which is  a coarse moduli space of elliptic curves with level-$H$ structure; that is, given any $L$ an extension of $K_N$, the $L$-points of $X_H(N)$ correspond to $\bar{L}$-isomorphism classes of elliptic curves over $L$ whose associated mod-$N$ representation on $N$-torsion points is contained in (a conjugate of) $H$. (Sometimes this is a fine moduli space, but I don't think this is important for this question). 
It is certainly not the case that, if $E/L$ has level-$H$ structure at $N$, then so too do all twists of $E$. For example, if $H$ were the subgroup $\left\{\left(\begin{array}{cc}1 & \ast \\ 0 & \ast \end{array}\right)\right\}$, corresponding to $X_1(N)$, then it is not true.
But this is true sometimes; e.g., if $H = \left\{\left(\begin{array}{cc}\ast & \ast \\ 0 & \ast \end{array}\right)\right\}$, the Borel subgroup; this follows from the slogan: if an elliptic curve has a rational cyclic isogeny, then so too do all twists. A similar argument sows this for $H$ being the normaliser of a split Cartan subgroup, viz $\left\{\left(\begin{array}{cc}\ast & 0 \\ 0 & \ast \end{array}\right)\right\} \bigcup \left\{\left(\begin{array}{cc}0 & \ast \\ \ast & 0 \end{array}\right)\right\}$. I'll say that these level structures are twist-invariant. 
This leads me to ask:

Is there a characterisation of the twist-invariant level structures? What is it about a subgroup $H$ that makes it twist-invariant?

[DeRap] : P. Deligne, M. Rapoport, ``Les schemas de modules de courbes elliptiques''.

REPLY [8 votes]: The criterion is that H contains -I.<|endoftext|>
TITLE: Measure-preserving maps from the square to the cube
QUESTION [19 upvotes]: There is a measure preserving map from the unit interval onto the unit cube that is Lipschitz of order 1/2, that is $|f(x)-f(y)| \leq A |x-y|^{1/2}$. By considering the image of small intervals, one can see that one could not have a smoother map.
Now consider maps from $[0,1]^2$ onto $[0,1]^3$ that preserve measure. By looking at the image of small balls we see that f can't be smoother than Lipschiz 2/3.
Does there exist a measure preserving map from $[0,1]^2$ onto $[0,1]^3$ that is Lipschitz 2/3?

REPLY [5 votes]: There seems to be a $\frac23$-Hölder (or $\frac23$-Lipschitz) onto map $[0,1]^2\to[0,1]^3$.
It was apparently found in this paper.
I don't know if it has been done elsewhere.
The space filling curves due to Hilbert, Peano and Sierpinski preserve measure and the cited article contains generalizations of these, so it seems quite possible that measure is preserved by some of the corresponding surfaces as well.
I have only been able to access the abstract.
It would be great if someone with full text access could check the exact results in the paper.

Here is Fig.12b of the cited paper,
showing their Hilbert surface, whose fractal dimension approaches $3$.
(*Image added by J.O'Rourke*):

 
 
 
![Fig12Hilbert][1]<|endoftext|>
TITLE: Taking direct sums in $K$-theory in Kirchberg-Phillips classification
QUESTION [7 upvotes]: A theorem by Kirchberg and Phillips states that two unital separable nuclear simple purely infinite $C^*$-algebras (so called Kirchberg algebras) satisfying the Universal Coefficient Theorem are isomorphic if and only if their topological $K$-theory groups are isomorphic.
Taking here direct sums of abelian groups at the $K$-theory side, what does it mean at the Kirchberg algebras side?

REPLY [4 votes]: The most concrete way of constructing Kirchberg algebras with given $K$-theory that I know of, is due to Rordam and Elliott-Rordam, and is given in a series of papers "Classification of Certain Infinite Simple C*-Algebras" (Rordam) and "Classification of Certain Infinite Simple C*-Algebras, II" (Elliott-Rordam). There, Kirchberg algebras are constructed as crossed products of either AF-algebras or AT-algebras (depending on whether there is torsion on $K_1$ or not) by endomorphisms, similar to how the Cuntz algebras are obtained from UHF-algebras. I suppose one could go through their construction and see how they get the AF/AT-algebras and the endomorphism from the given $K$-theoretical data (which in general will also include the class of the unit), and then try to see if there is any way of relating the algebras one gets for two different choices, and the algebra one gets for the sum of the $K$-groups. One complication is that all the AF and AT-algebras involved are simple (so taking the sum of the AF/AT-algebras won't work either).<|endoftext|>
TITLE: Eigenvalues of principal minors Vs. eigenvalues of the matrix
QUESTION [5 upvotes]: Say I have a positive semi-definite matrix with least positive eigenvalue x. Are there always principal minors of this matrix with eigenvalue less than x?
(Here "semidefinite" can not be taken to include the case "definite" -- there should be a
zero eigenvalue.)

REPLY [3 votes]: Say the matrix $S$ has unitary orthonormal eigenvectors $u_1,u_2,\dots u_n$ with eigenvalues 
$$\lambda_1\leq \lambda_2\leq \dots \leq \lambda_{n-1}\leq \lambda_n$$
I think the smallest (resp. largest) eigenvalue of a principal minor is $\leq \lambda_2$ (resp. strictly $\geq \lambda_{n-1}$). 
A sufficient condition for this is to have ortho-normal eigenvectors in $S$ and in its minors (true in a positive definite
world with symmetric matrices). 
For any chosen coordinate $i$, the idea is to consider a linear combination of $v=au_1+bu_2$ such that $v_i=0$ as suggested above
(if I understood well the comment of T. Tao). I think one can prove that $\frac{|Sv|}{|v|}\in [\lambda_1,\lambda_2]$,
where $|v|^2=a^2+b^2$ (to verify it, write $Sv=a\lambda_1 u_1+b\lambda_2 u2$ and observe $|Sv|^2= \lambda_1^2a^2 +
\lambda_2^2b^2 $. In other words, the multiplication by $S$ ''stretches'' $v$ by something between $\lambda_1$ and $\lambda_2$.
Now, we can see how the minor $S^i$ ($S$ without line and column $i$) acts on $v^i$ ($v$ without position $i$). Column $i$ does nothing to $v$ in the calculation of $Sv$, and so, $S^iv^i$ is the same as vector $Sv$ but without 
the (non-zero) value of $Sv$ at coordinate $i$. By forgetting this coordinate, $S^iv^i$ has an even smaller 2-norm than $Sv$. Vector 
$v^i$ is ``stretched'' via the $S^i$ multiplication by something $\leq \lambda_2$. As such, $S^i$ can not have all eigenvalues larger than $\lambda_2$, this would mean that it
''stretches'' any $(n-1)-$dimensional vector (including $v^i$) by more than $\lambda_2$.
Finally, the inequalities are not strict because $a$ or $b$ above can be zero. The very basic matrix 
$\begin{pmatrix}
1 & 0 \\
0 & 0 \\
\end{pmatrix}$
has eigenvalues 0 and 1 but the top-left minor has eigenvalue 1.<|endoftext|>
TITLE: Branched Regular Cover over 4-times punctured sphere
QUESTION [5 upvotes]: This is probably trivial but has been bothering me all day.
Suppose $f:\Sigma_g\to \mathbb{S}^2$ is a $g+1$ fold branched conformal map with $\Sigma_g$ a connected genus $g$ surface and $f$ having $4$ branch points (all of order $g+1$).
Is it true that $f:\Sigma_g^* \to (\mathbb{S}^2)^*$ (the cover obtained by removing the branch points) is a regular cover (i.e. so the deck group acts transitively on the fibers of $f$)?
I thought this was obvious but then realized I didn't know how to prove it  (admittedly this is a bit outside my area of expertise).  Part of the problem is that I don't have a good example of a branched map from a connected surface which is not a regular cover away from the branch points.
Edit
Based on some comments of Sebastian.  It would be enough for my purposes to prove that the conformal structure of $\Sigma_g$ was uniquely determined by the cross-ratio of the branch points. However, its not clear to me whether this can be shown without knowing that the cover is regular...

REPLY [6 votes]: The answer is no:
The monodromy group of the cover, as a permutation group on the fiber $f^{-1}(b)$ of a non-branch point $b$, is generated by four elements $s_1,s_2,s_3,s_4$ with $s_1s_2s_3s_4=1$. By Riemann's existence theorem, the question is equivalent to the following, with $n=g+1$:

Let $s_1,s_2,s_3,s_4$ be $n$-cycles in the symmetric group on $n$ letters with $s_1s_2s_3s_4=1$. Does this imply that the group generated by $s_1, s_2,s_3, s_4$ has order $n$?

This does not hold for $n=4$: Take for example $s_1=(1, 2, 3, 4)$, $s_2=(1, 3, 4, 2)$, $s_3=s_2^{-1}$, $s_4=s_1^{-1}$. Then $s_1s_2s_3s_4=1$, but $s_1s_2=(2, 4, 3)$.
This works for any $n\ge4$: Simply take $s_1$ an $n$-cycle and $s_2$ another $n$-cycle which is not a power of $s_1$, and $s_3$, $s_4$ the inverses of $s_2$ and $s_1$ as above.
(Note: I use the right action of permutations, so in $s_1s_2$, one first applies $s_1$, and then $s_2$ afterwards.)<|endoftext|>
TITLE: On local Rankin-Selberg L-functions for Steinberg representations
QUESTION [9 upvotes]: This is a basic question concerning the local Langlands correspondence for $GL_2(\mathbb{Q}_p)$, in particular the compatibility with tensor products . I apologize if this is too basic but I hope someone with more experience can answer it quickly.
Let $G=GL_2(\mathbb{Q}_p)$, $B \subset G$ the subgroup of upper triangular matrices and $St_G$ be the Steinberg representation of $G$; this is the unique irreducible quotient of the representation $Ind_B^G(|\cdot|^{1/2},|\cdot|^{-1/2})$ (normalized induction).
According to the local Langlands correspondence for $G$, we can attach to $St_G$ a Weil-Deligne representation $(V=\mathbb{C}^2,\sigma:W_F \rightarrow GL(V),N \in End(V))$: in this case the inertia acts trivially, $\sigma(Fr) = \begin{pmatrix} p^{-1/2} & 0 \\\ 0 & p^{+1/2} \end{pmatrix}$ as a representation of $W_F$ and $N=\begin{pmatrix} 0 & 1 \\\ 0 & 0 \end{pmatrix}$.
Now consider the equality of $L$-factors $L(St_G \times St_G,s)=L(\sigma \otimes \sigma,s)$. According to Jacquet (Automorphic Forms on GL(2), Part II, p. 23), we have $L(St_G \times St_G,s)=(1-q^{-s-1})^{-1}$, an $L$-function of degree $1$. On the RHS, the tensor product of Weil-Deligne representations is defined by $(V,\sigma,N) \otimes (V',\sigma',N')=(V\otimes V',\sigma \otimes \sigma',N \otimes 1 + 1 \otimes N')$. But, if I am not mistaken, $\dim_\mathbb{C} Ker(N \otimes 1 + 1 \otimes N')=2$ so that the RHS seems to be of degree $2$! Where did I go wrong?

REPLY [9 votes]: I apologize for answering my own question, but I think that the answer might be of some interest to others.
Since I was getting no answers I asked Professor Jacquet directly today. He explained that the expression for $L(St_G \times St_G,s)$ in the book cited above is not correct. The correct $L$-function appears in Proposition 1.4 of his paper A relation between automorphic forms of GL(2) and GL(3), joint with S. Gelbart.  With the expression obtained in that paper we do have the expected equality $L(St_G \times St_G,s)=L(\sigma \otimes \sigma,s)$.<|endoftext|>
TITLE: Non-constructive existence proofs without AC?
QUESTION [13 upvotes]: Hi everyone,
This is a question I have been asking from long, but none of my colleagues could ever answer me:
It is a well-known fact that the axiom of choice (AC) allows one to prove the existence of some set with some property $P$, though we cannot exhibit such a set: for instance, one knows that there exists a well-ordering on $\mathbb{R}$, but cannot define any though.
In formal terms, this means the following: working is the ZFC system, one can find a (first-order) logical property $P(x)$ (with one free variable) such that
$$\vdash (\exists x) (P(x)),$$
yet there is no logical property $Q(x)$ such that
$$\vdash (\exists! x) (P(x) \wedge Q(x))$$
Apparently that is the very phenomenon why many people are dubious about the relevance of AC. Yet I have seen nowhere the statement (and even less the proof) that such a phenomenon would not occur in ZF...
So, is it true that, whenever ZF proves the existence of a set having some property, it is also possible to define some non-ambiguous such set?
Regards,

REPLY [12 votes]: If ZF is consistent, then the answer is no.
ZF proves that there is an $x$ such that $P(x)$: either $x$ is an $L$-generic Cohen real or there is no $L$-generic such real.  (Here, by $L$ I mean the constructible universe of Gödel.) 
In a universe with $L$-generic Cohen reals, $P$ holds only of them, but in a universe without any $L$-generic Cohen reals, $P$ holds of everything.
But if ZF is consistent, then it cannot prove that there is a definable such object satisfying $P(x)$, because we may consider the
forcing extension $L[c]$ obtained by adding an $L$-generic Cohen real. In such a model $L[c]$, which still satisfies ZF, the predicate $P$ holds
only of the $L$-generic Cohen reals, but no such real is definable in $L[c]$ because the forcing is almost homogeneous, and consequently all ordinal
definable objects of $L[c]$ are in the ground model $L$.
Since these are also ZFC models, they also serve to establish the claim you had made about ZFC having this phenomenon. Indeed, there can be no (parameter-free) definable well-ordering of the reals in $L[c]$, since in this case there would be a definable $L$-generic Cohen real, namely, the least one to appear in the well ordering. Meanwhile, in $L[c]$ there is a $c$-definable well-ordering of $\mathbb{R}$, of complexity $\Delta^1_2(c)$, just as in $L$ there is a parameter-free $\Delta^1_2$ definable well ordering of $L$. 

But actually, I've realized that any example from the ZFC case can be transformed to the ZF case by this procedure. In other words, suppose that $P$ has your property for ZFC. Let $P'(x)$ assert "either $P(x)$ or there are no instances of $P$".  So ZF proves the existence of $x$ for which $P'(x)$. But there can be no $Q$ that ZF proves to define such an $x$, since such a $Q$ would also ZFC-provably define an $x$ for which $P(x)$, which we had assumed was not the case. 
Indeed, as mentioned by Rodrigo in the comments, any subtheory of ZFC at all proves $\exists x\ P'(x)$, but no such theory can prove that there is a definable such $x$, since we assumed that ZFC itself does not prove that there is a definable $x$ for which $P(x)$.<|endoftext|>
TITLE: Wedge Product of Lie Algebra Valued One-Form
QUESTION [6 upvotes]: I've been reading about the formal structure of gauge theories and am a little confused by the notation. Could someone clarify this for me?
Suppose that $A$ is a Lie algebra valued 1-form corresponding to the gauge potential of a $U(1)$ gauge theory. Then several sources define the field strength tensor by
$$F=dA+A\wedge A$$
Surely in this case $A\wedge A$ is simply zero, since $\mathfrak{u}(1)=\mathbb{R}$ and the wedge product is an alternating construction? If this is true, why is this term included? Or am I missing something odd about Lie algebra valued 1-forms?

REPLY [15 votes]: There's something about the notation you should know before you get confused when trying to do non-abelian gauge theory. The second term in the field strength should involve a combination of the wedge product of forms and the Lie bracket: the field strength (in the case of an arbitrary gauge group $G$ with Lie algebra $\mathfrak{g}$) should actually be
$$F = dA + \tfrac{1}{2}[A \wedge A],$$
where if $\omega$ is a $\mathfrak{g}$-valued $k$-form and $\eta$ is a $\mathfrak{g}$-valued $p$-form, then
$$[\omega \wedge \eta](X_1, \dots, X_{k+p}) = \sum_{\sigma \in S_{k+p}} (-1)^{\text{sgn}(\sigma)} [\omega(X_{\sigma(1)}, \dots, X_{\sigma(k)}), \eta(X_{\sigma(k+1)}, \dots, X_{\sigma(k+p)})]$$
for any $k + p$ vector fields $X_1, \dots, X_{k+p}$. In particular, if $A$ is a $\mathfrak{g}$-valued $1$-form, then
$$[A \wedge A](X_1, X_2) = [A(X_1), A(X_2)] - [A(X_2), A(X_1)] = 2[A(X_1), A(X_2)].$$
So in components, the field strength is given by
$$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu + [A_\mu, A_\nu],$$
which is the form you'll see most frequently in the physics literature.
When the gauge group $G$ is abelian (e.g. in ${\rm U}(1)$ gauge theory), the Lie bracket on $\mathfrak{g}$ is trivial so that $[A \wedge A] \equiv 0$ and the field strength is just the exterior derivative of the gauge potential: $F = dA$.<|endoftext|>
TITLE: Characterizing Hessians among symmetric bilinear tensors
QUESTION [15 upvotes]: I apologize in advance if this is somewhat elementary, but:

Let $(M,g)$ be a compact Riemannian manifold. Is there a "characterization" of which symmetric bilinear tensors $B\in Sym^2(M)$ are Hessians of functions $f\colon M\to \mathbb R$?
In other words, for which $B\in Sym^2(M)$ there exists $f\colon M\to\mathbb R$ such that $B(X,Y)=Hess \ f(X,Y)$, where $Hess\ f(X,Y)=g(\nabla_X \nabla f,Y)=X(Y(f))-\mathrm df(\nabla_X Y)$?

Obviously, since $M$ is assumed compact, any continuous function will attain a min and a max, so a symmetric bilinear form that is a candidate to Hessian cannot be definite. This maybe pretty naive, but I cannot think of any other property that distinguishes Hessians among bilinear symmetric tensors in this context... For example, it doesn't seem to be the case that $B$ has to be parallel, or satisfy any other sort of similar "nice" conditions.
Note that even if we impose the extra condition that $B$ is the Hessian of a Morse function, I don't quite see if the topology of $M$ will impose any properties on $B$ via Morse theory, since $B$ does not know if it is at a (future) critical point or not, right?

REPLY [2 votes]: Given $f : N \to \mathbb R$, there is its derivative $Df : TN \to \mathbb R$, which has its dual gradient $\nabla f: N \to TN$, which you can covariantly differentiate $c \circ D\nabla f : TN \to TN$, and this map is dual to the Hessian $Hf : TN \oplus TN \to \mathbb R$.  From this perspective there's two things to determine.  Here $c : T^2 N \to TN$ is the connection, in the Ehresmann formalism. 
1) Your Hessian is adjoint to a bundle map $TN \to TN$, is this the covariant derivative of a vector field on $N$? 
2) If the answer to (1) is yes, among the solution vector fields from (1) is there a vector field which is the gradient of a real-valued function on $N$?  
(2) Has the traditional cohomological answer so I'll focus on (1). 
A bundle map $TN \to TN$ is the covariant derivative of a vector field $N \to TN$ means that you could write the vector field as a type of holonomy integral, by adding to your parallel transport an integral of the bundle map $TN \to TN$. So there will be a local triviality condition on this holonomy integral, as well as a global triviality condition.<|endoftext|>
TITLE: Generalization of the equilateral triangle?
QUESTION [6 upvotes]: I consider points in the two-dimensional plane.
An equilateral triangle is a set of three points in the plane which are equidistant.
Suppose now I have $n$ points $x_1,...,x_n$. What is the configuration which minimizes:
$$H(x):=\sum_{i,j} (a-\|x_i-x_j\|^2)^2$$
where $a$ is positive real number.
Clearly, if $n=3$, one recovers the equilateral triangle. Could you draw the solution for larger $n$ ? What is the value of $H_{min}$ as a function of $n$?
Thanks!

REPLY [9 votes]: First, let's assume $a=1$; for other values we can scale a solution with $\sqrt{a}$. 
So we want to minimize $H=\sum_{i,j} (1-\|x_i-x_j\|^2)^2$.
I globally optimized the problem numerically for $n=4$ and $n=5$ and obtained as solutions
for $n=4$, the square with side length $\sqrt{\frac{2}{3}}\approx0.8165$, giving $H_4=\frac{2}{3}$
 
for $n=5$, the regular pentagon with side length $\sqrt{\frac{5-\sqrt{5}}{6}}\approx0.6787$, giving $H_5=\frac{5}{3}$

So in these cases the "generalizations of the equilateral triangle" are the regular polygons. This might be an indication that the regular polygons could be solutions to the optimization problem for other $n$. In any case the regular polygon yields an upper bound. To find the value of $H$ for the regular polygon one just has to find the optimal side length and then plug in the values. Perhaps surprisingly, $H^{C_n}$ takes a particular simple form:
$$H^{C_n}=\frac{n(n-3)}{6}.$$
So we know $H^n_{min}\leq H^{C_n}$ and this bound is sharp for $n\in \{3,4,5\}$.
For larger $n$, there are different solutions that obtain the same minimum as the regular polygon. For example distributing the points evenly on the vertices of the equilateral triangle gives $$H=\left\lfloor\frac{\binom{n-1}{2}}{3}\right\rfloor,$$ which is the same as $H^{C_n}$ above if $n$ is a multiple of $3$.
Let me give you two pictures of solutions that are not the regular $n$-gons, but obtain the same minimum:
for $n=6$:

for $n=7$:

I think minimizing this function is interesting also in dimensions other than $d=2$, even $d=1$, but mainly $d>2$. Maybe one can exploit the symmetry in the polynomial $H$ in some way. I don't know the references and the connection to explosives that Ryan Budney mentions in his comment.<|endoftext|>
TITLE: Solvable models in quantum mechanics
QUESTION [6 upvotes]: Is there anyone who studied on the book "Solvable Models In Quantum Mechanics" by Albeverio? I don't succed in understanding the proof of page 116 about the eigenvalues of the Hamiltonian with point interactions.
In particular I don't understand the proof of the fact that eigenvalues conrespond to the zero of the determinant of the matrix of the Hamiltonian with N point interxations:
$$A(k)=\bigg[\bigg(\alpha_j-\frac{ik}{4\pi}\bigg)\delta_{jj'}-\tilde{f}(y_j-y_{j'})\bigg]_{jj'}$$
where
$\tilde{f}(x)$ is $\tilde{f}(x)=\frac{e^{ik|x|}}{4\pi|x|}$ se $x\neq 0$ and $0$

REPLY [3 votes]: I don't know if this will satisfy, but here is a `physics proof' of the above formula. The wavefunction satisfies the Lippmann--Schwinger equation (with units mass $m=1/2$ and $\hbar=1$)
$$\psi_{\mathbf{k}}(\mathbf{r}) = e^{i\mathbf{k}\cdot\mathbf{r}} - \int d^3\mathbf{r}'\frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{4\pi|\mathbf{r}-\mathbf{r}'|} V(\mathbf{r}')\psi_{\mathbf{k}}(\mathbf{r}'),$$
where $V(\mathbf{r})=\sum_i V_i(\mathbf{r})$ is the potential due to the scatterers. Now a single point scatterer has the (angle independent) scattering amplitude
$$f_i^{(1)}=-\frac{a_i}{1+ia_ik},$$
where $a_i$ is the scattering length (i.e. the `effective range' of the potential is zero). Recall that the scattering amplitude is defined by the asymptotic behavior as $r\to\infty$
$$\psi_{\mathbf{k}}(\mathbf{r}) \to e^{i\mathbf{k}\cdot\mathbf{r}} + f(\theta,\phi)\frac{e^{ikr}}{r}$$
Imagine iterating the Lippmann--Schwinger equation. Since we know the answer for a single scatterer, we can sum up repeated scattering off the same scatterer in the resulting (Born) series to give the scattering amplitude above, only keeping track of when we switch to a different scatterer
$$\psi_{\mathbf{k}}(\mathbf{r}) = e^{i\mathbf{k}\cdot\mathbf{r}}+4\pi\sum_i \tilde f(\mathbf{r}-\mathbf{r}_i) f_i^{(1)}\psi_{\mathbf{k}}(\mathbf{r}_i),$$
where $\tilde f$ is the function you defined above.  $\psi_{\mathbf{k}}(\mathbf{r}_i)$ on the right hand side should be understood in the asymptotic sense i.e. as one approaches the $i^{\text{th}}$ scatterer. This is for the scattered wave. If you want a bound state, it should be a solution of the homogenous equation for the wavefunction at each of the $N$ scatterers i.e. without any incoming plane wave. Then the appropriate determinant must vanish, which is just the condition stated in your question (with the identification $\alpha_i=-(4\pi a_i)^{-1}$).
I believe this is called the Foldy--Lax method in scattering theory.<|endoftext|>
TITLE: Connections between topos theory and topology
QUESTION [14 upvotes]: What are some "applications to" / "connections with" topology that one could hope to reasonably cover in a first course on topos theory (for master students)? I have an idea of what parts of the theory I would like to cover, however, I would love some more nice examples and applications. Of course, I have some ideas of my own, but am open to suggestions. Thank you!
P.S.
I am also interested in perhaps learning about some new connections myself, which would be out of reach for such a course, so feel free to leave these as well, qualified as such.

REPLY [2 votes]: I will copy and paste the description of chapter 2 and 3 of my master thesis.
Chapter 2 plays an important motivational role in the thesis and is aimed at pointing out geometric characteristics of a Grothendieck topos. For a topological space $X$ one can consider the category of its open sets $\mathcal{O}(X)$ that is a complete Heyting algebra, so that we have a functor $$ \text{Spaces} \to \text{cHa}^{\text{op}}. $$ We study properties of this functor concluding that there is a huge subcategory of spaces (sober ones) that embeds into $\text{cHa}^{\text{op}}$ via this functor. For a notational motivation we call Locales the opposite of cHa,
$$\text{SobSpaces} \hookrightarrow \text{Locales} $$
So locales  naturally are generalized (sober) spaces. 
In Chapter 2 we present the notion of localic topos that is a Grothendieck topos on a locale and we prove that there is an equivalence of category between the category of locales and the category of localic toposes
\begin{matrix}  \text{Locales} & \leftrightarrows & \text{LocToposes.} &   \end{matrix} 
This equivalence is the precise sense in which a localic topos is a generalized topological space, that is the same in which its associated locale is a generalized topological space.
A generic Grothendieck topos has not this fascinating property, there is not a topological space from which it comes from, but precisely in this rift one can collocate Barr's theorem.
Chapter 3 is devoted to proving Barr's theorem that we can formulate right now:

Any Grothendieck topos is covered by a localic boolean one.

This theorem states that not any Grothendieck topos is geometric but not far from it there is an other Grothendieck topos that is not only geometric (better say localic) but it is also boolean.
This is the first and most naive interpretation of Barr's theorem.<|endoftext|>
TITLE: Rigorous numerical integration
QUESTION [12 upvotes]: I need to evaluate some (one-variable) integrals that neither SAGE nor Mathematica can do symbolically. As far as I can tell, I have two options:
(a) Use GSL (via SAGE), Maxima or Mathematica to do numerical integration. This is really a non-option, since, if I understand correctly, the "error bound" they give is not really a guarantee.
(b) Cobble together my own programs using the trapezoidal rule, Simpson's rule, etc., and get rigorous error bounds using bounds I have for the second (or fourth, or what have you) derivative of the function I am integrating. This is what I have been doing.
Is there a third option? Is there standard software that does (b) for me?

REPLY [4 votes]: VNODE-LP is a C++ package with which you should be able to achieve the desired task:

VNODE-LP is a C++ package for computing bounds on solutions in IVPs for ODEs. In contrast to traditional ODE solvers, which compute approximate solutions, this solver tries to prove that a unique solution to a problem exists and then computes bounds that contain this solution. Such bounds can be used to help prove a theoretical result, check if a solution satisfies a condition in a safety-critical calculation, or simply to verify the results produced by a traditional ODE solver.

This package is a successor of the VNODE package of N. Nedialkov (mentioned above in the comments by Gilead).
It is free and can be downloaded here:
http://www.cas.mcmaster.ca/~nedialk/vnodelp/<|endoftext|>
TITLE: Invertibility of a certain matrix indexed by the Hamming cube
QUESTION [13 upvotes]: For reasons which the margin of this page is too small to hold, I have been reading parts of a recent paper by O. Selim
On submeasures on Boolean algebras, arXiv 1212.6822v3
and in Section 7 the following technical lemma is given (Lemma 7.5 in the paper)
Lemma (paraphrased) Let $S$ be the set of non-empty subsets of some fixed finite set $F$, and consider the matrix $A:S\times S\to {\mathbb Q}$  where
$$ A_{I,J} =  1 \hbox{ if $I\cap J\neq \emptyset$, and }
    A_{I,J} =  0 \hbox{ if $I\cap J = \emptyset$.} $$
Then $A$ is invertible.
Selim gives a proof by induction that the columns of $A$ are linearly independent, but he says "we could not find a particularly enlightening proof". So my question is this: do we have a more conceptual argument to show this (real, symmetric) matrix is invertible?

[EDIT/UPDATE 2012-03-07: this was poorly phrased on my part; I was hoping to find some explanation that involved the lattice or group stucture on $\{0,1\}$, and which took advantage of the very particular structure of this matrix, although I am grateful for all answers received so far. In some sense I wanted to know: "what is the pattern?" or "what is the underlying algebraic mechanism?" -- the matrix is defined in terms of some incidence or order structure, so does that give some way to interpret invertibility of this matrix as part of a more general result? (I do not mean a result like "a matrix with non-zero determinant is invertible".)
Benjamin Steinberg's answer comes closest, at present, to what I was hoping for, but Benjamin Young's answer is also very suggestive and helpful.
I suspect this will be routine for several MO regulars, but hope it is not too elementary or "too localized".

[older comments/thoughts, left here for context]
My vague thoughts are that one could view $A$ as the corner of a square matrix indicated by the power set of $F$, and then perhaps do some kind of Fourier transform on the group $\{0,1\}^{|F|}$. Or perhaps there is some kind of Möbius inversion at work here?
While I'm here, a question on terminology: the matrix $A$ is of course the adjacency matrix of a certain graph whose vertex set is $S$. Does this graph have an established name?

REPLY [8 votes]: For a vast generalization, see Exercise 3.96(a) of Enumerative Combinatorics, vol. 1, second ed. To get the posted problem, take $L$ to be the boolean algebra of all subsets of $F$ (ordered by inclusion), and set $F(u,s)=1$ if $u\neq\emptyset$ or $s=\emptyset$, and otherwise $F(u,s)=0$. (Note that I am using $F$ in two different ways: one is Yemen's use, and the other is the use in EC1.) Then in the row of the matrix $F(s\wedge t,s)$ indexed by $\emptyset$, every entry is 0 except in the column indexed by $\emptyset$. 
Hence the determinant remains the same if we remove the row and column and indexed by $\emptyset$, but this gives the matrix $A$.<|endoftext|>
TITLE: Dimension of Specht Modules $S^\lambda$
QUESTION [7 upvotes]: In the study of representation theory of $S_n$, we know that the irreducible characters of $\chi_\lambda$ of $S_n$ are indexed by partitions $\lambda \vdash n$. There are several methods in determining the dimension of each $S^\lambda$, $f^\lambda$. One of the method is by considering
$ f^\lambda = \frac{n!}{\prod \text{hook length}} $
Let $\lambda' \vdash n$ be a partition obtained by taking 'transpose' in Ferrer's diagram of $\lambda$. For example, if $\lambda = (5,4,1)$, then $\lambda' = (3,2,2,2,1)$. Using the formula of $f^\lambda$ above, we thus have $f^\lambda = f^{\lambda'}$.
After some observations, I found out that when $n \geq 8$, then the dimension of $S^\lambda$ is unique up to transpose. In other words,
"Given any $\lambda \vdash n$, then there exists no other $\alpha \vdash n$ such that $f^\lambda = f^{\alpha}$ except when $\alpha = \lambda$. "
Is the above result well-known or established by anyone?
Thanks for the help!

REPLY [2 votes]: An even stronger counter-example is presented in Example 8 in my paper.
It shows that $\lambda = (8,5,4)$ and $\mu=(7,7,2,1)$
both have the same multi-set of hook values. This of course implies $f^\lambda=f^\mu$.
What is interesting though is that the corresponding Ehrhart functions for the posets defined by $\lambda$ and $\mu$
are different.
Perhaps you can salvage your conjecture, and ask if for $|\lambda|>8$, the Ehrhart polynomial is unique (up to transposition). The value of $f^\lambda$ can be recovered from the leading coefficient of the Ehrhart polynomial.<|endoftext|>
TITLE: Random reals and strongly meager sets
QUESTION [13 upvotes]: Adding a single Cohen real makes the set of reals from the ground model strong measure zero (see this question).
The notion of strong measure zero sets has its dual concept in the category branch -- strongly meager sets. A set $X\subseteq \mathbb{R}$ is strongly meager if for any null set $Y$ there exists $t\in  \mathbb{R}$ such that $(t+X)\cap Y=\varnothing$. 
One can see duality of these notions due to Galvin-Mycielski-Solovay Theorem which states that a set $X\subseteq  \mathbb{R}$ is strong measure zero if and only if for any meager set $Y$ there exists $t\in  \mathbb{R}$ such that $(t+X)\cap Y=\varnothing$. 
Random real forcing is dual to Cohen forcing in the sense of measure and category. Therefore it makes sense to ask, whether:

The set of reals from generic model $ \mathbb{R}\cap V$ is strongly meager after adding a single random real?

I have heard that the answer is affirmative, but I have not been able to find any published proof. 
Note that $\mathbb{R}\cap V$ is meager after adding a random real (see this question).

REPLY [2 votes]: As I have written above the affimative answer itself was known to many people including T. Bartoszyński. The following proof is due to T. Weiss (my advisor).
Proof. We follow closely the proof and notation of Lemma 3.2.42 from [1]. Let $A$ be a Borel measure zero set in $M[r]$, where $r$ is a random real over $M$. There exists $\dot{A}\subseteq 2^{\omega}\times 2^{\omega}$ measure zero set coded in $M$, such that $\dot{A}_{r}=A$ (notation: $\dot{A}_{r}=\{y\colon \left<r,y\right>\in \dot{A}\}$).
Then 
$$\dot{A}\subseteq\bigcap_{m\in\omega}\bigcup_{n\geq m} [s_{n}]\times[t_{n}]$$ 
where $s_n, t_n\in 2^{<\omega}$, $\sum_{n=0}^{\infty}\frac{1}{2^{2|s_{n}|}}<\infty$ and we can assume that $|t_{n}|=|s_{n}|$ for any $n\in\omega$.
Let $z\in 2^{\omega}\cap M$ and $f\in\omega^\omega$ be increasing. Then
$$\mu(\{x\colon\left<x,x_f+z\right>\in [s]\times[t]\})\leq \frac{2^{f^{-1}(|s|)}}{2^{|s|+|t|}}$$
(where $x_f\in 2^{\omega}$ such that $x_{f}(n)=x(f(n))$). By induction on length $|s_{n}|$ we define an increasing function $f_{A}\in\omega^{\omega}$ such that 
$$\sum_{n=0}^{\infty}\frac{2^{f_{A}^{-1}(|s_{n}|)}}{2^{2|s_{n}|}}<\infty.$$
It is easy to see that such function exists as for any $\varepsilon>0$ we can find $N_{\varepsilon}\in\omega$ such that $\sum_{n\geq N_{\varepsilon}}\frac{1}{2^{2|s_{n}|}}<\varepsilon$.
Notice also that $\left< x,x_{f}+z\right>\in [s]\times[t] $ if and only if $\left<x,x_{f}\right>\in[s]\times [t+z]$.
The set
$$H_{z}=\{x\in\left<x,x_{f_A}+z\right>\in\bigcap_{m\in\omega}\bigcup_{n\geq m} [s_{n}]\times[t_{n}]\}$$ has measure zero and is coded in $M$ for every $z\in 2^{\omega}\cap M$. Thus $r\notin H_{z}$ and $\left<r,r_{f_{A}}+z\right>\notin \dot{A}$. This implies that $r_{f_{A}}\notin A+z$ for every $z\in 2^{\omega}\cap M$, so $(2^{\omega}\cap M)+A\neq 2^{\omega}$ and so $2^{\omega}\cap M$ is strongly meager. 
$\square$
References
[1] T. Bartoszyński, H. Judah, Set thoery: on the structure of the real line, A K Peters, 1995<|endoftext|>
TITLE: ERA, PRA, PA, transfinite induction and equivalences
QUESTION [8 upvotes]: I'm quite sure I don't understand very well the links between proof theoretical ordinals of theories, the axioms of transfinite induction and the objects a theory can prove to exist.
For instance I'm considering Peano Axioms ($\mathbf{PA}$), of proof theoretic ordinal $\epsilon_0$, Primitive Recursive Arithmetic ($\mathbf{PRA}$) of proof theoretic ordinal $\omega^\omega$ and Elementary Recursive Arithmetic ($\mathbf{ERA}$), which is a fragment of $\mathbf{PRA}$.
I was wondering if $\mathbf{PRA}+TI\{\alpha\in\epsilon_0\}$ (where $TI$ stands for transfinite induction) was equivalent in some sense to $\mathbf{PA}+TI\{\alpha\in\epsilon_0\}$ or/and to $\mathbf{ERA}+TI\{\alpha\in\epsilon_0\}$ ?
And more generally, if it was true that for any set $A$ of (countable) ordinals such that $\epsilon_0 \subset A$, $\mathbf{ERA}+TI\{\alpha\in A\} = \mathbf{PRA}+TI\{\alpha\in A\} = \mathbf{PA}+TI\{\alpha\in A\}$ ?
Any enlightenment would be most welcome :)
Thanks in advance.

REPLY [9 votes]: This entirely depends on what exactly you mean by $TI$, as there are several options (I actually do not understand what the $\{\alpha\in\epsilon_0\}$ part of the notation is supposed to mean either, but I will assume it just means transfinite induction up to $\epsilon_0$):

$TI\{\alpha\in\epsilon_0\}$ is the schema
$$\forall x\,(\forall y\prec x\,\phi(y)\to\phi(x))\to\forall x\,\phi(x),$$
where $\phi$ is an arbitrary formula, and $\prec$ the standard ordering of type $\epsilon_0$. It is easy to see that transfinite induction implies ordinary induction over a weak base theory (say, $I\Delta_0$), hence in this case, $I\Delta_0+TI\{\alpha\in\epsilon_0\}=\mathrm{PA}+TI\{\alpha\in\epsilon_0\}$ (and the same holds for any base theory in between).
$TI\{\alpha\in\epsilon_0\}$ is the same schema restricted to formulas of bounded complexity $\Gamma$. Typically used choices for $\Gamma$ include $\Pi^0_2$, $\Pi^0_1$, or open formulas in the language of PRA or EA (also called ERA or EFA). In all these cases, $\mathrm{PRA}+TI\{\alpha\in\epsilon_0\}$ is strictly weaker than $\mathrm{PA}+TI\{\alpha\in\epsilon_0\}$, since the former theory can be axiomatized by formulas of bounded complexity, and no consistent set of formulas of bounded complexity can imply full ordinary induction (which is equivalent to the full uniform reflection schema). In the case where $\Gamma$ are open EA-formulas, $\mathrm{EA}+TI\{\alpha\in\epsilon_0\}$ is likewise strictly weaker than $\mathrm{PRA}+TI\{\alpha\in\epsilon_0\}$. On the other hand, if $\Gamma\supseteq\Pi^0_1$, then $TI\{\alpha\in\epsilon_0\}$ implies $I\Sigma_1\supseteq\mathrm{PRA}$ over a weak base theory.
$TI\{\alpha\in\epsilon_0\}$ is the second-order induction axiom
$$\forall X\,\forall x\,(\forall y\prec x\,y\in X\to x\in X)\to\forall x\,x\in X.$$
Then one needs to include some comprehension schema in the base theory to make any sense, and its strength determines the strength of the $TI\{\alpha\in\epsilon_0\}$. In particular, if we take at least $\Sigma^0_1$-comprehension, we are in the same situation as in 1. If we take recursive comprehension, it is the same as 2 with $\Gamma=\Delta^0_1$.<|endoftext|>
TITLE: Permutation character of the symmetric group on subsets of certain size
QUESTION [9 upvotes]: The symmetric group $S_n$ acts on $[n]:=\{1,\ldots,n\}$, thereby inducing an action on the set $$\wp_k(n)=\{\: A\subseteq[n] \::\: \#A=k \:\}$$ of subsets of cardinality $k$, simply by $$(g,A)\mapsto g(A)=\{\: g(a) \::\: a\in A \:\}.$$
This finally yields an action on $V_k:=\mathbb C[\wp_k(n)]$.
By Lemma 4 in this paper (arXiv:0903.2864), the character of $V_k$ is given as
$$\tag1 \sum_{r=0}^k \chi_{(n-r,r)}$$
where $\chi_\lambda$ is the irreducible character of $S_n$ corresponding to the partition $\lambda\vdash n$. There is no proof of Lemma 4 in the above reference, but the author says that the result is due to Frobenius. I would like to have a reference of $(1)$ - it doesn't have to be the original paper of Frobenius, in fact I would prefer a more recent work which also has a proof.

REPLY [6 votes]: A classical construction of the Specht modules of $S_n$ says that $\chi_{(n-k,k)}$ is present in the character $\pi_k$ of the action $S_n$ on $V_k$, with multiplicity 1. Indeed $M^\lambda:=V_k$ is the permutation module arising along the way of constructing the Specht module $S^\lambda$ for the partition $\lambda=(n-k,k)$. 
Moreover, it is easy to see that the centralizer of the action of $S_n$ on $V_k$ in the full matrix algebra of $\binom{n}{k}\times\binom{n}{k}$ is commutative, and has dimension $k+1$ (The centralizer is spanned as an algebra by the 0-1 matrices corresponding to the orbits of $S_n$ on the ordered pairs of $k$-subsets - this is a general fact about permutation representations of finite groups; here these matrices are symmetric, and thus it's a commutative algebra). Thus $\pi_k$ is a sum of $k+1$ irreducible characters, each of them with multiplicity 1. At this moment we know two of them, namely $\chi_{(n-k,k)}$ and $\chi_{(n)}$ (the latter is there, as it's the trivial character, present in every permutation character).
There is a description (see e.g. Volume 2 of the Richard Stanley's book) of irreducible characters arising in $M^\lambda$, for any $\lambda$. Namely, 
$$M^\lambda=S^\lambda\oplus \oplus_{\mu\triangleright\lambda} 
K_{\lambda\mu} S^\mu,$$
where $\triangleright$ stands for the dominance partial ordering on the partitions of $n$, and the $K_{\lambda\mu}$'s are famous Kostka numbers (in our case they are all 0 or 1). Using this, one can easily complete the proof of (1).

PS. In Stanley's book, this question is Example 7.18.8 on p.355, Volume 2.<|endoftext|>
TITLE: Character table entries and sums of roots of unity
QUESTION [9 upvotes]: It is well-known that the entries of the character table of a finite group are sums of roots of unity. 
Question: Is the converse true? Explicitly, given $z\in \mathbb{Z}[\mu_\infty]$, can I find a finite group $G$ and irreducible  character $\chi$ with $z=\chi(g)$ for some $g\in G$?
It's certainly true if I dropped the irreducibility; in this case one can take $G$ to be abelian, and just hit each summand of $z$ one at a time and take a direct sum. Also noteworthy is that if one has a single $G$ one can compose with a surjection to $G$, so it will be true for infinitely-many groups if it is for one. 
A related question (possibly less trivial): If it's YES, is there a natural "minimal" subclass of finite groups that suffice for this purpose?
If it's "NO," are the obstructions completely understood? 
This is partially meant as a (hopefully) easier relative to a previous question: A Realization Problem for Character Tables. I apologize in advance if it is trivial (one way or the other), as I fear it may be.

REPLY [6 votes]: I think that a clearer way to construct the desired example is not to let the acting group be $S_n$, but take the cyclic group $C = C_n$ instead. Thus, start with the group $G = (C_m)^n$, as before, and let $C$ act by cycling the direct factors of $G$. Let $\lambda$ be a linear character of $G$ whose kernel is exactly the product of all but the first cyclic factor of $G$ (so the restriction of $\lambda$ to the first factor of $G$ is faithful). Only the identity of $C$ stabilizes $\lambda$, so working in the semidirect product $\Gamma = GC$, the stabilizer of $\lambda$ is $G$, and it follows that the induced character $\chi = \lambda^\Gamma$ is irreducible.
Now the restriction $\chi_G$ is a sum of $m$ linear characters $\lambda_i$, where the kernel of $\lambda_i$ is the product of all but the $i$th cyclic factor of $G$. Now given any list of $n$ $m$th roots of unity $\varepsilon_i$, we can choose an element $g$ of $G$ such that $\lambda_i(g) = \varepsilon_i$, and thus $\chi(g) = \sum\varepsilon_i$, as wanted.<|endoftext|>
TITLE: When is the projective model structure cartesian? When is the internal hom invariant?
QUESTION [9 upvotes]: If M is a sufficiently nice model category and D is a small category then there are two natural model structures we can impose on the functor category $Fun(D,M)$ where the weak equivalences are the levelwise weak equivalences.  

The projective model structure, which is characterized by the fibrations being the levelwise fibrations, and
The injective model structure, which is characterized by the cofibrations being the levelwise cofibrations. 

Here "nice" means cofibrantly generated in the projective case and combinatorial in the injective case. Now I am interested in the case that $M = sSet$, the category of simplicial sets with the usual Kan-Quillen model structure. In other words (replacing D by its opposite) I am interested in the projective and injective model structures on simplicial presheaves. 
Let us say that a model category is a cartesian model category if it is cartesian closed and it satisfies the pushout-product axiom. This is the same as saying the product is a Quillen adunction of two variables. In particular it implies that if $A$ is cofibrant and $X$ is fibrant, then the functors:
$$ Hom(A, -) $$
$$ Hom(-, X) $$
are part of Quillen adjunctions. (Here "Hom" is the inner hom). In particular the assignment $$ A,X \mapsto Hom(A,X)$$
sends weak equivalences to weak equivalences, provided the As are cofibrant and the Xs are fibrant.
A catchy way to summarize this last observation is to say that the derived functor of internal hom is homotopically meaningful. 
Now the category of simplicial presheaves is cartesian closed, i.e. it has products and an internal hom. In section 2 of Rezk's paper "A Cartesian presentation of $(\infty,n)$-categories" (arXiv:0901.3602), he reviews these model categories and states that the injective model structure is always cartesian. 
So this raises some questions about the projective model structure:


When is the projective model structure a cartesian model structure? There are examples where the injective and projective model structure agree (e.g. D = pt), so presumably there are less severe conditions one can impose on D to ensure this happens? 
Is there an illuminating example for how the projective model structure can fail to be cartesian?
(Main question) Setting aside the question of cartesian-ness, we can also ask about whether the internal hom is invariant under weak equivalences, always assuming the source is cofibrant and the target is fibrant. This is the question I am most interested in. It is of course implied by cartesian-ness of the projective model structure, but a priori seems weaker. 


I tried to construct a counter example to 3 by looking at the case $D = (0 \to 1)$, the free-walking arrow. However in this case I was thwarted by the fact that simplical sets is a right proper model category. In this specific easy case (even though the injective and projective structures differ) the internal hom is invariant. 
There are two classes of D which interest me the most. First there are combinatorial sorts of categories which show up often in the theory of higher categories. I am thinking now of things like $D = \Delta, \Theta_n$ or Segal's $\Gamma$. The other case I am interested in is when $D$ is something like the site of smooth manifolds. The general setting might be out of reach, but hopefully something can be said in these cases. 
There are also a variety of related bonus questions:


What happens if we localize our model category? Does the invariance of internal hom persist? Does cartesian-ness?. 
What can we say when simplicial sets is replaced with another nice cartesian model category? What if this model category is right proper?

REPLY [4 votes]: Just a remark about making the projective model structure on simplicial presheaves (on a category $C$) cartesian.
In the projective model structure, representable objects are cofibrant, (and in some sense all cofibrant objects are built from them).  So a helpful condition to impose might be: 

finite products of representable functors are projective cofibrant. 

This is clearly a necessary condition.  Perhaps it is also sufficient?  
One way to satisfy this is to assume that $C$ itself has finite products, so finite products of representables are representables.  This sounds like the result of Sinan Yalin  that D. White quotes.  
Your $(0\to 1)$ example satisfies this property.  
Another example which satisfies the condition is $C=$ discrete group $G$.  The free functor is $G$ as a $G$-set, and $G\times G$ has a free $G$-action, so is cofibrant.  (Though it is not itself a representable functor.)  The projective model structure on  $\mathrm{Psh}(G,\mathrm{sSet})$  is in  fact cartesian.  
(I should add that the fact that the injective model structure on simplicial presheaves is cartesian is not my observation: it is clear from Jardine's papers on the Joyal-Jardine model structures for simplicial (pre)sheaves, for instance.)<|endoftext|>
TITLE: Relationship between the derivative of a matrix and its eigenvalues
QUESTION [6 upvotes]: Is there any relationship between the derivative of a matrix and its eigenvalues? If, for example, the derivative is strictly positive definite, can I say that the eigenvalues are strictly increasing?
In particular, my matrix is
$$-\frac{ik}{4\pi} I + \mathrm{diag}[a_1,\ldots,a_n]+A(k)$$ where $I$ is the identity matrix, $\alpha_i$ are real constants and
$$ [A(k)]_{jj}=0 $$ 
$$ [A(k)]_{jl}= -\frac{e^{ik|y_j-y_l|}}{4\pi|y_j-y_l|}\quad \text{for } j\neq l $$
with each $y_i$ a point in $\mathbb{R}^3$.

REPLY [7 votes]: The relation you are looking for is in the article "On Eigenvalues of Matrices Dependent on Parameter" by P.Lancaster (1964), theorem 5.
It states, that for any matrix $A$:
$$
\frac{\mathrm{d} \lambda^{(j)}_t}{\mathrm{d} t} = \frac{ y_t^{(j)T} A'_t x^{(j)}_t }{ y_t^{(j)T} x^{(j)}_t }
$$
For real parameter $t$ and $y^{(j)}_t$, $x^{(j)}_t$ being left and right eigenvectors of $A_t$ corresponding to j-th eigenvalue $\lambda^{(j)}_t$.
If $A_t$ is additionally symmetric, then $y^{(j)}_t = x^{(j)}_t$ and it can be chosen real-valued. If also $A'_t$ is positive definitive, then the right side of equation is strictly positive and so is the derivative of eigenvalue.
In your particular case in the book you mentioned $k$ is set: $k=i\chi$ and $\chi$ is real-valued parameter.<|endoftext|>
TITLE: what are mutations of sheaves all about?
QUESTION [5 upvotes]: Suppose I have a smooth projective variety $X$  and a semi-orthogonal decomposition of its bounded derived category of coherent sheaves $D^b(X)$. Then I can apply right or left mutations to the full asmissible triangulated subcats that make up the semi-orth decomposition. Algorithmically it is clear to me what happens, but what is the spirit of such a transformation? what is the geometry behind it? in particular I would be curious to understand this in the case of full exceptional sequences and when all subcats in the semi-orth decomposition are generated by sheaves, line or vector bundles.

REPLY [6 votes]: I wouldn't say that in general there is some special geometry behind mutations. This just explains that whenever you have one sod, you have many of them. 
However, in some cases there is some meaning. For example, if $D(X) = < A, {}^\perp A >$ then the left mutation functor $L_A:{}^\perp A \to A^\perp$ is isomorphic to the composition of the Serre functor of $D(X)$ and the inverse Serre functor of ${}^\perp A$:
$$
L_A = S_X\circ S_{{}^\perp A}^{-1}.
$$
In the particular case, when ${}^\perp A$ is generated by one exceptional object, one has $L_A = S_X$.<|endoftext|>
TITLE: Is there a conceptual reason why topological spaces have quotient structures while metric spaces don't?
QUESTION [16 upvotes]: Of the mathematical objects that I am familiar with, it is normally the case that the product of 2 objects is an object of the same type and that an equivalence relation on an object induces a quotient object of the same type.
I think I have some understanding as to why the product of 2 fields is not a field, because a field is not an algebra in the universal algebra sense. But I don't see a reason as to why an equivalence relation on a metric space fails to induce a quotient structure, apart from the fact that it just doesn't work.

REPLY [14 votes]: One way to get around the lack of quotients is to work in the category of Uniform Spaces. This is more general than a metric space, but still gives you a way to talk about how "close" two points are. Uniform spaces are also cool because they simultaneously generalize metric spaces and topological groups. I believe their study was popular back in the 60s and 70s for instance because this category of uniform spaces has much nicer categorical properties than either of the categories it generalizes. For instance, with uniform maps it's complete and cocomplete, as can be seen here.
Even if your maps are just continuous maps, you still get quotients as can be seen here. Isbell's book is a great reference for this subject.
Side note: I like Eric Wofsey's answer a lot and I wonder if it's secretly related to this answer. The wikipedia page points out the connection between uniform spaces and pseudo-metrics but for categorical things it's probably easy to work with the former than the latter.<|endoftext|>
TITLE: Formal writing: numbers under 10
QUESTION [8 upvotes]: I've been tasked with proofreading an Engineering/Mathematics thesis paper. I was always told that numbers under 10 should be spelled out (one, two, three, ...) but I was wondering if this rule holds in math and science fields as well. This paper understandably uses a lot of numbers in it and switching back and forth between numbers written out in words and numerals seems inconsistent at best and somewhat confusing at worst.

REPLY [2 votes]: I believe that the rule here is to spell out numbers used for counting (unless they require three or more words to be spelled out), and use digits otherwise. Just check the examples in the answers above to see that they all fit this rule. As one more example: "There are five theorems in the manuscript labeled Theorem 5".<|endoftext|>
TITLE: Lattice in motion group
QUESTION [5 upvotes]: Let $\Gamma$ be a discrete cocompact subgroup of the euclidean motion group
$$
G={\mathbb R}^d\rtimes O(d).
$$
Let $\phi:G\to O(d)$ the projection homomorphism.
Is it true that $\phi(\Gamma)$ is finite?

REPLY [5 votes]: The answer is yes. This is a theorem of Bieberbach (see Corollary (8.26) of the book "Discrete Subgroups of Lie Groups" by M.S.Raghunathan (Springer Ergebnisse Tract).<|endoftext|>
TITLE: Is displacement controled by stable norm?
QUESTION [8 upvotes]: Let $T^n$ be the $n$-dimensional torus and $g$ be a Riemannian metric on $T^n$. Let $\tilde g$ be the induced metric on the universal covering; using suitable coordinates, $\tilde g$ is therefore a $\mathbb{Z}^n$-periodic metric on $\mathbb{R}^n$ (I shall conflate the lattice $\mathbb{Z}^n$ with the fundamental group of $T^n$ in the sequel).
Let $d$ be the distance induced by $\tilde g$ and $t:\mathbb{Z}^n\to (0,+\infty)$ be the function defined by $t(\gamma)=d(0,\gamma(0))$. Recall that the stable norm $\Vert\cdot\Vert_S$ is a norm on $\mathbb{R}^n$ defined by the property that for any $\gamma\in\mathbb{Z}^n$,
$$ \Vert\gamma\Vert_S = \lim_k \frac{t(\gamma^k)}{k} $$
Question: is it true that for all $\gamma\in\mathbb{Z}^n$, we have
$$t(\gamma)\le \Vert \gamma \Vert_S+2\mathrm{diam}(g)?$$
If not, does some similar control hold? The formula could depend on $n$ but not on the metric $g$. 
A pointer to good literature on this kind of metric Riemannian geometry would already be much appreciated.
Important edit: if needed, I am ok with the additional, very strong assumption that the sectional curvature of $g$ is bounded above by some positive $\varepsilon=\varepsilon(n)$. The formula for a lower bound on $\Vert\cdot\Vert_S$ can depend upon $\varepsilon$ (explicitely).
As for motivation, I need this kind of control for a project of showing some constraints on Riemannian metrics on the torus $T^n$ by using quantitative versions of Milnor's argument in his paper on the growth of fundamental groups and volume of Riemannian manifolds.
Happy ending edit: while I was not able to complete the argument proposed in the accepted answer, with Stéphane Sabourau we managed to get a good enough bound, by passing to a finite cover not increasing diameter too much, see https://arxiv.org/abs/1707.07876

REPLY [8 votes]: If you allow an additive term $C(n)diam(g)$ rather than $2diam(g)$, then yes, the statement is true. In the paper D.Burago, "Periodic metrics", Adv. Soviet Math. 9, (1992), 205-210, he proves that for every periodic metric on $\mathbb R^n$ there is a constant $C$ such that
$$
 | d(x,y)-\|x-y\|_{st} | \le C
$$
for all $x,y\in\mathbb R^2$. In this non-invariant formulation, $C$ depends not only on the metric but also from a particular choice of coordinates used to identify the torus with the standard $T^n$. However, if  $x$ and $y$ are from one orbit of $\mathbb Z^n$, then going through the proof shows that one can take $C$ of the form $C(n)diam(g)$.
Unfortunately the paper is published in an obscure place and I can no longer find it on Google Books. But the proof is rather short and I think I can write down all details here if needed.
Added later. I'm adding a proof outline upon a request from comments.
Fix $p\in \mathbb R^n$ and $v\in\mathbb Z^n$. Define $f:\mathbb Z_+\to\mathbb R$ by $f(m)=d(p,p+mv)$. Note that $ \|v\|_{st} = \lim_{k\to\infty} f(k)/k$. The function $f$ satisfies the following properties:
(1) $f(m_1+m_2)\le f(m_1)+f(m_2)$
(2) $f(2m)\ge 2f(m)-C$,
where $C$ depends only on the metric (and in fact equals $C(n)diam(g)$). These properties immediately imply that $k\|v\|_{st}\le f(k)\le k\|v\|_{st}+C$ with the same $C$. For $k=1$, this gives us the desired estimate on $d(p,p+v)=f(1)$.
The property (1) is just the triangle inequality. The property (2) is based on the following topological lemma:
Lemma. Let $s:[0,1]\to\mathbb R^n$ be a continuous path. Then there is a collection of disjoint intervals $[a_i,b_i]\subset[0,1]$, $i=1,2,\dots,k\le (n+1)/2$, such that
$$
 \sum (s(b_i)-s(a_i)) = \frac{s(1)-s(0)}2 .
$$
Let me derive property (2) from the lemma. Apply the lemma to a shortest path $s$ connecting $p$ to $p+2mv$. This gives us a collection of at most $n/2$ vectors $v_i=s(b_i)-s(a_i)$ in $\mathbb R^n$ whose sum equals $mv$. Let $p_0=p,p_1,\dots,p_k$ be points such that $p_i-p_{i-1}=v_i$. Then $p_k=p+mv$, and each pair of points $p_{i-1},p_i$ is a (possibly non-integer) parallel translation of $s(a_i),s(b_i)$. Due to periodicity, this implies that $d(p_{i-1},p_i)\le d(s(a_i),s(b_i))+C_1$ for some constant $C_1$ depending on $g$. I will address this dependence later. Therefore $f(m)=d(p,p_k)$ is bounded by a constant $C_1n$ plus the length of the parts of $s$ covered by the intervals $[a_i,b_i]$.
Consider the complement intervals $[0,a_1]$, $[b_i,a_{i+1}]$, $[b_k,1]$. The corresponding vectors $s(a_1)-s(0)$ etc, also add up to $mv$. Hence the above argument applies to these intervals as well and implies that $f(m)$ is no greater than $C_1n$ plus the remaining part of the length of $s$. One of these two parts is no greater that half of the total length of $s$, hence the result.
The problem is how to control the constant $C_1$ by the diameter only. To do so, one moves the division points $s(a_i)$, $s(b_i)$ to nearby lattice points (losing at most the diameter on each move), then the required parallel translations preserve the distance and the dependence on the metric goes away. But the new collection of vectors no longer sums up to $mv$. So we have to choose those "nearby" lattice points wisely, so that the accumulated error is bounded by the diameter times $C(n)$. This (hopefully) can be done as follows: approximate $s$ by a sequence of lattice points (with distances between neighboring ones at most twice the  diameter), then apply the lemma to the Euclidean broken line connecting these points, then move each division point to the next vertex of this broken line. The error in the sum of the resulting vectors has bounded stable norm and this should imply a bound on the metric distance. (I have not worked through this detail yet.)
Proof of lemma. It was discussed in this answer but the Google Book link there died since that (probably some copyright maniac killed it, now it looks as if the volume was never digitized).  The proof it the following, Consider the standard unit sphere $S^n$ of points $x\in\mathbb R^{n+1}$ with $\sum x_i^2=1$. To each point $x\in S^n$, associate a partition $0=t_0\le t_1\le\dots t_{n+1}=1$ of $[0,1]$ such that $t_i-t_{i-1}=x_i^2$, and define $f(x)\in\mathbb R^n$ by
$$
 f(x) = \sum sign(x_i) (s(t_i)-s(t_{i+1}) .
$$
The resulting function $f:\mathbb S^n\to\mathbb R^n$ is continuous and odd (i.e. $f(-x)=f(x)$). Therefore $f(x)=0$ for some $x$. Now let $[a_j,b_j]$ be the segments of the partition $\{t_i\}$ associated to this $x$, whose corresponding coordinates $x_i$ are positive. If there are too many of them, take the negative ones instead.<|endoftext|>
TITLE: Topological characterization of the closed interval $[0,1]$
QUESTION [19 upvotes]: This question is related to question 92206 "What properties make $[0, 1]$ a good candidate for defining fundamental groups?" but is not exactly equivalent in my opinion.  It is even suggested in one of the answers to 92206 that "there is nothing fundamental about the unit interval," but i would like to know what is fundamental about the unit interval.  I have learned some answers from the answers to 92206, but i wonder if there is more.  Another related question: "Topological Characterisation of the real line".
The question is:

What would be a "natural" topological characterization of the closed interval $[0,1]$?

Motivating questions (please do not answer them):

Why is all the algebraic topology built around it?
Why does it appear (in the guise of Lie groups) in the study of general locally compact topological groups (Gleason-Yamabe theorem)?
Is it really necessary to define it algebraically as a subset of a field ($\mathbb R$) just to use it as a topological space?

I am mostly interested in the significance of $[0, 1]$ for the study of Hausdorff compacts (like metrizability, Urysohn's lemma).
I have some ideas and have asked already on Math.StackExchange, but decided to duplicate here.
One purely topological way to define $[0, 1]$ up to homeomorphism would be to define path connectedness first: $x_1$ and $x_2$ are connected by a path in a topological space $X$ if for every Hausdorff compact $C$ and $a,b\in C$, there is a continuous $f\colon C\to X$ such that $f(a)=x_1$ and $f(b)=x_2$. Then it can be said that in every Hausdorff space $X$ with $x_1$ and $x_2$ connected by a path, every minimal subspace in which $x_1$ and $x_2$ are still connected by a path is homeomorphic to $[0, 1]$.
Maybe in some sense it can be said that $[0, 1]$ is the "minimal" Hausdorff space $X$ such that every Hausdorff compact embeds into $X^N$ for some $N$, but i do not know if this can be made precise.
In one of the answers to 92206 it was stated that $([0, 1], 0, 1)$ is a terminal object in the category of bipointed spaces equipped with the operation of "concatenation."  This is the kind of answers i am interested in.  As 92206 was concerned mostly with the fundamental group and tagged only with [at.algebraic-topology] and [homotopy-theory], i am asking this general topological question separately.

REPLY [4 votes]: Consider the class of all Hausdorff compacts with distinct points (i.e. which have more than $1$ point) that are absolute retracts in the class
of Hausdorff compacts.
Then $[0,1]$ is up to homeomorphism the only member of this class that embeds into every other.<|endoftext|>
TITLE: Symplectic block-diagonalization of a real symmetric Hamiltonian matrix
QUESTION [7 upvotes]: Given a $2n\times 2n$ real, symmetric, Hamiltonian matrix $W$ (anticommutes with the symplectic metric), is there an orthogonal, symplectic matrix $R$ such that $R^\top WR$ is block-diagonal? 
Being symmetric, for sure one can diagonalize $W$ by an orthogonal matrix. However, we only want block-diagonalization. This relaxation may let me make $R$ symplectic. I am looking for an explicit construction of $R$ in terms of i.e., SVD's of $W$'s blocks.
Let $J$ be the symplectic form 
$$
  J= \left(\begin{array}{cc}
      0&1\\\
      -1&0
    \end{array}\right).
$$
Given that $W$ is symmetric ($W^\top=W$) and Hamiltonian ($WJ=-JW$), it has block form
$$
   W=\left(\begin{array}{cc}
       x&y\\\
       y&-x
     \end{array}\right)
$$
with $x=x^\top$ and $y=y^\top$. $R$ can be parametrized as
$$
  R=\left(\begin{array}{cc}
      c&s\\\
      -s&c
    \end{array}\right)
$$
both the symplectic and orthogonal conditions boil down to $cc^\top+ss^\top=I$ and $cs^\top-sc^\top=0$. 
I am looking for $c$ and $s$ such that
$$
  R^\top W R=D
$$
where
$$
  D=\left(\begin{array}{cc}d&0\\\
0&-d
\end{array}\right).
$$
Hint: Writing $WR=RD$ yields equations
\begin{align}
   xc-ys&=cd\\\
 xs+yc&=-sd
\end{align}
Defining $t=sc^{-1}$ and equating the $d$'s in previous equations one can write (assuming $c$ is nonsingular!)
$$
   t y t-tx-xt-y=0.
$$
This is a particular form of the Riccati equation. It admits a "simple" solution if $y\geq0$ and $y+xy^{-1}x\geq0$ and from here one may be able to find a solution. Unfortunately I cannot assume these.
Note: Numerical tests suggest that this is always possible. Namely, for any $W$ that I generate, it seems that given the orthogonal diagonalization $W=O^\top D O$ there is always a permutation matrix $P$ such that $OP$ is symplectic. Furthermore, the resulting diagonalization is always strictly diagonal, not only block-diagonal.

REPLY [6 votes]: Yes, you can always do this.  The right way to understand this is to look at the complex symmetric matrix $X = x + i y$.  You want to act on this by a complex matrix $C = c + is$ where $C$ is unitary, by the rule, $X\mapsto C^{T}XC$, and you want to know whether you can arrange that $C^{T}XC$ is a real, diagonal matrix.  
You can do this, in two stages:  First, look at the Hermitian symmetric matrix $H = {\bar X} X = {\bar H}^T$.  When you act by $C$, the matrix $H$ will transform as $H\mapsto {\bar C}^{T}HC$.  Thus, one can choose $C$ to that $H$ is diagonal and hence real.  In particular, one is reduced to the case in which $X=x+iy$ satisfies $xy-yx=0$.  Thus, $x$ and $y$ are commuting symmetric matrices and hence they can be simultaneously diagonalized by a real orthogonal matrix.  Thus, you can now assume that $x$ and $y$ are diagonal.
Second, let $C$ be diagonal and unitary and let it act on a diagonal $X$ by $X\mapsto C^{T}XC$.  Since the diagonal elements of $C$ are unit complex numbers, we can clearly choose them so that the entries of $C^{T}XC$ are real and nonnegative, which is what you wanted.<|endoftext|>
TITLE: Hyperbolic groups with infinitely generated commutator subgroups
QUESTION [10 upvotes]: I am trying to get a sense of how often the commutator subgroup $[G,G]$ of a (Gromov) hyperbolic group $G$ is infinitely generated.
Remarks: 

$[G,G]$ is infinitely generated if is $G$ is free noncyclic, and of course $[G,G]$ is finitely generated when $G$ has finite abelianization.  
There are examples when a hyperbolic group has an finitely generated, normal (infinite) subgroup (of infinite index), obtained via various versions of the  Rips constructions, or Morse theory considerations of Bestvina-Brady and Brady, see  here . 

More generally, how often is the kernel of a surjection $G\to\mathbb Z$ infinitely generated (or infinitely presented)? Here is a specific:
Question.
Suppose the abelianization of $G$ has rank $>1$, so that there are infinitely many surjections $G\to \mathbb Z$. Is it true that there are infinitely many surjections $G\to\mathbb Z$ whose kernel is not finitely generated?

REPLY [3 votes]: The question to determine, given an f.g. group $G$, which homomorphisms to abelian groups (esp. cyclic groups) have a f.g. kernel, is addressed in detail in the paper: Bieri, W. Neumann, Strebel, A geometric invariant of discrete groups, Invent.math. 90, 451477 (1987). An older reference is Neumann, W.D.: Normal subgroups with infinite cyclic quotient. Math. Sci. 4, 143-148 (1979). This is not specific to the setting of hyperbolic groups (and in particular does not supersede Agol's answer).<|endoftext|>
TITLE: On the $L^1$-norm of certain exponential sums
QUESTION [14 upvotes]: I am stuck with an elementary-looking problem, which does not belong to my usual field of research so I eventually decided to ask it on MO.
Let $S$ be a finite set of integers. For $P$ a subset of $S$, I note
$$s_P = \sum_{s \in P} s$$
the sum of elements in $P$.
I assume that $S$ satisfies the following property
$$ (*) \ \ \ \ \ \  \forall P,Q \subset S, \ \ \ s_P=s_Q  \Longrightarrow P=Q$$
In other words, $(*)$ means that the set $R=$ {$ s_P,\  P \subset S$} has
maximal cardinality:
$$ (*) \ \ \ \ \ \  |R| = 2^ {|S|}$$
(Is there a name for the property $(*)$? additively free maybe?)
Now for $x \in [0,1]$ a real number, I consider the exponential sum
$$f(x) = \prod_{s \in S} (1+e^{2 i \pi s x}) = \sum_{r \in R} e^{2 i \pi r x}.$$
(Seen as a function of $e^{2i \pi x}$ on the unit circle of $\mathbb C$, $f$ is the Fourier
transform $\widehat{\chi_R}$ of the chatacteristic function $\chi_R$ of $R \subset \mathbb Z$.) I am interested in upper bounds for the $L^1$ norm of $f$,
$$||f||_1 = \int _{x=0}^1 |f(x)| dx.$$
A very simple application of Cauchy-Schwarz gives
$$||f||_1 < \sqrt{|R|} = \sqrt{2}^{|S|}$$
My question is:

When $|S|$ goes to $\infty$, can we prove an asymptotically better upper bound for $||f||_1$ than the Cauchy-Schwarz estimate above?
For example, is there a positive real
$\alpha < \sqrt{2}$ such that $||f||_1 < \alpha^{|S|}$ for all $S$ satisfying $(*)$ ? (Edit: same question for sets $S$ satisfying the stronger property $(*2)$ below).


Example: when $S=${$ 1,2,4,\dots,2^{n-1}$}, then $|S|=n$ and $R=${$0,1,2,3,\dots,2^n-1$}, so $S$ satisfies $(*)$, then $f(x)$ is essentially (up to multiplication by a complex of modulus 1)
the  Dirichlet kernel $D_{2^{n-1}}(x)$ and a famous result (of Dirichlet?) says that $$||f||_1 \sim \log (2^{n-1}) \sim n = |S|,$$ so in this case we get a much better bound than the Cauchy-Schwarz bound, and which is essentially optimal by the Littlewood conjecture (now a theorem, saying that $||\widehat{\chi_R}||_1 >> \log |R|$).
Edit: This suggests to restrain ourselves at first to sets $S$ satisfying the stronger property
$(*2)  \ \ \ $ every element of $S$ is a power of $2$.
Non-Example: removed as pointless after Noam's comment; replaced with this Remark:
Without the hypothesis $(*)$, (for $f$ defined as a product over $S$ as above, or as a sum over $R$ if $R$ is interpreted as a multiset),
the Cauchy-Schwarz bound $||f||_1 \leq \sqrt{2}^{|S|}$ does not hold any more, and one can not in general improve on the trivial bound $||f||_1 \leq 2^{|S|}$ as shown in Noam's comment
below.

But in general I am stuck. I thought that the product expression of $f(s)$ might help,
but I was not able to use it in  a clever way.
Numerical evidence is not very conclusive but does not seem to point toward a polynomial bounds in $|S|$. Any advice, reference, intuition, conjecture, solution (proof or disproof) welcome. I am not even sure that my tags are right, please feel free to modify them. PS: my motivation comes form Galois representations.

REPLY [10 votes]: As Noam correctly mentioned, the upper bound has to be exponential. However, the exponent can, indeed, be improved. 
One (fairly cheap) trick is to look at $|f(x)|^2=2^{|S|}\prod_S(1+\cos 2\pi sx)$. If we can show that the product is bounded from above by $e^{-c|S|}$ outside a set of measure $e^{-c|S|}$, we can get away with Cauchy-Schwarz. Now, $\log(1+y)\le y-cy^2$ for $y\in[-1,1]$ with some $c>0$. Since $2\cos^2 y=1+\cos 2y$, we see that the square term will give us a fixed linear in $|S|$ push down, so it remains to show that for every $a>0$, the set of $x$ for which $\left|\sum_{s\in S} \cos 2\pi sx \right|$ is at least $a|S|$ has exponentially small measure (we'll need to use that twice: once for $x$ and once for $2x$). 
This is actually pretty easy if we recall the reverse inequality $\log(1+x)\ge x-Cx^2$ for $x\in[-\frac 12,\frac 12]$, say. Note that the unique representation as a sum property implies that
$$
\int_0^1 \prod(1+t\cos 2\pi sx)\,dx=1
$$
for every $t\in\mathbb R$. Assuming that $\sum_{s\in S} \cos 2\pi sx>a|S|$ on a set $E$, choose $t\in(0,\frac 12)$ so small that $b=at-Ct^2>0$. Then, clearly, the product is at least $e^{b|S|}$ on $E$ and the desired exponential bound on the measure of $E$ follows at once. The other set where $\sum_{s\in S} \cos 2\pi sx<-a|S|$ is treated the same way using small negative $t$.
This all is extremely crude, of course. However, it answers the original question in the affirmative, so I'll stop here :).<|endoftext|>
TITLE: What are the sub $C^*$-algebras of $C(X,M_n)$?
QUESTION [6 upvotes]: Let $X$ be a locally compact Hausdorff topological space, denote by $M_n$ the $C^*$-algebra of complex $n\times n$ matrices, by $C_0(X,M_n)$ the $C^*$-algebra of continuous functions on $X$ with values in $M_n$ vanishing at infinity, and by $C_b(X,M_n)$ the $C^*$-algebra of bounded continuous functions on $X$ with values in $M_n$. Does there exist a description of all sub $C^*$-algebras of $C_0(X,M_n)$ or $C_b(X,M_n)$? I am most interested in the (easier?) case where $X$ is compact and sub $C^*$-algebras of $C(X,M_n)$ containing the unit of $C(X,M_n)$.
For the case $n=1$, i.e., for $C_0(X)$, you can find the answer here: What is the commutative analogue of a C*-subalgebra?.

REPLY [6 votes]: Every $C^*$-subalgebra $A$ of $C_0(X,M_n)$ has irreps of dimension $\leq n.$ (Just because every irrep of $A$ can be continued to an irrep of $C_0(X,M_n)$). Such $C^*$-algebras are called $n$-subhomogeneous. 
There is a complete (rather complicated) description of such $C^*$-algebras in algebraic topological terms: 
Vasilʹev, N. B. "$C^∗$-algebras with finite-dimensional irreducible representations."
Uspehi Mat. Nauk 21 1966 no. 1 (127), 135–154. 
Much easier is to describe the $C^*$-algebras with all irreps of the same dimension $n$. They are exactly the algebras of $C_0$-sections of vector bundles over l.-c. hausdorff spaces with fiber $M_n$ and group $PU_n,$ see
Tomiyama, Jun; Takesaki, Masamichi
Applications of fibre bundles to the certain class of $C^∗$-algebras.
Tôhoku Math. J. (2) 13 1961 498–522.<|endoftext|>
TITLE: Is the evaluation of polynomial functors appropriately continuous?
QUESTION [10 upvotes]: I'd like a nice proof of the following fact.
Let $C$ and $D$ be categories, and let $\mathbf{Cat}/(C\times D)$ be the usual (1-categorical) slice category whose objects are triples $(X,F\colon X\to C, G\colon X\to D)$, $$C\xleftarrow{F}X\xrightarrow{G}D,$$
and whose morphisms are honest commuting triangles. That is, $$\mathrm{Hom}((X,F,G),(X',F',G'))=\{f\colon X\to X' \mid F'\circ f=F, G'\circ f=G\}.$$ 
Given $(X,F,G)$ as above, let $F^{\ast}\colon \mathbf{Set}^C\to \mathbf{Set}^X$ be the pullback functor and let $F_{\ast}$ be its right adjoint; similarly for $G$. Given a $C$-set $\delta\colon C\to \mathbf{Set}$, we will be interested in the continuous (polynomial-like) operation of moving it over to a $D$-set as $G_{\ast}F^{\ast}\delta$.
At this point we collect everything in sight into a functor, which might be called the evaluation functor $$K\colon (\mathbf{Cat}/(C\times D))^{op}\times \mathbf{Set}^C\to \mathbf{Set}^D$$ defined on objects as
$$K((X,F,G),\delta):= G_{\ast}F^{\ast}\delta.$$
I want to show that $K$ preserves limits. I have convinced myself that this should work using a very low-level argument, but I am looking for efficient avenues by which to think about this, rather than small-minded plug and chug.  A reference would be greatly appreciated. (Similarly, it seems true that the evaluation functor $L\colon (\mathbf{Cat}/(C\times D))\times \mathbf{Set}^C\to \mathbf{Set}^D\ $ given by $L((X,F,G),\delta)=G_!F^*\delta\ $ preserves colimits.) 
I'd also be happy with an analogous theorem about polynomial functors in a locally cartesian closed category.

REPLY [3 votes]: Ok, this follows trivially from the construction of limits in ${\bf Cat}$. 
Let $I\ $ be a small category, let $X\colon I\to{\bf Cat}\ $ denote a functor, and let $Y=\text{colim}_{i\in I}X_i\ $ be its colimit with inclusion maps $q_i\colon X_i\to Y$. Let $d,e\colon Y\to {\bf Set}\ $ be any functors. The question posed here reduces to the question of whether the following function is an isomorphism:
$$Hom_{Y-{\bf Set}}(d,e)\to \lim_{i\in I}\ Hom_{X_i-{\bf Set}}(q_i^{\ast}d,q_i^{\ast}e)$$
It indeed is an isomorphism because the category $Y-{\bf Set}\ $ of functors $Y\to{\bf Set}\ $ is the limit of the categories $X_i-{\bf Set}\ $, and we know that the objects and hom-sets of a limit in ${\bf Cat}$ are computed pointwise.
We now prove that the above observation implies that the functor $K$ above preserves limits. It suffices to fix each variable in the domain of $K$, because a limit in a product of categories is the product of the limits. Since $g_{\ast}$ and $f^{\ast}$ are right adjoints, $K$ is continuous in the second variable. It suffices to show that if $\delta\colon C\to{\bf Set}\ $ is any functor and $(Y,F,G)=\text{colim}_{i\in I}(X_i,F_i,G_i)\ $ in ${\bf Cat}/(C\times D)$ then the map $$G_{\ast}F^{\ast}\delta\to\lim_{i\in I}(G_i)_{\ast}(F_i)^{\ast}\delta$$ is an isomorphism of $D$-sets. 
Let $\epsilon\colon D\to{\bf Set}\ $ be a $D$-set. Using the Yoneda imbedding and the $(G^{\ast},G_{\ast})\ $ adjunction, it suffices to show that the function
$$Hom_{Y-{\bf Set}}(F^{\ast}\delta,G^{\ast}\epsilon)\to\lim_{i\in I}Hom_{X_i-{\bf Set}}(F_i^{\ast}\delta,G_i^{\ast}\epsilon)$$
is a bijection, which was the observation above.<|endoftext|>
TITLE: FIltered colimits of truncated objects in $\infty$-topoi
QUESTION [6 upvotes]: The bare question:
Let $\mathcal{C}$ be an $\infty$-topos, and let $\tau_{\leq 0}\mathcal{C}$ be the subcategory of 0-truncated objects (which is the nerve of an ordinary Grothendieck topos: see HTT 6.4.1.3).

Does the inclusion $\tau_{\leq 0}\mathcal{C} \hookrightarrow \mathcal{C}$ preserve filtered (or at least directed) ($\infty$-)colimits?

Motivation:
Let $Aff_\mathbb{C}$ the Grothendieck site of complex affine schemes. We can then consider the topos of sheaves, $Shv(Aff_\mathbb{C})$, and the $\infty$-topos of $\infty$-stacks, $Shv_\infty(Aff_\mathbb{C})$. The nerve of the first is equivalent to the subcategory of 0-truncated objects in the second.
Given a scheme $X$ and a closed subscheme $Y$ in it defined by a sheaf of ideals $\mathcal{I}$, we can construct the so-called formal completion of $X$ along $Y$ as the directed colimit $X_Y^{\mbox{^}} = \mathrm{colim}\:V(\mathcal{I}^n)$. Typically this is done in $Shv(Aff_\mathbb{C})$. Since the homotopy theory in the latter is trivial, it is also the homotopy colimit of the same diagram. But how does this play with the inclusion $Shv(Aff_\mathbb{C}) \hookrightarrow Shv_\infty(Aff_\mathbb{C})$? Is $X_Y^{\mbox{^}}$ still the homotopy colimit of the same diagram in $Shv_\infty(Aff_\mathbb{C})$?
An affirmative answer to my last question would be enough for me, but I suppose it is a natural question to ask whether this holds for general formal schemes —i.e., sheaves that are locally formal completions as above.

REPLY [10 votes]: I believe the answer is YES and, more generally, that $\tau_{\leq n}\mathcal{C}\subset\mathcal{C}$ preserves filtered colimits for any $\infty$-topos $\mathcal{C}$. For the $\infty$-topos of $\infty$-groupoids this is well-known. This implies the result in any presheaf $\infty$-topos since colimits and truncations are computed objectwise. Finally, if the result is true in $\mathcal{C}$ and $\mathcal{D}\subset\mathcal{C}$ is a left exact localization, then the result is true in $\mathcal{D}$ as well because left exact functors preserve $n$-truncated objects (HTT, Prop. 5.5.6.16).<|endoftext|>
TITLE: Applications of line graphs
QUESTION [5 upvotes]: I am trying to collect a few examples of applications of line graphs in sciences other than mathematics. To be more precise: I am thinking of models where there is a clear conceptual added value in switching the paradigm from a description focused on agents (nodes) to a description focused on relations (edges). 
I know: put it like this it sounds very much like I am thinking of anthropological/sociological models, and indeed they are the most natural I could come up with$^{*}$, but I do believe that similar ideas appear elsewhere, too, modulo considering graph operations/decorations that allow for more flexible descriptions.
Inter/transdisciplinary examples would be particularly appreciated.
$^{*}$ It is not by chance that line graphs were so much pushed by Frank Harary after all, of all possible graph theorists :)

REPLY [4 votes]: Consider a wireless adhoc network that can be modeled as a graph $G=(V,E)$.  Two links $e, f \in E$ cannot be simultaneously active if they are incident to the same node. This type of interference is called primary interference.  The graph that models this interference is the line graph of $G$.  More details about this application can be found in my paper (at http://link.springer.com/article/10.1007%2Fs11276-013-0680-z) and the references therein.<|endoftext|>
TITLE: How to see the geometry and arithmetic of tannakian fundamental groups?
QUESTION [8 upvotes]: The etale fundamental group is an inverse limit of automorphism groups of finite etale coverings. We can see the geometry of etale fundamental group very well from etale coverings just like topologically fundamental group. But the tannakian fundamental groups are defined as the automorphism of a fiber functor of a tensor category, this is definition abstract for me. 
I want ask that how to see the geometry of tannakian fundamental groups? 
What is the relation between first etale cohomology and pro-unipotent fundamental group?
I would like to extend my question. If we consider a hyperbolic curve $X$ over a local field $K$ with valuation ring $R$, there is an natural Galois action (i.e., outer Galois action) of $G_{K}$ on the etale fundamental group $\pi_{1}(X)$. From this outer Galois action, we can understand some geometry of $X$ and their reduction. For example, there is a good reduction criterion in terms of pro-$l$ fundamental groups (Oda, Tamagawa).
My question is: Dose there exist some similar Galois actions or criterions for Tannakian fundamental groups? or dose there exist some anabelian type theorems for Tannakian fundamental groups?

REPLY [5 votes]: To answer your second question, for any nilpotent neutral Tannakian category $\mathcal{C}$, (i.e. one in which every object is an iterated extension of the unit object $\underline{1}$), with fibre functor $\omega$ and assocaited pro-unipotent group scheme $G=G(\mathcal{C},\omega)$, there is an isomorphism
$\mathrm{Lie}(G^\mathrm{ab})^*\cong \mathrm{Ext}^1_\mathcal{C}(\underline{1},\underline{1})$
So in our case, taking $\mathcal{C}$ to be the category of unipotent lisse $\mathbb{Q}_\ell$-sheaves on some variety over an algebraically closed field of characteristic $\neq\ell$, we can recover the first étale cohomology as the dual of the abelianisation of the pro-unipotent étale fundamental group, exactly as one might expect from the Hurewiz theorem.<|endoftext|>
TITLE: Rational points on surfaces of general type
QUESTION [13 upvotes]: The weak Lang conjecture asserts that rational points on a variety of general type defined over $\mathbb{Q}$ are not Zariski dense (same replacing $\mathbb{Q}$ with a number field). This one is proved in dimension $1$, and in dimension $2$ it would in particular mean that the rational points of surface of general type are contained in the union of a finite number of rational and elliptic curves, plus a finite set of "isolated points".
It seems natural for me that a surface of general type has "less points" because the polynomials that define it are of "higher" degree. However, is there some more precise evidence for this conjecture ? 
And in which particular cases is the conjecture known to be true?
I am mostly interested in the case of surfaces, but higher dimension case is also interesting.

REPLY [9 votes]: A Theorem of Faltings states that any proper subvariety of an abelian variety has finitely many rational points provided this subvariety does not contain a translate of any non-trivial proper abelian subvariety:
G. Faltings, The general case of S. Lang’s conjecture, pages 175–182 of Barsotti Symposium
in Algebraic Geometry (Abano Terme, 1991), Perspect. Math. 15, Academic Press, San Diego,
CA, 1994.
Harris and Silverman use this to show that if C is curve of genus $\ge 2$ that is neither hyperelliptic nor bielliptic, then the set of rational points on $C^{(2)}$ is finite. Here $C^{(2)}$ is the symmetric square of $C$. 
J. Harris and J. H. Silverman, Bielliptic curves and symmetric products, Proceedings of the
American Mathematical Society 112 (1991), no. 2, 347–356.
My understanding is that if $C$ has genus $3$ (say a plane quartic) then $C^{(2)}$ is a surface of general type. 
I admit however that symmetric powers of curves are really special and probably not what the OP is hoping for.<|endoftext|>
TITLE: Inductive tensor product and smooth functions
QUESTION [16 upvotes]: Given complete, locally convex Hausdorff vector spaces $E$ and $F$, let
$$ E \otimes_i F, \qquad E \otimes_\pi F$$ denote the (completed) inductive and projective tensor products respectively.  The inductive (resp. projective) tensor product has the universal property for separately (resp. jointly) continuous bilinear maps out of $E \times F$.  By the universal property, there is a continuous linear map
$$E \otimes_i F \to E \otimes_\pi F.$$
If $E$ and $F$ are Frechet, then this map is an isomorphism.  This is because separate and joint continuity coincide in this case.  I am interested in knowing if this map is an isomorphism in situations where the hypotheses on $E$ are strengthened and the hypotheses on $F$ are relaxed.
More specifically, I am interested in the case $E = C^\infty(M)$, the nuclear Frechet space of smooth functions on a compact manifold $M$.  In this case,
$$C^\infty(M) \otimes_\pi F \cong C^\infty(M,F),$$ the space of smooth functions on $M$ with values in $F$.  So I would like to know what hypotheses, if any, are needed on $F$ to ensure that
$$C^\infty(M) \otimes_i F \to C^\infty(M,F)$$ is an isomorphism.  For example, is it an isomorphism if $F$ is an $LF$-space, that is, an inductive limit of Frechet spaces?

REPLY [14 votes]: If you mean by LF-space a strict inductive limit of Frechet spaces (as it was done by Dieudonne and Schwartz) I think the answer is yes.
Here is what I believe could be made a proof: Since the inductive tensor product
respects inductive limits you have $C^\infty(M) \otimes_i F = \lim C^\infty(M) \otimes_i F_n$ if $F= \lim F_n$ is a strict LF-space. Since $F_n$ is a closed topological
subspace of $F$ you get that $C^\infty(M) \otimes_i F_n = C^\infty(M) \otimes_\pi F_n$
and this implies that $C^\infty(M) \otimes_i F = \lim C^\infty(M) \otimes_\pi F_n$ is a dense topological subspace of $C^\infty(M) \otimes_\pi F$. On the other hand, it is complete
because it remains strict (as for as I remember this is due either to Dieudnne-Schwartz or to Köthe).

EDIT: Sorry, I confused the Frechet case with the DF case -- only in the latter case
$\lim E \otimes_\pi F_n$ is a topological subspace of $E\otimes_\pi \lim F_n$.
For strict LF-spaces the answer to your question was already given in Grothendieck's thesis (part I, page 47): If E is a proper (non-normable) Frechet space and $F=\lim F_n$ is a strict inductive limit with a strictly increasing sequence $F_n$ then
$\lim E\otimes_\pi F_n$ is NEVER a topological subspace of $E \otimes_\pi \lim F_n$.
On the other hand, if the nuclear Frechet space $E$ has a continuous norm (as it is the case for $C^\infty(M)$) and $F=\lim F_n$ is a strict LF-space then 
$\lim E\otimes_\pi F_n = E\otimes_\pi F$ holds algebraically: 
The latter space is the space of all continuous linear operators $T$ from $E'$ to $F$ and the former of those operators with values in some $F_n$. But $E'$ contains a bounded set
(the polar of the $0$-neighborhood corresponding to the continuous norm) which is total, i.e., its linear span is dense in $E'$. For any continuous $T$ the bounded set $T(B)$ is contained in some $F_n$ (by the so called regularity of strict LF-spaces) and since $F_n$ is closed in $F$ one gets from the continuity $T(E') \subseteq \overline{[T(B)]} \subseteq F_n$.

There might be non-strict inductive limits where $\lim E \otimes_\pi F_n = E\otimes_\pi\lim F_n$ holds topologically but I do not know an example since both, the algebraic and the topological, conditions are very restrictive. If $F$ is the dual of a nuclear Frechet space $X$ then the arguments above should show that the algebraic coincidence is equivalent to 
$L(X,E) = LB(X,E)$, the space of all operators from mapping some $0$-neighborhood of $X$ into a bounded subset of $E$. This situation has been investigated by Vogt,
J. Reine Angew. Math. 345 (1983), 182–200, and it implies Ext$^1(E,X)=0$. On the other
hand, I believe results of Grothendieck and Vogt yield that the topological coincidence
implies Ext$^1(X,E)=0$. If $X$ and $E$ are both power series spaces these conditions contradict each other.

In conclusion: The topological equality $C^\infty(M) \otimes_i F = C^\infty(M) \otimes_\pi F$ seems very unlikely whenever $F$ is not a Frechet space.<|endoftext|>
TITLE: Differences between various categories of surface embeddings in 4-space  
QUESTION [8 upvotes]: This is a very naive question, but I'm trying to understand the difference between the various categories when it comes to embedding surfaces in 4-dimensional manifolds.  The situation I'd really like to understand is when we have a surface neatly and properly embedded in the 4-ball $D^4$, so that its boundary lies in $\partial D^4 = S^3$ as a link.  I'd like to understand some of the differences between assuming everything in the setup is just continuous, versus when we assume it is either locally-flat, PL locally-flat, or smooth. 
For example, if we have a PL-locally flat surface, after arbitrarily small PL-isotopy we can arrange so it has a movie presentation with only max, min, and saddle points.  So it seems like we can treat such an embedded surface much as we would a smooth surface with Morse function.  Intuitively it seems that the theories of smoothly embedded surfaces and PL locally-flat surfaces should be equivalent in some sense, though I've never seen it dealt with in the literature.  In particular it seems like we should be able to smooth PL locally-flat surfaces and vice versa, so that equivalence under the different types of isotopy is preserved.
Where I get really murky is when we drop the PL or smooth condition and consider topological locally-flat embeddings.  Here things can be quite different, as there are knots which bound topologically locally-flat disks, but whose smooth slice genus is arbitrarily high.  It seems like there's an awful lot more freedom once we drop smoothness, or even just the PL condition from PL locally-flat, even though we can use embedded PL locally-flat disks to $\varepsilon$-approximate any topological embedding.

With that rambling out of the way, my main questions are he following:

1) Are the smooth and PL locally-flat situation really that "close" to each other?  Can surfaces in one category be approximated by surfaces in the other, so that equivalence under smooth and PL-isotopy is preserved?  Or are their hidden subtleties I'm missing?
2) Is there anyway to understand the problems we encounter when we drop to the topologically locally-flat situation?  For example, whenever the locally-flat condition is explained, the example of the cone over a nontrivial knot is invoked to illustrate a situation where it fails.  Is there a similar example to understand (even just a single situation) where a topologically locally-flat embedding fails to be PL locally-flat or smooth? 

Of course anything else which might help me understand the situation better would be greatly appreciated as well.

REPLY [4 votes]: Regarding (1), yes there's a unique (up to smooth isotopy) smoothing of a PL locally-flat embedding.  You can similarly ensure that if your original surfaces are close in some PL compact-open sense, that their smoothings can be made to be close in the smooth C^1 compact-open sense. 
Regarding (2), yes, there are knots that are topologically locally-flat slice, yet they are not smoothly slice.  A knot with a trivial Alexander polynomial does the job.  These examples aren't as "concrete" as your cone example as they depend on theorems of Freedman which involve infinite constructions.<|endoftext|>
TITLE: Is every topological (resp. Lie-) group the isometrygroup of a metric space (resp. Riemannian manifold)?
QUESTION [20 upvotes]: The isometry group of a metric space is a topological group (with the compact open topology). The isometry group of a Riemann Manifold is a Liegroup. (Thm. of Steenrod-Myers)
So, is every topological group isomorphic (in the category of topological groups) to the isometry group of a metric space?
And what about the differentiable version: Is every Liegroup isomorphic (in the category of Liegroups) to a isometry group of a Riemann manifold?
Edit: Benjamin Steinberg gave a reference, which fully answers the question in the topological case.
Ryan Budney gave an idea how to realize at least every compact lie group as the isometry group of a Riemannian manifold. (This is proven in THE ISOMETRY GROUPS OF MANIFOLDS
AND THE AUTOMORPHISM GROUPS OF DOMAINS, by RITA SAERENS and WILLIAM R. ZAME)
So for me, still open is the question about the non-compact case: Is even every non-compact Lie group realizable as the isometrygroup of a riemannian manifold?

REPLY [8 votes]: This seems to be completely answered in the topological category by Thm 1.4 of http://arxiv.org/pdf/1202.3368v3.pdf. 
Edit
If X is a topological space which is not Dieudonne complete (meaning its topology cannot be given by a complete uniformity), then it seems theorem 7.24 of http://books.google.com/books?id=v3_PVdvJek4C&pg=PA35&lpg=PA35&dq=free+topological+group+dieudonne+complete&source=bl&ots=QIt0C2TCjN&sig=d4BJjO2h3zM4jDLoTzijrDVYt3w&hl=en&sa=X&ei=VIw8UeuBJpO30QHek4C4AQ&ved=0CC4Q6AEwAA  and Thm 1.4 of the paper above shows that the free topological group on X is not the isometry group of a metric space. Googling shows completely regular spaces exist which are not Dieudonne. 
I believe that any polish group or locally compact group is an isometry group of a metric space by the paper I linked. 
Also the author of the first paper has shown that every Lie group is the isometry group of another Lie group with respect to some proper metric. http://arxiv.org/pdf/1201.5675v2.pdf<|endoftext|>
TITLE: Chern numbers via Euler characteristics?
QUESTION [12 upvotes]: Let $X$ be a space good enough to have a fundamental class, and $E$ a complex vector bundle on $X$.  Let $P$ be some polynomial expression, and say I want to evaluate $P(c_i(E)) \cap [X]$. 

Is there some naturally associated space, $\mathbf{P}(X, E)$ such that (e.g.) the topological Euler characteristic $\chi(\mathbf{P}(X, E)) = P(c_i(E)) \cap [X]$?

REPLY [2 votes]: Let $B = Gr_{\dim E}(\infty)$ denote the classifying space for $\dim E$-bundles. Your $P(c_i(E))$ defines a class on $B$, which (if $P$ has integer coefficients, which you better've meant it to!) can pretty obviously be represented by an actual cycle, $\mathcal P \subseteq B$.
Assume the classifying map $c : X \to B$ is transverse to $\mathcal P$. Then $c^{-1}(\mathcal P)$ is the space you're looking for. Too bad it's finite!
Not quite. I want $c$ to be not just transverse to $\mathcal P$, but to meet it positively (unless you're okay with $\chi$ of a "negative point" being $-1$). Something like, $X$ should be algebraic, and the classifying map should be algebraic, some condition like $E$ being globally generated.<|endoftext|>
TITLE: Is there an editors checklist for mathematics?
QUESTION [20 upvotes]: I always have trouble with editing math papers for publication.  I know there are plenty of checklists for English exposition but is there one for math specific exposition errors?

REPLY [4 votes]: In the same spirit as above: look at the video where Jean-Pierre Serre explains in English how to write mathematics - badly!
http://www.dailymotion.com/video/xf88b5_jean-pierre-serre-writing-mathemati_tech#.UTsRq44zVHI
Often hilarious, but so true...<|endoftext|>
TITLE: Kodaira dimension of symmetric products of curves
QUESTION [21 upvotes]: What is the Kodaira dimension of symmetric products of curves? That is, given a projective smooth, connected complex curve $C$, what is the Kodaira dimension of $C^{(d)}=C^d/\mathfrak S_d$?
When $d> g$, the genus of $C$, then $C^{(d)}$ is a bundle in projective spaces over the Jacobian of $C$, hence all pluriforms on $C^{(d)}$ vanish on the fibers of this fibration and $\kappa(C^{(d)})=-\infty$ in this case.
(Said in another way, $C^{(d)}$ is uniruled.) 
Is something known for $2\leq d\leq g-1$? (This question is prompted by this other post.) I suspect that the answer will strongly depend on fine properties of the curve $X$ (gonality, Brill-Noether properties) and there might not be a general and neat answer.
Perhaps surprisingly, the analogous question in higher dimensions is quite different since if $X$ is a projective smooth connected variety of dimension $>1$, the Kodaira dimension of (any desingularization) of $X^{(d)}$ is equal to $d \kappa(X)$, where $\kappa(X)$ is the Kodaira dimension of $X$ (D. Arapura, S. Archava, Kodaira dimension of symmetric powers, Proc. AMS, 2003).
[Edited to correct an incorrect assertion, as pointed out by @pbelmans]

REPLY [23 votes]: Let $C$ be a smooth projective connected complex curve of genus $\geq 2$. Let me show that $C^{(d)}$ is of general type if $1\leq d\leq g-1$.
Equivalently, one needs to show that the image $W_d$ of $C^{(d)}$ in the jacobian $J(C)$ is of general type, because $C^{(d)}\to W_d$ is birational. If $W_d$ were not of general type, then, by [Ueno, Classification of algebraic varieties I, Theorem 3.10], there would be a non-trivial abelian variety $A\subset J(C)$ such that $W_d$ is stable by translation by $A$ (this is the argument in MP's comment above). But then, $W_{g-1}$ would also be stable by translation by $A$. Now choose a point $x$ outside of $W_{g-1}$ and consider the orbit $A.x$ : it is a positive-dimensional variety avoiding $W_{g-1}$. This is a contradiction because $W_{g-1}$ is an ample divisor: the theta divisor.<|endoftext|>
TITLE: A formal group law over oriented bordism
QUESTION [10 upvotes]: My question is related to the following question by Mark Grant here on math overflow:
Formal group law of unoriented cobordism
There it is stated that $MO_*$ has a formal group law $F_0$, universal for formal group laws in characteristic $2$ which satisfy $F(x,x) = 0$.
Of course this formal group law is classified by a map $MU_* \to MO_*$ and in view of the model of $MU$ as unitary ring spectrum and $MO$ as orthogonal ring spectrum this map is indeed induced by a canonical map of (unitary) ring spectra
$$c: MU \to MO,$$
see also the answer by Neil Strickland here: Which cohomology theories are real- and complex-orientable? .
Now I have two questions:
Q1) Does the map $c$ factor over $MSO$ ?
Q2) If so, can one say anything about the corresponding formal group law over $MSO$?
In view of how I think one can construct the map $MU \to MO$ I think it should be obvious that the map factors through $MSO$, since the map $U(n) \to O(2n)$ factors over $SO(2n)$.
But I have not been able to find a reference that gives the example of $MSO$ being complex oriented.
So I guess Q2) is more interesting, already for the simple reason that the ring $MSO_\ast$ is much more delicate than the ring $MO_\ast$. 
Moreover (as Johannes Ebert also pointed out in a comment in the link above) real oriented theories come equipped with a map from $MO$ and hence split in Eilenberg-MacLane spectra, whereas $MSO$ behaves quite differently.
In addition, I have one last question:
Supposing Q1) and Q2) are true, then as I have already mentioned that $MSO$ is complex oriented. In particular we have a notion of abstract chern classes for $MSO$-cohomology. Moreover we have a canonical map $MSO \to H\mathbb{Z}$. The composite $MU \to H\mathbb{Z}$ is just the usual orientation of integral cohomolgy, in particular the abstract chern classes of this complex oriented theory are just the usual chern classes, but since oriented real bundles have pontryagin classes I ask myself:
Q3) Is there any way to view pontryagin classes as special case of abstract chern classes?
(It seems not to be the case for the cohomology theory $MSO$ which I would have thought of first).

REPLY [8 votes]: MSO is certainly complex-oriented
The resulting formal group law satisfies $[-1]_F(x)=-x$
Let $R_*$ be the universal example of a graded ring with a formal group with the above property, so there is a natural map $R_*\to MSO_*$.  I think this is injective, and it becomes an isomorphism after inverting $2$.
Let $U$ be the set of all positive integers not of the form $2^j-1$, let $U_0$ be the subset of all even integers, and put $U_1=U\setminus U_0$.  One can show that $R_*$ has a polynomial generator $a_i$ of degree $2i$ for each $i\in U$, and the only relations are $2a_i=0$ for $i\in U_1$. 
The additive structure of $MSO_*$ is known by old work of Wall and Atiyah.  I don't know a really convincing interpretation in terms of formal groups.  All the torsion is killed by $2$, and there is a lot of stuff in odd degrees.<|endoftext|>
TITLE: The existential theory of the reals
QUESTION [5 upvotes]: Some definitions of the existential theory of the reals (ETR) allow a real closed field and some definitions allow only rational numbers as coefficients of polynomials. Which one is correct? Will the answer to the question have an effect on the PSPACE proof by J. Canny? For example, ETR is not in PSPACE, if numbers in real closed field are allowed as coefficients?

REPLY [4 votes]: In order for this to be a computational problem in the first place, you have to fix a representation of the coefficients by finite strings (which in particular implies that the field is countable). The answer will in general depend on the representation.
For the most obvious case, if the coefficients are taken from the field of real algebraic numbers, and are represented in a common way (minimal polynomial + an isolating interval or a BKR sign condition), then the problem is equivalent to the one with rational coefficients, because we can just plug the definitions of the coefficients into the formula.<|endoftext|>
TITLE: Is the space of gradient-like vector fields contractible?
QUESTION [6 upvotes]: Let $M$ be a compact manifold (without boundary) and let $f:M\to \mathbb{R}$ be a fixed Morse-function.

Question: Is the space $GVect(M,f)$ of all gradient-like vector-fields for $f$ contractible or even convex?

To eliminate any misunderstanding, this is the definition of gradient-like vector fields that I'm using:

Definition: A vector field $X\in Vect(M)$ is called gradient-like for $f$ if
   1.  $\forall q\in M\setminus Crit(f):\; df(q)X(q)<0$.
   2. For each critical point $p\in Crit(f)$ of $f$, $\exists$ Morse coordinate chart $\phi:U\to\mathbb{R^n}$ (i.e. $\phi(p)=0$ and $f\circ\phi^{-1}(x)=f(p)-\sum_{i=1}^k x_i^2+\sum_{i=k+1}^n x_i^2$ for $x\in\phi(U)$, $k=index(p)$) such that $$\phi_*X(x)=d\phi(\phi^{-1}(x))X(\phi^{-1}(x))=(2x_1,\ldots,2x_k,-2x_{k+1},\ldots,-2x_n)^T$$  for $x\in\phi(U)$ (i.e. in this Morse-chart $X$ coincides with the negative gradient of $f$ w.r.t. the euclidean metric on  $\mathbb{R^n}$).

Thank you in advance for any insights.

REPLY [6 votes]: $GVect(M,f)$ is a convex set in the space of all vector fields. Thus it is contractible.
Edit, since you asked for details: I assume that the Morse charts $\Phi$ are the same for all $X$. 

Let $X,Y$ be gradient-like for $f$, then:

1. $df(q)(t.X(q)+(1-t).Y(q)) = t.df(q)(X(q)) + (1-t).df(q)(Y(q)) < 0$ for $0\le t\le 1$.

2. In the Morse charts the two vector fields look the same.
Second edit: Okay, you want the Morse chart to depend on the vector field.
Maybe, your definition of a gradient-like vector field is equivalent to:
a) At a critical point $p$ we have $X(p)=0$. 

b) Near each critical point $p$, the function $-X(f)= -df(X)$ is Morse with critical point $p$ a minimum. Then you can use $-df(X)$ as the square distance from $p$ for a Riemannian metric, and the Morse chart construction with respect to that Riemannian metric leads to a Morse chart as you require.

c) Off the critical points, we have $df(X)<0$ (i.e. X is transversal to the level sets and points towards lower level sets).<|endoftext|>
TITLE: Does the cubic planar graph with 6 3-faces and 6 7-faces have a name?
QUESTION [5 upvotes]: There is exactly one cubic planar graph with six 3-faces and six 7-faces (and no other faces). Surely it must have a name. What is it called?
Here is a picture of the graph embedded on the plane with a point at infinity:

(source)
A slightly more general question: How can I find out what the names are of semi-famous graphs? A graph is semi-famous if it has an established name but is not easily found in a standard textbook on the subject.
I did try the House of Graphs by searching using their drawing tool, but found nothing.

REPLY [2 votes]: Sage has search facilities for graphs with specified properties, and "knows" a large number of "named" graphs. As you work within a system running Python, you can do much more much easier, compared to what you get just by searching online databases using a www interface. The manual of the corresponding part is here.<|endoftext|>
TITLE: Similarities between Post's Problem and Cohen's Forcing
QUESTION [31 upvotes]: Remark: I have since learned that G.H. Moore addresses this question in the third reference listed at the end of this post, beginning on p. 157 in which he cites a letter from Kreisel to Gödel dated 4/15/63, followed by the parenthetical note: Thus Kreisel saw an analogy between forcing and Friedberg's [1957] priority argument. The following page suggests other logicians agree that forcing was implicit in priority arguments from recursion theory (p. 158) and mentions Kunen (who is cited in one of the answers here). On the same page, it is mentioned that Kreisel claimed he had a form of forcing in his interpretation of intuitionism in the 1961 paper "Set-theoretic problems suggested by the notion of potential infinity." However, Moore contends that Cohen was the first to use forcing and related ideas in Set Theory. Nevertheless, I am grateful for the responses thus far, and would warmly welcome further clarification regarding my question below from other experts in the area of Set Theory and/or Mathematical Logic.

Background: Paul Cohen began to think deeply about the Continuum Hypothesis in 1962, and published his proof of its independence (in two parts) by the following year. Of course, there were earlier mathematical moments that led to his "discovery of forcing," including his familiarity with Skolem's work (in particular, the Löwenheim–Skolem theorem) and a desire to think in terms of "decision procedures." I will include a few relevant references at the end of this question, including a retrospective/introspective piece by Cohen himself.
My question concerns an occurrence in 1957, when Richard Friedberg provided a solution to Post's Problem. First, allow me to transcribe an excerpt from Cohen's talk at the 2006 Gödel centennial:
At that time there was great interest among Raymond [Smullyan] and some other people about the Post Problem. And that’s a problem which could have interested me; it had a mathematical flavor to it. But I never thought about it, and occasionally we’d have coffee and I’d hear these people talk about it. But one day, someone came to my office and said, “This problem’s been solved.” And I said, “Really?” “Yes, here’s the letter. I can’t believe it’s true!” And he gave it to me and I read it. I went to the blackboard, took some chalk, and I said, “Well, it seems right.” This is the proof by Friedberg – and so that was my only contact with logic at that point. But I still never lost this idea of somehow thinking about the foundations of mathematics: trying to find some kind of inductive technique for simplifying propositions; perhaps leading to a decision procedure, when impossible.
This talk is summed up in the introduction to a re-printing of "Set Theory and the Continuum Hypothesis" (Cohen, 2008) in which the remarks corresponding to the above excerpt are:
A small group of students were very interested in Emil Post's problem about maximal degree of unsolvability. I did dally with the thought of working on it, but in the end did not. Suddenly, one day a letter arrived containing a sketch of the solution by Richard Friedberg (Friedberg, 1957), and it was brought to my office. Amidst a certain degree of skepticism, I checked the proof and could find nothing wrong. It was exactly the kind of thing I would like to have done. I mentally resolved that I would not let an opportunity like that pass me again.
I find the last sentence of this latter quotation rather interesting, particularly since it concerns a time five years before Cohen's work on ~CH officially commenced. A quick check of Wikipedia gives a problem statement and solution (i.e., the priority method) that sound remarkably similar, at least on the superficial level, to Cohen's subsequent work with forcing. Unfortunately, work on Turing degrees falls well outside of my bailiwick.
Question: Can someone who specializes in Set Theory or Mathematical Logic comment on the similarities between Post's problem/the priority method and ~CH/Cohen's forcing? In particular, is there reason to believe that what was Cohen's "only contact with logic" by 1957 would have contributed in a meaningful (mathematical) way to his work half a decade later?

References:
Cohen, P. (2002). The discovery of forcing. Rocky Mountain Journal of Mathematics, 32(4).
Kanamori, A. (2008). Cohen and set theory. The Bulletin of Symbolic Logic, 351-378. 
Moore, G. H. (1988). The origins of forcing, Logic Colloquium ’86. Studies in Logic and the Foundations of Mathematics, North-Holland, Amsterdam, 143-173.

REPLY [18 votes]: The primary difference between forcing arguments in set theory and priority constructions is that the latter care about the complexity of the generic filter in a way the former do not. In particular, forcing in set theory involves hitting every dense set in a given model, and so the generic cannot possibly be a member of that model (unless the poset is trivial), whereas if you frame priority constructions as forcing arguments the goal is to hit a fixed countable set of dense sets with a generic which is computable (or close to that) as a set of conditions (the set coded by the generic, on the other hand, won't be computable).
(Let me write this out explicitly, in the case of Friedberg-Muchnik: the poset in question is the set of pairs $(p, \rho)$, where $p$ is the set built so far (i.e., $p\in 2^{<\omega}$) and $\rho$ is the collection of restraints imposed so far (so $\rho$ is a finite subset of $\omega\times(\omega\cup\lbrace -1\rbrace$) with each element of $\omega$ occurring as the left component of at most one element of $\rho$). The poset is ordered in the following way: $(p, \rho)\le(q, \pi)$ if 

$\vert p\vert\ge\vert q\vert$ and $\forall n<\vert q\vert, q(n)\le p(n)$ (this is the c.e. condition - we're not allowed to remove elements from the set we're building); 
for all $(n, m)\in\pi$ with $m>-1$, either $p\upharpoonright m+1=q\upharpoonright m+1$ or for some $k<n$ and $j>-1$ we have $(k, j)\in \rho-\pi$ (this is the missing condition Francois pointed out, which stipulates that restraints can only be violated by the actions of higher-priority requirements); and 
$$ \forall n\in dom(\rho), \rho(n)\not=\pi(n)\implies \exists m < n[(\rho(m)=-1\vee m\not\in dom(\rho))\wedge \pi(m)\downarrow>-1).$$ (This just says that $\rho$ "could occur from $\pi$ by injury.") 

Note that conditions in this poset can be coded by natural numbers, so it makes sense to talk about a set of conditions - i.e., a filter - being computable. Now the requirements in the Friedberg-Muchnik argument can be represented by dense sets (which are not computable) in this poset, two for each $\Phi_e$, and the theorem turns into "there is a computable filter through this poset generic for this collection of dense sets." In fact, I believe the poset described above is "universal" for finite injury arguments, in the sense that each finite injury argument can be dealt with by coming up with an appropriate list of dense subsets of this poset and then arguing that there is a computable filter which is generic for that collection of sets, but I'm not sure about that.)
A possible stronger analogy: priority arguments are like forcing axioms. Forcing axioms say "for such-and-such a poset and collection of dense sets, there is a generic filter already in the model." For example, Martin's Axiom says that for any poset $\mathbb{P}$ with the c.c.c. and any collection $\mathcal{D}$ of $< 2^{\aleph_0}$-many dense sets, there is already a $\mathcal{D}$-generic filter. By analogy, the punchline of a priority argument is often "for this particular poset and these dense sets, there is a computable generic filter." In both cases, we start with the Rasiowa-Sikorski Theorem ("we can hit countably many dense sets") and try to strengthen it: in the set-theoretic case, by enlarging the class of dense sets, and in the computability-theoretic case, by restricting the collection of filters we consider. In fact, this is an analogy I've spent a lot of time thinking about over the course of the last year; it hasn't helped me understand priority arguments, but it has helped me understand forcing axioms.
That said, I do think of priority constructions as a type of forcing; I just realize that this isn't necessarily convincing.

Now let me briefly address the question of what role, if any, Friedberg/Muchnik (or related earlier work by Kleene/Post) played in Cohen's development of forcing. On the one hand, in his paper "The Discovery of Forcing" (which you cite), Cohen describes his process as mostly self-contained, but at one key juncture bolstered by reading Goedel's monograph on $L$. No reference is made to computability theory, and indeed the words "Friedberg," "Muchnik," "Post," "priority," and "recursion" appear nowhere in the article. ("Kleene" appears once, but only in reference to the fact that Kleene's tome on mathematical logic did not include anything especially relevant to Cohen's project.) On the other hand, I seem to recall an article in which Cohen described his picture of forcing as involving an adapting oracle, which he connected to computability theory - which would suggest a real influence. But I can't track down that citation at present; all I can find is a comment by Chad Groft on Terry Tao's blogpost http://terrytao.wordpress.com/2010/03/19/a-computational-perspective-on-set-theory/ that says Cohen explained forcing in this way in his later years. But Cohen may have changed his intuition about forcing over time, so that it is quite possible that Cohen came to view his forcing arguments as related in spirit to recursion theory, while not actually having drawn any inspiration from the subject originally.<|endoftext|>
TITLE: What goes wrong for the Sobolev embeddings at $k=n/p$?
QUESTION [28 upvotes]: For $u\in W^{k,p}(U)$, where $U\subseteq\mathbb{R}^n$ is open and bounded with $C^1$-boundary, we have the celebrated Sobolev inequalities:
If $k < n/p$ then $u\in L^q(U)$ for $q$ satisfying $\frac{1}{q}=\frac{1}{p}-\frac{k}{n}$,
If $k > n/p$ then $u$ lies in a particular Hölder space.  
It is also known that this doesn't work for the borderline case $k=n/p$ (which is related to the Sobolev conjugate $p^*\to \infty$ as $p\to n$), although one would expect/hope for $u\in L^\infty(U)$.
**An exception as Denis mentions: it works for $(k,p)=(n,1)$  via the fundamental theorem of calculus.
**As a counterexample, the function $u(x)=\log\log(1+\frac{1}{|x|})$ with domain $U=int(\mathbb{D}^n)$ lies in $W^{1,n}$ but not in $L^\infty$ (for $n>1$). Instead, apparently the result we end up with is that the functions lie in "the space of functions with bounded mean oscillation".
1) Is there an intuitive / deeper reason as to what goes wrong?
2) Is there some sort of geometric reason why the Sobolev embedding theorem has to fail for $k=n/p$? I've been told that this critical case seems to arise often in geometry/topology.

REPLY [18 votes]: I'll take a stab. In the following we consider the case $W^{1,n}$ in $\mathbb{R}^n$. My short answer is that under rescaling by factor $\lambda$, derivatives scale by $\lambda$ and volumes by $\lambda^{-n}$, so integrating derivatives to the $n$ won't change under rescaling. The following examples illustrate how this affects embeddings. 
As for no Holder continuity, look at a smooth bump function $\varphi$ supported on $B_1$ with $|\nabla \phi| < 2$. The rescalings $\varphi(x/\epsilon)$ have arbitrarily bad modulus of continuity, but bounded $W^{1,n}$ norm, since (key point) the derivative to the $n$ (~$\epsilon^{-n}$) grows exactly like the volume of support (~$\epsilon^{n}$) decays. This says that we cannot control the modulus of continuity by the $W^{1,n}$ norm. (As expected, these functions have unbounded $W^{1,p}$ norm for $p > n$.)
As for not embedding into $L^{\infty}$, the way I would try to see how things could go wrong is take a function $\psi$ positive, supported on $B_2$, with $\psi \equiv 1$ on $B_1$ and $|\nabla \psi| < 2,$ and add dyadic rescalings together. Consider
$$u(x) = \sum_{i} h_i\psi(2^{i}x)$$
for some $h_i$ we will choose to give bounded $W^{1,n}$ norm but unbounded height of $u$. Note that $|\nabla (h_{i}\psi(2^{i}x))|$ grows like $h_i2^{i}$ and they are supported on disjoint dyadic rings of volume going like $2^{-in}$. Thus, to get bounded $W^{1,n}$ norm we want
$$\sum_{i} h_i^{n} < C.$$
Again, the key point is that volume decays with the same power that the derivatives of rescalings to the $n$ grows.
To give unboundedness we just want
$$\sum_{i} h_i = \infty.$$
The canonical example of such a sequence is $h_i = 1/i$. Ultimately this is just the same example as you gave since $\sum_{i=1}^k 1/i$ ~ $\log(k)$ ~ $\log\log(2^k)$ is the size of $u$ at $r = 2^{-k}$, but it shows how this example naturally arises.<|endoftext|>
TITLE: How do you prove that every curve of constant width is convex?
QUESTION [5 upvotes]: Cross-posted from math.stackexchange:
Let C be a simple closed plane curve and let D be its interior. Recall that the width of C in a direction θ is the distance between two supporting lines for D which are perpendicular to θ. A curve is said to have constant width if its width is the same in every direction; see the Reuleaux polygons for nontrivial examples.
It is often stated (for instance in "The Enjoyment of Math" by Rademacher and Toeplitz) that curves of constant width are necessarily convex. Does anyone know how to prove this? I can't find any references, and a proof eludes me.

REPLY [4 votes]: Say that the compact set $K \subset \mathbb{R}^2$ of constant width $d$ is not convex; then there is some support line that touches two points, say $A$ and $B$ on $\partial K$. The parallel line that supports $K$ "from the other side" touches at least one point $C \in \partial K$. The distance between these two lines is the constant width $d$.
But either the distance $\overline{AC}$ or $\overline{BC}$ (or both) is strictly larger than $d$, so the width in the direction orthogonal to that line is larger than $d$ and $K$ has not constant width.<|endoftext|>
TITLE: What do formal group laws of height $\geq 3$ look like?
QUESTION [18 upvotes]: By the classification of formal groups in characteristic $p$, we know that isomorphism classes of connected smooth $1$-dimensional formal groups, equivalently group scheme structures on $\operatorname{Spec} \bar{\mathbb 
F}_p[[x]]$, are in a bijection, defined by the height invariant, with $\mathbb N \cup \infty$ .
I understand the formal groups of height $1$ and $\infty$ quite well - they are the germs of the additive and multiplicative groups. There are extremely simple explicit formulas for the group laws, or rather the group laws of some example.
I understand the formal groups of height $2$ fairly well. They are the germs of supersingular elliptic curves. This gives a procedure to compute the power series, but not a very easy one.
What about formal groups of other heights? Is it possible to give an explicit formula for the coefficients of the power series? How difficult are they to compute? Can the power series be taken to be algebraic functions?

REPLY [23 votes]: A few years ago I computed some formal group laws over ${\mathbb F}_2$ of heights 2, 3, and 4. I've just put the resulting pictures online:

height 2 
height 3
height 4

I find the patterns fascinating because there is a fractal element to them (patterns are repeated at different scales). I also wonder if one can define a limit that captures the large scale look of these pictures. 
As for the mathematical question whether & how higher height formal group laws occur in nature, you might like Jan Stienstra's "Formal group laws arising from algebraic varieties". He computes the formal Brauer group of a K3 surface. If I recall this correctly that can give you formal group laws up to height 10.<|endoftext|>
TITLE: Non-existence of ergodic measures
QUESTION [6 upvotes]: Good afternoon.
Can anybody give me an example of a continuous map $T:X\to X$ defined on a Polish space $X$  which admits an invariant Borel probability measure but no ergodic Borel probability measure? Typically, the space $X$ could be a separable infinite-dimensional Banach space, or the space $\mathbb N^{\mathbb N}$.
Thanks in advance.

REPLY [12 votes]: The existence of such an example is prevented by the ergodic decomposition theorem, which asserts that every $T$-invariant measure on a standard probability space $(X,\mathcal{B},m)$ can be expressed as a (possibly uncountably infinite) convex combination of ergodic $T$-invariant measures by means of an integral over the set of ergodic $T$-invariant measures on $(X,\mathcal{B},m)$. In particular, since the integral must have a nonzero outcome the set of such measures is nonempty.
This theorem is relatively technical to prove and seems to be left unproved or even unstated in standard textbooks on ergodic theory. For example, Walters (p.34) describes the theorem without stating it formally, referring instead to the original work of V. I. Rokhlin (Selected topics in the metric theory of dynamical systems, Uspekhi Mat. Nauk. 4 (1949) 58--127). Petersen (p.81) mentions the result in passing but does not provide a reference. Aaronson's book on infinite ergodic theory gives a proof of the ergodic decomposition theorem for probability spaces in the case where $T$ is invertible (p. 62--64) and sets the case of general $T$ as an exercise; Aaronson attributes the result to von Neumann (who discovered it independently of Rokhlin) and cites Zur Operatorenmethode in der Klassischen Mechanik, Ann. Math. 33 (1932) 587--642.
Terence Tao's weblog has a nice discussion of the result here.<|endoftext|>
TITLE: Is the $n$-th prime $p_n$ expressible as the difference of coprime $A, B$ such that the set of prime divisors of $AB$ is $\{p_1, \dots, p_{n-1}\}$?
QUESTION [11 upvotes]: We define recursively 
$p_1=2,p_2=3$
and 
$$p_{n}= \min_{(A,B)\in F_{n-1}}|A-B| $$
Where
$$
\begin{split}
F_n=\{(A,B) |&\gcd (A,B)=1,\quad |A-B| \not =1,
\\\
&\text{both $A$ and $B$ are products of powers of $p_i$ for $i\le n$},
\\\
&\text{for each $i\le n$, either $p_i |A$ or $p_i |B$}\}
\end{split}
$$
Is always $p_n$ the $n-th$ prime?

REPLY [6 votes]: You can find some amusing (I might say amazing) papers in this area by searching for "primes at a glance" and " primes at a (somewhat lengthy) glance". 
In the first paper they give (along with many other interesting things) what they "believe to be a complete list" of all pairs of integers $B,L$ with

$N=B+L$
$B \ge |L|$
$\gcd(B,L)=1$
if $p \le\sqrt{N}$, then $p$ divides $BL.$
if $Q | BL$ and $p \lt q$ then $p | BL$

For $N \gt 1$ this proves $N$ to be prime as it rules out any proper divisors. Such a presentation provides an at a glance proof that $p$ is prime. For $N=31$ the presentations range from 
$31=2^4+3\cdot5$ 
to 
$31=2^3\cdot7\cdot11\cdot17\cdot23-3\cdot5^2\cdot13^2\cdot19$
The first primes for which they give no solutions are $541,547$
the last for which they do is
$2521=19\cdot43\cdot37\cdot2\cdot3^2\cdot5\cdot29^2\cdot41\cdot47^2-7\cdot11^3\cdot13\cdot17^2\cdot23^2\cdot31$
Without condition 5 there are many solutions. $88711$ is the product of $7,19,23,29$ and $72930$ is the product of $2,3,5,11,13,17$ so we can certainly find positive coprime integers with $1=88711x-72930y.$ Then $31=88711s-72930t$ is a difference of coprime values for $s=31x+72930$ and $t=31y+88711$ You can always do that.But probably not with $st$ having all prime factors below 31 ( in which case the prime factors would split the same way, given that none of them divide $31$.)<|endoftext|>
TITLE: what is Deligne's cohomological descent (and what are some examples)
QUESTION [8 upvotes]: As far as I understand Deligne's far reaching generalisation of Čech cohomology is called cohomological descent and is used to endow any variety with a (mixed) Hodge structure.
Again, AFAIU, the idea is to resolve the variety, then take intersections of the exceptional pieces, resolve those and so on.
As you can see I don't understand it well, so can someone please help? Also, what is the most ridiculously easy example one can have? (I guess the thing I'm interested in most is very simple examples illustrating the possible behaviours)
The only places I know that discuss this are an SGA, Brian Conrad's notes and Peters-Steenbrink.

REPLY [17 votes]: As you said, it is a generalization of Cech theory. The standard example to understand first is a divisor $D=\bigcup D_i$ with simple normal crossings.  A resolution of singularities is obtained by simply taking a disjoint union of components $X_0= \coprod D_i$. Let $\pi_0:X_0\to D$ be the obvious map. The cohomology of $X_0$ with your favourite coefficients will not (usually) be the same as $D$. So you want to correct this by adding in "higher simplicies". Let $X_1$ be the disjoint union  $\coprod D_i\cap D_j$. This has a pair of "face" maps to $X_1\to X_0$. We can continue this process to get a (strict) simplicial object
$$ \ldots X_1\rightrightarrows X_0\to D$$
with an augmentation to $D$. Given a sheaf $F$ on $D$, we can pull it back to a get a collection of sheaves $F_i$ on $X_i$ with various structure maps. We can define the cohomolgy
$H^i(X_\bullet, F_\bullet)$ by taking (for example) by compatible injective resolutions, applying  $\Gamma$, and taking cohomology of the total complex of the resulting double complex. Now the point is that the machinery of descent tells you that
$$H^i(D, F)\cong H^i(X_\bullet, F_\bullet)$$
or if you prefer, there is a spectral sequence relating $H^*(X_p, F_p)$ and $H^*(D,F)$.
This reduces down to a Mayer-Vietoris type sequence 
$$\ldots H^i(D,F)\to H^i(D_1, F)\oplus H^i(D_2,F)\to H^i(D_1\cap D_2, F)\ldots$$
when $D$ has two components. Why is this good? Because for certain things, e.g.constructing mixed Hodge structures, it's better to replace the singular space by a bunch of smooth spaces.

As per request, I'm expanding my comment, although this may be a bit too concise. If $X_\bullet\to X$ and $Y_\bullet\to X$ are two
simplicial resolutions, take the fibre product to get simplicial scheme 
$Z_n =\coprod_{i+j=n}X_i\times_X Y_j$ dominating both. However, it will be singular. You can build simplicial resolution $\tilde Z_\bullet$ of this inductively.
First resolve singularites of $Z_0$ to get $\tilde Z_0$. For the next step, choose a resolution of $Z_1$ which maps to $\tilde Z_0\times_{X}\tilde Z_0$ etc.<|endoftext|>
TITLE: Versions of large cardinals with target model in a generic extension
QUESTION [5 upvotes]: (I ran into this question while thinking about the (Strong) Inner Model Hypothesis: see http://www.jstor.org/stable/4093051 or http://arxiv.org/abs/0711.0680.)
A measurable cardinal is a cardinal $\kappa$ such that there is an elementary embedding $j: V\rightarrow M\subseteq V$ with $M$ an inner model of $V$ and $crit(j):=\min\lbrace \alpha\in ON: j(\alpha)\not=\alpha\rbrace=\kappa$. 
Now we can allow the target model of $j$ to live, not inside $V$, but inside some set-generic extension of $V$ as follows. Say that $\kappa$ is outer-measurable if there is some poset $\mathbb{P}\in V$, some $G$ which is $\mathbb{P}$-generic over $V$, and some transitive inner model $M$ of $V[G]$ such that there is an elementary embedding $$ j: V\rightarrow M\subseteq V[G]$$ with $crit(j)=\kappa$.
In general, given any large cardinal property $(\*)$ defined in terms of elementary embeddings, we can define outer-$(\*)$-ness to be the property $(\*)$ where the target model $M$ is allowed to be an inner model of some set-forcing extension of $V$, rather than $V$ itself. My questions, then, are:

Is there a large cardinal property $(\*)$ such that we can have an outer-$(\*)$ cardinal which is not $(\*)$?
Is there a large cardinal property $(\*)$ such that the consistency strength of an outer-$(\*)$ cardinal is weaker than the consistency strength of a $(\*)$-cardinal?

I suspect that the answer to the second question is "no;" I have no idea about the first question.
[EDIT: Thanks to Joel for pointing out that my now-removed claim that "measurable=outer-measurable" is wrong.]

REPLY [3 votes]: If there is a precipitous ideal $I$ on $\omega_1$---a hypothesis equiconsistent over ZFC with the existence of a measurable cardinal---then after forcing with $P(\omega_1)/I$, we get an elementary embedding $j:V\to M\subset V[G]$ with critical
point $\omega_1$. Thus, on this hypothesis, $\omega_1$ is outer measurable in the sense you have described. This answers question 1 and also shows that your assertion that


ZFC proves that a cardinal is outer-measurable only if it is measurable.


is not correct if these large cardinal hypotheses are consistent. Similar phenomenon arise with analogues of many of the other large cardinal concepts,
such as supercompactness and others.  Matt Foreman has particularly emphasized the richness of these generic large
cardinal concepts, and he has explored this topic extensively. Some of this material is in his chapter in the Handbook of Set Theory.
Perhaps you were thinking of the related assertion, which I believe is correct (but this should be confirmed by inner-model theorists), namely, if $\kappa$ is outer measurable, then there is an inner model where $\kappa$ is measurable. That is, if $\kappa$ is the critical point of a generic embedding, then I believe one can still construct the canonical inner model $L[\mu]$ in which $\kappa$ is measurable. 
Finally, you seem to indicate that you believe that the concept of outer-measurability is not first-order expressible in ZFC.  But I think it is expressible: $\kappa$ is outer measurable if and only if there is a partial order $\mathbb{P}$ forcing that there is a $V$-ultrafilter $\mu$ on $\kappa$ such that $\text{Ult}(V,\mu)$ is well-founded.<|endoftext|>
TITLE: Maximum number of edges in a planar graph
QUESTION [6 upvotes]: What is the maximum number of edges a planar subgraph of $K_n$ can have? Is there a simple way to calculate this if not, are there some values of n for which this is easier to find out?

REPLY [3 votes]: Proposition: If $\Gamma(E,V)$, is a planar graph (no multigraph) then
$|E| \le 3 |V| - 6$. 
Proof:
  Let us note that this does not work for a multigraph where more than one
    edge could be attached to the same two vertices. Imagine a figure (below)
of two vertices and 5 segments attached to the two vertices with no intersections
other that the ends of the segments. This figure 
    has $|E|=5$ and $|V|=2$, while the inequality $5 \le 0$ is false.

Of course $K_n$ is not a multigraph but it is good to be aware of counter-examples.
We first assume that the faces are all triangles and show the inequality 
    \begin{equation}
  2|E| \ge 3|F|   \quad \quad  (1)
\end{equation}
    For example for one face $|E|=3$ and
    $F=2$ so the equality $6=6$ is achieved. However for two faces
    (for example a rectangle with a diagonal) we have  $|E|=5$ and
    $|F|=3$, here the inequality $10 > 9$ is strict.
    We do this by induction over $|F|$ . Let us introduce a new face $|F|$ by
adding one more vertex and two more edges on the boundary of the graph (note that since all faces are triangular we can not add faces inside the graph). We need to show
    that $2| |E| +2| \ge 3| |F|+1|$. 
    \begin{equation}
  2 | |E| + 2| = 2 | E| + 4 \ge 3|F|  + 4 \ge 3|F| + 3 \ge 3| |F|+1|.
\end{equation}
Now we use Eulers equation $|V|-|E|+|F|=2$, from which $|F|=2+|E|-|V|$ and
    replacing $|F|$ in equation (1)
    \begin{equation}
  2|E| \ge 3 (2 + |E| - |V|)    
\end{equation}
    then
    \begin{equation}
  |E| \le 3|V| - 6.
\end{equation} 
Now what happens if a face is not a triangle. We need to add edges until
    making it a triangle, use equation $|E'| \le 3|V'| -6$ which is valid
    for triangles then remove the edges and find that for the new graph
    $|E| \le 3|V| - 6$ is a valid inequality. After adding edges to make all faces 
    triangles we have $|E'| \le 3|V'| -6$ where $|E'|$ and $|V'|$ are the number of edges and
    vertices of the new triangulated graph. When we remove one edge 
    which is common to two triangular faces, we end up with a quadrilateral.
    The graph has one less edge without removing any vertex. In general,
    we remove edges but no vertices to go from the triangulated to the original
  graph, so $|E| < |E'|$ and $|V|= |V'|$. That is,
  \begin{equation}
  |E| < |E'| \le 3 | V'| - 6 = 3 |V| - 6,
  \end{equation}
  from which the proposition is shown.
Now, a $K_n$ graph has $n$ vertices so, $|E| \le 3n - 6$.
However the number of edges of $K_n$ can be exactly counted. Put the vertices
in a unit circle equally spaced. That is, on the $n$ complex roots of the
equation $z^n -1 = 0$. Each vertex can visit exactly $n-1$ other vertices, 
for a total of $n(n-1)$. But each edge was counted twice (from $v_i$ to $v_j$ and
from $v_j$ to $v_i$) so the exact maximum is $n(n-1)/2$.
It is interesting that this serves to show the inequality $3n-6 \le n(n-1)/2$. . Use gnuplot or any graph tool to illustrate this. Of course the only solution is for $n=3$, where we have a single triangle. After that the inequality is strict.<|endoftext|>
TITLE: Periodic point-free maps and free ultrafilters.
QUESTION [6 upvotes]: Let $X$ be a set and $u$ be a free ultrafilter on $X$. We can consider a topology on $X$ by declaring every element of $u \cup \{\emptyset \}$ to be open. 
El'kin's original motivation for looking at this topology is that $X$ is a space without isolated points which can't be split into disjoint dense subspaces. Another quirky feature: if $X$ is say $\mathbb{R}$ then translations are not continuous. This last fact makes me wonder: are there any periodic-point free continuous self maps at all on $X$ with this topology? In other words:

Is there a free ultrafilter $u$ on a set $X$ and a periodic point-free onto map $f: X \to X$ such that: $f^{-1}(A) \in u$, for every $A \in u$? Is there at least one such fixed-point free map?

REPLY [8 votes]: Although the question has already been thoroughly answered, it might be worthwhile to point out that the result can be separated into the combinatorial "meat", which doesn't involve ultrafilters, and a small corollary where the meat is fed to the ultrafilter problem.  The meat is the following theorem, which is, if I remember correctly, explicit in the early references; it is essentially proved (though not stated) in Joel Hamkins's answer here.  Given any set $X$ and any function $f:X\to X$, there is a partition of $X$ into four disjoint (possibly empty) sets $X=A_0\sqcup A_1\sqcup A_2\sqcup A_3$ such that $A_0$ is the set of fixed points of $f$ and each of the other three $A_i$'s is disjoint from its image under $f$.  Once one has this result, one immediately sees that any ultrafilter on $X$ must contain one of the four $A_i$'s and if that $A_i$ isn't $A_0$ then the ultrafilter can't be $f$-invariant as in the question.<|endoftext|>
TITLE: Do eigenfunctions of elliptic operator form basis of $H^k(M)$?
QUESTION [10 upvotes]: We know that the eigenfunctions of the Laplacian on a compact manifold $M$ form a countable basis of $H^1(M)$ and $L^2(M)$. 
If $L$ is a $2k$-order elliptic operator, do the eigenfunctions of $L$ form a basis for $H^k(M)$? References/more detail would be appreciated. Thanks.
(Crossposted from https://math.stackexchange.com/questions/324777/do-eigenfunctions-of-elliptic-operator-form-basis-of-hkm due to lack of replies)

REPLY [14 votes]: The answer is no, if the Fredholm index of $L$, which is the integer
$\mathrm{ind}\, L=\dim\mathrm{ker}\, L- \dim \mathrm{ker} \, L^\ast$,
is negative. (Regard $L$ as an unbounded operator on $L^2(M)$ with domain $D(L)=H^{2k}(M)$.) A proof by contradiction goes as follows. If the eigenvectors formed a basis of $H^k(M)$, then their span would also be dense in $L^2(M)$. The elements of $\mathrm{ker}\, L^\ast$, i.e. of the null space of the adjoint operator, are orthogonal to the range of $L$, which contains the eigenvectors with non-zero eigenvalues. The assumption $\mathrm{ind}\, L<0$ implies that there exists a non-zero element of $\mathrm{ker}\, L^\ast$ orthogonal to all eigenvectors. Thus we obtain the contradiction that the span of the eigenvectors is not dense in $L^2(M)$.
I have no explicit example of a scalar elliptic operator of even order with negative index. Because of $\mathrm{ind}\, L^\ast= -\mathrm{ind}\, L$ it would suffice to find an operator with non-zero index. By the Atiyah-Singer index theorem, this is a topological condition on the manifold $M$ and on the principal symbol of the operator $L$.
You received the answer "yes" to your question on SE under the additional assumption that $L$ is self-adjoint and coercive. It was explained there how this follows using the spectral theorem and using, what I am accustomed to call, Gelfand triplets. As a reference for the latter I mention the book: Wloka, Partial Differential Equations. Note that self-adjointness implies that the index is zero.<|endoftext|>
TITLE: How long should one wait for a report before asking about its status?
QUESTION [8 upvotes]: I apologize if this question is too soft or if its answers would be too subjective for this site. However, I would find it highly useful to have such a question answered on this site, and I believe that others might feel this way, too.
I have certainly found a related question (see 13 months and not even one report. what would you do?) quite helpful.
My question is essentially already stated in the title. When submitting a paper, there is obvisouly some time that one simply has to wait patiently. Although I have experienced this a few times by now, I am still completely clueless about the etiquette in this case.
After how many months (or years?) do you think it is appropriate to write to the editor and kindly ask about the status of the paper?

REPLY [10 votes]: As others have mentioned, possible parameters are too wide for a one-size-fits-all answer. But here are some data points:

My personal default is to wait six months before inquiring into a paper's status. In other words, I have decided that not hearing anything in six months is pretty much never a surprising turn of events to me.
For a long paper (say over 40 pages), I might extend this somewhat, but probably not longer than nine months. For a very short paper (say under 8 pages), I might inquire at four months; but knowing me, I'd probably just wait until six months anyway.
Surely these durations can be decreased if one is in the position where a paper accepted or not will have a significant impact upon one's career. I think it's reasonable to write after three months, politely ask about the status of the paper, and add that you're going on the job market at time X and would be very grateful to know the decision on the paper by then.
Above all else, word your inquiry with the understanding that both the editors and referees are volunteering their time. Authors have the right to a timely evaluation of their paper (that comes with the agreement not to submit elsewhere while it's being evaluated), and sometimes the editors/referees just need a gentle nudge to notice that it's been a long time. But (needless to say) implying malice, laziness, or incompetence is extremely unlikely to make anything better. (Even if it's the third time around asking about a paper, when those nouns probably do in fact apply!)<|endoftext|>
TITLE: Applications of n-dimensional crystallographic groups
QUESTION [9 upvotes]: I would like to know what are the applications of the theory of $n$-dimensional crystallographic groups (aka space groups) 
1) in mathematics
2) outside of mathematics,
besides the applications to $2$-dimensional and $3$-dimensional crystallography (or related fields like chemistry, or physics, of crystals).
One possible application of $4$-dimensional space groups is already reported in the wikipedia article I linked to (see "Magnetic groups and time reversal").

REPLY [5 votes]: The following are applications in the theory of $p$-groups: 
Space groups have been used by 

Felsch, Neubüser, Plesken: Space groups and groups of prime-power order. IV: Counterexamples to the class-breadth conjecture. Journal London Math. Soc. (2), 24 (1981) 113-122 

to construct counterexamples to the class-breadth conjecture for $p=2$. Recall the the conjecture claims $\text{class} \le \text{breath} + 1$ for $p$-groups $P$ where the breath $b$ is defined such that $p^b$ is the maximal size of the conjugacy classes of $P$. In their counterexamples $P=S/2^kT$ where $S$ is a space group, $T$ the translation subgroup and $k$ a carefully choosen integer. 
Space groups those point groups are $p$-groups are also the core in proving the celebrated coclass conjectures of Leedham-Green and Newman (see the book Leedham-Green, McKay: The structure of groups of prime power order, 2002). I don't know enough to tell details, but it's striking that the series of papers that contain the proof are titled "Space groups and groups of prime-power order" (I-VIII).<|endoftext|>
TITLE: When does End(M) consist entirely of zero, zero divisors, and units?
QUESTION [5 upvotes]: Let $R$ be a commutative ring (with $1$) such that every non-zero divisor in $R$ is a unit (see Rings in which every non-unit is a zero divisor for various stabs at what these are called). Let $M$ be a finitely generated $R$-module. 
My question: 

Can we conclude that every non-zero divisor in $\mathrm{End}(M)$ is a unit?

For example: When $M$ is a free module, we have $\mathrm{End}(M) \cong R^{n \times n}$ is a matrix algebra and so every non-zero divisor is a unit (see Do these matrix rings have non-zero elements that are neither units nor zero divisors?).
A few notes:

Even matrix algebras can misbehave when $R$ is non-commutative. Also, if $R$ has a non-zero divisor which is not a unit, then $\mathrm{End}(R)=R^{1 \times 1}=R$ has a non-zero divisor which is not a unit (thus the assumptions on $R$).
If $V$ is an infinite dimensional vector space (over some field). Then $\mathrm{End}(V)$ has non-zero divisors which aren't units (thus the "finitely generated" assumption).
If the answer to the question is "No" (I'd like a counter-example), then I would like to know under what circumstances $\mathrm{End}(M)$ does have this property. What assumptions on the module will force this to hold? (e.g. This holds when $M$ is a ??? module -- like flat or projective or something.) Or what assumptions on the ring will force this to hold for all modules? (e.g. This obviously holds when $R$ is a field.)
Motivation and background: I've been working with a few undergraduates on a related problem. We tripped over this question and I have no idea whether this is true or not (I'm not a ring theorist and have a feeling this may be a difficult question to entirely resolve). 

Thanks for reading my question! We (my group of undergrads and myself) appreciate your help!

REPLY [7 votes]: Counterexample: $R=k[[x,y]]/(x^2,xy)$ and $M = R/(x) \cong k[[y]]$.

REPLY [7 votes]: Here is a small positive result: If $R$ is Artinian, $End_R(M)$ is an $R$-algebra which is f.g. as $R$-module, hence Artinian, hence classical (as I call these rings, following Lam).
On the other hand, as soon as the Krull dimension of $R$ is $\geq 1$, $M = R/p$ for a non-maximal prime ideal $p$ will give a counterexample, as in Graham's answer.
Added: I think if $R$ is Noetherian (hence semilocal by a theorem of Faith, see Lam's Lectures on Modules and Rings theorem 8.31 p. 283), the assertion is true for $M$ f.g. projective (= f.g. flat).
Because then $R \simeq \prod_{i=1}^{n} R_i$ with indecomposable semilocal $R_i$ with corresponding idempotents $e_i$, $End_R(M) \simeq \prod End_{R_i}(e_i M)$ and $e_iM$ is f.g. projective over $R_i$. We can thus reduce to the indecomposable case, and here f.g. projectives are free (this remains true without "f.g.", says Hinohara), so we are done by the OP's second link.<|endoftext|>
TITLE: zero-dimensional completely regular space with $\sigma$-complete clopen algebra
QUESTION [6 upvotes]: Suppose $X$ is a zero-dimensional completely regular space (clopen sets form a base) such that the Boolean algebra of clopen sets is a $\sigma$-complete Boolean algebra.  Must $X$ be basically disconnected?  That is, must every cozero set in $X$ have open closure?
I thought this was known, but I can't find it. The point is to show that such $X$ is strongly zero-dimensional, i.e., $\beta(X)$ is zero-dimensional, so that every cozero set is a countable union of clopens.
It should be noted that the analogous statement for extremally disconnected spaces is true: A zero-dimensional space is extremally disconnected (open sets have open closures) if and only if the Boolean algebra of all its clopen sets is a complete Boolean algebra. The proof is an easy exercise.

REPLY [5 votes]: The answer to the posted question is probably no.  Thanks to the above comment by @Joseph Van Name, the claim "yes" is posed as problem 6K in the Porter-Woods book.  Grant Woods has told me in a recent email that this matter came to his attention some time ago, and he concluded that the problem statement in the book should have stated "strongly zero-dimensional" in the assumption. 
A complete answer to the posted question will require a counterexample.  This may not be so well-known.  This will require a zero-dimensional space which is not strongly zero-dimensional, whose Boolean algebra of all clopen sets is $\sigma$-complete.  Perhaps a modification of the Dowker construction would work (see Porter-Woods Problem 4V, or Gillman-Jerison Problem 16M, or Engelking Example 6.2.20).<|endoftext|>
TITLE: A counterexample to a conjecture of Nash-Williams about hamiltonicity of digraphs?
QUESTION [16 upvotes]: Maybe I am missing something, but found potential counterexample to a conjecture
of Nash-Williams.
According to HAMILTONIAN DEGREE SEQUENCES IN DIGRAPHS

The outdegree and indegree sequences of digraph $G$ are
$d_1^+ \le \cdots \le d_n^+$ and $d_1^- \le \cdots \le d_n^-$.
Note that the terms $d_i^+$ and $d_i^-$ do not necessarily corresponds
to the degree of the same vertex of $G$.
Conjecture 1 (Nash-Williams). Suppose that $G$ is a strongly connected digraph
on $n \ge 3$ vertices such that for all $i < n/2$
(i) $d_i^+ \ge i + 1$ or $d_{n-i}^- \ge n - i$,
(ii) $d_i^- \ge i + 1$ or $d_{n-i}^+ \ge n - i$,
Then $G$ contains a Hamilton cycle.
The potential counterexample is $G$ on $6$ vertices with edges:
[(0, 3), (0, 5), (1, 4), (1, 5), (2, 3), (2, 4), (3, 0), (3, 2), (3, 4), (3, 5), (4, 0), (4, 1), (4, 3), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4)]

$G$ is strongly connected and by inspection the degree sequences satisfy the hypotheses
for $i \in [1,2]$ (both degree sequences are $[2, 2, 2, 4, 4, 4]$).    
Nonhamitlonicity was shown using exhaustive search, sage 5.6 and Max Alekseyev's
hamiltonian cycle counting pari program.    

Is this really a counterexample to the conjecture of Nash-Williams?

Drawing of $G$:

REPLY [4 votes]: I realize this question was asked seven years ago and hasn't had a comment in four years, but I just came across it and thought it might be worth sharing what I've learned.
As @HughThomas mentions, since $\{0,1,2\}$ is an independent set, the question boils down to whether the bipartite digraph between $\{0,1,2\}$ and $\{3,4,5\}$ has a Hamiltonian cycle.  I wondered whether your example could be generalized for all $n=4k+2$.  To generalize your example, we take sets $X=\{x_1, x_2, \dots, x_{2k+1}\}$ and $Y=\{y_1, y_2, \dots, y_{2k+1}\}$.  We make $X$ an independent set and add all possible edges inside $Y$.  The idea is to come up with a bipartite digraph $D$ between $X$ and $Y$ such that every vertex has indegree and outdegree at least $k+1$, but $D$ has no Hamiltonian cycle (as this would give a digraph having no Hamiltonian cycle with a degree sequence $[k+1,k+1,\dots,k+1,3k+1,3k+1,\dots,3k+1]$ satisfying Nash-Williams condition).
As it turns out D. Amar and Y. Manoussakis (see Theorem 1.7 and Fig. 1 in On the Meyniel condition for Hamiltonicity in bipartite digraphs by J. Adamus, L. Adamus, A. Yeo) proved that if $D$ is a bipartite digraph with $2k+1$ vertices in each part and every vertex has indegree and outdegree at least $k+1$, then $D$ has a Hamiltonian cycle unless $D$ is exactly your digraph on 6 vertices! (that is, the important part between the sets $\{0,1,2\}$ and $\{3,4,5\}$)
Note that if $n=4k$ you can create a bipartite digraph between $X=\{x_1, x_2, \dots, x_{2k}\}$ and $Y=\{y_1, y_2, \dots, y_{2k}\}$ in which every vertex has indegree and outdegree at least $k$ and there is no Hamiltonian cycle, but the resulting degree sequence will be $[k,k,\dots,k,3k-1,3k-1,\dots,3k-1]$ which just barely fails the Nash-Williams condition.
So my guess is that the conjecture is safe for all $n\neq 6$.<|endoftext|>
TITLE: Trace Class Functions on locally compact groups
QUESTION [12 upvotes]: Let $G$ be a locally compact subgroup, $\mu$ a Haar-measure.
For $f \in L^1(G)$, and for $\pi$ a unitary, topology irreducible, representation of $G$ on 
an Hilbert space $H_\pi$, it is customary to define a continuous operator $\pi(f)$ on $H_\pi$ by $$\pi(f) v = \int_G f(g) \pi(g)(v) d\mu(g).$$
Let us say here that $f \in L^1(G)$ is of trace class if for every  unitary, absolutely irreducible representation $\pi$ of $G$, the operator $\pi(f)$ is of trace class. 
(side question: is there a name outside for this notion?). 
I have many question about this notions, but let me give one:

Is it true that the space of trace class functions on $G$ is dense in $L^1(G)$? If not, 
  is it true for a separable type I locally compact group ? 

I know of certain larges classes of group for which it is true. Compact group and abelian locally compact groups are trivial examples, since all irreducible $\pi$'s are finite dimensional. Real and $p$-adic Lie group are other examples since smooth compact support functions are trace class (IIRC) with the usual meaning of smooth ($C^\infty$ in the real case, locally constant in the $p$-adic case), and are dense. Yet this is proved in a rather ad hoc way, for example in the real case (Duflo-Labesse) 
by showing that such a function $f$ is the convolution of
two such functions $f_1$ and $f_2$, so that $\pi(f)=\pi(f_1)\pi(f_2)$ and $\pi(f_1), \pi(f_2)$ are Hilbert-Schmiddt. Also the adelic case (necessary for the global trace formula) 
follows from the real and $p$-adic case. 
Yet I don't know if this is true in general, or at least for a large abstract class of locally compact group (defined with a property such as type I, not as a list of example such as "real Lie groups"), with if possible a uniform proof.
If the question is too difficult, or the answer is no, does that help if
we ask the same question for functions that are reduced trace class, i.e. such that $\pi(f)$ is of trace class for all irreducible $\pi$'s in the reduced spectrum (i.e. in the support of the regular representation). 
PS: Clearly, the condition of being a trace class does not change if we replace $L^1(G)$ by $C^\ast(G)$ with its natural norm. Hence one of the tag.

REPLY [5 votes]: The following is a suggestion how to prove a weak variant of the OP's original question affirmative.

Theorem: Let $G$ be a unimodular, seperable, type-I group.
  For every element $\phi$ in $C_c^\infty(G)$ (in the sense of Bruhat), the operator $\pi(\phi)$ is trace class for almost every unitary representation $\pi$ (with respect to the Plancherel measure).

Proof: The Plancherel theorem states for $\phi \in C_c^\infty(G)$, that 
$\pi( \phi \ast \phi^{\ast})$ 
is a trace class operator for a.e. $\pi$. The Dixmier-Malliavin theorem allows us to write general $f$ as a sum of convolution product $S \ast T^*$. A convolution product is the linear combination of four positive functions (polarisation identity), i.e., 
$$4 \cdot S \ast T^\ast = (S+T) \ast (S+T)^\ast - (S-T) \ast (S-T)^\ast$$
$$ - i(S-iT) \ast (S-iT)^* + i(S+iT) \ast (S+iT)^*.$$
Applying that  $ \pi ( (S+T) \ast (S+T)^\ast ), \dots$ are trace class by Plancherel for a.e. $\pi$ yields the theorem. Q.e.d.
The following lemama should follow rather easily from the equivalent result of Lie groups:

Lemma: $C_c^\infty(G) \subset L^1(G)$ is a dense embedding. 

Discussion of Dixmier-Malliavin:
Every smooth, compactly supported functions is a finite sum of convolution products. This is due to Dixmier and Malliavin, originally proven for a Lie groups only. 
Writing general $G$ as an projective limit of Lie groups, we obtain a notion of smooth, compactly supported functions on $G$. So the proof theorem Dixmier-Malliavin extends to locally compact groups. See also my Phd thesis http://webdoc.sub.gwdg.de/diss/2012/palm/palm.pdf, where I discuss this in chapter 5, in particular Theorem 5.2.1. Approximation via Lie groups is explained in chapter 4.
Discussion of the Plancherel theorem: A necessity of the Plancherel theorem (check e.g. http://www.encyclopediaofmath.org/index.php/Unitary_representation) is that the $tr\; \pi (f\ast f^*)$ is well-defined for every $f \in L^1(G) \cap L^2(G)$ and a.e. $\pi$ with respect to the Plancherel measure $\mu$. 
The Plancherel theorem states then for $G$ be a unimodular separable locally compact group of type I
$$ \int\limits_G  |f(g)|^2 d g = \int\limits_{unitary \; dual} tr\; \pi( f \ast f^*) d \mu(\pi).$$
There exist extensions, which drop unimodular, type I or seperable as stated in the above link, but don't know how the extensions look like, so that's why I worked with rather special $G$ here.
You can furthermore put a (Fell-)topology on the unitary dual and so on, but I don't know wheter $\pi \mapsto \pi(\phi)$ will then be a continuous function!? Even a smooth structure via the approximation-by-lie-groups-business. 
Using Dixmier-Malliavin plus the polarization identity you obtain for $f \in C_c^\infty(G)$  (Theorem 6.2.6, pg.83 in my thesis) the following variant of the Plancherel theorem
$$ f(1) = \int\limits_{unitary \; dual} tr\; \pi( f) d \mu(\pi).$$
For non-type-I groups, the Plancherel measure will not be unique (?), so I am sceptical about a direct generalization omitting that notion. I think that neither being unimodular nor seperable is a crucial hypothesis.
Discussion of reductive groups:
For reductive groups over local fields, you can replace a.e. w.r.t. to Plancherel measure by every smooth, admissible representation, but I guess you are well aware of that. Some reps are not tempered and do not contribute to the Plancherel formula, so this far more. The notion of smooth in the sense I indicate above coincides with the usual one in the non-archimedean world/totally disconnected groups. Be careful, general Lie groups are not type I. For the adelic points of a reductive group over a global field, you need an argument along the lines of Is a reductive adelic group a Type I group?, but the situation is similar to that of reductive local groups.<|endoftext|>
TITLE: B(H) as a direct sum of a closed two sided ideal and a subalgebra
QUESTION [10 upvotes]: Let $B(H)$ is the C*-algebra of all bounded linear operators on 
Hilbert space $H$. Are there a closed two-sided ideal $I$ and a 
subalgebra $A$ of $B(H)$ such that $B(H)=I \oplus A$ (direct sum I and A)?

REPLY [11 votes]: Let me complement Bill's answer. It is basically the same idea. 
Gramsch and Luft described the lattice of closed ideals of $B(H)$ where is $H$ a non-separable Hilbert space.

B. Gramsch, Eine Idealstruktur Banachscher Operatoralgebren, J. Reine Angew. Math. 225 (1967),
  97–115.
E. Luft, The two-sided closed ideals of the algebra of bounded linear operators of a Hilbert space, Czechoslovak Math J. 18 (1968), 595–605. 

They proved that for a non-separable Hilbert space $H$ with density character ${\rm dens}\; H$ (this is the minimal cardinality of a dense subset) all the closed ideals of $B(H)$ are of the form $\{0\}$, $K(H)$ (the compact operators) and  
$$K_\lambda(H) = \{T\in B(H)\colon \mbox{dens }T[H]< \lambda\}$$
where $\lambda\leqslant\kappa^+$. Certainly, $$B(H) = K_{({\rm dens}\; H)^+}(H).$$ 
Hence, we are interested in the case only where $\lambda\leqslant  {\rm dens}\; H$. 
Fix an orthonormal basis for $H$ and identify operators on $H$ with matrices with respect to this basis. Consider the Banach space $\ell_\infty^\lambda({\rm dens}\; H)$ of all bounded complex-valued functions on the cardinal ${\rm dens}\; H$ with at most $<\lambda$ non-zero entries, endowed with the sup-norm. 
If you intersect the ideal $K_\lambda(H)$ with the diagonal masa (that is, $\ell_\infty({\rm dens}\; H)$) then you'll get precisely $\ell_\infty^\lambda({\rm dens}\; H)$. The diagonal masa is complemented because $\ell_\infty({\rm dens}\; H)$ is an injective Banach space. Consequently, if $K_\lambda(H)$ was complemented in $B(H)$ then $\ell_\infty^\lambda({\rm dens}\; H)$ would be complemented in $\ell_\infty({\rm dens}\; H)$. This is however a contradiction because $\ell_\infty^\lambda({\rm dens}\; H)$ is not injective (see this paper for a discussion of this space and its injectivity-like properties).
Proof of non-injectivity of $\ell_\infty^\lambda(\kappa)$: Assume $\ell_\infty^\lambda(\kappa)$ is injective. Manifestly, it contains $c_0(\kappa)$. Then by Rosenthal's theorem 

H.P. Rosenthal, On relatively disjoint families of measures, with some applications
  to Banach space theory, Studia Math., 37 (1970), 13–36.

it would contain a copy of $\ell_\infty(\kappa)$ and by Pełczyński's decomposition method, it would be isomorphic to $\ell_\infty(\kappa)$; a contradiction.
EDIT 05.05.15: This is actually an ancient observation by Pełczyński and Sudakov:

A. Pełczyński and V. N. Sudakov, Remark on non-complemented subspaces of the space $m(S)$, Colloq. Math. 19 (1962), 85-88.

EDIT: As Bill pointed out we have to be careful why this is indeed impossible. Under GCH this is evident but life would be too easy with GCH. Actually, to see this is a result in ZFC, one has to tweak the argument in Paragraph d) on page 12 of the hyperlinked paper by replacing $\mathbb{N}$ with $\lambda$ and $\aleph_1$ with $\lambda$. Then the whole proof carries over. (Frankly, I learnt it from one of the authors of this paper some time ago and presumed that it is well-known. My apologies for that.)<|endoftext|>
TITLE: Mayer-Vietoris sequence in homology with local coefficients
QUESTION [6 upvotes]: Background. I'm trying to compute some homology groups using a Mayer-Vietoris argument, but I really need local coefficients.
Question 1. What does the Mayer-Vietoris sequence look like when using local coefficients?
Consider an open cover $X = U \cup V$ with inclusion maps
$$\begin{array}{ccccc}
& & U \cap V & &\\
& i \swarrow & & \searrow j &\\
U & & & & V\\
& k \searrow & & \swarrow l &\\
& & X & &\\
\end{array}$$
and a coefficient module $M$ on $X$. (Assume all four spaces are path-connected if needed.) I believe that the Mayer-Vietoris sequence takes the form
$$\ldots \to H_n(U \cap V; (ki)^*M) \to H_n(U;k^* M) \oplus H_n(V;l^*M) \to H_n(X;M) \to$$
$$\to H_{n-1}(U \cap V; (ki)^*M) \to \ldots$$
where $k^*M$ denotes the restriction of $M$ to $U$ along the inclusion $k \colon U \to X$. Is that correct?
Question 2. Are there good references for homology with local coefficients, and in particular the Mayer-Vietoris sequence in that context?
Sections 5.3 and 5.4 of Lecture Notes in Algebraic Topology by J. Davis and P. Kirk are a good start, especially Theorem 5.13 and the remark afterwards.
Question 3. Are there good references that treat local coefficients as functors from the fundamental groupoid $\Pi_1(X) \to Ab$ and describe homology with local coefficients in that context?
I wouldn't mind reducing the problem to the case of path-connected spaces, but I feel like the argument would be cleaner without such reductions or choices of basepoints.

REPLY [7 votes]: Whitehead's "Elements of Homotopy Theory", in particular chapter VI, seems to have everything you ask for. (Note that the Mayer-Vietoris sequence is a formal consequence of excision and the long exact sequence of a pair; see section 2.3 of Hatcher's book).<|endoftext|>
TITLE: Characteristic subgroup of nilpotent group that is not invariant under powering
QUESTION [13 upvotes]: I want an example of a nilpotent group $G$, a characteristic subgroup $H$, and a prime number $p$ such that:

$G$ is $p$-powered, i.e., every element of $G$ has a unique $p^{th}$ root in $G$.
$H$ is not $p$-powered. In this case, it would mean that there exists an element of $H$ that does not have any $p^{th}$ root in $H$.

Note that there do not exist any abelian examples, because in an abelian $p$-powered group, the $p^{th}$ root map is an automorphism and hence preserves every characteristic subgroup.
The intuitive reasoning for why there shouldn't be any examples: even though the $p^{th}$ root map is not an automorphism, it seems possible generally to find automorphisms that behave a lot like this map to a sufficient extent that characteristic subgroups must be preserved under taking $p^{th}$ roots. 
The intuitive reasoning for why there may well be examples: nilpotent groups can in principle rigidify a lot of powering structure.
I don't know which side to believe.
PS: There are certainly characteristic subgroups of non-nilpotent groups with this property. For instance, let $G = GA^+(1,\mathbb{R})$ (the identity component of the affine group of degree one over the reals, i.e., maps of the form $x \mapsto ax + b, a > 0, a,b \in \mathbb{R}$), and let $H$ be the subgroup comprising those elements where $a$ is rational. $H$ is characteristic in $G$, $G$ is powered over all primes, and $H$ is not powered over any prime.

REPLY [5 votes]: There exists such example (thus also disproving Conjecture 4.1.28 here).
Indeed there exists a nonzero nilpotent Lie algebra $\mathfrak{g}$ with rational coefficients whose automorphism group $A$ is unipotent (references (1,2) below). Therefore $\mathfrak{g}$ admits an $A$-invariant $\mathbf{Q}$-defined proper subalgebra $\mathfrak{l}$ such that the induced $A$-action on $\mathfrak{g}/\mathfrak{h}$ is trivial. 
Let $G$ is the corresponding unipotent $\mathbf{Q}$-group ($G_K$ can be defined by endowing $\mathfrak{g}_K$ with the group law defined by the Baker-Campbell-Hausdorff formula) and $L_\mathbf{Q}$ the corresponding subgroup, so that $\mathrm{Aut}(G_\mathbf{Q})=A_{\mathbf{Q}}$. Then $L_\mathbf{Q}$ is a characteristic subgroup of $G_\mathbf{Q}$ and the action of $\mathrm{Aut}(G_\mathbf{Q})$ on $G_\mathbf{Q}/L_\mathbf{Q}$ is trivial. It follows that every subgroup of $G_\mathbf{Q}$ containing $L_\mathbf{Q}$ is also a characteristic subgroup. So, if we define $H\subset G_\mathbf{Q}$ as the inverse image of an infinite cyclic subgroup of $G_\mathbf{Q}/L_\mathbf{Q}$, then $H$ is not $p$-divisible (hence not $p$-powered) for any prime $p$, while $G_\mathbf{Q}$ is $p$-powered for all primes $p$. 
For those who rather like Lie groups: if we perform the same construction in the real case, we get $H\subset G_\mathbf{R}$, where $G_\mathbf{R}$ is a simply connected nilpotent Lie group, $H$ is a closed proper subgroup closed under the whole group of continuous group automorphisms of $G_\mathbf{R}$.

References
((1)) (link, MR link) JL Dyer, A nilpotent Lie algebra with nilpotent automorphism group, Bulletin of the
American Mathematical Society 76 (1970) 52–56
((2)) (MR link) Favre, Gabriel
Une algèbre de Lie caractéristiquement nilpotente de dimension 7. (French) 
C. R. Acad. Sci. Paris Sér. A-B 274 (1972)<|endoftext|>
TITLE: Quotient of trivial bundles
QUESTION [8 upvotes]: Suppose you are on a manifold. Suppose you have a trivial bundle and a trivial subbundle of it. If you divide this trivial bundle with its trivial subbundle, do you get a trivial bundle as a quotient? The answer is generally no. What are the conditions on the manifold (s.c. etc.) and bundles (codim e.g.) to get a trivial quotient?

REPLY [6 votes]: If the dimension of the base $X$ of the bundles is less than the difference $n-k$ of the fiber dimensions then the quotient bundle is trivial. To see this it is enough to consider the case $k=1$ and then induct. An embedding of a trivial rank one bundle in a trivial rank $n$ bundle amounts to a map $X\to S^{n-1}$. If the dimension of $X$ is less than $n-1$ then such a map is nullhomotopic.<|endoftext|>
TITLE: Baer's criterion for projective modules
QUESTION [8 upvotes]: Let $R$ be a commutative ring. If necessary, assume that $R$ has any convenient properties you like.

Is there some $R$-module $Q$ such that an $R$-module $P$ is projective if and only if $\hom_R(P,Q) \to \hom_R(P,Q')$ is surjective for all quotients $Q'$ of $Q$?
If yes, can $Q$ be chosen as a cogenerator?
What happens when we restrict to finitely generated $P$?

Neither $Q=R$ nor $Q=\hom_{\mathbb{Z}}(R,\mathbb{Q}/\mathbb{Z})$ work. This question is a follow-up of math.SE/325495.

REPLY [6 votes]: I think the paper "Whitehead Test Modules" by Jan Trlifaj (Trans. AMS 348 (1996), 1521-1554) and the references in it, answer your question positively for perfect rings, but show that a negative answer is consistent with ZFC+GCH if $R$ is not perfect.
In Lemma 2.4 he seems to prove that (if $R$ is not perfect) then for any cardinal $\kappa$, it is consistent for there to be a non-projective module $M$ such that $\operatorname{Ext}^1(M,N)=0$ for all modules $N$ of cardinality less than $\kappa$, which contradicts the property you want $Q$ to have if $\kappa$ is larger than the cardinality of $Q$. But I'm not a set theorist, and may be misunderstanding something.
For the restriction to finitely generated $P$, I assume you want $Q$ also to be finitely generated? Otherwise you could just take $Q$ to be a free module of infinite rank.<|endoftext|>
TITLE: $\prod_{n=1}^{\infty} n^{\mu(n)}=\frac{1}{4 \pi ^2}$
QUESTION [36 upvotes]: When I tested this in Mathematica, I had expected it to say it did not converge.  However, I got this:  
$$\prod_{n=1}^\infty n^{\mu(n)}=\frac{1}{4 \pi ^2}$$
Note: this is the reciprocal of (3) zeta-regularized product over all primes.  
This indicates there is a slight skew toward the square-free numbers that have an odd number of factors.
I haven't been able to find this in the literature, but my guess is that it is already known.
Can someone point me in the right direction?
Edit: We have been told it is a bug. Mathematica V.10 does not produce a symbolic answer. 
Edit 2 We can make this work by creating a function $f(k)$ that returns the $k$-th square-free number (in the constructive? order).  We can find a few ideas from A019565.  
$$\prod_{n=1}^{\infty}f(n)^{\mu(f(n))}$$
$$\sum_{n=1}^\infty \mu(f(n))$$
At the steps when $n>2$ is a power of $2$, sum$=0$ and prod$=1$.
Also, at those steps, $f(n)$ is the primorial with $p_k$ as the greatest factor, with $k=\frac{\log(n)}{\log(2)}$  
There is a revised question over here. It is a variation to Edit 2.

REPLY [12 votes]: This is a discussion following from David Speyer about Cesaro summability of the series. 
This series is not Cesaro summable in any order. 
In "The General Theory of Dirichlet Series" by Hardy and Riesz. Chapter 4.
Let $\kappa\geq 0$. $0\leq \lambda _0<\lambda_1<\cdots<\lambda_n\rightarrow\infty$, and write
$$
A_{\lambda}^{\kappa}(w) = \sum_{\lambda_n \textrm{<}w}(w-\lambda_n)^{\kappa}a_n.
$$
\bf Summability $(\lambda,\kappa)$. \rm If $A_{\lambda}^{\kappa}(w)\sim Aw^{\kappa}$ as $w\rightarrow\infty$, then we say that $\sum a_n$ is summable $(\lambda,\kappa)$ to sum $A$.
The second consistency theorem(we do not need first consistency theorem here) states that 
If $\sum a_n$ is summable $(l,\kappa)$ where $l_n=e^{\lambda_n}$, then it is summable $(\lambda,\kappa)$ to the same sum. In particular, Cesaro summability of order $\kappa$ implies $(\log n, \kappa)$ summability to the same sum. 
We consider the formula
$$
\sum_{n=1}^{\infty}a_ne^{-\lambda_ns}=\frac{1}{\Gamma(\kappa+1)}\int_0^{\infty}s^{\kappa+1}e^{-s\tau}A_{\lambda}^{\kappa}(\tau)d\tau.$$
This is originally valid if $\sigma> max(0,\sigma_c)$ where $\sigma_c$ is the abscissa of convergence of the left series. On the other hand, the right allows an analytic continuation of the function represented by the series on the left up to $\sigma>0$. 
What we use here is $a_n = \mu (n)\log n$ and  $\lambda_n = \log n$ . 
Hence, assuming Cesaro summability of any order $\kappa$, will allow 
$$-\frac{\zeta'(s)}{\zeta^2(s)}$$
to have an analytic continuation to $\sigma>0$ which contradicts the fact that $\zeta(s)$ has zeros on the critical line.<|endoftext|>
TITLE: Pullbacks as manifolds versus ones as topological spaces
QUESTION [16 upvotes]: My question is: Does the forgetful functor F:(Mfd) $\to$ (Top) preserve pullbacks?
Detailed explanation is following.
A pullback is defined as a manifold/topological space satisfying a universal property: a manifold/topological space $Z$ (more precisely a diagram $Y \overset{g}{\leftarrow}Z\overset{g'}{\rightarrow}Y'$) is said to be a pullback if for any diagram $Y \overset{h}{\leftarrow}W\overset{h'}{\rightarrow}Y'$, there is a unique smooth/continuous map $u:W \to Z$ such that $h=g \circ u$ and $h'=g' \circ u$.
Now suppose that we have a manifold $Z$ which is a pullback of the diagram $Y \overset{f}{\rightarrow}X\overset{f'}{\leftarrow}Y'$ in (Mfd). Then I wonder whether the manifold $Z$ is a pullback in (Top) or not. 
Notes:

(Top) is the category of topological spaces and continuous maps.
(Mfd) is the category of smooth manifolds and smooth maps.
It is well known that a pullback is given by a submanifold $$Y \times_X Y' = \lbrace (y,y') \in Y \times Y'\mid f(y)=f'(y') \rbrace$$ of $Y \times Y'$ if either $f$ or $f'$ is a submersion. But we do not assume that a pullback is of this form. (If a pullback is always given by this form in (Mfd), then this implies that the forgetful functor preserves pullbacks.)
I am NOT asking about an example of a pullback in (Top) which is not a manifold. Sorry for my confusing question.

REPLY [18 votes]: Here's a counterexample.  Let $X=Y=\mathbb{R}$, and let $Y'$ be a point.  Let $f:\mathbb{R}\to\mathbb{R}$ be a smooth map such that $f^{-1}(\{0\})=\{1,1/2,1/3,\dots\}\cup\{0\}$, and let $f'$ map $Y'$ to $0$.  Then the topological pullback is just the space $W=\{1,1/2,1/3,\dots\}\cup\{0\}$.  But since manifolds are locally connected, any map from a manifold to $W$ factors through $Z$, a space with the same points as $W$ but with the discrete topology.  Thus $Z$ is the pullback in manifolds, but the canonical map $Z\to W$ is not a homeomorphism.
An interesting question is whether this is the only way to get a counterexample.  That is, can you prove the following: A pair of smooth maps has a pullback in manifolds iff every path component of the topological pullback is a manifold.
EDIT: I believe you can get a path-connected counterexample by taking a pair of maps for which the topological pullback is a topologist's sine curve which has been connected up so that if you "unglued" the two halves of it, it would be homeomorphic to $\mathbb{R}$.  That is, it should look kind of like a letter P, except that as the loop of the P approaches the vertical line, it wobbles up and down infinitely like a topologist's sine curve.  Then any map from a manifold in should factor through $\mathbb{R}$, since manifolds are locally path-connected.<|endoftext|>
TITLE: What are some triangulations of Grassmannians?
QUESTION [17 upvotes]: A while ago I heard that there was no known triangulation of the Grassmannian of 3-planes in 6-space.
To believe a statement like that, you have to be a little bit ungenerous about what you mean by "known."  For instance if something is cut out by some polynomial equations and inequalities in R^n, I think there is an algorithm for writing down a simplicial complex homeomorphic to it.  But it is not a very satisfying one, maybe even computationally infeasible in some sense.
Real projective space is easy to triangulate.  Complex projective space is of the form (torus) x (simplex) divided by a simple equivalence relation on the boundary, so probably a triangulation can be cooked up from there.  And there's a beautiful 9-vertex triangulation of CP^2 which is probably not of this form:
http://www.math.brown.edu/~banchoff/howison/newbanchoff/publications/pdfs/9Vertex.pdf
The real Grassmannian of 2-planes in n-space could probably be triangulated by refining the partition by oriented matroids.  But for 3-planes and higher, this partition is too crazy.
What are some other triangulations of Grassmannians?  Is the claim about 3-planes in 6-space accurate?  Is there a reference for 2-planes in n-space?

REPLY [3 votes]: I came across a weaker problem recently, namely finding a simplicial set (or complex) homotopy equivalent to $BO(n)$ was what I needed. The answer is in
http://annals.math.princeton.edu/wp-content/uploads/annals-v158-n3-p04.pdf
that states that certain "realizations of MacPhersonians" (they are simplicial
complexes) are homotopy equivalent to Grassmannians $BO(n)$.
As for triangulation, I would bet on the approach via Schubert cell decomposition. In the case of $BO(n)$ it is not regular, but there is an analog in the case of "oriented Grassmanian" $BSO(n)$, a double cover of $BO(n)$ ($BSO(n)/Z_2=BO(n)$). 
In https://circle.ubc.ca/bitstream/handle/2429/21381/UBC_1979_A6_7%20J96.pdf?sequence=1 it is shown, that incidence nubers between cells of dimension $r$ and $r-1$ are $\pm 1$ or $0$.
As far as I understand, they implicitly even prove that every face is regular. However, there is something still needed to prove that the cell structure is regular, if that is true.
If that is true, then barycentric subdivision (normal subdivision) of the Schubert decomposition of $BSO(n)$ will be simplicial set homeomorphic to $BSO(n)$. As simplicial sets are closed under taking quotients, we get a simplicial set for $BO(n)$ as well. In the end, any simplicial set $X$ can be converted into a simplicial complex by taking certain double subdivision $B_{\ast}Sd(X)$ where $Sd$ is the normal subdivision (it takes into account degenerate simplices) and $B_*$ is its variant that ignores the degenerate simplices.<|endoftext|>
TITLE: Are Thom spectra MU, MSO and K-theory spectra KU, KO modules over some truncations of the sphere spectrum?
QUESTION [13 upvotes]: The Thom spectrum MO is a module over the ring spectrum π≤0S=HZ, where S is the sphere spectrum.
In particular, MO is equivalent to the Eilenberg-MacLane spectrum Hπ*(MO).
On the other hand, MU and MSO are not modules over π≤0S, because they have nontrivial k-invariants.
I wonder if the above result about MO can be generalized to other Thom spectra like MU and MSO
by considering higher truncations π≤kS of the sphere spectrum.
Such a result would be interesting because it is related to the question of representing Thom spectra
as (weak) simplicial objects in the k-category of symmetric monoidal (k+1)-groups (k-groupoids with invertible objects), in the sense that
interpreting symmetric monoidal (k+1)-groups as stable homotopy k-types and taking the homotopy colimit should give back the Thom spectrum under consideration.
Such models are interesting because they are more strict then (say) Segal's Γ-spaces.
A negative answer would necessarily preclude the existence of such models because a stable homotopy k-type
is a module over π≤kS, and the homotopy colimit of a simplicial diagram of modules over π≤kS in the category of spectra is a again a module over π≤kS.
Is MU or MSO a module over the ring spectrum π≤kS for some k>0?  Same question for KU and KO.
(Incidentally, the above result for MO implies that MO can be represented as a simplicial abelian group.
I wonder if there is a geometric model for MO as a simplicial abelian group along the lines of Galatius-Madsen-Tillmann-Weiss theorem,
i.e., n-simplices should be related to unoriented n-manifolds.
Such a model would necessarily make explicit use of properties of unoriented manifolds as opposed to oriented or stably complex manifolds.)

REPLY [24 votes]: For any $k$ and any $0<n<\infty$, $K(n)\wedge \pi_{\leq k}S=0$.  Indeed, this is true for any spectrum with finitely many homotopy groups, since $K(n)\wedge H\mathbb{Z}=0$ and any such spectrum has a finite filtration into Eilenberg-MacLane spectra.  If a spectrum $M$ admits a $\pi_{\leq k}S$-module structure, then $M$ is a retract of $\pi_{\leq k}S\wedge M$, and so $K(n)\wedge M$ is a retract of $K(n)\wedge\pi_{\leq k}S\wedge M=0$ and hence $K(n)\wedge M=0$.  But this is not true for $n=1$ and all of your examples (for $MSO$, you must work at an odd prime).<|endoftext|>
TITLE: How to triangulate a math reference?
QUESTION [19 upvotes]: Today I had the following question: "Is the category of continuous functors a cartesian closed subcategory of ${\bf Cat}\ $?" Suppose I want to find an existing reference for this claim. In that case, 
I need to find the result in the literature somehow!
And for me, that's always hard. I look through some of my category theory books, I look on nlab, I check out wikipedia, and I can't find what I'm looking for. Does that mean the result is not in those places? Absolutely not. I usually take it to mean that I don't have an efficient search strategy. Maybe the result is sitting right there, disguised as "Any lextensive category with strongly orthogonal orbits is quintessentially monoidal", or something that I'm simply not able to "see". Or maybe it's there in plain site, but I just missed it. Why can't we avoid this stupid human error?
The present MO question really has three parts, in ascending order of coolness but descending order of concreteness.

Does anyone know a reference for my category theory claim in quotes above?
Does anyone know of a nice strategy for triangulation in math research? I'd like some way to search for "continuous functors" and "cartesian closed", but  wouldn't you know it --  doing that in google returns useless results. How can one use books, the web, etc. to perform the kind of research I'm talking about: finding what's known about your question? E.g. suppose the answer is in a work of Johnstone or Kelly. What techniques would I use to realize that fact, given that I don't know it to begin with? 
I'd like to hear ideas about the proper structure for the world of mathematical theorems. I'm not looking for an answer such as "It's ${\bf Prop}$, the category of propositions," or anything so simple. Instead, I'm looking for a "strongly searchable" structure in which to store mathematical literature. By strongly searchable, I mean a structure that enables the kind of triangulation I discuss above in 2. This is of course an open-ended question, but perhaps someone has a good idea.

REPLY [8 votes]: I find myself doing searches like this a lot. In the pre-google days I imagine my method would be basically depth first search through the references, but I don't claim this is optimal. Anyway, I know David knows what DFS is, but I'll describe what I mean in detail anyway because it's the basis for what I really do. First, I'll try to remember where I've read something like this before or start googling till I find something in the ballpark of the result I want. If that paper gives references to other sources for background material (e.g. textbooks, etc) then I'll go look them up. I keep tracing back the chain of references till I am pretty sure the result is not there. Then I go up the chain of papers till I find another avenue which seems fruitful and check it out. Because I keep digital copies of almost all papers and books it's easy to search within them for the keywords. The benefit of doing it this way is that if the terminology has changed at some point in history, and if the papers I'm tracing back along are well-written, then I'll be made aware of which keywords to search for in older documents.
Since we live in a world with google, I often supplement this process above by adding keywords or phrases to my original google search to pare down the number of hits. I never thought of it like this before, but this is the triangulation part of the search. The idea is that if a paper is going to contain the result I want then it will also probably contain this phrase. If I know it should contain the definition of a foobar then I might add the phrase "a foobar is" or "is called a foobar" to the googling. I usually add enough phrases, one step at a time, till the number of results is down to between 5 and 30 hits.
Obviously, there are lots of choices being made in which phrases to add, so I have to estimate probabilities on the fly to guess at which choice is going to be most likely to pare down the results without throwing away the paper that contains the reference I want. I guess this would be the hardest part for a machine to emulate. I make my estimates based on all the papers I've read and the way in which authors usually write. Computers can do things like this nowadays, e.g. the algorithms used to summarize the news. I keep multiple tabs open with different google search results so I can try several of the most likely searches. The tabs usually have some overlap but also some unique results. In the end I don't always get the original place the result appeared, but I can find a place where it's there as a lemma or something. And if I really care about the original place then I can follow the references in that paper and usually find it, or copy the verbiage of that lemma into google to see where the author got it from.<|endoftext|>
TITLE: Topology, the board game
QUESTION [9 upvotes]: Edit: I am reposting this question fom math.stackexchange.com; there may be some professors here who have more experience teaching topology.
This is a math education question that I've been thinking of when taking and teaching topology.
For a few years now I've had an idea for a board game that could help teach students topology. However, I've had trouble working out the specifics, and wondered if the community would be able to help. Of course, this may be closed for being off topic, but if so, I'll post a link where we could continue the conversation elsewhere for those interested.
Setup: Basically like battleship. The playing board would be squares of transparent paper (or chessboards, go boards, etc.). 

Players secretly place markers (which are circles) indicating where gardens are (or mines, etc.) on their paper, hidden from the other player. 
A card is then flipped over selecting a topology. 
Players then send five agents to the other player's board; the agents are points. If the points are placed directly in the garden, the player gets his opponents garden. If not, each agent has the option of moving (in a topological path) or of setting off a bomb. Bombs explode to form an open set; any agents or gardens caught in the bomb perish (so you may have to sacrifice your agents). Players describe the shape of the bomb, and their opponent tells them if they've hit the hidden gardens.
At the end of the round, players gain one point for each garden they possess and lose two points for each agent lost.

Possible topologies include:
-Discrete topology: Bombs can take any shape, but agents cannot move.
-Indiscrete topology: Agents can move anywhere, but the only possible bomb is a total nuke.
-Finite complement topology: Agents can still move anywhere, but bombs can miss the agents.
-Dictionary order: Most interesting if agents aren't allowed to move through each other.
-Product topology: Each direction is one of discrete/indiscrete/finite complement/standard
-Metric topology: In metric topology, we require bombs to be formed of metric balls. Then we have the standard metric, the square metric, etc.
-Torus topology: Identify opposing edges (could do other surfaces, use orientation perhaps)
-Subspace topology: Players place an overlay on their boards marking out a subset (like 
topologist's sine curve, etc.) and then flip over another topology to combine with the overlay.
Now, I think this could be made more interesting. Possible variants could include that agents don't find gardens when placed in them until they do a "search" which means they can detect gardens in a compact connected set containing them (but the test only detects if there is at least one garden, so if the only compact set in the whole space is the space itself, the test is always positive).
I'm sure you all could think of many improvements and better rules. I think this could really help people learning topology for the first time. I would have loved to had it for my students this last semester. What ideas do you have to incorporate other parts of topology (like connectedness or algebraic topology) and how could scoring and set up be improved? In other words, how could this be made playable with real strategies? I don't want this to end up like Quidditch. Thanks, and happy Boxing Day!
Edit: I forgot to mention, I don't know whether it would be better to do analog (paper and pen) or discrete (pegs on a go board).

REPLY [3 votes]: This isn't quite topology, but Mike Shulman and John Baez have some interesting things to  say about "gamification".
http://golem.ph.utexas.edu/category/2012/06/the_gamification_of_higher_cat.html<|endoftext|>
TITLE: What can we say about this generalization of simply-connectedness?
QUESTION [8 upvotes]: Let $S$ be a connected scheme. We say that $S$ is simply connected if every smooth and proper morphism $X \to S$ of relative dimension $0$ has a section. This is equivalent to the standard definition, since finite etale morphisms are smooth and proper morphisms of relative dimension $0$, and, after reducing to the connected case, havinga  section is equivalent to being trivial.
Now say that $S$ is $d$-trivial if every smooth and proper morphism $X \to S$ of relative dimension $d$ has a section. So being $0$-trivial is the same as being simply connected.
We might extend the definition to call $\infty$-trivial schemes that are $d$-trivial for all $d$.

For what numbers $d>0$ can we say something about which schemes are $d$-trivial?

This question was originally inspired by this excellent question in which it is proven that $\operatorname {Spec} \mathbb Z$ is $1$-trivial but not $6$-trivial. The apparent difficulty of further progress on $\operatorname{Spec} \mathbb Z$ led me to wonder if other cases, such as varieties over $\mathbb C$, might be easier. In particular:

For what $d$ is $\mathbb P^1_\mathbb C$ $d$-trivial? $\mathbb P^n_\mathbb C$?

REPLY [6 votes]: Already $\mathbb{P}^2_{\mathbb{C}}$ is not $1$-trivial, e.g., the relative Proj of the symmetric algebra of the sheaf of relative differentials (the projectivized tangent bundle) cannot have any section.  
$\textbf{Edit}.$  Also you can use Serre's construction to construct similar locally free sheaves of rank $2$ over $\mathbb{P}^1\times \mathbb{P}^1$.  This raises a red flag in considering "$d$-simple" as a generalization of "simply connected".  For every generalization of simply connected that I know of, a product of two "simply connected" varieties is again "simply connected".  Yet that fails for "$1$-simple".<|endoftext|>
TITLE: A Universal Elliptic Curve
QUESTION [6 upvotes]: I'm working through Deligne's "Formes modulaires et representations l-adiques" paper and I find one of his constructions particularily ambigious. I'm hoping someone can give me a bit of clarification so I can work out a careful proof of the proposition that follows it (2.2.i - below) for myself. (The paper is available in English at http://abel.math.harvard.edu/~jay/writings/deligne-l-adic.pdf , page 3)
Easy formalism: Let $e_1, e_2$ and $1, i$ be ordered bases for the $\mathbb{R}$-vectorspaces $\mathbb{R}^2$, and $\mathbb{C}$ respectively. Let $X = \textrm{Hom}^+(\mathbb{R}^2, \mathbb{C}$) denote the set of $\mathbb{R}$-vectorspace isomorphisms that preserve the given orientations (defined by $e_1\wedge e_2 > 0$ and $1\wedge i > 0$). $X$ is naturally given a complex structure.
Deligne says that on this space one arranges a universal exact sequence:
$$0 \rightarrow \mathbb{Z}^2_X \rightarrow \mathbb{G}_a \rightarrow E_0 \rightarrow 0$$
Here, the first term is the constant sheaf with stalk $\mathbb{Z}^2$ on $X$. Although Deligne doesn't explicitly state what $E_0$ is in this sequence, two lines later (Prop 2.2.i) he uses $E_0$ to denote an elliptic curve over $X$, and says that the pair $(X, E_0)$ satisfies a particular universal property (classifying elliptic curves over analytic spaces with some additional data - see the end of post).
$\textbf{Questions:}$
1.) In particular, I'm not sure how to interpret $\mathbb{G}_a$ in this context (all he says is that this is the additive group). What is it? Is it a sheaf for us? The trivial vector bundle on $X$ with fiber $\mathbb{C}$?
2.) What is this an exact sequence of? Perhaps sheaves of abelian groups? In which case, is there a way we interpret the last term $E_0$ as a sheaf of analytic sections of an elliptic curve over $X$ or something? Somehow $E_0$, whatever it is, has to naturally correspond to an elliptic curve over $X$.
3.) What is the map $\mathbb{Z}^2_X \rightarrow \mathbb{G}_a$?

Prop 2.2.(i) The functor which associates to each analytic space $S$ the set of isomorphism classes of elliptic curves $f: E \rightarrow S$ with identity section denoted $e$, equipped with 
(A) isomorphisms $e^*(\Omega_{E/S}) \cong \mathbb{G}_a$ 
(B) isomorphisms $R^1f_*\mathbb{Z}_E \cong \mathbb{Z}^2$ that induce the map $1$ on the second exterior powers
is representable by the analytic space $X$ equipped with a universal elliptic curve $E_0$ 
http://abel.math.harvard.edu/~jay/writings/deligne-l-adic.pdf (Page 3)

REPLY [7 votes]: Every point of $X$ gives you an embedding of $\mathbb Z^2$ as a lattice in $\mathbb C$ (via the canonical inclusion $\mathbb Z^2 \subset \mathbb R^2$), and hence by quotienting an elliptic curve. This produces a universal diagram 
$$ X \times \mathbb Z^2 \to X \times \mathbb C \to E,$$
where $E$ is a family of elliptic curves over $X$. The analytic sections of this diagram over $X$ give you Deligne's short exact sequence, you can think of it just as a sequence of sheaves of abelian groups.<|endoftext|>
TITLE: finite index subgroup of a fuchsian group
QUESTION [5 upvotes]: Given G, a fuchsian group and a finite sub set A of G. Does there exist a finite index subgroup H in G such that inter section of A with H is empty?

REPLY [12 votes]: On Yves' request, here is a proof of residual finiteness of non-finitely generated Fuchsian groups. (I do not know a reference for this: In general, people do not like working with infinitely generated groups, so, it is possible that nobody bothered writing a proof, even though many people could. The result could be in Beardon's or Maskit's book though, I did not check.) A side remark: Residual finiteness of Fuchsian groups was a partial motivation for this question. 
A Fuchsian group is a discrete group $\Phi$ of isometries of hyperbolic plane $H^2$. Then $\Phi$ contains an index 2 subgroup $\Gamma$ which preserves orientation on the hyperbolic plane and, hence, $\Gamma < PSL(2, R)$. Since residual finiteness is a commensurablity invariant, it suffices to consider the case of discrete subgroups of $PSL(2,R)$. Let $O=H^2/\Gamma$  be the quotient orbifold. Its singular set is a discrete subset $\Sigma=\{x_i: i\in I\}$ of the underlying space $X$ of $O$. I will assume that $\Gamma$ is not finitely generated, thus, $O$ is noncompact.  
Let $T\subset X$ be an infinite locally finite, 1-ended properly embedded tree (with geodesic edges) which contains $\Sigma$. Let $U$ be a small neighborhood (with smooth boundary) of $T$ in $X$ which admits a  retraction to $T$. Every boundary component of $U$ is contractible. Let $O_U$ be the suborbifold of $O$ supported by $U$. Thus, by Seifert-van Kampen theorem for orbifolds applied to this situation, the fundamental group $\Gamma$ of $O$ splits as a free product 
$$
\pi_1(O_U) * \prod^*_i \pi_1(S_i)
$$ 
where $S_i$'s are components of $X\setminus U$, and $\prod^*$ means free product. Now, since each surface $S_i$ is noncompact, by a theorem of Whitehead, it is homotopy-equivalent to a graph (see the discussion here). Hence, $\Gamma \cong  \pi_1(O_U) * F$, where $F$ is a free group. Lastly (again by Seifert-van Kampen theorem applied to $O_U$), the group $\pi_1(O_U)$ is a free product of finite cyclic groups $Z_{n_i}$, where $n_i$ is the order of the singular point $x_i\in \Sigma$. Thus, $\Gamma$ is the free product of countably many cyclic groups (some of which might be finite  and some infinite). 
The converse is also true: If $\Gamma$ is a countable product of cyclic groups $C_i$, then $\Gamma$ is isomorphic to a discrete subgroup of $PSL(2, R)$. Namley,  make each $C_i$ act 
isometrically on $H^2$ with the fundamental domain $D_i$, so that the sets $cl(H^2 -D_i)$ are pairwise disjoint. Now, take the subgroup $\Phi$ of $PSL(2, R)$ generated by the subgroups $C_i$. By Klein combination theorem (also known as the "ping-pong argument") $\Phi$ is discrete (its fundamental domain is the intersection of $D_i$'s) and isomorphic to $\Gamma$. 
Now, suppose that $\Gamma$ is a group which is a countable free product of residually finite groups $\Gamma_i, i\in {\mathbb N}$. Let $S\subset \Gamma$ be a finite subset not containing the identity. Then 
$$
S\subset \prod^*_{j\in J} \Gamma_j$$
where $J\subset I$ is a finite subset. Hence, $S$ projects bijectively to a subset $P$ in the quotient
$$
\Lambda=\Gamma/\langle \langle \bigcup_{i\notin J} \Gamma_i\rangle \rangle. 
$$
Now, $\Lambda$ is residually finite (as a finite free product of residually finite groups). Hence, $\Lambda$ contains a finite-index subgroup $\Lambda'$ disjoint from $P$. Taking preimage of $\Lambda'$ under the epimomorphism $\Gamma \to \Lambda$, we obtain a finite index subgroup in $\Gamma$ disjoint from $S$. This argument, of course, applies to free products of cyclic groups, thereby proving residual finiteness of non-finitely generated Fuchsian groups.<|endoftext|>
TITLE: Projectives in the category of discrete G-modules 
QUESTION [5 upvotes]: If $G$ is a profinite group, then the category $Mod(G)$ of discrete $G$-modules has sufficiently many injectives (Neukirch, Schmidt, Wingberg: Cohomology of Number Fields, 2.6.5). 
Since the cited book says nothing on projectives within this category, I guess $Mod(G)$ doesn't have sufficiently many projectives. Can you give me an example of a discrete $G$-module $M$ such that there is no epimorphism  $P \to M$ with $P$ projective ?  If there are little projectives, is it even possible to classify them ? 
Added: Note that a discrete $G$-module $M$ is an abelian group (with the discrete topology) with a continuous $G$-action $G \times M \to M$ (N-S-W, 1.1.5). For example $\mathbb{Z}$ with trivial $G$-action is a discrete $G$-module.

REPLY [2 votes]: Here is a link a proof that the category $\text{Mod}(G)$ does not have enough projectives, when $G$ is an infinite profinite group. I think the proof is fairly elementary. There might be a few details missing though - sorry about that.<|endoftext|>
TITLE: How large is this "algebra" of defining graphs for Right-angled Artin groups?
QUESTION [7 upvotes]: As part of my research, I have been trying to construct a spherical space at infinity for every right-angled artin group. I've been able to work it out for a certain class of defining graphs. I'd like to know how large this class is (I.e. does it consist of all graphs satisfying a certain invariant equation?), but I have very little graph theoretical background. 
The class of graphs is invariant under disjoint union, join (connecting every vertex in one graph to every vertex in another), and the operation of identifying a single vertex in two disjoint graphs.
The "generating set" consists of all graphs not containing an induced square or an induced subgraph consisting of two triangles identified along a single edge (equivalently, it is the set of all graphs where every four-cycle bounds a tetrahedron). This set includes all trees and all graphs without 4-cycles, as well as all complete graphs.
My hope is that all or many hyperbolic 3-manifold groups virtually embed into RAAG's with such a defining graph. Some or all surface subgroups embed into such RAAG's (since any RAAG with a defining graph with an induced 5-cycle contains a surface subgroup). Thanks for your help!
Edit: I forgot to mention that RAAG's are subgroup separable iff their defining graphs have diameter at most 2 and contain no induced square.

REPLY [3 votes]: This is not a complete answer, but after tlking to some people at a comference about this, the only virtual embeddings we could think of explicitly are related to reflection groups.
Andreev's theorem (or Rivin's extension) characterizes right-angled hyperbolic polyhedra; in particular, they have trivalent boundary graphs (I think that the only other characteristic they have is that they lack prismatic circuits of size less than four). The reflection group is a right angled coxeter group whose defining graph is the dual graph of the polyhedron's boundary graph. There is a finite orbifold cover that is a manifold, and so these manifolds virtually embed into these right angled coxeter groups (one of the comments mentioned this kind of embedding).
Since the defining graphs are dual to trivalent graphs, they consist of triangulations of surfaces, which contain nothing but diamonds. 
For my research, RACG's and RAAG's have identical subdivision rules, so diamonds may be more common than I thought.<|endoftext|>
TITLE: Concentration of sum of pairwise squared Euclidean distances of random vectors
QUESTION [7 upvotes]: Let $X_1, \ldots, X_n $ be independent random vectors in $B(0, D) \subset R^d$ ($\ell_2$ ball of radius $D$ centered at the origin). I am trying to find the concentration of the following quantity around its expectations:
$$f(X_1, \ldots, X_n) = \frac{1}{n^2} \sum_{1 \leq i,j \leq n} \|X_i - X_j \|_2^2$$
Using McDiarmid's inequality, I can show that 
$$ \Pr(|f(X_1,\ldots,X_n) - \mathbb{E} f(X_1,\ldots,X_n)| \geq \epsilon) \leq 2 \exp\left(- \frac{2n\epsilon^2}{16D^4}\right). $$
This tells me that with high probability, $\epsilon \sim O(1/\sqrt{n})$. 
Usually sum of $n$ independent random variables concentrate with the similar dependence on $n$ (that is, $O(1/\sqrt{n})$) unless I use Berstein's inequality. And it is believed that in some sense, among functions of independent random variables, sums are the least concentrated.
In my situation, the function is a sum of $n^2$ non-independent random variables (if you think of $Z_{ij} = \| X_i - X_j \|_2^2$ as a random variable). So I believe that the concentration should be better than $O(1/\sqrt{n})$.
Does anybody have any idea of how such a guarantee can be achieved? Or can anybody show that the $O(1/\sqrt{n})$ is the best I can hope for?

REPLY [5 votes]: You're right, you can get a better concentration inequality with $\epsilon \sim O(1/n)$ considering that's a sum of $n^2$ dependent random variables. You first need to use decoupling for $U$-statistics that relates the dependent setting to the independent one as it is shown in  [1]. In your case, you can show that there exist a constant $C$ such that
$ P\left(\left| \frac{1}{n^2} \sum_{1 \leq i \ne j \leq n} \|X_i - X_j \|_2^2
 - \mathbb{E}f\right| \ge t\right) \le \\
\qquad \qquad C P\left(C\left|\frac{1}{n^2} \sum_{1 \leq i \ne j \leq n} \|X_i^{(1)} - X_j^{(2)} \|_2^2
 - \mathbb{E}f\right| \ge t\right) , $
where $X_i^{(1)}, X_i^{(2)}$ are independent copies of $X_i$.
Then you can apply the McDiarmid inequality on the sum of $n(n-1)$ independent terms and get an inequality achieving the rate $O(1/n)$.
Moreover, as your U-statistic is symmetric you can reverse this inequality to get a lower bound and convince you that you can't do better.
[1] de la Peña, V. H., & Montgomery-Smith, S. J. (1995). Decoupling inequalities for the tail probabilities of multivariate U-statistics. The Annals of Probability, 806-816. Avialable : http://arxiv.org/pdf/math/9309211v2.pdf<|endoftext|>
TITLE: Existence of different knots in $RP^3$ having the equivalent liftings in $S^3$
QUESTION [8 upvotes]: I'm looking for the answer to following question. Do exist different knots in $RP^3$ which have equivalent liftings in $S^3$ under covering $p:S^3\rightarrow RP^3$?

REPLY [4 votes]: If either knot is hyperbolic, this is not possible. 
First, consider the non-null homologous case. 
Let $K_1,K_2\subset \mathbb{RP}^3$ be two knots, such that their preimages $K_1',K_2'\subset S^3$ are isotopic to the hyperbolic knot $K$ (Remark: if $K_1$ is hyperbolic, then so is $K$ and therefore $K_2$ by the geometrization theorem). The covering translation induces involutions $\iota_{1,2}:S^3 \to S^3$ such that $\iota_{i}(K)=K$. Restricting to the hyperbolic space $S^3\backslash K$, we get involutions $\iota_i:S^3\backslash K \to S^3\backslash K$, $i=1,2$, which are isotopic to (fixed-point free) hyperbolic isometries. The hyperbolic isometry restricts to the torus $T=\partial\mathcal{N}(K)$ as an isometry (e.g. taking a horotorus representative). Each isometry is determined by its action on $T$. There are 3 possible fixed-point free involutions of a torus. Let $\mu, \lambda\subset T$ be representatives of the meridian and longitude. The involutions rotate around $\mu$ (preserving $\mu$), or rotate around $\lambda$, or rotate around $\mu\lambda$. The rotation around $\mu$ is not possible, since this would extend to an involution of $S^3$ with fixed-point set $K$, which is impossible by the Smith conjecture. Thus, $\iota_1$ and $\iota_2$ must induce the other two involutions. But then $\iota_1\circ\iota_2$ must be the forbidden involution, a contradiction. So there is a unique involution preserving $S^3\backslash K$, and therefore at most one isotopy class of knot in $\mathbb{RP}^3$. 
In the null homologous case, there are knots $K_1,K_2\subset \mathbb{RP}^3$ such that the preimage is isotopic to a two-component link $L$. There are two fixed-point free involutions of $S^3$ preserving $L$ and exchanging its components, and isotopic to isometries of the hyperbolic metric on $S^3\backslash L$ which exchanges the two cusps. Thus, they generate a dihedral group in the isometries of the hyperbolic knot complement, which must be finite order. Since both isometries preserve the meridians of $L$, this may be extended to a dihedral group action on $S^3$, generated by two fixed point free involutions. However, any such (smooth) action is conjugate to a group of isometries of $S^3$ by the orbifold theorem, which gives a contradiction, since the only fixed-point free involution is the antipodal map. 
I think the general case should follow in a similar fashion, but one would have to consider how the involutions permute the JSJ decomposition.<|endoftext|>
TITLE: Is there an analog of 'Directed Graph' for topological spaces?
QUESTION [9 upvotes]: In a directed graph, certain paths are distinguished: Those that follow the edge direction on each edge. I was wondering if one could (or does) formulate a notion of a 'Directed Topological Space', on which certain paths are distinguished. Conceptually, we could take a space like R^2 \ 0 and say that counterclockwise paths are the 'directed paths'.
I had two ideas: First, if you have a smooth Riemannian manifold, then a smooth vector field X gives you a 'Direction': The directed paths being those whose tangent vector v satisfies g(v,X) > 0 everywhere.
Alternatively, with an arbitrary topological space, you could say that a 'Directed' space comes equipped with a sorting of homotopy classes of paths into 'Forward', 'Backward', and 'Trivial' (the last consisting of just trivial loops), where the compositions 'Forward'*'Forward' = 'Forward', 'Forward'*'Trivial' = 'Forward', etc. and the reverse of a 'Forward' path is always a 'Backward' path (note that you wouldn't have any results involving 'Foward'*'Backward' or 'Backward'*'Forward').
So I suppose my question is: Is this idea studied at all? Has anyone determined this (or perhaps another generalization of 'Directed') to be a useful notion?
Thanks.

REPLY [11 votes]: There is quite a growing number of researchers nowadays in so-called "Directed Algebraic Topology" (it even has its own Wikipedia page). It can be seen as part of Applied Algebraic Topology, since it arose out of an attempt to model concurrent systems in Computer Science. 
I can recommend the papers of Fajstrup, Raussen and Goubault (linked to from the Wiki page) and Jeremy Gunawardena on homotopy and concurrency as good introductions to the subject of directed homotopy.

REPLY [2 votes]: Partially ordered sets are closely related to directed graphs, and partially ordered spaces have been studied in general topology quite a bit since there is a strong connection between ordered sets and topologies. I realize that this may not be exactly what you had in mind since you may have been concerned mainly with paths on such spaces, but I will answer this question since ordered topological spaces are very interesting and very natural. Besides, one may also consider paths on ordered topological spaces as order preserving continuous maps from $[0,1]$ to your space $X$. Furthermore, I suppose that you can probably generalize this notion to a "locally ordered topological space" or a "locally preordered topological space" so that you can consider $\mathbb{R}^{2}\setminus\{0\}$ with a notion of going clockwise and counterclockwise. I will briefly outline some basic ordered topological spaces here.
If $X$ is a topological space, then the specialization ordering on $X$ is the preordering $\leq$ on $X$ such that $x\leq y$ if and only if $x\in\overline{\{y\}}$. Furthermore, the specialization ordering on $X$ is a partial ordering if and only if $X$ is a $T_{0}$-space. It follows from the definition that $X$ is a $T_{1}$-space if and only if the specialization ordering $\leq$ coincides with equality. Thus the specialization ordering is next to useless for spaces satisfying separation axioms from $T_{1}$ and above. However, the specialization ordering is very important for analyzing spaces that do not satisfy the $T_{1}$-separation axiom.
As another example of an ordered topological space, an Alexandroff space is a topological space where the arbitrary intersection of open sets is open. The category of Alexandroff spaces is isomorphic to the category of all preordered sets where the morphisms between preordered sets are the order preserving maps. If $X$ is an Alexandroff space, then the specialization ordering gives you a preordering on $X$. If $X$ is a preordered set, then the collection of all downwards closed sets are precisely the closed sets in an Alexandroff topology on $X$. These correspondences give you the equivalence between the category of Alexandroff spaces and the category of partially ordered sets.
Also, people have endowed partially ordered sets with other topologies such as the Lawson topology and Scott topology (see the book Continuous Lattices and Domains or the book the Compendium of Continuous Lattices) for more details. The Scott topology almost never satisfies higher separation axiom, but it turns out that the Lawson topology is compact and Hausdorff on a large class of ordered sets called continuous lattices.
Other ordered spaces that satisfy higher separation axioms are also useful.
For instance, one can use ordered compact Hausdorff spaces in order to study distributive lattices. We say that a topological space $X$ with a partial order $\leq$ is totally order disconnected if whenever $x\not\leq y$, then there is an upwards closed clopen set $U$ with $x\in U,y\not\in U$. A Priestley space is a compact totally order disconnected topological space. It turns out that the category of Priestley spaces is equivalent tot the category of all ordered topological spaces.
In my own personal research I have studied connections between least upper bounds in posets and topological spaces, and this research is quite similar to the Scott topology on a topological space. In my research, I have shown that a certain category of zero-dimensional ordered topological spaces is isomorphic to a category certain of pairs of the form $(X,\mathcal{A})$ where $\mathcal{A}$ is a collection of subsets of $X$ with least upper bounds.<|endoftext|>
TITLE: Properties of ring epimorphisms that are true only over commutative rings 
QUESTION [6 upvotes]: I'm interested in knowing/collecting some properties of epimorphisms of rings (with identity) that are true over commutative rings but are false in the non-commutative case.
Example: I learned from MO that if $R \hookrightarrow S$ is an epimorphism of commutative rings then $S/R$ is a torsion left $R$-module. But there are counter-examples to this property if $R$ is permitted to be non-comutative.

REPLY [4 votes]: For Artinian commutative rings $R$ all ring epimorphisms $R\to S$ are surjective. 
In general, this is false if $R$ is non-commutative. For, Isbell has constructed an epimorphism $R \to S$ where $A$ is finite (hence Artinian) and $S$ infinite. I'll have a look if I can find Isbell's example later on.<|endoftext|>
TITLE: What about schemes built up out of graded rings?
QUESTION [8 upvotes]: Toen-Vaquié construct a category of schemes relative to some complete cocomplete closed symmetric monoidal category $C$. Affine schemes correspond by definition 1:1 to commutative monoid objects in $C$. For $C=\mathsf{Ab}$ we get the usual category of schemes, where affine schemes correspond to commutative rings. For $C=\mathsf{Set}$ one gets one of the various definitions of schemes over $\mathbb{F}_1$. One of the drawbacks of this quite general theory is that schemes are defined via their functors on commutative monoid objects, without any geometric incarnation.
Question. What happens when $C$ is the category of graded abelian groups (equipped either with the usual symmetry, or with the twisted symmetry)? Here affine schemes correspond to (graded) commutative rings. Is there any connection with the usual Proj construction? Is there any more geometric interpretation of these schemes? For example one might hope for a fully faithful functor into the category of locally ringed spaces. Is this category of schemes something new at all?

REPLY [10 votes]: It is my impression (without being very careful about it) is that the C-geometry (for C the category of graded abelian groups) is the same as geometry over $BG_m$, i.e., geometry of schemes with 
an action of the multiplicative group. This should follow from the identification of $C$ with representations of $G_m$. (In particular there is a close relation with Proj - except we don't throw away the "irrelevant ideal", we just consider the stacky quotient of a homogeneous variety in affine space by $G_m$.)
More generally, suppose your category $C$ is identified by a generalized Tannakian reconstruction theorem with $QCoh(X)$, the (complete cocomplete closed symmetric monoidal) category of quasicoherent sheaves on $X$, a quasicompact stack with affine diagonal. Then $C$-schemes should be identified with schemes over $X$, so that $C$-geometry just means geometry relative to the base $X$. Thus for example $C$ could be $R-mod$ for a commutative ring $R$ ($X=Spec(R)$), or $Rep(G)$ for an affine group scheme $G$ ($X=BG$).<|endoftext|>
TITLE: Stokes theorem for manifolds without orientation?
QUESTION [10 upvotes]: In textbooks Stokes' theorem is usually formulated for orientable manifolds (at least I couldn't find any version not using orientability). Is Stokes theorem: $\int\limits_{M}d\omega=\int\limits_{\partial M} \omega$ also true for non-orientable manifolds?
Are there any references, you can tell me?
Regards.

REPLY [10 votes]: This is mostly an elaboration on Peter Michor's answer, but I think more deserves to be said, including a major caveat. A reference which treats this material more fully and explicitly than his book cited above is Theodore Frankel's The Geometry of Physics. 
From the "double-cover" perspective, I believe the usual differential forms are the ones in the $+1$ eigenspace of the covering map, which de Rahm calls paire ("even"). The ones in the $-1$ eigenspace ("odd forms" impaire) are called "pseudo-forms" by Frankel, although Frankel doesn't explicitly work in terms of the double cover. See section 2.8.e of Frankel for his definitions and section 3.4 for a discussion of integration, including a statement of Stokes' theorem for odd forms. (I had trouble finding such a statement in Michor's book...)
The caveat I want to mention is that like the familiar even forms, odd forms do require some orientation data to integrate over a submanifold (as in Stokes' Theorem). It's just different data. Whereas an even form is integrated over an oriented submanifold, an odd form is integrated over a transverse-oriented submanifold. That is, if $N^k \subset M^n$ and $\omega$ is an even $k$-form, then you need to choose an orientation on the tangent bundle of $N$ before you can integrate $\omega$ over $N$. If $\omega$ is an odd form, then you need to choose an orientation on $N$'s normal bundle in order to integrate (a fair number of arbitrary need to be made to specify such a thing, including, at least locally, a choice of normal bundle, but they turn out not to matter).
This treatment of orientation means that, for example, an odd $(n-1)$ form on an $n$-manifold can naturally represent a "flux": integrate over a hypersurface to determine the rate of transport through that hypersurface per unit time -- the transverse orientation tells you which direction of flow you consider to be "positive".
But the case that comes out most naturally is when you have an odd $n$-form on an $n$-manifold. In this case, when you integrate over an $n$-submanifold, the normal bundle is 0-dimensional, so there is a canonical choice of transverse orientation (between "$+$" and "$-$", just take "$+$"!), so no orientation data is really needed to integrate. This is analogous to the fact that no orientation data is needed to integrate an even 0-form over a 0-submanifold (i.e. to add up the values of a function at a set of points!). 
The upshot is that a volume form or other nice measure is naturally an odd form, because the volume/measure of a submanifold certainly doesn't depend on any choice of orientation. Whereas when you try to treat volume as integration over an even form, you have to fix an orientation arbitrarily.<|endoftext|>
TITLE: Bijective-equivalent collections of proper classes in set theory
QUESTION [9 upvotes]: In ZFC set theory (or better in NBG set theory, where the language is more flexible with proper classes), we have that every unbounded class of ordinal numbers is a proper subclass of the class On of all ordinals and that every such proper class has a unique bijection (by the enumeration function) with the proper class On. More generally, every proper class that is (class) well-orderable (that is, that is well-orderable, with every descending section being a set) is also bijective with On.
So that, every proper class is bijective with On iff V (the class of all sets) is (class) well-orderable (this is known as the global choice axiom). So, if we do not have the global choice axiom there is no bijection between On and V, and we are left with at least two bijective-equivalent collections of proper classes.
Question 1: Does ZFC (or NBG) prove more that that concerning the bijective-equivalent collections of proper classes  ?
Question 2: If we pass to a set theory whose language allows for quantification over proper classes (like KM, the Kelley-Morse set theory), do we know more about these bijective-equivalent collections of proper classes ?.
Gérard Lang

REPLY [4 votes]: I will give a partial answer for models of $NBG$ in which local choice ($AC$) holds, namely:

Theorem. There is a model $M$ of $ZFC$ whose natural expansion $(M,\cal{A})$ to a model of $NBG$ (where $\cal{A}$ is the collection of all parametrically definable classes of $M$) has at least 3 bijection-inequivalent classes, namely: $Ord$ (the class of ordinals), $\cal{P}$$(Ord)$ (the class of all subsets of ordinals), and $V$ (of class of sets).
Moreover, it is a theorem of $NBG$ plus $AC$ that there is a bijection between $V$ and $\cal{P}(\cal{P}$$(Ord))$.

Proof. Let $M$ be a model of $ZFC$ which has a definable class of pairs with no no choice function (see, e.g., the answers given by Hamkins and myself to this MO question).
It is clear that there is no injection from $V$ into $Ord$ that is canonically coded in $\cal{A}$.
We next observe that there is no injection from $V$ into $\cal{P}$$(Ord)$ that is coded in $\cal{A}$. This is because there is an obvious definable linear ordering of $\cal{P}$$ (Ord)$  (by viewing each subset of ordinals as a binary sequence), and the existence of such an injection would contradict the fact that there is no definable choice function on a class of pairs.
Finally, to show that there is no injection from $\cal{P}$$(Ord)$ into $Ord$ in our model of $NBG$, we take advantage of the axiom of choice within $M$. More specifically, recall that in the presence of the axiom of choice, every set is definable from a subset of ordinals, the idea being: given any set $x$, let TC($x$) be the transitive closure of {$x$}. Then by AC there is bijection $f$ between TC($x$) and some ordinal $\alpha$. This allows us to copy the $\in$ relation on TC($x$) to a binary relation $r$ on $\alpha$, and then we can use a canonical pairing function on ordinals to code $r$ as some subset $s$ of ordinals. Then $x$ is definable from $s$ via "$x$ is the top element of the transitive collapse of $r$".
Note that the above procedure gives rise to a definable surjection $F$ from $\cal{P}$$(Ord)$ onto $V$: given a subset $s$ of ordinals, first decode it as a binary relation $r$ on ordinals, and then ask whether $r$ is an extensional, well-founded relation with a top element. If no, let $F(s)$ be $0$, and if yes, then let $F(s)$ be the top element of the transitive collapse of $r$.
Now if there exists a definable injection $F$ from $\cal{P}$$(Ord)$ into $Ord$, then in light of the fact that $\cal{P}$$(Ord)$ would then be definably well-ordered, the above map $F$ could be "inverted" by an injection $G$ from $V$ into $\cal{P}$$(Ord)$, where $G(x)$ is the first $s$ such that $F(s)=x$. This, in turn would enable us to definably well-order $V$, contradiction.
Finally, to verify the last claim of the theorem: the coding explained above allows us to define an injection $H$ from $V$ into $\cal{P}(\cal{P}$$(Ord))$, namely $H(x)$ is the collection of all subsets $s$ of $\kappa$, where $\kappa$ is the cardinality of the transitive closure of {$x$}, such that when $s$ is canonically decoded as a relation $r$ on $\kappa$, then $r$ is an extensional well-founded relation with a top element such that $x$ is equal to the top element of the transitive closure of $r$.

Therefore, since the class version of the Schröder-Bernstein theorem holds in models of $NBG$, there is a bijection between $V$ and $\cal{P}(\cal{P}$$(Ord))$ in models of $NBG$ in which the (local) axiom of choice holds.<|endoftext|>
TITLE: Sums of Squares
QUESTION [13 upvotes]: Every prime $p = 4k + 1$ can be uniquely expressed as sum of two squares, but for which
integers $x$ is $x^2 + y^2 =$ some prime $p$?  Stated differently, does the square of
every positive integer appear as one of the squares in the representation of some
prime $p$?

REPLY [6 votes]: This is a special case of Bateman–Horn conjecture, which in this case states that for given $y\in\mathbb{N}$ the polynomial $p(x)=x^2+y^2$ assumes prime values for infinitely many $x\in\mathbb{N}$, more specifically,
$$\#\{x\leq N: p(x)\text{ is prime}\}\sim\frac{1}{2}\prod_{p\nmid y}\frac{p-1-(-1)^{\frac{p-1}{2}}}{p-1}\cdot\frac{N}{\ln N},$$
thus the asymptotics should depend on $y$, but only by a multiplicative constant.
However the only proved case of Bateman-Horn (at least as far as I know) is for one linear polynomial, ie Dirichlet theorem.<|endoftext|>
TITLE: Does the Manin-Drinfeld theorem hold over number fields?
QUESTION [17 upvotes]: The Manin-Drinfeld theorem has various equivalent statements. Let $\Gamma$ be a congruence subgroup of $\mathrm{SL}_2(\mathbb{Z})$. Then:

for any congruence subgroup $\Gamma$ of $\mathrm{SL}_2(\mathbb{Z})$, the subgroup of $\operatorname{Jac}(X(\Gamma))$ generated by the cusps is finite;
if $c_1, c_2$ are cusps and $f \in S_2(\Gamma)$ is a Hecke eigenform, the integral $\int_{c_1}^{c_2} f(z) \mathrm{d}z$ is a linear combination of the periods $\Omega_+(f)$ and $\Omega_-(f)$ with coefficients in the field generated by the Hecke eigenvalues of $f$;
the natural map $H^1_c(Y(\Gamma), \mathbb{Q}) \to H^1(Y(\Gamma), \mathbb{Q})$ has a unique Hecke-equivariant section.

I'm interested in extensions of this to the context of arithmetic quotients of $\mathrm{GL}_2(\mathbb{A}_K)$, where $K$ is a number field. The first statement only makes sense if $K$ is totally real, but the second and third can be formulated for any $K$. Are they true in this generality? 
(I have a translation of a 1978 paper by Kurcanov which gives the proof for $K$ an imaginary quadratic field; and I believe there is a more recent paper of Kurcanov that covers CM fields, but I can't read Russian and there doesn't seem to be an English translation.)

REPLY [2 votes]: Along the line Venkataramana brought up, it's mentioned in section 7.6 of the paper Harmonic analysis in weighted $L_2$ spaces by Franke that if compactly-supported cohomology in the 3rd statement is replaced by the cuspidal part of cohomology (which is by definition a Hecke-summand but only defined over $\mathbb{C}$) and we ask about rationality, then when $G=Res^K_{\mathbb{Q}}GL_n$ for a number field $K$ an affirmative answer was given by Clozel (and Franke proved there the rationality of the $\{P\}$-part of cohomology for any class $\{P\}$ of associate parabolic subgroups). It's also remarked there Harder and Clozel seemed to be the first to mention this rationality generalizes Manin-Drinfel'd.<|endoftext|>
TITLE: Factoriality: local or global?
QUESTION [13 upvotes]: Let $X$ be an algebraic variety.  I have read the following definitions:

$X$ is factorial if every Weil divisor on $X$ is Cartier.
$X$ is locally factorial if all its local rings are unique factorisation domains.

I am comfortable with each of these.  However, I have also seen it stated (in this question) or implied (in this answer to a related question) that these are equivalent, and I'm not sure about this.
For example, consider the local model for a threefold node, $\mathrm{Spec}\left( \mathbb{C}[x,y,z,w]/(xy - wz) \right)$.  Its local ring at the origin is not a UFD, as $xy = wz$.  This corresponds to the existence of non-Cartier divisors like $\{x = z = 0\}$.
But this singularity occurs in varieties which do not have non-Cartier divisors.  Consider, for example, the quintic threefold given by projectivising the equation ($xy - wz +$ generic terms up to order five) $=0$.  I know for a fact that this does not have non-Cartier divisors, yet it has a node.
So, are 1. and 2. really equivalent?  If so, the higher-order terms in the above example must restore factoriality of the local ring, right?  I'd been under the impression that such higher-order terms don't change the local structure; this might be where I've gone wrong.

REPLY [16 votes]: It's a standard result in commutative algebra that every noetherian integral domain is a UFD if and only if every prime ideal of height 1 is principal. When applied to the local rings of X this gives exactly the equivalence above.<|endoftext|>
TITLE: Can the expansion of a large integer in all bases consist of almost all zeroes?
QUESTION [18 upvotes]: Let $n$ be a positive integer. Given an integer base $b\ge 2$, let $C_b(n)$ be the number of non-zero digits in the expansion of $N$ in base $b$. Further, let $M(n)=\max\{C_b(n):b\ge 2\}$ be the maximum of all $C_b$. Obviously, for each $n$ only $b\le n$ count, and $M(n)\le 1+\log_2 n$. 
I would like to understand the relation between the $C_b(n)$ for different bases $b$. It seems reasonable to conjecture (and I'm sure stronger conjectures exist) that if $b, b'$ are multiplicatively independent, then $C_b(n)+C_{b'}(n)>\varepsilon \log n$ for some $\varepsilon(b,b')>0$ and all $n$. However, this seems to be out of reach. The best I have found are estimates of the form $C_b(n)+C_{b'}(n)\ge \frac{\log\log n}{\log\log\log n}$, using Baker's Theorem.

My question is:

Can we do better by considering more bases? In particular, is it true that there is $\varepsilon>0$ such that $M(n) \ge \varepsilon \log n$ for all $n$?

The special case in which $n$ is a power would already be interesting, and is perhaps a little easier:

Does there exist $\varepsilon>0$ such that $M(2^k) > \varepsilon k$ for all $k$?


This question came up in studying the absolute continuity of the convolution of certain self-similar measures, a question at the interface of fractal geometry/dynamics/number theory.

REPLY [7 votes]: A full answer to your question seems tough, because for any fixed value of $\epsilon$, only finitely many bases are available to establish the bound.  However, there is an elementary partial answer to your second question:  The digits of $2^{n^2}$ in base $2^n-1$ are exactly the $n$th row of Pascal's triangle.  A variation of this idea yields $M(2^k) = \Omega(\sqrt{k})$.  This suggests the following strategy for general $n$:  Choose a base $b$ so that $n = d+\prod_i (b+c_i)$ with $d \ll n$ and $0 < c_i \ll b$, in such a way that the leading digits of $n$ in base $b$ are the same as the coefficients of the polynomial $p(x) = \prod_i (x+c_i)$.  I have not done a calculation, but maybe this could give you $M(n) = \Omega((\log n)^\alpha)$ for some $\alpha > 0$.<|endoftext|>
TITLE: Splitting of a Short-Exact Sequence of Lie Groups
QUESTION [8 upvotes]: Let $G$ be a Lie group (possibly disconnected). Consider the natural short-exact sequence $$1\rightarrow G_0\rightarrow G\rightarrow\pi_0(G)\rightarrow 1,$$ where $G_0$ is the identity component of $G$, and $\pi_0(G)\cong G/G_0$ is the component group. Under what conditions does this sequence admit a splitting, so that $G$ is a semidirect product of $G_0$ and $\pi_0(G)$?. Useful answers might include (but need not be limited to) some types of Lie group $G$ for which a splitting exists. I would appreciate any and all references and suggestions.
Thanks!

REPLY [2 votes]: There is a crossed module associated to this situation, which will be trivial if $G \to \pi_0(G)$ splits on the level of gropus. Thus an obstruction for the existence of a splitting is the non-triviality of the crossed module.
In some more detail, the crossed module is given by the universal covering $\tilde{G_0}\to G_0$, composed with the inclustion $G_0\to G$. The action is given by lifting the conjugation action of $G$ on $G_0$ to $\tilde{G_0}$ (note that each automorphism of $G_0$ lifts in a unique way to its simply connected cover). This defines the smooth crossed module $\tilde{G_0} \to G$. This has an obstruction class in the group cohomology $H^3(\pi_0(G),\pi_1(G))$, where the action of $\pi_0(G)$ on $\pi_1(G)$ in induced by the conjugation in $G$.
The purpose of the obstruction class is twofold. On one hand, it is an obstruction for the existence of the splitting. On the other hand, it allows you to reduce the question to simply connected $G_0$ in case that the obstruction vanishes, since then there exists a $H$ with $H_0$ simply connected and $\pi_0(H)=\pi_0(G)$ and $G\to\pi_0(G)$ splits if $H\to \pi_0(H)$ does so. Details on the latter arguing can be found in Theorem III.8 of http://arxiv.org/abs/math/0504295 (I hope that I am not overseeing some smoothness issues here).<|endoftext|>
TITLE: Probing a manifold with closed curves
QUESTION [5 upvotes]: Since fedja's excellent comment on Joseph's question on probing a manifold with geodesics remained uncommented (especially by topologists), I'd like to make a question out of it:

Conjecture: Given an orientable 2-dimensional manifold and two
  closed curves on it which intersect
  transversally in exactly one
  point. Then the two curves cannot be
  homotopic.

(An immediate consequence of this would be that living on a surface with two such curves, one would know, that it is not homeomorphic to the sphere.)

How to proof this conjecture (if it's true)?

REPLY [7 votes]: I wanted to give a somewhat "low-tech" answer to this, here's the best I came up with:
Call the curves $C_1$ and $C_2$ and the surface that they lie in $\Sigma$.  Take a small open neighborhood $N$ of $C_1 \cup C_2$.  What does $N$ look like?  Well, it's a neighborhood of $C_1$ stuck to a neighborhood of $C_2$ in an operation called a "plumbing" (which is the same operation as sticking together two strips of paper to make a cross).
The neighborhood of $C_i$ is a cylinder that has been plumbed to itself a few times (one plumbing for each self-intersection of $C_i$).
Now if $C_1$ can be homotoped to coincide with $C_2$ inside $\Sigma$ then it can be homotoped to $C_2$ inside the space $S$ that I make by crushing $\Sigma \setminus N$ to a point.
The idea is to convince yourself that $S$ is a torus with a finite number of points identified (one point for each boundary circle of $N$), and that $C_1$ wraps once around one direction of the torus while $C_2$ wraps once around the transverse direction.<|endoftext|>
TITLE: Subgroups of $SL_2(F)$ generated by unipotent elements
QUESTION [5 upvotes]: I am interested in the following problem : given a finite field $F$ and two unipotent elements $g_1,g_2\in\mathrm{SL}_2(F)$ which do not commute, what can we say about the subgroup they generate? More specifically, is there an additional condition under which it is known that they generate all of $\mathrm{SL}_2(F)$? 
For $F=\mathbf F_p$ it is easily seen that this is always the case. For $F=\mathbf{F}_{p^2}$ I convinced myself a while ago (using lenghty computations of products of $g_1,g_2$) that they generate either the whole of $\mathrm{SL}_2(F)$ or a subgroup conjugated to $\mathrm{SL}_2(\mathbf{F}_p)$. I would be glad to see a more enlightening proof of this fact, and even more if it pointed at generalizations to other algebraic groups than $\mathrm{SL}_2$.

REPLY [5 votes]: A few complements to Geoff's answer and Agol's comment:
1) Dickson's theorem is worked out in detail (somewhat lengthy) in Gorenstein's
Chapter 2, page 44+, as Theorem 8.4.   His section 8 is devoted to a study of this matrix group and its projective quotient.  The treatment is elementary but not conceptual.   In the exceptional case of the field of 9 elements, the proper subgroup you might generate is of order 120 and has the simple group of order 60 as a quotient.
2) Larger matrix groups are definitely more complicated, though the theme of generation by unipotent matrices (geometrically, transvections) is in any case prominent in the study of such classical groups.  For instance, when you look at $3 \times 3$ matrices over a finite field, you easily find two unipotent Jordan blocks (with a single block just above the diagonal) which fail to commute and only manage to generate a subgroup of upper triangular matrices.   
3) As Agol comments, you still need to get your arbitrary non-commuting unipotent matrices to be conjugate to a pair treated in Dickson's theorem.    What Agol suggests is in the more sophisticated but also more conceptual framework of Chevalley groups (as in Steinberg's old Yale lectures), or more generally semisimple linear algebraic groups.   Looking at $\mathrm{SL}_2$ in this spirit, there are two Borel subgroups containing a typical maximal torus such as the group of diagonal matrices (and any two distinct Borels intersect in such a torus).   In turn, the unipotent subgroups (say upper and lower unitriangular matrices) suffice to generate the entire group.   Such a unipotent group is cyclic over the prime field, but otherwise elementary abelian (which creates complications for applying Dickson's theorem).    Moreover, the quotient of the full group by a Borel subgroup is a projective line, on which the torus has precisely two fixed points, etc.    This is where Agol's comment comes from.
4) Most of this Chevalley group formalism is well encoded in the more elementary theory of split BN-pairs, where it's still fairly easy to see conceptually how the generation by unipotents works.   See Steinberg's lecture notes (now online at his UCLA homepage) or Carter's 1968 book for results in this direction.<|endoftext|>
TITLE: Models of Determinacy
QUESTION [5 upvotes]: Today we have that $L(\mathbb{R}) \models AD$ (assuming there are $\omega$ many Woodin cardinals and a measurable above them all). I was wondering what other models of $AD$ might look like and if it is possible to characterize them using inner model theory. For instance the $HOD^{L(\mathbb{R})}$ is a certain mouse, $\mathcal M_{\omega}$, which has information about its iteration strategy, namely it is the direct limit of a directed system containing all mice which are properly small, full and are $\Game^{\mathbb{R}} \Pi^1_1$-iterable (this last condition turns out to be the same as $\Sigma_1^{L(\mathbb{R})}$ and these statements are absolute between $V$ and $L(\mathbb{R})$ by Martin and Steel, see Cabal books). 
Can we similarly characterize all of $L(\mathbb{R})$ as a certain type of mouse? More generally can one characterize models of $AD$ by just saying that they are the mice satisfying some basic properties? I am reading the CMIP of Steel (i.e I'm still stuck in the 90's) so I don't know how far Core Models have reached today but can one construct a Core Model $K$ for $\omega$ many Woodin Cardinals and a measurable above them and then somehow derive another core model from it which would satisfy determinacy? Thx.

REPLY [6 votes]: Certainly, $L(\mathbb R)$ is not a mouse (rather, a weasel) over a countable set, and the only way I see of thinking of it as a mouse and still capturing all the reals is making it a mouse over $\mathbb R$, in which case the answer is trivial. There is interesting work on $\mathbb R$-premice, but I do not think this is what you had in mind here. 
More relevantly, if determinacy holds, then $L(\mathbb R)$ is a derived model, which I think is how you should think of it. There is a key relationship between models of determinacy and models with limit many Woodin cardinals, given by the derived model theorem. This is significant result. It says, on the one hand, that a symmetric extension of any model of choice with limit many Woodin cardinals results in a model of $\mathsf{AD}^+$. On the other hand, by an appropriate use of Prikry forcing, one can weave together $\mathsf{HOD}$-like models inside models of $\mathsf{AD}^+$, and recover a model of choice with limit many Woodin cardinals. This is how several of the key equiconsistencies between versions of determinacy and appropriate instances of large cardinals have been proved. 
Derived models of (fine-structural) mice are special, of course, and there is a significant body of work on their properties. (And I think we can still say much more.)
A good reference for both topics ($\mathbb R$-premice and derived models of mice) is the paper

John R. Steel. Derived models associated to mice. In Computational prospects of infinity. Part I. Tutorials, pp. 105–193, Lect. Notes Ser. Inst. Math. Sci. Natl. Univ. Singap., 14, World Sci. Publ., Hackensack, NJ, 2008. MR2449479 (2009m:03082). 

(A preprint is available at John's page, where you will also find preprints of papers presenting the derived model construction in detail.)
Work on $\mathbb R$-premice leads to the notion of $K(\mathbb R)$, which indeed gives us a nontrivial way of thinking of some models of determinacy as mice, as we can think of "nice" models of $\mathsf{AD}^+$ as initial segments of $K(\mathbb R)$. (The precise meaning of niceness here is technical, and probably not too illuminating at this point. I mean that appropriate versions of the mouse set conjectures hold. John's paper covers this carefully.)
Work on the core model induction gives us that not just from large cardinals, but also from many strong combinatorial statements, we get many mice (over countable sets) inside $L(\mathbb R)$ (or inside general models of determinacy), and the fact that determinacy holds in $L(\mathbb R)$ can be thought of as a consequence of the presence of these mice. The key notion is the concept of a mouse operator, and a good introduction to this topic is the beginning of the monograph in preparation by John and Ralf:

Ralf Schindler and John R. Steel. The core model induction.

For the gap between CMIP and these papers, you may want to look at John's paper in the Handbook, a preprint of which can also be found at his page:

John R. Steel. An outline of inner model theory. In Handbook of set theory. Vols. 1, 2, 3, pp. 1595–1684, Springer, Dordrecht, 2010. MR2768698. 

You are probably interested in this general area, so you may enjoy knowing that we posted notes of many of the talks and background material at the page for the first conference on the core model induction and hod mice, Münster, July 19-August 06, 2010.
For the mouse set conjectures in particular, you want to read Grigor's thesis (posted at the conference site), or the preprint he wrote based on it, and available at his webpage:

Grigor Sargsyan. A tale of hybrid mice.<|endoftext|>
TITLE: Is there an explicit expression for the imaginary part of some non-trivial zero of zeta,
QUESTION [11 upvotes]: Is there an explicit expression for the imaginary part of some non-trivial zero of zeta, in terms of well-known constants, such as say $\gamma$ or $\pi$ say ?

REPLY [3 votes]: There is no explicit value for the imaginary part of the n-th zero.   However it satisfies a simple transcendental equation for each n,  whose solution is well approximated by the Lambert function.   See  LeClair and Franca  on arXiv,   math.NT<|endoftext|>
TITLE: Mean minimum distance for N random points on a unit square (plane)
QUESTION [9 upvotes]: A previously posted question "mean minimum distance for N random points on a one-dimensional line" produced an elegant answer: for a line of length L, the expected minimum distance (between random points) = L/(N^2/1), or simply 1/(N^2-1) for a line of unit length. 
I am wondering what the corresponding value would be for the unit plane? I am not a mathematician, but my intuition suggests the answer is simply 1/(N -1), given N points on a unit square (plane). 
Am I correct?

REPLY [5 votes]: This is a continuation to Douglas Zare's answer. I believe his lower bound of $1/n\sqrt{2}$ is the correct asymptotic behaviour. To see that, notice that all in inequalities in the calculation of the lower bound tend to equalities as $n\to \infty$, with the possible exception of the first inequality.
But in fact, the first inequality also tend to an equality when $n\to \infty$. This is because the error term is the measure of the overlap between discs of radius $x$ around the $k$ points and for $x=O(1/n)$, which is the relevant order of magnitude, this overlap is negligible - it's $O(k^2/n^4)$ whereas the measure of the discs is $\Theta(k/n^2)$.<|endoftext|>
TITLE: Uniformly integrable sequence such that a.s. limit and conditional expectation do not commute
QUESTION [13 upvotes]: Hi,
Could anyone give an example such that:
$$Y_i \rightarrow Y_{\infty}, \text{a.s.},$$
and $Y_i$'s are uniformly integrable.
But $\mathbb{E}(Y_i|\mathcal{G})$ does not converge a.s. to $\mathbb{E}(Y_{\infty}|\mathcal{G})$ for some sub-$\sigma$ algebra $\mathcal{G}$.
Thanks,
John

REPLY [13 votes]: Construct (on a suitable product of probability spaces) two independent sequences $(X_n)_{n\in\mathbb N}$ and $(Z_n)_{n\in\mathbb N}$ of positive random variables such that $E(X_n) \to 0$ but
$X_n$ does not converge a.s., $E(Z_n)=1$ and for each $\omega\in\Omega$ there is $n\in\mathbb N$ such that $Z_m(\omega)=0$ for $m\ge n$. Then $Y_n=X_nZ_n$ converges to $0$
(even point-wise) and in $\mathscr L_1$ (because of the positivity and $E(Y_n)=E(X_n)E(Z_n)$) but for the $\sigma$-algebra 
$\mathscr G =\sigma (X_n:n\in\mathbb N)$ you can pull out the measurable factor to obtain  $E(Y_n | \mathscr G)= X_n E(Z_n|\mathscr G) =X_n E(Z_n)=X_n$.

Although this is known it is not as well-known as it should be (a colleague once showed me a book on probability theory where the a.s. convergence of the conditional expectations was claimed).<|endoftext|>
TITLE: Applications of topological chiral homology and factorization algebras (aka higher Hochschild cohomology)
QUESTION [13 upvotes]: I recently heard a talk about these topics and found them very interesting.
The talk was centered on the formal structure and didn't really focus on examples.
So my question is: what is your favorite application of topological chiral homology? (or its other variants and specialisations)

REPLY [12 votes]: It might be worth pointing out that topological chiral homology is a specialization to the topological (or locally constant) setting of a construction that originated (AFAIK) in conformal field theory, namely the functor of conformal blocks for a vertex algebra and its derived formulation
as chiral homology, due to Beilinson and Drinfeld.
As such there are loads of concrete applications and calculations.
Perhaps the best known is the proof of the Verlinde formula, calculating the dimension of spaces of nonabelian theta functions on a Riemann surface by realizing them as conformal blocks and using the gluing properties of conformal field theory to reduce to a calculation in the representation theory of loop groups. (Recently Gaitsgory gave a derived enhancement of this calculation,
showing that the full chiral homology of the integrable level k vertex algebra of a loop group is still the space of nonabelian theta functions).
Perhaps my favorite application still is the one that motivated Beilinson and Drinfeld to introduce chiral homology in the first place. Namely, they wished to construct special $D$-modules on the moduli of $G$-bundles on a Riemann surface which were eigen-objects for Hecke functors, with eigenvalues particular $G^\vee$-local systems on the surface. (This is the central goal of the geometric Langlands program). They achieved this goal when the local system is a so-called "oper" by showing first how to use the Feigin-Frenkel theorem (a loop algebra version of the Harish Chandra isomorphism for the center of the enveloping algebra) to solve a local version of the problem.
In other words, they showed that Feigin-Frenkel provide representations of the loop algebra which are eigenobjects for Hecke functors. But then the mechanism of conformal blocks (or chiral homology) allowed them to globalize this construction on a Riemann surface, resulting in the desired global objects. The in-depth examination that resulted of what vertex algebras really mean geometrically led to the notion of factorization algebras and chiral homology which are now becoming so influential, thanks to work of (alphabetically) Costello, Francis, Gaitsgory, Gwilliam, Lurie, Rozenblyum and others.
The Beilinson-Drinfeld idea of chiral homology captures in a profound geometric way the adelic structure of moduli spaces of bundles on curves. The recent work of Gaitsgory and Lurie (as well as Rozenblyum, Barlev, Zhu and others) greatly furthers the Beilinson-Drinfeld dream of realizing the entire geometric Langlands correspondence on a Riemann surface as the "integral" (chiral homology) over the surface of a simpler local statement. 
In short, chiral homology is providing a formula for the "path integral" over a Riemann surface, telling us how to integrate more categorical but simpler local invariants to get global invariants. 
For an introduction to more recent developments (and in particular the migration of factorization algerbas outside of the confines of CFT or TFT), I recommend Costello's ICM address (about recovering the Witten genus through factorization algebras).
Also in a sense made precise I think by Ayala and Rozenblyum, 
all the values of any extended TFT (in the sense of the cobordism hypothesis) are given by calculating chiral homology of an appropriate local quantity (i.e., a higher category is a kind of "partially defined E_n algebra"), so in principle the examples are everywhere!<|endoftext|>
TITLE: The four-square theorem from the Gauss-Legendre three-square theorem
QUESTION [7 upvotes]: I've been studying some proofs of the four-square theorem. Some of them are pretty clear. However, I came across a statement that the four-square theorem can be easily derived from Gauss-Legendre three-square theorem.
Hard as I tried, I couldn't find out how to do it.
I was hoping someone could give me some idea or point out some article.
Thank you!

REPLY [8 votes]: Also, the Gauß/Legendre theorem in question implies the following (improved) version of the four-squares theorem:
Every positive integer is the sum of four POSITIVE squares unless it belongs to the set
$A \cup B$
where
$A$ $=$ {$1, 3, 5, 9, 11, 17, 29, 41$}
and
$B$ $=$  {$2\cdot 4^{m}: m \in \mathbb{Z}^{+}$} $\cup$ {$6\cdot 4^{m}: m \in \mathbb{Z}^{+}$} $\cup$ {$14\cdot 4^{m}: m \in \mathbb{Z}^{+}$}.
Proof (d'après Prof. J. H. Conway in "The sensual quadratic form" [page 140]). By Gauß/Legendre and the well-known result on numbers that can be written as a sum of two squares it follows that every natural number of the form $8k+3$ (or $8k+6$) is the sum of three positive squares. Multiplying by $4$, we see that the same conclusion applies to natural numbers of the forms $32k+12$ and $32k+24$. Then, one can show that any integer $>49$ which is not a multiple of $8$ is the sum of four positive integers by subtracting an square so as to obtain a number of one of the aforementioned forms (for instance, from a number of the form $8k+2$ subtract $2^{2}$ in order to obtain a number of the form $8k+6$...). The proof is completed by checking the numbers up to $49$ and verifying that a number $n$ divisible by $8$ is the sum of four positive squares only if $\frac{n}{4}$ is. QED.<|endoftext|>
TITLE: Reference for a nice proof of "undetermined coefficients"
QUESTION [12 upvotes]: I'm teaching an honors differential equations class and have been using linear algebra heavily.  I thought it would be interesting to include a proof of the method of undetermined coefficients along the lines described below, which I figured is already known.  Nonetheless, Google turns up no proof other than the obvious one by integration, which is significantly messier and uses no linear algebra, so perhaps I'm wrong.

What is the (standard, best, earliest) reference to the following proof?

Note: I have made a conscious decision not to ask this on math.stackexchange since, based on my last question asking about a reference there, I don't think it will even be noticed.
Theorem: Let $L$ be a constant-coefficient $n$th order linear differential operator with characteristic polynomial $p(\lambda)$, and let $y(t) = t^m e^{qt}$ for some integer $m \geq 0$ and $q \in \mathbb{C}$.  Suppose that $q$ is a $k$-fold root of $p(\lambda)$.  Then the equation $Lx = y$ has a solution of the form $t^k f(t) e^{qt}$, where $f(t)$ is a polynomial of degree $m$.
Proof outline:

$y$ is a solution of the equation $L_2 y = 0$, where $L_2 = (\tfrac{d}{dt} - q)^m$.  The pair of equations $Lx = y, L_2 y = 0$ can be written as a system of $n + m + 1$ first-order equations via the linear transformation $\vec{z} = (x, x', \dots, x^{(n - 1)}, y, y', \dots, y^{(m)})$, for which the coefficient matrix is:
$$M = \left(\begin{array}{c|c}
A & \begin{smallmatrix} 0 & \cdots & 0 \\ \vdots & & \vdots \\ 1 & \cdots & 0 \end{smallmatrix} \\ \hline
0 & C
\end{array}\right)$$
where $A$ is the companion matrix of $L$ and $C$ is the companion matrix of $L_2$.
It is easily verified that:

The eigenvalues of $M$ are the same as those of $A$, with an additional $m + 1$ repetitions of $q$; each eigenvalue $\lambda$ has a one-dimensional eigenspace, spanned by $(1, \lambda, \dots, \lambda^{n - 1}, p(\lambda), \lambda p(\lambda), \dots, \lambda^m p(\lambda))$.
Every Jordan chain for $A$, namely, vectors $\vec{v}_i$ associated with an eigenvalue $\lambda$ such that $(A - \lambda I) \vec{v}_i = \vec{v}_{i - 1}$, upgrades to vectors $\vec{u}_i$ with the same property for $M$, by attaching $m + 1$ zeros to the end.

It follows that $M$ has a generalized eigenbasis consisting of upgrades of a generalized eigenbasis for $A$, together with $m + 1$ additional generalized eigenvectors filling out a Jordan chain of length $k + m + 1$ for eigenvalue $q$.
A generalized eigenvector $\vec{v}_h$ of eigenvalue $\lambda$ at the top of a Jordan chain produces a fundamental solution to $\vec{z}' = M\vec{z}$ via the matrix exponential:
$$e^{\lambda t} \sum_{i = 0}^{h - 1} \frac{1}{i!} t^i \vec{v}_{h - i}.$$
The general solution is therefore the general linear combination of these.  Taking the first components (which we may assume are nonzero only for the eigenvectors (indexed $\vec{v}_1$ in the above sum)), we get a solution of $Lx = y$.  The terms coming from "upgrades" are part of the homogeneous solution to $Lx = 0$, and the remaining terms are some linear combination of $t^k e^{qt}, \dots, t^{k + m} e^{qt}$, as claimed.

REPLY [2 votes]: Hi Ryan and everybody,
Besides the very beautiful proof by Tao, a very nice and easy linear algebra approach to the undetermined coefficients method can be found in C. C. Ross ``Why the method of undetermined coefficients works'', Am. Math. Monthly 98 (1991), pp. 747-749.
I also suggest R. C. Gupta ``Linear differential equations with constant coefficients: a recursive alternative to the method of the undetermined coefficients'', Int. J. Math. Educ. Sci. Technol. 27 (1996), pp. 757-760.
A very easy formula (quite good for the students) that gives a particular solution of the equation $$L(\frac{d}{dt})y=t^m e^{q t}, \ q \in \mathbb C,$$ 
where $L$ is a nth order linear differential operator with constant coefficients, can be found in O. R. B. de Oliveira ``A formula substituting the undetermined coefficients and the annihilator methods'', Int. J. Math. Educ. Sci. Technol. 44 (3), 2013, pp. 462-468, http://dx.doi.org/10.1080/0020739X.2012.714496. The proof of the formula is in p. 465 and it is short and very elementary. 
The idea is the following: the given ode has a particular solution $y=f(t)e^{q t}$ where $f$ is a polynomial that satisfies the following ode with constant coefficients
$$\frac{p^{(n)}(q)}{n!}f^{(n)} + \cdots + \frac{p'(q)}{1!}f' + \frac{p(q)}{0!}f= t^m,$$
where $p=p(\lambda)$ is the characteristic polynomial associated to $L$. The short and elementary proof of this formula, with some examples, is over there. Using it, we find a particular solution through solving a triangular linear system.
The same formula shows that if $q$ is a root of multiplicity $k$ of $p(\lambda)=0$ then $e^{qt},\ldots, t^{k-1}e^{qt}$ are solutions of the homogeneous equation $Ly=0$, since we have $p(q)=\cdots=p^{(k-1)}(q)=0$ and $\frac{d^l}{dt^l}(t^j)=0$ if $l>j$. Furthermore, in this case the formula boils down to
$$\frac{p^{(n)}(q)}{n!}f^{(n)} + \cdots + \frac{p^{(k)}(q)}{k!}f^{(k)} = t^m,$$
which shows that $f^{(k)}$ is a polynomial of order $m$. Integrating it $k$-times we are able to choose $f(t)=t^kg(t)$, where $g$ is a polynomial of order $m$.
Oswaldo R. B. de Oliveira<|endoftext|>
TITLE: Group G hasn't all conditions of Lie group
QUESTION [9 upvotes]: Is there a group $G$ with the property that $G$ is a smooth manifold, the multiplication map of $G$ is smooth, but the inversion map of $G$ is not smooth?

REPLY [5 votes]: Every Čech-complete paratopological group is a topological group. That means that for Čech-complete groups you do not have to require the continuity of the inverse, continuity of multiplication suffices. Every manifold is Čech-complete. Using the affirmative answer to Hilbert’s fifth problem we get that every paratopological group on a manifold is actually a Lie group uniquely determined by the topological group structure.
In the spirit of Martin let me give a correct (I hope I have not forgotten anything) definition which is even wronger than the definition without the inverse:

A topological space $G$ with a function $\cdot\colon G^2\to G$ is called an $n$-dimensional Lie group if and only if

$G$ is second-countable
There exists an injective, open continuous map $\iota\colon \mathbb{R}^n\to G$
For every $g\in G$ the map $x\mapsto g\cdot x$ is continuous and surjective
There exists $e\in G$ such that $x\mapsto x\cdot e$ is the identity
For every $g\in G$ the map $x\mapsto x\cdot g$ is continuous
$\cdot$ is associative<|endoftext|>
TITLE: Does erosion mix faster than a riffle shuffle?
QUESTION [18 upvotes]: It is a famous result of Aldous and Diaconis1 that 

seven shuffles are necessary and suffice to approximately 
  randomize 52 cards.2

Here the shuffles are the standard riffle shuffle, where the deck 
of cards is cut
approximately in half and then the two halves "riffled" (interleaved) together.
The randomization is measured by the total variation distance
to the uniform distribution.
(References below.)
I was exploring a rough model of soil erosion that can
be viewed (with some distortion) as a 3D riffle shuffle.
Let me explain it in 2D.
Rather than a linear deck of $n$ cards, suppose the cards are arranged
in a square $\sqrt{n} \times \sqrt{n}$ matrix $M$.
Below I show a $6 \times 6$ matrix of cards/numbers.
The matrix is cut into two halves $A$ and $B$, 
and then reassembled by randomly selecting
items from matrix entries on either side of the cut interface.
Note that here there is a (perhaps significant?) difference
from riffle shuffling, which selects from the top of both
halves, rather than selecting from either side of the cut.
Just as in the Aldous/Diaconis work, the probability of
taking from $A$ is $|A|/(|A|+|B|)$, and similarly from $B$.
The "interface" from which an entry might be selected drills
down the columns, i.e., any entry that is exposed above might
be selected.
The process is illustrated below in several snapshots,
leading to a reassembled matrix $M'$.

     


One might view the "cards" in the matrix as units of soil or rock, 
and the halving cut
as a fissure through which water carries away the soil and deposits
in a new matrix $M'$.
My question is simple:

Q. Does this 2D shuffle, or its 3D analog, mix the $n$ elements
  faster than a riffle shuffle of $n$ items in a linear deck of cards?

In 3D, the $n$ items in are in a $(n^{\frac{1}{3}})^3$ cube.
My hunch answer to Q is Yes, because the interface from
which the items are drawn is "larger" in some sense.
But given the complexity of the Aldous/Diaconis analysis,
I do not trust my intuition.
Thanks for insights or references!

(Detail added for clarity.) The matrix $M'$ is filled row-by-row, left-to-right.  Here
is an accounting of the above example.

 $B$ is chosen. Among $(19,20,21,22,23,24)$, $24$ is selected and placed in the $(1,1)$ cell of $M'$.
 $A$ is chosen. Among $(13,14,15,16,17,18)$, $16$ is selected and placed in the $(1,2)$ cell of $M'$.
 $A$ is chosen. Among $(13,14,15,10,17,18)$, $17$ is selected and placed in the $(1,3)$ cell of $M'$.
 $A$ is chosen. Among $(13,14,15,10,11,18)$, $10$ is selected and placed in the $(1,4)$ cell of $M'$.
 $A$ is chosen. Among $(13,14,15,4,11,18)$, $15$ is selected and placed in the $(1,5)$ cell of $M'$.
 $B$ is chosen. Among $(19,20,21,22,23,30)$, $19$ is selected and placed in the $(1,6)$ cell of $M'$.
 Etc., next filling the $(2,1)$ cell of $M'$ with $20$, and so on.


1
David Aldous and Persi Diaconis, 
"Shuffling cards and stopping times." 
American Mathematical Monthly 93, 1986, 333-48.
(JSTOR link)
2
David Austin,
"How Many Times Do I Have to Shuffle This Deck?"
MAA Feature Column. (MAA link)

REPLY [6 votes]: I didn't understand the exact details of the shuffling method, so I cannot completely answer the question. However, it seems like the entropy you introduce in each shuffle is linear in $n$ (as in riffle shuffle). Since the entropy of a uniform permutation is $\Theta(n \log(n))$, you need at least $\Theta(\log(n))$ shuffles, so the mixing cannot be faster then riffle shuffle by more than a multiplicative constant.<|endoftext|>
TITLE: An expander (?) graph
QUESTION [13 upvotes]: For a prime $p$, consider the graph on the vertex set ${\mathbb F}_p$, in
which every vertex $z$ is adjacent to $z\pm 1$ and also to $z^{-1}$ (unless
$z=0$). I was told that this graph is known to be an expander, but the person
who told me this couldn't recall where exactly this graph has been studied.
Does anybody know the reference? Thanks!

REPLY [18 votes]: Here is Theorem 4.4.2 in Lubotzky's lovely book "Discrete groups, expanding graphs and invariant measures". On the projective line over $\mathbb{F}_p$, connect $z$ to $z\pm 1$ and to $-\frac{1}{z}$; this is a family of 3-regular expander graphs. The proof is indeed based on Selberg's 3/16-theorem.<|endoftext|>
TITLE: Where was the arithmetic zeta function of a scheme first defined?
QUESTION [8 upvotes]: Let $X$ be an arithmetic scheme, that is, a scheme of finite type over the integers. We denote the set of closed points of $X$ by $|X|$. For every $x\in|X|$, write $N(x)$ for the cardinality of the residue field $\kappa(x)$.
The arithmetic zeta function of $X$ is defined as
$$\zeta_X(s)=\prod_{x\in|X|}\frac{1}{1-N(x)^{-s}}.$$
This definition (up to a change in variable) can be found in

A. Grothendieck, Formule de Lefschetz et rationalité des fonctions $L$, Séminaire Bourbaki  279 (1964), 41-55.

Grothendieck attributes this definition to Weil, but as far as I know, Weil only defined the Hasse-Weil zeta function: if $X$ is a smooth projective variety over $\mathbb{F}_q$ and $N_r=|X(\mathbb{F}_{q^r})|$, then
$$Z_X(t)=\exp\left(\sum_{r=1}^\infty N_r(X)\frac{t^r}{r}\right).$$
Of course, it is easy to show these two functions satisfy
$$\zeta_X(s)=Z_X(q^{-s}),$$
but Weil did not address the notion of the zeta function of a scheme, at least not in the original paper:

A. Weil, Numbers of solutions of equations in finite fields, Bull. Amer. Math. Soc. 55 (1949), 497-508.

Where was the zeta function of an arithmetic scheme first defined? If anyone knows the actual paper in which this first appears, that would be optimal. (Of course, it's very possible that this definition was well-known but unpublished for some time; that would be an acceptable answer too.)

REPLY [5 votes]: I accidentally found the answer to this question in the book "The Abel Prize. 2003-2007 The First Five Years", page 74:

"A first lecture on zeta and L-functions in the setting of the theory
  of schemes (of finite type over $\mathrm{Spec}(\mathbb{Z})$) was given
  by Serre in [S112, 64(1965)]."

The paper is, of course,

Jean-Pierre Serre, Zeta and L functions (1965) Arithmetical Algebraic Geometry (Proc. Conf. Purdue Univ., 1963) pp. 82-92 Harper & Row, New York [also in Collected Papers, vol. II, n° 64)]

This confirms Tim Dokchitser comment.
On a side note, the concept of scheme didn't even exist at the time of Weil's paper!<|endoftext|>
TITLE: When does the direct image functor nicely push past the power/exists functor?
QUESTION [5 upvotes]: Let $D$ and $E$ be toposes and let $f_{\ast}\colon D\to E$ be the direct image part of a geometric morphism $(f^{\ast},f_{\ast})$ between them. Considered as categories, we have (covariant) power-object endofunctors on each:
$$P_D\colon D\to D \hspace{.5in} P_E\colon E\to E$$
where, for a morphism $\phi$ in $D$ we have $P_D(\phi)=\exists_\phi$, sending a sub-object of the domain to its image under $\phi$.
I'm trying to construct a natural transformation $$A_f\colon\ f_{\ast}\circ P_D\to P_E\circ f_{\ast}$$
of functors $D\to E\ $.
Question: For what geometric morphisms $(f^\ast,f_\ast)$ is such a natural transformation $A_f$ guaranteed to exist?
For example, such a thing exists in the case of change-of-base morphisms between slice toposes of ${\bf Set}$. If $q\colon X\to Y$ is a function, it induces a logical morphism $\Pi_q\colon {\bf Set}/X\to {\bf Set}/Y$. In this case the natural transformation $$A^~_{\Pi^~_q}\colon\Pi_q\circ P_{{\bf Set}/X}\to P_{{\bf Set}/Y}\circ \Pi_q$$
exists. It acts fiberwise on $Y$; for each $y\in Y$ it sends a $q^{-1}(y)\ $-indexed collection of subsets to their product.
How to construct this map $A_f$ in general? I wanted to use what would generalize to the morphism $f_{\ast}f^{\ast}\Omega_E\to\Omega_E\ $ induced by the mono-part of the epi-mono factorization for $\Omega_E\to f_{\ast}f^{\ast}\Omega_E$, where $\Omega_E$ is the subobject classifier in $E$. But while such a map does exist for all geometric morphisms, and can be used to construct the components of my desired $A_f$, I couldn't see how to prove that the naturality squares for $A_f$ commute. I showed it in the ${\bf Set}$ case using a basic set-theoretic argument.
So what made it work for slice toposes of ${\bf Set}$? Was it very specific to that case? Was it that these change-of-base functors are logical morphisms, or that they're essential geometric, or does such an $A_f$ always exist?

REPLY [9 votes]: A different way to describe the same answer that Zhen and Todd arrived at is to work in the internal logic of $E$.  That way we may pretend that $f_*$ is the global sections functor $\mathrm{Hom}(1,-) : D \to \mathrm{Set}$, as long as we treat $\mathrm{Set}$ constructively.  Then we have the components of a putative natural transformation
$$ \mathrm{Hom}(1,P A) \to P(\mathrm{Hom}(1,A)) $$
which, under the universal property of power objects $ \mathrm{Hom}(1,P A) \cong \mathrm{Sub}(A)$, sends a subobject $S\rightarrowtail A$ in $D$ to the set of all global sections $1 \to A$ which factor through it.  The naturality square for $p:A\to B$ requires that if we take the direct image subobject $p_!(S)$, then a global section of $B$ factors through $p_!(S)$ just when it lifts to some global section of $A$ factoring through $S$.  It's easy to see that this is the same as asking that $1\in D$ be projective, which is equivalent to saying that the global sections functor $f_* = \mathrm{Hom}(1,-)$ preserves epimorphisms.

REPLY [5 votes]: Since every power object is an internal Heyting algebra, and $f_*$ preserves the structure of internal Heyting algebras, there are trivial examples of such natural transformations corresponding to the constants $\top$ and $\bot$. Of course, this is uninteresting. 
Let me write $P^{\mathcal{D}}$ and $P^{\mathcal{E}}$ for the respective contravariant power object functors. Since $f_*$ preserves monomorphisms, there is a canonical comparison morphism $f_* \Omega_{\mathcal{D}} \to \Omega_{\mathcal{E}}$; since $f_*$ preserves products, there is a canonical natural morphism $f_* (Y^X) \to (f_* Y)^{f_* X}$; and so there is a canonical natural morphism $f_* P^{\mathcal{D}} X \to (f_* \Omega_{\mathcal{D}})^{f_* X} \to P^{\mathcal{E}} f_* X$. So there is an interesting canonical natural transformation $\theta : f_* P^{\mathcal{D}} \Rightarrow P^{\mathcal{E}} f_*$. 
Now allow me to argue using generalised elements. Let $T$ be an arbitrary object of $\mathcal{E}$, and let $p : X \to Y$ be a morphism in $\mathcal{D}$. Given a generalised element $t : T \to f_* P_\mathcal{D} X$, what is $\theta_X \circ t : T \to P_{\mathcal{E}} f_{\ast} X$, and what is $f_* \exists_p \circ t : T \to f_* P_{\mathcal{D}} Y$? Let $t' : f^* T \to P_\mathcal{D} X$ be the left adjoint transpose of $t$, and let $A' \rightarrowtail X \times f^* T$ be the subobject classified by $t'$. It is clear that $\theta_X \circ t$ is just the classifying morphism for the pullback of $f_* A' \rightarrowtail f_* X \times f_* f^* T$ along $f_* X \times T \to f_* X \times f_* f^* T$. Also, by naturality, $f_* \exists_p \circ t$ must be the right adjoint transpose of $\exists_p \circ t' : f^* T \to P_\mathcal{D} Y$, which is none other than the classifying morphism for the image of the composite $A' \rightarrowtail X \times f^* T \to Y \times f^*T$.
This suggests the crucial criterion is that $f_*$ preserve epimorphisms (and hence, epi–mono factorisations) – and this automatic for all base change morphisms for slices over $\textbf{Set}$ because $\textbf{Set}$ and its slices have the axiom of choice. So assume $f_*$ preserves epimorphisms. If we write $A \rightarrowtail f_* X \times T$ for the subobject classified by $\theta_X \circ t$, $B' \rightarrowtail Y \times f^* T$ for the image of $A' \rightarrowtail X \times f^* T \to Y \times f^* T$, and $B \rightarrowtail f_* Y \times T$ for the subobject classified by $\theta_Y \circ f_* \exists_p \circ t$, then the preservation of epi–mono factorisations implies that $f_* B'$ remains the image of $f_* A'$ under $f_* X \times f_* f^* T \to f_* Y \times f_* f^* T\ $; but epi–mono factorisations are stable under pullback in a topos, hence $B$ is the image of $A$ under $f_* X \times T \to f_* Y \times T$. Thus, we have
$$\theta_Y \circ f_* \exists_p \circ t = \exists_{f_{\ast} p} \circ \theta_X \circ t$$
for all generalised elements $t : T \to f_* P_{\mathcal{D}} X$, and thus $\theta$ is also a natural transformation $f_* P_{\mathcal{D}} \Rightarrow f_* P_{\mathcal{E}}$.<|endoftext|>
TITLE: The "interplay" between additive and multiplicative structure in a field
QUESTION [8 upvotes]: A field is an ordered triple $(F, +,\cdot)$ of a set $F$ and binary operations $+,\times$ on $F$ such that $(F,+)$ and $(F\backslash 0,\times)$ are abelian groups satisfying the distributive laws
$\forall a,b,c \in F:\begin{cases} a\times(b+c)=(a\times b)+(a\times c)\newline (a+b)\times c= (a\times c)+(b\times c) \end{cases}$.
Someone mentioned that the "interplay" between the additive structure $(F,+)$ and the multiplicative structure $(F\backslash 0,\times)$ in a field is still "not well understood". Another friend mentioned that the distributive laws fully characterize the relationship between $+,\times$ by definition, but that these laws are "not fully understood".
What is this deep understanding? What is known and what is unknown? Restrict this question to specific fields if necessary, like finite.
EDIT: I should clarify that this came up during a discussion of Wang's attacks on hash functions in cryptography.

REPLY [11 votes]: This answer is just adding flesh to @Frank Thorne's earlier answer. He noted that the idea of "addition and multiplication interacting" comes up in additive combinatorics. Perhaps the most obvious instance of this is in the study of the sum-product phenomenon (SPP).
Roughly speaking the SPP asserts that any subset of a field $F$ must ``grow'' under either addition or multiplication. The way one makes this rough statement precise depends on the field in question. Let me consider two instances:
Suppose $F=\mathbb{R}$. In this situation the central conjecture is due to Erdos and Szemeredi:

For every $\varepsilon\in (0,1)$, there exists $c>0$, such that for $A$ a finite subset of $\mathbb{R}$,
  $\max(|A+A|, |A\cdot A|) \geq c |A|^{2-\varepsilon}$.

This conjecture is still open, however progress has been made. The strongest statement is (I believe) due to Solymosi, but it's also worth mentioning the work of Elekes. With a very simple argument, he connected SPP to questions in incidence geometry in the plane and to the  idea of the crossing number in $\mathbb{R}^2$ to prove:

There exists $c>0$, such that for $A$ a finite subset of $\mathbb{R}$,
  $\max(|A+A|, |A\cdot A|) \geq c |A|^{5/4}$.

One last remark - another way of thinking about the Erdos-Szemeredi conjecture is this: it says that a set $A$ of real numbers cannot simultaneously be both a geometric progression and an arithmetic progression (since, by results of Freiman and others, these are the classes of sets that do not grow under multiplication and addition, respectively).
Suppose $F=\mathbb{Z}/p\mathbb{Z}$. In this situation, the central result is due to Bourgain, Katz and Tao:

For every $\delta>0$ there exists $\varepsilon>0$ and $c>0$, such that for $A$ a finite subset of $\mathbb{Z}/p\mathbb{Z}$ with $|A| < p^{1-\delta}$, we have
  $\max(|A+A|, |A\cdot A|) \geq c |A|^{1+\varepsilon}$.

The statement is slightly different to that in $\mathbb{R}$ because it is clear that sets that are almost as large as the field itself cannot possibly grow.
This result has been generalized in various ways to arbitrary finite fields. However in this more general setting one has to deal with the presence of finite subfields (again this does not crop up in $\mathbb{R}$), and so statements tend to be slightly technical. There is also a wealth of work giving values for $\varepsilon$ when $\delta$ is, say, $\frac12$, as well as a lot of work connecting this result to geometry over finite fields (in the spirit of the work of Elekes).<|endoftext|>
TITLE: Multisimplicial geometric realization
QUESTION [13 upvotes]: Does anyone know a reference or proof for the following?  Let $k\geq 1$ and let $X$ be a space.  There is a $k$-fold multisimplicial set whose simplices in degree $n_1,\ldots,n_k$ are the maps $\Delta^{n_1}\times\cdots\times\Delta^{n_k}\to X$.  There is a map from the geometric realization of this multisimplicial set to X.  I'd like to know that this map is a weak equivalence, or at least a homology isomorphism.

REPLY [2 votes]: [This answer is mostly a long comment to Peter May's answer.]
Edit: I have corrected some arrows which were pointing the wrong way.$\newcommand{\real}[1]{\lvert #1\rvert}
\newcommand{\Map}{\operatorname{Map}}
\newcommand{\Top}{\mathrm{Top}}
\newcommand{\sSet}{\mathrm{sSet}}
\newcommand{\diag}{\operatorname{diag}}
\newcommand{\To}{\longrightarrow}
\newcommand{\DeltaOp}{\Delta^{\mathrm{op}}}$
I was unable to prove the map $X\to\real{Y}$ that Peter May uses in his answer is always a weak equivalence. Unfortunately, the answer Peter links to gives no details. The result would hold if $Y$ were Reedy cofibrant, but it does not seem to be.
So here is an alternative method which proves directly that the map $\real{SY}\to X$ is a weak equivalence, thus answering the question. It avoids any use of realizations of (multi) simplicial spaces, which are problematic by not always being homotopy invariant. Perhaps there is an easier method. In any case, I want to make clear that Peter's idea to use the multi-simplicial space $Y$ is absolutely fulcral. I am essentially dealing with some technical details.
Let $X$ be a space. Recall that the object $Y:(\DeltaOp)^{\times(k-1)}\to\Top$ from Peter's answer is a multi-simplicial space whose space of $(n_2,\cdots,n_k)$-simplices is the space of maps $\Map(\Delta^{n_2}\times\ldots\times\Delta^{n_k},X)$. Let $cX$ be the constant $(k-1)$-fold multi-simplicial space with value $X$. Then we have an obvious map $cX\to Y$ which is objectwise a homotopy equivalence — as Peter remarks — due to the contractibility of products of simplices.
Let $S:\Top\to\sSet$ be the singular complex functor taking a space $Z$ to the simplicial set whose set of $n$-simplices is the set of maps $\Top(\Delta^n,Z)$. Applying $S$ objectwise to $cX$ and $Y$, we get a map $S(cX)\to SY$ (where $SY$ actually stands for $S\circ Y$, and similarly for $cX$). Since $S$ preserves weak equivalences, then $S(cX)\to SY$ is objectwise a weak equivalence. Now we conclude the map on realizations $\real{S(cX)}\to\real{SY}$ is an equivalence. This follows from the fact that any functor $(\DeltaOp)^{\times(k-1)}\to\sSet$ is Reedy cofibrant: this is a straightforward generalization of proposition 15.8.7 of Hirschhorn; alternatively, it is a particular case of the main theorem (proposition 3.15) of the article "Reedy categories and the $\Theta$-construction" by Charles Rezk and Julia Bergner.
Let us now view functors $(\DeltaOp)^{\times(k-1)}\to\sSet$ as $k$-fold multi-simplicial sets instead. As Peter observes, $SY$ is then just the $k$-fold multi-simplicial set described in the question. So it suffices to prove that the natural map $\real{SY}\to X$ is a weak equivalence. But we already saw above that $\real{S(cX)}\to\real{SY}$ is a weak equivalence. Moreover, $\real{S(cX)}=\real{\diag(S(cX))}=\real{SX}$ as Tom Goodwillie observes in a comment above (together with the simple fact that $\diag(S(cX))=SX$). But $\real{SX}$ is well-known to be weakly equivalent to $X$ (the case $k=1$ of the question). Finally, this weak equivalence $\real{SX}\to X$ is the composite of the maps
$$ \real{SX}=\real{S(cX)}\To\real{SY}\To X $$
and the first map is a weak equivalence. In conclusion, the second map $\real{SY}\To X$ is also a weak equivalence, as desired.<|endoftext|>
TITLE: Are there results from gauge theory known or conjectured to distinguish smooth from PL manifolds?
QUESTION [15 upvotes]: My question begins with a caveat: I sometimes spend time with topologists, but do not consider myself to be one.  In particular, my apologies for any errors in what I say below — corrections are encouraged.
My impression is that manifold topologists like to consider three main categories of (finite-dimensional, paracompact, Hausdorff) manifolds, which I will call $\mathcal C^0$, $\mathrm{PL}$, and $\mathcal C^\infty$, corresponding to manifolds whose atlases have transition functions that are, respectively, homeomorphisms, piecewise-affine transformations, and diffeomorphisms.  The latter two categories obvious map faithfully into the first, and a theorem of Whitehead says that every $\mathcal C^\infty$ manifold admits a unique PL structure.
These categories are not equivalent in any reasonable sense.  The generalized Poincare conjecture is true in $\mathcal C^0$, true (except possibly in dimension $4$) in $\mathrm{PL}$, and false in many dimensions including $7$ in $\mathcal C^\infty$.  $\mathcal C^0$ is the realm of surgery and h-cobordism.  In $\mathcal C^\infty$, and in particular in $4$ dimensions, there is a powerful tool called "gauge theory", which provides the main technology used to prove examples of homeomorphic but not diffeomorphic manifolds.
By definition, gauge theory is that part of PDE that studies connections on principal $G$-bundles for Lie groups $G$.  The most important gauge theories for distinguishing between the $\mathcal C^\infty$ and $\mathcal C^0$ worlds are Donaldson Theory (which studies the moduli space of $\mathrm{SU}(2)$ connections with self-dual curvature) and the conjecturally equivalent Seiberg–Witten Theory (which studies an abelian gauge field along with a matter field, and which I understand less well).  Another important gauge theory that I understand much better is (three-dimensional) Chern–Simons Theory, whose PDE picks out the moduli space of flat $G$ connections; for example, counting with sign the flat $\mathrm{SU}(2)$ connections on a 3-manifold is supposed to correspond to the Casson invariant.
My impression, furthermore, has been that the categories $\mathcal C^\infty$ and $\mathrm{PL}$ are in fact quite close.  There are more objects in the latter, certainly, but in fact many of the results separating $\mathcal C^0$ from $\mathcal C^\infty$ in fact separate $\mathcal C^0$ from $\mathrm{PL}$.  A side version of my question is to understand in better detail the distance between $\mathcal C^\infty$ and $\mathrm{PL}$.  But my main question is whether the technology of gauge theory (possibly broadly defined) can be used to separate them.  A priori, the whole theory of PDE is based on smooth structures, so it would not be unreasonable, but I am not aware of examples.

Are there gauge-theoretic invariants of smooth manifolds that distinguish nondiffeomorphic but $\mathrm{PL}$-isomorphic manifolds?

Of course, a simple answer would be something like "For any $X,Y \in \mathcal C^\infty$, the inclusion $\mathcal C^\infty \hookrightarrow \mathrm{PL}$ induces a homotopy equivalence of mapping spaces $\mathcal C^\infty(X,Y) \to \mathrm{PL}(X,Y)$."  This would explain the impression I have that $\mathcal C^\infty$ and $\mathrm{PL}$ are close — if it is true, it is not something I recall having been told.

REPLY [9 votes]: First, it follows from the work of Kirby and Siebenmann that in dimensions $\le 6$ PL and DIFF categories are equivalent. In particular, if you are working in dimension 4 (where gauge-theoretic invariants are mostly used) then the answer to your your question is negative. Starting in dimension 7, there are smooth manifolds which are PL-equivalent but not diffeomorphic. Milnor's exotic 7-spheres are the first examples of such manifolds. The smooth structures on Milnor's spheres are distinguished via index and 1st Pontryagin class. Whether you consider such invariants gauge-theoretic or not, depends on how broadly you interpret gauge theory. For instance, characteristic classes of smooth manifolds can be defined via differential forms, i.e., Chern-Weil theory, or as indices of some elliptic operators. Does this qualify as gauge theory? (You can lift forms from, say, tangent bundle to the principal bundle- frame bundle, if you so desire.) The point of usage of gauge theory in dimension 4 is that the "traditional" topological invariants turned out to be insufficient, so one considers spaces of connections satisfying some differential equation (like self-duality) and uses such spaces to derive some smooth invariants of 4-manifolds. As far as I know, nobody used this viewpoint in higher (i.e., at least 7) dimensions, since there was no need for it.<|endoftext|>
TITLE: What is the difference between an automorphic form and a modular form?
QUESTION [49 upvotes]: This is more of a question about terminology than about math.
The term "automorphic form" is clearly a generalization of the term "modular form." What is not clear is exactly which generalization it is. Many sources use the term in different ways.
Any classical holomorphic modular form for $\mathrm{SL}_2(\mathbb{Z})$ is called a modular form, and usually (but not always) so are modular forms for congruence subgroups. Often, "automorphic form" is used when one considers either other Fuchsian groups, forms on groups other than $\mathrm{GL}_2$, or non-holomorphic forms (such as Maass forms and real analytic Eisenstein series). Alternatively, Diamond and Shurman define an "automorphic form" in Section 3.2 as being like a modular form but possibly meromorphic instead of holomorphic.
As another example, Miyake's book Modular Forms writes on p.114, "Automorphic functions and automorphic forms for modular groups are called modular functions and modular forms, respectively," and his definition of "modular group" seems to coincide with that of congruence subgroup.
The Princeton Companion to Mathematics writes in section III.21, "automorphic forms, which are generalized versions of the classical analytic functions called modular forms [III.61]," but it does not specify what the generalization is. The book contains other, similar statements (in III.61, "And indeed, automorphic forms, which are generalizations of modular forms").
So what exactly is the* definition of modular form as opposed to automorphic form? Since there is likely no "right" answer, what I really want to know is what is the history and what are the different conventions and the relations between them.

REPLY [6 votes]: There are many aspects to the relation between modular forms and automorphic forms. First let me recall what a modular form of weight $2k$ for, say $SL(2,{\mathbb Z})$ is. This is a function on the upper half plane, which is $holomorphic$, has an equivariance property 
$$f(\frac{az+b}{cz+d})=(cz+d)^{2k}f(z)$$ for 
$\begin{pmatrix} a & b \cr c & d\end{pmatrix}\in SL(2,{\mathbb Z})$ 
 and at infinity (the "cusp" for $SL(2,{\mathbb Z})$ in the fundamental domain) has the Fourier expansion 
$$f(z)= \sum _{k=0}^{\infty} a(n)q^n,$$  where $q=e^{2\pi i z}$ for $z$ in the upper half plane. For example, the Ramanujan $\Delta$ is such a modular form. But, $1/\Delta (z)$, while it is still holomorphic on the upper halp plane, has its Fourier expansion at infinity starting from $q^{-1}$ and is hence not a modular form.  
We can convert a modular  form into a function on the group $SL(2,{\mathbb R})$ by taking the related function $F_f(g)= (ci+d)^{-2k}f(g(i))$. This is a function on $SL(2,{\mathbb R})$ on which the centre of the enveloping algebra acts by scalars (translating the holomorphy condition on $f$), is $SL(2,{\mathbb Z})$ invariant (translating the equivariance for the modular form $f$),  and has moderate growth on the fundamental domain for $SL(2,{\mathbb Z})$ in $SL(2,{\mathbb R})$. 
This can be generalised to any congruence subgroup of the modular group, and any weight etc, and you get $K$-finite , $Z$ finite functions on $G$, which are $\Gamma$ invariant. These are called automorphic forms. If you do this construction for the inverse of $\Delta$ you get most properties, but the moderate growth, and hence you do not get an automorphic form.  
On the other hand, any function on $G/\Gamma$ (say, $G$ semi-simple and $\Gamma$ an arithmetic group) which is $K$ finite, $z$ finite and has moderate growth (along with its derivatives ) on a fundamental domain, is defined to be an automorphic form. So even for $SL(2,{\mathbb R})$ there are many more automorphic forms than those arising from modular forms (those from modular forms generally have infinite component of the discrete series (this keeps track of the weight $k$ of the modular form)).<|endoftext|>
TITLE: One question about the quandle
QUESTION [5 upvotes]: Given a finite quandle $Q$, for any knot $K$ one can associate an invariant, i.e. the number of proper colorings $p(K)$. Let us consider the inverse $K^{-1}$ and mirror image $K'$ of $K$. My queston is, is there any relation between $p(K^{-1})$ $(p(K'))$ and $p(K)$? In general, with an appropriate finite quandle, can we distinguish a knot from its inverse (mirror image) by counting the number of proper colorings?
It seems that the number of proper colorings is exactly the number of homomorphisms from the knot quandle of $K$ to the quandle $Q$ (trivial homomorphism corresponds to trivial coloring). Hence my question concerns the relation between the knot quandle of $K^{-1}$ $(K')$ and the knot quandle of $K$. I only know that the knot quandle is a complete invariant for unoriented knots. 
Any hints are welcome.

REPLY [3 votes]: Answers to some of your questions can be found in this paper:
  Quandle Colorings of Knots and Applications
  http://arxiv.org/abs/1312.3307<|endoftext|>
TITLE: Does there exist a regular graph of degree 4, diameter 3, 32 vertices?
QUESTION [6 upvotes]: Does there exist a regular graph of degree 4 having its diameter equal to three and its number of vertices equal to 32? What I know is http://www-ma4.upc.es/~comellas/delta-d/desc_g/desc_g3.html
( the second graph).

REPLY [9 votes]: Here's an adjacency list, it's 4-regular, so there should be 4 vertices per line.   I'm sure it will be easier for you to get the graph that for me to figure out the software interface here.
14    3    9   24
 5    8   25    3
30   16   21    7
 1    0   29   30
15   28   23   11
28   16   26    1
31   19   10   23
20   27    2   29
15   14    1   22
 0   13   17   28
 6   21   22   25
17   13   18    4
20   24   26   22
 9   11   25   20
 8    0   31   19
21    4    8   24
 5   31   20    2
 9   11   30   27
31   11   26   30
 6   28   14   27
13   16   12    7
10   24    2   15
12   10    8   27
26   29    4    6
12   21    0   15
10   13    1   29
12    5   23   18
19   17    7   22
 5   19    4    9
23    3    7   25
17    2    3   18
16   18    6   14

There appear to be many such graphs.

(Added by J.O'Rourke).
Here is the graph above, but with vertices labeled from $1$ to $32$ rather than $0$ to $31$:<|endoftext|>
TITLE: Is there a list of Kazhdan-lusztig polynomials?
QUESTION [9 upvotes]: When studying the combinatorics and representation of Coxeter groups, I often find it irksome to compute KL- and R- polynomials from scratch on Maple or Sage. The time it consumes to generate a complete list for all pairs of elements of a Coxeter group soon becomes unbearable as the magnitude of the group increases. 
Surprisingly, the only one I could find is a small one by Mark Goresky(link text) obtained in 1996. Is there a more comprehensive list of such polynomials?
It seems meaningful work to generate such lists, compared with calculating the next biggest known prime number.

REPLY [11 votes]: I have placed data files containing the polynomials for $S_n$ with $n\in \{4,5,6,7,8,9\}$  here.  The corresponding file for $S_{10}$ is a few gigabytes as a plain text file; I could certainly send that if you're interested.
This belongs as a response to Jim Humphrey's post, but I don't think I have the reputation for that, so: Patrick Polo's result (for which you can also find a combinatorial proof 
 by Caselli ) is a wonderful, important result:  Given a polynomial $f(q)$ of degree $d$ with constant term $1$ and nonnegative integer coefficients, Polo constructs a pair of permutations $x$ and $w$ living in some $S_N$ for which $P_{x,w}(q) = f(q)$.  But $N = 1 + d + f(1)$.  So even though $1+14q + 60q^2 + 96q^3 + 43q^4 + 4q^5$ appears as a Kazhdan-Lusztig polynomial for a pair of permutations in $S_{10}$, Polo's construction returns permutations in $S_{224}$.  There's still a lot to be learned about what can happen for small $S_n$.  Theory is going to be a crucial guide in studying these polynomials.  But studying the data as one would do in the physical sciences is also, I think, going to play an important role.<|endoftext|>
TITLE: Characters of p-groups
QUESTION [7 upvotes]: Berkovich mentioned the following result of Mann in his book on p-groups:
The number of nonlinear irreducible characters of given degree in a p-group is divided by p-1.
Do you know any reference for this statement? Also, I would like to ask you if a similar assertion holds for conjugacy classes.

REPLY [2 votes]: As already explained by Geoff, the number of conjugacy classes of size $>1$ is divisible by $p-1$. A reference is:  
A. Mann: Groups with few class sizes and the centralizer equality subgroup. Israel J. Math. 142(2004), 367-380. 
The result (with a proof quite similar to Geoff's) is on page 376 (after Corollary 18).<|endoftext|>
TITLE: Does a Riemannian manifold with bounded geometry admit an isometric proper embedding into Euclidean space with uniformly thick tubular neighborhood
QUESTION [18 upvotes]: Suppose $(M,g)$ is an open Riemannian manifold with bounded geometry, i.e., the injectivity radius is $\ge \epsilon>0$ and each iterated covariant derivative of curvature is bounded with respect to $g$.
Question: Does there exist an embedding into some high dimensional $\mathbb R^N$ with the following properties:

$\bullet$ isometric

$\bullet$ proper (i.e., compact sets have compact inverse images)

$\bullet$ The normal tubular neighborhood contains a uniformly thick disk in each fiber.
The Gauss equations give a relationship between the (normal bundle valued) second fundamental form (or shape operator); but by embedding arc-length parameterized curves one can show that this is not enough to force focal points uniformly away. 
There is a related question 
(here), but it does not answer this question.
By taking a product with another Euclidean space one can then have embedding where the normal bundle is trivial. Then one can use such embedding to carry over results valid on $\mathbb R^n$ to Riemannian manifolds with bounded geometry. 
Edit: Many thanks for the comments and Anton for a very succinct answer.
I end with a wild guess:
Maybe, one can characterize the Riemannian manifolds of bounded geometry admitting such embeddings as those whose ends are asymptotically flat times something compact (by which I mean an bundle)?

REPLY [23 votes]: It does not hold for hyperbolic plane.
It follows since the volume growth of the hyperbolic plane is exponential, while volume growth of $\mathbb{R}^N$ is polynomial.<|endoftext|>
TITLE: Principal maximal ideals in Z[x]/(F)
QUESTION [15 upvotes]: Is there some irreducible $F \in \mathbb{Z}[x]$ such that $\mathbb{Z}[x]/(F)$ has no principal maximal ideal? Equivalently, is it possible that the $1$-dimensional integral domain $\mathbb{Z}[x]/(F)$ has no prime element?
The maximal ideals have the form $(p,f)$, where $p \in \mathbb{Z}$ is a prime and $f \in \mathbb{Z}[x]$ is a monic polynomial such that $f \bmod p$ is an irreducible factor of $F \bmod p$. In particular, $F$ should be reducible modulo every prime number. See here for a classification of biquadratic polynomials with this property. But this does not suffice, as $F=x^4+1$ shows. This is irreducible, reducible modulo every prime number, but $(2,x+1)=(x+1)$ in $\mathbb{Z}[x]/(x^4+1)$. A better candidate seems to be $F=x^4-10x^2+1$, I have checked $(p,f) \neq (f)$ for some primes $p$.
I suspect that the question is connected with the class group of the curve $V(F) \subseteq \mathbb{A}^1_{\mathbb{Z}}$?
Background: The question is equivalent to the question if $\mathbb{Z}[x]$ wins in the game of integral domains, which is a simplification of the game of rings by Will Sawin.

REPLY [7 votes]: Let $A$ be an order in the ring of integers $O_K$ of a number field $K$.  We claim that there are infinitely many principal maximal ideals $P$ of $A$.  By using localization at rational primes, we have a bijection between the sets of maximal ideals of $A$ and $O_K$ with residue characteristic relatively prime to $N = [O_K:A]$ via $P \mapsto P \cap A$ and $P' \mapsto P'O_K$.  
In this way, we see that it is harmless to replace $A$ by a sub-order, so we may assume $A = \mathbf{Z} + M O_K$ for an integer $M > 0$. For $x \in 1 + MO_K$, we have $A \cap xO_K = xA$. Indeed, if $y \in O_K$ and $xy = c + Mt$ for $t \in O_K$ then we have to show that $y \in \mathbf{Z} + M O_K$, but this is clear since $xy \equiv c \bmod M O_K$ and $x \equiv 1 \bmod M O_K$.  Hence, if $xO_K$ is a prime ideal of $O_K$ then $xA$ is a prime ideal of $A$, so it suffices to construct infinitely many maximal ideals $P$ of $O_K$ admitting a generator congruent to 1 modulo $M O_K$.  
This latter formulation does not mention the order at all, and is a special case of the more general fact that for any nonzero ideal $J$ of $O_K$ whatsoever, $O_K$ has infinitely many principal maximal ideals $P$ admitting a generator $x \equiv 1 \bmod J$.  The existence of infinitely many such $P$ follows from the method of proof of the "abelian" case of the Chebotarev Density Theorem (using generalized ideal class characters in the role of Dirichlet characters in the proof of Dirichlet's theorem on primes in arithmetic progressions). So tacitly here we are using the basic analytic properties of $L$-functions attached to characters of generalized ideal class groups (which can be proved in various ways, such as using $\zeta$-functions and class field theory if one wants to be ahistorical).<|endoftext|>
TITLE: Can one compare monads arising from homotopy equivalent adjunctions?
QUESTION [7 upvotes]: Suppose we have a Quillen adjunction $L\colon {\mathcal C} \leftrightarrow {\mathcal D}: R$. For convenience let us assume that all objects of $\mathcal C$ are cofibrant and all objects of $\mathcal D$ are fibrant, so that $L$ and $R$ are homotopy functors. Let $(L', R')$ be another Quillen adjunction. Let's suppose that we have a natural transformation $\alpha_L\colon L \to L'$. It induces a natural transformation $\alpha_R\colon R' \to R$. Let us assume that $\alpha_L$ is a weak equivalence. If I am not mistaken, it follows that $\alpha_R$ is a weak equivalence.
Let $T$ and $T'$ be the monads $RL$ and $R'L'$ respectively. As far as I see, in general there is no map of monads between $T$ and $T'$.
 Question : Are the homotopy categories of $T$ algebras and $T'$-algebras equivalent? 
 Remark : The adjunction $(L, R)$ induces a Quillen adjunction between $T$-algebras and $\mathcal D$. This adjunction is sometimes (quite often?) a Quillen equivalence. In this case one may say that $T$ satisfies homotopy descent. It seems that if $T$ satisfies homotopy descent, then so does $T'$, and in this case Ho$(T-\mathrm{alg})$ and Ho$(T'-\mathrm{alg})$ are equivalent, since both are equivalent to Ho$(\mathcal D)$. I want to know if there is a direct way to compare algebras over $T$ and $T'$, without using descent.

REPLY [4 votes]: Hi Greg, I'll give an alternative, elementary, starting point 
towards an answer.  Dylan, there is a question about using the bar 
construction that seems maybe to be missing from your answer, and I don't 
understand your statements about dealing with multiplicative structure or
with uniqueness. Also, if you will forgive a little teasing, if your 
favorite tool is a sledgehammer, then every nail looks like a rock.
I'll use the two-sided bar construction from "The geometry of
iterated loop spaces", [12] on my web page; see [112] 
for a modernized discussion and some category theory relevant to the
question.   For a $T$-algebra $A$, define a $T'$-algebra $F(A)$ 
by $F(A) = R'B(L,T,A)$, where $T$ acts on the right of $L$ via $\epsilon R$.
This is a $T'$-algebra since $R'D$ is a $T'$-algebra for any $D\in \mathcal D$.
Similarly, For a $T'$-algebra $A'$, define a $T$-algebra $G(A')$ by 
$G(A') = RB(L',T',A')$. Of course, $F$ and $G$ are functors.
I assume that we have natural weak equivalences
$$\gamma'\colon B(R'L, T, A) \to R'B(L,T,A)$$
and 
$$ \gamma\colon B(RL', T', A') \to RB(L',T',A').$$
We expect the second to come from a weak equivalence of the general
form $|RX|\to R|X|$ for a simplicial object $X\in \mathcal C$, and similarly
for the first.  With $R = \Omega$ and $\mathcal C$ the category of based spaces, 
proving that there is such a weak equivalence $\gamma$ was the hardest thing 
technically in [12].  It is something like this that I find missing in your sketch,
Dylan, but if we can see directly that $B(R'L, T, A)$ is a $T'$-algebra (say by
commuting $T'$ past $B$) and similarly for $B(RL',T',A')$, then perhaps that is 
not needed here.
Since $T = RL$, we have the chain of natural weak equivalences (ignoring 
algebra structure)
$$ A \leftarrow B(T,T,A) \leftarrow B(R'L,T,A) \to F(A) $$
where the middle arrow is induced by $\alpha\colon R'\to R$, which I'll assume
given (rather than taking a conjugate). Similarly,
we have the chain of weak equivalences
$$ A' \leftarrow B(T',T',A') \rightarrow B(RL',T',A') \to G(A'). $$
Therefore $F(G(A'))\simeq A'$ and $G(F(A))\simeq A$.  It remains to prove
that these equivalences (or related ones) are equivalences of $T'$-algebras 
and $T$-algebras.  I haven't had time to try.<|endoftext|>
TITLE: On the category of virtual species
QUESTION [9 upvotes]: In Foncteurs analytiques et espèces de structures, Joyal defines virtual species, as a (quotient) of formal differences of functors $F,G:\mathbb{B}\rightarrow \mathsf{Set}$, and then proceeds to show that these form a (commutative) ring.  The quotienting operation is that of equivalence of species.
Has anyone instead studied what might be best termed the category of virtual species, where the quotient is not carried out?  The objects would be formal differences of species, and the arrows would be equivalences, basically "unpacking" the quotient groupoid.
As this seems relatively straightforward, I would expect that it has been carried out before.  The point here is that various isomorphisms (like $A+B \simeq B+A$) have computational content, and thus it is useful to work in a setting where these are representable.  When studying the computational content of $A+M=B+M$, one is quite naturally led to looking at virtual species as the right place to ask that question -- if only it were done categorically rather than algebraically!

REPLY [10 votes]: I had tried to do something like this around 1994, and then again a little later as I will explain in a moment. The first time around, I had tried formalizing this in the context of species valued in a category of "virtual sets" (finite sets, to avoid an Eilenberg swindle), guided by Joyal's pretty observation that the implication "$A + C \cong B + C$ implies $A \cong B$" definitely has computational meaning as you say: starting with a given bijection $f: A + C \to B + C$, feed back the outputs of $f$ that land in $C$ as inputs, and iterate to obtain a bijection $g: A \to B$. This operation is natural in the sense fully explained in Joyal-Street-Verity's paper, Traced Monoidal Categories. (Essentially the same observation appears in Conway and Doyle's paper, Division by Three, if I remember correctly.) 
One can categorify the taking of formal differences, starting with the category of finite sets and bijections and applying the tortile category construction in Traced Monoidal Categories. It's pretty simple in this case and one winds up with the compact closed category of oriented 1-cobordisms (so objects are oriented compact 0-manifolds, i.e., multisets of +'s and -'s, and morphisms are diffeomorphisms of oriented compact 1-manifolds with boundary). I tried developing a calculus of virtual species as functors from $\mathbb{B}$ to this category, but at some point the project petered out. Unfortunately I cannot recall exactly where it petered out at this remove in time, but I suspect it had to do with the fact that to get the full richness of the usual theory of species uses good properties of $\mathrm{Set}$ such as cartesian closure, and that some of these properties don't translate well to the category of virtual sets (the category of oriented 1-cobordisms). For example, taking the negative of a set should be a duality functor (a contravariant equivalence), but cartesian closed categories that are self-dual collapse to posets. 
A few years later I had another crack at it (while I was hanging out as a visitor at U. Chicago, around 1996), but this time I was more interested in formalizing virtual linear species (i.e., differences of $\mathrm{Vect}$-valued species), since the primary application in the Joyal paper centered on linear species and particularly the structure of the Lie species. The basic idea was to use differential $\mathbb{Z}_2$-graded spaces modulo quasi-isomorphism as the receiving category, thinking of $(V_0, V_1, d)$ as representing a formal difference $V_0 - V_1$. This approach seemed much more promising, and is implicit in my Notes on the Lie Operad, which you can find here if you are interested. However, I was never satisfied with those notes and never tried to publish them -- I should return to this, and particularly the model category of $\mathbb{Z}_2$-graded chain complexes as an environment for virtual linear species. (Perhaps someone else has fleshed it all out in the meantime?)<|endoftext|>
TITLE: Rigorous numerics for maxima and minima (one variable)
QUESTION [7 upvotes]: Let $f:\mathbb{R}_0^+\to \mathbb{R}$ be defined by some combination of the four basic operations and square roots. (The argument of square-roots is assumed is to be non-negative, and the value of square roots is defined to be non-negative as well.) How can one find the global maximum and minimum of $f$ numerically, with full rigor?
(A rigorous numerical solution has to be given as an interval, e.g., [1.500001,1.500002],
within which the maximum (say) lies.)
Rather convincing but not quite rigorous method:
Plot $f'(x)$, see roughly where the zeroes are, and narrow them down using some standard
(e.g. SAGE, mathematica). Then verify this (non-rigorous) datum rigorously simply by
checking that $f'(x)$ has different signs at $x=x_0-\epsilon$ and at $x=x_0+\epsilon$,
where $x_0$ is an alleged zero of $f'(x)$. Compare the values of $f(x)$ at all $x_0$, and also at $x=0$ and as $x\to \infty$.
There is really only one way in which this procedure isn't rigorous: there is no guarantee that there aren't any zeroes of $f'(x)$ we have missed. (If $f$ is a polynomial, this can be easily dealt with, but $f$ is not necessarily a polynomial.)
Are there any (free) programs out there that for the minimum and maximum of $f$ rigorously?

REPLY [4 votes]: You can convert your problem into a multivariate one, with polynomial equality constraints, by introducing new variables for each square root and division. Thus you end up with the problem of minimizing a polynomial function on a semialgebraic set (add various necessary conditions for the optimum, too, e.g. what you get from $f'(x)=0$; this won't need more variables to be introduced), which you indeed can do rigorously, i.e. you could do even more: find your minimum and maximum as real algebraic numbers, by quantifier elimination. 
If you're lucky then necessary conditions for the optimum give you a zero-dimensional ideal to deal with, with zeros being (potential) optima.
Well, this is in theory; in practice, you might run into problems, and instead opt for using a sum of squares-based approach of polynomial optimization, which turns into solving a sequence of semidefinite programming problems of increasing size, with optimal values approximating the real optimum with increasing precision.<|endoftext|>
TITLE: Solid angles of a tetrahedron
QUESTION [13 upvotes]: This is a problem I have had for a while. For a triangle, the side opposite the largest angle has the largest length (and similarly for smallest angle). For a tetrahedron, the question is whether the face opposite the largest solid angle (trihedral angle - the area of the spherical triangle with angles equal to the dihedral angles incident on one vertex) has the largest area?
I can prove that this is true for tetrahedra coming from sphere packings (each vertex $v_i$ has a weight $r_i$ and the edge lengths are equal to $r_i+r_j$, etc.) but this excludes degenerations of a tetrahedron such as the one going to a box with two diagonals.
I suspect this is true but have no proof. Perhaps someone else does?

REPLY [5 votes]: One can show that  the face opposite the largest solid angle has the largest perimeter. This follows directly from here (see the  answer of Greg Egan with an explicit linear relation).<|endoftext|>
TITLE: Is there a general theory of fiber theorems?
QUESTION [26 upvotes]: Here are three vague theorems rolled up in one.

Let $X$ and $Y$ be sufficiently nice topological spaces and $f:X \to Y$ a sufficiently nice surjection. If for each $y \in Y$, the fiber $f^{-1}(y) \subset X$ is (topologically, homotopically, homologously) equivalent to a point, then $f$ induces a (topological, homotopic, homological) equivalence between $X$ and $Y$.

There are many such theorems (Vietoris-Begle for homology, Smale for homotopy), and generalizations abound. For instance, instead of requiring fibers to be homotopy-equivalent to points, you can ask that they be $n$-connected for some $n$; under this weaker hypotheses one can still extract the fact that the homotopy groups of $X$ and $Y$ agree up to dimension $n-1$.
This suggests a fairly general template: take a nice map $f:X \to Y$, and assume that each fiber $f^{-1}(y)$ is equivalent to a point under [insert topological invariant $I$ here]. Then, $X$ and $Y$ are $I$-equivalent.

Question: Is there a general theorem like the one suggested above?

For instance, one might ask: if each fiber has the Euler characteristic of a point, must we obtain $\chi(X) = \chi(Y)$? The reason I thought to ask is because I would like to know if such a theorem holds for simple homotopy equivalence (see here for instance). Even if nothing like the super-general theorem above can hold, I'd still be interested in knowing if one has the following precise, restricted version:

Let $X$ and $Y$ be connected CW complexes and $f:X \to Y$ a cellular surjection. If for each $y \in Y$, the fiber $f^{-1}(y) \subset X$ is simple homotopy equivalent to a point, then $X$ and $Y$ are simple homotopy equivalent.

If this is true, where can one find this theorem in the literature? I scoured Marshall Cohen's book on Simple Homotopy Theory but could not find it.

REPLY [10 votes]: Edit: I have added some definitions and details to my answer.
In the most general form I can find, your third question is a consequence of two results regarding cell-like maps and fine homotopy equivalences. It is closely related to and makes use of Chapman's celebrated theorem on topological invariance of Whitehead torsion.
Firstly, a cell-like map is a map which is proper, and whose fibres are all cell-like spaces. A compact metric space is called cell-like if it is "shape contractible" in the sense of having trivial (Borsuk) shape. Secondly, a map $f:X\to Y$ is a fine homotopy equivalence if, for any open cover $\mathcal{U}$ of $Y$, $f$ has a homotopy inverse $g$, and corresponding homotopies to the two identity maps which both follow paths never leaving some open in $\mathcal{U}$. Finally, recall that ANR abbreviates absolute neighbourhood retract. Importantly, and to avoid piling up qualifiers, all ANRs in this answer will be implicitly assumed metric and separable. A good example of a (metric separable) ANR is a locally finite, countable CW-complex.
The main theorem in the article "Mappings between ANRs that are fine homotopy equivalences" by William Haver implies that any cell-like map between locally compact ANRs is a (proper) fine homotopy equivalence. In that article, a cell-like map would be called a proper $UV^\infty$-map instead.
On the other hand, corollary 3.2 in the article "The homeomorphism group of a compact Hilbert cube manifold is an ANR" by Steve Ferry implies that any proper fine homotopy equivalence between locally compact ANRs is a simple homotopy equivalence. The proof of this result by Ferry uses the theory of Hilbert cube manifolds, via the following facts: (1) a proper fine homotopy equivalence between Hilbert cube manifolds is approximable by homeomorphisms (as proved by Ferry in the article cited above), and (2) the product of a locally compact ANR with the Hilbert cube is a Hilbert cube manifold (by a result of Robert Edwards). Finally, it uses Chapman's results on the topological invariance of Whitehead torsion.
In any case, putting together the two results stated above, we conclude the following.

Any cell-like map between locally finite, countable CW-complexes is a simple homotopy equivalence.

To relate this to the actual question, first observe that simple homotopy type is not defined for every space. In the greatest generality I am aware of, it is definable for locally compact ANRs. Then being simple homotopy equivalent to a point is equivalent to being a compact contractible ANR. Since such a space is necessarily cell-like, we obtain the following answer to your last question.

Let $f:X\to Y$ be a proper map between locally finite, countable CW-complexes. Assume each fibre of $f$ is a contractible ANR, i.e. has trivial simple homotopy type (since the fibres are compact). Then $f$ is a cell-like map. By the above result, $f$ is a simple homotopy equivalence.

That is the most general statement I can find. If you only care about the case of finite CW-complexes, then the result admits a formulation with fewer classifiers, and was also proved a few years earlier. For example, it is stated explicitly as a theorem in page 17 of the article by Lacher cited below.

If $f:X\to Y$ is a cell-like map between compact ANRs, then $f$ is a simple homotopy equivalence. In particular, if $f:X\to Y$ is a map between finite CW-complexes whose fibres are all contractible ANRs, then $f$ is a simple homotopy equivalence.

Finally, two nice overviews of the theory of cell-like maps and related topics are:

the 1978 ICM address of Robert Edwards which describes many interesting connections with the topology of manifolds;
the article "Cell-like mappings and their generalizations" by Lacher, which gives a thorough exposition of the theory of cell-like maps.

Moreover, the first part of the article by Lacher contains an interesting exposition of theorems of Vietoris-Begle type, much like the ones mentioned in the question. It might be a good place to start looking for a general theory of such results.<|endoftext|>
TITLE: On the convexity of element-wise norm 1 of the inverse
QUESTION [8 upvotes]: Question first asked on math.stackexchange here: https://math.stackexchange.com/questions/317209/on-the-convexity-of-element-wise-norm-1-of-the-inverse
On the convexity of element-wise norm 1 of the inverse
Let us define $\|A\|_1$ the element wise norm 1 of a matrix $A \in \mathbb{R}^{n \times m}$ as $$\|A\|_1=\sum_{i,j} |A_{i,j}| $$
Obviously, this function is convex over $\mathbb{R}^{n \times m}$. Is it true that the function $f:S^n_{++} \longrightarrow \mathbb{R}$, defined as $f(A) = \|A^{-1}\|_1$, is convex? Here we denote with $S^n \supset S^n_+ \supset S^n_{++}$ respectively the set of Symmetric, Positive Semidefinite (PSD) and Positive Definite (PD) $n \times n$ matrices.
Some observations:
Unfortunately matrix inversion is convex with respect to $S_+$, while $\|.\|_1$ is non decreasing with respect to the cone $R^{n \times m}_+$ and not with respect to $S_+$ (it is easy to find counterexamples). Hence theorems on combination of convex functions are not applicable.
I have tried to formulate function $f$ as $max\{ \ trace(M_i A^{-1}) \ \}_{i\in\mathcal{I}}$ where the $M_i\in S^n$ live in a family of matrices with elements equal to 1 or -1 so as to cover  all the possible combinations of signed sums of elements of $A^{-1}$ but unfortunately not all such $M_i$ are PSD and therefore not all $trace(M_i A^{-1})$ are convex in $A$. Therefore I was not able to define $f$ as the pointwise max of convex functions.
I have also tried to consider
$$
\|A^{-1}\|_1= \sum_{i,j}=\frac{|\det A_{\hat{\imath}\hat{\jmath}}|}{detA}
$$
where $\det A_{\hat{\imath}\hat{\jmath}}$ is the minor associated to the $n-1 \times n-1$ sub-matrix obtained by eliminating row $i$ and column $j$ from $A$. But I have no intuition on how to go further...
Any thought?

REPLY [3 votes]: The function that you have is convex for unitarily invariant norms, but for the (basis dependent) elementwise absolute value, it can clearly break as a trivial counterexample below shows.
\begin{equation*}
 X = \begin{pmatrix}
 2 & 0 & 0\\\\
 0 & 1 & 0\\\\
 0 & 0 & 1
\end{pmatrix},\qquad Y = \begin{pmatrix}
 10  &    9 &    5\\\\
     9 &   10 &    5\\\\
     5 &    5 &    4
\end{pmatrix}
\end{equation*}
Now, simply define $Z = 0.5(X+Y)$, and consider $g(H)=\|H^{-1}\|_1$ as your function. 
Then, we see that $g(Z) = 3.1692$, while $0.5(g(X)+g(Y)) = 3$, clearly violating convexity.<|endoftext|>
TITLE: Is there a direct proof that affine schemes are fppf quasi-compact?
QUESTION [11 upvotes]: Let $A$ be a (commutative) ring. A family $(B_i)_{i\in I}$ of $A$-algebras is said to be an fppf cover if it satisfies three properties: (1) each $B_i$ is flat as an $A$-module, (2) each $B_i$ is finitely presented as an $A$-algebra, and (3) the family is faithful, meaning that any $A$-module map $M\to N$ is an isomorphism if (and only if) each of the induced maps $B_i\otimes_A M \to B_i\otimes_A N$ is.
It is a theorem that for every fppf cover $(B_i)_{i\in I}$, there is a finite subfamily $J\subseteq I$ such that $(B_j)_{j\in J}$ is an fppf cover. In other words, affine schemes are quasi-compact in the fppf topology. The standard proof uses algebraic geometry: combine the fact that any flat finitely presented map of affine (say) schemes is open with the fact that affine schemes are quasi-compact in the Zariski topology.
My question is whether anyone knows a "direct" proof of this fact. Ideally it would not use the theorem that flat finitely presented morphisms are open or maybe not even any algebraic geometry at all. You might also hope that it works in some more general context, for instance for maps of (commutative?) monoids in any symmetric monoidal category.

REPLY [5 votes]: Since the ring map $A \to B_i$ is of finite presentation, the image of $\text{Spec}(B_i)$ in $\text{Spec}(A)$ is constructible, see Theorem Tag 00FE, usually known as Chevalley's theorem. Then $\text{Spec}(A)$ is the union of these constructible sets. A constructible set is open in the constructible topology on $\text{Spec}(A)$, see Section Tag 08YF. Moreover, the constructible topology is compact (ibid). Thus we need only finitely many indices $i$.
Probably not the answer you were looking for. Bork, bork, bork! (I was going to run this answer through the Swedish chef translator, but that would probably annoy the powers that be. Chicken in the basket, two points!)<|endoftext|>
TITLE: Is the set of the absolutely continuous functions a Borel set of the space of the continuous functions?
QUESTION [6 upvotes]: Does anyone knows whether the set of the absolutely continous functions $F :[0,1]\to \mathbb{R}^d$ of the form $$F(t)= a + \int_0^tf(s) ds$$ where $f$ is an integrable function is a Borel set of the Banach space $C$ of the continuous funtions $$F : t\in [0,1] \to F(t)\in \mathbb{R}^d$$ with the norm of the uniform convergence ?

REPLY [2 votes]: With a little more machinery we can give a very short proof.  A theorem of Lusin and Souslin states that if $X,Y$ are Polish, $\Phi : X \to Y$ is continuous, $A \subset X$ is Borel and $\Phi|_A$ is injective, then $\Phi(A)$ is Borel.  (See for instance Theorem 15.1 of Kechris's Classical Descriptive Set Theory.)  Taking $A = X = \mathbb{R} \times L^1([0,1])$, $Y = C([0,1])$, and considering the map $(a,f) \mapsto a + \int_0^\cdot f$ which is continuous and injective and whose image is the absolutely continuous functions, we have the result.<|endoftext|>
TITLE: Qustions on R.Bryant's papaer "Calibrated embeddings in the special Lagrangian and coassociative cases"
QUESTION [5 upvotes]: I am reading the paper "Calibrated embeddings in the special Lagrangian and coassociative cases" by R.Bryant (here the link: http://arxiv.org/abs/math/9912246) and there are certain things that are unclear to me.

Bryant defines on page 11 in is paper the set $V_{n}(\mathcal{I},\pi) =$ {$E\in V_{n}(\mathcal{I})|D_{u}\pi : E \rightarrow T_{\pi(u)}M$ is injective} and claims that $V_{n}(\mathcal{I},\pi) \subset Gr_{n}(TF)$ is a submanifold of codimension $np$, where $p$ is the codimension of $G \subset SO(n)$ in $SO(n)$. My question is: why is this so? I tried to use several chart representations or the implicite function theorem but without any succes.
On page 12 there are the subspaces $\mathfrak{h}_{k}$ defined. Then he computes that $H(E _ {k}) = E + ( \mathfrak{h} _ {k} ) _ {u}$. Why does this hold?
How does he show that $SU(n)$ is reglarly presented in $SO(2n)$?

I hope that some of you have the answers to some of my questions.
Best regards
Mario

REPLY [13 votes]: I'm afraid that that article does not do a lot of details in the introductory Section 0, just because more complete explanations were already available in earlier articles of mine. Here are some brief answers.  Right now, I don't have the time to write out the explanations in greater detail.

This is a consequence of the claim, proved in the reference [Br$_2$], that the condition of being torsion-free is $np$ first order PDE for a section $\sigma:M\to S=F/G$ that defines a $G$-structure (where $n$ is the dimension of the manifold and $p$ is the codimension of $G$ in $\mathrm{SO}(n)$).  The hypothesis of strong admissibility implies (indeed, it is equivalent to the condition) that all of these equations are captured by the condition of closure of the differential forms associated to the $G$-structure, which is exactly the condition that an $n$-plane $E$ be an integral element of $\mathcal{I}$.
If you unwind the definitions, you will see that the spaces ${\frak{h}}_k$ were defined so as to make this equation true.  This computation is best carried out up on $F$, where one can write out the definition of $\hat \alpha$ in a coframing of $F$ given by the structure equations and compute $d\hat\alpha$ explicitly.  (When I have more time, maybe I can put in a brief description of this computation.)
I didn't claim to give the proof that $\mathrm{SU}(n)\subset\mathrm{SO}(2n)$ is regularly presented for all $n$, I just said that it can be proved.  You can get an indication of how the proof goes by looking a little further along in the paper where, on pages 13–14, I do (briefly) give the argument for $n=3$, the case of most interest in the article.<|endoftext|>
TITLE: projection of sobolev spaces onto cones
QUESTION [5 upvotes]: Consider the Sobolev space $W^{k,p}(\Omega)$ for $k\in \mathbb N$, $p\in [1,\infty]$ and some open domain $\Omega\subset \mathbb R^n$ $^*$. Then it is known that $W^{k,p}(\Omega)$ is an ordered Banach space, and indeed a lattice-ordered Banach space if $k=1$, but not a Banach lattice because the norm is not monotone on the positive cone $W^{k,p}(\Omega)_+$. Furthermore, it is known that in the reflexive range $W^{k,p}(\Omega)$ is isomorphic, but in general not lattice isomorphic to an $L^p$ space. It is also clear that the positive cone of $W^{k,p}(\Omega)$ is a closed subset, hence at least in the reflexive range each element of $W^{k,p}(\Omega)$ can be projected onto $W^{k,p}(\Omega)_+$.
That said: 

Does anybody have an idea of how said orthogonal projection looks like?

$^*$ I would be happy already with an answer in the case of $n=1$, $k=1$, $p=2$, $\Omega$ bounded.

REPLY [5 votes]: I will consider the case $k = 1$ and $p = 2$ (some arguments may generalize to $k \in \mathbb{N}$).
Let us use the norm $\|u\|^2 = \|u\|^2 + \|\nabla u\|^2$ in $H^1(\Omega)$ (both are $L^2$-norms).
The associated scalar product is denoted by $(\cdot,\cdot)$.
We denote by $K = \{v \in H^1(\Omega) : v \ge 0\}$ the positive cone in $H^1(\Omega)$.
Then, the projection of $u \in H^1(\Omega)$ onto $K$ is given by the minimizer $v$ of $\frac12 \, \| u - v \|^2$ over $K$.
Now, one can show, that there is a Lagrange multiplier $\lambda \in (H^1(\Omega))'$.
This multiplier lies in the polar cone $K^\circ$, that is
$$
 \langle \lambda, w \rangle \le 0 \quad\mbox{for all } w \in K
$$
and is orthogonal to $v$, i.e., $\langle \lambda , v \rangle = 0$.
Moreover, we have the equation
$$
(v - y, w) + \langle \lambda, w \rangle = 0 \quad\mbox{for all } w \in H^1(\Omega),
$$
which couples all involved quantities. This equation is the weak formulation of a PDE.
Note that these optimality conditions are nothing more than an equivalent reformulation of the projection inequality
$$
 (v - y, w - v) \ge 0 \quad\mbox{for all } w \in K.
$$
The inclusion $\lambda \in K^\circ$ can be interpreted as non-positivity of $\lambda$.
Moreover, it should be possible to show (at least in the case $n = 1$, you are interested in), that $\lambda$ is represented by some negative measure (along the lines that negative distributions are essentially measures).
Then, the complementarity $\langle \lambda , v \rangle = 0$ means, that $\lambda$ is only strictly negative, where $v = 0$ (i.e. where the projection is active).
As you can see,
the calculation of the projection requires the solution of a nonlinear (even non-differentiable) PDE.
This PDE is just the optimality condition (i.e., the projection inequality) of minimizing the distance in the Dirichlet energy.
Hence, there is no easy formula.
Finally, let me comment on the case $u \in H^2(\Omega)$.
That is, we are projecting a more regular element (but still w.r.t. the $H^1$-norm).
Then one can show (see, e.g., the book by Kinderlehrer and Stampacchia - I can provide an exact reference, if needed),
that (under some regularity of $\Omega$)
again $v \in H^2(\Omega)$ and $\lambda \in L^2(\Omega)$.
Due to this regularity, on can interpret the above relations pointwise and obtain
$\lambda(x) \ge 0$ and $v (x) \, \lambda(x) = 0$ for almost every $x \in \Omega$.
Finally, the PDE can be written as
$$
\max\Big( {-\Delta(u-v)} + (u-v), \; v \Big) = 0 \quad\mbox{almost everywhere in }\Omega.
$$
Again, one can see, that there is no easy (explicit) formula for the projection.
I found this 'high regularity case' interesting, because its similar to the case of projecting a $H^1$-function w.r.t. the $L^2$-norm, where the projection is again $H^1$.<|endoftext|>
TITLE: Proof theory and primitive roots
QUESTION [8 upvotes]: I have had this question on my mind for two decades. We know, after Heath-Brown, that one out (say) of 3, 5, 7 is a primitive root mod p for infinitely many primes p. We just don't know which one. (We presume each is, or some GRH goes wrong per Hooley.)
Now if I invoke proof theory, it must be in terms of the proof of the result, not just that it has been proved. The result is a disjunction "3 is" or "5 is" or "7 is", and a proof according to constructive canons ought to be rendered down to a proof of one such statement. 
So we can assume there is an obstruction to making the proof constructive in any facile way. What I'd like to ask is for a friendly logician to point out where the obstruction would be, and if possible name it as some standard issue.
The structure of the proof is in two parts. One analytic part uses Chen's Goldbach technique. It basically locates primes p of type 2qr + 1, if I recall correctly, with q and r primes in certain ranges by magnitude and I suppose in some arithmetic progressions. And an algebraic part, which is at the level of Lagrange's theorem. I have always assumed the trouble is with the second part. Which is relatively so much shallower, prompting my wish to know more.

REPLY [12 votes]: It's the Pigeonhole Principle in the final step which is not constructively valid, not Heath-Brown's main result which, after inspection, doesn't show any of the tell tale signs of non-constructiveness.
Let $P_t$ be the set of all primes for which $t$ is a primitive root. Heath-Brown shows that if $q$, $r$, $s$ are three non-zero integers which are multiplicatively independent and that $q$, $r$, $s$, $-3qr$, $-3qs$, $-3rs$ and $qrs$ are not squares, then $P_q \cup P_r \cup P_s$ is infinite (and indeed that $\left|(P_q \cup P_r \cup P_s) \cap\lbrace1,\dots,x\rbrace\right| \gg x/(\log x)^2$). It is not constructively valid to draw from this the conclusion that one of $P_q$, $P_r$, $P_s$ is infinite. However, one can draw the more negative conclusion that $P_q$, $P_r$, $P_s$ are not all three finite.
One can see this using a Brouwerian counterexample which is analogous to the above situation. Let $A$ be the set of all $n$ for which there is a string of $333$ consecutive $3$'s in the first $n$ digits of $\pi$, and let $B$  be the complement of $A$. This is legitimate since we can always compute the first $n$ digits of $\pi$ to determine whether $n \in A$ or $n \in B$. Clearly $A \cup B = \lbrace1,2,3,\dots\rbrace$ is infinite. However, we cannot assert that $A$ or $B$ is infinite without knowing whether the digits of $\pi$ do or do not contain $333$ consecutive $3$'s. In the same way that GRH allows us to say that all three sets $P_q$, $P_r$, $P_s$ are infinite, the well-known conjecture that $\pi$ is normal allows us to assert that $A$ is infinite and $B$ is finite, but we don't know yet.
I haven't gone through Heath-Brown's argument in sufficient detail to assert without doubt that it is completely constructive but if there is a non-constructive part it must be hidden somewhere in some of the results he cites and not in the paper itself. In the paper, Heath-Brown explicitly computes an asymptotic lower bound on $\left|(P_q \cup P_r \cup P_s) \cap \lbrace1,\dots,x\rbrace\right|$. The argument is not straightforward but it is not convoluted and looks completely constructive. Instead of considering all primes, he considers a smaller but still large set of well-behaved primes $p$ for which he can break down the three multiplicative subgroups mod $p$ generated by $q$, $r$, $s$ into a handful of cases. He then computes upper bounds for the number of well-behaved primes up to $x$ falling into a case where none of $q,r,s$ are primitive roots. Adding these up and subtracting the result from a lower bound on the number of well-behaved primes up to $x$ gives $\left|(P_q \cup P_r \cup P_s) \cap \lbrace1,\dots,x\rbrace\right| \geq Cx/(\log x)^2$ where $C$ is a constant that may depend on $q, r, s$. Since the multiplicative subgroups of mod $p$ generated by $q$, $r$, $s$ are finite subgroups of a finite group, the various cases are decidable and it is fine to use the law of excluded middle to break into cases that way.<|endoftext|>
TITLE: Are all free groups linear, i.e., admit a faithful representation to GL(n,K) for some field K ?
QUESTION [22 upvotes]: All free groups of finite or infinite countable rank are subgroups of the free non-abelian group $F_2$, which is linear. However, a free group of infinite uncountable rank will not be a subgroup of $F_2$. Is it linear, too ?
This might easily follow from model theory, but I could not find a proof in the literature so far.

REPLY [10 votes]: Linearity of uncountable free groups 
can be easily deduced from linearity of free groups of countable rank using basic set theory and algebra only, without using compactness theorem, or ultraproducts, or non-trivial arguments of Misha Kapovich. Here is the simple proof.
Proposition. Let $K$ be a field, $n\ge 2$, and $S$ a subset of ${GL}_n(K)$. Suppose all elements of all matrices from $S$ are distinct and algebraically independent in $K$. Then $S$ is a free subset of ${GL}_n(K)$.
Proof.  Suppose not. Then there exist  distinct matrices $A_1,\dots,A_m$ from $S$ and a non-trivial  reduced group word $w(x_1,\dots,x_m)$ such that 
$w(A_1,\dots,A_m)=E$.  
For any field $F$ and  matrix $C=(c_{ij})$ in $GL_n(F)$,
the inverse of $C$ can be found using a formula of the form
$C^{-1}=(f_{ij}(c_{11},\dots,c_{nn})),$
where all $f_{ij}$ are certain rational functions over $\mathbb Z$ in $n^2$ variables,
which do not depend on the choice of $F$ and $C$.
Therefore,  for any field $F$ and matrices $C_1,\dots,C_m$
 in $GL_n(F)$, where $C_k=(c^k_{ij})$, we have
$$w(C_1,\dots,C_m)=
(g_{ij}(c^1_{11},\dots,c^1_{nn},\dots,c^m_{11},\dots,c^m_{nn})),
$$
for some  rational functions $g_{ij}$  over $\mathbb Z$ 
in $mn^2$ variables; these $g_{ij}$ do not depend on the choice of
$F$ and  $C_1,\dots,C_m$.
Let $A_k=(a^k_{ij})$. Since $w(A_1,\dots,A_m)=E$, 
the algebraic independence of all $a^k_{ij}$ 
implies the $g_{ij}\equiv 0$ for 
$i\ne j$, and $g_{ij}\equiv 1$ for $i=j$, for any field of characteristic $p$,
where $p={\rm char}(K)$ (zero or prime).
Hence $w(C_1,\dots,C_m)=E$
for any field $F$ of characteristic $p$ and matrices $C_1,\dots,C_m$  in $GL_n(F)$.
Hence, for any field $F$ of characteristic $p$, in  $GL_n(F)$ there is no free subgroup of rank $\ge m$. 
This contradicts to the known fact that
free subgroups of countable rank embed into
$GL_n(\mathbb Q)$ and $GL_n({\bf F}_q(t))$, for any prime $q$. $\square$
Corollary. If $\lvert K\rvert=\kappa>\aleph_0$
then the free group of rank $\kappa$ embeds
into ${GL}_n(K)$.
Proof.  Since $\kappa>\aleph_0$, 
any transcendence basis $B$ of $K$
is of power $\kappa$. Then there is
a subset $S$ of $M_n(K)$ of power $\kappa$ such that all elements of matrices from $S$ belong to $B$ and are distinct.
Since the elements of $B$ are algebraically independent,
for $A\in S$ we have ${\rm det}(A)\ne 0$ 
and so $A\in {GL}_n(K)$. By Proposition, $S$ is free.  $\square$
EDITED VERSION. In his answer
Misha Kapovich proved, using his 
non-trivial geometric arguments, that
for any field $K$ of uncountable power $\kappa$, the free group of rank $\kappa$ embeds into $SL_n(K)$, for any $n\ge 2$. In my answer I gave an elementary proof of a weaker result, with $GL$ instead of $SL$. However, my proof can be easily modified to the case of $SL$. Here is this modified proof.
For a non-singular square matrix $A$ denote by $\bar A$ the
matrix  obtained from $A$ by replacing any element $a$
in the first row of $A$ with $a/{\rm det}(A)$. Clearly, 
${\rm det}(\bar A)=1$, and $\bar A=A$ if ${\rm det}(A)=1$.
Proposition. Let $K$ be a field, $n\ge 2$,
and $S$ a subset of ${GL}_n(K)$. 
Suppose all elements of 
all matrices from $S$ are distinct and algebraically independent in $K$. 
Then ${\bar S} =\{{\bar A}: A\in S\}$ is a free subset of ${SL}_n(K)$.
Proof. Suppose not. Then there exist  distinct matrices 
$A_1,\dots,A_m$ from $S$ such that 
$w({\bar A}_1,\dots,{\bar A}_m)=E$
for some  non-trivial  reduced group word $w(x_1,\dots,x_m)$.
As above, for any field $F$ and matrices $C_1,\dots,C_m$
 in $GL_n(F)$, where $C_k=(c^k_{ij})$, we have
$$w({\bar C}_1,\dots,{\bar C}_m)=
(g_{ij}(c^1_{11},\dots,c^1_{nn},\dots,c^m_{11},\dots,c^m_{nn})),
$$
for some  rational functions $g_{ij}$  over $\mathbb Z$ 
in $mn^2$ variables; these $g_{ij}$ do not depend on the choice of
$F$ and  $C_1,\dots,C_m$.
Let $A_k=(a^k_{ij})$. Since $w({\bar A}_1,\dots,{\bar A}_m)=E$, 
the algebraic independence of all $a^k_{ij}$ 
implies the $g_{ij}\equiv 0$ for 
$i\ne j$, and $g_{ij}\equiv 1$ for $i=j$,
for any field of characteristic $p$ , where $p={\rm char}(K)$  (zero or prime). 
Hence 
$w({\bar C}_1,\dots,{\bar C}_m)=E$
for any field $F$ of characteristic $p$ and matrices $C_1,\dots,C_m$  in $GL_n(F)$. In particular, 
$w(C_1,\dots, C_m)=E$
for any field $F$ of characteristic $p$ and matrices $C_1,\dots,C_m$  in $SL_n(F)$.
Hence, for any field $F$ of characteristic $p$, in  $SL_n(F)$ there is no free subgroup of rank $\ge m$. This contradicts to the fact that
free subgroups of countable rank embed into $SL_n(\mathbb Q)$ and $SL_n({\bf F}_q(t))$, for any prime $q$.
 $\square$
Corollary. If $\lvert K\rvert=\kappa>\aleph_0$
then the free group of rank $\kappa$ embeds
into ${SL}_n(K)$. 
Proof.  Since $\kappa>\aleph_0$, 
any transcendence basis $B$ of $K$
is of power $\kappa$. Then there is
a subset $S$ of $M_n(K)$ of power $\kappa$ such that all elements of matrices from $S$ belong to $B$ and are distinct.
 Since the elements of $B$ are algebraically independent,
for $A\in S$ we have ${\rm det}(A)\ne 0$ 
and so $A\in {GL}_n(K)$. 
Clearly, $\lvert \bar S\rvert=\kappa$.
By Proposition, $\bar S$ is 
free in $SL_n(K)$.  $\square$<|endoftext|>
TITLE: What can be expressed in and proved with the internal logic of a topos?
QUESTION [30 upvotes]: The title of this post expresses what I really want, which is to learn how to wield the internal logic of a topos more effectively. However, to bring it down to earth, I'll ask a few basic questions about the topos $\mathcal{Grph}$ of directed graphs, whose underlying category is 
$$\mathcal{Grph}:={\bf Set}^{\mathcal{G}}.$$ 
Here $\mathcal{G}$ is the indexing category for graphs, with two objects $A$ and $V$ and with two morphisms $src,tgt\colon A\to V\ $.
What can I express in the internal logic of the topos $\mathcal{Grph}$, say using the Mitchell-Benabou language or the Joyal-Kripke semantics, or what have you? How can I use this logic to prove things? What kinds of things can't be said or proven in this way?
Below are some concrete questions. In each case when I ask "Can I...", I mean "To what degree can I use the internal language and logic of $\mathcal{Grph}$ to...". 

Can I express that a graph $X$ is finite, (respectively complete, discrete)? 
Can I take a graph $X$ and produce the
paths-graph $Paths(X)\ $, whose
vertices are those of $X$ but whose
arrows are all finite-length paths
in $X$?  
Can I express that $Paths$ is a monad, i.e. produce some morphism $X\to Paths(X)\ $ and another $Paths(Paths(X))\to Paths(X)\ $ with some properties?
Can I take a graph $X$ and produce the subgraph $L\subseteq X$ consisting of all vertices and arrows that are involved in a loop? (That is, an arrow $a\in X(A)$ is in $L$ iff there exists a path $P$ of length $n$ in $X$ such that $a\in P$ and $P$ is a loop: $P(0)=P(n)\ $. A vertex $v\in X(V)$ is in $L$ if it is the source of an arrow in $L$.)
Can I prove that if a graph has no loops and finitely many vertices then it has finitely many paths?
Can I do something else in $\mathcal{Grph}$ that might be fun and informative?

Maybe this post reflects a basic misunderstanding of how to think about the internal logic of a topos. If so, please set me straight. Also, a quiz question might be nice -- something of the form "see if you can express/prove this in $\mathcal{Grph}$: __."

REPLY [29 votes]: In general the internal language of a topos can only express those statements that make sense in every topos. In essence, this limits you to something like bounded Zermelo set theory, without global membership.
The right way to use the internal language of a particular topos, such as your topos of directed graphs, is to enrich the general internal language of toposes with new primitive types and new axioms. If you are lucky you may be able to add just axiom and define the new types (i.e., the new types can be characterized in the internal language). Let us see what these may be in the case of the topos of directed graphs.
Because we are dealing with a presheaf topos we can tell in advance that the (covariant) Yoneda embedding $y : \mathcal{G} \to \mathbf{Set}^\mathcal{G}$ will give us something important. Indeed, $y(V)$ is the graph with one vertex and no arrows, while $y(A)$ is the graph with two vertices and one arrow in between. Let me write $V$ and $A$ instead of $y(V)$ and $y(A)$, respectively. We might call $V$ "the vertex" and $A$ "the arrow". We call the objects of our topos "graphs", obviously.
Simple calculations reveal that, for a given graph $G$:

$G \times V$ is the associated discrete graph on the vertices of $G$.
$G^V$ is the associated complete graph on the vertices of $G$.
$G \times A$ is the following graph: for each vertex $g$ in $G$ we get two vertices $(g,s)$ and $(g,y)$ in $G \times A$ (think of them as "$g$ as a source" and "$g$ as a target"), and for each arrow $a : g \to g'$ in $G$ we get an arrow $a : (g,s) \to (g',t)$ in $G \times A$. This probably means something to graph theorists, I would not be surprised if they have a name for it.
$G^A$ is the associated "graph of arrows": the vertices of $G^A$ are pairs of vertices $(g,g')$ of $G$; and for each arrow $a : g \to h$ we get an arrow $a : (g,g') \to (h',h)$ in $G^A$. This makes more sense once you compute the global points of $G^A$: they correspond precisely to the arrows in $G$. Also, it is helpful to think of the vertices of $G^A$ as "potential arrows of $G$".

We can already answer some of your questions:

A graph $G$ is discrete when the projection $G \times V \to G$ is onto.
A graph $G$ is complete when the canonical map $G \to G^V$ is onto.

You would like to have the graph of paths $\mathsf{Path}(G)$ of a given graph $G$. I think you've described the wrong gadget, i.e., what you should be looking for is a graph whose global points are the paths in $G$, but there will be many other "potential" things floating around. We have so far not used the fact that there are two morphisms $s, t : A \to V$. These allow us to form "generic paths of length $n$" $P_n$ as pullbacks: $P_1 = A$, $P_2 = A \times_V A$ is the pullback of $s : A \to V$ and $t : A \to V$, and so on. With a little bit of care we should be able to form the object of "generic paths" $P$, equipped with a concatenation operation that turns it into a monoid. I am going to naively guess that the vertices of $P$ are pairs of natural numbers $(k,n)$ with $k < n$ and that arrows are of the form $(k,n) \to (k+1,n)$. But this needs to be checked, and in any case it should be possible to define the "correct" $P$ internally. The graph $\mathsf{Path}(G)$ that you are looking for  ought to be the dependent sum $\sum_{p : P} G^p$ (and this looks a lot like a polynomial functor). The monoid structure on $P$ should give you a monad.
Regarding cyclic paths (you call them loops): if I am not mistaken the internally projective graphs are those graphs whose in- and out-degrees are all 1, in other words the cycles and the infinite path stretching in both directions. This should help with getting a grip on cyclic paths. That a graph $G$ is internally projective can be expressed in the internal language as "every $G$-indexed family of inhabited graphs has a choice function", i.e., these are the objects that satisfy the axiom of choice, internally.
The vertex $V$ is a subobject of the terminal object $1$, which is the graph with a single vertex and a single arrow. Thus, there is a corresponding truth value $v \in \Omega$, which is a kind of "intermediate" truth value. We can define a closure operator $j : \Omega \to \Omega$ (a modality) by $j(p) = (v \Rightarrow p)$. This modallity should be called "vertex-wise". Indeed, if $H \hookrightarrow G$ is a subgraph of $G$ then its $j$-closure $\bar{H} \hookrightarrow G$ is the subgraph of $G$ induced by the vertices of $H$. Ah, but this is then the same as the compleement of the complement of $H$, so we see that $j$ is just the double negation closure. (I hope I am doing this right, I am speaking off the top of my head.)
If I am correct, then we can define $V$ in the internal language, using an axiom:

Axiom: there is a truth value $v \in \Omega$ such that $(v \Rightarrow p) = \lnot\lnot p$ for all $p \in \Omega$.

Then $V = \lbrace * \in 1 \mid v \rbrace$. We still have to do something about $A$, though.
In any case, my experience with internal languages is that they are well worth using. It takes a bit of effort, thoough, to figure out the optimal way of setting up the internal language of a particular topos. The general idea is to introduce as few new types as possible, characterize them with suitably chosen axioms, and figure out what other useful axioms are valid in your topos.<|endoftext|>
TITLE: On the naturality of the bar construction
QUESTION [10 upvotes]: Let $X$ be a based space. Then the Moore loop space $MX$  is defined to be the topological monoid whose points are based loops $[0,a] \to X$ where $a \ge 0$ is allowed to vary. Composition is gotten by concatenating loops.
Since $MX$ is a topological monoid, we can form the bar construction $BMX$.
This is geometric realization of the simplicial space
$$
[n] \mapsto (MX)^{\times n}
$$ 
where the face and degeneracy maps are defined using: projection, multiplication, and
insertion of the identity element.
If $X$ is a connected based CW complex, then there is a homotopy equivalence
$$
BMX \simeq X  .
$$
A standard way to prove this is to construct a quasi-fibration $EMX \to BMX$ whose
fiber at the basepoint is identified with $MX$ up to homotopy equivalence, 
in which $EMX$ is contractible. The quasi-fibration
is functorially associated with $MX$. However, it seems to me that the equivalence
$BMX \simeq X$ gotten in this fashion depends on a contractible space of choices. In particular it doesn't seem to be natural.
Question: 
Is there a zig-zag of  natural transformations 
$$
BMX = f_0(X)  \leftarrow f_1(X) \to f_2(X) \leftarrow \cdots \to f_n(X) =  X
$$
which yields a chain of equivalences when $X$ is a connected CW complex?

REPLY [10 votes]: I wrote down two quick and simple solutions in Lemmas 14.3 and 15.4 on pages 84 and 90 of 
"Classifying spaces and fibrations ([15] on my web page).  The first is the evident zigzag
of natural weak equivalences 
$$   X \leftarrow B(PX,MX,\ast) \rightarrow  B(\ast,MX,\ast) = BMX,   $$
where $PX$ is the Moore path space.
The left arrow is obtained from an obvious map of simplicial spaces to the constant simplicial
space at $X$:  Just compose loops and paths and evaluate at the end point.  The second is 
induced from $PX\to \ast$.
The second starts with a very slight modification of what Charles wrote down. For a point 
$u=(t_0,\cdots, t_p)\in \Delta^p$ with 
$0\leq t_i\leq 1$ and $\sum t_i = 1$, let $u_{i} = t_0 + \cdots + t_{i-1}$. For $\gamma_i\in MX$
of length $a_i$, $1\leq i\leq p$, define
$$ \xi |(\gamma_1,\cdots,\gamma_p),u| = (\gamma_1\cdots\gamma_p)(\sum_{1\leq i\leq p} u_i a_i). $$
This modification makes the compatibility check with the simplicial identities a bit more trivially trivial. To see that $\xi$ is a weak equivalence, take its ordinary loops. There is a natural inclusion $\Omega X \to MX$, which is a weak equivalence; as for
any grouplike monoid, there is a natural weak equivalence $MX \to \Omega BMX$; and we have 
$\Omega \xi\colon \Omega BMX\to \Omega X$. The composite of these three maps is the identity 
map $\Omega X\to \Omega X$, hence $\Omega \xi$ is a weak equivalence.  If $X$ is connected, it
follows that $\xi$ is a weak equivalence.<|endoftext|>
TITLE: Consistency of many Erdos cardinals
QUESTION [5 upvotes]: Does anyone have a reference for the consistency strength of saying that the Erdos cardinal $\kappa(\alpha)$ exists for all $\alpha$?  Or of various weakenings about how far such cardinals extend into $\text{Ord}$?
This comes up while looking at the consistency strength of sharps of sets of ordinals, so any reference on that front would also be welcome.

REPLY [7 votes]: I agree with the last sentence of Bill Mitchell's comment. But here is something closer than the cardinals mentioned. In [1] the notion of "almost Ramsey" cardinal was coined. Such a cardinal $\kappa$ is required to be $\alpha$-Erdos for all $\alpha<\kappa$. Then $V_\kappa$ is a model of what you are after (but in fact this is still not exact as there are many other $\gamma <\kappa$ for which this is true too. Almost Ramseys also get an outing in [2].
If you are interested in sharps alone, then, eg just for sharps for reals, $\kappa \rightarrow (\omega_1)^{<\omega}_{2}$ is already overkill: one really just needs for any function $f$ homogeneous sets of arbitrarily large but countable length, all of which have the same "type".  This has been investigated closely in [3].  Similar considerations would hold for sharps of other sets of ordinals.  The least inner model in which every set has a sharp, $L^\#$ say, is too thin to contain any Erdos cardinals (other than trivially $\kappa(\delta)$ for $\delta< \omega_1^{L^\#}$).  
[1] J. Vickers & P.Welch "Elementary Embeddings of an inner model into the Universe" JSL vol 66, 2001.
[2] A. Apter & P. Koepke "Making All cardinals almost Ramsey" Archive for Math. Logic, vol 47, 2008.
[3] J. Baumgartner & F.Galvin "Generalized Erdos cardinals and $0^\#$", Ann. of Math. Logic, vol. 15 , 1978.<|endoftext|>
TITLE: Was the early calculus inconsistent?
QUESTION [21 upvotes]: This question does NOT concern the RIGOR, or lack thereof, of the early calculus. Rather the question is of its CONSISTENCY. 
George Berkeley wrote in 1734 with reference to the early calculus that such a method is "a most inconsistent way of arguing, and such as would not be allowed in Divinity". This passage is quoted by William Dunham in 2004. Dunham concludes: "Bishop Berkeley had made his point. Although the results of the calculus seemed to be valid ... none of this mattered if the foundations were rotten". See page 72 of http://books.google.co.il/books?id=QnXSqvTiEjYC&source=gbs_navlinks_s 
On the other hand, Peter Vickers in 2007 challenged "The ubiquitous assertion that the early calculus of Newton and Leibniz was an inconsistent theory" at http://philsci-archive.pitt.edu/3477/ (soon to appear in book form at Oxford University Press), and concluded that this only holds in a limited sense and "can only be imputed to a small minority of the relevant community". 
Was the early calculus consistent as far as most practitioners were concerned, as Vickers contended, or was it a most inconsistent way of arguing, as did Berkeley and Dunham?
Note 1. Berkeley claimed that calculus was based on an inconsistency that can be expressed in modern notation as $(dx\not=0)\wedge(dx=0)$.  Thus he was using the term "inconsistent" in much the same sense it is used in modern logic.
Note 2. For a closely related thread, see https://math.stackexchange.com/questions/445166/is-mathematical-history-written-by-the-victors
Note 3. There is a related thread at the history SE: https://hsm.stackexchange.com/questions/3301

REPLY [10 votes]: Coming back to the B.Berkeley critics, there is a common denominator of all known getarounds, both the two mainstream ones (Wstrass and NSA) and exotic ones like the SDG interpretation. 
That is, one considers an extension - call it $R^+$ - of the true reals R and a map 
$R^+ \to R\cup \{\infty\}$ - call it the valuation map. For instance, 
1) $R^+$ consists of all convergent infinite real sequences and the valuation map is the 
`` taking the limit'' map
2) $R^+$ is a nonstandard extension of $R$ and the valuation map is the ``standard part'' map ($\infty$  for infinitely large objects)
3) Nilpotent or any other applicable exotics.
It occurs that the evaluation map cannot be a homomorphism, it always lacks something. For instance the value of a non-0 infinitesimal is 0, the value of its inverse is $\infty$, but $0\cdot \infty=1$ makes little sense in $R$. 
This is I believe the only sound way to view the medieval controversies around infinitesimals. That is, accept that a non-0 infinitesimal is not equal to the real number 0, it just has the value 0. Maybe, a devoted scholar of Leibnizz, Euler, etc. (although there is no much of etc. after Euler!) can find a support of this point of view. 
Obviously, a modern mathematician would ask for either a concrete mathematically defined model of both $R^+$ and the valuation map - and the two mainstream such models are listed above, with perhaps more yet to come under category 3 - or at least to set it up in the form of calculus of propositions, with rigorous rules of inference albeit w/o a fixed interpretation of objects. 
Vladimir Kanovei<|endoftext|>
TITLE: Smooth function algebra on cartesian product and beyond 
QUESTION [5 upvotes]: Short question:
Let $M$ and $N$ be smooth manifold, with appropriate smooth function algebras
$C^\infty(M,\mathbb{R})$ and $C^\infty(N,\mathbb{R})$. 
Can we express the smooth function algebra of the cartesian product manifold
in terms of $C^\infty(M,\mathbb{R})$ and $C^\infty(N,\mathbb{R})$? 
I know it is neither (equivalent to) 
$C^\infty(M,\mathbb{R}) \oplus C^\infty(N,\mathbb{R})$ nor
$C^\infty(M,\mathbb{R})\otimes_{\mathbb{R}}C^\infty(N,\mathbb{R})$.
...
A more general question is,if there is a general rule to get from categorical
constructions on manifolds to constructions on the appropriate smooth function algebra.
Maybe this boils don to the question whether or not the functor $C^\infty(\cdot,\mathbb{R})$
from smooth manifolds to ass. comm. unitary $\mathbb{R}$-algebras preserves (co)limits.
That's indeed the deeper question.
...
P.S.: I tagged it in particular as algebraic-geometry related, do to the category theory
related part...

REPLY [4 votes]: If $M$ and $N$ are compact, then $C^\infty(M\times N)=C^\infty(M)\bar\otimes_{i}C^\infty(N)$, the completed injective tensor product which coincides with the completed projective tensor product, since the locally convex spaces involved are nuclear.
Edit: This also holds for for non-compact $M,N$; see [Treves: Topological Vector Spaces, Distributions, and Kernels, Page 530]. 
If the manifolds are not compact, the same holds for the space of smooth functions with compact support.<|endoftext|>
TITLE: Does there exist a half-integer weight theta function which is is equivalent to 1 modulo 4?
QUESTION [8 upvotes]: Given a quadratic form $F$ in $n$ variables, there is an associated theta function $\theta_F(z) = \sum_{m \in \mathbb{Z}} q^{F(m)}$, which is a modular form of weight $n/2$.  Letting $F(m) = m^2$ gives a theta function $\theta_F(z) = 1 + 2 \sum_{n = 1}^\infty q^{n^2} \equiv 1 \pmod{2}$ with weight $1/2$.  
Does there exist a theta function $\theta_F(z) \equiv 1 \pmod{4}$ with half-integer weight?

REPLY [6 votes]: This is nothing like a complete answer (EDIT-now it seems to be--see below), but it may suggest a fruitful attack, based on the comment (see below for a version incorporated into this answer) that I made earlier. Your question is related to my question 124243 on this site about a certain ring $M$ consisting of the mod $2$ reductions of elements of $\mathbb{Z}[[x]]$ that are the Fourier expansions of modular forms for $\gamma_0 (N)$. Suppose first that $N$ is an odd prime $p$.
Theorem(?) The element $g=x+x^4+x^9+x^{16}+\dots$ of $\mathbb{Z}/2[[x]]$ does not lie in $M$.
Proof(?)  In the notation of my question, and the other question of mine that it links to, let $h=g+g^2+r$. Then
$$ h(1+h)=(g+g^4)+(r+r^2)=f1+(f1+fp)=fp . $$
(Again, see below). Now I conjecture in my question that in the field of fractions of $M$, $fp$ has a zero of order $p$, a zero of order $1$ and $p+1$ poles counted with multiplicity. If this is true then both $h$ and $h+1$ have $\frac{p+1}{2}$ poles, counted with multiplicity. Furthermore one of them must have a zero of order $p$, giving a contradiction.
Now I'm reasonably sure that this conjecture is correct, is known to the experts, and follows from results proved by Igusa in the 1950's. Unfortunately no expert has yet responded to my question. I also think that similar techniques will show that $g$ is not the mod $2$ reduction of any modular form for $\gamma_0 (N)$ when $N$ is odd. But the theory, if I understand correctly, is much harder when the level is divisible by the characteristic, and if a proof is to be given based on the idea of my comment this case has to be considered.
Edited comment--Suppose that $F$ is a positive definite quadratic form over $\mathbb{Z}$ in $s$ variables where $s$ is odd. Here's an idea for showing that the power series $\theta_F$ in $\mathbb{Z}[[x]]$ attached to $F$ (I'll defy convention and write $x$ instead of $q$) cannot be congruent to $1$ mod $4$. Choose $S$ so that $s\times S=8m-1$ and let $G$ be a direct sum of $S$ copies of $F$ and one of the one variable 
form $z^2$. Then if $\theta_F$ is $1$ mod $4$, $\theta_G$ is $1+2x+2x^4+2x^9+2x^{16}+\dots$ mod $4$. Now (see Schoeneberg's paper from the 1930's for example) there is an $N$ such that $\theta_G$ is the "expansion at infinity" of a weight $4m$ modular form, $\phi$, for $\gamma_0 (N)$. Then $(1/2)(\phi-(E_4)^m)$ is a 
weight $4m$ modular form for $\gamma_0 (N)$ whose expansion at infinity lies in $\mathbb{Z}[[x]]$, and has mod $2$ reduction equal to $g=x+x^4+x^9+x^{16}+\dots$.  There are perhaps reasons to believe this can't happen--at any rate it appears to be an approachable problem of a mainstream variety.
One more remark. In my proof? given above I use the fact that the mod $2$ reduction $f_1$ of the expansion at infinity of the cusp form delta is $x+x^9+x^{25}+\dots$. This lovely if well-known fact comes from the infinite product formula for delta and Jacobi's triple product identity.
EDIT---MARCH 24
I now believe I have an argument showing that $g$ is not the mod $2$ reduction of a modular form for any $\gamma_0 (N)$, completing the answer to your question. I'll dispense with any ideas of Igusa, and instead use Hecke operators and classical results of Gauss on ternary quadratic forms. The idea is this. For each odd prime $q$ there is a "Hecke operator" $T_q$ on $\mathbb{Z}/2[[x]]$.
If $h$ is the reduction of a modular form for $\gamma_0 (N)$ then the $T_q(h)$ span a finite dimensional subspace of $\mathbb{Z}/2[[x]]$. Now let $h=g^{11}$. I'll use this last result to show that $h$ cannot be the reduction of a modular form.
To this end, fix $K$ and primes $p_1,...,p_K$ each of which is $5$ mod $8$. Then choose primes $q_1,...,q_K$ each of which is $7$ mod $8$ so that the Legendre symbol $(q_i/p_j)$ is $-1$ if $i=j$ and is $1$ otherwise. Now the coefficient of $x^{p_j}$ in $T_{q_i}(h)$ is the coefficient of $x^{p_j}(q_i)$ in $h$. By the last paragraph of the comments following my answer to MO question 28462-why are there usually an even number of representations as a sum of $11$ squares--(an answer based on Disquisitiones Arithmeticae)--, this coefficient is $1$ if $i=j$ and $0$ otherwise.
So the $T_{q_i} (h)$ are linearly independent. Since $K$ may be chosen arbitrarily large, the criterion of the last paragraph gives the result.<|endoftext|>
TITLE: Analogue of Tate or Hodge conjecture for varieties over $\mathbb Q_p$
QUESTION [16 upvotes]: I've been learning about p-adic Hodge theory recently (I'm a beginner), and I've been wondering about the following question the past couple of weeks.  Sorry for the long setup, it's mainly background; the question is down towards the bottom.
Background: For a field $k$, and a second field $K$ of characteristic 0, we can consider the $K$-linear category $Mot(k)_{K}$ of motives defined by smooth projective varieties over $k$ (pure motives) up to homological or numerical equivalence (let's assume homological equivalence equals numerical equivalence for each Weil cohomology we consider).  Then this is a semi-simple abelian category (due to Jannsen), and there are realization functors to various categories of vector spaces extending the functor sending a variety $X$ to its cohomology groups (e.g. its etale cohomology groups $H^i(X,\mathbb{Q}_\ell)$ for $\ell \not= p$ in which case $K = \mathbb{Q}_\ell$). 
In the case when $k = \mathbb{C}$ and we take $K = \mathbb{Q}$, we can consider the singular cohomology realization $Mot(\mathbb{C})_\mathbb{Q} \rightarrow {Vect}_{\mathbb{Q}}$ sending $X$ to $H_{sing}^*(X,\mathbb{Q})$.  One way of stating the Hodge conjecture is that when we consider the extra structure of the Hodge decomposition on $H^*_{sing}(X,\mathbb{Q}) \otimes \mathbb{C}$ and therefore consider the enriched realization $Mot(\mathbb{C})_{\mathbb{Q}} \rightarrow {Hdge}_{\mathbb{Q}}$ to pure $\mathbb{Q}$-Hodge structures, this functor is fully faithful.
Similarly, if $k$ is a finite field or a number field, we can take $K = \mathbb{Q}_\ell$, we can consider the functor $Mot(k)_{\mathbb{Q}_{\ell}} \rightarrow GalRep(\mathbb{Q}_\ell)$ which sends $X$ to its $\ell$-adic cohomology $H^i(X_{\overline{k}},\mathbb{Q}_\ell)$ as an $\ell$-adic Galois representation.  Then the Tate conjecture is that this functor is fully faithful.  In the number field case, there are some other conjecturally fully faithful enriched realization functors discussed in Ch. 7 of Andre's book "Une Introduction Aux Motifs".
Motivation: Now in $p$-adic Hodge theory, we deal with varieties over a complete discrete valuation field (for example $k = \mathbb{Q}_p$), and study the extra structures which one obtains on the de Rham cohomology of such a variety from Hodge theory, crystalline cohomology, and etale cohomology.  To me it seems that one is trying to endow the de Rham cohomology of a variety $X$ with as much "natural" structure as possible, at least enough to determine the Galois representation $H^i(X_{\overline{k}},\mathbb{Q}_p)$, but possibly one can add more.  So I wonder the following:

Let $k$ be a finite extension of $\mathbb{Q}_p$.  Is it reasonable to believe that there exist a characteristic-0 field $K$ and a fully faithful functor $Mot(k)_K \rightarrow \mathcal{C}$, where $\mathcal{C}$ is the category of $K$-vector spaces with some "natural" extra structure, for example that considered in $p$-adic Hodge theory?  If not, why not?  If so, is there a candidate for $\mathcal{C}$?

EDIT: I want to add an additional condition: that when one takes the forgetful functor $\mathcal{C} \rightarrow Vect(K)$, the composition $Mot(k)_K \rightarrow Vect(K)$ is the functor defined by some Weil cohomology (e.g., de Rham cohomology or $p$-adic etale cohomology).  
My impression from the example given by user ulrich in the comment to 
Should the etale cohomology of a smooth projective variety (over rationals) be semi-simple; why?
is that we CANNOT take $K = \mathbb{Q}_p$ and take the functor sending $X$ to $H^*(X_{\overline{k}},\mathbb{Q}_p)$ , because the the $p$-adic cohomology of $X$ might not be semi-simple, for example when $X$ is an elliptic curve with non-split multiplicative reduction.  But perhaps there's another possible $\mathcal{C}$?
EDIT: As wccanard explained in a comment (now deleted), the example above doesn't actually show that the $p$-adic etale cohomology functor isn't fully faithful, since the subobject of $H^1(X,\mathbb{Q}_p)$ which is not split is not "motivic".  However, there are other reasons why $p$-adic etale cohomology can't work as explained in wccarnard's answer below.

REPLY [7 votes]: This is not an answer, it's just a note that some of the more standard examples of functors don't seem to work because they're not fully faithful. This was originally in the comments but it clogged them up so I deleted them and moved them here.
The observation I want to make is simply that if $k=\mathbb{Q}_p$ then I think that none of the following work in general: $K=\mathbb{Q}_\ell$ and the functor is $\ell$-adic etale cohomology; $K=\mathbb{Q}_p$ and the functor is $p$-adic etale cohomology; and finally $K=\mathbb{Q}_p$, the target category is $K$-vector spaces equipped with a filtration, a Frobenius and a monodromy operator in the usual $p$-adic Hodge theory waym and the functor is $D_{pst}$.
The reason none of these work is the following. Fix an elliptic curve over $\mathbb{F}_p$ with $a_p=0$ (assume such a curve exists). Then there are uncountably many lifts of this curve to $\mathbb{Q}_p$, so there are two non-isogenous lifts. The motives attached to the $h^1$ of these curves will, I imagine, have no non-zero maps between them. However the functors described above are all controlled by the special fibre so produce isomorphic linear algebra objects. In the $D_{pst}$ case what is happening is that the Frobenius has trace zero and the $Fil^1$ can't be either of the eigenspaces so it must be another line; however all other lines give isomorphic filtered $\phi$-modules. 
The reason this doesn't contradict the Hodge conjecture is that you can take coefficient field equal to $\mathbb{Q}$ in that setting, which is much smaller, and then the filtration (which is defined once you tensor up to $\mathbb{C}$ gives you far more information.<|endoftext|>
TITLE: Which E_∞-spaces are homotopy colimits of k-truncated E_∞-spaces?
QUESTION [5 upvotes]: This question is closely related to my previous question about modules over truncated sphere spectra, in particular, it has the same motivation.
Recall that every space (or ∞-groupoid) can be represented as the homotopy colimit of some simplicial diagram of 0-truncated spaces, i.e., sets considered as homotopy 0-types.
For example, if X is a Kan simplicial set, then the homotopy colimit of X considered as a functor X: Δop → Set → Space is equivalent to X itself.
For connective spectra (i.e., group-like E∞-spaces) a more complicated picture emerges:
homotopy colimits of simplicial diagrams of 0-truncated connective spectra (i.e., abelian groups) are precisely Eilenberg-MacLane spectra (of connective chain complexes of abelian groups).
More generally, homotopy colimits of simplicial diagrams of k-truncated connective spectra for some k>0 are modules over
the k-truncation π≤kS of the sphere spectrum S.
In particular, not every connective spectrum can be represented in this way.
(Presumably, being a module over π≤kS is also a sufficient condition, but I haven't checked any details to claim this.)
I wonder what happens in the intermediate case of (not necessarily group-like) E∞-spaces.
Which E∞-spaces can be represented as homotopy colimits of k-truncated E∞-spaces for some k>0?  What if we require the diagrams to be simplicial, i.e., indexed by Δop?
In particular, if some E∞-spaces cannot be represented in this way, what tools do we have to detect this?
Specifically, I'm interested in the answer for the case of E∞-spaces coming from connective spectra like MU, MSO, KU, or KO.

REPLY [12 votes]: If you don't group complete, then free $E_{\infty}$-spaces are $1$-truncated.
Consequently, for $k > 0$, the answer is "all $E_{\infty}$-spaces". When $k=0$, you'll
get those which are homotopy equivalent to simplicial commutative monoids.<|endoftext|>
TITLE: stopping time expectation for gambler's ruin
QUESTION [5 upvotes]: 2 players A and B start with x & y dollars respectively, and they bet against each other 1 dollar each time by tossing a fair coin.
I let $X_n = x + \sum_{i=1}^{n}\xi_i$ where $\xi_i$ are i.i.d. with $P(\xi_i=\pm 1)=\frac{1}{2}$.
Let $\tau_0 = \inf{\{n:X_n = 0}\}$, $\tau_{x+y} = \inf{\{n:X_n = x+y}\}$, and $\tau=\tau_0\wedge \tau_{x+y}$, which are all stopping times w.r.t. the martingale $(X_n)$.
I know that $E[\tau]=xy$, and $E[X_\tau]=x$.
Am i able to find $E[\tau^2]$ by using any moment generating function in this case? 
Any suggestion is welcome! Thanks!

REPLY [9 votes]: You can continue to play the same "game" that lead to the probabilities for hitting $0$ and $x+y$ and the expectations for $\tau$ and $X_\tau$. Just construct martingales to which you can apply the optional stopping theorem. Note that $X_n$, $X_n^2-n$, $X_n^3-3nX_n$, $X_n^4-6nX_n^2+n(3n+2)$, are all martingales. Applying the optional stopping theorem to the first two yields
$P(X_\tau=0)=\frac{y}{x+y}$, $P(X_\tau=x+y)=\frac{x}{x+y}$, $E(\tau)=xy$, as you already know.
With the next one you can show that
$E(\tau X_\tau)=\frac{1}{3}xy(2x+y)$.
Conditioning on $X_\tau=0$ and $X_\tau=x+y$ gives
$E(\tau X_\tau)=\frac{y}{x+y}E(\tau X_\tau|X_\tau=0) + \frac{x}{x+y}E(\tau X_\tau|X_\tau=x+y)$,
which tells you that
$E(\tau|X_\tau=x+y) =\frac{1}{3}y(2x+y)$, $E(\tau|X_\tau=0)=\frac{1}{3}x(x+2y)$,
i.e. now you know the expectations of the hitting times conditional on player A winning or losing.
Applying the optional stopping theorem to the fourth martingale will give you the second moment of the hitting time...
You can construct martingales that are higher order polynomials in $X_n$ and $n$, or, since after a while this gets a little bit tedious, you might want to put them together and use mgf's, to describe the law of $\tau$ (conditionally on hitting 0 and $x+y$). Just observe that
$E(\exp(\lambda \xi_i))=\cosh(\lambda)$,
so
$M_n(\lambda) = \frac{\exp(\lambda X_n)}{\cosh(\lambda)^n}$
is a martingale.<|endoftext|>
TITLE: ramified quaternion algebras
QUESTION [5 upvotes]: I'm trying to better understand the connection between the concepts of ramification of a field extension, and ramification of a quaternion algebra.  I'm also trying to build a better understanding of the motivation behind this term.  Bear with me as I summarize my understanding of the stuff...
If $K$ is a field with valuation $v$, and $L$ is a finite extension of it, the definition of ramification depends on whether $v$ is Archimedean.  If $v$ is Archimedean, then it corresponds to an embedding $\sigma:K\rightarrow\mathbb{C}$, and also $v$ extends to a valuation $w$ on $L$.  We say the extension is ramified here iff there is a nontrivial $\tau\in Gal(L,K)$ such that $w\tau=w$ (turns out to be independent of choice of $w$, hence well-defined). If $v$ is not Archimedean, then $v$ corresponds to a prime ideal $\mathcal{P}$ of $R_K$, and there are prime ideals $\mathcal{P}_i$ of $R_L$ such that $\mathcal{P}R_L=\Pi_{i=1}^r\mathcal{P}_i^{e_i}$, and we have that the extension is ramified here iff $\exists i: e_i>1$.
If $A$ is a quaternion algebra over the field $K$, and $v$ is a valuation on $K$, let $A_v:=A\otimes_KK_v$, and then we have (after a good amount of work) that $A_v\cong M_2(K_v)$ or $A_v$ is the unique division algebra over $K_v$.  And the definition of $A$ being ramified at $v$ is that $A_v$ is the division algebra and not the other one.
Now, behind these definitions, there is a lot of other theory one would use to actually determine these things.  Let's stick to number fields for simplicity.  For a number field as an extension of $\mathbb{Q}$, the infinite places are very simple, just have to do with whether an embedding is totally real.  And for the finite places you've got Kummer's theorem: look at when the minimal polynomial of the primitive element has roots modulo $p$, which I find very enlightening because it gives you an explicit description of how $(p)$ splits in the number field.  In the case of the quaternion algebras, any $A$ has a Hilbert symbol $(\frac{a,b}{K})$, and you can figure out where it ramifies by looking at $a$ and $b$.  The infinite places depend on the sign of the images of $a$ and $b$ under the embedding, and the finite places depend on whether or not $a$ and $b$ (and sometimes $-a^{-1}b$) are in $\mathcal{P}$ and $\mathcal{P}^2$.
So I get all that, but what exactly are we trying to measure here?  The infinite places seem more obvious: we are looking at whether or not there are embeddings in the bigger field that look the same in the smaller field.  But why should it be so important that the prime factorization of a prime in the bigger ring has repetition?  Why is it not called "ramified" just for the prime to split apart into several primes?  Also, for $A$ to ramify at a finite place $v$, does this imply some kind of factorization of lifts of prime ideals?  $A$ doesn't have a "ring of integers," but it does have an order, does this happen there?
I apologize for the vagueness of my questions, I'm just looking for some good intuition about this from someone who knows it very well.  My intention is to understand enough of this for now to use it for something else, but I would like to come back to it periodically until I've mastered it.

REPLY [8 votes]: If $F/{\mathbf Q}$ is a quadratic field then all but finitely many places $v$ of ${\mathbf Q}$ are unramified in $F$, and we could interpret what that means in a couple of ways: prime ideal factorization (for nonarchimedean $v$), extensions of absolute values (any $v$), or base extension by ${\mathbf Q}_v$ (any $v$). Let's use the last way: we look at ${\mathbf Q}_v \otimes_{\mathbf Q} F$. First suppose $v$ is nonarchimedean. For unramified $v$, the tensor product is ${\mathbf Q}_v \times {\mathbf Q}_v$ (if $v$ splits) or it is an unramified quadratic extension of ${\mathbf Q}_v$ (if $v$ is inert). For ramified $v$, the tensor product is a ramified quadratic extension of ${\mathbf Q}_v$ in the sense of ramified extensions of complete discretely valued fields. Next suppose $v$ is the archimedean place of ${\mathbf Q}$. We call $v$ unramified in $F$ when ${\mathbf Q}_v \otimes F$ is ${\mathbf R} \times {\mathbf R}$ ($F$ is real quadratic), and call it ramified in $F$ when ${\mathbf Q}_v \otimes F$ is $\mathbf C$ ($F$ is imaginary quadratic).
When $B$ is a quaternion algebra over ${\mathbf Q}$ and $v$ is a place of ${\mathbf Q}$, the base extension ${\mathbf Q}_v \otimes_{\mathbf Q} B$ is the matrix algebra ${\rm M}_2({\mathbf Q}_v)$ except for finitely many $v$, when it is a division algebra (something subtle). Think of ${\rm M}_2({\mathbf Q}_v)$ as being a noncommutative analogue of ${\mathbf Q}_v \times {\mathbf Q}_v$, and this is a reason for calling ${\rm M}_2({\mathbf Q}_v)$ the unramified case while the division algebra case is called ramified since those are the finitely many peculiar cases. The situation with quaternion algebras doesn't have anything like the inert case from quadratic fields, so "split" and "unramified" are used as synonyms (we say $v$ is "split" or "unramified" in $B$ when ${\mathbf Q}_v \otimes_{\mathbf Q} B$ is a matrix algebra, and "nonsplit" or "ramified" otherwise).
An analogy with quadratic extensions $L$ of $K :={\mathbf C}(X)$, where all residue fields are algebraically closed, is useful to bear in mind. For a discrete valuation $v$ on $K$ that is trivial on the constants, there is a notion of ramified or unramified for $v$ in $L$. It could be defined in terms of how the prime ideal associated to $v$ in the localization ${\mathcal O}_v$ of $K$ decomposes in the integral closure of ${\mathcal O}_v$ in $L$. Or it could be defined in terms of the base extension of $L$ by the completion of $K$ at $v$,$K_v \otimes_{K} L$: if this is $K_v \times K_v$ then we call $v$ unramified or split in $L$ (the two words are synonyms here) and otherwise this base extension is a quadratic ramified extension of $K_v$ (in the sense of ramified extension of a complete discretely valued field) and we then call $v$ ramified or nonsplit in $L$.<|endoftext|>
TITLE: Are all dinilpotent groups solvable, i.e., groups G=AB with nilpotent subgroups A, B ?
QUESTION [5 upvotes]: The Wielandt-Kegel theorem states that all finite dinilpotent groups G=AB are solvable.
This results extends to many infinite groups, e.g., to finitely-generated linear groups.
In general however, I think there will be a counterexample. I have searched the literature,
but could not find an answer. Is there a (nice, not too pathological ) counterexample  ?

REPLY [5 votes]: As Derek Holt mentioned, the question is open (if either $A$ or $B$ is not Abelian). Ito's theorem has been generalized several times by N. S. Chernikov and others, for example, see  Chernikov, N. S.
On groups that are factorizable by two subgroups with Chernikov commutants. Ukraïn. Mat. Zh. 52 (2000), no. 3, 396--402; translation in 
Ukrainian Math. J. 52 (2000), no. 3, 457–463. A typical result showing the current state  of the problem:
 A residually finite group $G$ factorized by two subgroups $A$ and $B$ with finite derived subgroups has a normal subgroup $H$ of finite index the second derived subgroup of which is finite and lies in the center $Z( H)$.
If you allow several (but finite number of) factors: $G=A_1A_2...A_n$, then $SL_n(\mathbb{Z})$, $n\ge 3$, is factorizable by cyclic groups (=boundedly generated), and Muranov proved that there are simple 2-generated groups that are boundedly generated. See Carter, D., Keller, G. (1983). Bounded elementary generation of $SL_n(O)$. Am. J. Math. 105(3):673–687, Muranov, Alexey, Diagrams with selection and method for constructing boundedly generated and boundedly simple groups. Comm. Algebra 33 (2005), no. 4, 1217–1258.<|endoftext|>
TITLE: Applications of higher-order reflection principles
QUESTION [8 upvotes]: Let $L_3$ be the third-order language of set theory with identity on the first sort. Variables $x$ are first-order, $y$ are second-order, and $z$ are third-order. In the style of Lévy and Bernays, we can extend the third-order theory $\mathrm{ZFC3}$ by reflection axioms of the form $\phi \rightarrow \exists x [\mathrm{Trans}(x) \wedge \mathrm{Rel}(\phi,x)]$ for each $L_3$ formula $\phi$ without free third-order variables, where $\mathrm{Trans}(x)$ says that $x$ is transitive and $\mathrm{Rel}(\phi,x)$ is the relativization of $\phi$ to $x$. Tharp's methods show both that the resulting theory $\mathrm{BL3}$ is consistent as long as first-order $\mathrm{ZF}$ is for each $n$ consistent with the existence of a $\Pi^2_n$-indescribable cardinal, and moreover that $\mathrm{BL3}$ proves that the $\Pi^2_n$-indescribable cardinals are stationary.

Does the latter fact have any significant applications that aren't already covered by weaker large-cardinal hypotheses, e.g. the unboundedness of the weakly compact cardinals?

We know that $\mathrm{BL3}$ becomes inconsistent if we drop the restriction that reflected formulas $\phi$ lack free third-order variables. On the other hand, work of Peter Koellner implies that $\mathrm{BL3}$ remains consistent relative to $\kappa(\omega)$ if we add reflection axioms in which $\phi$ has the form $\forall y_1 \exists u_1 \dots \forall y_k \exists u_k \psi$ where $\psi$ is a positive $L_3$ formula without higher-order quantifiers and each $u_i$ is an $L_3$ variable. Here an $L_3$ formula is said to be positive iff it is built using $\vee$, $\wedge$, $\exists$, and $\forall$ from atoms of the forms $x_n = x_k$, $x_n \neq x_k$, $x_n \in x_k$, $x_n \notin x_k$, $x \in y$, $x \notin y$, and $y \in z$.

Does the resulting theory $\mathrm{BL3}\prime$ prove anything significant not already proved by $\mathrm{BL3}$ or by $\mathrm{ZF}$ with unboundedly many weakly compacts?

Now get $\mathrm{BL3}^\ast$ from $\mathrm{BL3}\prime$ by adding reflection axioms in which $\phi$ can be any positive $L_3$ formula. It is not hard to show that $\mathrm{BL3}^\ast$ is consistent relative to a $2$-strong cardinal. But is there any reason to care about this theory?

Does $\mathrm{BL3}^\ast$ prove anything significant not already proved by $\mathrm{BL3}\prime$, by $\mathrm{BL3}$, or by $\mathrm{ZF}$ with unboundedly many weakly compacts?

Using a result of Rupert McCallum, I could continue asking such questions about extensions of the higher-order analogs of $\mathrm{BL3}$. But I fear I've bored you already.

REPLY [8 votes]: I view these kinds of hypotheses as on a continuum stretching from
weak compactness up through all the levels of the indescribability
hierarchy, second order, third order and so on. This hierarchy of
indescribability continues transfinitely via the strongly
unfoldable cardinals, which can be thought of simply as the
transfinite-order indescribable cardinals.
The basic phenomenon here is that as you rise in this hierarchy, you can prove the consistency of
stronger and stronger fragments of the proper forcing axiom. The
situation is that, although one seems to need a supercompact
cardinal to get the consistency of the full proper forcing axiom,
nevertheless weaker versions of PFA can be shown consistent from
these much weaker large cardinals in the vicinity of your
hypotheses.
For example, in my article J. D. Hamkins, T. Johnstone, The proper and
semi-proper forcing axioms for forcing notions that preserve $\aleph_2$ or $\aleph_3$,
we prove the following:
Theorem. The following are equiconsistent over ZFC:

There is an unfoldable cardinal.
PFA($\aleph_2$-preserving) + PFA($\aleph_3$-preserving) + $\text{PFA}_{\aleph_2}$ + $2^\omega=\aleph_2$.
$\text{PFA}_{\frak{c}}$.

A more local version of this result is:
Theorem. If $\kappa$ is $(\theta+1)$-strongly unfoldable, with
$\kappa\leq\theta$, and $0^\sharp$ does not exist, then the PFA
lottery preparation of $\kappa$ forces the principles
PFA($\aleph_2$-preserving$\cap V_\theta$),
PFA($\aleph_3$-preserving$\cap V_\theta$) and $PFA_{\frak{c}}(V_\theta)$.
This is very close to your third-order situation, if you take
$\theta=\kappa+1$ or $\theta=\kappa+2$; the cases of $\theta=\kappa+n$ are essentially the finite order indescribibility hypotheses. The point is that as you
rise to higher orders here, you can get the consistency of greater
fragments of the PFA axiom, applying it to larger partial orders.
Since $\kappa$ is totally indescribable if and only if it is
$(\kappa+m)$-strongly unfoldable for every finite $m$, we similarly
obtained the following:
Corollary. If $\kappa$ is totally indescribable and $0^\sharp$
does not exist, then the PFA lottery preparation of $\kappa$
forces PFA(size$\lt\beth_\omega$ +$\aleph_2$-preserving) and
PFA(size$\lt\beth_\omega$ + $\aleph_3$-preserving).
In addition, Neeman and Neeman and Schimmerling established the
following theorem:
Theorem.(Neeman, Neeman+Schimmerling)

PFA(c-linked) is equiconsistent over ZFC with the
existence of a $\Sigma^2_1$-indescribable cardinal.
If the existence of a $\Sigma^2_1$-indescribable cardinal is consistent with
ZFC, then so is PFA($\mathfrak{c}^+$-c.c.).

Our paper has further references and similar results.<|endoftext|>
TITLE: Is there a database somewhere for sharing translations of mathematical works?
QUESTION [24 upvotes]: When doing my language exams for my doctorate, I requested to translate articles relevant to my interests that had not previously been translated, and to be able to do them at home, with the agreement that they would be much more difficult than what you might do on the spot in an exam setting.  This was a very enlightening process for me when I did my French exam, and I expect it will be again when I start the German one next week.  But I was hoping to make more of a contribution to the community with these translations, rather than them just sitting in a file somewhere at the school.
Does anyone know of a database for downloading and voluntarily submitting translations of math articles?  I can't find anything fitting that description online.  I expect such a thing would want to be moderated.  But my translations are (and will be) good, I'm doing them very carefully, and if they weren't good my professors wouldn't accepted them.
In the meantime, if anyone is interested in an English translation of an article by Weil called "Sur la théorie des formes différentielles attachées à une variété 
analytique complexe," feel free to contact me and I'll send it.  It's actually a letter he wrote to de Rham in 1946.
Also, if anyone is interested in an English translation of a 1977 article by Volker Schneider called "Elliptische Fixpunkte und Drehfaktoren zur Modulgruppe in Quaternionenschiefkörpern über reellquadratischen Zahlkörpern," I should have that done in a couple of weeks.
...but there's gotta be a better way of making one's translations available to those who would benefit from them.
UPDATE:
Following a suggestion from @AdeelKhan here is a link where the translation mentioned can be found, as well as a translation of "Elliptische Fixpunkte und Drehfaktoren zur Modulgruppe in Quatemionenschieikörpern über reellquadratischen Zahlkörpern" (Volker Schneider, 1977).
http://joequinn.ws.gc.cuny.edu/articles/
UPDATE 2:
My grad school hosted website no longer exists, so that link won't get you anywhere.

REPLY [20 votes]: Put it on arxiv.org, in the math.HO category ("HO" stands for "History and Overview").  I've seen (and read) contributions there that are no more than translations of Euler from Latin to English.  You'll want to write it up like a paper, with an explanatory paragraph providing context before the actual translation.  This plan seems to provide everything you want, since arxiv.org is indexed by search engines, available to everyone, and should last a long time.

REPLY [10 votes]: I do not know of such database, but I like the idea of having one very much. Then we can ask all
students passing the language exam to contribute to it:-)
Mathematical papers are very rarely translated from French and German to English.
(Most papers of Gauss and Riemann are not available in a good English translation). Unlike Russian
papers: almost all main Russian journals are translated cover-to-cover, at least this was the case
few years ago. Presumably people think that if you understand the alphabet then you can read
mathematics written in this alphabet in any language:-)
Meanwhile, before this database is created, you can submit your translation to the arXiv.
People do submit translations to the arxiv. Search for example on the name: Euler.<|endoftext|>
TITLE: How many proofs of the Weil conjectures are there?
QUESTION [20 upvotes]: I hope this this is not seen as too much as jumping on the band-wagon, but here goes.
Deligne's proof of the last of the Weil conjectures is well-known and just part of a huge body of work that has lead to prizes, medals etc (wink wink). The other conjectures were proved by Dwork and Grothendieck. According to Wikipedia, Deligne gave a second proof, and then mentions three more proofs. However, it is unclear from what I read as to which of the conjectures they all prove - namely all of them, or just the Riemann hypothesis.

So how many proofs of the complete set of Weil conjectures are there?

REPLY [40 votes]: I guess that "just the Riemann hypothesis" means the statement about the eigenvalues of Frobenius acting on the $H^i$ of a smooth proper variety, and "all of then" means the full theory of weights : definition of mixed sheaves, how the 4 operations affect weights etc (from which you get the hard Lefschetz theorem, the decomposition theorem...).
So here is my understanding :

Deligne's first proof : just the Riemann hypothesis.
Deligne's second proof : the full package for $\ell$-adic complexes, using the same kind of ideas as his proof but improving them.
Laumon's proof with the $\ell$-adic Fourier transform : It allows to simplify parts of Deligne's second proof, but does not replace all of it. More precisely, Laumon gives a shorter proof of theorem 3.2.3 of Deligne's "La conjecture de Weil II" that doesn't use section 2 of that paper, but he still needs results from section 1 (the calculation of the monodromy of lisse sheaves on curves in 1.8 and the fact that "most" irreducible lisse sheaves are pure proved in 1.5.1). Once you have theorem 3.2.3, it is pretty easy to deduce the rest of the "full package" from it, so Laumon doesn't say anything about that.
Katz's proof : Katz gives a simpler proof of theorem 3.2.3 of Deligne's Weil II paper, or rather of a weakening of it that is enough to deduce the Riemann hypothesis and hard Lefschetz. It is not enough to deduce the full package immediately, although this can be done without too much pain. I am not very familiar with Katz's proof. It seems to be different from Deligne's and Laumon's proofs but to use the same kind of techniques as Deligne's proof. My impression is that it is more concrete that Deligne's proof.
(Note : I am not convinced that the difference between "just the Riemann hypothesis" and "the full package" is so big. The reason is that, if you know the Riemann hypothesis, then you should be able to deduce the full package for complexes of geometric origin. But global Langlands tells us that, over a smooth curve over a finite field, every irreducible lisse sheaves is of geometric origin up to twisting by a rank 1 sheaf. So in a way everything is of geometric origin. The situation is very different over a number field, of course.)
Kedlaya's proof : It is supposed to be modeled on Deligne's second proof modulo the simplification of Laumon, but of course the actual technical are different because Kedlaya uses p-adic coefficients. Also, he got the Riemann hypothesis but not the full package, because at the time the theory of relative p-adic coefficients was not fully developed. But now it is !
Abe and Caro's proof : See http://arxiv.org/abs/1303.0662
They develop the theory of weights for overholonomic D-modules with Frobenius structure on varieties over finite fields and obtain a full Weil II package for them (and even the Asterisque 100 extension, ie the theory of weights for perverse sheaves). They say in the introduction that their method is independent from Kedlaya's method.<|endoftext|>
TITLE: Is the primitive element theorem a cohomological statement?
QUESTION [5 upvotes]: If $K$ is a field, then as is well known every finite separable extension $L$ of $K$ is of the form $L=K(\alpha)$ for some $\alpha \in L$.
A similar statement can be made about an extension of discrete valuation rings with separable residue field extension.
These statements very much resemble the statement "every projective module over a principal ideal domain $A$ is free". This last statement can be interpreted as the vanishing of a certain cohomology group. Now my question is: can the primitive element theorem be interpreted as the vanishing of a cohomology group?
Thank you!

REPLY [12 votes]: The vanishing of the cohomology group $H^1(Spec(R),GL_n)$ doesn't actually say that all projectives of rank $n$ are free; it says only that all projectives of rank $n$ are isomorphic.  Combining this with the observation that at least one such projective is free, we get that they're all free.
But in the case of field extensions, it is not true that all finite separable extensions  are isomorphic, even though they're all generated by primitive elements.  Therefore, I think the analogy you're seeing is largely illusory.<|endoftext|>
TITLE: Categorical description of the second K-group
QUESTION [9 upvotes]: Let $\mathcal{P}$ be a (small) exact category. Without delving into any homotopy theory, we can provide characterisations of $K_0(\mathcal{P})$ and $K_1(\mathcal{P})$ as plain categorical constructions:
$K_0(\mathcal{P})$ is the free abelian group on the objects of $\mathcal{P}$ under the relations $[P] = [P'] + [P'']$ if $0 \to P' \to P \to P'' \to 0$
$K_1(\mathcal{P})$ is the free abelian group on the pairs $(P, \alpha)$, where $P$ is an object and $\alpha$ one of its automotphisms, under suitable relations.

Does $K_2(\mathcal{P})$ admit a similar description?


My question has been answered (and then some!). It has been pointed out however that I was slightly mistaken: the $K_1(\mathcal{P})$ given above only coincides with the standard one ($\pi_2(Q \mathcal{P})$) when short exact sequence of $\mathcal{P}$ splits. So I actually refer to the `Bass $K_1$'.

REPLY [9 votes]: Algebraic generators and relations for Quillen's K-group $K_n(P)$ are given in this paper: "Algebraic K-theory via binary complexes".  Those you mention don't give the Quillen K-group $K_1(P)$ in general, but just for Quillen-exact categories where every short exact sequence splits; the group they give is known as Bass' $K_1$.

REPLY [6 votes]: The $K_1$ group you describe is the automorphism $K_1$, which is in general not isomorphic to Quillen's $K_1$ of an exact category. It coincides with Quillen's when exact sequences in $\mathcal P$ split. For a description by generators and relations of $K_1$ of any exact category see:
Nenashev, A.
K1 by generators and relations. (English summary) 
J. Pure Appl. Algebra 131 (1998), no. 2, 195–212. 
generators are pairs of short exact sequences on the same objects, and relations are given by $3\times 3$ diagrams. A generalization of this result to all $K_n$ is given in
Grayson, Daniel R.
Algebraic K-theory via binary complexes. 
J. Amer. Math. Soc. 25 (2012), no. 4, 1149–1167.<|endoftext|>
TITLE: Are there results in "Digit Theory"?
QUESTION [19 upvotes]: Results about numbers that are related to their decimal representation are usually confined to recreational mathematics. There I have seen mainly questions about individual numbers, like finding a  number that is the sum of the cubes of its digits. But I have not yet seen a systematic study of such questions.
A result that prompted me to asked this question was the following from Kurt Hensel's "Zahlentheorie". There he shows that the exponent of highest power of a prime $p$ that divides the number $n$ is $$\frac{s_{n-1} - s_n + 1}{p - 1},$$ where $s_n$ is the sum of the digits in the representation of $n$ with base $p$. From this he derives that the highest power of $p$ that divides $n!$ is $$\mu_n = \frac{n - s_n}{p - 1}.$$ This is a different and much simpler formula than the representation $\mu_n = \sum_{k = 0}^\infty \left\lfloor \frac{n}{p^k} \right\rfloor$ by Legendre, which I usually see in the literature.
I would like to know whether there are more results of this type. To be more precise, I define the not-yet-existing field of "Digit Theory" as the study of number-theoretic questions with help of the positional representation of numbers, and the study of the positional representation itself.

So are there already systematic treatments of "Digit Theory" or of parts of it? Are there other interesting results like that one above?

REPLY [8 votes]: James Maynard has a survey paper Digits of primes. In Primes with restricted digits he shows that for any digit $d\in\{0,1,\dotsc,9\}$ there are infinitely many primes that do not have $d$ in their decimal expansion. This might be the deepest known result in Digit Theory.<|endoftext|>
TITLE: Hyperfiniteness of CCR algebra
QUESTION [6 upvotes]: Hi, It is known that the double commutant of the CCR algebra in it's GNS space 
with respect to some quasi-free states are always type III factors.
My question is;  Will some of them be hyperfinite factors?

REPLY [5 votes]: Yes, Araki-Woods showed they're always ITPFI factors and ITPFI factors are hyperfinite. See the following:
Araki-Woods, A classification of factors
It's a bit of a monster paper, the stuff on CCR algebras is near the end.<|endoftext|>
TITLE: "Wild" solutions of the heat equation: how to graph them?
QUESTION [30 upvotes]: It has long been known that the Cauchy initial-value problem for the
classical heat equation on $\mathbb{R}$ (or $\mathbb{R}^n$) doesn't
have unique solutions, without additional assumptions.  In particular:

Theorem. There exist nontrivial $C^\infty$ functions $u :
    (-\epsilon, \epsilon) \times \mathbb{R} \to \mathbb{R}$ satisfying
      the heat equation
  $$\partial_t u - \partial_x^2 u = 0$$
  with the initial condition $u(0,x) = 0$.

These functions must somehow represent disturbances arriving from
infinity in finite time.  They seem pretty weird to me and I think a
picture would help with intuition.

I would like to see a graph of such a function.  Where can I find one,
  or how can I generate it?

There are some explicit examples known.  For instance, Rosenbloom and
Widder [1] give the following:
$$u(t,x) = \int_0^\infty e^{-y^{4/3}} y \cos(\sqrt{3} y^{4/3}) 
(e^{xy} \cos(xy + 2ty^2) + e^{-xy} \cos(xy-2ty^2))\,dy$$
as well as
$$v(t,x) = \int_{a-i\infty}^{a+i\infty} e^{st + x\sqrt{s} - s^{2/3}}\,ds.$$
Unfortunately, the integrals are oscillatory and don't look so nice to
approximate numerically.
Of course, there are various uniqueness theorems that give us
qualitative information about what these nasty solutions must look
like.  Tychonoff's uniqueness theorem says $u(t,x)$ must grow faster
than $e^{cx^2}$ near $x = \infty$, and Widder's theorem says that $u$
must be unbounded above and below for arbitrarily small $t$.  But I'd really like to have an explicit picture to look at.
[1] Rosenbloom, P.C. and Widder, D.V.  A temperature function which
vanishes initially.  American Mathematical Monthly
65(8):607-609, 1958.

REPLY [15 votes]: Here's another plot (using Maple 17, and the first 21 terms) of the solution Carlo referred to.<|endoftext|>
TITLE: Modular interpretation of nebentypus
QUESTION [8 upvotes]: Recall that for a subgroup $\Gamma \subset SL_2(\mathbb{Z})$ a modular form $f$ of weight $k$ is a holomorphic function from the upper-half plane into the complex numbers such that for any 
$\begin{pmatrix}
a & b \\ 
 c& d
\end{pmatrix}\in \Gamma$
and any $z$ in the upper half-plane, we have
$f(\frac{az+b}{cz+d}) = (cz+d)^k f(z)$
and $f$ is meromorphic (or holomorphic, depending on taste) at the cusps, which can be formulated as a growth condition, which I won't make precise here. 
For many subgroups $\Gamma$, we have a modular interpretation of this. For example, for $\Gamma = SL_2(\mathbb{Z})$, a modular form corresponds to  a section of $\omega^{\otimes k}$ on the (compactified, if we want holomorphic at the cusps) moduli stack of elliptic curves over $\mathbb{C}$. For $\Gamma = \Gamma_0(N) = \{\begin{pmatrix}
a & b \\ 
 c& d
\end{pmatrix}\in SL_2(\mathbb{Z}) | c \equiv 0 \mod N\}$, we consider instead elliptic curves with a fixed subgroup of order $N$. As this all makes sense over $\mathbb{Z}$ (or, at least, $\mathbb{Z}[\frac{1}{N}]$) instead of $\mathbb{C}$, we also get integral versions of the rings of modular forms. 
But in some sense, this seems to be just the tip of the iceberg. Many functions which show some "modular behaviour" are not modular forms in the sense above. In particular, this is true for many kinds of $\Theta$-functions, where both the aspects of a nebentypus and a half-integral weight come in. 
Let $\chi: (\mathbb{Z}/N)^\times \to \mathbb{C}$ be a character. Then a modular form of weight $k$, level $N$ (or level group $\Gamma_0(N)$) with nebentypus $\chi$ is a holomorphic function $f$ on the upper half-plane such that for any
$\begin{pmatrix}
a & b \\ 
 c& d
\end{pmatrix}\in \Gamma_0(n)$
and any $z$ in the upper half-plane, we have
$f(\frac{az+b}{cz+d}) = \chi(d)(cz+d)^k f(z)$
and $f$ is meromorphic (or holomorphic) at the cusps.
Is there any modular interpretation for modular forms with nebentypus? Are there integral versions of rings of modular forms with nebentypus? And how about the story when we have only half-integral weight?

REPLY [11 votes]: Yes, there is such an interpretation, which is standard (cf. Delinge-Rapoport or Katz-Mazur).
You begin by interpreting the modular forms of level $\Gamma_1(N)$ and weight $k \in \mathbb N$ as section of $\omega^k$ (plus condition at infinity) on the moduli space
$Y_1(N)$ (over $\mathbb C$) of all elliptic curves $E$ together with what is called "a $\Gamma_1(N)$-level structure" and is simply a point $P$ on your elliptic curve of order exactly $N$.
Then you define an action of $(\mathbb Z/N\mathbb Z) ^\ast$ on $Y_1(N)$ simply by $x \cdot (E,P)=(E,xP)$.
Note that since $x$ is invertible modulo $N$, $xP$ is still of exact order $N$. This action
induces an action on the sheaf $\omega$, hence on the sections of $\omega^k$, hence
(putting under carpet the condition at infinity) hence on the space of modular forms.
This actions is the same as the action of the diamond operators.
Hence, to get your space with Nebentypus $\epsilon$, you just take the $\epsilon$-part for that action.
An other way to say the same thing is to say that a modular form with Nebentypus $\epsilon$
of weight $k$ and level $N$ is a law, which to any triple $(E/S,w,P)$
where $S$ is a scheme over $\mathbb C$, $E$ an elliptic curve over $S$, $w$ an invariant differential form on $E$, $P$ a point in $E(S)$ of order $N$ attaches an element of $\Gamma(S,{\cal O}_S)$. The law being subject to those three conditions:
(a) it commutes with base change $S' \rightarrow S$. (b) Multiplying $\omega$ by $\lambda$ multiplies the results by $\lambda^{-k}$, for $\lambda$ in $\Gamma(S,{\cal O}_S)$.
(c) Multiplying $P$ by $x \in (\mathbb Z/N\mathbb Z)^\ast$ multiplies the result by $\epsilon(x)$.
Of course all of this has an interpretation over $\mathbb Q(\zeta_N)$ and furnishes an algebraic version of your space of modular forms. You can also go to integers if you like.<|endoftext|>
TITLE: Have we ever proved any non-solvable case of reciprocity without the Langlands program ?
QUESTION [24 upvotes]: The reciprocity of the title is the following not completely well-posed problem:
Fix $P(X)$ a monic irreducible polynomial of degree $n$, with coefficients in $\mathbb Z$. "Describe" 
(in some sense) the set of prime $p$ such that $P \mod p$ has a given reduction (for example is irreducible, or on the opposite is a product of $n$ linear factors, etc.).
As is well-known, when one root (hence all roots) of $P$ happens to be a cyclotomic integer (in particular in the very special case where $P$ is quadratic), then the problem has a precise solution, called Artin's reciprocity law, which is the core of class field theory. The sets of primes $p$ such that $P$ has a given reduction are then given
by congruences modulo fixed integers satisfied by $p$, that is those sets are union of
"set primes in artithemtic sequence". The Galois group of $P$ is abelian in this case.
As the opposite of the complexity spectrum, there are polynomial $P(X)$ of degree $n \geq 5$, whose Galois group is big, say $S_n$, or $A_n$ in particular not solvable. In this case, it is my understanding that our only hope to find a reciprocity law in general is by making huge progress in the Langlands program. 

But
  is it possible, in some special case (for polynomials $P$ of Galois group $S_n$ say satisfying
  somme assumtions, or for an explicit family of polynomials of variable degree $n \geq 5$,
  or even for just one explicit polynomials) where some kind of reciprocity law has been worked out (even only partially) by some method that does not involve the Langlands program (i.e. modular forms, automorphic forms, etc.) ?

REPLY [6 votes]: This question gives me the chance to advertise a result contained in http://arxiv.org/abs/1201.2124 which characterizes primes which are completely split in torsion fields extensions $K(E[N])/K$ of elliptic curves over number fields. Sorry for being self-referential.
Let $K$ be a number field, $E$ an elliptic curve over $K$, and $N$ an integer $>0$. For a finite prime $\mathfrak{p}$ of $K$ with residue field $k_\mathfrak{p}$, denote by $a_\mathfrak{p}$ the trace of $E \text{ mod } \mathfrak{p}$, and by $\Delta_\mathfrak{p}$ the discriminant $a_\mathfrak{p}^2-4|k_\mathfrak{p}|$ of the characteristic polynomial $x^2-a_\mathfrak{p}x+|k_\mathfrak{p}|$.
$\textbf{Theorem}.$ There exists a universal family of polynomials $\{\mathcal{P}_D(x)\}_{D\leq 0}$ satisfying the following property. Let $\mathfrak{p}$ be a prime of good reduction for $E$ which does not divide $N$, and for which $E\text{ mod }\mathfrak{p}$ is not special* if $N=2$. Then $\mathfrak{p}$ splits completely in $K(E[N])/K$ if and only if both conditions below are satisfied:
i) $N^2$ divides $\Delta_\mathfrak{p}$, and $\mathcal{P}_{\Delta_\mathfrak{p}/N^2}(\;j_E\;)\equiv 0\text{ mod }\mathfrak{p}$;
ii) $a_\mathfrak{p}\equiv 2 +\dfrac{\Delta_\mathfrak{p}}{N}\text{ mod }N^*$;
where $N^*=N$ if $N$ is odd, and $N^*=2N$ otherwise.

*this condition, not explained here, avoids only finitely many $\mathfrak{p}$.
If $D$ is a negative discriminant, the polynomial $\mathcal{P}_D(x)$ is monic with integer coefficients. Its roots are the $j$-invariants of complex elliptic curves with CM by an order containing the imaginary quadratic order of discriminant $D$. Moreover $\mathcal{P}_0(x)=0$ and $\mathcal{P}_D=1$ if $D$ is not a discriminant.
The proof of the result is via local methods and relies on the fact that if the ring of $k_\mathfrak{p}$-endomorphisms of $E\text{ mod }\mathfrak{p}$ is a quadratic order, then the action of $\text{Frob}_\mathfrak{p}$ on $E[N](\bar K)$ is equivalent to the action of $\text{Frob}_\mathfrak{p}$ on $\tilde E_\mathfrak{p}[N](\bar K)$, where $\tilde E_\mathfrak{p}$ is the Deuring lifting of $E\text{ mod }\mathfrak{p}$. The evaluation of the polynomials $\mathcal{P}_D(x)$ at the $j$-invariant $j_E$ of $E$ enters in condition i) in order to identify the correct lifting of $E\text{ mod }\mathfrak{p}$ (for this to work in the supersingular case one has to make some observations).
The Theorem was well known if $\mathfrak{p}$ is an ordinary prime for $E$. The fact that the above formulation remains true for supersingular primes (infinitely many when $K$ is real) is perhaps the novelty.
Since the methods used in the proof are rather antique, I realize that the result might be not so interesting to experts. But at least its statement gives an idea of how a reciprocity law in a non-solvable context might look like.
Adelmann in his book "The Decomposition of Primes in Torsion Point Fields" treats the same problem. He employs modular polynomials to characterize complete split primes.<|endoftext|>
TITLE: Coboundaries and Gluing in Cech Cohomology - Intuition?
QUESTION [13 upvotes]: I'm trying to develop an intuition for Cech cohomology geometrically, but am currently failing. A lot of people seem to say that the groups $H^n$ measure obstructions to gluing local sections to get global ones. However I don't see how this works, and I think it's because I don't understand coboundaries.
I see that $H^0$ is the group of global sections. Moreover $B^1\setminus 0$ tells us all the 1-cochains arising from the failure of 0-cochains to glue to give global sections. This suggests to me that we might want to study $B^1$, but instead we quotient by it to get $H^1$. Why?
I know that $Z^1$ tells us all the 1-cochains that agree on overlaps, and obviously $B^1$ is a subset of $Z^1$. But I don't know what $Z^1$ modulo $B^1$ is meant to tell me. I especially don't see how it has anything to do with gluing things. As far as I can tell it just informs me about the consistent 1-cochains which don't have anything to do with the failure to glue 0-cochains together. What am I missing?
Edit: it has just occurred to me that maybe the thing I don't understand is the meaning of gluing here. If I have the wrong idea about that could somebody please explain what is meant by it? Many thanks!

REPLY [6 votes]: Regarding higher cohomology, I was about to write another comment, but the comments are getting a bit long, so let say a few words here. Cech theory is really a generalization of simplicial cohomology, as Liviu explained, and this is really the best way to understand where the formulas come from. But if you find that unhelpful, let me work out a basic  example.
Suppose $X$ is a $C^\infty$ or complex manifold, and let $\mathcal{O}_X$ denote either the sheaf of complex valued $C^\infty$ functions or holomorphic functions. A complex line bundle
is a manifold $L$ of the same  type with a map $L\to X$ which makes it locally a product.
In other words we have isomorphisms $\sigma_i:\mathbb{C}\times U_i\cong L|_{U_i}$. The first components of $\sigma_i^{-1}\sigma_j$ gives a Cech $1$-cocycle of $g_{ij}\in \mathcal{O}^*(U_{ij})$. If you wish to eliminate the choice of trivialization ($\sigma_i$), you need to work  with cohomology class. 
The basic topological invariant of $L$ is its first Chern class
$c_1(L)\in H^2(X,\mathbb{Z})$ which measure the failure of $L$ to be (topologically) a product. The first question is what does $H^2$ mean? The second is what is $c_1$? For 1, you can triangulate $X$, and use simplicial cohomology, or you can use Cech cohomology. The fact that the formulas are similar, means that the group may be similar. In fact, they are the same. For 2, we note that in the $C^\infty$ case, $H^1(X,\mathcal{O}_X)=0$. We can attempt
lift back to this trivial group by taking the normalized logarithm
$\frac{1}{2\pi \sqrt{-1}}\log g_{ij}$. However, because the logarithm is "multvalued", there is no guarantee that this is a cocycle. So take the Cech boundary to get
$$c_1(L) = \frac{1}{2\pi \sqrt{-1}}(\log g_{ij}+\log g_{jk}+\log g_{ki})$$
When the cohomology class defined by this vanishes, then $L$ is topologically trivial.<|endoftext|>
TITLE: Combining van der Waerden's theorem with Ramsey's theorem
QUESTION [10 upvotes]: Consider positive integers $c$, $k$, and $s$. Does there exist some $N = N(c,k,s)$ such that the following holds? 

Take any $c$-coloring of the $k$-tuples of integers in $[1,N]$. Then there is an arithmetic progression of length $s$ such that all $k$-tuples of it have the same color. More precisely, there exists a set $S \subseteq [1,N]$ such that $|S| = s$, $S$ is an arithmetic progression, and all tuples in $S^k$ have the same color.


If $k=1$, then this is van der Waerden's theorem. If I don't care for an arithmetic progression (and am happy with just a set), then this is Ramsey's theorem. This is asking for the best of both theorems. 
I was unable to find such a statement in my searches. I thought this might be a consequence of the Hales-Jewett Theorem, but could not show this. It seems like a first principles approach combining proofs of van der Waerden and Ramsey's theorem might work. But I was wondering if this is already known.
Thanks!

Update
User Wei Wang has refuted this statement below with a simple counterexample. Exampled added here for clarity.
Suppose $N(2,3,4)$ existed. Simply color all three tuples in $[1,N]$ that form an AP blue, and remaining tuples red. Consider any AP $S = (s_1, s_2, s_3, s_4)$. The tuple $\{s_1, s_2, s_3\}$ is blue and $\{s_1, s_3, s_4\}$ is red.
User quid in his answer below has given a reference showing that a variant of my question is true. If I understand correctly, it is the case that there are k APs $S_1, S_2, \ldots, S_k$ such that tuples in $S_1 \times S_2 \times \cdots \times S_k$ have the same color. One can also ensure that these APs have the same minimum difference. (One can get a lot more, as explained in quid's post.) All in all, I think his post gives "more than an answer" to my question.

REPLY [2 votes]: Simple simultaneous generalizations of Ramsey's and van der Waerden's Theorem are available. The first was found by Bergelson and Hindman in their only paper published in Combinatorica. The second, which is similar but incompatible and perhaps cleaner, is available in my paper with Samet, Corollary 3.15. 
Say that a set of natural numbers is AP if it contains arbitrarily long arithmetic progressions.
This corollary asserts that the Baumgartner-Taylor partition relation holds for AP sets:
$$
\mathrm{AP}\longrightarrow \lceil \mathrm{AP}\rceil^k_n
$$
for all $k$ and $n$.
For $n=2$, this corollary asserts the following: For each AP set $A$ and each finite edge-coloring of the complete graph with set of vertices $A$, there is an AP set $M$ of vertices and a partition of $M$ into finite sets such that all edges among distinct pieces of this partition are of the same color.
Using the compactness theorem, one may obtain a finite version of this theorem.<|endoftext|>
TITLE: infimum of the Calabi energy in a given Kahler class
QUESTION [5 upvotes]: Given a compact Kahler manifold $M^n$ and a Kahler class $\Omega$. We have the Calabi energy (or Calabi functional) 
$$\textrm{Ca}(\omega)=\int_Ms^2(\omega)\omega^n,\qquad \forall \omega\in\Omega.$$
Here $s(\omega)$ is the scalar curvature with respect to the metric $\omega$. My question is whether or not there is an explict expression of
$$\textrm{inf}_{\omega\in\Omega}\textrm{Ca}(\omega).$$
Here by "explicit" I mean that if it can be expressed in terms of some well-known or familar quantities in Kahler geometry.
According to my best knowledge, A.D. Hwang and Xiuxiong Chen have a result that $\textrm{Ca}(\omega)$ has a lower bound in terms of the Calabi-Futaki functional evaluted at the extremal vector field and the equality holds iff there exists an extremal metric in $\Omega$. So according to their result, $\textrm{inf}_{\omega\in\Omega}\textrm{Ca}(\omega)$ also has this as its lower bound.

REPLY [7 votes]: If $\Omega$ is a rational Kähler class, so $M$ is projective algebraic, then Donaldson was the first to prove the Chen-Hwang lower bound that you stated. In fact, in this case he proved more, namely that
$$\inf_{\omega\in\Omega}\mathrm{Ca}(\omega)\geq \sup_{\mathfrak{X}}\Psi(\mathfrak{X}),$$ where $\mathfrak{X}$ is any test configuration for $(M,\Omega)$, and $\Psi(\mathfrak{X})$ is a suitable normalization of the Futaki invariant of $\mathfrak{X}$ (see the paper of Donaldson for definitions). The point is that if $(M,\Omega)$ is K-unstable, then the supremum is strictly positive (by definition), and so you conclude that there is no constant scalar curvature Kähler metric in the class $\Omega$.
In the same paper (p.455) Donaldson conjectures that equality above should always hold. This would be the sharp lower bound that you are looking for, and it is still open in general. Very recently, as a consequence of work of Chen, Donaldson, Sun, Tian and Li, it was shown that equality holds when $M$ is Fano, $\Omega$ is the anticanonical class, and $(M,\Omega)$ is K-stable or K-semistable (i.e. the case when the supremum on the RHS is zero).
Therefore there is no general "simple" formula for the lower bound of the Calabi energy, and this is related to the difficult open problem of relating existence of constant scalar curvature Kähler to algebro-geometric stability. See for example these surveys for more on this problem.
By the way, Donaldson's inequality was inspired by a similar identity proved by Atiyah-Bott for holomorphic vector bundles on algebraic curves, with the infimum of the Calabi energy replaced by the infimum of the $L^2$ norm of the curvature of all compatible unitary connections. This identity was recently extended by A.Jacob to all holomorphic vector bundles on compact Kähler manifolds.<|endoftext|>
TITLE: When are isometry groups of hyperbolic 3-manifolds finite?
QUESTION [6 upvotes]: If $M$ is a finite volume hyperbolic 3-manifold, then its isometry group is finite. I believe this is also true for geometrically finite 3-manifolds. What is the most general condition on a hyperbolic 3-manifold that is known to ensure that its isometry group is finite?

REPLY [6 votes]: Here is the detailed answer. First, you have to assume that your hyperbolic manifold is complete and has finitely generated fundamental group, otherwise you will get no answer except for the tautological one. Thus, you are dealing with a finitely generated torsion free Kleinian group G in SO(3,1). If such a group is elementary then the normalizer of the group is not discrete, consider for instance the case of an abelian discrete group of isometries. 
Thus, let us exclude elementary groups G. Then the normalizer is discrete (consider its action on the limit set of G). Now, if G is also geometrically finite, then the normalizer is a finite extension of G. To see this, consider convex core of your hyperbolic manifold and observe that its thick part is compact. In the geometrically infinite case, there is one example you should be aware of: Take finite volume hyperbolic 3-manifold which fibers over the circle and take M to be its infinite cyclic cover associated with the fiber. Then the isometry group of M contains infinite cyclic group. Covering Theorem of Thurston and Canary in addition to the positive solution of tameness conjecture by Agol, Calegari and Gabai shows that such M's are the only examples where the isometry group is infinite.  Moreover, the isometry group in this case is a finite extension of an infinite cyclic group. (The latter is immediate.)
A side note: If you consider higher dimensional hyperbolic manifolds, the situation in geometrically finite case is essentially the same except you have to assume that the limit set is not contained in a round sphere of codimension 2. In geometrically infinite case, there are more examples one has to exclude and there is no even conjectural classification of exceptions.<|endoftext|>
TITLE: Rational points on a sphere in $\mathbb{R}^d$
QUESTION [21 upvotes]: Call a point of $\mathbb{R}^d$ rational if all its $d$ coordinates are rational numbers.

Q1.
  Are the rational points dense on the unit sphere $S :\; x_1^2 +\cdots+ x_d^2 = 1$, i.e. does $S$ contain a dense set of rational points?

This is certainly true for $d=2$, rational points on the unit circle.

Q2.
  If (as I suspect) the answer to Q1 is Yes, is there a sense in which the rational coordinates are becoming arithmetically more complicated with larger $d$, say in terms of their height?

If $x= a/b$ is a rational number in lowest terms (i.e. gcd$(a,b)=1$), then the height of $x$ is $\max \lbrace |a|,|b| \rbrace$.
This is far from my expertise.  No doubt this is known, in which case a pointer would suffice.  Thanks!

(Added, 22Mar13).  I just found this reference. 

Klee, Victor, and Stan Wagon. Old and new unsolved problems in plane geometry and number theory. No. 11. Mathematical Association of America, 1996. p.135.

REPLY [12 votes]: We (with Keith Merrill) recently wrote a paper further quantifying the density of rational points on $S^n$, see http://arxiv.org/abs/1301.0989. Hope it helps!<|endoftext|>
TITLE: Silly question about mixing
QUESTION [9 upvotes]: Let $T$ be a measure-preserving transformation on a probability space $(\Omega,\mathcal B,\mu)$. Assume that for any pair of measurable sets $A,B\in\mathcal B$ with $\mu(A), \mu(B)>0$, one can find $N$ such that $T^{-n}(A)\cap B\neq\emptyset$ for all $n\geq N$. Is $T$ mixing with respect to $\mu$? This is likely to be a simple exercise. So, apologies, and thanks in advance.

REPLY [3 votes]: The authors left out that $T$ is ergodic in Lemma 2.4.  If $T$ is ergodic and $\liminf_{n\to \infty}\mu (T^nA\cap A) > 0$ for all sets $A$ of positive measure, then $T$ is lightly mixing. In other words, $\liminf_{n\to \infty}\mu (T^nA\cap B) > 0$ for all sets $A,B$ of positive measure. Proof: Suppose there exists sets $A,B$ of positive measure such that $\liminf_{n\to \infty}\mu (T^nA\cap B)=0$. There exists a sequence $n_k$ such that $\lim_{k\to \infty} \mu (T^{n_k}A\cap B)=0$. Since $T$ is ergodic choose, $\ell$ such that $\mu (A\cap T^{\ell}B)>0$. Let $A'=A\cap T^{\ell}B$. Thus, $\mu (T^{n_k+\ell}A'\cap A')\leq \mu (T^{n_k+\ell}A\cap T^{\ell}B) = \mu (T^{n_k}A\cap B)$. The last equality holds by measure preservance. Therefore, $\lim_{k\to \infty} \mu (T^{n_k+\ell}A'\cap A')=0$.<|endoftext|>
TITLE: Recovering $\sum_{n \leq x} a(n)$ from $\sum_{n \leq x} a(n)e^{-n/x}$
QUESTION [8 upvotes]: In the theory of automorphic forms and multiple Dirichlet series, we often take inverse Mellin transforms of Dirichlet series to come up with Tauberian theorems, like the Ikehara Tauberian method. In particular, from
$ \displaystyle \sum_n \frac{a(n)}{n^s}$, we might expect to use Mellin to get asymptotics for $\displaystyle \sum_{n \leq x} a(n)$.
Sometimes, the series doesn't behave well enough to integrate. A common way around that seems to be to multiply by the gamma function to get better decay. But this has the side effect of 'smoothing' the sum to get asymptotics for sums that look like $\displaystyle \sum_{n \leq x} a(n) e^{-n/x}$.
For a while, I just accepted that sometimes we need to accept smoothed sums. But I began to wonder how hard it would be to recover the regular, un-smoothed sum. So I'm prompted to ask:

Suppose we can calculate $F(x) := \displaystyle \sum_{n \leq x} a(n) e^{-n/x}$ for any given $x > 1$. For a fixed $N$, can one recover $\displaystyle \sigma_N := \sum_{n \leq N} a(n)$ from $F(x)$? In particular, it is ever possible to recover $\sigma_N$ by calculating $F(x_i)$ for a suitably chosen, but finite set of $x_i \in \mathbb{R}$?

REPLY [8 votes]: In many applications, you can use the fact that
$$ e^{-n/x}=1 + O\Big(\frac{n}{x}\Big)$$
uniformly for $n\le x$ and therefore
$$ \sum_{n\le x} a_n e^{-n/x} =  \sum_{n\le x} a_n+ O\left( \frac{1}{x} \sum_{n\le x}n \ a_n\right). $$
This works well for sequences $a_n$ generated by $L$-functions near the $1$-line. For instance, if
$$  \sum_{n\le x} a_n \asymp (\log x)^k$$
for some $k$.<|endoftext|>
TITLE: More expanders?
QUESTION [5 upvotes]: Having received several exhausting answers to my recent question about
the expansion properties of a certain graph, I now wonder whether anything is
known on the following graphs of a similar nature:
1) The graph on ${\rm GF}(p)$ with $z$ adjacent to $-z$ and also to $gz$,
where $g$ is a fixed primitive root mod $p$.
2) The graph on ${\rm GF}(2^n)$ with $z$ adjacent to $z+e$ and also to $gz$,
where $e$ is a fixed non-zero element, and $g$ is a generating element of
${\rm GF}(2^n)$.
3) The graph on $({\mathbb Z}/2^n{\mathbb Z})^\times$ (odd residue classes
mod $2^n$) with $z$ adjacent to $z^{-1}$ and also to $z+2$.
Are these (families of) graphs known to be good expanders? Can one
investigate them using Selberg's 3/16-theorem or other "standard" tools
used to study the graph my original question concerned with?

REPLY [7 votes]: Freddie Manners is right: graphs (1) and (2) are not expanders
for any choice of $g$.
For (1), he already showed this by exhibiting large vertex sets with
$O(1)$ neighbors.  For (2) we prove it below by contructing vectors $v$
orthogonal to the all-$1$ vector for which the Rayleigh quotient
$\langle Av, v \rangle / \langle v, v \rangle$
is within $o(1)$ (indeed $O(1/n)$) of the graph degree,
thus proving that there is no spectral gap.
Graph (3) is probably an expander,
because the maps taking $z$ to $z^{-1}$ and $z+2$
generate a congruence subgroup of index 3 in ${\rm PSL}_2({\bf Z})$
(as Serre notes in the very last section of A Course in Arithmetic),
and the graph vertices can be identified with an orbit of cusps of a modular curve
of level $2^n$, so one should be able to use the $3/16$ bound on that curve;
but I'll leave that to the folks who actually know these techniques.
For (2): let $k$ be a finite field of $2^n$ elements, and consider the graph
where each $z$ is adjacent to $gz$ $-$ and thus also to $g^{-1} z$ $-$
and to $z+e$.
Fix a nontrivial homomorphism $\epsilon$ from $(k,+)$ to the group
$\lbrace\pm1\rbrace$; the usual choice is $\epsilon(x) = (-1)^{{\rm Tr}(x)}$.
Then for each $c \in k$ we have a homomorphism
$\epsilon_c: (k,+) \rightarrow \lbrace\pm1\rbrace$ defined by
$\epsilon_c(x) = \epsilon(cx)$,
and these $\epsilon_c$ form an orthonormal basis for the Euclidean space of
real-valued functions on $k$ with inner product
$$
\langle f, g \rangle := \frac1{2^n} \sum_{x\in k} f(x) g(x).
$$
So we're seeking a linear combination of the $\epsilon_c$ with $c \neq 0$
that is almost fixed by the adjacency matrix $A$ in the sense that
$\langle Av, v \rangle / \langle v, v \rangle = 3 - O(1/n)$.
Write $A = A_1+A_2$
where $A_1$ is induced by translation by $e$ 
and $A_2$ is induced by multiplication by $g^{\pm 1}$.
Then for each $c \in k$ we have $A_1 \epsilon_c = \epsilon(ce) \epsilon_c$
and $A_2 \epsilon_c = \epsilon_{gc} + \epsilon_{g^{-1}c}$.
We shall take $v = \sum_{i=1}^{n-1} \epsilon_{g^i c}$ for some $c\neq 0$,
so $\langle v, v \rangle = n-1$ and $\langle A_2 v, v \rangle = 2n-4$.
We next show that $c$ can be chosen so that $v$
is an eigenvector of $A_1$, corresponding to the eigenvalue $1$.
Because $\lbrace c : \epsilon(ce) = +1 \rbrace$ is a
$({\bf Z}/2{\bf Z})$-subspace of $k$ of codimension $1$,
the same is true of $\lbrace c : \epsilon(g^i ce) = +1 \rbrace$
for each $i$, so the intersection of these $n-1$ subspaces
has positive dimension.  Choosing nonzero $c$ in this intersection
makes $A_1 v = v$ as claimed.  Then
$$
\langle A v, v \rangle =
\langle A_1 v, v \rangle + \langle A_2 v, v \rangle =
(n-1) + (2n-4) = 3n-5 = (3-O(1/n)) \langle v, v \rangle,
$$
QED.
Exercise: Adapt this technique to show that the graph on
${\bf Z} / p {\bf Z}$ where each $z$ is connected to
$z \pm 1$ and $g^{\pm 1} z$
is not an expander either for any $g \in ({\bf Z} / p {\bf Z})^*$.
(That's what I first thought graph (1) was when I quickly read the question.)<|endoftext|>
TITLE: The minimal norm of a shifted stochastic matrix
QUESTION [5 upvotes]: Given a row-stochastic matrix $M$ with singular values $\sigma_{1} \geq \cdots \geq \sigma_{n}$, I am looking for an upper bound on the expression
$$\min_{\alpha} \left\| M- \frac{\alpha}{n}J_{n} \right\|_{2}$$ 
where $J_{n}$ is the matrix with all ones.
It is not hard to see that if $M$ is doubly stochastic, the above expression is exactly $\sigma_{2}$ (as the singular vector of the largest singular values is the vector of all ones), for $\alpha =1$. Can you find a similar bound when $M$ is only row stochastic?
Thank you.
Edit: Suppose we take $a, b$ to be the left and right singular vectors corresponding to the largest singular value $\sigma_{1}$. Then, 
$$\| M- \frac{\alpha}{n}J_{n}-\sigma_{1}ab^T+\sigma_{1}ab^T \|_{2} < \sigma_{2}+\sigma_{1}\left(\sqrt{1-\frac{\left< a,e\right> ^{2}\left< b,e\right> ^{2}}{N^{2}}}\right)$$ 
for 
$$\alpha = \frac{\sigma_{1}\left< a,e\right> ^{2}\left< b,e\right>^{2}}{N^{2}}$$
For a doubly stochastic matrix, this bound is tight (as the first singular vectors are $\frac{1}{\sqrt{N}}e$). What can we say, for example, of $\left< a,e\right> \left< b,e\right>$, when $\sigma_{1}$ is not 1, but very close to it?

REPLY [3 votes]: Some experimentation led me to the following results:

$\alpha=1$ yields the minimum value even for the operator-2 norm (certainly provable directly too)
$\|M-ee^T/n\|_2$ can be larger than $\sigma_2 + \cdots + \sigma_n$
$\|M-2ee^T/n\|_2 = \sigma_1(M)$ (obviously, since $M$ is rs)

So the only reasonable bound is: $\|M-ee^T/n\|_2 \le c_n\sigma_1(M)$, where the constant $c_n$ depends on the dimensionality $n$. As of now, I am using $c_n=1$, but have not paid much thought to what the best constant would be (need to dig up some singular value inequalities for that).
Here is an example matrix for which $\|M-ee^T/n\|_2 \ge 0.9\sigma_1(M)$.
\begin{equation*}
 M = \frac{1}{1000}\begin{pmatrix}
   10  &  24  &  966\\\\
   410 &  576 &   14\\\\
   529 &  362 &  109
 \end{pmatrix}.
\end{equation*}<|endoftext|>
TITLE: Good uses of Siegel zeros?
QUESTION [27 upvotes]: The short version of my question goes: What is known to follow from the existence of Siegel zeros?
A longer version to give an idea of what I have in mind: The "exceptional zeros" of course first cropped up in the work of Deuring and Heilbronn on the Gauss class number problem, the finitude of discriminants $D$ where $h(D) = 1$ (or $h(D) = c$ in general) is shown to follow from the failure of the RH (Deuring-Mordell) or, more generally, the failure of the GRH (Heilbronn). Since the finitude also follows (with good bounds) on the assumption of the GRH (Hecke-Landau) this settles the result unconditionally. (A more careful analysis by Heilbronn-Linfoot gives explicit bounds so that $h(D) = 1$ can happen at most 10 times.) Siegel later crystallised out the peculiar case of the exceptional real zero and proved the result in a very sharp form.
Linnik then used these ideas to show that $P(a,q) \ll q^L$, $P(a,q)$ being the least prime $\equiv a \pmod q$ and $L$ an explicit constant. Again this was known to follow on the GRH (with $L < 2 + \varepsilon$), hence the result is unconditional.
Iwaniec [Conversations on the Exceptional Character, p. 118 ] mentions on the other hand that Siegel zeros can be exploited to give even stronger results than what is available on the (presumably true) GRH; for example it follows that $L < 2$, hence improving on the bound known on the GRH. This is related to earlier work by Heath-Brown showing that the existence of exceptional zeros implies both that $L < 3$ (which is better than what is known unconditionally, although not better than the $L < 2 + \varepsilon$ that follows on the GRH) and the existence of infinitely many twin primes.
The results of Heath-Brown and Iwaniec, however, assumes an infinite number of exceptional zeros, to increasing modulus $q$, to derive results stronger than what is known unconditionally (or even on the GRH). What I would like to know is if similar results are proved in the literature working only from a single exceptional zero (or even from the weaker assumption of the mere existence of any zeros off the critical line, like Deuring and Heilbronn did)? Putting $\Theta =$ supremum of the real parts of the zeros, we have of course that $1/2 < \Theta$ (i.e. the failure of the RH) would mess up the approximate formulas for different prime counting functions (e.g. by a theorem of Schmidt we have $\Pi(x) - {\rm li} (x) = \Omega_\pm(x^{\Theta-\varepsilon})$, thus $= \Omega_\pm(x^{1/2})$ in case RH fails), but as long as the mere existence of any zero off the line is assumed (not some horrifying disaster like $\Theta = 1$), I know of no results stronger than the unconditionally known irregularites in the approximation (e.g. from Littlewood's theorem). The assumption of a Siegel zero of course is an assumption of such a special failure of the GRH on the other hand, so perhaps I'm missing some rather obvious examples?

REPLY [5 votes]: I just ran across this paper. I don't know if it is a "good" use or not. It is in the spirit of Heath-Brown I presume.
Lithuanian Mathematical Journal January 2010, Volume 50, Issue 1, pp 43-53
L. Germán, I. Kátai, On multiplicative functions on consecutive integers
http://link.springer.com/article/10.1007%2Fs10986-010-9070-8
From the abstract:
Concerning the Liouville function $\lambda$, we find an upper estimate for 
${1\over x}|\sum_{n\le x}\lambda(n)\lambda(n+1)|$ under the unproved hypothesis that $L(s,\chi)$ have Siegel zeros for an infinite sequence of L-functions.<|endoftext|>
TITLE: On Lipschitz embeddability of certain compact metric spaces into $\mathbb{R}^n$
QUESTION [7 upvotes]: Suppose that $K$ is a compact metric space which is homeomorphic to a subspace of $\mathbb{R}^n$. Does there exist $f:K\rightarrow \mathbb{R}^n$ which is one-to-one and Lipschitz?

REPLY [9 votes]: No. There is a length metric counter-example in dimension 3. See Theorem 1 in this paper.
Let me briefly explain the construction here. For a large $n$, consider a unit segment $[p_nq_n]\subset \mathbb R^3$ and let $U_n$ be its neighborhood of radius $2/n$. Inside $U_n$, consider a "chain" formed by $n$ small solid tori. Each solid torus is a $(1/10n)$-neighborhood of a circle of radius $1/n$, and each of these circles is linked to the next one. The first link in the chain is nearby $p_n$ and the last one is nearby $q_n$. In each of these solid tori introduce a conformal Riemannian metric as follows. At the middle circle, the conformal factor is very small, so that the length of this circle equals $1/n^2$. And the conformal factor equals 100 almost everywhere between the middle circle and the boundary of the torus. So despite the fact that the middle circle is very short, the distances between points outside the torus are no shorter than the Euclidean distances. It follows that the distance between $p_n$ and $q_n$ in the resulting Riemannian metric is at least 1.
Repeat this constriction for $n=10,100,1000,\dots$, choosing the segments $[p_nq_n]$ so that they converge to some $[pq]$ (but the neighborhoods $U_n$ are disjoint). This yields a length metric on $\mathbb R^3$ which defines the standard topology (i.e., the identity is a homeomorphism). Restrict it to a compact ball containing all the segments of the construction.
There is no Lipschitz homeomorphism from this ball with this metric to $\mathbb R^3$. Indeed, suppose that $f$ is a 1-Lipschitz homeomorphism. Consider the image of the neighborhood of $[p_nq_n]$. The images of the chained circles have lengths (and hence diameters) at most $1/n^2$, Since they are linked, each next circle is within distance $1/n^2$ from the previous one. So the endpoints of the chain are mapped to points at distance at most $2/n$ or so from each other: $|f(p_n)-f(q_n)|\le 2/n$. It follows that $f(p)=f(q)$, a contradiction.<|endoftext|>
TITLE: Possible to find a set of log-concave functions with log-concave sums?
QUESTION [5 upvotes]: While the set of log-convex functions is closed under addition, the set of log-concave functions is not. Yet if $f$ is log-concave, $\ln(k f) = \ln(k)+\ln(f)$, with $k \in \mathbb{R}^+$ constant, is concave. This suggests one can find a set of log-concave functions whose sums are still log-concave (possibly closed under addition). 

I am interested in the concavity/convexity of $\ln(b+s)$, where $b,s$ are functions on $\mathbb{R}^+$, increasing, positive, convex and log-concave. 

Example: I found $\ln(x^2+x^\beta)''<0, \forall x>0$ for $\beta<6$  (numerical result, perhaps not exact). 

Are there any results that could help delineate this set of functions? Perhaps for polynomials?

Some suggestions and more detailed description here
The initial motivation for this problem comes from population dynamics.

REPLY [5 votes]: In fact $\ln(x^2 + x^\beta)$ is concave for $x > 0$ iff $3-2\sqrt{2} \le \beta \le 3 + 2 \sqrt{2}$. 
This comes from writing
$$ \dfrac{d^2}{dx^2} \ln(x^2 + x^\beta) = \dfrac{x^2}{(x^2 + x^\beta)^2} \left(-\beta x^{2\beta - 4} + (\beta^2 - 5 \beta + 2) x^{\beta - 2} -2\right)$$ 
and noting that the discriminant of $-\beta t^2 + (\beta^2 - 5 \beta + 2) t - 2$ with respect to $t$ is $(\beta^2 - 6 \beta + 1) (\beta - 2)^2$.
EDIT: Somewhat more generally, $\ln(x^\alpha + x^\beta)$ is concave for $x > 0$ iff
$\alpha=\beta$ or  $(\alpha-\beta)^2 - 2 (\alpha + \beta) + 1 \le 0$.<|endoftext|>
TITLE: Forcing Diamond
QUESTION [11 upvotes]: It is well known that adding a subset of a regular cardinal $\kappa$ with partial functions of size $< \kappa$ forces $\Diamond_\kappa$.  One can also see that if $S \in V$ is a stationary subset of $\kappa$, then $Add(\kappa,1)$ also forces $\Diamond(S)$.  Question: If $G \subseteq Add(\kappa,1)$ is generic, do we get $\Diamond(S)$ for all stationary $S \in V[G]$, $S \subseteq \kappa$?

REPLY [13 votes]: The answer is yes, and the construction is essentially the same as for $\diamondsuit$.    Let $\dot S$ be a name for the stationary set, and define a diamond sequence $\langle A_\alpha\mid\alpha\in S\rangle$ as follows:  Let $f$ be the $Add(\kappa,1)$ function.   Suppose $\alpha\in S$.  Then let $\gamma$ be least such that $f\upharpoonright\gamma\Vdash\alpha\in\dot S$ and let $A_\alpha=\{\nu<\alpha\mid f(\gamma+\nu)=1\}$.
In answer to the comment, let $\dot C$ be a name for a club set $C$ and $\dot x$ the name for a subset of $\kappa$.  Let $C'$ be the set of $\nu\in C$ such that $f\upharpoonright\nu$ decides the value of $\dot x\cap\nu$ and forces $\nu$ is a limit point of $\dot C$ (and hence is in $\dot C$).   Then it is forced that $\dot C'$ is club in $V[\dot f]$, so there is a dense set of $q$ such that, setting $\eta=\operatorname{domain}(q)$, $q\Vdash\eta\in\dot C'$ and $q$ does not force $\eta\notin\dot S$.   Then let $q'\le q$ be of minimal length forcing $\eta\in\dot S$, and let $q''$ be $q'$ with the known value of $\dot x\cap \eta$ tacked onto the end.  Then $q''$ forces that $\dot A_\eta= \dot x\cap\eta$.<|endoftext|>
TITLE: Generalizations and limitations of Quillen's F-isomorphism theorem
QUESTION [8 upvotes]: Quillen proved in 1971 ("The Spectrum of an Equivariant Cohomology Ring: I,II")  for a large class of groups $G$ including 

compact Lie groups 
groups of finite virtual cohomological dimension 
compact topological groups with a finite number of conjugacy classes of elementary abelian $p$-subgroups

that the map 
$$\text{res}: H^\ast(BG,\mathbb{F}_p) \to \varprojlim_E H^\ast(BE,\mathbb{F}_p)$$
where $E$ runs through the elementary abelian $p$-subgroups of $G$ (ordered by inclusion and conjugacy) is an F-isomorphism (i.e. it has finite kernel and for each $x$ in the RHS, a $p$-power of $x$ is in the image). 
Question 1: Has this theorem been generalized to other classes of groups in the meantime ? 
Conversely, I'm also interested in counterexamples to the theorem, i.e. 
Question 2: What is an example of a topological group $G$ s.t. the map above is no F-isomorphism ?

REPLY [2 votes]: This might be a rather different direction than you're interested in, but analogues of Quillen's theorem have been studied for graded cocommutative Hopf algebras.  These generalizations were originally motivated by computations in stable homotopy theory that used finite subalgebras of the Steenrod algebra.  Wilkerson ("The cohomology algebras of finite dimensional Hopf algebras") proved that an analogue of Quillen's theorem holds for all finite subalgebras of the mod 2 Steenrod algebra, but gave counterexamples for more general connected graded cocommutative Hopf algebras, including subalgebras of the mod $p$ Steenrod algebra for odd $p$.  Palmieri ("A note on the cohomology of finite dimensional Hopf algebras") gave a generalization of the theorem that holds for any finite-dimensional connected graded cocommutative Hopf algebra, but Palmieri uses a very general class of algebras in the place of elementary abelian subgroups, to the point that the theorem essentially becomes purely formal.<|endoftext|>
TITLE: Six operations for (quasi)-coherent sheaves
QUESTION [9 upvotes]: Can someone point me to a reference with an overview of what Grothendieck's six operations formalism looks like for schemes and (quasi)-coherent sheaves (or derived category objects with (quasi)-coherent cohomology sheaves)?  Do I have to read Residues and Duality? I'm particularly curious about what the two shriek functors look like.  Are there distinguished triangles associated to a closed immersion and its open complement?  What kind of theorems about commutation of pushforwards with pullbacks are true?

REPLY [3 votes]: Well, you could read SGA.  But my two favorite sources for this material are here for an abstract treatment that doesn't (as far as I remember) talk specifically about quasi-coherent sheaves, and here for a considerably longer but readable treatement that does.
(More precisely, the first reference has a section that basically takes various properties of quasi-coherent sheaves as axioms and proceeds from there.  If you're willing to accept these axioms without working through all the geometry, that's probably the reference you're looking for.)<|endoftext|>
TITLE: On Determinants of Laplacians on Riemann Surfaces
QUESTION [27 upvotes]: History of the Formula: In their famous paper "On Determinants of Laplacians on Riemann Surfaces" (1986), D'Hoker and Phong computed the determinant of the Laplacian $\Delta_n^+$ on the space $T^n$ of spinor/tensor fields on the compact Riemann surface $M$ (for simplicity in this question I'm assuming $n\geq0$). Their result reads:
$$\det(\Delta_n^+)=\mathcal{Z}_{n-[n]}(n+1)\cdot e^{-c_n\mathcal{X}(M)}.$$
Where $\mathcal{Z}_{n-[n]}(s)$ are two Selberg zeta functions, and $c_n$ is a constant which is explicitly computed in the article above. 
In their exposition there are for sure some trivial mistakes. One comes from an error in the reference paper "Fourier coefficients of the resolvent for a Fuchsian group" (1977) by Fay, and it has been considered in the review paper "Geometry of String Perturbation Theory" (1988) of D'Hoker and Phong. But it doesn't lead to any change in the above considered formula. 
Another trivial mistake is a misuse of the volume formula for an hyperbolic Riemann surface, this has been first pointed out by Bolte and Steiner, in their (unpublished) paper: "Determinants of Laplace-like Operators on Riemann Surfaces" (1988). In this article the authors use a new way to compute the same quantity $\det(\Delta_n^+)$, but they arrive to a different result. Calling $\frac{1}{2}\Delta_n^+$ what D'Hoker and Phong called $\Delta_n^+$ (according to Bolte and Steiner), it reads:
$$\det(\frac{1}{2}\Delta_n^+)=\mathcal{Z}_{n-[n]}(n+1)\cdot e^{-k_n\mathcal{X}(M)}.$$
Where the constant $k_n$ has been explicitly computed, and it turns out to be different from $c_n$.
Unfortunately the mistakes pointed out above are not enough to account for the difference between $c_n$ and $k_n$. Bolte and Steiner suspect that the error in D'Hoker and Phong could come from a missing factor $2$ in the definition of $\Delta_n^+$.
The end of the story seems to be the article "Notes on determinants of Laplace-type operators on Riemann surfaces" (1990) of Oshima. In this article the author recalls the correction of some mistakes in the original paper of D'Hoker and Phong, moreover he points out a conceptual mistake made by Bolte and Steiner. Finally, simply assembling previous results, he provides the following formula:
$$\det(\frac{1}{2}\Delta_n^+)=\mathcal{Z}_{n-[n]}(n+1)\cdot e^{-l_n\mathcal{X}(M)}.$$
Again $l_n$ is given explicitly, and it turns out to be different from both $c_n$ and $k_n$. 
Question: Is it really the end of the story? It seems to me that the whole situations is quite messy, for example no one seems to point out new specific mistakes in the original article of D'Hoker and Phong (there must be some, according to Oshima). In specific, do the experts agree on the validity of the formula provided by Oshima? Are there other successive articles on the same topic?
Note: I added the tags "Arithmetic Geometry" and "Analytic Number Theory" because the laplacians $\Delta_n^+$ are conjugated to the Maass Laplacians $D_n$ acting on automorphic forms of weight $n$ on $M$. So I suspect an answer could come from people in arithmetic as well.
Thank you very much for reading all this!

REPLY [2 votes]: A fairly complete treatment is given by Christian Grosche:
Grosche, C., Path integrals, hyperbolic spaces and Selberg trace formulae, Singapore: World Scientific. xi, 280 p. (1996). ZBL0883.58003.
(the citation tool gives a reference to the first edition, the second edition came out in 2013. See, in particular, Chapter 14 in the Second Edition (I don't have the first).<|endoftext|>
TITLE: Existence of non-split vector bundles on smooth projective varieties
QUESTION [17 upvotes]: Question. Is it known/easy to see that every smooth projective variety $X$ (over an algebraically closed field), except for the point and $\mathbb{P}^1$, has a vector bundle which is not a direct sum of line bundles? 

I have a result which trivially shows the above fact in positive characteristic, but I don't know whether I should state it as a corollary as it might be well-known or obvious for reasons I don't see.

REPLY [17 votes]: Yes, it is true that (over an algebraically closed field)  the only positive-dimensional smooth projective variety on which every algebraic vector bundle splits as a sum of line bundles is $\mathbb P^1$.
More generally Ballico has proved that if every algebraic vector bundle splits as a sum of line bundles on a reduced but maybe reducible positive-dimensional  projective variety , then that variety is a chain of $\mathbb P^1$'s.
For holomorphically inclined readers, here is a related result in complex geometry.<|endoftext|>
TITLE: A question on normal closures of elements in free groups.
QUESTION [11 upvotes]: Let $F$ be a free group of finite rank, and $p, b \in F$, where $b$ is a root element (i.e. not a proper power). 
I have a case where $p^{n_k} = V_{n_k}^{-1}b^{-1} V_{n_k} \cdot U_{n_k}^{-1}b U_{n_k}$, for some $n_k \in \mathbb{Z}$ ...i.e. some powers of $p$ are products of two conjugates of $b$ and $b^{-1}$. 
What can be said about $p$ and $b$ ? 
Some immediate implications:
By Karras-Magnus-Solitair, since $b$ is root, one-relator group $ < F \ | \ b >$ is torsion-free, so if a proper power of $p$ is in $ncl(b)$ then $p\in ncl(b)$. 
Also, by going to an abelianization of $F$, it is clear that $p\in \[ F, F \]$. 
I was hoping that $p$ is conjugate to $b$...by Magnus, if we can show that $b$ is also in the the normal closure of $p$ then that would be the case. 
I also have a somewhat related general question. Say, if we have $b$ as normal root of $p$ i.e. $p \in ncl(b)$, so that $p = \displaystyle{\prod_{i=1}^{n}} T_i^{-1} b^{\epsilon _i} T_i$, where $\epsilon _i = \pm 1$. Clearly the above product is not unique. 
Is there a notion of associating to $p$ a minimal integer $n_p \geq 1$  so that $p$ can be decomposed to the product $p = \displaystyle{\prod_{i=1}^{n_p}} T_i^{-1} b^{\epsilon _i} T_i$?
For example if $n_p=1$ then $p$ and $b$ are conjugate.  
Apologies if I missed something obvious or the question(s) doesn't make sense. 
Thanks!

REPLY [11 votes]: This is just to flesh out the details of my comment above.  I think we can show that $n_k=1$ or $p=1$.
After conjugating (and simplifying notation slightly), your equation easily becomes
$p^n=[w,b]$
(for $w=vu^{-1}$). The stable commutator length of $p$ is equal to the infimum of $cl(p^n)/n$ over all $n>0$, so in your case we have
$\mathrm{scl}(p)\leq 1/n$.
But a theorem of Duncan--Howie asserts that scl in a free group is always at least 1/2, so $n\leq 2$.  (This is all explained in Danny Calegari's book scl.)  Suppose therefore than $n=2$.
In this case, your equation can be realized by a map of the fundamental group of the surface $\Sigma$ of Euler characteristic -1, which has presentation $\langle p,w,b|p^2=[w,b]\rangle$.
A theorem of Lyndon asserts that every such map factors through $H_1(\Sigma)$ (though Lyndon didn't say it like that, of course).  Because $p$ is torsion in $H_1(\Sigma)$ and free groups are torsion-free, it follows that $p=1$.
I'll add some proper references when I have more time.<|endoftext|>
TITLE: How much of a variety can be reconstructed from codimension-zero data?
QUESTION [14 upvotes]: This is an improved version of this question. Maybe it should be an edit -- I'm not sure what the MO convention is.
I'm curious, more or less, how much information one can get out of the derived category of coherent sheaves on a variety which are trivialized outside a finite set. Specifically, suppose $X$ is a smooth, proper variety of finite type over $\mathbb{C}$. Consider the category $\mathcal{C}=\text{Coh}(X)_{\text{codim }0}$ of coherent sheaves which are trivial (in the sense of being isomorphic to a power of the constant sheaf) outside a finite set of $\mathbb{C}$-points of $X$. This is a symmetric monoidal category with inner Homs. Suppose we do not know $X$, but know the category $\mathcal{C}$, together with derived data (the derived category of complexes whose cohomology is in $\mathcal{C}$, together with derived tensor products, Homs and most importantly global sections). Suppose in addition we are given the set of points of $X$ together with their formal neighborhoods, and know for every sheaf in $\mathcal{C}$ how it can be patched out of a globally trivial sheaf together with patching data at some finite set of formal neighborhoods. 
How much information about $X$ does the category $\mathcal{C}$ contain? In particular can $X$ be recovered from $\mathcal{C}$?

REPLY [7 votes]: Let $F$ be a sheaf which is trivial on $X - S$. This means that $j^*F = O_{X-S}^{\oplus n}$ for some $n$, where $j$ is the embedding of $X - S$ into $X$. Note that by Hartogs theorem one has $j_*O_{X-S} = O_X$, so by adjunction one has a morphism $F \to j_*j^*F = O_X^{\oplus n}$ which is an isomorphism on $X - S$. So, this means that such $F$ is a modification of the trivial bundle in the finite number of points. In other words, $Coh(X)_{codim 0}$ is the extenion closed abelian subcategory of $Coh(X)$ generated by $O_X$ and all structure sheaves of points.
EDIT. Let us see what the information we have in this category. First, we have the cohomology of $O_X$. Second, we have a bunch of structure sheaves of points, and this part of the category depends only on the dimension of the variety. Finally, we have multiplication maps
$$
Ext^\bullet(O_X,O_S)\otimes Ext^\bullet(O_S,O_X) \to Ext^\bullet(O_X,O_X)
$$
and
$$
Ext^\bullet(O_X,O_S)\otimes Ext^\bullet(O_S,O_X) \to Ext^\bullet(O_S,O_S).
$$
Since $S$ is $0$ dimensional the first factor lives in degree $0$ and the second in degree $n = \dim X$, so the only thing we have are the maps 
$$
Hom(O_X,O_S)\otimes Ext^n(O_S,O_X) \to Ext^n(O_X,O_X)
$$
and
$$
Hom(O_X,O_S)\otimes Ext^n(O_S,O_X) \to Ext^n(O_S,O_S).
$$
Now let $S= x$ be a point. By Serre duality these maps correspond to the maps
$$
Hom(O_X,O_x)\otimes Hom(O_X,\omega_X) \to Ext^n(O_X,\omega_X\otimes O_x)
$$
and
$$
Hom(O_X,O_x)\otimes Hom(O_x,O_x) \to Hom(O_X,O_x).
$$
So, the second map contains no information, while the first gives the canonical morphism 
$$
X \to P(H^0(\omega_X)^*) = P(H^n(O_X)).
$$
So, on one hand, if $H^n(O_X) = 0$ you cannot reconstruct anything. On the other hand, if the canonical class is very ample, by reconstrcting the anticanonical image you reconstruct all $X$. I am not quite sure what goes on in the intermediate cases.<|endoftext|>
TITLE: Bounding the second derivative of the log-determinant
QUESTION [6 upvotes]: I'm trying to use the log-determinant to regularize an optimization problem. To make the argument work, I need to bound the second derivative of the log-determinant.
I need to prove that $\text{Tr}\left( \left(A^{-1} B\right)^2\right) \geq 1$ whenever:

$A$ is positive semidefinite
$B$ is symmetric, has zeroes on the diagonal, and has at least one entry which is $\geq 1$.
The diagonal entries of $A$ are 1.

From messing around a bit it seems like in fact there are always unit vectors $v, w$ such that $v^T A^{-1} B w \geq 1$---for example, this is true if B has only a single large entry, or if A and B commute. That would imply the conclusion and seems like it should be easy enough to confirm or deny, but I'm stumped. 
Any thoughts?

REPLY [5 votes]: Here is an argument based on convex optimization.
Consider the case where $B$ has a one on the diagonal and the rest of the matrix is arbitrary. Without loss of generality, we can assume that $b_{11}=1$. Thus, we can write the general matrix $B=E+C$, where $E=e_1e_1^T$ with $e_1=(1,0,\ldots,0)$ being the first canonical basis vector, and $C$ is a symmetric matrix with $C_{11}=0$.
(Note: If the entry $\ge 1$ is an off-diagonal, then we can write $B=e_ie_j^T+e_je_i^T$, and run an argument similar to the one below---this one is somewhat more tedious so I did not work it out).
To save on typing, first define the notation:
$\DeclareMathOperator{\vect}{vec}$
\begin{equation*}
  Z = A^{-1},\quad M = Z \otimes Z,\quad e=\vect(E)\quad c=\vect(C).
\end{equation*}
Then, the trace under question is:
\begin{equation*}
  \mbox{trace}(Z(E+C)Z(E+C)) = e^TMe + c^TMc + 2c^TMe.
\end{equation*}
Actually, we have $(Kc)^TMc$ as the second term, where $K$ is the commutation matrix, but after some simplifications, it will turn out that we can drop $K$.
Now, we minimize $c^TMc + 2c^TMe$ subject to $c^Te=0$ (and the constraint that $Kc=c$ to ensure symmetry, but this constraint can be eliminated, so I've dropped it). 
Introduce the Lagrange multiplier $\nu$ corresponding to the constraint $c^Te=0$, the 
first-order optimality conditions for this convex optimization problem are:
\begin{equation*}
  2Mc + 2Me - \nu e = 0,\qquad c^Te=0.
\end{equation*}
Thus, we have
\begin{eqnarray*}
  Mc &=& (\nu/2) e - Me\\\\
  c  &=& (\nu/2) M^{-1}e - e\\\\
  &\implies& c^Te = (\nu/2) e^TM^{-1}e - e^Te\\\\
  &\implies& \nu=2\qquad\text{since}\ e^TM^{-1}e=1.
\end{eqnarray*}
Thus, in particular, an optimum $c$ must satisfy
\begin{equation*}
  c = M^{-1}e - e,
\end{equation*}
from which it follows that $e^TMc =  1 - e^TMe$ and $c^TMc=0-c^TMe$. Thus, the objective is
\begin{equation*}
  c^TMc + 2c^TMe= c^TMe = 1 - e^TMe,
\end{equation*}
from it immediately follows that
\begin{equation*}
  \mbox{trace}(Z(E+C)Z(E+C)) = e^TMe + c^TMc + 2c^TMe \ge 1.
\end{equation*}<|endoftext|>
TITLE: Change of variables formula for Riemann integration and Lebesgue Integration
QUESTION [6 upvotes]: I've put this question on math.SE for a while without getting any answers. I thought it must be a rather trivial question for MO so that I didn't put it here. But I do want to get some help anyway (before being closed for whatever reason). 
In the setting of Riemann integration, we have the following change of variables formula:

Let $[a,b]$ be a closed interval, and let $\phi:[a,b]\to[\phi(a),\phi(b)]$ be a continuous monotone increasing function. Let $f:[\phi(a),\phi(b)]\to{\Bbb R}$ be a Riemann integrable function on $[\phi(a),\phi(b)]$. The $f\circ\phi:[a,b]\to{\Bbb R}$ is Riemann-Stieltjes integrable with respect to $\phi$ on $[a,b]$ and 
  $$
\int_{[a,b]}f\circ\phi\ d\phi=\int_{[\phi(a),\phi(b)]}f. \tag{1}
$$

In the setting of Lebesgue integration, we have the following: 

Let $(X,{\mathcal B},\mu)$ be a measure space, and let $\phi:X\to Y$ be a measurable morphism from $(X,{\mathcal B})$ to another measurable space $(Y,{\mathcal C})$. Define the pushforward $\phi_*\mu:{\mathcal C}\to[0,+\infty]$ of $\mu$ by $\phi$ by the formula
  $\phi_*(E):=\mu(\phi^{-1}(E))$. If $f:Y\to[0,+\infty]$ is measurable, then
  $$
\int_Y f\ d\phi_*\mu=\int_X(f\circ\phi)\ d\mu. \tag{2}
$$

My question is: how can I interpret (1) in terms of (2)?

A naive attempt is letting $X=[a,b]$ and $\phi$ be the measurable morphism from $[a,b]$ to $[\phi(a),\phi(b)]$ in (1). But the result doesn't match (2) at all.

REPLY [7 votes]: If $\phi:X\to Y$ is a $C^1$ diffeomorphism between open subsets of euclidean space, then the change of variables formula reads
$$\int_Y f(y)\, d\lambda(y)=\int_X (f\circ\phi)(x)|\det\phi'(x)|\,d\lambda(x).$$
This means that the Lebesgue measure is the push-forward, $\lambda=\phi_* \mu$, of the absolutely continuous measure $\mu=|\det\phi'|\,\lambda$.<|endoftext|>
TITLE: Does the poset of free factors of a free group form a lattice?
QUESTION [5 upvotes]: Let $F_n$ denote a free group of rank $n$. The set of its free factors is partially ordered by inclusion. Recall that a psoet is called a lattice if any two elements have a smallest upper bound and a greatest lower bound. 
Is this true for this poset?

REPLY [8 votes]: ORIGINAL ANSWER, ADDRESSING A SLIGHTLY DIFFERENT QUESTION: There is a closely related poset for which greatest lower bounds and least upper bounds indeed exist. Instead of an individual free factor $A$, first consider its conjugacy class $[A]$. Then, instead of individual conjugacy classes of free factors $[A]$, consider a "free factor system" (in the language of Bestvina-Feighn-Handel): a finite set $\mathcal{F} = \{[A_1],\ldots,[A_k]\}$ such that there exists a free factorization of the form $F_n = A_1 * \cdots * A_k * B$ where $B$ may or may not be trivial but all the $A_i$'s are nontrivial.
The partial ordering $\mathcal{F}\sqsubset \mathcal{F}'$ is defined by requiring that for each $[A] \in \mathcal{F}$ there exists $[A'] \in \mathcal{F}'$ such that $A$ is conjugate to a subgroup of $A'$. 
The Kurosh Subgroup Theorem can be translated into the statement that this poset has greatest lower bounds. The greatest lower bound of $\mathcal{F}$ and $\mathcal{F}'$ is 
$$\mathcal{F} "meet" \mathcal{F}' = \{[A \cap A'] \, | \, [A] \in \mathcal{F}, [A'] \in \mathcal{F}', A \cap A' \ne 1\}
$$
(I don't know how to get the "meet" operator in this version of TeX).
CORRECTION: You have to allow the "trivial" free factor system in order for this meet to be well-defined, because it is possible that the definition above produces the emptyset. In which case I should have left out the condition "$A \cap A' \ne 1$" in my definition of the meet.
The meet operator can then be extended to an operator on arbitrary sets of free factor systems, and using this one gets least upper bounds too: the least upper bound of $\mathcal{F}$ and $\mathcal{F}'$ is the meet of all free factor systems $\mathcal{F}''$ (including the improper free factor system $\{[F]\}$) such that $\mathcal{F} \sqsubset \mathcal{F}''$ and $\mathcal{F}' \sqsubset \mathcal{F}''$. This $\mathcal{F}''$ is called the ``free factor support'' of $\mathcal{F}$ and $\mathcal{F}'$.
ADDITION, ADDRESSING THE ORIGINAL QUESTION: Now that I've had a chance to think more about the Kurosh subgroup theorem myself, I realize that the answer to the original question is just "yes".
First, for greatest lower bound: the intersection of any collection of free factors of the finite rank free group $F_n$ is a free factor. For two free factors $A,B$ this is a consequence of the Kurosh subgroup theorem which says that the set of nontrivial intersections of $A$ with conjugates of $B$ is a finite set $\{C_1,\ldots,C_K\}$ with the property that there exists a free factorization $A = C_1 * \cdots * C_k * D$ where $D$ may be trivial; so $A \cap B$, if nonempty, is one of the $C$'s, and is therefore a free factor of $A$, and a free factor of a free factor of $F_n$ is a free factor of $F_n$. In general, for any collection of free factors, write them in a sequence, intersect each initial segment of the sequence, and use the fact that in a finite rank free group, a free factor strictly included in another has smaller rank (proved by abelianizing).
Then, for least upper bound: given any collection $\mathcal A$ of free factors, the intersection of all free factors (including the improper free factor $F_n$) containing each element of $\mathcal A$ is the least a free factor containing each element of $\mathcal A$.<|endoftext|>
TITLE: A question of compactness in the geometry of numbers
QUESTION [5 upvotes]: Given a star body  $S \subset \mathbb{R}^n$ with the origin as interior point, the critical determinant of $S$---usually denoted as $\Delta(S)$---is the infimum of the determinants of all lattices that intersect $S$ only
at the origin. 
The quantity ${\rm vol}(S)/\Delta(S)$ is a linear invariant of compact, star bodies. If we restrict its domain to the space $\mathcal{K}_0^n$ of convex bodies in $\mathbb{R}^n$ that contain the origin in their interior, then it is a continuous linear invariant. 
The question: Is there a nice little proof or a reference for the statement that the the sublevel sets
$$
\{ K \in \mathcal{K}_0^n : {\rm vol}(K)/\Delta(K) \leq C \}
$$
are compact and hence the functional must attain a global minimum?
I think this is true (because of  Macbeath's compactness theorem and the fact that $\Delta(K)^{-1}$ blows up if the origin is near the boundary of $K$), but have not really checked thoroughly thinking that is must be widely known although I cannot find the explicit statement in Cassel or Lekkerkerker.
Addendum. Sergei has given an excellent reason why this is not to be found in Cassel or Lekkerkerker ... However, much work has gone into giving lower bounds of the functional 
$K \mapsto {\rm vol}(K)/\Delta(K)$. For example, the Minkowski-Hlawka theorem says it is bounded below by $1$ (or $\zeta(n)$ to be more accurate). Is it known whether this functional attains a (global) minimum on $\mathcal{K}_0^n$ ?

REPLY [6 votes]: It is not compact unless you allow the origin at the boundary (or maybe impose some kind of uniform strict convexity).
Consider a rectangle $K=[-1,1]\times[-\delta,1]$ in the plane, where $\delta$ is positive and goes to 0. If $K$ intersects some lattice only at the origin then so does $-K$, by symmetry. But $K\cup -K$ is a convex symmetric set of area 4, hence $\Delta(K)=\Delta(K\cup -K) \ge 1$ by Minkowski's theorem.
Therefore the ratio $vol/\Delta$ is bounded for all such $K$, but they have no limit in $\mathcal K^2_0$ as $\delta\to 0$.<|endoftext|>
TITLE: A set of positive measure with cardinality less than that of the continuum?
QUESTION [8 upvotes]: Is it consistent with ZFC that there is a subset of $[0,1]$ whose cardinality is less than that of the continuum but which has positive Lebesgue measure?
Obviously not given CH.  And, given ZFC, there is such a subset iff there is a subset of full measure that has cardinality less than that of the continuum.  Moreover, I think it follows from the consistency of ZFC with the non-existence of Sierpinski subsets of $[0,1]^2$ that it is consistent with ZFC that there is a subset of $[0,1]$ whose cardinality is less than that of the continuum and which has positive outer Lebesgue measure.

REPLY [16 votes]: No. It is a famous exercise that if $X\subset\mathbf{R}$ has positive measure then $X-X$ contains an interval. It follows that $X$ has cardinality continuum.<|endoftext|>
TITLE: Passage from the moduli functor to the functor of points of the coarse moduli space
QUESTION [6 upvotes]: Let $F: (Sch)^{o}\to (Set)$ be a functor that admits a coarse moduli $Y$ (a scheme). We can consider $Y$ as a representable functor $h_{Y}: (Sch)^{o}\to (Set)$. Is there a direct way to produce $h_Y$ from $F$ (maybe by taking sheafification of $F$ in some topology or something like that)?
The same question for a Deligne-Mumford stack $F: (Sch)^{o}\to (Groupoids)$ and its coarse moduli space (a scheme or an algebraic space). Is it produced as taking $\pi_0$ of $F$ and then sheafification or something like that?

REPLY [9 votes]: It is not true that one can obtain the coarse moduli space of a functor or stack $F$ by sheafifying $F$ or $\pi_0(F)$, so probably the answer to both of your questions is NO.
I'll take the following definition of a coarse moduli space of a functor or stack $F: (Sch)^{op} \rightarrow \mathrm{Gpoid}$: It's an algebraic space $X$ together with a natural transformation $F \rightarrow X$ such that

$X$ is universal for maps from $F$ to algebraic spaces (a bit stronger than being universal to schemes), and
For any algebraically closed field $k$, the natural map $F(k) \rightarrow X(k)$ is a bijection.   

For this definition (which I think is the standard one), the coarse moduli space is typically NOT obtained by sheafifying $\pi_0(F)$.  For example, take $F$ to be the Deligne-Mumford stack $[\mathbb{A}^1/(\mathbb{Z}/2)]$, for $\mathbb{A}^1 = \mathrm{Spec \ }\mathbb{C}[x]$ and $\mathbb{Z}/2$ acting in the obvious way by $x \mapsto -x$.  Then the coarse moduli space of $F$ is given by $X = \mathbb{A}^1$, and the map $F \rightarrow X$ is induced by the map $x \mapsto x^2$ from $\mathbb{A}^1$ to $\mathbb{A}^1$.  But $\mathbb{A}^1$ is not isomorphic to the sheafification of $\pi_0(F)$ for the etale topology (and probably not for any subcanonical topology, although I'm not sure how to prove this at the moment).
To see why, let $Y$ be the etale sheafification of $\pi_0(F)$, and let $T_0 F$, $T_0 X$, and $T_0 Y$ be the tangent spaces to these functors at 0 (by which I mean the set of lifts of $0 \in F(\mathbb{C})$ to $F(\mathbb{C}[\epsilon])$, where $\epsilon^2 = 0$).  If $X$ were given by the etale sheafification of $\pi_0(F)$, there would be a natural map $Y \rightarrow X$ (induced from the projection $F \rightarrow X$) which would be an isomorphism on tangent spaces.  But one sees that the map $T_0 F \rightarrow T_0 X$ is the zero map since this is true for the map $x \mapsto x^2: \mathbb{A}^1 \rightarrow \mathbb{A}^1$.  On the other hand, the map $T_0 X \rightarrow T_0 Y$ is non-zero, because $\mathbb{C}[\epsilon]$ has no non-trivial etale covers and so $T_0 Y = T_0(\pi_0(F))$, which is isomorphic to $\mathbb{C}/(\mathbb{Z}/2)$ with the projection map $T_0 F \rightarrow T_0 Y$ identified with the quotient map $\mathbb{C} \rightarrow \mathbb{C}/(\mathbb{Z}/2)$.  
If you look at the proof of the existence of coarse moduli spaces, you'll see that it's done by first considering the case of quotienting an affine scheme by a finite flat groupoid, where the coarse moduli space is given by Spec of the invariants.  One then carefully glues together the coarse moduli spaces on open patches.  This construction makes me suspect that there is no direct (i.e., global) construction of the coarse moduli space, at least not one that is easy to work with.<|endoftext|>
TITLE: Geometric description of the Deligne-Mumford stacks
QUESTION [6 upvotes]: It is well known that a one-dimensional smooth Deligne-Mumford stack (over $\mathbb{C}$) could be described as a collection of its "stacky" points (finitely many) on its coarse moduli space with the orders of their stabilizers. As wccanard noted in the comments, here we need to assume that all but finitely many points of the stack have the trivial automorphisms group.

Is there an analogous description for two-dimensional DM stacks? For $\dim > 2$ DM stacks? Maybe for stacks with some additional restrictions (for example, assume that its coarse moduli is a smooth scheme)? By analogous description I mean a collection of closed subschemes on its coarse moduli space with the orders of their stabilizers. Is there an example of two non-isomorphic DM-stacks for which such descriptions coincide?
Is there some geometric description for one-dimensional DM 2-stacks (n-stacks)?

REPLY [3 votes]: For question 2, I've never seen a definition of DM 2-stack, so I don't know where to start.  For question 1, I have a tentative counterexample even for the case where both the stack and the coarse moduli space are smooth.
The étale local picture near a stacky point is that we take an affine space, quotient by a finite group $G$, and take the coarse moduli space to get an affine space.  By the Chevalley-Shepard-Todd theorem, if you want your coarse moduli space to be smooth, it is necessary and sufficient that $G$ be a complex reflection group, acting by complex reflections.  In order to find a pair of suitable non-isomorphic stacks for which labelings of the coarse space by orders of groups does not distinguish them, we need to find two complex reflection groups satisfying the following conditions:

They have the same order.
The orders of stabilizers of affine subspaces of a given dimension form identical sets of positive integers.
There is an isomorphism of the coarse moduli spaces taking the images of suitably labeled subspaces to each other.

The relevant information is in the big table at the Wikipedia page on complex reflection groups.
I will confine myself to rank 2 groups, because then the nontrivial subspace stabilizers are precisely the reflections.  For some reason, I'm unable to think clearly right now (this may have something to do with the awkward wording of condition 3), so I'm going to make an assumption about transitivity that might not be true.
Assumption: For each reflection order in an irreducible complex reflection group $G$, all reflection hyperplanes of that order in the source affine space are mapped to a single hyperplane in the target.  Equivalently, irreducible complex reflection groups act transitively by conjugation on the quotient of the set of reflections of any fixed order by the corresponding hyperplane stabilizers.
If the assumption is true, we only need to find two rank 2 complex reflection groups of the same order, whose reflections have the same order.  For example, the exceptional groups of order 144 listed as numbers 7 and 14 in the Wikipedia table work, as well as several other exceptional groups matched with the semidirect products $G(m,p,2)$.<|endoftext|>
TITLE: Is it consistent with ZFC that there is a translation-invariant extension of Lebesgue measure that assigns nonzero measure to some set of measure less than c?
QUESTION [6 upvotes]: It is consistent with ZFC (but not ZFC+CH, of course) that there is a subset $A$ of nonzero outer Lebesgue measure that has cardinality less than $c$.  There will then be an extension of Lebesgue measure that assigns non-zero measure to $A$ and there will be a translation-invariant extension of Lebesgue measure that assigns zero measure to $A$ (since there is an extension that assigns zero to all sets of cardinality less than $c$, using the uncountable cofinality of $c$ and the fact that any such set has null inner measure).  
Question: Is it consistent with ZFC that there be a translation-invariant extension of Lebesgue measure that assigns nonzero measure to some set of cardinality less than $c$?  
If yes, then it will be consistent with ZFC that there be a translation-invariant extension of Lebesgue measure which has a set of null measure whose complement has cardinality less than $c$, which will be rather amusing, I think.
(There are two kinds of proofs I've seen of the fact that every set of nonzero Lebesgue measure has cardinality $c$.  One kind depends on there being a closed subset of nonzero measure and then a bunch of bisections.  That won't work for extensions of Lebesgue measure.  The other kind depends on the continuity of convolutions of characteristic functions, which then depends on the $L^1$-continuity of translation, which then depends on approximation by characteristic functions of intervals, and that won't work either.)

REPLY [14 votes]: There can be no translation invariant extension of the Lebesgue measure which gives a set of cardinality less than continuum positive measure. Suppose that $\nu$ is a translation invariant extension of the Lebesgue measure with $\nu(A)>0$ for some set $A$ of cardinality less than continuum. Take note that $\mathbb{R}$ is a vector space over $\mathbb{Q}$. Let $B$ be the vector subspace of $\mathbb{R}$ generated by $A$. Then $B$ has cardinality less than continuum as well, so the quotient group $\mathbb{R}/B$ has cardinality continuum. Therefore there is an uncountable set $C\subseteq\mathbb{R}$ such that $c+B\neq d+B$ for distinct $c,d\in C$. Clearly $c+A\neq d+A$ for distinct $c,d\in C$. In particular, by translation invariance, $\mu(c+A)>0$ for all $c\in C$. However, this means that $\mathbb{R}$ has uncountably many pairwise disjoint sets of positive $\nu$-measure. This contradicts the $\sigma$-finiteness of $\nu$.<|endoftext|>
TITLE: Seifert genus of the lift of a knot in its cyclic branched covers
QUESTION [6 upvotes]: I was wondering if there are any known examples of knots $K$ in $S^3$ with Seifert genus $g$ so that the lift of $K$ sitting inside its $n$-fold cyclic branched cover bounds an embedded surface of genus less than $g$? 
If $g_n(K)$ is the smallest possible genus of an embedded surface bounding the lift of $K$ in its $n$-fold cyclic branched cover, what happens to $g_n(K)$ if we allow $n$ to grow? (Assume $K$ is not fibered.)

REPLY [15 votes]: A non-trivial (and much more general) result of Gabai implies that $g_n(K)=g_1(K)$ for all $n$. This is encapsulated in the phrase ``Gromov norm equals Thurston norm". Roughly, the Gromov norm represents the minimal genus of an immersed Seifert surface, whereas the Thurston norm represents the minimal genus embedded Seifert surface. It follows that the minimal genus embedded Seifert surface realizes the minimal genus over all immersed Seifert surfaces. 
Gabai's proof is interesting and enlightening. He constructs a taut foliation of the knot complement with a minimal genus Seifert surface as a leaf, making use of the theory of taut sutured manifolds that he develops in the papers. The euler class of this foliation gives a lower bound on the genus of immersed Seifert surfaces, and realized by the embedded Seifert surface.<|endoftext|>
TITLE: Overconvergent/infinitesimal site, base change and six operations
QUESTION [7 upvotes]: This question is about 6 operations formalism for 'crystalline' cohomology theories - more specifically the infinitesimal cohomology of smooth $\mathbb{C}$-varieties, and the overconvergent cohomology of varieties in positive characteristic. I'm not really sure of a good reference for the former, for the latter I'm talking about le Stum's site theoretic approach to rigid cohomology, as developed in his paper "The Overconvergent Site".
In particular, I'm interested in whether or not one might hope for a 6 operations formalism for the cohomology of the overconvergent site. One thing that suggests to me that this is not the 'right' set-up to get 6 operations is that I think I can convince myself that for any Cartesian square of $k$-varieties ($k$ a perfect field of characteristic $p>0$)
$$ \begin{matrix} X' & \overset{g'}\rightarrow &X \\
f'\downarrow &&\downarrow f \\
Y' & \overset{g}{\rightarrow} &Y\end{matrix}$$
the induced base change morphism
$$ g^*_{\mathrm{An}^\dagger} \mathbb{R}f_{\mathrm{An}^\dagger*}E \rightarrow \mathbb{R}f'_{\mathrm{An}^\dagger*}g'_{\mathrm{An}^\dagger}^*E$$
is an isomorphism. This seems 'wrong' to me - one should only expect such a base change for a proper morphism $f$. My question is whether or not this should be taken seriously as a reason why 6 operations with good 'topological' properties won't exist in this context, or is there something I've missed? 
I'd also be interested to know whether or not 6 operations has been worked out for the infinitesimal site of $\mathbb{C}$-varieties (or varieties over any alg. closed char $0$ field). In the introduction to another paper of his, "Constructible $\nabla$-modules on curves", le Stum says that  constructible sheaves have a definition in terms of the infinitesimal site (due to Deligne, but unpublished), but what about complexes and six operations? Is there anything known in this direction?
Also, just for reference, the argument that we have base change for any Cartesian square is basically just combining the paragraph before 1.4.2 of http://perso.univ-rennes1.fr/bernard.le-stum/Publications_files/OverconvergentSite.pdf with 7.28.1 of http://stacks.math.columbia.edu/tag/04IT. One can also see it in terms of realisations.
EDIT: It may be worth making clear that I am familiar with both Caro's theory of 6 operations for overholonomic $F\text{-}\mathcal{D}^\dagger$-modules, and the theory of algebraic $\mathcal{D}$-modules in char $0$. I am more interested in whether or not, for example in char $0$, one can do all this purely in terms of the infinitesimal site.

REPLY [7 votes]: The six operations formalism includes a good theory of support: for any closed immersion $i:Z\to X$ with complement $j:U=X-Z\to X$, there are short exact sequences of the form $$0\to j_{!}j^\ast(F)\to F\to i_\ast i^\ast(f)\to 0$$
This is not compatible with a base change formula for arbitrary morphisms, and explain why working with too big sites is not an option (at least without further work).
For the $p$-adic version, this paper of D. Caro explains how to get the six operations for overholonomic $F$-complexes of $\mathcal{D}^\dagger$-modules. In characteristic zero, the following books are standard references:
M. Kashiwara $\mathcal{D}$-modules and microlocal calculus,
Translations of Mathematical Monographs, 217. Iwanami Series in Modern Mathematics. American Mathematical Society, Providence, RI, 2003.
R. Hotta, K. Takeuchi, T. Tanisaki, $\mathcal{D}$-modules, perverse sheaves, and representation theory, Progress in Mathematics, 236. Birkhäuser Boston, Inc., Boston, MA, 2008.
Note also that there is a way to produce the six operations formalism out of any reasonable cohomology theory (such as algebraic de Rham cohomology, or Berthelot's rigid cohomology), using Morel-Voevodsky's homotopy theory of schemes and mixed motives; see the last chapter of this paper (in particular, Examples 17.2.21, 17.2.22, and 17.2.23).<|endoftext|>
TITLE: Alternative Almost Complex Structures
QUESTION [8 upvotes]: Originally posted on Maths Stack Exchange.

Let $V$ be a real vector space. An almost complex structure on $V$ is a map $J : V \to V$ such that $J^2 = -\mathrm{id}_V$. An almost complex structure gives $V$ the structure of a complex vector space by defining $(a+bi)v = av + bJ(v)$. The idea behind this definition is that $J$ represents multiplication by $i$. What if you instead consider multiplication by other complex numbers?
Let $n > 2$ be a fixed positive integer and let $J : V \to V$ be such that $J^n = -\mathrm{id}_V$ which I will refer to as an alternative almost complex structure. One can regard $J$ as trying to capture multiplication by $\zeta = \exp\left(\frac{\pi i}{n}\right)$. In particular, as $\{1, \zeta\}$ is linearly independent over $\mathbb{R}$, it is a basis for $\mathbb{C}$ as a real vector space. One can then endow $V$ with a complex structure by defining $(a + b\zeta)v = av + bJ(v)$.

Do alternative almost complex structures give rise to the same results as the standard almost complex structures? In particular, in the case where we extend them to bundle endomorphisms of the tangent bundle of a manifold.
If not, what fails? If so, is dealing with almost complex structures easier than dealing with alternative almost complex structures?


As Henry T. Horton asks about in a comment to the original post, I am interested in the integrability of alternative almost complex structures but also whether there are any difficulties when combining with other structures such as a symplectic form or a Riemannian metric.

REPLY [5 votes]: Wait: if we ask for $J^3=−I$, then $J=−I$ is one solution, and that doesn't determine an almost complex structure, so Vladimir's answer doesn't quite cover all possible cases. If $J^k=−I$ and $k$ is even, then clearly Vladimir's answer is ok: $J$ commutes with a unique almost complex structure, and the whole story reduces to the story of that almost complex structure.<|endoftext|>
TITLE: Conceptual structuralism and continuum hypothesis
QUESTION [11 upvotes]: In Ferefman's paper 'Is the Continuum Hypothesis a definite mathematical problem?', he argues that within the philosophy of conceptual structuralism, the continuum hypothesis is not a definite mathematical problem.
However, Koellner wrote a response to this paper in his paper 'Feferman on the Indefiniteness of CH'. In this, he proposes that it is possible to reach the conclusion that the continuum hypothesis is a definite mathematical problem via conceptual structuralism though he presents no argument for it. Instead, he relates this to the views of Martin and Isaacson.
Does anyone know where I can find an argument that the continuum hypothesis is a definite mathematical problem by the view of a conceptual structuralist?

REPLY [11 votes]: Koellner is probably referring to the paper

Daniel Isaacson, The reality of mathematics and the case of set theory, in: Zsolt Novak and Andras Simonyi (eds), Truth, Reference, and Realism, Central European University Press, 2011. JSTOR
(author pdf).

Isaacson appeals to Kreisel's argument that the second-order categoricity of set theory makes the continuum hypothesis a definite problem.<|endoftext|>
TITLE: Machin-like formulas for logarithms
QUESTION [8 upvotes]: I found this math puzzle blog post
http://fredrikj.net/blog/2013/03/machin-like-formulas-for-logarithms/
which I'm reposting here with permission.  I'm setting this to community wiki to minimize the perception that I'm leeching for internet points, but if this question is somehow inappropriate then I would take the blame and not the original author.

Machin-like formulas express $\pi$ as an integer combination of arctangents evaluated at reciprocals of integers. The most famous is
$$\pi = 16 \arctan \frac{1}{5} - 4 \arctan \frac{1}{239}$$
which historically has been used for many record computations of $\pi$, including Machin’s own accomplishment of breaking the 100-digit barrier in 1706. (Nowadays the more efficient Chudnovsky algorithm is usually used, but Machin-like formulas still hold up quite well.)
For a Machin-like formula
$$\pi = \sum_{i=1}^N A_i \arctan \frac{1}{B_i}$$
one can define its efficiency as
$$e = \sum_{i=1}^N \frac{1}{\log B_i}.$$
A smaller $e$ roughly corresponds to a shorter computation time. Larger $B_i$ give faster convergence of each arctangent series, but formulas with larger $B_i$ usually involve more terms, so there is a tradeoff.
By considering hyperbolic arctangents, one can obtain rapidly converging representations for logarithms of integers. For example, here is a simple Machin-like formula for $\log 2$ (it’s easy to find others):
$$\log 2 = 2 \operatorname{arctanh} \frac{1}{5} + 2 \operatorname{arctanh} \frac{1}{7}.$$
Having efficient formulas of this kind for the logarithms of small integers is very useful in some applications.
I recently noticed by accident (while looking at the output from a quick brute force search for Machin-like formulas with PSLQ in mpmath) that there is a set of particularly efficient Machin-like formulas that allow computing $\log 2$, $\log 3$ and $\log 5$ simultaneously from just three hyperbolic arctangents:
$$\log 2 = 14 \operatorname{arctanh}\frac{1}{31} + 10 \operatorname{arctanh}\frac{1}{49} + 6 \operatorname{arctanh}\frac{1}{161}$$
$$\log 3 = 22 \operatorname{arctanh}\frac{1}{31} + 16 \operatorname{arctanh}\frac{1}{49} + 10 \operatorname{arctanh}\frac{1}{161}$$
$$\log 5 = 32 \operatorname{arctanh}\frac{1}{31} + 24 \operatorname{arctanh}\frac{1}{49} + 14 \operatorname{arctanh}\frac{1}{161}$$
This trivially also allows one to compute the logarithm of any integer of the form $2^i 3^j 5^k$.
Here is a challenge: for a positive integer $n$, what is the most efficient set of hyperbolic arctangents that gives you the logarithms of all integers up to $n$ (or equivalently just the primes up to $n$) simultaneously? Can you find a more efficient set for $n = 5$?
Note that if $n$ gets large and we already have a basis for the integers up to $n-1$, we can just add
$$\log(n) = \log(n-1) + 2 \operatorname{arctanh} \frac{1}{2n-1},$$
so it might be enough to consider smallish $n$.

Note that for $x \gt 1, \text{ arctanh}\frac{1}{x} = \frac{1}{2} (\log (x+1) - \log (x-1)).$ So, it makes sense to consider denominators $x$ so that $x-1$ and $x+1$ are smooth, which explains the denominators of $31$, $49$, and $161$ above. Also, the following system is not as good
$$\begin{eqnarray}\log 2 &=& 2 \text{ arctanh} \frac{1}{5} + 2 \text{ arctanh} \frac{1}{7} \newline \log 3 & = & 4 \text{ arctanh} \frac{1}{5} + 2 \text{ arctanh} \frac{1}{7} \newline \log 5 &=& 4 \text{ arctanh} \frac{1}{5} + 4 \text{ arctanh} \frac{1}{7} + 2 \text{ arctanh} \frac{1}{9} \end{eqnarray}$$ 
since we want larger denominators instead of smaller so that we can compute $\operatorname{arctanh} \frac{1}{b}$ more rapidly. A measure of the cost of computing $\sum a_i \operatorname{arctanh} \frac{1}{b_i}$ is $\sum \frac{1}{\log b_i}$.

REPLY [5 votes]: The "efficiency" of the set of denominators $\lbrace 31, 49, 161 \rbrace$ is $0.744$. It is better to use $\lbrace 251, 449, 4801, 8749\rbrace$, which lets you compute the logs of the first $4$ primes for an efficiency of $0.573$. Those are the largest numbers I could find which are sandwiched between $7$-smooth numbers. Using the first $5$, $6$, or $7$ primes didn't improve this, as they resulted in slightly larger efficiencies, e.g. $0.601$ for $\lbrace 28799,57121,62425,74359,246401,388961,672281 \rbrace$, the $7$ largest numbers I could find so that adding or subtracting $1$ produces a $17$-smooth number. 
$$\begin{eqnarray} \operatorname{arctanh}\frac{1}{251} &=& \frac{1}{2}(\log 2 + 2 \log 3 - 3 \log 5 + \log 7) \newline
\operatorname{arctanh}\frac{1}{449} &=& \frac{1}{2}(-5\log 2 + 2 \log 3 + 2 \log 5 - \log 7) \newline
\operatorname{arctanh}\frac{1}{4801} &=& \frac{1}{2}(-5\log 2 -\log 3 - 2\log 5 + 4 \log 7) \newline
\operatorname{arctanh}\frac{1}{8749} &=& \frac{1}{2}(-\log 2 -7\log 3+4\log 5+\log 7). \end{eqnarray}$$
By inverting this system we get
$$\begin{eqnarray}\log 2 &=& 144~a(251) + 54~a(449)-38~a(4801)+62~a(8749)  \newline \log3&=& 228~a(251)+86~a(449)-60~a(4801)+98~a(8749) \newline\log5 &=& 334~a(251) + 126~a(449)-88~a(4801)+144~a(8749)  \newline \log 7 &=& 404~a(251)+152~a(449)-106~a(4801)+174~a(8749) \end{eqnarray}$$
where $a(n) = \operatorname{arctanh}\frac{1}{n}$.
It is plausible that it would be better to use a small set of primes, but not the smallest ones.<|endoftext|>
TITLE: Conjugacy depth of free groups
QUESTION [11 upvotes]: This question is related to this one but for some reason I did not see a connection before Ben McReynolds asked me that question. 
 Question  What is the smallest function $f(n)$ such that for every two  non-conjugate words $u,v$ from the free group $F_2$ of length $\le n$ there exists a finite homomorphic image of $F_2$ of order $\le f(n)$ where $u,v$ are not conjugate?
The function $f(n)$ can be called the conjugacy depth function of $F_2$. The ordinary depth function intruduced by Bou-Rabee "serves" the word problem in a similar way. It is known and not difficult to deduce from the fact that $F_2$ is linear that the depth function of the free group is polynomial (the best known estimate is due to Kassabov and Matucci). Is the conjugacy depth function of the free group bound by a polynomial? It would be so if the answer to the that question is "yes" because matrices with different traces are not conjugate and if two matrices in $SL_3(Z[x_{11},...,y_{33}])$ have different traces, then these two matrices will have different traces in some small finite factor. As Ben McReynolds pointed out to me, if one uses nilpotent factors of $F_2$ to estimate the conjugacy depth growth,one gets exponential estimates. 
 Update.  One does not need work of Marshall Hall, Hempel, Stallings or  Stebe to prove conjugacy separability of $F_2$. One does not need linear groups as well. Here is a proof. Let $u,v$ be cyclically reduced words in $F_2$, $u$ is not a cyclic shift of $v$, $n=|v|\ge |u|$. Consider a finite 4-regular graph $(V,E)$ containing exactly one cycle $C$ of length $n$ (exercise: construct such a graph). Label the edges of $C$ by the generators of $F_2$ and their inverses so that $C$ spells $v$. Extend the labeling of $C$ to a lebeling of $(V,E)$ by the generators of $F_2$ and their inverses to obtain a complete automaton (no two edges having the same tail have the same label). Then each generator of $F_2$ induces a permutation of $V$. So the labeling induces a homomorphism of $F_2$ into the symmetric group $S_{V}$. The images of $u$ and $v$ under this homomorphism are not conjugate because the image of $v$ has a fixed point while the image of $u$ has not.

REPLY [11 votes]: The question, as far as I know, is open. Sean Lawton, Lars Louder, and I are just finishing up a paper that addresses the complexity of the depth function for conjugacy separability. At present, we can prove that if you fix a word in the free group, then distinguishing the associated conjugacy class from another conjugacy class in finite quotients is polynomial in the word length of the varying class. The degree of the polynomial depends on the fixed word. We give two algorithms. One is geometric and one is algebraic.
We tie the complexity of this function with linear representations of groups as Mark mentioned above. The case of n=3 is not special and having any linear representation of the free group that sends conjugacy classes to distinct traces suffices for a polynomial upper bound. I hesitate to say too much as to where I lean but I would guess that the complexity of the depth function for conjugacy in free groups is super-polynomial. We provide a decent amount of evidence for this in the forthcoming paper. 
We also prove that for groups like SL(n,Z), which are known by work of Stebe to not be conjugacy separable for n at least 3, have a rather dramatic failure of conjugacy separability. Essentially every infinite word will have infinitely many powers that are conjugate to other words that are not conjugate to the power in SL(n,Z). These other words are also primitive (not powers of another word). This investigation also leads to a general method for addressing the depth function for conjugacy on free groups. 
With one of my students, Mark Pengitore, we are currently writing a paper on these conjugacy depth functions for other classes of groups; virtually polycyclic for instance.<|endoftext|>
TITLE: Homology in the $A_\infty$ World
QUESTION [18 upvotes]: This question is turning out to be a little long so let me start off with the headline.  Given a differential graded algebra $A$, we can recover $A$ from its homology $HA$ if we know "the" $A_\infty$-structure of $HA$.  Vaguely, I would like to know:

Question 0: Can this be done "functorially" in some sense?

Ok, now for the long version.
For a dg algebra $A$, the association $A\mapsto HA$ gives a functor from the category of dg algebras into the category of graded algebras.  Kadeishvili's theorem states that there is a unique(ish) $A_\infty$-structure on the homology $HA$ with certain nice properties.  In this way we can think of the association $A\mapsto HA$ as having values in the category of $A_\infty$-algebras.  Unfortunately, there seems to be two problems when trying to make this into a functor:

 Given a dg morphism $f:A\to B$ the induced graded morphism $Hf:HA\to HB$ may not be an $A_\infty$-morphism
 One can get an $A_\infty$-morphism $f_*:=q\circ f\circ j:HA\to HB$ where $j:HA\to A$ and $q:B\to HB$ are $A_\infty$-quasi-isomorphisms, but then the association $f\mapsto f_*$ is no longer functorial.

This brings us to: 

Question 1: Is there any way to fix this?  Eg., can we somehow view homology as an $A_\infty$-functor or some other sort of "functor up to homotopy"?    

Similarly, for a dg $A$-module $M$ there is an $A_\infty$-$HA$-module structure on $HM$ having nice properties.  

Question 2: Can we recover the category of dg $A$-modules from the category of $A_\infty$-$HA$-modules, i.e., is there a functor $A$-mod$\to$ $HA$-$A_\infty$-mod (or better yet in the other direction) giving some sort of equivalence?

References also would be much appreciated.

REPLY [4 votes]: The following answer does not address question 1 in full, but it shows, I think, what one would need to think through.
Take a dg algebra $(A,d)$ over a commutative ring $k$. Specify a splitting of the cocycles as cohomology plus coboundaries:
$$  \mathrm{ker} (d) = HA \oplus \mathrm{im}(d) $$
(such a splitting exists provided that $HA$ is projective), and let $i\colon HA\to A$ be the resulting inclusion. We can then construct canonically
(i) An $A_\infty$ structure on $HA$ with differential $\mu^1=0$;
(ii) an $A_\infty$ morphism $\mathcal{I} \colon HA \to A$ whose first term is the given inclusion $i$.
So $HI \colon HA \to HA$ is the identity map.  This is Kadeishvili's construction.
These structures are defined by explicit recursive formulae. As such, they already have some desirable functoriality properties. For instance, if a group $G$ acts by automorphisms on $A$, and if the summand $i(HA)$ is $G$-invariant, then the $A_\infty$ data will be $G$-equivariant. 
Now suppose we specify in addition a splitting of $A$ as $\mathrm{ker} (d) \oplus A'$. We then have a projection $p\colon A \to HA$, and this extends canonically to 
(iii) an $A_\infty$ morphism $\mathcal{P}\colon A\to HA$, with
(iv) a nullhomotopy of $\mathcal{P}\circ \mathcal{I}- \mathrm{id}_{HA}$.
Moreover, there exists a nullhomotopy of $\mathcal{I}\circ \mathcal{P}-\mathrm{id}_A$, but I'm not sure how canonical this nullhomotopy is. A reference for these assertions is Paul Seidel's book Fukaya categories and Picard-Lefschetz theory, chapter 1. In general, $A_\infty$ quasi-isomorphisms induce quasi-equivalences of their module-categories, and this gives an affirmative answer to question 2.
Now take a dg morphism $f\colon A \to B$, and suppose we're given splittings of $A$ and of $B$ as coboundaries plus cohomology plus complement and that $f$ respects these summands. Then we can construct an $A_\infty$ morphism
$$  \mathcal{H}f = \mathcal{P}_B \circ (Hf) \circ \mathcal{I}_A \colon A\to B,  $$ 
as indicated in the question. Under composition $g\circ f$ of splitting-respecting dg maps, there is a homotopy $\mathcal{H}g\circ \mathcal{H}f \simeq \mathcal{H}(g\circ f)$. The homotopy comes from the existence of a homotopy $I_B \circ P_B \simeq \mathrm{id}_B$.` 
So we get a functor from the category of dga with splittings to the category of $A_\infty$-algebras and homotopy classes of morphisms. Presumably, if one can establish just how canonical the homotopy $I_B \circ P_B \simeq \mathrm{id}_B$ is, one can sharpen the functoriality statement.<|endoftext|>
TITLE: The existence of meromorphic functions on Riemann surfaces
QUESTION [9 upvotes]: In Miranda's book on algebraic curves and Riemann surfaces, Miranda writes:

It is a basic and highly nontrivial
  result that a compact Riemann surface
  has nonconstant meromorphic functions
  on it [...] The theory involved in
  producing meromorphic functions for an
  unknown compact Riemann surface is
  rather technical analysis and
  functional analysis. After one has
  access to meromorphic functions,
  however, the theory is completely
  algebraic, or at least can be made to
  be so.

I've seen this claim a number of other places as well. It seems unnatural to use real analysis to prove a theorem about Riemann surfaces, which are geometric/algebraic objects. Is there genuinely no purely geometric/algebraic way to realize an abstract Riemann surface as a branched cover of the Riemann sphere?

REPLY [7 votes]: This deep fact is essentially the same as the uniformization theorem. The problem is
how to construct at least one holomorphic or meromorphic form with prescribed singularity.
All known proofs use some Analysis, and none of them is simple. Once you have Uniformization,
it is easy to construct holomorphic forms.
A good modern proof (in full generality) is contained in the book of J. Hubbard, Teichmuller Theory.
The complexity of the proof depends on how exactly you define a Riemann surface.
A Riemann surface is a 1-dimensional complex manifold.
A 1-dimensional complex manifold is separable (has a countable base). This fact is a part of the
modern statement of the Uniformization theorem.
However, if you include this separability to the DEFINITION of a Riemann surface, the proof substantially simplifies. Especially simple proof (assuming separabiity)
can be found in Goluzin's book Geometric theory
of functions of a complex variable.
All Riemann surfaces arising in real life are easily seen to be separable (give me an example if I am wrong), so there is no real harm if we include this in the definition:-)
But with or without this separability condition, all proofs of existence of holomorphic forms
or of the uniformization use Analysis. Complex or real, I don't see a sharp distinction between
these two.<|endoftext|>
TITLE: Are Lurie's operads special SMCs?
QUESTION [15 upvotes]: In Higher Operads, Higher Categories, Leinster nicely characterizes operads among monoidal categories (as PROs). Roughly speaking, a monoidal category comes from an operad if its set of objects is freely generated as a monoid by some generating set, and if every morphism is a monoidal product of morphisms with one output. But for the purposes of this question, all I need to say is that Leinster characterizes operads as special monoidal categories. Symmetric operads are again special symmetric monoidal categories.
In Higher Algebra, Lurie defines the notion of symmetric monoidal category (SMC) in a really interesting way: it is a coCartesian fibration $p\colon C\to Fin_{\ast}$, where $Fin_{\ast}$ is the category of pointed finite sets, such that the $n$ different "inert" maps $\langle n\rangle\to\langle 1\rangle\ $ induce an isomorphism $p^{-1}\langle n\rangle\cong (p^{-1}\langle 1\rangle)~^n$. But instead of defining a (symmetric colored) operad as a special kind of SMC, Lurie seems to relax the above coCartesian condition. An operad to Lurie is a functor $p\colon C\to Fin_{\ast}$, with coCarteisan lifts guaranteed only for certain arrows downstairs, e.g. over "inert" morphisms. While he does include other conditions not present in the SMC definition, we don't see directly how an operad is a special kind of SMC, even though both are kinds of functors $p\colon C\to Fin_{\ast}$.
While I really like Lurie's discussion (2.1.1) of operads, I find his definition (2.1.1.10) a bit opaque. I was expecting to see that an operad is a monoidal category with a certain extra (PRO-like) condition, but I don't see that in his definition. 
Question: Is there a nice way to characterize Lurie's operads as special SMCs, i.e. as coCartesian fibrations $p\colon C\to Fin_{\ast}$ satisfying a PRO-like condition? How does that condition manifest in the 1-truncated case?
PS. While Lurie works with $\infty$-categories, I'm only interested in the 1-truncated case. I just appreciate the parsimony in his definition of SMC, and was surprised to see it evaporate in his definition of operad.

REPLY [10 votes]: Given a symmetric monoidal category one can construct its underlying operad (well, symmetric colored operad, but I won't keep mentioning this). This operad has the same objects as the SMC. The operads arising this way can be characterized, so that we can regard a SMC as a special kind of operad.
Conversely, given an operad, one can construct its symmetric monoidal envelope, a symmetric monoidal category whose objects are not just the objects of the operad but rather formal (commutative) products of them. The SMCs arising this way can be characterized, as David mentions in the question, so that we can also regard operads as special SMCs.
Now, when operads and SMC are implemented by Lurie's definitions, the direction in which it is easy to go is that of regarding a SMC as an operad. If the SMC is given by a functor $p : C \to Fin_{\ast}$, then regarded as an operad it is the same functor, and whether we view it as a SMC or as an operad its objects are the objects of the category $p^{-1}\langle 1 \rangle$.
To go in the other direction, we need to change the functor: given an operad $p : C \to Fin_{\ast}$, its symmetric monoidal envelope will be a different functor $q : D \to Fin_{\ast}$. The objects of the envelope, which are the objects of $q^{-1}\langle 1 \rangle$, will be all the objects of $C$, not just the objects of the subcategory $p^{-1}\langle 1 \rangle$.
Lurie constructs the symmetric monoidal envelope in section 2.2.4 of Higher Algebra.<|endoftext|>
TITLE: Division by 3 on elliptic curve
QUESTION [10 upvotes]: There is a classical theorem (cf. Theorem 4.1 on Husemoller "Elliptic Curve" book) that states conditions on the coordinates of a point P in an elliptic curve to be twice another point. The result is the following:
Theorem: Let $E$ be an elliptic curve defined over a field $K$ by the equation
$$
E:y^2=(x-\alpha)(x-\beta)(x-\gamma),\quad\mbox{with $\alpha,\beta,\gamma\in K$}. 
$$
For $(x',y')\in E(K)$ there exists $(x,y)\in E(K)$ with $2(x,y)=(x',y')$ if and only if $x'-\alpha$, $x'-\beta$ and $x'-\gamma$ are squares.
I would like to know if there is something similar for 3. That is, condition on the coordinates $(x',y')\in E(K)$ such that there exists $(x,y)\in E(K)$ with $3(x,y)=(x',y')$.

REPLY [8 votes]: In this paper by Ed Schaefer and myself, there is a reasonably explicit description of `3-descent' on an elliptic curve. There is a somewhat more detailed version of the paper here.
Assuming that your curve $E$ over $K$ is such that the Galois action on the points of order 3 is transitive, you look at the field $L = K(P_0)$, where $P_0$ is any point of order 3 on $E$. Then $P_0 = (\xi,\eta) \in E(K)$ and you can consider the tangent line $y = \lambda x + \mu$ at this point $P_0$. The map
$$E(K) \longrightarrow L^\times/L^{\times 3}, \qquad P \longmapsto f(P) = y(P) - \lambda x(P) - \mu$$
(modulo $L^{\times 3}$, where $L^{\times 3}$ means the subgroup of cubes) has kernel $3 E(K)$. This means that $P$ is divisible by 3 in $E(K)$ if and only if ($P = O$ or) $f(P)$ is a cube in $L$.
This is in fact the correct generalization of Joe's reply (for $m=3$) to the case when you don't have full rational 3-torsion: $f$ is the correctly scaled function with divisor $3(P_0) - 3(O)$, and working over $L$ basically means that we look at all the $f$'s for the various choices of $P_0$ at the same time.
When the Galois action is not transitive, you may have to consider two such functions $f$ (such that the orbits of the corresponding points generate $E[3]$ and have total cardinality not a multiple of 3).<|endoftext|>
TITLE: Sum of two tangent bundles of $S^{2n}$
QUESTION [10 upvotes]: I was wondering if the sum $TS^{2n}\oplus TS^{2n}$ is a trivial bundle?
The same is true for spheres of odd dimension (one can find a nowhere zero section of the second bundle, add it to the first, the first becomes trivial and the rest of second bundle plus trivial bundle of rk 2 is trivial too).
It seems that one should take $2n$ sections $v=(v_1,\dots,v_{2n})$ of $TS^{2n}$ (for example projections of coordinate vector fields from $\mathbb R^{2n+1}$ to $S^{2n}$), $u=(u_1,\dots,u_{2n})$ for the second part and than perturb a little u=u+av, v=v+bu.
Nevertheless I can not prove that it works. From the other point of view I see no reasons for this bundle to be non-trivial.

REPLY [20 votes]: Yes. Let $V$ be a real vector bundle whose base is a $d$-dimensional manifold or cell complex, and whose fibers are $r$-dimensional. Then (1) if $r>d$ then $V=W\oplus \epsilon$ where $\epsilon$ is a trivial rank one bundle, and (2) if $r>d+1$ then the rank $r-1$ bundle $W$ is determined up to isomorphism by $V$. In particular stably trivial bundles of rank greater than $d$ are trivial.<|endoftext|>
TITLE: When does a modular form satisfy a differential equation with rational coefficients?
QUESTION [25 upvotes]: Given a modular form $f$ of weight $k$ for a congruence subgroup $\Gamma$, and a modular function $t$ with $t(i\infty)=0$, we can form a function $F$ such that $F(t(z))=f(z)$ (at least locally), and we know that this $F$ must now satisfy a linear ordinary differential equation
$$P_{k+1}(T)F^{(k+1)} + P_{k}(T)F^{(k)} + ... + P_{0}(T)F = 0$$
Where $F^{(i)}$ is the i-th derivative, and the $P_i$ are algebraic functions of $T$, and are rational functions of $T$ if $t$ is a Hauptmodul for $X(\Gamma)$.
My question is the following:

given a modular form $f$, what are necessary and sufficient conditions for the existence of a modular function $t$ as above such that the $P_i(T)$ are rational functions?


For example, the easiest sufficient condition is that $X(\Gamma)$ has genus 0, by letting $t$ be a Hauptmodul.
But, this is not necessary, as the next condition will show.
Another sufficient condition is that $f$ is a rational weight 2 eigenform. I can show this using Shimura's construction* of an associated elliptic curve, and a computation of a logarithm for the formal group in some coordinates (*any choice in the isogeny class will work).
Trying to generalise, I have thought of the following: if $f$ is associated to a motive $h^i(V)$ of a variety $V$, with a pro-representable Artin-Mazur formal group $\Phi^i(V)$ of dimension 1, then we can construct formal group law a-la Stienstra style, and get a logarithm using the coefficients of powers of a certain polynomial. This makes the logarithm satisfy a differential equation with rational functions as coefficients. Since the dimension is 1, the isomorphism back to "modular coordinates" will be a single modular function $t$, and this answers the question positively.
This was the original motivation for the question - a positive answer is weaker, but maybe suggests the existence of associated varieties to rational eigenforms. 
Putting non-eigenforms aside, since I'm not interested as much in them, we are left with non-rational eigenforms. We can try to perform the same Stienstra construction, but this time we get that the galois orbit of $f$ is associated to a "formal group law" of a motive with dimension greater than one. This will make for an interesting recurrence for the vector of the galois orbit, but not necessarily for each form individually, as the isomorphism of formal groups laws (between Stienstra's and those with the modular forms as logarithm) might scramble them together. Maybe not, and this solves might the question.
I realise this last paragraph might be difficult to understand, for the wording is clumsy, and the mathematical notions are even worse. If you're really interested in this, I'd be happy to elaborate.

REPLY [5 votes]: Let $K$ be the algebraic closure of the differential field $\mathbb{C}(T)$. 
Let $\partial$ denote differentiation w.r.t. $T$.  Now $\mathbb{C}(T)[\partial] \subseteq K[\partial]$ are rings of differential operators. Your function $F$ is a solution of $L(F)=0$ where $L \in K[\partial]$ is the differential operator $L = P_{k+1} \partial^{k+1} + \cdots + P_0 \partial^0$. The rings $\mathbb{C}(T)[\partial]$ and $K[\partial]$ (multiplication = composition) satisfy all properties of a Euclidean domain except commutativity. In particular, one can define an LCLM (least common left multiple) which behaves just like an LCM in Euclidean domains.
Let $L_1,\ldots,L_d$ be the conjugates of $L$ over $\mathbb{C}(T)$, obtained by applying ${\rm Gal}(K/\mathbb{C}(T))$ to $P_{k+1},\ldots,P_0$. Now let $M = {\rm LCLM}(L_1,\ldots,L_d)$. Then $M \in \mathbb{C}(T)[\partial]$ and $M$ is right-divisible by $L$. In particular $M(F)=0$.
In summary: Any function $F$ that satisfies a linear differential operator $L$ with algebraic-function coefficients will also satisfy a linear differential operator $M$ with rational-function coefficients. In Maple you can find $M$ with the command DEtools[LCLM](L, `and conjugates`);<|endoftext|>
TITLE: Stone–Čech Compactification of $\mathbb{Z}$ with Fürstenberg Topology
QUESTION [20 upvotes]: The Stone–Čech Compactification of $\mathbb{N}$ as a discrete space has been extensively studied and can be represented using ultrafilters.
Consider $X=(\mathbb{Z},\mathcal{T})$, where $\mathcal{T}$ is the Fürstenberg topology generated by arithmetic sequences. Equipped with this exotic topology, $X$ is a topological ring, metrizable, and totally disconnected. 
Since $X$ is metrizable, it is Tychonoff and the map from $X$ to its image in $\beta X$ (its compactification) is a homeomorphism. 
Has $\beta X$ been studied? Is there a straightforward description analogous to the compactification of $\mathbb{N}$ with the discrete topology?

REPLY [7 votes]: Topologically speaking $\mathbb{Z}$ with the topology mentioned above is just (homeomorphic to) the space of rational numbers. The space $\beta\mathbb{Q}$ has been studied a lot (not as much as $\beta\mathbb{N}$), for example by Eric van Douwen in Remote Points (MR link), freely available here at DMLPL.<|endoftext|>
TITLE: computing the order of the image of 0 under the modular parametrization map for an elliptic curve
QUESTION [5 upvotes]: Let $E/\mathbb{Q}$ be an elliptic curve of rank 0, with modular parametrization $\phi: X_0(N) \to E$.  Let $\Omega_0$ be the least positive real period.  A paper I'm reading (Yoshida, Some variants of the congruent number problem I) seems to use the following line of reasoning. Say $\Omega_0/L(E/\mathbb{Q}, 1) = c$. We have that $\Omega_0/c = L(E/\mathbb{Q}, 1) = -I(0)$, so $I(0)$ must be a torsion point of exact order $c$.  
I'm confused because from Cremona's tables, there are elliptic curves (for example 11a3) with torsion subgroup of order 5 but $c = 25$.  So something is wrong here... Any idea what it is?  I'm most likely just misinterpreting the paper or incorrectly generalizing (he was only doing this for a few specific elliptic curves).

REPLY [5 votes]: Here is an expanded version of g6hq's answer. Your question is indeed sensitive to the Manin constant of the modular parametrization $X_0(N) \to E$. This in turns depends on whether the elliptic curve $E$ is the so-called "strong Weil curve" in its isogeny class, which by definition means that the kernel of $\phi_* : J_0(N) \to E$ is connected.
For example when $E=11a1=X_0(11)$, then $\phi$ is an isomorphism so everything is fine. But when $E=11a3=X_1(11)$, the modular parametrization $\phi : X_0(11) \to X_1(11)$ of least degree has degree 5 and in fact satisfies $\phi^* \omega_E = c_E \omega_f$ where $f=f_E \in S_2(\Gamma_0(11))$ is the newform associated to $E$, with a non-trivial Manin constant $c_E=5$. Thus the torsion point $I(0)$ has, in fact, order $c/c_E=5$.
In general, given any elliptic curve $E/\mathbf{Q}$, there is an optimal parametrization $\phi : X_1(N) \to E$. Such a parametrization conjecturally satisfies $\phi^* \omega_E = \omega_f$ (Stevens' conjecture). But the class of the divisor $[0]-[\infty]$ in the Jacobian of $X_1(N)$ is defined only over $\mathbf{Q}(\mu_N)^+$, not over $\mathbf{Q}$.<|endoftext|>
TITLE: Area of a lattice polygon in terms of its width
QUESTION [5 upvotes]: Let $M$ be a lattice polygon on a plane (i.e. its vertices are integer points $(i,j)\in\mathbb Z^2$).
Let us define lattice width in a direction $v=(m,n)\in\mathbb Z^2$ as $w_v(M)=\max\limits_{x,y\in M} v\cdot(x-y)$.
Suppose the $minimal$ lattice width of $M$ equals $d$. It is clear that the area of $M$ should be proportional to $d^2$.
One can prove an inequalities $area\geq d^2/4$ as is written in comments.
But it seems far from the best one.
So, the question is what is the best $\alpha$ such that $area(M)\geq \alpha d^2$ for each $M$ with minimal lattice width equals $d$. From my comment below one can extract that $\alpha\leq 3/8$
Added: F. Petrov gave a link to the proof of the fact, but there is no complete proof in the Internet so I rewrote it in the appendix in http://arxiv.org/abs/1306.4688

REPLY [3 votes]: $\alpha=3/8$ is sharp according to, say this article, the authors refer to 
[L. Fejes-Toth and E. Makai, Jr., On the thinnest non-separable lattice of convex
plates, Studia Sci. Math. Hungar. 9 (1974), 191–193.]<|endoftext|>
TITLE: bijection between number of partitions of 2n satisfying certain conditions with number of partitions of n
QUESTION [9 upvotes]: Suppose $\lambda = (\lambda_1,\lambda_2,.....,\lambda_k)$ is a partition of $2n$ where $n \in \mathbb N$ satisfying the following conditions:
(1) $\lambda_{k} = 1$.
(2) $\lambda_{i} - \lambda_{i+1} \leq 1$ for every $i\leq k-1$.
(3) In the partition $\lambda$, the number of odd parts in odd places & the number of odd parts in even places are equal. Here a part $\lambda_{i}$ is said to be in even place if ${i}$ is even, whereas $\lambda_{i}$ is said to be in odd place if ${i}$ is odd. $\lambda_{i}$ 's are called parts of $\lambda$ and $\lambda_{i}$ is called an odd part if it is odd & is called even part if it is even.
Now the question is to give a bijection between number of partitions of $2n$ satisfying the above conditions and number of partitions of $n$.

REPLY [3 votes]: This is the bijection indicated in the answer by Matt Fayers, described more explicitly. Note that I needed to correct the wrong definition of "conormal nodes" that I had found in the document mentioned in my comment to that answer. 
Represent partitions by their Young diagram, laid out on a checkerboard pattern that makes the top-left square a black one. Condition (3) means that the diagram covers equally many black and white squares, while conditions (1) and (2) mean that all nonzero column lengths are different.
The procedure below will successively add $1$ black square, then $2$ white squares, then $3$ black squares, and so on with alternating colours and increasing numbers, until at some point a step adds squares to every nonempty column. At that point, if there are now $k$ nonempty columns (and the last step added $k$ squares), remove $k,k-1,\ldots,2,1$ squares respectively from the successive columns (which compensates for all the the squares that had been added); this makes the column lengths (no longer necessarily distinct but) all even (this follows from the colour of the last square added to each column), and dividing them all by $2$ gives the desired partition (or its conjugate). The crux of the definition of the procedure is of course to specify exactly which $i$ squares to add in step $i$; this is done as follows.
From the theory of $2$-quotients, it follows that for any Young diagram, if $d$ is the number of black squares it contains minus the number of white squares, then the number of places (cocorners) where a white square can be added minus the number of places (corners) where a white square can be removed equals $2d$, and then the number of places where a black square can be added minus the number of places where a black square can be removed equals $1-2d$. An elementary proof would be to check it for $2$-cores (staircase diagrams), and check that the mentioned differences do not change when adding a domino to the diagram. In the above description, where the number $d$ takes successive values $0,1,-1,2,-2,\ldots$, one sees that at the point where one must add $i$ squares of some colour, the number of addable squares of that colour exceeds the number of removable squares of that colour by $i$ (so in particular one cannot have a shortage of candidate squares). Now pair up each such removable square with an addable square which it "cancels", as follows. Order all these addable and removable squares of the correct colour cyclically, from the top-right to the bottom left along the border of the diagram, and then "warp around" back to the start. If a removable square pair is immediately followed in this order by an addable square then the two pair up; continue pairing up with the same rule, but ignoring already paired-up squares for the "immediately followed" condition. All removable squares are eventually paired up, and $i$ addable squares remain unpaired. It is these $i$ squares that will be effectively added in this step.
(One can avoid talking about a "wrap around" cyclic order, in which case some number $l$ of removable squares may be left unpaired; then one only adds the $i$ lowest unpaired addable squares and $l$ addable squares before them are not added, although they remained unpaired. However I find it more intuitive to say that every removable square knocks out precisely one addable square, and without cycling there are precisely $l$ unmatched removable squares, which are all below all unmatched addable squares.)
The proof that each step is reversible is easy. Each square that was added has become removable; therefore the pairing for the reverse direction proceeds by having each addable square pair up with a removable square immediately preceding it in the cyclic order (again ignoring previously paired-up squares), and it is the unpaired removable squares that in the end are effectively removed. 
Of course one must also show that the construction in both directions preserves the "distinct nonzero column lengths" condition; this is also easy. In the forward direction, one could only make two column lengths equal if both columns had addable squares of the correct colour, but only the rightmost one of them was actually added. But that means the leftmost one must have been paired up with a removable square; but such pairs are never in the same column, and the presence of the rightmost addable square makes such a pairing impossible. The argument in the oppsite direction is quite similar.<|endoftext|>
TITLE: Quadratic forms and $p$-adic integers
QUESTION [15 upvotes]: I want to prove a result on equivalences of quadratic forms over $\mathbb{Q}_p$, with a control on the height of the change-of-basis matrix.
(I am more generally interested in hermitian forms over division algebras over local fields, but for the purposes of this question the simplest case seems the most awkward.)  I need the following result, which I can prove in a tedious way:

Let $p$ be a prime not equal to $2$ and $M \in \mathrm{M}_n(\mathbb{Z}_p)$.  If there exists $A \in \mathrm{M}_n(\mathbb{Q}_p)$ such that $M = AA^t$ then there exists $B \in \mathrm{M}_n(\mathbb{Z}_p)$ such that $M = BB^t$.

(The difference between $A$ and $B$ is that $B$ has integer entries, while $A$ might not.)
I can prove this as follows: the quadratic form represented by $M$ can be diagonalised over $\mathbb{Z}_p$, so we may assume that $M$ is diagonal.
We can also assume that its diagonal entries are contained in $\{ 1, p, u, pu \}$ for some fixed non-square unit $u$.
Since $M = AA^t$ this quadratic form is equivalent to the standard one over $\mathbb{Q}_p$.
Thus $\det M$ is a square and it has Hasse invariant $+1$.
This translates into conditions mod 4 on how many times $p$, $u$ and $pu$ appear (the conditions are different depending on whether $p$ is 1 or 3 mod 4).
Thus there is a finite list of matrices of size at most 4 which have to be checked, and for each of these you can exhibit explicitly a suitable $B$.
Going through this list is quite tedious.  Is there a more elegant proof of the result, or has it been written down somewhere already?

REPLY [5 votes]: Let $V$ be the $n$-dimensional quadratic space over $\mathbb Q_p$ corresponding to the sum of $n$ squares.  The matrix $M$ can be viewed as the Gram matrix for a $\mathbb Z_p$-lattice $L$ on $V$.  Your condition on $M$ is equivalent to saying that $L$ is $\mathbb Z_p$-integral; so it must be inside a $\mathbb Z_p$-maximal lattice $K$ on $V$.  Now, any two $\mathbb Z_p$ maximal lattices on $V$ are isometric.  In particular, $K$ is isometric to the $\mathbb Z_p$-lattice $I_n$ ($I_n$ is maximal because $p$ is odd).  Therefore, $I_n$ must contain an isometric copy of $L$, which is the same as saying that there exists an $B \in M_n(\mathbb Z_p)$ such that $M = BB^t$.<|endoftext|>
TITLE: What is the smallest cardinality a topology can have which is c.c.c but not separable (in ZFC)?
QUESTION [5 upvotes]: What is the smallest cardinality a topology can have which is c.c.c but not separable (in ZFC)?

REPLY [3 votes]: This may be helpful for you:

Let $X$ be a space with $|X|= \aleph_1$, let $\tau_X= \lbrace U: X\setminus U \text{ is countable } \rbrace$. This space is CCC, but not separable.

Proof: $X$ is not separable: for any countable set $A \subset X$, clearly, $U=X\setminus A$ is open and $U \cap A=\emptyset$.
$X$ is CCC: if $X$ has uncountable disjoint open sets $\lbrace \cal U_\xi: \xi \in \aleph_1\rbrace$. Pick one open set, for example, $U_0$. Because $X \setminus U_0$ is uncountable, it is a contradiction with $U_0$ is open.<|endoftext|>
TITLE: Why is a proper, affine morphism finite?
QUESTION [7 upvotes]: Can you tell me, how to prove that every proüper and affine morphism between locally Noetherian schemes is finite?
Every help will be appreciated.

REPLY [13 votes]: There is an elementary proof of the result "universally closed + affine $\Rightarrow$ integral" that I learnt from Olivier's paper "Going up along absolutely flat morphisms." In fact, it's so simple, I can present it here.
Observation 1: Say $\phi:A \to B$ is an injective ring map that is closed on $\mathrm{Spec}$. Then $\phi^{-1}(B^\ast) = A^\ast$.
(This proof was edited and corrected to reflect xuhan's comment.)
Proof: Fix $a \in A$ with $\phi(a) \in B^\ast$. We must show $a \in A^\ast$ or, equivalently, $a$ is non-zero in the residue field $\kappa(\mathfrak{p})$ of $A$ at any prime $\mathfrak{p} \in \mathrm{Spec}(A)$. Note that the last statement is clearly true if $\mathfrak{p}$ lies in the image $Z$ of $\mathrm{Spec}(\phi)$. So it suffices to show $Z = \mathrm{Spec}(A)$. By closedness, $Z = V(I)$ for some ideal $I \subset A$ (set-theoretically). Localizing at any prime $\mathfrak{p}$ shows $I \subset \mathfrak{p}$ by the injectivity hypothesis. Then $I$ is contained in all primes of $A$, so it contains only nilpotents, and hence $Z = \mathrm{Spec}(A)$.
Observation 2: Say $\phi:A \to B$ is an injective ring map, and $\phi[T]:A[T] \to B[T]$ is closed on $\mathrm{Spec}$. Then $\phi$ is integral.
Proof: Fix some $f \in B$, and consider the surjective map $B[T] \to B[\frac{1}{f}]$ given by $T \mapsto \frac{1}{f}$. If we write $C \subset B[\frac{1}{f}]$ for the image of the composite $A[T] \to B[T] \to B[\frac{1}{f}]$, then $C \to B[\frac{1}{f}]$ is an injective ring map that is closed on $\mathrm{Spec}$. The image of $T$ in $C$ becomes a unit in $B[\frac{1}{f}]$, and hence must be a unit on $C$ by Observation 1, so we can write $f = \sum_{i=0}^n a_i \big(\frac{1}{f}\big)^i$ in $B[\frac{1}{f}]$ for $a_i \in  A$. Clearing denominators shows that $f \in B$ satisfies a monic polynomial over $A$.
Observation 2 + killing the kernel shows:
Theorem: If $\phi:A \to B$ is a ring map that is universally closed on $\mathrm{Spec}$, then it is integral.<|endoftext|>
TITLE: Units of MO and MU
QUESTION [13 upvotes]: Real (or complex) cobordism is described by a symmetric ring spectrum MO (or MU respectively) as explained in examples 2.8 and 2.9 here. Associated to such a ring spectrum $R$, we have a unit spectrum $GL_1(R)$ (see for example chapter 22 in the book by May and Sigurdsson. 

Is there a nice description or interpretation of the units $GL_1(MO)$ or $GL_1(MU)$? 

Let me "define" what I mean by nice by giving an analogy. For the units of $K$-theory, we can identify $GL_1(KU) = BU_{\otimes} \times \mathbb{Z}/2\mathbb{Z}$ and we know that this is the classifying space for virtual vector bundles of virtual dimension $\pm 1$, similarly for $KO$.

REPLY [13 votes]: This is really just a comment, but a bit too long. Thanks Tyler.  Ulrich, that is a good question.  I once worked hard to understand the zeroth space of MU (and more simply BP), one prime at a time, calculating Dyer Lashof operations.  I threw away my notes because the answers I was getting seemed unhelpful.  I seem to recall a paper of Steve Wilson that considers the Hopf ring of BP_0, but my memory may be failing me.  The problem is that these spaces are very  large and seem to have nothing like the economical and intuitive description of their K-theory analogues.  It could be a good problem to try to understand what these infinite loop spaces really look like and what they mean geometrically.  A short modern summary of early work on GL_1(R), alias FR, in general is in [114] on my web page. There is work in progress by John Lind that makes honest the idea that BGL_1R is a classifying space for a reasonably concrete kind of ``principal GL_1R-bundle''.<|endoftext|>
TITLE: Why are all these families of polynomials finally log-concave?
QUESTION [5 upvotes]: This started when I was examining certain families of unimodal polynomials, i.e. $\sum_{k=0}^n a_kx^k$ where $a_0\le a_1\le\cdots \le a_k\ge\cdots \ge a_n$.
(Notation: in the following, the $a_k$ depend on $n$, but writing $a_k$ instead of $a_{k,n}$ makes it easier to read.)
If for a given $n$, we display the coefficients of $f_n=f_n(x)=\sum_{k=0}^n a_kx^k$  as points $(k,a_k)$ defining a graph (say, with linear segments between) and normalize it such that it maximally fits into the rectangle with corners $(\pm1,0)$ and $(\pm1,1)$, then, as $n$ grows, for many of the "classical" families, the shapes of these graphs ressemble more or less a "Gaussian bell curve" with width $\sigma$ proportional to $1/\sqrt{n}$. So it makes sense to look at the "log graph" $\lbrace(k,\ln|a_k|)\rbrace$ instead, which we will again scale to fit into the same rectangle. This scaling "filters" out $\sigma$, and even though the coefficients normally don't yield exactly Gaussian distributions (thus no "Central limit theorem" here), 

for many, if not all, of the classical families of polynomials, as $n\to\infty$, these log graphs converge to a well-defined limit curve, i.e. to the graph of a function $L:[-1,1]\to\mathbb [0,1]$.

(Note that we will extend this "log graph" definition to even and odd polynomials by displaying only the non-zero coefficients.)
An easy example is $f_n(x)=(1+x)^n$, i.e. the binomial coefficients. By Stirling's formula, we find for the limit function, after scaling, 
$L(x)={\ln\left[\left(\dfrac{1+x}2\right)^{\frac{1+x}2} \left(\dfrac{1-x}2\right)^{\frac{1-x}2}\right]} \Bigl/ {\ln(\dfrac12)}=\dfrac1{\ln4}\left(\ln\dfrac4{1-x^2}+x\ln\dfrac{1-x}{1+x}\right),$ 
which is the red curve about half way between the parabola $y=1-x^2$ (blue - would correspond to a "Gaussian bell curve" for the $a_k$) and the upper half of the unit circle (pink). ![Click here for a picture.]


Or consider the Legendre and both kinds of Chebyshev polynomials (ignoring the zero coefficients). Again by Stirling's formula, for each of the three families, we obtain the same limit curve 

$L(x)=\dfrac{(x+3) \ln(x+3)-(1-x) \ln(1-x)-2 (x+1) \ln(x+1) }{4\ln(1+\sqrt{2})}.$ 
This is the green curve above, which has its maximum at $x=\sqrt{2}-1$ and $\lim\limits_{x\to1}L(x)=\dfrac{\ln2}{\ln(1+\sqrt{2})}\approx0.7864397.$ The convergence is numerically rather rapid.

More generally, it seems in fact that all Jacobi polynomials $P_n^{(\alpha,\beta)}(x)$ for fixed $\alpha,\beta\in\mathbb R$ yield this same limit curve as $n\to\infty$, independently of $\alpha$ and $\beta$. 
Doing the same with the Hermite polynomials (again ignoring the zero coefficients) or the Bell polynomials, the convergence is much slower, as displayed below, so the limit curve is more difficult to guess due to limited computer power. I think for both the limit is nothing but the green line $y=\dfrac{1-x}2$.



If we take the above classical polynomials but shift their zeros by a constant, e.g. in PARI using $\mathtt{subst(pollegendre(n,x),x,x+1)}$, or more generally, replacing $x$ by a certain fixed polynomial, it seems like the curves converge quite rapidly, the result depending of course of this fixed polynomial. The next illustration shows e.g. what happens for the Chebyshev polynomials with $x$ replaced by $x^2-10x-10$. The $n=250$ curve showed here has no visible difference with the $n=125$ curve.



The same sort of convergence happens if $f_n$ is defined by its zeros, all real, with fixed proportions of zeros randomly chosen in certain fixed intervals; e.g. in the next image for $n=200$, where $f_n$ is of degree $2n$ with $n$ zeros randomly taken in $[-50,0]$ and $n$ zeros in $[0,1]$.


So far this is nothing very spectacular, but it is an interesting observation that in all these cases, the limit curve is concave.
This still seems to hold if $f_n$ is any "combination" of the above, as defined more formally below, including the possibility for the parameters, like $\alpha$ and $\beta$ for the Jacobi polynomials, to also depend on $n$.
On the other hand, it all breaks down if we admit e.g. the cyclotomic polynomials $\Phi_n(x)$. Well, unlike the "classical" families of polynomials, these cannot be recursively defined by, say, a three-term recurrence, and I'm wondering if this recursiveness is linked to the log-concavity of "$\lim\limits_{n\to\infty}f(n)$". So the question in its general form:

Let $f_n=P(g_1,...,g_k)$, where $P$ is a fixed multivariate polynomial and $g_1,...,g_k$ are polynomials (in $x$ or in fixed polynomials of $x$, as above) belonging to the classical families (meaning, to any family defined by a three-term recurrence) and depending on $n$. Then, as $n\to\infty$, does the limit curve of the log graph of $f_n$'s coefficients always exist, and moreover, is it always concave?

REPLY [11 votes]: I can explain a lot of what you are seeing.
(1) If $f(x)$ has negative real roots, then the coefficients of $f$ are always log concave. Proof: Let $f(x) = \prod (x+\lambda_i)$. Then the coefficients of $f$ are the elementary symmetric polynomials $e_k(\lambda_1, \lambda_2, \ldots, \lambda_n)$. The elementary symmetric polynomials obey Newton's inequalities.
(2) Set $s(a)$ and $t(a)$ be concave functions. I don't know how to make the bounds in what I'm about to say rigorous, so I'll just say everything in the approximate sense. Suppose that $f_n(x) = \sum_{k=0}^n f_{k,n} x^k$ is a family of polynomials with $f_{k,n} \approx e^{n \cdot s(k/n)}$ and that $g_n(x)$ is a similar family of polynomials with $g_{k,n} \approx e^{n \cdot t(k/n)}$. Set $h_n(x) = f_n(x) g_n(x)$. Then the coefficient of $x^{2k}$ in $h$ is
$$\sum_{\ell=0}^{2k} f_{\ell,n} g_{2k-\ell,n} \approx \sum_{\ell=0}^{2k} e^{n \cdot ( s(\ell/n) + t((2k-\ell)/n) )}$$
$$ \approx \exp( n \cdot \max_{\ell} (s(\ell/n) + t((2k-\ell)/n) )\approx \exp(n \max_{0 \leq a \leq 2k/n} s(a) + t(2k/n -a)) $$
If $s$ and $t$ are concave, then the function 
$$u(b) = \max_{0 \leq a \leq 2b} s(a) + t(2b-a)$$
is also concave. I think that's what you're seeing when you take "fixed proportions of zeros randomly chosen in certain fixed intervals": There is some limit curve for zeroes chosen uniformly in one interval, and choosing multiple intervals combines them by the above rule.
(3) Let $F(x,y) = \sum_n f_n(x) y^n$. If the singularities of $F$ are not too complicated, then there are good tools to extract the asymptotic behavior of the $f_{k,n}$, and those methods will "often" give convex curves of the sort you describe. I'll be more precise about what I mean by often.
$\def\CC{\mathbb{C}}$Let $\bar{U}$ be the set of $(x,y) \in \CC^2$ where $\sum f_{k,n} x^k y^n$ converges absolutely. Assume that $\bar{U}$ contains a neighborhood of $(0,0)$, and let $U$ be the interior of $\bar{U}$. Whether or not a point $(x,y)$ is in $U$ depends only on $(|x|, |y|)$. Let $D$ be the image of $U$ under $(x,y) \mapsto (\log |x|, \log |y|)$. Then $D$ is a convex set which obeys the property that $(u,v) \in D$, with $u' \leq u$ and $v' \leq v$ imply $(u', v') \in D$. See section 2 of these notes for much more.
For a positive number $a$, let $s(a) = - \sup_{(u,v) \in D} (au+v)$. Then it is often true that $\log f_{k,n} \approx n s(k/n)$. The function $\sup_{(u,v) \in D} (au+v)$ is (up to sign conventions) called the Legendre transform of the boundary of $D$.
What do I mean by often?
If we have a sequence $k_n$ with $k_n/n \to a$, it is always true that $\sup_{n \to \infty} \frac{1}{n} \log |f_{k_n,n}| \leq s(a)$. Proof: Choose $r>s(a)$. Then there is a point $(u,v) \in D$ with $au+v = -r$. Then 
$$f_{k_n,n} = \int_{|x| = u, |y|=v} F(x,y) x^{-k_n} y^{-n} \frac{dx dy}{x y}$$
An easy bound gives $f_{k_n, n} \leq C \exp (- k_n u - n v) = C \exp(- ((k_n/n) u +v) )$ for some constant $C$.
Conversely, suppose the following conditions hold: There is a single point $(u,v) \in \partial D$ where $au+v$ achieves its maximum. There is an open set $\Omega$ in $\CC^2$ containing the closed polydisc $\{ |x| \leq e^u,\ |y| \leq e^v \}$ such that $F$ extends to a meromorphic function on $\Omega$, with single simple pole along a divisor $\Delta \subset \Omega$. There is a single point $(x,y)$ in $\Delta$ with $(\log |x|, \log |y|) = (u,v)$, and this point is a smooth point of $\Delta$. With all these hypotheses (and perhaps some I have forgotten), $(1/n) \log |f_{k_n,n}| \to s(a)$. See Pemantle and Wilson, part 1. See also this paper where Pemantle and Wilson provide 20 applications of their method, including many of the examples you give.
If $F$ extends to a meromorphic function on some $\Omega$, but the pole set is more complicated, you need to read Pemantle and Wilson's later papers. See especially 2, 3.<|endoftext|>
TITLE: Maximal chain of 1s in binary strings
QUESTION [5 upvotes]: Let $S$ be the set of $2^n$ binary $n$-bit strings. For every $x\in S$, let $f(x)$ is the maximal chain of bits 1 in $x$. So Can we find a good upper bound of $$F(n)=\frac{\sum_{x\in S}f(x)}{2^n}$$
Of course, $O(1)\le F(n) \le O(n)$. I think the upper bound is a constant or $O(\log n)$. Can anyone help me?

REPLY [3 votes]: There is an easy way to get a good upper bound. 
The probability that there is a streak of length $k$ is at most the expected number of streaks of $1$s length $k$, which is at most the expected number of all-$1$ substrings of length $k$ (which may overlap). It is easy to get the last expected value. There are $n-k+1$ possible substrings of length $k$, so the expected number of all-$1$ substrings of length $k$ is $(n-k+1)2^{-k} \lt n /2^k$. For $k = \lceil \log_2 n \rceil + c$ this gives us an upper bound of $1/2^c$ for the probability that there is a streak of length $\lceil \log_2 n \rceil + c$. So, the average excess over $\lceil \log_2 n \rceil$ is at most $1$, and the average length of the longest streak is at most $\lceil \log_2 \rceil + 1$.
Of course, it's not clear that this upper bound is good until you get a lower bound which is close to this. I think the value should be something like $(\log_2 n) -1$. (Actually, my guess is that the difference from $\log_2 n$ is not asymptotically constant, but fluctuates depending on the fractional part of $\log_2 n$.)<|endoftext|>
TITLE: Mathematical Paper That Just Links Two Different Fields of Sciences
QUESTION [18 upvotes]: I have a soft question that is interesting for me in some aspects. I appreciate your answers and comments about it. 
Four years ago, one of my friends in MIT, in the biology lab, had working on neuroscience and specially he worked on Deja-Vu phenomenon. When he asked me about writing a program with Matlab for simulating this phenomenon with a network of cells that they want simulate the Sinc function, I found that there are many good theorems in graph theory that can be useful for his research. When I suggested him this idea, he found it very interesting. 
My question are about this event in a little bit different way. Is it possible that we publish a paper in some mathematical journals that:
1) The only new thing in the paper is relation between a real phenomena and a field of mathematics that is well known. For example, we just model the controversy with bandwidth problem, and no more things just using the theorems that proved for bandwidth problem.
2) This paper does not have new theorems as like as theorems that are common in mathematical papers. This paper just use mathematical theorems in its direction.
Also, do we have some mathematical journals that publish such a papers? And if yes, is there some evidences for this type of publication?
Maybe someone think about the Hilbert Spaces and quantum mechanic. But, in my view, this is not the case. We use Hilbert spaces to model some aspects of quantum mechanics and we get some new results and theorems in quantum mechanic. If we want to think this relation, the paper only must be contain the modeling of quantum mechanic by Hilbert spaces and no more.
Briefly, suppose we found a connection between a real phenomenon and a field of mathematics that can be acceptable or a new view point for analysis the phenomenon. For example, if we found a relation between Darwin's evolutionary theory and a game on graph, is it possible that we can publish such a results as a paper in a mathematical journal? And what kind of mathematical journal is good for this work?
Sorry me for long question.

REPLY [5 votes]: Here's a practical example:
This paper (now considered a classic in the field with 120+ citations) connected the Thurston-Neilsen surface classification theorem with the problem of mixing in fluids.
Boyland, Philip L., Hassan Aref, and Mark A. Stremler. "Topological fluid mechanics of stirring." Journal of Fluid Mechanics 403 (2000): 277-304

REPLY [3 votes]: Further journals devoted to papers linking mathematics with other scientific areas are the following:
Mathematical medicine and biology;
Bulletin of mathematical biology;
Computers and mathematics with applications;
Journal of mathematical imaging and vision;
Mathematical geology;
Mathematical geosciences.<|endoftext|>
TITLE: Quotient of a reductive group by a non-smooth subgroup
QUESTION [5 upvotes]: This is a continuation of my question Quotient of a  reductive group by a non-smooth central finite subgroup.
Let $G$ be a smooth, connected, reductive $k$-group over a field $k$ of characteristic $p>0$.
Let $H\subset G$ be a $k$-subgroup, not necessarily smooth.
Question 1: Does the quotient $G/H$ exist as a $k$-variety?
I am interested in the following special case.
Let $H^{\rm mult}$ denote the largest quotient of $H$ which is a $k$-group ($k$-group scheme) of multiplicative type.
Set $H_1=\ker[H\to H^{\rm mult}]$.
I assume that $H_1$ is smooth, connected and semisimple. 
Question 2:  Does the quotient $G/H$ exist as a $k$-variety under this assumption? (I do not assume that $H^{\rm mult}$ is smooth.)
All comments and references are welcome!

REPLY [2 votes]: As xuhan points out, SGA3 provides a fairly comprehensive view of what can be said about quotients involving group schemes.   The price of this thoroughness is of course a lengthy technical treatment from which you might have trouble extracting just the amount of information you need.   (SGA3 is currently available online
here.)   In any case, it would be helpful if xuhan expands the comment here into a full answer.
Two later expositions may be worth consulting for your specific questions, since they are written in different styles:
Demazure-Gabriel, Groupes algebraiques, Chap. III, section 3.3.
Jantzen, Representations of Algebraic Groups (2nd ed., AMS, 2003), I.5 (especially the early sections of this chapter).   Note that Jantzen follows closely the Demazure-Gabriel treatment, but he tries to make his own version as self-contained as possible.   Moreover, his study of group schemes is slanted particularly toward work with reductive group schemes over fields of prime characteristic.<|endoftext|>
TITLE: A direct proof of the Harer-Zagier recursion enumerating the ways to paste a 2n-gon to get a genus g surface? 
QUESTION [18 upvotes]: In a 1986 paper, Harer and Zagier proved the recursion:
$$(n+1)e(g,n)=(4n-2)e(g,n-1)+(2n-1)(n-1)(2n-3)e(g-1,n-2)$$
where e(g,n) is the number of ways of grouping sides $S_1...S_{2n}$ of a 2n-gon into n pairs, such that, after identifying pairs with the corresponding orientation one gets an orientable surface of genus g.
Their proof is really indirect, it involves pages of calculations. 
Is there a simple proof of this fact?

REPLY [11 votes]: How about 

Goulden, I. P.; Nica, A.
  A direct bijection for the Harer-Zagier formula. 
  J. Combin. Theory Ser. A 111 (2005), no. 2, 224–238.

or one of the references therein?

REPLY [8 votes]: http://arxiv.org/abs/0712.2448
Gluing of Surfaces with Polygonal Boundaries
 E. T. Akhmedov, Sh. Shakirov
By pairwise gluing of edges of a polygon, o produces two-dimensional surfaces with handles and boundaries. In this paper, we count the number ${\cal N}_{g,L}(n_1, n_2, n_L)$ of different ways to produce a surfac of given genus $g$ with $L$ polygonal boundaries with given numbers of edges $n_1, n_2, >..., n_L$. Using combinatorial relations between graphs on real two-dimensional surfaces, we derive recursive relations between $\cal N_{g,L}$. We show that Harer-Zagier numbers appear as a particular case of ${\cal N}_{g,L}$ and deriv a new explicit expression for them. Comments: 7 pages, 9 figures
It seems proposes quite elementary proof.
The key idea that they found some generalization which is more easy to prove.<|endoftext|>
TITLE: Unbounded representations of groups
QUESTION [7 upvotes]: Let $H$ be a Hilbert space and $G$ be a finitely generated group. Let $\pi:G\rightarrow GL(H)$ be a representation.
A map $c:G\rightarrow H$ is called cocycle if $c(gh)=π(g)c(h)+c(g)$ for all $g,h$ in the group $G$. A cocycle  $c$ is proper if the map $g\mapsto c(g)$ is proper, i.e., for every constant $K$ the number of elements $g$ in the group such that $||c(g)|| \lt K$ is finite.
Q1: Does there always exists a representation (bounded or not) of a group on a Hilbert space which admits a proper cocycle?
the answer to this question is in the comment of Mikael.
Q2: Does there always exists a representation (by bounded operators) of a group on a Hilbert space which admits a proper cocycle?
Here we only assume that $||\pi(g)||<\infty$, but $\pi$ is not necessarily uniformly bounded.

REPLY [10 votes]: Q1 and Q2 have a positive answer for all countable groups (conversely a discrete uncountable group cannot bear any proper function). Let $\mu$ be a proper function from $G$ to the positive reals, and view it as a discrete measure on $G$. Assume in addition that $\mu$ grows reasonably, and more precisely satisfies an equality of the form $\mu(gh)\le C_g\mu(h)$ with $C_G\>0$ (e.g. fix a proper subadditive length $|\cdot|$ and define $\mu(g)=|g|+1)$. 
Then the action of left action of $G$ on itself induces a well-defined left regular representation $\pi$ of $G$ on $\ell^2(G,\mu)$, which is bounded ($\|\pi(g)\|\le C_g^{1/2}$).
Let $e$ be the unit in $G$ and $\delta_g$ the Dirac function at $g\in G$.
Define $b$ as the coboundary $b(g)=\delta_e-g\delta_e=\delta_e-\delta_g$. Then $\|b(g)\|\ge \mu(g)-\|\delta_e\|$, which is proper; thus $b$ is a proper cocycle.<|endoftext|>
TITLE: Characterise all pairs of n/m stars that have the same inner radius
QUESTION [6 upvotes]: Geometry, algebra, and examples
Let n and m be integers, with 2 ≤ m < n/2. Consider the bounding polygon of an n/m star (that is, a star with n points each of which connects to the two points ±m away) inscribed in the unit circle. Such a bounding polygon has 2n points, n on the unit circle and n on an inner circle with:
    $Inner Radius = \frac{\cos{\left(\frac{\pi m}{n}\right)}}{\cos{\left(\frac{\pi (m-1)}{n}\right)}}$  
    
    
    
E.g., inner radius of 5/2 star is ½(3–√5) ≈ 0.381966, that of 8/2 star is √(2–√2) ≈ 0.765367, and that of 8/3 star is √(1–√½) ≈ 0.541196. 
I wish to characterise the different stars that have the same inner radius. Two series of pairs are known: ∀ integer i ≥ 2, stars (6i–2)/i and (18i–6)/(6i–2) have the same inner radius, as do (6i–4)/i and (18i–12)/(6i–3). Proof that these pairs do match is boringly elementary, given the identity Cos[θ] Cos[φ] = ½Cos[θ+φ] + ½Cos[θ–φ], and at jdawiseman.com.
But are there any other non-trivial matches, perhaps another series, perhaps sporadic? 
(I know that any other matches must have inner radius > 0.998122, and of course strictly < 1. So any other matches shown as a graphic smaller than about 2k pixels across must look like a circle — practical rather than proper progress.)
The Question

Characterise all pairs of stars, n₁/m₁ and n₂/m₂ (all integer), such that the stars have the same inner radius. It is known that there are two series of such pairs of stars, {(6i–2)/i, (18i–6)/(6i–2)} and {(6i–4)/i, (18i–12)/(6i–3)}. Are there any other series? Are there any sporadic matches?

REPLY [7 votes]: [Expanding some on my comment of a few days ago]
This is a special case of a problem that was solved by Gerrit Bol in 1936 [B];
that nearly-forgotten result was rediscovered, using slightly different
methods, by Bjorn Poonen and Michael Rubinstein [PR].  (As it happens
I used such coincidences in my own work a few years ago [E].)
They find all ways that three diagonals of a regular polygon can meet 
at a point: there are several infinite families (comprising algebraically "trivial" 
solutions that are not always geometrically obvious, plus the four "nontrivial"
families of Table 3 on page 12), and 65 sporadic solutions (Table 4 on page 13).
If two stars have the same inner and outer radii then we can rotate them
so they share an inner vertex; then the outer vertices are contained
among the vertices of a regular polygon, and the shared inner vertex
is on at least four diagonals $-$ indeed at least five if we include
the line of symmetry (and double the order of the regular polygon
if necessary).  Four infinite families (all symmetrical) of such
quintuple intersections are listed in Table 6 (page 16), and a
finite computation limits further sporadic solutions to denominators
18, 24, and 30 (pages 15-16).  If you've already computed far enough
to find any sporadic solutions then the infinite families must account
for everything else.
References
[B] Gerrit Bol: Beantwoording van prijsvraag no. 17, 
Nieuw Archief voor Wiskunde 18 (1936), 14-66.
[PR] Bjorn Poonen and Michael Rubinstein:
The Number of Intersection Points Made by the Diagonals of a Regular Polygon,
SIAM J. Discrete Math 11 (1998), 135-156
(http://www-math.mit.edu/~poonen/papers/ngon.pdf).
[E] Noam D. Elkies: 
On some points-and-lines problems and configurations, 
Periodica Mathematica Hungarica 53 #1-2 (2006), 133-148
(arXiv:MG/0612749).<|endoftext|>
TITLE: $f^3,f^2$ are the cube and quadratic of f respectively and both infinite differentiable on $R$,how to show so is $f$
QUESTION [25 upvotes]: Let $f$ be a real function with domain R.
If $f^2$ and $f^3$ are both infinitely differentiable on R,
how to prove $f$ is infinitely differentiable on R?
I have been thinking about this problem for a long period, but I 
I can not find an accurate proof. So if somebody can help me,
I will appreciative this very much.

REPLY [17 votes]: Posted here because Criterion for smooth functions got closed when I was typing all this. Look at the comment chain there if the opening phrases make no sense to you :-)
Since I couldn't recall the name of the theorem, let me at least try to recall the $\bar\partial$ proof of it :-). Also, if I remember things right (no warranty on that), the proof in Monthly is one of the flawed ones, so at the very least be extremely careful when reading it. The proof referred to by Iosif is correct,  brilliant, and short too, works in much more general setting, but still has one small drawback: looking at it, it is not so easy to understand what happens when only some finite smoothness is assumed, i.e., what is the implication of $f^2,f^3\in C^N$ with some finite $N$. Note that this implication cannot be that $f\in C^N$ because $f(x)=0$ for $x<0$, $f(x)=x^{N/2+\delta}$ for $x\ge 0$ satisfies the requirements but is only $C^{N/2}$ smooth. However, it is, indeed, possible to show that $f\in C^{rN}$ for some fixed $r>0$ (though I'll not try to optimize $r$ in the argument below)
We need three pieces of standard machinery. I'll present the cheapest versions that will work for us though almost everything can be refined quite a bit.
The first one is that a function $f(x)$ on an interval $I\subset\mathbb R$ is in $C^N$ essentially if and only if there exists a uniformly bounded family of functions $f_\varepsilon(z)$ ($\varepsilon\in(0,1)$) defined and analytic in the strips $D_\varepsilon=I\times(-\varepsilon,\varepsilon)$ such that $|f-f_\varepsilon|\le\varepsilon^N$ on $I$.
To get one direction, just define $f(x+iy)=\sum_{k=0}^{N-1}\frac{f^{(k)}(x)}{k!}(iy)^k$ formally extending the Taylor series of $f$ to the complex plane from the nearest point. Then $(\bar\partial f)(x+iy)=\frac 12(\partial_x+i\partial_y)f(x+iy)=\frac 12\frac{f^{(N)}(x)}{(N-1)!}(iy)^{N-1}$, which is bounded by $C\varepsilon^{N-1}$ in $D_\varepsilon$. To get an analytic approximation, it is enough just to subtract the solution of the $\bar\partial$ problem that is obtained by the convolution of $\chi_{D_\varepsilon}\bar\partial f$ with $\frac 1{\pi z}$ as usual.
To get the other direction, let us notice that $|f_{2\varepsilon}-f_{\varepsilon}|\le C\varepsilon^N$ on $I$, so, due to the uniform boundedness and something like the Hadamard 3 lines lemma (or, if you prefer, crude estimates for the harmonic measure),
we get $|f_{2\varepsilon}-f_{\varepsilon}|\le C\varepsilon^{N(1-q)}$ in $D_{q\varepsilon}$ for any $q\in(0,\frac 12)$. Thus, decomposing
$$
f=f_1-\sum_{\varepsilon\in E} (f_{2\varepsilon}-f_{\varepsilon})
$$
where $E=\{2^{-k},k=1,2,3,\dots\}$ and applying the Cauchy estimate for the $m$-th derivative, we conclude that the formally differentiated $m$ times series on the right is dominated by 
$$
Cm!\sum_{\varepsilon\in E}(q\varepsilon)^{-m} \varepsilon^{N(1-q)}\,,
$$
which is a convergent series when $m<N(1-q)$, so if we choose $q<N^{-1}$, we shall get $f\in C^{N-1}$. 
Note that we were a bit sloppy here: we went from $N$ to $N-1$ and the "three line lemma" works the way we want only if we are separated from the endpoints of $I$, but since the smoothness is a local property and we are ready to lose even some portion of $N$ in our count of derivatives, such sloppiness can be easily tolerated.
The second piece of machinery is the computation of the Laplacian 
$$
\Delta \log(|h|^2+\rho^2)=4\frac{\rho^2|h'|^2}{(|h|^2+\rho^2)^2}
$$
and the Green formula
$$
\int_{\mathbb D}\Delta u(z)\log(1/|z|)\,dA(z)=\int_{\mathbb T} u(z)dm(z)-2\pi u(0)
$$
where $\mathbb D=\{z:|z|<1\}$, $\mathbb T=\{z:|z|=1\}$, $dA$ is the area measure and $dm$ is the length measure.
It follows immediately that if $h$ is a bounded analytic function in the unit disk, then
$$
\int_{\frac 12\mathbb D}|h'|^2\chi_{\{|h|\le\rho\}}\le C\log\frac{\|h\|_\infty^2+\rho^2}{\rho^2}\rho^2\,.
$$
Rescaling to $\varepsilon\mathbb D$ and covering $D_{\varepsilon/3}$ by $\approx \varepsilon^{-1}$ disks of radius $\varepsilon/2$, we conclude that for any analytic function $h:D_\varepsilon\to\mathbb C$ that is bounded by a constant, we have
$$
\int_{D_{\varepsilon/3}}|h'|^2\chi_{\{|h|\le\rho\}}\le C\varepsilon^{-1}\log\frac{1+\rho^2}{\rho^2}\rho^2\,.
$$
The last piece of machinery is the bound for the uniform norm of the convolution $w*\frac 1{\pi z}$ in a bounded domain. Split $\frac 1{\pi z}=K_r+K^r$ where $K_r$ is the part in the disk  $r\mathbb D$ and $K^r$ is the part outside the disk (we assume that the function is truncated beyond the diameter of the domain). Then $\|K_r\|_{L^1}\le Cr$ and $\|K^r\|_{L^2}\le C\sqrt{\log(1/r)}$ for small $r$. Hence, we get
$$
\left\|w*\frac 1{\pi z}\right\|\le Cr\|w\|_{L^\infty}+C\sqrt{\log(1/r)}\|w\|_{L^2}.
$$
for every $r\in(0,\frac 12)$, say.
Now everything is ready for the "punchline". Assume that $f^2$ and $f^3$ are $C^{N}$ on $I$ with derivatives up to $N$ bounded by $1$, say. Fix $\varepsilon>0$. Construct  their bounded analytic approximations in $D_{\varepsilon}$ with error $\varepsilon^N$ and call them $h$ and $g$ respectively. Then $|g^2-h^3|\le C\varepsilon^N$ on the real line, so if $q<1/N$, this estimate (with $N(1-q)>N-1$ instead of $N$) spreads to $D_{q\varepsilon}$. Thus, replacing $\varepsilon$ by $q\varepsilon$, we can safely assume that $|g^2-h^3|\le\delta=C\varepsilon^{N-1}$ in $D_{\varepsilon}$. 
How would a normal person recover $f$ grom $g\approx f^3$ and $h\approx f^2$? One possibility is just to say $f\approx g/h$. This would be perfect if we didn't have the small (or even $0$) denominator problem, so let us take $F=\frac{g\bar h}{\max(|h|^2,\rho^2)}$ instead with some $\rho\gg \delta$ to be chosen later and put $f_\varepsilon=F-(\bar\partial F)*\frac 1{\pi z}$. We just need to do two things now: to estimate $|F-f|$ on $I$ and to bound the convolution in the uniform norm.
Estimate of $|F-f|$
Case 1:  $|h|<\rho$. In this case $|f|\le \sqrt(|h|+\delta)\le \sqrt{\rho+\delta}$ and 
$\frac{|g||h|}{\rho^2}\le \frac{\sqrt{\rho^3+\delta}}{\rho}\le C\sqrt{\rho}$, provided that 
$\rho^3\ge \delta$. Thus $|F-f|\le C\sqrt{\rho}$.
Case 2: $|h|\ge\rho$. In this case $h=f^2+O(\delta)$, $g=f^3+O(\delta)$, so $|f|\ge \sqrt{\rho-O(\delta)}\approx \sqrt\rho$, so  $|f|^3\ge \rho^{3/2}\gg \delta$ and
$$
F=\frac gh=\frac{f^3+O(\delta)}{f^2+O(\delta)}=f+O\left( |f|\left(\frac{\delta}{|f|^3}+\frac{\delta}{|f|^2}\right) \right)=f+O(\delta/|f|^2)=f+O(\delta/\rho)\,,
$$
whence $|F-f|\le \frac\delta\rho$ in this case.
Thus, we get $|F-f|\le C(\sqrt{\rho}+\frac{\delta}{\rho})$ whenever $\rho^3\ge\delta$.
Estimate of the convolution
Notice that $w=\bar\partial F=\rho^{-2}g\bar{h'}\chi_{\{|h|\le\rho\}}$. Also, we have already seen that $|h|\le\rho$ implies $|g|\le C\rho^{3/2}$ if $\rho^3\ge\delta$, so we get
$$
|w|\le C\rho^{-1/2}|h'|\chi_{\{|h|\le\rho\}}\,. 
$$ 
Thus, $\|w\|_{L^2}^2\le  C\varepsilon^{-1}\log\frac{1+\rho^2}{\rho^2}\rho$ in $D_{\varepsilon/3}$. Also, since $h$ is bounded in $D_\varepsilon$, we have $|h'|\le C/\varepsilon$ and $\|w\|_{L^\infty}\le C\rho^{-1/2}\varepsilon^{-1}$  in $D_{\varepsilon/3}$.
Hence, by the last standard estimate, we get the bound
$$
Cr\rho^{-1/2}\varepsilon^{-1}+C\sqrt{\varepsilon^{-1}\rho\log\frac{1+\rho^2}{\rho^2}}\sqrt{\log(1/r)} 
$$
for the $L^\infty$ norm of $w*\frac 1{\pi z}$. If we choose $r=\rho$, this becomes $C\varepsilon^{-1}\sqrt\rho\log(1/\rho)$.
Now it is easy to see that with $\rho^3=\delta=C\varepsilon^{N-1}$, all our bounds are at most $C\varepsilon^{\frac N6-2}\log(1/\varepsilon)$, so we have shown that we have at least $\frac N6-3$ continuous derivatives (and there is still some room for improvement).<|endoftext|>
TITLE: Relatively numerically trivial divisor
QUESTION [6 upvotes]: Hi,
Let $f : X \rightarrow Y$ a projective morphism of quasi-projective algebraic varieties over $\mathbb{C}$. Assume that $X$ is smooth, that $Y$ is normal and that: 
$$\textbf{R} f_* \mathcal{O}_X = \mathcal{O}_Y $$ 
(here $ \textbf{R} f_* $ denotes the derived push-forward of $f$). Let $E$ be a Cartier divisor on $X$ such that $E$ is numerically trivial on every fiber of $f$. Is there an integer $m$ such that $mE = f^* F$ for some Cartier divisor $F$ on $Y$ ?
In case $Y$ is a point, I think it is true, because $E$ numerically trivial implies that $mE$ is a deformation of $\mathcal{O}_X$ for some $m>>0$. One concludes with the equality:
\begin{equation*}
T_{\mathrm{Pic}^0(X), \mathcal{O}_X} = H^1(X,\mathcal{O}_X),
\end{equation*}
where $T_{\mathrm{Pic}^0(X), \mathcal{O}_X}$ is the tangent space to the neutral component of the Picard scheme at $\mathcal{O}_X$.
I guess that if $f$ is flat, the same argument should work with the "relative Picard scheme". However I am interested in the case where $f$ is not flat. 
Many thanks in advance!

REPLY [2 votes]: Hi,
So the result is true and a proof has been nicely explained to me by Yoshinori Namikawa.
First, let's recall the following "classical" fact (whose proof can be found as lemma $7.7$ in the book Fourier-Mukai and Nahm transforms in Geometry and Mathematical Physics, for instance):
Lemma :
Let $f : Y \rightarrow X$ be a proper morphism of varieties such that that 
$$ \mathrm{R} f_* \mathcal{O}_X = \mathcal{O}_Y. $$ 
Let $E$ be a Cartier divisor on $Y$. Then $E$ is the pull back of a Cartier divisor on $X$ if and only if for all $x \in X$, there is a neighborhood $U$ of $x$ in $X$ such that $E$ restricted to $f^{-1}(U$) is trivial.
Let $x \in X$, and let $U$ be a contractible neighborhood of $x$ in $X$. Since $\mathrm{R} f_* \mathcal{O}_{f^{-1}(U)} = \mathcal{O}_U$, the exponential exact sequence shows that:
$$ \mathrm{Pic}(f^{-1}(U)) = H^2(U,f^{-1}(U), \mathbb{Z}).$$
Let $F_x$ be the reduced scheme underlying $f^{-1}(x)$. As the open $U$ is contractible, we have $H^2(f^{-1}(U), \mathbb{Z}) = H^2(F_x, \mathbb{Z})$. So we only have to prove that $E$ is homologically trivial on $F_x$. Denote by $\mathrm{Pic}^h(F_x)$ (resp. $\mathrm{Pic}^{\tau}(F_x)$ the group of homologically trivial line bundles (resp. numerically trivial line bundles) on $F_x$. Since $F_x$ is reduced, a theorem of Matsusaka (see Kleiman : Towards a numerical theory of ampleness for a proof) tells us that the quotient group:
$$ \mathrm{Pic}^{\tau}(F_x)/\mathrm{Pic}^h(F_x)$$
is of finite order. Hence some multiple of $E$ is homologically trivial on $F_x$, which conclues the proof.<|endoftext|>
TITLE: Reverse mathematics below RCA
QUESTION [11 upvotes]: I'm sure this is a fairly basic question, but I can't seem to find a solid answer:
My primary question is: Is there a reasonably nice subsystem of second-order arithmetic corresponding essentially to "primitive recursive comprehension?" I'm interested only in $\omega$-models - that is, I don't care about how much induction the system allows. In fact, I'd prefer it to have full induction, so that I know the weakness of the system lies squarely in its comprehension axioms. I do mean "comprehension" here: I'd like this system to be a two-sorted system, like $RCA$ itself, so e.g. $PRA$ is not what I'm looking for - although maybe it can be tweaked into a satisfactory system in an easy way?
My secondary question is: assuming a positive answer to the first question, what are some natural statements which are equivalent (over this base system) to $RCA$ (or $RCA_0$)?
(I'm almost certain this is written up nicely somewhere easily accessible, and my google-fu is simply failing me; if this is the case, and this question is therefore inappropriate for mathoverflow, please feel free to close it.)

REPLY [12 votes]: Surprisingly, this kind of base system hasn't gotten much attention in the reverse mathematics literature. Note that the system $\mathsf{RCA}_0^*$ proposed by Simpson in §X.4 of Subsystems of Second Order Arithmetic is a weakening of $\mathsf{RCA}_0$ in precisely the opposite direction: namely $\mathsf{RCA}_0^*$ plus primitive recursion is equivalent to $\mathsf{RCA}_0$ [D. R. Hirschfeldt, R. A. Shore, Combinatorial principles weaker than Ramsey's theorem for pairs, JSL 72 (2007), 171–206; PDF].
I implicitly used a system like you want in A variant of Mathias forcing that preserves ACA0 [AML 51 (2012), 751–780; arXiv:1110.6559]. This is a function-based system of the form $\mathfrak{N} = (\mathbb{N},\mathcal{N}_1,\mathcal{N}_2,\ldots)$ where $\mathbb{N}$ is the underlying set and each $\mathcal{N}_k$ is a set of functions $\mathbb{N}^k\to\mathbb{N}$ which together form an algebraic clone: each $\mathcal{N}_k$ contains all the constant functions, the projections
$\pi_i(x_1,\dots,x_k) = x_i,$ and if $f \in \mathcal{N}_\ell$ and $g_1,\dots,g_\ell
 \in \mathcal{N}_k$ then the superposition $f(g_1(x_1,\dots,x_k),\dots,g_\ell(x_1,\dots,x_k))$ belongs to $\mathcal{N}_{k}$. And this algebraic clone is closed under primitive recursion: there are distinguished $0 \in \mathbb{N}$ (zero) and $\sigma \in \mathcal{N}_{1}$ (successor) such that for any $f \in \mathcal{N}_{k-1}$ and $g \in \mathcal{N}_{k+1}$ there is a unique $h \in \mathcal{N}_k$ such that $$h(0,\bar{w}) = f(\bar{w}) \quad\text{and}\quad h(\sigma(x),\bar{w}) = g(h(x,\bar{w}),x,\bar{w})$$ for all $x, \bar{w} \in \mathbb{N}.$ Note that the uniqueness requirement on $h$ is crucial since this is the only form of induction in this system. Strangely, one needs to assume dichotomy $x \dot- y = 0 \lor y \dot- x = 0$ to avoid a few bizarre models. Over this base system, $\Delta^0_1$-comprehension boils down to what I called uniformization: 

If $f \in \mathcal{N}_{k+1}$ is such that $\forall \bar{w}\,\exists x\,{f(x,\bar{w}) = 0},$ there is a $g \in \mathcal{N}_k$ such that $\forall\bar{w}\,{f(g(\bar{w}),\bar{w}) = 0}.$

And arithmetic comprehension boils down to what I called minimization: 

For every $f \in \mathcal{N}_{k+1}$ there is a $g \in \mathcal{N}_k$ such that $\forall x,\bar{w}\,{f(x,\bar{w}) \geq f(g(\bar{w}),\bar{w})}.$

(The above description is semantic but it is straightforward to formalize the above as a multi-sorted system.)
Similar systems do appear in proof theory (as described in Carl's answer) but only a few are restricted to second-order types. Kohlenbach proposed similar systems in various places. The closest I've seen is what he called $\mathsf{PRA}^2$ in Things that can and things that cannot be done in PRA [APAL 102 (2000), 223–245; PDF]. He doesn't actually give a fully detailed description in that paper, but if you chase references you see that it is exactly as the system I described above except that closure under superposition and primitive recursion are ensured by type-2 functionals $\mathbf{S}^k_\ell:\mathcal{N}_k\times\mathcal{N}_\ell^k\to\mathcal{N}_\ell$ and $\mathbf{R}_k:\mathcal{N}_{k-1}\times\mathcal{N}_{k+1}\to\mathcal{N}_{k}$. Uniformization is equivalent to the AC0,0-qf scheme in Kohlenbach's paper. The system assumes quantifier-free induction, which is probably not stronger than the formally weaker form of induction in the system I described above.<|endoftext|>
TITLE: What can be proved about the Ramanujan conjecture using elementary means?
QUESTION [25 upvotes]: The Ramanujan conjecture states that the coefficients $\tau(n)$ in the identity
$$q\prod_{m=1}^\infty(1-q^m)^{24}=\sum_{n=1}^\infty\tau(n)q^n$$
satisfy the inequality $|\tau(n)|\leq d(n)n^{11/2}$, where $d(n)$ is the number of divisors of $n$. A positive answer to the conjecture followed (non-trivially) from Deligne's proof of the Riemann hypothesis for varieties over finite fields, conjectured by Weil.
One can interpret the coefficient $\tau(n)$ combinatorially as follows. Let $(a_n)$ be the sequence $1,1,1,\dots,1,2,2,2,\dots,2,3,3,3,\dots,3,4,\dots$, where each number occurs 24 times. Let $\rho(n)$ be the number of ways of writing $n-1$ as a sum $x_1+\dots+x_r$ of $r$ terms of the sequence for some even number $r$, and let $\sigma(n)$ be the number of ways of writing $n-1$ as a sum $x_1+\dots+x_s$ of $s$ terms of the sequence for some odd number $s$. Then $\tau(n)=\rho(n)-\sigma(n)$. It is easy to check that both $\rho(n)$ and $\sigma(n)$ grow very fast -- at least at a rate $\exp(c\sqrt{n})$ (and I'm almost sure that that is the correct order of magnitude, but haven't carefully checked so don't want to claim it as a fact). So Deligne's result tells us that there is a huge amount of cancellation: the number of representations as an even sum is almost exactly half the total number of representations. 
Of course, it tells us something considerably more precise than that, and it seems clear that elementary methods are unlikely to be sufficient for this more precise result. But what if we just ask for a proof that $\tau(n)$ grows at most polynomially? Is that easy to show? If it is, then a much more general result ought to be true. For example, suppose we take an arbitrary non-decreasing sequence $a=(a_1,a_2,a_3,\dots)$ of positive integers such that $cr\leq a_r\leq Cr$ for every $r$. Define $\rho_a(n)$ to be the number of ways of writing $n$ as a sum of an even number of terms of this sequence and $\sigma_a(n)$ as the number of ways of writing $n$ as a sum of an odd number of terms. Must $\tau_a(n)=\rho_a(n)-\sigma_a(n)$ grow at most polynomially? If so, then what can be said about the degree of this polynomial in terms of $c$ and $C$? 
Note that we have the identity
$$\prod_{r=1}^\infty(1-q^{a_r})=\sum_n\tau_a(n)q^n\ .$$
(A small remark is that we don't quite recover Ramanujan's $\tau$ function when the sequence takes each positive integer 24 times, because in this combinatorial context there is no 
motivation for taking the initial $q$ and shifting everything by 1.)
However, in this more general problem, we don't obtain a modular form when we substitute $q=e^{2\pi iz}$. As far as I can see, this rules out not just Deligne's proof methods but also earlier methods such as those of Rankin that gave weaker bounds. However, I don't know my way around this literature: is a result like the one suggested known? Is it even true?
Edit: As Garth Payne points out, it isn't true if all the $a_r$ are odd, since then the parity of the number of terms you need to pick depends on the parity of $n$. So for my question to make sense I need to add some extra condition. I don't really care what that condition is, but one possibility is to take $C$ to be less than 1, but to modify the growth condition to $cn-u\leq a_n\leq Cn+u$ for some $c>0$, $C<1$ and $u$. Or we could insist that each $a_r$ occurs in the sequence an even number of times. Basically, any condition that rules out this kind of example would leave a question that would interest me. 
Further edit: Maybe better than those two conditions is just to assume that the density of terms of the sequence that are even is bounded below the whole time.

REPLY [5 votes]: It seems very unlikely. There are plenty of meromorphic modular functions (as in Payne's answer) whose coefficients fail to satisfy polynomial growth, for example,
$$q^{\frac{M-1}{24}} \cdot \frac{\eta(\tau)}{\eta(M \tau)}
= \prod_{n \nmid M} (1-q^n).$$
If one multiplies this by any power of $\eta(\tau)$, the result (a meromorphic modular form) still has a pole at the cusp above $0$, and the coefficients will exhibit similar behavior. So it would seem that very regular sequences will still cause you problems. The fact that the functions above exhibit coefficient growth of the flavour $\exp(c \sqrt{n})$ [at least in arithmetic progressions] is not obvious, but I imagine such a proof is incidental for your purposes.<|endoftext|>
TITLE: Constructing a linear ODE for a product of two holonomic functions without introducing additional singularities
QUESTION [5 upvotes]: A function $f$ is called holonomic if it satisfies some linear differential equation with polynomial coefficients $$p_n(x) f^{(n)}(x)+\dots+p_1(x)f'(x)+p_0(x)f(x)=0.$$ Now if $f,g$ are holonomic then so are their sum and product. To obtain a differential equation for $h=fg$, first observe that
$$ h^{(k)} = \sum_{i=0}^k {k\choose i} f^{(i)} g^{(k-i)}.$$ Since each $f^{(i)}$ is a linear combintation of $f,f',\dots,f^{(n-1)}$ with rational coefficients (where $n$ is the order of the ODE satisfied by $f$), and analogously each $g^{(i)}$ is a linear combination of $g,g',\dots,g^{(m-1)}$, then $h,h',h'',\dots$ span a finite-dimensional vector space of dimension at most $d=mn$, and hence there exists a nontrivial relation of the form
$$ r_d(x)h^{(d)}+\dots+r_1(x)h'+r_0(x)h=0$$with $r_i(x)$ rational functions.
Now suppose that the ODE which $f$ satisfies had singular points $u_1,\dots,u_n$, while the equation of $g$ had singular points $w_1,\dots,w_m$. The ODE for $h$ might have additional singular points besides $u_1,\dots,u_n,w_1,\dots,w_m$. For example, if
$$(x-a)f''(x)+cf(x)=0\\\\ (x-b)g''(x)+dg(x)=0$$ then $h=fg$ satisfies an ODE of order 4 with leading coefficient $$(x-a)^3 (x-b)^3 \biggl((c-d)x-(bc-ad)\biggr)h^{(4)}+\dots$$ (I calculated this using C.Mallinger's GeneratingFunctions Mathematica package).
Is it possible to construct a holonomic ODE for $h$ (in the above example and in general) without introducing additional singular points?

REPLY [4 votes]: Hi Dima, 
The operator you obtain by the algorithm you describe has minimal possible order. You pay for the minimality of the order by having (in general) a nonminimal degree of the polynomial coefficients. You can turn the minimal-order operator into a minimal-degree operator if you are willing to pay the price of a higher order. There is in general no operator which has both order and degree as small as possible.
As long as you are only interested in the singularities, i.e., you are only concerned about avoiding extra factors in the leading coefficient, you can use a desingularization algorithm to turn the minimal order operator into one which has no unnecessary singular points. For differential operators, this is a classical technique, explained for example in the ODE book of Ince (Section 16.4). For recurrence operators, there is an algorithm by Abramov and van Hoeij (http://www.math.fsu.edu/~hoeij/papers/issac99/new.pdf). 
If you are interested in minimizing the degree not only of the leading coefficient, but of all the polynomial coefficients in the operator simultaneously, see my joint paper with Chen, Jaroschek, and Singer, to appear on this year's ISSAC (http://www.risc.uni-linz.ac.at/people/mkauers/publications/jaroschek13.pdf).
Best regards,
Manuel<|endoftext|>
TITLE: More on Kunen's inconsistency result
QUESTION [7 upvotes]: I would like to suggest another argument for Kunen's inconsistency result, and I wonder to know if the argument is correct. I am also interested to see, if the proof is correct, which part of the argument uses AC.
Theorem.
There is no non-trivial elementary embedding $j: V \rightarrow V.$
Proof. Assume on the contrary that there is a non-trivial elementary embedding $j: V \rightarrow V$ with $\kappa=crit(j).$ Iterate it to get $\langle \langle M_n: n\leq \omega \rangle, \langle j_{n,m}: n\leq m\leq \omega  \rangle \rangle$ where $\langle M_{\omega}, \langle j_{n, \omega}: n\leq \omega  \rangle \rangle =dirlim \langle \langle M_n: n<\omega \rangle, \langle j_{n,m}: n\leq m< \omega  \rangle \rangle.$ Note that $j=j_{0,1}$ and that for all $n<\omega, M_n=V.$ Let $\kappa_n = j_{0, n}(\kappa), n\leq \omega.$ Then $\kappa_{\omega}=sup_{n<\omega}\kappa_n.$ Also we have $M_{\omega}=V$ and we get a contradiction, since $\kappa_{\omega}$ must be regular in $M_{\omega}=V.$

REPLY [10 votes]: This is an interesting idea, but I don't believe that your
argument succeeds.
Notice first that if it were correct, then it would also refute
the existence of $I_1$ rank-into-rank cardinals, that is, $j:V_{\lambda+1}\to V_{\lambda+1}$, since
$\lambda=\kappa_\omega$ and the singularity of this cardinal is
revealed in $V_{\lambda+1}$.
Some people might object that you haven't said what you mean by iterating $j$, but this is not actually a problem, for there is a natural way to do this. Any class embedding $j:V\to M$ can be iterated, since the
definition of $j$ can be extended to any proper class $A$ by
defining $j(A)=\bigcup_\alpha j(A\cap V_\alpha)$. This way, we get
$j(j)$, which will be an elementary embedding from $M$ to $j(M)$.
Thus, if we start with $j:V\to V$, we may indeed iterate
$j$ to form a system $V\to V\to V\to\cdots$ as you describe.
The problem is that you claimed that the direct limit of this system is $V$, but I
don't see why this should be true; this is a gap in
your proof. It may seem reasonable to suppose that the
direct limit of a system of structures, all of which are the same,
is again that same structure, but in fact there are counterexamples to
this general principle. (For a quick example, embed the discrete order
$\mathbb{Z}$ into itself by stretching by a factor of two; the
direct limit of the iteration of this process is the dense order
$\mathbb{Q}$.)
Indeed, let me argue directly that the direct limit is definitely
not $V$. The reason is that the direct limit model has only
$\lambda$ many subsets of $\lambda$, where
$\lambda=\kappa_\omega$. This is simply because every subset of
$\lambda$ in the direct limit is born at some stage $M_n$ as a
subset of $\kappa_n$, having the form $j_{n,\omega}(A)$ for some
$A\subset\kappa_n$. But there are only $\lambda$ many $A$ like
that, and so the limit model is missing most subsets of $\lambda$.
Note that the limit of the iteration is well-founded, by
essentially the usual Kunen argument. Namely, if it were
ill-founded, then we may consider the least $\gamma$ with
$j_{0,\omega}(\gamma)$ in the ill-founded part of the limit. By
pushing $\gamma$ into $M_n$, we may conclude on the one hand that
$j_{0,n}(\gamma)$ must be least sent into the ill-founded part,
but on the other hand we may also drop down to smaller ordinal
born at that stage still in the ill-founded part, making a
contradiction. (To formalize this precisely, one may avoid the
complications caused by $j$ being a proper class simply by
chopping off $j$ to the set $j\upharpoonright V_\alpha$, such that
the iteration whose iteration is ill-founded.) This argument relies fundamentally on the fact that the iteration of $j$ in $M_0$ produces the same limit as the iteration of $j(j)$ in $M_1$. 
Finally, if on the other hand you had formed your iteration by using the same
embedding $j_{n,n+1}=j$ at each stage, then this is clearly
ill-founded, since the threads starting with $\kappa$ at each
stage form a descending sequence of ordinals in the limit.<|endoftext|>
TITLE: Finite vertex-transitive graphs that look like infinite vertex-transitive graphs
QUESTION [13 upvotes]: For a vertex-transitive graph $G$ and a positive integer $d$, and let $G(d)$ be the subgraph induced by all vertices of $G$ within distance $d$ of some given vertex $v$ (since $G$ is vertex-transitive, this doesn't depend on $v$). 
Given an infinite (but locally-finite) graph $G$, does there always exist a finite vertex-transitive graph $G_d$ such that $G(d)$ is isomorphic to $G_d(d)$ (intuitively, a neighborhood of radius $d$ in the infinite graph "looks like" a neighborhood of radius $d$ in the finite graph)? If so, are there any good methods for constructing such a graph (given some nice representation of $G$, say as a Cayley graph of some infinite group). It seems that if $G$ is the Cayley graph of some group (which is true by Sabidussi's theorem), then it might be possible to take some presentation of this group and add some extra relations so it becomes finite, but I haven't been able to make this method work.
One example of a graph $G$ where such a family of graphs $G_d$ exists are lattice graphs $\mathbb{Z}^m$ (where points are connected if they are at unit distance); these are similar to toroidal grid graphs $(\mathbb{Z}/d\mathbb{Z})^m$. I can also construct other such graphs $G$ by "adding edges" to $\mathbb{Z}^m$ (by say, connecting all points at distance < 5). Barring a positive answer to the above question, are there any other graphs $G$ where such a family exists?

REPLY [8 votes]: Several comments. 
First, it is wrong to interpret the Sabidussi theorem as claiming that vertex transitive graphs are necessarily Cayley graphs. Sabidussi's theorem contains an additional condition of existence of a simply transitive action. Without this assumption the claim is false. Eskin, Fisher and Whyte recently proved that so-called Diestel-Leader graphs are not even quasi-isometric to any Cayley graph of a finitely generated group in spite of being vertex transitive.
Second. The group of isomorphisms of vertex transitive graphs which satisfy your approximation property must necessarily be unimodular, so that any vertex transitive graph whose group of isomorphisms is not unimodular provides a counterexample. Such graphs do exist; probably the simplest is the so-called "grandfather graph". The Diestel-Leader graphs used by Eskin et al. also have non-unimodular group of isomorphisms.    
Third. As it was pointed out by HW, for edge-labelled Cayley graphs finitely presented groups which are not residually finite provide a counterexample (it is, indeed, less clear what happens for unlabelled Cayley graphs). Note also that your approximation is way stronger than the one required in the definion of sofic groups. In particular, existence of such approximation would imply that the considered group is sofic. Whether any finitely generated group is sofic is currently unknown, but most people believe that there is no reason for that.<|endoftext|>
TITLE: Cancellation law for $M^n\times \mathbb R= N^n\times \mathbb R$.
QUESTION [7 upvotes]: Assume $M^n$ and $N^n$ are null bordant, i.e. each can be realized as boundary of an $n+1$ dimensional manifold. Suppose $M^n \times \mathbb R$ is homeomorphic to $N^n\times \mathbb R$. Is there any example shows that $M$ is NOT homeomorphic to $N$?

REPLY [18 votes]: [Edit: I have added some details and a more explicit example by Milnor.]
I will present a couple of examples verifying the conditions required in the question.$\newcommand{\RR}{\mathbb{R}}
\newcommand{\TT}{\mathbb{T}}
\newcommand{\ZZ}{\mathbb{Z}}$
$H$-cobordant manifolds
We will make fundamental use of the following fact: if $M$ and $N$ are $h$-cobordant, closed, smooth manifolds of dimension greater than 3, then $M\times\RR$ is diffeomorphic to $N\times\RR$. [Aside: A similar statement holds for topological manifolds; simply replace diffeomorphism with homeomorphism.] Here are a couple of quick references for this fact, both making use of the $s$-cobordism theorem:

In a related mathoverflow question, Oscar Randal-Williams provides a fairly direct argument in a comment to one of the answers.
The article "Contact manifolds with symplectomorphic symplectizations" states the result for smooth manifolds as corollary 2.5. [The proof described there applies also to topological manifolds.]

[Further aside: Incidentally, the converse also holds — although we will not make use of it. In another mathoverflow question, Igor Belegradek remarks in his answer and the ensuing comments that if $M\times\RR$ is homeomorphic to $N\times\RR$, then $M$ and $N$ are $h$-cobordant.]
So it suffices to find closed smooth manifolds $M$ and $N$ which are $h$-cobordant, yet not homeomorphic. As required by the question, we also want $M$ to be null-cobordant; then $N$ will be null-cobordant as well. As a minor bonus, since $M$ and $N$ are smoothly $h$-cobordant, $M\times\RR$ is actually diffeomorphic to $N\times\RR$.
An example by Farrel and Hsiang
One such example is given by Farrel and Hsiang in their article "$H$-cobordant manifolds are not necessarily homeomorphic". There, the manifold $M$ is a product $M=L\times\TT^n$, where $L$ is any 3-dimensional lens space such that $\pi_1(L)\simeq\ZZ/p^2$ for some prime $p$, and $\TT^n=(S^1)^{\times n}$ is a torus of dimension $n\geq 3$. Note that $M$ is null-cobordant since the torus $\TT^n$ is null-cobordant. Unfortunately, the methods used by Farrel and Hsiang do not yield an explicit description of the manifold $N$.
Milnor's examples
A more explicit example is given in the celebrated article "Two complexes which are homeomorphic but combinatorially distinct" by John Milnor. Milnor essentially proves that we can take $M=L(7,1)\times S^{2n}$ and $N=L(7,2)\times S^{2n}$ for any $n>0$, where $L(p,q)$ denotes the lens space of type $(p,q)$. This is only partially stated as theorem 4 in that article. By the way, observe that $M$ and $N$ are null-cobordant since spheres are null-cobordant.
I will briefly describe how to conclude from Milnor's article that the preceding $M$ and $N$ meet our requirements. In section 2, Milnor sketches a proof that $M$ and $N$ are $h$-cobordant. At the end of section 4, in the proof of corollary 2, he uses the fact that $L(7,1)$ and $L(7,2)$ have distinct Reidemeister torsions, and concludes that $M$ and $N$ also have distinct Reidemeister torsions, and thus are not diffeomorphic. But more is true: since the Reidemeister torsions of $M$ and $N$ differ, $M$ and $N$ are not homeomorphic. This follows from the topological invariance (i.e. invariance under homeomorphisms) of Reidemeister torsion, which is itself a simple consequence of the topological invariance of Whitehead torsion — proved by Chapman after the publication of Milnor's article.
In general, the results in sections 2 and 4 of Milnor's article actually prove that we can take $M=K\times S^n$ and $N=L\times S^n$, where:

$K$, $L$ are parallelizable, smooth, closed $k$-manifolds;
$K$, $L$ are homotopy equivalent;
$n$ is even and $n\geq k$;
the Reidemeister torsions of $K$ and $L$ are defined (for some complex-valued representation of their common fundamental group), and have different absolute values.

For instance, we can take $K$ and $L$ to be 3-dimensional lens spaces, and not just $L(7,1)$ and $L(7,2)$. The homotopical classification and Reidemeister torsions of lens spaces are well-known: see http://www.map.mpim-bonn.mpg.de/Lens_spaces.<|endoftext|>
TITLE: How many polynomial Morse functions on the sphere?
QUESTION [27 upvotes]: Let $f$ be a homogeneous polynomial of degree $d$ in $n$ variables. Restricted to the unit sphere $S^{n-1}$, it might or might not be a Morse function.
If $f$ is a Morse function of degree $1$, you get everyone's favorite Morse function on the sphere, with two critical points, corresponding to the minimal CW decomposition.
If $f$ is a Morse function of degree $2$, you get everyone's second-favorite Morse function, corresponding to the minimal regular CW decomposition with two points, two lines, two faces, and so on.
If $f$ has degree $d\geq 3$, there are a lot more possibilities. How many?
The space of degree $d$ homogeneous polynomials in $n$ variables can be identified with $\mathbb R^N$ where $N={\left( \begin{array}{c}n+d-1 \\ n-1\end{array}\right)}$. The ones that are Morse functions form an open subset.

About how many connected components does this open subset have for large $d$ and/or $n$?

REPLY [29 votes]: I've been trying to answer this question for several years and it turned out to be really  hard, even for the $2$-sphere.   Below I will discuss this case. 
First of all  one should  ask   what  is the number $m(k)$ of topological types  of (stable) Morse functions on $S^2$  with  precisely $k$ saddle points. (such a function has $2k+2$ critical points.)  I showed that the generating series
$$ x(t) := \sum_{k\geq 0} \frac{m(k)}{(2k+1)!} t^{2k+1}, $$
is the inverse  of an elliptic integral; see  this paper.  More precisely $x(t)$ is the inverse of the function
$$ x\mapsto t(x)=\int_0^x \frac{ds}{\sqrt{s^4/4-s^2-2sx+1}} ds. $$
This fact   leads to a positive answer to a question of V.I. Arnold   who conjectured that
$$\log m(k)\sim 2k\log k $$
as $k\to \infty $. I refer you to  this paper for details.   This shows that $m(k)$ grows rather fast as $k\to \infty$.
Any   polynomial $P$ of degree $d$ in $\newcommand{\bR}{\mathbb{R}}$ on $\bR^n$   can  be uniquely decomposed  as a sum
$$ P= \sum_{0\leq j+2k\leq d} r^{2k} H_{j}, \;\; r^2= (x_1^2+\cdots +x_n^2), $$
where $H_{j}$ is a darmonic polynomial of degree  $j$.      On $\bR^3$ the space of degree $d$ hormonic polynomials has dimension  $2d+1$.   If we denote by $U_d$ the subspace of $C^\infty(S^2)$ consisting of the restrictions to $S^2$   of the  polynomials of degree $\leq d$  we deduce that
$$\dim U_d=\sum_{0\leq k\leq d} (2k+1)=(d+1)^2. $$
Denote by $K_d$ the expected number of critical points of a random function in $U_d$. I showed  that 
$$ K_d\sim C\dim U_d\sim Cd^2 $$
as $d\to \infty$ where $C$  is a certain explicit constant; see this paper and this paper.   
It turns out that the   number of critical points of a random function in $U_d$ is     highly concentrated around its mean $K_d$, i.e., the probability that the number of critical points of a random function  in $U_d$ is far from the mean  $K_d$ is extremely small as $d\to\infty$.  In more precise technical terms, the variance of the  (random) number of critical points of a (random) function in $U_d$  has the same size  as $K_d$, which makes the standard deviation  of size $\sqrt{K_d}$, much, much smaller than $K_d$ for $d$ large.   
I personally believe, based on some empirical evidence,   that the mean  is close to the  maximum number of critical points    in the sense  that if  we denote by $\mu_d$ the maximum number of critical points of a  Morse function in $U_d$, then $\mu_d \sim C'' d^2$ as $d\to\infty$.
My guess is that the number of topological types of functions in $U_d$  as $d\to \infty$ is roughly
$$ \sum_{k=1}^{K_d/2} m(k), $$
where I recall that $m(k)$ denotes the number of topological types  of Morse functions with $k$ saddle points, i.e., $2k+2$ critical points.<|endoftext|>
TITLE: Quasinilpotent elements of group C-star algebras
QUESTION [11 upvotes]: If $G$ is a discrete torsion-free group, can its (reduced or full) group C-star algebra contain non-zero quasinilpotent elements? I've seen various examples in the group von Neumann algebra setting (usually in the context of finding quasinilpotent generators of II-1 factors) but if I remember correctly these usually don't belong to the reduced group C-star algebra.
Can we get such examples for free groups, for instance?
(My reason for assuming torsion-free is that if $G$ contains a finite non-abelian subgroup $H$, then the C-star algebra generated by H will contain a non-trivial matrix algebra and hence will contain nilpotent elements.)

REPLY [10 votes]: Prompted by Douglas Somerset's answer to look harder, I've found a paper of Behncke that also gives what I need, according to MathSciNet, by proving a stronger result:
MR0283582 (44 #813)
Behncke, H.
Nilpotent elements in group algebras. (Russian summary)
Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 19 (1971) 197-198.

"The aim of this note is to prove the following theorem: The group algebra $L^1(G)$ of a locally compact group $G$ has nilpotent elements if and only if $G$ is non abelian."

(I haven't yet got to the library to check the proof; in particular, I don't know if it bootstraps from Kaplansky's result as mentioned in the other answer, or mimics the proof in a different setting, or uses something more closely related to the group structure.)<|endoftext|>
TITLE: easter problem - egg shapes
QUESTION [5 upvotes]: Inspired by an exceptionally silly article in today's newspaper I pose the following "egg parametrization problem".

Give an explicit function $ f(x,y,t) : \mathbb{R}^2\times I \to \mathbb{R}$ such that for each $t$ from interval $I$ the solution set of equation $f(x,y,t) = 0$ looks like an egg. 

I'm looking for function that provides most of the various egg shapes found in nature.

REPLY [10 votes]: Here is a nice Mathematica Demonstration (incidentally colored according to the Riemann zeta function), with the shape based on the mathematics of egg shapes described at an egg curves website www.mathematische-basteleien.de:
           
One of the neat constructions explained there is the Gardener's Egg:<|endoftext|>
TITLE: Diagonalizing a Complex Symmetric Matrix
QUESTION [16 upvotes]: For a symmetric matrix M with complex entries, I want to diagonalize it using a matrix A, such that
$AMA^T = D$, where  D is a diagonal matrix with real-positive entries.
Question 1: When can this be done?
Question 2: Is $A$ unitary, i.e., is $A^\dagger A = 1$ ?
Question 3: How do I construct $A$?
The question is motivated by Majorana masses of fermions, which are complex symmetric matrices, and need to be diagonalized as above to get the physical masses. Obviously masses need to be positive and the basis-rotation by $A$ must preserve probabilities, and needs to be unitary.

REPLY [3 votes]: Since no one said it, without the condition of question 2 this can always be done. Choose a vector $v$ such that $v^T M v\neq 0$. If this can't be done then the matrix is symmetric and symplectic, hence zero, hence diagonal. If it can, scale $v$ so the product is positive real, restrict to the kernel of $v^T M$, and apply induction.<|endoftext|>
TITLE: Morphisms $\mathrm{SL}_n(\mathbb Z) \to \mathrm{SL}_m(\mathbb Z)$
QUESTION [5 upvotes]: From a result Obtained by O. Schreier and B. L. van der Waerden [Math. Sem. Univ. Hamburg 6, 303- 322 (1928)], one can show that for two fields $\mathbb F$ and $\mathbb G$, and integers $n>m>2$, the only group homomorphism $\mathrm{SL}_n(\mathbb F) \to \mathrm{SL}_m(\mathbb G)$ is the trivial homomorphism.
My question is: Is there a non trivial group homomorphism $\mathrm{SL}_n(\mathbb Z) \to \mathrm{SL}_m(\mathbb Z)$ with $n>m>2$?
Thanks.

REPLY [5 votes]: $SL_n(\mathbb Z)$ is generated by elementary matrices. In general let $E_n(R)$ be the subgroup of $SL_n(R)$ generated by elementary matrices. Let's show that if $m\lt n$ and $n\ge 3$ then the only homomorphism $E_n(R)\to GL_m(F)$ is the trivial one, if $F$ is a field. Wlog $F$ is algebraically closed.
Let $T_n(R)\subset E_n(R)$ be the group of strictly upper-triangular matrices. It is generated by elementary matrices $e_{ij}(a)$ where $1\le i\lt j\le n$ and $a\in R$. Let's first show that every homomorphism $h:T_n(R)\to GL_m(R)$ takes $e_{1n}(a)$ to a scalar matrix. This is by induction on $n$. The element $e_{1n}(a)$ is central in $T_n(R)$, so the image of $h$ is contained in the centralizer of $h(e_{1n}(a))$. If $h(e_{1n}(a))$ is not a scalar, then by conjugating we can assume the image of $h$ is contained in $GL_{m_1}(F)\times GL_{m_2}(F)$ for some positive $m_1$ and $m_2$ adding up to $m$. (EDIT: I have changed the next few sentences to correct an error.) Or rather, that the image of $h$ is contained in the subgroup of $GL_m(F)$ preserving $F^{m_1}\times 0\subset F^{m_1}\times F^{m_2}=F^m$. By induction on $n$ the images of $e_{1,n-1}(a)$ under the resulting homomorphisms to $GL_{m_1}(F)$ and $GL_{m_2}(F)$ are central. (To begin the induction, if $n=3$ then $m_1$ and $m_2$ are both $1$ so this step still is OK.) So $h$ takes $T_n(R)$ into some conjugate of $T_m(F)$. Since $n>m$, it follows that $e_{1n}(a)$, being an $(n-1)$-fold commutator in $T_n(R)$, to the identity after all. 
In particular a homomorphism $E_n(R)\to GL_m(F)$ must take $e_{1n}(a)$ to a scalar matrix. Of course more generally it must take $e_{ij}(a)$ to a scalar matrix. Since $n\ge 3$, we can express $e_{ij}(a)$ as the commutator of $e_{ik}(a)$ and $e_{kj}(1)$, and it follows that $e_{ij}(a)$ goes to the identity.<|endoftext|>
TITLE: Which group extensions are normal?
QUESTION [6 upvotes]: Consider a central extension
$$1 \longrightarrow \mathbb{Z} \longrightarrow G \longrightarrow Q \longrightarrow 1$$
with Euler class $\zeta \in H^2(Q;\mathbb{Z})$.  Let $Q'$ be a normal subgroup of $Q$ and let $\zeta' \in H^2(Q';\mathbb{Z})$ be the restriction of $\zeta$.  Finally, consider some $\zeta'' \in H^2(Q';\mathbb{Z})$ such that there exists some $n \geq 1$ with $n \cdot \zeta'' = \zeta'$.  Corresponding to $\zeta''$, there is a central extension
$$1 \longrightarrow \mathbb{Z} \longrightarrow G'' \longrightarrow Q' \longrightarrow 1$$
which fits into a commutative diagram
$$\begin{array}{ccccccccc}
1 & \longrightarrow & \mathbb{Z} & \longrightarrow & G'' & \longrightarrow & Q' & \longrightarrow & 1\\
  & & \downarrow \times n & & \downarrow & & \downarrow & & \\
1 & \longrightarrow & \mathbb{Z} & \longrightarrow & G & \longrightarrow & Q & \longrightarrow & 1
\end{array}$$
The group $G''$ is thus a subgroup of $G$.
Question : What assumptions can I place on $\zeta''$ which would ensure that $G''$ is a normal subgroup of $G$?

REPLY [3 votes]: As a set, $G' = Q' \times \mathbb{Z}$, with the group operation given by $(p,x)(q,y) = (pq, x + y + \zeta'(p,q))$, and $G'' = Q' \times n\mathbb{Z}$ (again, as a set) with group operation $(p,na)(q,nb) = (pq,n(a + b + \zeta''(p,q)) = (pq,na + nb + \zeta'(p,q))$. 
Now, let $p \in Q'$, consider the element $(p,na) \in G''$, and conjugate it by an arbitrary element of $G$, say $(q,x)$ where $q \in Q$. We have
$(q,x)(p,na)(q,x)^{-1} = (qp,x + na + \zeta(q,p))(q^{-1}, - x - \zeta(q,q^{-1}))$
The first coordinate is then $qpq^{-1}$ which is in $Q'$ since $Q'$ is normal in $Q$ and $p \in Q'$. The second coordinate is
$x + na + \zeta(q,p) - x - \zeta(q,q^{-1}) + \zeta(qp,q^{-1})$
So the condition you need is exactly that for any $p \in Q'$ and any $q \in Q$, $\zeta(q,p) - \zeta(q,q^{-1}) + \zeta(qp,q^{-1})$ is divisible by $n$. I'm pretty sure this condition is necessary and sufficient (even if it's not exactly a condition on $\zeta''$, as requested).<|endoftext|>
TITLE: Functorality of universal central extension
QUESTION [11 upvotes]: I'm looking a reference (or quick proof) for the following fact, which doesn't appear in the standard sources I've consulted (for instance, Milnor's book on algebraic k-theory).
Let $\phi : G \rightarrow G'$ be a homomorphism between perfect groups.  Set $A = H_2(G)$ and $A' = H_2(G')$, and let
$$1 \longrightarrow A \longrightarrow \tilde{G} \longrightarrow G \longrightarrow 1$$
and
$$1 \longrightarrow A' \longrightarrow \tilde{G}' \longrightarrow G' \longrightarrow 1$$
be the universal central extensions.  Then there exists a unique homomorphism $\tilde{\phi} : \tilde{G} \rightarrow \tilde{G}'$ lifting $\phi$.  Also, the restriction $A \rightarrow A'$ of $\tilde{\phi}$ to $A$ is the map $A \rightarrow A'$ induced by $\phi$.

REPLY [11 votes]: Two sources come to mind, though there may be more recent ones.   These are both available online now.
1) A concise exposition of central extensions is given by Steinberg in Section 7 of  his 1967-68 Yale lectures on Chevalley groups, distributed in mimeographed form and currently linked on his UCLA homepage here.   Item (ix) at the bottom of page 76 is what you need.  The proof is elementary, and standard, but note an obvious typo on page 77: the first handwritten symbol $\psi$ should have been $\pi$. 
Originally Steinberg developed these ideas mainly in order to justify lifting of projective representations of finite Chevalley groups to ordinary linear representations of such groups of "universal" type.   But along the way he worked out in considerable generality the algebraic analogues of topological connectedness and simple connectedness: the former imitated by a group being perfect (equal to its derived group) and the latter by a group being a universal central extension of a given group (possible only when the given group is perfect).
2) An even more concise account can be found in the important 1968 paper by Calvin Moore here.  See his statement in the middle of page 10.
As indicated by the title Group extensions of $p$-adic and adelic linear groups, Moore came to these ideas from a rather different direction.   But his approach and Steinberg's soon merged in the work of Matsumoto and others on the Congruence Subgroup Property in the case of Chevalley (split) groups over number fields.
P.S. Concerning your last line, the formalism for the algebraic "fundamental group" varies (homology vs. cohomology), but I think the treatments of central extensions given by Steinberg and others yield immediately what you want.   Work on the CSP tends to emphasize the cocycle approach here, featuring Steinberg cocycles in the special cases of interest.   (There is some exposition with references in the last part of my 1980 Arithmetic Groups, Springer Lecture Notes 789.)<|endoftext|>
TITLE: Interpretability and consistency strength
QUESTION [7 upvotes]: I have heard there is some fairly recent result showing that whenever theories $T$ and $T'$ have the same consistency strength, then each can interpret the other.  I suppose it refers to first order theories, and I do not know exactly what kind of interpretability it uses or what measure of consistency strength.
Can anyone give me the result, or point me in a good direction?

REPLY [3 votes]: John Steel spoke at the EFI series at Harvard concerning his ideas on The triple helix, which has to do in part with the interplay of large cardinal strength and the arithmetic interpretability hierarchy. This is not exactly the claim you mention, but there is a family resemblance.<|endoftext|>
TITLE: Is it true that the geodesics on SO(n) and SU(n) are closed?
QUESTION [6 upvotes]: I mean for the bi-invariant metric (but actually any metric would work). In this metric geodesics are translates of 1-parameter subgroups so we need only to show that $exp(t X)$ for any X in the lie algebra is a closed curve. Then we can use the standard forms (like the Jordan form) for matrices. My source of doubt comes from the fact that these groups don't seem to be on the list of Riemannian manifolds with periodic geodesic flow.

REPLY [10 votes]: Clearly there are geodesics which are not periodic. Take the maximal torus of say $SO(4)$, and let 
$$ 
X = \begin{pmatrix} J & 0 \\
 0 & \alpha J \end{pmatrix} 
$$
be in block diagonal form, where $J = \begin{pmatrix} 0 & 1  \\ -1 & 0 \end{pmatrix}$, with $\alpha$ irrational. Then the geodesic generated by $X$ will be dense in the set of 2x2 block diagonal elements of $SO(4)$, but is not the whole set, hence can't be closed.<|endoftext|>
TITLE: Do different Dehn fillings produce homeomorphic 3-manifolds ?
QUESTION [8 upvotes]: Hi, everyone.  I am interested in the dehn filling and Hyperbolic 3-manifold. 
Suppose M be an orientable compact 3-manifold with one torus boundary and int(M) admit a 
hyperbolic structure.  Thurston proved that almost all the dehn fillings produced 
hyperbolic 3-manifolds. 
My question is:
Among all the hyperbolic dehn fillings,  is there possible that two different hyperbolic dehn fillings produce same 3-manifold, for some M? Is it true for all the one cusp hyperbolic 3-manifolds? 
If the answer to the question is positive, is there an theorem describing this thing?
Any answer and reference are welcome! Thanks!

REPLY [4 votes]: To add even more details to Richard Kent's wonderful answer, the question of cosmetic surgeries showed up as Problem 1.81 on Kirby's list. 
Although there are no non-trivial knot complements in $S^3$, that admit non-trivial $S^3$ surgeries the Blieler, Hodgson and Weeks paper that Richard mentions gives knot complements in lens spaces that have this property. In earlier work, Gabai and Berge independently found infinitely many examples of hyperbolic knot complements in $S^1 \times D^2$ that admit at least one (and sometimes two) non-trivial $S^1 \times D^2$ fillings. 
A good reference for some recent advances on the cosmetic surgery problem is Yi Ni and Zhongtao Wu's work here. For knots in $S^3$, they are able place some restrictions on what pairs of rational slopes can admit, truly cosmetic (sometimes called purely cosmetic) surgeries. 
In particular, they have a succinct statement of the cosmetic surgery conjecture that they attribute to the authors of Kirby's problem list. Namely:
If $M \not\cong S^1 \times D^2$, then for any knot $\gamma \subset M$, $M-\gamma$ does not admit any truly cosmetic fillings.<|endoftext|>
TITLE: Automorphism classes of the free group
QUESTION [5 upvotes]: As is well known, the conjugacy classes of the free group $F_2$ are parametrised by cyclically reduced words, up to cyclic permutation. In particular, it's easy to tell whether two elements of $F_2$ are conjugate.
What about the automorphism classes of $F_2$? For $u,v\in F_2$ write $u\sim v$ if there is an automorphism of $F_2$ mapping $u$ to $v$. Is there a similarly simple description of a representative from each $\sim$ class?
For example, if $F_2 = \langle x,y\rangle$ then $xyxyxy\sim xxx$ while $xyxyxyx\sim x$.
In particular, is it easy to tell whether $u\sim v$ for general words $u$ and $v$?

REPLY [5 votes]: "Let $w_1$ and $w_2$ be elements of a free group $F$. Then it is decidable 
whether there is an automorphism of $F$ carrying $w_1$ into $w_2$."
(R.C.Lyndon, P.E.Schupp, Combinatorial Group Theory, Chapter I, Prop.4.19)
Is this an answer on your question? I think it's hard to get something more specific.<|endoftext|>
TITLE: Naive question about the representation theory of algebraic groups and hopf algebras
QUESTION [6 upvotes]: I have been learning some representation theory and have some questions about the following pattern:
 Instance 1:  If we have a finite group $G$ and a field $k$, a representation of $G$ over $k$ consists of a finite dimensional $k$-vector space $V$ and a homomorphism $G \to GL(V)$. We have a Hopf algebra $kG$ and an equivalence between representations of $G$ over $k$ and finite dimensional $kG$-modules.
 Instance 2: If we have a Lie group $G$, a representation of $G$ consists of a finite dimensional $\mathbb{R}$-vector space $V$ and a smooth homomorphism $ G \to GL(V) $. Let $\mathfrak{g}$ be the lie algebra of $G$. Then we have a Hopf algebra $U(\mathfrak{g})$ and an equivalence between representations of $G$ and finite dimensional $U(\mathfrak{g})$-modules.


 Question:  If $G$ is an algebraic group over some field $k$, is there a Hopf algebra floating around?


Thanks!

REPLY [2 votes]: There is a direct way to define group algebras for different classes of groups (locally compact groups, Lie groups, algebraic groups, etc.). 
For an (affine) algebraic group $G$ one should take the algebra ${\mathcal R}(G)$ of polynomials (i.e. regular functions) on $G$ and then consider the dual algebra ${\mathcal R}^\star(G)$ of linear functionals on ${\mathcal R}(G)$ with the natural topology (in this case the topology of poitwise convergence on ${\mathcal R}(G)$). Then ${\mathcal R}^\star(G)$ becomes a (topological) group algebra for $G$: each  "polynomial representation" of $G$ corresponds bijectively to a continuous representation of ${\mathcal R}^\star(G)$. At the same time both ${\mathcal R}(G)$ and ${\mathcal R}^\star(G)$ are Hopf algebras in the monoidal category of stereotype spaces, see details in: S.S.Akbarov, Pontryagin duality in the theory of topological vector spaces and in topological algebra, Journal of Mathematical Sciences, 113(2): 179-349, 2003 (Theorem 10.12).<|endoftext|>
TITLE: Iterates converging to a continuous map
QUESTION [6 upvotes]: I have no doubt that the following observation is quite well known. Let $\varphi:[0,1]\to [0,1]$ be a continuous map. Assume that the iterates $\varphi^n$ converge pointwise to some continuous map $\varphi_\infty$. Then the convergence is in fact uniform. However, I was unable to locate a reference. Does anybody know where I can find it written? Thanks in advance.

REPLY [8 votes]: I don't know a reference but maybe the following proof is shorter than yours.
By continuity, $\varphi\circ\varphi_\infty=\varphi_\infty$. Hence $\varphi$ is identity on the set $I:=\varphi_\infty([0,1])$. Hence $I$ is the set of fixed points of $\varphi$. And it is compact and connected. Then there are two cases: either $I$ is a nontrivial interval $[a,b]$ or it is a single point $\{a\}$.
In the first case, we have $f(x)<x$ for all $x>b$ and $f(x)>x$ for all $x<a$, otherwise there is a fixed point outside $I$. There is an $\varepsilon>0$ such that $f(x)>a$ for all $x\in(b,b+\varepsilon)$ and $f(x)<b$ for all $x\in(a-\varepsilon,a)$. Let $U=(a-\varepsilon,b+\varepsilon)$. Then $\varphi(U)\subset U$ and for every $x\in U$ we have
$$
|\varphi^n(x)-x|<\delta(\varepsilon)
$$
where $\delta(\varepsilon)\to $ as $\varepsilon\to 0$ ($\delta(\varepsilon)$ is the maximum of the diameters of the sets $\varphi([a-\varepsilon,a])$ and $\varphi([b,b+\varepsilon])$)  For every $x\in[0,1]$, the iterations $\varphi^n(x)$ eventually get into $U$, hence
$$
 \bigcup_n \varphi^{-n}(U) = [0,1] .
$$
By compactness, this implies that $[0,1]$ is covered by finitely many sets $f^{-n}(U)$ and hence there exist $n$ such that $\varphi^n([0,1])\subset U$. After this $n$, all iterations stay within the distance $\delta(\varepsilon)$ from the limit. Since $\varepsilon$ can be arbitrarily small, this implies uniform convergence.
In the second case (when $I$ is a single point $a$), similarly observe that $f(x)<x$ for $x>a$ and $f(x)>x$ for $x<a$, and the same holds for $\varphi^2$, $\varphi^3$, etc. This implies that if e.g. $x>a$ and $f(x)<a$ then all iterations $\varphi^n(x)$ lie between $x$ and $f(x)$. This easily implies that there is an arbitrarily small neighborhood $U$ of $a$ such that $f(U)\subset U$. Then uniform convergence follows similarly to the first case.<|endoftext|>
TITLE: Fubini-Study metric for an infinite dimensional Hilbert space
QUESTION [5 upvotes]: Hi,
Can one define a Fubini-Study metric/Kaehler metric on the projective space of an infinite dimensional Hilbert space, i.e. using the formula $\partial \bar{\partial} \log |Z|^2$? 
This should be very well-known to the experts. Anyhow I don't have much experience with infinite dimension and worried that something may go wrong. I appreciate any comments or references.

REPLY [3 votes]: Maybe it's easier to see that the definition extends if you don't use a formula:
The unit sphere inherits a Riemannian metric from the Hilbert space in the standard manner and since it is invariant under the circular symmetry $(e^{i\theta},x) \mapsto  e^{i\theta} x$, it will
project down to a Riemannian metric on the complex projective space.<|endoftext|>
TITLE: Complete intersection space curves
QUESTION [5 upvotes]: Fix two positive integers $d, e$ and assume $d>e$. Is it true that a general degree $e$ curve which lies in a complete intersection of a degree $e$ and a smooth degree $d$ surface in $\mathbb{P}^3$ is a plane curve (in the sense the radical ideal of the curve contains a linear polynomial)? 
There is another way of formulating this problem. Let $H$ be the Hilbert flag scheme of pairs $(C_1, C_2)$ with $C_1 \subset C_2$, $C_2$ is a complete intersection of a degree $e$ and a smooth degree $d$ surface in $\mathbb{P}^3$ and $C_1$ is of degree $e$.
The question is does there exists an irreducible component $H'$ of $H$ such that a generic element of $pr_1(H')$ is a plane curve in the sense described above (Here $pr_1$ is the natural projection map onto the first coordinate)? 
In the above question we can assume $C_1$ is a local complete intersection curve in $\mathbb{P}^3$.

REPLY [3 votes]: The answer is no in general. The simplest example is for $e=3$ (because a line and a conic in $\mathbb{P}^3$ are plane curves). There are plenty of twisted cubic curves lying on a smooth cubic surface in $\mathbb{P}^3$ and all lie in a complete intersection: take a general quartic passing through the twisted cubic.
If you want to be more precise, we can see a smooth cubic surface $S\subset \mathbb{P}^3$ and view it as the blow-up $\pi\colon S\to \mathbb{P}^2$ of six points in general position. An hyperplane section of $S$ corresponds, by adjunction, to the  anti-canonical divisor $-K_S$, which corresponds, by ramification formula, to a cubic of $\mathbb{P}^2$ passing through the six points. Take then a line of $\mathbb{P}^2$ not passing through the six points. Its intersection with the anti-canonical divisor is $3$, hence it is a cubic curve in $\mathbb{P}^3$. Then, take a curve of degree $11$ passing through the $6$ points with multiplicity $3$, and its strict transform on $S$, added to the twisted cubic, will be the complete intersection of $S$ with a quartic. If you wonder if such curve of degree $11$ exists, just take union of lines and conics.<|endoftext|>
TITLE: Bijection from $\mathbb{R}$ to $\mathbb{R^2}$
QUESTION [13 upvotes]: Importantly, I am looking for a constructive proof (which does not rely on the Cantor–Bernstein–Schroeder theorem). Motivated by this discussion.

REPLY [13 votes]: Here is a bijection that uses the decimal (or binary, whatever) expansion of reals. (Even though I think the approach using continued fractions is more canonical.)

Let $\alpha:\mathbb Z\times \mathbb Z\to \mathbb Z$ and $\beta:[0,1)\times [0,1)\to [0,1)$ be bijections.
Let $f:\mathbb R \to \mathbb Z \times [0,1) $ be the natural bijection $x\mapsto (\lfloor x\rfloor, x-\lfloor x\rfloor)$.
Together, $f$, $\alpha$, $\beta$ will define bijections $$ \mathbb R\times \mathbb R \to^f (\mathbb Z\times [0,1)) \times (\mathbb Z\times [0,1)) \simeq \mathbb Z^2 \times [0,1)^2 \to ^{\alpha,\beta} \mathbb Z \times [0,1) \to^{f^{-1}}\mathbb R$$

The main part is of course the definition of $\beta$.  [EDIT: This is not my construction; I am not sure where I first read it.  In his book on the real numbers, Oliver Deiser gives a very similar construction (blocking zeroes instead of nines) and calls this Julius König's trick.  König's wikipedia page mentions it but omits the details.]
Represent each real number $x\in [0,1)$
as a sequence of DIGITS, where each DIGIT is either in $\{0,\ldots,8\}$ or is of the form $10*(10^k-1)+i$ with $k\ge 1$ and $i\in \{0,\ldots,8\}$ (i.e., in $\{90,\ldots, 98; 990,\ldots, 998; 9990,\ldots, 9998; \ldots\}$.   
For example, the number 0.0129449956$\dots$ would be represented by $(0,1,2,94,4,995,6,\ldots)$.   
Given a pair $(x,y)$, $\beta$ interleaves these two representations.<|endoftext|>
TITLE: The octonions on a bad day
QUESTION [23 upvotes]: We can define the algebra of quaternions $\mathbb H$ over any field $k$, and depending on the arithmetic of $k$ it is either a division algebra or a matrix algebra.
We can also define the algebra of octonions $\mathbb O$ over any field $k$, and if over $k$ the $8$-ary quadratic form $Q=x_0^2+x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2+x_7^2$ is anisotropic, then $\mathbb O$ is again a division algebra —a non-associative one, but oh well.

What happens to $\mathbb O$ if $k$ is such that $Q$ is isotropic?

The classical structure theory of non-commutative Jordan algebras tells us that $\mathbb O$ is, over any field, a direct product of simple flexible power-associative algebras coming from a rather restricted list: simple commutative Jordan algebras, quasiassociative algebras, and flexible quadratic algebras with nondegenerate norm forms (Shafer's book on non-associative algebra explains all this, which is —I'd say— mostly forgotten nowadays) but I am pretty sure one can be very specific about what comes out in the case of octonions. In other words, one can probably find something playing the role of «matrix algebra» in the statement about quaternions.
N.B. All this is over fields of characteristic zero.

REPLY [23 votes]: $\DeclareMathOperator\Tr{Tr}
$Suppose that $k$ is a field with $\operatorname{char}(k) \neq 2$.  Let's agree that an "octonion algebra" over $k$ is an 8-dimensional unital $k$-algebra $A$, endowed with a quadratic form $N: A \rightarrow k$, whose associated bilinear form $T(x,y) = N(x+y) - N(x) - N(y)$ is nondegenerate, and which satisfies $N(xy) = N(x) N(y)$ for all $x,y \in A$.
When $x \in A \setminus k$, the trace $\Tr(x) = T(x,1)$ and norm $N(x)$ are determined by the algebra structure: $x$ is the root of a quadratic polynomial with coefficients $-\Tr(x)$ and $N(x)$.  Hence the algebra structure determines the quadratic form $N$ in a convenient way.  One can talk about the "isomorphism class of an octonion algebra" while carrying around the quadratic form or not; it doesn't really matter.
It is an old theorem (Jacobson?  Albert?  I can't recall … check "The Book of Involutions" too) that the isomorphism class of the octonion algebra $k$ is determined by the isomorphism class of the quadratic form $N$.  Now, essentially by the Cayley–Dickson doubling process, the norm form $N$ is a Pfister form, i.e., $N$ is isomorphic to $\langle1,-a\rangle \otimes \langle1,-b\rangle \otimes \langle1,-c\rangle$ for some $a,b,c \in k$.
It is a fact about Pfister forms that when they are isotropic, they are split.  So if the norm form represents zero, then $N$ is isomorphic to $\langle1,-1\rangle \otimes \langle1,-1\rangle \otimes \langle1,-1\rangle$ and the octonion algebra is isomorphic to the split octonion algebra over $k$.
In this level of generality, the isomorphism classes of octonion algebras over $k$ are classified up to isomorphism by the Galois cohomology $H^3(k, \mu_2)$; you can see such a cohomology class too from the Pfister form perspective.  The Pfister form $\langle1,-a\rangle \otimes \langle1,-b\rangle \otimes \langle1,-c\rangle$ depends only on the square-classes of $a, b, c$, giving three classes in $H^1(k, \mu_2)$ whose cup product is the element of $H^3(k, \mu_2)$ classifying the octonion algebra.<|endoftext|>
TITLE: Simple field extension and rational points
QUESTION [6 upvotes]: Let $F$ be an infinite field and $f$ a homogeneous form on $F$ such that $f$ has no non-trivial zero in $F$. Let $F'$ be a finite extension of $F$ such that $f$ has a non-trivial zero in $F'$. Is it true there exists a simple extension of $F$ of the form $F(\alpha)$ contained in $F'$ which contains a non-trivial zero of $f$?

REPLY [5 votes]: I believe the following is a negative example: Let $s,t,u,v$ be variables over $\mathbb F_p$. Set $F=\mathbb F_p(s,t,u,v)$ and $F'=F(\sigma,\tau)$ with $\sigma^p=s$, $\tau^p=t$.
Set $$f(X,Y,Z)=(X^p-sZ^p)u+(Y^p-tZ^p)v.$$
Then $f(\sigma,\tau,1)=0$.
We show that any solution of $f=0$ over $F'$ has this form up to a scalar factor: Let $x,y,z\in F'$ with $f(x,y,z)=0$. As $F'=\mathbb F_p(u,v,\sigma,\tau)$, we get
$$x^p-sz^p, y^p-tz^p\in\mathbb F_p(u^p,v^p,\sigma^p,\tau^p)=\mathbb F_p(u^p,v^p,s,t),$$
hence
$$A(u^p,v^p,s,t)u+B(u^p,v^p,s,t)v=0$$
for rational functions $A,B$ over $\mathbb F_p$ with $x^p-sz^p=A(u^p,v^p,s,t)$ and $y^p-tz^p=B(u^p,v^p,s,t)$.
This implies $A(u^p,v^p,s,t)=0$, for otherwise $u$ were a rational function in $u^p$. For the same reason $B(u^p,v^p,s,t)=0$. We get $x^p-tz^p=0=y^p-tz^p$, and the claim follows.<|endoftext|>
TITLE: Are there any nontrivial ring homomorphisms $M_{n+1}(R)\rightarrow M_n(R)$?
QUESTION [27 upvotes]: Let $R$ be a finitely generated ring with identity, $M_n(R)$ the set of $n\times n$ matrices. Are there any nontrivial ring homomorphisms $M_{n+1}(R)\rightarrow M_n(R)$? This should be an elementary question in abstract algebra. But even if $R$ is a field, I couldn't get a quick (negative) proof. Any comments are welcomed.
RMK: If we view the natrual map $M_{n}(R)\rightarrow M_{n+1}(R)$ as a ring homomorphism, we will not require that a ring homomorphism preserves identities.

REPLY [5 votes]: Since it is not clear that $R$ must be commutative (none of the tags eliminate this possibility), the finite generation hypothesis (which I interpret as finitely generated as an algebra over a field) is not  significant, in view of the existence of rings for which all finitely generated projective modules are cyclic and free. 
The standard dense subalgebra of the Cuntz C*-algebra is such an example, and is  finitely generated as an algebra over the complexes, and satisfies $M_n R $ isomorphic to $R$ for all $n$, yielding unital maps (isomorphisms) $M_{n+1}R \to M_n R$.<|endoftext|>
TITLE: How did Takagi prove Kronecker's Jugendtraum for Q(i)?
QUESTION [13 upvotes]: In Noah Snyder's historical undergraduate thesis on Artin L-Functions, it mentions that Takagi proved Kronecker's Jugendtraum in the case of Q(i) in his doctoral thesis. Since I don't know how to get access to Takagi's thesis, does anyone have any idea how he did this?
I know how to prove complex multiplication (e.g. as in Silverman's book) by first assuming class field theory. But Takagi was working before class field theory had been demonstrated. So how did he do it?
I understand how one can directly prove the reciprocity law directly for extensions generated by torsion points on elliptic curve. The nontrivial part is this: how does one prove that an arbitrary abelian extension of an imaginary quadratic field arises in such a manner?
Note that this question has everything to do with my previous question.

REPLY [23 votes]: Takagi's goal is the following: 

show the existence of sufficiently many cyclic extensions defined 
  by division values of sn $u$ (the lemniscatic sine);
prove that each abelian extension of ${\mathbb Q}(i)$ is contained in 
  the compositum of these fields.

Step 1 is analogous to the construction of the fields of $p^n$-th roots
of unity (in particular, proving the irreducibility of the corresponding
cyclotomic polynomials and determining their dirscriminants),  whereas
Step 2 is the analogue of the theorem of Kronecker-Weber.
Step 1
Takagi (thesis, Sect. 6) constructs the following cyclic extensions
of ${\mathbb Q}(i)$:

For each Gaussian prime $\mu$ such that $p^h$ is the largest power
  of the prime $p$ that divides $(N\mu-1)$, there exists a cyclic 
  extension unramified outside $\mu$ with degree $p^h$.
For each Gaussian prime $\pi$ with odd prime norm $p \equiv 1 \bmod 4$
  and each integer $\lambda \ge 1$ there exists a cyclic extension of 
  prime power degree $p^\lambda$ unramified outside $\pi$.
For each prime number $q \equiv 3 \bmod 4$ and each integer 
  $\lambda \ge 1$ there exist $q^\lambda +1$ cyclic extensions with
  degree $q^\lambda$ unramified outside $q$. 
For each Gaussian prime $\mu$ with $N\mu-1 = 2^{h+2} u$, where
  $u$ is an odd integer, there is a cyclic extension of degree $2^{h+2}$
  unramified outside $(1+i)\mu$ for which the subfield of degree $2^h$ 
  is unramified outside $\mu$.
For each integer $\lambda > 1$ there are $2^\lambda +1$ cyclic
  extensions of degree $2^\lambda$ unramified outside $1+i$. 

Takagi also proves the decomposition law in these extensions.
Step 2
In Sect. 9, Takagi begins proving the analogue of the 
theorem of Kronecker-Weber. His approach is the one used by Hilbert
in his proof of the theorem of Kronecker-Weber: Given an abelian
extension $L/{\mathbb Q}(i)$, we write $L$ as a compositum of cyclic 
extensions of prime power degree. If the odd Gaussian prime $\mu$ is 
ramified in $L$, form the compositum of $L$ and the cyclic extension 
$K/{\mathbb Q}(i)$ unramified outside $\mu$ constructed in Step 1 and 
show that the compositum $KL$ contains a cyclic extension 
$M/{\mathbb Q}(i)$ unramified at $\mu$ such that $L$ is contained 
in $KM$; the case of wild ramification requires a careful investiagtion 
of the ramification subgroups. 
By induction, this process reduces the proof to cyclic extensions that 
are unramified everywhere. There are various ways of proving that 
such extensions do not exist:

We can use Takagi's Lemma, according to which every cyclic extension 
of ${\mathbb Q}(i)$ that is normal over ${\mathbb Q}$ is a cyclotomic 
field. This is how Takagi proves this claim.  
By Hilbert's Theorem 94, unramified cyclic extensions of prime
degree $\ell$ of a number field $K$ exist only if $K$ has class
number divisible by $\ell$.
Minkowski's bounds show that ${\mathbb Q}(i)$ does not 
admit any nontrivial unramified extension.

P.S. Takagi's thesis is contained in his collected papers. I also would like to refer you to a beautiful article by Cox and Hyde on The Galois theory of the lemniscate.<|endoftext|>
TITLE: On closed simple curve with curvature at most 1
QUESTION [17 upvotes]: I am looking for the reference to the following theorem.
I have to apply a similar statement, and it would be nice to trace the source.
Please note, I know few proofs in fact it is Problem 3 in my collection of exercises. 

Theorem. Let $\gamma$ be a closed simple plane curve with curvature at most 1. Then the  disc bounded by $\gamma$ contains a circle of radius $1$.

Here is an illustration for those who are lazy to read.

REPLY [7 votes]: Here is summary of letter from Yurii Nikonorov, it gives complete answer.
The statement is called Pestov--Ionin theorem:

Пестов Г., Ионин В. О наибольшем круге, вложенном в замкнутую кривую // Докл. АН СССР. – 1959. – 127, № 6. – С. 1170–1172.

An alternative proof via curve shortening flow was given by Pankrashkin:

Pankrashkin, Konstantin An inequality for the maximum curvature through a geometric flow. Arch. Math. (Basel) 105 (2015), no. 3, 297–300.<|endoftext|>
TITLE: Local concentration of measure on Erdos-Rényi graph
QUESTION [6 upvotes]: Let $G_n=(V_n,E_n)$ be an Erdos-Rényi random graph, precisely the vertex set is $V_n=(1,\dots,n)$ and the edge set is $E_n=(ij\in\mathcal{P}_2(V_n)\ |\ \epsilon_{ij}=1)$ where $(\epsilon_{ij})_{ij}$ are i.i.d. random variables with distribution $\textrm{Bernoulli}(c/N)$ .
It is known that $(G_n)_{n\in\mathbb{N}}$ locally converges to tree. I'm trying to prove this fact "by hands". The part which gives me some troubles is the concentration of measure. I state precisely my problem.
Fix $r\in\mathbb{N}$ and $(T,o)$ a finite rooted tree with at most $r$ generations. For any vertex $v\in V_n$ consider $B_{G_n}(v,r)$, the rooted sub-graph of $G_n$ induced by the vertices at graph-distance $\leq r$ from $v$. Consider the following random variable
$$M_n:=\frac{1}{n}\sum_{v=1}^{n}\chi(B_{G_n}(v,r)\equiv(T,o))\ ,$$
where $\chi(A)$ is the indicator function of the event $A$. And consider its expectation
$$\mathbb{E}[M_n]=\frac{1}{n}\sum_{v=1}^{n}\mathbb{P}(B_{G_n}(v,r)\equiv(T,o))\ .$$
I would like to prove that
$$|M_n-\mathbb{E}[M_n]|\xrightarrow[n\to\infty]{}0\ \ a.s.$$
Any idea? I tryed to consider a Doob martingale, exposing vertex by vertex, and to bound its differences in order to apply Azuma-Hoeffding inequality. But did not manage to find a useful bound. Can you help me?
Edit. I write more clearly the way I've tried. Fix $n\in\mathbb{N}$.
Define the filtration which at each step let you know the subgraph of $G_n$ induced by the first $i$ vertices:
$$\mathcal{F_1}:=(\Omega,\emptyset),\ \ \mathcal{F_i}:=\mathcal{F_{i-1}}\cup\sigma(\epsilon_{i1},\epsilon_{i2},\dots,\epsilon_{i(i-1)})\ \ \forall i=2,\dots,n.$$
Then define the following Doob martingale:
$$A_i:=\mathbb{E}[M_n|\mathcal{F_i}]\ \ \forall i=1,\dots,n$$
noticing that $A_n=M_n$ and $A_0=\mathbb{E}[M_n]$.
Now if one finds a $c_n>0$ such that $n\ c_n^2\xrightarrow[n\to\infty]{}0$ (fast enough) and
$$|A_i-A_{i-1}| < c_n\ \ \forall i=2,\dots,n$$
then by the Azuma-Hoeffding inequality one obtains that
$$\mathbb{P}(|M_n-\mathbb{E}[M_n]|>t)\ \leq\ 2\ \exp(-\frac{t^2}{2\ n\ c_n^2})\ \ \forall t>0$$
which allows to conclude thanks to Borel-Cantelli lemma.
The problem is bounding $|A_i-A_{i-1}|$.
Edit2. Let $d$ be the maximum degree of the tree $T$. Note that in general if two graphs $G,\tilde{G}$ differ only for one edge (i.e. $G$ contains a given edge $ij$ while $G'$ does not), then the two sums
$$\phi_G:=\sum_{v=1}^n\chi(B_{G}(v,r)\equiv(T,o))\ \ ,\ \ \phi_{\tilde{G}}:=\sum_{v=1}^n\chi(B_{\tilde{G}}(v,r)\equiv(T,o))$$
may differ at most for $2 \sum_{l=0}^r d^l$ terms (this is an upper bound for the number of vertices $v$ which can be reached from $i$ or $j$ by a walk of lenght $l\leq r$ completely made of vertces with degree $\leq d$).
Now what happens to $|\phi_G-\phi_{\tilde{G}}|$ if instead the graphs $G,\tilde{G}$ differ for one vertex (i.e. for some edges attached to a given vertex $i$)? I fear the previous bound explodes becoming unuseful in the Azuma-Hoeffding inequality... Am I right?
Maybe is there a way to exploit the fact that in a Erdos-Rényi graph the edges are not too much (precisely $|E_n|/n\xrightarrow[n\to\infty]{}c/2\ a.s.$)?

REPLY [2 votes]: Why not use second moment? For fixed $r$, the correlation between the events that $B_{G_n}(v,r)$ is a tree and the same event for $v'\neq v$ is small, of order $f(c)/n$. Try $r=2$ to see what I mean. You can even (since you need only a lower bound) throw into the random variables the event that the maximal degree in $B_{G_n}(v,r)$ is bounded by some function
$g(c,r)$ that grows fast enough.
This argument may  be not strong enough for a.s. limits though.<|endoftext|>
TITLE: Meromorphic continuation of a Dirichlet series associated to an irrational number
QUESTION [9 upvotes]: Let $\theta>0$ be irrational, and define
$$\zeta_\theta(s) = \sum_{n=1}^\infty \frac{\lfloor n \theta \rfloor}{n^s}.$$
This converges to an analytic function on $\Re(s)>2$. Does $\zeta_\theta(s)$ have a meromorphic continuation to $\mathbb{C}$?
Some motivation: I'm interested in zeta functions associated to distance functions on $\mathbb{R}^n$: For a distance function $d$, consider 
$$\zeta_d(s)=\sum_{\mathbf{m}\in \mathbb{Z}^n}' d(\mathbf{m})^{-s}.$$ 
If $d$ is smooth outside $0$, one can show $\zeta_d$ has a meromorphic continuation. If $d(x,y)=\max\{|y|,|x|/\theta\}$, then $\zeta_\theta(s)$ is $d(x,y)^{-s}$ summed over the integral points in the sector $\{(x,y)\ |\ y\theta>x>0\}$. Meromorphic continuation of $\zeta_\theta$ and $\zeta_{\theta^{-1}}$ would then imply the meromorphic continuation of $\zeta_d$.
Based on numerical evidence, I conjecture that for $\theta$ a quadratic irrational, $\zeta_\theta(s)$ has a meromorphic continuation to (at least) a region $\Re(s)>\sigma_0$, where $\sigma_0<0$, and that its poles in $\Re(s)\geq 0$ occur at $s=2,1$ and a subset of $it_0\mathbb{Z}$ for some $t_0\in \mathbb{R}$. A more tentative conjecture is that this is true for all real algebraic numbers.
EDIT: George Lowther gave a nice answer below, but I've since learned that this question was answered by Hecke in 1922:
E. Hecke, Uber Analytische Funktionen und die verteilung von zahlen mod
eins, Hamburg. Math.Abh., 1(1922) 54 - 76

REPLY [7 votes]: Yes, the conjecture is true -- $\zeta_\theta(s)$ extends to a meromorphic function on $\mathbb{C}$ for all real quadratic irrationals $\theta$ with poles at $s=2$, $s=1$ and in the infinite set of vertical lines $\lbrace -2n+i\mathbb{R}\colon n\in\mathbb{Z}_{\ge0}\rbrace$. I'm not sure about the more tentative conjecture for arbitrary algebraic numbers. Also, more simply, it will extend to a meromorphic function over $\Re(s) > 0$ whenever $\theta$ cannot be approximated too well by rationals (in particular, if $\theta$ has irrationality measure 2 which, by the Thue-Siegel-Roth theorem, includes all algebraic irrationals). Conversely, if an irrational $\theta$ can be approximated too well by rationals then there will not be a meromorphic extension, which will be the case outside of a meagre subset of $\mathbb{R}$. I can give a rough proof of these statements now.
[Edit: As Pieter pointed out in the comments, my initial answer carelessly missed out a rather important term, so was not correct. I've updated the proof sketch to deal with this term, and restricted to quadratic irrationals for the meromorphic extension. The proof is a bit sketchy, as I haven't justified that we can rearrange the order of the various summations/integrals. As they are only conditionally convergent, this would need to be fleshed out to make it more rigorous but, thinking about this a bit more, it looks like it all works out with no problems.]
First, let $\lbrace x\rbrace = x - \lfloor x\rfloor$ denote the fractional part of $x$, and write
$$
\tilde\zeta_\theta(s)=\sum_{n\ge1}\frac{\lbrace n\theta\rbrace-1/2}{n^s},
$$
which converges, at least, on $\Re(s) > 1$.
Then, the function $\zeta_\theta$ defined in the question is
$$
\zeta_\theta(s)=\theta\zeta(s-1)-\frac12\zeta(s)-\tilde\zeta_\theta(s),
$$
where $\zeta$ is the Riemann zeta function. So we have an extension to $\Re(s) > 1$. In the case where $\theta$ is rational then $\lbrace n\theta\rbrace-1/2$ is periodic in $n$, so $\tilde\zeta_\theta$ is a linear combination of Dirichlet L-functions and has a meromorphic extension to $\mathbb{C}$. Henceforth, I will only consider irrational $\theta$. Setting $F(x)=\sum_{1\le n\le x}(\lbrace n\theta\rbrace-1/2)$, the equidistribution theorem states that $F(x)/x\to0$ as $x\to\infty$. Rearranginging the expression for $\tilde\zeta_\theta$ gives
$$
\tilde\zeta_\theta(s)=\sum_{n\ge1}\frac{F(n)}{n}n^{1-s}\left(1-(1+1/n)^{-s}\right)
$$
If $\lvert F(n)/n\rvert$ is bounded by some $\delta > 0$ then this expression will be bounded by $\delta\zeta(s)$ for all $s > 1$. So, by equidistribution, it follows that $(s-1)\tilde\zeta_\theta(s)\to0$ as $s\to1$ (on $s > 1$). If a meromorphic extension existed in a neighbourhood of 1 then this would imply that $\tilde\zeta_\theta$ is bounded near 1. However, if $\theta=p/q$ is rational ($p,q$ coprime) then $F(x)/x\to-1/(2q)$ so, if $x$ is too closely approximated by rationals then there will be a slight bias in $F(x)/x$ and $\tilde\zeta_\theta(s)$ will not be bounded close to 1. I think that the existence of rational approximations $p_n/q_n$ with $q_n^2\log\lvert\theta-p_n/q_n\rvert\to-\infty$ is enough for this to happen.
Now, suppose that $\theta$ has irrationality measure no greater than some finite $\gamma$, so that there are only finitely many rational approximations $\lvert\theta-p/q\rvert\le q^{-\gamma-\delta}$ for any $\delta > 0$. Equivalently, there is a constant $C_\delta > 0$ such that $n^{\gamma-1+\delta}\lvert n\theta-m\rvert\ge C$ for all positive integer $n$ and integer $m$. Then, the discrepancy of the set $\left\lbrace \lbrace k\theta\rbrace\colon 1\le k\le n\right\rbrace$ is bounded by $Kn^{-1/(\gamma-1)+\delta}$ (for a constant $K$). This means that $F(n)/n$ is bounded by $2Kn^{-1/(\gamma-1)+\delta}$ and, hence, the expression above for $\tilde\zeta_\theta(s)$ has summand going to zero at rate $O(n^{-s-1/(\gamma-1)+\delta})$. So, $\tilde\zeta_\theta(s)$ is an analytic function on $\Re(s) > (\gamma-2)/(\gamma-1)$. In particular, for algebraic irrationals, $\gamma=2$ and we get a meromorphic extension of $\zeta_\theta(s)$ to $\Re(s) > 0$ with simple poles at $s=2$ and $s=1$.
I now attempt to extend to the whole of $\mathbb{C}$ via Hurwitz's formula which, for $\Re(s) > 0$ and irrational $x$ in the unit interval, gives
$$
\begin{array}{l}
&\sum_{n\ge1}e^{2\pi inx}n^{-s}=(2\pi)^{s-1}\Gamma(1-s)\left(e^{i\pi(1-s)/2}\zeta(1-s,x)+e^{i\pi(s-1)/2}\zeta(1-s,1-x)\right)
\end{array}
$$
where $\zeta(s,x)=\sum_{n\ge0}(n+x)^{-s}$ is the Hurwitz Zeta function.
Plugging in the Fourier series $\lbrace x\rbrace=1/2+\sum_{k\not=0}(i/(2\pi k))e^{2\pi ikx}$, applying Hurwitz's formula, and being careless about when the sums converge/commute (it can be made more rigorous) gives,
$$
\begin{array}{rl}
\tilde\zeta_\theta(s)&=\sum_{n\ge1}\sum_{k\not=0}\frac{ie^{2\pi ink\theta}}{2\pi kn^s}\cr
&=(2\pi)^{s-2}\Gamma(1-s)\sum_{k\not=0}\frac{i}{k}\left(e^{i\pi(1-s)/2}\zeta(1-s,\lbrace k\theta\rbrace)+e^{i\pi(s-1)/2}\zeta(1-s,1-\lbrace k\theta\rbrace)\right)\cr
&=(2\pi)^{s-2}\Gamma(1-s)\sum_{k\not=0}\frac{i}{k}\left(e^{i\pi(1-s)/2}\zeta(1-s,\lbrace k\theta\rbrace)-e^{i\pi(s-1)/2}\zeta(1-s,\lbrace k\theta\rbrace)\right)\cr
&=2\sin((s-1)\pi/2)(2\pi)^{s-2}\Gamma(1-s)\sum_{k\not=0}k^{-1}\zeta(1-s,\lbrace k\theta\rbrace)\cr
&=2\sin((s-1)\pi/2)(2\pi)^{s-2}\Gamma(1-s)\sum_{n\ge0}\sum_{k\not=0}k^{-1}(n+\lbrace k\theta\rbrace)^{s-1}.
\end{array}
$$
Split up the summation as
$$
\sum_{n\ge1}n^{s-1}\sum_{k\not=0}k^{-1}\left(1+\lbrace k\theta\rbrace/n\right)^{s-1}+\sum_{k\not=0}k^{-1}\lbrace k\theta\rbrace^{s-1}.
$$
The first term can be handled easily. Use the fact that $\theta$ has irrationality measure 2, so that $\left\lbrace\lbrace k\theta\rbrace\colon k=1,\ldots,n\right\rbrace$ has discrepancy $O(n^{-1+\delta})$. As the function $x\mapsto(1+x/n)^{s-1}$ has bounded variation of size $O(1/n)$ over the unit interval, the summation over $k$ converges and is of size $O(1/n)$. Therefore, the sum over $n$ converges on $\Re(s) < 1$, showing that the first term converges to an analytic function on $\Re(s) < 1$.
The final summation over $k$ is more problematic. Assuming $\theta$ is a quadratic irrational, and using $\bar z$ to denote the conjugate of $z\in\mathbb{Q}(\theta)$, it can be rewritten (up to order of summation) as
$$
(\bar\theta-\theta)\sum_{0 < z < 1}\frac{z^{s-1}}{\bar z-z}=(\bar\theta-\theta)\sum_{j=0}^\infty\sum_{0 < z < 1}{\bar z}^{-j-1}z^{s+j-1}.
$$
Here, I have substituted in $z=\lbrace k\theta\rbrace$, and the summation is over all $z\in\mathbb{Z}+\mathbb{Z}\theta$ lying in the unit interval. For each bounded region for $s$ on which we want to extend we only actually have to expand out a finite number of terms in the sum over $j$ above, up until $\Re(s)+j-1$ is positive. In that case, the remainder is of order $\bar z^{-j-1}z^{s+j-1}=O(k^{-1-j})$ and has uniformly convergent sum. So, we just need to show that each term in the sum over $j$ extends to a meromorphic function.
For the moment, I'll restrict to the case where $\mathbb{Z}+\mathbb{Z}\theta$ is a number ring. For example, $\theta=\sqrt{d}$ for squarefree $d\not\cong1$ (mod 4) or $\theta=(1+\sqrt{d})/2$ for squarefree $d\cong1$ (mod 4). Let $\eta$ be the fundamental unit of $\mathbb{Z}[\theta]$ lying in the unit interval. Then, each $z$ in the unit interval can be written uniquely as $\eta^ny$ for $y$ in $[\eta,1)$ and nonnegative integer $n$. Writing $\epsilon=\bar\eta\eta=\pm1$, the term for a fixed $j$ in the expansion above is (again, being a bit careless about order of summation),
$$
\begin{array}{rl}
\sum_{0 < z < 1}{\bar z}^{-j-1}z^{s+j-1}&=\sum_{\eta\le z < 1}\sum_{n=0}^{\infty}(\bar\eta ^n\bar z)^{-j-1}(\eta^nz)^{s+j-1}\cr
&=\sum_{\eta\le z\le1}\sum_{n=0}^\infty{\bar z}^{-j-1}z^{s+j-1}\epsilon^{n(j+1)}\eta^{n(s+2j)}\cr
&=\left(1-\epsilon^{j+1}\eta^{s+2j}\right)^{-1}\sum_{\eta\le z < 1}{\bar z}^{-j-1}z^{s+j-1}.
\end{array}
$$
Now that the summation has being restricted to $z\ge\eta$, the term $z^{s+j-1}$ will be bounded, and the sum is guarateed to converge. To see this, substitute back $z=\lbrace k\theta\rbrace$ and $\bar z=k(\bar\theta-\theta)+\lbrace k\theta\rbrace$. The summand is of size $O(k^{-1-j})$, so the sum converges absolutely for $j\ge1$. For $j=0$ it is $(\bar\theta-\theta)^{-1}\lbrace k\theta\rbrace^{s-1}/k + O(k^{-2})$ for $\lbrace k\theta\rbrace\ge\eta$. As $x^{s-1}$ has finite variation over the interval $[\eta,1)$, the sum for $j=0$ also converges.
Finally, the term $(1-\epsilon^{j+1}\eta^{s+2j})^{-1}$ is meromorphic on $\mathbb{C}$ with zeros on the vertical line $-2j+i\mathbb{R}$. So, $\zeta_\theta$ does extend with poles in the claimed set.
In the argument above, it was assumed for simplicity that the module $M\equiv\mathbb{Z}+\mathbb{Z}\theta$ is a number ring. The generalization to arbitrary quadratic irrationals is easy enough. The only use of the fact that $M$ is a number ring was to ensure that $\eta M\subseteq M$ and $\eta^{-1}M\subseteq M$. However, there will always exist some $r > 0$ such that $\eta^{\pm r}M\subseteq M$ and the argument above follows through unchanged with $\eta^r$ in place of $\eta$. To see that such an $r$ does exist, let $R$ be the ring of algebraic integers in $\mathbb{Q}(\theta)$, and let $a,b$ be nonzero integers with $bR\subseteq aM\subseteq R$. Then, multiplication by $\eta$ gives a permutation of the finite quotient $R/bR$, which has finite order $r > 0$. So, $\eta^{\pm r}x-x\in bR\subseteq aM$ for all $x\in R$, from which $\eta^{\pm r}aM\subseteq aM$ follows.<|endoftext|>
TITLE: A “mother of all groups”? What kind of structures have "mother of all"s?
QUESTION [37 upvotes]: For ordered fields, we have a “mother of all ordered fields”, the surreal numbers $\mathbf{No}$, a proper-class “field” which includes (an isomorphic copy of) every other ordered field as a subfield. So, I wondered: does a similar mother object exist for other kinds of object, like groups? That is, is there some “meta-group” that's a proper-class “group” which every other group is a subgroup? If so, how could it be constructed?
My attempt at a construction was this. It is based on Cayley's theorem, that every group is isomorphic to some group of permutations. In particular, the symmetric groups on $n$ letters, $S_n$, can be considered as “mother groups” for all finite groups of order $n$ or less: every such group is embedded isomorphically as a subgroup of $S_n$. It is also possible to see how this holds for infinite groups as well: there are groups $\mathrm{Per}_{\kappa}$ for infinite cardinals $\kappa$ of all permutations (bijective functions) of a set of cardinality $\kappa$ that act as mother groups for all groups of cardinality $\kappa$ or less.
So what I imagine was the following idea. We start with a class of sets such that:

they are ordered by inclusion,
every cardinality is represented exactly once

We here use the initial ordinals of cardinals, interpreted as sets in the usual way (we assume the axiom of choice here). Then we consider building up on each initial ordinal $I_{\alpha}$ (which is indexed with ordinals $\alpha$ such that as we run up from 0 through to $\omega$, but not reaching $\omega$, $I_{\alpha}$ is the finite ordinal $\alpha$, then $I_{\omega}$ is $\omega$, the initial ordinal of cardinal $\aleph_0$, then $I_{\omega+1}$ is $\omega_1$, the initial ordinal of cardinal $\aleph_1$, etc.) its corresponding group of permutations, $\mathrm{Sym}(I_\alpha)$. We now imagine the class-sized mega-union of all these $\mathrm{Sym}(I_\alpha)$, i.e. 
$$\mathbf{SYM} = \bigcup_{\alpha \in \mathbf{On}} \mathrm{Sym}(I_\alpha)$$.
Next, we proceed to define a composition of permutations in $\mathbf{SYM}$. Let $p$ and $q$ be two such permutations. If $\mathrm{dom}(p) = \mathrm{dom}(q)$, then their composition $p \diamond q = p \circ q$ – the usual composition. However, if $\mathrm{dom}(p) \ne \mathrm{dom}(q)$, then we have to first extend one or the other permutation. Define, for $\mathrm{dom}(p) < \mathrm{dom}(q)$,
$$p^q: \mathrm{dom}(q) → \mathrm{dom}(q)$$
$$p^q(a) = \begin{cases}p(a),&\ \mathrm{if}\ a \in \mathrm{dom}(p)\\
               a,&\ \mathrm{otherwise}\end{cases}$$
.
Then, if $\mathrm{dom}(p) < \mathrm{dom}(q)$, $p \diamond q = p^q \circ q$, and if $\mathrm{dom}(q) < \mathrm{dom}(p)$, $p \diamond q = p \circ q^p$. We then define an equivalence relation ~ on permutations in $\mathbf{SYM}$ such that two are equivalent if one can be extended to the other in the fashion above. Then an operation between equivalence classes of composition can be defined by taking two representative permutations and composing. Of course, this runs into a difficulty since we cannot collect these equivalence classes together as they themselves are proper classes. But we can take the representative defined on the lowest possible $I_\alpha$. Denote this lowest representative of a permutation $p$ by $\mathrm{low}(p)$. Now we should have a “mother of all groups”, given by the class of all $\mathrm{low}(p)$ for every $p \in \mathbf{SYM}$, with composition defined by taking the low of the composition as defined before.
My questions are: does the above construction make any sense? If not, where's the flaw? If so, is there a way to get around the obvious use of the axiom of choice that I mentioned? One thing I notice about rejecting choice is that then the cardinals are not necessarily totally-ordered any more, so then however we'd go about choosing representatives, we'd run into the problem where we could not “nest” them together and so could not extend the permutations so as to be able to perform the composition. Does this mean the existence of the mother group depends on the axiom of choice? Also, what about “mother objects” of other types? I suspect that in a manner analogous to the above, we can construct a “mother of all rings” by the corresponding Cayley's theorem analogue for rings (rings are isomorphic to rings of endomorphisms of abelian groups). So this makes me wonder: what kind of conditions are required for some kind of structure to have a “mother structure” of proper-class size? Does any structure have one or just some?
Another thing I notice here is that this “mother group” seems not to include all proper-class “super” groups, only “normal”, i.e. set, groups, whereas, I think, No includes all proper-class “super” ordered fields as well. Which makes me wonder: what kind of criteria are needed to ensure the existence of a “true mother” proper-class version of a structure that includes all other such structures including other proper-class ones as subsets/classes? What do ordered fields have that groups don't, and what else besides ordered fields share this property?

REPLY [4 votes]: For finitely presented simple groups, there are the existentially closed groups. These groups are groups $M$ for which any equation or inequality definable over $M$ has a solution in $M$. These have many remarkable properties, including the following, which is Theorem 1.8 in G. Higman, E. Scott, "Existentially Closed Groups", Oxford University Press (1988):

Let $M$ be an existentially closed group. Then $M$ cannot be finitely generated, $M$ contains every finitely presented simple group (and hence every finite group), and $M$ is simple.

One can (thankfully!) show that existentially closed groups exist, and that there even are $2^{\aleph_0}$ countable existentially closed groups! Indeed, analogously to how Hall's universal group is locally finite and contains every finite group, it can be shown that there exist locally finitely presented countable existentially closed groups. As an added bonus, all countable groups can be embedded in a countable existentially closed group.<|endoftext|>
TITLE: Strassen's algorithm
QUESTION [7 upvotes]: I am reading Landsberg's "Tensors: Geometry and Applications". Here he mentions tensor formulation of Strassen's algorithm and shows that the rank of Strassen's matrix multiplication tensor is $7$ and $7$ is the lower bound for any such tensor and hence one needs $7$ multiplications for $2 \times 2$ matrix multiplication. My question is the following: is $7$ the absolute lower bound among all algorithms or just those that can be formulated via tensors?

REPLY [5 votes]: Quid's reply already gives you a complete answer for the $2\times 2$ case; however, there is an additional consideration that adds to the picture of the optimal exponent of matrix multiplication. 
One can prove that, for any $t$, if there exists a straight-line program that multiplies $n\times n$ matrices in $O(n^t)$ ops (without any bilinearity assumption or restriction on the operations made: divisions, additions, products of three more terms, everything is allowed), then there is a bilinear algorithm (i.e., one that can be "formulated with tensors") that achieves the same asymptotic complexity $O(n^t)$. So one does not lose generality by restricting to bilinear algorithms, when looking for the optimal exponent of matrix multiplication.
I had this same doubt myself a few years ago, and I found a proof of this statement in Borodin-Munro, The computational complexity of algebraic and numeric problems.<|endoftext|>
TITLE: Intersection of conjugates of subgroups in free groups
QUESTION [8 upvotes]: I am looking for a reference for the following
Fact 1: if $A$ and $B$ are finitely generated subgroups of infinite index in a finitely generated free group $F$ then there exists $f \in F$ such that $fAf^{-1} \cap B=\{1\}$.
I know several proofs for this, but surely this should be classical, so a reference would be desirable. Modulo a known result of W. Neumann (Neumann, Walter D.
`On intersections of finitely generated subgroups of free groups.' Groups—Canberra 1989, 161–170,
Lecture Notes in Math., 1456, Springer, Berlin, 1990), this fact would immediately follow from the following
Fact 2: with the assumptions of Fact 1, $F$ cannot be covered by finitely many double cosets of the form AgB, $g \in F$.
I proved a generalization of Fact 2 for hyperbolic groups some time ago, but I do not know of an  explicit (classical) reference for it.

REPLY [2 votes]: I'm sorry I don't have a reference but here is a constructive proof of Fact 1 just using basic algebraic topology: Let $F$ be generated by $a_1$ and $a_2$ (the higher rank case is an easy generalization). Let $B$ be the figure eight with basepoint $b_0$. Let $X$ and $Y$ be covers corresponding to $A$ and $B$ with base points $x_0$ and $y_0$. Since $A$ and $B$ are finitely generated and of infinite-index, there exists infinite graphs that are isomorphic, with labelings, to the images of the lifts of all cyclically reduced words which start with $a_1$ (and similarly $a_1^{-1}$) in the universal cover. Call $I_X$ this infinite graph corresponding to $a_1$ (including the initial $a_1$ edge). Call $I_Y$ the infinite graph corresponding to $a_1^{-1}$ in $Y$ but do not include the edge $a_1^{-1}$ in $I_Y$ (so it does not include the initial $a_1^{-1}$ edge).
Take the graph $X$ and remove the infinite piece $I_X$ to obtain $X_0$. Take $Y$ and remove the infinite piece $I_Y$ to obtain $Y_0$. Glue $X_0$ to $Y_0$ in the obvious place. This gives a new graph $Z$ which is a cover of $B$. A path from $x_0$ to $y_0$ in $Z$ is an $f$ you want.<|endoftext|>
TITLE: Identity involving Fresnel integrals
QUESTION [5 upvotes]: In the paper E. Mehlum, Appell and the apple (nonlinear splines in space), Technical
Report No. 1676 (1981), Central institute for industrial research, Oslo (reproduced in the book Mathematical Methods for Curves and Surfaces, pages 365–384, Vanderbilt University Press, 1995) we can find the following interesting identity $$S^2(x)+C^2(x)=\sum\limits_{n=0}^\infty \frac{(-1)^n2^{2n}x^{4n+2}}{(2n+1)(4n+1)!!},$$
involving Fresnel integrals $$S(x)=\int\limits_0^x \sin{(t^2)}dt,\;\;\;
C(x)=\int\limits_0^x \cos{(t^2)}dt.$$ In the paper the identity is proved indirectly as a by product of a solution of an intricate problem. The author also mentions that the identity 
can be deduced directly from the definition of the Fresnel integrals, but does not provide any details of such deduction. How this identity can be proved? I suspect we can use $$\int\limits_0^x dy\int\limits_0^y dz\;\cos{(y^2-z^2)}=\frac{1}{2}(C^2(x)+S^2(x))$$ and the Taylor-MacLaurin series representation of the cosine function.

REPLY [6 votes]: It seems easier to integrate over the unit square, giving
$$ C^2(x) + S^2(x) = \int_0^x \int_0^x \cos(y^2-z^2) \;\mathrm{d}y\;\mathrm{d}z. $$
Using the Taylor series for cosine, and then normalizing the integral by setting $s = xy$, $t = xz$, gives
$$ C^2(x) + S^2(x) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{4n+2} \int_0^1 \int_0^1 (s^2-t^2)^{2n} \; \mathrm{d}s \; \mathrm{d}t. $$
So it is sufficient to prove that
$$ \int_0^1 \int_0^1 (s^2-t^2)^{2n} \;\mathrm{d}s\; \mathrm{d}t = \frac{2^{2n}(2n)!}{(2n+1)(4n+1)!!}. $$
This can be done immediately by Mathematica (although not Maple, to my surprise). Alternatively, here is a short proof using binomial coefficients. Expand the left-hand side using the binomial theorem and perform both integrations to get the equivalent identity
$$ \sum_{m=0}^{2n} \binom{2n}{m} \frac{(-1)^m}{(2m+1)(4n-2m+1)} =  \frac{2^{2n}(2n)!}{(2n+1)(4n+1)!!}. $$
Using partial fractions on the left-hand side this is equivalent to
$$  \sum_{m=0}^{2n}\binom{2n}{m} \frac{(-1)^m}{2m+1} = \frac{2^{2n}(2n)!}{(4n+1)!!} $$
which follows from a more basic integral, namely
$$ \int_{0}^1 (1-u^2)^{2n} \;\mathrm{d} u = \frac{2^{4n}}{4n+1} \binom{4n}{2n}^{-1}. $$<|endoftext|>
TITLE: sum of three cubes and parametric solutions
QUESTION [14 upvotes]: The first paragraph in the following link asserts that the equation $x^3+y^3+z^3=2$ has finite many parametric solutions over $\mathbb{Q}$, i.e., there are finite many polynomial triples $(x(t),y(t),z(t))$ with $x(t),y(t),z(t)\in\mathbb{Q}[t]$ satisfying the equation $x^3+y^3+z^3=2$.
Question: What might be the exact evidence for such an assertion?
Edit: Complementary materials on this problem:
Segre, Beniamino. "A note on arithmetical properties of cubic surfaces." Journal of the London Mathematical Society 1.1 (1943): 24-31.
Bremner, Andrew. "On diagonal cubic surfaces." manuscripta mathematica 62.1 (1988): 21-32.
The first paper states that a genus 0 curve on such diagonal cubic surface must be the complete intersection with another surface. The second one states that a genus 0 curve corresponding to a parametric solution should have some unusual properties at infinity. Although they are quite strong confinements, it is still ambiguous that why such a statement(finite many parametric solutions) tends to be reasonable.

REPLY [17 votes]: Seeing that the link is to a 1996 announcement that I posted to 
sci.math.research, I suppose I should explain.  Yes, the formulation 
in that announcement is not clearly stated $-$ one doesn't polish 
USENET posts like a published paper, and can't even edit after the fact 
(as is possible on mathoverflow) to correct blatant typos like the stray 
"+1" in the formula "(x,y,z)=(1+6t^3+1,1-6t^3,-6t^2)".
As it happens here there is a
 published paper
that appeared only a few years later:

Elkies, Noam D.: Rational points near curves and small nonzero 
  $|x^3-y^2$| via lattice reduction, 
  Lecture Notes in Computer Science 1838 (proceedings of ANTS-4, 2000; 
  W.Bosma, ed.), 33-63 (arXiv:math.NT/0005139).

but the relevant section (3.2) doesn't address parametrizations of
$x^3+y^3+z^3=2$.  The answer to the present question is that
D.Burde is basically right: the correct statement was, and still is,
that all known solutions of $x^3 + y^3 + z^3 = 2$ in ${\bf Q}[t]$
come from the identity
$$
(1+6t^3)^3 + (1-6t^3)^3 + (-6t^2)^3 = 2
$$
by permuting  $x,y,z$ and substituting some polynomial for $t$.
The substitution need not be linear, but nonlinear substitutions like
$(x,y,z) = (1+6t^6, 1-6t^6, -6t^4)$ give no new $(x,y,z)$ solutions either.
I don't think any method is known that would prove that there are
no other nonconstant solutions, or that there's no nonconstant solution in 
${\bf Q}[t]$ to $x^3+y^3+z^3=d$ unless $d$ is a cube or twice a cube.
All that can be said is that if there were such a solution that had
small enough degree and coefficients then it would have turned up in 
searches for integral solutions such as the searches described in that ANTS-4 paper
and also on this page.<|endoftext|>
TITLE: Are subfunctors of left exact functors also left exact?
QUESTION [6 upvotes]: Consider the direct image functor $f_*: Sh(X) \rightarrow Sh(Y)$, let $X$ and $Y$ be topological spaces, let $f: X \rightarrow Y$ be a continuous map, let $G \in Sh(X)$ be a sheaf. I was reading this course on sheaves:
http://bit.ly/14IBBTZ.
On page 1 they define this subpresheaf $f_!G \subset f_*G$ then on page 2 in the first line they remark that: 
"As $f_!$ is a subfunctor of $f_*$ it is left exact".
I wanted to ask, are subfunctors of left exact functors also left exact?

REPLY [20 votes]: Here's a counterexample with additive functors on abelian categories.  If $A$ is an abelian group, let $F(A)$ denote the subgroup of elements that are divisible by $2$.  It is easy to see that $F:Ab\to Ab$ is an additive functor, and $F$ is a subfunctor of the identity.  But $F$ is not left exact because it does not preserve kernels.  For instance, $F$ sends the exact sequence $$0\to \mathbb{Z}\to\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}$$ to $$0\to2\mathbb{Z}\to\mathbb{Q}\to\mathbb{Q}/\mathbb{Z},$$ which is not exact.<|endoftext|>
TITLE: Darboux like theorem for non-degenerate 3-forms in 6-manifolds
QUESTION [6 upvotes]: we know Darboux theorem for higher-symplectic geometry is not correct in general, 
but is there any Darboux like theorem for non-degenerate 3-forms in 6-manifolds?

REPLY [13 votes]: This depends on what you mean by 'Darboux-like'.  It is certainly not true that a closed nondegenerate 3-form on a 6-manifold is necessarily locally equivalent to one of the 'flat' models, so there is no direct analog of the Darboux' theorem in this case.  
As I remark in my article "Remarks on the geometry of almost complex $6$-manifolds" (Asian Journal of Mathematics 10 (2006), 561–606, also available on the arXiv as arXiv:math/0508428), the closed $3$-forms of elliptic type in dimension $6$ essentially depend on 4 arbitrary functions of 6 variables (modulo diffeomorphism).  
However, there is an analog if you are willing to consider something stronger: If $\phi\in\mathcal{A}^3_+(M^6)$ is a $3$-form of elliptic type on an oriented $6$-manifold $M$, then there is a unique $J(\phi)\in\mathcal{A}^3_+(M^6)$ with the property that the complex $3$-form $\Upsilon = \phi+i \ J(\phi)$ is decomposable and hence of type $(3,0)$ with respect to an almost complex structure $J_\phi$ on $M$ that induces the given orientation of $M$.  All this is algebra.  However, now, if one adds the hypothesis that $d\Upsilon = 0$, which is, of course, the same as $d\phi = d\bigl(J(\phi)\bigr)=0$, then one has that there always exist local complex functions $z^1,z^2,z^3$ such that $\Upsilon = dz^1\wedge dz^2\wedge dz^3$.  In particular, $\phi = \mathrm{Re}(dz^1\wedge dz^2\wedge dz^3)$, so this is a sort of Darboux-like theorem; it's just that you need more hypothesis than the closure of the original form.
There is a similar result for nondegenerate $3$-forms of hyperbolic type.  This is the case in which the form $\phi$ can be written locally as $\phi = \phi_+ + \phi_-$ where each of $\phi_\pm$ is decomposable while $\phi_+\wedge\phi_-\not=0$.  (These two summands are unique up to permutation.)  In this case, the 'Darboux-like' theorem is that $\phi$ can be put in normal form if and only if $d\phi_+=d\phi_-=0$, which is stronger than $d\phi=0$ (which is not sufficient by itself).
Finally, there is the case of nondegenerate $3$-forms of what might be called 'nilpotent type' (which is not a stable type in Hitchin's sense, but is nondegenerate in the sense described by the OP).  A $3$-form $\phi$ on $M^6$ is nondegenerate of nilpotent type if and only if each point lies in some open set $U$ on which there exists a coframing $\omega^1,\ldots,\omega^6$ for which
$$
\phi = \omega^4\wedge\omega^2\wedge\omega^3
      +\omega^5\wedge\omega^3\wedge\omega^1
      +\omega^6\wedge\omega^1\wedge\omega^2.
$$
For such a $\phi$, the conditions that it can locally be put in this form with $\omega^i = dx^i$ for some coordinates $x = (x^1,\ldots,x^6)$ consist of two things:  First, the condition $d\phi=0$, which is clearly necessary; second, the condition that $3$-plane field $D\subset TM$ defined by $\omega^1=\omega^2=\omega^3=0$ (which is well-defined by $\phi$) should be Frobenius.  It is not hard to show that these necessary conditions are also sufficient, so this is the 'Darboux-like' normal form theorem in this case.
Note the interesting fact that, in each of these three cases, the 'Darboux-like' conditions are all first order equations on $\phi$.  This does not continue in higher dimensions.  In dimension $7$, the two stable types of $3$-forms each have examples that are flat to first order but not flat to second order, so the 'Darboux-like' theorems in this case turn out to involve a mixture of first and second order conditions.<|endoftext|>
TITLE: conservation law and generalized Symplectic Monge-Ampere equation arising from 3-variables
QUESTION [14 upvotes]: If we have a Jacobi PDE system with conservation law $\theta \in \Omega^1(M)$ such that $d \theta$ is non-degenerate 2-form , then we know this fact that it can be written as symplectic 2D Monge-Ampere equation. so if we extend our Jacobi PDE system to 3-forms instead of 2-forms 
and assume conservation law $\theta \in \Omega^2(M)$  such that 3-form $d \theta$ is non-degenerate 3-form, then the new Jacobi PDE system can be written locally as the generalized  3D Symplectic Monge-Ampere equation arising from 3-forms  ?
PS: Firstly, I try to explain more the question and will try to solve it for 2D symplectic monge ampere equation and Jacobi plane and after solve the question for 3D, but I am still looking for an affirmative answer.
Let me explain the Jacobi PDE system view of 2D symplectic Monge Ampere equation .In fact by using Darboux Theorem we can give a nice answer to this question in 2D case . So here, First I prove the following proposition:
But before I try to define Jacobi PDE system for readers as follows
Definition : We call following system as Jacobi PDE system
$a_1+b_1\frac{\partial h_1}{\partial x_1}-c_1\frac{\partial h_1}{\partial x_2}-d_1\frac{\partial h_2}{\partial x_2}+e_1\frac{\partial h_2}{\partial x_1}+f_1\left (\frac{\partial h_1}{\partial x_1}\frac{\partial h_2}{\partial x_2}-\frac{\partial h_1}{\partial x_2}\frac{\partial h_2}{\partial x_1} \right )=0$
$a_2+b_2\frac{\partial h_1}{\partial x_1}-c_2\frac{\partial h_1}{\partial x_2}-d_2\frac{\partial h_2}{\partial x_2}+e_2\frac{\partial h_2}{\partial x_1}+f_2\left (\frac{\partial h_1}{\partial x_1}\frac{\partial h_2}{\partial x_2}-\frac{\partial h_1}{\partial x_2}\frac{\partial h_2}{\partial x_1} \right )=0$
and we can correspond each equation in Jacobi PDE system as symplectic 2-form $\omega$. 
So we correspond a Jacobi PDE system to two 2-from $\omega_1, \omega_2$ ,also it is clear that the combination of $\omega_1$ and $ \omega_2$ are also Jacobi PDE equation, so we can correspond each Jacobi PDE system as $\prod=<\omega_1, \omega_2>$.
Proposition: Let $\prod=<\omega_1,\omega_2>$ be a Jacobi PDE system with a conswerwation law $\theta \in \Omega^1(M)$ , such that $d\theta=a\omega_1+b\omega_2$ is non-degenerate 2-form , then locally the Jacobi PDE system can be written as Monge-Ampere equation.
The reason is: In fact the symplectic Monge-Ampere equation have the following form
$\hat{a}+b\frac{\partial^2\varphi}{\partial x_1^2}-d\frac{\partial^2\varphi}{\partial x_2^2}-c\frac{\partial^2\varphi}{\partial x_2\partial x_1}+e\frac{\partial^2\varphi}{\partial x_1\partial x_2}+\check{a}\frac{\partial\varphi}{\partial x_1}+\tilde{a}\frac{\partial\varphi}{\partial x_2}+$
$f(\frac{\partial^2\varphi}{\partial x_1^2}\frac{\partial^2\varphi}{\partial x_2^2}-\frac{\partial^2\varphi}{\partial x_2\partial x_1}\frac{\partial^2\varphi}{\partial x_1\partial x_2})=0$
where the coefficients are smooth functions of $x$ and $\frac{\partial\varphi}{\partial x}$ 
.
By the following trick, we reduce the symplectic Monge-Ampere equation into a Jacobi PDE system and vise versa. In fact by substitution $h_1=\frac{\partial\varphi}{\partial x_1}$, and $h_2=\frac{\partial\varphi}{\partial x_2}$ and taking following compatibility condition
$\frac{\partial h_2}{\partial x_1}=\frac{\partial h_1}{\partial x_2}$ we get the following Jacobi PDE system:
$\frac{\partial h_2}{\partial x_1}-\frac{\partial h_1}{\partial x_2}=0$ 
$a+b\frac{\partial h_1}{\partial x_1}-c\frac{\partial h_1}{\partial x_2}-d\frac{\partial h_2}{\partial x_2}+e\frac{\partial h_2}{\partial x_1}+f(\frac{\partial h_1}{\partial x_1}\frac{\partial h_2}{\partial x_2}-\frac{\partial h_1}{\partial x_2}\frac{\partial h_2}{\partial x_1})=0$
where $a=\hat{a}+\breve{a}h_1+\tilde{a}h_2$ .
Therefore the corresponding 2-forms of this system are:
$\omega_1=dx_1\wedge du_1+dx_2\wedge du_2$
$\omega_2=a(x,u)dx_1\wedge dx_2+ b(x,u)du_1\wedge dx_2+c(x,u)du_1\wedge dx_1+$
$d(x,u)du_2\wedge dx_1 +e(x,u)du_2\wedge dx_2+f(x,u)du_1\wedge du_2$
So the non-degenerate 2-form $d\theta$ , determines a symplectic structure on $M$ .Moreover by applying Darboux theorem, locally there exists a canonical coordinate system for $d\theta $, say $(x_1,x_2,u_1,u_2)$, such that : $d\theta=dx_1\wedge du_1+dx_2\wedge du_2$.
Now, let $\omega'$ be a 2-form, such that $<\omega', d\theta>$ is a local basis for $\prod$
. Then $\omega'$ has the same form as $\omega_2$ as above. Therefore the Jacobi PDE system $d\theta$ and $\omega'$ can be written as the 2D symplectic Monge-Ampere equation as above.
So if we continue this method for 3D symplectic monge ampere equation, we first define 3D Jacobi PDE system(this is my definition for compatibility of 3D Monge-Ampere equation and generalized Jacobi PDE system, please check it yourself again)
Definition (3D Jacobi PDE system): We call following system(4 PDE equations) as a 3D Jacobi PDE system ($k=1,2,3,4$)
$a^k(x,h(x))det\pmatrix{\frac{\partial u_1}{\partial x_1} & \frac{\partial u_1}{\partial x_2}&\frac{\partial u_1}{\partial x_3}\cr \frac{\partial u_2}{\partial x_1} & \frac{\partial u_2}{\partial x_2}&\frac{\partial u_2}{\partial x_3}\cr \frac{\partial u_3}{\partial x_1}&\frac{\partial u_3}{\partial x_2}&\frac{\partial u_3}{\partial x_3}\cr} +\sum_{i,j}b^k_{ij}(x,h(x))det \pmatrix{\frac{\partial u_i}{\partial x_1} & \frac{\partial u_i}{\partial x_2}\cr \frac{\partial u_j}{\partial x_1} & \frac{\partial u_j}{\partial x_2}\cr} -$
$\sum_{ij}k^k_{ij}(x,h(x))\pmatrix{\frac{\partial u_i}{\partial x_1} & \frac{\partial u_i}{\partial x_3}\cr \frac{\partial u_j}{\partial x_1} & \frac{\partial u_j}{\partial x_3}\cr}+\sum_{ij}c^k_{ij}(x,h(x))\pmatrix{\frac{\partial u_i}{\partial x_1} & \frac{\partial u_i}{\partial x_3}\cr \frac{\partial u_j}{\partial x_2} & \frac{\partial u_j}{\partial x_3}\cr}+$
$\sum_i m_i^k (x,h(x))\frac{\partial u_i}{\partial x_1}-\sum_i n_i^k (x,h(x))\frac{\partial u_i}{\partial x_2}+\sum_i l_i^k (x,h(x))\frac{\partial u_i}{\partial x_3}+e^k(x,h)=0$
, my computation show that by reducing 3D symplectic monge ampere equation to Jacobi PDE system we have four 3-forms. So if we assume  $\prod=<\omega_1,\omega_2,\omega_3,\omega_4>$ be a 3D Jacobi PDE system with a conserwation law $\theta \in \Omega^2(M)$ such that $d\theta=a\omega_1+b\omega_2+c\omega_3+d\omega_4$ be a non-degenerate 3-form. In fact $\omega_1,\omega_2,\omega_3,\omega_4$ have the following 3-forms
$\omega_1=dx_1\wedge dx_2 \wedge du_1+dx_3\wedge dx_2\wedge du_3$
$\omega_2=dx_3\wedge dx_2 \wedge du_2+dx_3\wedge dx_1\wedge du_1$
$\omega_3=dx_1\wedge dx_2 \wedge du_2-dx_3\wedge dx_1\wedge du_3$
$\omega_4=du_1\wedge du_2 \wedge du_3+\sum_{ij}\sum_{k=1}^n a_{ij}^k du_i\wedge du_j\wedge dx_k+$
$\sum_i\sum_{j,k}b_{jk}^i du_i\wedge dx_j \wedge dx_k+dx_1\wedge dx_2 \wedge dx_3$
In fact, $\omega_1,\omega_2, \omega_3$ come of the following compatibility conditions:
$\frac{\partial h_1}{\partial x_2}=\frac{\partial h_2}{\partial x_1}$
$\frac{\partial h_2}{\partial x_3}=\frac{\partial h_3}{\partial x_2}$
$\frac{\partial h_1}{\partial x_3}=\frac{\partial h_3}{\partial x_1}$
So we obtain , the Jacobi PDE system $\prod=<\omega_1,\omega_2,\omega_3,\omega_4>$ can be written as following system:
3-D symplectic monge Ampere equation =
$A=a(x,h(x))[\frac{\partial^2\varphi}{\partial x_1^2}\frac{\partial^2\varphi}{\partial x_2^2}\frac{\partial^2\varphi}{\partial x_3^2}+\frac{\partial^2\varphi}{\partial x_1 \partial x_2}\frac{\partial^2\varphi}{\partial x_2 \partial x_3}\frac{\partial^2\varphi}{\partial x_3 \partial x_1}+\frac{\partial^2\varphi}{\partial x_2 \partial x_1}\frac{\partial^2\varphi}{\partial x_3 \partial x_2}\frac{\partial^2\varphi}{\partial x_1 \partial x_3}]-$
$(\frac{\partial^2\varphi}{\partial x_1^2}\frac{\partial^2\varphi}{\partial x_2 \partial x_3}\frac{\partial^2\varphi}{\partial x_3 \partial x_2}+\frac{\partial^2\varphi}{\partial x_2^2}\frac{\partial^2\varphi}{\partial x_1 \partial x_3}\frac{\partial^2\varphi}{\partial x_3 \partial x_1}+\frac{\partial^2\varphi}{\partial x_3^2}\frac{\partial^2\varphi}{\partial x_1 \partial x_2}\frac{\partial^2\varphi}{\partial x_2 \partial x_1})+$
$b(x,h(x))[\frac{\partial^2\varphi}{\partial x_1^2}\frac{\partial^2\varphi}{\partial x_2^2}-\frac{\partial^2\varphi}{\partial x_1 \partial x_2}\frac{\partial^2\varphi}{\partial x_2 \partial x_1}]+c(x,h(x))[\frac{\partial^2\varphi}{\partial x_1^2}\frac{\partial^2\varphi}{\partial x_3^2}-\frac{\partial^2\varphi}{\partial x_1 \partial x_3}\frac{\partial^2\varphi}{\partial x_3 \partial x_1}]+$
$d(x,h(x))[\frac{\partial^2\varphi}{\partial x_2^2}\frac{\partial^2\varphi}{\partial x_3^2}-\frac{\partial^2\varphi}{\partial x_2 \partial x_3}\frac{\partial^2\varphi}{\partial x_3 \partial x_2}]+e(x,h(x))[\frac{\partial^2\varphi}{\partial x_1^2}\frac{\partial^2\varphi}{\partial x_2 \partial x_3}-\frac{\partial^2\varphi}{\partial x_1 \partial x_2}\frac{\partial^2\varphi}{\partial x_1 \partial x_3}]-$  
$f(x,h(x))[\frac{\partial^2\varphi}{\partial x_2^2}\frac{\partial^2\varphi}{\partial x_1 \partial x_3}-\frac{\partial^2\varphi}{\partial x_1 \partial x_2}\frac{\partial^2\varphi}{\partial x_2 \partial x_3}]+g(x,h(x))[\frac{\partial^2\varphi}{\partial x_3^2}\frac{\partial^2\varphi}{\partial x_1 \partial x_2}-\frac{\partial^2\varphi}{\partial x_1 \partial x_3}\frac{\partial^2\varphi}{\partial x_2 \partial x_3}]+$
$h'(x,h(x))\frac{\partial^2\varphi}{\partial x_1^2}-i(x,h(x))\frac{\partial^2\varphi}{\partial x_2^2}+j(x,h(x))\frac{\partial^2\varphi}{\partial x_3^2}+k(x,h(x))\frac{\partial^2\varphi}{\partial x_1 \partial x_2}-$
$l(x,h(x))\frac{\partial^2\varphi}{\partial x_2 \partial x_3}+m(x,h(x))\frac{\partial^2\varphi}{\partial x_1 \partial x_3}$
$n(x,h(x))\frac{\partial \varphi}{\partial x_1}+o(x,h(x))\frac{\partial \varphi}{\partial x_2}+p(x,h(x))\frac{\partial \varphi}{\partial x_3}+q(x,h(x))=0$
&
$\frac{\partial^2\varphi}{\partial x_1\partial x_2}=\frac{\partial^2\varphi}{\partial x_2\partial x_1}$ 
$\frac{\partial^2\varphi}{\partial x_1\partial x_3}=\frac{\partial^2\varphi}{\partial x_3\partial x_1}$ 
$\frac{\partial^2\varphi}{\partial x_2\partial x_3}=\frac{\partial^2\varphi}{\partial x_3\partial x_2}$ 
But automatically we have last three equations, because $h_i\in C^1$ so $\varphi\in C^2$ and by substituting these equations in first equation $A$ we get the generic 3D Monge-Ampere equation.

REPLY [5 votes]: I think that you should be careful to define your terms, but let me guess:  A Jacobi PDE system for three unknown functions $h_1,h_2,h_3$ of three independent variables $x_1,x_2,x_3$ is a set of three partial differential equations, each of which can be written as a linear combination of the minors (of any rank, including $0$ and $3$) of the Jacobian matrix $J_x(h) = \frac{\partial h}{\partial x}$ with coefficients that are explicit functions of the $6$ variables $x_1,x_2,x_3,h_1,h_2,h_3$.  In other words, on $M=\mathbb{R}^6$ (or some open subset) with coordinates $x_1,x_2,x_3,h_1,h_2,h_3$, you have specified three $3$-forms $\Upsilon_i$ ($i=1,2,3$), and a graph 
$$\Gamma_u = \bigl\lbrace\bigl(x,u(x)\bigr)\ \mid\ x\in D\subseteq\mathbb{R}^3\bigr\rbrace
$$ solves your system with $h=u(x)$ for some function $u:D\to\mathbb{R}^3$ if and only if the pullback of each of the $\Upsilon_i$ to $\Gamma_u$ vanishes.  To get a decent theory, you'll also need to impose some kind of nondegeneracy condition, such as the condition that, for a 'generic' pair of tangent vectors $v_1,v_2\in T_pM$, the $1$-forms $\theta_i(v) = \Upsilon_i(v_1,v_2,v)$ are linearly independent at $p$.
Now, it turns out that such a system is never equivalent to a so-called symplectic Monge-Ampère system, which is locally described on a $6$-manifold $M$ by a choice of a symplectic $2$-form $\Omega$ on $M$ and a nonzero $3$-form $\Upsilon$ that is $\Omega$-primitive, i.e., such that $\Omega\wedge\Upsilon=0$.  (In this case, the $3$-manifolds you want to study are the $\Omega$-Lagrangian submanifolds $L\subset M$ to which $\Upsilon$ pulls back to be the zero $3$-form.  
The reason is that for Jacobi PDE systems as above (that satisfy the nondegeneracy condition), the general solution depends locally on three arbitrary functions of $2$-variables while, for the symplectic Monge-Ampère systems as defined above, the general solution depends locally on two arbitrary functions of $2$-variables.
On the other hand, you could consider a generalization of 'symplectic Monge-Ampère system' in which you have a differential ideal $\mathcal{I}$ on a $6$-manifold that is generated by a nondegenerate $2$-form $\Omega$ and a nonvanishing $\Omega$-primitive $3$-form $\Upsilon$.  The condition that they generate a differential ideal is just that there exist a $1$-form $\alpha$ and a function $a$ such that 
$$
d\Omega = \alpha\wedge\Omega + a\ \Upsilon.
$$
(If $a\equiv0$, then one finds that $d\alpha=0$, so that, at least locally, one can scale $\Omega$ to make it be closed, and you are back in the symplectic Monge-Ampère case.  However, if $a\not=0$, you cannot do this.)  In this generalized situation, the general solution will depend on two arbitrary functions of $2$ variables, just as in the symplectic Monge-Ampère case.  It would now make sense to think of conservation laws as closed $3$-forms in $\mathcal{I}$, and there are interesting question to ask there.<|endoftext|>
TITLE: Exterior and symmetric  powers of external tensor products of representations    
QUESTION [10 upvotes]: Let us assume that $\pi: G\to Aut(V)$  and $\rho : K\to Aut(W)$ are two finite-dimensional representations of two Lie groups $G$ and $K$, and   consider the representation
  $\pi\hat{\otimes}\rho : G\times K\to  Aut(V\otimes W)$, the so called external tensor product of $\pi$ and $\rho$, given by
$$
(\pi\hat{\otimes}\rho)(g, k)(v\otimes  w):= \pi(g)v\otimes \rho(k)w,
$$
for any $g\in G$, $k\in K$, $v\in V$ and $w\in W$. For the second exterior power of the representation  $\pi\hat{\otimes}\rho$ is known the following 
isomorphism:
$$
\Lambda^{2}(\pi\hat{\otimes}\rho)=(\Lambda^{2}\pi \  \hat{\otimes}\ Sym^{2}\rho) \ \oplus  (Sym^{2}\pi \ \hat{\otimes} \ \Lambda^{2}\rho).
$$
Similarly, for the second symmetric power it holds that
$$
Sym^{2}(\pi\hat{\otimes}\rho)=(Sym^{2}\pi \  \hat{\otimes}\ Sym^{2}\rho) \ \oplus  (\Lambda^{2}\pi \ \hat{\otimes} \ \Lambda^{2}\rho).
$$
I would like to understand  how these formulas can be  generalized for exterior and symmetric powers of bigger degree.  For example, what we can say about 
$$
\Lambda^{3}(\pi\hat{\otimes}\rho), \ \Lambda^{4}(\pi\hat{\otimes}\rho),  \ Sym^{3}(\pi\hat{\otimes}\rho), \ Sym^{4}(\pi\hat{\otimes}\rho), \ \dots \ ?
$$
For the isomorphisms above, you can see for example the link
An isomorphism of 2-Schur modules
Thank you!

REPLY [11 votes]: One has
$$
Sym^k(\pi\otimes\rho) = \bigoplus_{|\alpha|=k} \Sigma^\alpha(\pi)\otimes\Sigma^\alpha(\rho),
$$
the sum is over all Young diagrams with $k$ boxes, $\Sigma^\alpha$ is the Schur functor. Similarly,
$$
\Lambda^k(\pi\otimes\rho) = \bigoplus_{|\alpha|=k} \Sigma^\alpha(\pi)\otimes\Sigma^{\alpha^T}(\rho),
$$
where $\alpha^T$ is the transposed diagram.<|endoftext|>
TITLE: The transcendence degree of $\mathbb R$ after adding a Cohen
QUESTION [11 upvotes]: Let $V\models\sf ZFC$, and let $V[r]$ be a generic extension obtained by adding one Cohen real, or equivalently $\omega$ Cohen reals.
It is clear that $\Bbb R^{V[r]}$ and $\Bbb R^V$ have the same cardinality, and that $\Bbb R^{V[r]}\setminus\Bbb R^V$ also have that cardinality.
Furthermore any generic real must be transcendental (I think, because we don't change the rationals), so $\Bbb R^{V[r]}$ is a transcendental extension of $\Bbb R^V$. But what's the transcendence degree of this extension?
It clearly cannot be finite, because adding one Cohen real adds $\omega$ pairwise generic Cohen reals, and the above argument gives us that they cannot be algebraically dependent. But is it $\aleph_0$, or is it $2^{\aleph_0}$? Maybe something else?

REPLY [2 votes]: I think the following argument might be easier than Hamkins answer. It also does not use AC. So suppose $g$ is a Cohen real. We want to show that there are continuum many mutually generic Cohen reals in $V[g]$.
In $V$, fix a canonical enumeration $F: 2^{<\omega} \to \omega$ such that if $|s|<|t|,$ then $F(s)<F(t).$ For any $t\in (2^\omega)^V$, define $g_t$ by $g_t(n)=g(F(t\restriction n)).$ Then $(g_t: t\in (2^\omega)^V)$ is as required.<|endoftext|>
TITLE: Infimum of a finite number of distances in the plane
QUESTION [9 upvotes]: Suppose one has a finite number of distances $d_1,\ldots,d_k$ on the Euclidean plane all of which metricize the usual Euclidean topology.
Define for each pair of points $x$ and $y$ in the plane
$$d(x,y) = \inf\left\lbrace d_{i_1}(x_0,x_1) + \cdots d_{i_l}(x_{l-1},x_l) \right\rbrace$$
where the infimum is taken over all finite sequences with $x_0 = x,x_1,\ldots,x_l = y$ and all possible choices of $i_1,\ldots, i_l$.
Question: Is it possible for $d(x,y)$ to be $0$ for a pair of distinct points $x \neq y$?
Also, I'm interested in the same question replacing $\mathbb{R}^2$ by an arbitrary Polish space.

REPLY [11 votes]: This is possible even on the real line.
There is a strictly increasing continuous function $f:[0,1]\to\mathbb R$ whose derivative is zero almost everywhere. It is a suitable sum of a series of Cantor functions. See, for example, Gelbaum and Omsted, "Counterexamples in analysis" (1964), Chapter 8, Example 30.
This function has the following property: for every $\varepsilon>0$ there is a collection of disjoint intervals $[a_i,b_i]\subset[0,1]$ with total length greater than $1-\varepsilon$ and total variation of $f$ less than $\varepsilon$:
$$
\sum (b_i-a_i)>1-\varepsilon, \qquad \sum (f(b_i)-f(a_i)) < \varepsilon .
$$
Now define metrics $d_1$ and $d_2$ on $[0,1]$ as follows: $d_1$ is the standard metric and $d_2$ is the pull-back of the standard metric by $f$, i.e. $d_2(x,y)=|f(x)-f(y)|$. You can go from 0 do 1 through points $a_1,b_2,a_2,b_2,\dots$ using the distance $d_2$ between $a_i$ and $b_i$ and $d_1$ between $b_i$ and $a_{i+1}$. Thus $d(0,1)<\varepsilon$ for every $\varepsilon>0$.

REPLY [5 votes]: The infimum can be zero as pointed out by Anton Petrunin.  Here's a construction on the interval $[0,1]$.
Consider a sequence of piecewise linear functions $f_n:[0,1] \to [0,1]$ each of which is strictly increasing defined by (see figure):  

$f_0(x) = x$ for all $x$.
$f_{n+1}$ coincides with $f_n$ on diadic numbers of denominator $2^{-n}$.
$f_{n+1}\left(\frac{2k+1}{2^{n+1}}\right) = f_n\left(\frac{k}{2^n}\right) + \frac{1}{3^{n+1}}$ for integer $k$.

Let $f = \lim\limits_{n \to +\infty}f_n$ and $d_1$ be the pullback of the standard Euclidean distance on $[0,1]$ under $f$.   One has that
$$d_1(0,1/2) = \frac{1}{3}$$
$$d_1(0,1/4) = d_1(1/2,3/4) = 1/9$$
$$d_1(0,1/8) = d_1(1/4,3/8) = d_1(1/2,5/8) = d_1(3/4,7/8) = 1/27$$
etc...
One can construct in similar fasion a distance $d_2$ satisfying 
$$d_2(1/2,1) = \frac{1}{3}$$
$$d_2(1/4,1/2) = d_2(3/4,1) = 1/9$$
$$d_2(1/8,1/4) = d_2(3/8,1/2) = d_2(5/8,3/4) = d_2(7/8,1) = 1/27$$
etc...
By considering diadic partitions one sees that the infimum is $0$ when one can use the distances $d_1$ and $d_2$.<|endoftext|>
TITLE: differential geometry using Robinson's infinitesimals?
QUESTION [9 upvotes]: Is there a detailed treatment of differential geometry using Robinson's infinitesimals?

REPLY [6 votes]: Somewhat belatedly we developed foundations for differential geometry using infinitesimal displacements here:
Nowik, T.; Katz, M. "Differential geometry via infinitesimal displacements." Journal of Logic and Analysis 7:5 (2015), 1-44.<|endoftext|>
TITLE: Random variables invariant under almost automorphisms.
QUESTION [9 upvotes]: Let $\Omega$ be a standard atomless probability space, we can assume $\Omega=(0,1)$ with Lebesgue measure. A bijection $f:\Omega/A_1\to\Omega/A_2$ is almost automorphism, if $P(A_1)=P(A_2)=0$, $f(A)$ is measurable if and only if $A$ is, and $P(f(A))=P(A)$ in this case. An almost automorphism preserves a (real-valued) random variable (r.v.) $X$ if $X(\omega)=X(f(\omega))$ for almost all $\omega$. 
The question is: Assume that every almost automorphism preserving $Y$ preserves also $X$. Does this imply that $X$ is $Y$-measurable, that is $X=g(Y)$ a.s. for some $g:R \to R$?

REPLY [4 votes]: Fix for convenience $\Omega = [0,1)$ with Lebesgue measure $\mu$.  We exhibit a family of continuum-many Borel functions $X_\alpha \colon [0,1) \to \mathbb{R}$, indexed by $\alpha \in (0, 1/2)$, such that

Each $X_\alpha$ has only trivial almost automorphisms, in the sense that each almost automorphism $f$ preserving $X_\alpha$ has $\mu$-null support (i.e., the set $\{r : r \neq f(r)\}$ is $\mu$-null).
When $\alpha \neq \beta$ we have that $X_\alpha$ is not in your sense $X_\beta$-measurable.

In particular we get a negative answer to the question.
For each $\alpha \in (0,1/2)$ put $I_\alpha = [0,\alpha)$ and $J_\alpha = [0,1) \setminus I_\alpha = [\alpha, 1)$.  Let $i_\alpha \colon [0,1) \to [0,1)$ be the obvious fixed-point-free involution flipping $I_\alpha$ and $J_\alpha$.  Note that since $\alpha < 1/2$ this involution distorts $\mu$ in the following sense: any $\mu$-positive Borel set $A$ contains a $\mu$-positive Borel set $A' \subseteq A$ such that $\mu(A') \neq \mu(i_\alpha[A'])$.  In particular, no restriction of $i_\alpha$ to a set of $\mu$-positive measure can be a $\mu$-preserving map.
Now define $X_\alpha$ by 

$X_\alpha(r) = r$ if $r \in I_\alpha$,
$X_\alpha(r) =  i_\alpha(r)$ if $r \in J_\alpha $.

Note that $X_\alpha(r) = X_\alpha(s)$ iff $r = s$ or $r = i_\alpha(s)$.  It follows that any almost automorphism preserving $X_\alpha$ is almost everywhere some restriction of $i_\alpha$, but by the previous paragraph it must in fact be almost everywhere the identity.  So each $X_\alpha$ has only trivial almost automorphisms.
Finally, suppose that $\alpha < \beta$ in $(0,1/2)$.  Put $R = [\alpha, \beta)$.  Note that for each $r \in R \subseteq J_\alpha$ we have $i_\beta(r) \in J_\alpha$ and in particular $X_\alpha(r) \neq X_\alpha(i_\beta(r))$.  Of course by construction we have $X_\beta(r) = X_\beta(i_\beta(r))$.  So if $g \colon \mathbb{R} \to \mathbb{R}$ is any function, for each $r\in R$ we must have at least one of $g(X_\beta(r)) \neq X_\alpha(r)$ or $g(X_\beta(i_\beta(r)) \neq X_\alpha(i_\beta(r))$ and in particular $g \circ X_\beta$ does not agree $\mu$-a.e. with $X_\alpha$.  In other words, $X_\alpha$ is not $X_\beta$-measurable.
A similar argument using $i_\alpha$ shows that $X_\beta$ is not $X_\alpha$-measurable.<|endoftext|>
TITLE: Invariance of the l.h.s. of Euler-Lagrange equation
QUESTION [12 upvotes]: Let $M^n$ be a smooth manifold equipped with a nondegenerate Lagrangian $L:TM\to\mathbb R$, $L=L(x,y)$, $x\in M$, $y\in T_xM$. The stationary points of the corresponding integral functional on curves are the solutions of the Euler-Lagrange equation, which in coordinates reads
$$
 \frac d{dt} \frac{dL}{dy_i}(x(t),\dot x(t)) - \frac{\partial L}{\partial x_i}(x(t),\dot x(t)) = 0, \qquad i=1,\dots,n .
$$
Consider a smooth curve $t\mapsto x(t)$ which is not stationary. Plugging it into the l.h.s. of the equation yields coordinates of a co-vector (from $T^*_{x(t)}M$) which depends on the curve but not on the coordinate system. The invariance of this co-vector can be seen e.g. from the first variation formula for the functional. 
Question: Is there a coordinate-free definition of this co-vector?
Actually I am interested only in the case when $L$ is the Lagrangian associated to a Finsler metric (i.e. $L$ is quadratically homogeneous).
Notes

The equation itself (i.e. the property that the co-vector is zero) has an invariant expression e.g. with  Hamiltonian formalism.
In the Riemannian case, the co-vector in question (for a unit-speed curve) corresponds to the geodesic curvature vector under the isomorphism between $TM$ and $T^*M$ defined by the metric. This can be defined invariantly via the Levi-Civita connection.

REPLY [16 votes]: There is a coordinate-free description using only natural objects on $TM$.  Here is one way to do it.
First, consider the basepoint submersion $\pi:TM\to M$.  For each $v\in TM$, the linear map $\pi'(v):T_v(TM)\to T_{\pi(v)}M$ is surjective, and the $\pi$-fiber through $v$ is equal to $T_{\pi(v)}M$, a vector space.  It follows that the kernel of $\pi'(v)$ is naturally isomorphic to $T_{\pi(v)}M$.  Call this isomorphism $\iota_v: T_{\pi(v)}M\to \mathrm{ker}\bigl(\pi'(v)\bigr)\subset T_v(TM)$, and let $\nu_v : T_v(TM) \to T_v(TM)$ be the nilpotent endomorphism $\nu_v = \iota_v\circ \pi'(v)$. 
Next, consider a Lagrangian $L:TM\to \mathbb{R}$, which I will assume to be smoothly differentiable. Define a $1$-form $\omega_L$ on $TM$ by 
$$
\omega_L(w) = dL\bigl(\nu_v(w)\bigr)
$$ 
for all $w\in T_v(TM)$.  Let $R$ be the vector field on $TM$ that is tangent to the fibers of $\pi$ and that is the natural radial vector field on each such (vector space) fiber, and set $E_L = dL(R) - L$.  (If $L$ is quadratic homogeneous on each $\pi$-fiber, then, by Euler's relation, $E_L = L$.)
Finally, if $\gamma:[a,b]\to M$ is a twice differentiable curve, with $\gamma':[a,b]\to TM$ its velocity vector and $\gamma'':[a,b]\to T(TM)$ the velocity vector of $\gamma'$, then, for each $t\in[a,b]$, consider the co-vector $\beta(t)\in T^\ast_{\gamma'(t)}TM$ defined by the rule
$$
\beta(t)(w) = d\omega_L(\gamma''(t),w)+ dE_L(w)
$$ 
for $w\in T_{\gamma'(t)}TM$.  Then $\beta(t)(w)=0$ for $w\in \mathrm{ker}\bigl(\pi'(\gamma'(t))\bigr)$, so $\beta(t) = \pi'(\gamma'(t))^\ast(\delta\gamma(t))$ for a unique co-vector $\delta\gamma(t)\in T^\ast_{\gamma(t)}M$.  
This assignment $t\mapsto \delta\gamma(t)$ is the canonical 'variation $1$-form' of the Lagrangian $L$ along $\gamma$.  It vanishes identically if and only if $\gamma$ satisfies the Euler-Lagrange equation for $L$.<|endoftext|>
TITLE: Binary expansion of squares
QUESTION [10 upvotes]: I came across the following simple question: what odd integer squares have exactly 3 ones in their binary expansion?
After looking at it for a while I convinced myself that the only solutions to $r^2 = 1+ 2^m + 2^n$ (with $n > m \ge 3$) are $n=2m-2$ (the "trivial" case) and $(m,n) = (4,5),(4,9)$ (the "sporadic" cases).  My first attempt was to analyze things 2-adically: write out the Taylor Series:
$f(x) = 1 - 2 \sum_{n=1}^{\infty} (-1)^n C_{n-1} x^n$, where $C_n = \frac{1}{n+1} \binom{2n}{n}$ is the $n$-th Catalan number, and $v_2(x) > 0$.  We have $f(x)^2 = 1 + 4x$, so that $a_m := \sqrt{1+2^m} = f(2^{m-2})$ (Note that $1+2^m$ is not a rational square when $m \ne 3$).  If such an integer $r$ exists, then we must have $r \equiv \epsilon(a_m + 2^{n-1}) \bmod {2^{n+1}}$, where $\epsilon = \pm 1, \pm (1+2^n)$ is one of the 4 square roots of 1 modulo $2^{n+1}$.  Also, we must have $r < 2^{(n+1)/2}$.  To get rid of the annoying cases, take all of this modulo $2^{n-1}$, where we find that the top $(n-3)/2$ bits of $a_m \bmod 2^{n-1}$ must be either all 0's or all 1's.  By writing out the first few terms of $f(x)$ I found that there's a run of $m-2$ 0's which corresponds to the trivial solution, and a run of $m-1$ 1's which corresponds to the sporadic solution (the condition that $m-1 \ge (n-3)/2$ limits the possible $m$'s and $n$'s to a small set).  However, I don't see any easy way to show that there are no long "runs" of 1's or 0's among the higher bits of $a_m$ (by the bits, I mean to write out $a_m = \sum_{n=0}^{\infty} b_{m,n} 2^n$, where $b_{m,n} \in \{0,1\}$).  Such a statement would be to the effect that we can't approximate $a_m$ too closely by small integers -- this is the sort of $p$-adic Diophantine approximation statement that I would think should be true, but I can't find it.
Absent that, another approach is via $S$-unit equations.  This show (see below) that for each $m$, there are only a finite number $n$'s for which $1+2^m + 2^n$ is a rational square, but I'd like to prove (with the exception of the sporadic solution) that there is exactly 1.  Since, as I mentioned above, $1+2^m$ is not a rational square when $m > 3$, denote by $K = K_m = \mathbb{Q}(\sqrt{1+2^m})$.  It's easy to see that the prime 2 splits in $K$: call $\frak{a},\frak{a}'$ the ideals above 2. Let $k>0$ denote the order of $\frak{a}$ in the ideal class group of $K$ and $\beta$ a generator of the principal ideal $\frak{a}^k$.  Also denote conjugation of $K/\mathbb{Q}$ by $\quad '$.  Then we have $r+\sqrt{1+2^m} = \epsilon \beta^t 2^u$, for some non-negative integers $t,u$ and $\epsilon \in K$ is a unit.  Note that we might need to look at $r-\sqrt{1+2^m}$ instead.  By taking norms we see that $2^{2n} = 2^{2u + kt}$, so that $2n = 2u+kt$.  By noting that $\epsilon \beta^t 2^u - \epsilon' {\beta'}^t 2^u = 2 \sqrt{1+2^m}$, we see that $u=0,1$.  For each of those values we then have an equation of the form $\alpha - \alpha' = \gamma$, where $\gamma$ is fixed, and $\alpha,\alpha'$ are $S$-units (here $S$ consists of the primes above 2), so we know that there are only a finite number of solutions.  Any suggestions as to how to proceed to show that my initial guesses are correct?

REPLY [17 votes]: This was solved in a paper of Szalay (in Indag. Math. in 2002), using lower bounds for approximations to $\sqrt{2}$ by rationals with denominators of the form $2^k$ (obtained by Beukers using Pad\'e approximations to $\sqrt{1-z}$).<|endoftext|>
TITLE: How often do two powers of 2 equal two powers of 10 (when summed)?
QUESTION [5 upvotes]: In thinking about a MathOverflow question pertaining to numbers whose decimal and binary digit sums are equal, I found myself asking:  

Are there any solutions in
  non-negative integers $(a,b,c,d)$ to
  the equation
$$2^a + 2^b = 10^c + 10^d$$
aside from the trivial solution
  $(0,0,0,0)$ and the nontrivial
  solutions $(2,4,1,1)$ and $(4,2,1,1)$?

(This is, in effect, the case $s=2$ to the other MO question regarding which numbers $s$ can be a simultaneous binary and decimal digit sum and if so, how often.  It seems likely that 20 is the only number whose digit sums are both 2, but that's basically what I'm asking here.)
In searching for other solutions, we may as well assume, for the sake of simplicity, that $a\le b$ and $c\le d$.  It's quickly clear that we can restrict to looking for positive solutions, and we can also dismiss the possibility that $a=b$.  That is, we can assume $0\lt a\lt b$ and $0\lt c \le d$.
The possibility that $c=d\ne 1$ is ruled out by Mihăilescu's proof of Catalan's conjecture:  If $2^a + 2^b = 2\cdot10^c = 2^{c+1}\cdot5^c$ (with $a\lt b$) we necessarily have $a=c+1$, leaving the equation $1 = 5^c - 2^n$, where $n=b-a$, whose only solution is $c=1$, $n=2$.  (Aside:  We don't actually need to invoke Catalan's conjecture.  See below.)
If now we restrict to $c\lt d$, it's immediately clear that we must have $a=c$.  After writing $n=b-a$ (as before) and $m=d-c$ and doing a little factoring, the problem reduces to what I take to be the core question:

Does the equation
$$1+2^n = 5^a(1+10^m)$$
have any solutions in positive
  integers $(a,m,n)$?

This is as far as I've gotten in any meaningful sense.  The question is obviously related to the Cunningham project.  It's easy to see that $a>0$ implies $n\equiv2\mod4$, so the aurifeuillian factorization 
$$2^{4k+2}+1 = (2^{2k+1}-2^{k+1}+1)(2^{2k+1}+2^{k+1}+1)$$
may or may not help.  (It does help avoid the use of invocation of Catalan's conjecture above:  Only one aurifeuillian factor is divisible by 5.)  It's also easy to see that
$$n\log2 + \log(1+1/2^n) = a\log5 + m\log10 + \log(1+1/10^m)$$
implies
$$(a+m)\log5 \approx (n-m)\log2$$
if $m$ and $n$ are large.  Finally, writing $5^a = (1+4)^a = 1+4a+16{a\choose2}+\cdots$, it may be possible to say something useful about the relationship of $a$ to $n$, but I don't see what.

REPLY [12 votes]: There's an elementary way of solving this (and similar equations). Let's start with
$$ 
1+2^n=5^a(1+10^m)
$$
which you want to solve in positive integers $n$, $a$, $m$. Clearly $m < n$ and $a < n$. As wccanard points out, $2$ is a primitive root modulo $5^a$ and so we obtain that $n$ is divisible by $2 \cdot 5^{a-1}$. In particular,
$$
2 \cdot 5^{a-1} \le n.
$$
Now let's use the fact that $m < n$ are reduce the equation modulo $2^m$. We obtain,
$$
5^a \equiv 1 \pmod{2^m}. 
$$
As in your question you write this as
$$
4a + 16 \binom{a}{2}+\cdots \equiv 0 \pmod{2^m}.
$$
This implies that $2^{m-2} \mid a$, and so 
$$
2^{m-2} \le a.
$$
The inequalities we now have show that the left-hand side of the equation is much bigger than the right-hand side as soon as the $n$ is large!
Appended by the OP:  With apologies to Samir if he had something slicker in mind, I thought I'd add my own elaboration on when $(1+2^n)$ becomes bigger than $5^a(1+10^m)$.
The first displayed inequality implies $5^a \le (5/2)n$ and the second, combined with $a \lt n$, gives $2^m \le 4a < 4n.$  Clearly, $1+10^m \lt 16^m = (2^m)^4\lt (4n)^4$.  This all gives
$$2^n \lt 1+2^n = 5^a(1+10^m) \lt (5/2)n(4n)^4 = 640n^5,$$
and it's easy to check that this implies $n\lt 35$.  
Knowing also that $n$ must be congruent to 2 mod 4 means we can finish the problem off by checking the factorizations of $1+2^n$ for $n=2,6,10,14,18,22,26,30,$ and $34$, which is easy enough to do.  A less crude estimate than $1+10^m \lt 16^m$ might shave off a few of the larger values of $n$, but it doesn't seem worth the effort.<|endoftext|>
TITLE: Would a closed universe with special relativity violate causality? Does the universe have to be simply connected?
QUESTION [12 upvotes]: This question may be more appropriate for physics.stackexchange.com, but it would be helpful to get feedback from experts in Minkowski geometry.
The classic twin paradox is a false thought experiment trying to disprove special relativity. Special relativity says that an observer at rest watching a passerby at a high velocity will see the other person's clocks running slower. However, relativity also says there is no preferred frame of uniform motion, so the passerby sees the clock of the original observer running slow. So if a twin leaves the earth at a high speed and returns, leaving her sister on earth the whole time, both should find the other younger, a contradiction.
This is resolved in flat, open spacetime by the fact that the passenger twin must accelerate to leave and return, and acceleration is a preferred frame in the sense that the laws of physics are different for an accelerating person.
Now, though, imagine a closed universe (I.e. one with closed geodesics). In this universe, two objects in relative uniform motion will encounter each other repeatedly. Even if a twin must accelerate to get into uniform motion, she will pass her sister on earth over and over again if she keeps on a fixed closed geodesic. Now both frames are equivalent, and so time is slower for both observers. Thus, past and future are not well-defined, and causality is violated. 
Is there something wrong with this reasoning? Is there any way to handle special relativity and a closed universe so the above paradox does not happen?
Edit: For geometrists, I'm really asking if the invariant quadratic form becomes poorly-behaved if we take a quotient of $\mathbb{E}^{3,1}$ by a proper action of a subgroup of the Lorenz group.
Edit: By closed universe, I meant a universe compact in space dimensions for some observer. By a poorly behaved quadratic form, I was just thinking of how proper time is the root of the quadratic form for events with time like separation, so any paradox with proper time is an inconsistency with the quadratic form.

REPLY [3 votes]: In addition to Igor Khavkine's comment. There are some more recent articles about twin paradox in compact spaces:
https://arxiv.org/abs/gr-qc/0101014 (The twin paradox in compact spaces, by John D. Barrow and Janna Levin).
https://arxiv.org/abs/0910.5847 (Time, Topology and the Twin Paradox, by J.-P. Luminet).<|endoftext|>
TITLE: Rookie questions about k3's
QUESTION [5 upvotes]: Hi everyone,
I am trying to go through parts of Saint-Donat's 1974 paper 'Projective Models of K3-surfaces', and have been stuck on a few claims for a while now - I'd appreciate some help explaining them.
Here is the set-up:  $X$ is a (smooth, projective, algebraic) K3 surface.  For a divisor $D$ on $X$ the Riemann-Roch theorem reads that
$ h^0(D) + h^0(-D) = 2 + \frac{1}{2}D^2 + h^1(D)$
Since the most troublesome thing here is $h^1(D)$, we should try to say something about it: the 'something' is the following:  if $D$ is effective then $h^1(D)=h^0(D, \mathcal{O}_D)-1$.
We furthermore have the following dichotomy for effective divisors $D$ without fixed component:  Either:
(1)  $D^2 > 0$, $h^1(D)=0$, and the generic member of $D$ is an irreducible curve $1+\frac{1}{2}D^2$ or
(2)$D^2=0$.  Then $D$ is linearly equivalent to $kE$ for some irreducible curve of (arithmetic) genus 1, and $k \geq 1$.  We furthermore have that $h^1(L)=k-1$ and that every divisor of $|D|$ is equal to a sum of the form $E_1 + ... +E_k$ with each $E_l \in |E|$.
Furthmore, a remark that will probably be relevant is the following: if $\Delta$ is an effective divisor on $X$, then $\dim |\Delta|=0$ if and only if $h^1(\Delta, \mathcal{O}_\Delta)=0$.  Furthermore, if $\Delta$ is connected and reduced, then we have $\Delta^2 = -2$.
Consider now an effective divisor $D$ satisfying $D^2 \geq 0$.  We write $D \sim D'+\Delta$ where $\Delta$ is the fixed part of $D$.  Decompose $\Delta$ into its connected reduced components $\Delta_1, ..., \Delta_N$.  Since $D'$ has no fixed part, we are in one case of the dichotomy; assume we are in the second.  The claim is that there exists one and only one $\Delta_i$ which satisfies $D' . \Delta_i>0$, and that this $\Delta_i$ in fact satisfies $\Delta_i .E=1$.  
Why is the claim true?  The author writes that it is an easy consequence of the Riemann-Roch theorem and the equation $h^1(D)=h^0(D, \mathcal{O}_D)-1$ given above.
Thanks a lot!

REPLY [3 votes]: As you said, it's a consequence of Riemann-Roch and the equation $h^1(D)=h^0(D,\mathcal O_D)-1$, which follows from taking cohomology of
$$0\to \mathcal O_X(-D)\to \mathcal O_X\to \mathcal O_D\to 0$$
for an effective divisor $D$.
Applying Riemann-Roch to $D=D'+\Delta$, we get
$$h^0(D)-h^1(D) = \frac{1}{2}D^2+2 = D'.\Delta-N+2,$$
where $h^2(D)=h^0(-D)=0$ because $D$ is effective.  Plugging in the formula for $h^1(D)$, this becomes
$$h^0(D)-h^0(D,\mathcal O_D)=D'.\Delta-N+1.$$
Now, $h^0(D)=h^0(D')=h^1(D')+2=h^0(D',\mathcal O_{D'})+1$.  Since $D'$ is equivalent to $k$ disjoint irreducible genus 1 curves (the fibers of the linear series $|D'|$), its structure sheaf has $k$ global sections.  The equation then becomes
$$k-h^0(D,\mathcal O_D)=kE.\Delta-N.$$
The $h^0$ term is the number of connected components of $D$, or equivalently $E\cup \Delta$, where $E$ is a generic fiber of the linear series.  Also, $E.\Delta = \sum_i E.\Delta_i$, where each summand is nonnegative because $E$ is movable.  To finish, we just do some casework, counting connected components for each incidence possibility.
If $E.\Delta=0$, then $D^2=-2N<0$, which contradicts the initial assumption that $D^2\geq 0$.
If $E.\Delta=1$, then we win.
If $E.\Delta=m>1$, then the only partition of $E.\Delta$ that gives us an integral value of $k$ is $E.\Delta_i=1$ for all $\Delta_i$, and this forces $k=1$ and $N=m$.  I couldn't rule out this case a priori, but if you look at Saint-Donat's remark 2.7.4, he assumes $k>2$ before making the claim, so all is well.<|endoftext|>
TITLE: How to work out a grammar if we know the language?
QUESTION [5 upvotes]: How could we work out a grammar if we know the language? How could we work out a grammar if we know the language that is restricted to a special kind like CFL or CSL? For example, we know $$L=\{a^nb^nc^n \mid n \in \mathbb{N}\}$$ How can we get the grammar? Is there any algorithm?
PS: Language here means at least the recursively enumerable one, or computably enumerable one.

REPLY [7 votes]: Theorem. There is no computable procedure which, given as
input a Turing machine program $e$ that enumerates a c.e. set that
happens to be a context-free language, outputs a context-free
grammar for that language.
Proof. Let us denote by $W_e$ the set enumerated by program $e$.
Suppose that there were such a computable procedure $e\mapsto
g(e)$, where $g(e)$ is a context-free grammar for $W_e$, if indeed
$W_e$ is a context-free language. (If $W_e$ is not context-free, then we do
not assume $g(e)$ is meaningful or even that it converges.)
We define a certain computable function $f$. For any program $e$, let $W_{f(e)}$ be the c.e. set defined as
follows. At first, we enumerate nothing into $W_{f(e)}$ until
$g(e)$ converges and outputs a context-free grammar. At this
point, if this grammar generates a non-empty language, then we
continue to enumerate nothing into $W_{f(e)}$ and thereby ensure that $W_{f(e)}$ is empty. Alternatively, if the language generated by the grammar $g(e)$ is empty, then we ensure that $W_{f(e)}=\{0\}$, containing a single
string. (Note that the emptiness problem for context-free grammars
is computably decidable.)
By the recursion theorem, there is a particular program $e$
such that $W_e=W_{f(e)}$. Since $W_{f(e)}$ is either empty or a
singleton, it follows that it is a context-free-language, and so
$g(e)$ is defined. But by construction, we have ensured that
$g(e)$ is a grammar for the empty language if and only if $W_e$ is
non-empty. And so for this program, $g(e)$ is not a grammar for
$W_e$. Contradiction. QED<|endoftext|>
TITLE: Moduli Spaces of Higher Dimensional Complex Tori
QUESTION [5 upvotes]: I know that the space of all complex 1-tori (elliptic curves) is modeled by $SL(2, \mathbb{R})$ acting on the upper half plane. There are many explicit formulas for this action.
Similarly, I have been told that in the higher dimensional cases, the symplectic group $Sp(2n, \mathbb{R})$ acts on some such space to give the moduli space of complex structures on higher dimensional complex tori. Is there a reference that covers this case in detail and gives explicit formulas for the action? 
In the 1-dimensional case, all complex tori can be realized as algebraic varieties, but this is not the case for higher dimensional complex tori. Does the action preserve complex structures that come from abelian varieties? 
Crossposted at https://math.stackexchange.com/questions/345713/moduli-spaces-of-higher-dimensional-complex-tori where it has been unanswered for awhile.

REPLY [7 votes]: The fact that all $1$-dimensional tori are projective means care is sometimes needed in making analogies with higher dimensional tori. This is one of those times. The natural 'moduli space' of all $d$-dimensional complex tori constructed by Will is not very nice when $d>1$. For example, the action of $GL_{2d}(\mathbb Z)$ on $GL_d(\mathbb C)\backslash GL_{2d}(\mathbb R)$ isn't properly discontinuous, so the resulting quotient space isn't Hausdorff (an observation due to Siegel).
A slightly better state of affairs is available if we begin by looking at the Siegel upper half-plane
$$ \mathcal H_d = \{ \tau \in M_d(\mathbb C) \colon \tau^t = \tau, \mathrm{Im}(\tau) > 0 \}. $$
The group $Sp(2d,\mathbb R)$ acts transitively on $\mathcal H_d$ via
$$ \begin{pmatrix}A & B\\C & D\end{pmatrix} \tau = (A\tau+B)(C\tau+D)^{-1} $$
(this is well-defined) and the isotropy subgroup of $iI_d$ is (essentially) $U(g)$. Thus we can view $\mathcal H_d$ as $Sp(2d,\mathbb R)/U(d)$. Now a point $\tau \in \mathcal H_d$ gives us a complex torus $\mathbb C^d/(\mathbb Z^d + \mathbb Z^d \tau)$ -- but this torus is not an arbitrary one: it comes with a "principal polarization". Moreover the natural action of $Sp(2d,\mathbb Z)$ on $\mathcal H_d$ (which, by the way, is properly discontinuous) preserves the isomorphism class of the corresponding torus as well as its principal polarization. Thus the points of the space $\mathcal H_d/Sp(2d,\mathbb Z)$ are in bijection with isomorphism classes of principally polarized $d$-dimensional complex tori (these are abelian varieties). Note that if $d=1$ we're reduced to the familiar modular curve $\mathcal H/SL(2,\mathbb Z)$ which parametrizes all $1$-dimensional tori.
For (much) more on this, see II.4 of Mumford's Tata Lectures on Theta I or van der Geer's chapter in The 1-2-3 of Modular Forms.<|endoftext|>
TITLE: is there an anyon structure analogous to spin structure for rank 2 bundle?
QUESTION [7 upvotes]: A spin structure on a Riemannian bundle of rank >2 is the lift of the structure group from $\text{SO}(n)$ to its universal cover $\text{Spin}(n).$ It may also be defined in the case $n=2$ as the lift of the structure group to a double cover of $\text{SO}(2)$, which is of course not the universal cover. 
So what about lifts to other covers of $\text{SO}(2)$. Does the lift to a three-to-one cover of $\text{SO}(2)$ have a topological obstruction living in $H^2(B,\mathbb{Z}/3)$? Is the concept of a lift to the universal cover of $\text{SO}(2)$ affected by the unusual statistics of anyons?
Searching the literature for spin structures, I found only the double cover case. And searching the archive for anyons yields nothing but solid state physics articles. So I would be happy just to have a reference to any work discussing this case.

REPLY [3 votes]: Here is a different explanation, in terms of exact sequences. There is an exact sequences:
$0 \to \mathbb Z/n \to SO(2) \to SO(2) \to 0 $
since $SO(2)$ is the $n$-fold cover of $SO(2)$. Taking cohomology, we get:
$H^1(B, SO(2) ) \to H^1(B, SO(2) ) \to H^2(B, \mathbb Z/n)$
Since $H^1(B,G)$ is the group of principal $G$-bundles for $G$ an abelian group, the exact sequence property means that the bundle lifts if and only if the class in $H^2(B,\mathbb Z/n)$ is zero. 
It is a bit more difficult to see why this method why the cohomology class is just the first Chern class of the associated complex line bundle modulo $n$.<|endoftext|>
TITLE: Hereditarily Countable Names and Proper Forcing
QUESTION [5 upvotes]: The 'hereditarily countable names' are as defined in Shelah's Proper and Improper Forcing, Chapter 3 Definition 4.1. Let $\mathbb{P}$ be a proper forcing notion and $\dot{Q}$ a $\mathbb{P}$-name such that $\Vdash_{\mathbb{P}}$ "$\dot{Q}$ is a proper forcing notion with set of elements $\check{\kappa}$ and maximal element $\check{0}$".
Let $A$ be the closure of {$\check{\alpha}$ : $\alpha < \kappa$} under the following 2 functions.
(1) given sequences ($p_n$ : $n \in \omega$) and ($\tau_n$ : $n \in \omega$), let $\tau$ be a name forced to be equal to: 
(i) $\tau_i$ where $i$ is the least $n$ satisfying $p_n \in \dot{G_\mathbb{P}}$, if such $i$ exists, and 
(ii) $\check{0}$, otherwise.
(2) given ($\tau_{m, n}$ : $m$, $n \in \omega$), let $\tau$ be: 
(i) the $\epsilon$-least element of $\dot{Q}$ such that for all $m \in \omega$, {$\tau_{m, n}$ : $n \in \omega$} is predense below $\tau$, if such an element exists, and
(ii) $\check{0}$, otherwise.
My question is, is it true that if $p \in \mathbb{P}$ and $\sigma$ is a $\mathbb{P}$-name such that $p \Vdash \sigma \in \dot{Q}$, then there is a $\mathbb{P}$-name $\tau \in A$ such that $p \Vdash \tau \le \sigma$, or even better, $p \Vdash \tau = \sigma$? If so, why is that?
Much thanks in advance.

REPLY [6 votes]: Counterexample:  Let $\kappa $ be uncountable and let $\mathbb P= \kappa$ be an antichain (with a special weakest element $0_{\mathbb P}$, and let $\mathbb Q$ be forced to be the same forcing. Let $\sigma$ be the $\mathbb P$-name of the generic element of $\mathbb P$. 
Now note that each "hereditarily countable" name uses (hereditarily) only countably many of the names $\check \alpha$, $\alpha<\kappa$.   More precisely:  For every subset $B\subseteq \{ \alpha: \alpha < \kappa\}$ define the family of HC-$B$-names naturally. (That is: all conditions and names appearing in the recursive construction must be in $B$, or HC-$B$-names, respectively.) 
Then show that the set of HC names which are an HC-$B$-name for some countable $B$ is closed under the two operations (1) and (2). 
Every HC-$B$-name for a condition is forced to be equal to the empty condition, or in $B$. 
If the empty condition forces $\tau\le \sigma$ for some HC-$B$-name $\tau$, for some countable $B$. Now choose a condition $p_0\notin B$; then $p_0$ forces that $\sigma=p_0$, but cannot force $\tau=p_0$. 
(I think the point of HC-names is this: If you have a countable set of names $\sigma_n$, and a condition $p$, then you can find a stronger condition $q$ and a set of HC names $\tau_n$, such that $q$ forces $\tau_n \le \sigma_n$  for all $n$. To prove this, let $N$ be a sufficiently large countable elementary model, and let $q$
be generic for $N$. Obtain $\tau_n$ from $\sigma_n$ by ignoring elements outside $N$.)<|endoftext|>
TITLE: injectivity radius of hyperbolic surface
QUESTION [6 upvotes]: Given a positive real number $l$. Does there exist a closed hyperbolic surface $X$ so that injectivity radius not less than $l$?

REPLY [8 votes]: Let $Y$ be a compact hyperbolic surface.  There are only finitely many closed geodesics in $Y$ whose lengths are less that $\ell$.  Since $\pi_1(Y)$ is residually finite, there is a normal subgroup of finite index in $\pi_1(Y)$ that does not contain anything in the conjugacy classes corresponding to these geodesics. The corresponding finite covering space $X$ of $Y$ has no closed geodesics of length less than $\ell$.<|endoftext|>
TITLE: Books on advanced galois theory
QUESTION [12 upvotes]: I have been studying galois theory on my own and find it very fascinating. I have gone through Ian Stewarts book: http://www.amazon.co.uk/Galois-Theory-Third-Chapman-Mathematics/dp/1584883936. I am not sure which book to study from next but I would like to learn about something along the lines of algebraic closures of Q and other related topics. 
Related to this, I would also like to learn more finite field theory. I would like books that are not too technical and provide some context since I am doing a self study.

REPLY [2 votes]: The book, Algèbre et théories galoisiennes, by Adrien and Régine Douady, discusses Galois theory vs. the topological theory of coverings, especially in the context of Riemann surfaces. It concludes by an introduction to the theory of dessins d'enfants.<|endoftext|>
TITLE: Is there an effective way to calculate K-theory using Morse functions?
QUESTION [16 upvotes]: Let $M$ be a compact manifold and let $f$ be a Morse function with exactly one critical point at each critical level.  Then one can recover a CW-complex with the homotopy type of $M$ from just the critical point data of $f$: simply go through the critical points of $f$ in order and attach an $n$-cell every time you pass through a critical point of index $n$.  
Hence it is possible to recover any homotopy theoretic information you could want from $M$.  In some cases it is possible to directly calculate topological invariants using Morse theory; for instance, one can recover the cohomology ring of $M$ using Morse cohomology.  
I'm wondering if there is a tool in the spirit of Morse cohomology which recovers the topological K-theory of $M$.  I'm looking for something more effective than "use a Morse function to build a CW-complex and calculate its K-theory."  For instance, there should be a natural Chern character map from "Morse K-theory" to Morse cohomology.
I haven't seen anything like this in the literature, but that certainly doesn't mean it isn't there.  Any ideas?

REPLY [2 votes]: Roland Voigt has submitted his thesis "Transport functions and Morse K-theory" at Universität Leipzig -- his advisor was Matthias Schwarz. He has not published his results so far and the thesis will only become available (automatically) once he has defended.
You can contact him, his email address is (Roland.Voigt@math.uni-leipzig.de).<|endoftext|>
TITLE: who invented projective space $\mathbb{P}^n$?
QUESTION [12 upvotes]: Who invented projective space $\mathbb{P}^n$ as an extension of the usual affine space $\mathbb{A}^n$?
Who was the first person to consider projective closure of
plane affine algebraic curves (curves in $\mathbb{A}^2$)? Was it the same person?

REPLY [2 votes]: One should also mention 
Karl Georg Christian von Staudt (1798 – 1867), a German mathematician.
His book "Geometrie der Lage (1847)" was a landmark in projective geometry. 
Staudt was the first to adopt a fully rigorous approach. Without exception his predecessors still spoke of distances, perpendiculars, angles and other entities that play no role in projective geometry.<|endoftext|>
TITLE: Methods for solving two variable recurrence
QUESTION [7 upvotes]: I have a recurrence 
$$f(i,j) = 1+ \frac{N-i}{N} f(i,j-1) + \frac{i}{N} f(i-1, j)$$
$$f(i,0)  = 0$$
$$f(0,j)  = j$$
I would like to compute $f(N,M)$ in terms of N and M. The system is defined for $0\leq i\leq N$ and $0\leq j\leq M$ where $N$ and $M$ are non-negative integers.  
I am familiar with techniques for solving single variable recurrences but cannot see how to extend them to this situation. Is there some generating function solution one can write down? Is it known how to solve this sort of recurrence?
By a hand wavy argument I believe the solution looks asymptotically roughly like $c\sqrt{NM}$ for some constant $c$, when $N > M$. 
[Fixed typo in question.]

REPLY [5 votes]: This problem has a probabilistic interpretation. Namely, if one considers the Markov chain on the integer points in the rectangle $[0,N]\times [0,M]$ with the transition probabilities
$$
p\bigl((i,j),(i,j-1)\bigr) = \frac{N-i}{N} \;, \qquad p\bigl((i,j),(i-1,j)\bigr) = \frac{i}{N} \;,
$$
then $f(i,j)$ is the expected number of steps until the chain attains the horizontal line $\{j=0\}$ where it is absorbed. Now, the $i$ component of this chain is the Markov chain on $[0,N]$ with the transition probabilities
$$
p(i,i) = \frac{N-i}{N} = p_i \;, \qquad p(i,i-1) = \frac{i}{N} = 1-p_i\;,
$$
which can be interpreted as the deterministic chain $i\to i-1$ with additional "holding" at each state. The distribution of this holding time at a point $i$ is geometric with the parameter $1-p_i$, i.e., the holding time is 0 with probability $1-p_i$, is 1 with probability $p_i(1-p_i)$, etc. Thus, the expectation of the holding time at point $i$ is $p_i/(1-p_i)$, and the expected vertical displacement of the original chain provided its horizontal component attains a point $t\in [0,n]$ is
$$
M_t=\frac{p_N}{1-p_N} + \dots + \frac{p_t}{1-p_t} = \frac{1}{N-1} + \dots + \frac{t}{N-t} \sim \int_0^t \frac{x}{N-x}dx = N\log\frac{N}{N-t}-t \;.
$$ 
In particular, if $t$ is close to $N$, then $M_t\gg N>M$, which means that our chain most likely becomes absorbed before it attains the vertical segment $\{i=0\}$. [More rigorously here one should also estimate the variance of the vertical displacement to make sure it does not differ much from $M_t$.]
Now the equation
$$
M = N\log\frac{N}{N-t}-t
$$
will give us the likely number of horizontal steps $t$ necessary to have vertical displacement $M$, i.e., to be absorbed. The total expected time before absorption will be therefore
$$
f(N,M)=M + t = N\log\frac{N}{N-t} \;.
$$ 
The above equation can be rewritten as
$$
\frac{M}{N} = -\log (1-t/N) - t/N \;.
$$
Under the assumption that $M \ll N$ (i.e., $t \ll N$) one can expand the log in the RHS, which gives
$$
\frac{M}{N} \approx \frac{t^2}{2N^2} \;, 
$$ 
whence $t\approx\sqrt{2MN}$ and $f(N,M)\approx\sqrt{2MN}$.<|endoftext|>
TITLE: Convolution in $\ell_p$ when $0<p<1$
QUESTION [6 upvotes]: Background
It is known that given real sequences $a = (a_n)_{n \in \mathbb Z} \in \ell_p$ and $b = (b_n)_{n \in \mathbb Z} \in \ell_q$, their convolution defined as
$$ a * b (n) = \sum_{k \in \mathbb Z} a_{n-k} b_k $$
is in $\ell_r$ if $1 \le p, q < \infty$ and $\frac 1 r = \frac 1 p + \frac 1 q -1 $.
Question
What happens when $0 < p, q < 1$? Obviously, since $a$ and $b$ are in $\ell_1$, their convolution $a * b$ is in $\ell_1$. Can we say better, i.e. $a*b$ is in $\ell_r$ for some $r < 1$?

REPLY [2 votes]: If $0< p<1$, then $(u+v)^p\leqslant u^p+v^p$ for any non-negative numbers $u$ and $v$. Consequently, working first on finite sums and then after taking the limit, we can see that
$$|a*b(n)|^r\leqslant \sum_{k\in\mathbb Z}|a(k)|^r\cdot |b(n−k)|^r$$
for any integer $n$ and any $r\geqslant \max\{p,q\}$. Then, by an application of Fubini-Tonelli's theorem, we get that $a*b\in\ell_{\max(p,q)}$. 
Angelo's remarks shows that this is optimal, because assuming that $a$ and $b$ are sequences of positive numbers, then 
$$a*b(n)\geqslant \max\{a(n)b(0),a(0)b(n)\}.$$<|endoftext|>
TITLE: Density of smooth functions in Sobolev spaces on manifolds
QUESTION [12 upvotes]: Hebbey defines the Sobolev space of functions on a Riemannian manifold (M,g) as the completion of smooth functions under the Sobolev norm. However, I have seen (elsewhere) that Sobolev spaces have been defined as the collection of locally integrable functions whose weak derivatives (w.r.t to the Levi-Civita connection) are in L^p. For compact manifolds, or for manifolds with bounded geometry and positive injectivity radius (like $\mathbb{R}^n$), I can see why these two definitions are equivalent. Are they equivalent for general noncompact manifolds?

REPLY [8 votes]: According to pages 14 and 15 of:

MR2343536  Reviewed Eichhorn, Jürgen Global analysis on open manifolds. Nova Science Publishers, Inc., New York, 2007. x+644 pp. ISBN: 978-1-60021-563-6; 1-60021-563-7 (Reviewer: Yuri A. Kordyukov)

The poof is given in:

MR1066741  Reviewed Eichhorn, Jürgen Elliptic differential operators on noncompact manifolds. Seminar Analysis of the Karl-Weierstrass-Institute of Mathematics, 1986/87 (Berlin, 1986/87), 4–169, Teubner-Texte Math., 106, Teubner, Leipzig, 1988. (Reviewer: Steven Rosenberg) 

The results are:
The closure of smooth functions with compact support, the closure of smooth function in the Sobolev space, and the Sobolev space are all different in general on open Riemannian manifolds. If the manifold is of bounded geometry (of order $k$) then all these spaces coincide up to Sobolev order $k+2$. This holds even for sections of vector bundles.
Thus on compact manifolds all these spaces coincide also.<|endoftext|>
TITLE: Homotopy equivalence from contractibility of fiber
QUESTION [7 upvotes]: Suppose $X$ and $Y$ are two $CW$ complexes and $f:X\rightarrow Y$ is a continuous surjection such that fiber of each point (i.e. $f^{-1}(y)$ for each $y\in Y$)  is contractible. Does it implies that $X$ and $Y$ are homotopy equivalent.
PS-1:By Whitehead's Theorem it will be enough to show that $f$ induces an isomorphism between all homotopy groups. 
PS-2:In question Equivariant Cohomology for actions with finite stabilizers there are some discussion regarding the above question but in terms of homology. If anybody thinks that my question can be a consequence of this discussion please explain the connection.

REPLY [12 votes]: In his paper
MR0087106 (19,302f)
Smale, Stephen
A Vietoris mapping theorem for homotopy. 
Proc. Amer. Math. Soc. 8 (1957), 604–610. 
Smale proved the following theorem:
Theorem : Let $X$ and $Y$ be connected, locally compact separable metric spaces.  Assume also that $X$ is locally contractible.  Consider a proper surjective continuous map $f : X \rightarrow Y$.  Assume that for all $y \in Y$, the space $f^{-1}(y)$ is contractible and locally contractible.  Then $f$ is a weak homotopy equivalence.
To see how this fits into your situation, remember that (for instance) finite CW complexes are locally compact and locally contractible.  So you need to impose conditions on the fibers to ensure that they are also locally contractible.<|endoftext|>
TITLE: On Prufer domains
QUESTION [5 upvotes]: Are there any Prufer domains that have an infinity of prime ideals but only one of those primes is not finitely generated?

REPLY [2 votes]: In $\mathbb{Z}+X\mathbb{Q}[X]$, all prime ideals are principal except $X\mathbb{Q}[X]$ which is not finitely generated. Also $\mathbb{Z}+X\mathbb{Q}[X]$ is a Bezout domain (A domain in which each finitely generated ideal is principal) and a Bezout domain is always Prufer.<|endoftext|>
TITLE: Ordinary Generating Function for Mobius
QUESTION [16 upvotes]: Is there any information known for the Ordinary Generating Function for Mobius?
$$
\sum_{n=1}^{\infty}  {\mu(n)}x^n
$$
I know that

It has radius of convergence 1.
Does not have limit as $x\rightarrow 1$. 

My question is 

Does it have limit as $x\rightarrow \textrm{exp}(i\theta)\neq 1$?
Is there a similar result for $\textrm{exp}(i\theta)\neq 1$ with 
$$
\sum_{n\leq x}\mu(n)=o(x)$$  i.e. is there a result of this type?
$$
\sum_{n\leq x}\mu(n)\textrm{exp}(in\theta) = o(x)$$

EDIT1: Aside from my questions, there are some known results for the OGF for Mobius. 
EDIT2: Found a reference for (III)
Reference: Jason P. Bell, Nils Bruin, Michael Coons "Transcendence of generating functions whose coefficients are multiplicative" available at http://www.ams.org/journals/tran/2012-364-02/S0002-9947-2011-05479-6/
P. Borwein, T. Erd´elyi, and F. Littman, Polynomials with coefficients from a finite set
available at http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P195.pdf
(I) This function has the unit circle as natural boundary. 
(II) This function is a transcendental function. 
(III) It is not bounded on any open sector($\{z:|z|<1, z=re^{i\theta}, \alpha<\theta<\beta\}$)of unit disc. 
This result (III) together with Baire Category answers my question 1 with a dense set of $\theta$'s, but as Prof Tao mentioned, there can be a possibility that it is bounded for some values of $\theta$.

REPLY [14 votes]: Mobius randomness heuristics suggest that $\sum_n \mu(n) (r e^{i\theta})^n$ does not converge to a limit as $r \to 1^-$ for any (or at least almost any) $\theta$.  If it did converge for some $\theta$, then we would have
$$\sum_n \mu(n) e^{in\theta} \psi_k(n) \to 0$$
as $k \to \infty$, where
$$ \psi_k(n) := (1-2^{-k-1})^n - (1-2^{-k})^n.$$
The cutoff function $\psi_k$ is basically a smoothed out version of the indicator function $1_{[2^k,2^{k+1}]}$.  The Mobius randomness heuristic (discussed for instance in this article of Sarnak) then suggests that the sum $\sum_n \mu(n) e^{in\theta} \psi_k(n)$ should have a typical size of $2^{k/2}$, and so should not decay to zero as $k \to \infty$.
Note from Plancherel's theorem that the $L^2_\theta$ mean of $\sum_n \mu(n) e^{in\theta} \psi_k(n)$ is indeed comparable to $2^{k/2}$.  This does not directly preclude the (very unlikely) scenario that this exponential sum is very small for many $\theta$ and only large for a small portion of the $\theta$, but if one optimistically applies various versions of the Mobius randomness heuristic (with square root gains in exponential sums) to guess higher moments of $\sum_n \mu(n) e^{in\theta} \psi_k(n)$ in $\theta$, one is led to conjecture a central limit theorem type behaviour for the distribution of this quantity (as $\theta$ ranges uniformly from $0$ to $2\pi$), i.e. it should behave like a complex gaussian with mean zero and variance $\sim 2^k$, and in particular it should only be $O(1)$ about $O(2^{-k})$ of the time, and Borel-Cantelli then suggests divergence for almost every $\theta$ at least.  Unfortunately this is all very heuristic, and it seems difficult with current technology to unconditionally rule out a strange conspiracy that makes $\sum_n \mu_n (re^{i\theta})^n$ bounded as $r \to 1$ for some specific value of $\theta$ (say $\theta = \sqrt{2} \pi$), though this looks incredibly unlikely to me.  (One can use bilinear sums methods, e.g. Vaughan identity, to get some nontrivial pointwise upper bounds on these exponential sums when $\theta$ is highly irrational, but I see no way to get pointwise lower bounds, since $L$-function methods will not be available in this setting, and there may well be some occasional values of $\theta$ and $k$ for which these sums are actually small.  But it is likely at least that the $\theta=0$ theory can be extended to rational values of $\theta$.)<|endoftext|>
TITLE: Quantized Enveloping Algebras at $q=1$
QUESTION [10 upvotes]: As is well-known, the quantized enveloping algebra $U_q(\frak{sl}_2)$ is not well-defined when $q=1$ because of the relation
$$
[E,F] = \frac{K-K^{-1}}{q-q^{-1}}.
$$
To address this problem, one has an alternative formulation of $U_q(\frak{sl}_2)$: One removes the offensive relation, introduces an extra generator $G$, along with extra relations, 
$$
[G,E] = E(qK+q^{-1} K^{-1}), ~~~~~~~~~ [G,F] = -(qK + q^{-1} K^{-1})F,
$$
and 
$$
[E,F] = G, ~~~~~~~~ (q-q^{-1})G = K - K^{-1}.
$$
and canonical extensions of the Hopf algebra maps. This new algebra can be shown isomorphic to $U_q(\frak{sl}_2)$, yet is well-defined at $q=1$. It is not equal to $U(\frak{sl}_2)$, but to a double cover of $U(\frak{sl}_2)$.
What happens in the general $U_q(\frak{g})$ case? Again $U_q(\frak{g})$ is not well-defined since we the relation
$$
[E_i,F_j] = \delta_{ij}\frac{K_i-K^{-1}}{q-q^{-1}}.
$$
Judging by the representation theory, there should exist a reformulation of $U_q(\frak{g})$, where the offensive relations are removed, $n$ new generators $G_i$ (where $n$ is the rank of $\frak{g}$) are introduced, along with an extra set of new relations, which I would guess as
$$
[G_i,E_j] = \delta_{ij}E_i(qK_i+q^{-1} K_i^{-1}), ~~~ [G_i,F_j] = -\delta_{ij}(qK_i + q^{-1} K_i^{-1})F_i,
$$
and
$$
[E_i,F_j] = \delta_{ij}G_i, ~~~~~~~ (q-q^{-1})G_i = K_i - K_i^{-1}.
$$
This should then be equal to $U_q[\frak{g}$, and is clearly well-defined when $q=1$. I would then guess that at $q=1$ we have an $n$-fold cover of $U(\frak{g})$, which one consider as justification for the $2^n$ representations types of $U_q(\frak{g})$. Am I OK in reasoning here?

REPLY [8 votes]: Even if strictly speaking it's not true, it's always usefull to remember that morally $$K_i=q_i^{H_i}:=\exp(H_i\log(q_i))$$ where $q_i=q^{d_i}$ for some integer $d_i$ attached to the Cartan matrix of $\mathfrak g$, and where $H_i$ is the $i$th Cartan generator of $\mathfrak g$. This explain why the classical limit of $K_i$ should be 1.
Now set 
$$G_i=\frac{K_i-K_i^{-1}}{q_i-q_i^{-1}}$$
Then using again this heuristic it's easliy seen that you have
$$\lim_{q\rightarrow 1} G_i= \lim_{q\rightarrow 1}\frac{q_i^{H_i}-q_i^{- H_i}}{q_i-q_i^{-1}}= H_i$$
It's now rather clear how to prove that it indeed work, since $G_i$ is precisely the R.H.S. of the relation you want to modify, so just replace it by
$$[E_i,F_j]=\delta_{i,j} G_i$$
and add the relaiton
$$(q-q^{-1})G_i=K_i-K_i^{-1}$$
which is nothing but the definition of $G_i$. That you get the same algebra is obvious because you did not really change anything. The former relation does not depend on $q$ anymore so gives you the appropriate relation at the classical limit, and the latter relation just become trivial.
Then following again the heuristic that $K_i$ should goes to 1 at the limit, take the quotient by the ideal generated by $K_i-1$, and you get the classical envelopping algebra.<|endoftext|>
TITLE: Is there an analog of determinant for linear operators in infinite dimensions as that of finite dimensions?
QUESTION [11 upvotes]: I am trying to find out the essence of what a determinant is. Besides, in finite dimensions, determinant is the kind of numerical invariant that determines the invertibility of a linear operator, but what about infinite case? Is there a similar invariant? Why if not?

REPLY [3 votes]: The answers above point out that one cannot define a determinant in a meaningful way on the algebra of bounded operators on a Banach space, unless finite-dimensional.  However, this does not preclude the possibility of doing this for suitable subclasses and this is precisely what Alexander Grothendieck did  in his celebrated (amongst functional analysts)
article "Théorie de Fredholm" (Bull. Soc. Math. vol. 84---freely available online).
This is one of those articles which changed the face of mathematics forever.  The closely related question of which operators have a  trace has been investigated in great detail, by, for example,  Albrecht  Pietsch and Hermann König.<|endoftext|>
TITLE: Are there non-categorical notions in topos theory?
QUESTION [15 upvotes]: Suppose that $\mathcal{T}$ is an abstract $2$-category we know is equivalent to the $2$-category of Grothendieck topoi via some equivalence $$\phi:\mathcal{T} \to \mathfrak{Top},$$ and let $E$ be an object of $T$. Can we recover the underlying category $\phi(E)$ without using $\phi$? 
I am asking because often properties of morphisms of topoi use that we know what topoi are (certain categories) and what maps between them are (certain pairs of adjoint functors), e.g. by referencing elements of the domain, or by saying one of the pairs of adjoint functors has a further adjoint with certain properties. But, a general principle of category theory is that one shouldn't care what  things are, just about the maps between them-and not really the maps, but how they are related, i.e. the category you get. If you have an equivalent category, then you should be able to make the same statements. So the $2$-category $\mathcal{T}$ should be enough. Hence, I ask, how does one recover $\phi(E)$? It would suffice have a completely categorical description of etale geometric morphisms (i.e. one not depending on the "evil" choice of a particular presentation of $\mathfrak{Top}$ as consisting of certain categories called topoi) because any topos is equivalent to the category of etale morphisms over itself.

REPLY [17 votes]: There is a Grothendieck topos $\textbf{Set}[\mathbb{O}]$ with the following universal property: for all Grothendieck toposes $\mathcal{E}$, the category $\textbf{Geom}(\mathcal{E}, \textbf{Set}[\mathbb{O}])$ of geometric morphisms $\mathcal{E} \to \textbf{Set}[\mathbb{O}]$ and "geometric transformations" (a misnomer; they actually code algebraic data!) is naturally equivalent to $\mathcal{E}$ itself. Such a $\textbf{Set}[\mathbb{O}]$ is called an object classifier.
We can construct $\textbf{Set}[\mathbb{O}]$ explicitly using the theory of classifying toposes: one presentation is as the presheaf topos $[\textbf{FinSet}, \textbf{Set}]$. Indeed, by Diaconescu's theorem, a geometric morphism $\mathcal{E} \to [\textbf{FinSet}, \textbf{Set}]$ is the same thing as a left exact functor $\textbf{FinSet}^\textrm{op} \to \mathcal{E}$, but any such is freely and uniquely determined by the image of $1$; this correspondence extends to 2-morphisms as well.
Addendum. To address Simon Henry's comments, here is an abstract construction of $\textbf{Set}[\mathbb{O}]$. It is known that the 2-category of Grothendieck toposes has tensors with small categories. Indeed, 
$$[\mathbb{C}, \textbf{Geom}(\mathcal{E}, \mathcal{F})] \simeq \textbf{Geom}([\mathbb{C}, \mathcal{E}], \mathcal{F})$$
but we know that $\textbf{Set}$ is a pseudo-terminal object in the 2-category of Grothendieck toposes, so we may take $\textbf{Set}[\mathbb{O}] = \textbf{FinSet} \otimes \mathcal{S}$, where $\mathcal{S}$ is any pseudo-terminal Grothendieck topos.<|endoftext|>
TITLE: New Geometric Methods in Number Theory and Automorphic Forms
QUESTION [17 upvotes]: The MSRI is organising a programme with the above title from Aug 11, 2014 to Dec 12, 2014. Here is a short description from their website : 

The branches of number theory most
  directly related to the arithmetic of
  automorphic forms have seen much
  recent progress, with the resolution
  of many longstanding conjectures. 
  These breakthroughs have largely been
  achieved by the discovery of new
  geometric techniques and insights. The
  goal of this program is to highlight
  new geometric structures and new
  questions of a geometric nature which
  seem most crucial for further
  development. In particular, the
  program will emphasize geometric
  questions arising in the study of
  Shimura varieties,  the $p$-adic
  Langlands program, and periods of
  automorphic forms.

Question Which new geometric structures, techniques and insights have been crucial for this recent progress ?

REPLY [20 votes]: To complement Joel's wonderful and (as far as I understand) very much on point answer, let me quote from the proposal for the parallel program on Geometric Representation Theory, which touches on several related themes:
Representation theory is the study of the basic symmetries of
mathematics and physics. The primary aim of the subject is to
understand concrete linear models for abstract symmetry groups. A
signature triumph of the past century is our understanding of compact
Lie groups. At the foundation, there is Cartan's classification of Lie
algebras and Borel-Weil-Bott's uniform construction of all
representations in the cohomology of line bundles on flag
varieties. Thus we have a list of every compact Lie group we could
ever encounter and every way in which it could appear concretely as a
matrix group. Furthermore, there is a deep combinatorial theory of
other key structures such as the tensor product of
representations. The ideas and results of this subject are the basic
input to diverse areas from number theory to quantum field theory.
Though mysteries still remain, the theory of compact Lie groups is a
representative model for what we would like to achieve with other
symmetry groups.  Because of their universal importance, symmetry
groups come in many different flavors: finite groups, Lie groups,
$p$-adic groups, loop groups, ad\'elic groups,...  and the list will
only increase with the discovery of new important structures.  For the
above examples, our understanding is still very coarse though the last
decades have witnessed breathtaking advances.  The Langlands program
along with its geometric spinoffs provide a visionary roadmap for
where the subject could go in the coming years.  In particular,
current developments such as the recent proof of the Fundamental Lemma
create great optimism that geometric techniques will have a deep
impact.
Geometric representation theory seeks to understand groups and
representations as a consequence of more subtle but fundamental
symmetries. A groundbreaking example of its success is
Beilinson-Bernstein's uniform construction of all representations of
Lie groups via the geometry of $\mathcal D$-modules on flag varieties.
The result is not difficult to state and prove but has the
Borel-Weil-Bott theorem and the Kazhdan-Lusztig multiplicity
conjectures as immediate consequences. A reasonable reaction is to
wonder what allows one to prove deep results about representations of
Lie groups with so little effort. One answer is that the true focus of
the Beilinson-Bernstein theory is not representations but rather the
symmetries of flag varieties. The geometric notion of $\mathcal
D$-module allows one to localize symmetries and to apply the
sheaf-theoretic techniques of algebraic geometry.  In this way, our
understanding of representations of Lie groups follows from that of
infinitesimal symmetries of algebraic varieties.
There are now many instances where difficult questions about
representations can be translated into more tractable questions about
geometry.  Other famous examples include the Deligne-Lusztig theory of
representations of finite groups of Lie type, the Springer theory of
representations of Weyl groups, the Kazhdan-Lusztig theory of modules
for Hecke algebras, and Lusztig and Nakajima's theory of
representations of Kac-Moody algebras and quantum groups via quiver
varieties. In the theory of automorphic forms, one of the fundamental
tools is the realization of representations of ad`elic groups in the
cohomology of Shimura varieties (in the case of number fields) and
Drinfeld modular varieties (in the case of function fields). The
recent proof of the Fundamental Lemma exploits the fact that computing
orbital integrals in $p$-adic groups can be reduced to calculating
cohomology of fibers of Hitchin's integrable system.
The twin modern goals of geometric representation theory are to
explore the above dictionaries and to discover new unexpected ones.
As an important consequence, the geometric realization of
representations often reveals deeper layers of structure in the form
of categorification. Categorification typically turns numbers (for
example, the coefficients of Kazhdan-Lusztig polynomials) into the
dimensions of vector spaces (in this case, the Ext groups of
intersection cohomology sheaves).  It is a primary explanation for
miraculous integrality and positivity properties in algebraic
combinatorics.  At the center of geometric representation theory is
Grothendieck's categorification of functions by $\ell$-adic
sheaves. An important example is Lusztig's theory of character
sheaves: it provides a uniform geometric source for the characters of
all finite groups of Lie type.  Another important example is the
theory of canonical bases: here categorification replaces
representation spaces with linear categories equipped with canonical
generating objects.  Another broad example is the geometric Langlands
program: it provides a categorification of the Langlands program in
the setting of function fields, providing new insights into many
classical constructions.
Geometric representation theory has close and profound connections to
many fields of mathematics, which we expect to play a significant role
in the program. Perhaps the most significant are to number theory, via
the theory of automorphic forms, L-functions and modularity. Much
current activity in the field is motivated either directly by problems
in number theory or by more tractable geometric analogues
thereof. Another major influence on the subject comes from physics, in
particular gauge theory, integrable systems, and recently topological
string theory.  There is a significant interaction with the theory of
$C^*$-algebras through the Baum-Connes conjecture, an instance of
which provides an organizing principle for representation theory of
real and $p$-adic groups. Finally, geometric representation theory is
closely entwined with very active areas in combinatorics  such as
Schubert calculus, its affine analogue, and the theory of
Macdonald polynomials.
The proposed program has two primary goals: 

To bring together researchers working in the
arithmetic and geometric Langlands programs so as to discover 
new relations between the objects appearing in the two
subjects.
To explore new principles and paradigms within 
geometric representation theory.

A particular emphasis will be placed on geometric methods in
representation theory over local fields. This is an area which has advanced
dramatically in the past decade. More importantly, it holds great promise for
future discoveries of interest to a diverse collection of researchers.<|endoftext|>
TITLE: Resolvent of Laplacian
QUESTION [7 upvotes]: Hello!
Let $(M,g)$ be a Riemannian manifold and $-\Delta$ the Laplacian on M (acting on smooth functions). Then the resolvent $R(\xi)$ of $-\Delta$ is a compact operator.
Is it possible to find for every $\epsilon>0$, a point in the resolvent set $\xi$, s.t. $\Vert R(\xi) \Vert\leq \epsilon$? 
Maybe it is very easy to prove, but I'm not so familiar with spectral theory. I hope you can help me.
Regards.

REPLY [7 votes]: This is mostly enhancing Nik Weaver's comments.  Suppose  that $M$ is compact of dimension $m$. If $m\geq 2$, then for a generic metric $g$ on $M$ the eigenvalues $\lambda_k$  of the Laplacian $\Delta_g$ are   simple. In general, for any $m$,  Weyl's spectral estimates  imply that
$$\lambda_k \sim C_m \left(\frac{k}{{\rm vol}_g(M)}\right)^{\frac{2}{m}}\;\;\mbox{as $k\to\infty$},  $$
where $C_m$ is an explicit universal constant that depends only on $m$. (Hat-tip to Marc Palm!)  In particular this shows that for $m\geq 2$ and a generic metric  we have
$$0<\lambda_{k+1}-\lambda_k =O(1). $$
Now Nik Weaver's argument shows that there exists $r_0>0$ such   for any $\xi\in [0,\infty)\setminus {\rm spec}\;(\Delta)$  we have  $\Vert(R(\xi)\Vert\geq r_0$.<|endoftext|>
TITLE: Is there non-simple-connected projective variety(over C) with trivial etale fundamental group?
QUESTION [5 upvotes]: As the title. Geometrically, is there a projective complex manifold(or more generally an projective algebraic variety) accepting only infinite nontrivial cover(which may not be projective)?
Thanks.

REPLY [11 votes]: There are several classes of spaces for which this question can be asked, here are the answers:

Compact complex-projective manifolds (also frequently called manifolds admitting flat complex-projective structures): These are n-manifolds admitting an atlas where transition maps are elements of $PGL(n+1, C)$. For such manifolds, either the fundamental group is trivial or it admits a nontrivial projective-linear representation (holonomy of the structure). It follows from Malcev's theorem, that the group always contains a finite-index subgroup (different from the original group). Thus, the answer to OP's question in this setting in NO.  
Complex-projective varieties. It was observed by Serre (and many others) that every finitely-presented group $\pi$ appears as fundamental group of such a variety. One does not need Simpson's paper for this (he was answering a much harder question). Here is the construction. Take finite simplicial complex with the given fundamental group. Embed it as a subcomplex of a standard affine k-simplex. Replace each face of this subcomplex with its complex-projective span. Take the union of these subspaces of $CP^k$. That's your variety. Then, as Ricky noted, the answer to OP's question is YES. 
Smooth complex-projective varieties. Then it becomes a well-known open problem. It is also open for compact Kahler manifolds. 
Compact complex manifolds. It was proven by Taubes in 1992 that every finitely-presented group is the fundamental group of such a 3-dimensional manifold. Thus, the answer again is, YES.<|endoftext|>
TITLE: Two questions about combinatorics journals
QUESTION [19 upvotes]: Hello,
I have two questions regarding combinatorics journals. I hope that this is the right place for such questions.

Which combinatorics/DM journals would you consider as the "top tier"?
I tried to look for an answer online, and found these two links:
http://www.scimagojr.com/journalrank.php?category=2607 and
Top specialized journals .
These somewhat contradict each other (especially regarding EJC), and I assume that the SJR ranking might not be identical to the general public opinion.
What exactly is the difference between Journal of Combinatorial Theory series A and Journal of Combinatorial Theory series B? Wikipedia states that "Series A is concerned primarily with structures, designs, and applications of combinatorics. Series B is concerned primarily with graph and matroid theory.", but this seems a bit vague. For example, JCTA does contain many papers concerning graph theory. I also heard that the journal split due to a disagreement between its founders (or editors?). Can this disagreement shed some light on the difference?

Many thanks,
Adam

REPLY [4 votes]: The closest thing to an official account of why JCT split into JCTA and JCTB may be found in Edwin F. Beschler's article Gian-Carlo Rota and the Founding of the Journal of Combinatorial Theory, J. Combin. Theory Ser. A 91 (2000), 2–4.

However, no single journal or set of editors, prestigious and hard-working as they might be, could overcome the tremendous diversity in hopes and aspirations for combinatorics. Competition for page allocations and scheduling soon led to intolerable strain on the editorial board and there developed a real threat to continuing cooperation and growth.

Thus it seems that the volume of submissions was a major driving factor behind the bifurcation into A and B. Theodore Motzkin agreed to be the chief editor of A, and W. T. Tutte agreed to be the chief editor of B.  The research interests of the chief editors undoubtedly played a role in the decision as to which papers went to which journal.  Some further information is provided in the editorial article Fifty years of the Journal of Combinatorial Theory, J. Combin. Theory Ser. A 144 (2016), 1–6.

The first thing that had to be done was to decide the fate of papers submitted to JCT, but which would only appear after the split into series A and B.  Together with the editors for JCTB, Bill Tutte and Dan Younger, it had to be decided which papers should be routed to JCTB and which to JCTA.  Typically papers on graph theory or matroids went to JCTB and the rest to JCTA.  This amounted to a rough split of about 40/60. Naturally, papers often did not fit obviously into one series or the other and judgements had to be made. This has remained a recurrent problem which we still face to this day.  Over the years the distinction based on graphs and matroids has become much less pronounced as, with the emergence of new subfields such as additive combinatorics, combinatorial commutative algebra and physical combinatorics, the field of combinatorics has grown significantly.<|endoftext|>
TITLE: Categories of recursive functions
QUESTION [20 upvotes]: I have a couple of conjectures on recursive functions, that I feel must have been proved or refuted by someone else, but I don't know where to look. In short:
1. The primitive recursive functions form a pseudoinitial small finite product category with natural number object.
2. The partial recursive functions form a pseudoinitial small regular category with natural number object.
The longer version is as follows.
In non closed Cartesian categories, the natural number objects should be defined in a more stable way: $N$ together with $0:1\to N$ and $s:N\to N$ is a natural number object if for each $f:X\to Y$ and $g:Y\to Y$ there is an $h:N\times X \to Y$ such that $h(0,x) = f(x)$ and $h(n+1,x) = g(h(n,x))$. This roughly means that the projection $N\times X\to X$ is a natural number object in the slice over $X$ for every object $X$.
A 0-cell in a 2-category is pseudoinitial if there is an up to isomorphism unique 1-cell to every other two cell.
The first conjecture in full is as follows: The 2-category of small finite product category with a chosen natural number object and finite product preserving functors which preserve the choice of natural number object and all natural transformations, has a pseudoinitial 0-cell. One of these pseudoinitial 0-cells is the category whose objects are powers of $\mathbb N$ and whose morphisms are primitive recursive functions.
The second conjecture says that there is a pseudoinitial object in the category of small regular categories with NNO. This time the category of recursively enumerable sets and recursive functions is such a pseudoinitial 0-cell.
The conjectures fail if non standard models of arithmetic can exclude primitive / partial recursive functions that exists in the standard model: categories of non standard recursive functions could be counterexamples.
Have you seen anything like this before? If so, am I right?

REPLY [6 votes]: Regarding question 1.: it's hard to find the statement spelled out in black and white in the literature. Apparently Gavin Wraith proved it in unpublished notes titled "Notes on arithmetic universes and Gödel incompleteness theorems" (1985). (And, one might imagine Joyal proved this as well in his unpublished work.) Happily, a crisp statement is provided in Alan Morrison's very fine Master's Thesis, Lemma 5.11: the functions representable in the initial Skolem theory (the initial object among Lawvere theories in which the generating object is a parametrized natural numbers object) are precisely the primitive recursive functions. He works through the details using a simple programming language of register machines, called PRIM. My understanding is that this is very closely modeled after Wraith's notes; see Maietti's article Joyal's arithmetic universes via type theory, especially her remark just after Definition 4.2. 
Hot off the press is Gödel’s Incompleteness after Joyal by Joost van Dijk and Alexander Gietelink Oldenziel. From the introduction: 

This article concerns an alternative proof of the first Gödel Incompleteness theorem in the language of category theory due to André Joyal, relying crucially on his newly introduced notion of 'Arithmetic Universe’. In 1973 Joyal lectured on his new proof, and a set of notes were circulated among a small group of workers in topos theory. Unfortunately, the proof has never been made publicly available. This document means to remedy this gap in the literature.<|endoftext|>
TITLE: Simply connected algebraic groups and reductive subgroups of maximal rank
QUESTION [13 upvotes]: Recall that a connected semisimple algebraic group $G$ over an algebraically closed field $K$ of arbitrary characteristic was defined by Chevalley to be simply connected if the character group $X(T)$ of a maximal torus $T$ in $G$ is "as large as possible": equal to the abstract weight lattice associated to the root lattice $\mathbb{Z} \Phi \subset X(T)$ (here $\Phi$ denotes the root system of $G$ relative to $T$).   Since all maximal tori are conjugate, this is independent of the choice of $T$. Fortunately this notion of "simply connected" agrees with the usual topological notion when $K = \mathbb{C}$.  A typical example is $\mathrm{SL}_n$.
The closed reductive subgroups $H$ of $G$ containing $T$ have been well studied, but one loose end still bothers me.  The connected component of the identity $H^\circ$ is generated by $T$ and certain pairs of opposite root groups for roots forming a subsystem $\Psi$ of $\Phi$ (Borel-Tits).   Typical examples are the derived groups of Levi subgroups in parabolic subgroups of $G$ (complementary there to the unipoent radical), which are always connected.   Imitating work of Borel and de Siebenthal (1949) on compact Lie groups, one gets these and sometimes more examples by taking root subsystems corresponding to proper subsets of vertices in the extended Dynkin diagram.   
Equally natural are the groups $H = C_G(s)$, centralizers of semisimple elements (not necessarily connected, unless $G$ is simply connected).   These have especially been studied in prime characteristic and have identity components which turn out to be of the type described above: work of Springer and Steinberg, Carter and Deriziotis, McNinch and Sommers. 
Assume for convenience that $G$ is both simply connected and almost-simple, thus of Lie type $A - G$.  

Is the semisimple derived group $H'$ of a connected reductive subgroup of $G$ including $T$ always simply connected?   (If so, is there a uniform proof?)

When $H$ is a Levi subgroup of some parabolic subgroup of $G$, this was proved by Borel-Tits (in the more complicated context of a field of definition).
Some experimental work with subgroups of Borel and deSiebenthal type suggests that the derived groups are also simply connected, but if so the reason is obscure.   Maybe the topological case in characteristic 0 is suggestive?
Example: Take $G$ to be of type $G_2$ (so it is both simply connected and adjoint), with short simple root $\alpha$ and long simple root $\beta$ (picture here). The six long roots form a subsystem of type $A_2$.   Since the original highest root is $3\alpha +2\beta$, its negative (call it $\gamma$) along with $\delta := \beta$ correspond to two vertices of the extended Dynkin diagram and span a subsystem $\Psi$ of type $A_2$ belonging to a subgroup $H$.   Here $H$ is already connected and almost-simple.   In fact it is simply connected: direct calculation shows that the two fundamental weights for $H$ are $-2\alpha -\beta$ and $-\alpha$, thus lie in the weight lattice (= root lattice) of $G$. 
ADDED: My formulation (and example) probably oversimplify the computations involved in most cases.   When $H'$, with a maximal torus $S := T \cap H'$, has lower rank than $G$ or has a root system with multiple irreducible components, the study of $X(S)$ tends to be more complicated.   This already shows up in the usual Levi subgroups.   For instance, the group $G_2$ has a standard Levi subgroup with derived group of type $\mathrm{SL}_2$ corresponding to either $\alpha$ or $\beta$, but the fundamental weight in $X(S)$ won't lie in $X(T)$.  I hoped one might see whether or not a uniform pattern exists without too much case-by-case work, but that may be unrealistic.
UPDATE: The comments (especially by Paul) help to clarify what is going on, though I'm left with the partial question: Is there any predictable pattern to the occurrence or non-occurrence of simply connected groups $H'$?  I was motivated to look more closely at an extreme case involving the simple (both adjoint and simply connected) group $G$ of type $E_8$.   By removing the vertex of the extended Dynkin diagram corresponding to $\alpha_2$ (Bourbaki numbering), one gets a subgroup $H= H'$ of type $A_8$ with the same maximal torus $T$ and character group $X(T)$.   Using Chevalley's classification, $H$ will be one of three nontrivial quotients of $\mathrm{SL}_9$ (as a group scheme).   Some tedious bookkeeping with roots and weights shows that $H$ is the "intermediate" group, whose group of rational points has a center of order 3 (when the characteristic is not 3).   Thus $H$ fails to be simply connected or adjoint, the former not obvious a priori but certainly consistent with results of McNinch-Sommers..

REPLY [2 votes]: If $H$ is the fixed points of an involution of $G$ the answer is known completely,
and $H$ is "almost always" not simply connected. This includes many subgroups of maximal rank (whenever the involution is inner).
Suppose $G$ (complex) is simple and simply connected, $H$ is the fixed points of an involution of $G$, and $H$ is semisimple. Then the fundamental group of $H$ is $\mathbb Z/2\mathbb Z$ except in the following cases: $Sp(2n)\subset SL(2n)$, $Spin(n)\subset Spin(n+1)$ ($n\ge 3$), $Sp(2p)\times Sp(2q)\subset Sp(2(p+q))$, 
$F_4\subset E_6$, and $Spin(9)\subset F_4$. The proof is a fairly simple, case-free, 
root/weight lattice argument, along the lines of some of the preceding discussion.
See http://www.ams.org/mathscinet-getitem?mr=2112326.
A closely related question is When is a finite dimensional real or complex Lie Group not a matrix group
It would be interesting, and probably not too difficult, to extend this to the case when $H$ is the fixed points of an automorphism of finite order, and is maximal.<|endoftext|>
TITLE: Is there a "mathematical" definition of "simplify"?
QUESTION [29 upvotes]: Every mathematician knows what "simplify" means, at least intuitively. Otherwise, he or she wouldn't have made it through high school algebra, where one learns to "simplify" expressions like $x(y+x)+x^2(y+1+x)+3(x+3)$. 
But is there an accepted rigorous "mathematical" definition of "simplify" not just for algebraic expressions but for general expressions, which could involve anything, like transcendental functions or recursive functions? If not, then why? I would think that computer algebra uses this idea.

REPLY [2 votes]: There is a theory called Kolmogorov Complexity(KC,also called algorithmic information) which has been initiated by Chaitin,Solomonoff,and Kolmogorov.Roughly speaking,an object is simple if it's KC is shorter,it is related to recursive function or computability theory,or uncomputabilty ,See Kolmogorov complexity and it's application by Ming Li and Vitanyi for the exact definition and examples.or http://www.scholarpedia.org/article/Algorithmic_complexity. It may be what you are looking for.<|endoftext|>
TITLE: A criterion for freeness over a local ring
QUESTION [8 upvotes]: Let $A=K[[X_1,\dots,X_n]]$ where $K$ is a field. Let $M$ be a finitely generated torsion-free $A$-module, such that

for all $k$, the $A[1/X_k]$-module $M[1/X_k]$ is free of rank $d$;
for every $i \neq j$, we have $M = M[1/X_i] \cap M[1/X_j]$.

Does this imply that $M$ is free? 
It certainly does if $n=1$ (easy), and also if $n=2$ (reduce $M$ modulo $X$), but things seem trickier if $n \geq 3$. 
This question comes up when trying to prove that some $(\varphi,\Gamma)$-modules over rings in several variables are free.

REPLY [8 votes]: No, this is false as soon as $n ≥ 3$. A second syzygy $M$ of the residue field $K$ gives a counterexample: each $M[1/X_i]$ is projective, hence free, and it is reflexive, so the second condition is satisfied. On the other hand the projective dimension of $K$ as a module is $n$, so $M$ can not be free.

REPLY [3 votes]: I think the answer is no.  Suppose $M = B$ is a finite ring extension of $A$, say with an isolated singularity, which is both Cohen-Macaulay except over the closed point of $A$ and normal.
Then $M$ cannot be free, since free modules are Cohen-Macaulay.  On the other hand, it is free away from the closed point by the Cohen-Macaulay hypothesis.  The second hypothesis follows from the S2 assumption.<|endoftext|>
TITLE: Which powers of the closed unit interval are homogeneous?
QUESTION [5 upvotes]: It is known that no finite power of the closed unit interval I is homogeneous, while the countable power, i.e., the Hilbert cube, is. It seems that a power is not homogeneous for every successor cardinal. But, is it homogeneous for every limit cardinal, or is this dependent on some of the more sensitive cardinal properties of the exponent?

REPLY [11 votes]: For $\kappa \geq \omega$, $I^\kappa$ is homeomorphic to $(I^\omega)^\kappa$, and product of homogeneous spaces is homogeneous, so $I^\kappa$ is homogeneous for every infinite $\kappa$.<|endoftext|>
TITLE: Is there a deep reason for the fecundity of involutions?
QUESTION [7 upvotes]: You might have come across the book of involutions in your travels. A colleague of mine asked whether there is a natural global reason (versus ad-hoc trickery) for considering involutions in mathematics. The above book provides many situations that suggest such a global perspective. Having witnessed the extraordinary power of certain involutions in operator algebras (e.g. in Tomita-Takesaki theory), I'd be interested in hearing about such a global perspective in summary from an expert. I'm aware that this question as I've asked it risks being trite...perhaps warranting the answer "it's the simplest nontrivial symmetry" but the existence of the above book might suggest otherwise:

Question: What are some "global" reasons for considering involutions in mathematics?

REPLY [3 votes]: It all boils down to the fact that $\mathbb{R}$ has two ends!
In all (most) mathematical processes there is some notion of direction: counting, moving (along a curve), mapping from one space into another, reading a formula from left to right... the geometry of $\mathbb{R}$ is present always in one way or other, and the flip of the negative and positive ends usually induces some sort of involution.<|endoftext|>
TITLE: Hilbert scheme of points on a surface as moduli space of semistable sheaves
QUESTION [8 upvotes]: It is well-known that $Hilb^n(X)$, the hilbert scheme of $n$ points on a smooth projective surface $X$, is isomorphic to $M_X(1,\mathcal O_X,n)$, the moduli space of rank one semistable sheaves with trivial determinant and second chern class $n$.  The canonical morphism in one direction sends a subscheme $Z\subset X$ to it's ideal sheaf $\mathcal I_Z$.  I was wondering how to go in the other direction.  Namely, given a semistable sheaf of rank 1, trivial determinant, and with second chern number $n$, how do I get an injection into $\mathcal O_X$?
A similar result holds for example for Hilbert schemes of curves on Calabi-Yau 3-folds, so an explanation which takes into account this case as well is preferable.

REPLY [11 votes]: By definition semistable sheaves are torsion-free.  Any torsion-free $F$ includes into its double dual, $F\to F^{\ast\ast}$.  The double dual is a reflexive sheaf, so any singularities occur in codimension 3.  In the surface case, we conclude $F^{\ast\ast}$ is a line bundle with trivial determinant, so must be $\mathcal O_X$.
More generally, in "Vector Bundles on Complex Projective Spaces" (Okenek-Schneider-Spindler) it is shown that any rank one reflexive sheaf on a smooth variety $X$ is necessarily a line bundle (Lemma 2.1.1.15).  So even in higher dimensional cases, it follows that $F^{\ast\ast} = \mathcal O_X$.
(In fact, OSS defines the determinant of a torsion-free sheaf $F$ of rank $r$ by $$\det F = (\Lambda^r F)^{\ast\ast},$$ and notes that the determinant is always a line bundle by the cited lemma.  In the rank one case, this is just the double dual, so we get a map $$F \to F^{\ast\ast} = \det F = \mathcal O_X.)$$<|endoftext|>
TITLE: Coefficients of real k-theory with coefficients
QUESTION [7 upvotes]: Question: Calculate the group $ \pi_{8k+2}(KO \wedge M\mathbb Z/l\mathbb Z) $.
Here $KO$ denotes the real k-theory spectrum and $M\mathbb Z/l\mathbb Z $ denotes a Moore Spectrum associated to the cyclic group of order $l$.
Assume for the following that $2$ divides $l$ (otherwise there are no problems).
Using the short exact sequence for Moore spectra one can easily calculated all the coefficients but $ \pi_{8k+2}(KO \wedge M\mathbb Z/l\mathbb Z) $ .
There one gets only a short exact sequence:
$$\mathbb Z/2\mathbb Z \hookrightarrow \pi_{8k+2}(KO \wedge M\mathbb Z/l\mathbb Z)
\twoheadrightarrow \mathbb Z/2\mathbb Z $$
Note that for $l=2$ we have:
$ \pi_{2}(KO \wedge M\mathbb Z/l\mathbb Z)\cong KO_{3}(\mathbb RP^2) $
But I'm neither able to calculate the latter group.
What techniques are used to solve problems like this?
Answer: (following the answer by Tom Goodwillie)
Idee: 
Use $S$-Duality to calculate $\pi_{2}(KO \wedge M\mathbb Z/l\mathbb Z)\cong KO_{3}(C_l)$
where $C_l$ denotes a Moore space to the cyclic group of order $l$ (and $C_2\simeq\mathbb RP^2$).
The $S$-dual of $C_l$ is $\Sigma^{-3}C_l$ (use that $C_l$ can be realized as the cofibre of
$S^1\to S^1$).
By $S$-duality we have: $KO_{3}(C_l) \cong KO^0(C_l)$.
The latter group can be calculated by comparing $C_l$ and $\mathbb RP^2$.
In the end we get that the coefficients are as follows:


Note that the extension splits for some $l$ which I find surprising.

REPLY [9 votes]: By $S$-duality the group $KO_3(\mathbb RP^2)$ is isomorphic to $KO^0(\mathbb RP^2)$. The latter group is not killed by $2$. Let $L$ be the nontrivial real line bundle on $\mathbb RP^2$. The bundle $L\oplus L$ on $\mathbb RP^2$ is nontrivial, even stably nontrivial, because its second Stiefel-Whiney class is not zero.<|endoftext|>
TITLE: Is there an algorithm to decide if an ideal contains monomials?
QUESTION [5 upvotes]: Let $I\subset k[x_1,\dots,x_n]$ be an ideal in a polynomial ring in commuting variables. Is there a procedure to decide if $I$ contains a monomial and possibly to find one? 
Gröbner bases come to mind, but for example, $$I = \langle x-y-z, y^4z^2+2y^3z^3+y^2z^4  \rangle = \langle  x-y-z,x^2y^2z^2\rangle$$ has $$\left\{x^4y^2-2x^3y^3+x^2y^4, y^4z^2+2y^3z^3+y^2z^4, x^4z^2-2x^3z^3+x^2z^4, x-y-z\right\} $$ as its universal Gröbner basis, which makes me pessimistic about Gröbner methods.  In this example, the symmetry can be broken by computing a primary decomposition which reveals the monomials, but what can be said in general?

REPLY [7 votes]: Computing colon ideals is pretty quick. You could colon out the
variables in order. If the ideal changes, record the variable that
worked, and go back to the beginning of the list.
Either you get to the unit ideal, in which case you found 
the lex-first monomial that's in the ideal, or you make it to the
end of the list, in which case there's no monomial in your ideal.
EDITS:

There's no need to loop back to the beginning of the list. Do $x_1$ to completion, then $x_2$, and so on.
Doing $x_i$ only requires a Gröbner basis for an elimination order for $x_i$, and the colon ideal will come with a Gröbner basis again, for free. So you don't even need a universal Gröbner basis, just $n$ elimination Gröbner bases.
You could also do this by coloning out $x_1 x_2 \cdots x_n$, which is very close to 
François Brunault's answer comment. But my understanding is that colon ideals are computed using the elimination of a new variable, so I doubt this is actually faster than what I'm suggesting.
You don't even need those full Gröbner bases, just ones that are Gröbner enough; see e.g. theorem 6.3 of http://arxiv.org/abs/dg-ga/9706003 where we use a criterion like this to compute a cohomology ring.

ANOTHER:
You could slice with random hyperplanes, and whenever you get isolated points, see if each of those points has some coordinate $=0$.<|endoftext|>
TITLE: Every continuous function is homotopic to a locally Lipschitz one
QUESTION [11 upvotes]: I would like to know for which category/class/set of metric spaces the following holds: for any two metric spaces $X$, $Y$, for any continuous function $f:X\to Y$ there exists a locally Lipschitz continuous function $g:X\to Y$ which is homotopic to $f$.
EDIT: One could also ask a class of metrizable topological spaces such that each one of them can be given a metric so that the above property holds. Actually, I am more interested in the underlying topological space than in the actual metric space.
In general, the metric spaces I am considering are complete and weakly separable (there exists a sequence $(\phi_h)$ of $1$-Lipschitz functions such that for any two point $x,\ y$ $d(x,y)=\sup_h|\phi_h(x)-\phi_h(y)|$).
I don't know if this is a known fact among experts or not; in that case, I apologize for the standard question and would ask only for a reference.
ADDENDUM: Although I also have an interest for the general question as it is posed above, I could try to highlight some classes of metrizable spaces I have particular interest in knowing if they fulfill the request or not: manifolds, singular spaces (which singularities are allowed), spaces which are manifolds outside a "small" (in some sense) set, compact manifolds of infinite dimension or manifolds modeled on some "nice" linear space (Banach, Hilbert, Fréchet, ...).
Thanks.

REPLY [3 votes]: Such approximation is possible under some mild assumptions about domain and range.
For the domain you want to have structure of a finite dimensional structure of a metric simplicial complex of locally bounded geometry. For example, a Riemannian manifold or Alexandrov space would do.  For the target you should impose some conditions implying local linear contractibility, for instance, a space which is locally CAT(k), where $k<\infty$ would suffice. The proof is based on barycentric maps of smplices, which you can find in the paper of Bruce Kleiner, "The local structure of length spaces of curvature bounded above", Math. Z. 1999. 
The construction of Lipschitz approximation is the same as cellular approximation in algebraic topology. First, approximate your map on the set of vertices. Then extend to simplices by induction on skeleta, using barycentric simplices as in Kleiner's paper. 
Some of this might even work if domain is infinite dimensional, but you would need to control the Lipschitz constant for the barycentric maps.<|endoftext|>
TITLE: Status of Beilinson conjectures?
QUESTION [12 upvotes]: (I hesitate to post that question here, but I received on answer on FB:) 
Does anyone know how the current status of work on them is? And how the possible generalizations etc. which one thinks currently on, look like?

REPLY [10 votes]: In addition to Andreas's excellent answer, we should also mention the Tamagawa number conjecture of Bloch and Kato, which predicts the undetermined rational factor arising in Beilison's conjectural description of the $L$-value. The Bloch-Kato conjecture was later reformulated and generalized by Fontaine and Perrin-Riou to the case of motives with coefficients in an arbitrary number field. Here are some references :
Bloch, Kato, L-functions and Tamagawa numbers of motives.
Fontaine, Perrin-Riou, Autour des conjectures de Bloch et Kato: cohomologie galoisienne et
 valeurs de fonctions L.
Colmez, Fonctions L p-adiques.
Kings, The Bloch-Kato conjecture on special values of L-functions. A survey of known results.
Flach, The equivariant Tamagawa number conjecture : A survey.
Gealy, On the Tamagawa Number Conjecture for Motives Attached to Modular Forms.
Bellaïche, An introduction to the conjecture of Bloch and Kato.<|endoftext|>
TITLE: Explicit formula for Bockstein hom in group cohomology of elementary abelian p-groups
QUESTION [5 upvotes]: Suppose $G$ is an elementary abelian $p$-group of rank n (for simplicity we can assume n=1). Denote by $\beta$ the well-known Bockstein boundary map from $H^1(G,\mathbb F_p)$ to $H^2(G,\mathbb F_p)$. I am looking for an explicit formula for $\beta(f)$ on $[g|h]$ if we know the value of $f$ on $G$.

REPLY [6 votes]: Write $G=\langle \sigma\rangle\cong \mathbb{Z}/p$. $f\in H^1(G,\mathbb{F}_p)$ can be taken as group homomorphism $f: G \to \mathbb{F}_p$. If $B$ denotes the bar resolution then $\beta(f)$ is represented by the cocycle 
$$B_2 \to \mathbb{F}_p,\;\; [\sigma^i,\sigma^j] \mapsto 
\begin{cases}f(\sigma) & , & i+j\ge p \newline 0 &,& i+j < p\end{cases}\qquad (0\le i,j < p)$$ 
For, $\beta(f)$ is the composition of the connecting homomorphism $\delta: H^1(G,\mathbb{F}_p) \to H^2(G,\mathbb{Z})$ that results from the short exact sequence $0 \to \mathbb{Z} \xrightarrow{\cdot p}\mathbb{Z}\to \mathbb{F}_p\to 0$ and the mod-p reduction $\rho: H^2(G,\mathbb{Z}) \to H^2(G,\mathbb{F}_p)$. 
Let $f(\sigma)=k \mod p$. $\tilde{f}: B_1 \to \mathbb{Z},\;[\sigma^i]\mapsto ik\;(0\le i< p)$ is a lift of $f$ and the homomorphism $B_2 \xrightarrow{\partial}B_1 \xrightarrow{\tilde{f}}\mathbb{Z}$ is 
$$[\sigma^i,\sigma^j] \mapsto \begin{cases}pk & , & i+j \ge p \newline 0 &,& i+j< p\end{cases}\qquad (0\le i,j < p)$$
Since $\delta(f)$ is represented by the cocycle $\frac{1}{p}(\tilde{f}\circ \partial)$, the explicit formula above follows.<|endoftext|>
TITLE: Why does the naive choice of homogeneous coordinate ring of a product of projective schemes not work?
QUESTION [10 upvotes]: Let $S$ be a graded ring with $A := S_0$. Set $X := \textrm{Proj} (S)$. Then the projective coordinate ring of  $X \times_A X$ is the graded ring $ \bigoplus_{n \geq 0} S_n \otimes_A S_n $, cf Hartshorne, Exercise II.5.11. (This matches with geometric intuition because this ring corresponds [at least when $S$ is nice enough, eg if $X$ is projective over $A$ and $S$ is the homogeneous coordinate ring of an embedding into $\mathbf P^m_A$] to the section ring of a very ample sheaf on $X \times_A X$ coming from the exterior product of very ample sheaves on $X$.)
However, there is another natural graded ring that we can construct here, namely $S \otimes_A S$ with the obvious grading coming from total degree. This ring is clearly not the same as the graded ring in the first paragraph above. Naively, though, by analogy to the affine case one might still expect that $\textrm{Proj}(S \otimes_A S) \cong X \times_A X$. Is this true? Probably this is not true, although I don't know how to show it; so a followup question is: what scheme does this produce?

REPLY [20 votes]: A ($\mathbb{Z}$-)grading on a ring $S$ is equivalent to an action of the group $\mathbb{G}_m$ on $U=\operatorname{Spec} S$.  This is an exercise in algebra; the statement is that the symmetric monoidal category of comodules over the Hopf algebra $A[t,t^{-1}]$ is equivalent to the category of graded $A$-modules.  The construction $\operatorname{Proj} S$ then corresponds to taking the quotient of $U\setminus U_0$ by the $\mathbb{G}_m$ action, where $U_0$ is the subscheme corresponding to the "irrelevant ideal".  This is really just the familiar fact that $\mathbb{P}^n$ is the quotient of $\mathbb{A}^{n+1}\setminus 0$ by rescaling.
What happens when we take a product of two such schemes $U$ and $V$ with corresponding graded rings $S$ and $T$?  I'm going to ignore $U_0$ and $V_0$ in this discussion to simplify notation (i.e., I'm going to pretend that $\operatorname{Proj} S$ is just $U/\mathbb{G_m}$).  Well, $U\times V$ will naturally have an action of $\mathbb{G}_m\times \mathbb{G}_m$; this corresponds to the natural bigrading on $S\otimes T$.  By taking the diagonal $\mathbb{G}_m\to\mathbb{G}_m\times\mathbb{G}_m$, we get an action of $\mathbb{G}_m$ on the product, and this corresponds to the "total" grading on $S\otimes T$.  But if we mod out $U\times V$ by the diagonal action of $\mathbb{G}_m$, we don't get $U/\mathbb{G}_m\times V/\mathbb{G}_m$; to get that, we would need to mod out the entire action of $\mathbb{G}_m\times \mathbb{G}_m$.  Alternatively, instead of modding out the whole action of $\mathbb{G}_m\times\mathbb{G}_m$ at once, we could first mod out the action of the antidiagonal $\alpha:\mathbb{G}_m\to\mathbb{G}_m\times\mathbb{G}_m$ given by $\alpha(t)=(t,t^{-1})$ and then mod out the action of the diagonal; this works since the antidiagonal and the diagonal together generate the whole group.  What do we get when we mod out the antidiagonal?  Well, it turns out that in this case the quotient will still be affine, and its coordinate ring can be found by taking invariants of the action on $S\otimes T$.  If $x\in S_n$ and $y\in T_m$, then the the antidiagonal action of $t\in\mathbb{G}_m$ sends $x\otimes y$ to $t^{n-m}x\otimes y$, so the invariants are exactly $\bigoplus S_n\otimes T_n$.  Thus the product $U/\mathbb{G}_m\times V/\mathbb{G}_m$ can be obtained as $\operatorname{Proj}(\bigoplus S_n\otimes T_n)$.<|endoftext|>
TITLE: IBN for algebraic theories
QUESTION [11 upvotes]: Let us say that a finitary algebraic theory $\tau$ has IBN (invariant basis number) if the free functor $F : \mathsf{Set} \to \mathsf{Mod}(\tau)$ reflects the isomorphism relation: If $S,T$ are sets with $F(S) \cong F(T)$, then $S \cong T$. When $\tau$ is the theory of $R$-modules for some ring $R$, then this is the usual IBN property of $R$ (at least when we restrict to finite sets).
If $\tau \to \sigma$ is a homomorphism and $\sigma$ satisfies IBN, then also $\tau$ satisfies IBN. Besides the classical example of vector spaces, this gives lots of examples for IBN theories (abelian groups, modules over commutative rings $\neq 0$, groups and Lie algebras (using abelianization), monoids, semigroups, quasigroups, loops, magmas, commutative variants of them, etc.). One can show IBN for (commutative) $R$-algebras, where $R \neq 0$ is a commutative ring. Benjamin Steinberg has remarked in the comments that $\tau$ has IBN when there is a $\tau$-module with finite cardinality $>1$. This gives lots of further examples.
A directed colimit of IBN theories $\tau = \mathrm{colim}_i \tau_i$ is IBN for finite sets and therefore IBN by E: If $F_\tau(S) \to F_\tau(T)$ is a homomorphism, it is given by a map $S \to F(T)$, which factors through some $S \to F_{\tau_i}(T)$. Similarily the other way round. That $F_\tau(S) \to F_\tau(T) \to F_\tau(S)$ is the identity, already holds for the factorizations if $i$ is big enough. Therefore it suffices to consider finitely presented theories. 
Now I have several questions:
A. Has the IBN property for algebraic theories in general been studied in the literature?
B. What are further interesting examples of IBN or $\neg$ IBN (beyond module categories)?
C. What about the infinitary theory of compact Hausdorff spaces? If $X,Y$ are sets such that their Stone-Čech compactifications $\beta(X),\beta(Y)$ are homeomorphic, does it follow $X \cong Y$? (answered by Benjamin Steinberg: Yes)
D. Do nontrivial commutative algebraic theories satisfy IBN? In other words, is the rank of a free module over a nontrivial generalized ring à la Durov well-defined? This should be crucial for the theory of generalized schemes, right?
E. Is there some algebraic theory which satisfies IBN for finite sets, but not IBN for arbitrary sets? (answered by Joseph Van Name: No).
F. When $|S| < \kappa$, then $F(S)$ is $\kappa$-presentable. Is there some $\tau$ which satisfies IBN, but $F(S)$ is $\kappa$-presentable for some $\kappa < |S|$?

REPLY [2 votes]: A. Has the IBN property for algebraic theories in general been studied in the literature?

Yes, mostly by many polish authors. See Gr\"atzer, universal algebra, Chap. 5.
For example, at pag. 198 you read the already cited result

Corollary. If an algebra has an infinite basis, then all bases are infinite and have the same cardinal number.

A sample from pag. 204: 

For $v$ and $v^*$-algebras, one can prove that any two bases have the same number of elements by using the same technique as for vector spaces, namely, by use of thef EIS property. It is rather surprising that, although the EIS property is not valid for $v^{**}$-algebras, the invariance of the number of elements in a basis can still be proved.<|endoftext|>
TITLE: The sum of same powers of all matrices modulo p
QUESTION [20 upvotes]: The following is a problem from our department algebra competition for
students:

Non-question.
   An experimental-math geek was trying to raise all matrices $17\times17$
  over the field with 17 elements to the power of 100, sum the returns,
  and observe the result, when his computer broke. Help him.

Probably, the most of us can calculate the sum without use of computer
(alternatively, Russian readers can find a solution
here).
The problem seems to become much
harder when we replace 100 by, say, 80. More generally, my question is the
following:

Question.
  What is the sum
  $$
> \sum_{A\in M_p(\mathbb F_p)}A^k,
> $$
  where $p$ is prime and $k$ is a multiple of $p-1$?

It is easy to show that the sum is a scalar matrix, but which one?
Note that the coincidence of the matrix size and the characteristic
makes trace useless.

REPLY [14 votes]: [corrections applied, per Ilya and Anton]
Consider the formal series
$$
f(x) = \sum_{k=0}^\infty (\sum_A A^k) x^k
$$
where $A$ runs through $p \times p$ matrices.  It is equal to $\sum_A (I - Ax)^{-1}$, a rational function with values in scalar matrices.  Thus, for some $d$, if the first $d$ coefficients vanish all coefficients must vanish.
To determine $d$, we need to bound the degree of the numerator and denominator of $f(x)$.  Note by Cramer's rule, each $(I - Ax)^{-1}$ is a matrix of degree $(p-1)$ polynomials divided by the degree $p$ determinant of $(I - Ax)$.
There are $p^p$ different denominators, i.e. polynomials of the form 
$$
\det(I - Ax) = c_p x^p + c_{p-1} x^{p-1} + \cdots + c_1 x + 1
$$
The $c_i$ are unconstrained, they appear at least once in case $A$ is the companion matrix with $-c_p,-c_{p-1},\ldots$ along the last column.  So we can write the sum as
$$
\sum_{c_p,\ldots,c_1} \frac{\text{something of degree $p-1$}}{c_p x^p + \cdots + c_1 x + 1} 
$$
There are $(p-1)$ linear denominators, $p (p-1)$ quadratic denominators, $p^2 (p-1)$ cubic denominators and so on, so the product of all the $c_p x^p + \cdots + 1$ has degree
$$
d = (p-1) + 2p(p-1) + 3p^2(p-1) + \cdots + p p^{p-1}(p-1)
$$
i.e. $d = p^{p+1} - \frac{p^p - 1}{p - 1}$.  So $f(x)$ is a polynomial of degree less than $d$ divided by a polynomial of degree $d$.
Ilya's answer shows that the first $p^{p+1} - p$ Taylor coefficients of $f(x)$ vanish, and when $p \geq 3$ this is larger than $d$ so all the remaining Taylor coefficients are vanish as well.  When $p = 2$ Will points out the sums are not always zero, in fact for $p=2$
$$
f(x) = \frac{x^5}{(x+1)^2 (x^2 + x + 1)} I
$$<|endoftext|>
TITLE: What is the relationship between these two notions of "period"?
QUESTION [16 upvotes]: The motivation for this question is to understand a recent theorem of Francis Brown which implies that all periods of mixed Tate motives over $\mathbb{Z}$ lie in $\mathcal{Z}[\frac{1}{2\pi i}]$, where $\mathcal{Z}$ is the $\mathbb{Q}$-span of the set of multiple zeta values (of positive integer arguments). My picture of mixed Tate motives is not very clear, and I would like to be able to relate their periods to something I understand better.
There is a survey article of Kontsevich and Zagier which defines a period as a complex number whose real and imaginary parts are given by convergent integrals of rational functions with rational coefficients, over domains in $\mathbb{R}^n$ cut out by finitely many polynomial inequalities with rational coefficients.

What is the relationship between the set of periods of mixed Tate motives over $\mathbb{Z}$ and the set of periods in the sense of Kontsevich/Zagier? Does one of these sets contain the other?

I would be interested to see examples of periods of one kind which are not periods of the other.

REPLY [3 votes]: Let me try to explain the relation, without going into technical details of inverting $2\pi i.$ 
There are  three Tannakian categories appearing here.

Conjectural abelian category of mixed motives over $\mathbb{Q}.$ Periods in the sense of Kontsevich-Zagier appear as matrix coefficients of the comparison isomorphism between de Rham and Betti fiber functors. Explicitly, they will look like integrals of rational functions over domains, defined by rational inequalities. For every period one can find a smooth proper algebraic variety $X$ and a pair of devisors $D_1, D_2$ so that the period equals to 
$$
\int_{\gamma}\omega
$$
for a cycle with boundary on $D_2$ and an algebraic form with poles on $D_1.$ Cohomology groups $$H^k(X\setminus D_1, D_2)$$ define  objects in the category of mixed motives.

This category is terribly big and complicated. It also has lots of simple objects, coming from cohomology of smooth proper varieties defined over $\mathbb{Q},$ like elliptic curves. 

Mixed Tate Motives over $\mathbb{Q}.$ This category is known to exist. It should be a subcategory of the category of mixed motives, generated by objects
$$H^k(X\setminus D_1, D_2),$$
where the mixed Hodge structure on cohomology is of mixed Tate type. Typical periods here are polylogarithms at rational values, like $Li_5\left(\frac{23}{7}\right).$ 

This category is smaller, but still big. All its simple objects are pure Tate motives $\mathbb{Q}(n),$ it has homological dimension one and $Ext^1(\mathbb{Q}(0), \mathbb{Q}(n))$ is finite dimensional for $n\ge 2.$ Notice that 
$$
Ext^1(\mathbb{Q}(0), \mathbb{Q}(1))=K_1(\mathbb{Q})_\mathbb{Q}=\mathbb{Q}^{\times}_{\mathbb{Q}},
$$ is not finite dimensional. This is the reason why this category is still very big. In particular, the graded Hopf algebra of periods is not finitely generated in any degree.
3.Finally, mixed Tate motives unramified over $\mathbb{Z}.$ This is a tiny subcategory of the previous one. It has the same simple objects and the same 
$Ext^1(\mathbb{Q}(0), \mathbb{Q}(n))$ for $n\geq 2,$ but 
$$
Ext^1(\mathbb{Q}(0), \mathbb{Q}(1))=K_1(\mathbb{Z})_\mathbb{Q}=0.
$$
This means that the graded Hopf algebra of its periods is finitely generated in each degree. This is (thanks to the theorem of Francis Brown) the algebra of motivic multiple zeta values.<|endoftext|>
TITLE: When are Brieskorn Manifolds Homeomorphic?
QUESTION [13 upvotes]: Let $a_0, \dots, a_n, b_0, \dots, b_n \in \mathbb{N}$ and consider two polynomials $f = \sum_{i = 0}^{n} z_{i}^{a_i}$, $g = \sum_{i = 0}^{n} z_{i}^{b_i}$. Given two Brieskorn manifolds $\Sigma(a_0, \dots, a_n) = f^{-1}(0) \cap S^{2n+1}$ and  $\Sigma(b_0, \dots, b_n) = g^{-1}(0) \cap S^{2n+1}$ (for sufficiently small spheres), under what necessary and sufficient conditions on the exponents can one conclude a homeomorphism 
\begin{align}
\Sigma(a_0, \dots, a_n) \cong \Sigma(b_0, \dots, b_n).
\end{align} 
NB: I do not assume that either of the Brieskorn Manifolds are necessarily integral homology $(2n-1)$-spheres, so the exponents are not necessarily pairwise coprime.
For $n = 1$, the manifolds are torus links. Therefore, $(a_0, a_1) = (b_0, b_1)$ (ignoring sign and order). 
For $n = 2$, the genera of the corresponding Seifert surfaces must be equal,
\begin{align}
g = \frac{1}{2}\left(  \frac{d}{\tau} - l \right) + 1 = \frac{1}{2} \left( \frac{d^{\prime}}{\tau^{\prime}} -  l^{\prime} \right) + 1 = g^{\prime},
\end{align}
where $l = \gcd(a_0,a_1) + \gcd(a_1,a_2) + \gcd(a_2,a_1)$, $d =\gcd(a_0,a_1)   \gcd(a_1,a_2)   \gcd(a_2,a_1)$ and $\tau = \gcd(a_0,a_1,a_2)$ and the primed version corresponding to $(b_0, b_1, b_2)$.
Is this equality also sufficient to imply homeomorphism (for $n = 2$)?
For the general case, is it necessary and/or sufficient that the sums of the GCDs of the subsets of the same size of the exponents sets be equal? 
Is there a general reference on the invariants of Brieskorn Manifolds for the general case?

REPLY [8 votes]: This is by no means a complete answer but, rather, a DIY suggestion:
Let $B$ be a $2k+1$-dimensional Brieskorn manifold. Then $B$ is $k-1$-connected. 
C.T.C. Wall wrote in "Classification problems in differential topology—VI. Classification of (s−1)-connected (2s+1)-manifolds" a complete list of invariants (up to diffeomorphism) for such manifolds (there are few exceptions, but one should be able to deal with them on case-by-case basis), assuming, of course, that $k\ge 2$, so the fundamental group is trivial. The invariants are (mostly) of homological nature, so you should compare them with computations done by Hirzebruch's and Milnor, to see if you get enough information from there to determine a complete list of diffeomorphism invariants for Brieskorn manifolds in terms of the parameters $a_i$. 
It is quite possible that such analysis was made by Alan Durfee in his 1971 thesis "Diffeomorphism classification of isolated hypersurface singularities". You may want to ask Durfee (he is at Mount Holyoke College) for a copy. I have no idea why his thesis was never published, but people refer to it quite a bit. 
Now, if $k=1$, then the situation is quite different and $B$ is a Seifert manifold. Topology of such manifolds is completely determined by "Seifert invariants" which were first computed by Milnor  here if there are no singular fibers (which reduces to computing genus of the base ond Euler number of the fibration) and, in general by Neumann and Raymond here, in terms of the parameters $a_i$, following an earlier paper by Neumann which I do not have access to. So, answering your question in this case, is still a DIY project, working through the formulae in the paper by Neumann and Raymond. 
Addendum: Here is the link to a scan of Neumann's thesis. Since it is a scan, it is harder to read, but, unlike the paper of Neumann and Raymond, it deals specifically with Brieskorn manifolds, not with complete intersections of such.  
Here is the description (taken fron Neumann's thesis) of a complete set of topological invariants of $\Sigma=\Sigma(a_0,a_1,a_2)$ from the vector $(a_0,a_1,a_2)$ in the generic case (for nongeneric cases see Corollary 9.2 in Neumann's thesis). This is not at all pretty (to say the least), but it is what it is. Define numbers
$$
d=gcd(a_0,a_1,a_2),
$$
$$
a_i'= \frac{1}{a_i} lcm(a_0,a_1,a_2),
$$
$$
t_i= gcd(a_j', a_k'), \{i,j,k\}=\{0,1,2\}
$$
$$
s_i= \frac{1}{d} gcd(a_j, a_k), \{i,j,k\}=\{0,1,2\}
$$
$$
g= \frac{1}{2}(d^2 s_0 s_1 s_2- d(s_0+s_1+s_2)) +1. 
$$
Genericity assumption: $t_0, t_1, t_2$ are all $\ne 1$. Now, find integers $\beta_i'$ so that
$$
0\le \beta_i'< t_i, \quad \beta_i'a_i' =  1 (mod\ t_j) 
$$
and set
$$
b= \frac{d}{t_0t_1t_2}(1- \sum_{i=0}^2 \beta_i'a_i')
$$
Then the tuple
$$
(g; b; \{ds_0(t_0, \beta_0'), ds_1(t_1, \beta_1'), ds_2(t_2, \beta_2')\})
$$
is a complete topological invariant of $\Sigma$. Here 
$$
ds_i(t_i, \beta_i')= (ds_i t_i, ds_i \beta_i'). 
$$
Topological meaning of some of the quantities in this tuple:

$g$ is the genus of the base-orbifold $O$ of the Seifert fibration on $\Sigma$. 
Under our genericity assumptions, the base-orbifold $O$ will have $3$ singular points of the orders
$$
ds_i t_i, i=0, 1, 2. 
$$
The numbers 
$$
ds_i \beta_i'
$$
define the second set of invariants for the Seifert fibration at the singular fibers. 
The number $b$ is responsible for the Euler number of the Seifert fibration (I did not bother to write a precise formula for the transition between these invariants, maybe it literally is the Euler number). 

Given how complex this description is, it is very likely that the complete sets of topological/smooth invariants in higher dimensions is much messier. 
Now, consider the special case where the numbers $a_0, a_1, a_2$ are pairwise coprime, and greater than $1$. Then 
$$
g=0, t_i=a_i, d=1, s_i=1, ds_it_i=a_i. 
$$
In particular, the numbers $a_i$ are orders of cone-points of the base-orbifold. In particular, for coprime numbers $a_i$, the vector $(a_0, a_1, a_2)$ is the complete topological invariant of $\Sigma$. This was the original comment made by myself and Bruno: We both missed the coprimality condition.  
In this setting, your "genus" equals $2$ (I still do not know why do you call it "genus"; I would write instead:
$$
\frac{1}{2}\left(\frac{d}{\tau} -l\right) + 1,
$$
then, at least in the coprime case it matches the genus of the base-orbifold.) Now, it is clear that this number is insufficient to determine the topology of $\Sigma$.<|endoftext|>
TITLE: Approximate number of primes below a given integer?
QUESTION [6 upvotes]: The problem of the complexity of the exact counting problem for primes is interesting. The best result we have about primes is that it is hard for TC0. But counting the number of witnesses to a TC0 function can be as difficult as sharpP. The best upper-bound is P by AKS which gives a sharpP upper-bound for the exact counting problem.
There is no known algorithm for exact counting better that P, otherwise we would be able to check if a number x is a prime by counting the number of primes before x-1 and x and comparing them. 
The problem can't be in P unless we can solve the problem of finding prime numbers in P which is an open problem: we can use a binary search to find the first prime number after x by finding the first number y>x where there are more primes before y than x.
Can we do better if we relax the question to approximate counting? I know that the question is not completely well-defined. We need to first clarify what we mean by "approximate", e.g., an absolute error or a relative error? etc. But I don't know what would be a good definition.
Given $n$, how well can we approximate the number of prime numbers below $n$ in polynomial time?

REPLY [10 votes]: The Lagarias-Odlyzko algorithm gives a method for counting the number of primes less than $n$ in time $O_{\epsilon} ( n^{1/2+\epsilon})$ time. Roughly speaking, this algorithm proceeds by expressing the prime counting function as an integral involving the Riemann zeta function and then approximating the integral with numerical integration. This is currently the best known result in this direction, however the Polymath4 project found an algorithm for computing the parity of the number of primes  less than $n$ in time $n^{1/2 -\delta}$ for some small $\delta > 0 $.<|endoftext|>
TITLE: Hypercohomology of a complex via Cech cohomology
QUESTION [12 upvotes]: Let $X$ be a reasonable topological space. If $\mathcal{F}$ is a sheaf of abelian groups then Cech cohomology gives us a method to compute the cohomology groups $H^p(X, \mathcal{F})$ - the main input being the sections $\mathcal{F}(U)$ for various open sets $U \subset X$.
I would like to have a similar procedure to compute hypercohomology of a finite complex $C^\cdot$ of abelian sheaves on $X$ (or coherent sheaves on a scheme). Is this possible, and is there a reference you can recommend? I couldn't find this in the Stacks project, which incidently explains how to compute cohomology of a complex (not hypercohomology) via a Cech argument.

REPLY [5 votes]: There is a nice treatment of it in chapter 1 of Brylisnki's Loop Spaces, Characteristic Classes and Geometric Quantization. In the Stacks projects, look for section 19.19.<|endoftext|>
TITLE: Irreducible Degrees and the Order of a Finite Group 
QUESTION [22 upvotes]: This is a question of aesthetics. 
For a finite group of order $n$, the proof that  the degree $d$ of a complex irreducible representation  divides $n$ goes by showing that the rational number $n/d$ is an algebraic integer. As an application of the fact that  $\mathbf Z$ is an integrally closed domain,  this proof is really spectacular. But I feel that this  is an indirect proof, not providing an insight into what is actually going on.
Being a statement of very basic nature perhaps 
  there are other 'natural'  or alternative ways of seeing why  this happens. I would be grateful if experts here can point out other proofs or explain what is happening in the traditional proof.

REPLY [6 votes]: Here is an exposition of the ``alternative hack'' to which David Speyer alluded.  For concreteness, I have fixed a group; the argument will go through in general, of course.
Let $g = (12) \in S_3$, a transposition in the symmetric group on three elements.  The conjugacy class of $g$ consists of all three transpositions.  Form $a = \frac{1}{3}\left[(12) + (13) + (23) \right] \in \mathbb{C}S_3$, the average of $g$'s conjugacy class.  It is clear that for any $x \in G$,
$$xax^{-1} = a$$
since conjugating $a$ will simply rearrange the terms in the sum.  Equivalently,
$$xa = ax,$$
and $a$ lies in the center of $\mathbb{C}S_3$.  It follows that if $\rho: S_3 \longrightarrow \mbox{Aut}(V)$ is an irreducible representation, the matrix
$$\rho(a) = \frac{1}{3}\left[\rho(12) + \rho(13) + \rho(23) \right]$$
commutes with any $\rho(x)$.  Schur's lemma tells us that $\rho(a)$ is a scalar matrix, that is, $\rho(a) = \lambda I$ for some $\lambda \in \mathbb{C}$.
Summarizing, for each irreducible representation $\rho$, we may define a class function $\psi_{\rho}$ which associates to any $g \in S_3$ the scalar $\lambda$ by which $a=\frac{1}{|S_3|}\sum_{x \in S_3} xgx^{-1}$ acts on $V$.
As it happens, $\psi_{\rho}(g)$ can be computed easily in terms of the character $\chi^{\rho}$: after all, every element conjugate to $g$ has the same trace; by linearity,
$\mbox{Tr}(\rho(a)) = \chi^{\rho}(a) = \chi^{\rho}(g)$.
It follows that
$$\psi_{\rho}(g) = \frac{\chi^{\rho}(g)}{\mbox{dim}V}.$$
A natural next step is to eliminate reference to a particular $g$ using an inner product:
$$\langle \psi_{\rho},\chi^{\rho} \rangle = \frac{1}{\mbox{dim}V}.$$
Expanding the definition of the inner product on class functions,
$$\mbox{Tr}\left[\frac{1}{|S_3|^2}\sum_{g \in S_3} \left(\sum_{x \in S_3} \rho(xgx^{-1}) \right) \rho(g^{-1}) \right] = \frac{1}{\mbox{dim}V},$$
and
$$\frac{1}{|S_3|^2}\sum_{g \in S_3} \sum_{x \in S_3} \chi^{\rho}(xgx^{-1}g^{-1}) = \frac{1}{\mbox{dim}V}.$$
We are led to consider the formal sum $d=\sum_{g,h \in S_3} ghg^{-1}h^{-1}$ mentioned in David Speyer's post.  This sum is invariant under any automorphism of $S_3$ (in particular inner automorphisms) and so lies in the center of the group algebra $\mathbb{C}S_3$.  Schur's lemma tells us that the image of $d$ under any irreducible representation $\rho$ is a scalar.  By the above we get
$$\sum_{g \in S_3} \sum_{h \in S_3} \rho(ghg^{-1}h^{-1}) = \left[\frac{|S_3|}{\mbox{dim}V}\right]^2I.$$
Considering now the regular representation $\mathbb{C}S_3$, we see that the element $d$ acts with rational number eigenvalues.  In other words, its characteristic polynomial splits completely over $\mathbb{Q}$.  But we can also see by inspection that $d$ acts by an integer matrix.  The characteristic polynomial is a monic integer polynomial and splits into linear factors, so its roots are integers.  Since $\mathbb{C}S_3$ contains every irreducible representation as a summand at least once, we see that each
$$\frac{|S_3|}{\mbox{dim}V} \in \mathbb{Z}.$$
In particular, $1$, $2$, and $1$ all divide $6$.<|endoftext|>
TITLE: Is the space of diffeomorphisms homotopy equivalent to a CW-complex?
QUESTION [30 upvotes]: Clarification: My question concerns the homotopy type of the space of $C^k$ diffeomorphisms with the compact-open $C^k$ topology, where $0< k \leq\infty$. I have stated my question below with $k=1$ for definiteness and simplicity. I am not particularly interested in any specific value of $k$.$\newcommand{\Diff}{\operatorname{Diff}}$$\newcommand{\RR}{\mathbb{R}}$
It is fairly well-known that the space $\Diff(M)$ of $C^1$ diffeomorphisms of a closed smooth manifold $M$ is homotopy equivalent to a CW-complex. Here are the relevant facts:

$\Diff(M)$ is a Banach manifold modelled locally on the space of $C^1$ vector fields on $M$.
Metrizable Banach manifolds have the homotopy type of CW-complexes, as shown by Palais.

[Remark: When $M$ is closed, the spaces of $C^k$ diffeomorphisms of $M$ (for $0 < k \leq\infty$) are all homotopy equivalent to each other via the natural inclusions. This can be shown by embedding $M$ smoothly in $\RR^N$, and then using smoothing operators defined by taking convolution with a mollifier.]
When we allow $M$ to not be compact, there are several common topologies on $\Diff(M)$. I am interested in the compact-open (or weak) $C^1$ topology. 

Questions: Is it known whether the space $\Diff(M)$ of $C^1$ diffeomorphisms with the compact-open $C^1$-topology is homotopy equivalent to a CW-complex when $M$ is a smooth manifold without boundary? Is there a known counter-example? Are there particular cases where the answer is known, for example if $M$ is the interior of a compact manifold?

Feel free to use instead $C^k$ diffeomorphisms and/or the compact-open $C^k$ topology for any $0 < k \leq\infty$.
I would also be interested in hearing about any known results related to this question: e.g. for spaces of embeddings of manifolds in the compact-open/weak topology when the source is not the interior of a compact manifold.
Edit: Allen Hatcher gave a very nice answer to my question. Afterwards, I also posted an answer very similar to Allen's, which I was writing when Allen posted his. A pertinent question still remains: (1) Does the result hold for the interior of a compact manifold? Here is a perhaps less pertinent question: (2) For $M$ without boundary, do the path components of $\Diff(M)$ have the homotopy type of a CW-complex?

REPLY [28 votes]: Here is an example where ${\rm Diff}(M)$ with the compact-open topology is not homotopy equivalent to a CW complex. Take $M$ to be a surface of infinite genus, say the simplest one with just one noncompact end.  I will describe an infinite sequence of diffeomorphisms $f_n:M\to M$ converging to the identity in the compact-open topology and all lying in different path-components of ${\rm Diff}(M)$.  Assuming this, suppose $\phi:{\rm Diff}(M) \to X$ is a homotopy equivalence with $X$ a CW complex.  The infinite sequence $f_n$ together with its limit forms a compact set in ${\rm Diff}(M)$, so its image under $\phi$ would be compact and hence would lie in a finite subcomplex of $X$, meeting only finitely many components of $X$.  Thus $\phi$ would not induce a bijection on path-components, a contradiction.
To construct $f_n$, start with an infinite sequence of disjoint simple closed curves $c_n$ in $M$ marching out to infinity, and let $f_n$ be a Dehn twist along $c_n$.  The $f_n$'s converge to the identity in the compact-open topology since the $c_n$'s approach infinity.  We can choose the $c_n$'s so that they represent distinct elements in a basis for $H_1(M)$ and then the $f_n$'s will induce distinct automorphisms of $H_1(M)$.  If two different $f_n$'s were in the same path-component of ${\rm Diff}(M)$ they would have to induce the same automorphism of $H_1(M)$ since any path joining them would restrict to an isotopy of any simple closed curve in $M$ (see the next paragraph below) and a basis for $H_1(M)$ is represented by simple closed curves.
If $g_t$ is a path in ${\rm Diff}(M)$ then the images $g_t(c)$ of any simple closed curve $c$ vary by isotopy since this is true as $t$ varies over a small neighborhood of a given $t_0$, so since the $t$-interval $[0,1]$ is compact, a finite number of these neighborhoods cover $I$ and the claim follows.
Remark: The $f_n$'s were chosen to be Dehn twists just for convenience.  Many other choices of diffeomorphisms would work just as well.  One can easily see how to generalize to higher dimensions.

REPLY [6 votes]: [Edit: Allen Hatcher posted an answer while I was writing this one. Both answers seem to use similar ideas. I will leave my answer here anyway.]
$\newcommand{\Diff}{\operatorname{Diff}}$$\newcommand{\ZZ}{\mathbb{Z}}$$\newcommand{\RR}{\mathbb{R}}$$\newcommand{\CC}{\mathbb{C}}$$\newcommand{\connsum}{\mathbin{\#}}$$\newcommand{\id}{\mathrm{id}}$$\newcommand{\set}[1]{\lbrace #1 \rbrace}$Reading the answer by Peter Michor gave me an idea for a counter-example, which I explain below. The question still remains as to whether the result is valid for the interior of a compact manifold.
Claim: The path components of a space homotopy equivalent to a CW-complex are all open.
A homotopy equivalence induces a bijection on path components. Consequently, if a space $X$ is homotopy equivalent to a space $Y$, and the path components of $Y$ are open in $Y$, then the path components of $X$ are open in $X$. Finally, observe that any CW-complex is locally path connected, and thus its path components are open.
Construction of the counter-example
It thus suffices to present a smooth manifold $M$ without boundary such that the path components of $\Diff(M)$ are not open. Define $M$ to be the open submanifold of $\RR \times S^1$ given by
$$ M = (\RR \times S^1) \setminus (\ZZ \times \set{1}) $$
where $1\in S^1 \subset\CC$. One can also view $M$ as a connected sum of infinitely many punctured spheres $P=S^2\setminus\set{(1,0,0)}$:
$$ M \cong \, \cdots \connsum P \connsum P \connsum P \connsum \cdots $$
Proof that the path component of $\id_M$ is not open in $\Diff(M)$
Pick a neighbourhood $U$ of $\id_M$ in $\Diff(M)$. I will describe a diffeomorphism $\varphi\in U$ which is not in the path component of $\id_M$ in $\Diff(M)$, thus concluding the proof. By definition of the compact-open topology on $\Diff(M)$, there exists a compact subspace $K$ of $M$ such that a given diffeomorphism of $M$ is in $U$ if it is the identity on $K$. Let $n$ be a positive integer large enough so that $K\subset [-n,n]\times S^1$. Now we define the required diffeomorphism $\varphi$ of $M$:
$$ \varphi(t,x) = \bigl( t , e^{i\cdot\theta(t-n)} x \bigr) $$
where $\theta:\RR\to\RR$ is any smooth function which is identically zero on $(-\infty,0]$, and equals $2\pi$ on $[1,+\infty)$. Viewing $M$ as the connected sum of infinitely many punctured spheres, the diffeomorphism $\varphi$ is the result of applying a Dehn twist at the junction cylinder joining the spheres numbered $n$ and $n+1$.
It is easy to see that $\varphi:M\to M$ is not even homotopic to the identity: in fact, the homomorphism induced by $\varphi$ on $\pi_1(M)$ (based at the point $(n,-1)$) is not conjugate to the identity. Here is a simple way to see this:

The group $\pi_1 M$ is free on infinitely many generators.
The homomorphism $\pi_1 \varphi$ coincides with the identity on all loops in $M$ contained in $(-\infty,n]\times S^1$. In particular, $\pi_1 \varphi$ fixes two distinct (actually, infinitely many) free generators of $\pi_1 M$.
The homomorphism $\pi_1 \varphi$ is not the identity homomorphism on $\pi_1 M$: a loop going once around the puncture $(n+1,1)\in\RR\times S^1$ is not fixed by $\pi_1 \varphi$.
In a free group, conjugation by a non-identity element can fix at most one of the free generators.<|endoftext|>
TITLE: dense orders are saturated
QUESTION [9 upvotes]: In a FOM msg of Fri May 21 19:59:44 EDT 1999, 
http://www.cs.nyu.edu/pipermail/fom/1999-May/003149.html
Simpson gave a short proof of the following theorem:
Theorem.  Let F be a real closed ordered field and suppose that 
the ordering of F is $\kappa$-dense, i.e., for every pair of sets $X,Y$ in
F with $X < Y$ and ${\rm card}(X \cup Y)\le \kappa$, there
exists z in F such that $X < z < Y$.  Then F is ($\kappa^+$)-saturated.
In other words, the saturation property wrt cuts implies the full 
model-theoretic saturation. 
It is an immediate corollary that the known field surreal numbers NO is ordered field-isomorphic to a class-size full set-saturated elementary extension R* of the reals R 
(under the global choice axiom), and hence NO admits a full superstructure which is also an elementary extension of the full superstructure over R . 
Simpson comments as follows:
"This result [the thm above - VK] may be well known to those who know it well.  But I
didn't know it before.  Can anyone supply a reference for it? "
As far as I can follow the FOM thread, nobody managed to give a reference. 
Neither can I now. 
Therefore, a question: will anybody be so kind as to offer a reference wrt the theorem above in a published paper.
Vladimir Kanovei
After reading the comments
First of all, I sincerely thank correspondents for taking time and trouble with this issue. 
In the rest, 
1) Regarding [Erdos ea 1955] - thankfully it is available from the Erdos site - theorem 2.1 there literally says that: 
for any ordinal $\alpha>0$, any two rcf of cardinality $\aleph_\alpha$ which are $\eta_\alpha$ sets are isomorphic. 
There is no saturation claim here, so one has still to work towards saturation from this result on. Yet most importantly the setup is restricted to the case of rcf of cardinality $\aleph_\alpha$ - which (and Erdos ea note it just after the theorem) is equivalent to CH being true on $\aleph_\alpha$. 
To conclude, [Erdos ea 1955] is not a solution to the problem. 
2) Regarding [Marker, Model theory] - thankfully, it's downloadable either. Ex 4.5.18 on p. 165 claims that a rcof F is $\kappa$-saturated iff the underlying order is, thus leaving aside the (how much simpler?) problem to prove the Simpson's theorem for orders (or just some type of orders) instead of fields. 
3) Regarding [Ehrlich] pre-1999 papers mentioned - thankfully, downloadable from Philip's www page. The AU 1988 paper contains a theorem on p. 12 saying that 
if $\aleph_\alpha$ is a saturation cardinal then a certain $\alpha$-fragment of No is a unique $\aleph_\alpha$-universally extending ordered field of cardinality $\aleph_\alpha$. 
Even if we take it as given that the universally extending property implies saturation in this context, this does not yield a proof of Simpson's theorem in its general context. 
With thanks again, I reproduce here Simpson's own proof, just for the sake of completeness of the discussion. 
Proof.  By Tarski's result on quantifier elimination for real closed
ordered fields, any subset of $F$ which is definable over $F$ allowing
parameters from $F$ is a finite union of intervals, all of whose
endpoints are in $F$.  But then Tychonoff's theorem 
[in a later msg, Simpson refers to the Rado selection theorem, which is better - VK] plus $\kappa$-density
of $F$ implies that any family of $\kappa$ such sets has the finite
intersection property.  Hence $F$ is $\kappa^+$-saturated.$\square$
Best regards
Vladimir Kanovei
PS. Regarding Hausdorff, his early papers on ordered sets have been reprinted in a brand-new Band 1a of Felix Hausdorff, Gesammelte Werke, with comments (including those on "Pantachies"), now available. 
A commented English translation of Hausdorff early papers, see 
Hausdorff on ordered sets, Volume 25 of History of Mathematics, Felix Hausdorff, 
Editor  J. Jacob M. Plotkin, AMS, 2005.

REPLY [7 votes]: Philip Ehrich and Emil Jeřábek have given useful references (and Dave Marker and I seem to have simultaneously written our responses).

This note is to point out that the result itself was originally established by Erdös, Gillman, and Henriksen; it appears as Theorem 2.1 of their paper below:

Erdös, P.; Gillman, L.; Henriksen, M. An isomorphism theorem for real-closed fields. 
Ann. of Math. (2) 61, (1955). 542–554. 

By the way, an analgous result is also true, surprisingly, for models of PA, and of ZFC. 

For PA, it is due to Jean-François Pabion, in his paper:
Saturated models of Peano arithmetic. J. Symbolic Logic 47 (1982), no. 3, 625–637. 
For ZFC, it is due to Schmerl and Kaufmann; see line 4, section 2 of the following paper (which also includes a proof of Pabion's result).
Kaufmann, M; J.H. Schmerl, Saturation and simple extensions of models of Peano arithmetic. Ann. Pure Appl. Logic 27 (1984), no. 2, 109–136.<|endoftext|>
TITLE: Quantum-Jimbo Algebras: Why Such Fuss About Roots of Unity?
QUESTION [14 upvotes]: Coming from a Lie algebraic background, I'm trying to branch onto quantum group theory. The divide I see all the time, is $q$ a root of unity or $q$ not a root of unity. I am wondering why is this? If I have understood, two reasons are:
(i) The representation theory of the two types is very different.
(ii) At roots of unity, one can quotient $U_q(\frak{g})$ to get a quasi-triangular Hopf algebra.
Do there exist other important ways in which these two types differ?

REPLY [4 votes]: Your first statement is certainly right on the money.  Away from roots of unity the representation theory behaves essentially the same as the representation theory of the classical Lie algebra, while at roots of unity it becomes nonsemisimple and behaves quite like the more complicated rep theory of Lie algebras over finite fields.  Your second statement is a little off though. $U_q({\mathfrak{g}})$ is quasitriangular (up to issues of completion) and indeed ribbon for all values of $q,$ and thus gives things like link invariants.  In fact what happens at roots of unity is that a quotient of a subcategory of the representation theory forms a modular category, which gives a TQFT and a three manifold invariant, which seems to replicate the theory one expects heuristically to get from Chern-Simons quantum field theory.  So the roots of unity are particularly challenging to understand and particularly interesting in terms of representation theory, topology and physics.<|endoftext|>
TITLE: Finite subgroups of $PGL(3,K)$ 
QUESTION [11 upvotes]: It is well-known that finite subgroups of $PGL_2(\mathbb{C})$ are cyclic groups, dihedral groups, A4, S4 and A5 and each of these groups occurs exactly once (up to conjugacy). These facts are classical.
If $K$ is an arbitrary field then theres are Beauville's notes about $PGL_2(K)$:
http://arxiv.org/abs/0909.3942
In dimension $n=3$ there also classical results about subgroups of $PGL(3,\mathbb{C})$. One can also show the connection between subgroups of $SU(3)$ and $PGL(3,\mathbb{C})$. See this discussion for details:
https://math.stackexchange.com/questions/42904/finite-subgroups-of-pgl3-c
So my questions is what can we say about finite subgroups of $PGL(3,K)$ where $K$ is not necessarily algebraicly closed? More concretely, knowing, for example, finite subgroups of $PGL(3,\mathbb{C})$ can one classify finite subgroups in $PGL(3,K)$ for any subfield $K\subset\mathbb{C}$ (up to conjugacy)?

REPLY [5 votes]: Just to complicate things a little, it is possible for a 3-dimensional representation to have Schur index 3. For example, the group of order 63 with centre of order 3 and presentation
$\langle x,y \mid x^7= y^9=1, y^{-1}xy = x^2\rangle$
has four such 3-dimensional complex characters. Their character rings are all equal to the cyclotomic field $F$ of $21$st roots of 1, which is an extension of ${\mathbb Q}$ of degree 12.
I don't know how easy it is to decide whether this representation can be written over some specified subfield $K$ of ${\mathbb C}$. Certainly $K$ must be an extension of $F$ of degree at least 3. One possible $K$ of minimal degree is the field of $|G|$-th roots of 1, but there are probably others.<|endoftext|>
TITLE: On the oscillation of the summatory totient about its average
QUESTION [5 upvotes]: Let
    $$R(x)=\sum_{n\leq x}\phi(n)-\frac{3x^2}{\pi^2}.$$
Montgomery has shown that $R(x)=\Omega_{\pm}(x\sqrt{\log\log x})$, which is the best known lower bound. It seems interesting therefore that
    $$\int_0^{\infty}\frac{R(x)dx}{x^2}=0,$$
because it tells us that the oscillations (which continue indefinitely) are particularly regular. 
I cannot find any references for this integral, so I am wondering if it is known. I would particularly like to find other work of this nature as I cannot prove anything about the rate of convergence of the improper integral (other than $o(1)$ as $X\rightarrow\infty$ where $X$ is the upper limit of integration).

REPLY [5 votes]: I am not sure if this result is explicitly mentioned in the literature, but it certainly is classical.
Let
$$R(x) = \sum_{n \leq x}{\varphi(n)} - \frac{3x^2}{\pi^2}, \qquad H(x) = \sum_{n \leq x}{\frac{\varphi(n)}{n}} - \frac{6x}{\pi^2}.$$
Then by partial summation,
$$\int^{x}_{0}{\frac{R(t)}{t^2} \: dt} = H(x) - \frac{R(x)}{x}.$$
A classical result of Chowla states that
$$H(x) - \frac{R(x)}{x} = O\left((\log x)^{-4}\right).$$
See Lemma 13 of S. Chowla, "Contributions to the analytic theory of numbers", Mathematische Zeitschrift 35:1 (1932), 279-299.
(If you have access to Springer Link then it is available here.)
From a cursory glance of Chowla's proof, the negative powers of a logarithm stem from the prime number theorem applied to the summatory function of the Möbius function, so it is likely that this bound could be improved with more modern estimates for this.
For what it's worth, I answered a question closely related to this here.<|endoftext|>
TITLE: How long can this string of digits be extended?
QUESTION [21 upvotes]: Consider a number $a_1a_2a_3a_4 \dots a_n$ in some base $b$, such that for each $k, 1\leq k \leq n$, the subnumber $a_1a_2\dots a_k$ is a multiple of $k$.
For instance $1836$ is such a number in base $10$, because $1$ is a multiple of $1$, $18$ is a multiple of $2$, $183$ is a multiple of $3$, and $1836$ is a multiple of $4$.
Let $N(b)$ be the maximum possible value of $n$ for base $b$.

How large is $N(b)$?

We might expect $N(b)$  to be about $eb$. Indeed, there are about $b^n/n!$ such numbers of length $n$, which by Stirling's approximation goes below $1$ sometime around $n=eb$.
The only lower bound I have is that $N(b)\geq b$. I don't have any upper bound at all.
This question is the result of a conversation with John Conway.

REPLY [11 votes]: Following links at the OEIS entry mentioned above takes one in a step or two to this page where there are posts (from 2005) with  Maple code and results out to base $23$. The values $N(b)$ for $2 \le b \le 23$ are reported to be 
$2, 6, 7,  10, 11, 18, 17, 22, 25,  26, 28, 35, 39,  38, 39, 45, 48, 48,  52, 53, 56, 58$ 
Note that $N(7) \gt N(8)$ and $N(14) \gt N(15)$ and $N(19)=N(20).$
The ratios $\frac{N(b)}{b}$ are
$1.0, 2.0, 1.75, 2.0, 1.833, 2.571, 2.125, 2.444, 2.5, 2.364, 2.333, 2.692, $$2.786, 2.533, 2.438, 2.647, 2.667, 2.526, 2.60, 2.524, 2.546, 2.522$
Based on the data so far one might feel somewhat safe speculating that  $2\lt \frac{N(b)}{b} \lt 3$ provided $b \gt 6.$ As far as I can see, little nothing is known for sure (including that $N(b)$ is finite althoughthat seems highly likely.)
For each fixed value of $b$ there is a tree of possibilities (if we use a formal root node for level $0$.) A node at level $k-1$ has at most $\lceil \frac{b}{k}\rceil$ children. It might (or might not) be worth looking at the distribution of leaf levels.<|endoftext|>
TITLE: Image of L^1 under the Fourier Transform
QUESTION [24 upvotes]: The Fourier Transform $\mathcal{F}:L^1(\mathbb{R})\to C_0(\mathbb{R})$ is an injective, bounded linear map that isn't onto.  It is known (if I remember correctly) that the range isn't closed, but is dense in $C_0$.  Everything I have read/heard says that the range is "difficult to describe".  
But, since $\mathcal{F}$ is injective and continuous, the image of $L^1$ must be Borel inside of $C_0$.  Is anything else known about its descriptive complexity?  If not, might this be an example of a natural set of high Borel rank?
This question may be relevant. Thanks for any insight/references.

REPLY [9 votes]: It seems that Mike's argument shows that in fact the range $R$ of the Fourier transform is $F_{\sigma\delta}$ in $C_0(\mathbb R)$. Define the $T_ng$ as above, 
$$T_ng(x)=\int_{\mathbb R} e^{-a_n\pi\vert t\vert}g(t) e^{2i\pi tx}dt\;.$$ Then a function $g\in C_0(\mathbb R)$ is in $R$ iff two things hold:
(i) all $T_ng$ are in $L^1$;
(ii) the sequence $(T_ng)$ is Cauchy in $L^1$.
Condition (i) can be written as follows:
$$ \forall n\;\exists N\in\mathbb N\; \left ( \forall d\in\mathbb R_+ \;:\;\int_{-d}^d \vert T_ng(x)\vert dx\leq N\right)$$
By dominated convergence, the condition under brackets is closed with respect to $g$; so (i) defines an $F_{\sigma\delta}$ subset of $C_0(\mathbb R)$.
Condition (ii) reads
$$\forall k\in\mathbb N\;\exists N \;\left(\forall p,q\geq N\; \forall d \;:\; \int_{-d}^d \vert T_pg(x)-T_qg(x)\vert dx\leq \frac 1k\right)$$
By dominated convergence again, the condition under brackets is closed wrt $g$, so (ii) defines an $F_{\sigma\delta}$ subset of $C_0(\mathbb R)$.
Altogether, $R$ is the intersection of two $F_{\sigma\delta}$ sets, hence an $F_{\sigma\delta}$ subset of $C_0(\mathbb R)$. I would be extremely surprised if it were better than that; i.e.   I  "conjecture" that it is not $G_{\delta\sigma}$.<|endoftext|>
TITLE: relative resolution of singularity
QUESTION [6 upvotes]: Is there a relative version of resolution of singularities (in characteristic 0)?
For finite type morphism $f:X\rightarrow S$ where $S$ is a variety over a field $k$ with char $k=0$(or more generally excellent scheme with residue field are characteristic 0?). Are there modifications $i:S' \rightarrow S$, $j:X'\rightarrow X$, and morphism $f':X'\rightarrow S'$ s.t. $fj=if'$, and $f'$ smooth, $i$ proper, surjective.
I'm not sure the above statement is a proper version of relative desingularities.
Thanks

REPLY [3 votes]: There are a bunch of very basic fundamental problems with doing this. 
Here is one example and you are invited to generalize this to get others: Let $X$ be a smooth surface and $f$ a surjective flat morphism to a smooth curve $S$. If $f$ is not already smooth, then it is not possible to make it smooth the way you suggest (or any reasonable way). 
This is one reason why moduli spaces of smooth objects are in general not compact or put it in another way why compactifying moduli spaces is a non-trivial question. (See $\overline {\mathfrak M}_g$ for instance).
To see that the above claim is true consider the following:

If $f$ has a singular fiber, then for any $i:S'\to S$ the base change morphism:
$f_{S'}:S'\times_S X\to S'$ will also have singular fibers. 
For any $f':X'\to X$ as in the question, the $fj=if'$ assumption means that $f'$ factors through $f_{S'}$, so in other words you want a $g:X'\to S'\times_S X$ modification such that $f'=f_{S'}g$ plus the rest of the requirements. 
$S'\times_S X$ is normal. This is because the fibers of $f$ are CM and hence so are the fibers of $f_{S'}$. Hence $S'\times_S X$ is $S_2$ and by a similar argument it is also $R_1$, so by Serre's criterion it is normal. 
There is no such $g$. If indeed $f$ had a singular fiber, then the same singular fiber appears as a fiber of $f_{S'}$. Being a curve its unique resolution is its normalization, so a $g$ like that would have to be a finite morphism of degree $1$. However, since $S'\times_S X$ is normal, Zariski's Main Theorem implies that $g$ would have to be the identity. (For simplicity assume that there is at least one non-unibranched singularity on one of the fibers).<|endoftext|>
TITLE: Fixing a proof of the systolic inequality for higher genus surfaces
QUESTION [5 upvotes]: I'm currently learning some stuffs about systolic inequalities. While reading the relevant sections (p329 to 340) in Berger's Panoramic View of Riemannian Geometry, I noticed a gap in one of the proofs (starting at page 331). The goal is to prove :

For any non simply connected compact surface $(M,g)$, $Area(M,g)\geq Sys(M,g)^2/2$.

Berger considers a periodic geodesic $c$ realizing the systole $L=Sys(M,g)$, pick a point $m$ on $c$ and claims that $Vol B(m,L/2)\geq L^2/2$, which is enough to show the result.
However in the proof of the claim, he invokes the following fact : for $r<L/2$, $B(m,r)$ is a topological disk. He says that this comes from the "very definition of the systole". But this claim is false, it is enough to consider a torus with a long thin finger glued at some point to see it (fig 7.26 on the page where the claim is made shows exactly this). In the paper "Systolic and inter-systolic inequalities", Gromov, facing the same kind of situation, just says "chop the fingers", while this is intuitively convincing I don't see a way to make it rigorous.
My question is the following : is this fixable ?
Thinking a little bit about it, it seems enough for the rest of the argument to show that all but one of the connected component of $M\backslash B(m,r)$ are disks, and that the systole $c$ doesn't meet the components which are disks. But I am not able to prove this at the moment.
I'm aware that another proof is available, through estimating the homological systole. But I like the proof on the next page of Berger that the systolic ration grows at least like the square root of the genus, which uses the same argument.
I should also say that I don't have access the article of Hebda to which Berger refers.

REPLY [3 votes]: You don't really need to reattach chopped off fingers to get the estimate.  The point is that for almost all (in the sense of Sard) $r$, the points $\gamma(-r)$ and $\gamma(r)$ are in the same connected component of the $r$-level curve of the distance function.  Applying the coarea formula and the triangle inequality then gives the estimate.<|endoftext|>
TITLE: Commutativity of Tor
QUESTION [5 upvotes]: Let $A$ be a commutative ring with $1$ and $M,N$ be $A$-modules. Can you give a quick proof that $\textrm{Tor}_i(M,N) \cong \textrm{Tor}_i(N, M)$ using derived categories?
In his Homological algebra book, Weibel proves this with an argument via a double complexes: the so-called "acyclic assembly lemma", and from what I understand this argument can be essentially reworded into the language of spectral sequences. Hartshorne's discussion (in "Residues and Duality") of derivatives of functors in two variables is quite short, but it's not clear to me if this result (commutativity of Tor) immediately follows from the relevant derived category formalism.

REPLY [13 votes]: Yes, the derived category of a commutative ring is symmetric monoidal $M\otimes_A^{\mathbb L}N=N\otimes_A^{\mathbb L}M$ and Tor is the homology of this tensor product.<|endoftext|>
TITLE: Minimal size of subsets $A,B$ in a finite group $G$ such that $AB=G$
QUESTION [12 upvotes]: Given a finite group $G$ with $N$ elements what are the "smallest" possible subsets
$A,B$ of $G$ such that $G=AB$ (ie. every element of $G$ is a product of an element in $A$
and an element in $B$)? 
We have of course $\sharp(A)\sharp(B)\geq N$
but can one always find two subsets $A,B$ of size roughly $\sqrt N$ with the property $G=AB$?
Since I do not know the optimal answer a good definition of "roughly" 
should be part of an answer (one can surely not do better than
$A,B$ of size $O(1)+\sqrt N$ but I guess this is much to optimistic.
It is perhaps more reasonable to hope for $\sqrt N$ times something bounded).
The optimal answer $A,B$ of size (at most) $1+\sqrt N$ works for cyclic groups: take 
$A=\mathbb N\left\lceil \sqrt N\right\rceil\cap \{0,\dots,N-1 \}$
and $ B= \{ 0,1,\dots,\left\lfloor \sqrt{N} \right \rfloor \} $ in $\mathbb Z/N\mathbb Z=
\{0,\dots,N-1\}$.
On the other hand, if $G$ has a (not-necessarily normal) subgroup $H$ of size roughly $\sqrt{N}$ one can 
consider $A=H$ and take for $B$ coset-representantives.
Combining the two constructions we get a nearly optimal result for abelian groups.
Is there for example a sequence of "bad" groups $G_i$ such that, say
$$\left((1+\sharp(A_i))(1+\sharp(B_i))-\sharp(G_i)\right)\Big/\sqrt{\sharp(G_i)}$$
is unbounded for every (sequence of) subsets $A_i,B_i\subset G_i$ 
such that $G_i=A_iB_i$? (The existence of such a sequence would of course
the hope for the existence of $A,B$ of size $O(1)\sqrt{\sharp(G)}$ (with $AB=G$)
in every finite group $G$.

REPLY [10 votes]: For square root of N times something bounded, this can be done with A=B (and the bound
being 4 over the square root of 3) according to
Kozma and Lev, Bases and decomposition numbers of finite groups
Arch. Math. (Basel) 58 (1992), 417-424
which seems to do as you say, namely find a subgroup of order about the square root of
N (via an appeal to CFSG).
It wouldn't surprise me if the O(1)+square root of N case is open.<|endoftext|>
TITLE: Can the sum of two roots of unity be a root of unity?
QUESTION [6 upvotes]: Let $p$ be prime, and $z_0, z_1, ..., z_{p-1}$ be all the $p$-th roots of unity, i.e. solutions of the equation $z^p = 1.$
Is it true or false that a combination of two (or more, in general) of the roots can give us another root of the same order? 
In mathematical terms, does there exist indices $i_1,i_2,...,i_s, j,$
such that
$z_j = \sum_{k=1}^s z_{i_k}?$
It seems to be that this is not possible, but I also don't have proof of that.
Thank you!

REPLY [18 votes]: If $p$ is intended to be prime, you should say so explicitly. I'll assume $p$ is prime. Then the subset sums are distinct except that the sum of all $p$th roots of unity is $0$, the sum over the empty set. Any coincidence of subset sums $\sum_{i \in I} \zeta_p^i = \sum_{j\in J} \zeta_p^j $produces a polynomial of degree at most $p-1$ with coefficients in $\lbrace-1,0,1\rbrace$ so that $\zeta_p$ is a root. This polynomial must be a multiple of the minimum polynomial for $\zeta_p$, the $p$th cyclotomic polynomial $\Phi_p(x) =1 + x + ... + x^{p-1}$. The only possibilities are scalar multiples corresponding to $1 + \zeta_p + ... + \zeta_p^{p-1} = 0$.

REPLY [10 votes]: It is known that for every positive integer $n$, the primitive $n$-th roots of unity are linearly independent over $\mathbb{Q}$ if and only if $n$ is square free.<|endoftext|>
TITLE: Name for algebra and its tensor products
QUESTION [14 upvotes]: Is there a name for the algebra (and its tensor products) given by generators $M_{j}$, $j \in \mathbb{Z}_{n}$ under the conditions $$M_{j} = (1 - M_{j-1})(1-M_{j+1})$$ where $j=1\implies j-1=n$ and $j=n\implies j+1=1$?
There is no restriction on the commutativity of $M_{j}$ and so is there a matrix representation of this algebra for all possible cases for $M_{j}$?


By tensor product I mean generalization of following extension to multiple indices:
For single index $$M_{j} = \prod_{\substack{i\in\{-1,0,1\}\\i\neq j}}(1 - M_{j+i})$$ which is same as above and for two indices $$M_{j,j'} = \prod_{\substack{i,i'\in\{-1,0,1\}\\i\neq j\wedge i'\neq j}}(1 - M_{j+i,j'+i'})$$ relation holds.
Update Like Lena I too thought it is similar to a Hecke algebra however since I am not familiar I could not pin down details.

REPLY [7 votes]: As requested, I elaborate on my comment.
First of all, let me make a change of variables $a_i=U_i-1$. The relations then become $a_i+1=a_{i-1}a_{i+1}$. 
For $n=2,3,4$ I used the Magma online calculator. The commands
F<x,y> := FreeAlgebra(RationalField(),2);
B := [x^2-y-1,y^2-x-1];
GroebnerBasis(B);
give the result
[
    x^2 - y - 1,
    x*y - y*x,
    y^2 - x - 1
]
so the algebra has a basis $1,x,y,yx$, so is four-dimensional. 
The commands
F<x,y,z> := FreeAlgebra(RationalField(),3);
B := [x*z-y-1,y*x-z-1,z*y-x-1];
GroebnerBasis(B);
give the result
[
    x*y*z - z*x*y + y*z - z*x - x + y,
    y*z*x - z*x*y - x*y + y*z - x + z,
    y*z^2 + y*z - z*x - x - z - 1,
    z^2*x - y*z + z*x - y - z - 1,
    z^3 - x*y + z^2 - x - y - 1,
    x^2 - z^2 + x - z,
    x*z - y - 1,
    y*x - z - 1,
    y^2 - z^2 + y - z,
    z*y - x - 1
]
so the algebra has a basis $1,x,y,z,xy,yz,zx,z^2,zxy$, so is nine-dimensional. 
A similar computation for $n=4$ gives a Gröbner basis which is a bit too long to format properly, and a basis for the algebra $1,x,y,z,t,xy,yz,zt,tx,tz,x^2,y^2,z^2,t^2,xyz,y^2z,yzt,z^2t,ztx,tx^2,txy,t^2x,t^3,txyz,t^2x^2$, so the algebra is 25-dimensional. 
For $n=5$, the calculator spits more and more elements as the degree grows, so it might even be that the Gröbner basis is infinite. However, if we consider the abelianisation of this algebra, the command
F<x,y,z,t,u> := PolynomialRing(RationalField(),5);
B := [x*z - y-1, y*t-z-1, z*u-t-1, t*x-u-1, u*y-x-1];
GroebnerBasis(B);
produces the result
[
    x - y*u + 1,
    y*t - z - 1,
    z*u - t - 1
]
which defines an infinite-dimensional algebra (since the associated monomial algebra is defined by the relations $x=0$, $yt=0$, $zu=0$), so the original algebra must for sure be infinite-dimensional. 
[Alternatively, we can note that for commuting elements $a_i$, the equation $a_{i-1}a_{i+1}=a_i+1$ is the celebrated "pentagon recurrence" related to cluster algebras of type $A_2$, and we have $a_{n+5}=a_n$ for all choices of $a_0,a_1$ for which the sequence is uniquely defined, so the corresponding abelianisation corresponds to something 2-dimensional geometrically, and the algebra is infinite-dimensional.]
Overall, it is not quite clear if we should hope that these algebras obey a nice pattern or are easily recognisable, but they do look interesting.<|endoftext|>
TITLE: Euler's mathematics in terms of modern theories?
QUESTION [21 upvotes]: Some aspects of Euler's work were formalized in terms of modern infinitesimal theories by Laugwitz, McKinzie, Tuckey, and others. Referring to the latter, G. Ferraro claims that "one can see in operation in their writings a conception of mathematics which is quite extraneous to that of Euler." Ferraro concludes that "the attempt to specify Euler's notions by applying modern concepts is only possible if elements are used which are essentially alien to them, and thus Eulerian mathematics is transformed into something wholly different"; see http://dx.doi.org/10.1016/S0315-0860(03)00030-2. 
Meanwhile, P. Reeder writes: "I aim to reformulate a pair of proofs from [Euler's] "Introductio" using concepts and techniques from Abraham Robinson's celebrated non-standard analysis (NSA). I will specifically examine Euler's proof of the Euler formula and his proof of the divergence of the harmonic series. Both of these results have been proved in subsequent centuries using epsilontic (standard epsilon-delta) arguments. The epsilontic arguments differ significantly from Euler's original proofs." Reeder concludes that "NSA possesses the tools to provide appropriate proxies of the inferential moves found in the Introductio"; see http://philosophy.nd.edu/assets/81379/mwpmw_13.summaries.pdf (page 6). 
Historians and philosophers thus appear to disagree sharply as to the relevance of modern theories to Euler's mathematics. Can one meaningfully reformulate Euler's infinitesimal mathematics in terms of modern theories?
Note 1. There is a related thread at Would Euler's proofs get published in a modern math Journal, especially considering his treatment of the Infinite?
Note 2. We challenged a reductionist view of Euler's infinitesimal mathematics in a recent article in The Mathematical Intelligencer.  Here we refute H. Edwards' reduction of Euler to an Archimedean framework.

REPLY [3 votes]: A somewhat delayed response is provided in our detailed study of Euler accepted for publication in Journal for General Philosophy of Science.
We apply Benacerraf's distinction between mathematical ontology and mathematical practice (or the structures mathematicians use in practice) to examine contrasting interpretations of infinitesimal mathematics of the 17th and 18th century, in the work of Bos, Ferraro, Laugwitz, and others. We detect Weierstrass's ghost behind some of the received historiography on Euler's infinitesimal mathematics, as when Ferraro proposes to understand Euler in terms of a Weierstrassian notion of limit and Fraser declares classical analysis to be a "primary point of reference for understanding the eighteenth-century theories." Meanwhile, scholars like Bos and Laugwitz seek to explore Eulerian methodology, practice, and procedures in a way more faithful to Euler's own.
Euler's use of infinite integers and the associated infinite products is analyzed in the context of his infinite product decomposition for the sine function. Euler's principle of cancellation is compared to the Leibnizian transcendental law of homogeneity. The Leibnizian law of continuity similarly finds echoes in Euler. 
We argue that Ferraro's assumption that Euler worked with a classical notion of quantity is symptomatic of a post-Weierstrassian placement of Euler in the Archimedean track for the development of analysis, as well as a blurring of the distinction between the dual tracks noted by Bos. Interpreting Euler in an Archimedean conceptual framework obscures important aspects of Euler's work. Such a framework is profitably replaced by a syntactically more versatile modern infinitesimal framework that provides better proxies for his inferential moves.<|endoftext|>
TITLE: Classification of geometric outer automorphisms of free groups
QUESTION [6 upvotes]: Good evening everyone,
an outer automorphism $[\phi]\in Out(F_n)$ is geometric if it is induced by a surface homeomorphism $h\colon M\stackrel{\cong}{\to}M$, where $M$ is a compact surface with nonempty boundary.
Bestvina and Handel gave a classification of those outer automorphisms that are induced by a pseudo-Anosov homeomorphism of a compact surface with connected boundary ('Train tracks and automorphisms of free groups', Annals, 1992):
Theorem: $[\phi]\in Out(F_n)$ is induced by a pseudo-Anosov homeomorphism of a compact surface with one boundary component if and only if each $[\phi]^l$ is irreducible and there is a conjugacy class $s\in \mathcal{C}(F_n)$ such that $[\phi] (s)=s$ or $[\phi] (s)=\overline{s}$.
Can we deduce from this result a complete classification of all geometric outer automorphisms of $F_n$? Or do their assumptions on the surface and the homeomorphism not allow for a corollary that treats the general case? (We obviously can't just drop 'pseudo-Anosov' and 'one boundary component' from the statement of the theorem.)

REPLY [2 votes]: I think the short answer is "No", you cannot deduce from this result a classification of all geometric outer automorphisms. 
I think it might eventually be possible to obtain a classification of geometric outer automorphisms from the more powerful "relative train track / lamination" machinery developed in the works of Bestvina, Feighn, and Handel, although that has not been done. Nonetheless one can deduce bits and pieces of such a classification. 
For instance, suppose that $\phi \in Out(F_n)$ has an attracting lamination $\Lambda$ with the property that the smallest free factor of $F_n$ that supports $\Lambda$ is the whole free group. In this case one can prove that $\phi$ is geometric if and only if there exists a finite $\phi$-invariant set of root-free conjugacy classes $c_1,...,c_k$ such that the smallest free factor of $F_n$ that supports $c_1,...,c_k$ is also the whole free group, $c_1,...,c_k$ are the only root-free conjugacy classes that are not attracted to $\Lambda$ under iteration of $\phi$, and a few other nondegeneracy conditions hold. The picture to keep in mind is that $c_1,...,c_k$ represent the boundary components of a surface on which $\phi$ is represented as a pseudo-Anosov homoemorphism with unstable lamination $\Lambda$. The "nondegeneracy" conditions I mentioned are needed to avoid counterexamples where, say, three of the $c's$ are identified to the same closed curve, and these conditions can be expressed in an intrinsic manner in terms of ``Nielsen theory'' which means the asymptotic behavior of automorphisms representing the outer automorphism $\phi$.
This statement can be found in a slightly different form in Proposition 2.38 of the paper Subgroup classification in $Out(F_n)$ by Handel and myself, and in this exact form in the soon-to-appear Part III of the expanded version "Subgroup decomposition in $Out(F_n)$".<|endoftext|>
TITLE: Criteria for ghost-Witt vectors: looking for history and references
QUESTION [7 upvotes]: I am looking for references (both of the readable and of the historical kind!) for the following result (which I formulate in one of its least general forms, so as not to complicate the discussion). I can prove all of it, and I have proven much of it on the internet (witt5, witt5f, Exercise 2.9.6 in Hopf Algebras in Combinatorics), but this isn't something I want to cite in a paper.

Theorem. Let $\left(b_1,b_2,b_3,...\right)$ be a sequence of integers. Then, the following assertions are equivalent:
Assertion $\mathcal C$: Every positive integer $n$ and every prime divisor $p$ of $n$ satisfy
  \begin{equation}
b_{n\diagup p}\equiv b_{n}\operatorname{mod}p^{v_{p}\left(  n\right)} ,
\end{equation}
  where $v_p\left(n\right)$ denotes the $p$-adic valuation of $n$ (that is, the greatest integer $m$ such that $p^m$ divides $n$).
Assertion $\mathcal D$: There exists a sequence $\left(x_1,x_2,x_3,...\right)$ of integers such that every positive integer $n$ satisfies $b_n = \sum\limits_{d\mid n}dx_{d}^{n\diagup d}$.
Assertion $\mathcal D^{\prime}$: Same as Assertion $\mathcal D$, but with "a sequence" replaced by "one and only one sequence".
Assertion $\mathcal E$: There exists a sequence $\left(y_1,y_2,y_3,...\right)$ of integers such that every positive integer $n$ satisfies $b_{n}=\sum\limits_{d\mid n}dy_{d}$.
Assertion $\mathcal E^{\prime}$: Same as Assertion $\mathcal E$, but with "a sequence" replaced by "one and only one sequence".
Assertion $\mathcal F$: Every positive integer $n$ satisfies $n\mid \sum\limits_{d\mid n}\mu\left(  d\right)  b_{n\diagup d}$, where $\mu$ denotes the Möbius function.
Assertion $\mathcal G$: Every positive integer $n$ satisfies $n\mid \sum\limits_{d\mid n}\phi\left(  d\right)  b_{n\diagup d}$, where $\phi$ denotes Euler's totient function.
Assertion $\mathcal H$: Every positive integer $n$ satisfies $n\mid \sum\limits_{i=1}^{n}b_{\gcd\left(  i,n\right)  }$.
Assertion $\mathcal I$: There exists a sequence $\left(q_1,q_2,q_3,...\right)$ of integers such that every positive integer $n$ satisfies $b_{n}=\sum\limits_{d\mid n}d\dbinom{q_{d}n\diagup d}{n\diagup d}$.
Assertion $\mathcal I^{\prime}$: Same as Assertion $\mathcal I$, but with "a sequence" replaced by "one and only one sequence".
Assertion $\mathcal J$: There exists a ring homomorphism from the ring $\mathbf{Symm}$ of symmetric functions in infinitely many variables over $\mathbb{Z}$ which sends $p_{n}$ (the $n$-th power sum symmetric function) to $b_{n}$ for every positive integer $n$.
Assertion $\mathcal K$: There exist two sets $M$ and $N$ and two maps $f:M\rightarrow M$ and $g:N\rightarrow N$ such that every positive integer $n$ satisfies
  \begin{equation}
\left\vert \operatorname*{Fix}\left(  f^{n}\right)  \right\vert <\infty
, \qquad \left\vert \operatorname*{Fix}\left(  g^{n}\right)
\right\vert <\infty \qquad\text{ and }\left\vert \operatorname*{Fix}\left(  f^{n}\right)
\right\vert -\left\vert \operatorname*{Fix}\left(  g^{n}\right)  \right\vert
=b_{n} ,
\end{equation}
  where $\operatorname*{Fix}\left(  h\right)  $ denotes the set of fixed points of any map $h:S\to S$ (for any set $S$).

I assume more results can be added to this. It should be noticed that each of the assertions $\mathcal D^{\prime}$, $\mathcal E^{\prime}$, $\mathcal I^{\prime}$ follows from the respective un-primed assertion due to $\mathbb Z$ being torsionfree, and that Assertion $\mathcal H$ is more or less a trivial reformulation of Assertion $\mathcal G$.
The above theorem can be seen as a generalization of the famous "necklace divisibility" which states that $n \mid \sum\limits_{d\mid n} \phi\left(d\right)q^{n\diagup d}$ for any positive integer $n$ and any integer $q$. Many similar divisibilites also follow from that theorem. The sequences $\left(b_1,b_2,b_3,...\right)$ which satisfy the equivalent assertions of the Theorem can be called "ghost-Witt vectors over $\mathbb Z$", though the real motivation of this notion comes not from considering sequences of integers but (more generally) families of elements of a commutative ring.
The above theorem is a kind of folklore, except for Assertions $\mathcal I$ and $\mathcal I^{\prime}$ which I have not seen anywhere (but they are sufficiently epigonal that I wouldn't expect them to be new). Lemma 9.93 in Hazewinkel's "Witt vectors. Part 1" (when will there finally be a part 2?) yields $\mathcal C\Longleftrightarrow\mathcal D$. The equivalence of $\mathcal D$, $\mathcal F$, $\mathcal G$ and $\mathcal H$ is stated as the Corollary on page 9 of Andreas Dress, Christian Siebeneicher, The Burnside ring of the infinite cyclic group and its relations to the necklace algebra, $\lambda$-rings and the universal ring of Witt vectors (where $\mathcal D$ appears in a generating function form), and a quick look at this paper makes me pretty confidence he proves their equivalence to $\mathcal K$ there. The equivalence between Assertions $\mathcal E$ and $\mathcal F$ follows from Möbius inversion and seems to be implicit in the papers mentioned. I am irked by the fact that I cannot find Assertion $\mathcal J$ explicit in literature, although it is clearly well-known. I have heard that Vladimir Arnold was studying sequences satisfying the equivalent assertions when he was doing discrete dynamical sequences, but I am not sure what a good keyword would be to search for.
Further equivalent assertions that could be added to the list are also welcome!
UPDATE: Further references found (thanks to Richard Stanley and Keith Conrad):

The equivalence $\mathcal C \Longleftrightarrow \mathcal D$ is Lemma 1 in Lars Hesselholt, Lecture notes on Witt vectors, 2005. It is credited to Dwork there, although it seems that the forms Dwork stated it in are rather far apart from the form I need.
Richard Stanley, Enumerative Combinatorics, volume 2, CUP 2001, exercise 5.2 a proves the equivalence $\mathcal C \Longleftrightarrow \mathcal F$. Some more of the above assertions appear in equivalent rewritings in that exercise.
The closest thing I could find to a proof of $\mathcal D \Longleftrightarrow \mathcal J$ in literature is Christophe Reutenauer, On Symmetric Functions Related to Witt Vectors and the
Free Lie Algebra, Advances in Mathematics, vol. 110, issue 2 (February 1995),
pp. 234-246.. He never does this equivalence explicitly, but his statement that "the $q_n$ freely generate $\Lambda$ over $\mathbb Z$" (as a commutative algebra) on p. 236, combined with the equality (2.3), yield it very easily.
The equivalence $\mathcal C \Longleftrightarrow \mathcal D \Longleftrightarrow \mathcal F$ goes back to Issai Schur, Arithmetische Eigenschaften der Potenzsummen einer algebraischen Gleichung, Compositio Mathematica, 4 (1937), pp. 432-444. There he also proves a statement that can be interpeted as a finitary version of $\mathcal C \Longleftrightarrow \mathcal J$; in fact, he is not considering the ring $\mathbf{Symm}$, but rather he works with actual power sums of the roots of a monic degree-$m$ integer polynomial. Unfortunately it's easier to reprove the equivalence $\mathcal C \Longleftrightarrow \mathcal J$ than derive it from Schur's results, but he clearly would have stated it if he would work with today's notations. Ironically Schur is one of the founding fathers of $\mathbf{Symm}$...
Here is a reference from Stanley's EC2 which I could not find: W. Jänischen (I suspect that it should be Jänichen with no "s"), Sitz. Berliner Math. Gesellschaft 20 (1921), pp. 23-29. If anyone can send me a scan I'd be very happy. I don't know if "Sitz. Berliner Math. Gesellschaft" is a standalone periodical or a part of "Archiv der Mathematik und Physik" (it used to be the latter at the beginning of the 20th century).
Several people refer to A. Dold, Fixed point indices of iterated maps, Inventiones mathematicae, 1983, Volume 74, Issue 3, pp. 419-435. While this is obviously related to assertion $\mathcal K$ (it is mostly about its continuous analoga), I can't see any part of the theorem being proven in this paper. But I've just quickly skimmed the paper.
Keith Conrad mentions Donald Knutson, $\lambda$-rings and the representation theory of the symmetric group. Since this text (one of the early good introductions into the representation theory of $S_n$) considers Witt vectors in relation to $\mathbf{Symm}$, it inevitably grazes the above theorem, but it seems to never actually state any part of it. Or have I missed something?

UPDATE 2: I just modified Assertion $\mathcal K$. The original version of this assertion was this: "There exists a set $M$ and two maps $f:M\rightarrow M$ and $g:M\rightarrow M$ such that every positive integer $n$ satisfies
$\left\vert \operatorname*{Fix}\left(  f^{n}\right)  \right\vert <\infty
$, $\left\vert \operatorname*{Fix}\left(  g^{n}\right)
\right\vert <\infty$ and $\left\vert \operatorname*{Fix}\left(  f^{n}\right)
\right\vert -\left\vert \operatorname*{Fix}\left(  g^{n}\right)  \right\vert
=b_{n}$."
While this original version is equivalent to the new version of Assertion $\mathcal K$, it is a rather unnatural statement (and the equivalence is ugly to prove as far as I can tell).

REPLY [8 votes]: This question is a bit hard to answer for a few reasons. As you say, most of the equivalences are implicit in the literature. So then it becomes a question of how explicit you want them to be stated, and that's a matter of personal taste. Also, some of them were known earlier in the p-typical case (going all the way back to Witt work in the 30s), although the same ideas will work in the "big" case. So whether such things count is also not clear.
That said, here are my reactions, although you probably know much of what I have to say. I believe all the references I mention are given in my paper "Basic geometry of Witt vectors, I".
For D=J, this is of course a consequence of the fact that big Witt vectors and lambda-rings are two sides to the same coin. So it will be at least implicit in any reference that mentions both of them, such as the chapter in Bourbaki, Hazewinkel's book, or Joyal's two papers in the Canadian Comptes Rendus. There is also Witt's unpublished note from 60s which appears in his collected works. From what I remember, he doesn't talk about lambda-rings, but he talks about Witt vectors and power series, so it's surely at least implicit there too. (I'm also interested in the question of who first realized that Witt vectors and lambda-rings are the same thing. I've spoken to Cartier and Serre about it, and neither is sure. This does appear in the Grothendieck-Mumford correspondence in Mumford's selected papers II page 692. See the letter to Mumford 31 August 1964. Perhaps Grothendieck was the first to realize that the two might be related.)
For D=D', this is of course a consequence of the fact that the ghost map is injective for torsion-free rings. So again, it will be implicit in just about any reference that mentions that fact. Aside from the references above, there is Bergman's chapter in Mumford's book. There are some exercises in the first edition of Lang's Algebra (1969?) which discuss big Witt vectors. Perhaps it's implicit there as well.
For C/E/E'=D/D'/J, I think of this as a special instance of the linear description of Witt vectors of lambda-rings. I believe this is in Joyal. (Update: It's only there in the p-typical case.) I'm not sure if it appeared anywhere earlier.
The equivalence of K and the others is, as you say, presumably just a disguised form of the description of W(Z) as the Burnside ring and hence will at least be implicit in Dress-Siebeneicher. That's probably the first appearance of W(Z)=Burnside; so I think it's unlikely to be anywhere earlier.<|endoftext|>
TITLE: Partitions-sum of divisors identity
QUESTION [13 upvotes]: A few years ago I first read about the marvelous Euler identity:
$\sum_{n\in\mathbb{N}}p(n)z^n=\prod_{k\geq1}\frac{1}{1-z^k}$,
where $p(n)$ is the number of partitions of $n$ ($p(0)=1$ by convention) and some of its beautiful consequences (like the pentagonal number theorem). Taking log of both sides of Euler identity and differentiating, the following nice recursive formula magically appears:
$np(n)=\sum_{k=0}^{n-1}p(k)\sigma(n-k)$,
where $\sigma(n)$ denotes the sum of the divisors of $n$. After some googling I found this identity quoted in a few places, but always without any reference. Since I am quite ignorant about the theory of partitions and related matters, I would like very much to know:
1) Who discovered this identity? Does it have a name?
and the much more interesting: 
2) Is there a proof without generating functions?
Thank you!

REPLY [5 votes]: Part 1.) R. Stanley mentions on page 59 (in the answer to exercise 24a, which is the identity mentioned in the OP) in Enumerative Combinatorics vol. 1 (exercise is 78a in later edition) the following somewhat inconclusive fact:

Some related results are due to Euler and recounted in $\S$303 of P.A. MacMahon, Combinatory Analysis, vol. 2, Cambridge University Press, 1916...

Searching the Euler archive showed a few similar results (but those were on the generating function of $\sigma(n)$ being related to the logarithmic derivative of the Euler generating function.<|endoftext|>
TITLE: $k[[x]]$ as a $(k[[x]])^p$ module for ugly fields
QUESTION [6 upvotes]: Suppose that $k$ is a field of characteristic $p$ such that $k$ is not a finite $k^p$-module.  For example, $k = \mathbb{F}_p(x_1, x_2, x_3, ...)$.  
Is it true that $k[[x]]$ is a free $(k[[x]])^p$-module?  We know that it is flat by a theorem of Kunz.  However, it is certainly not finite, so flat is not the same as free.
The naive thing to try (in terms of a basis) is to do the following.  Choose {$\lambda_i$} a basis for $k$ over $k^p$.  Consider the set {$\lambda_i x^j$} for $0 \leq j \leq p-1$.  If $k$ is a finite $k^p$-vector space, this set is easily seen to be a basis for $k[[x]]$ over $(k[[x]])^p$.  
However, because of our (ugly) field, we can consider the power series:
$$\lambda_0 + \lambda_1 x^1 + \lambda_2 x^2 + ... + \lambda_n x^n + ...$$
where say $\lambda_0, \lambda_1, ...$ runs over some countably infinite subset of the {$\lambda_i$}.  It is easy to see that this cannot be written as a finite $(k[[x]])^p$-linear combination of subset of the $\lambda_i x^j$ (where again $0 \leq j \leq p-1$).
But of course, maybe there's some clever way to choose a basis that actually does work?

REPLY [5 votes]: The answer to the question is no. 
Let $R = k[[x]]$, $S = k[[y]]$, and $f:R \to S$ is the absolute Frobenius. We must show that $S$ is not a free $R$-module. By assumption on $k$, we know that $S$ must have infinite rank if it were free. On the other hand, $S$ is $x$-adically complete (since $x^p = y \in S$). Hence, the claim follows from:
Lemma: The $R$-module $M := R^{\oplus I}$ is $x$-adically complete if and only if $I$ is finite.
Proof: The "if" direction is clear. For the reverse, note that direct summands of complete $R$-modules are complete. By replacing $I$ with a subset, we may assume $I = \mathbf{N}$. Now consider the sequence $m_n := \sum_{i=1}^n x^i \cdot e_i \in M$, where $e_i$ is the $i$-th basis vector. This sequence is Cauchy (clear); I claim it has no limit in $M$. If there was a limit $m = \sum_{i=1}^N a_i \cdot e_i \in M$ with $a_i \in R$, then the neighbhourhood $m \in m + x^k M$ would contain infinitely many $m_n$ for any fixed $k$, but staring at the coefficient of $e_{N+1}$ shows this is false as long as $k \geq N+2$.<|endoftext|>
TITLE: When is Ad(pi) an irreducible representation ?
QUESTION [7 upvotes]: For a finite group $G$ with a representation $\pi:G\to GL_n(\mathbb C)$ one can define the adjoint representation $Ad$ as the non-trivial summand in $End(\pi)$, i.e.
$End(\pi)=\pi\otimes \pi^{\vee}=1\oplus Ad(\pi)$.  
For example, if $\pi$ is self-dual then $\pi\otimes \pi=Sym^2(\pi)\oplus \bigwedge^2(\pi)$, and for $n>2$ it is possible to show that $Ad(\pi)$ is reducible. 
Are there any known examples when $Ad(\pi)$ is irreducible ?

REPLY [5 votes]: Derek Holt has given a bunch of examples for $n>2$. In the case $n=2$, $Ad \pi$ is irreducible
if and only if it does not contain a character $\chi$ (since it has dimension 3), that is if and only if the two condition holds:
$Hom_G(\pi,\pi)$ has dimension $1$, and $Hom_G(\pi,\pi(\chi))$ has dimension $0$ for every non-trivial character $\chi$. Now the first condition is automatic by Schur's lemma (since $\pi$ is assumed irreducible) and the second one is equivalent to $ \pi \not \simeq \pi(\chi)$ for 
every non-trivial character $\chi$. Now if for instance you take a group which has no non-trivial character $\chi$ (that is a group whose abelianization is trivial, a.k.a a prefect group), and an irreducible representation $\pi$ of dimension $2$ of that group, you get an example as you asked for, that is such that $Ad \pi$ is irreducible. For example if consider a subgroup $G$ of $Gl_2(\mathbb C)$ is isomorphic to $A_5$, and $\pi$ the canonical representation, then you get an example. (Edit: actually, "2.A_5", see the comments below)<|endoftext|>
TITLE: Model structure on the category of small $A_\infty$ categories, hocolims.
QUESTION [11 upvotes]: I strangely could not find a reference for this. What are some (if any) model structures on the category of small $A_{\infty}$ categories, with weak equivalences quasi-equivalences. Same question in the case underlying chain complexes are unbounded, ungraded. 
Related question, let $M$ denote the category of small $A_\infty$ categories, $M^D$ category of $D$ shaped diagrams with $D$ Reedy with fibrant constants. Think of $M^D$ as a homotopical category with weak equivalences objectwise quasi-equivalences, similarly with $M$. Is the functor
$$colim M^D \to M$$ homotopical? It is not homotopical in the dg world, but it seems it might be in the $A_\infty$ world. If it is not homotopical and the answer to question one is "none" does the derived functor $hocolim: ho M^D \to ho M$ exist?

REPLY [9 votes]: Look at Lefevre-Hasegawa PhD Thesis (unfortunately in French) at http://arxiv.org/abs/math/0310337, where this is done. For a reference in English, you can look there: http://math.unice.fr/~brunov/HomotopyTheory.pdf , where you have to restrict to the the Koszul operad $As$ of associative algebras and then consider "mutliple objects", ie $A_\infty$-categories.  
Strictly speaking, you cannot have a model category structure on $A_\infty$-categories since not all the equalizers exist. But "all" the other axioms hold. In this case, weak equivalences are $\infty$-quasi-isomorphisms (does this correspond to your quasi-equivalences?), cofibrations are $\infty$-monomorphisms (the first component map is a monomorphism) and fibrations are $\infty$-epimorphisms (the first component map is an epimorphism). In this model category structure, everybody is fibrant and cofibrant. (I have an extended version of loc. cit. including this, that I can share.)<|endoftext|>
TITLE: Reals with integer powers bounded away from integers?
QUESTION [6 upvotes]: Do there exist real numbers whose integer powers are bounded away from integers?  More precisely, for an arbitrary  constant 0 < $\epsilon$ < 1/2, does there exist real x such that for all positive integer n,
$ \epsilon  < (x^n)$ mod 1 < $1-\epsilon $?
Pisot numbers more or less provide the opposite of this behavior, and at first I briefly thought that applying Hurwitz' Theorem to the logs might help...but ln(3)/ln(2) isn't rational, and $2^n$ and $3^n$ seem to keep a minimum unit distance apart. ^_^

REPLY [6 votes]: We should be able to construct such a real number $x$. Let $\epsilon\in(0,1/2)$ be fixed. If $S\subset \mathbb{R}_{>0}$ and $r>0$, we write $S^r$ for the set of positive $r$-th powers of elements of $S$. Recursively define a sequence of intervals $I_n$ of the form $[N_n+\epsilon,N_n+1-\epsilon]$ where the $N_n$ are positive integers. Let $I_1=[N_1+\epsilon,N_1+1-\epsilon]$ where $N_1$ chosen large enough that $N_1(1-2\epsilon)>2$. Once $I_n$ has been defined, $I_n^{(n+1)/n}$ has length at least $2$, so we can find an integer $I_{n+1}$ so that $[I_{n+1}+\epsilon,I_{n+1}-\epsilon]\subset I_n^{(n+1)/n}$. Then the intervals $I_n^{1/n}$ are nested and have diameter going to 0, and we choose $x$ to be the unique element in the intersection of $I_n^{1/n}$. This $x$ is constructed in such a way that $x^n\in[I_n+\epsilon,I_n+1-\epsilon]$ for all $n$.

REPLY [6 votes]: Such numbers exist. Here is a way to construct one of size close to $n$, for $n\ge 2$:
Let $a_1 = n + \frac{1}{2}$. Inductively, for each $k > 1$ let $a_k = (\lfloor a_{k-1}^k\rfloor+\frac{1}{2})^{1/k}$. Then $a = \lim_{k\rightarrow\infty} a_k$ exists and is between $n$ and $n+1$.
More precisely, we have $|a_k-a_{k-1}| = |a_k^k-a_{k-1}^k|/(\sum_{i=0}^{k-1}a_k^ia_{k-1}^{k-1-i}) < \frac{1}{2kn^{k-1}}$, so $|a-a_k| < \sum_{j>k} \frac{1}{2jn^{j-1}} < \frac{1}{kn^k}$. By the same argument, if we let $\alpha_k = \inf_{j\ge k} a_k$ then we have $|a-a_k| < \frac{1}{k\alpha_k^k}$, and since $\alpha_k > a-\frac{1}{kn^k}$ we have
$|a^k-a_k^k| < |a-a_k|k(a+\frac{1}{kn^k})^{k-1} < \frac{(a+\frac{1}{kn^k})^{k-1}}{(a-\frac{1}{kn^k})^{k}} < \frac{e^{1/n^k}}{n}.$
Thus the number $a$ we've constructed will satisfy your condition with $\epsilon = \frac{1}{2}-\frac{e^{1/n}}{n}$.
Example: For $n = 10$, we have $a = 10.51174467290...$, and the smallest fractional part of $a^k$ for $k$ from $1$ to $100$ is $0.452...$. The largest is $0.543...$.<|endoftext|>
TITLE: Intuition behind the spectral density of random matrices
QUESTION [10 upvotes]: Hi,
I have read that the spectral density of an NxN random matrix consisting of iid random variables with zero mean and unit variance converges as N goes to infinity to the uniform distribution on the unit disc.  I was wondering if there is an intuitive way of understanding why this should be the case.

REPLY [14 votes]: I don't know of a fully intuitive derivation, but there are some informal arguments that give the circular law with a relatively small amount of calculation.
Let $M$ be a matrix where the entries are iid with mean zero and variance one.  One can begin with the determinant formula
$$ \log |\det( M - z )| = \sum_{j=1}^n \log |\lambda_j - z|.$$
The circular law suggests that the eigenvalues $\lambda_j$ should be uniformly distributed in the disk of radius $\sqrt{n}$, so one should be proving something like
$$ \log |\det( M - z )| \approx \frac{1}{\pi} \int_{|w| \leq \sqrt{n}} \log |w-z|\ dw.$$
A routine calculation (e.g. using Jensen's formula, or the fundamental solution for the Laplacian) reveals that the RHS is equal to $n \log |z|$ when $|z| \geq \sqrt{n}$ and $\frac{1}{2} n \log n - \frac{1}{2} n + \frac{1}{2} |z|^2$ for $|z| \leq \sqrt{n}$.  So heuristically, the circular law is equivalent to the approximations
$$ |\det(M-z)| \approx |z|^n$$
for $|z| \geq \sqrt{n}$ and
$$ |\det(M-z)| \approx n^{n/2} e^{-n/2} e^{|z|^2/2}$$
for $|z| \leq \sqrt{n}$.  Here one should interpret the $\approx$ symbol rather loosely (in particular, polynomial factors in $n$ should be considered negligible).
However, by the Leibniz formula for determinants and the iid mean zero variance one nature of the entries (which makes all the covariances between the terms in the Leibniz formula vanish), one can easily compute that
$$ {\bf E} |\det(M-z)|^2 = \sum_{j=0}^n |z|^{2j} \frac{n!}{j!}.$$
(This type of calculation goes back to an old paper of Turan.)
For $|z| \gg \sqrt{n}$, the $|z|^{2n}$ term on the RHS dominates, while for $|z| \ll \sqrt{n}$, the RHS is most of the Taylor series for $n! e^{|z|^2}$.  The claim then morally follows from Stirling's approximation.
(Incidentally, my recent paper with Van Vu on local versions of the circular law  basically proceeds by making the above argument rigorous; see also recent work of Bourgarde, Yau, and Yin.  The idea of controlling the spectrum of an iid matrix through its log-determinant goes back to the early work of Girko.)
--
Note also that without too much calculation, one can see that the limiting law of the spectrum should be invariant with respect to rotations around the origin (especially if one assumes that the entries of the iid matrix are similarly invariant, e.g. they are complex gaussian).  From the matrix inequality $\sum_{j=1}^n |\lambda_j|^2 \leq \hbox{tr}(M M^*)$ and the law of large numbers we also see that the typical size of an eigenvalue $\lambda_j$ should be of the order of $\sqrt{n}$.  These facts fall well short of the full circular law but are certainly consistent with that law.

REPLY [6 votes]: Minor initial comment: You probably mean "uniform distribution in the unit circle", rather than on the unit circle --- the eigenvalues $\lambda_{k}$ ($i=1,2,... n$), scaled by $\sqrt{n}$, uniformly fill the interior of the unit circle in the complex plane. So what is commonly called the "circle law" is perhaps more appropriately referred to as the "disc law".
You ask for a simple/intuitive derivation. The easiest I know is given by Eric Kostlan in On the spectra of Gaussian matrices. When the real and imaginary parts of each matrix element are independent normally distributed, then it is rather easy to show that the absolute values squared of the eigenvalues, $\mu_{k}=|\lambda_{k}|^{2}$,  have independent $\chi^2$ distributions with $k=2,4\ldots 2n$ degrees of freedom. 
The large-$n$ asymptotics of the $\chi^2$ distribution then implies that $\mu$ is uniformly distributed in the interval $(0,n)$. It thus follows that the eigenvalues uniformly fill a disc in the complex plane of radius $\sqrt{n}$.<|endoftext|>
TITLE: Surreals and NSA: some foundational issues
QUESTION [5 upvotes]: Surreals and NSA: some foundational issues. 
A. 
Leaving aside the whole internal machinery of surreals (with funny questions like is $\omega$ an entire number and if yes is it odd or even, simple, a factorial, etc.), it is a principal foundational achievement of surreals that they give concrete, well-defined examples of saturated fields, in particular, 
1) a concrete, well-defined countably saturated rcof, 
2) a concrete, well-defined set-size saturated rcof (No itself), by necessity of class size. 
The saturation properties in this context were first explicitly observed probably by Ehrlich in 1980s, but they follow from the $\eta_\alpha$-properties (obvious by construction) and a result (CK, Ex 5.4.4 on p. 369, later reproved by Simpson) which likely was "of common knowledge" in model theory in 1970s if not earlier. 
Note that Hausdorff studied his $\eta_\alpha$ fields most successfully as "pantachies", that is, linearly ordered subsets of a certain partial order of R^N, and with a heavy dose of the axiom of choice - which does not yeild any single, well-defined, concrete example. 
Thus the countably saturated rcof which emerges as a certain initial part of No is probably the closest thing to the notoriously inconsistent "infinitaire pantachie" of DuBoisReymond known so far. 
(As a marginal remark, a countably saturated rcof cannot be Borel - even cannot have a Borel set as the domain and Borel relation as the order - so it cannot be too concretely well-defined.)
B. 
In the context of foundations of infinitesimals, the surreals have a major defect: they are just a rcof, w/o an adequate subsystem of "surnatural" numbers, which is a sine qua non for any full-scale treatment of infinitesimals.   
C. 
Nonstandard extensions of R commonly denoted by R* do not suffer from this, of course, and moreover, along with a related system of hyperintegers N* they are equipped with the asterisk of the whole $\omega$-high superstructure over R. 
Yet for a long time they used to have an own foundational issue: unique, concretely defined examples of R* were not known. 
Such examples of OD (ordinally definable) models R* of any amount of saturation were first defined by Kanovei-Shelah (JSL, 2004) - including 
(i) 
the full set-size saturated R* of class-size, 
and in fact 
(ii)
an ordinal-definable full set-size saturated elementary extension of the whole set universe of ZFC considered in detail in Kanovei-Reeken, Nonstandard analysis axiomatically, Springer 2004. 
D. 
As any two full set-size saturated rcof (both of class size, of course) are isomorphic under any suitable class theory with GC (global choice) by means af a standard application of the b&f method, No and the ground rcof structure of R* as in (i) above are isomorphic (first observed by Ehrlich). 
This means that, postfactum, the surreals No do indeed contain (under GC) a suitable system of surintegers and do allow the whole system of real functions and much more - simply inherited from R* as in (i) by means of the mentioned isomorphism. 
This leads to the following problems of foundational importance, yet to be solved.
Problem 1. Note that both surreals and R* of type (i) are OD, well-defined ZFC classes, but the isomorphism between them is not such - it is an application of GC. So we ask: does there exist a ZFC-well-defined isomorphism between them? 
Problem 2. Sharper, does there exist a ZFC-well-defined adequate system of surintegers, satiafying PA at least and preferably making (No,surintegers) to be an elementary extension of (R,N) ?
Problem 2 can be answered in the affirmative both by affirmatively answering Problem 1, and by means of an own surrealistic construction.

REPLY [4 votes]: Installment 2.
Having addressed a historical point in the first installment of my response to Vladimir (see below), I now turn to the first of his two interesting questions. To begin with, Theorem 20 of my The Absolute Arithmetic Continuum and the Unification of All Numbers Great and Small, The Bulletin of Symbolic Logic 18 (1) 2012, pp. 1-45—reads as follows:
In NBG [which I take to include global choice] there is (up to isomorphism) a unique structure (R, R*,* ) such that Axioms A–E [Keisler’s (1976) axioms for saturated hyperreal number systems] are satisfied and for which R* is a proper class; moreover, in such a structure R* is isomorphic to No. Such a structure is in fact (to within isomorphism) the unique model of Axioms A–E whose existence can be established in NBG without additional assumptions.
Vladimir’s first question regards the relation between the hypperreal number system of Theorem 20 and the full set-saturated hyperreal number system developed in his and Reeken’s important treatise on nonstandard analysis. I am skeptical that Vladimir’s first question admits an affirmative answer. In fact, I propose the following
Conjecture: Without global choice, one can not prove that the Kanovei-Reeken full set-saturated hyperreal number systems is isomorphic to the one of Theorem 20. 
In NBG with global choice, No is (up to isomorphism) the unique homogeneous universal ordered field, i.e. it contains an isomorphic copy of every ordered field whose universe is a set or proper class of NBG, and every isomorphism between subfields of No, whose universes are sets, can be extended to an automorphism. The proof that No contains an isomorphic copy of every ordered field of power On uses global choice as does the proof of homogeneity. It is not clear to me how one can prove either of these results about the Kanovei-Reeken system in the Kanovei-Reeken framework. Of course, I may not be thinking creatively enough; moreover, it might be possible that Vladimir's first question would have a positive answer for some pared down version of No, but based on a private exchange with Vladimir as well as his question, it's No of Theorem 20 he has in mind.
I will address Joel’s question about the omnific integers as well as Dave’s and Emil’s informative answers in a further installment
First Installment:
I hope to return soon (perhaps after I finish my taxes) to address some of the interesting questions raised by Vladimir. In some cases I will expand on the answers I provided Vladimir in response to his recent private letter to me, responses which turned out to be incorporated into the motivation and formulation of some of his questions. At that time, I will also explain why the nonnegative portion of the omnific integers referred to by Joel is not a full model of PA and make a few points about them as well.
For the time being, I merely wish to correct misconceptions about Hausdorff’s great writings on $\eta_{\alpha}$-orderings that one might be apt to walk away with after reading Vladimir’s remarks. Since I treat these matters with some care in Section 8 of my paper, 
The Absolute Arithmetic Continuum and the Unification of All Numbers Great and Small, The Bulletin of Symbolic Logic 18 (1) 2012, pp. 1-45
I refer interested parties to that paper for the requisite definitions and details. Even greater detail will be found in a forthcoming work of mine entitled From du Bois-Reymond’s Infinitary Pantachie to the Surreal Numbers.
To begin with, in Hausdorff's first great paper on ordered sets of 1906, he introduces the idea of an $\eta_{\alpha}$-ordering in precisely the way we use it today and he continued to use it in the same fashion in all of his subsequent writings. The definition is given on page 132 of the original paper and on page 150 of the wonderful recent English translation by Plotkin. As I explain in my aforementioned BSL paper, Hausdorff was motivated to introduce the idea of an $\eta_{1}$-ordering to characterize the order type of his very insightful reconfiguration of Paul du Bois-Reymond's flawed conception of an infinitary pantichie. In fact, he proves:
HAUSDORFF 1 [1907]: Infinitary pantachies exist. If P is an infinitary pantachie, then P is an $\eta_{1}$-ordering of power $2^{\aleph_{0}}$; in fact, P is (up to isomorphism) the unique $\eta_{1}$-ordering of power $\aleph_{1}$, assuming (the Continuum Hypothesis) CH.
In his investigation of 1907, Hausdorff also raises the question of the existence of a pantachie that is algebraically a field, but he only makes partial headway in providing an answer. However, in 1909 he returned to the problem and provided a stunning positive answer. Indeed, beginning with the ordered set of numerical sequences of the form r, r , r, …, r, … where r is a rational number, and utilizing what appears to be the very first algebraic application of his maximal principle, Hausdorff proves the following little-known, remarkable result.
HAUSDORFF 2 [1909]. There is a pantachie H of numerical sequences of real numbers indexed over the natural numbers (with operations suitably defined) that is an ordered field. Any such pantachie is, in fact, a real-closed ordered field.
Writing before Artin and Schrier [1926], Hausdorff of course does not refer to H as real closed; but he essentially establishes H is real-closed by showing it is the union of a chain of ordered fields, each of which admits no algebraic extension to a more inclusive ordered field.
Thus, contrary to what Vladimir contends, Hausdorff does not formulate his theory of  $\eta_{\alpha}$-orderings in terms of pantachies, but rather in the manner we know and love; moreover, he uses the special case of an $\eta_{1}$-ordering to characterize the order type of his pantachies. 
I'm delighted to see that Vladimir has edited his remarks on Hausdorff, presumably in light of the above remarks.<|endoftext|>
TITLE: Fixed point theorems
QUESTION [42 upvotes]: It is surprising that fixed point theorems (FPTs) appear in so many different contexts throughout Mathematics: Applying Kakutani's FPT earned Nash a Nobel prize; I am aware of some uses in logic; and of course everyone should know Picard's Theorem in ODEs. There are also results about local and global structure OF the fixed points themselves, and quite some famous conjectures (also labeled FPT for the purpose of this question).
Many results are so far removed from my field that I am sure there are plenty of FPTs out there that I have never encountered. I know of several, and will post later if you do not beat me to them :)
Community wiki rules apply. One FPT per answer, preferably with an inspiring list of interesting applications.

REPLY [8 votes]: I forgot who proved it, but the statement is nice and very easy to prove: A function $f:X\to X$ is fixed point free if and only if there is a partition of $X$ into three subsets s.t. $f$ maps each of the three subsets into the union of the other two.  An immediate application is that if $f$ is fixed point free on a set $X$ then so is its continuous extension to a function on $\beta X$.<|endoftext|>
TITLE: Is the Segre embedding projectively normal?
QUESTION [5 upvotes]: Consider the Segre embedding
$$
\mathbb P^n\times \mathbb P^m \hookrightarrow \mathbb P^N.
$$
It seems to me that from the definition it is clear that 
$$
H^0(\mathbb P^N, \mathscr O_{\mathbb P^N}(1)) = H^0(\mathbb P^n, \mathscr
O_{\mathbb P^n}(1))\otimes H^0(\mathbb P^m, \mathscr O_{\mathbb P^m}(1)), 
$$
but is it obvious that 
$$
H^0(\mathbb P^n\times \mathbb P^m, {\mathscr O_{\mathbb P^{N}}(1)}|_{\mathbb P^n\times \mathbb P^m}) = 
H^0(\mathbb P^n, \mathscr O_{\mathbb P^n}(1))\otimes H^0(\mathbb P^m, \mathscr
O_{\mathbb P^m}(1))? 
$$
More generally:
Question: Is the Segre embedding projectively normal?
EDIT Note that strictly speaking the above condition is only linearly normal, I did mean to ask projective normality, which in this case follows from linear normality. (See J.C. Ottem's answer).

REPLY [6 votes]: Yes. The Segre embedding $i:P:=\mathbb P^n \times \mathbb P^m\to \mathbb P^N$ is defined by the sections of the line bundle $O_P(1,1):=pr_1^*O_{\mathbb P^n}(1)\otimes pr_2 O_{\mathbb P^m}(1)$ on $\mathbb P^n \times \mathbb P^m$ and by definition of $i$ we have $i^*O_{\mathbb P^N}(1)=O(1,1)$. In other words, the equality $$H^0(\mathbb P^n \times \mathbb P^m,i^*O_{\mathbb P^N}(k))=H^0(\mathbb P^n \times \mathbb P^m,O(k,k))$$ holds for any $k\in \mathbb Z$, and so $H^0(\mathbb P^n \times \mathbb P^m,i^*O(k))$ decomposes as the tensor product  $H^0(\mathbb P^n, \mathscr O_{\mathbb P^n}(k))\otimes H^0(\mathbb P^m, \mathscr
O_{\mathbb P^m}(k))$ by Kunneth's theorem. Note however, that the definition of projectively normal requires that the map $$H^0(\mathbb P^N,O(k))\to H^0(P,O(k,k))$$is surjective for all $k\ge 0$ not just for $k=1$ (in which case the embedding is called linearly normal). This is true in our case, since global sections of $O(k,k)$ are polynomials in sections of $O(1,1)$. Of course, when $k=2$, the kernel of the above map is generated by the quadrics defined as the $2\times 2$ minors defining the Segre embedding.<|endoftext|>
TITLE: Is it possible to get another math PhD
QUESTION [15 upvotes]: My undergrad being in a field other than mathematics (and not in the US), I couldn't get into the Princeton/Harvard/MIT tier of universities for my math PhD. I settled for a lesser program and got my PhD. I did land a postdoc at a half decent institution, but 4 years later, I can't even get search committees to look at my TT applications. I am no genius, but I don't think my publication list is that uncompetitive. 
So I have two questions:

Do search committees not want to put a blot in their faculty list by interviewing/taking 
applicants who don't have Harvard/Princeton (and similar) etc on their CVs? 
Is it possible to completely re-do my PhD? I mean, can I apply simply to math PhD programs again? Do I stand a chance at the top tier universities, especially since I am still below 30? I know extremely well that a PhD from Princeton/Stanford etc is not guarantee of a TT position, but it obviously improves the odds. And if I apply to PhD programs, should I hide the fact that I already have a math PhD?

REPLY [24 votes]: I agree with item 2 of Alexandre Eremenko's answer. Lower tier departments who care more about how they appear to the dean rather than the actual quality of the research do sometimes go after Ph.D.'s from top schools because it looks better. But departments that have actual aspirations of having strong research will look at anyone who has a strong track record, regardless of where that person got a Ph.D.
Here are some thoughts:
1) Get to know as many strong senior mathematicians as possible, especially those who work in your field or closely related ones. Try to make sure they are familiar with your work by giving seminars at their schools, attending and giving talks at conferences, posting your papers on arxiv, and sending a PDF of your paper to people you think might be interested (but don't overdo this).
These people are invaluable in at least two ways: 1) Putting in a good word for you, whether in a formal letter or informally by email or phone. 2) Providing useful guidance to you on which departments are hiring and where you stand the best chance.
Who writes strong letters for you matters at least as much as where you got a Ph.D.
2) When visiting a department that you think you have a shot at getting a job at, don't be afraid to ask them what the prospects are and what they are looking for. Getting a better sense of how each department chooses how to hire is invaluable.
3) Maintain good and friendly relations with absolutely everyone in your field. For example, when you write a paper, cite people generously, including anyone who has done anything related to your work, whether you used the work or not in your paper. Showing your appreciation for others' work by citing them is a good way to build a lot of goodwill.<|endoftext|>
TITLE: What is the length of the shortest law of $S_n$?
QUESTION [16 upvotes]: What is the length of the shortest word $w\in F_2$ such that $w(x,y)$ is trivial for every $x,y\in S_n$?
There is a simple argument showing that we must have $\ell(w)\geq n$. See here for instance. It seems likely however that $\ell(w)$ must be super-polynomial in $n$.
What is the state of the art for lower bounds?
(As for upper bounds, the paper "Identical relations in symmetric groups and separating words with reversible automata" by Gimadeev and Vyalyi finds an upper bound of $\exp(\sqrt{n}\log n)$.)

REPLY [15 votes]: As far as I know, the state of the art gives some improvements to those bounds, but they are not huge. For a better lower bound: given a nontrivial element $w \in F_2$ of word length $\ell$, using a result of Buskin (Economical separability in free groups, Sib. Math. J., 50 (2009), 603-608) there exists a subgroup, $H$, of index $\ell/2+2$ that does not contain $w$. By looking at the action of $F_2$ on $F_2/ H$ we get a representation of $F_2$ into $S_{\ell/2+2}$, that does not kill $w$. Therefore, in order for $w$ to be trivial in any representation of $F_2$ into $S_n$ we must have that $n \leq \ell /2 + 2$, or $2(n-2) \leq \ell$.
There are also better upper bounds known (see, for instance, Asymptotic growth and least common multiples in groups (me and Ben McReynolds), Bulletin of the LMS (2011)).
Your question is equivalent to quantifying residual finiteness of free groups (the non-normal case), for which the precise answer is still unknown (the best known bounds are from the papers above).<|endoftext|>
TITLE: Is the unitary matrix group path-connected?
QUESTION [5 upvotes]: Is the group $\mathcal{U} (n)$ of all $n\times n$ unitary matrices over $\mathbb{C}$ a (local) path-connected space? 
If so, what are the connected components of the unitary matrix group $\mathcal{U}(n)$? Is the number of components finite? What is the representative for each component? Is each components closed in norm topology?

REPLY [9 votes]: Every matrix Lie group is a smooth manifold, hence it is path-connected if and only if it is connected.
And  $U(n)$ is compact and connected as a topological space (any unitary matrix can be diagonalized by a unitary matrix, this gives a path from it to the identity). It is not simply-connected, though.
We have $\pi_1(U(n))\simeq \mathbb{Z}$.<|endoftext|>
TITLE: What metatheory proves $\mathsf{ACA}_0$ conservative over PA?
QUESTION [13 upvotes]: Simpson's book shows  $\mathsf{ACA}_0$ is conservative over $\mathsf{PA}$ in the natural way by model theory using definable subsets.  Of course, $\mathsf{ACA}_0$ being conservative over PA is interesting even apart from consistency strength.  So one might not be explicit about the metatheory.  And I am sure it is not a strong one.  
But I am curious to know the exact logic of this argument.

REPLY [18 votes]: The conservativity of $\mathrm{ACA}_0$ over $\mathrm{PA}$ is provable in $I\Delta_0+\mathit{SUPEXP}$ by a cut elimination argument. It is not provable in $I\Delta_0+\mathit{EXP}$, since $\mathrm{ACA}_0$ has superexponential speedup over $\mathrm{PA}$ (a result attributed to Solovay, Pudlák, and Friedman).
EDIT: Let me briefly sketch a proof of the speedup theorem here.
Theorem:

$I\Delta_0+\mathit{EXP}\nvdash\mathrm{Con}_\mathrm{PA}\to\mathrm{Con}_{\mathrm{ACA}_0}$.
For any $k$, there are true $\Sigma^0_1$-sentences $\phi$ such that $l_\mathrm{PA}(\phi)\ge2_k^{l_{\mathrm{ACA}_0}(\phi)}$, where $l_T(\phi)$ denotes the smallest Gödel number of a $T$-proof of $\phi$, and $2_k^x$ is the $k$-times iterated exponential of $x$. Consequently, the ($\Sigma^0_1$-)conservativity of $\mathrm{ACA}_0$ over $\mathrm{PA}$ is not provable in $I\Delta_0+\mathit{EXP}+\mathrm{Th}_{\Sigma^0_2}(\mathbb N)$.

Proof:
For 1, a variant of Parikh’s theorem shows that if $I\Delta_0+\mathit{EXP}\vdash\mathrm{Con}_\mathrm{PA}\to\mathrm{Con}_{\mathrm{ACA}_0}$, there is a constant $k$ such that
$$\tag{$*$}I\Delta_0\vdash p\le\log^{(k)}(x)\land\mathrm{Proof}_{\mathrm{ACA}_0}(p,\ulcorner\bot\urcorner)\to\exists q\le x\,\mathrm{Proof}_{\mathrm{PA}}(q,\ulcorner\bot\urcorner).$$
Working in $\mathrm{ACA}_0$, one can define a cut $I(n)$ consisting of those $n$ such that there exists a truth-predicate for $\Sigma^0_n$-sentences satisfying Tarski’s definition (the inductive clauses can be described by an arithmetical formula, hence $I$ is $\Sigma^1_1$-definable). Using the method of shortening of cuts, let $J(n)$ be a cut $\mathrm{ACA}_0$-provably closed under $\omega_k$ and included in $I$. We have
$$\mathrm{ACA}_0\vdash(I\Delta_0+\Omega_k+\mathrm{Con}_\mathrm{PA})^J.$$
Let $K(n)\iff J(2_k^n)$. Then $K$ is a cut closed under multiplication, and we have
$$\mathrm{ACA}_0\vdash(I\Delta_0+\mathrm{Con}_{\mathrm{ACA}_0})^K$$
using $(*)$, which contradicts a suitable version of Gödel’s incompleteness theorem.
For 2, let $\psi$ be the $\Pi^0_1$-sentence
$$\forall s\in\Sigma^0_1\forall x,p\,\bigl(p\le\log^{(k)}(x)\land\mathrm{Proof}_{\mathrm{ACA}_0}(p,s)\to\exists q\le x\,\mathrm{Proof}_{\mathrm{PA}}(q,s)\bigr)$$
expressing that $\mathrm{ACA}_0$ has no speedup faster than $2_k^x$ over $\mathrm{PA}$ (for $\Sigma^0_1$-sentences). By the same argument as in 1, we show
$$\mathrm{ACA}_0+\psi\vdash(I\Delta_0+\mathrm{Con}_{\mathrm{ACA}_0+\psi})^K,$$
hence $\mathrm{ACA}_0\vdash\neg\psi$ by Gödel’s theorem. In particular, $\psi$ is false, hence $\mathrm{ACA}_0$ has speedup faster than $2_k^x$ over $\mathrm{PA}$ for some $\Sigma^0_1$-sentences $\phi$.
One can construct an explicit such sentence, as well as avoid the appeal to the $\Sigma^0_1$-soundness of $\mathrm{ACA}_0$ at the end of the proof, by defining
$$\vdash\phi\leftrightarrow\exists x,p\,\bigl(p\le\log^{(k)}(x)\land\mathrm{Proof}_{\mathrm{ACA}_0}(p,\ulcorner\phi\urcorner)\land\forall q\le x\,\neg\mathrm{Proof}_{\mathrm{PA}}(q,\ulcorner\phi\urcorner)\bigr)$$
using self-reference, and working with $\psi=\neg\phi$.
Notice that there is nothing particularly specific to $\mathrm{PA}$ in the proof. In general, if $T$ is a consistent sequential theory in a finite language, axiomatized by finitely many axioms and axiom schemata (allowing for all formulas with parameters, such as the induction schema of $\mathrm{PA}$), and $T^+$ is its “second-order” conservative extension with first-order comprehension axioms, and the schemata of $T$ replaced with the corresponding $\Pi^1_1$-sentences, then the speedup theorem holds with $T$ and $T^+$ in place of $\mathrm{PA}$ and $\mathrm{ACA}_0$. Particular examples of such theories are $\mathrm{ZF(C)}$ and $\mathrm{GB(C)}$.<|endoftext|>
TITLE: Jacobi sums on tori
QUESTION [6 upvotes]: The Jacobi sum of $n$ multiplicative character $\chi_1,\dots,\chi_n$ on a finite field
$\mathbb  F_q$ is defined as 
$$J(\chi_1,\dots,\chi_n) = \sum_{x_1,\dots,x_n \in \mathbb F_q, x_1+\dots+x_n=1} \chi_1(x_1) \chi_2(x_2) \dots \chi_n(x_n).$$ One also considers the variant:
$$J_0 (\chi_1,\dots,\chi_n) = \sum_{x_1,\dots,x_n \in \mathbb F_q, x_1+\dots+x_n=0} \chi_1(x_1) \chi_2(x_2) \dots \chi_n(x_n)$$
It is well-known (see e.g. Ireland and Rosen) that one can compute the complex modulus
of $J(\chi_1,\dots,\chi_n)$ and $J_0(\chi_1,\dots,\chi_n)$ (when the $\chi_i$ are "general" in some precise sense, $|J(\chi_1,\dots,\chi_n)|=q^{n-3/2}$, and the other cases are not hard to determine as well).
The definition of the Jacobi sum can be rewritten in a more compact way using the maximal diagonal torus $T$ of $Gl_n$: for $\chi=(\chi_1,\dots,\chi_n)$ a character of the maximal torus,
$$J(\chi) = \sum_{x \in T(\mathbb F_q), \ tr\  x = 1} \chi(x)$$
$$J_0(\chi)=  \sum_{x \in T(\mathbb F_q), \ tr\  x = 0} \chi(x)$$
Now it is clear that the definition above of $J(\chi)$ makes sense when $T$ is replaced by any
subtorus of $Gl_n$ defined over $\mathbb F_q$, not necessarily maximal or split. My question is

Is it possible to determine, or at least estimate, $J(\chi)$ for general character of
  $T$ when $T$ is a general torus of $Gl_n$ defined over $\mathbb F_q$? 

In  particular, by how much (if anything) is it possible to improve on the trivial bound
$|J(\chi)| \leq \sum_{x \in T(\mathbb F_q), \ tr\  x = 1} 1$?
If the general question is too hard, here is one case I am especially interested: $T$ is the torus of $Gl_4$ of diagonal matrices $(x,y,y^{-1},x^{-1})$. So in this case, $T$ is still 
a split torus, but is not maximal (actually it it is the maximal torus of the symplectic group $Sp_4$ seen as a torus of $Gl_4$ through the natural inclusion). So in this case, in down-to-earth terms, $J_0(\chi)=\sum_{x,y \in \mathbb F_q, x + y + x^{-1} + y^{-1} =0} \chi_1(x) \chi_2(y)$.

REPLY [3 votes]: For general tori this problem gets really weird. For instance, given any polynomial equation $f(x_1,..,x_n)$ defined over $\mathbb F_p$, I can encode the point-counting problem $|\{x_1,\dots,x_n  \in \mathbb F_q | x_1,\dots x_n \neq 0, f(x_1,\dots,x_n)=0 \}|$ as one of these. This point-counting problem is usually at least as hard as counting solutions to $f$ which is a tough problem that we can usually only approximate the answer to.
Indeed, just consider a diagonal matrix where each monomial in $x_1,\dots,x_n$ occurs with multiplicity congruent modulo $p$ to its coefficient in $f$. As $x_1$ and $x_n$ vary, this defines a torus. Choose a trivial character of this torus, and you recover the point-counting problem.
Or choose a nontrivial, generic one - it shouldn't make the problem any easier.
Of course, in this specific case, Felipe Voloch has the solution.<|endoftext|>
TITLE: Do all subtraction-free identities tropicalize?
QUESTION [19 upvotes]: If you take a subtraction-free rational identity like $(xxx+yyy)/(x+y)+xy=xx+yy$ and replace $\times$,$/$,$+$,$1$ by $+$,$-$,min,$0$, do you always get a valid min,plus,minus identity like min(min($x+x+x,y+y+y$)$-$min($x,y$),$\:x+y$)$\ =\ $min($x+x,y+y$)?

REPLY [18 votes]: It suffices to show that whenever $F$ is a function $\mathbb R_{\geq 0}^k\to\mathbb R_{\geq 0}$ defined using $\times,/,+,1$, and $f,g$ is the corresponding tropicalization $\mathbb R^k\to\mathbb R$, for all real $x_1,\dots,x_k$ we have $$F(\exp(-\beta x_1),\dots,\exp(-\beta x_k))^{1/\beta}\to \exp(-f(x_1,\dots,x_k)).$$
as $\beta\to+\infty$.
This follows by structural induction on the formula defining $f$, using the following lemma.
Lemma: Let $F$ be one of the operations $\times,/,+$, let $f$ be the corresponding tropical operation, and let $G_1$ and $G_2$ be functions $\mathbb R_{> 0}\to\mathbb R_{> 0}$ such that
$G_1(\beta)^{1/\beta}$ tends to some limit $e^{-x_1}$ as $\beta\to\infty$, and likewise $G_2(\beta)^{1/\beta}\to e^{-x_2}$. Then
$$F(G_1(\beta),G_2(\beta))^{1/\beta}\to \exp(- f(x_1,x_2)).$$
as $\beta\to+\infty$.
Proof: The operations $\times$ and $/$ are easy - the only non-trivial step is
$$(G_1(\beta)+G_2(\beta))^{1/\beta}\to e^{-\beta \min(x_1,x_2)}.$$

REPLY [17 votes]: Yes.  Replace $x$ with $e^{-Na}$, $y$ with $e^{-Nb}$, etc. Then take the log, then divide by $N$. One gets a new identity where $\times$ is replaced by $+$, $/$ by $-$, $1$ by $0$, and $u+v$ by $-\ln (e^{-N u} + e^{-N v}) / N= \min(u,v) - \ln\left( 1+ e^{-N |u-v|}\right)/N = \min(u,v) + O(1/N)$. Then take the limit as $N$ goes to $\infty$. You now have a tropcal identity.
This fits with the idea of tropical geometry as the limit of classical algebraic geometry as variables get very large.<|endoftext|>
TITLE: Monotonicity of Loewner ellipsoid?
QUESTION [10 upvotes]: Given two $0$-symmetric convex bodies $K \subset L \subset \mathbb{R}^n$, is it true that the Loewner ellipsoid of $K$ is contained in the Loewner ellipsoid of $L$?
I have just finished proving a lemma stating that the Loewner ellipsoid depends continuously on parameters and the proof is a bit more elaborate than I first expected.  I suddenly realized that I was unconsciously assuming that the Loewner ellipsoid is not monotone (otherwise the lemma would be trivially true), but that I did not have a ready example showing that this was the case.
I profit to ask a second question: is there a reference for the fact that the Loewner ellipsoids of a continuous family of convex bodies form a continuous family?
Rephrase in terms of normed or Finsler bundles and Euclidean structures if you want to be very rigorous.
My proof of this fact involves "looking under the hood" at the proof of uniqueness of the Loewner ellipsoid and using Berge's maximum theorem for set-valued maps. It's natural (after all this is just a problem in mathematical programming), but I was expecting a triviality.

REPLY [16 votes]: No, the Loewner ellipsoid is not monotone w.r.t. inclusion. Let $K$ be a square, whose Loewner ellipsoid is its circumcircle. Let $L$ be any other ellipse through the four vertices of $K$. The Loewner ellipsoid of $L$ is $L$ itself but it does not contain the circle. (I assume that the Loewner ellipsoid is the minimal circumscibed one. If you meant the maximal inscribed one, just consider the dual bodies.)
The continuity of Loewner ellipsoid follows from its uniqueness via a standard compactness argument. Its volume is continuous for trivial reasons. If the Loewner ellipsoid $E(K)$ were discontinuous at some convex body $K$, then by compactness there would be a sequence $K_i\to K$ such that $E(K_i)$ converge to some ellipsoid other than $E(K)$ but of the same volume, contrary to the uniqueness.<|endoftext|>
TITLE: Homotopy excision and homotopy pushout
QUESTION [5 upvotes]: I have three related questions.
I understand homotopy pushouts via the standard model structure on the diagrams - and taking the derived functor of the pushout. 
I'm not sure, but I believe that in classical algebraic topology, we use implicitly at least three model structures in Top: Quillen structure, Hurewicz structure and mixed structure.
So, we obtain (at least) three notions of homotopy pushouts, right? 
First question: Is there a theorem relating these three types of homotopy pushouts? 
Second question: Can homotopy excision be formulated as a result about homotopy pushouts? If so, can we formulate homotopy excision as a result of which (of those three) kind of homotopy pushout?
Third question: Do you know a good reference on this stuff? (Specially, references about homotopy excision using homotopy pushouts).
Remark: "Homotopy excision" is the following result: If $(X,A,B)$ is a excisive triad, such that $(A,C)$ is $(n-1)$-connected and $(B,C)$ is $(m-1)$-connected (with $n\geq 2 $ and $m\geq 1 $), Then
$(A,C)\to (X,B) $ is a $(m+n-2)$ equivalence.
It seems clear that all homotopy pushouts, at the HUrewicz model, can be viewed as a excisive triad... And, then, we can formulate the result...
I don't know about the other structures.
But, even in the Hurewicz model structure, the reformulated version seems to be weaker (since I can't prove that all excisive triad is equivalent to a homotopy pushout - I mean, if $(X, A, B) $ is a excisive triad, I can't prove that X is the homotopy pushout of $A\cap B\to A $ along $A\cap B \to B $).

REPLY [5 votes]: There are old-fashioned classical ways to think about excision, which can easily be translated
into model theoretical language of homotopy pushouts as desired.  Any excisive triad can be approximated up to weak equivalence of triads by a CW triad, where a CW triad (X;A,B) is just a CW complex X that is the union of subcomplexes A and B (theorem on page 77 of "A concise course in algebraic topology"). A CW triad is not excessive, but it is homotopy equivalent to the obvious double mapping cylinder, which is excessive (lemma on page 78).  Note that  excision in any homology theory is obvious for a CW triad since A/A cap B is isomorphic as a CW complex to X/B. The Blakers-Massey theorem is given an easy homotopical proof from this CW approximation result on page 85, opus cit.  No originality is claimed: I'm pretty sure I learned the idea from Boardman (probably unpublished). 
Chapter 17 of "More concise algebraic topology" has a reasonably thorough treatment of the three model structures and their comparison.  The mixed model structure is due to Cole.
(However `More Concise' unfortunately has a mistake, taken from a different paper of Cole, in the proof of the axioms for the Strom model structure; a nice paper of Barthel and Riehl,
"On the construction of functorial factorizations for model categories" explains and corrects the mistake.)<|endoftext|>
TITLE: Closed totally disconnected subspaces
QUESTION [18 upvotes]: It is a remarkable property of uncountable compact metric spaces that each of them contains a homeomorphic copy of the Cantor set. In general, one cannot expect containment of Cantor cubes (in particular, in the case of scattered compact spaces) but instead different totally disconnected subspaces occur quite often. My question is the following

Let $K$ be a compact Hausdorff space. Does $K$ contain a closed, totally disconnected subspace having the same cardinality as $K$?

REPLY [7 votes]: EDIT: I tried to improve the presentation. I hope that it is a bit more readable now.
I think that the following construction gives a consistent counter-example. It uses the combinatorial principle $\diamondsuit$, which implies CH (the continuum hypothesis) and is independant from ZFC. 
I'll work with ordinals, and since $\alpha<\omega_1$ (for instance) may be seen either as a point or as a an initial segment in $ \omega_1 $, I'll use the interval notation when I refer to the latter.
Start with a circle bundle over $\omega_1$, that is, a space $X$ with a continous map $\pi$ onto $\omega_1$, such that the fiber over each point $\pi^{-1}(\{\alpha\})$ is a circle. (Of course, $\omega_1$ is endowed with the order topology.) Say that a subset of $X$ is bounded if it is contained in $\pi^{-1}([0,\alpha])$ for some $\alpha<\omega_1$, and unbounded otherwise. 
Nyikos showed in his article "The theory of non-metrizable manifolds" in the Handbook of set theoretic topology (Example 6.17) that using $\diamondsuit$, we can construct a circle bundle over $\omega_1$ with the following properties:
1) The underlying set of $X$ is $\omega_1\times\mathbb{S}^1$, but the topology is not the product topology.
2) $\pi^{-1}([0,\alpha])$ is homeomorphic to $[0,\alpha]\times\mathbb{S}^1$ with the usual product topology for any $\alpha<\omega_1$. In particular, $X$ is locally compact. We may fix an homeomorphism $\psi_\alpha:\pi^{-1}([0,\alpha])\to [0,\alpha]\times\mathbb{S}^1$ for each $\alpha<\omega_1$.
3) If $E\subset X$ is closed, then either $E$ is bounded, or $E$ contains $\pi^{-1}(C)$ for $C$ a closed and unbounded subset of $\omega_1$. In particular, in the latter case $E$ is not totally disconnected since it contains copies all the fibers (which are circles) above the member of $C$.
(Note: Actually, Nyikos builds a bundle over the long ray ${\mathbb{L}}_+$ which is even a surface, but we take only the restriction to $\omega_1$.)
Now, define a circle bundle $\pi':Y\to\omega_{\omega_1}$ as follows. The underlying set of $Y$ is $\omega_{\omega_1}\times\mathbb{S}^1$. $X$ will be included in $Y$ as the union of the fibers above the $\omega_\alpha$, for $\alpha<\omega_1$. 
The topology "between" $\omega_\alpha$ and $\omega_{\alpha+1}$ is the usual product topology. That is, if $\omega_\alpha < \gamma < \omega_{\alpha+1}$, given $x\in\mathbb{S}^1$, take $\beta,\beta'$ with $\omega_\alpha\le\beta<\gamma<\beta'< \omega_{\alpha+1}$ and an open $O\subset\mathbb{S}^1$ containing $x$, then $(\beta,\beta')\times O$ is a neighborhood of $\langle\gamma,x\rangle$. 
We now define the neighborhoods of $\langle\omega_{\alpha},x\rangle$. 
In $X$, choose a neighborhood $U$ of $\langle \alpha,x\rangle$, and denote by $U^\beta$ the intersection $U\cap \pi^{-1}(\{\beta\})$. Set $V^{\omega_\beta}=U^\beta$, and
if $\omega_\beta<\gamma<\omega_{\beta+1}$, set $V^{\gamma}=U^{\beta+1}$. That is: $V^\gamma$ is equal to the intersection of $U$ with the fiber over $\beta+1$. Then a neighborhood of $\langle\omega_{\alpha},x\rangle$ is given by the union of $\{\gamma\}\times V^{\gamma}$ for all $\gamma$ greater than some $\gamma'$. 
Then $Y$ has the following properties:
2') For any $\alpha<\omega_{\omega_1}$, $(\pi')^{-1}([0,\alpha])$ is homeomorphic to $[0,\alpha]\times\mathbb{S}^1$ with the usual product topology (and thus $Y$ is locally compact). To see this, define the homeomorphism $\phi:(\pi')^{-1}([0,\alpha])\to [0,\alpha]\times\mathbb{S}^1$ by setting $\phi(\langle\omega_{\alpha},x\rangle)=\psi(\langle\alpha,x\rangle)$, and for $\omega_\alpha<\gamma<\omega_{\alpha+1}$, set $\phi(\langle\gamma,x\rangle)=\psi(\langle\alpha+1,x\rangle)$, where $\psi$ is defined in 2) above. 
3') As 3) with $\omega_{\omega_1}$ rather than $\omega_1$.
Now, a bounded subset of $Y$ is contained in some $[0,\alpha]\times\mathbb{S}^1$, and has thus cardinality $\max\{|\alpha|,\omega_1\}<\omega_{\omega_1}$ (since $2^\omega=\omega_1$ by CH). Taking the one-point compactification yields a compact space such that a closed set of cardinality $\omega_{\omega_1}$ cannot be totally disconnected.
I hope that I did not miss something. 
I don't know whether another construction can be done in ZFC, but this particular one needs at least something more than ZFC+CH, because Nyikos space $X$ would have to contain a copy of $\omega_1$ (which is impossible, and thus the space does not exist) in a model of ZFC+CH due to Eisworth and Nyikos.<|endoftext|>
TITLE: Status of (Global) Langlands Conjecture for $GL_2$ over $\mathbb{Q}$
QUESTION [24 upvotes]: Apologies if this question has already been dealt with on MO. I am wondering about the status of the global Langlands conjectures for $GL_2$ over the rational numbers. How close is humanity to the proof of these conjectures? 
I guess Langlands picture is related to (or, should I say, includes?) the Taniyama-Shimura, Fontaine-Mazur, and Serre Conjectures. There has been great progress on these latter conjectures recently. Even assuming these conjectures are settled, how much closer are we to understanding the Langlands program for $GL_2$? 
It would be helpful if somebody points to a paper where a modern and precise version of the global Langlands conjecture for $GL_2$ is stated. For the local Langlands conjecture, we have the nice paper of Vogan. Is there an analogous paper for the global theory?

REPLY [2 votes]: Additionally to what Joel was saying, Langlands conjectured the existence of a universal group $\widehat{G}$ (depending on the number field only) whose 2-dim'l representation correspond to automorphic representation of GL(2) in a suitable way. These would include automorphic forms with the archimedean factors not being algebraic, e.g., even Maass forms with Laplace eigenvalue $\neq 1/4$. From what I understand, Arthur suggests a definition of $\widehat{G}$ here:  http://www.claymath.org/cw/arthur/pdf/automorphic-langlands-group.pdf. The situation is similar to case (1c) described above.<|endoftext|>
TITLE: Numbers integrally represented by a ternary cubic form
QUESTION [8 upvotes]: Given integers $a,b,c,$ and cubic form
$$ f(a,b,c) = a^3 + b^3 + c^3 + a^2 b  - a b^2 + 3 a^2 c - a c^2  + b^2 c - b c^2  - 4 a b c $$
$$ f(a,b,c) =
\det  \left(  \begin{array}{ccc}
  a  &  b & c  \\\
  c & a + c & b + c \\\
   b + c   &  b + 2 c  & a + b + 2 c
\end{array} 
  \right)  .  $$ 
what primes $p$ can be integrally represented as
$$  p = f(a,b,c)? $$
(A): I think it is  all primes $(p| 11) = -1 ,$ and all $p = u^2 + 11 v^2$ in integers, but not any $q = 3 u^2 + 2 u v + 4 v^2.$ 
Note that, if $-p$ is represented, so is $p.$ 
(B): I also suspect that if  prime $q = 3 u^2 + 2 u v + 4 v^2$ and $f(a,b,c) \equiv 0 \pmod q,$  then all three $a,b,c \equiv 0 \pmod q,$ and  $f(a,b,c) \equiv 0 \pmod {q^3}.$ Checked correct for $q=3,5.$ Maybe I will do a few more. 
Note that if $f$ integrally represents both $m,n$ then it represents $mn.$ That is because $f(a,b,c) = \det(aI + b X + c X^2),$ where
$$ X =  \left(  \begin{array}{ccc}
  0  &  1 & 0  \\\
  0 & 0 & 1 \\\
   1  &  1  & 1
\end{array} 
  \right)    $$ Then $X^3 = X^2 + X + I$ and $X^4 = 2 X^2 + 2 X + I.$ 
If all suspicions are correct, we can correctly describe all numbers integrally represented by this polynomial: positive or negative are unimportant, most prime factors are unimportant, all that matters is that every exponent of a prime factor $q = 3 u^2 + 2 u v + 4 v^2$ must be divisible by 3.
I should have done this last time: most of the class field part has already been done, by Hudson and Williams (1991), Theorem 1 and Table 1 on page 134. You get my version of the polynomial by negating their variable $x.$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=  

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
       p           a           b           c 
       2           0           1           1
       7           0         -11           6
      11           0          -3           2
      13           0          -1           2
      17          -1           0           2
      19           1           2           4
      29           0          -7           4
      41           0           3           2
      43           0           4          -1
      47           0           5          -2
      53           0           1           4
      61           0          46         -25
      73           2         -36          19
      79           0           3           4
      83           0          24         -13
     101          -1          12          -6
     103           0          15          -8
     107           1          -9           5
     109           1           2           6
     127           1          -2           4
     131           1           7          -3
     139           1          -6           4
     149          -1           4           2
     151           0         -20          11
     163           0           5           2
     167          -1           1           5
     173           0           6          -1
     193           1         -52          28
     197           0           9          -4
     199          -1           5           1
     211           0         -12           7
     227          -2           0           5
     233           0         -16           9
     239           0          -6           5
     241           0          -4           5
     257           0          -1           6
     263           2           4           9
     269          -1           0           6
     271           2           8          -3
     277           1          -7           5
     281           0           2           7
     283          -1           2           6
     293          -1          -8           6
     307           2          -1           6
     311           0           5           6
     337          -2           5           2
     347           1           7           5
     349           0          19         -10
     359          -1           9          -3
     373           2           5          10
     397           1          -1           7
     401           0         -68          37
     409           3         -77          41
     419           0          -7           6
     421           0           7           2
     431           1         -14           8
     439           0           8          -1
     457           0           1           8
     461           0          -2           7
     479           1          -8           6
     491           0           7           4
     499           0          13          -6
     503          -1         -36          20
     523           0           9          -2
     541           2         -12           7
     547           1         -11           7
     557          -1          25         -13
     563          -2         -11           8
     569           0           8           1
     571           1          -3           7
     587           0         -29          16
     593           3         -25          13
     599          -1           0           8
     601           0           7           6
     607           0          11          -4
     613           0           4           9
     617           2          -1           8
     659           0           8           3
     673           0          -6           7
     677           0         -17          10
     683          -1           4           8
     701           2          13          -6
     733           1          10          -2
     739          -1          14          -6
     743          -2           1           8
     757           0          81         -44
     761          -1           8           2
     769           0         -25          14
     773          -1           7           5
     787           2           5          12
     809          -1         -10           8
     811          -4           0           7
     821          -1           3           9
     827           2          10           7
     853           0         -11           8
     857          -2           3           8
     863           0           9           2
     877          -2         -15          10
     883           0         -14           9
     887           2          -3           8
     907           0          -5           8
     911           0           8           7
     919           0          -2           9
     929           1           7          11
     937           3           8          14
     941           3          -1           9
     953          -1           6           8
     967           1          13          -5
     991           1         -35          19
     997          -3           7           3

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Primes represented by $x^2 + 11 y^2$ and then by $3 x^2 + 2 x y + 4 y^2,$ both up to $1000.$
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primego
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2 
1  0  11
Discriminant  -44

Modulus for arithmetic progressions? 
11
Maximum number represented? 
1000
           p        p  mod 11 
          11           0
          47           3
          53           9
         103           4
         163           9
         199           1
         257           4
         269           5
         311           3
         397           1
         401           5
         419           1
         421           3
         499           4
         587           4
         599           5
         617           1
         683           1
         757           9
         773           3
         863           5
         883           3
         907           5
         911           9
         929           5
         991           1


    0    1    3    4    5    9

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ 
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primego
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2 
3  2  4
Discriminant  -44

Modulus for arithmetic progressions? 
11
Maximum number represented? 
1000
           p        p mod 11
           3           3
           5           5
          23           1
          31           9
          37           4
          59           4
          67           1
          71           5
          89           1
          97           9
         113           3
         137           5
         157           3
         179           3
         181           5
         191           4
         223           3
         229           9
         251           9
         313           5
         317           9
         331           1
         353           1
         367           4
         379           5
         383           9
         389           4
         433           4
         443           3
         449           9
         463           1
         467           5
         487           3
         509           3
         521           4
         577           5
         619           3
         631           4
         641           3
         643           5
         647           9
         653           4
         661           1
         691           9
         709           5
         719           4
         727           1
         751           3
         797           5
         823           9
         829           4
         839           3
         859           1
         881           1
         947           1
         971           3
         977           9
         983           4


    1    3    4    5    9

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ 

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
joro asked about high powers being represented primitively. It turns out that the polynomial is not divisible by 8 unless $a,b,c$ are all even. This, despite the fact that 2 is represented. I believe this happens for all the (unrepresented) primes $q = 3 u^2 + 2 u v + 4 v^2$ as well, in the strongest manner: the polynomial is not divisible by $q$ itself unless $a,b,c$ are. I thought there might be trouble with the prime 11, but no. Anyway, here are some prime powers represented primitively, where $47 = 36 + 11$ and $53 = 9 + 44:$
           7           0           1           2
          49           1          -1           3
         343           6           4           5
        2401         -11          -3           9
       16807         -11          30          -8
      117649         -19          75         -29
      823543          -2        -117          82
     5764801         162          43          12
    40353607         205         -64         186



          11          -1           1           1
         121          10          15          16
        1331         -10          -2           7
       14641          12          28           9
      161051           1          25          59
     1771561          53         -78          70
    19487171          37          46         300
   214358881         171        -210         460




          13           1           3           3
         169          10          17          18
        2197          -4           3          10
       28561         -15          -8          24
      371293           8          71          34
     4826809         -54          98          77
    62748517        -257         125         167



          47           1           3           5
        2209          10          12           3
      103823         108         181         202
     4879681         104          32         153
   229345007        -128         319         432




          53          -1           1           3
        2809          10          23          24
      148877         100         163         170
     7890481         100          18         187
   418195493         342        -308         451

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

REPLY [13 votes]: Your conjectures are correct.  So was the "someone else at MSRI [who]
muttered something about norm forms" (mentioned in earlier edits
of the question), except for the part about laughing at you.
As you in effect note, $f(a,b,c)$ is the norm $N_{K/{\bf Q}}(a+bx+cx^2)$,
where $x$ is one of the roots of $x^3-x^2-x-1 = 0$
and $K$ is the cubic number field ${\bf Q}(x)$.
This field has discriminant $-44$,
and ${\bf Z}[x]$ is the full ring of integers $O_K$
(equivalently, the field discriminant of $K/{\bf Q}$
equals the polynomial discriminant of $x^3-x^2-x-1$;
to check this in gp, compute
poldisc(x^3-x^2-x-1)
nfdisc(x^3-x^2-x-1)

and observe that both return $-44$).  Now for (A),
you already know that $x^3-x^2-x-1$ has at least one root modulo
any prime $q$ unless $q$ is represented by the nonprincipal quadratic
form $3u^2+2uv+4v^2$ of discriminant $-44$.  (For other $q$:
there's a triple root for $q=2$, a double and a simple root for $q=11$,
three distinct roots for $q=u^2+11v^2$, and one simple root for odd
$q$ not congruent to a square $\bmod 11$.)  Equivalently, $K$ has
an ideal of norm $q$ unless $q = 3u^2+2uv+4v^2$.  But $O_K$ is
a principal ideal domain, so once there's an ideal of norm $q$
then it has a generator $a+bx+cx^2 \in O_K$, and then $q=f(a,b,c)$
(or $q=f(-a,-b,-c)$ if we chose $a+bx+cx^2$ of norm $-q$).  The discriminant of $K$
is small enough that one can check unique factorization by hand using
the Minkowski bound; nowadays this exercise can also be done routinely
on the computer, e.g. in gp
K = bnfinit(x^3-x^2-x-1); K.cyc

(This functionality happens to be one of the "Usage examples" in the current
Wikipedia
page on gp.)
[EDIT In fact this $K$ happens to be one of the handful of number fields
whose Minkowski bound is so tight that nothing needs to be checked!
The discriminant $\Delta_K = -44$ is small enough in absolute value that 
the bound
$$
\frac4\pi \frac{3!}{3^3} \left|\Delta_K\right|^{1/2} = 1.8768\ldots
$$
is less than $2$, which means every ideal $I$ has a nonzero element
of norm $\pm \left|I\right|$ and is thus automatically principal.  TIDE]
(B) Translating the factorization of $x^3-x^2-x-1 \bmod q$
into the factorization of the ideal $(q)$ in $O_K$, we see that
if $q = 3u^2+2uv+4v^2$ then $(q)$ remains prime in $O_K$,
and thus that $q \mid N_{K/{\bf Q}}(a+bx+cx^2)$ iff $q \mid a+bx+cx^2$.
For $q=2$ the ideal $(q)$ is the cube of $(1+x)$, so $8 \mid f(a,b,c)$
iff $a,b,c$ are all even.  Any power of a prime $q$ other than
those of the form $3u^2+2uv+4v^2$ can be represented primitively by $f$,
even $q=11$ (for which $(q)$ factors as $(2+x)(3-2x)^2$).  If we do not
care about primitivity then we can also represent all powers of $2$,
and all powers of $q^3$ for $q = 3u^2+2uv+4v^2$.
By multiplicativity this also proves the final conjecture:
the nonzero $n \in {\bf Z}$ that are represented by $f$
are precisely those whose $q$-valuation is a multiple of $3$
for all primes $q = 3u^2+2uv+4v^2$.<|endoftext|>
TITLE: A question on Iwasawa theory
QUESTION [5 upvotes]: Let $p$ be an odd prime. Let $K_{\infty}$ be the field extension of $\mathbb{Q}$ generated by all $p^n$-th roots of unity. Let $M/K_{\infty}$ be the maximal abelian $p$-extension of $K_{\infty}$ unramified outside $p$. Denote its Galois group by $\mathcal{X}$. Then the  Iwasawa main conjecture concerns the structure of $\mathcal{X}^{+}$ (here, $\pm$ means the $\pm1$-eigenspace of complex conjugation).  My question is whether one can  say anything about the structure of $\mathcal{X}^-$? 
I don't find the answer in the basic textbook (e.g. Lang's Cyclotomic fields & Washington's Introduction to cyclotomic fields). I wonder whether $\mathcal{X}^-$ is a free $\mathbb{Z}p[[\Gamma]]$-module where $\Gamma$ is the Glois group of $K_{\infty}/\mathbb{Q}(\zeta_{p})$?
Can somebody tell me the answer?
Thank you very much.

REPLY [2 votes]: Thank you very much.
Kummer-Vandiver conjecture will imply $X^+=0$, not just $X^+\sim 0$. Recently, I found some connection between Kummer-Vandiver conjecture and the freeness of $\mathcal{X}^-$ as a $\mathbb{Z}p[[\Gamma]]$-module. That is why I asked about $\mathcal{X}^-$.<|endoftext|>
TITLE: Whitney stratifications
QUESTION [14 upvotes]: Many results on characteristic classes of singular varieties (as well as other singularity-theoretic constructions) make use of a so-called "Whitney stratification" of the variety under consideration, which from what I understand is a filtration of the variety into smooth locally closed subvarieties subject to some techinical conditions (the technical conditions may be found here: http://en.wikipedia.org/wiki/Whitney_conditions). However, other than merely reading the bare-bones definitions of these technical conditions, I have never seen these technical conditions explicitly used. Moreover, I have never once seen a concrete example of a Whitney stratification. As such, an elucidation of the technical conditions and why they are useful along with a concrete example of a Whitney stratification (perhaps say a Whitney umbrella) would be greatly appreciated. Thanks!

REPLY [11 votes]: Here is an example:
Let $V = \lbrace y^2 = t^2 x^2 + x^3\rbrace \subset \mathbb R^3$. Then the singular set of $V$ is the whole $t$-axis.
Let $Y$ be the $t$-axis and $X = V - Y$.
Now set $X_1 = X \cap \lbrace x > 0 \rbrace$,
$X_2 = X \cap \lbrace x < 0 \rbrace \cap \lbrace t>0 \rbrace$, and
$X_3 = X \cap \lbrace x < 0 \rbrace \cap \lbrace t<0 \rbrace$.
I wish I could draw pictures but I don't know how to do that here. It's fairly easy though. 
Notice that, $X_1$, $X_2$, $X_3$ and $Y$ are smooth submanifolds of $\mathbb R^3$, they are disjoint, and their union is $V$. Thus we have stratified $V$ into submanifolds. 
You can now verify that this stratification of $V$ is an $(a)$-regular stratification. However, it is not a $(b)$-regular (Whitney) stratification. Verify that $X_2$ and $X_3$ are not $(b)$-regular over $Y$.
Let $Z$ be the origin and $W = Y - Z$. Now, $X_1$, $X_2$, $X_3$, $W$ and $Z$ are disjoint, smooth submanifolds of $\mathbb R^3$ and their union is $V$. This stratification is $(b)$-regular or what is called a Whitney stratification of $V$. 
I hope the example helps visualizing $(a)$-regular and $(b)$-regular stratifications.
You can read `Notes on topological stability' by Mather for properties of Whitney conditions (yes it is published now after more than 40 years).
http://www.ams.org/journals/bull/2012-49-04/S0273-0979-2012-01383-6/S0273-0979-2012-01383-6.pdf
Another good reference is the Ph.D. thesis of David Trotman titled `Whitney stratifications: faults and detectors'.<|endoftext|>
TITLE: The category theory of $(\infty, 1)$-categories
QUESTION [16 upvotes]: There are many proposed models for the theory of $(\infty, 1)$-categories and it has now been shown that many of these theories have Quillen-equivalent model categories, i.e. that they are equivalent as homotopy theories. But what about as category theories?
For example, suppose I have a homotopical category $\mathcal{C}$ and a homotopically initial object $A$ in $\mathcal{C}$, in the sense of [Dwyer, Kan, Hirschhorn, and Smith, Homotopy limit functors on model categories and homotopical categories]. That means, there are endofunctors $F$ and $G$ on $\mathcal{C}$ such that $\Delta A$ is weakly equivalent to $F$, $G$ is weakly equivalent to $\textrm{id}_{\mathcal{C}}$, and there is a natural transformation $\alpha : F \Rightarrow G$ such that $\alpha_A : F A \to G A$ is a weak equivalence. It is straightforward to show that $A$ descends to an initial object in $\operatorname{Ho} \mathcal{C}$, but what about in, say, the hammock localisation $L^H \mathcal{C}$ – is the hom-space $L^H \mathcal{C}(A, Z)$ contractible for every object $Z$ in $\mathcal{C}$? 
More generally, how are the notions of homotopy limit/colimit, adjunction, Kan extension etc. in the different theories (where defined) related? Pointers to relevant literature would be much appreciated.

REPLY [12 votes]: Browsing through the old unanswered questions, I've come across this one, which happily can be partially answered now by the work of Riehl and Verity (Zhen Lin will be aware of this, which is why I'll CW it, but the average mathematician using the site might not be).
The first step is to identify what "a category theory" is. In ordinary category theory, we have categories, functors, bimodules, and natural transformations. Perhaps the term "bimodule" (or "profunctor" or "distributor") is unfamiliar in this context, but a bimodule from $C$ to $D$ is just a functor $C^\mathrm{op} \times D \to \mathsf{Set}$, and they compose via a coend, analogously to tensoring ordinary bimodules between rings. Categories and functors form a 2-category; so do categories and bimodules. The relationship is that functors are included into bimodules, by sending the functor $F: C \to D$ to the representable bimodule $D(1,F)$, and moreover this bimodule has a right adjoint given by the corepresentable bimodule $D(F,1)$. Such a structure is called a proarrow equipment. It's essential to have the bimodules in addition to the functors in order to express things like what it means to be a pointwise Kan extension.
This much was known. The next step is to try to extract a proarrow equipment from an arbitrary model of $\infty$-categories, and then to check whether various notions such as (co)limits and Kan extensions agree with the notions that have already been defined in ad hoc ways in various models. The idea of course is that the role of "functors" should be played by appropriate $\infty$-functors and the role of "bimodules" should be played by appropriate 2-sided $\infty$-presheaves. Riehl and Verity do this in a uniform manner for some (but not all) models by setting up an axiomatic framework of $\infty$-cosmoi. From an $\infty$-cosmos, one can extract a homotopy 2-category of $\infty$-categories and $\infty$-functors, and then the bimodules are defined using certain weak comma objects which are guaranteed to exist by the $\infty$-cosmos axioms. This is similar to a familiar construction using comma objects, but Riehl and Verity had to weaken the universal property of the comma object in a novel way in order for it to exist in this setting. And it turns out that all the categorical notions formulated this way agree with those which had already been formulated in more ad hoc ways.
Since the proarrow equipment associated to an $\infty$-cosmos is defined in a uniform way, Riehl and Verity are then able to formulate simple criteria for equivalence of these equipments. And they find that all of the models of $\infty$-categories which actually form $\infty$-cosmoi are indeed equivalent. This includes those models which form model categories which are simplicial with respect to the Joyal model structure, including quasicategories, complete Segal spaces, naturally marked simplicial sets, and Segal categories. But, for example, simplical categories and homotopical categories are not among the models which can be compared using the $\infty$-cosmos framework. But it is still natural to approach these comparisons by extracting proarrow equipments from these models.
I've perhaps overstated the role of the proarrow equipment in this story. I think Riehl and Verity regard it as an open question to determine just how much "category theory" is correctly encoded in terms of the proarrow equipment. (I've also glossed over the fact that Riehl and Verity use a slightly weaker, and more-complicated-to-define, but to my mind no less natural structure called a virtual equipment.) But I suspect that even concepts which can't be expressed in terms of proarrow equipments will be expressible in terms of $\infty$-cosmoi, using the addtional structure available which I haven't gone into, and Riehl and Verity have actually constructed equivalences at the $\infty$-cosmos level, so the models they consider should be equivalent in a strong sense.
References: Notes and video from Riehl's YTM lectures. The notes contain fuller references, laid out in the introduction.<|endoftext|>
TITLE: Genus one fibered links
QUESTION [12 upvotes]: It is well-known that the only genus one fibered knots are the trefoil and the figure-eight. On the other hand, there exist infinitely many fibered links for any fixed higher genus.
My question is about what happens if we go to more boundary components: fixing the number of boundary components, are there infinitely many genus one fibered links? 
Another related question: the monodromy of a fibered genus one link corresponds, after collapsing the boundary, to an element of $SL_n(\mathbb Z)$. For example the trefoil yields $((0,1),(-1,1))$ and the figure-eight $((2,1),(1,1))$. 
Which conjugacy classes of $SL_n(\mathbb Z)$ can be obtained as monodromies of many-components genus one fibered links?

REPLY [2 votes]: Agol's comment works to give 3 component fibered links with genus 1 fiber, but we can dial it back a bit further.   
The connect sum of a Hopf band and its mirror is a 3-component link of unknots with genus 0 fiber.  Moreover one component is 0-framed so Dehn twists along it will give an infinite family of 3 component fibered links with genus 0 fiber.  So to get an infinite family of 2 component links with genus 1 fiber, you can just plumb on another Hopf band along an arc joining the other two boundary components.  This plumbing will be disjoint from the twisting.
I think the Alexander polynomials of these links will distinguish them, perhaps even the Conway polynomials.  
It would be interesting to see the Hopf stabilization equivalences in these families.<|endoftext|>
TITLE: At what point would an elementary generalization of Bertrand's Postulate be interesting?
QUESTION [13 upvotes]: I know that in 1952 Jitsuro Nagura was able to show that there is always a prime between $k$ and $\frac{6k}{5}$ for $k > 24$.
At what point would an improvement on Nagura's result be interesting?  If an approach could show for example that for any $k$, there is a specific value $X$ which could be calculated such that for all $x \ge X$, there is a prime between $kx$ and $(k+1)x$, would this be interesting?
Or, does the Prime Number Theorem provide us enough insight that short of a proof of Legendre's Conjecture, elementary results are not very interesting at this time?

REPLY [2 votes]: Once again, being new here I tried to comment your answer, but obviously don't have 50 points yet.
Here is the original text of my comment with my mistake on value of x
"As a check on your equations I set k=100 x=2478 and should therefore expect one prime between 2478 and 2478 + 2478/100 = 2502.78.(for k=100, x must be greater than 2202.65 as detailed by your equations - this condition is met for x=2478).But there are no primes between 2478 and 2502.78. (Though 2503 is prime) Is the domain of x for your answer the positive integers? Then everything checks out okay. Or as you warned 'up to error'?"
Assigning x=24.78 is the correct value.
@quid: Note the prime is between kx and (k+1)x, so with my corrected value of x there should be a prime between 2478 < (some prime) < 2502.78, but the next prime is 2503.
@stefan kohl: understood and my calculation and answer/comment also gave x > 2202.65. My mistake was not putting the correct value for x<|endoftext|>
TITLE: Sharpness of the Sobolev embedding theorem
QUESTION [5 upvotes]: We know that $W^{k,p}\hookrightarrow C^{k-\lfloor\frac{n}{p}\rfloor-1,\gamma}(\bar{\Omega})$ with $kp>n,\gamma=\lfloor\frac{n}{p}\rfloor+1-\frac{n}{p}$, where $n$ is the dimension of $\Omega$, $\Omega$ is a bounded domain in $\mathbb{R}^n$ with $C^1$ boundary.
From Wikipedia. This can also be seen in C.L.Evan's "Partial Differential Equations".
However, when $n/p$ is an integer, the theorem does not state anything more about $\gamma=1$. Is there any counterexample to 
$W^{k,p}\hookrightarrow C^{k-\frac{n}{p}-1,1}(\bar{\Omega})$ when $n/p$ is an integer? I don't know how to construct it, Thanks for your attention!.

REPLY [3 votes]: This response is closely related to my answer Here.
For the case $W^{2,n}$ we automatically get Holder regularity by applying Sobolev and then Morrey. However, we don't get Lipschitz. The following example "integrates" the counterexample to boundedness of $W^{1,n}$ functions.
Take a function $\psi$ supported on $B_2$ with $\psi$
linear and nonconstant on $B_1$ and $|\nabla \psi| < c$, and add dyadic rescalings together as follows. Consider
$$u(x) = \sum_{i=1}^{\infty} h_i\psi(2^ix)$$
for some $h_i$ we will choose to get bounded $W^{2,n}$ norm but unbounded derivative.
Note that $|D^2(h_i\psi(2^ix))|$ grows like $h_i2^{2i}$ and they are supported on disjoint dyadic rings of volume going like $2^{-in}$. Thus, to get bounded $W^{2,n}$ norm we want
$$\sum_i h_i^n2^{in} < C.$$
To give unboundedness of the derivative we want
$$\sum_i h_i2^{i} = \infty.$$
The natural choice for $h_i$ is $\frac{2^{-i}}{i}$, which gives the counterexample.
Remark: This function has size $~ 2^{-k}\sum_{i=1}^k \frac{1}{i}$ ~ $2^{-k}|\log\log(2^{-k})|$ at $|x| = 2^{-k},$ so a more explicit example might look something like $|x||\log\log(|x|)|$, which looks almost like a cone away from $0$ but the slope gets unboundedly high near $0$.<|endoftext|>
TITLE: Reconciling two notions of geometric quantization.
QUESTION [14 upvotes]: Let $(M,\omega)$ be a compact symplectic manifold and $(L,\nabla)$ a prequantum line bundle.  There are two schemes to quantize this data:

Choose a polarization $P$ of $M$ and define the quantum Hilbert space to be sections of $L$ that are parallel along $P$.  This space admits an action of the Poisson algebra of smooth functions on $M$.
Choose a compatible almost complex structure.  This gives a $Spin^c$ structure and define the quantum space to be the index of the corresponding Dirac operator twisted by $L$.  Here there is no action of $C^\infty(M)$.  When a Lie group $G$ acts on everything, the index is an element of the representation ring of $G$.

In the case that $M$ is Kähler, $L$ a holomorphic line bundle, and $P = T^{0,1}M$, the first method gives the space of holomorphic sections of $L$ and the second method gives the index of $\bar\partial_L + \bar\partial_L^*$.
The first method is more tied to physics and may seem a little ad-hoc from a mathematical point of view.  The second seems more natural mathematically (since it fits in well with symplectic reduction and the index of an elliptic operator is more well-behaved mathematically than the kernel of $\nabla$ along $P$).  But it seems to be a very weak notion of quantization since there is no action of $C^\infty(M)$.  .
How can these two viewpoints be reconciled? Should the second version be viewed as just a more natural mathematical construction?  Is there a nice way to tie it back to the physics?

REPLY [5 votes]: Klaas Landsman has been proposing that geometric quantization is best understood as taking values in KK-cocycles. (A quick review of this idea is for instance around p. 134 of his student's thesis available here.) If one remembers the KK-cocycle before passing it through the Baum-Connes assembly map to compute its index (which is supposedly an iso anyway...), then this means remembering the module structure that you are asking for. 
From this and various other perspectives it looks like regarding geometric quantization as a map into KK-theory is a rather attractive idea.
A collection of relevant references by Landsman, his students, and related articles is here: http://ncatlab.org/nlab/show/geometric+quantization+by+push-forward#References<|endoftext|>
TITLE: What is the purpose of section 3 of BBD?
QUESTION [15 upvotes]: I am not quite sure that this question is appropriate for Mathoverflow, yet I would be deeply grateful for any hint: what happens in section 3 of Beilinson A., Bernstein J., Deligne P., Faisceaux pervers//Asterisque 100, 1982, 5-171? Is there any statement that is important for the following sections of this treatise? I do not know French; yet this does not prevent me from understanding all the other sections.
Upd. I wonder: does there exist a 'plan' of BBD?

REPLY [7 votes]: I'm not an expert at BBD, but the introduction explains that chapter 3 is supplementary technical information that can be skipped.  For example, if you want to work with $\mathbb{Z}$-coefficients, or the filtered derived category, you might find chapter 3 useful.  As far as a "plan" of BBD is concerned, the introduction has a brief description of the contents of each chapter, but if you are looking for some kind of "tree" of lemmata leading to main results like the decomposition theorem, I haven't seen one.
I only found 3 references to chapter 3 in the rest of the book:

The proof of Proposition 2.1.23 uses section 3.2.
Part 2.2.19 gives a proof of Proposition 2.1.23 that does not use results of chapter 3.
The first page of chapter 4 mentions section 3.3 in an inessential way.<|endoftext|>
TITLE: A Fraïssé class without the strong amalgamation property.
QUESTION [5 upvotes]: I am trying to find some examples of Fraïssé classes that do not have the strong amalgamation property. Anyone?

REPLY [6 votes]: The following paper gives an example showing why distributive lattices have amalgamation but not strong amalgamation.
E Fried, G Grätzer.
Strong amalgamation of distributive lattices.
Journal of Algebra,
Volume 128, Issue 2, 1 February 1990, Pages 446–455.
doi:10.1016/0021-8693(90)90033-K
Their example is the following. 
Let A and B be the 4-element Boolean algebras A = {s0,s1,s2,a} and B = {s0,s1,s2,b} with s0 bottom and s1 top. Let S = {s0,s1,s2} be the intersection of A and B. S is a distributive lattice, indeed a sublattice of both A and B. Since both a and b are complements of s1, the 4 element Boolean algebra D = {s0,s1,s2,c} where c identifies a and b, provides an amalgamation of A and B wrt S. But S is not the intersection (formally: pullback) of A and B in D (as both A->D and B->D are isomorphisms), therefore D is not a strong amalgamation. Moreover, it follows from D being the pushout of (S->A,S-B) that no other amalgamation of A and B wrt S is strong.<|endoftext|>
TITLE: Homology groups of divisible and powered (nilpotent) groups
QUESTION [5 upvotes]: (1) Suppose $\pi$ is a set of primes and $G$ is a $\pi$-divisible nilpotent group, i.e., for any $g \in G$ and $p \in \pi$, there exists $x \in G$ such that $x^p = g$. Is it necessary that all the homology groups of $G$ for the trivial group action with coefficients in $\mathbb{Z}$ are also $\pi$-divisible groups? I am in particular interested in the second homology group $H_2(G;\mathbb{Z})$, which is the Schur multiplier of $G$. Note that the result is true for $H_1(G;\mathbb{Z})$, which is the abelianization of $G$.
I believe that the following example shows that the result is false for $H_2(G;\mathbb{Z})$, but I'm not sure whether I am applying results correctly. The example group $G$ that I have in mind is defined as follows. Let $UT(3,\mathbb{Q})$ denote the group of $3 \times 3$ upper triangular matrices over the rational numbers that have 1s on the diagonal. Let $Z$ be an infinite cyclic subgroup of $UT(3,\mathbb{Q})$ generated by any non-identity element in its center. Let $G = UT(3,\mathbb{Q})/Z$. Then, $G$ is divisible by all primes, on account of being a quotient of $UT(3,\mathbb{Q})$, which is divisible by all primes. $G$ admits $UT(3,\mathbb{Q})$ as a stem extension group and the base of the extension is $Z$, isomorphic to $\mathbb{Z}$, which is not divisible by any prime. The base of any stem extension is a quotient of the Schur multiplier, so the Schur multiplier (which I haven't computed precisely) should not be divisible by any prime.
(2) Suppose $\pi$ is a set of primes and $G$ is a $\pi$-powered nilpotent group, i.e., for any $g \in G$ and $p \in \pi$, there exists unique $x \in G$ such that $x^p = g$. Is it necessary that all the homology groups of $G$ for the trivial group action with coefficients in $\mathbb{Z}$ are also $\pi$-powered groups? I am in particular interested in the second homology group $H_2(G;\mathbb{Z})$, which is the Schur multiplier of $G$. Note that the result is true for $H_1(G;\mathbb{Z})$, which is the abelianization of $G$ (the proof of this uses that $G$ is nilpotent).
Can anything be said in general, in the non-nilpotent case?

REPLY [3 votes]: For $G=UT(3,\mathbb{Q})/Z$, I computed $H_2(G,\mathbb{Z})=\mathbb{Z}$, which confirms that $G$ is a counterexample to (1). (This doesn't contradict Adam's answer, since $G\supseteq  \mathbb{Q}/\mathbb{Z}$ isn't uniquely divisible)
The essential steps in my computation are (coefficients are always $\mathbb{Z}$ which are suppressed): 

The LHS spectral sequence of the central extension $\mathbb{Q}/\mathbb{Z} \hookrightarrow G \twoheadrightarrow \mathbb{Q}^2$ yields $H_3(G)=0$ and $H_2(G) = E^3_{2,0} \le H_2(\mathbb{Q}^2)=\mathbb{Q}$. 
The LHS of the central extension $\mathbb{Q} \hookrightarrow UT(3,\mathbb{Q}) \twoheadrightarrow \mathbb{Q}^2$ yields $H_2(UT(3,\mathbb{Q}))=\mathbb{Q}^2$ 
The LHS of the central extension $\mathbb{Z} \hookrightarrow UT(3,\mathbb{Q}) \twoheadrightarrow G$ has 
$$E^2=\begin{array}{cccc}
|0 & & & & \newline 
|\mathbb{Z} & \mathbb{Q}^2 & & \newline 
|\underline{\mathbb{Z}} & \underline{\mathbb{Q}^2} & \underline{H_2(G)} & \underline{0} \newline 
\end{array}$$ 
Since $UT(3,\mathbb{Q}) \twoheadrightarrow G$ induces an isomorphism on the abelianization, we have $E^3_{01}=0$. Hence $d^2: H_2(G)\to \mathbb{Z}$ is epi. For $K = \ker d^2$ there is a s.e. sequence $\mathbb{Q}^2 \hookrightarrow H_2(UT(3,\mathbb{Q})) \twoheadrightarrow K$. Tensoring yields $K\otimes \mathbb{Q}=0$. If $K \neq 0$ there is $\mathbb{Z} \hookrightarrow K$ (since $K\le H_2(G)\le \mathbb{Q})$ and tensoring gives the contradiction $\mathbb{Q} \hookrightarrow K\otimes \mathbb{Q}=0$. Hence $K=0$ and $H_2(G)\cong \mathbb{Z}$ as claimed.<|endoftext|>
TITLE: What is known about the strong Arnold conjecture?
QUESTION [11 upvotes]: Here are the two versions of Arnold's conjecture on Hamiltonian orbits:

Let $(M,\omega)$ be a closed symplectic manifold, and let $H: \mathbb{R/Z} \times M \to \mathbb{R}$ be a nondegenerate Hamiltonian.  Then the number of $1$-periodic orbits of the vector field $X_H$ defined by $\omega(X_H, \cdot) = dH$ is bounded below by 

the sum of the rational Betti numbers of $M$ (weak version)
the minimum number of critical points of a Morse function on $M$ (strong version).


The weak version has been settled using Floer homology.  My question is:

Has anyone made any progress on the strong Arnold conjecture?

I have asked a couple of experts in person, but they didn't know of anything (see also Tim Perutz's answer to this question).
Also, I wonder if anyone can tell me the right intuition for the strong conjecture.  The only idea I have for why the strong conjecture should be true is that when $H$ is time-independent, each critical point of $H$ is a $1$-periodic orbit of $X_H$.
Edit about the version of the conjecture without assuming $H$ nondegenerate: Thanks to Thomas's comment below I looked up Arnold's original statement of the conjecture and realized that it is different than the one I wrote above (which I have never seen in print).  Call the conjecture above the nondegenerate Arnold conjecture; here is the original, "possibly degenerate" Arnold conjecture from Mathematical methods of classical mechanics:

Let $(M, \omega)$ be a closed symplectic manifold, and let $H: \mathbb{R}/\mathbb{Z} \times M \to \mathbb{R}$ be a Hamiltonian.  Then the number of $1$-periodic orbits of $X_H$ is bounded below by the minimal number of critical points of a smooth function on $M$.

There is also a weak version (not from Arnold's book) where the bound is replaced by $1$ plus the cuplength of $M$.
Some progress has been made toward the strong version of this conjecture.  In "On analytical applications of stable homotopy", Yuli Rudyak proves that if $(M,\omega)$ is a closed symplectic manifold such that $\omega|_{\pi_2(M)} = 0$ and $c_1|_{\pi_2(M)} = 0$ and the Lusternik-Schnirelmann category of $M$ is $\dim M$, then the strong version of the "possibly degenerate" Arnold conjecture holds.  In "On the Lusternik-Schnirelmann category of symplectic manifolds and the Arnold conjecture", John Oprea and Rudyak eliminate the LS hypothesis by showing that $\text{LS}(M) = \dim M$ whenever $M$ is a closed symplectic manifold such that $\omega|_{\pi_2(M)} = 0$.

REPLY [3 votes]: Assume dim$(M) \geq 6$ and $M$ is simply connected.
By cancelling Morse critical points for a Morse function $f$ on $M$ one can get the number of critical points equal to the minimum number of generators needed in $C_* (M)$ to generate $H_* (M)$. I.e. the sum of the Betti numbers plus 2 for each torsion generator. This is also the number of cells in a minimal cell structure (CW-complex) for $M$.
This means that if one can define Floer homology with $\mathbb{Z}$ coefficients and the PSS isomorphism works with $\mathbb{Z}$ coefficients, then the strong Arnold conjecture is true.
I believe this is the case for e.g. monotone symplectic manifolds.<|endoftext|>
TITLE: Local Norm Mapping for Abelian Varieties
QUESTION [6 upvotes]: Let $A/K$ be an abelian variety defined over a nonarchimedean local field $K$ of characteristic $0$ and let $L$ be a finite extension of $K$. Consider the norm map $$A(L)\xrightarrow{N_{L/K}}A(K)$$ I want to know if the following statement is true - 
If $A_{0}(K)$ is the subgroup of $A(K)$ specializing to the connected component of the Neron model, then $$\bigcap_{L}N_{L/K}A(L)\cong A_{0}(K)$$ where the intersection is over all finite unramified extensions $L$. 
When $A$ has non-degenerate reduction over $K$ (i.e. the dual variety ${A}^{\vee}$ has good reduction), this follows immediately from Tate Local Duality and Corollary 4.4 in Mazur's 1972 paper 'Rational Points of Abelian Varieties with Values in Towers of Number Fields.' Does it also hold when $A$ does not have non-degenerate reduction over $K$?

REPLY [2 votes]: One doesn't need Tate local duality to analyze the good reduction case, and in general the answer is affirmative.
First, let's review the general nonsense for norm maps with commutative group functors (to sidestep representability issues). For any finite etale map of schemes $f:S' \rightarrow S$ and any commutative functor $F$ on the category of $S$-schemes there is a natural "norm" map of group functors $N: f_{\ast}(F_{S'}) \rightarrow F$, where $F_{S'}$ denotes the restriction of $F$ to the category of $S$-schemes (represented by base change when $F$ is representable).  Indeed, if $T$ is an $S$-scheme then $f_{\ast}(F_{S'})(T) = F(T \times_S S')$ is identified with the group of global sections of the $f_T$-pullback of the restriction of $F_T$ to the small etale site over $T$, so to define $N$ on $T$-points we simply use the natural trace map $(f_T)_{\ast} \circ f_T^{\ast} \rightarrow {\rm{id}}$ on the category of abelian sheaves on the small etale site of $T$. (This trace map is defined more generally using just that $f$ is quasi-finite flat and separated and finitely presented, but later we will need that $f$ is finite etale and the definition of the trace map is much simpler in the case of finite etale $f$ anyway.)
For example, taking $F$ to be the functor of points of the Neron model $\mathcal{A}$ of $A$ over $O_K$ and $f$ to correspond to $O_K \rightarrow O_L$, the above norm map on $O_K$-points is precisely the natural norm map $A(L) \rightarrow A(K)$ due to the universal mapping property of Neron models and the compatibility of Neron models with unramified base change.  The advantage of this geometric interpretation is that the norm map of smooth finite type $O_K$-groups 
$$N_{A,L/K}:{\rm{R}}_{O_L/O_K}(\mathcal{A}_{O_L}) \rightarrow \mathcal{A}$$
is an $O_K$-form of the addition map $\mathcal{A}^{[L:K]} \rightarrow \mathcal{A}$.  More generally, for any $f$ and $F$ as in the preceding paragraph with $f$ of constant degree $d$, the "norm" map $f_{\ast}(F_{S'}) \rightarrow F$ becomes the addition map $F^{\oplus d} \rightarrow F$ after passing to an etale cover of $S$ that splits $S'$ into a disjoint union of $d$ copies of $S$ (a fiber power of $S'$ over $S$ is such a cover). Indeed, the formation of the norm commutes with base change on $S$, and in the case of a split covering it is clear from the construction that the norm is identified with the componentwise addition map.
Now we see that by taking $[L:K]$ divisible by the order of the finite etale component group of the special fiber $\mathcal{A}_0$, $N_{A,L/K}$ lands inside the relative identity component $G := \mathcal{A}^0$ (open complement of the closed union of the finitely many non-identity components in the special fiber) since such an assertion is local for the etale topology on $O_K$ and hence it can be checked by computing with the $O_K$-form of the map given by componentwise addition $\mathcal{A}^{\oplus [L:K]} \rightarrow \mathcal{A}$.  Hence, we may replace $\mathcal{A}$ with $G$ and reduce to checking that the norm map
$$N_{G,O_L/O_K}:{\rm{R}}_{O_L/O_K}(G_{O_L}) \rightarrow G$$
is surjective on $O_K$-points for any smooth $O_K$-group $G$ of finite type with connected fibers.  
The kernel $G' = \ker N_{G,O_L/O_K}$ is a form of $G^{[L:K]-1}$ for the etale topology on $O_K$, so it is also smooth of finite type with connected fibers. Hence, for any $O_K$-point $g$ of $G$, the $N_{G,O_L/O_K}$-pullback of $g:{\rm{Spec}}(O_K) \rightarrow G$ is a smooth $O_K$-scheme $X$ of finite type that is a $G'$-torsor for the etale topology, and we just need to show that $X(O_K)$ is non-empty.  Since $X$ is smooth and $O_K$ is henselian (e.g., complete), such non-emptiness holds provided that the special fiber $X_0$ has a rational point over the residue field $k$ of $O_K$.  
We have come down to the task of showing that any torsor for a smooth connected finite type group scheme over $k$ must have a $k$-point.  So far there is no number theory in any of this, just general arithmetic geometry over discrete valuation rings.  But finally we bring in finiteness of $k$ to apply Lang's theorem which asserts exactly that such torsors always have a rational point when the ground field $k$ is finite.<|endoftext|>
TITLE: Intuition for Levi-Civita connection via Hamiltonian flows
QUESTION [11 upvotes]: A Riemannian metric on a manifold $X$ defines a function on the symplectic space $T^*X$ whose Hamiltonian flow gives geodesics. Is there a similar interpretation of the Levi-Civita connection?

REPLY [12 votes]: The intuition is that the Levi-Civita connection corresponds to the linearization of the geodesic flow plus a simple projective-geometric construction:
Let $c(t)$ be an orbit of the geodesic flow (projecting down to a geodesic), consider the vertical subspaces $V(t)$ along $c(t)$ and bring them back to the tangent space of the cotangent bundle over the point c(0) by using the differential of the flow. You get a family of (Lagrangian) subspaces $l(t) := D\phi_{-t}(V(t))$ that is "fanning" or "regular". 
Now forget you ever had a geodesic flow: all that you need is the curve of subspaces. A bit of differential projective geometry shows that you also get a second curve $h(t)$ of (Lagrangian subspaces) in $T_{c(0)}(T^*M)$that is transversal to $l(t)$. The subspace h(0) is the horizontal subspace of the connection and  $T_{c(0)}(T^*M) = l(0) \oplus h(0)$ is the decomposition into vertical and horizontal subspaces.
I think this is nicely written in this paper ;-) 
However, this is classical: it's just the geometry behind the Schwartzian derivative. There are plenty of references in the paper if you're interested in this.  
Addendum on the geometry of the Schwartzian derivative.
Since the Schwartzian derivative has many "standard" geometric interpretations, for completeness sake I'll sketch the (new?) one that I'm refering to. I will do it in a way just a bit more conceptual than was done by Duran and myself in the paper. Actually, it's just a "high brow" version of what was done in the paper, but I will do it here just for curves on the projective line. The reader may amuse himself/herself at extending this to curves on the Grassmannian of $n$-planes in $\mathbb{R}^{2n}$ and checking against the paper.
Consider the action of the linear group $GL(2;\mathbb{R})$ on the projective line and lift it to an action on its cotangent bundle. The moment map of this action takes values on the set of nilpotent matrices. In fact, if you take away the zero section, the moment map takes value in the coadjoint orbit formed by $2 \times 2$ nilpotent matrices of rank one. In higher dimensions, one needs to take away a bit more than the zero section and "rank one" is replanced by "the kernel and the image of the matrix (seen as linear transformation) are the same". 
We will also need a neat thing about the geometry of the tangent space of the projective line: there is a canonical way to identify non-zero tangent vectors to non-zero covectors. In fact, the tangent space of the projective line at a line $\ell$ is the space of linear maps between $\ell$ and the quotient space $\mathbb{R}^2/\ell$, which is the tensor product $(\mathbb{R}^2/\ell) \otimes \ell^* $. If the map is invertible, its inverse is a map from $\mathbb{R}^2/\ell$ to $\ell$ and, therefore, an element of  $ \ell \otimes (\mathbb{R}^2/\ell)^*$, which is the cotangent space at $\ell$. 
Now consider a curve $\gamma(t)$ on the projective line whose derivative never vanishes. We lift the  curve to the curve $\dot{\gamma}(t)$ on the tangent bundle and use the isomorphism described above to obtain a curve $\Gamma(t)$ on the cotangent of the projective line. Use the moment map to obtain a curve $F(t)$ of nilpotent matrices. Note that everything we have done is projective-equivariant. 
Finally we come to the little miracle: the time derivative of $F(t)$ is a curve of reflections $\dot{F}(t)$ (i.e., $\dot{F}(t)^2 = I$) whose -1 eigenspace is the curve of lines $\gamma(t)$ and whose $1$-eigenspace defines a "horizontal curve" $h(t)$ equivariantly attached to $\gamma(t)$. This is the construction that yields the Levi-Civita connection (and what is behind the formalisms of Grifone and Foulon for connections of second order ODE's on manifolds).
Differentiate $F(t)$ a second time to find the Schwartzian derivative. Geometrically, it just describes how the curve $h(t)$ moves with respect to $\gamma(t)$. For comparison, recall that the curvature of a connection is obtained by differentiating (i.e., bracketing) horizonal vector fields and projecting onto the vertical bundle.<|endoftext|>
TITLE: When does $ZFC \vdash\ ' ZFC \vdash \varphi\ '$ imply $ZFC \vdash \varphi$?
QUESTION [16 upvotes]: Being a new member, I am not yet sure whether my question will be taken as a research level question (and thus, appropriate for MO). However, I have seen similar questions on MO, couple of which led me asking mine, and I seem to not be able to find many resources except discussion on FOM and MO. So, any references to resolve the question and fix my possible confusion would be appreciated.
As the title suggests, I want to understand the relation between $ZFC \vdash \varphi$ and $ZFC \vdash\ 'ZFC \vdash \varphi'$. Let me give my motivation (and some partial answers) asking this question so that what I'm trying to arrive at is understood.
We know that if $ZFC \vdash \varphi$, then $ZFC \vdash\ 'ZFC \vdash \varphi'$ for we could write down the Gödel number of the proof we have for $\varphi$ and then check that the formalized $\vdash$ relation holds. I believe even more can be checked to be true for this provability predicate (Hilbert-Bernays provability conditions).
Is the converse true in general? Not necessarily. (Just to make sure that it will be pointed out sooner if I am doing any mistakes, I will try to write down everything unnecessarily detailed using less English and more symbols!)
Let us assume only that $ZFC$ is consistent (However, I am not assuming the formal statement $Con(ZFC)$, that is $\ 'ZFC \nvdash \lceil 0=1 \rceil'$). Then, it is conceivable that $ZFC \vdash\ \ 'ZFC \vdash \lceil 0=1 \rceil'$ but $ZFC \nvdash 0=1$. It might be that in reality ZFC is consistent but $\omega$-inconsistent.
Indeed, if I am not missing a point, it is consistent to have this situation:
$ZFC \vdash Con(ZFC) \rightarrow Con(ZFC+\neg Con(ZFC))$ (Gödel)
$ZFC \vdash Con(ZFC) \rightarrow\ \exists M\ M \models ZFC+\neg Con(ZFC)$ (Gödel)
$ZFC \vdash Con(ZFC) \rightarrow\ \exists M\ M \models\ ZFC+\ 'ZFC \vdash \lceil 0=1 \rceil'$
$ZFC \vdash Con(ZFC) \rightarrow\ \exists M\ M \models\ ZFC+\ 'ZFC \vdash\ 'ZFC \vdash \lceil 0=1 \rceil\ '\ '$ (Soundness and the second provability condition here)
$ZFC \vdash Con(ZFC) \rightarrow Con(ZFC+\ 'ZFC \vdash \neg Con(ZFC)\ ')$
So we cannot hope to have $ZFC \vdash\ 'ZFC \vdash \varphi'$ implying $ZFC \vdash \varphi$ for an arbitrary formula without requiring an additional assumption. At least, we know this  for $\varphi: 0=1$ (this is not because of the consistency argument above, but because consistency and $\omega$-inconsistency of ZFC is a possibility).
If you believe that ZFC's characterization of natural numbers coincides with what we have in mind and agree that ZFC should not be $\omega$-inconsistent, then you might want to throw in the assumption $Con(ZFC)$.
Now imagine a universe where $Con(ZFC)$ holds but all the models of ZFC is $\omega$-nonstandard and believe $\neg Con(ZFC)$. I do not know whether this scenario is even possible (which is another question I am wondering) but if it is possible, then it would be the case that $'ZFC \vdash \neg Con(ZFC)\ '$, by completeness since $\neg Con(ZFC)$ is true in all models. Then if the implication in title (or should I say, an informal version of it: $V \models ZFC \vdash \varphi$ implies $V \models \varphi$) held, then $\neg Con(ZFC)$ which contradicts our assumption that there are models at all. The point is arbitrary models of ZFC may not be sufficient to have existence of ZFC-proofs implying existence of actual proofs.
However, if we add a stronger assumption $\psi$ that there is an $\omega$-model, then whenever we have an arithmetic sentence $\varphi$, if
$ZFC \vdash\ 'ZFC \vdash \lceil \varphi \rceil'$
then
$ZFC+\psi \vdash \exists M\ \omega^M=\omega \wedge M \models ZFC+\ \varphi$
and because $\omega$ in the model is the real one, by taking care of quantifiers one by one we can deduce $ZFC+\psi \vdash \varphi$. Thus, existence of an $\omega$-model solves our problem for arithmetical sentences. I cannot see any reason to make this work for arbitrary sentences without strengthening the assumption. Here is a thought:
We know, by the reflection principle, that we can find some limit ordinal $\alpha$ such that $\varphi \leftrightarrow \varphi^{V_{\alpha}} \leftrightarrow V_{\alpha} \models \varphi$. Thus, if we could make sure somehow that $V_{\alpha}$ is a model of ZFC while we reflect $\varphi$, then we would be done. But I could not modify the proofs of reflection in such a way that this can be done and am not even sure that this could be done.
My question to MO is to what extent (and under which assumptions) can we get the implication in the title?
Edit: After reading Emil Jerabek's answer, I realized I should clarify some details.
Firstly, I want to treat ZFC only as a formal system (meaning that if you are claiming some assumption $\psi$ does what I want, I want to have a description of how that proof would formally look. This is why I kept writing all the leftmost $ZFC \vdash$'s all the time). Then, it is clear by the above discussions that even if we could prove $ZFC \vdash \varphi$ within our system, we may not prove $\varphi$ without additional assumptions on our system.
One solution could be that our system satisfies the "magical" property that whenever we have $ZFC \vdash \exists x \in \omega\ \varphi(x)$, say for some arithmetic sentence, then we have $ZFC \vdash \varphi(SSS...0)$ for some numeral. This of course is not available by default setting for we know that the theory $ZFC$+ $c \in \omega$ + $c \neq 0$ + $c \neq S0$ +... is consistent if $ZFC$ is consistent. Thus, that magical property seems like an unreasonably strong assumption.
To make my question very precise, what I want is some assumption $\psi$ so that for some class of formulas, whenever I have $ZFC \vdash ZFC \vdash \varphi$, then $ZFC + \psi \vdash \varphi$. For arithmetic sentences, existence of an $\omega$-model is sufficient.
I agree that $\Sigma^0_1$ soundness should be sufficient for arithmetic sentences if what you mean by $\Sigma^0_1$ soundness is having $ZFC \vdash ZFC \vdash \varphi$ requiring (maybe even as a derivation rule, attached to our system!) $ZFC \vdash \omega \models \phi$, where $\phi$ is the translated version of $\varphi$ into the appropriate language, since I can again go through quantifier by quantifier and prove the sentence itself, that is $ZFC \vdash \varphi$.
However, I see no reason why $\Sigma^0_1$ soundness should be enough for arbitrary sentence $\varphi$. It seems to me that what we need is some structure for which we have the reflection property that formal truth in the structure is provably equivalent to $\varphi$ and that structure being model of all the ZFC-sentences used in the ZFC-proof of $\varphi$.
I believe existence of $\Sigma_n$-reflecting cardinals which are inaccesible is more than sufficient for sentences up to $n$ in the Levy hiearchy. By definition of those, we have the equivalence $\varphi \leftrightarrow V_{\kappa} \models \varphi$ and then provability of $\varphi$ in ZFC implies $V_{\kappa} \models \varphi$. However, I am not sure if we had to go that far. ${}{}$

REPLY [13 votes]: With regard to your sub-question,


Now imagine a universe where $\text{Con}(\text{ZFC})$ holds but all the models of $\text{ZFC}$ are  $\omega$-nonstandard and believe $\neg \text{Con}(\text{ZFC})$. I do not know whether this scenario is even possible...


Indeed this is possible, assuming that
$\text{ZFC}+\text{Con}(\text{ZFC})$ is consistent. The reason is
that by the incompleteness theorem, if this theory is consistent,
then there is a model of
$\text{ZFC}+\text{Con}(\text{ZFC})+\neg\text{Con}(\text{ZFC}+\text{Con}(\text{ZFC}))$.
In this model, we have that $\text{Con}(\text{ZFC})$ holds, but
there is no model of $\text{ZFC}+\text{Con}(\text{ZFC})$, and so
all models of $\text{ZFC}$ in this model satisfy
$\neg\text{Con}(\text{ZFC})$. In particular, all models of
$\text{ZFC}$ in this model are also $\omega$-nonstandard from the
perspective of this model.<|endoftext|>
TITLE: Does closed Alexandrov space admit a bi-Lipschitz embedding into $\mathbb R^N$?
QUESTION [7 upvotes]: As the title says.
Let $A^n$ be an $n$-dimensional closed Alexandrov space. Does it admit a bi-Lipschitz embedding into Euclidean space $\mathbb R^N$ for sufficiently large $N$?
I know there are some spaces that do not admit such an embedding; for example, a theorem by Pansu says that: 

The Heisenberg group
  equipped with the Carnot-Caratheodory distance does not biLipschitz embed into $\mathbb R^n$,
  for any $n$.

REPLY [2 votes]: In my preprint "Bi-Lipschitz embeddings of SRA-free spaces into Euclidean spaces" (https://arxiv.org/abs/1906.02477) I prove the quantitative version of this statement.

Theorem 2.
  For $n \in \mathbb{N}$, $k < 0$ and $R > 0$ there exist $D > 0$ and $N \in \mathbb{N}$ satisfying the following.
  For every $n$-dimensional Alexandrov space of curvature $\ge k$ and every $x \in X$
  there exists an embedding $\phi:B_R(x) \rightarrow \mathbb{N}^N$ which 
  bi-Lipschitz distortion does not exceed $D$.
For $n \in N$ there exist $D_0 > 0$ and $N_0 \in \mathbb{N}$ such that 
  every $n$-dimensional Alexandrov space of non-negative curvature 
  allows an embedding into $\mathbb{N}^{N_0}$ which 
  bi-Lipschitz distortion does not exceed $D_0$.<|endoftext|>
TITLE: Locally flat non-smooth discs
QUESTION [6 upvotes]: There are many knots (e.g. the $P(-3,5,7)$-pretzel knot) that are topologically, but not smoothly slice; "topologically" slice means that there is a locally flat embedding of a disc into the 4-space, such that the disc's boundary is the knot, and smoothly slice means that there is such an embedding that is smooth.
Smooth slice discs are (conjecturally) always ribbon discs, so it is fairly easy to visualise them (see e.g. a ribbon disc for 6_1). Or, using Morse theory, every smooth slice disc may be drawn as a movie, i.e. a sequence of Reidemeister moves and $a$ births, $b$ saddle moves and $c$ deaths from a diagram of the knot to a diagram of the unknot, such that $b = a + c$.
But how can one visualise the disc embedded in a locally flat way which is bounded by a topologically, but not smoothly slice knot? Do such discs e.g. have movies in which some additional non-smooth moves feature?

REPLY [3 votes]: You could read the appendix of this paper
http://people.mpim-bonn.mpg.de/teichner/Papers/Alexander.pdf
which is very nice.  It basically says that for an Alexander polynomial one knot you push a Seifert surface into the 4-ball and then surger it to a disc.  The surgery discs need Freedman-Quinn disc embedding though.  This cannot be visualised so easily, if at all, since the grope and tower height raising used to construct convergent towers in the proof go over many parallel copies of dual surfaces or grope arms.<|endoftext|>
TITLE: Counter-example to faithfully flat descent
QUESTION [10 upvotes]: I am looking for a counter example to the fact that a faithfully flat morphism is
an effective descent morphism for the category of quasi-coherent sheaves
when one forgets the quasi-compact hypothesis. If possible, I'd like a counter-example
on the descent of morphism aspects. In other words, I am looking for
a morphism of scheme $X' \rightarrow X$, faithfully flat but not quasi-compact,
and two quasi-coherent sheaves $F$ and $G$ on $X$, such that, calling $X''=X' \times_X X'$,
$F'$, $F''$, $G'$, $G''$ thepull-back of $F$,$G$ to $X'$, $X''$, the sequence of modules
$$0 \rightarrow Hom_X(F,G) \rightarrow Hom_{X'}(F',G') \rightarrow Hom_{X''}(F'',G'')$$
is not exact (here the last morpheme is as it should the difference between the morphism 
induced by the two projections $X'' -> X'$. I wrote it this way because I am not sure how to do a double arrow in mathoverflow's tex:-)

REPLY [12 votes]: I think the following should be a counter-example.
Let $X=\text{Spec}(\mathbb{C}[T])$ be the affine line and $X'=\bigsqcup_{x\in\mathbb{C}}\text{Spec}(\mathbb{C}[T]_{(T-x)})$ be its faithfully flat cover built from all local rings at closed points. Let $F=\mathcal{O}_{X}$ and $G=\bigoplus_{x\in\mathbb{C}}\mathcal{O}_X/(T-x)\simeq\bigoplus_{x\in\mathbb{C}}\mathbb{C}_x$ be two quasi-coherent sheaves on $X$.
Then, there is a natural quotient morphism $u':F'\to G'$ given by $\mathbb{C}[T]_{(T-x)}\to \mathbb{C}[T]/(T-x)$ on $\text{Spec}(\mathbb{C}[T]_{(T-x)})$. This morphism obviously satisfies the descent condition, as the only non-trivial conditions to check are on connected components of $X''$ that are isomorphic to the generic point of $X$, and $G''$ is zero on these components. However, this morphism does not come from a morphism $u:F\to G$ because $u(1)$ should have a non-zero component in every factor of the direct sum, which is impossible.<|endoftext|>
TITLE: Inertia subgroup in the ordinary reduction case when $p=2$
QUESTION [5 upvotes]: Dear MO,
Let $K/\mathbb{Q}_2$ be a finite extension, and let $E/K$ be an elliptic curve with good ordinary reduction, and such that $\mathbb{Q}_2(j(E))=K$. Let $\rho:\operatorname{Gal}(\overline{K}/K)\to \operatorname{GL}(2,\mathbb{Z}_2)$ be the $2$-adic representation associated to $T_2(E)$, the $2$-adic Tate module of $E$, and let $I_K$ be the image of the inertia subgroup via $\rho$. The theory tells us that there is a $\mathbb{Z}_2$-basis $\{P,Q\}$ of $T_2(E)$, such that $I_K$ is a subgroup of 
$$ \left\{\left( \begin{array}{cc} \ast & \ast \\\  0 & 1\\\ \end{array}\right) \right\} \subset \operatorname{GL}(2,\mathbb{Z}_2),$$
and, moreover, if $\sigma\in I_K$, then the upper-left corner is $\chi_K(\sigma)$, where $\chi_K$ is the $2$-adic cyclotomic character $\chi_K:\operatorname{Gal}(\overline{K}/K)\to (\mathbb{Z}_2)^\times$. Let $\Phi$ be the image of $\chi_K$.
My question is the following: is $I_K$ necessarily a conjugate in $\operatorname{GL}(2,\mathbb{Z}_2)$ of a subgroup of the form
$$ B(\Phi,\Psi)=\left\{\left( \begin{array}{cc} a & b \\\  0 & 1\\\ \end{array}\right) : a\in \Phi, b \in \Psi \right\} \subset \operatorname{GL}(2,\mathbb{Z}_2),$$
for some fixed additive subgroup $\Psi$ of $\mathbb{Z}_2$? If so, what is the reason?
This would rule out some weird subgroups of $\operatorname{GL}(2,\mathbb{Z}_2)$ to appear as inertia subgroups. For example, can the following group $I$ appear as an inertia subgroup $I_K$ in $\operatorname{GL}(2,\mathbb{Z}/8\mathbb{Z})$?
$$I=\left\langle \left( \begin{array}{cc} 5 & 2 \\\  0 & 1\\\ \end{array}\right) \right\rangle =\left\{ \left( \begin{array}{cc} 1 & 0 \\\  0 & 1\\\ \end{array}\right),\ \left( \begin{array}{cc} 5 & 2 \\\  0 & 1\\\ \end{array}\right),\ \left( \begin{array}{cc} 1 & 4 \\\  0 & 1\\\ \end{array}\right),\ \left( \begin{array}{cc} 5 & 6 \\\  0 & 1\\\ \end{array}\right)\right\}\subset \operatorname{GL}(2,\mathbb{Z}/8\mathbb{Z}).$$
Edit: The example group $I$ may appear as inertia in $\operatorname{GL}(2,\mathbb{Z}/8\mathbb{Z})$ if we do not assume that $K$ is the minimal field of definition of $E/K$, i.e., $K=\mathbb{Q}_2(j(E))$, for trivial reasons. For instance, suppose $E / \Bbb Q_2$ has inertia subgroup $B((\mathbb{Z}/8\mathbb{Z})^\times,\mathbb{Z}/8\mathbb{Z})$ in $\mathbb{Q}_2(E[8])/\mathbb{Q}_2$, and let $K$ be the fixed field $\Bbb{Q}_2(E[8])^I$, where $I$ is as in the example above modulo $8$. Then the inertia subgroup for $E/K$ in $\operatorname{GL}(2,\mathbb{Z}/8\mathbb{Z})$ is as indicated in the example.
Thank you!

REPLY [4 votes]: The assumption that $K$ is not the minimal field of definition is no assumption at all. By Krasner's Lemma, the $E[8]$-torsion is locally constant in any family of elliptic curves over $K$, so simply deform your example slightly (e.g. by $2$-adically peturbing the coefficients of the Weierstrass equation) so that the $j$-invariant of the resulting curve cuts out $K$.<|endoftext|>
TITLE: Is the group of integer points of ${\rm SO}(n,1)$ maximal?
QUESTION [8 upvotes]: That is, is it true that there does not exist a lattice in $G = {\rm SO}(n,1)$ which contains the group of integer points of $G$ as a proper subgroup (obviously then of finite index)? if such a lattice exists, what is the maximal possible index? I know that ${\rm SL}(n,\mathbb{Z})$ is a maximal arithmetic subgroup of ${\rm SL}(n,\mathbb{R})$ so the answer is yes if $n=2$, what is known for larger $n$? this is probably related to hyperbolic manifolds of minimal volume but I can't locate an appropriate reference... any suggestions will be appreciated :)

REPLY [4 votes]: Here is a proof for even $n$, I am not quite sure about odd $n$ since $G=SO(n,1)$ is not an adjoint group and weak approximation might fail in this case (I do not know enough about this, you would need to ask experts, like Andrei Rapinchuk). First, note that the commensurator of $G({\mathbb Z})$ in $G({\mathbb R})$ is $G({\mathbb Q})$ (as it has to preserve the set of rational lines in the light cone: They are the fixed points of unipotent elements of $G({\mathbb Z})$). 
Claim. Let $G=SO(n,1)$, $n$ is even. Then $G({\mathbb Z})$ is a maximal subgroup of $G({\mathbb Q})$ and, hence, in $G({\mathbb R})$. 
Proof. It is known that 
$$
G({\mathbb Z}_p)= G({\mathbb Q}_p)\cap SL_{n+1}({\mathbb Z}_p)$$ `
is a maximal compact subgroup of 
$G({\mathbb Q}_p)$ for all $p$. By weak approximation (since $G$ is an adjoint group), $G({\mathbb Q})$ is dense in $G({\mathbb Q}_p)$, which implies that $G({\mathbb Z})$ is dense in $G({\mathbb Z}_p)$. If $G({\mathbb Z})$ is not a maximal subgroup of $G({\mathbb Q})$ and $G({\mathbb Z})<\Gamma< G({\mathbb Q})$ is a larger discrete subgroup, then $\Gamma$ is not contained in $G({\mathbb Z}_p)$ for some prime $p$. The closure of $\Gamma$ in $G({\mathbb Q}_p)$ is compact (since $\Gamma$ is a finite extension of $G({\mathbb Z})$) and strictly contains $G({\mathbb Z}_p)$, which contradicts maximality of  $G({\mathbb Z}_p)$. qed<|endoftext|>
TITLE: quasi conformal, area preserving homomorphisms of the disc
QUESTION [8 upvotes]: Restricting a quasi-conformal homeomorphism of the disc to the boundary gives a surjective homomorphism from $QC(D^2)$ (quasi-conformal homeos of $D^2$) to $QS(S^1)$ (quasi-symmetric homeos of the circle).   Surjetivity here follows, for example, from the existence of natural extensions of QS homeos like Douady-Earle.
The same restriction-to-the-boundary map from the group of area preserving smooth diffeos (symplectomorphisms if you will) to Diff$(S^1)$ is also surjective -- one way to see this is to construct a by-hand extension defined in a collar neighborhood and then use Moser's theorem (here), and details of perhaps another approach have been written up here.
My question is: what are the possible "boundary values" of area preserving QC homeos of the disc?  My guess is that the map from area preserving QC homeos to $QS(S^1)$ is not surjective... perhaps someone with a good understanding of symplectic homeos has an idea of what the image is.

REPLY [7 votes]: Claim. A QS homeomorphism $f$ of the circle extends to a QC area preserving map of the disk if and only if $f$ is BL (bi-Lipschitz). 
Proof.  

Suppose that $f: S^1\to S^1$ is a QS map which admits an area-preserving quasiconformal extension $F: D^2\to D^2$. Then it immediately follows from the definition of quasiconformality that $F$ has to be (globally) bi-Lipschitz. In particular, $f$ was bi-Lipschitz to begin with. 
Suppose now that $f$ is a BL homeomorphism of the unit circle.  I claim that it extends to a BL area preserving homeomorphism $F$ of the disk.  By looking at the BL flow from the identity to $f$, we see that the problem reduces to the case of Lipschitz vector fields $v$ tangent to the circle which we have to extend to Lipschitz divergence-free vector fields $V$ on the disk.  By removing a point on the circle, we reduce the question to the case when $v=u'$, where $u$ is a $C^{1,1}$ function on the circle (maybe minus a point).  Now, extend $u$ to a harmonic function $U$ on the disk. Then the gradient $V$ of $U$ has zero divergence and is Lipschitz.  Furthermore, $V$ will be Lipschitz at the boundary (except maybe for one point), to it will extend to the remaining point as well.  Qed

I was assuming here that $f$ is orientation-preserving.  However, by composing with a symmetry of the disk, the general case reduces to this one, provided that you allow QC maps to reverse orientation.<|endoftext|>
TITLE: Proving that a space cannot be delooped.
QUESTION [13 upvotes]: Suppose we have some pointed connected topological space $X$. How can we determine if there exists a space $BX$, called delooping of $X$, such that its space of based loops $\Omega BX$ is homotopy equivalent to $X$? More generally, when can we deloop a space $n$ times?
Abstract nonsense
There is a general theory answering this question, going back to J.D. Stasheff's 1963 papers "Homotopy Associativity of H-Spaces. I, II" and vastly developed later. The basic answer is that $X$ should carry an action of $A^\infty$ operad, corresponding to composition of loops. This composition is not associative (at least with an obvious definition), but associativity fails in a well-behaved way --- up to coherent higher homotopies. If $X$ is an iterated loop space $X = \Omega^n Y$, then it carries $n$ compatible actions of $A^\infty$, because there are roughly $n$ really different ways to compose two $n$-dimensional spheres in $Y$: a choice of composition corresponds to a map $S^n \to S^n \vee S^n$, which can be obtain by contracting some base-pointed equator on $S^n$, and there are $n$ really different ways to choose it. Composition of $n$-dimensional spheres is also not associative, and different choices of composition do not commute, but all these statements fail up to coherent higher homotopies, encoded in an action of so-called $E_n$ operad on $X$ ($E_1 = A^\infty$).  In this case $X$ can be delooped up to $n$ times. An example of explicit construction is given in "The Geometry of Iterated Loop Spaces" of J.P. May: there is a natural right action of $E_n$ operad on the $n$-th suspension functor $\Sigma^n$, so we can consider "derived tensor product" $\Sigma^n \otimes_{E_n} X$. It is a geometric realization of simplicial space $$\Sigma^n X \leftarrow \Sigma^n E_n X \leftleftarrows \Sigma^n E_n E_n X \leftleftarrows \dots$$
(last term shoul have 3, not 2 arrows, but how can you do this with MO TeX?)
The problems
It's all nice and neat, but how can you determine if a space can have an action of $E_n$? In all examples I'm familiar with this action is supplied with the definition of $X$, because $X$ is defined as an iterated loop space, or a group, or have a natural action of something that is close to $E_n$, like little disks in some manifold $E_M$, or is a representing object for some cohomology theory. Further problem is, even if we find some group structure on $X$, it doesn't necessarily mean that we have captured all structure. For example, a group can be noncommutative, but still admit higher homotopies, assembling into the action of $E_n$ or even $E_\infty$. A simple example is the infinite-dimensional projective unitary group $PU(H)$, where $H$ is some separable Hilbert space. Kuiper's theorem states that $U(H)$ is contractible. Its center is $U(1)$ and we have a fiber sequence 
$$U(1) \to U(H) \to PU(H)$$
This means that $PU(H)$ is $K(\mathbb{Z},2)$ and an infinity-loop space. While it may be that these higher commutativity homotopies can be written down explicitly, it is not at all obvious that they exist in the first place. If $X$ is an infinity-loop space, we can at least hope that after several iterations of taking loops we will get some periodicity, or an obviously commutative group (naive, but is there much choice?). If $X$ is $n$-loop space for finite $n$, then even this hope is futile.
While I'm not expecting that there is some way, besides finding a certain $E_n$ action,  to prove that $X$ is $n$-loop space (but I still would be glad to hear any different methods), I do hope that it is possible to prove that $X$ is not $n$-loop. Possibly there are some homotopic or homological invariants that would be obstructions to finding such a structure. However, I can only think of obvious ones: $\pi_1$ should be abelian. Perhaps some homotopical or homological operations should also vanish for $n$-loop spaces, but I don't know how to describe them.
An iterative check for the existence of $A^\infty$ structure can be performed based on Milnor's theorem. If $X$ has $A^\infty$ action, then it is homotopy equivalent to a topological group $G$, and we get "projective space bundles"
$$G \to G^{\ast n} \to B_n$$
Here $B_n$ is some space, and $G^{\ast n}$ is n-th iterated join. For $G=S^1$ or $G=S^3$ this gives standard Hopf bundles over complex and quaternionic projective spaces. A theorem of Adams asserts that such bundles do not exist for spheres of other dimensions (except for $\mathbb{O}P^1$ and $\mathbb{O}P^2$, $G=S^7$). However, his methods seem to rely on $G$ being a sphere, and I am not sure how to prove or disprove the existence of a fibration with fiber $X$ and total space $X^{\ast n}$ for a general space $X$.
Question
Are there different ways to prove that $X$ is $n$-loop space, besides finding an explicit $E_n$ action? More importantly, how can one prove that such action does not exist? What homological/homotopical invariants and operations can be used to disprove its existence, or prove that some known $E_n$ structure can not be lifted to a $E_{n+k}$ structure?

REPLY [17 votes]: The general question is indeed hard, but in view of Anton's side question to Dylan about compact Lie groups, I feel compelled to advertise a beautiful old result of John Hubbuck that everyone should know: If $X$ is a connected finite CW complex and a homotopy commutative $H$-space, then $X$ is homotopy equivalent (as an $H$-space) to a torus.  The reference is:
J.R. Hubbuck. On homotopy commutative H-spaces. Topology 8 1969 119–126.  Thus the original question trivializes unless $X$ is infinite dimensional.<|endoftext|>
TITLE: Centralizers of nilpotent elements in semisimple Lie algebras
QUESTION [6 upvotes]: Let $G$ be a connected, simply-connected, complex, semisimple Lie group with Lie algebra $\frak{g}$, and let $\xi\in\frak{g}$ be a nilpotent element. I am interested in understanding the structure of $$C_{\mathfrak{g}}(\xi)=\{\eta\in\mathfrak{g}:[\xi,\eta]=0\}, \quad C_G(\xi)=\{g\in G:\mathrm{Ad}_\mathfrak{g}(\xi)=\xi\},$$ and $\pi_0(C_G(\xi))=C_G(\xi)/C_G(\xi)_0$. I would appreciate any references you suspect would give useful structural information. Also, I would welcome any advice and suggestions.
Thanks!

REPLY [6 votes]: This determination of component groups goes back to Elashvili and Alexeevskii, but has been improved somewhat in a 1998 IMRN paper by Eric Sommers and a later joint paper by him and George McNinch here.   Your set-up is essentially equivalent to studying the same problem for a semisimple algebraic group and its Lie algebra in arbitrary characteristic, but good characteristic (including 0) is essential for getting uniform results.
In particular, the situation for nilpotent elements of the Lie algebra and unipotent elements of the group is essentially the same, by Springer's equivariant isomorphism between the two settings   The classes/orbits and centralizers correspond nicely in good characteristic.
P.S. Concerning structural information on the centralizers, you can also consult Roger Carter's 1985 book on characters of finite groups of Lie type.   There he includes a lot of details about the classes and centralizers in your question over an algebraically closed field.   Since there are only finitely many unipotent classes or nilpotent orbits (the same number in good characteristic), his tables provide a clear overview.    There is less detail about exceptional types in the book by Collingwood-McGovern on nilpotent orbits, but it provides the full Dynkin-Kostant theory over $\mathbb{C}$. Fine points of structure are also treated extensively in the newer AMS book by Martin Liebeck and Gary Seitz, in arbitrary characteristic (including good and bad primes).<|endoftext|>
TITLE: Is there a "right" proof of Riemann's Theta Relation?
QUESTION [17 upvotes]: Let $\theta$ denote the usual Jacobi Theta function (with auxiliary parameter $\tau = i$, for simplicity), i.e.
$$
\theta(z) = \sum_{n \in \mathbb{Z}} \exp(-\pi (a + n)^2 + 2 \pi i n z) \ .
$$
I'm interested in Riemann's quartic theta relation, which is (barring mistakes):
$$
\theta(x_1) \theta(y_1) \theta(u_1) \theta(v_1) = \frac{1}{2} \sum_{\eta \in \lbrace 0, 1/2, i/2, (1+i)/2 \rbrace } c_\eta \theta(x + \eta) \theta(y + \eta) \theta(u+\eta) \theta(v+\eta)
$$
where $c_\eta$ are some exponential factors, and $x_1$, $y_1$, $u_1$, $v_1$ are certain linear functions of the variables $x$, $y$, $u$, $v$ as follows:
$$
x_1 = (x + y +  u + v) / 2
$$
$$
y_1 = (x + y - u - v) / 2
$$
$$
u_1 = (x - y + u - v) / 2
$$
$$
v_1 = (x - y - u + v) / 2 \ .
$$
(This is taken from Mumford's "Tata lectures on theta I", chapter 1, section 5.)
There are many variations.  They're all fairly straightforward to prove with bare hands and the above formula; that is precisely the line taken in Mumford's book and every other reference I've seen.  However, I've been promised (somewhere) that every fact about special functions should have a "nice" interpretation coming from representation theory, and in particular from their interpretation as matrix coefficients (here, of the Heisenberg group).  Is there a clean proof along those lines?
To put it another way, if I only told you some properties of $\theta$ and its variants -- being an eigenfunction of certain operators, periodicity conditions and so on -- but not its formula, is there an enlightening reason to expect something like the Riemann theta formula to hold?
As usual, apologies if I've missed this in a standard reference.

REPLY [5 votes]: I think there is indeed a "right" proof of Riemann's quartic theta relation. Namely, as a corollary of Mumford's first and second "fundamental identities" (described in 6.4, 6.5 of Tata III). Mumford shows us that there is a simple machinery (thetas with quadratic forms) producing an infinite number of theta relations. This abundance of theta relations is not a totally obvious fact (at least not until one finds the fundamental identities).  
Riemann's quartic theta relation is apparently one very remarkable identity, or at least as remarkable as the $4\times 4$ rational symmetric positive definite and orthogonal matrix $$A=\frac{1}{2}\begin{pmatrix} 1 & 1 & 1 & 1 \cr 1 &1 &-1&-1 \cr 1&-1&1&-1 \cr 1&-1&-1&1 \end{pmatrix}.$$
Let us displace the question of "proving" riemann's quartic relation with confirming the identity, i.e. as a physical process of displacing and shuffling thetas. Then we should look maybe for a reason why $A$ is so remarkable. Or, at least see where else $A$ occurs.  
But let's agree on what a theta function is. This requires replacing the upper half plane $\mathbb{H}$ consisting of $z=x+iy$ with $y>0$ with Siegel's upper half space $\mathbb{H}_g$ consisting of symmetric $g\times g$ complex matrices $T$ whose imaginary part is positive definite. An element $T\in \mathbb{H}_g$ yields a hermitian form on $\mathbb{C}^g$, by defining $T[z]:={}^tzTz$. Likewise for any $g\times h$ matrix $N$ we set $T[N]={}^t N T N$. It's also useful for a ring $R$ to let $R(g,h)$ denote all $g\times h$ matrices over $R$. 
Computing from Mumford's second identity requires (explicit) coset representatives for the quotient $$\mathbb{Z}(g,4){}^t A / \mathbb{Z}(g,4){}^tA \cap \mathbb{Z}(g,4).$$ But the rational matrix $A$ is peculiar enough to have $A={}^tA=A^{-1}$ leading the coset representatives to take the form $(\eta, \eta, \eta, \eta)$, where $\eta$ varies over $\frac{1}{2}\mathbb{Z}^g / \mathbb{Z}^g$.
Identifying the coset representatives with the diagonal in $$\frac{1}{2}\mathbb{Z}^g / \mathbb{Z}^g \times \frac{1}{2}\mathbb{Z}^g / \mathbb{Z}^g  \times\frac{1}{2}\mathbb{Z}^g / \mathbb{Z}^g   \times\frac{1}{2}\mathbb{Z}^g / \mathbb{Z}^g  $$ is perhaps reason enough to affirm $A$'s singularity. The identification of this diagonal with these cosets may in fact have some other root system interpretation (in sense of Jeff Harvey's answer involving string things).   
But in a sense, after Mumford, it is the matrix $$A=\begin{pmatrix} 1 &1 \cr 1&-1 \cr \end{pmatrix}$$ which is responsible for riemann's quartic relation. This is Theorem 7.4 in Tata III. 
Now a comment regarding the OP's secondary question concerning, for lack of a better expression, a priori knowledge of holomorphic functions on the complex torus $\mathbb{C}^g / \mathbb{Z}^g + T \mathbb{Z}^g$, for $T \in \mathbb{H}_g$.
At the beginning of the theory, by whatever divine interference, we define a theta function $\theta:\mathbb{C}^g \times \mathbb{H}_g \to \mathbb{C}$ by the expression 
$$\theta(z,T)=\sum_{n \in \mathbb{Z}^g} \exp \pi i(T[n]+2{}^tnz).$$ In maintaining one's honesty (and i can only speak for myself) it is worth thinking about the necessity of our condition $im(T)>0$ in guaranteeing that $\theta(z,T)$ is a holomorphic function in $z$. One should also derive the periodicity and quasi-periodicity relations of $\theta(z,T)$ with respect to translations by  $\mathbb{Z}^g$ and $T\mathbb{Z}^g$ in $\mathbb{C}^g$.
From the definition we find ourselves with a holomorphic function on $\mathbb{C}^g$ periodic under translations by $\mathbb{Z}^g$ and satisfying the quasi-periodicity relation $$f(z+Tm) =\exp \pi i(-T[m] - 2{}^t mz) f(z).$$ The following observation is already contained on pp.121 of Tata I.
If $f:\mathbb{C}^g \to \mathbb{C}$ is any entire function satisfying the quasiperiodicity relations of $\theta(z,T)$, then $f= c\theta(z,T)$ for some constant $c\in \mathbb{C}$. Indeed $\mathbb{Z}^g$-periodicity permits a fourier expansion $$f(z)=\sum_{n\in \mathbb{Z}^g} c_n \exp 2\pi i {}^tn z.$$ Now quasiperiodicity permits us to derive recursive relations amongst the coefficients $c_n, \\ n\in \mathbb{Z}^g$. Explicitly, we can relate $c_{n}$ to $c_{n + \epsilon_k}$, where $\epsilon_k$ is the $k^{th}$ unit vector. It's amusing to think of how running around the lattice generated by $\mathbb{Z}^g T$ varies the cofficients in the fourier expanision.
So knowing that $f$ is a quasiperiodic holomorphic function already yields the formula for the theta function $\theta(z,T)$ -- whose usual formula we should recognize already as a fourier expansion. But I also know nothing about $\theta(z,T)$ as eigenfunction.
Now concerning "promises" relating the Heisenberg representations to facts on theta functions. From what I can tell, the only promise is Stone-von Neumann-Mackey's theorem on irreducible unitary representations (which are faithful on the centre) of the Heisenberg group: all such representations are equivalent. So given two representations (and there are many), there is an "intertwining" among them. But for some facts, one suffers. In 7.4 Tata III there is a description of riemann's quartic relation as arising from such an intertwining. However the Heisenberg group is defined over the finite adeles, i.e. $Heis(2g, \mathbb{A}_f)$. I would like to know where triality, spinors, or $so(8)$ interact in $\mathbb{A}_f$.<|endoftext|>
TITLE: How many vertices can a convex polytope have?
QUESTION [16 upvotes]: One has an $n$-dimensional convex polytope $P$ represented by an intersection of half-spaces:
\begin{equation}H_i = \{ (x_1,x_2, \ldots,x_n) \in \mathbb{R}^n \mid \sum_{j=1}^n a_{ij} x_j \ge a_{i0}, \ a_{ij} \in \mathbb{R} \}, \ i = \overline{1,k} \end{equation}
\begin{equation}P = \{ X \in \mathbb{R}^n \mid X \in \bigcap_{i=1}^k H_i \} \end{equation}
Then let's say that $(v_1,v_2,\ldots,v_n) \in P$ is a vertex if we can find at least $n$ subspaces (out of $k$) such that 
\begin{equation}\sum_{j=1}^n a_{\tilde i j} v_j = a_{\tilde i 0}, \quad \text{ for some } \tilde i \in \{ 1,2, \ldots , k\} 
\end{equation}
Now let's get to the point. It is known that we must have at least $k = n+1$ subspaces in order to construct a closed convex polytope. In this initial case it'd be a simplex having $n+1=|V|$ vertices (here $V$ is a set of vertices). 
We can obviously introduce any number of somewhat trivial additional subspaces and stay at $|V|=n+1$ vertices, but I am interested in the maximum number of vertices for any given number of subspaces $k>n$ in $n$-dimensional space.

$\mathbb{R}^2$ case: since restrictions are geometrically half-planes one can draw lines on paper and see what happens. The simplest closed figure is a triangle, it has 3 vertices and needs a minimum of 3 restrictions to exist. Now every additional restriction can add no more than 1 additional vertice (this is obvious in 2-dimensional space). This means that for $k \ge 3$ restrictions there can exist no more than $k$ vertices: $\max |V| = k.$
$\mathbb{R}^3$ case: similarly the simplest closed figure is a triangular pyramid, it has 4 vertices and needs a minimum of 4 restrictions to exist. Now it gets a little more complicated with additional restrictions, but what follows from fiddling with Euler characteristic is that for $k \ge 4$ restrictions there can exist no more than $(2k-4)$ vertices: $\max |V| = 2k-4.$
$\mathbb{R}^n$ case: I only understand clearly that $\max_{k = n+1} |V| = n+1.$ What happens after introducing additional restrictions is hard to conceive.
How can I generalize it? I hoped to draw some conclusions from Dehn–Sommerville equations but I am not quite experienced enough and maybe it's a completely wrong way to look at this problem...

REPLY [20 votes]: Using $k$ half-spaces, the polytope has at most $k$ facets. For a fixed number of facets, the number of vertices is maximized, for example, by the dual polytope of the cyclic polytope with $k$ vertices. More generally, by the dual polytope of any neighborly polytope with $k$ vertices.
This maximum number of vertices is equal to
$${k-\lceil n/2 \rceil \choose \lfloor n/2 \rfloor} + {k-\lfloor n/2 \rfloor - 1 \choose \lceil n/2 \rceil - 1}.$$

REPLY [13 votes]: What you call $n$ in your posting is often called $d$ in the literature, but I will stick with 
your notation.  So $n$ is the dimension.  Let $V$ be the number of vertices, and $k$ the number
of facets.
Then $V = \Theta( k ^ {\lfloor n/2 \rfloor} )$.
More precisely, the maximum $V$ is given by McMullen's Upper Bound Theorem, 
realized by duals of cyclic polytopes. Cyclic polytopes maximize the number
of facets for a fixed number of vertices, so their duals maximize the number of vertices
for a fixed number of facets.

    


See, e.g.,

"Basic Properties Of Convex Polytopes."
  Martin Henk, Jürgen Richter-Gebert, and Günter M. Ziegler.
  Handbook of Discrete and Computational Geometry, Chapter 16. CRC Press. 2004.
  (Citeseer link)<|endoftext|>
TITLE: Detecting homotopy nontriviality of an element in a torsion homotopy group
QUESTION [7 upvotes]: I have a map, constructed geometrically, $S^4 \to S^3$. I suspect that it is a representative for the generator $\eta_3\in \pi_4(S^3) \simeq \mathbb{Z}_2$, but I am not 100% sure ($\eta_3$ is defined as the suspension of the Hopf map $S^3 \to S^2$). I'd like to be able to detect the fact that my map is homotopically nontrivial in a geometric or combinatorial way. Is this possible? I imagine there is some sort of index one can calculate, but unfortunately the terms one looks these things up in are very generic (forms, integration, index etc).

REPLY [22 votes]: How about thinking about framed cobordism, which in this case gives an isomorphism between $\pi_4(S^3)$ and the group of cobordism classes of normally framed 1-manifolds in $S^4$.  Since your map is constructed geometrically, it probably has a regular value.  Pull this back to a collection of disjoint circles in $S^4$, forming a trivial link, with framings of their normal bundles. Since $\pi_1SO(3)$ has order 2, just count up the number of circles with nontrivial framing to see whether this number is odd or even.<|endoftext|>
TITLE: When does the sheaf cohomology of a topological space vanish?
QUESTION [10 upvotes]: The question is in the title. A more precise formulation is:
Let $X$ be a topological space. When does $H^i(X,F) = 0$ for all $i > 0$ and all abelian sheaves $F$ on $X$?
The obvious example is a discrete space. I'd be happy with a characterization of compact Hausdorff topological spaces $X$ satisfying the above property.
Edit: Following Georges Elencwajg's answer, I would like to clarify that these spaces will be quite pathological from the viewpoint of classical topology. Nevertheless, I do not know a single example which does satisfy the above vanishing property and is not discrete. For example, does the Cantor set have this property?

REPLY [10 votes]: 0)  I would guess that the compact spaces you are looking for are extremely rare.   
1)  For example the extremely simple contractible space $I=[0,1]$ is not suitable:
Consider the inclusion $j\colon U=(0,1)\hookrightarrow I
$ and take on $I$ the sheaf$j_!(\mathbb Z_U)$,  the constant sheaf $\mathbb Z_U$ on $U$    extended to $I$ by zero.
Claim:  $H^1(I,F)\cong\mathbb Z$
Proof of claim:
Consider the open embedding  $i:F=I\setminus U\hookrightarrow X$ and the short exact sequence of sheaves on $I$ (see Hartshorne's Algebraic Geometry, Exercise I.19, page 68)
: $$0\to j_!\mathbb Z_U\to \mathbb Z_I\to i_*\mathbb Z_F\to 0$$ Taking the corresponding long exact sequence in cohomology we get the fragment $$0\to \Gamma(I,j_!\mathbb Z_U) \to \Gamma(I,\mathbb Z_I)\to \Gamma(I,i_*\mathbb Z_F)\to H^1(I,j_!\mathbb Z_U) \to    H^1(I,\mathbb Z_I)     $$
Since $\Gamma(I,j_!\mathbb Z_U)=0$ and 
 $H^1(I,\mathbb Z_I)=H^1_{\operatorname {singular}}(I,\mathbb Z)=0$ the above fragment becomes $$0\to 0\to \mathbb Z \to \mathbb Z^2\to H^1(I,j_!\mathbb Z_U) \to   0 $$   so that  $H^1(I,j_!\mathbb Z_U)\cong \mathbb Z\neq 0$
2) There is a very similar statement in scheme theory saying that $H^1(\mathbb A^1_k,j_!(\mathbb Z_U))=\mathbb Z$, where now $U$ is the complement of two rational points on the affine line $\mathbb A^1_k$: see Hartshorne's Algebraic Geometry, Exercise III 2.1      
3) Of course on afffine schemes, quasi-coherent sheaves have zero cohomology in positive degree, but that is not a purely topological statement and as shown in the example 2) above does not apply to arbitrary sheaves of abelian groups: even theorems by Serre necessitate some hypotheses!<|endoftext|>
TITLE: Does any lower bound on proofs of FLT improve Shepherdson 1965?
QUESTION [11 upvotes]: In 1965 Shepherdson proved that FLT is independent of the fragment of PA that uses only open induction  and signature $0,S,+\times$. Indeed $2x+1\neq 2y$ is independent of that fragment.  Schmerl gives a good general criterion for independence from that fragment in ``Diophantine equations in a fragment of number theory'' in the book Computation and Proof Theory, Springer Lecture Notes in Mathematics Volume 1104, 1984, pp 389-398.
Is FLT currently known to be independent of any larger fragment of PA?

REPLY [4 votes]: Leszek Kołodziejczyk has devised a method how to extend some type of Shepherdson-like models of IOpen into models of Buss’s theory $T^0_2$ (a weak subsystem of $I\Delta_0+\Omega_1$). In particular, he has shown that $T^0_2$ does not prove that $x^3+y^3=z^3$ has no nontrivial solution.<|endoftext|>
TITLE: How unique is a conformal compactification?
QUESTION [15 upvotes]: I'm trying to understand the term "conformal compactification" which is often used in physics. I reckon that most places take this to mean a (sometimes specific) compact conformal completion. That is, a conformal compactification of a manifold $M$ is a compactification $\tilde{M}$ in which all conformal transformations defined locally extend globally. Firstly, is this correct?
Now for my main question. Obviously not all compactifications are conformal compactifications (take the Alexandroff compactification of Minkowski space for example). But what about the other direction? Are all conformal completions compact? I can't think of a counterexample, but then my intuition about conformal completions is pretty shaky. I hoped that some combination of Liouville's theorem and perhaps the Hopf-Rinow theorem would help, but I'm not sure it does. 
Finally how unique are conformal compactifications? I'd like to think that for any manifold there's only one conformal compactification of the same dimension, given by the usual Penrose construction. But I can't find any references to help me start to get some intuition.
If anyone has any hints or suggestions for good literature I'd be very pleased to hear them! Many thanks!

REPLY [12 votes]: For Lorentzian manifolds, the conformal completion need not be compact. A typical example is the universal covering of the $d$-dimensional anti-de Sitter space-time (the maximally symmetric solution of the vacuum Einstein equations with negative cosmological constant) - its conformal boundary is diffeomorphic to $\mathbb{R}\times\mathbb{S}^{d-2}$.
Several examples of conformal completions of different space-times (i.e. time oriented Lorentzian manifolds) such as above can be found in Chapter 5 of S. W. Hawking and G. F. R. Ellis, "The Large Scale Structure of Space-Time" (Cambridge, 1973). As for the uniqueness of the procedure of conformal completion, there is of course (at least) the freedom to multiply the metric in the conformal completion by a positive smooth function thereof. There is an extensive discussion on the ambiguities in the definition of a conformal completion by Robert Geroch in F. P. Esposito and L. Witten (eds.), "The Asymptotic Structure of Space-Time" (Plenum, 1977), pp. 1-105. The book "General Relativity" by R. M. Wald (Chicago University Press, 1984) also discusses conformal completions to some length in Chapter 11.<|endoftext|>
TITLE: Endomorphisms in Category O and Schubert Classes
QUESTION [5 upvotes]: Let $\mathfrak{g}\supset \mathfrak{b}\supset \mathfrak{h}$ be a complex semisimple Lie algebra, with choice of Borel and Cartan subalgebras, $W$ the Weyl group. 
W. Soergel's 'Endomorphismensatz' allows for the identification of $End_{\mathcal{O}_{0}}(P(w_{0}))$ with the algebra of coinvariants $\mathbb{C}[\mathfrak{h}^{\ast}]/\mathbb{C}[\mathfrak{h}^{\ast}]^{W}_{+}$, a finite dimensional quotient of a polynomial algebra, equipped with a $W$-action. Here $\mathcal{O}_{0}$ is the block of the BGG category $\mathcal{O}$ corresponding to the trivial central character.
Moreover, it is a classical result (due to Borel?) that we can identify $\mathbb{C}[\mathfrak{h}^{\ast}]/\mathbb{C}[\mathfrak{h}^{\ast}]^{W}_{+}$ with the cohomology algebra of the flag variety of $\mathfrak{g}$, and that there is a basis of this cohomology algebra given by the Schubert basis $\lbrace S_{w}\rbrace_{w\in W}$, where $S_{w}$ is the class of the corresponding Schubert cell.
Question(s): 1) does anyone know to which morphisms in category $\mathcal{O}$ the Schubert classes correspond to under the above identifications? 

If yes; is there a 'nice' intrinsic (in terms of category $\mathcal{O}$) description of these morphisms that would give a 'canonical' description of the Schubert classes?
(rubbish, vague question) if no; would this be an interesting/valuable thing to know? (ie, are there any immediate applications?)

2) Is there a way to see the $W$-action on the endomorphism algebra in category $\mathcal{O}$?
Also, replace 'Schubert class' by 'first Chern class of tautological bundles' in the above questions; is anything known in this case?
If this is standard material then my apologies; any references/directions would be much appreciated. In particular, any references for Soergel's work (in English/French) would be particularly appreciated.
Cheers, George

REPLY [3 votes]: 1) I believe there is such a description, though I think its pretty debatable whether it is likely to tell you very much about Schubert calculus.
Category $\mathcal{O}$ has a nice collection of objects called "tilting modules"; these are distinguished by have a Verma and dual Verma filtration (actually all of these are self-dual); the indecomposables are indexed by the elements of the Weyl group (look at the lowest elements whose associated Verma or dual Verma appears in the filtration). 
The self-dual projective $P(w_0)=T(e)$ is an example of a tilting module.  
Furthermore, these all have graded lifts in the graded version of category $\mathcal{O}$; in particular, there's a way of grading the Hom spaces between these objects so that the endomorphisms of $P(w_0)$ become $H^*(G/B)$ with the homological grading.  If you choose the gradings correctly, the Hom spaces $Hom(T(e),T(w))$ and $Hom(T(w),T(e))$ have lowest degree $\ell(w)$ and dimension 1 in that degree.  I believe the Schubert class for $w$ is (up to scalar) the composition of elements from these Hom spaces in lowest degree.
I won't give a detailed proof, but the point is that from Soergel's work you can identify $Hom(T(e),T(w))\cong Ext^\bullet(\mathbb{C}_{G/B},\mathbf{IC}_{S_w})$ and $Hom(T(w),T(e))\cong Ext^\bullet(\mathbf{IC}_{S_w},\mathbb{C}_{G/B})$ with composition being Yoneda product.  This shows that any map $T(e)$ to $T(e)$ that factors through $T(w)$ is a sum of Schubert classes for $S_{w'}$ with $w'>w$ (it also shows the claim I made about Hom spaces).  Thus, $S_w$ is (up to scalar) the only such element of degree $\ell(w)$.
2) Yes, if you're willing to think about $\mathrm{End}(P(w_0))$ via its canonical isomorphism with the center of category $\mathcal{O}$.  It's the induced action on the center of a categorical braid group action.  See Section 3 of this paper of Stroppel: http://arxiv.org/abs/math.RT/0608234<|endoftext|>
TITLE: Locally free extension of locally free sheaf
QUESTION [11 upvotes]: Given a coherent sheaf $F$ on a smooth variety $X$, we know that $F$ is locally free on an open subset $U$ in $X$ outside a codimension two subset. Say the rank is $k$.
Is  there a locally free sheaf $E$ of rank $k$ on $X$ together with a map $E\rightarrow F$, at least on $U$?
In other words, does the restriction $F_U$ admit a locally free extension on $X$?
Usually coherent extensions are subsheaves of $j_*F_U$, but I dont know if one can have a locally free extension of the locally free $F_U$, on $X$.

REPLY [15 votes]: Since you assumed that $X$ is smooth (less would be enough, but you need at least $S_2$), $G=j_*F_U$ is reflexive, that is $G^{\ast\ast}=G$ where $G^{*}=\mathscr Hom_X(G,\mathscr O_X)$ is the dual.
If two reflexive sheaves agree on an open set with a codimension $2$ complement, then they agree, so your question is equivalent to asking that $G$ be locally free.
If the rank of $F$ is $1$, then $G$ corresponds to a divisor and again since $X$ is smooth it is Cartier and hence $G$ is locally free.
If the rank of $F$ is at least $2$, then the locus where a reflexive sheaf is not locally free is at least $3$-codimensional (see Lemma 1.1.10 in Vector Bundles on Complex Projective Spaces, by Okonek, Schneider, Spindler). 
So what you want can be done on curves and surfaces, but not necessarily on varieties of dimension $\geq 3$. An alternative way to get $G$ is to take the double dual of $F$, that is, $G=F^{\ast\ast}$. In other words, your question is whether that is locally free.
To complete the picture look at Example 1.1.13 in ibid. That shows that reflexive sheaves of rank $\geq 2$ are not always locally free.
To summarize:

If $k=1$ or $\dim X\leq 2$ then the sheaf you are looking for is $j_*F_U$
Otherwise, if $j_*F_U$ is locally free you've got it, if it is not, then there is no such sheaf.<|endoftext|>
TITLE: Cohomology of configuration spaces
QUESTION [13 upvotes]: Let $F(X,n)$ be the configuration space of ordered $n$-tuples of distinct points in $X$, where $X$ is a smooth manifold. Is there a procedure for computing the Poincare polynomial of $F(X,n)$? I am particularly interested in the case where $X$ is a 2-dimensional torus.
If $X$ is a smooth, projective, complex algebraic variety (for example an elliptic curve), Burt Totaro (in his paper "Configuration spaces of algebraic varieties") uses the Leray spectral sequence for the inclusion $F(X,n)\to X^n$  to find an explicit DGA whose cohomology is isomorphic to the cohomology ring of $F(X,n)$.  But it is not clear from this description how to compute the Betti numbers.

REPLY [2 votes]: It is perhaps worth pointing out that the virtual Poincare polynomial of $F(X,n)$ is easy to compute, at least when $X$ is a complex variety. This is treated in Section 2 of Fulton and MacPherson's paper on compactifications of configuration spaces, and the result is that the virtual Poincare polynomial of $F(X,n)$ is just 
$$
P(X)\cdot (P(X)-1)\cdot \dots (P(X)-i+1).
$$
Here the virtual Poincare polynomial of $X$ is given in terms of a signed sum over the weight graded pieces of cohomology of $X$ with compact supports. 
I think that this should allow some simple calculations, such as the Betti numbers for the (complex) braid arrangements being given by unsigned Stirling numbers of the first kind.<|endoftext|>
TITLE: Perimeter/Neighborhood of a graph on grid
QUESTION [5 upvotes]: Hello,
I have a $\sqrt{n}\times\sqrt{n}$ lattice graph $G=(V,E)$ i.e. vertices on said 2-dim integer lattice, and two vertices have an edge if their $L_1$ distance is one.
Now I want to claim something like this:
For any partition of $V$ into $V_1, V_2$ with $n/4\leq |V_1| \leq n/2$, there are at least $\sqrt{n}$ edges
of type $v~w$ where $v\in V_1$ and $w\in V_2$.
It seems quite easy, and I was wondering if there is a simple crisp known proof for it. It seems some geometry might help e.g. to claim that perimeter of subset $S$ of area $n/4$ is minimized when $S$ is just one rectangle etc.
Any reference for extensions to higher dimension would be helpful too. Thank you!

REPLY [2 votes]: The $n$-dimensional case is solved in Bollobás and Leader, Edge-isoperimetric inequalities in the grid, Combinatorica 11 (1991), no. 4, 299–314.  In two dimensions the optimal sets are small squares, then "half-space" rectangles, then complements of small squares.  The range you specify is exactly the range where rectangles are best, and your guess is correct: there are almost exactly $\sqrt n$ edges in the smallest edge boundary.
Ryan O'Donnell's suggestion to consider the vertex isoperimetry problem will give the answer up to a constant factor (because the degrees are bounded), but curiously the extremal sets are very different: always sets roughly of the form $x+y \leq r$.<|endoftext|>
TITLE: Localization sequence for K^0(X)
QUESTION [5 upvotes]: Suppose $X$ is any quasi-projective variety. let $K^0(X)$ denote the Grothendieck group of locally free sheaves.
Suppose $U$ is an open subset of $X$.
Is there a localization sequence:
$$
 K^0(X)\rightarrow K^0(U)\rightarrow 0.
$$
I saw that this exists for group of coherent sheaves $K_0(X)$.
If $X$ is not smooth then do we still have the localization sequence for $K^0$.
thanks.

REPLY [4 votes]: An element of $K_0(U)$ is represented by a perfect complex $F^.$ in the derived category of $U$-modules.  As long as $X$ and $U$ are quasi-compact and quasi-separated, the class $[F^.]$ lifts to $K_0(X)$ if and only if $F^.$ is the restriction (in the derived category) of a perfect complex on $X$.  (This is the "key proposition" of Thomason and Trobaugh's paper on "Higher Algebraic K-Theory of Schemes and of Derived Categories".)  So any perfect complex that doesn't lift gives a counterexample to the surjectivity on $K_0$.

REPLY [3 votes]: There is a localization sequence, as given in the reference mentioned in Angelo's comments. However the map is not always surjective. Perhaps an easy example is $X=Spec(R)$ where $R=k[x,y]_{(x,y)}/(xy)$ and $U=X$ minus the closed point. Then $K^0(X)=\mathbb Z$ since $R$ is local, but $K^0(U)=\mathbb Z^2$ since $U$ is two disjoint points.<|endoftext|>
TITLE: Question about topological monoid maps
QUESTION [7 upvotes]: Let Mon be the category of topological monoids. I am happy to work with the model structure mentioned here: 
Model Structure/Homotopy Pushouts in topological monoids?. 
I'm looking for a reference for the following statement, which I believe to be true.
 Suppose  $X$ and $Y$ are topological monoids, with $X$ cofibrant, and $Y$ group-like. 
Assertion: The map of spaces
$$
\text{monoid-maps}(X,Y) \to \text{maps}_\ast(BX,BY)
$$
is a weak homotopy equivalence. 
My guess is that one could prove this by induction,
using the fact that $X$ is a retract of an object given by attaching free things. For example, here is a verification of the statement when $X = FU$, the
free monoid on the points of a based space $U$. In this case, 
$$
\text{monoid-maps}(FU,Y)  = \text{maps}_\ast(U,Y) ,
$$
whereas
$$
\text{maps}_*(BFU,BY) \simeq \text{maps}_\ast(\Sigma U,BY)
$$
using, say James theorem $FU \simeq \Omega\Sigma U$. Since $Y$ is group-like, we
have $Y \simeq \Omega BY$ and we get 
$$
\text{maps}_*(\Sigma U,BY) \simeq \text{maps}_\ast(U,Y) 
$$
verifying the assertion in this special case.
More generally, it seems to me that if $X = \text{colim}(X_0 \leftarrow FA \to FB)$
where $(B,A)$ is a cofibration pair (with the colimit taken in topological monoids), 
and if the assertion is true for $X_0$ then it 
is also true for $X$ using the above and by noting that (i) function spaces convert pushouts in the domain to pullbacks and (ii) the classifying space functor preserves homotopy pushouts.  This would then give the inductive step. 
Added Later: I'd like to reiterate that I'm really looking for a decent reference.

REPLY [3 votes]: EDIT 2: Sorry about the confusion, I will try to be careful now, and I'll put some comments at the end to clear up the business about adjoints etc. Here's what we're going to do.
We'd like to show that
$$
\text{Map}_{\text{Mon}}(X, Y) \rightarrow \text{Map}_{Spaces_*}(BX, BY)
$$
is an equivalence when $X$ is cofibrant and $Y$ is grouplike. The strategy will be to consider the string
$$
\text{Map}_{\text{Mon}}(X,Y) \rightarrow \text{Map}_{\text{Spaces}_*}(BX,BY) \rightarrow \text{Map}_{\text{Mon}}(\Omega'BX, \Omega'BY) \rightarrow 
$$
$$
\text{Map}_{\text{Mon}}(X, \Omega'BY) \cong \text{Map}_{\text{Mon}}(X,Y)
$$
where $\Omega'$ denotes the Moore loop space (strictly associative multiplication), and show that the composite is homotopic to the identity. This clearly reduces down to showing that the third map is a weak equivalence, since I may as well have started the string with the inverse of the last equivalence ($\Omega' BY$ and $Y$ are interchangeable).
So we want to show that
$$
\text{Map}_{\text{Mon}}(\Omega'BX, \Omega'BY) \rightarrow \text{Map}_{\text{Mon}}(X, \Omega'BY)
$$
is a weak equivalence when $X$ is cofibrant ($Y$ can be arbitrary).
We will prove this by showing that these two spaces represent the same functor on the homotopy category of spaces, the fundamental fact being that the natural map $\Omega'B Y \rightarrow \Omega'B\Omega'B Y$ is a weak equivalence.
Given a map $K \rightarrow \text{Map}_{\text{Mon}}(X, \Omega'BY)$ we get a map $K \rightarrow \text{Map}_{\text{Mon}}(\Omega'BX, \Omega'B\Omega'BY)$, and composing with the natural weak equivalence gives us a map $K \rightarrow \text{Map}_{\text{Mon}}(\Omega'BX, \Omega'BY)$. One can check that composition gives back the original map, up to homotopy. This proves surjectivity of
$$
\text{Map}_{h\text{Spaces}}(K, \text{Map}_{\text{Mon}}(\Omega'BX, \Omega'BY)) \rightarrow \text{Map}_{h\text{Spaces}}(K, \text{Map}_{\text{Mon}}(X, \Omega'BY))
$$
to get injectivity just note that we can recover the original value by applying $\Omega B'$ and using the weak equivalence again.
This completes the proof.

In my "second edition" I erroneously stated that there was a left adjoint to the inclusion of group-like topological monoids into monoids. Ricardo gracefully explains why this can't be true. However, it is a consequence of the above argument that there is a homotopical left adjoint, namely $\Omega'B$. (We've shown that the image of $\Omega'B$ is a homotopically reflective subcategory, and so all that's left is to note that the essential image is all group-like topological monoids.)
I've also erased the original answer, since it seems silly now that we have this one. 
A reference for the argument I gave that basically showed that $\Omega'B$ was a localization functor can be found in Higher Topos Theory Proposition 5.2.7.4. Lots more love can be found throughout Higher Algebra; one can, for example, deduce a similar statement about $\mathbb{E}_1$-spaces in general, which actually implies this result since the inclusion of monoids into $\mathbb{E}_1$-spaces is an equivalence of $\infty$-categories.
Hopefully this is all correct now! Let me know if there's still errors.<|endoftext|>
TITLE: Diagrammatic proof of unique prime decomposition of knots
QUESTION [9 upvotes]: Consider a knot to be a diagram in a plane--- i.e. a drawing of a finite connected planar graph (loops and multiple edges allowed) whose vertices are 4-valent with cyclic ordering for the incident edges of each vertex, two non-consecutive of which are called "over" and two "under" (with an additional connectivity condition to ensure there is one component)--- modulo Reidemeister moves. If we can find a disc in the plane such that a representative for the knot K intersects the boundary of the disc at two points, and intersecting K with a disc closes to give a non-trivial knot (and the same for intersecting K with the complement of the disc), then the resulting sub-knot is called a connect summand of K. A knot with no connect-summands is called prime. 
Schubert (1949) proved that every knot has a unique prime decomposition, up to permutation of the factors. The proof is geometric and uses Seifert surfaces. See for example Theorem 7.12 of Burde-Zieschang. There is also a smooth proof-- see e.g. Ryan Budney's paper. Recent work of Korablev concerns prime decomposition of virtual knots, where there is no Seifert surface in sight of course, but the proof is also geometric, and involves modifications of thickened surfaces in which the virtual knot is embedded.

Question: Is there a purely diagrammatic proof for unique prime decomposition of knots? (no topology or geometry, definitely no Seifert surfaces, diagrammatic algebra and combinatorics only). Who is it due to? 

Unique prime decomposition is a fairly elementary property of knots, so it feels like there should be a diagrammatic proof- but it is also a global property, whereas the diagrammatic approach is local/quantum, so I don't know. There is a weakly analogous problem which I'm looking at for coloured knots, in which there is an easy diagrammatic proof (but the statement is also much weaker in that context).
Note that quantum knot invariants are diagrammatic (Jones polynomial, HOMFLYPT, Kontsevich invariant), and we know that these see a lot, so a direct proof of unique prime decomposition using the Jones polynomial, for example, would qualify.

REPLY [6 votes]: I am fairly certain that there's no known diagrammatic proof of the uniqueness of prime decompositions. That answers the "who is it due to?" question.
Of course, a diagrammatic argument may still be out there, waiting to be discovered. But, echoing Ryan's comment, I expect that line of argument to be very difficult, for the following reason.
Suppose you're looking at a diagram $D(K)$ of some composite knot. Perhaps the diagram even shows $K$ to be a connected sum in some fashion. The crux of what you need to show is that there is no alternate prime decomposition besides the one you see. In other words, you would need to show that any prime decomposition of $K$ is visible in some (suitably nice) diagram.
At present, the Jones polynomial and its relatives are only known to place strong restrictions on diagrams for certain classes of knots and links. These include alternating links (where everything is easiest) and, to a smaller extent, adequate and semi-adequate links. It's conjectured that if a semi-adequate knot is composite, every semi-adequate diagram must also be composite -- and that conjecture is probably within reach. (See this paper by Ozawa, as well as Problem 10.6 in this book.) However, any solution is likely to involve essential surfaces, which you want to avoid. Furthermore, the semi-adequate setting is essentially the limit of where current knowledge about Jones-type invariants can reach diagrammatic information.
Without these invariants, you find yourself looking at problems like the additivity of crossing number, which are known to be devilishly hard. In particular, if the additivity conjecture does get solved, the solution would surely involve something beyond purely diagrammatic methods.<|endoftext|>
TITLE: Are there refuted analogues of the Riemann hypothesis?
QUESTION [56 upvotes]: The classical Riemann Hypothesis has famous analogues for function fields and finite fields which have been proved.  It has by now very many analogues, many of them still open.  Are there important analogues that are now known to be false?

REPLY [12 votes]: It is some time ago that I read it, so I hope I recall the details correctly but:
in Eisenstein series and the Riemann zeta function (Automorphic Forms, Representation Theory and Arithmetic, Springer-Verlag, Berlin-Heidelberg-New York (1981) 275-301)
Don Zagier shows that the RH is equivalent to the unitarity of a certain (non-admissible) $SL(2, \mathbb{R})$ representation. Next he gives a second construction of this representation in terms of the adeles over $\mathbb{Q}$. Then a rather surprising punch line follows: he applies the same construction to the adeles over $\mathbb{Q}(\sqrt{2})$, thus obtaining a similar represenation and shows that this representation is not unitary (by exhibiting an invariant subspace that is not a direct summand).
Although it is more a variant of an equivalent reformulation of the RH that is shown false than a variant of the RH itself, at least it goes to show, as Zagier points out, that any proof of the RH should exploit some very special properties of the rationals not shared by just any global field.  
EDIT: I just found an online version of the paper and it turns out I remembered it wrongly. Apparently the unitarity of the mentioned representation is not equivalent to RH but strictly stronger. In particular it would also imply that the zeroes of zeta are simple (which is believed to be even harder than RH according to the answer to this MO question: Are the nontrivial zeros of the Riemann zeta simple?.) Apparently this harder conjecture is known to be false over fields other than $\mathbb{Q}$ and it is this fact that is used to construct the counterexample to the unitarity of the analogous representations coming from other fields (also, to my surprise, $\mathbb{Q}(\sqrt{2})$ is mentioned nowhere explicitly). So not an answer to the original question after all, I'm sorry. Still the article is quite interesting.<|endoftext|>
TITLE: Which p-adic algebraic groups are type I?
QUESTION [20 upvotes]: It was proved by Jacques Dixmier (Sur les représentations unitaires des groupes de Lie algébriques, Annales de l'institut Fourier, 7 (1957), p. 315-328, doi: 10.5802/aif.73, MR 20 #5820, Zbl 0080.32101)
that algebraic groups over the reals, are type I. Is a similar result known for algebraic groups over non-archimedean local fields (possibly of characteristic 0)? I am only aware of the result by Bernstein
http://www.math.tau.ac.il/~bernstei/Publication_list/publication_texts/bernstein-P-tame-FAN.pdf
that reductive algebraic groups over non-archimedean local fields, are type I.

REPLY [5 votes]: Theorem 1.2.3 (p18) of
MOHAMED HACHMI SLIMAN
Théorie de Mackey pour les groupes adéliques
Astérisque, tome 115 (1984) Link at SMF site
asserts that the $F$-points of linear algebraic groups over characteristic zero local fields $F$ are of type 1.  The author states the result in a slightly more general setting that allows for some covering groups.<|endoftext|>
TITLE: Surfaces filled densely by a geodesic
QUESTION [38 upvotes]: Which smooth, closed surfaces $S \subset \mathbb{R}^3$ have no
  single geodesic $\gamma$ that fills $S$ densely?

Say a geodesic $\gamma$ "fills $S$ densely" if the closure of the set of points
through which $\gamma$ passes equals $S$.
Some examples:

 A sphere: every geodesic is a great circle.
 Zoll surfaces, as discussed here:
"Surfaces all of whose geodesics are both closed and simple."
 An ellipsoid.


(Image from GeographicLib.)


 A torus generally has many geodesics that fill the surface.



(Image by John Oprea)


My assumption is that almost all surfaces have geodesics that fill them.
Is this known, under any interpretation of "almost all"?
I would also be interested in extending the list of exceptional surfaces
beyond {sphere, Zoll, ellipsoid}.
Thanks for pointers!

Answers Summary (18Apr2013):

 (Robert Bryant, Mikhail Katz)
Any surface of revolution with poles has no dense geodesic.
This holds for convex or nonconvex surfaces of revolution.
 (Robert Bryant)
There are generalizations of
Liouville surfaces (due to Goryachev-Chaplygin and to Dullin-Matveev) that have no dense geodesic.
 (Misha Kapovich)
Every surface may be perturbed by gluing on "focusing caps" so 
that it has dense geodesics.
 (Keith Burns) Guess:
There is always a dense geodesic on a closed Riemannian surface of genus 
$\ge 2$.

REPLY [4 votes]: I feel there is some implicit idea here but nobody formulated it, so let me try:
a smooth closed surface has no filling geodesic if and only if its geodesic flow is integrable
What do you think about this conjecture ?
For me it seems probable since integrable flow have their Arnold tori that don't fill up all the space: for example, for the ellipsoid, the picture of the space that a geodesic fills in is a projection of an Arnold tori (an annulus).
All the examples stated above are integrable, and Burns guess is nice because surfaces of big genus can't have an integrable flow (Kozlov theorem, if I am not wrong...)
I am really excited about this conjecture.<|endoftext|>
TITLE: Does "induction" for a functor algebra imply it is initial?
QUESTION [7 upvotes]: I originally posted this question to MSE but there were no answers except for a partial one from me, so I'm trying again here. I'm an undergrad, not a researcher, so forgive me (and correct me!) if my terminology is nonstandard, or my ideas confused.

By "induction" I mean "no proper subalgebras". My thinking goes like this:

For natural numbers, recursion and induction are in some sense the same thing. In particular, given a recursive definition of $f$ you would prove its totality roughly by saying "if I can define $f$ on $1\dots n$, then I can define it on $n+1$", i.e. by induction.
The proper categorical notion of recursion is initial algebras – in particular, for $F(X) = 1\sqcup X$ an initial algebra is a natural number object, the property of being initial being precisely what you need to define functions by recursion.
An initial algebra automatically has a notion of induction: initial objects have no proper subobjects, so if you have some subobject that is closed under the algebra operations, this means precisely that the inclusion is an algebra homomorphism, and therefore an isomorphism.
I'd really like to go the other way, but in general the implication "$I$ is initial $\implies$ every mono into $I$ is an iso" cannot be reversed (its dual has a counterexample in $\mathbf{Set}$, in that every epi into $0$ is an iso but $0$ is not terminal).

Are there circumstances where an algebra having no proper subalgebra means it is initial?

Observations that have been made since posting the question:

There is an algebra $1 \sqcup \mathbb Z/n\mathbb Z \to \mathbb Z/n\mathbb Z$ that has no proper subalgebras, so induction works for it. Nevertheless it is not initial. The problem is that the algebra satisfies equations, since $n$ successors give the identity, so there aren't morphisms to algebras that don't satisfy those equations. This suggests that the algebra needs to be "free" in some sense.
In light of the above, I might decide I'm interested in quasi-initial objects, since it is at least true that any morphism from $\mathbb Z/n\mathbb Z$ that is an algebra homomorphism is unique. It turns out that in any category with equalisers, if an object has no proper subobjects then it is quasi-initial, and of course a category of algebras has equalisers if its underlying category does.

What's the "freeness" condition that I want? What can I do to show that any morphisms exist from an object with no proper subobjects?

REPLY [5 votes]: It looks as though you might have already observed this yourself, but suppose $F: C \to C$ is an endofunctor and $C$ has equalizers. Then if $F$ has a weakly initial algebra $X$ (meaning that for every $F$-algebra $Y$ there exists an $F$-algebra map $X \to Y$), then $X$ is initial if and only if every $F$-subalgebra of $X$ is all of $X$. For, if $f, g: X \to Y$ are two $F$-algebra maps, then their equalizer is an $F$-subalgebra of $X$, and its being equal to $X$ would force $f = g$. 
This is not a completely idle observation however, because one sometimes has formal ways to construct weakly initial $F$-algebras. Suppose for instance that $C$ is finitely complete and cartesian closed and $F$ carries a structure of $C$-enriched functor. Then, if the end 
$$\int_{c \in C} c^{c^{F(c)}}$$ 
exists (meaning an object universal with respect to dinatural maps to the functor $G: C^{op} \times C \to C$ defined by $G(d, c) = c^{d^{F(c)}}$), it is a weakly initial $F$-algebra. For instance, when $C = Set$ and $F(c) = 1 + c$, we have 
$$\int_{c} c^{c^{1 + c}} \cong \int_c c^{c^c \times c} \cong \int_c (c^c)^{(c^c)}$$ 
which is the set of dinatural transformations $c^c \to c^c$. This is isomorphic to $\mathbb{N}$ conceived as the set of Church numbers $\lambda f: f^{(n)}$ where $f$ is of variable type $c^c$. 
If you are interested specifically in natural numbers objects, then there is a remarkable theorem due to Peter Freyd: 
Theorem: In a topos $E$, the following are equivalent for a structure $(N, o: 1 \to N, s: N \to N)$: 

$(N, o, s)$ is a natural numbers object; 
The maps $o: 1 \to N, s: N \to N$ are coproduct injections (that witness $N$ as a coproduct $1 + N$) and $N \to 1$ is the coequalizer of the pair $(1_N, s): N \to N$; 
$s$ is monic, the subobjects $s: N \to N$ and $o: 1 \to N$ are disjoint (have the initial object as their pullback), and any subalgebra of the algebra structure $(o, s): 1 + N \to N$ on $N$ is all of $N$. 

The proof is not at all easy, but you can find it in Johnstone's Sketches of an Elephant, section D.5.<|endoftext|>
TITLE: References on techniques for solving equations with discontinuous functions such as floor and ceiling?
QUESTION [6 upvotes]: Here I describe the sort of reference I'm after with a motivating example. I am not seeking solutions to my equations on this forum; I'm quite happy to do that myself. Rather, I'm asking for some good references on solving equations of this sort.
I had an equation

$2 \lfloor a \rfloor_{c} - a = d - c$

with $a,c,d \in {\mathbb Z} $, and where $\lfloor n \rfloor_{k}$ is my notation for $k \lfloor n/k \rfloor$ — essentially the floor function down to the nearest multiple of $k$.
I wished to solve for $a$. Now, I didn't know how to tackle this algebraically, as the usual technique of bringing the $a$'s together does not seem to be available. However, I had some notion of what form a solution was likely to take. After some guesswork and experimentation with Maxima, I found the solution:

$a = 2 \lceil d \rceil_{c} - (c + d)$

This appears to be correct, but I have not yet found a way to prove this — but that's not my question.
This approach is very unsatisfactory to me. I would much rather solve the problem algbraically.
I'd like to know if there are any recommended references, either books or on-line, about techniques that can be used to solve equations involving the floor ($\lfloor \cdot \rfloor$), ceiling ($\lceil \cdot \rceil$), fraction-part, and similar functions, either in ${\mathbb Z}$, ${\mathbb Q}$ or ${\mathbb R}$.
Beyond my particular equation of interest, I'd be interested to learn how to tackle this sort of equation more generally.
(In case you're interested why I was looking at this equation: I have recently encountered the remarkable Stern diatomic sequence. The equation in question is related to the successor function on ratios of consecutive terms; I wished to find the inverse function.)

REPLY [3 votes]: Expressions formed by composing polynomials and the integer-part operator are refered to in numerous papers by the not very google-friendly name ``generalized polynomials''. The problem of determining whether a generalized polynomial equation has integer solutions includes Hilbert's Tenth problem, and is therefore effectively unsolvable. On the other hand there are some interesting results on the distribution of values of generalized polynomials, which you might find relevant:

Bergelson and Leibman's paper ``Distribution of values of bounded generalized polynomials'' available here.
Leibman's paper ``A canonical form and the distribution of values of generalized polynomials'' available here.

There are related papers on Leibman's website and also by Haland and McCutcheon. Leibman's paper  gives a cannonical form for generalized polynomials that helps to grasp what values the gp can assume mod 1. His paper is a follow-up to the  Bergelson-Leibman paper, in which Bergelson shows very roughly speaking that every bounded generalized polynomial can be thought of as a matrix power map composed with a piecewise-polynomial function. Bergelson shows how tools from Ergodic Theory and Lie Theory can be brough to bear on the study of generalized polynomials.
Incidentally, the problem of which equations $g=0$ are identities (i.e. hold for all integer values of the variables), where g is a gp, is also effective unsolvable by reduction to Hilbert's Tenth Problem: Let $f$ be any polynomial with integer coefficients. Then the equation 
$$\lfloor \sqrt{2}f(\bar{x})\rfloor+\lfloor- \sqrt{2}f(\bar{x})\rfloor+1=0$$ is an identity if and only if $f$ has no integer zeros.<|endoftext|>
TITLE: Does the vanishing of the Poisson bracket on $S(\mathfrak{g})^{\mathfrak{g}}$ inspire the disover of Duflo's isomorphism theorem?
QUESTION [6 upvotes]: For any finite dimensional Lie algebra $\mathfrak{g}$, we know that the universal enveloping algebra $U(\mathfrak{g})$ is a deformation of the symmetric algebra $S(\mathfrak{g})$. In fact let's define
$$
U_t(\mathfrak{g}):=\text{T}(\mathfrak{g})/(X\otimes Y-Y\otimes X-t[X,Y]).
$$
Then $S(\mathfrak{g})=U_0(\mathfrak{g})$ and $U(\mathfrak{g})=U_1(\mathfrak{g})$. Moreover we have the symmetrization map
$$
I_{PBW}:S(\mathfrak{g})\longrightarrow U_t(\mathfrak{g})
$$
which pulls back the product on $U_t(\mathfrak{g})$ to a product on $S(\mathfrak{g})$. We call it the star product and denote it by $*_t$.
Obviously $*_t$ is different from the original product on $S(\mathfrak{g})$. In fact we can prove that the first order deformation of the product is governed by the $\textit{Poisson bracket}$ on $S(\mathfrak{g})$. More precisely the Poisson bracket is defined to be $ \text{{a,b}} := c^k _ {ij}   X_k \cdot \partial^i a \cdot \partial^j b$  ( $c^k _ {ij}$ is the structure constant of $\mathfrak{g}$ )  and we can prove that
$$
a *_t b= ab+\frac{t}{2}\text{{a,b}}+O(t^2).
$$
Furthermore, we have the following result

The Poisson bracket vanishes on the invariant subalgebra $S(\mathfrak{g})^{\mathfrak{g}}$. This is almost the definition.
The  symmetrization map $I_{PBW}$ maps $S(\mathfrak{g})^{\mathfrak{g}}$ isomorphically (as vector spaces, not as algebras) onto the center  $Z(U(\mathfrak{g}))$. 
(Duflo's Isomorphism Theorem) We can precompose a map $D: S(\mathfrak{g})\rightarrow S(\mathfrak{g})$ such that the composition restrict to $S(\mathfrak{g})^{\mathfrak{g}}$ is an $\textit{algebraic isomorphism}:~S(\mathfrak{g})^{\mathfrak{g}}\rightarrow Z(U(\mathfrak{g}))$. 

The Duflo's Isomorphism Theorem is of course highly non-trivial and we can refer to Calaque and Rossi's book http://math.univ-lyon1.fr/~calaque/LectureNotes/LectETH.pdf, as well as well as many other resources, for further discussions.
I usually wonder that (maybe historically, maybe not) how people could expect that there is an algebraic isomorphism between $S(\mathfrak{g})^{\mathfrak{g}}$ and  $Z(U(\mathfrak{g}))$. The thing we can notice is that the first order deformation, which is the Poisson bracket, vanishes. We know that it is a necessary condition (at least it should vanish in the second Hochschild cohomology) to find an algebraic isormorphism.
My question is: Does the vanishing of the Poisson bracket plays an important role in finding and proving Duflo's isomorphism theorem? Or it is just an literally first step?

REPLY [8 votes]: My question is: Does the vanishing of
  the Poisson bracket plays an important
  role in finding and proving Duflo's
  isomorphism theorem? Or it is just an
  literally first step?

Let $A_0$ be a Poisson algebra and $A$ a deformation quantization of $A_0$ (assume we are in a context when it exists). 
Assume you have a quantization map $Q:A_0\to A$, by which I mean a section of the classical limit map $A\to A/(\hbar)=A_0$. 
Then for any two elements $a,b\in A_0$, $[Q(a),Q(b)]=\hbar\{a,b\}+O(\hbar^2)$. 
Hence if you want to have $Q(ab)=Q(a)Q(b)$ you must at least assume that $\{a,b\}=0$. 
My (non-)answer to your question is then: 

the vanishing of the Poisson bracket is a necessary requirement 
  if you want a statement of Duflo-type. It is just a first step. 

The actual history comes from the Harish-Chandra isomomorphism. 
Duflo noticed that the original formula could be written for any Lie algebra, without any use of roots and similar stuff specific to the semi-simple case.<|endoftext|>
TITLE: Does the derivative of log have a Dirac delta term?
QUESTION [15 upvotes]: Dirac writes down the following formula on page 61 of his "Principles of quantum mechanics":
$\frac{d}{dx}\log x = \frac{1}{x} -i\pi\delta(x)$, see http://adsabs.harvard.edu/abs/1947pqm..book.....D for the exact reference (but no text). What is the best way of formalizing this to a mathematician's satisfaction?

REPLY [5 votes]: Here is another context where something like this shows up, though I don't know how or if it's related to your question. Consider the simple left $\mathscr{D}$ module on $\mathbf{A}^1$ generated by $\text{log}x$, and similarly for $1/x$ and $\delta(x)$
$$\mathscr{D}\cdot\text{log}x\ =\  \mathscr{D}/\mathscr{D}\partial x\partial \ = \ k[x]\log x +k[x,x^{-1}],$$
$$\mathscr{D}\cdot 1/x\ =\  \mathscr{D}/\mathscr{D}\partial x \ = \ k[x,x^{-1}],$$
$$\mathscr{D}\cdot\delta(x)\ =\  \mathscr{D}/\mathscr{D}x \ = \ k[\partial]\delta(x)$$
where the global sections of $\mathscr{D}$ is the algebra $k[x,\partial]$ with $[x,\partial]=1$. There is an exact sequence of $\mathscr{D}$ modules
$$0 \ \longrightarrow \ \mathscr{D}\cdot 1/x \ \longrightarrow \ \mathscr{D}\cdot\log x\ \longrightarrow \ \mathscr{D}\cdot \delta(x)\ \longrightarrow \ 0$$
which in plain language just means that the subspace $k[x,x^{-1}]$ is preserved by $x$ and $\partial$.
Note that to get this $``\log=1/x+\delta"$ principle to work you really have to work with the whole $\mathscr{D}$ modules, not just consider elements of them. Indeed, the element $\log x\in\mathscr{D}\cdot\log x$ just satisfies $\partial \log x=1/x$.<|endoftext|>
TITLE: Tame morphism from a curve to $\mathbb{P}^1$
QUESTION [18 upvotes]: Let $k$ be an algebraically closed field of characteristic $p\ge 0$. Let $C$ be a smooth projective curve over $k$. Is it possible to find a map $C \to \mathbb{P}^1$ that is tamely ramified at every point of $C$, i.e. such that the ramification index at every point of $C$ is prime to $p$?
A result of Fulton says that, if $k$ is (algebraically closed) of characteristic $p\ne 2$, then it is possible to find a morphism $C \to \mathbb{P}^1$ that is a simple cover: only double points may appear and at most one in every fiber. (This is theorem 8.1 in "Hurwitz schemes and the irreducibility of moduli of algebraic curves", Ann. of Math. 90, 1969. He says it is classical and dates back to Severi.)
Fulton's result gives a positive answer for fields of characteristic $p\ne 2$. But what about characteristic 2? Does the result still hold? I would already be interested in answers in particular cases (elliptic curves for instance).
EDIT: I added the hypothesis that the field is algebraically closed in order to focus on what I am really interested in. Still, I would also appreciate comments on how relevant this hypothesis is.

REPLY [4 votes]: In [Stefan Schroer, Curves with only triple ramification], the author gives lower bounds on the dimension of the subset of the moduli space of curves for which the question has a positive answer.<|endoftext|>
TITLE: What are the possible motivic Galois groups over $\mathbb Q$?
QUESTION [38 upvotes]: Let $E$ be a motive over $\mathbb Q$. (I should precise, that by a motive I mean
here a pure motive over $\mathbb Q$, with coefficients in $\mathbb Q$, that I see here as a conjectural object which exists in a world where standard conjectures and anything else you can expect is true, in the spirit of Grothendieck, and like in Serre's paper in Motives, PSPM 55, volume 1 - a paper which unfortunately does not seem to be online except partially on Amazon). 
Recall that the motivic galois group of $E$, denoted $G_{M(E)}$, is defined as the tensor-automorphic group
of the functor "Betti realization" from the category $M(E)$, defined as the smallest Tannakian sub-category of the category of motives containing $E$, to the category of $\mathbb Q$-vector spaces. 
An equivalent (under the conjectures of Tate and Hodge)
and more concrete (at least for me) definition is as follows (see the same paper of Serre):
let $H_b(E)$ be the Betti realization of $E$ (a finite-dimensional $\mathbb Q$-vector space),
and let $\rho_\ell$ be the $\ell$-adic realization of $E$, which is a continuous semi-simple
representation $\rho_\ell$ of $\Gamma_{\mathbb Q}$ (the absolute Galois group of $\mathbb Q$) over $H_b(E) \otimes \mathbb Q_\ell$. Then $G_{M(E)}$ is the algebraic subgroup $Gl(H_b(E))$ such that $G_{M(E)} \otimes \mathbb Q_\ell$ is the Zariski-closure of the image of $\rho_\ell$, for every prime number $\ell$.
In any case, $G_{M(E)}$ is a reductive group. My question is 

What reductive groups over $\mathbb Q$ arise as $G_{M(E)}$ for some pure motive $E$? 

This question is asked by Serre in the section 8 of his paper. At that time there was not much known on it apparently (it was 20 years ago): Serre notices that such a group $G$ must have an element $\gamma \in G(\mathbb R)$ such that $\gamma^2=1$ and whose centralizer is a maximal compact subgroup of $G(\mathbb R)$. This necessary conditions implies that $Sl_2$ is not 
a motivic Galois group. On the positive side, I believe that symplectic group $GSp_{2n}$
are known to be motivic Galois group (attached to $E=$ an abelian variety over $\mathbb Q$ of dimension $n$, sufficiently generic), and I know a few other examples like those attached to
elliptic curves with complex multiplication.
There has been a lot of work and progresses on motives since then, and I am asking to know
what progresses have been made in the direction of this question. I would be very surprised
if it had been solved completely (even assuming the standard conjectures + Hodge conjecture + Tate conjecture), but I would be interested in any partial results:

For example, what if we just try to classify the connected reductive group which arise as neutral component
  of a $G_{M(E)}$? those groups up to isogeny ? those groups after extension to the algebraic closure of $\mathbb Q$?  Conjectural answers (conjectural in the sense that they are not proved even assuming the conjectures assumed above) are also more than welcome.

Even a list of example of groups that are provably or possibly motivic Galois group which goes beyond the small list I have given would be very useful. And of course, any references more recent than Serre's could be very useful.

REPLY [12 votes]: I realise that I'm a bit late to this party, but maybe it is still interesting to you:
Under the Hodge conjecture the motivic Galois group coincides with the Mumford–Tate group. These have been classified by S. Patrikis: http://arxiv.org/abs/1302.1803.<|endoftext|>
TITLE: Homotopy classes of maps to Lie groups
QUESTION [13 upvotes]: In Physics one often encounters maps from a certain manifold $M$ to a Lie group $G$. For example, in gauge theories, this gives a gauge transformation, wich is a symmetry of a theory. It is then important to give a homotopy classification of such maps. One of the physical motivations is that maps non-homotopic to constant identity give examples of "large gauge transformations", which can turn out to be not exactly symmetries of the theory, and the requirement for them to be a symmerty leads to certain quantisation conditions, for example the quantization of level $k$ in non-abelian Chern-Simons theory.
Typical manifolds are $\mathbb{R}^n$, which is trivial, $S^n$, which reduces to $\pi_n(G)$. 
The general problem of classification of homotopy classes $[M,N]$ is surely very hard, but I thought there may be some results when $N=G$ is a Lie group (for me it seems that this fact brings a group structure to the set of classes via pointwise multiplication of representatives). Probably I am not really good at looking for references, so I decided to post a question here:
Does anybody know a reference where the set $[M,G]$ is discussed for $G$ a Lie group?
(I know that for $U(1)$ it is $[M,K(\mathbb{Z},1)]=H^1(M,\mathbb{Z})$, pretty usefull in physics, but thats all.)

REPLY [3 votes]: Notice that since $G$ is a topological group, we have 
$$\pi_0[X,Y] = \pi_1 [X, BG]$$
Here $[X,Y]$ denotes the mapping space with standard compact-open topology, $BG$ is the classifying space for group $G$ (a connected space, such that its based loop space is homotopy equivalent to $G$). Since $G$ is a connected group, we have $\pi_0(BG) = \pi_1(BG) = 0$. For $k>1$ all $\pi_k(BG)$ are abelian and $\pi_k(BG) = \pi_{k-1}(G)$.
There exists a so-called Federer spectral sequence, which is a homological second quadrant spectral sequence, having
$$E^2_{p,q} = H^{-p}(X,\pi_{q}(Y))$$
for $p+q\geqslant 0$ and $E^2_{p,q} = 0$ otherwise. Here (if $\pi_1(Y)$ acts trivially on higher homotopy groups of $Y$) $H^{-p}$ groups are singular cohomology of $X$ with coefficients in corresponding (abelian) homotopy groups of $Y$. Under suitable finiteness conditions this sequence converges to $\pi_{p+q}([X,Y],f)$, where $f\colon X\to Y$ is some specific mapping, considered as a base point in mapping space (the spectral sequence depends on path component). Sufficient convergence conditions include $X$ being a finite CW-complex. Since we are interested in the loop space $\Omega [X, BG]$, we should consider $f\colon X \to BG$ that is a trivial map. Also $\pi_1(BG) = 0$, so all conditions of the stated theorem are satisfied.
Obviously, if $\dim X = n$, then to calculate $\pi_i([X,Y])$ for $i\geqslant k$ we need to know only first $n+k$ homotopy groups of $Y$, so in principle you can find anything you want to know about the mapping space. There is also a standard problem that a spectral sequence generally gives not exactly the groups it converges to, but only some graded groups associated to the corresponding filtration.
H. Federer, A study of function spaces by spectral sequences, Trans. Amer. Math. Soc. , 82 (1956) pp. 340–361<|endoftext|>
TITLE: Explicit 2-Cocycles of G=Z2×Z2xZ2 over U(1)
QUESTION [7 upvotes]: We know that group cohomology $H^2(G,U(1))$ consists of 2-cocycles $\beta(A,B)\in U(1)$ corresponding to elements in the group $H^2(G,U(1))$, where $A\in G,B \in G$. Note that $\beta(A,B)$ satisfies 2-cocycles conditions: $\frac{\beta(A,B)\beta(AB,C)}{\beta(A,BC)\beta(B,C)}=1$, with $A,B,C\in G$. 
For example,
(1)$H^2(Z_2,U(1))=Z_1$
(2)$H^2(Z_2^2,U(1))=Z_2$
(3)$H^2(Z_2^3,U(1))=Z_2^3$
(4)$H^2(Z_n^k,U(1))=Z_n^{k(k-1)/2}$
(5)$H^2(Z_n \times Z_m,U(1))=Z_{gcd(n,m)}$.
We know (1)$H^2(Z_2,U(1))$ has a 2-cocycle $\beta(A,B)=1$ (up to a 2-coboundary term), this corresponds to the unique element in $Z_1$.
Questions: What are the explicit forms of 2-cocycles $\beta(A,B)$ for the cases of (2)$H^2(Z_2^2,U(1))$,(3)$H^2(Z_2^3,U(1))$? 
The answer should look like:
For (2), $\beta(A,B)=\beta_1^{n_1}$ with $n_1\in \{ 0,1\}=Z_2$, with $\beta_1$ as a generator of 2-cocycles. 
For (3), $\beta(A,B)=\beta_1^{n_1}\beta_2^{n_2}\beta_3^{n_3}$ with $n_1,n_2,n_3\in \{ 0,1\}=Z_2$, with $\beta_1,\beta_2,\beta_3$ as generators of 2-cocycles.
Similarly, any answer for explicit 2-cocycles for (4)$H^2(Z_n^k,U(1))$ and (5)$H^2(Z_n \times Z_m,U(1))$? 
Any comments, concise/short reference or better understanding will be helpful. I am a physicist, on a modest level trying to absorb this http://arxiv.org/abs/hep-th/0001158. Thank you so much.

REPLY [7 votes]: It turns out that by playing around the $U(1)$ form of 2-cocylces, I manage to provide some answers to (2) and (3) and partially (4).
For (2)$H^2(Z_2^2,U(1))=Z_2$, 
the 2-cocycles are $\beta(b,c)=\beta_1^{n_1}=\exp({i\pi}n_1(b_1 c_2))$, with $b=(b_1,b_2)\in Z_2^2$, $c=(c_1,c_2)\in Z_2^2$. Here $b_1,b_2,c_1,c_2,n_1\in\{0,1\}=Z_2$.
More generally, $H^2(Z_n^2,U(1))=Z_n$, 
the 2-cocycles are $\beta(b,c)=\beta_1^{n_1}=\exp({i2\pi}\frac{n_1}{n}(b_1 c_2))$, with $b=(b_1,b_2)\in Z_n^2$, $c=(c_1,c_2)\in Z_n^2$. Here $b_1,b_2,c_1,c_2,n_1\in\{0,1,\dots,n-1\}=Z_{n}$.
For (3)$H^2(Z_2^3,U(1))=Z_2^3$, 
the 2-cocycles are $\beta(b,c)=\beta_1^{n_1}\beta_2^{n_2}\beta_3^{n_3}$. 
Explicitly,
$\beta_1^{n_1}=\exp({i\pi}n_1(b_2 c_3))$,
$\beta_2^{n_2}=\exp({i\pi}n_2(b_1 c_3))$,
$\beta_3^{n_3}=\exp({i\pi}n_3(b_1 c_2))$,
 with $b=(b_1,b_2,b_3)\in Z_2^3$, $c=(c_1,c_2,c_3)\in Z_2^3$. Here $b_1,b_2,b_3,c_1,c_2,c_3,n_1,n_2,n_3\in\{0,1\}=Z_2$.
Thus, similarly, we can partially answer (4) for $H^2(Z_n^3,U(1))=Z_n^3$, 
the 2-cocycles are $\beta(b,c)=\beta_1^{n_1}\beta_2^{n_2}\beta_3^{n_3}$. 
Explicitly,
$\beta_1^{n_1}=\exp({i2\pi}\frac{n_1}{n}(b_2 c_3))$,
$\beta_2^{n_2}=\exp({i2\pi}\frac{n_2}{n}(b_1 c_3))$,
$\beta_3^{n_3}=\exp({i2\pi}\frac{n_3}{n}(b_1 c_2))$,
 with $b=(b_1,b_2,b_3)\in Z_n^3$, $c=(c_1,c_2,c_3)\in Z_n^3$. Here $b_1,b_2,b_3,c_1,c_2,c_3,n_1,n_2,n_3\in\{0,1,\dots,n-1\}=Z_{n}$.
The 2-cocycles $\beta(b,c)$ above, have the satisfactory properties: 
(i) all satisfy 2-cocycles conditions $\frac{\beta(A,B)\beta(AB,C)}{\beta(A,BC)\beta(B,C)}=1$.
(2) A generator correspond to an element of $H^2(G,U(1))$ for the given $G$. 
(3) Any two generators $\beta_i \neq \beta_j$ are not different simply by a 2-coboundary. i.e. $\beta_i \neq \beta_j \frac{\gamma(b)\gamma(c)}{\gamma(bc)}$ for any 1-cochian $\gamma(g)\in U(1)$.
Maybe the above answer implies it will not be too difficult to fully answer (4) and (5).<|endoftext|>
TITLE: Does there exist an infinite non-commutative division ring with finite center?
QUESTION [6 upvotes]: Does there exist an infinite non-commutative division ring with finite center?

REPLY [14 votes]: Yes.
The following construction is attributed to Hilbert in "A first course in noncommutative rings" by T.Y. Lam., p. 217.
Let $k$ be a field with an automorphism $\sigma$.  Write $D = k((x,\sigma))$ for the noncommutative ring of formal Laurent series $\sum_{i = n}^\infty a_ix^i$ with twisted multiplication rule $xa = \sigma(a) x$ for $a\in k$.  Then $D$ is a division ring.  If $k_0$ is the fixed field of $\sigma$, then either
$Z(D) = k_0$ or $Z(D) = k_0((x^s))$
depending on whether $\sigma$ has infinite order or finite order $s$, respectively.  
Thus all we need to do is choose $\sigma$ and $k$ such that $k_0$ is a finite field and $\sigma$ has infinite order.  This can be done by choosing say the Frobenius automorphism of $\overline{\mathbb{F}}_p$.<|endoftext|>
TITLE: What is about J. v. Neumann's "Continuous geometry"?
QUESTION [13 upvotes]: I am curious about von Neumann's "Continuous geometry", but found no recent text or survey on it. Does anyone know the book and would be so nice to share their impression, and if/how the concept of such geometries fits into contemporary tries to generalize geometry?

REPLY [5 votes]: In plane projective geometry, there are "subspaces" of dimension 0 (points) dimension 1 (lines) and dimension 2 (the whole space).  We may also add dimension -1 (the empty set).  Axioms may be given concerning incidence among these various subspaces.  There is a "duality" that switches points and lines; the whole space and the empty set; reverses inclusion.
In $n$-dimensional projective geometry, there are "subspaces" of dimension $-1,0,1,\dots,n$.  Axioms may be given concerning incidence again.  And duality.
The idea of continuous geometry is to allow subspaces of dimensions in more general totally ordered sets.  The main example is $[0,1]$.  Still there are axioms concerning incidence.  And duality.
As noted in the other answers, it did not turn out to be an important idea.<|endoftext|>
TITLE: Reconciling Lusztig's results with the Langlands philosophy
QUESTION [25 upvotes]: Let $\boldsymbol{G}$ be a reductive group over a finite field $\mathbb{F}_q$, $G = \boldsymbol{G}(\mathbb{F}_q)$, $W = \mathrm{W}(\mathbb{F}_q)$ the Witt vectors over $\mathbb{F}_q$, and $K = \mathrm{Frac}(W)$ its fraction field. I'll abuse notation by also writing $\boldsymbol{G}$ for the corresponding (unramified) reductive group over $K$.
When $\boldsymbol{G}$ has connected center, Lusztig proves the following:    
Theorem. There is a canonical bijection between (isomorphism classes of) irreducible $\overline{K}$-representations of $G$ and special conjugacy classes in $\widehat{\boldsymbol{G}}(\overline{K})$ stable under $g \mapsto g^q$.
Here $\widehat{\boldsymbol{G}}$ is the dual group of $\boldsymbol{G}$. The condition of being special is a condition of Lusztig relating to special representations of Weyl groups through the Springer correspondence.
On the other hand, one could write a Langlands-type statement as follows: irreducible $\overline{K}$-representations of $G$ should correspond to $\widehat{\boldsymbol{G}}(\overline{K})$-conjugacy classes of $L$-parameters for $G$.
One possible definition of an $L$-parameter is that of a  homomorphism over $\mathrm{Gal}(\overline{\mathbb{F}_q}/\mathbb{F}_q)$ of the form
$$\varphi \colon \langle \mathrm{Frob}_q \rangle \times \mathrm{SL}_2(\overline{K}) \to {}^L \boldsymbol{G}(\overline{K}), $$
where ${}^L \boldsymbol{G} = \widehat{\boldsymbol{G}} \rtimes \mathrm{Gal}(\overline{\mathbb{F}_q}/\mathbb{F}_q)$ is the Langlands dual group of $\boldsymbol{G}$, and where we require $\mathrm{Frob}_q$ to have semisimple image in $\widehat{\boldsymbol{G}}(\overline{K})$, and the restriction of $\varphi$ to $\mathrm{SL}_2(\overline{K})$ to be algebraic.
Of course, in this situation, the data of an $L$-parameter $\varphi$ is equivalent to that of a particular kind of $\widehat{\boldsymbol{G}}(\overline{K})$-conjugacy class in ${}^L \boldsymbol{G}(\overline{K})$. However, Lusztig's conditions on the conjugacy classes do not appear anywhere. For instance, when $\boldsymbol{G}$ is split, an $L$-parameter is simply a $\widehat{\boldsymbol{G}}(\overline{K})$-conjugacy class in $\widehat{\boldsymbol{G}}(\overline{K})$.
Instead, we might want to add an extra condition which would pin down the image of $\mathrm{Frob}_q$ to lie in $\widehat{\boldsymbol{G}}(K)$, in order to parallel the following result, pertaining to the semisimple part of the correspondence:  
Theorem. There is a canonical bijection between (isomorphism classes of) irreducible semisimple $\overline{K}$-representations of $G$ and semisimple conjugacy classes in $\widehat{\boldsymbol{G}}(\mathbb{F}_q)$.
However, such a modification would not account for the unipotent part, nor would it account for the condition of being special.
It seems, then, that this notion of $L$-parameter is the wrong one. What is the fix?

REPLY [12 votes]: The way I like to think about this is that a Langlands parameter for the group $G({\mathbb F}_q)$ should be the "restriction to inertia" of a tame Langlands parameter for the group $G(K)$.
That is, a tame Langlands parameter (say, over ${\mathbb C}$) for $G(K)$ should be a
pair $(\rho,N)$, where $\rho$ is a map $W_K \rightarrow \hat G({\mathbb C})$ that factors through the tame quotient of $W_K$ and $N$ is a nilpotent "monodromy operator", that is, a nilpotent element of the Lie algebra of $\hat G$ that satisfies a certain commutation relation with $\rho$.
The tame quotient of $W_K$ is generated by the tame inertia subgroup $I_K$ and a Frobenius element $F$; local class field theory identifies $I_K$ with the inductive limit of the groups
${\mathbb F}_{q^n}^{\times}$, and conjugation by Frobenius acts on this by raising to $q$th powers.
I don't have the details in front of me, but if I recall correctly the Deligne-Lusztig parameterization involves several choices (for instance, an identification of $\overline{\mathbb F}_q^{\times}$ with a suitable space of roots of unity in ${\mathbb C}$.)  My understanding is that if one unwinds these choices, they amount to a choice of topological generator $\sigma$ for the inductive limit of the ${\mathbb F}_{q^n}^{\times}$.  
Thus, if one starts with a Langlands parameter $(\rho,N)$ for $G(K)$ and restricts $\rho$
to inertia, this restriction is determined by $\rho(\sigma)$, which is a semisimple element of $\hat G({\mathbb C})$ that is conjugate to its $q$th power.  The
pair $(\rho(\sigma),N)$ should be the Langlands parameter for the group $G({\mathbb F}_q)$.  
There should then (roughly) be a compatibility between depth zero local Langlands and the Deligne-Lusztig parameterization, as follows: let $K$ be the kernel of the reduction map
$G(W) \rightarrow G({\mathbb F}_q)$, and let $\pi$ be an irrep of $G(K)$ with
Langlands parameter $(\rho,N)$.  Then the $K$-invariants $\pi^K$ of $\pi$ are naturally a
$G({\mathbb F}_q)$-representation, and $\pi^K$ should contain the representation
of $G({\mathbb F}_q)$ corresponding to $(\rho(\sigma),N)$ via Deligne-Lusztig.  You should take this with a bit of a grain of salt, as I haven't thought the details through carefully.  But it should be correct on a "moral" level, at least.
For $GL_n$ this falls under the rubric of the so-called "inertial local Langlands correspondence".  For more general groups the picture is more conjectural, but there
are ideas along these lines in the paper of DeBacker-Reeder on the depth zero local Langlands correspondence.<|endoftext|>
TITLE: Are virtual cubulated groups cubulated?
QUESTION [10 upvotes]: Suppose $G$ has a finite index subgroup $N$ such that $N$ acts properly and cocompactly on a CAT(0)-cube complex. Does $G$ also act properly and cocompactly on a CAT(0)-cube complex?
Edit: After searching the web a little I found that the answer to this question is no, in general. There exists for example a 3-dimensional torsion-free crystallographic group that does not act freely and cocompactly on a CAT(0)-cube complex. (see Example 16.11 in 'THE STRUCTURE OF GROUPS WITH A QUASICONVEX HIERARCHY', by Dani Wise).
So my new question becomes: what conditions can one impose on $G$, such that the answer to the original question is yes? For example, what if $G$ is hyperbolic (as suggested by HW, in comments below)?

REPLY [9 votes]: The answer is 'yes' when your group $G$ is word-hyperbolic.  This can be deduced from Sageev's theorem.  I'll explain this here, but a good reference is Hruska--Wise's paper 'Finiteness properties of cubulated groups' (arXiv:1209.1074v2).
The first step is to notice that $G$ admits a proper, but not cocompact, action on a cube complex.
Lemma: Suppose $|G:N|<\infty$ and $N$ acts properly on a CAT(0) cube complex $X$.  Then $G$ acts properly on $X^{|G:N|}$.
The proof of this is the usual messing around with the induced representation, so I'll leave it as an exercise.  The one extra fact about this action that we will need is that the new hyperplane stabilizers are precisely the $G$-conjugates of the hyperplane stabilizers in $N$.
When $G$ is word-hyperbolic, the Schwarz--Milnor Lemma implies that $X$ is Gromov-hyperbolic.  We also have the following facts about hyperplane stabilizers in $G$.

Hyperplane stabilizers in $N$ are codimension-one subgroups.  Since this is a coarse property, the same is true of hyperplane stabilizers in $G$.
Hyperplanes in $X$ are convex.  Since quasigeodesics in Gromov-hyperbolic spaces are uniformly close to geodesics, it follows that hyperplane stabilizers are quasiconvex in $N$, and hence in $G$.

Associated to any finite collection $\lbrace H_i\rbrace$ of codimension-one subgroups in a group $G$, Sageev constructed a CAT(0) cube complex on which $G$ acts by isometries.  In this case, we will take $\{H_i\}$ to be a set of $G$-conjugacy representatives for the hyperplane stabilizers in $G$.
Sageev's theorem: If the $\{H_i\}$ are all quasiconvex then $G$ acts cocompactly on the associated cube complex.
By 1 and 2 above, Sageev's theorem applies.  It remains to prove that $G$ also acts properly.  This is a matter of making sure that we have chosen a large enough collection of hyperplane stabilizers. 
The Hruska--Wise paper contains some useful criteria for checking that the action is proper.   Roughly speaking, their Theorem 5.4 says that, as long as the axis of every infinite-order element crosses some hyperplane, the action is proper.  But, indeed, if $g\in G$ has infinite order then it shares an axis with $g^{|G:N|}\in N$, and this certainly crosses a hyperplane, since the action of $N$ on $X$ was proper.
This completes the proof in the word-hyperbolic case (modulo filling in some details).
Finally, I'll just mention that much of the purpose of the Hruska--Wise paper is to generalize these ideas to the relatively hyperbolic setting.  In particular, something similar should be true there, subject to suitable restrictions on the action of the parabolic subgroups.<|endoftext|>
TITLE: how to visualize the class number of an imaginary quadratic field?
QUESTION [20 upvotes]: Let me detail the title of the question. I'm trying to give students an intuition of what the class number is.
Let $K=\mathbb{Q}(\sqrt{-d})$, with $d>0$ a square-free integer, be a quadratic imaginary field. Let $\mathcal{O}_K$ be its ring of integers. It is of the form $\mathbb{Z}[\tau]$ with $\tau=\sqrt{-d}$ or $\tau=\frac{1+\sqrt{-d}}{2}$ depending on the value of $d$ mod $4$. 
So let us think of $\mathcal{O}_K$ as the lattice of $\mathbb{C}$ generated by $1$ and $\tau$. Then ideals should correspond to sublattices of $\mathcal{O}_K$ and two of them should define the same class in the class group if one can pass from one to another by multiplying by an element of $\alpha$, isn't it? 
Could anybody help me to make this analogy precise? For instance, how can one see that $\mathbb{Q}(i)$ has class number 1 but $\mathbb{Q}(\sqrt{-5})$ doesn't just by looking at the corresponding lattices? The (non-equivalent) decompositions
$2.3=(1+\sqrt{5}i)(1-\sqrt{5}i)$ 
suggest to consider the lattices $\mathbb{Z}\cdot 2+\mathbb{Z}\cdot(1+\sqrt{5}i)$ and $\mathbb{Z}\cdot 3+\mathbb{Z}\cdot(1-\sqrt{5}i).$ Is that what I have to do?
Thanks!

REPLY [10 votes]: I think that showing the lattices $(1, \sqrt{-5})$ and $(2, 1+\sqrt{-5})$ is a good idea. As Felipe Voloch says, not all lattices are ideals, but the fact that these lattices are ideals can be shown visually: You just need to check that $\sqrt{-5} I \subset I$ and, of course, multiplication by $\sqrt{-5}$ means multiplication by $\sqrt{5}$ and rotation by $\pi/2$. And it is geometrically obvious that $(2, 1+\sqrt{-5})$ is not principal -- it does not have a rectangular fundamental domain. (You could contrast this with $(2, 1+\sqrt{-1})$, which is principal.)
Beyond that, if you are going to talk about reduced forms, you could plot the roots of $a z^2+bz+c=0$ for all reduced forms of some discriminant and show how they lie in the fundamental domain. For example, here are the $23$ reduced forms of discriminant $-647$ (with real and imaginary axes transposed to fit nicely on the page).

If you want to play with this example, here are the coefficients in $(a,b,c)$ form
{{12, 5, 14}, {12, -5, 14}, {13, 9, 14}, {13, -9, 14}, {12, 11, 16}, {12, -11, 16}, {9, 1, 18}, {9, -1, 18}, {8, 5, 21}, {8, -5, 21}, {7, 5, 24}, {7, -5, 24}, {6, 1, 27}, {6, -1, 27}, {6, 5, 28}, {6, -5, 28}, {4, 3, 41}, {4, -3, 41}, {3, 1, 54}, {3, -1, 54}, {2, 1, 81}, {2, -1, 81}, {1, 1, 162}}<|endoftext|>
TITLE: textbooks on asymptotic expansions 
QUESTION [5 upvotes]: Where can I find a readable textbook or lecture notes on asymptotic expansions ?

REPLY [7 votes]: There is a very large literature on asymptotic expansions, including books. What is the best books
depends on your needs.
A comprehensive (advanced) book oriented at physicists and applied mathematicians is
MR0499926 Dingle, R. B. Asymptotic expansions: their derivation and interpretation. Academic Press [A subsidiary of Harcourt Brace Jovanovich, Publishers], London-New York, 1973.
Another good book for physicists/engineers is
MR1721985 Bender, Carl M.; Orszag, Steven A. Advanced mathematical methods for scientists and engineers. I. Asymptotic methods and perturbation theory. Springer-Verlag, New York, 1999.
Books more oriented at pure mathematicians is 
MR0435697 Olver, F. W. J. Asymptotics and special functions. Computer Science and Applied Mathematics. Academic Press, New York-London, 1974.
and (an older book)
MR0115035 Ford, Walter B. Studies on divergent series and summability & The asymptotic developments of functions defined by Maclaurin series. Chelsea Publishing Co., New York 1960 x+342<|endoftext|>
TITLE: When is the module of Kahler volume forms torsion-free?
QUESTION [11 upvotes]: Let $R$ be a commutative algebra over a field $k$.  Denote the $R$-module of Kahler differentials by $\Omega^1_kR$; this is the $R$-module generated by symbols of the form $da$, $a\in R$, and relations
$$ d(\lambda a+b)=\lambda da+db,\;\;\; dab =adb+bda,\;\;\; d1=0, \;\;\; \forall a,b\in A, \lambda\in k$$
The $R$-module of $i$-forms is the $i$th exterior power $\Omega^i_kR:= \Lambda^i_k\Omega^1_kR$.
When $R$ has Krull dimension $n$, call $\Omega^n_kR$ the module of volume forms.  When $R$ is the coordinate ring of a smooth affine variety $X$ of dimension $n$, then elements of $\Omega^nR$ determine volume forms on $X$.
For singular varieties, the behavior of $\Omega^nR$ can be weirder. For example, the simple cusp in the plane
$$R=\mathbb{C}[x,y]/x^3-y^2$$
has a volume form $3ydx-2xdy$ which is non-zero, but
$$y(3ydx-2xdy) = 3x^3dx-2xydy = xd(x^3)-xd(y^2)=0$$
So, the module $\Omega^1_kR$ of volume forms here has a torsion element.
My question is:
Under what conditions on $R$ is $\Omega^n_kR$ torsion-free?

REPLY [6 votes]: This is an interesting question! When $n=1$ there is a conjecture by Berger that the module of differential is torsion-free iff the curve $X$ is non-singular. Googling ``Berger conjecture module of differentials" should give you some references. Bernd Ulrich's first paper (by Mathcsinet) was on this topic, so you can look there also. I do not know anything about higher dimensions but it is plausible that torsion-freeness of $\Omega^n_k$ imposes strong conditions on $R$, as tensor products of non-projective modules tend to have torsions.<|endoftext|>
TITLE: Connected groupoids and action groupoids
QUESTION [5 upvotes]: It is written in Wikipedia http://en.wikipedia.org/wiki/Groupoid, that any connected groupoid $A\rightrightarrows X$ is isomorphic to an action groupoid  $G\ltimes X$ coming from a transitive action of some group $G$ on $X$. I do not understand how to construct such a group $G$, and would be grateful for an explanation or a reference.
I think  that in general one cannot recover $G$ from the action groupoid $G\ltimes X$. Indeed, if $G$ acts simply transitively on $X$, then the action groupoid is given by the equivalence relation $X\times X$ on $X$, hence does not depend on $G$, provided that ${\rm Card}(G)={\rm Card}(X)$. Is this correct?
This question is a version of my question at Math Stack Exchange
to which I got no answers.

REPLY [3 votes]: Here is my stackexcnage answer. 
Another way of looking at this is to use the equivalence of categories between covering morphisms of a groupoid $P$ and actions of $P$ on sets. (Recall that a covering morphism $p:G \to P$ is a groupoid morphism having unique path lifting. Not necessarily unique path lifting gives a fibration of groupoids.) Given an operation of $P$ on a set $X$ then the corresponding covering morphism may be written $P \ltimes X$, an action groupoid, and thought of as a semidirect product because it is a special case of the semidirect product for an action of a groupoid $P$ on a groupoid $H$. For this one needs a morphism of groupoids $\omega: H \to Ob(P)$, where the latter is thought of as a discrete groupoid, and an element $w: x \to y$ in $P$ gives a morphism of groupoids $w_*: \omega^{-1}(x) \to \omega^{-1}(y)$. One has to be precise on conventions to get all this right, which I won't do here. 
So a groupoid $G$ has a representation as an action groupoid whenever you  are given a covering morphism $ G \to P$. This is closely related to Omar's answer, of course. 
I'll add that more details of these ideas are in my book Topology and Groupoids. 
Addition: May 19, 2013 Here is a version of Sam's nice argument but in the language of covering morphisms. 
If $G$ is a group then its universal cover $p: T \to G$ is a covering morphism of groupoids such that $T$ is connected and has trivial vertex groups; this is determined by the action of $G$ on itself by left multiplication. 
The set of objects  of $T$ is bijective with the set $G$ and $T$ is the indiscrete groupoid  (also called tree groupoid) on its set of objects. (Note that if $S$ is a generating set for $G$ then $p^{-1}(S)$ is a graph, namely the Cayley graph of $(G,S)$.) 
Now let $A$ be a connected groupoid with $X$ as its set of objects. Let $T$ be the indiscrete groupoid on $X$. Then for any object $x$ of $A$,  $A$ is isomorphic to $A(x) \times T$. But if $G$ is as above, then 
$$1 \times p: A(x) \times T \to A(x) \times G$$
is a covering morphism.  
The next question is whether this argument can illuminate the case $A$ is a $\Gamma$-groupoid.<|endoftext|>
TITLE: What is the correspondence between combinatorial problems and the location of the zeroes of polynomials called?
QUESTION [11 upvotes]: (From MSE)
In the wikipedia article on the Italian-born American mathematician and philosopher Gian-Carlo Rota, it is stated that the one combinatorial idea he would like to be remembered for  

"... is the correspondence between combinatorial problems and problems of the location of the zeroes of polynomials."

Also, a refence [1] is given for this quote. Upon reading through the interview, though, I didn't discover any more about this correspondence, nor did I find a lot by searching for it on the web.
Question 1: What is this correspondence called? 
I am also interested in how these two (which seem to me) disparate problems in mathematics relate to one another, so:
Question 2: How does this correspondence work? Any references? 
Finally:
3 more questions: To what extent has this correspondence been developed any further since Rota's discovery? Are there any other connections between zeroes of polynomials and combinatorics? I know algebraic geometry is concerned with the study of zeroes of polynomials, so is there any connection between (algebraic) combinatorics and algebraic geometry? 
Reference: 
[1]  http://web.archive.org/web/20070811172343/http://www.rota.org/hotair/rotasharp.html

REPLY [3 votes]: I'm not sure why such an old question suddenly bubbled up to the (my) first page after several years, but I won't let that stop me answering it.
As explained above, Rota's critical problem is usually phrased as finding certain (integer) values where the characteristic polynomial of something (often a graph, or representable matroid) is non-zero.
But I think that describing these problems/results as "location of zeros of polynomials" is a bit disingenuous - graph colouring, like the 4-colour theorem, is essentially a combinatorial subject and the chromatic polynomial and its roots played no role in the eventual proof of the 4CT. Saying the "any planar graph is 4-colourable" is logically equivalent to "4 is not a zero of the chromatic polynomial" but I'd argue that it in this case the equivalence is superficial.
Subjects that really use the non-integer real and/or complex roots of polynomials seem to better meet the "spirit" of the question (though anyone else is free to disagree). Here the obvious candidate application is the study of the partition function of the Potts model, where statistical physicists specifically wish to compute the (limiting curves of) complex roots of these partition functions for various families of graphs. These curves provide information about phase transitions in the underlying Potts model.
The chromatic polynomial is a special limiting case of this partition function and the entire partition function is equivalent to the Tutte polynomial, so these connections somehow seem more fundamental.<|endoftext|>
TITLE: Nash Embedding Theorems for Pseudo-Riemannian Manifolds?
QUESTION [13 upvotes]: Are there analogs of the Nash Embedding Theorems for Pseudo-Riemannian Manifolds?

REPLY [5 votes]: Not clear where you are headed with your concise question, 
but if you have any interest in Lorenzian manifolds as instances of pseudo-Riemannian manifolds,
then this might be of interest, especially for the theorem of Campbell:
"The embedding of General Relativity in five dimensions."
Carlos Romero, Reza Tavakol, Roustam Zalaletdinov.
General Relativity and Gravitation.
March 1996, Volume 28, Issue 3, pp 365-376. (Springer link.)

Abstract.
  We argue that General Relativistic solutions can always be locally embedded in Ricci-flat 5-dimensional spaces. This is a direct consequence of a theorem of Campbell (given here for both a timelike and spacelike extra dimension, together with a special case of this theorem) which guarantees that any $n$-dimensional Riemannian manifold can be locally embedded in an $(n+1)$-dimensional Ricci-flat Riemannian manifold. [...]

And there are many papers in some sense following, e.g.: "The embedding of space–times in five dimensions with nondegenerate Ricci tensor,"
F. Dahia and C. Romero, J. Math. Phys. 43, 3097 (2002). (AIP link.)<|endoftext|>
TITLE: Is there an elliptic surface over $Y(1)$?
QUESTION [5 upvotes]: Actually I have a few related questions.
Here, by $Y(1)$ I mean the affine $j$-line $\text{SL}_2(\mathbb{Z})\backslash\mathcal{H}$.
I know $Y(1)$ is only a coarse moduli space, so there isn't a universal elliptic curve over it, but does there exist an elliptic surface over it such that the fiber above every point is the elliptic curve corresponding to that point?
Now, I know that above the open set $Y(1) \setminus \{i, e^{2\pi i/3}\}$, there is such an elliptic surface. My second question is: Does there exist an open cover of $Y(1)$ such that above each open set in the cover, there exists an elliptic surface with the above property? (I'm not asking for the surfaces to glue.)
Complex analytically, one can consider the space:
$$\mathcal{H}\times\mathbb{C}$$
On this space we have an action (on the right) by $\mathbb{Z}\times\mathbb{Z}$, acting via $$(\tau,x)\cdot(a,b) = (\tau, x + a\tau + b),$$
and an action (on the left) by $\text{SL}_2(\mathbb{Z})$, acting via
$$\gamma\cdot(\tau,x) = (\gamma\tau, x)$$
where $\gamma$ acts on $\mathcal{H}$ by fractional linear transformations. Since intuitively, $\mathbb{Z}\times\mathbb{Z}$ acts on $\mathcal{H}\times\mathbb{C}$ "discretely", so the quotient $\mathcal{H}\times\mathbb{C}/\mathbb{Z}\times\mathbb{Z}$ should be a complex manifold, and is essentially an elliptic surface over $\mathcal{H}$.
If $\Gamma \subset \text{SL}_2(\mathbb{Z})$, then we may also try to form the quotient 
$$\mathbb{E}(\Gamma) := \Gamma\backslash\mathcal{H}\times\mathbb{C}/\mathbb{Z}\times\mathbb{Z}.$$
Intuitively, if $\Gamma$ has no elliptic elements, then it should act "discretely" on $\mathcal{H}\times\mathbb{C}/\mathbb{Z}\times\mathbb{Z}$ and the quotient ought to be a manifold. Hence, if $\Gamma = \Gamma(2)$ (ie, matrices congruent to the identity mod 2), then since $\Gamma(2)$ has no elliptic elements, shouldn't $\mathbb{E}(\Gamma(2))$ be a (complex) manifold? If it is, then it's a complex manifold above $Y(2) := \Gamma(2)\backslash\mathcal{H}$, which is again only a coarse moduli scheme, and hence has no universal elliptic curve. In this case, my third question is: is $\mathbb{E}(\Gamma(2))$algebraic? and if it is, can someone describe heuristically how it's different from a universal elliptic curve over $Y(2)$, if one existed?
The context for these questions comes from me trying to understand why Katz's definition of modular forms for $\Gamma(N)$ (in his paper on $p$-adic modular forms) is properly a generalization of the analytic definition of modular forms. In particular, I'm trying to understand why they must give holomorphic functions on $\mathcal{H}$. I see why this must be the case when $N \ge 3$, since then you have a universal elliptic curve, and in this case holomorphicity is a result of the required compatibility with base change, but in the case of $N = 1$ and $N = 2$, I'm still a little confused. Relevant references would be appreciated as well.
thanks,

will

REPLY [6 votes]: 1 The last exercise of Hartshorne chapter 4 section 4 proves that there is no elliptic surface over $\mathbb A^1$ with nonconstant $j$ invariant. Obviously this is a special case.
2 No as well. Put the fiber in Weirstrass form near $j=0$. Then one can write it as $y^2=x^3-g_2x-g_3$, where $g_2$ and $g_3$ are both functions of $j$ that are well-defined at $j=0$. Furthermore clearly $g_3$ is nonvanishing at $j=0$ and $g_2$ is vanishing at $j=0$. Then computing the $j$ invariant near $j=0$ as $g_2^3/(4g_2^3-27g_3^2)$ up to a constant, we see it vanishes to at least third order, an obvious falsehood.
3 Yes, you're thinking of the Legendre family $y^2=x(x-1)(x-\lambda)$. The Legendre family is universal up to a quadratic twist - every family with full level two structure is a pullback of Legendre, up to a quadratic twist. So it is as close an approximation to the universal elliptic curve over $Y(2)$ as you can get in the category of schemes.<|endoftext|>
TITLE: In Szemerédi's Regularity Lemma, how many blocks are in the partition?
QUESTION [9 upvotes]: Suppose $G = (V,E)$ is a directed graph.
For sets $A$ and $B$ of vertices of $G$,
let $d(A,B) = |(A \times B) \cap E| / (|A||B|)$ denote the edge density between $A$ and $B$,
and say that the pair $A,B$ is $\epsilon$-regular if
$$ |d(X,Y) - d(A,B)| \lt \epsilon $$
whenever $X \subseteq A, Y \subseteq B$, $X$ contains more than an $\epsilon$-fraction of the vertices of $A$, and $Y$ contains more than an $\epsilon$-fraction of the vertices of $B$.
An equipartition is a partition with the sizes of blocks of the partition differing pairwise by at most 1.
Szemerédi's Regularity Lemma can then be stated as:

For any $\epsilon > 0$, there exists $M(\epsilon)$ such that for every graph $G=(V,E)$ there is some $k\le M(\epsilon)$ and an equipartition $V = V_1 \cup \ldots \cup V_k$ in which each block $V_i$ contains at most
  $\lceil \epsilon |V|\rceil$ vertices,
  and having the property that for all but at most $\epsilon k^2$ of the pairs $(i,j)$, the pair $V_i, V_j$ is $\epsilon$-regular.

Do we have any bounds or asymptotics for how $M(\epsilon)$ behaves as a function of $\epsilon$?
I vaguely recall having read a comment that $M(\epsilon)$ is likely to be extremely large, making the Regularity Lemma only useful for truly large graphs.  But I have not been able to find this assessment again, so a pointer would be appreciated.  (I did check Terence Tao's exposition again, and some of the more obvious references.)

REPLY [7 votes]: In addition to Gowers' breakthrough paper, a recent paper addressing the bounds in Szemerédi's regularity lemma, and its weak and strong variants is: 
D. Conlon and J. Fox, Bounds for graph regularity and removal lemmas, Geom. Funct. Anal. 22 (2012), 1191-1256.
It is shown that the bound on the number of parts for Szemer\'edi's regularity lemma grows as at least a tower of $2$'s of height $\Omega(\epsilon^{-1})$, for the Frieze-Kannan weak regularity it grows as $2^{\Theta(\epsilon^{-2})}$, and for the strong regularity lemma of Alon, Fischer, Krivelevich, and Szegedy it grows as a wowzer in a power of $1/\epsilon$. The wowzer function is the next function in the Ackermann hierarchy after the tower function. 
Update (April 2014): Moshkovitz and Shapira recently gave a simple proof of a bound which is a tower of height a power of $1/\epsilon$ in the paper entitled: 
A Short Proof of Gowers' Lower Bound for the Regularity Lemma.
Laszlo Miklos Lovasz and I recently give a bound on the tower height in a version of Szemer\'edi's regularity lemma which is tight up to an absolute constant factor in the paper entitled:
A tight lower bound for Szemerédi's regularity lemma.<|endoftext|>
TITLE: HIgher Homotopy Groups and Representation Theory
QUESTION [11 upvotes]: Let $G$ be a compact Lie group, and $g$ its associated Lie algebra.
In what ways do the higher homotopy groups $\pi_{n}(G)$ with $n>1$ appear in the representation theory of $G$? 
As an example, note that there is at least one place where the fundamental group is significant, namely in distinguishing representations of $g$ from those of $G$ itself.  Indeed, if $\tilde{G}$ is the simply connected universal cover of $G$ then all representations of $g$ can be integrated to representations of $\tilde{G}.$  However, if $G=\tilde{G}/H$ ($H$ a discrete subgroup of $\tilde{G}$) then a necessary and sufficient condition for a representation of $g$ to integrate to a representation of $G$ is that $\pi_{1}(G)=H$ acts trivially in the representation of $\tilde{G}$.

REPLY [2 votes]: When $n = 2,3$:
For any simply-connected finite-dimensional Lie group $G$, $\pi_2(G) = 1$ is trivial.  So it can have no applications.
If $G$ is simply-connected, then $\pi_3(G) = \mathrm{H}_3(G,\mathbb Z)$ by Hurewicz's theorem.  This is a copy of $\mathbb Z$ whenever $G$ is simple.  Classes here can be used to construct central extensions of the loop group of $G$.
I do not know of applications of higher $\pi_n$.<|endoftext|>
TITLE: geometric interpretation of Lie bracket
QUESTION [34 upvotes]: On page 159 of "A Comprehensive Introduction To Differential Geometry Vol.1" by Spivak has written:
We thus see that the bracket $[X,Y]$ measures, in some sense, the extent to
which the integral curves of $X$ and $Y$ can be used to form the "coordinate
lines" of a coordinate system. If $X$ and $Y$ are two vector fields in a neighborhood of p, then for sufficiently small $h$ we can
(1) follow the integral curve of $X$ through $p$ for time $h$ ;
(2) starting from that point, follow the integral curve of $Y$ for time $h$;
(3) then follow the integral curve of $X$ backwards for time $h$ ;
(4) then follow the integral curve of $Y$ backwards for time $h$.

Question:
Before reading this book I thought that $\mathcal{L}_{X}Y=[X,Y]$ calculates changes of $Y$ along Integral curve of $X$.But in this Figure, the integral curves of both vector fields are used. I'm confused. Can someone help me?
Thanks.

REPLY [3 votes]: One view of the Lie bracket that is intermediate between the other views expressed is in terms of the symmetries of the double tangent bundle.  It may not be the most clean geometric picture, but it is very useful when you want to think about other problems -- for example it gives a very clean formulation of torsion, the Riemann curvature tensor, and the Frobenius theorem on integrability of sub-bundles (to foliations). 
Given a function between manifolds $f : M \to N$ there is an induced map of tangent bundles
$$Tf : TM \to TN$$
where $TM$ is the space of pairs of points, one point in $M$, and the other in the tangent space.
$$TM = \{ (p,v) : p \in M, v \in T_p M \}$$
$$Tf(p,v) = (f(p), Df_p(v))$$
this gives a nice formulation of the chain rule: $T(f \circ g) = Tf \circ Tg$. 
From this perspective, the double tangent bundle is the space
$$T^2 M = \{ (p,v,w,y) : p \in M, v \in T_p M, (w,y) \in T_{(p,v)} TM \}$$
The latter condition requires that $w \in T_p M$.  Although it is maybe not obvious, there is a canonical involution of $T^2 M$ given by $\iota(p,v,w,y) = (p,w,v,y)$. 
From this perspective, if $v,w : M \to TM$ are vector fields, you compute $Tv \circ w$ and $\iota \circ Tw \circ v$ and observe that these are taking values in the same fibers of $T^2 M \to TM$, so you can subtract them:
$$Tv \circ w - \iota \circ Tw \circ v$$
Moreover, one can check these are in the "vertical" fiber.  Specifically, take the bundle projection map $\pi : TM \to M$, $\pi(p,v) = p$ then $Tv \circ w - \iota \circ Tw \circ v$ is taking values in $ker(T\pi)$, which one can identify with the tangent spaces of $M$. Call the identification $\alpha$, then you get the formula
$$[w,v] = \alpha \left( Tv \circ w - \iota \circ Tw \circ v \right).$$
I don't claim this is how you want to always think about the Lie bracket, but it is a very formally useful device, and this perspective can make computations of things like the Riemann curvature tensor a bit simpler.  The integrability condition for the Frobenius theorem has a simple statement from this perspective, as well.  If you have a sub-bundle of $TM$, it has a "tangent bundle" that is a sub-bundle of $T^2 M$.  The Frobenius integrability statement has the relatively elegant formulation that this "tangent bundle" must be invariant under $\iota$.  This is close to formally equivalent to the statement that tangent vector fields have Lie brackets that must be tangent, but I find it more pleasant -- and easier to compute, i.e. use as an obstruction.<|endoftext|>
TITLE: Examples of stationary set preserving forcings that are not semiproper?
QUESTION [5 upvotes]: A notion of forcing $P$ is called stationary set preserving iff each stationary subset of $\omega_1$ remains stationary in $V^P$. It is standard to show that semiproper (and of course proper) notions of forcing are stationary set preserving.
On the other hand Shelah realized that assuming the Semiproper Forcing Axiom (which is consistent relative to a supercompact cardinal) each stationary set preserving notion is already semiproper.
So the question naturally arises if there is a nice example of a notion of forcing which is stationary set preserving but not semiproper?

REPLY [5 votes]: Philipp Doebler and me showed in  https://ivv5hpp.uni-muenster.de/u/rds/doebler_sch.pdf  that the forcing to increase u2 (which is stationary set preserving) is semi-proper iff all stationary set preserving forcings are semi-proper. So that forcing provides a canonical example.<|endoftext|>
TITLE: A basis of the symmetric power consisting of powers
QUESTION [5 upvotes]: I have asked this question on math.se, but did not get an answer - I was quite surprised because I thought that lots of people must have though about this before:
Let $V$ be a complex vector space with basis $x_1,\ldots,x_n\in V$. Denote by $v_1\odot\cdots\odot v_k$ the image of $v_1\otimes\cdots\otimes v_k$ in the symmetric power $\newcommand{\Sym}{\mathrm{Sym}}\Sym^k(V)$. It is well-known that the Elements $v^{\odot k}$  for $v\in V$ generate this space (see, for instance, this answer on math.se), so they must contain a basis. 
In other words, let $N=\binom{n+k-1}k$, then there must be $v_1,\ldots,v_N\in V$ with 
$$\mathrm{Sym}^k V = \mathbb Cv_1^{\odot k} \oplus \cdots \oplus \mathbb C v_N^{\odot k}.$$
I am looking for an explicit description of such a basis. Is such a description known? Is there maybe even a "nice" or somewhat "natural" choice for the $v_i$ as linear combinations of the $x_i$?

REPLY [8 votes]: I would look up a book on the calculus of finite differences in a multivariate setting.
The claim here is to show that for any multi-index $\alpha=(\alpha_1,\ldots,\alpha_n)$
of length $k$ one can express the multiple derivative at zero
$$
\left(\frac{\partial}{\partial t}\right)^\alpha \ 
(t_1x_1+\cdots+ t_n x_n)^k
$$
as a linear combination of finite difference analogous expressions
which should only involve the evaluation of $(t_1x_1+\cdots+ t_n x_n)^k$
at integer points $(t_1,\ldots,t_n)$ with nonnegative coordinates adding up
to $k$. This is the same as the above candidate basis considered by you and Peter.
I don't know if there exists a multivariate analogue of the Newton series. If so
then this would immediately imply the wanted statement.

Edit:
Apparently there is such a formula due to Lascoux and Schutzenberger, see Theorem 9.6.1 page 148 in the book
"Symmetric functions and combinatorial operators on polynomials" by Alain Lascoux.
Another source on the web is here. It also has the required property here which is that the number of finite differences taken is the same as the degree of the multiplying Schubert polynomial.

Edit: @Jesko you're right it is a bit more complicated than what I said. Also, the Lascoux-Schutzenberger formula might not be the simplest to use here.
First note that expressions
$$
\prod_{i=1}^{n-1} (x_i-x_n)^{\beta_i}\ \times\ (kx_n)^{k-|\beta|}\ ,
$$
where $\beta$ ranges over multiindices with $n-1$ components and length $|\beta|\le k$,
form a basis.
Now you get the latter as derivatives
$$
\left(\frac{\partial}{\partial t}\right)^{\beta} \ 
\left(t_1x_1+\cdots+ t_{n-1} x_{n-1}+\left(k-\sum_{i=1}^{n-1}t_i\right)x_n\right)^k
$$
at $t=0$.
Call $f(t_1,\ldots,t_{n-1})$ the polynomial function to be hit with derivatives.
One has a multivariate Newton expansion for it:
$$
f(t)=\sum_{m} (t-a)^m \partial^m f(a_{11},a_{21},\ldots,a_{n-1,1})
$$
as follows.
Here $a$ stands for a matrix of indeterminates $(a_{ij})$ with $1\le i\le n-1$
and $1\le j\le d$, with $d$ high enough.
Let $\partial_{ij}$ denote the divided difference operator acting on functions of these indeterminates
as
$$
\partial_{ij} g=\frac{1}{a_{i,j+1}-a_{ij}}\left(
g({\rm argument\ with\ }a_{i,j+1}\ {\rm and}\ a_{ij}\ {\rm exchanged})- g
\right)\ .
$$
The notation $m=(m_1,\ldots,m_{n-1})$ is for a multiindex with nonnegative entries.
We also write the corresponding operator
$$
\partial^m = \prod_{i=1}^{n-1} \left(\partial_{i, m_i}
\cdots\partial_{i,2}\partial_{i,1}\right)
$$
noting that finite difference operators concerning different groups of variables commute.
Finally 
$$
(t-a)^m=\prod_{i=1}^{n-1} \left((t_i-a_{i,m_i})\cdots(t_i-a_{i,2})(t_i-a_{i,1})\right)\ .
$$
The formula basically amounts to applying Newton's univariate formula in each coordinate direction separately.
Now use this with the choice $a_{i,j}=j-1$, then take the beta derivative in the $t$'s
and that should be it.<|endoftext|>
TITLE: The structure of small components in random graphs with a given degree sequence
QUESTION [6 upvotes]: Background and definitions
Consider a random graph on $n$ vertices with a nicely behaved degree sequence.  That is, letting $d_i(n)$ denote the number of vertices of degree $i$, suppose that for all $i$, there exists a constant $\lambda_i$ such that $d_i(n)/n \to \lambda_i$ as $n \to \infty$.
Let $G$ be a randomly chosen graph with this degree sequence.  Molloy and Reed proved that as $n \to \infty$, if $\sum_{i = 0}^{\infty} i(i - 2)\lambda_i > 0$, then w.h.p. $G$ has a giant component, while if $\sum_{i = 0}^{\infty} i(i - 2)\lambda_i < 0$, then w.h.p. $G$ does not have a giant component.
Moreover, the same authors gave a formula for the limiting size of the giant component in the former case.
My question concerns the "small" components in the case where a giant component exists.  In the case of an Erdos-Renyi random graph $G_{n,p}$, the structure of these components is well understood: if $p = cn/2$ for some $c > 1$, then with high probability the graph is the union of the giant component, small tree components, and a growing but relatively small number of unicyclic components. (For a precise statement of this, see Theorem 6.11 of Bollobas's Random Graphs.)
My question
For which degree sequences is it true that with high probability the small components are mostly trees with a comparatively small number of unicyclic components?  In some cases, it's easy to see that no small component can be a tree: for example, in the case of an $r$-regular graph for $r \geq 2$, $G$ cannot contain a tree, because every tree has a vertex of degree $1$.  (However, for $r \geq 3$, the results in the second paper linked to above show that w.h.p. the giant component consists of the entire graph, so in this case the desired property holds vacuously.)  So, for which degree sequences is it "non-trivially" true that most small components are trees?
(N.B.: In the first paper linked to above, Molloy and Reed showed that in the "subcritical" case in which there is no giant component, w.h.p. every component contains at most one cycle.)
The motivation
I'm teaching a course on complex networks using M.E.J. Newman's Networks: An Introduction.  In the chapter on random graphs with a given degree sequence (see SEction 13.7), the author claims that for essentially any given degree sequence, almost all of the small components are trees.  However, the argument given is far from rigorous, and I am rather skeptical of the conclusion.

REPLY [3 votes]: I think the answer to your question is: It is "non-trivially" true that most small components are trees for all degree sequences where the giant component is not the whole graph. 
The answer comes from the two papers you provided. Theorem 2 of the second paper gives what they call a "Discrete Duality Principle" which basically says that if you remove the giant component, then the remaining graph looks like a random graph on the remaining vertices with degree sequence given by these $\lambda_i'$s (which they provide to you). 
The $\lambda_i'$s will satisfy the $Q(\mathcal{D}) < 0$ condition,
so you can apply Theorem 1(b) from the first paper to conclude that the random graph with that degree sequence is mostly tree components (as you pointed out).
I hope this is right, I don't know anything about this other than the two papers you provided.<|endoftext|>
TITLE: when constant scalar curvature implies Einstein?
QUESTION [6 upvotes]: Assume that $(M^n,g)$ is an $n$ dimensional ($n \geq 3$) closed Riemannian manifold with constant scalar curvature and $Ric_g$ nonnegative. Then is $g$ Einstein?

REPLY [5 votes]: As an example where this does hold, for $\omega$ a Kähler metric of constant scalar curvature with $\pi c_1(M) = \lambda [\omega]$, then $\omega$ is Kähler-Einstein. This is Proposition 2.12 in Tian's "Canonical metrics in Kähler Geometry".<|endoftext|>
TITLE: Giving $Top(X,Y)$ an appropriate topology
QUESTION [13 upvotes]: I am not sure if its OK to ask this question here.
Let $Top$ be the category of topological spaces. Let $X,Y$ be objects in $Top$. 
Let $F:\mathbb{I}\rightarrow Top(X,Y)$ be a function (I will denote the image of $t$ by $F_t$). Let $F_{*}:X\times \mathbb{I}\rightarrow Y$ be the function that sends $(x,t)$ to $F_t(x)$. 
Is there a topology on $Top(X,Y)$ such that $F$ is continuous iff $F_{*}$ is continuous ?
Motivation: In the definition of a homotopy $F$ from $f$ to $g$ (for some $f,g\in Top(X,Y)$) it is tempting to think of $F$ to be a "path" (as in the definition of $PY^X$) from $f$ to $g$. Now I really wanted to see if $F$ could be thought of as real path from $f$ to $g$ in $Top(X,Y)$. More precisely, I wanted to know whether $F_{*}:\mathbb{I}\rightarrow Top(X,Y)$ that sends $(x,t)$ to $F_t(x)$ is a path or not. Note that $F_{*}(0)=f,F_{*}(1)=g$,thus if $F_{*}$ is continuous it would be a path from $f$ to $g$ in $Top(X,Y)$.
Hence, I still think that the case when $\mathbb{I}$ is the unit interval is still of some intrest.

REPLY [20 votes]: Dear Ronnie,
The example you refer to (April 22) uses the idea of a placement of a body B in an environment E, and notes that a path in the space of placements corresponds uniquely to a placement of B in the space of paths, because both correspond to a lower-order map from IxB to E itself. These correspondences are invertible, as well as smooth, recursive, etc.(i.e. they preserve whatever structure characterizes the ambient category of spaces). But the transformations are far from banal because it is the smoothness of functionals on the map spaces that the sought structure controls, and composing paths or placements with possible functionals yield new quantities of very diverse needed kinds. This is described for example in Article V, Session 30, and Session 31 of Conceptual Mathematics (Lawvere & Schanuel). My paper about Volterra's functionals (published in the
RCM Palermo in 2000 and downloadable from my Buffalo homepage) discusses the unsuitability for functional analysis (as well as for homotopy theory) of the attempt to characterize continuity or cohesion using open sets or other contravariant structure. That unsuitability was pointed out by Hurewicz who formulated the
necessary exponential laws and inspired Fox (BAMS 1945) to make use of the compact open construction that had
been defined by Lefschetz in 1942. Hurewicz himself introduced, in Princeton lectures in the late 1940's,
the k-spaces; these k-spaces seemed to be an adequate repair of the difficulty and so led to later studies
with various refinements by Kelley, Steenrod, and (of course) yourself. However, an equally serious flaw in
the traditional definition of continuity had already been revealed by Peano's construction of space-filling
curves, suggesting that a return to Frechet's covariant conception based on paths or on Eilenberg's singular figures should be considered. This is almost trivially the basis for the success of exponentials in simplicial sets and similar categories, including those studied in synthetic differential geometry; but the analysis given by Johnstone as a prelude to his important topological topos (which of course still contains the Peano pathologies) reveals that there are still significant foundational issues involved in the appropriate definition of a monoid of continuous reparameterizations of paths; these issues were emphasized by Grothendieck in his Tame Topology and have now been partially addressed by the o-minimal theorists.
(Hurewicz used the k to abbreviate 'kompakt erzeugte')<|endoftext|>
TITLE: central/critical/special values of L-functions terminology
QUESTION [17 upvotes]: I have a question about the terminology for special values
of L-functions.  Is the following a correct description of
standard usage:
Suppose L(s) is an L-function which satisfies a functional
equation relating $s$ to $w+1-s$, where $w$ is the (motivic)
weight.  ADDED LATER:  I am assuming the L-function is motivic,
otherwise (please correct me if I am wrong) there is nothing
special about the value at any integer.
1) If $m$ is an integer then $L(m)$ is a special value of
the L-function.
2) If $m$ is an integer and neither $m$ nor $w+1-m$ is a pole
of a $\Gamma$-factor of the L-function, then $m$ is a critical point
and $L(m)$ is a critical value of the L-function.
3) $L(\frac{w+1}{2})$ is the central value of the L-function.
4) If $\frac{w+1}{2}$ is not an integer, then the central value
is not a special value.
I am pretty sure 2) is correct, unless Deligne's notion of critical
point is not the only one.  I am also pretty sure 3) is correct,
since the central point of the functional equation is pretty
unambiguous.  It is 1) and 4) that I am hoping the experts can
clarify.

REPLY [13 votes]: I think everything you write is correct, and moreover very clear. For example, the value $\zeta(1/2)$ is not a special value of the Riemann Zeta function: $w=0$ in this case. On the contrary, if $E$ is an elliptic curve, $w=1$ and $L(E,(w+1)/2)=L(E,1)$ is a special and the central value of $E$, and it is the value of interest for the Birch-Swinnerton Dyer.
There is a dichotomy between the $L$-function of motives with even, and with odd, motivic weights, and a Tate twist won't change in which world your $L$-motive is, because it adds an even
integer to the weight. 
The Bloch-Kato's conjecture aims at understanding the order $n$ (and then the value of
the $n$-th derivative) of an $L$-function $L(M,s)$ at any point $s \in \mathbb Z$, in terms of algebraic informations on the motive $M$ giving rise to that $L$-function. 
Proving that the order $n$ of $L(M,s)$ at some $s \in \mathbb Z$ is at least what the Bloch-Kato conjecture predicts is completely elementary (provided we have the functional equation for $L(M,s)$ at every $s \in \mathbb Z$ except when $s$ is the central value. When $w$ is even, 
this means that the lower bound of Bloch-Kato for the order of $L(M,s)$ at any integer $s$
is known. It remains to prove the upper-bound, and then to compute precisely the value of the
suitable derivative, which of course is not an easy task. When $w$ is odd, both proving the BK lower bound and proving the BK upper bound on the order
of $L(M,s)$ at the central point $(w+1)/2$ are open. That makes this point the center of all attentions. Example as above: $L(E,1)$ for $E$ an elliptic curve, or an abelian variety.
One can complete this lexicon with:
5) The near central points are defined, in the case $w$ even only, as follows:
 the point $w/2$ and $w/2+1$.

REPLY [8 votes]: How are you normalizing your $L$-function? I ask because in some fields, it's conventional to normalize everything by shifting so the functional equation always takes $s$ to $1-s$. If the original motivic weight is odd, this means that what I would call critical values may sometimes get moved to half-integers rather than integers. (This is why I personally dislike that normalization.) 
If your $L$-function is the $L$-function of a motive and you don't do any strange shiftings (so your usage is consistent with the motives literature), then (2) and (3) are unambiguously correct, (4) is a consequence of (1), and (1) accords with how I use the term but I've never seen it written down as a formal definition.
E.g. if the $L$-function is $L(f, s) = \sum_{n \ge 1} a_n(f) n^{-s}$ for $f = \sum_{n \ge 1} a_n q^n$ a modular cusp form of weight $k$, then the functional equation sends $s$ to $k - s$, the special values are at $s \in \mathbb{Z}$, the critical values are at $\{1, \dots, k-1\}$, and if $k$ is odd then the central value at $s = k/2$ is not a critical or special value.<|endoftext|>
TITLE: What is the homotopy fiber of the map from a space to its James construction?
QUESTION [6 upvotes]: Given a pointed space X, let $J(X)$ denote its James construction. There is a natural inclusion $X\rightarrow J(X)$ which can equivalently be described as the unit map $\eta_X:X\rightarrow \Omega \Sigma X$.
The homotopy fiber of the counit map $\epsilon_X:\Sigma \Omega X\rightarrow X$ is known to be $\Omega X * \Omega X$. Is there an analogous result for the homotopy fiber of $\eta_X$ in terms of $X$?

REPLY [2 votes]: This might not be a satisfactory answer, but let me point out that S.-c. Wong constructed a "little cube" model for the homotopy fiber of $\Omega^{k-1}E^n : \Omega^{k-1}\Sigma^{k}X \to \Omega^{k-1+n}\Sigma^{k+n} X$ in this paper in 1994.
Note that May's little cube model for $\Omega\Sigma X$ is essentially the James construction.<|endoftext|>
TITLE: Heat Kernel Asymptotics with low regularity
QUESTION [6 upvotes]: Let $M$ be a smooth manifold with Riemannian metric $g$, which is not smooth but only continuous.
Question: Is there still an asymptotic expansion of the heat kernel of the form
$$ p_t(x, y) \sim (4 \pi t)^{-n/2} e^{-\frac{1}{4t}d(x, y)^2}\sum_{j=0}^\infty t^j \Phi_j(x, y)$$
where the correction terms $\Phi_j$ are only continuous (or some other regularity that makes sense)?
Assume the following "mild form" of irregularity of $g$: Assume $g$ is smooth everywhere except at a submanifold $N$, where it is still smooth in direction of $N$ and only non-smooth in directions transversal to $N$, but maybe still Lipschitz-continuous.

REPLY [5 votes]: The Varadhan's asymptotics
$t \ln p_t(x,y) \to_{t \to 0} -\frac{d^2(x,y)}{4}$
is known to hold in Lipschitz Riemannian manifolds. It was proved by James Norris in his paper:
Heat kernel asymptotics and the distance function in Lipschitz Riemannian manifolds, Acta Mathematica, Volume 179, Issue 1, pp 79-103
This Varadhan's asymptotics was later generalized by K.T. Sturm,  [J. Math. Pures Appl. 75 (1996), 273–297] and Ramirez  [Comm. Pure Appl. Math. 54 (2001), 259–293] to even more singular spaces. 
Concerning the full  Minakshisundaram–Pleijel expansion for the heat kernel on singular spaces, I am not aware of general results. This is still an active domain of research.  There is an interesting example that I like, that comes  from a measurable Riemannian structure on the Sierpinski gasket. See for instance the short summary
Measurable Riemannian structure and its heat kernel analysis on the Sierpinski gasket
It shows that weird and interesting things can happen.<|endoftext|>
TITLE: When does one obtain different 3-manifolds by pasting two tori? 
QUESTION [6 upvotes]: Consider a compact solid torus $T$ and a diffeomorphic copy of it $T' \subset T$ embedded in the interior of $T$ in such a way that it makes two turns around the central circle of $T$.

I would like to consider the compact boundaryless 3-manifolds $M_f$ obtained from the difference $M = T \setminus \mathring{T'}$ of $T$ and the interior of $T'$ by gluing the two torus boundaries $T_1$ and $T_2$ of $M$ with a diffeomorphism $f$.
If two diffeomorphisms $f,f'$ are isotopic then $M_f$ and $M_{f'}$ are diffeomorphic.   
Question:  Is this the only way one can have $M_f$ diffeomorphic to $M_{f'}$?  If not, is there a characterization of when one obtains diffeomorphic manifolds?  Is there a way to know what Thurston geometry the manifold $M_f$ will have?
I've spent some time trying to calculate the fundamental group of the resulting manifolds via Van Kampen's theorem but I didn't get very far.

REPLY [5 votes]: Here's a way to get different gluing maps giving the same glued-up manifold.
There's an annulus embedded in M that meets each boundary in a closed curve (on the outer torus $T_1$ this closed curve runs twice around the longitudinal direction).  There's a self-diffeomorphism of $M$ supported near the annulus that restricts to a Dehn twist $D_1$ on the closed curve in $T_1$ and to a Dehn twist $D_2$ on the closed curve on $T_2$.  It follows that if $f : T_1 \rightarrow T_2$ is a gluing map, then for $f' = D_2 \circ f \circ D_1^{-1}$ we have $M_f = M_{f'}$.<|endoftext|>
TITLE: Doubling space without Besicovitch covering theorem?
QUESTION [10 upvotes]: A metric space is doubling if any ball of radius $2R$ can be covered by $N$ balls of radius $R$ and $N$ is fixed once forever.

Is there an example of complete length-metric space which is doubling, but the Besicovitch covering theorem does not hold?

REPLY [10 votes]: The Besicovitch covering theorem fails for example in the Heisenberg group, see 
[ E. Sawyer and R. L. Wheeden,
Weighted inequalities for fractional integrals on Euclidean and homogeneous spaces, 
Amer. J. Math. 114 (1992), no. 4, 813–874. http://www.jstor.org/stable/2374799 ] 
The following was more of a comment to Misha's question.
It is easier to see (without assuming the space to be a length-space)  that the Lemma 4.2.2. in http://www.math.wustl.edu/~sk/books/root.pdf is not true for general complete doubling metric spaces nor is the Besicovitch covering theorem:
Take for instance a space $X = \mathbb{N} \cup\{0\}$ with the distance
$$d(0,j) = 2^{-j} \text{ for } j \ne 0$$ and $$d(i,j) = 2^{-j}+2^{-i} \text{ for }0 \ne i\ne j \ne 0.$$
To see that this does not satisfy the Lemma nor the Besicovitch covering theorem consider for any $k \in \mathbb{N}$ the collection $\{B(j,2^{-j}+2^{-k}) ~:~ j = 1, \dots, k-1 \}$.<|endoftext|>
TITLE: Is rigour just a ritual that most mathematicians wish to get rid of if they could? 
QUESTION [73 upvotes]: "No". That was my answer till this afternoon! "Mathematics without proofs isn't really mathematics at all" probably was my longer answer. Yet, I am a mathematics educator who was one of the panelists of a discussion on "proof" this afternoon, alongside two of my mathematician colleagues, and in front of about 100 people, mostly mathematicians, or students of mathematics. What I was hearing was "death to Euclid", "mathematics is on the edge of a philosophical breakdown since there are different ways of convincing and journals only accept one way, that is, proof", "what about insight", and so on. I was in a funny and difficult situation. To my great surprise and shock, I should convince my mathematician colleagues that proof is indeed important, that it is not just one ritual, and so on. Do mathematicians not preach what they practice (or ought to practice)? I am indeed puzzled! 
Reaction: Here I try to explain the circumstances leading me to ask such "odd" question. I don't know it is MO or not, but I try. That afternoon, I came back late and I couldn't go to sleep for the things that I had heard. I was aware of the "strange" ideas of one of the panelist. So, I could say to myself, no worry. But, the greatest attack came from one of the audience, graduated from Princeton and a well-established mathematician around. "Philosophical breakdown" (see above) was the exact term he used, "quoting" a very well-known mathematician. I knew there were (are) people who put their lives on the line to gain rigor. It was four in the morning that I came to MO, hoping to find something to relax myself, finding the truth perhaps. Have I found it? Not sure. However, I learned what kind of question I cannot ask!  
Update: The very well-known mathematician who I mentioned above is John Milnor. I have checked the "quote" referred to him with him and he wrote 

"it seems very unlikely that I said that...".

Here is his "impromptu answer to the question" (this is his exact words with his permission):

Mathematical thought often proceeds from a confused search for what is true to a valid insight into the correct answer. The next step is a careful attempt to organise the ideas in order to convince others.BOTH STEPS ARE ESSENTIAL. Some mathematicians are great at insight but bad at organization, while some have no original ideas, but can play a valuable role by carefully organizing convincing proofs. There is a problem in deciding what level of detail is necessary for a convincing proof---but that is very much a matter of taste.
The final test is certainly to have a solid proof. All the insight in the world can't replace it. One cautionary tale is Dehn's Lemma. This is a true statement, with a false proof that was accepted for many years. When the error was pointed out, there was again a gap of many years before a correct proof was constructed, using methods that Dehn never considered.
It would be more interesting to have an example of a false statement which was accepted for many years; but I can't provide an example.

(emphasis added by YC to the earlier post)

REPLY [10 votes]: Another MO question about rigor got me thinking about this old question again.  One valuable feature of rigor, which I don't think has been said explicitly in the other answers, is that rigor allows me to be confident, in the privacy of my own study, that my argument is correct.
Much has been said by philosophers of mathematics about how a person working in isolation can easily make mistakes without realizing it, and how a proof is of little value unless it is absorbed by the mathematical community at large. All that is true, but should not be allowed to obscure the fact, which is nearly unique to mathematics, that because there is such a thing as mathematical rigor, I have the ability to verify on my own that an argument that I've come up with on my own is objectively correct.  In the natural sciences, for example, a hypothesis that I come up with has to be checked against the empirical facts, and that might cost millions of dollars.
Euler was mentioned in another answer.  Someone of Euler's caliber can, for example, manipulate divergent series without rigorously defined rules and not get into trouble, because he knows of many ways to sanity-check his calculations.  However, what is a mere mortal to do?  If rigorous definitions and proofs are not clearly laid out, then the average mathematician or student has no reliable way of telling whether their calculation with (say) infinite series yields a correct conclusion or is nonsense.  They have to ask an expert and accept the verdict of the expert.
Let me emphasize that I am focusing on the value of rigor in the setting of a single individual working alone.  The Atiyah quote in the other answer I mentioned above emphasizes the importance of rigor in allowing knowledge to be objectively codified and transferred to other people.  I fully agree that rigor plays a vital role here, but I am saying something more.  Rigor also plays a vital role when I am sitting quietly at my desk trying to come up with new mathematics.  It lets me tell when I have a complete solution and when I have an incomplete solution.  It gives me confidence that my rigorously proven lemma can be used as a solid foundation for further investigation.
Rigor therefore contributes to making mathematics more "democratic." I don't deny that the mathematical community has hierarchies and non-democratic features. Nevertheless, it is rigor that makes it possible for an independent researcher to build something of permanent value with limited contact with the larger mathematical community.  Even for those who are "plugged in" to the main community, a large proportion of creative mathematics is initially generated by individuals coming up with their own ideas privately, and testing and validating them before socializing them.  Rigor plays an absolutely fundamental role in guiding and shaping this private thinking process.  It lets me have a very good sense ahead of time, before I say anything to my colleagues, whether my argument is going to be accepted.  If I'm on reasonably good terms with my colleagues then I know that either they will be convinced, or they will point out my mistake and I will agree that I erred.  Without rigor, there is no way I can enjoy this kind of confidence.
I sometimes wonder what it would have been like to try to do research in analysis in the days before calculus was put on a rigorous footing.  It's hard for me to imagine. I think I would always be unsure whether my arguments were really correct.  People who pooh-pooh rigor have, I think, been "spoiled" by the fact that nowadays everything can be made as rigorous as anyone cares to make it.  Next time someone tries to downplay the importance of rigor, ask them what they think it would be like to do research in mathematics in the absence of rigor.  I'm reminded of the famous quote by Kant, "The light dove, cleaving the air in her free flight, and feeling its resistance, might imagine that its flight would be still easier in empty space."<|endoftext|>
TITLE: Untwisting Heegaard diagrams
QUESTION [6 upvotes]: Most Heegaard diagrams contain many rectangles, for instance from loops that circle around one of the handle disks. You can always `twist' a Heegaard diagram to get more and more rectangles (as in page 5 of this paper). Can one reverse this process to eliminate rectangles?
My real question is, is there a systematic way of constructing Heegaard diagrams without rectangles? 
I am interested in Heegaard diagrams without prismatic circuits of length $\leq 4$; in particular, such diagrams can have no rectangles. I'm interested to know how common such Heegaard diagrams are, so I'm asking this question as a first step.

REPLY [8 votes]: I'm sorry to say that this condition is quite rare in practice. If the splitting has high enough distance in the curve complex, any pair of curves from the two disk sets will intersect a lot, resulting in many rectangles. Furthermore, high-distance splittings are "generic."
The word "generic" can be made precise in two ways. By the work of Joseph Maher, random walks in the mapping class group result in a high-distance splitting with probability approaching 1. In a different direction, the work of Lustig and Moriah implies that high-distance splittings are "generic" in a measure-theoretic sense.
Here are the references:
Maher: http://dx.doi.org/10.1112/jtopol/jtq031
Lustig-Moriah: http://arxiv.org/abs/1002.4292
Update: Actually, I am becoming convinced that only very low distance splittings can have rectangles.
Lemma: Let $S$ be a Heegaard splitting surface of genus $g \geq 5$, with Hempel distance $\geq 2$. Then any Heegaard diagram for $S$ contains rectangles. 
In other words, for genus $g \geq 5$, any Heegaard diagram without rectangles must come from a weakly reducible splitting. I strongly suspect this is true in every genus.
Proof: Let $\alpha_1, \ldots, \alpha_g$ and $\beta_1, \ldots, \beta_g$ be the curves of any Heegaard diagram for this splitting. By hypothesis, every $\alpha_i$ intersects every $\beta_j$. Now, cut $S$ along all the $\alpha_i$. We get a sphere with $2g$ holes. The maximal number of disjoint, non-parallel arcs in this surface is $6g-6$. On the other hand, since every $\alpha_i$ intersects every $\beta_j$, there are at least $g^2$ remnants of the $\beta$ curves in this sphere. Since $g^2  > 6g-6$, when $g \geq 5$, some of these arcs must run in parallel. QED<|endoftext|>
TITLE: Obtaining conditional probabilities as pushforwards of [0,1]
QUESTION [5 upvotes]: It is standard that every Borel probability measure on a polish space $X$ can be obtained as pushforward of the uniform measure $\lambda$ on $[0,1]$ along an almost-everywhere-defined Borel-measurable function $d: [0,1] \to X$ . (In fact, $d$ can always be taken to be continuous on a measure-$1$ $G_\delta$ subset. But this is not important for what follows.) My question is whether there is a "conditional" version of this result, along the following lines. I first formulate this below as a precise question, and then discuss possible variations.
I shall state the question as a property of the category whose objects are polish probability spaces (i.e., a polish space together with a Borel probability measure) and whose maps are almost-everywhere-defined measure-preserving Borel-measurable functions modulo almost-everywhere equality. Note that, in terms of this category, the fact stated at the start asserts that $[0,1]$ (with $\lambda$ as its probability measure) is a weakly initial object (i.e., there exists at least one map from $[0,1]$ to every object $X$.)  
Precise question. Given a map $f: X \to Y$, does there exist a map $e : Y \times [0,1] \to X$ such that $f \circ e = \pi_1$ (where $\pi_1$ is first projection)? (Here, $Y \times [0,1]$ is the topological product  with product Borel measure.) 
Remarks and variations 

This is a conditional version of the initial fact in the following sense. Given such an $e$, the function mapping $y \in Y$ to the pushforward of $\lambda$ along $e(y,-): [0,1] \to X$ is a disintegration of $f$, thus giving conditional probability measures $P(-|f(x) = y)$.
Category-theoretically, the question asks if every projection $\pi_1: Y \times [0,1] \to Y$ is a weakly initial object in the slice category over $Y$. 
For the application I have in mind, a weaker result would suffice. It is enough for there to exist $d: [0,1] \to Y$ and $e:[0,1] \times [0,1] \to X$ such that $f \circ e = d \circ \pi_1$.
I am not that fussy about the precise choice of category. One might, e.g., generalise to analytic or standard probability spaces. In general, I'm interested in pointers to any result at all that is similar in nature to the one discussed.

REPLY [4 votes]: I'am not sure this is what you want, but there is a way to represent regular conditional probabilities as some kind of pushforward-measure. Let $\kappa$ be a probability kernel from $Y$ to $X$. That is $\kappa:Y\times\mathcal{B}(X)\to[0,1]$ is measurable as a function of $Y$ and a probability measure as a function of $\mathcal{B}(X)$. Then there is a measurable function $f:Y\times[0,1]\to X$ such that $\kappa(y,\cdot)$ is the distribution of $f(y,\cdot)$. This can be found as Proposition 10.7.6 in Bogachev's Measure Theory.<|endoftext|>
TITLE: Gelfand representation and functional calculus applications beyond Functional Analysis
QUESTION [7 upvotes]: I think it is fair to say that the fields of Operator Algebras, Operator Theory, and Banach Algebras rely on Gelfand representation and functional calculus in a crucial way.

I am curious about applications of these techniques beyond Functional Analysis. Do you know any?

I am aware of this thread and that thread. I am not primarily interested in interactions of that theory with Algebraic Geometry and Differential Geometry. Also, rather than general ideas, I am looking for concrete examples at the research level. Non research level ones are welcome too as Gelfand's proof of Wiener's lemma, for instance, would be a great fit if I did not know it already.
Thank you.
Edit: by Gelfand representation, I refer to this. By functional calculus, I mean either the holomorphic, or  the continuous, or the Borel functional calculus.
Edit: in view of the comments below, I will try to say it slightly differently. The tools I mentioned and linked to above belong in Functional Analysis. I am aware of the categorical meaning of Gelfand transform. Out of pure curiosity, I would like to collect results and problems from other areas of Mathematics to which these techniques apply. As Fourier Analysis is such another area even though it has a nontrivial intersection with Functional Analysis, I think Gelfand's proof of Wiener's lemma is a canonical example as it raised the interest in Banach Algebras, as far as I know. The answers by Asaf and Alain Valette are good examples of what I am hoping for.

REPLY [2 votes]: Very late answer but here is one small application of holomorphic functional calculus in time series/discrete-time stochastic process. Very basic stuff but maybe something a pure mathematician doesn't see everyday.
A baseline time series statistical model is the ARMA (Auto-Regressive Moving-Average) model, which is described by stochastic difference equation
$$
x_t = \sum_{i = 1}^p  \phi_i x_{t-i} + \sum_{j = 0}^q \psi_j \epsilon_{t-j}, 
$$
where $\{ \epsilon_t \}$ is a white noise process, a sequence of independent identically distributed random variables. The statistician is interested in finding a solution of the form 
$$
x_t =  \sum_{i = 0}^{\infty} \theta_i \epsilon_{t-i}.
$$ 
Such a solution necessarily lies in the Banach space $X$ of sequences of bounded (in the $L^2$-norm) random variables. So we are trying to solve the equation
$$
(I - \sum_{i = 1}^p \phi_i B^i) x_t = (\sum_{j = 0}^q \psi_j B^j) \epsilon_t = 
\psi(B)\epsilon_t, 
$$
in $X$, where $B$ is the backward-shift on $X$. If the polynomial $1 - \sum_{i = 1}^p \phi_i z^i$ doesn't vanish on the unit disk, then it has a holomorphic inverse $\theta$ and the desired solution is
$$
x_t = \theta(B)  \psi(B) \epsilon_t. 
$$
The causality/no roots in unit disk condition is always assumed in selecting the model and usually the first thing one checks after fitting data.<|endoftext|>
TITLE: Age of Stochasticity?
QUESTION [25 upvotes]: One user on MSE made an interesting question, which was unanswered so I suggested him to post it here but he refused for personal reasons and said I could ask it here.
The question is this:

Today I came across D. Mumford's 1999
  article The Dawning of the Age of
  Stochasticity, which is quite
  remarkable even after more than a
  decade. The title already indicates
  the theme, but I copy the abstract for
  the convenience of the reader:

For over two millennia, Aristotle's logic has ruled over the thinking of
    western intellectuals. All precise
    theories, all scientific models, even
    models of process of thinking itself,
    have in principle conformed to the
    straight-jacket of logic. But from its
    shady beginnings devising gambling
    strategies and counting corpses in
    medieval London, probability theory
    and statistical interference now
    emerges as better foundations for
    scientific models, especially those of
    the process of thinking and as
    essential ingredients of theoretical
    mathematics, even foundation of
    mathematics itself. We propose that
    this sea change in our perspective
    will affect virtually all of
    mathematics in the next century.

In the article he proposes a new
  approach to mathematical science,
  putting random variables and
  stochasticity into foundations of
  mathematics (rather than building them
  upon measure theory), especially in
  theory of differential equations and
  artificial intelligence.
I am wondering how is this program
  going? I know something about
  stochastic differential equations from
  finance, and I know probability theory
  is fundamental to machine learning and
  artificial intelligence.
However, it seems to me stochasticity
  is still far from the foundations of
  mathematics, and much mathematics is
  still ruled by logic. Of course as an
  undergraduate maybe I am just too far
  from the frontier. 
So can someone tell me how is this
  program going? Is it really some
  advantage in this new approach Mumford
  proposed?
Thanks very much!

REPLY [14 votes]: Here's an example of something that I think Mumford might advocate in the foundations of mathematics: Solovay's model.
The axiom of choice is generally accepted by mathematicians, but it has always suffered from the nagging problem that it violates certain intuitions we have.  Almost all these counterintuitive  consequences of the axiom of choice are related in one way or another to the existence of non-measurable sets.  (See this related MO question for more information, in particular Ron Maimon's answer.)  Solovay's model shows that we can come close to having our cake and eating it too: We can simultaneously have the axioms "all Lebesgue sets are measurable" and the axiom of dependent choice.  The former pretty much eliminates all the probabilistic paradoxes while the latter gives us almost all of the "desirable" consequences of the axiom of choice.
The reason that I think this is the sort of thing Mumford might advocate is that Mumford's discussion of Freiling's theorem shows that he really wants to preserve probabilistic intuition even at the expense of jettisoning a well-accepted axiom.  In the paper he suggests getting rid of the power-set axiom, but my guess is he was probably not familiar with Solovay's model at the time, and if he were, he would have been favorably disposed towards it. 

EDIT: In particular, in Solovay's model, all the following hold: (1) the axioms of ZF, including powerset; (2) all sets are Lebesgue measurable (which is most of what we need to capture probabilistic intuitions); (3) Freiling's axiom of symmetry; (4) the continuum hypothesis in the form "every uncountable subset of $\mathbb R$ can be put into 1-1 correspondence with $\mathbb R$."  The only price one pays is that the axiom of choice has to be weakened to dependent choice. (Thanks to Ali Enayat for pointing this out.)  My view is that Freiling's argument shows only that probabilistic intuition is incompatible with full-blown AC (which is something we knew already); the continuum hypothesis is a red herring.

For more information about the practical impact of adopting Solovay's model and some speculation on why it hasn't already been adopted widely, see this MO question and Andreas Blass's answer to this MO question.<|endoftext|>
TITLE: Intersection of all normalizers
QUESTION [7 upvotes]: This is probably standard for group-theorists:
Let $G$ be a finite group. Is it true that the intersection of all normalizers of subgroups equals the center?
If so, where do I find a proof? What about the same question for infinite groups?
The original question can be reformulated as follows: Let $G$ be a finite group and fix $g\in G$. Assume that for every $x\in G$ there exists a natural number $k(x)$ such that
$$
gxg^{-1}=x^{k(x)},
$$
does it follow that $k(x)=1$ for all $x$? One gets the reformulation by applying the original statement to cyclic groups. It suffices to consider cyclic groups, as an element that normalizes all cyclic groups, normalizes every subgroup.

REPLY [2 votes]: See R. Baer. He Proved: N(G)=E if and only if Z(G)=E.<|endoftext|>
TITLE: Normal subgroups of finite index in free groups
QUESTION [13 upvotes]: Hi all,
This is a question about the groups $H_{n,s}$ introduced by Völklein in his book "Groups as Galois groups", §7.1, and defined as follows: let $N$ be the intersection of all normal subgroups of index $\le n$ in $F_s$, the free group on $s$ generators, and define $H_{n,s} = F_s / N$.

Have these groups been studied? Do they have a name? Is it possible to compute their orders, at least in some cases?

For example for $s=1$, then $H_{n,1}$ is the cyclic group of order $lcm(1, 2, \ldots, n)$.
In Völklein's book, these are introduced primarily to avoid talking about profinite groups (the inverse limits of the $H_{n,s}$, with fixed $s$, is the free profinite group of $s$ generators). 
Any information you may have on these will be great appreciated.
Thanks!
Pierre

REPLY [12 votes]: I am preparing a paper with Ian Biringer, Martin Kassabov, and Francesco Matucci, where we study the growth of the index of the intersection of all normal subgroups of index at most $n$ in a given group. We call this the study of intersection growth of the group. In your notation, for the free group of rank $s$, $F_s$, and every natural number $n$, the intersection growth function, $i_{F_s}(n)$, is defined to be the order of $H_{n,s}$. As general motivation for studying this growth, for a general group $\Gamma$, we show that the growth of $i_\Gamma(n)$ determines the dimension of the profinite completion of $\Gamma$.
This paper (which we may split into two) has some examples worked out: we have precise calculations for this growth for nilpotent groups and certain arithmetic groups. In the case of a rank $s$ free group, we found the lower bound $e^{n^{s-2/3}}$ (which we compute by finding the precise growth when one only intersects maximal subgroups).

REPLY [10 votes]: Yes, you can compute their orders in a few easy cases. For example, if $p$ is prime, then $H_{p,s}$ is elementary abelian of order $p^s$, and  $H_{p^2,s}$ is homocyclic abelian of order $p^{2s}$. I expect it would not be too hard to describe the structure when $n=p^3$, or when $n=pq$ for distinct primes $p,q$.
For small $n$ you could compute $H_{n,s}$ directly. For example, when $n=6$, it has order 972.
But it would be very difficult to compute the order more generally. Some interesting quotients of $H_{n,s}$ have been studied. For example, if we take $s=2$, $n=60$, and let $K$ be the intersection of the kernels of homomorphisms of $F_2$ onto $A_5$, then $F_2/K$ is a direct product of 19 copies of $A_5$.<|endoftext|>
TITLE: Slope of classical modular forms
QUESTION [5 upvotes]: A Celebrated theorem of Coleman says that any overconvergent eigenform of weight $k$ and slope $< k-1$ is classical (here $k \geq 2$ and the level is $\Gamma_1(N) \cap \Gamma_0(p)$). This result is almost optimal as the slope of a classical eigenform of weight $k$ is $\leq k-1$. This seems to be well known and probably easy, but I cannot find a proof. So the question is:
why is the slope of a classical modular eigenform of weight $k$ less or equal to $k-1$?
Thank you very much!
Robert

REPLY [7 votes]: I assume that you are talking of an eigenform $g$ for $U_p$ of level $\Gamma_1(N) \cap \Gamma_0(p)$, with $p$ prime to $N$ (otherwise the result may be false: the slope may be infinite).
For a low-level proof, there are two cases to consider. The most important (because it is generic in a $p$-adic family sense) is when $g$ is old at $p$. That is, when $g$ is a linear combination of the form $f(z)$, $f(pz)$, for $f$ a form of level $\Gamma_1(N)$, eigenform for $T_p$, say with eigenvalue $a_p$. Then one can compute the matrix of $U_p$ in the basis $f(z)$, $f(pz)$. This is not hard since $T_p h$ is  $U_p h $ plus an explicit  scalar
times $h(pz)$, by the very definitions of the Hecke operators $T_p$ and $U_p$, and one find
that that the characteristic polynomial of $U_p$ is $X^2 - a_pX + p^{k-1} \epsilon(p)$, where $\epsilon$ is the nebentypus of $f$. (you can remove this term if $f$ is of level $\Gamma_0(N)$, say). Then the eigenvalue of $U_p$ on $G$ has to be a root $\alpha$ and $\beta$ of this polynomial, but since $a_p$ is an algebraic integer, so are the roots $\alpha$ and $\beta$; in particular $v_p(\alpha) \geq 0$ and $v_p(\beta) \geq 0$.
Since $\alpha \beta = p^{k-1} \epsilon(p)$, and $\epsilon(p)$ is a root of unity,
$v_p(\alpha)+v_p(\beta)=k-1$, and this implies $v_p(\alpha), v_p(\beta) \leq k-1$.
(with possible equality only when one of those two slope is $0$, which is equivalent to $v_p(a_p)=0$.) For the case of a new form, a computation shows that the slope is always $(k-2)/2$.
Edit: I have removed the first sentence and the last paragraph which were non-sense.<|endoftext|>
TITLE: A question about primitive recursive functions
QUESTION [5 upvotes]: I have a question about primitive recursive functions. Maybe it's trivial, if it is I will move it into math.stackexchange.
Is there a primitive recursive function $f$ which is a bijection of $N$ onto $N$ such that $f^{-1}$ is not primitive recursive ?

REPLY [3 votes]: More concisely, it is not far from the truth to say that the purpose of (adding) the minimalisation operation for general recursion is to define inverse functions. That this is Difficult is put to practical use in many methods of encryption.<|endoftext|>
TITLE: Efficient (divergent) summation for sum of zetas at negative arguments?
QUESTION [6 upvotes]: In a question in MSE (see bottom of my own answer) I'm considering the following series, depending on a parameter m:
$$ L(m) = -\zeta(1m)/1 - \zeta(2m)/2 - \zeta(3m)/3 - \ldots $$
where I want to make sense of that sums at negative m.    
For the numerical evaluation I use a customized version of the Noerlund-summation ($ \mathfrak N $) (with the software Pari/GP), but which converges only poorly and even with 128 terms I get not much more than some 15 correct digits.
However, at least for $m=-1$ I get a convincing guess using mathematica (at wolfram alpha) such that
$$L(-1) \underset{\mathfrak N}{=} 1- \log( \sqrt{2 \pi}) \qquad \small \text{ // guessed from 15 digits }$$
For fractional m between $0 \gt m \gt -1$ the summation still works in principle but with even less reliable digits precision. Here are some guesses for $B(m) = \exp(L(m))$
$$ \small \begin{array} {r|lr}
 m & B(m) (\text{ using } \mathfrak N ) \\
\hline
 -1 & 1.0844375514192275466 & \qquad (=\exp(1)/ \sqrt{2 \pi} = 1-A_{-1})\\
 -1/2 & 1.2904007518681174634 \\
 -1/3 & 1.48044921903 \\
 -1/4 & 1.65184851943 \\
 \end{array} $$
Unfortunately for $m \lt -1$ my procedures seem to be completely useless.
So I'd like to ask here:
Q: Which (efficient) procedure can I use to get meaningful evaluations for $L(m)$ at negative m ?  
[update]
To respond to @joro's computation: I did also a Borel-summation. The result for $L(-1)$ was $$L(-1) \sim 0.08106146679532725821967026359438236013860...$$     
I proceeded this way. My function to be summed at -1 is
$$ L(m) = - \sum_{k=1}^\infty \zeta(km)/k \qquad \text{ at } m=-1$$
The Borel-transform is
$$ \mathfrak B L(-1) = - \sum_{k=0}^\infty \zeta(-1-k)/(1+k) \cdot x^k/k! $$
and we define 
$$ B_0(x) = - 1/x \sum_{k=1}^\infty \zeta(-k) \cdot x^k/k! $$
(where the index is also conveniently adapted)      
Then the Borel-sum is computed by the integral
$$ L(-1) \underset{\mathfrak B}{=}\int_0^\infty \exp(-t) B_0(t) dt $$
Now using the software Pari/GP and the sumalt-procedure for $B_0(x)$ we are still confined to small x, so the integral cannot be evaluated at high values of t . But the $B_0(x)$ can be expressed in a closed form using only the exponential, which I denote here as $B_1(x)$:
$$ B_1(x)= \left(\frac 12- \frac 1x  -{1 \over 1-\exp(x)}\right) \cdot \frac 1x $$
This integral can now be evaluated numerically by Pari/GP with a far better interval:
$$ L(-1) \underset{\mathfrak B}{\sim}\int_{1e-20}^{1e6} \exp(-t) B_1(t) dt $$
and gives the above value to about 30 digits precision (I don't think it is a rational value).        
Unfortunately, I cannot generalize that transformation to closed form for sequences of $\zeta(1m)/1,\zeta(2m)/2 , \zeta(3m)/3, ...$ where a negative $m$ is different from $-1$, say $m=-1/2$ or $m=-3$ ...
[/update]

REPLY [2 votes]: This is a partial answer for $m=-1$, proving that Gottfried Helms' guess is correct. 
What we need is the following: 
When $$B_1(x)= \left(\frac 12- \frac 1x  -{1 \over 1-\exp(x)}\right) \cdot \frac 1x, $$
$$\int_0^\infty \exp(-t) B_1(t) dt {=}1- \log( \sqrt{2 \pi})  $$
I have used the same idea for this question in MSE: https://math.stackexchange.com/questions/340718/references-to-integrals-of-the-form-int-01-left-frac1-log-x-frac/342072#342072
The idea is considering the following integral:
$$F(s)=\int_0^\infty \left(\frac 12- \frac 1t  -{1 \over 1-\exp(t)}\right) \cdot t^{s-1} e^{-t} dt$$
This is well-defined if $\textrm{Re}(s)>-1$. In particular for $s=0$. 
This integral can be treated term by term if we have $s$ with sufficiently large real part (absolute convergence).
The result is 
$$\frac 12 \Gamma(s) - \Gamma(s-1) + \Gamma(s)(\zeta(s)-1)$$ 
Since the integral defines analytic function on $\textrm{Re}(s)>-1$, the result of integral should be analytic continuation of the function $\frac 12 \Gamma(s) - \Gamma(s-1) + \Gamma(s)(\zeta(s)-1)$.
Now the result for $s=0$ follows if we take limit $s\rightarrow 0$ in that expression.<|endoftext|>
TITLE: Can nuclearity be determined by tensoring with a single C*-algebra?
QUESTION [25 upvotes]: A C*-algebra is nuclear if the algebraic tensor product $A\odot B$ ($B$ is any other C*-algebra) admits a unique C*-norm. This definition requires testing the condition for nuclearity with `all' C*-algebras. But is there a C*-algebra $B$ which is good enough for all separable $A$s? More precisely,

Does there exist a C*-algebra $B$ such that, given a separable C*-algebra $A$, $A$ is nuclear if and only if there is a unique C*-norm on $A\odot B$?

Presumably, $B$ has to be non-separable and highly non-nuclear. Is it known for $B$ being the Calkin algebra?

REPLY [2 votes]: Somewhat related to this, a C*-algebra $A$ is nuclear if and only if $(A^{**}, C^*(\mathbb{F}_\infty))$ is a nuclear pair (that is, has a unique tensor norm), where $A^{**}$ is the enveloping von Neumann algebra of $A$. Not quite what you're after, but related (nuclearity is decided if there is at most one norm not on $A$, but $A^{**}$, tensored with a single other algebra).
This follows first from the fact that $A$ is nuclear if and only if $A^{**}$ is injective, a von Neumann algebra $M$ is injective if and only if it has the WEP, and of course Kirchberg's seminal result [1] that a C*-algebra $A$ has the WEP if and only if $(A, C^*(\mathbb{F}_\infty))$ is a nuclear pair (also mentioned by Prof. Ozawa above). Pisier in his recent book [2] gives a somewhat simpler proof than Kirchberg of this characterization of the WEP, assuming you're familiar with the many interesting nuances of completely positive maps.
Kirchberg's result opened the door for a number of variations on the theme - in fact there are a few other C*-algebras (and operator spaces) you could take in place of $C^*(\mathbb{F}_\infty)$. For instance

$C^*(\mathbb{F}_n)$ for any $n\ge 2$
$C^*(\mathbb{Z}_2 * \mathbb{Z}_3)$ ($*$ denoting the free product)
$C^*(SL_2(\mathbb{Z}))$
the "non-commutative $n$-cube operator system" $NC(n) = \overline{\text{span}}(\{1\}\cup\{u_i\}_{i=1}^n)$, with $u_i$ the generators of $C^*(*_{i=1}^n \mathbb{Z}_2)$
"Farenick's operator space" $J = \mathbb{C}^5/\text{span}((1, 1, -1, -1, -1))$ (the operator space quotient being considered here).

See [3] and [4].
Also on a side note, it appears recent work in quantum complexity theory has refuted the Connes embedding problem (I wish I had a reference, but it's outside my ken), so I take it $B = \mathcal{Q}(\ell_2)$ is definitively insufficient.
[1] - E. Kirchberg. "On non-semisplit extensions, tensor products
and exactness of group C*-algebras". In: Inventiones
mathematicae (1993)
[2] - Gilles Pisier. "Tensor Products of C*-Algebras and Operator Spaces". Cambridge University Press, 2020
[3] - Douglas Farenick, Ali S. Kavruk, Vern I. Paulsen, Ivan G. Todorof, "Characterizations of the Weak Expectation Property". 2013. Retrieved from https://arxiv.org/pdf/1307.1055.pdf.
[4] - Douglas Farenick, Vern I. Paulsen "Operator System Quotients and Tensor Products". 2011. Retrieved from https://arxiv.org/pdf/1010.0380.pdf.<|endoftext|>
TITLE: Prescribing the Lie derivative of the metric?
QUESTION [5 upvotes]: This is a question that arises from my research problem. Suppose $(M,g)$ is a compact Riemannian manifold with boundary and $g$ is smooth up to the boundary (if you like, take $M$ to be diffeomorphic to a closed ball). Let $h$ be a symmetric traceless 2-tensor on $M$ (in my case this is actually the traceless Ricci $\stackrel\circ {Ric}$). 
Now, I have shown that $\int_M \langle h, \mathcal L_Xg\rangle dV=0$ for any vector field $X$ on $M$. I wonder does it follow that $h=0$? (Well, may be not...) 
Naturally, one would try to see if there is $X$ such that $\mathcal L_X g=h$. While this seems plausible, I have no idea when it will hold. 
In case this is not always possible, instead one ask: Let $\epsilon>0$ and $p\ge 1$, does there exists functions $\lambda, \mu$ on $ M$ with $\lambda\ge 1$, and a vector field $X$ such that
  $\|\mathcal L_X g -(\lambda h+ \mu g)\|_{L^p(M)}<\epsilon$?
Well, if they don't hold in general, can we impose conditions on $(M,g)$ so that one of them hold?

REPLY [9 votes]: If $h\in S^2(TM^*)$ is a symmetric $(0,2)$-tensor such that $\int_M \langle h,\mathcal L_X g\rangle =0$ for all vector fields $X$ on $M$, then $h$ need not be zero. In fact, the above is precisely equivalent to $\delta h=0$, where $$\delta=\nabla^*\big|_{S^2(TM^*)} \colon S^2(TM^*)\to \Omega^1(M)$$ is the divergence operator, i.e., the formal $L^2$-adjoint of the covariant derivative $\nabla \colon \Omega^1(M)$ $\to S^2(TM^*)$.
The above claim follows from an infinitesimal version of Ebin's slice theorem for the pull-back action of the diffeomorphism group $\mathcal D:=Diff(M)$ on $S^2(TM^*)$, see this paper of Berger and Ebin, JDG '69. The statement I am referring to is the $L^2$-orthogonal decomposition: $$S^2(TM^*)=\ker \delta\oplus Im \ \delta^*.$$
Given any element $g\in S^2(TM^*)$, e.g., a Riemannian metric, the space $\ker\delta$ is the tangent space to the slice at $g$ to the $\mathcal D$-action and $Im\ \delta^*=T_g \mathcal D(g)$ is the tangent space to the orbit of $g\in S^2(TM^*)$. The latter is precisely formed by tensors of the form $\mathcal L_X g$, where $X$ is some field on $M$. In fact, consider $\phi_g\colon\mathcal D\to S^2(TM^*)$, $\phi_g(\eta)=\eta^*(g)$, and let $\eta_t\in\mathcal D$ be a curve of diffeomorphisms, with $\eta_0=id$ and $\dot\eta_0=X$. Then $$\mathrm d\phi_g(id)X=\frac{\mathrm d}{\mathrm dt}\phi_g(\eta_t)\big|_{t=0}=\frac{\mathrm d}{\mathrm dt}\eta_t^*(g)\big|_{t=0}=\mathcal L_X g=2\delta^*(X^b),$$ where $X^b=g(X,\cdot)$ is the $1$-form dual to the field $X$. The last equality follows from $\delta^*(\omega)=\tfrac12\mathcal L_{X_\omega}g$, where $X_\omega$ is the vector field dual to the $1$-form $\omega$. Note the above line also proves that $Im \ \delta^*=T_g\mathcal D(g)$.
Thus, going back to your original question, the above result of Ebin and Berger tells you that your symmetric traceless tensor $h$ is geometric, in the sense that it is tangent to the slice of the $\mathcal D$-action on $S^2(TM^*)$, or, equivalently, $L^2$-orthogonal to the tangent space to the $\mathcal D$-orbit. In some sense, this means that it ``descends'' to an object in the quotient space of tensors modulo diffeomorphisms (where, e.g., the moduli space of Riemannian metrics should live).
The above observations also answer your question regarding what symmetric $(0,2)$-tensors are of the form $\mathcal L_Xg$; namely, they are precisely the ones in $Im \ \delta^*$. Hope this helps...<|endoftext|>
TITLE: How to understand Chern-Simons action
QUESTION [24 upvotes]: Hi all. The question I have should be a rather simple one, but I just can't think it through.
So the Chern-Simons action reads
\begin{equation}
S = \int_M {\rm tr} (A\wedge dA + \frac{2}{3} A\wedge A \wedge A)
\end{equation}
where $M$ is 3-fold, and similarly for higher dimensional manifold.
Now, my question is:
*since $A$, the connection 1-form is only defined patch by patch, what do we really mean by doing the integration? *
It would be understandable if I write
\begin{equation}
S = \int_M {\rm tr} \left[(A-A_0)\wedge d(A-A_0) + \frac{2}{3} (A-A_0)\wedge (A-A_0) \wedge (A-A_0)) \right ]
\end{equation}
where $A_0$ is some reference connection, since $A-A_0$ is globally defined 1-form valued in ${\rm Lie}G$.
I see that under gauge transformation (or different chart),
\begin{equation}
CS(A^g) - CS(A) = d\alpha(A,g) + Q(g)
\end{equation}
where $Q(g)$ is closed. But I don't know how I can infer the validity of doing the integration from this gauge transformation.
Thank you!

REPLY [17 votes]: Here is a mid 1970s  point of view, courtesy of Atiyah-Patodi-Singer.
Suppose  you have a  complex vector bundle $E$   of rank $r$   over   a smooth manifold $M$.   A polynomial  function $P$ on the space of  $r\times r$ matrices is called invariant if $P(T AT^{-1})=P(A)$ for any  $r\times r$ complex matrix  $A$ and any invertible $r\times r$ matrix $T$.   If you look at 
$$ \Delta(x)=\det(1+ xA)=\sum_{k=0}^r c_k(A) x^k, $$
then the coefficient  $c_k(A)$ is a homogeneous invariant polynomial function of degree $k$. For example
$$c_1(A)={\rm tr}\; A,\;\;c_r(A)=\det A. $$
To  a connection $\nabla$ on $E$ with curvature $F(\nabla)$, we can associate the  degree $2k$ form on $E$
$$ c_k(\nabla) = c_k\bigl(\; F(\nabla)\;\bigr), $$
where in the above equality one thinks of $F(\nabla)$ as an $r\times r$-matrix whose entries are $2$-forms.  For example 
$$ c_1(\nabla)= {\rm tr}\; F(\nabla)= F_{11}(\nabla)+\cdots +F_{rr}(\nabla). $$
Chern-Weil theory proves two things:

The form $c_k(\nabla)$ is closed.
If $\nabla^1$, $\nabla^0$ are two  connections on $E$, then there exists  a canonical form  of   degree  $(2k-1)$,  called the transgression form and denoted by $Tc_k(\nabla^1,\nabla^0)$,     which satisfies 

$$ d Tc_k(\nabla^1, \nabla^0)= c_k(\nabla^1)-c_k(\nabla^0). $$
In other words, the cohomology class determined by $c_k(\nabla)$ is independent of $\nabla$. This cohomology class is  the $k$-th Chern class of $E$.
Suppose now that $\dim M= 2k-1$.  Then, on account of dimension, $c_k(\nabla)=0$, yet $Tc_k(\nabla^1,\nabla^0)$  is a   top degree form  well defined for any choices of $\nabla^0,\nabla^1$.
Suppose additionally that $E$ is trivial and we have fixed a trivialization. Then we can choose $\nabla^0$ to be the trivial connection on $E$    and then we set
$$ CS_k(\nabla):= Tc_k(\nabla,\nabla^0). $$
The usual  Chern-Simmons theory is a special case of this construction when $k=2$, i.e., $E$ is a trivial complex vector bundle of rank $r\geq 2$ over a $3$-manifold.<|endoftext|>
TITLE: What are the invariant Pseudo-differential operators on a Lie group?
QUESTION [11 upvotes]: It is well-known that (left) $G$-invariant differential operators on a Lie group $G$, has an algebraic description, i.e. universal enveloping algebra of the Lie algebra of the group.  
On the other hand $\Psi$Do are generalization of the differential operators on general smooth manifold.
My question is that:
Does any algebraic description for the $G$-invariant $\Psi$DOs on the Lie groups (or more general on homogeneous spaces $G/H$) exist ? 
Thanks

REPLY [2 votes]: There is the following description of $G$ invariant pseudodifferential operators on a Riemannian homogeneous space $G/H$: The Schwartz kernels are smooth outside the diagonal and conormal with respect to the diagonal; this is equivalent to Beals' commutator characterization mentioned in Pedro Lauridsen Ribeiro's answer. Moreover, the full geometric symbols are, modulo symbols of order $-\infty$, invariant under the symplectic action of $G$ on $T^*(G/H)$.
Geometric symbols are defined by pulling back Schwartz kernels under the exponential map of the Levi-Civita connection to a neighbourhood of the zero section of the tangent bundle $T(G/H)$, and then taking Fourier transforms in the fiber variable. See here for more details of and references to the geometric pseudodifferential calculus. That principal symbols must be invariant is clear from the transformation behaviour under diffeomorphisms (special case of Egorov's theorem, or FIO calculus). Lemma 6.2 in the  paper in Math. Z., of which I am a coauthor, gives the invariance of the full geometric symbol if the diffeomorphism is an isometry, e.g. an action on $G/H$ by a group element.
The description discards smoothing operators. This is reasonable if one adopts, from Sato's microlocal analysis, the view that pseudodifferential operators operate on microfunctions.
The above characterization of invariant pseudodifferential operators is not algebraic. However, I think that it is simple and straightforward. I don't know if there is a characterization of $G$ invariant microdifferential (pseudodifferential) operators in the setting of algebraic analysis which is more algebraic.<|endoftext|>
TITLE: Elementary proof for identity involving sums of binomials
QUESTION [6 upvotes]: Is there an elementary proof of this identity?
$$n  + 1 - \sum_{k=1}^{n} k^{k-1} \binom{n}{k} \frac{(n-k)^{n+1-k}}{n^{n}} =1 + \sum_{k=1}^n \frac{n!}{(n-k)!n^k}\;?$$
The term on the right is the average time to find a duplicate birthday from the classic Birthday Problem and the sum on the left is a special case of this sum.  As a sanity check, if you compute the answers numerically (exactly) for small $n$ they are exactly the same.
The numerator of the sum on the left is $n^{n+1}-Q(n)n^n$ (Q(n) is called Ramanujan's function by Knuth) according to A219706 and A063169.  This immediately gives the identity as the term on the right is $1+Q(n)$.  Am I just missing a simple direct proof of their equality too?

REPLY [16 votes]: After canceling $1$'s and clearing denominators, the identity can be rearranged to this one: 
$$n^{n+1} = \sum_{k=1}^{n} \binom{n}{k} k^{k-1} (n-k)^{n-k+1} + \sum_{k=1}^n \binom{n}{k} n^{n-k} k!$$ 
and now we proceed to give a bijective proof. The left side counts data of the form 

Endofunction $f: S \to S$ on an $n$-element set $S$, plus a distinguished point $s$ of $S$. 

Draw the endofunction as a directed graph, where an edge is drawn from $x$ to $y$ if $f(x) = y$. This graph can be decomposed into connected components, and there are two disjoint possibilities: 

Case 1: either $s$ belongs to a cycle (this includes the case where $f(s) = s$), or 
Case 2: removal of the edge from $s$ to $f(s)$ produces a disconnection of the connected component of $s$. 

Let us count the possibilities for the first case. The cycle to which $s$ belongs defines a $k$-element set, say, and gives a structure of linear order on this set, namely the ordering $s, f(s), \ldots, f^{k-1}(s)$. There are $k!$ ways of giving such a linear order, and this order describes the restriction of $f$ to the cycle containing $s$. The remainder of $f$ is just given by some function $S \backslash \mathrm{cycle}(s) \to S$ on the complement, of which there are $n^{n-k}$ many. Thus the second sum of the asserted identity counts the possibilities for case 1. 
For case 2, consider the set of elements $x$ such that $f^{(j)}(x) = s$ for some $j \geq 0$. This defines some $k$-element subset, and the graph of the restriction of $f$ to this subset is described as a rooted tree with root at $s$. The number of such rooted tree structures on a $k$-element set is $k^{k-1}$, by Cayley's theorem. Then, the rest of $f$ is given by its restriction to the complement of this $k$-element set (notice it takes the complement to itself) -- there are $(n-k)^{n-k}$ possibilities here -- plus knowledge of the element $f(s)$ which belongs to this $(n-k)$-element set. Thus the first sum on the right side of the identity counts the number of possibilities for case 2, and we are done.<|endoftext|>
TITLE: Is the derived double dual of a quasi-coherent sheaf a sheaf?
QUESTION [13 upvotes]: Let F be a quasi-coherent sheaf on a smooth projective variety X over an algebraically closed field and $D_X(?)=RHom(?,\omega _X[n])$ the dualizing functor. Is it the case that $D_X(D_X(F))$ is a sheaf? (i.e. $\mathcal H^n(D_X(D_X(F)))=0$ for $n\ne 0$). If so is there a convenient reference?

REPLY [16 votes]: For coherent sheaves, by using resolutions, one checks that the assertion in question is true: for any perfect complex $K$, one even has $D_X^2(K) \simeq K$. However, for larger quasicoherent sheaves $F$, the value $D_X^2(F)$ fails to even be a sheaf; I give an example below in Corollary 5.
Fix a countable set $I$, and a PID $A$ with at least two non-zero irreducible elements $p,q \in A$ such that $(p,q) = (1)$. We begin with the following basic observation that lies at the heart of the argument:
Lemma 1: One has $Hom_A(\prod_I A,A) \simeq \oplus_I A$ via the natural map in $D(A)$.
Proof: See Is it true that, as $\Bbb Z$-modules, the polynomial ring and the power series ring over integers are dual to each other?.
This Lemma shows that $\prod_I A$ is not projective. More precisely, we get:
Lemma 2: The complex $K := RHom_A(\prod_I A,A)$ has non-zero $H^1$.
Proof: If $k$ is a residue field of $A$ at a maximal ideal (ex: $k = A/(p)$), then $k$ is $A$-perfect, so $- \otimes_A k$ commutes with all (homotopy) limits and colimits. This gives $K \otimes_A k \simeq Hom_k(\prod_I k, k)$, which is a vector space of uncountable dimension. On the other hand, by Lemma 1, if we assume that $H^1(K) = 0$, then we get $K \otimes_A k \simeq \oplus_I k$, which is a vector space of countable dimension; so we conclude $H^1(K) \neq 0$.
Now we globalize. Set $X = \mathbf{P}^1_k$ over some field $k$. All complexes below are considered as living in the derived category $D(O_X)$ of all $O_X$-modules on $X$, and similarly for functors. (I prefer this to using $D_{qc}(O_X)$ as the description of products in the latter is more subtle.) The global analogue of Lemma 2 is: 
Lemma 3: The complex $K := RHom_X(\prod_I O_X,O_X)$ has a non-zero $H^1$.
Proof: This does not formally follow from Lemma 2 as $\prod_I O_X$ is not quasi-coherent. However, we can argue as follows: (a) first show that $H^0(K) := Hom_X(\prod_I O_X,O_X) \simeq \oplus_I O_X$ via the natural map, and (b) show that $K \otimes_{O_X} k \simeq Hom_k(\prod_I k,k)$ is a vector space of uncountable dimension, where $k$ is the residue field at $0$ viewed as an $O_X$-complex. Then (a) and (b) immediately imply the claim by the argument of Lemma 2. Moreover, (b) is clear since $k$ is perfect over $O_X$, so it remains to show (a). Consider the natural map $\eta:\oplus_I O_X \to Hom(\prod_I O_X,O_X)$. Clearly $\eta$ is injective. For surjectivity, fix some $f:\prod_I O_X \to O_X$. We want to show that $f$ has finite support, i.e., that $f$ factors through a projection to finitely many components of the product on the left. For any affine open $U \subset X$, the map $f(U)$ has finite support by Lemma 1. A glueing argument then finishes the proof.
Writing a product as the dual of a sum then gives:
Lemma 4: The complex $RHom_X(RHom_X(\oplus_I O_X,O_X),O_X)$ has non-zero cohomology in degree $0$ and $1$.
Proof: This follows formally from Lemma 3 as $RHom_X(-,O_X)$ carries coproducts to products.
Here is the promised example:
Corollary 5: On $X = \mathbf{P}^1_k$, if $D_X = RHom_X(-,\omega_X[1])$, then the functor $D_X^2$ carries the sheaf $\oplus_I O_X$ to a complex with non-zero $H^0$ and $H^1$.
Proof: As $\omega_X$ is invertible, one has $D_X = RHom_X(-,O_X) \otimes \omega_X[1]$, and hence $D_X^2 = RHom_X(RHom_X(-,O_X),O_X)$, so the claim follows form Lemma 4.<|endoftext|>
TITLE: Solutions to $\binom{n}{5} = 2 \binom{m}{5}$
QUESTION [25 upvotes]: In Finite Mathematics by Lial et al. (10th ed.), problem 8.3.34 says:

On National Public Radio, the Weekend Edition program posed the
  following probability problem: Given a certain number of balls, of
  which some are blue, pick 5 at random.  The probability that all 5 are
  blue is 1/2.  Determine the original number of balls and decide how
  many were blue.

If there are $n$ balls, of which $m$ are blue, then the probability that 5 randomly chosen balls are all blue is $\binom{m}{5} / \binom{n}{5}$.  We want this to be $1/2$,
so $\binom{n}{5} = 2\binom{m}{5}$; equivalently,
$n(n-1)(n-2)(n-3)(n-4) = 2 m(m-1)(m-2)(m-3)(m-4)$.
I'll denote these quantities as $[n]_5$ and $2 [m]_5$ (this is a notation for the so-called "falling factorial.")
A little fooling around will show that $[m+1]_5 = \frac{m+1}{m-4}[m]_5$.
Solving $\frac{m+1}{m-4} = 2$ shows that the only solution with $n = m + 1$ has $m = 9$, $n = 10$.
Is this the only solution?
You can check that $n = m + 2$ doesn't yield any integer solutions, by using the quadratic formula to solve $(m + 2)(m  +1) = 2(m - 3)(m - 4)$.  For $n = m + 3$ or $n = m + 4$, I have done similar checks, and there are no integer solutions.  For $n \geq m + 5$, solutions would satisfy a quintic equation, which of course has no general formula to find solutions.
Note that, as $n$ gets bigger, the ratio of successive values of $\binom{n}{5}$ gets smaller; $\binom{n+1}{5} = \frac{n+1}{n-4}\binom{n}{5}$
and $\frac{n+1}{n-4}$ is less than 2—in fact, it approaches 1. So it seems possible that, for some $k$, $\binom{n+k}{5}$ could be $2 \binom{n}{5}$.
This question was previously asked at Math StackExchange, without any answer, but some interesting comments were made. It was suggested that I ask here.

REPLY [13 votes]: This isn’t a complete answer, but the problem “reduces” to finding the finitely many rational points on a certain genus 2 hyperelliptic curve. This is often possible by a technique
involving a reduction to finding the rational points on a finite set of rank 0 elliptic curves—see for example “Towers of 2-covers of hyperelliptic curves” by Bruin and Flynn in Trans. Amer. Math. Soc. 357 (2005) #11 4329–4347.
In your case, the curve is $u^2= 9t^6+16t^5-200t^3+256t+144$. There are the following 16
rational points: $(t,u)$ or $(t,-u)= (0,12)$, $(1,15)$, $(2,12)$, $(4,204)$, $(-1,9)$, $(-2,36)$, $(-4,180)$, and $(7/4,411/64)$. If these are the only rational points then the only non-trivial solution to your equation is $n=10$, $m=9$. To see this suppose that $$n(n-1)(n-2)(n-3)(n-4)=2m(m-1)(m-2)(m-3)(m-4).$$ Let $$y=(n-2)^2\text{ and }x=(m-2)^2.$$ Squaring both sides we find that $$y(y-1)(y-1)(y-4)(y-4)=4x(x-1)(x-1)(x-4)(x-4).$$ Suppose $y$ isn’t $0$. Then $4x/y$ is $t^2$ for some rational $t$ with $$(y-1)(y-4)=t(x-1)(x-4).$$ We replace $y$ by $4x/t^2$ in this equation and find that $$(t^5-16)x^2-(5t^5-20t^2)x+(4t^5-4t^4)=0.$$ So this last quadratic polynomial in $x$ has a rational root and its
discriminant is a square. This gives the hyperelliptic curve above. Note that the case $n=10$, $m=9$ of your problem corresponds to the point $(7/4,411/64)$ on this curve.
EDIT:
More generally one can look for rational $m$ and $n$ with $[n]_5= 2[m]_5$. If $(t,u)$ is a
rational point on the hyperelliptic curve with $t$ non-zero, set $x=(5t^5-20t^2+t^2u)/(2t^5-32)$ and $y=4x/t^2$. Then if $x$ is a square in $\mathbb Q$, one gets such an $m$ and $n$ with $m=2+\sqrt x$ and $n=2+\sqrt y$. Joro’s points lead in this way to the solutions
$(n,m)=(10,9)$, $(10/3,5/3)$, and $(78/23,36/23)$, the last one being rather unexpected. (And as François notes, each $(n,m)$ gives a second solution $(4-n,4-m)$.) Perhaps these solutions and
the trivial solutions with $m$ and $n \in \{0,1,2,3,4\}$ are the only rational solutions.<|endoftext|>
TITLE: Thompson's group T
QUESTION [6 upvotes]: Does there exist a non trivial homomorphism from Thompson's group T to a linear group?

REPLY [12 votes]: No: T is infinite, finitely presented and simple.  Fg linear groups are residually finite, by Mal'cev's theorem. QED.<|endoftext|>
TITLE: Optimality of p-Lebesgue Differentiation Theorem for Sobolev Functions
QUESTION [7 upvotes]: This is the third question in a series whose purpose has been to flesh out an example of the optimality of the p-Lebesgue differentiation theorem for Sobolev functions.  This theorem says that for $f \in W^{1,p}_{loc}(\mathbb{R}^N;\mathbb{R})$, 
$\lim_{r\to0} \frac{1}{r^{N+p}}\int_{B(0,r)} |f(x+h)-f(x)-\nabla f(x)h|^p\;dx = 0$
for $\mathcal{L}^N$ almost every $x\in \mathbb{R}^N$.
EDIT: The previously mentioned lower bound is true, but it is not infinite, and thus does not serve as a counterexample to the question at hand.  Therefore, the real question is about the optimality of the p-Lebesgue differentiation theorem for Sobolev functions.  Can one find $f \in W^{1,1}$ or $f \in W^{1,p-\epsilon}$ (and $f \notin W^{1,p}$) such that the above limit is plus infinity on a set of positive measure?

REPLY [5 votes]: It depends on the values of $p$ and $N$ - sometimes there will be counterexamples and sometimes there will not.  For instance, any function $f \in W^{1,q}$ for $q>N$ is classically differentiable almost everywhere, and hence the above result is true for all $1\leq p < \infty$.  When $1\leq q < N$, then the result holds for all $1\leq p \leq q^\ast$, and this is optimal.
To see these claims, check Evan's proof of differentiability of $W^{1,q}$ for $q>N$ and observe that it is a clever use of Morrey's embedding to show that
$\frac{|f(y)-f(x)-\nabla f(x)(y-x)|}{|y-x|} \leq C \frac{1}{r^N}\int_{B(x,r)}|\nabla f(y)-\nabla f(x)|^q\;dy$,
and the right hand side goes to zero by the classical Lebesgue differentiation theorem for $\nabla f \in L^q$.
A similar proof allows one to use the Sobolev-Gagliardo-Nirenberg embedding to let $f \in W^{1,q}$ and consider the function $u(y)=f(y)-f(x)-\nabla f(x)(y-x)$ which has average value on a ball $\frac{1}{\alpha(N)r^N} \int_{B(x,r)} u(y)\;dy= \frac{1}{\alpha(N)r^N} \int_{B(x,r)} f(y)\;dy - f(x)$ to deduce that
\begin{align*}
\lim_{r\to 0}\frac{1}{r^{N+q^\ast}}\int_{B(x,r)} |f(y)-\frac{1}{\alpha(N)r^N}\int_{B(x,r)}f(y)dy-\nabla f(x)(y-x)|^{q^\ast}\;dy =0,
\end{align*}
and thus, writing
\begin{align*}
\frac{1}{r^{N+q^\ast}}&\int_{B(x,r)} |f(y)-f(x)-\nabla f(x)(y-x)|^{q^\ast}\;dy \newline
 &\leq \frac{C}{r^{N+q^\ast}}\int_{B(x,r)} |f(y)-\frac{1}{\alpha(N)r^N}\int_{B(x,r)}f(y)dy-\nabla f(x)(y-x)|^{q^\ast}\;dy \newline
&\;\;+\left[\frac{\tilde{C}}{r^{N+1}}\int_{B(x,r)}|f(y)- f(x)-\nabla f(x)(y-x)|\;dy \right]^{q^\ast}
\end{align*}
and applying the standard theorem on $L^q$ differentiability the second term tends to zero.  Thus, there is a possible improvement of differentiability theorems for Sobolev functions.  To see that this is optimal, just take a function $f \in W^{1,q}$ and $f \notin L^p$ of any open set (for $p>q^\ast$.  Then the above integral is plus infinity.  For example, consider the function
$g_\epsilon(x) = |x|^{1-\frac{N}{q}+\epsilon}$.
Then $\int_{B(0,r)} |g_\epsilon|^p\;dx = |S^{N-1}|\int_0^1 r^{p(1-\frac{N}{q}+\epsilon)+N-1}\;dr$,
and $|g|^q$ will be not be integrable if 
$p(1-\frac{N}{q}+\epsilon)+N-1 \leq -1$,
which is equivalent to 
$p \geq \frac{Nq}{N-q-\epsilon q}$.
Now for any $p>\frac{Nq}{N-q}$, we can find an $\epsilon>0$ small such that
$p \geq \frac{Nq}{N-q-\epsilon q}$
and necessarily $g_\epsilon \in W^{1,q}$ by construction, since
\begin{align*}
\int_{B(0,r)} |\nabla g_\epsilon|^q\;dx \leq C \int_0^1 r^{-N+\epsilon q+N-1}\;dr <\infty
\end{align*}
Now, let $f_\epsilon(x):= \sum_n \frac{1}{2^n} g_\epsilon(x-x_n)$
for $\{x_n\}$ be a dense sequence of some bounded open set, and then we have
$\int_{B(x,r)}|f_\epsilon(x+h)-f_\epsilon(x)-\nabla f_\epsilon(x)h|^p\;dh = +\infty$,
for every $x$ in this set, since $f \notin L^p$ and the result is demonstrated.<|endoftext|>
TITLE: Additive Combinatorics - reference request
QUESTION [6 upvotes]: Let $A$ be a finite set of integers with $|A \hat{+} A| \leq K|A|$, where the $\hat{+}$ denotes restricted sumset: the set of all $a_1 + a_2$ with $a_1, a_2 \in A$ and $a_1 \neq a_2$. 
Claim: $|A + A| \leq (K + o(1))|A|$, where $o(1)$ denotes a quantity tending to 0 as $|A|$ tends to $\infty$.
Sketch proof: Let $A'$ be the set of all $a \in A$ for which $2a$ cannot be represented as $a_1 + a_2$ with $a_1, a_2 \in A$, $a_1 \neq a_2$. Note that $2A' = (A + A) \setminus (A \hat{+} A)$, so it suffices to show that $A'$ is small. But $A'$ has no 3-term arithmetic progressions. But by standard results (Freiman, Ruzsa...) a set $A'$ with no 3-term progressions has $|A' + A'| \gg |A'| \log^c |A'|$. Thus indeed $|A'| = o(|A|)$. 
I know I've seen this argument in the literature, and my question is simply: where?

REPLY [7 votes]: For a somewhat similar argument, see Proposition 2.5 from Alon's 1987 paper ``Subset sums", available here.<|endoftext|>
TITLE: Needless axiom for Grothendieck topologies?
QUESTION [7 upvotes]: Hi,
The first axiom for a Grothendieck (pre)topology on a category $C$ says that for every object $X\in C$, the family consisting of just the identity $1_X : X\to X$ should be a covering family.
Why is this axiom needed? Obviously a functor $F : C^{opp}\to Sets$ will satisfy the sheaf condition with respect to this family, so nothing seems to be gained by adding this covering family...
Am I missing something?
Thanks!

REPLY [9 votes]: The only important axiom in order to define a notion of sheaf is the stability under pullback. There is a proposition in SGA4 saying that if you have a family of sieves only satisfying the pullback stability condition then a presheaf is a sheaf with respect to this family if and only if it is a sheaf with respect to the topology it generates. Hence to some extend the identity axiom is useless, as well is the locality axiom.
The reason why this two other axioms are important is that when you have all three axioms then you can prove that if two topologies defines the same notion of sheaf then they are equal: adding new covering will reduces the number of sheaf, and this is the case only if the three axioms are satisfied.<|endoftext|>
TITLE: Lecture on Fractals for Middle School Students
QUESTION [5 upvotes]: I'm going to have a one-hour lecture for middle school students next Monday. It will be about fractals. The students know virtually nothing about this subject. 
I'll show some fractal images and a few short films which I've found on youtube, discuss on the ways a fractal can be constructed, and introduce a software. But I'll need other things, too:

Some serious mathematical content.
Some questions to propose to the students for further study.
Could you please help me with these?

Thanks in advance.

REPLY [2 votes]: A very accessible application of fractals can be found in Richard Taylor's "Order in Pollock's chaos" (an article in the November 2002 issue of Scientific American), which is based on the Richard Taylor, Adam Micolich, and David Jonas's "Fractal analysis of Pollock's drip paintings" (a paper in a June 1999 issue of Nature).  The Nature paper has the following abstract:

Scientific objectivity proves to be an essential tool for determining the fundamental content of the abstract paintings produced by Jackson Pollock in the late 1940s. Pollock dripped paint from a can onto vast canvases rolled out across the floor of his barn. Although this unorthodox technique has been recognized as a crucial advancement in the evolution of modern art, the precise quality and significance of the patterns created are controversial. Here we describe an analysis of Pollock's patterns which shows, first, that they are fractal, reflecting the fingerprint of nature, and, second, that the fractal dimensions increased during Pollock's career.

From the Scientific American article:

The painting is scanned into a computer. It is separated into its different colored patterns, then covered with a computer-generated mesh of identical squares. The computer analyzes which squares are occupied and which are empty. This is done for different mesh sizes. The patterns were found to be fractal over the entire size range.<|endoftext|>
TITLE: Inverse of a totally unimodular matrix
QUESTION [9 upvotes]: A unimodular matrix $M$ is a square integer matrix having determinant $+1$ or $−1$.
A totally unimodular matrix  (TU matrix) is a matrix for which every square non-singular submatrix is unimodular. A totally unimodular matrix need not be square itself. Obviously, any totally unimodular matrix has only $0$, $+1$ or $−1$ entries.
Now suppose a $n\times n$ non-singular matrix $A$ is totally unimodular. Can we prove that
$A^{-1}$ is also totally unimodular? Or if it is not correct, can we have a counterexample?

REPLY [14 votes]: The answer is yes, because if $B=A^{-1}$, then we have an equality between minors:
$$B(I,J)=\pm\frac{A(J^c,I^c)}{\det A},$$
for every subsets $I,J\subset[[1,n]]$ of same cardinals. This formula generalizes that giving the entries of $A^{-1}$ in terms of minors of $A$. The $\pm$ sign is not essential to prove the stability of the TU class under inversion.<|endoftext|>
TITLE: What does a mathematician expect from mathematics education? 
QUESTION [7 upvotes]: Consider that my question is not a personal and/or subjective question. Perhaps, you have hired a mathematics educator in your department and you are interested in finding a way to communicate with him/her and vice versa. Or not, for some reasons you are looking at one of the writings of those peoples, and more often than not you get disappointed since you can hardly find what you are looking for. What do you expect from mathematics education as a discipline?
Addendum: After being revised by others it occured to me that my description had the potential of being interpreted as rude or offensive. Thus I decided to accept the revision done. Furthermore, to be less argumentative , it would be great to include in your answers/comments some good examples of the ideas from mathematics education that you have found useful, and perhaps an explanation of how they were useful for you (there are one or two below). I hope this would be a posstive way to go.

REPLY [20 votes]: I recall working with a reasonably reputable mathematics educator once, teaching a calculus class. At one point it became evident that the guy wasn't comfortable summing a geometric series. One thing I expect from mathematics education is that mathematics educators know some mathematics. Saying this any less bluntly would lose too much of the sentiment I'm trying to convey, so there it is.
As a mathematician, I'd personally like to see work in mathematics education that helps me teach what it is that I actually do as a mathematician. I was introduced by Ken Appel to some of the ideas of Hy Bass on mathematics education, e.g. the "granularity" concept which asserts that at different levels of sophistication mathematicians allow different jump sizes in their arguments. Awareness of granularity made explicit like this really has changed the way I organize my undergraduate course material and for me was revolutionary.
Other ideas of Bass that I'd like to see followed up include the idea of a common structure problem set. Such an idea might help to get a large chunk of a difficult aspect of the mathematical aesthetic into the curriculum. 
In general, I'd like to see mathematics education address how the practices of the best mathematicians can be brought to the graduate and undergraduate population in universities, and how we can bring more of mathematics itself as experienced by mathematicians to our students. I'm more interested in this than this than studies of how to improve calculus course assessment, for example. I've always been frustrated that nobody seems to study the learning approaches of successful mathematicians rather than average students. I'd personally like to see more of our best practices being studied and propagated. This last paragraph is a bit ignorant, I know, but it's my honest impression.<|endoftext|>
TITLE: Can every $\mathbb{Z}^2$ disk be pinball-reached?
QUESTION [32 upvotes]: Let every point of $\mathbb{Z}^2$ be surrounded by a mirrored disk of radius $r < \frac{1}{2}$,
except leave the origin $(0,0)$ unoccupied by a disk.

Q. Is it the case that every disk can be hit by a lightray emanating from the origin
  and reflecting off the mirrored disks?

Lightrays are composed of (infinitely thin) segments, and reflect off the disks
with angle of incidence equal to angle of reflection.
For example, here is one way (of many ways)
to hit the $(0,2)$ disk when $r = \frac{1}{4}$ with two reflections;
it clearly cannot be reached directly, with zero reflections:



I believe the answer to my question Q is Yes, but I would be
grateful for confirmation from the dynamical systems experts.
(Forgive me if I have not learned sufficently from my previous, related question, 
"Pinball on the infinite plane.")
It occurs to me it might be interesting to color the disks according to the minimum number
of reflections needed to hit each...

REPLY [3 votes]: Permit me to draw attention to a new expository survey related to this question:

Alex Wright. "From rational billiards to dynamics on moduli spaces." Apr. 2015.
  (arXiv abstract.)

He discusses the "Wind Tree" model, also known as the Ehrenfest model,
which is essentially the model I was considering, but with square (or rectangular)
"trees" rather than disk trees:<|endoftext|>
TITLE: Which hard mathematical problems do you have to solve to earn bitcoins ?
QUESTION [17 upvotes]: A virtual currency called bitcoins has been in the news recently.  It is said that in order to "mine" bitcoins, you have to solve hard mathematical problems.
Now, there are two kinds of mathematical problems.  The difference is best explained by the following beautiful quotation from Langlands :

[T]here is an appealing fable that I learned from the mathematician Harish-Chandra, and that he claimed to have heard from the French mathematician Claude Chevalley. When God created the world, and therefore mathematics, he called upon the Devil for help.  He instructed the Devil that there were certain principles, presumably simple, by which the Devil must abide in carrying out his task but that apart from them, he had free rein.  Both Chevalley and Harish-Chandra were, I believe, persuaded that their vocation as mathematicians was to reveal those principles that God had declared inviolable, at least those of mathematics for they were the source of its beauty and its truths.  They certainly strived to achieve this. If I had the courage to broach in this paper genuine aesthetic questions, I would try to address the implications of their standpoint.  It is implicit in their conviction that the Devil, being both mischievous and extremely clever, was able, in spite of the constraining principles, to create a very great deal that was meant only to obscure God’s truths, but that was frequently taken for the truths themselves.  Certainly the work of Harish-Chandra, whom I knew well, was informed almost to the end by the effort to seize divine truths.

Question  Which kind of mathematical problems do you have to solve in order to mine bitcoins ?
Let me clarify that I'm not interested in mining, only in knowing whether the problems are divine or devilish.

REPLY [5 votes]: Another way to earn bitcoin is not to mine them, but to have them wired to you from other bitcoin accounts. The transactions are signed using ECDSA. So if you solve the discrete logarithm problem in an elliptic curve (for bitcoin this is the curve Secp256k1), then you potentially can earn many bitcoins (illegally of course...)
Personally I find this problem much more beautiful than finding lot of zeroes in the SHA-256 hash function.<|endoftext|>
TITLE: Will quantum computing kill cryptography ?
QUESTION [14 upvotes]: I apologize as this question is not really mathematical, and therefore perhaps not
well-suited for this site. Please feel free to close it if you think it is not. My reason
for asking it here is that I am not satisfied (that is not convinced in any sense)
by many discussions relative to that question I have seen on various forums (in particular some devoted to bitcoins),
So, the basic fact is that currently used method of cryptography, based on prime factorizations or elliptic curves, would not work anymore when a quantum computer is available, as proved by Shor. My main question is:

Do we know other cryptographic algorithms that would work in a world were quantum computers
  exist ?

If yes, will they be easy to implement quickly when quantum computers appear ? If not, are people working on this ? have we hope to find such algorithms anytime soon ? Is there some theoretical obstruction to the existence of such algorithm ?

To be honest, let me give more argument to close the question by indicating my motivation for asking this question, which is not mathematical. I am curious about the real-world implications of quantum-computers in particular on bitcoins. Cryptography is currently used in many transaction using real-world currencies and, by design, in all transactions using bitcoins. If cryptography became unusable because of the appearance of quantum computers,
either for ever or for a sufficiently long period of time (in years), this 
would likely have enormous implication on the economy and the real world. 
To be sure, real money have worked for centuries without cryptography and if needed, 
one could go back to this. But cryptography and anonymity seems embedded in bit coin in a fundamental way, so would the appearance of quantum computers doom bitcoin?

REPLY [13 votes]: There is a very good book that you can find your answer there completely. This book's name is:
"Post-Quantum Cryptography" by "Daniel J. Bernstein, Johannes Buchmann and Erik Dahmen". 
As a part of this book, today we know that these cryptosystems can be broken by quantum computers:
$1)$ RSA public key encryption
$2)$ Diffie-Hellman key-exchange
$3)$ Elliptic curve cryptography
$4)$ Buchmann-Williams key-exchange
$5)$ Algebraically Homomorphic
and these cryptosystems (and also with some variants) are safe:
$1)$ McEliece public key encryption
$2)$ NTRU public key encryption
$3)$ Lattice-based public key encryption
Also, the good cryptosystems is not usable today because of the storage space problem and complexity. We have some limit on quantum computers that help us to design some good cryptosystems. There are some problems that if we have very large quantum computer and the best quantum algorithm,still we need exponential time for solving them. For example, searching among very large database to find special data, is very hard problem for quantum computer. We can prove that if we have $N$ cases that there is only one suitable case, the best quantum algorithm need $O(N^\frac{1}{2})$ to solve it. Also, it is proved that there is not better result. So, we can hope that we can find some efficient algorithms against the power of quantum computer and quantum algorithms.

REPLY [7 votes]: There is a web site and conference series on post-quantum cryptography, leading up to Bernstein et al's book mentioned by Shahrooz.  See:

http://pqcrypto.org/<|endoftext|>
TITLE: Examples of applications of the Freyd-Mitchell embedding theorem.
QUESTION [11 upvotes]: The Freyd-Mitchell embedding theorem states the following:

Let $\mathcal{A}$ be a small abelian category. There exists a unital ring $R$ and a full, faithful and exact functor $F\colon\mathcal{A}\rightarrow R\mathcal{Mod}$, where $R\mathcal{Mod}$ is the (abelian) category of (left) $R$-modules.

I am searching for (elementary and non-elementary, useful and non-useful) applications of this theorem. I am also interested in applications, for which one does not necessarly need the embedding theorem.
The only "real" application I know at the moment is all about diagram chasing: 

Checking exactness of diagrams in arbitrary abelian categories can be done by checking those in the categories of $R$-modules, where one can apply element-wise diagram chasing. 

(This can be made precise, what is e.g. done in Freyd's "Abelian Categories - An Introduction to the Theory of Functors".)

REPLY [13 votes]: An interesting class of examples comes from the following construction, also due to Freyd.  Let $\mathcal{T}$ be a small triangulated category.  Form a new category $\mathcal{A}$, with one object $I(u)$ for each morphism $u$ in $\mathcal{T}$.  The morphisms from $I(u:A\to B)$ to $I(v:C\to D)$ are the pairs $(f:A\to C,g:B\to D)$ such that $gu=vf$, modulo those for which $gu=vf=0$.  There is a functor $J:\mathcal{T}\to\mathcal{A}$ given by $J(X)=I(1_X)$.  One can show that $\mathcal{A}$ is abelian, that $J$ is full and faithful, and that the essential image of $J$ is the subcategory of projective objects, which is the same as the subcategory of injective objects.
I think that the embedding theorem is much less obvious for these abelian categories than it is for the more usual examples.
As a special case, we can take $\mathcal{T}$ to be the category of finite spectra in the sense of stable homotopy theory.  The embedding theorem, combined with the above construction, tells us that $\mathcal{T}$ embeds in the category of $R$-modules for some $R$ (which is not unique).  Freyd also conjectured more specifically that in this case the stable homotopy functor gives a full and faithful embedding of $\mathcal{T}$ in the category of modules over the ring of stable homotopy groups of spheres (and he proved that many interesting consequences would follow from that).  This conjecture is still wide open half a century later.<|endoftext|>
TITLE: Does a nonlinear additive function on R imply a Hamel basis of R?
QUESTION [5 upvotes]: A function is additive if $f(x+y) = f(x) + f(y)$.  Intuitively, it might seem that an additive function from R to R must be linear, specifically of the form $f(x) = kx$.  But assuming the axiom of choice, that is wrong, and the proof is rather simple: you just take a Hamel basis of $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and then you define your function f to be different in at least two distinct elements of the basis.
But my question is this: if there is no Hamel basis of $\mathbb{R}$, then must $f$ be linear?  To put it another way, does ZF + the existence of a nonlinear additive function imply the existence of Hamel basis of $\mathbb{R}$?
I checked the Consequences of the Axiom of Choice Project, a database of choice axioms and their relationships here, and it said that it didn't know.

REPLY [13 votes]: To my knowledge this is an open problem.
If one looks at Herrlich The Axiom of Choice, there is a diagram (7.23, p. 156) of implications related to non-measurable sets (which include discontinuous solution to the Cauchy functional equation problem), one can see that this is pretty far down below the existence of a Hamel basis.
Had it been known to be equivalent, an arrow back would be there -- and the book is not that old.
I doubt Herrlich would have omitted a reference to such fact, had it been known, and the last time I looked around, I couldn't find anything newer which proved anything related to that (and I did look around several times when answering a couple of questions).<|endoftext|>
TITLE: Upper bound on order of finite subgroups of GL_n(Z_p)?
QUESTION [9 upvotes]: Fix a prime $p$ and integer $n>1$, along with the ring $R$ of integers in a finite extension of the field $\mathbb{Q}_p$ (for example $R = \mathbb{Z}_p$).  

Is there an upper bound $C(n,p)$ on the orders of finite subgroups of $\mathrm{GL}_n(R)$?  Or can finite subgroups be arbitrarily large?

Probably this question has been resolved one way or the other in the literature but I don't recall a relevant source.

REPLY [10 votes]: Yes, for any fixed $p$-adic field $K$ the supremum of orders of finite subgroups of $\operatorname{GL}_n(K)$ is finite and can be explicitly bounded above.  There is a beautiful discussion of this tucked away somewhere in Serre's Lie Algebras and Lie Groups.  (What I say in the following is almost entirely derived from that, so you would do at least as well just to go back to the source.)
The first observation is that the answer is the same or $K$ as for its ring of integers $R$: a simple compactness argument shows that any finite subgroup of $\operatorname{GL}_n(K)$ can be conjugated into $\operatorname{GL}_n(R)$.  Then the idea is to show that for sufficiently large $k$, there is no torsion in the kernel of the natural surjective map $\operatorname{GL}_n(R) \rightarrow \operatorname{GL}_n(R/\pi^k R)$: this comes down to analysis of torsion in formal groups, which explains why it shows up in Serre's LALG.  
In fact Chapter 4 of these notes is devoted to precisely this and closely related questions, for instance including a proof of Selberg's Theorem: if $K$ is any field of characteristic zero and $G$ is a finitely generated subgroup of $\operatorname{GL}_n(K)$, then $G$ has a finite index torsionfree subgroup.  But the proof here is kind of antithetical to 
your precise question: one reduces to the case in which $K$ itself is finitely generated and applies a theorem of Cassels: any such field can be embedded in $\mathbb{Q}_p$ for some odd $p$.  
My notes give (standard) explicit upper bounds in the case of $\operatorname{GL}_n(\mathbb{Q}_p)$. 
Okay, but wait: I know how to do the general case too.  You want to combine the above arguments with:

Proposition: Let $K$ be a finite extension of $\mathbb{Q}_p$ with ramification index $e$, and let $R$ be its ring of integers, with maximal ideal $\mathfrak{m}$.  Let $F(X,Y)$ be any formal group law (of any finite dimension; here $X$ and $Y$ are vector variables) over $R$, with associated "standard" $K$-analytic Lie group $G^1 = F(\mathfrak{m})$.  Then the exponent of any 
  finite subgroup of $G^1$ divides $p^{\gamma_p(e)}$, where for any $m \in \mathbb{Z}^+$, $\gamma_p(m) = \lfloor \log_p \left( \frac{pm}{p-1} \right) \rfloor $.

(This is Proposition 9 from this paper of myself and Xavier Xarles.  It was well known to just about anyone who had worked in the area, but we couldn't find it in the literature, and in fact our paper was cited at least once for precisely this result.)
So I think that this does exactly what you want: let me know if I'm mistaken.  
Finally, I find it striking that the answer is completely different for local fields of positive characteristic: it is not possible to bound torsion in formal groups in this setting -- somehow $e = \infty$ in the above setup -- and Selberg's Theorem is false.  If I am not mistaken, there are indeed arbitrarily large finite subgroups of $\operatorname{GL}_n(\mathbb{F}_q((t)))$ for all prime powers $q$ and all $n \geq 2$.<|endoftext|>
TITLE: How we do actually compute the topological index in Atiyah-Singer?
QUESTION [24 upvotes]: This is migrated by math.stackexchange as I did not receive an answer. I do not know if it is too naive for this site. 
I am taking a lectured class in Atiyah-Singer this semester. While the class is moving on really slowly (we just covered how to use Atiyah-Singer to prove Gauss-Bonnet, and introducing pseudodifferential operators), I am wondering how practical this theorem is. The following question is general in nature:
Suppose we have a PDE given by certain elliptic differential operator, how computable is the topological index of this differential operator? If we give certain boundary conditions on the domain (for example, the unit circle with a point removed, a triangle, a square, etc), can we extend the $K$-theory proof to this case? I know the $K$-theory rhetoric proof in literature, but to my knowledge this proof is highly abstract and does not seem to be directly computable. Now if we are interested in the analytical side of things, but cannot compute the analytical index directly because of analytical difficulties, how difficult is it to compute the topological index instead? It does not appear obvious to me how one may compute the Todd class or the chern character in practical cases. 
The question is motiviated by the following observation: 
Given additional algebraic structure (for example, if $M$ is a homogeneous space, $E$ is a bundle with fibre isomorphic to $H$) we can show that Atiyah-Singer can be reduced to direct algebraic computations. However, what if the underlying manifold is really bad? What if it has boundaries of codimension 1 or higher?How computable is the index if we encounter an analytical/geometrical singularity?(which appears quite often in PDE). 
On the other hand, suppose we have a manifold with corners and we know a certain operator's topological index. How much hope do we have in recovering the associated operator by recovering its principal symbol? Can we use this to put certain analytical limits on the manifolds(like how bad an operator on it could be if the index is given)?

REPLY [18 votes]: This isn't really going to be an answer, but it's too long to be a comment and I think it will be helpful.
First, as the other answerers pointed out it is a good idea to avoid boundary value problems and singularities at first.  The usual index formula is not correct in this setting (even once you manage to formulate it correctly): there is a mysterious error term called the "eta invariant."  The eta invariant is generally even harder to compute than the Todd class because it is non-local and it depends on a choice of Riemannian metric.  It is defined in terms of the eigenvalues of your operator, and actually working these out requires quite a lot of symmetry so that you can do harmonic analysis.
Second, I don't think the real point of the index theorem is to numerically compute all of the integrals involved (though it's nice that you can in principle).  It is often useful enough just to know that it is even possible to compute the index in terms of local data.  For instance, this immediately implies that any invariant which can be expressed as the index of an operator is multiplicative under coverings, something which might not always be obvious.  Also, if you understand the geometric structures which produced your operator then you can often refine the Atiyah-Singer integrand; for instance, it is useful to know that the Euler characteristic is the integral of curvature terms and that the signature of a 4-manifold is the integral of Pontryagin classes.
Third, in applications the goal is often to find geometric reasons why the index of an operator is $0$, and sometimes you can do this even if you don't know exactly what the integrand looks like.  This sometimes happens in Seiberg-Witten theory, for instance, where the index of the Dirac operator tells you the dimension of a certain moduli space and thus you're pretty happy if it happens to be $0$.
Finally, I would argue that the actual cohomological formula for the index is sort of beside the point.  The index theorem should really be viewed as a version of Poincare duality in K-theory, and all those messy characteristic classes are the price we pay for trying to stuff the theorem into ordinary cohomology.  Working with index invariants at the level of K-theory is usually much easier and you can pick up on much more refined index invariants that don't necessarily have a local formula (such as the Clifford index in real K-theory).<|endoftext|>
TITLE: Tannaka duality for C*-algebras?
QUESTION [10 upvotes]: Tannaka-Krein
duality shows
how to recover a group $G$ from its category $\mathbf{Rep}(G)$ of finite-dimensional
complex representations and the forgetful functor $F:\mathbf{Rep}(G)\to
\mathbf{Vect}_{\mathbb{C}}$.
On the other hand, there is Takesaki's
theorem
which shows how to recover a (separable) $C^*$-algebra $A$ from its representation theory. We
have just started looking into the details of this and trying to reformulate this in categorical
terms. It seems tantalizingly similar to the Tannaka-Krein reconstruction
theorem. In particular, it seems that Takesaki secretly also considers natural
transformations from the forgetful functor $F:\mathbf{Rep}(A)\to\mathbf{Hilb}$ to itself.
So, the question is:

Can Takesaki's duality theorem indeed be formulated in categorical terms similar to Tannaka-Krein duality? Where can we read about it?

We're mostly interested in the unital case. Follow-up question:

Consider the category of unital $C^*$-algebras and unital completely positive maps. Is there a "nice" description of its opposite as a concrete category?

REPLY [10 votes]: This isn't an answer but a long comment. Tannaka-Krein duality will mislead you about how hard you should expect this result to be. "Tannaka-Krein duality for rings" is actually very easy and looks like this. 
Theorem: Let $R$ be a ring. Then $R$ is isomorphic to the endomorphism ring of the forgetful functor $\text{Mod-}R \to \text{Ab}$. (Recall that the category of functors from a category into an $\text{Ab}$-enriched category is $\text{Ab}$-enriched, so endomorphism monoids are rings.)
Proof. This functor is represented by the right $R$-module $R_R$, so its endomorphism ring is the endomorphism ring of $R_R$ by the Yoneda lemma. By a second application of the Yoneda lemma (!), the endomorphism ring of $R_R$ is $R$. $\Box$
(And of course if you forget the forgetful functor then you cannot recover $R$ at all, only its Morita equivalence class.) 
"Set-theoretic Tannaka-Krein duality for monoids" is also very easy and says that a monoid $M$ is isomorphic to the endomorphism monoid of the forgetful functor $\text{Set-}M \to \text{Set}$; the proof is identical. Tannaka-Krein duality itself is hard because we are considering linear representations of groups, which a priori should only let us at best recover the group algebra; we need more structure (e.g. the tensor product) to recover the group itself.<|endoftext|>
TITLE: Connection between codata and greatest fixed points
QUESTION [11 upvotes]: This is, I'm afraid, another question that MSE couldn't answer.
It's easy to see how inductively-defined data types correspond to least fixed points. Let's take the natural numbers as an example, whose constructors are $0 : \mathbb N$ and $s : \mathbb N \to \mathbb N$. Define the operation $F(X) = \{0\} \cup \{ s(n) : n \in X \}$, which applies the constructors to all elements of $X$. The Knaster–Tarski fixed point theorem says that the least fixed point of $F$ is $\bigcap\{ X : F(X) \subseteq X \}$.
What does that set mean? Well, it's the intersection of all sets such that applying the constructors doesn't give you anything new. So, since applying the constructors throws in $0$, $0$ must be in there. Since applying the constructors throws in every successor, every successor must be in there. The minimality property ensures that there are no more elements in there, so induction holds for this set. So it's the natural numbers.
Okay, great. Something similar is going to work for other inductive datatypes too. This indicates we can consider inductively defined datatypes as $\subseteq$-least fixed points for the set operation that applies all the constructors.
But what about coinductive data types? I've heard it said, though I'm not sure anywhere reliable, that codata corresponds to greatest fixed points. Let's go look at Knaster-Tarski again: the greatest fixed point is $\bigcup \{ X : X \subseteq F(X) \}$. What does this tell us? Well, promisingly, it says that every element is either $0$ or a successor. Certainly all of the natural numbers are in this greatest fixed-point. From what I know about coinduction, I'm expecting to get the natural numbers plus a unique fixed point of the successor.
But why should the successor have a unique fixed point? Why can't we have two, violating coinduction? Certainly this depends on what $s$ is, precisely, but all we asked of $s$ when we were doing the inductive definition was that it was injective ("free" in some sense), which feels like a much lighter condition.
How do I ensure the right number of fixed points for the constructors? How do I recover the coinduction principle? Or is it just that the order for which these constructions are LFPs and GFPs respectively isn't subset inclusion?

REPLY [4 votes]: Following Andrej's recommendation in comments, I went and looked up Vicious Circles by Barwise and Moss. There, codata can indeed be modelled by greatest fixed points in the subset-inclusion order, but working in a set theory with an antifoundation axiom.
In particular, we have (for example) that the greatest fixed point in set-theory-with-foundation for the $F$ I gave above (with $s$ interpreted as ordinal successor) is $\omega$, which is lacking a unique fixed point for $s$, but with a suitable antifoundation axiom it is $\omega \cup \Omega$ where $\Omega$ is the unique solution to $x = \{x\}$, so also a fixed point of $s$.
The antifoundation axiom is playing a key role here: not only does it ensure the existence of the necessarily ill-founded objects that accompany codata, they also include a uniqueness condition so that coinduction and corecursion can be properly defined. If I have that every element of the "conatural numbers" is either zero or the successor of a conatural number, I can recursively unfold this into a set picture which will necessarily be some chain of successor applications that is either infinite or ends in 0 – crucially, by antifoundation, the length of the chain determines uniquely the identity of the element, just as required by the coinduction principle.
The conclusion is that greatest fixed points may or may not exist in various contexts, but it's the antifoundation axiom which ensures that they are the right thing with regards to coinduction and corecursion principles.<|endoftext|>
TITLE: How many distinct eigenvalues does a random graph have?
QUESTION [5 upvotes]: It is well-known that a random graph a.e. has diameter 2. It is also well-known that the number of distinct eigenvalues of a graph is at least the diameter plus one.
But what is known about the expected number of distinct eigenvalues of a random graph?

REPLY [12 votes]: In this recent paper of Erdos, Knowles, Yau, and Yin, it is shown that in the bulk of the spectrum, the spacing between eigenvalues of an Erdos-Renyi graph on $n$ vertices obeys GOE statistics asymptotically.  This implies that most of the eigenvalues are simple (i.e. $n-o(n)$ of the $n$ eigenvalues are simple) asymptotically almost surely, so that the number of distinct eigenvalues is $n-o(n)$ a.a.s..  It is very likely that in fact a.a.s. all of the $n$ eigenvalues are simple; Van Vu and I have some preliminary unpublished results in this direction but we are still working on the full problem.<|endoftext|>
TITLE: Internal Day convolution
QUESTION [7 upvotes]: Let me recall that any small category $\mathbb{A}$ enriched in a complete and cocomplete symmetric monoidal closed category $\mathbb{V}$ admits embedding (the Yoneda embedding):
$$y_\mathbb{A} \colon \mathbb{A} \rightarrow \mathbb{V}^{\mathbb{A}^{op}}$$
into a complete and cocomplete $\mathbb{V}$-enriched category. And:

this embedding preserves any monoidal structure $\langle \otimes_\mathbb{A}, I_\mathbb{A} \rangle$ defined on $\mathbb{A}$; the induced structure $\mathbb{V}^{\mathbb{A}^{op}} \otimes \mathbb{V}^{\mathbb{A}^{op}} \rightarrow \mathbb{V}^{\mathbb{A}^{op}}$ is given by the Day convolution:
$$\langle F, G \rangle \mapsto \int^{B \in \mathbb{A}, C \in \mathbb{A}} F(B) \otimes G(C) \otimes \hom(-, B \otimes_\mathbb{A} C)$$
the induced structure on  $\mathbb{V}^{\mathbb{A}^{op}}$ is always monoidal (bi)closed
more generally, any promonoidal structure (i.e. a weak monoidal object in the bi-category of $\mathbb{V}$-enriched distributors) defined on $\mathbb{A}$ induces a (bi)closed monoidal structure on $\mathbb{V}^{\mathbb{A}^{op}}$; moreover there is a bijective correspondence between (bi)closed monoidal structures defined on $\mathbb{V}^{\mathbb{A}^{op}}$ and promonoidal structures defined on $\mathbb{A}$

It seems (though I have not checked details) that all of the above carry to the context of categories internal to any finitely cocomplete locally cartesian closed category $\mathbb{C}$. For any $\mathbb{C}$-internal category $\mathbb{A}$ there exists a fibred fully faithful embedding:
$$y_\mathbb{A} \colon \mathbb{A} \rightarrow \mathbb{C}^{\rightarrow^{\mathbb{A}^{op}}}$$
where $\mathbb{C}^{\rightarrow^{\mathbb{A}^{op}}}$ is the (complete and cocomplete cartesian closed) fibration of $\mathbb{A}^{op}$-indexed diagrams in the fundamental (i.e. codomain) fibration on $\mathbb{C}$.
This observation is so blatantly obvious that it must have been made (and written somewhere) before. What are the references?
Much of the work has been done in "Cosmoi of Internal Categories" by Ross Street. However, I have not found any statement about transporting monoidal structures from $\mathbb{A}$ to $\mathbb{C}^{\rightarrow^{\mathbb{A}^{op}}}$ in the paper.

REPLY [2 votes]: As you say, I expect that this has been known to experts for a long time, but it follows as a special case of Theorem 11.22 in my paper Enriched indexed categories, since when $V$ is the self-indexing of $S$, small $V$-categories include internal $S$-categories, and large $V$-categories are essentially fibrations over $S$.<|endoftext|>
TITLE: H^d[U(1)^n,U(1)] of the Borel cohomology and Chern-Simons theory
QUESTION [9 upvotes]: Firstly I apologize that I am a physicist, with a relatively unrigorous math training. My approach of the problem can be Feynman style. Below $Z$ is the integer $\mathbb{Z}$, and $U(1)$ Abelian group is the same as $\mathbb{R}/\mathbb{Z}$. gcd stands for greatest common divisor. And $Z_1$ is the same as the group $0$.
The question is about $H^d[U(1)^n,U(1)]$ of the Borel cohomology. My questions have two parts (Q1) and (Q2).
My first question (Q1) is whether my guess below is correct:
$
\begin{cases}
H^3[U(1),U(1)]=Z,  \newline
H^3[U(1)\times U(1),U(1)]=(Z)^3 \text{(to be checked)} \newline
H^3[U(1)\times U(1)\times U(1),U(1)]=(Z)^7 \text{(to be checked)} \newline
\end{cases}
$
My second question (Q2) is that the above result seems to be inconsistent with the `universal coefficient theorem' and some facts about $H^d[U(1)^n,Z]$ and $H^d[U(1)^n,U(1)]$.

(Q1)
To calculate $H^d[U(1),U(1)]$ directly from the algebraic definition is very tricky to me since $U(1)$ has infinite uncountable many
elements.
Here, I will use a unrigorous physical argument (if you go against it, fine, no problem, but I will say Richard Feynman may do this) to calculate it by first calculating
$H^d[ \Pi_i Z_{n_i},U(1)]$, and then let $n_i\to \infty$. Below I will use the unproven and unrigorous $\lim_{n\to \infty}Z_n=Z$ (which is not true in general).
I start with the known fact:
$
\begin{cases}
H^3(Z_n,U(1))=Z_n \newline
H^3(Z_n\times Z_m,U(1))=Z_n \times Z_m \times Z_{gcd(n,m)}  \newline
H^3(Z_n \times Z_m\times Z_o,U(1))=Z_n \times Z_m  \times Z_o \times Z_{gcd(n,m)}\times Z_{gcd(n,o)} \times Z_{gcd(m,o)} \times Z_{gcd(n,m,o)}
\end{cases}
$      
What I had obtained is:
$
\begin{cases}
H^3[U(1),U(1)]=Z,  \newline
H^3[U(1)\times U(1),U(1)]=Z\times Z\times Z=(Z)^3 \text{(to be checked)} \newline
H^3[U(1)\times U(1)\times U(1),U(1)]=Z\times Z \times Z \times Z \times Z \times Z \times Z=(Z)^7 \text{(to be checked)} \newline
\end{cases}
$
My first question is whether my result is correct (Not the method).

(Q2) My second question is that the above result seems to be inconsistent with the 
(a)`universal coefficient theorem' 
$
\begin{align}
\ \ \ \ H^d(X,M)
\\
\simeq  H^d(X,Z)\otimes_{Z} M \oplus
\text{Tor}_1^{Z}(H^{d+1}(X,{Z}),M)  ,
\end{align}$
and 
(b)the following known facts:
 $H^d[U(1),U(1)]=
\begin{cases}
U(1) & \text{ if } d=0, \newline
Z_1  & \text{ if } d=0 \text{ mod } 2,\ \ d>0\newline
Z  & \text{ if } d=1 \text{ mod } 2.
\end{cases}$
$H^d[U(1),Z]=
\begin{cases}
Z  & \text{ if } d=0 \text{ mod } 2,\newline
Z_1  & \text{ if } d=1 \text{ mod } 2.
\end{cases}
$
This shows that $H^d[U(1),U(1)]=H^{d+1}[U(1),Z]$.
The inconsistency is
\begin{eqnarray}
H^3(U(1),U(1)) &=&[H^3(U(1),Z) \otimes U(1)] \times \text{Tor}^Z_1[H^4(U(1),Z),U(1)] \newline
&=&[Z_1\otimes U(1)] \times \text{Tor}^Z_1[Z,U(1)]=Z_1 \times Z_1 =Z_1
\end{eqnarray}
so it contradicts to $H^3(U(1),U(1))=Z$.
Surprisingly, the inconsistency already happens to the known fact H3[U(1),U(1)]=Z  !
The same for $H^3(U(1)\times U(1),U(1))=Z_1$ instead of $(Z)^3$.
The question is why $H^3(U(1),U(1))=Z$ is inconsistent here from the known facts (a)(b). Also the inconsistency at other $H^3(U(1)^n,U(1))$ with $n=2,3,\dots$,etc.
PS. The fact that $\text{Tor}^Z_1[Z,U(1)]=Z_1$ seems to tell me something nontrivial contrary to the naive $\text{Tor}^Z_1[Z_n,U(1)]=Z_n$ at $n \to \infty$. Suppose that $\text{Tor}^Z_1[Z,U(1)] \to Z$ instead, everything seems to match. This (Q2) was the main reason why I had asked this silly question: https://mathoverflow.net/questions/128203/torsion-product-torr-1-closed

[new update on April 27, 2013]
I provide some connections between group cohomology and Chern-Simons theory in the 2nd answer below. And add a question:
(Q3)
whether there is a symmetry breaking picture, such that one can obtain the result of 
$H^3[\mathbb{Z}_p^n,U(1)]$ of a discrete $\mathbb{Z}_p^n$ group from a large continuous group, say, from $U(1)^n$ broken down to $\mathbb{Z}_p^n$?
So that, for example, this guessed $H^3[U(1)^n,U(1)]$ broken down to a subgroup picture works
\begin{equation}
H^3[U(1)^n,U(1)] (\text{guessed})\to H^3[\mathbb{Z}_p^n,U(1)]
\end{equation}
as 
\begin{equation}
\mathbb{Z}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)} (\text{guessed})\to \mathbb{Z}_p^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}
\end{equation}
On the other hand, we know the fact that however 
\begin{equation}
H^4(B(U(1)^n),\mathbb{Z}) \to H^3[\mathbb{Z}_p^n,U(1)]
\end{equation}
broken down from 
\begin{equation}
\mathbb{Z}^{n+\frac{1}{2}n(n-1)} \to \mathbb{Z}_p^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}
\end{equation}
does not match when the cohomology group breaks down from $\mathbb{Z}^{n+\frac{1}{2}n(n-1)} \to \mathbb{Z}_p^{n+\frac{1}{2}n(n-1)}$ by the gauge group of $U(1)^n$ Chern-Simons theory  breaks down to $\mathbb{Z}^n_p$ Chern-Simons theory. 

[new update on May 12, 2013]
I would like to address Mariano's inquiry below: The Cohomology I mean here is Borel Group Cohomology. See for example, http://arxiv.org/pdf/1106.4772v6.pdf, p.26, Appendix D: Group cohomology and p.44, Appendix J: Calculations of group cohomology, Sec 4. Some useful tools in group cohomology. 
See also: http://arxiv.org/pdf/1110.3304.pdf.
For Segal and Mitchison's $\mathcal{H}^d_{SM}(G; Z)$
These two Ref show: 
\begin{equation}
\mathcal{H}^d_{SM}(G; Z)=H^d(BG; Z),\;\;
\mathcal{H}^d_{SM}(G; U(1))=H^{d+1}(BG; Z)
\end{equation}
Also this question: The relationship between group cohomology and topological cohomology theories
If it helps, I can change all my $H^d[G,U(1)]$ above to $\mathcal{H}^d[G,U(1)]$

REPLY [3 votes]: Let me myself comment briefly the calculation of $H^d[U(1)^n,U(1)]$ from another angle, the Chern-Simons theory.
It is known that the relation between Chern-Simons action $CS(A)$:
\begin{equation}
e^{2\pi i CS(A)}=\exp\Bigl[ 2\pi i k   \int_{X_3} A \wedge dA\Bigr]
\end{equation}
as a partition function of $U(1)$ Chern-Simons action $CS(A)$ on a 3-manifold $X_3$.
For a compact gauge group $G$ the Chern-Simons actions are classified by
an element [Ref:D-W] 
\begin{equation}
k\in H^4(BG,\mathbb{Z})
\end{equation}
The  CS actions $CS(A)$ for 
a compact gauge group $G$ are in one-to-one correspondence
with the elements of the cohomology 
group $H^4 (B{ G}, \mathbb{Z})$ of the classifying space
$B{ G}$ with integer coefficients $\mathbb{Z})$. 
For $G=U(1)$, CS actions $CS(A)$ is classified by
\begin{equation}
H^4(B(U(1)),\mathbb{Z}) \cong \mathbb{Z} 
\end{equation}
, and the element $\mathbb{Z}$ is simply the  integer $k$ (level $k$)
appearing in the expression for the $U(1)$ Chern-Simons action on the 3-manifold $X_3$.
For $K_{n\times n}$ Chern-Simons action $CS(A)$ with $U(1)^n$ gauge group:
\begin{equation}
e^{2\pi i CS(A)}=\exp\Bigl[ 2\pi i K_{ij}   \int_{X_3} A_{i} \wedge dA_{j}\Bigr]
\end{equation}
For $G=U(1)^n$, CS actions $CS(A)$ is classified by
\begin{equation}
H^4(B(U(1)^n),\mathbb{Z}) \cong \mathbb{Z}^{n+\frac{1}{2}n(n-1)}
\end{equation}

Let's shortly sketch the proof of $H^4(B(U(1)^n),\mathbb{Z}) \cong \mathbb{Z}^{n+\frac{1}{2}n(n-1)}
$
\begin{equation}
H^d(BU(1),\mathbb{Z}) \simeq  \begin{cases}
                 \mathbb{Z} & \text{if $d\in$ even} \newline
                 0 & \text{if $d\in$ odd}
                 \end{cases}
\end{equation}
For $H^d(B(U(1)^n),\mathbb{Z})$ with $n>1$, we can use  the Kunneth
formula, because the classifying space of the product group 
$U(1)^n$ is the same as the product of the the factorization of classifying spaces. 
That is, $B(U(1)^n) = B(U(1)^{n-1}) \times BU(1)$.
The calculation is analogue to $H^3[\mathbb{Z}_p^n,U(1)]$ case.
It is easier to show that for discrete abelian group:
\begin{equation}
\begin{cases}
H^3[\mathbb{Z}_p,U(1)]=\mathbb{Z}_p,  \text{(confirmed)}\newline
H^3[\mathbb{Z}_p^2,U(1)]=(\mathbb{Z}_p)^3 \text{(confirmed)} \newline
H^3[\mathbb{Z}_p^3,U(1)]=(\mathbb{Z_p})^7 \text{(confirmed)} \newline
H^3[\mathbb{Z}_p^n,U(1)]=\mathbb{Z_p}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}  \text{(confirmed)} \newline
\end{cases}
\end{equation}
Since the group $\mathbb{Z}$ is torsion free, however, the terms 
due to torsion products $Tor$ vanish in this case (thanks to the discussion: https://mathoverflow.net/questions/128203/torsion-product-torr-1-closed).
To summarize, the contribution of $H^4(B(U(1)^n),\mathbb{Z})$, comes from 
$n$ terms of the form $H^4(BU(1))\simeq \mathbb{Z}$. These
label the different CS actions of diagonal $K_{n\times n}$ matrix type:
\begin{equation}
\end{equation}
\begin{equation}
{\sum_{i=1}^n {K_{(ii)}} \epsilon^{\kappa\sigma\rho} {A_\kappa^{(i)}} \partial_{\sigma} A^{(i)}_{\rho} }
\end{equation}
In addition, there are $\frac{1}{2} n(n-1)$ terms 
of the form $H^2(BU(1)) \simeq \mathbb{Z}$ which label
the CS actions of off-diagonal $K_{n\times n}$ matrix type: 
\begin{equation}
{\sum_{i\neq j}^n {K_{(ij)}} \epsilon^{\kappa\sigma\rho} {A_\kappa^{(i)}} \partial_{\sigma} A^{(j)}_{\rho} }
\end{equation}
So overall, $H^4(B(U(1)^n),\mathbb{Z}) \cong \mathbb{Z}^{n+\frac{1}{2}n(n-1)}
$ (q.e.d.)

Importantly, this classification includes the case of finite gauge groups $H$. 
The isomorphism for a finite gauge groups $H$.
\begin{equation}                           
H^d(B{H},{\mathbb{Z}}) \simeq H^d({H},{\mathbb{Z}})  ,
\end{equation}
Moreover, the universal coefficient theorem(UCThm) shows the isomorphism 
\begin{equation}
H^{d} (H, \mathbb{Z})   \simeq   H^{d-1} ({ H}, U(1))    
\qquad \forall  n>1.
\end{equation}
We have then:
\begin{equation}
H^4 ({BH}, \mathbb{Z})  \simeq  H^3 ({ H}, U(1)) 
\end{equation}

It is this relation: $\begin{equation}
H^4 ({BH}, \mathbb{Z})  \simeq  H^3 ({ H}, U(1)) 
\end{equation}$
allure me from this finite gauge groups $H$ result,
\begin{equation}
\begin{cases}
H^3[\mathbb{Z}_p,U(1)]=\mathbb{Z}_p,  \text{(confirmed)}\newline
H^3[\mathbb{Z}_p^2,U(1)]=(\mathbb{Z}_p)^3 \text{(confirmed)} \newline
H^3[\mathbb{Z}_p^3,U(1)]=(\mathbb{Z_p})^7 \text{(confirmed)} \newline
H^3[\mathbb{Z}_p^n,U(1)]=\mathbb{Z_p}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}  \text{(confirmed)} \newline
\end{cases}
\end{equation}
to prompt the guessed result of:
\begin{equation}
\begin{cases}
H^3[U(1),U(1)]=\mathbb{Z},  \newline
H^3[U(1)^2,U(1)]=(\mathbb{Z})^3 \text{(to be checked)} \newline
H^3[U(1)^3,U(1)]=(\mathbb{Z})^7 \text{(to be checked)} \newline
H^3[U(1)^n,U(1)]=\mathbb{Z}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}  \text{(to be checked)} \newline
\end{cases}
\end{equation}
However, the result
$
H^4(B(U(1)^n),\mathbb{Z}) \cong \mathbb{Z}^{n+\frac{1}{2}n(n-1)}
$
already tells me the right way to view this $U(1)^n$ gauge symmetry C-S theory should be classified by $H^4(B(U(1)^n),\mathbb{Z})$
as
\begin{equation}
\begin{cases}
H^4(B(U(1)),\mathbb{Z})=\mathbb{Z},  \newline
H^4(B(U(1)^2),\mathbb{Z}) \cong \mathbb{Z}^{3}  \newline
H^4(B(U(1)^3),\mathbb{Z}) \cong \mathbb{Z}^{6} \newline
H^4(B(U(1)^n),\mathbb{Z}) \cong \mathbb{Z}^{n+\frac{1}{2}n(n-1)} \newline
\end{cases}
\end{equation}
This is what I should ask in (Q1).
So, instead, what I should really ask is, apart from (Q1)(Q2):
(Q3)
whether there is a symmetry breaking picture, such that one can obtain the result of 
$H^3[\mathbb{Z}_p^n,U(1)]$ of a discrete $\mathbb{Z}_p^n$ group from a large continuous group, say, from $U(1)^n$ broken down to $\mathbb{Z}_p^n$?
So that, for example, this guessed $H^3[U(1)^n,U(1)]$ broken down to a subgroup picture works
\begin{equation}
H^3[U(1)^n,U(1)] (\text{guessed})\to H^3[\mathbb{Z}_p^n,U(1)]
\end{equation}
as 
\begin{equation}
\mathbb{Z}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)} (\text{guessed})\to \mathbb{Z}_p^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}
\end{equation}
We know that however 
\begin{equation}
H^4(B(U(1)^n),\mathbb{Z}) \to H^3[\mathbb{Z}_p^n,U(1)]
\end{equation}
broken down from 
\begin{equation}
\mathbb{Z}^{n+\frac{1}{2}n(n-1)} \to \mathbb{Z}_p^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}
\end{equation}
This does not work, where we cannot simply replacing  $\mathbb{Z}$ to $\mathbb{Z}_p$ by symmetry breaking from $U(1)$ to $\mathbb{Z}_p$. It has been known that one may need a symmetry breaking of C-S action with a nonAbelian continuous group broken down to subgroup $Z_p^n$ to produce the all known elements of $H^3[\mathbb{Z}_p^n,U(1)]$. 
So that goes back to my guessed proposal:
\begin{equation}
\begin{cases}
H^3[U(1),U(1)]=\mathbb{Z},  \newline
H^3[U(1)^2,U(1)]=(\mathbb{Z})^3 \text{(to be checked)} \newline
H^3[U(1)^3,U(1)]=(\mathbb{Z})^7 \text{(to be checked)} \newline
H^3[U(1)^n,U(1)]=\mathbb{Z}^{n+\frac{1}{2}n(n-1)+\frac{1}{3!}n(n-1)(n-2)}  \text{(to be checked)} \newline
\end{cases}
\end{equation}
PS. I have to apologize what I had mentioned may be intriguing, the question turns out to overlap different math fields. Instead of directly answering the questions (Q1)(Q2), I now address the questions differently.

Ref:
[Ref:D-W]:Robbert Dijkgraaf, Edward Witten, Topological Gauge Theories and Group Cohomology, Commun. Math. Phys. 129 (1990), 393<|endoftext|>
TITLE: How unique are extensions of TQFTs to lower dimension?
QUESTION [13 upvotes]: Say I have an "ordinary" TQFT $F$ of dimension $n$, assigning groups or vector spaces to closed $(n-1)$-manifolds and linear maps to cobordisms. Consider the different ways $F$ can be obtained from a TQFT "extended one step" which assigns categories to manifolds of dimension $n-2$ (often derived categories of algebras or dg / $A_{\infty}$ algebras).
Is there expected to be any uniqueness to these extensions of $F$? For example, are there cases where you can extend the same $F$ two ways, but the corresponding (derived) categories associated to a codimension-2 manifold aren't equivalent?
This question is motivated partly by the situation in bordered Heegaard Floer homology; the dg algebra associated to a surface depends on a choice of parametrization, but these distinct algebras end up having equivalent derived categories (of type D or type A modules).
I'd be happy with a toy example in lower dimensions or anything illustrating this uniqueness holding or not holding- maybe in some restricted context. I'd also be curious to know if there's an $F$ which doesn't extend at all (maybe this is basic knowledge for the experts...)
Thanks!

REPLY [11 votes]: The question of which tqfts extend is a very interesting one. To make the question more mathematically precise, we can fix the target n-categories and ask for the tqfts to extend with respect to those targets. Then I can give precise answers.  
In general there are both existence and uniqueness issues, even in the n=2 case. That case is pretty instructive, and it doesn't get much simpler by increasing dimensions. 
It is well known that a 2D non-extended (oriented) tqft in vector spaces is the same thing as a commutative Frobenius algebra. Now we can ask that this "extends to points" using our favorite target 2-category, like linear categories or the 2-category of algebras, bimodules, and maps. There turns out to not be much difference between these two choices, so I will work with the latter. If you prefer the former, than you should just think of the category of modules associated to the algebra. It doesn't really effect what I am going to say.
Now a 2D extended (oriented) tqft in algebras, bimodules, and maps is the same thing as a non-commutative symmetric Frobenius algebra which is fully-dualizable. Over a perfect field fully-dualizable is the same as finite dimensional and semisimple. 
Now if A is such a fully-dualizable Frobenius algebra, then the center Z(A) will be a commutative Frobenius algebra and this is the algebra corresponding the the non-extended part of the 2D TQFTs. Since A is semisimple, this commutative algebra is semisimple. 
Thus, not every 2D TQFT extends to points (at least with the usual target categories). An explicit counter example is given by the non-semisimple commutative Frobenius algebra $k[x]/x^{n+1}$, where the trace is given by picking off the $x^n$-coefficient. 
Moreover we also see that uniqueness is an issue. For example consider working over the real numbers. Then the algebra $\mathbb{R}$, (with trivial trace) is a semisimple Frobenius algebra. The center is, of course, also $\mathbb{R}$. But we also have the quaternion algebra $\mathbb{H}$, which is also a semisimple Frobenius algebra (the trace is projection onto the real line in $\mathbb{H}$). The center of $\mathbb{H}$ is also $\mathbb{R}$, and so this gives an example of two extended tqfts which have the same underlying non-extended 2D tqft. Note that $\mathbb{H}$ and $\mathbb{R}$ are not Morita equivalent, and so in particular these are genuinely different extended tqfts. 
Different extended 2D tqfts can have the same underlying non-extended 2D tqft.<|endoftext|>
TITLE: Embeddings of finite groups into GL(n,Q_p)
QUESTION [11 upvotes]: This question is inspired by some interesting comments on this recent question.
Fix an integer $n \geq 1$ and a finite subgroup $G$ of $\mathrm{GL}_n(\mathbf{C})$. It is known that there are infinitely many primes $p$ such that $G$ embeds into $\mathrm{GL}_n(\mathbf{Q}_p)$. This follows from the Cassels Embedding Theorem : every finitely generated field of characteristic 0 embeds into $\mathbf{Q}_p$ for infinitely many primes $p$ (see Cassels's book Local Fields or Cassels's article An embedding theorem for fields. Bull. Austral. Math. Soc. 14 (1976), 193-198; see also Chapter 4 of Pete L. Clark's notes). 
Representation theoretic arguments actually show that $G$ occurs in $\mathrm{GL}_n(\overline{\mathbf{Q}})$ and thus in $\mathrm{GL}_n(K)$ for some number field $K$. From this and Chebotarev's density theorem, we deduce that the set $S(G)$ of primes $p$ such that $G \hookrightarrow \mathrm{GL}_n(\mathbf{Q}_p)$ has positive density.
Does the set $S(G)$ have a natural density?
Is the set $S(G)$ a Galoisian set of prime numbers in the sense of this question?

REPLY [9 votes]: $\def\Gal{\mathrm{Gal}}$
$\def\Res{\mathrm{Res}}$
$\def\GL{\mathrm{GL}}$
$\def\F{\mathbf{F}}$
$\def\Q{\mathbf{Q}}$
Edited to include more details.
Let $K$ be a field whose characteristic is prime to the order of $G$.
The algebra $K[G]$ is a product of matrix algebras over division rings.
Given any absolutely irreducible character $\chi$ with coefficients in $L/K$, the question of whether $\chi$ may be realized over $L$ is equivalent to whether the corresponding division algebra $D/K$ splits over $L$.
Suppose that $K = k$ is a finite field. Then, by Weddeburn's theorem, there are no non-trivial division algebras and irreducible characters with values in $k$ have models over $k$.
Suppose that $K$ is a number field. For all but finitely many places $v$ of $K$, $D \otimes_K K_v$ is a matrix algebra. So, at least away from finitely many places, irreducible characters with values in $K$ have models over $K_v$.
Example: Let $G = Q_8$ be the quaternion group, and let $\chi$ denote the absolutely irreducible faithful character of degree $2$. Then $\chi$ is valued in $\Q$. However, the corresponding quaternion algebra $D$ is ramified at $2$ and $\infty$. Hence $G$ does not have a faithful representation over either $\mathbf{R}$ nor $\mathbf{Q}_2$, but it does for $\Q_p$ and all odd primes $p$, because $D \otimes \Q_p = M_2(\Q_p)$ in those cases.
In the example above, we see that the obstruction arises for finite places only at the prime $2$ which divides $|G|$. Let us show that there is no obstruction for any $v$ such that the residue characteristic is finite and prime to $|G|$. In such cases, there is a bijection between absolutely irreducible characters in characteristics zero and $p$ given by reduction modulo $p$. Let $\chi$ be an irreducible character with values in a local field $E$. It preserves a lattice $\mathcal{O}$, and gives rise to an irreducible character over the finite field $k$, which has a model over $k$ by the remarks above. Hence (because the characters are in bijection) it suffices to prove the following: any representation:
$$G \rightarrow \GL_n(k)$$
with image $H$ admits a lift to $\GL_n(W(k))$, since $W(k)[1/p]$ will be necessarily be a subfield of $E$. (Not surprisingly, we see that all representations have models over unramified extensions when $p \nmid |G|$, since the characters are all valued in $\Q(\zeta_{m})$ where $m$ is the exponent of $G$.)
There is a projection $\GL_n(W(k)) \rightarrow \GL_n(k)$; let $\Gamma$ denote the inverse image of $H$. Since $H$ has order prime to $p$ and the kernel of $\Gamma \rightarrow H$ is pro-$p$, by Schur-Zassenhaus there is a splitting $H \rightarrow \Gamma$ which gives the required lift.
Suppose now that $\eta$ is a general (genuine) character of $G$ over a local field $K/\Q_p$ with values generate the field $L/K$, and suppose that
$p$ does not divide $|G|$. We prove that $\eta$ has a model over $L$. As noted above, $L/K$ is unramified, so in particular is Galois and $\Gal(L/K)$ is cyclic.
The result is true for 
irreducible characters from the discussion above.
Suppose that $\chi$ is an irreducible constituent of $\eta$. Since $\Gal(L/K)$ fixes $\eta$,
it follows that the $[L:K]$ distinct characters $\sigma \chi$ occur inside $\eta$.
Hence
$$\eta = \phi + \sum_{\Gal(L/K)} \sigma \chi,$$
where $\eta$ is a genuine character of lower dimension. Hence it suffices to note that
$$\bigoplus \sigma \chi$$
is defined over $K$, because if $V/L$ is a realization of $\chi$, then the above is 
$\mathrm{Res}_{L/K}(V)$,  where $L/K$ is thought of as a $[L:K]$-dimensional vector
space in the usual way. The result follows by induction.
Example: Suppose that $p \equiv -1 \mod 4$, and let $\chi$ be a faithful character of $\mathbf{Z}/4\mathbf{Z}$. Then $K = \Q_p(\chi)$ is unramified over $\Q_p$ of degree $2$, and the representation $\chi + \sigma \chi$ is sends a generator to $i \in K$ thought of as a vector space over $\Q_p$. If one chooses the basis $\{1,i\}$ of $K/\Q_p$, this is just the matrix:
$$\left( \begin{matrix} 0 & 1 \\\ -1 & 0 \end{matrix} \right).$$
Conclusion: Hence, at least for primes $p$ not dividing $|G|$, there is an injection from $G$ to $\GL_n(\Q_p)$ if and only if there is a faithful character $\eta$ with
values in a field $K$ which has a prime of norm $p$. Since $K$ will be abelian and ramified only at primes dividing $|G|$, this is equivalent to asking that $K$ split completly at $p$. So the set $S(G)$ (up to primes dividing $|G|$) is the union of primes which split completely in some finite number of fields determined by the faithful characters of degree $n$. If one restricts to a fixed character $\eta$, then $S_{\eta}(G)$ is indeed Galoisian.
Providing that $\eta$ has at least one faithful character of degree $n$, then $S(G)$ has rational positive density, answering 1. As for 2, it is not strictly Galoisian according to the definition of the previous question, since that required that the set be the set of primes
 which split completely in a single field. For example, one can take
$$G = \mathbf{Z}/12 \mathbf{Z}$$
and $n = 2$. Then $G$ is a subgroup of $\mathbf{GL}_2(\Q_p)$ for $p > 3$ if and only if  $p \equiv 1,5,7 \mod 12$, but not $11 \mod 12$, and this is not the set of primes which splits completely in any field $L$. In this case, if $\chi$ is a faithful character of $G$ of degree $1$, then $\chi + \chi^5$ and $\chi + \chi^{-1}$ are faithful characters with values in $\Q(\sqrt{-1})$  and $\Q(\sqrt{-3})$ respectively.
If $G$ has a faithful character of degree $n$ with values in $\Q$, then we see that $G$ embeds into $\GL_n(\Q_p)$ for all but finitely many primes $p$. The converse is quite possibly false, however; one could imagine $G$ having faithful characters of degree $n$ with values in $\Q(\sqrt{2})$, $\Q(\sqrt{3})$, and $\Q(\sqrt{6})$ but not in $\Q$ (although I don't have an example off the top of my head.)<|endoftext|>
TITLE: $H^1(X,O_X)$,  holomorphic $1$-forms, and $b_1(X)/2$ for normal $X$.
QUESTION [6 upvotes]: Suppose $X$ is a normal projective variety over $\mathbb C$. In the case $X$ is smooth according to Hodge theory $h^1(X,O(X))$ is the dimension of the space of holomorphic $1$-forms on $X$ and this number is equal as well to the half of the first Betti number $b_1(X)/2$ . 
I would like to know what happen in the case when $X$ is singular and normal.
1) Is there some relation (equality or inequality) between $h^1(X,O(X))$ and $b_1(X)$? For example does $b_1(X)=0$ imply $h^1(X,O(X))=0$?
2) Suppose that $h^1(X,O(X))=n$ is it true that there is a canonical $n$-dimensional space of $1$-forms on $X$, holomorphic outside of its singularities? (if yes, can something be said about their behaviour at singularities?)
Is there some pedagogical reference treating these questions?

REPLY [12 votes]: (Although I have pretty much "retired" from Mathoverflow, I will answer this, since the answer is nice but probably not all that well known.)

Theorem. If $X$ is complex normal projective variety, then  it is still true that $b_1(X)=2h^1(\mathcal{O}_X)$.

Proof. Let $\pi:\tilde X\to X$  be a desingularization. Since $X$ is normal, the fibres of $\pi$ are connected. Therefore $\pi_*\mathbb{Z}$ (with analytic topology) is connected.
It follows that $H^1(X,\mathbb{Z})\to H^1(\tilde X, \mathbb{Z})$ is injective since it can be identified with the edge map for Leray. Therefore the, a priori mixed, Hodge structure on $H^1(X)$ is pure of type $\lbrace (1,0),(0,1)\rbrace$. Consequently 
$$b_1(X)=2\dim [H^1(X)]^{(0,1)}=2 \dim im[H^1(X,\mathcal{O}_X)\to H^1(\tilde X, \mathcal{O}_{\tilde X}] = 2h^1(\mathcal{O}_X)$$

Regarding your question 2, you can construct an Albanese map $X\to Alb(X)$ to the torus
associated to the dual Hodge structure on $H_1(X)$. The space of $1$-forms on $Alb(X)$ will pullback to a space of the required dimension on $X$. 

To address your comments: a general reference for mixed Hodge structures is the book by
Peters and Steenbrink (although this may be bit a heavy). And yes, the argument does work in the Moishezon case, and a bit more generally. I guess that  I may as well admit the above argument was extracted from paper in Duke from 1990; this contains some more details and elaborations as well.<|endoftext|>
TITLE: K-theory of monoidal categories
QUESTION [6 upvotes]: I am novice in the algebraic K- theory and don' t know if this is the right place for the following questions. So some people might consider them as basic questions.
Consider an exact monoidal category and its  K- groups as introduced by Quillen.
Do $K_n(\mathcal  C)$ have a ring structure? Clearly for n=0 the Grothendieck groups  have a ring structure induced by the tensor monoidal functor of $\mathcal C$. So bassically the question can be reformulated if the same tensor functor induces also a ring structure in the higher K - groups.
Next natural question would be if  a monoidal functor $ F: \mathcal C \rightarrow D$ induces a ring homomorphism at any level of the K- groups, $F^*_n:K_n(\mathcal C)\rightarrow K_n(\mathcal D)$?
I guess these are basic things but unfortunately I don't know any text where they are treated.

REPLY [5 votes]: (I'm not sure the term exact monoidal category is a standard one in the literature, so I'll just assume I know what you mean by it.)
Yes, $K_n$ commutes with products, as some have mentioned, but the tensor monoidal functor $\otimes : \mathcal C \times \mathcal C \to \mathcal C$ is not exact, so it doesn't induce a map $ K_n \mathcal C \times K_n \mathcal C \to K_n \mathcal C$.  Rather, the functor $\otimes : \mathcal C \times \mathcal C \to \mathcal C$ is bi-exact, in the sense that it is an exact functor in each of its two variables, separately.  (That's what makes $K_0 \mathcal C$ into a ring.)
The result is that the collection of abelian groups $ K_n \mathcal C$, indexed by $n$, forms a graded ring, with products $K_m \mathcal C \otimes K_n \mathcal C \to K_{m+n} \mathcal C$.  Assuming the tensor product is commutative up to natural isomorphism, the ring is commutative in the graded sense that $ x y = (-1)^{mn} y x$.
The actual construction is intricate, the main idea being to construct a map $K \mathcal C \wedge K \mathcal C \to K \mathcal C$ of spectra.  See, for example, Waldhausen's approach in section 9 of Algebraic K-theory of generalized free products. I, II. Ann. of Math. (2) 108 (1978), no. 1, 135–204.  It uses the Q-construction of Quillen, but could be simplified by using the S-construction of Segal that Waldhausen developed extensively.
My paper with Gillet, The loop space of the Q-construction, Illinois Journal of Mathematics, 31 (1987) 574-597, gives another approach that avoids both spectra (i.e., delooping) and the phony multiplication trap that Steve mentioned.
PS: My paper Algebraic K-theory via binary complexes, Journal of the American Mathematical Society, 25 (2012) 1149-1167, gives an algebraic description of the elements of the higher K-groups that converts the construction of the product into a simple exercise involving tensor product of chain complexes.<|endoftext|>
TITLE: What are the most important open problems in algebraic combinatorics?
QUESTION [8 upvotes]: I have seen the paper of Stanley http://math.mit.edu/~rstan/pubs/pubfiles/116.pdf
but it is quite old and many of the problems are solved. I would like to know the two or three biggest open problems in algebraic combinatorics.

REPLY [2 votes]: One of the oldest standing open problems in algebraic combinatorics is Foulkes' conjecture; for some history and nice reformulations of the problem, see
On Foulkes' conjecture 
by William F. Doran IV in Journal of Pure and Applied Algebra (August 1998), 130 (1), pg. 85-98.<|endoftext|>
TITLE: The notion of multiplicity in algebraic geometry
QUESTION [7 upvotes]: Let $A$ be a commutative ring. Let $f\in A\setminus\{0\}$ and $I\subseteq A$ any ideal. I would like to define the multiplicity of $f$ at $I$ as 
$$\mu_f(I):= \max\{\, d\ge 0 \mid f\in I^d\,\},$$
where $I^0:= A$. In the case where $A$ is Noetherian and either local or an integral domain, the Krull Intersection Theorem (Eisenbud, Corollary 5.4) implies that $\mu_f(I)$ is well-defined. Main scenario: $A$ is the local ring of a locally Noetherian scheme $X$ at some point $P$, $I$ is the corresponding maximal ideal and $f$ is locally representing a Cartier divisor on $X$. 
I have only seen this in Hartshorne, Page 388, for surfaces, but I do not see why the definition should be limited to surfaces. In general, I only know the following definition of geometric multiplicity, for locally Noetherian schemes $X$ and points $P\in X$ of codimension one: 
$$\bar\mu_f(P):=\mathrm{length}_{\mathcal O_{X,P}}(\mathcal O_{X,P}/(f))$$
Does this coincide with the above definition? If yes, why is $\bar\mu$ so prominent? After all, $\mu$ is more general.

REPLY [4 votes]: In addition to two good answers, maybe one case the question has a positive answer is when $(R,m)$ is a regular local ring and $f$ is a nonzero element. For a Noetherian local ring $A$ and ideal $I$ which is primary to the maximal ideal, let $e(I, A)$ denote the Hilbert-Samuel multiplicity of $A$ with respect $I$. Then $e(m/(f),R/(f)) = \hbox{ord} (f) e(m, R) = \hbox{ord} (f)$ where $\hbox{ord} (f) = \sup \{ i \mid f \in m^i \}$. Here $e(m, R) = 1$ since $R$ is a regular local ring.
The special case is when the dimension of a ring (not necessarily regular) $R$ is $1$ and $f$ is a non zero-divisor (but not a unit) of $R$. Assume that $R$ is Cohen-Macaulay (domain or reduced implies Cohen-Macaulyness in dim $1$). Then we have that length $(R/(f)) = e((f), R) \ge \hbox{ord} (f) e(m,R)$ where the last inequality follows from Theorem 14.10 in Matsumura's commutative ring theory. A positive answer to your question implies an equality in the formula and $e(m,R) = 1$. The latter condition is equivalent to $R$ being regular if $R$ is unmixed. Note that the associated graded ring is a polynomial ring over the residue field if a ring is regular. In particular, it is a domain. I believe that once $R$ is a regular local ring, then the equality follows.
I believe in Hartshorne's algebraic geometry book, a surface is a nonsingular (locally regular) projective surface over an algebraically closed field. This probably is a reason why it works well. I hope someone can add some geometric point of view.<|endoftext|>
TITLE: Generalized Categories for "Higher Homotopy Groupoids"
QUESTION [13 upvotes]: I was thinking about the definition of higher homotopy groups $\pi_n$ of a topological space in comparison to the common extremely formal fundamental groupoid construction of $\pi_1$. I'd like to be able to do a similar construction in higher dimensions, and so I was wondering whether a principle similar to the following has been explored. Alas, the idea of these higher groupoids being a category fails for obvious reasons. We'll call this thing a groupesque. I won't give a definition of a groupesque in general - one of my questions is whether such a definition already exists - but I will describe the structure of the "homotopy groupesque." (The name is very conscious - the description that follows is grotesque.)
Throughout, consider all $n$-spheres, including $n$-spheres which are the boundary of $(n+1)$-disks, as being based spaces with a canonically chosen basepoint. Also, I may forget to apply the prefix "homotopy class of" to the word "map"; I apologize, and clarify ahead of time that every map in this post is actually considered up to homotopy.
The setup for the problem is quite long, but I imagine that a reader with an answer won't need to read much of it before they know exactly what I'm asking.

In the fundamental groupoid, our objects are points and maps are homotopy classes paths, maps $I^1\to X$ with appropriate boundary. In particular, our hom-sets are associated, as in any category, to pairs (a,b) of points. Equivalently, they are associated to maps $S^0=\partial I^1\to X$. This latter description has the convenient property that $\hom(a,b)$ is the set of homotopy classes of maps, relative the boundary, which "restrict to (a,b)" in the sense just described.
In the homotopy groupesques $\Pi_n$, we'd like our morphisms to be classes of maps $D^n\to X$ relative the boundary, and we'd like hom-sets to be associated to homotopy classes of maps $S^{n-1}\to X$, so that $\hom(f)$ is the set classes of maps $D^n\to X$ with $f$ as their boundary map (up to homotopy).
The "automorphism groups" of this groupesque in the traditional sense should end up being the hom-groups of constant map; this is nice, since these maps factor through maps from spheres.
But we still have no concept of composition, so when should there be a composition map on $\hom(f)\times\hom(g)$, and which hom-group does it land in? Note that the maps $f:S^{n-1}\to X$ that we're associating hom-sets to are actually elements of some automorphism group of $\Pi_{n-1}$, the previous homotopy groupesque. Basically, we'll say that we we can compose maps at $f$ with maps at $g$ is they lie in the same automorphism group, and hence if we can compose $f$ and $f$ as maps in $\Pi_{n-1}$. (This is where the basedness of all our spheres comes into play. We can compose them if and only if they map the baspoint to the same point.) In particular, composition runs $\hom(f)\times\hom(g)\to\hom(f\star g)$.
Note that this doesn't correspond in any obvious way with the case $n=1$: composition of those maps has nothing to do with a basepoint. This is an acceptable chimera.
Inductively, we can now check that the "automorphism groups" are genuinely what I just claimed they are, but this is where it gets interesting: there is always a composition $\hom(f)\times\hom(f)$ for any $f$, but if it happens to land in $\hom(f)$, then this implies that $f\star f=f$, which is if and only if $f$ is homotopically trivial.
Note that since maps no longer have a "source and target," it is not clear what an inverse is at all in general. However, when $f$ is homotopically trivial and we have a monoid structure this is clear. In that case, we can see that inverses to exist, and that our group structure corresponds to that of the higher homotopy group.
One last note: be sure that you see that $\Pi_1\neq\Pi$! The first homotopy groupesque is not the fundamental groupoid! In particular, hom-sets are associated to homotopy classes of maps $S^0\to X$, so we basically ended up with the skeleton of $\Pi$.

Finally, the questions:


Does this even make sense? There are so many places I could've gone astray while formulating this in the first place, that it's possible that something went wrong and these "groupesques" do not encode the homotopy groups at all.

Is this a higher category? I've thought and thought and I just can't seem to get it this object into that framework. The major issue is that even with an $\infty$-category structure, $n$-morphisms are attached to a pair of $(n-1)$-morphisms. It's not clear where "pairs" of anything comes into the picture here. (The higher-category approach seems to give "groups of maps from the $n$-torus" maybe, but not from the sphere.)

Do groupesques as a generalization of groupoids have a name? Is there a broader generalization of categories where hom-sets are associated to a single thing (rather than two), and some rule describes when a composition is defined and where it lands? (A traditional category is then one where the "single thing" is a pair of objects, and the "rule" checks whether the target of the first equals the source of the last.)

REPLY [17 votes]: Here is some background. In 1965 I noticed that the proof of the van Kampen theorem for the fundamental groupoid seemed to generalise to dimension 2, but there was a lack of a suitable gadget, a homotopy double groupoid. I also noticed that a proof due to J.F. Adams that any map $ S^r \to S^n$ for $ r < n $ is inessential (7.6.1 of Topology and Groupoids) should have algebraic consequences, but again there was no appropriate algebraic gadget. Nine years later we had found out a lot about double groupoids and crossed modules, but were still lacking the homotopy double groupoid! Then Philip Higgins and I did a strategic analysis which went as follows: 

J.H.C. Whitehead had a subtle theorem on $\pi_2(X \cup_\lambda e^2_\lambda,X,x)$ as a free crossed $\pi_1(X,x)$-module. This was an example, maybe the only then  example, of a $2$-dimensional universal property in homotopy theory. (here is a link to an  exposition of Whitehead's proof). 
If our conjectured $2$-dimensional van Kampen theorem was to be any good it should have Whitehead's theorem as a corollary. 
Whitehead's  theorem was about relative homotopy groups. 
So we should look for homotopy double groupoids in a relative situation, i.e. a space $X$ and a subspace $A$. 
The simplest way of doing this we could think of was to consider 


as in the above diagram maps of a square into $X$ which takes the edges into $A$ and the vertices to a subset $C$ of $A$ and to take homotopy classes of these rel vertices of the square. The proof that this works and gives  a strict double groupoid is not entirely trivial! To my knowledge, this was the first example of a strict higher homotopy groupoid. 
To our delight, this went swimmingly, and we were able to prove a $2$-d van Kampen theorem which had Whitehead's theorem as a Corollary. In fact we computed for example $\pi_2(X \cup_f CA,X,x)$ with Whitehead's theorem the case $A$ (not now a subspace) was a wedge of circles. 
R. Brown amd P.J. Higgins, ``On the connection between the second
relative homotopy groups of some related spaces'', Proc.
London Math.  Soc. (3) 36 (1978) 193-212.
Another surprise was that the submitted paper was asked to be withdrawn, in order not to embarrass the editor and two international authorities! A request for more information got another referee and a request (order?) to cut the paper by one third.  So the final paper had no pictures, and some slicker arguments. 
We also managed to work out the results for filtered spaces, and so all dimensions, and these were published in JPAA, 1981. See the book on Nonabelian algebraic topology. 
I emphasise that the basic methods were cubical, and we were unlikely  to conjecture let alone  prove these results using globular or simplicial methods. 
I like to think that these methods fulfill the dreams of the topologists of the early 20th century to find higher dimensional versions of the nonabelian fundamental group, since the nonabelian nature of the fundamental group was known to be useful in geometry and analysis.  
Over to you, reader, to get such  applications! 
Added 28 April: The contrast with what are  called in the literature "fundamental higher groupoids'' is that:(i)  those do not generalise the usual fundamental groupoid since they are not strict, and are just  singular complexes; (ii)  while they do satisfy some version of what is called the "small simplex theorem" that  does not  directly imply  strict colimit theorems in higher dimensions of a nonabelian type; (iii) the versions of higher groupoids we have worked with are strict structures, are defined non trivially for filtered spaces or $n$-cubes of spaces, and satisfy  nonabelian  colimit theorems with consequences not so far obtainable by other means. See for example the nonabelian tensor product of groups. 
These ideas relate  to and are aimed at relative homotopy theory, and $n$-adic homotopy theory, and I hope are seen in low dimensions as relevant to geometric group theory and  to geometric topology. 
The point is that one needs to evaluate what different approaches do and do not do, to compare and contrast. 
See also the question and answer to Infinity-categories vs Kan complexes 
Aug 5, 2014. The diagram there suggests  how convenient cubical methods are for multiple compositions, compared with simplicial or globular methods. 
July 7, 2014: A presentation   I gave to a workshop at the IHP,  Paris, June 5, 2014, entitled "Intuitions for cubical methods in nonabelian algebraic topology" is available on my preprint page. 
See also this stackexchange answer on homotopical excision in dim 2. 
Aug 5, 2014 A point should be made about the book on Nonabelian Algebraic Topology referred to above, and the foundations of algebraic topology, and in particular of homology theory. A key idea there is that of formal sums of geometric elements, an idea introduced by Poincaré. A definition of boundary then allows the notion of cycle, and boundary.  The more geometric method, as in this book, is that a "chain" is defined for a filtered space, and in dimension > $1$ is an element of a relative homotopy group. The "composition" of such chains is the composition in the relative homotopy groups. However cubical homotopy groupoids are used to prove many crucial properties of such chains.<|endoftext|>
TITLE: Compactness in Sobolev spaces
QUESTION [5 upvotes]: I was wondering whether the set $\lbrace f\in H_0^1(\Omega)|\|f\|_{L^\infty(\Omega)}\leq 1\rbrace$ is compact in $H_0^1(\Omega)$ or not. Here $\Omega$ is a convex domain in $\mathbb{R}^3$ with Lipschitz boundary.
Thanks.

REPLY [21 votes]: No.  As a general rule, in order to obtain compactness in some norm, one needs control of a higher regularity than what is associated to that norm, in order to shut down an "escape to frequency infinity".  For instance, $H^1_0$ has one degree of regularity, so one needs to control a norm involving more than one derivative in order to obtain compactness.  (But in many cases one only needs an epsilon more regularity; for instance, Arzela-Ascoli tells us that equicontinuity, which can be thought of as an infinitesimal amount of regularity, is enough (together with some additional hypotheses) to obtain compactness in the uniform norm, which has zero degrees of regularity.)
In this specific case, a concrete counterexample can be obtained by considering the unit cube $\Omega = [0,1]^3$ and the sequence $f_n := \frac{1}{n} \sin(2\pi n x) \sin(2\pi y) \sin(2\pi z)$, which  goes to zero in $L^\infty$ but stays away from zero in $H^1_0$, and so cannot have any $H^1_0$ convergent subsequence.<|endoftext|>
TITLE: Representing immersions from a surface into 3-space 
QUESTION [9 upvotes]: Let $\mathbb T^2=(S^1)^2$ be the 2-torus, for convenience. $\def\Imm{\operatorname{Imm}}$
Consider the Frechet manifold of immersion $\Imm(\mathbb T^2, \mathbb R^3)$ and the smooth mapping
$R:\Imm(\mathbb T^2,\mathbb R^3)\to C^\infty(\mathbb T^2,\mathbb R^3)$ which is given by
$$
R(f) := \partial_x f\times \partial_y f = f_x\times f_y,
$$
where $\times$ denotes the vector product. 
Note that $R(f)$ is the unit normal to $f(\mathbb T^2)$ times the the density function of the area measure, parameterized.

Question: Roughly, given $g\in C^\infty(\mathbb T^2,\mathbb R^3)$, can one solve $R(f)=g$ for $f$?
Can one reconstruct $f$ from $R(f)=g$ up to translations? Are there local solutions in    $\mathbb R^2$?
Of course one expects that the image $R$ is at best a codimension 6 submanifold of $C^\infty(\mathbb T^2,\mathbb R^3)$ because of the two periodicity conditions. 

The tangent mapping of $R$ is $T_fR.h = f_x\times h_y + h_x \times f_y = f_x\times h_y - f_y\times h_x$. This is not elliptic in $h$.
The symbol of $T_fR$ is 
$$
\sigma(T_fR)(\xi,\eta) = \begin{pmatrix} 0 & \gamma & -\beta \\\\ 
                                         -\gamma & 0 & \alpha \\\\ 
                                         \beta & -\alpha & 0  \end{pmatrix}
\quad\text{ where }\quad 
\begin{matrix} \alpha = f^1_y.\xi - f^1_x.\eta \\\\
\beta =  f^2_y.\xi - f^2_x.\eta \\\\
\gamma =  f^3_y.\xi - f^3_x.\eta 
\end{matrix}.
$$

Easier question: Is $T_fR$ injective? Is it surjective onto a codimension 6 linear subspace?

The interest in this problem comes from the fact that variants of $R$ like 
$f\mapsto |R(f)|^{-1/2}.R(f)$ pull back the flat $L^2$-metric on $C^\infty(\mathbb T^2,\mathbb R^3)$ to a weak Riemannian metric on $\Imm(\mathbb T^2,\mathbb R^3)$ which is invarint under the action of the diffeomorphism group of $\mathbb T^2$. 
The related mapping for closed curves has been used successfully here.
For the diffeomorphism group of $\mathbb R$ it has been used here.
The related problem for the unit normal for convex surfaces is classically called the Minkowski problem and there are solutions available, see
[JFM 34.0649.01 Minkowski, H.
Volumen und Oberfläche. (German)
Math. Ann. 57, 447-495 (1903)] and 
[MR0478079 (57 #17572)
Pogorelov, Aleksey Vasilʹyevich.
The Minkowski multidimensional problem. 
Translated from the Russian by Vladimir Oliker. Introduction by Louis Nirenberg. Scripta Series in Mathematics. V. H. Winston & Sons, Washington, D.C.; Halsted Press [John Wiley & Sons], New York-Toronto-London, 1978. 106 pp.]

REPLY [15 votes]: I may have to enter this as a sketch and fill in details later, but I thought that I'd go ahead and get the main ideas out there.
The first thing to notice is that the given problem is equivalent to the problem of solving $\rho(f)=g$ where $\rho:\mathrm{Imm}(T^2,\mathbb{R}^3)\to\Omega^2(T^2,\mathbb{R}^3)$ is the differential operator
$$
\rho(f) = f_x\times f_y\ \ dx{\wedge}dy
$$
taking values in (nonvanishing) $\mathbb{R}^3$-valued $2$-forms on $T^2$.  The advantage of considering $\rho$ instead of $R$ is that $\rho$ is well-defined, independent of coordinates on $T^2$.
The second thing to notice is that the inner product on $\mathbb{R}^3$ is something of a red herring.  The cross product is really just the wedge product followed by the orientation-and-metric induced canonical isomorphism between the second exterior power of $\mathbb{R}^3$ and $\mathbb{R}^3$ itself. 
Let's undo all this, and consider, instead, a $3$-dimensional vector space $V$ and a surface $\Sigma$ and define an operator $\rho:\mathrm{Imm}(\Sigma,V)\to\Omega^2\bigl(T^2,\Lambda^2(V)\bigr)$ that, in local coordinates $(x,y)$ on $\Sigma$, has the expression
$$
\rho(f) = f_x{\wedge}f_y\ \ dx{\wedge}dy.
$$
Now we want to ask the question "Given a nonvanishing $\Lambda^2(V)$-valued $2$-form $g$ on a surface $\Sigma$, when can it be written in the form $g = \rho(f)$ for some immersion $f:\Sigma\to V$ and in how many ways?"  This question is interesting both locally and globally.  Locally, it is three (nonlinear) first-order equations for three unknowns, but, as we shall see, it is highly degenerate, even when $g$ is 'generic', so that, for example, there is no hope of writing it in Cauchy form.
One distinct advantage of formulating the equation in this way is that its full equivariance is manifest.  Not only does $\rho$ not depend on a choice of local coordinates on the surface, if $L(v) = Av + b$ is any invertible affine transformation of $V$, then one has $\rho(L\circ f) = \Lambda^2(A)\bigl(\rho(f)\bigr)$.  In particular, replacing $g$ by $\Lambda^2(A)(g)$ does not change the problem, which shows that one can use the geometry of linear transformations of $V$ to uncover invariants of $g$ that will determine properties of the solution.
For example, in the (very degenerate) case in which $g$ is a $2$-form that takes values in a line in $\Lambda^2(V)$, one can write $g = e_1{\wedge}e_2\ \Phi$ where $e_1$ and $e_2$ are linearly independent and $\Phi$ is (nonvanishing) $2$-form on $\Sigma$.  Then the only possible solutions $f$ to $\rho(f)=g$ are to have $f$ immerse $\Sigma$ into a $2$-plane $P$ parallel to the subspace spanned by $e_1$ and $e_2$ in such a way that it pulls back the volume form on $P$ dual to $e_1{\wedge}e_2$ to be $\Phi$.  This is now essentially one equation for two unknowns and is always locally solvable, though there may well be no global solution.  This is a trivial special case, though, and doesn't give you any sense of the general case.
On this first pass, I will avoid discussing the singular and intermediate cases and go  directly to the 'generic' case, the case in which the projectivization of $g$, i.e., $[g]:\Sigma\to \mathbb{P}\bigl(\Lambda^2(V)\bigr)\simeq\mathbb{RP}^2$ is an immersion.  In this case, the problem reduces to either an elliptic or a hyperbolic problem, depending on the geometry of the mapping $g$.  
Example:  Before considering the general case, it's worth pausing to consider a simple example to illustrate where the analysis will take us:  Let $\Sigma$ be the $xy$-plane and consider the two maps $f_\pm:\Sigma\to\mathbb{R}^3$ defined by 
$$
f_\pm = \bigl(x,y,\tfrac12(x^2\pm y^2)\bigr).
$$
These have $g_\pm = \rho(f_\pm) = (1,0,x){\wedge}(0,1,\pm y)\ dx{\wedge}dy$.  We want to find the immersions $f:\Sigma\to\mathbb{R}^3$ that satisfy $\rho(f)=g_\pm$. Well, such an $f$ will have to satisfy
$$
df = (1,0,x)\ \xi + (0,1,\pm y)\ \eta = (\xi,\eta,\ x\xi{\pm}y\eta\ )
$$
for some $1$-forms $\xi$ and $\eta$ that satisfy $\xi\wedge\eta=dx\wedge dy$ and are such that $d(df)=0$.  Now, this latter equation implies $d\xi=d\eta=d(x\xi{\pm}y\eta)=0$, and so, assuming that a local solution is defined on a simply-connected open set $U\subset\Sigma$, one can then write $\xi = dp$ and $\eta = \pm dq$ for some functions $p$ and $q$ on $U$.  The remaining equations on $p$ and $q$ then become $0=d(x\xi{\pm}y\eta)=dx\wedge dp + dy\wedge dq$ and $dp\wedge dq = \pm dx\wedge dy$.  This is two PDE on $p$ and $q$ that can be written as a single Monge-Ampère equation by setting $p = u_x$ and $q = u_y$ (which works because $d(p\ dx + q\ dy) = 0$) and then the remaining equation becomes $u_{xx}u_{yy}-u_{xy}^2 = \pm 1$, the classic Monge-Ampère equation.  This is, of course, elliptic for $g_+$ and hyperbolic for $g_-$.  (We will see how this is reflected in the general analysis below.) The general solution to $\rho(f)=g_\pm$ is then
$$
f = (u_x,\ \pm u_y,\ xu_x+yu_y-u )
$$
where $u$ satisfies $u_{xx}u_{yy}-u_{xy}^2 = \pm 1$.
Now, I'm going to do the general analysis in terms of the moving frame and exterior differential systems, simply because that's the way I understand it and am most comfortable computing.  Of course, one can avoid this, but I'm too lazy (and short on time) to do that now.
So fix a nondegenerate $g\in \Omega^2\bigl(\Sigma,\Lambda^2(V)\bigr)$.  I'm going to define a bundle $P_g\to\Sigma$ whose elements will consist of the quadruples $(x;e_1,e_2,e_3)\in\Sigma\times V\times V\times V$ such that the $e_i$ are a basis of $V$ and, moreover, $e_1\wedge g(x) = e_2\wedge g(x) = 0$.  On $P_g$, there exists a nonvanishing $2$-form $\Phi$ such that $g = e_1{\wedge}e_2\ \Phi$.  This $\Phi$ is semi-basic for the projection to $\Sigma$ and hence is a multiple of any nonvanishing $2$-form on $\Sigma$ pulled up to $P_g$.  Note that $P_g$ is a right $G_1$-bundle over $\Sigma$, where $G_1\subset\mathrm{GL}(3,\mathbb{R})$ is the subgroup that consists of matrices $A\in \mathrm{GL}(3,\mathbb{R})$ that are $(2,1)$-block upper triangular, i.e., $A^3_1=A^3_2=0$.
Now, one has the usual structure equations $de_a = e_b\ \eta^b_a$, which satisfy $d\eta^a_b = - \eta^a_c\wedge\eta^c_b$. (The summation convention always in force.  I'll use the index ranges $1\le a,b,c \le 3$ and $1\le i,j,k \le 2$.)  Since $g$ is a vector-valued $2$-form on a surface, one has $dg=0$, and expanding this using the structure equations yields
$$
0 = dg = e_1{\wedge}e_2\ \bigl(d\Phi + (\eta^1_1+\eta^2_2)\wedge\Phi\bigr) 
         + e_3{\wedge}e_2\ \eta^3_1\wedge\Phi + e_1{\wedge}e_3\ \eta^3_2\wedge\Phi,
$$
so $\eta^3_1\wedge\Phi=\eta^3_2\wedge\Phi=0$.  Now, the hypothesis that $[g]:\Sigma\to\mathbb{RP}^2$ be an immersion is equivalent to $\eta^3_1{\wedge}\eta^3_2$ being nonvanishing, Consequently, $\Phi = \lambda\ \eta^3_1{\wedge}\eta^3_2$ for some nonvanishing function $\lambda$ on $P_g$.  One easily computes that, for $A\in G_1$,
$$
R_A^*\lambda = \left(\frac{A^3_3}{A^1_1A^2_2-A^2_1A^1_2}\right)^2\lambda,
$$
so the sign of lambda is constant on the fibers of $P_g$ and the equation $\lambda=\pm1$ defines a subset $P_g'\subset P_g$ that will be a right $G_2$-bundle over $\Sigma$, where $G_2\subset G_1$ is the subgroup consisting of those $A\in G_1$ that satisfy $A^3_3 = \pm(A^1_1A^2_2-A^2_1A^1_2)$.  
I'll say that $g$ is of elliptic (resp. hyperbolic) type if $\lambda = +1$ (resp. $\lambda=-1$) on $P_g'$.  (The reasons for these designations will become apparent below.)
Since $\Phi = \pm \eta^3_1{\wedge}\eta^3_2$ on $P_g'$, the structure equations plus the identity $0=d\Phi + (\eta^1_1+\eta^2_2)\wedge\Phi$ now show that $(\eta^3_3-\eta^1_1-\eta^2_2)\wedge\eta^3_1{\wedge}\eta^3_2=0$, so $\eta^3_3=\eta^1_1+\eta^2_2+b^1\eta^3_1+b^2\eta^3_2$ for some functions $b^1$ and $b^2$ on $P_g'$.  The structure equations now show that the equations $b^1=b^2=0$ define a $G_3$-subbundle $P_g''\subset P_g'$, where $G_3\subset G_2$ is the subgroup consisting of those $A\in G_2$ with $A^1_3=A^2_3=0$.  (This is the last reduction I will need to do to complete the calculation.)
Our goal now is to find $1$-forms $\eta^1$ and $\eta^2$ on $P_g''$ satisfying the equations
$$
d(e_1\ \eta^1 + e_2\ \eta^2) = 0\qquad\text{and}\qquad 
e_1{\wedge}e_2\ \eta^1{\wedge}\eta^2 = g = \pm e_1{\wedge}e_2\ \eta^3_1{\wedge}\eta^3_2\ .
$$
If we can do this, then, at least locally, $e_1\ \eta^1 + e_2\ \eta^2 = df$ for some $V$-valued function $f:\Sigma\to V$ such that $\rho(f) = g$, and this will solve our problem.
Of course, the second equation is just $\eta^1\wedge\eta^2 = \pm\ \eta^3_1{\wedge}\eta^3_2$, while using the structure equations to expand the first equation yields
$$
0 
= e_1 (d\eta^1 + \eta^1_1{\wedge}\eta^1+ \eta^1_2{\wedge}\eta^2)
 +e_2 (d\eta^2 + \eta^2_1{\wedge}\eta^1+ \eta^2_2{\wedge}\eta^2)
 +e_3 (\eta^3_1{\wedge}\eta^1 + \eta^3_2{\wedge}\eta^2) 
$$
Looking at the $e_3$-coefficient in this equation and applying Cartan's Lemma, we see that there must exist $h^{ij}=h^{ji}$ such that $\eta^i = h^{ij}\ \eta^3_j$, which, substituted into the second equation, says that $\det(h) = \pm 1$.  
Now, let $S_\pm$ denote the surface in the $3$-dimensional space of symmetric $2$-by-$2$ matrices that consists of the matrices with determinant $\pm 1$.  ($S_+$ is a hyperboloid of 2 sheets, while $S_-$ is a hyperboloid of 1 sheet.)  Let $h:P_g''\times S_\pm\to S_\pm$ be the projection onto the second factor.  Set $\eta^i = h^{ij}\eta^3_j$ as $1$-forms on $P_g''\times S_\pm$ and consider the pair of $2$-forms
$$
\Upsilon^i = d\eta^i + \eta^i_1{\wedge}\eta^1+ \eta^i_2{\wedge}\eta^2.
$$
The formulae derived so far show that $\Upsilon^i = \pi^{ij}\wedge\eta^3_j$, where $\pi^{ij}=\pi^{ji}$ are $1$-forms defined by the relations
$$
\begin{pmatrix}\pi^{11}&\pi^{12}\\\\
\pi^{21}&\pi^{22}\end{pmatrix} 
= dh + \begin{pmatrix}\eta^1_1&\eta^1_2\\\\
\eta^2_1&\eta^2_2\end{pmatrix} h 
+ h\begin{pmatrix}\eta^1_1-\eta^3_3&\eta^2_1\\\\
\eta^1_2&\eta^2_2-\eta^3_3\end{pmatrix}.
$$
Moreover, the equation $\det(h)=\pm1$ and the normalizations made so far imply that $\mathrm{tr}(h^{-1}\pi)=0$.  
These equations show that, on the quotient manifold $Q = (P_g''\times S_\pm)/G_3$, which is a bundle over $\Sigma$ with $2$-dimensional fibers, there is an ideal $\mathcal{I}$ that is locally generated by a pair of $2$-forms that pull back up to $P_g''\times S_\pm$ to be independent linear combinations of the $\Upsilon^i$, and that the integral surfaces of this ideal that are transverse to the fibers of $Q\to\Sigma$ represent solutions to our problem.  When $\det(h)=1$, the above relations show that this is an elliptic ideal, while, when $\det(h)=-1$, this ideal is hyperbolic.  In either case, the ideal is involutive, with Cartan characters $s_1=2$, $s_2=0$.
The Cartan-Kähler Theorem now applies, at least when $g$ is real-analytic, to show that there are local integral manifolds transverse to the fibers of $Q\to\Sigma$.  In the hyperbolic case, a smooth version of the Cartan-Kähler theorem guarantees local solvability while, in the elliptic case, there are existence theorems (basically coming from the theory of pseudo-holomorphic curves) that guarantee existence of local solutions.  Thus, the equations are locally solvable if $g$ is smooth.
Anyway, that's a sketch of an analysis of the nondegenerate case.  The point is that the original determined system of three equations for three unknowns is not formally integrable, but, in the generic case, it can be prolonged to either an elliptic or a hyperbolic system that is involutive and hence formally integrable.
One should still do the case when the rank of the differential of $g$ is $1$, and one should look for a way to reinterpret the system $\mathcal{I}$ as a single scalar equation of Monge-Ampere type, to which more standard PDE methods would apply.  There may even be a way to linearize this equation as either an elliptic or hyperbolic single scalar equation, but that would require more thought than I have had time to put into it so far.<|endoftext|>
TITLE: General theory of left-exact localization?
QUESTION [5 upvotes]: A left-exact localization of a category is a reflective subcategory such that the reflector preserves finite limits.  There are several prominent examples of such localizations, such as sheafification, and localization of module categories.  Is there a general theory of such localizations?  
I don't have any particular type of result in mind, but given the prominence of the two examples I mentioned, it seems like the topic of left-exact localizations must have been studied for its own sake.

REPLY [6 votes]: Just to tie this one up, the cited result in Borceux (Prop 5.6.1) says the following:
Proposition: Let $\mathcal C$ be a finitely-complete category and let $L \mathcal C$ be a reflective subcategory. Let $\mathcal W$ be the class of morphisms inverted by the reflector $r: \mathcal C \to L\mathcal C$. Then $r$ is left exact if and only if $\mathcal W$ is stable under base change.
This continues to hold in the $\infty$-categorical context. Here is a proof adapted from Borceux which works $\infty$-categorically.
Proof: The "only if" direction is clear; we prove "if". First, since $i$ preserves terminal objects we have that the terminal object of $L\mathcal C$ is the terminal object of $\mathcal C$ and in particular is $\mathcal W$-local, so $L$ preserves terminal objects.
So it will suffice to show that $L$ preserves pullbacks. To this end, it is sufficient to show that if we have a natural transformation from the pullback square on the left to the one on the right below, and if the components $B \xrightarrow \sim B'$, $C \xrightarrow \sim C'$, and $D \xrightarrow \sim D'$ are all in $\mathcal W$, then so is the map $A \to A'$.
$\require{AMScd} \begin{CD} A @>>> B\\ @VVV  @VVV\\ C @>>> D \end{CD} \qquad \Rightarrow \qquad\begin{CD} A' @>>> B'\\ @VVV  @VVV\\ C' @>>> D' \end{CD}$
By pullback-stability and 2/3, the map $B \times_{D'} C \xrightarrow \sim A'$ is in $\mathcal W$. So by 2/3 it will suffice to show that the map $A \to B \times_{D'} C$ is in $\mathcal W$. By pullback-stability and 2/3, the map $D \xrightarrow \sim D \times_{D'} D$ is in $\mathcal W$, so it will suffice by pullback-stability to show that the following two squares are pullbacks:
$\begin{CD} A @>>> B @>>> D \\ @VVV  @VVV @VV^\sim V \\
B \times_{D'} C @>>> B \times_{D'} D @>>> D \times_{D'} D \end{CD}$
This can be seen using the following two diagrams:
$\begin{CD} A @>>> B \\ @VVV  @VVV \\
B \times_{D'} C @>>> B \times_{D'} D @>>> B \\ @VVV @VVV @VVV \\
C @>>> D @>>> D' \end{CD}
\quad \begin{CD} B @>>> D \\ @VVV  @VVV \\
B \times_{D'} D @>>> D \times_{D'} D @>>> D \\ @VVV @VVV @VVV \\
B @>>> D @>>> D' \end{CD}
$
In each case we argue that the bottom-right square and the composite of the lower two squares is a pullback, so the bottom-left square is a pullback. Since the composite of the left two squares is also a pullback, it results that the top-left square is a pullback.

REPLY [5 votes]: This is my bibliography on the subject. Indeed Bourceux was quite relevant in this topic.


Borceux, Sheaves of algebras for a commutative theory, Ann. Soc. Sci. Bruxelles Sér. I 95 (1981), no. 1, 3–19
Borceux and Kelly, On locales of localizations, J. Pure Appl. Algebra, Volume 46, Issue 1, 1987, Pages 1-34.
Borceux and Veit, On the Left Exactness of Orthogonal Reflections
J. Pure Appl. Algebra, 49 (1987), pp. 33-42.
Borceux, Subobject Classifier for Algebraic Structures.
J. Algebra, 112 (1988), pp. 306-314.
Veit, Sheaves, localizations, and unstable extensions: Some counterexamples. J. Pure Appl. Algebra, Volume 140, Issue 2, July 1991, Pages 370-391.
Borceux and Quinteiro. A theory of enriched sheaves. Cahiers de Topologie et Géométrie Différentielle Catégoriques 37.2 (1996): 145-162.
Garner and Lack, Lex Colimits. J. Pure Appl. Algebra. Volume 216, Issue 6, June 2012, Pages 1372-1396.<|endoftext|>
TITLE: Who first observed that Conjugate Gradient for Symmetric Positive Definite linear systems is a Krylov method?
QUESTION [7 upvotes]: Conjugate gradient was originally presented in the 50's before the modern understanding of Krylov subspaces (and the resulting iterative methods) was fully realized.  As such, the method was derived using different tools and language.  My question is, who was the first person to observe that Conjugate Gradient is a Krylov subspace method?  I have heard Youssef Saad given this credit, but when I search for references to back up the assertion, I cannot find relevant references.  I figure someone must know or that perhaps the understanding came gradually and cannot be attributed to any one author (or even a small group).

REPLY [4 votes]: Krylov Subspace Methods for Solving Large Unsymmetric Linear Systems, Y. Saad, Mathematics of Computation 37, 105-126 (1981).

The purpose of the present paper is to
  generalize the conjugate gradient
  method regarded as a projection
  process onto the Krylov subspace K„.
  We shall say of a method realizing
  such a process that it belongs to the
  class of Krylov subspace methods. It
  will be seen that these methods can be
  efficient for solving large
  nonsymmetric systems.<|endoftext|>
TITLE: What are the best known lower and upper bounds for the second Chebyshev function $\psi(x)$
QUESTION [5 upvotes]: I was reading through Jitsuro Nagura's proof that there is always a prime between $x$ and $\frac{6x}{5}$ when $x \ge 25$.
In the paper, he uses the following bounds for the second Chebyshev function $\psi(x)$:
$$1.086x > \psi(x) > 0.916x - 6.954$$
If I apply the better upper bound from Rosser & Schoenfeld, 1962 of:
$$1.03883x > \psi(x)$$
Then Nagura's proof shows that there is always a prime between $x$ and $\frac{8x}{7}$ when $x \ge 34$.
Is this the best upper and lower bound for $\psi(x)$:  
$$1.03883x > \psi(x) > 0.916x - 6.954$$
Does anyone know of any results that improve on these bounds?
Thanks,
-Larry

REPLY [4 votes]: The most recent results on bounds for $\psi(x)$ are from this year:
Sharper estimates for Chebyshev's functions $\vartheta$ and $ψ$, February 2013.

In this article we present some
  improved results for Chebyshev's
  functions $\vartheta$ and $\psi$ using
  the new zero-free region obtained by
  H. Kadiri and the first
  $10^{13}$ zeros of the Riemann zeta
  function on the critical line calculated by
  Xavier Gourdon. The methods in the
  proofs are similar to those of the
  Rosser-Shoenfeld papers on this
  subject.<|endoftext|>
TITLE: Intuitionistic logic as quantization of classical logic?
QUESTION [13 upvotes]: A classically trained mathematician is more likely to be familiar (at least anecdotally) with an area of mathematical physics such as deformation quantization than with intuitionistic logic.  It is helpful in any sense (philosophical or mathematical) to think of intuitionistic logic as a quantization of classical logic?  Has anyone explored such an approach?  Note that the idea itself of comparing the passage from classical to intuitionistic logic to denying commutativity is not new; see Richman's "Interview with a constructive mathematician" at http://www.ams.org/mathscinet-getitem?mr=1400617 . ͏ ͏ ͏

REPLY [2 votes]: First, let me acknowledge Nik Weaver's objection.  However, there is a notion of quantization in error correction coding. This notion might be compared with numerical methods that establish a policy by which some rational number is fixed to represent the real number of a given calculation.  That is, an admissible value is accepted in relation to some actual value.  So, Margaret Friedland's notion will be accepted for this response.
The initial objections to classical logic by Brouwer and others had to do with the definiteness of working with finite systems that differed from infinite systems.  So, the sense of Margaret Friedland's notion would not seem to be the direct comparison to be made.
And, in that same context, the original posted question seems badly construed.
Insightfully, however, Ms. Friedland thought the issue might lie with semantics.
The following remarks are from personal unpublished researches.  As a philosophical matter, I reject logicism.  The focus of my investigations had been the sign of equality and identity.
When Leibniz introduced the principle of identity of indiscernibles in "Discourse on Metaphysics", he did so by invoking geometric intuitions,

"What St. Thomas affirms on this point
  about angels or intelligences ('that
  here every individual is a lowest
  species') is true of all substances,
  provided one takes the specific
  difference in the way that geometers
  take it with regard to their figures."
-Leibniz

My personal view on this is that numerical identity relies on geometric -- or, more precisely, topological -- notions.  So, I interpret Leibniz' remarks along the line of Cantor's intersection theorem for non-empty, nested closed sets of vanishing diameter.
The semantic sense of Leibniz' remarks are to be found in another quote from one of his papers on logic (the name of which escapes me at this moment).  Although somewhat deprecated in mathematical logic, the paradigm singular term in classical logic is the notion of a name.  This is what Leibniz says about names,

"All existential propositions, though true,
    are not necessary, for they cannot be
    proved unless an infinity of propositions
    is used, i.e., unless an analysis is
    carried to infinity.  That is, they can
    be proved only from the complete concept
    of an individual, which involves infinite
    existents.  Thus, if I say, "Peter denies",
    understanding this of a certain time, then
    there is presupposed also the nature of
    that time, which also involves all that
    exists at that time.  If I say "Peter
    denies" indefinitely, abstracting from
    time, then for this to be true -- whether
    he has denied, or is about to deny --
    it must nevertheless be proved from the
    concept of Peter.  But the concept of
    Peter is complete, and so involves infinite
    things; so one can never arrive at a
    perfect proof, but one always approaches
    it more and more, so that the difference
    is less than any given difference."
Leibniz

This, too, can be related directly to topological considerations in the guise of uniformities and uniform spaces.
Semantically, the identity relation is represented by the diagram or diagonal of the Cartesian product of a model domain.  Under the received view, logical identity is not given by a metric interpretation.  It is in the metrization lemma found in Kelley's "General Topology" wherein a system of relations containing the diagonal (say, a uniformity) and meeting certain other conditions generate a pseudometric.  In other words, it is in the theory of uniformities where the original Leibnizian conception and modern semantical notions coincide.
It is for this reason that I am personally inclined to view the topological designations of open and closed sets (in particular, closed sets) in the context of the kind of "quantization" mentioned by Margaret Friedland.
For completeness with respect to the preceding remarks, let me observe that both Frege and Russell included descriptivist theories of naming in their logical analyses.  Kleene reports that the eliminability of descriptions had been established in 1934 by Hilbert and 
Bernays.  Robinson had been critical of Russellian description theory and discusses the use descriptions in relation to model diagonals in his paper "On constrained denotation".
Along similar lines, the logical interpretation of Leibniz' principle of identity of indiscernibles seems to have been established by the time of Kant.  Kant criticizes Leibniz' application of that principle and asserts that identity associated with appearances is based on geometric notions.  With respect to modern description theory, this notion of numerical identity in relation to geometry can be found in another critic of Russell -- P. F. Strawson discusses the matter in his book, "Individuals".
Let me reiterate that these are only personal views, and, that they are non-standard by every account.<|endoftext|>
TITLE: Can I use both of setbuilder notations in one article?
QUESTION [5 upvotes]: I am sorry if this question is not for mathoverflow. I asked the same question in tex stackexchange and got an answer that this question is out of topic. If it's the same here, please let me know.
There are two setbuilder notations, the vertical bar and the colon. In some cases, it is better to make a choice. For example, compare
$$\{f \mid f\colon M \rightarrow N \text{ is continuous}\}$$
and
$$\{f : f\colon M \rightarrow N \text{ is continuous}\}.$$
Also compare
$$\{x : \lvert x \rvert = 1\}$$
and
$$\{x \mid \lvert x \rvert = 1\}.$$
Can I use both the vertical bar and the colon in one article to make it more readable?
Or should I choose one notation to be consistent?

REPLY [6 votes]: As the author, such things are generally up to you.  It would not be the first time an author chose clarity over consistency of notation.  I think many readers appreciate such tradeoffs, although I am sure there is someone who doesn't.  If you are going to use inconsistent notation, it is polite to make a footnote or something about why you have done so, but this case is probably transparent enough to make that unnecessary.
If using both really bothers you, you could use \big to do things like $\{ x\ \big\vert\ \lvert x\rvert = 1\}$, which is slightly more readable than when all the symbols are the same size.<|endoftext|>
TITLE: Fibered knot with periodic homological monodromy
QUESTION [11 upvotes]: It is well-known that there exist pseudo-Anosov automorphisms of surfaces that act trivially on the homology: they form the Torelli group. Similarly there exists pseudo-Anosov automorphisms that act periodically. 
On the other hand, given a fibered knot with pseudo-Anosov monodromy, the Alexander polynomial is the characteristic polynomial of the homological monodromy, and, at the same time, its degree is twice the genus of the fiber. This implies that no monodromy of a fibered knot lies in the Torelli group. 
My question is: can the homological monodromy of a knot be periodic and the geometric monodromy pseudo-Anosov? My guess is no, and this is probably known, but I could not find a reference nor a proof.
--edit--
Now I think that the answer is yes, see comment below!
An improved question could then be: can the homological monodromy of a knot be periodic, the geometric monodromy be pseudo-Anosov, and the fiber surface not contain an unlinked zero-framed link?

REPLY [5 votes]: There exists an infinite family of quasipositive fibre surfaces of genus 3 with the same Alexander polynomial as the torus knot T(2,7). This answers Pierre's improved question, since quasipositive surfaces do not contain any essential unlinked annuli.
Let us first recall that the fibre surface S' of T(2,7) is obtained from the fibre surface S of the torus link T(2,6) by a single positive Hopf plumbing operation. The plumbing operation is determined by a relatively embedded arc J in S. Given that the genus of S is two, there exists an infinite family of pairwise non-isotopic embedded intervals in the relative homology class of J. Plumbing positive Hopf bands along those instead of J gives rise to an infinite family of quasipositive fibred knots with the same Seifert matrix. All of them have pseudo-Anosov monodromy, since there is only one fibred knot of genus 3 with periodic monodromy of order 14: the torus knot T(2,7).<|endoftext|>
TITLE: What is a higher derived constructible sheaf
QUESTION [6 upvotes]: Suppose $X$ is a topological space and $k$ some discrete coefficient field. Let's define the category of "$\infty$-local systems on $X$" to be DG representations of the ring $C_*(\Omega X,k)$ of chains of loops on $X$. (I think this is equivalent, at least when $\text{char}(k)=0$, to locally constant sheaves where "sheaf" is defined in an appropriate higher-topos sense). 
I would like some construction of "$\infty$-constructible sheaves" which includes both this category and also constructible complexes of sheaves for some fixed stratification of $X$. Do topologists know of one? Is there an analogue of the Riemann-Hilbert correspondence for $X$ a complex algebraic manifold?

REPLY [6 votes]: In the appendix of "Higher Algebra" (http://www.math.harvard.edu/~lurie/papers/HigherAlgebra.pdf), Jacob Lurie describes constructible sheaves on a stratified space as representations of an exit path $\infty$-category. So one option is to take representations of this exit path $\infty$-category valued in the stable $\infty$-category (or appropriate DG category) of complexes of vector spaces. When the stratification is trivial, this recovers the usual notion of $\infty$-local system (as representations of the fundemental $\infty$-groupoid).
I would also be interested to hear about a Riemann-Hilbert Correspondence in this generality. In the local systems setting, there is Aaron Smith's thesis: http://repository.upenn.edu/cgi/viewcontent.cgi?article=1462&context=edissertations.<|endoftext|>
TITLE: Generating Random Young Tableaux: A peculiar probability identity
QUESTION [14 upvotes]: In the paper by Greene, Nijenhuis and Wilf, an algorithm is proposed for generating uniformly random Young tableaux of shape $\lambda$. The algorithm is to uniformly randomly pick a starting cell, and then do a hook walk algorithm until it terminates at one of the edges of the tableau. Another way of looking at this is to fix a starting cell $(a,b)$, and then start a random hook walk, so that one gets a path 
$$(a,b)=(a_1,b_1)\rightarrow(a_2,b_2)\rightarrow\cdots\rightarrow (a_k,b_k)=(\alpha,\beta)$$
where $(\alpha,\beta)$ is the terminal cell. This defines a probability of hitting terminal cell $(\alpha,\beta)$ given starting cell $(a,b)$. 
A peculiar observation is made on page 108, that:
$$P(\ (\alpha,\beta) \ |\ (a, b)\ ) = P(\ (\alpha,\beta) \ | \ (\alpha, b)\ ) \cdot P(\ (\alpha,\beta)\  |\ (a,\beta) \ )$$
In other words the probability of reaching the terminal point is the product of probabilities of starting in the row and column of the terminal point, which amounts to perpetually staying within the respective hooks. The authors point out that they have no obvious direct explanation of this fact.

Has anyone come up with an explanation more recently?

REPLY [11 votes]: This is a technical question for those very familiar with the proof.  Basically, GNW show by induction that the quantity on the l.h.s. has a product formula, with some hook numbers inside.  Since the probabilities add up to 1, this gives a recurrence relation for the number of standard Young tableaux (SYT), mimicking the branching rule.  This in turn allows one to prove the hook-length formula (HLF).  
Now, why does the probability $P( (α,β) | (a,b) )$ has a product formula is a bit of a miracle.  The way I see it, GNW aimed to prove HLF by induction, the summations were getting complicated, and they engineered the hook-walk to help keeping track of all paths they were counting.  Arguably, the miracle of the product formula is exactly the same as the miracle of the HLF.  The latter is a classical mysterious result, about which I can (but won't) go at length, but one thing is clear: there is no known philosophical reason why the ratio $n!/|SYT(\lambda)|$ has to be a product of $n$ "nice" integers, other than the fact that it's true.  
To finish on a positive note, let me mention that while there is no formal procedure how to obtain a hook-walk from a HLF, the idea of constructing it is extremely robust.  Roughly speaking, a variation on the HLF often enough can be derived via a hook-walk style proof (see a probably incomplete list below).  In each case we get the a miraculous product formula for the probability of a hook walk stopping at a given corner.  
1) in the shifted case (Sagan, 1980)
2) in the complementary case (GNW, 1984)

3) in the tree case (Sagan, Yeh, 1989)

4) in the q-case (Kerov, 1993)

5) in the q,t-case (Garsia, Haiman, 1998)

6) in the Han's tree formula case (Sagan, 2009)

7) in the weighted case (Ciocan-Fontanine, Konvalinka, Pak, 2011)

8) in the weighted complementary case (Konvalinka, 2011)

9) in the weighted shifted case (Konvalinka, 2011)<|endoftext|>
TITLE: Automorphisms of Generic Abelian Varieties
QUESTION [8 upvotes]: Automorphism groups of elliptic curves are very well understood. Of course, every elliptic curve has the automorphism $[-1]$ of order $2$. If we are over a (algebraically closed) field, this is the only non-trivial automorphism iff the $j$-invariant of our elliptic curve is neither $0$ nor $1728$. Over $\mathbb{C}$ the only other possibilities for automorphism groups are $\mathbb{Z}/4$ and $\mathbb{Z}/6$ occuring for $E = \mathbb{C}/\Lambda$ where $\Lambda$ is the square lattice or the "honeycomb" lattice respectively. 
In particular, the generic elliptic curve has automorphism group $\mathbb{Z}/2$, i.e. the locus of elliptic curves with bigger automorphism group in the moduli space of elliptic curves has codimension bigger than $0$. 
I am interested in the analogous question for abelian varieties. To force all the automorphism groups to be finite, it seems sensible to consider (principally) polarized abelian varieties. 

Let $\mathcal{A}_g$ be the moduli space of (principally) polarized complex abelian varieties of dimension $g$. What is the codimension of the locus of such varieties with automorphism group bigger than $\mathbb{Z}/2$ ?

Likewise, one can ask the same question not about arbitrary principally polarized abelian varieties, but such equipped with an endomorphism and level structure. So, in addition to the polarization on our (complex) abelian variety $A$ we also want a ring homomorphism $\mathcal{O}_F \to End(A)$ (for a fixed, say, quadratic imaginary field $F$), compatible with the Rosatti involution, and a level structure (plus some determinant/trace-condition). For a generic such $A$, its automorphism group may contain now $\mathcal{O}_F^\times$, depending on the level structure.

Let $Sh_g$ be the moduli space of such complex abelian varieties with PEL-structure (with the choices of $F$ and level implicit) of dimension $g$. Let $G$ be the smallest automorphism group of a point in $Sh_g$. What is the codimension of the locus of points of $Sh_g$ having automorphism group bigger than $G$?

REPLY [4 votes]: I will answer the first formulation of the question: As ulrich said, the answer is that the codimension is $g-1$, given by $\mathcal{A}_{g-1}\times X(1)$ mapping quasi-finitely to $\mathcal{A}_g$. 
To see this, let us work in Siegel space $\mathbb{H}_g$ with $Sp_{2g}(\mathbb{R})$ acting in the usual way. Having an extra automorphism is equivalent to being a stacky point in $\mathcal{A}_g$, which means that in $\mathbb{H}_g$ you are fixed by an element $t\in SP_{2g}(\mathbb{Z})$. It is easy to see that $t$ must have finite order. We will prove that for any finite order element $t$, the fixed set of $t$ in $\mathbb{H}_g$ is codimension at least $g-1$. 
To see this, first note that the claim is invariant under conjugating $t$, as this simply translates its fixed set. $t$ generates a finite, hence compact group. Thus we can conjugate $t$ inside of the standard maximal compact group $U(g)$. Moreover, we can further conjugate $t$ to lie in the standard maximal torus $T$ in $U(g)$, that is the diagonal matrices whose elements have unit norm. Thus we can write $$t=D(t_1,t_2,\dots,t_n)$$ where each $t_i$ is a root of unity. 
Now the fixed set of $t$ is easiest to see in the bounded model $B_g$ - which is biholomorphic to $\mathbb{H}_g$ in a $Sp_{2g}(\mathbb{R})$ equivaraint way so we are free to compute there instead.
In this model, $t$ multiplies $z_{ij}$ by $t_it_j$. Thus the fixed set is those matrices $Z$ that have non-zero entry $z_{ij}$ precisely when $t_it_j=1$. 
The problem is now purely a combinatorial one of trying to maximize the number of un-ordered pairs $i,j$ such that $t_it_j=1$. It is almost immediate to see that we may as well change all the  $t_i$'s that aren't $1$ to $-1$ and this will in fact increase the dimension of the fixed set. If we have $a$ 1's and $g-a$ -1's then we get a fixed locus of codimension $a(g-a)$ which is minimized when $a=1$ or $a=g-1$ (which are equivalent of course as $t$ and $-t$ give the same fixed set. This completes the proof.
It seems likely to me that by looking for which elements in $Sp_{2g}(\mathbb{Z})$ are conjugate to $t$ above one should be able to prove that $\mathcal{A}_{g-1}\times X(1)$ gives the unique component of maximal codimension, but I've not gone through and tried to do this.
Comment: A similar analysis should work for your second question, but I don't offhand know the uniformizing spaces.<|endoftext|>
TITLE: What is flexible about flexible algebras?
QUESTION [11 upvotes]: A possibly non-associative algebra is flexible if it satisfies the identity $$(xy)x=x(yx).$$ This is clearly a very weak form of associativity —and obviously an associative algebra is flexible— but it is also a weak form of commutativity —a commutative algebra is clearly flexible— and in practice it plays both of these roles.
Flexibility is such a weak condition that it is difficult to imagine getting much out of it, but then one reads the works of Albert, Schafer, Jacobson and several others, and sees it put to use with such extraordinary artistry and skill that one ends up reflecting on the excessiveness of our standard hypotheses of commutativity and associativity :-)
But I have not found so far an explanation for the choice of the term:

Do you know why flexible algebras are called flexible?

REPLY [3 votes]: I browsed through the 1948 paper by Albert where he introduced the notion of a flexible algebra. It is presented as a property that gives a certain flexibility to how algebraic operations are to be carried out. Like all good names, it is so natural it hardly needs an explanation.<|endoftext|>
TITLE: Can Inequivalent Topologies Have Same Sheaves/Cohomology?
QUESTION [6 upvotes]: Let $C$ be a fixed category, and let $T_1$ and $T_2$ be two Grothendieck (pre)topologies on $C$. We say $T_1$ is subordinate to $T_2$ if every covering in $T_1$ has a refinement in $T_2$. We say $T_1$ and $T_2$ are equivalent if $T_1$ is subordinate to $T_2$ and $T_2$ is subordinate to $T_1$. We know that if $T_1$ is subordinate to $T_2$, then any sheaf for the topology $T_2$ is also a sheaf for $T_1$. Is the converse true? In particular, if the categories of sheaves are the same, does that imply the topologies are equivalent? At least, if we know that $T_1$ is subordinate to $T_2$ but not equivalent to it, do we know that there is a sheaf for $T_1$ that is not a sheaf for $T_2$? How about if both topologies are subcanonical?
If $T_1$ is subordinate to $T_2$, we have a morphism of topologies $g: T_1 \to T_2$. If $g_*$ is exact (so the cohomology of all sheaves for $T_2$ is the same in the two topologies) does it imply that the two topologies are equivalent? Thank you.

REPLY [8 votes]: Does this hold water?
If the sheaves are the same, they have the same subobject classifier, which is the sheaf of closed sieves, so both topologies have the same notion of closed sieves, so the same notion of covers, hence they are equivalent.

REPLY [5 votes]: I would think so. 
Two pretopologies are equivalent when they generate the same topology, that is, when the have the same sieves. If $A$ is an object of $C$, a sieve on $A$ is a subfunctor of the functor $h_A$ represented by $A$; and a subfunctor $S$ of $h_A$ is a sieve with respect to a topology $T$ if and only if the induced morphism $S^{\rm sh} \to h_A^{\rm sh}$ of the sheafifications is an isomorphism. Of course the category of sheaves determines the sheafifications; so if two topologies define the same sheaves, they admit the same sieves, so they are equivalent.<|endoftext|>
TITLE: A Model where Dedekind Reals and Cauchy Reals are Different
QUESTION [30 upvotes]: Is there a model where Dedekind reals and Cauchy reals are different? I'd appreciate if someone can refer me to any related work in case such a model exists.

REPLY [2 votes]: A good reference for this is Fourman and Hyland's 1970 paper on sheaf models for analysis,

Fourman M.P., Hyland J.M.E. (1979) Sheaf models for analysis. In: Fourman M., Mulvey C., Scott D. (eds) Applications of Sheaves. Lecture Notes in Mathematics, vol 753, doi:10.1007/BFb0061823

which I managed to find online here:
https://www.dpmms.cam.ac.uk/~martin/Research/Oldpapers/analysis79.pdf<|endoftext|>
TITLE: A finely open set, not open up to polar set?
QUESTION [6 upvotes]: I already asked this on M.SE, but get no answers.
Is there a (simple) example of a finely open set (i.e. w.r.t. the fine topology in potential theory) $O$ in $\mathbb R^n$, $n \ge 2$, which is not open up to a polar set (i.e. zero capacity), i.e., there does not exist a polar set $M$, such that the symmetric difference of $O$ and $M$ is open?
I found some examples of finely open sets (e.g. constructed using the Lebesgue spine), but all were "almost" open (in the above sense).
I use the following definition of the ($H_0^1$)-capacity:
\begin{equation*}
\operatorname{cap}(A) = \inf\big( \|\nabla v\|_{L^2(\Omega)}^2 : v \in H_0^1(\Omega) \text{ and } v \ge 1 \text{ on a neighbourhood of } A\big).
\end{equation*}
Here, $\Omega$ may be $\mathbb R^n$ or a bounded, open set.

REPLY [3 votes]: I like the following abstract construction, which relies on a connection between quasi-continuity and the fine topology and also works in the setting of $p\ne 2$. It is debatable whether the resulting example is “simple”. The standard reference for the required techniques is Adams, Hedberg: Function spaces and potential theory, 1995.
Let $(B_k)$ be a sequence of open balls contained in the open unit ball $D\subset\mathbb{R}^n$ such that $A:=\bigcup_{k} B_k$ is dense in $\overline{D}$, but such that $\operatorname{cap}(A)<\operatorname{cap}(D)$. This can be arranged by centering the balls in a countable dense subset of $D$ and letting the radii decrease sufficiently rapidly. [As we are in the Hilbert space setting $p=2$, it is here where we need $n\ge2$.] Let $K:=\overline{D}\setminus A$. Note that $K$ is compact and has no interior points, but is necessarily still large in measure.
By Theorem 11.3.2 (in Adams, Hedberg), there exists a nontrivial nonnegative Sobolev function $u$ supported in $K$. In fact, let $w\in H^1(\mathbb{R}^n)$ be the capacitary extremal of $A$. Then $0\le \tilde{w}\le 1$ and $\tilde{w}=1$ quasi-everywhere on $A$, where $\tilde{w}$ is the quasi-continuous representative of $w$. As $\operatorname{cap}(A)<\operatorname{cap}(D)$, we cannot have $w=1$ a.e. on $D$. Let $\varphi\in C^\infty_c(\mathbb{R}^n)$ be such that $\varphi(x)>0$ for all $x\in D$ and $\varphi(x)=0$ for all $x\notin D$. Set $u := \varphi(1-w)$. Then $u\in H^1(\mathbb{R}^n)$ has the desired properties.
It follows that $U := \{x\in\mathbb{R}^n: \tilde{u}(x)>0\}$ is quasi-open and (possibly after removing a polar set) $U\subset K$. Let $O$ be the fine interior of $U$. Then $O\subset U$ and $U\setminus O$ is polar; see Section 6.4 and Proposition 6.4.12 (in Adams, Hedberg). As $u$ is nontrivial, both $U$ and $O$ have positive measure.
Consequently $O$ is a nontrivial finely open set that has no Euclidean interior (as it is contained in $K$). Let $V$ be an open set. If $V\cap O$ is nonempty, then $V\cap A$ has positive measure and clearly $V\cap A\subset V\setminus O$. Otherwise [i.e. if $V\cap O=\emptyset$] the set $O\setminus V=O$ has positive measure. It follows that the symmetric difference of $O$ and $V$ cannot be polar.
So $O$ is not equivalent (up to a polar symmetric difference) to any open set.<|endoftext|>
TITLE: AC and Krull's theorem equivalence
QUESTION [5 upvotes]: It is well known that the axiom of choice can be used to prove Krull's theorem which states that every ring has a maximal ideal. However, i heard once that Krull's theorem is equivalent to the AC (or to Zorn's lemma). Is that true?
So, we suggest each ring (perhaps, each commutative ring) has a maximal ideal and now we need to build some ring to prove the AC (Zorn's lemma, Zermelo theorem etc). Could anybody explain how to do that?

REPLY [6 votes]: Originally the proof was given by Hodges, from 1979 [1].
Banaschewski gave a new proof of the theorem in 1994 [2].

Bibliography:

Hodges, W., Krull implies Zorn. Journal of the London Mathematical Society 2 (1979), 285-287.
Banaschewski, B., A New Proof that “Krull implies Zorn”. Mathematical Logic Quarterly 40 (1994), 478-480.<|endoftext|>
TITLE: Algebras with finite essential arity
QUESTION [6 upvotes]: We are talking about algebras in the universal algebraic sense, that is, a set that $A$ is equipped with a set $F$ of finitary operations on $A$.

Definition: An algebra $(A,F)$ is said to have finite essential arity if there exists a number $n \in \mathbb{N}$ such that each of its term functions depends on at most $n$ variables.

I am mostly interested in the case where $A$ is finite. I have read some results about this,  but I am not sure what exactly is known about these kind of algebras. 
I am particularly interested in identities that hold (or don't hold) for algebras of this kind. I know a few conditions on identities that imply finite essential arity, but not so many that are required or even equivalent. Thus, I would be very interested in examples of this kind.
To ask one more specific question that goes in the same direction: Are there (finite) algebras with finite essential arity that are not strongly abelian? What would be an example for that?

REPLY [4 votes]: A finite algebra in a finite language has finite essential arity if and only if it is strongly nilpotent. This is proved in Section 4 of:
Keith A. Kearnes and Emil W. Kiss, Finite algebras of finite complexity. Discrete Math. 207 (1999), 89--135.<|endoftext|>
TITLE: Simplest complex curves with isomorphic Jacobian
QUESTION [5 upvotes]: Hi. 
What are the simplest examples to have in mind of non-isomorphic smooth (complex) algebraic curves with isomorphic jacobian variety?

REPLY [3 votes]: Slightly tangential -- the following paper of Bjorn Poonen offers some interesting computational perspectives on curves of genus at least $2$ which is as Francesco mentions, the lowest possible genus where such examples exist. This seems like it might be relevant.
http://www-math.mit.edu/~poonen/papers/ants2.pdf
I think the following is an interesting question to ask over a fixed number field $K$: Can one make tables of genus $2$ curves over $K$ ordered in some lexicographic fashion by the invariants? Perhaps the curves would be modelled by singular plane curves $y^2 = f(x)$ where $f$ has degree $5$ or $6$. The first pair $(X,Y)$ with isomorphic Jacobian variety to occur in this table might satisfy the request that the pair $(X,Y)$ be the "simplest" such pair to be defined over $K$. 
Poonen notes that not every not every hyperelliptic curve has a model of the form $y^2 = f(x)$ over
$k$ a field, because the quotient of the curve by its hyperelliptic involution might be a twist of $\mathbb{P}^1$ i.e. birational over $k$ to a conic in $\mathbb{P}^2$ without a $k$-rational point. However when $g$ is even, it seems to turn out (I'm not sure why) that the $k$-form of $\mathbb{P}^1$ has a $k$-point and the hyperelliptic curve has a model $y^2 = f(x)$, so for $g = 2$ we should be okay.<|endoftext|>
TITLE: Are finite index subgroups of inertia closed?
QUESTION [7 upvotes]: Let $K$ be a finite extension of the $p$-adic numbers. $G_K$ be its absolute Galois group and $I_K$ the inertia subgroup. Are finite index subgroups of $I_K$ closed in its profinite topology?
By a result of  Nikolov/Segal it suffices to show, that inertia is topologically finitely generated.

What has been proved in the direction of this or the analogous question on wild inertia?

Do the results of Jannsen/Koch/Wingberg on the finite generator rank of the local Galois group help? I couldn't find anything on this and considering its importance, it surely would be mentioned somewhere, if it were always known. So maybe someone has some conditional results?

REPLY [7 votes]: Here is another way to see that the wild inertia group $P_K=\mathrm{Gal}(\bar K|T)$ (where $T$ is the maximal tamely ramified extension of $K$ (a finite extension of $\mathbf{Q}_p$) in $\bar K$, an algebraic closure of $K$, is not finitely generated (as a profinite group).  One doesn't need class field theory, only Kummer theory and the fact that a pro-$p$ group $G$ is finitely generated if and only if the $\mathbf{F}_p$-space $G/D$ is finite, where $D$ is the closed subgroup generated by all commutators and all $p$-th powers in $D$.
In the case at hand, $P_K/D=\mathrm{Gal}(T(\root p\of{T^\times})|T)$, and it can be easily seen that the $\mathbf{F}_p$-space $T^\times/T^{\times p}$ is infinite.<|endoftext|>
TITLE: Entropy of edit distance
QUESTION [13 upvotes]: The edit or Levenshtein distance between two strings is the minimum number of single character insertions, deletions and substitutions to transform one string into another.   If we take random binary strings $A$ and $B$ of the same length $n$, is it known what the entropy of the edit distance between $A$ and $B$ is?
Update.  I would be happy with a proof that the entropy is asymptotically  $c \log n$ for some (unknown) $c>0$ as suggested by Anthony Quas (or even $(\Theta(\log{n})$ which potentially saves having to prove convergence).

REPLY [3 votes]: This is not a complete answer and may not tell you anything you don't already know, but it's too long for a comment.  
You can get an estimate on the number of words at an edit distance of $k$ from a given word of length $n$ by using the arguments in the proof of Lemma 2.6 in this paper.  (The lemma is on page 10 and its proof is on page 28.)
The idea is that if a word $w$ is at an edit distance of $k$ from a word $v$, then one can get from $v$ to $w$ via the following steps:

insert $k$ copies of the symbol 'e' (for 'edit') into the word $v$;
one by one, go through the symbols 'e' and either change the symbol immediately before it, delete the symbol immediately before it, or insert a symbol immediately before it, and then delete the 'e'.

Step 1 gives a word of length $n+k$ with $k$ occurrences of the symbol 'e', so there are ${n+k\choose k}$ possibilities after this step.  Then step 2 gives at most $3^k$ possible words for each of those possibilities, so for a fixed word $v$ of length $n$ one obtains the bound
$$
\#\{w \mid \text{edit distance from $v$ to $w$ is $k$}\} \leq 3^k {n+k \choose k}.
$$
For small $k$ this is a reasonable bound; in particular when $k\ll n$ this is not much larger than ${n\choose k}$, the number of words a Hamming distance of $k$ away.  The problem is that for larger values of $k$ the procedure described above succumbs to a lot of overcounting, so it's not clear what the actual bound should be.<|endoftext|>
TITLE: Probabilities of a random walk exiting a set
QUESTION [5 upvotes]: Let $F$ be a finite connected set in a graph (soon to be the Cayley graph of a group) and $\mathrm{Ex}_x^F$ be the function on the vertices in $F^c$ which are neighbour to vertices in $F$ defined as follow $\mathrm{Ex}_x^F(y)$ is the probability that the first time a random walker starting at $x$ exits $F$ is through $y$. To make sure this is defined, suppose the graph is transient.
$\textbf{Question}$: Assume the (transient) graph is the Cayley graph of an amenable group (for some finite generating set). Let $s$ be a neighbour of $e$. Does there exist a sequence of finite connected sets (containing $e$ and $s$) $\{F_i\}$ so that $\mathrm{Ex}_e^{F_i}(y) - \mathrm{Ex}_s^{F_i}(y) \overset{i}{\to} 0$ in $\ell^1$ norm?
The hypothesis about amenability is motivated by 

the above is clearly false in the free group and most probably false in any hyperbolic group. 
its more plausible as the Cayley graph of amenable groups (by Kesten) are those where $P^{(n)}(e,e)^{1/n} \to 1$, i.e. they tend to return "often" to where they were. Thus the random walks have bigger chance to "fuse", i.e. if a random walker starting at $s$, is at some time where the random walker starting at $e$ was, then the exit probabilities will be the same.

REPLY [7 votes]: Say that a graph $G$ has the Liouville property if all bounded (discrete-)harmonic functions on it are constant.

If it is possible to couple random walks on $G$ started from any two starting points in such a way that they almost surely coincide after some (random) time, then the graph is Liouville. The reason for that is that one can write the value of a bounded harmonic function $f$ at $x$ as the expected value of $f(X_t)$ where $X$ is a random walk on $G$ and $t$ is any stopping time; it random walks from $x$ and $y$ couple with high probability by time $t$ this gives an upper bound on $|f(x)-f(y)|$ which shows that $f$ has to be constant.
If your graph admits non-constant bounded harmonic functions, then the TV distance between harmonic measures from $e$ and $s$ (which is another way of naming what you are interested in)  cannot go to $0$ for all pairs $(e,s)$. Indeed, if $f$ is a non-constant bounded harmonic function and $f(e) \neq f(s)$, then writing $f$ as the expected exit value one gets a lower bound on the TV distance in terms of $|f(e)-f(s)|$.
On the other hand, assume that the $\ell^1$ norm you are interested in does not go to $0$. It means that there exists a sequence $g_i$ of functions bounded by $1$ in absolute value, each defined on $F_i^c$ and such that the integrals of $g_i$ through your two exit distributions differ by at least some $\delta>0$. Taking the harmonic extension $f_i$ of $g_i$ inside $F_i$ one gets $|f_i(s) - f_i(e)| \geq \delta$. Letting $i$ go to infinity and using a diagonal argument one gets a bounded harmonic function $f$ defined on the union of the $F_i$ (which I am assuming to be the whole graph?) and such that $|f(s)-f(e)|>0$, i.e. it is non-constant.

So, the answer to your question is positive iff the graph is Liouville. Now for the link with amenability: it is not true that every amenable group is Liouville, and a counterexample is given by the lamplighter group on $Z^3$ (or on any transient amenable group for that mattter). One natural non-constant harmonic function is the following: given any $(\omega,x)$ on the LL, where $\omega$ is the lamp configuration and $x$ the location of the walker, define $f((\omega,x))$ to be the probability, for a random walk $(\omega_t,x_t)$ started there, that the lamp at the origin $\omega_t(0)$ is eventually on. Note that the state of that lamp changes only finitely many times because the underlying random walk is transient. 
This function is clearly harmonic and bounded. To see that it is non-constant, just take $x$ very large: the probability that $x_t$ ever visits the origin will be very small, so with high probability the eventual state of the lamp at the origin will be the same as its state at time $0$, ie its state on $\omega$: $f((\omega,x))-\omega(0) = o_{x\to\infty}(1)$. So some places $f$ will be close to $0$ and some places it will be close to $1$.<|endoftext|>
TITLE: A_infinity structure on cohomology and the weight filtration
QUESTION [12 upvotes]: Let $X$ be a complex algebraic variety.  The rational cohomology of $X(\mathbb{C})$ carries a canonical filtration called the weight filtration.  It also carries a canonical equivalence class of $A_\infty$ structures (extending the cup product ring structure).  
How do these two structures interact? Is the weight of a higher multiplication $m_n(x_1, \cdots, x_n)$ determined by the weights of the $x_i$?  If the answer is no in general, then can one choose a particular nice $A_\infty$ structure within the equivalence class for which the answer is yes?

REPLY [6 votes]: This was meant to be a comment before it got too long.
One of the ways to define a minimal $A_\infty$ model of a cdga (more generally, an $A_\infty$-algebra) $A$ is Merkulov's recipe (see Merkulov http://arxiv.org/pdf/math/9809172.pdf or Chuang and Lazarev http://arxiv.org/pdf/0802.3507.pdf for a summary). All we need is а decomposition $A=W\oplus K$ into subcomplexes with the differential of $A$ vanishing on $W$ (so $W$ can be identified with $H^*(A)$), $K$ acyclic, and a contracting homotopy $h:K\to K$ satisfying $h^2=0$. This decomposition is called the Hodge decomposition since it becomes one when $A=\mathcal{E}^*(M)$, the algebra of complex-valued differential forms on a K\"ahler manifold $M$, in which case $W$ is the subspace of harmonic forms. Note that $W$ needn't be a subalgebra. Note also to find such a decomposition it suffices to find a splitting $A$ into cocycles and some complement and then splitting off the cohomology inside the cocycles, see Chuang, Lazarev, ibid., p. 5.
So if $A$ is some kind of cochain algebra associated to a complex algebraic variety and we take any mixed Hodge structure on $K$ compatible with $h$ and the differential and such that the multpplication $A\otimes A\to A$ is a map of mixed Hodge structures (the latter condition is probably too strong and may be relaxed in some way). Applying Merkulov's formula we get higher multiplications $m_n$ on $H^*(A)$ which are maps of mixed Hodge structures of degree $0$. These maps will be linear combinations of compositions of maps constructed from maps of the form $(0,h)$, multiplications in $A$ and projections to $W$ along $K$. In particular the weight of $m_n(x_1,\ldots,x_n)$ will be the sum of the weights of $x_1,\ldots, x_n$. Of course there is no reason this $A_\infty$ structure should be compatible with morphisms of varieties. I would guess it is possible to choose the splittings so that it is  but this would require some more work.<|endoftext|>
TITLE: Deriving Konig's Lemma directly from Infinite Ramsey's Theorem for triples
QUESTION [8 upvotes]: Let KL denote König's Lemma (for trees over $\mathbb{N}$), and RT(3) denote the
Infinite Ramsey Theorem for triples over $\mathbb{N}$ (notation as in Simpson's
book Subsystems of second order arithmetic, 2ed, 2009, see definitions below).

Question:  Is there a simple direct derivation of KL from RT(3)?

KL, RT(3), and ACA$_0$ are equivalent over RCA$_0$, and Simpson's book
has proofs (in RCA$_0$) of the following implications:
 ACA$_0$ $\implies$ KL $\implies$ RT(3) $\implies$ ACA$_0$.
I am 
looking for a direct "simple" proof of RT(3) $\implies$ KL (formalizable
in RCA$_0$ and without having to go through an intermediate ACA$_0$).
Definitions

Here KL (König's Lemma) is the statement "Every finitely
branching infinite tree over $\mathbb{N}$ has an infinite branch", where
(a) $T$ is a tree over $\mathbb{N}$ if $T$ is a set of finite sequences
from $\mathbb{N}$ such that any initial segment of any sequence in $T$ is in $T$,
(b) $T$ is finitely branching if for any sequence $u \in T$ there are
at most finitely many sequences in $T$ of length $\operatorname{length}(u)+1$
which extend $u$,
and (c) a branch of a tree is a maximal linearly ordered subset of it
(under the relation $u \preceq v$ iff $u$ is an initial segment of $v$).
RT(3) (Infinite Ramsey Theorem for triples) is the statement
"If $F$ is a finite set and $f \colon [\mathbf{N}]^3 \to F$, then there is
an infinite subset $H$ of $\mathbb{N}$ such that $f$ is constant on $[H]^3$",
where $[X]^n$ denotes the set of all $n$-element subsets of $X$ (for any set $X$).

REPLY [7 votes]: Here is an attempt...
The function $\Delta:[\mathbb{N}^{< \omega}]^2 \rightarrow \mathbb{N}$ is given by $\Delta(a,b)$ is the largest natural number $i$ such that $\langle a(0), a(1) \ldots a(i-1)\rangle = \langle b(0), b(1) \ldots b(i-1)\rangle$. If $a =b$ then this is the length of $a$, and if $a$ is an initial segment of $b$ then this is again the length of $a$. The point is that $\Delta(a,b)$ is the length at which $a$ and $b$ `branch'.
Now, let $T \subseteq \mathbb{N}^{< \omega}$ be a finitely branching infinite tree. Define the function $f: [T]^{3} \rightarrow 3$ by $f(a,b,c) = 0$ if $|\{ \Delta(a,b) , \Delta(a,c), \Delta(c,b)\}| = 1$, that is, if all three of them branch off at the same point. Otherwise, $f(a,b,c) = 1$ if $a< b < c$ (in the lexicographic order on $\mathbb{N}^{< \omega}$) and $\Delta(a,b) > \Delta(b,c)$. If not, let $f(a,b,c) = 2$. If $H$ is an infinite homogenous set for $f$, as $T$ is finitely branching, $f[H]$ cannot be $\{0\}$. Hence, it must be either $\{1\}$ or $\{2\}$. We observe that in either case, if $\{a,b,c\}$ is a subset of $H$, then if $a < b < c$ is their order lexicographically, then it is not possible that $\Delta(a,b) = \Delta(b,c)$, as this implies that $\Delta(a,b) = \Delta(b,c) = \Delta(c,a)$.
Now, suppose $f[H] = \{1\}$. Let $a < b < c < d$ be in $H$ then $\Delta(a,b) > \Delta(b,c) > \Delta(c,d)$. It follows that $H$ cannot contain an infinite lexicographically increasing sequence. Assuming that we can get an actual decreasing sequence $\langle a_i\rangle$ in $H$, we can choose a branch through $T$ by taking the union of all those elements of $T$ which are initial segments of cofinitely many of the $a_i$. This is a computable procedure because the first $n$ digits are fixed $a_{n+1}$ onwards.
So to pick this infinite decreasing sequence, do this naively, namely, if you have made the choices upto $i$, let $a_{i+1}$ be any element in $H$ which is lexicographically below $a_i$. To find such an an element you only need to look at more elements than the length of $a_i$ (here again we use the fact that if $a,b,c$ are distinct elements of $H$ then $|\{ \Delta(a,b) , \Delta(a,c), \Delta(c,b)\}| = 2$, so if you look at more elements than the length of $a_i$, you must find two elements which branch off from  $a_i$ at the same point. If neither of them is smaller than $a_i$, then we have a situation like this: $a_i < b < c$ and $\Delta(a_i, b) = \Delta(a_i, c) < \Delta(b,c)$, which is not possible by the homogeniety of $H$ with respect to $f$).
The proof for the case of $f[H] = \{2\}$ is similar.<|endoftext|>
TITLE: Do quasi convex hyperbolic subgroups remain quasi convex after adding redundant generators?
QUESTION [6 upvotes]: We know now that hyperbolic 3-manifolds virtually embed into right-angled Artin groups as quasiconvex subgroups. Also, quasiconvexity depends on the generating set.
I have been constructing a space at infinity for right-angled artin groups where the boundaries of quasiconvex hyperbolic subgroups will embed in a natural way; but to construct it, I've been forced to add all redundant generators of the form $b=a_1...a_k$, where all if the $a_i$ are distinct commuting generators in the standard generating set for the RAAG (they can also be unversed of generators). This is just adding in all 'diagonal' generators, and corresponds in free abelian groups to changing from an $L^1$ metric to an $L^{\infty}$ metric.
My question is, will this make some quasi convex subgroups no longer convex? In the $\mathbb{Z}^2$ case (generated by $a,b$), I think that the subgroup generated by $a$ is quasi convex before adding the generators $ab$, $ab^{-1}$, etc. So my final question is, will non-elementary hyperbolic quasi convex subgroups stay quasi convex after adding in these generators?
Edit: the easiest example of a non-elementary hyperbolic subgroup of a RAAG is the following: take the pentagon RAAG generated by $a,b,c,d,e$. Then the subgroup generated by $ab^{-1},bc^{-1},cd^{-1},de^{-1}$ is a two-holed torus group with standard relations.

REPLY [6 votes]: As this is too long for a comment, I'll post it as an answer.
As you seem to be interested in CAT(0) cube complexes, I want to point out that there is, in fact, a ninth type of quasiconvexity, which turns out to be the 'right' notion in this context.
Definition: Suppose that a group $G$ acts properly discontinuously and cocompactly by isometries on a CAT(0) cube complex $X$.  A subgroup $H$ is called combinatorially convex cocompact (CCC) if it acts properly discontinuously and cocompactly on a convex subcomplex $Y\subseteq X$.
(Note: this terminology is not standard---people often just write 'quasiconvex'---but I think this is less confusing and more descriptive.)
Haglund proved that this is equivalent to an $H$-orbit's being quasiconvex in the 1-skeleton of $X$ [1].  Therefore, in the special case where $G$ is a RAAG (equipped with its standard generating set) and $X$ is the universal cover of its Salvetti complex, this coincides with Misha's definition 1.
To justify that this is the 'right' definition in preference to definition 1, let me point out that it's clear that a CCC subgroup of a CCC subgroup is again CCC, whereas for definition 1 the transitivity of quasiconvexity doesn't even make sense (since the subgroup isn't equipped with a generating set).
[1]  Haglund, Frédéric, 'Finite index subgroups of graph products.', Geom. Dedicata 135 (2008), 167–209.<|endoftext|>
TITLE: What is the effect of adding 1/2 to a continued fraction?
QUESTION [30 upvotes]: Is there a simple answer to the question "what happens to the continued fraction expansion of an irrational number when you add 1/2?"  A closely related question is "what happens to such an expansion when you multiply by 2?"
Remarks:  I'm not sure what qualifies for an answer.  The motivation comes from wanting to better understand the equivalence relation on integer sequences generated by tail equivalence (which is generated by adding integers and taking reciprocals) and closure under doubling/halving.  It is known that this Borel equivalence relation is not hyperfinite, so the answer cannot be too simple.
Edit:  The answers are not really what I am asking for.  It is clear there is some recursive procedure for doing this, just like there is a recursive procedure for taking a square root of a decimal expansion.  I'm looking for something which one might call a "closed form".  For instance, if you start with a periodic expansion, adding 1/2 produces a new periodic expansion.  Is there a simple transformation on the initial and periodic parts which corresponds to adding 1/2?  For instance, can this be done with a finite state automaton?  An authoritative "there is no such nice answer" would actually be an acceptable answer.

REPLY [11 votes]: Kind of a late response, but since no one else mentioned it I think it is worth pointing out that the procedure for computing homographies/linear fractional transformations on continued fractions (mentioned in Douglas Zare's answer) was also described explicitly in terms of finite state automata by George N. Raney, in "On continued fractions and finite automata", Mathematische Annalen 206:4, 1973. (This was a little after Gosper's memo, but neither one cites the other, so I suppose Gosper and Raney came up with these ideas independently?)
Formally, Raney showed how to interpret homographies $\phi(x) = \frac{ax + b}{cx + d}$ (e.g., the map $x \mapsto x + 1/2$, taking $a=2,b=1,c=0,d=2$) as states of finite state transducers. The transitions executed by these transducers are of the form
$$\phi_1 \underset{I:O}{\longrightarrow} \phi_2$$
where $\phi_1$ and $\phi_2$ are states corresponding to a pair of homographies, and $I,O \in \{L,R\}^*$ are binary words describing a piece of the continued fractions of the input and output, respectively. Note that this is based on an encoding of continued fractions $x = [a_0;a_1,a_2,\dots]$ as binary sequences $R^{a_0}L^{a_1}R^{a_2}\cdots$ (which, incidentally, is closely related to the Stern-Brocot/Calkin-Wilf representations of the rationals).
I'm mostly ignorant as to how far Raney's approach has been taken (and curious myself). However, one significant follow-up work was by Pierre Liardet and Pierre Stambul in "Algebraic Computations with Continued Fractions" (1998), where they showed how to unify Gosper's and Raney's approaches by building transducers to interpret general "bihomographies"/"biquadratic fractional transformations" $\phi(x,y) = \frac{axy + bx + cy + d}{exy + fx + gy + h}$.

UPDATE:
I went ahead and derived the Raney transducer for computing $x \mapsto x + 1/2$.

There are four states $q_1,\dots,q_4$, corresponding to the homographies
\begin{aligned}
q_1(x) &= (2x + 1)/2 = x + 1/2 \\
q_2(x) &= 4x \\
q_3(x) &= x/4 \\
q_4(x) &= 2x/(x+2)
\end{aligned}
A total of 16 transitions, which can be grouped into eight transitions
\begin{aligned}
q_1 &\overset{R:R}{\longrightarrow} q_1 \\
q_1 &\overset{LL:LR}{\longrightarrow} q_2 \\
q_1 &\overset{LR:RLL}{\longrightarrow} q_3 \\
q_2 &\overset{R:RRRR}{\longrightarrow} q_2 \\
q_2 &\overset{LR:RR}{\longrightarrow} q_4 \\
q_2 &\overset{LLR:RL}{\longrightarrow} q_1 \\
q_2 &\overset{LLLR:RLLL}{\longrightarrow} q_3 \\
q_2 &\overset{LLLL:L}{\longrightarrow} q_2
\end{aligned}
together with a dual set of eight transitions:
\begin{aligned}
q_3 &\overset{L:LLLL}{\longrightarrow} q_3 \\
q_3 &\overset{RL:LL}{\longrightarrow} q_1 \\
q_3 &\overset{RRL:LR}{\longrightarrow} q_4 \\
q_3 &\overset{RRRL:LRRR}{\longrightarrow} q_2 \\
q_3 &\overset{RRRR:R}{\longrightarrow} q_3 \\
q_4 &\overset{L:L}{\longrightarrow} q_4 \\
q_4 &\overset{RR:RL}{\longrightarrow} q_3 \\
q_4 &\overset{RL:LRR}{\longrightarrow} q_2 \\
\end{aligned}

In case you are interested, here is a little Haskell implementation, which shows how to use this machine to add 1/2 to the continued fractions representing the golden ratio and $e$.<|endoftext|>
TITLE: motivating geometric representation theory
QUESTION [22 upvotes]: I am wondering if there is a good motivation for geometric representation theory from within the questions of classical representation theory.
In other words, I'd be curious to see something using geometry that is "meatier" than, say, just using geometric techniques to construct the exceptional isomorphisms between low-dimensional Lie groups --- but something that can still be stated in the framework of classical representation theory (unlike, say, the Borel-Weil theorem and friends).

REPLY [6 votes]: Consider triples $(\lambda,\mu,\nu)$ of dominant weights of $G$ such that the irrep $V_\nu$ occurs in $V_\lambda \otimes V_\mu$. Then this space of triples is closed under addition.
Proof. An intertwiner can be identified with a $G$-invariant section of the $(\lambda^*,\mu,\nu)$ equivariant line bundle over $(G/B)^3$. Tensoring two such sections together, we get a third, which is again nonzero because $(G/B)^3$ is reduced and irreducible.
(Moreover, this monoid is finitely generated, also not hard to prove with this approach.)<|endoftext|>
TITLE: A divergent series related to the number of divisors of of p-1
QUESTION [6 upvotes]: Let $d(n)$ denote the number of divisors of $n$. Is it known that the series
$$\sum_{p \text{ prime}} \frac{1}{d(p-1)}$$
diverges?
This would follow immediately from the Sophie Germain Conjecture. Indeed, if there are infinitely many primes of the form $2p+1$ ($p$ a prime), then infinitely many terms of the series are equal to $1/4$, so the series doesn't even satisfy the most basic requirement for convergence! So, surely there must be a direct proof?

REPLY [4 votes]: Answering my own question, because I totally overlooked the following ridiculous idea:
Obviously $d(n)\leq n$ for every $n$. Thus $d(p-1)\leq p-1 < p$, so $1/d(p-1) > 1/p$ and the divergence follows from the divergence of $\sum 1/p$ (if one is willing to assume that).<|endoftext|>
TITLE: What is the best *general triangle*?
QUESTION [7 upvotes]: During courses on geometry it is sometimes necessary to draw a triangle on the blackboard that can easily be recognized as a general triangle. It must not be rectangular and must not have two or more equal angles. Further all angles should be less than $\pi/2$. Has anybody optimized this old problem of geometry-teachers?

REPLY [9 votes]: Since the question is reopened, I convert my comments into an answer. 
The "general triangle" must have angles 75,60,45 degrees (as in Carl Dettmann's comment above), and so there is only one such triangle up to similarity. Here is a "psycho-mathematical" proof. 1) The triangle should not be obtuse (otherwise it is not a general triangle but an obtuse one). 2) The difference between any two angles should be at least 15 degrees (otherwise for an untrained eye of a student the triangle will look too isosceles). 3) The biggest angle should be at most 75 degrees (at least 15 degrees apart from 90), otherwise the triangle looks "too right". 4) The sum of angles must be 180 degrees (otherwise the 5th postulate would be broken). Parts 1),2),3), 4)  immediately imply the claim.
PS There are many general triangles on the hyperbolic plane and none on the sphere.<|endoftext|>
TITLE: Is there an algebra for divergent series summation operators?
QUESTION [13 upvotes]: Let $D$ denote a divergent series and let $C$ denote a convergent series. 
Furthermore, let $s : $ { Series } $\to$ $\mathbb{C}$  be a regular, linear divergent series operator, which is either one of these operators: 
(the hyperlinks will direct you to the wiki page of the relevant summation method, not the person who invented/discovered it) 

 Borel  summation
Abel summation
Euler summation
Cesàro summation
Lambert summation
Ramanujan summation
Summing the series by means of Analytic continutation 
Some  Regularization  method

I am wondering if there is any meaningful way to answer the following questions (Assuming $D_1 , D_2$ are summable with $s$):

What does $s(D_1 + D_2)$ equal? Is it always equal to $s(D_2 + D_1)$ ? How does it relate to $s(D_1)$ and $s(D_2)$ ?
What does $s(D_1 \cdot D_2) $ equal? Is it always equal to $s(D_2 \cdot D_1)$ ? How does it relate to $s(D_1)$ and $s(D_2)$ ?
What happens when we add convergent series into the mix? And what if we're summing linear combinations of $n$ convergent and $m$ divergent series? 

Do the results differ for different summation methods, listed above? 
(This question was migrated from MSE. I also asked a somewhat similar question on MO once.)

REPLY [8 votes]: I'll try to offer as much knowledge as I can on this topic, which might not be that much. 
Firstly your linear form $s$ is defined on a strict linear subspace $L$ of the vector spaces of all formal series. Secondly you seem to mix formal numeric series and formal power series, which are distinct rings.
1) 2) By definition of the ring of formal (numeric or power) series over $\mathbb C$, the addition and product are commutative, so $s(D1+D2)=s(D2+D1)$ and $s(D1\cdot D2)=s(D2\cdot D1)$. Now if $s$ is $\mathbb C$-linear then $s(D1+D2)=s(D1)+s(D2)$. This is not true anymore for the product, even for convergent series as the product of two conditionnally convergent series might only be divergent. If both are absolutely convergent then the Cauchy-product formula shows that $s(D1\cdot D2)=s(D1)s(D2)$. This formula for convergent series is related to 3) below.
3) Every method you mention agrees with the usual sum for convergent series (except maybe Ramanujan's which I don't know about). So adding finitely many convergent series to the mix does not change anything by linearity. By the way the "summation by analytic continuation" is often refferred to as Mittag-Leffler's summation. Be careful though that it is not well defined due to the fact that most analytic continuations yield multivalued functions.
I think I should mention a nice paper by Lyubich which tries to axiomatically define what should be a coherent summation method for numeric series. The main property is that, following the afore-mentionned axioms, the sum $1+1+1+\cdots$ will never be assigned a finite value, i.e. it does not belong to $L$.
Also as far as I know the regularization process in QFT first transform a "series" of infinite quantities into a formal power series over $\mathbb C$ which may be (is) divergent, as explained somehow in this thread. In some cases the latter is a Borel-summable power series. See this survey (in French) by J.-P. Ramis for more details regarding this topic.
You might also want to learn more about Borel-summation for Gevrey power series through the works of Ramis, Sibuya, Balser and others and also through the mould/alien calculus developed by Écalle. An example is presented in my answer here.<|endoftext|>
TITLE: The divisor bound for binary quadratic forms
QUESTION [5 upvotes]: The divisor bound asserts that for a large integer $n\in \mathbb{Z}$, the number of divisors of $n$ is at most $n^{o(1)}$ as $n\rightarrow\infty$. See here for a discussion of proofs using elementary analysis.
This  MO topic asked the same question with $\mathbb{Z}$ replaced by the ring of integers in a number field.
My question is whether a similar bound holds in the case of the number of representations by a   binary, positive, integral quadratic form.
More precisely, let $Q(x,y) = a x^2+ b x y + c y^2$ be a binary quadratic form with $a,b,c$ integers and discriminant $\Delta = b^2 - 4 a c <0$.
For any positive integer $n$, let $r_Q(n)$ be the number of integer solutions $(x,y)$ of the equation
$$n = Q(x,y).$$

Question: For $n$ large, is it known that $r_Q(n) = n^{o(1)}$?

Any references would be helpful.
When $\Delta$ is a fundamental discriminant, then one can use the theory of quadratic fields and the divisor bounds therein to deduce the same for our case. When $\Delta$ is a general discriminant, the cases seem tricky (especially when the associated fundamental discriminant $\Delta_0$ is $1 \mod 4 $).
From D.A.Buell, Binary Quadratic Forms, Section 4.4, it seems that the classical theory of quadratic forms doesn't impose any condition on $\Delta$. In fact it seems to proceed along the lines of factoring $n$ into prime factors, getting at most $2$ classes of forms which represent each prime (depending on $\Delta$, of course) and composing them to obtain the classes of forms representing $n$. Then, within each class, it is up to the modular group to generate more solutions (the orbit of $(\pm 1,0)$ under $SL_2\mathbb{Z}$). The cardinality of this orbit seems to be bounded by an absolute constant. 
In this approach, the answer to the question seems to be obtained by an analysis parallel to that of the divisor function. Sounds reasonable?

REPLY [6 votes]: Let $r_Q^*(n)$ be the number of primitive representations of $n$ by $Q$. Let $\mathcal{C}$ be 
a set of representatives of the classes of binary quadratic forms of discriminant $\Delta$. We can assume that $Q\in\mathcal{C}$. It is known that $\sum_{Q'\in\mathcal{C}}r_{Q'}^*(n)$ equals the number of residue classes $b\pmod{2n}$ such that $b^2\equiv\Delta\pmod{4n}$, see e.g. Page 66 in Zagier: Zetafunktionen und quadratische Körper (Springer 1981). By the Chinese Remainder Theorem, we can calculate this as a product of local densities, in fact this is an instance of Siegel's mass formula. Estimating crudely we get
$$ \sum_{Q'\in\mathcal{C}}r_{Q'}^*(n) \ll_\Delta 2^{\omega(n)},$$
where $\omega(n)$ is the number of prime factors of $n$, whence $r_Q^*(n)\ll_{\Delta,\epsilon} n^\epsilon$ readily follows. Finally,
$$ r_Q(n)=\sum_{d^2\mid n}r_Q^*(n/d^2) \ll_{\Delta,\epsilon} n^\epsilon. $$<|endoftext|>
TITLE: Exponent of the cohomology of a product of groups
QUESTION [7 upvotes]: Suppose $G$, $H$ are finite groups and $M$ is a module over $G\times H$. 
Question: Is the exponent of $H^i(G\times H,M)$ a divisor of $lcm(|G|,|H|)$ for $i> 0$ ? 
The Künneth formula answers the question affirmatively if $M$ is trivial or, more generally, if one of the groups acts trivially on $M$. But I don't know what to expect if $M$ is non-trivial.

REPLY [5 votes]: For any finite group $\Gamma$, if $I$ is the augmentation ideal of ${\mathbb Z}\Gamma$, then $H^1(\Gamma,I)\cong {\mathbb Z}/|\Gamma|{\mathbb Z}$, which gives a counterexample if you take $\Gamma = G\times H$ for any $G$ and $H$ whose orders are not coprime.<|endoftext|>
TITLE: Truncation of BG?
QUESTION [5 upvotes]: Let $G$ be a topological group. In some cases, e.g. when $G$ is discrete or when the spaces $G^n$ are locally contractible and the coefficients are discrete, the cohomology of the classifying space $BG$ is the group cohomology of $G$. So, for simplicity, let us assume that $G$ is discrete.
My question: is there a nice explicit space $B_{\leq k}G$ that is functorial in $G$, such that $H^n(B_{\leq k} G, M) = H^n (BG, M)$ for $n \leq k$ and $0$ for $n>k$? Here $M$ is a $G$-module.
For example, for $G = S^1$, $BG = CP^\infty$ and a possible choice for $B_{\leq 2} S^1$ is $CP^1$. (If it simplifies the question, feel free to assume $M = \mathbf Z$.)
Thank you.
(Edit: In response to Ralph's comment, let us assume $G$ is discrete to simplify, but an answer for non-discrete groups would be interesting, too.)

REPLY [7 votes]: There are several functorial models for $BG$, see for example [Adem, Milgram: Cohomology of Finite Groups, Chapter II] where 
$$BG = \coprod_{i=0}^\infty \sigma^i \times G^i/(\text{relations})$$
with $\sigma^i=\lbrace (x_1,...,x_i)\mid 0\le x_1 \le \cdots \le x_i \le 1\rbrace$ the standard $i$-simplex.  
Now assume $G$ is discrete. Then $$B_nG := \coprod_{i=0}^n \sigma^i \times G^i/(\text{relations})$$ 
is the n-skeleton of $BG$ and depends functorially on $G$. Since the cohomology in degree $\lt n$ of a CW complex is determined by the $n$-skeleton, we obtain $H^k(B_nG;M)=H^k(BG;M)$ for $0 \le k < n$ and $H^k(B_nG;M)=0$ for $k> n$ and all (local) coefficients $M$. 
In general $H^n(B_nG;M)\neq H^n(BG;M)$, but there is an exact sequence: Write $BG=EG/G$ $(EG$ is described explicitely in Adem-Milgram) and let $E_nG$ be the $n$-skeleton of $EG$. Then the following sequence is exact (Cartan, Eilenberg: Homologica Algebra, XVI, §9, Appl. 1): 
$$0 \to H^n(BG;M)\to H^n(B_nG;M)\to H^n(E_nG;M)^G \to H^{n+1}(BG;M)\to 0$$
Remark: In case that $G$ is not discrete, the above has to be adjusted accordingly. For example, if $G$ is (topologically) a $m$-dimensional CW complex then $B_nG$ is the $n(m+1)$-skeleton of $BG$.<|endoftext|>
TITLE: What is the Q-construction, metaphysically?
QUESTION [12 upvotes]: An exact (small) category $P$ is an environment in which we make sense of the "put-together"-edness of objects via (short) exact sequences. It seems like the K-theory of an exact category encodes the high order relations of how objects fit together, but I can't see how the $Q$-construction is the natural medium for this. 
$\bullet$ $0 \to B \to A \to C \to 0$ means that $A$ is put together in some way from $B$ and $C$. Letting $E$ denote the category of short exact sequences in $P$ (with obvious morphisms), we see that this holds in the large as well: If $p$ (resp. $q$) is the projection functor $E \to P$ that sends a s.e.s. to the first (resp. last) object in the sequence, then $(K_i(p), K_i(q)): K_i(E) \to K_i(P) \oplus K_i(P)$ is an isomorphism.
$\bullet$ Moreover, results like devissage, localisation, and resolution are also easily seen  to be reassurances that we are distilling a very sensible notion of put-together-edness.

Can anyone offer a (non-circular) reason as to why the Q-construction is right?

I don't see the sense or utility considering a category in which a morphism from $X$ to $Y$ is an isomorphism of $X$ with an (admissable) subquotient of $Y$.

REPLY [9 votes]: 1)  One wants the $Q$-construction to have the property that $\pi_1(QP)=K_0(P)$.
2)  To get this, one wants, for any covering space of $BQP$, that $K_0(P)$ acts naturally on the fiber over $0$.
3)  To get this, one wants to associate to any monomorphism $i$ in $P$ a morphism $i_!$ in $QP$ and to any epimorphism $j$ in $P$ a morphism $j^!$ in $QP$ in a way that satisfies certain simple properties; the statement of the properties, and the proof that they suffice to get this result, is in Quillen's Algebraic K-Theory I (Theorem I).
4)  Quillen's $Q$-construction is the Universal construction yielding such $i_!$ and $j^!$.  (The proof is in QUillen's paper, immediately preceding Theorem 1.)
5)  Therefore, there's a sense in which Quillen's $Q$-construction is the natural first guess for what should work.  (Of course the "naturality" of this guess appears only in hindsight; a lot of other people failed to find this construction.)
PS.  After you work through the constructions, you see that this is another way to see the same thing:  For any object $A$ in your category $P$, you want to associate the $K_0$-class $[A]$ to some loop in $BQP$.  The simplest thing to hope for is two canonically defined maps from $0$ to $A$ in $QP$, which together give you your loop.  Quillen's construction provides those two maps (recognizing $0$ as a quotient of both $0\subset A$ and $A\subset A$) in the simplest possible way.<|endoftext|>
TITLE: What is the doubling dimension of convex functions?
QUESTION [6 upvotes]: I am interested in the complexity of convex functions, specifically the "doubling dimension" of the class of convex functions defined on a compact subset of Euclidean space, when compared using the $L^\infty$ metric. 
Formally, let $F$ be the class of convex functions mapping $[0,1]^n$ to $[0,1]$. For $f,g \in F$, define the distance between them as $d(f,g) = sup_{x \in [0,1]^n} |f(x) - g(x)|.$ My question is: Is $(F,d)$ a doubling metric space, and if so what is its doubling dimension as a function of $n$?
Recall that a metric space is doubling if there exists an integer $M$ such that every ball of finite radius $r$ can be covered by at most $M$ balls of radius $r/2$. For such a space, the doubling dimension is defined to be $\log_2 M$. 
Observe that, had I replaced "convex" with "linear" in my statement of the question, the answer would have been "yes, and the doubling dimension is $O(n)$." I'm hoping for a similar answer for convex functions. Given that this seems like a fairly natural question, I would guess the answer is already known, or at least obvious to anyone working in functional analysis.

REPLY [4 votes]: No the space is not doubling.
Take a strongly convex function, say $f(x)=x^2$.
It is sufficient to show that there is $N(\varepsilon)$ which goes to $\infty$ as $\varepsilon\to0$, such that $\varepsilon$-neighborhood of $f$ contains $N(\varepsilon)$ points on distance $>\varepsilon$ from each other.
To see this take any finite collection of distinct points $\{x_i\}$ in $[0,1]$ and note that for all small $\varepsilon>0$ there is a function $f_i$ in the $\varepsilon$-neighborhood of $f$
such that $f_i(x_i)=f(x_i)+\varepsilon/2$ 
and $f_i(x_j)=f(x_j)-\varepsilon/2$ 
for all $i\ne j$.<|endoftext|>
TITLE: Isotopy classes on the disk and mapping tori
QUESTION [8 upvotes]: Is the following true?
"The conjugacy classes of two homeomorphisms of the n-times punctured disk have isotopic representatives iff the associated mapping tori are homeomorphic."
By conjugacy class of a homeomorphism is meant the homeomorphism up to (orientation preserving) topological changes of coordinates. The isotopies are not required to keep the boundary pointwise fixed. Thus the question may be rephrased as:
Is it true that, up to changes of coordinates and isotopies, two homeomorphisms of the punctured disk are the same iff their mapping tori are homeomorphic? The "only if" is true; it is the "if" that puzzles me.

REPLY [5 votes]: As I said in the comment above, knots in $D^2\times S^1$ which have a Dehn filling giving $D^2\times S^1$ were classified by Gabai and Berge. Gabai proved that knots in $D^2\times S^1$ giving back $D^2\times S^1$ are either cables or 1-bridge braids (note that the original statement was given before the knot complement problem, so Gabai states that the knot could also lie in a ball, but now this is known to not occur). I didn't check, but I believe the cable case will not work: even though there are many solid torus Dehn fillings, the link complement has an infinite image of the mapping class group in the mapping class group of the boundary, so I believe these all give equivalent cablings (although I didn't check this). 
For 1-bridge braids in a solid torus with solid torus surgery, Gabai gave a partial classification, and Berge gave a complete classification. Gabai shows that the other 1-bridge braid will have the same winding number (Corollary 3.3), so the same number of strands. For some of these examples, Dehn filling gives back a 1-bridge braid in the solid torus of the same type. But Berge shows that most examples will give back a different braid. 
To be explicit, I took the braid from Figure 8 of Gabai's paper, which is a braid on 10 strands. I input this into SnapPy, which shows that the braid is hyperbolic, and $(63,1)$ Dehn filling on cusp $0$ gives a manifold with fundamental group $\mathbb{Z}$, which therefore must be $D^2\times S^1$. Moreover, the symmetry group is $\mathbb{Z}/2^2$, and there is a $\mathbb{Z}/2$ subgroup which preserves the cusps, and acts as an elliptic involution, therefore preserving slopes. So the two 1-bridge braids must be inequivalent, but have the same complement.<|endoftext|>
TITLE: Integer dynamics hitting infinitely many primes
QUESTION [11 upvotes]: I am wondering if there are any rigorous results telling that some dynamical system hits infinitely many primes (except for the case when orbits are just arithmetic progressions). To make it specific, is there a polynomial $f(x)$ such that its iterations $f^n(x)$ are prime for infinitely many $n$ given that the integer $x$ is fixed. A simple observation here is that if $f(x) = 2x + 1$ then asking if iterations of $f(1)$ contain infinitely many primes is equivalent to the Mersenne primes conjecture. 
P. S.
This question seems to be related to the Bunyakovsky conjecture, so maybe somebody knows about any partial results in this direction?

REPLY [7 votes]: In 1947, American William . H. Mills proved that there is a real number $A$, greater than 1 but not an integer, such that integer part of
$$
A^{3^n} 
$$
is prime for all $n =1, 2, 3, \ldots$. The numnber $A$ is known as the Mill's constant. Its value is unknown, but if the Riemann hypothesis is true then,
$$
A \approx 1.3063778838630806904686144926\ldots
$$
[1] Mills, W. H. (1947) A prime representing function. Bull. Amer. Math. Soc., 53: 604: MR 8, 567.
[2] [Mill's constant]1<|endoftext|>
TITLE: History of the high-dimensional volume paradox
QUESTION [30 upvotes]: Inscribe an $n$-ball in an $n$-dimensional hypercube of side equal to 1, and let $n \rightarrow \infty$. The hypercube will always have volume 1, while it is a fun folk fact (FFF) that the volume of the ball goes to 0.
I first learnt of this in relation to Gromov. In the story I heard, he used to ask incoming students to compute the distance $(\sqrt{n}-1)/2$ from a hypercube corner to the ball, and observe them to see if they realized that the volume of the hypercube is concentrated in its corners.

Is this story correct? And is this the origin of this FFF? I could imagine a situation where several people noticed this at different times, but where the fact did not become "viral" until much more recenttly.

REPLY [44 votes]: A related (and to me, when I first saw it, much more surprising) Fun Fact:  Divide the n-dimensional cube in half in each of $n$ dimensions, to create $2^n$ smaller cubes of edge length 1/2.  Inscribe a ball in each of these subcubes, and then construct the smallest ball tangent to each of those (and centered at the center of the original cube) like so:
      (source)
What happens to the diameter of the central ball as $n$ gets large?
This question received much attention at an algebraic K-theory conference in Boulder in the early 1980s, where each new arrival was presented with a multiple choice problem:  Without stopping to compute, is the limit $-1$, $0$, $1/2$, $1$, $10$ or $\infty$?  You were allowed to choose any three answers out of six, and place a bet on whether the right answer was among them.  I can report that an overwhelming majority of algebraic K-theorists reason thusly: the answer can't be negative and can't be greater than 1 (the ball, after all, is obviously contained inside a box of side 1!); therefore it's safe to bet on the set 
$\lbrace 0,1/2,1 \rbrace $.  Feel free to make money off this.<|endoftext|>
TITLE: Can nonstandard analysis be used to prove results in constructive or computable analysis?
QUESTION [12 upvotes]: Nonstandard analysis is a useful tool which can be used to prove a number of results in analysis.
Question

Can it also be used to prove results in computable or constructive analysis?
If so, what are some examples?  (They don't need to be ground-breaking.)

Motivation
There seems to be this analogy involving small worlds and big worlds (model is probably a more accurate term).
computable math

small: computable real numbers
big: real numbers

nonstandard analysis:

small: standard real numbers
big: nonstandard real numbers

This analogy is quite common in logic (ground model vs forcing extension for another example).
Can statements about the computably of finite objects be moved to the "computability" of nonstandard finite objects, and then transferred to the computability of standard infinite objects?
I am aware of Sam Sanders' program to connect Bishop-style constructive analysis with nonstandard analysis, but I am not aware (possibly mistakenly) that it has been used to prove statements in computable/constructive mathematics.
Possible examples


Can one use nonstandard analysis to show that the supremum of a computable function $f$ on $[0,1]$ is computable uniformly from $f$?  (The corresponding finitary statement about finite functions of rationals is clearly true.)

What about the computability of the Riemann integral?

REPLY [14 votes]: Nonstandard Analysis (NSA) can be used to prove results in computable/constructive analysis; The central notion is $\Omega$-invariance, defined as follows.  
[As usual, the set $N$ consists of the standard/finite/natural numbers; The set ${^{\star}}N$ is an end-extension of $N$ and $\Omega={^{\star}}N\setminus N$ consists of the infinite/nonstandard numbers. For a standard formula or function $A(n)$ defined on $N$, the object $^\star A(n)$ is defined on $^\star N$.  Let $R$ be the set of real numbers.]  
($\Omega$-invariance) 
1) For a standard bounded formula $\varphi(n,m)$, and an infinite number $\omega\in \Omega$, the formula $^\star\varphi(n,\omega)$ is 
$\Omega$-invariant if 
$$(\forall n\in N)(\forall \omega'\in\Omega)[^\star\varphi(n,\omega)\leftrightarrow {^\star}\varphi(n,\omega')].$$
2) For a standard function $f:N\times N\rightarrow N$, and an infinite number $\omega\in \Omega$, the function $^\star f(n,\omega)$ is 
$\Omega$-invariant if 
$$(\forall n\in N)(\forall \omega'\in\Omega)[^\star f(n,\omega)= {^\star}f(n,\omega')].$$
3) For a standard function $F:R\times N\rightarrow R$, and an infinite number $\omega\in \Omega$, the function $^\star F(x,\omega)$ is 
$\Omega$-invariant if 
$$(\forall x\in R)(\forall \omega'\in\Omega)[^\star F(x,\omega)\approx {^\star}F(x,\omega')].$$
Note that $\Omega$-invariance is essentially "independence of the choice of infinitesimal".
Now, $\Omega$-CA is the comprehension axiom for $\Omega$-invariant formulas as follows: For all $\Omega$-invariant $^\star\varphi(n,\omega)$, we have $(\exists X^s \subset N)(^\star\varphi(n,\omega) \leftrightarrow n\in X^s)$.  Here, the superscript $^s$' refers to the fact that $X^s$ is a standard set.  
Recently, Antonio Montalbán and me showed the following: 
1) $^\star$RCA$_0+\Omega$-CA is a conservative extension (in the standard language) of RCA$_0$.  Here, $^\star$RCA$_0$ is a nonstandard version of $\text{RCA}_0$. 
2) In $^\star I\Sigma_1$, $\Omega$-CA implies $\Delta_1^0$-comprehension.   
3) $^\star$RCA$_0+\Omega$-CA proves that for every $\Omega$-invariant $^\star F(x,\omega)$, there is a standard $G:R\rightarrow R$ such that $(\forall x\in R)(^\star F(x,\omega)\approx {^\star}G(x)$. 
If I am not mistaken, the previous three observations answer your question regarding computable analysis:  As long as one produces $\Omega$-invariant functions, the results are computable.  (There is/should be some analogue to the Gaifman-Dimitracopoulos theorem here.)   
One can refine the above to 'constructive analysis', but explaining that would take up too much space.    
One a philosophical note, one might argue that most/all of the infinitesimal calculus used throughout physics is $\Omega$-invariant, and therefore computable.  
Three final remarks: 
0) There are a number of analogies one can use to compare NSA and constructive/computable analysis.  To me, these analogies are quite helpful/insightful.  Jason is right in pointing out his analogy.  Not everyone seems to agree on this, however.  
1) The above view of NSA is called 'Robinsonian'.  The same definitions etc. can be made in Nelson's `internal' framework without any problem.  
2) Somehow, mathoverflow does not parse '*' very well: one has to use '\star'.<|endoftext|>
TITLE: Noether-Lefschetz over finite fields
QUESTION [10 upvotes]: The classical Noether-Lefschetz theorem asserts the following: Over the complex numbers, a very general surface $S\subset \mathbb{P}^3$ has Picard number 1 (that is, $Pic(S)\simeq \mathbb Z$), provided that $\mbox{deg }S\ge 4$.
Over finite fields (or even countable fields), the corresponding statement does not make much sense, because here `very general' refers to the surface being chosen outside a countable union of closed proper subsets of the parameter space of surfaces. Still, I think makes sense to ask:

What evidence is there for the
  Noether-Lefschetz statement over
  countable fields? I.e., is there a sense
  in which the set of
  surfaces with low Picard number
  constitutes a 'large' proportion of
  the surfaces in $\mathbb{P}^3$ e.g.,
  over finite fields with many elements?

REPLY [6 votes]: There is a Noether-Lefschetz statement, but it's not quite the same as the one that holds over large fields and global fields.
Let the Tate-Picard rank be the formula for the (geometric) Picard rank implied by the Tate conjecture.
Let $n \geq 4$ be fixed, let $\mathbb F_q$ any large finite field, and let $X$ be a random smooth surface in $\mathbb P^3$ of degree $n$. I believe ne can show that:
If $n$ is odd, then with probability $1-o(1)$, the Tate-Picard rank of $X$ is $2$.
If $n$ is even, then with probability $1/2+o(1)$, the Tate-Picard rank of $X$ is $1$, and with probability $1/2+o(1)$, the Tate-Picard rank of $X$ is $3$.
Conditional on the Tate conjecture, this gives the distribution of the Picard rank. Otherwise, this is an upper bound for the Picard rank.
Here's a sketch of a proof:
Let $\mathcal F$ be the sheaf on the moduli space of smooth surfaces of degree $n$ in $\mathbb P^3$ whose stalk at a point is the primitive $2$nd cohomology of the corresponding hypersurface. Then $\mathcal F$ is lisse of rank $n^3-4n^2+6n-3$
Step 1: We compute the monodromy group of $\mathcal F$. This is probably in the literature somewhere (I know it is over $\mathbb  C$, which implies it at least for sufficiently large characteristic), but it's not hard.
By Poincare duality, it is contained in $O(n^3-4n^2+6n-3)$. By Larsen's conjecture, it contains $SO(n^3-4n^2+6n-3)$ if and only if the invariant subspace of its $8$th tensor power is $105$-dimensional. By Deligne's equidsitribution theorem, this happens as long as the average of the eight power of the number of points on the hypersurface, divided by q, minus q+1+1/q, is 105. This can be easily checked by exchanging the order of summation as soon as $n\geq 4$. Finally, one can use vanishing cycles near a nodal singularity to check that the determinant character is nontrivial.
Step 2: Compute the Tate-Picard rank in terms of the characteristic polynomial of Frobenius.
The Tate conjecture says that over $\mathbb F_{q^n}$, the cohomology classes that are generated by cycles are exactly the $q^n$-eigenspace of $\operatorname{Frob}_{q^n}$. So for $X$ defined over $\mathbb F_q$, the $\lambda$-eigenspace of $\operatorname{Frob}_{q}$ is generated by algebraic cycles defined over $\mathbb F_{q^n}$ if and only if $\lambda^n = q^n$, i.e. $\lambda$ is $q$ times an $n$th root of unity. 
Thus the Tate-Picard rank is defined to be the sum over all $n$th roots of unity $\zeta$ of the multiplicity of the eigenvalue $\zeta q$ of $$\operatorname{Frob}_{q}$, plus $1$ (as we pass from primitive cohomology to all of $H^2$).
Because the characteristic polynomial of Frobenius is a polynomial of degree $n^3-4n^2+6n-3$ with integer coefficients, the eigenvalue $\zeta q$ cannot appear unless the degree of $\mathbb Q(\zeta)$ is at most $n^3-4n^2+6n-3$, so we need only check a finite set of eigenvalues.
Step 3: By Deligne's equidistribution theorem, the distribution of the number of eigenvalues of Frobenius that are $q$ times a root of unity of bounded degree converges as $q$ goes to $\infty$ to the distribution of the number of eigenvalues of a matrix in $O(n^3-4n^2+6n-3)$ that are roots of unity of bounded degree.
This would follow immediately from Deligne's equidistribution theorem if the number of such eigenvalues were a semicontinuous function, which it isn't. However, it is upper semicontinuous - the set where of matrices with at least $r$ eigenvalues in a given finite set is closed. Moreover, the boundary of this closed set has Haar measure $0$, hence by standard measure theory, the convergence of the measure of this set follows from the convergence of the measure of continuous functions.
Step 4: Calculate this distribution for a random element of $O(N)$.
If $N$ is odd, then any orthogonal matrix has eigenvalues $\lambda_1, \lambda^{-1}, \lambda_2,\lambda_2^{-1},\dots, \pm 1$, with the last eigenvalue equal to the determinant. The eigenvalues $\lambda_1,\lambda_2$,etc. have probability $0$ of taking each eigenvalue, so the expected number of low degree roots of unity is $1$, for a Tate-Picard rank of $2$. This happens for $n$ even.
If $N$ is even, then an orthogonal matrix has eigenvalues $\lambda_1, \lambda^{-1}, \lambda_2,\lambda_2^{-1},\dots, \lambda_{N/2},\lambda_{N/2}^{-1}$ if its determinant is $1$ and $\lambda_1, \lambda^{-1}, \lambda_2,\lambda_2^{-1},\dots, 1,-1$ if its determinant is $-1$. Again, the variable eigenvalues have probability $0$ of attaining any specific value, so the number of root of unity eigenvalues is $0$ half the time and $2$ half the time, for a Tate-Picard rank of $1$ half the time and $3$ half the time.<|endoftext|>
TITLE: Distinguishing 3-manifolds by homologies of covers
QUESTION [13 upvotes]: In a blog post on ldtopology, a recent arxiv posting of Lins-Lins is discussed. The main argument of that paper is difficult to algorithmically distinguish two 3-manifolds and to that end the authors provide the challenge of showing two specific manifolds are not homeomorphic. 
Nathan Dunfield implemented a computation of the index 6 subgroups of the fundamental groups of the two manifolds. In the comments of the aforementioned blog post, he showed this computation distinguishes these two sets of subgroups by abelianizing the subgroups. In particular, this indicates the two manifolds are not homeomorphic.   
My first question is, "Can this trick always be used?", or more specifically: 
Let $M_1$ and $M_2$ be non-homeomorphic (finite volume) hyperbolic 3-manifolds 
and let $S_{i,k}$ be the set of degree $k$ covers of $M_i$. Finally, let $H_{i,k}$ be the sets of integral first homology groups for the manifolds in $S_{i,k}$. 
Is there a known pair of non-homeomorphic (finite volume) hyperbolic 3-manifolds $M_1$ and $M_2$ such that for all $n$, the sets of $H_{1,n}$ and $H_{2,n}$ are identical, i.e. there is a bijection between them?
As a second question:
In the absence of a pair of hyperbolic examples, is there a known pair of 3-manifolds with non-isomorphic fundamental groups with this property?

REPLY [3 votes]: To atone for my comment above, I'm going to write out the Sol
example. A Sol lattice takes the form $\mathbb Z^2\rtimes \mathbb Z$,
where the action is hyperbolic. The rank of the abelianization is
always 1, though the amount of torsion varies. The profinite
completion is $\hat{\mathbb Z}^2\rtimes \hat{\mathbb Z}$.
We can consider the group $\mathbb Z^2$ together with the action
as a module over $\mathbb Z[t,t^{-1}]$. One invariant is the
characteristic polynomial $P(t)$. Fixing this, we can consider the
group as a module over $\mathbb Z[t,t^{-1}]/P(t)$. For hyperbolic
actions, this ring is an order in a real quadratic number field.
Moreover, every real quadratic number field has a unit, and thus
has an order of this form. We can give $\mathbb Z^2$ the structure
of a rank 1 projective module over this ring; and if the ring has
nontrivial class group, there are non-isomorphic ways to do this.
So this gives rise to two Sol lattices that are not isomorphic,
but are pretty similar, having the same characteristic polynomial.
The obstruction to their isomorphism is global, so when we take
profinite completions and are dealing with modules over local
rings, where all projectives are free, the groups become
isomorphic. Thus their finite index subgroups are in bijection
preserving the quotients of corresponding groups; and preserving
the abelianizations. The action on the finite quotient group on
the homology of the subgroup is trivial on the torsion-free part,
because it is always just $\mathbb Z$, and matches the
corresponding action in the profinite group, so does not
distinguish the two discrete groups.
Similarly, it seems to me that there are pretty good candidates for
hyperbolic three-manifolds with isomorphic profinite completions.
Then these would have a bijection between finite index subgroups
preserving the abelianizations and quotient groups, answering the
original question. However, the action of the finite quotients on
the abelianizations might distinguish them, although that is a
subtle invariant. The idea is to take a imaginary quadratic number
field with appropriate class group and form nonisomorphic
rank 2 projective modules. For the right choice of projective
modules, the automorphism groups are not isomorphic as group
schemes and, I think, as discrete groups. Over the number field
all projective modules are free, so the groups of rational
automorphisms are equal, and thus the lattices are commensurable.
Over the ring of integers in a local field, all projective modules
are free, so the congruence completions are isomorphic. The
congruence subgroup property does not apply to these groups, but
the congruence kernel is attached to the rational group, so pretty
much the same for the two groups. Compatibility between the
congruence completion and the congruence kernel seems to me to be
the big problem in checking this candidate. Or maybe you could use
two $(3,1)$ quadratic forms in the same genus.<|endoftext|>
TITLE: Riemannian manifolds with small geodesics and bounded curvature
QUESTION [6 upvotes]: Let $(M,g)$ be a compact riemannian manifold with sectional curvature $|K_g| \leq 1$. A lemma due to Klingenberg asserts that then either the injectivity radius $i_g \geq \pi$ or $(M,g)$ contains a geodesic loop $\gamma$ of length $L(\gamma) = 2i_g$.
Let $n = 2$.  Suppose for a moment that $i_g < \pi$ and $M$ is orientable. Note that $\gamma$ is automatically simple. (I am note sure how often this occurs. An example, however, is a thin long cylinder with two balloon-like 2-spheres attached.) Then $\gamma$ cuts $M$ into two pieces. 
Intuition says that these two pieces should be rather large, as the manifold cannot curve together very quickly due to the curvature condition. By another Klingenberg lemma, known as the "Long Homotopy Lemma", any null-homotopy $\gamma_t$ of $\gamma_0 = \gamma$ must sweep through some curve $\gamma_{t_0}$ of length $L(\gamma_{t_0}) \geq \pi$. (This lemma holds for all $n$ and non-orientable manifolds. It is Exercise 1 in Chapter 10 of do Carmo's text book.)
To some extend, this is the kind of theorem I am looking for. It asserts that away from $\gamma$, $(M,g)$ must be slightly larger (on the scale of the curvature bound) than near $\gamma$. However, I'm looking for some result describes the geometry of $M$ away from $\gamma$ in a "more global" fashion. For instance: Is there any estimate of the diameter or the volume of the pieces that $M$ is cut into?  
In view of the examples given by horse with no name,
one should assume that $M$ has $g(M) \neq 0$ (or that $M$ has some other non-vashing characteristic class in the case that $n > 2$). 
As pointed out in the comments and Anton Petrunin's answer, $M$ should either be a sphere or  $\gamma$ must also be 
assumed to cut $M$ in half.

REPLY [2 votes]: It seems that you want $M$ to be a sphere (although you do not mention it).
The natural example is a tube with curvature $-1$ closed by two spherical cups with curvature $1$.
(This is a surface of revolution.)
This example should have the minimal area and minimal diameter 
among the spheres with given length of closed geodesic.
But at the moment I do not see a proof.
Some rough estimates will follow from the Klingenberg's result.
Take the neighborhood $U$ of your closed geodesic of size near $e^{1/\ell}$ where $\ell$ is the length of the geodesic.
Note that $U$ has to be a cylinder otherwise you end in singular point. 
It implies that diameter has to have order $e^{1/\ell}$.  
Rough bounds for the area follow Rbega comment. They are not far from the area of the surface of revolution, somewhere in the range between $4{\cdot}\pi$ to $(4+2{\cdot}\sqrt{2}){\cdot}\pi$.<|endoftext|>
TITLE: Visualize Fourth Homotopy Group of $S^2$
QUESTION [16 upvotes]: I know $\pi_4(S^2)$ is $\mathbb{Z}_2$. However, I don't know how to visualize it. For example, it is well known that $\pi_3(S^2)=\mathbb{Z}$ can be understood by Hopf Fibration. Elements in $\pi_3(S^2)=\mathbb{Z}$ can be understood as describing the number of links of the $U(1)$ fibers in $S^2$. 
So, do we have similar picture for $\pi_4(S^2)$? And, do we have similar topological invariant as links in previous case?

REPLY [2 votes]: The most explicit and geometric generator for $\pi_4(S^2)$ is probably the achiral genus-1 Lefschetz fibration $f : S^4 \to S^2$ with two Lefschetz singularies (one positive and one negative, with parallel vanishing cycles). In my knowledge, this Lefschetz fibration was first considered by Matsumoto ("On 4-manifolds fibered by tori", Proc. Japan Acad. 58 (1982), 298-301), and it is also described in Example 8.4.7 in the Gompf-Stipsicz book "4-Manifolds and Kirby Calculus".
I don't have now a proof of the homotopic non-triviality of this map, but I give this as a challenge!<|endoftext|>
TITLE: Infinity-categories vs Kan complexes
QUESTION [14 upvotes]: It is known (cf. Lurie's book Higher Topos Theory for instance) that higher ($\infty$-) category, in particular topological higher ($\infty$-) groupoids are "better" defined as weak Kan complexes, aka quasi-categories. Let me recall that a simplicial space $X_\bullet$ is weak Kan if every map $\Lambda_i^n\longrightarrow X$ (where $\Lambda_i^n$ is the horn) can be extended to a map $\Delta^n\longrightarrow X$ for all $0< i < n$. 
My problem is the following. Let us consider the fundamental $(\infty,0)$-groupoid $\pi_{\leq \infty}X$ of a nice space $X$: $X$ is the set of objects, morphisms between objects are maps $f:[0,1]\longrightarrow X$, $2$-morphisms are homotopies between morphisms, and so on. I am quite confused about how does one, "geometrically", see $\pi_{\leq \infty}X$ as a simplicial space; for an element of $\pi_2X$, for instance, has four edges, and hence four different face maps.

REPLY [11 votes]: I suspect your confusion arises in part because homotopies of paths are continuous maps $I^2 \to X$, while 2-morphisms in $\pi_{\lt \infty} X$ are continuous maps $\Delta^2 \to X$.  That is, 2-morphisms are not strictly the same as homotopies of paths.
The dictionary between the two structures is not too bad:

A 2-morphism in $\pi_{\lt \infty} X$ is a homotopy of paths, where either the beginning or the end is fixed (or perhaps it is a homotopy to or from a constant path).
A continuous map $I^2 \to X$ can be viewed as a composite of two 2-morphisms in $\pi_{\lt \infty} X$.  You end up using the diagonal $(0,0) \to (1,1)$ in the square to separate the two 2-morphisms, because the simplicial structure of $\Delta^1 \times \Delta^1$ has a diagonal 1-simplex.

I personally find it rather magical that among the sixteen 2-simplices in $\Delta^1 \times \Delta^1$, a pair of them pops out as non-degenerate - it is a rewarding computation.
More generally, higher dimensional cubes, viewed as products of the simplicial set $\Delta^1$, have canonical decompositions into nondegenerate simplices.  The corresponding higher homotopies are composites of homotopies with various pieces held constant.
I think this should account for the discrepancy you see in the number of face maps.<|endoftext|>
TITLE: For what spaces is the Hardy-Littlewood maximal operator of strong type $(p,p)$ if and only if $p > p_0 > 1$?
QUESTION [7 upvotes]: (This is essentially a continuation of my previous question, here.)
Let $(X,d,\mu)$ be a metric measure space, i.e. $\mu$ is a Borel measure on the metric space $(X,d)$. Further assume (though you can remove this assumption if you like) that each ball has positive finite volume. Let $M$ be the uncentred Hardy-Littlewood maximal operator, given by
$$Mf(x) := \sup_{B \ni x} \frac{1}{\mu(B)} \int_B |f(y)| \; d\mu(y)$$
with the supremum taken over balls containing $x$.
In my previous question, I asked when $M$ is not of weak type $(1,1)$, and was presented with an example ($\mathbb{R}^2$ with Gaussian measure) which isn't of weak type $(1,1)$, but which is of strong type $(p,p)$ for all $p>1$. (See Sjögren ('83), and Forzani, Scotto, Sjögren, Urbina ('02).)
Now I'm interested in the following question: are there any metric measure spaces $(X,d,\mu)$ for which $M$ is of strong type $(p,p)$ if and only if $p > p_0$ (or $p \geq p_0$) for some $p_0 > 1$ (note the strict inequality here)? In other words, do we ever have some but not all midpoint strong type estimates?
I know that there are maximal operators out there for which this is true, but I'm specifically interested in the uncentred Hardy-Littlewood maximal operator (and to some extent the centred version, though we have to assume some regularity of the measure here to ensure measurability).

REPLY [2 votes]: It was proved by Alex Ionescu that the uncentered Hardy Littlewood maximal function is restricted weak type $(2,2)$ and not strong type $(p,p)$ for $p<2$ for hyperbolic spaces in http://arxiv.org/pdf/math/0007200v1.pdf<|endoftext|>
TITLE: Classification of groups in which the centralizer of every non-identity element is cyclic
QUESTION [10 upvotes]: In which classes of groups is it feasible to classify those groups in which the centralizer
of every non-identity element is cyclic?

REPLY [11 votes]: If $G$ is a finite group in which the centralizer of every nonidentity element is cyclic, then as Geoff Robinson pointed out, the Fitting subgroup $F = F(G)$ is cyclic, and $G/F$ is cyclic. One can say more, however. Suppose that $G$ is not itself cyclic, so $F < G$. Let $x$ be an element whose image modulo $F$ generates $G/F$, so $x \ne 1$. Let $C = C_G(x)$, so $C$ is cyclic, and since $x \in C$, we have $FC = G$. Since both $F$ and $C$ are cyclic, their intersection centralizes both $F$ and $C$, so is central in $G$. If this intersection contains a nontrivial element $y$, then $G =C_G(y)$, so $G$ is cyclic, contrary to assumption. Thus $F \cap C = 1$, and so $G$ can be constructed as a semidirect product of $F$ acted on by $C$.
One can say still more. Suppose some nonidentity element $c \in C$ commutes with some nonidentity element $f \in F$. Then $C_G(f)$ contains $F$ and also contains $c$. Since $c \not\in F$, we see that $C_G(f)$ strictly contains $F$. This is impossible, however, because $C_G(f)$ is cyclic, and yet $F$ is its own centralizer in $G$. This shows that the action of $C$ on $F$ is "Frobenius", and in particular, $|C|$ divides $|F|-1$ so $|F|$ and $|C|$ are coprime. 
Conversely, if we start with an arbitrary finite cyclic group $F$, we can build all possible finite groups $G$ satisfying the cyclic-centralizer assumption and such that $F(G) = F$. Do this as follows. Let $d$ be the GCD of all of the numbers $p-1$ for primes $p$ dividing $|F|$, and let $C$ be any cyclic group of order dividing $d$. Then $C$ has a unique Frobenius action on $F$ and the semidirect product $G$ of $F$ by $C$ will have the desired property. The key to checking that $G$ does have this property is that every element of $G$ either lies in $F$ or is conjugate to an element of $C$, so it is enough to check that centralizers of nonidentity elements of $F$ and of $C$ are cyclic. These centralizers are respectively $F$ and $C$.

Later edit: Geoff Robinson points out that in general, the condition that all Sylow subgroups of a finite group are cyclic does not imply that $|G:F(G)|$ and $
|F(G)|$ are coprime. It is interesting, however, that $|G:G'|$ and $|G'|$ must be coprime. It follows that every group with all Sylows cyclic can be constructed as a semidirect product of cyclic groups with coprime orders. Without the centralizer-cyclic condition, however, the corresponding action need not be Frobenius.<|endoftext|>
TITLE: Positively curved manifold with almost extreme diameter
QUESTION [9 upvotes]: Suppose $M$ is a 1-connected closed manifold with sectional curvature $\ge 1$. So the diameter $D$ of $M$ satisfies
$$
D \le \pi
$$
When equality holds $M$ is isometric to round sphere. In fact this rigidity holds under the assumption of $Ric \ge n-1$. Hence it is natural to ask what happens to almost extreme case. i.e. when $$D \ge \pi -\epsilon $$Under the Ricci assumption only, the manifold is not necessary sphere by a counter example of M. Andersen. However with extra assumption: sectional curvature is bounded from below, Perelman proved it is homeomorphic to a twist sphere. (Is Perelman the first one prove this?)
By Grove-Shiohama's Diameter Sphere Thereom, under the assumption sectional curvature $\ge 1$ and $D\ge \pi-\epsilon$, $M$ is a twist sphere. (It also follows from Perelman's theorem above)
My question is: Is $M$ diffeomorphic to the standard sphere? 
Either Perelman's proof or Grove-Shiohama's Diameter sphere theorem uses the 'soft' approach, i.e. It is not by convergence argument, nor a Lipschitz distance between $M$ and $S^n$ is derived.
Actually, I suspect that it is not Gromov-Hausdorff close to the round sphere, as one might round off two tips of $S^2/\mathbb Z_p$, where $\mathbb Z_p$ acts on $S^2$ by rotation along the $z$-axis.
What if one assume there is also an upper bound $K$ on sectional curvatures, i.e. $$1\le sec(M) \le K$$

REPLY [3 votes]: I will answer the last question.
Namely, let me show that if $1\le \mathrm{sec}\,M \le K$ and $\mathrm{diam}\, M>\pi-\varepsilon$ for sufficiently small $\varepsilon>0$ then it has to be diffeomorphic to $\mathbb S^m$.
Asssume contrary, then there is a sequence of $m$-dimensional manifolds $M_n$ such that $1\le \mathrm{sec}\,M_n \le K$ and $\mathrm{diam}\, M_n\to\pi$ as $n\to \infty$.
Let's pass to a converging sequence $M_n\to A$.
Note that $A$ is $m$-dimensional and $1\le \mathrm{sec}\, A \le K$ in the sense of Alexandrov and $\mathrm{diam}\, A=\pi$. 
It follows that $A$ is isometric to $\mathbb S^m$
Finally note that $M_n$ is diffeomorphic to $A$ for all large $n$;
the later was essentally proved by Cheeger in his thesis.<|endoftext|>
TITLE: algebraic de Rham cohomology of singular varieties
QUESTION [15 upvotes]: Hi,
Is there a simple example of an (affine) algebraic variety $X$ over $\mathbb C$ where
the $H^*_{dR}(X/\mathbb C) = H^*(\Omega^\bullet_{A/\mathbb C})$ differs from the singular cohomology $H^*_{sing}(X(\mathbb C)^{an},\mathbb C)$?
Such an example has to be singular (by a theorem of Grothendieck), but I am having a hard time finding one. The case $xy = 0$ doesn't work (both cohomology theories give the same answer).
Thanks!
EDIT: I would like a reduced example if possible...

REPLY [3 votes]: Let's look at the cuspidal cubic curve $C: x^3 = y^2$. I claim $H^1_{dR}(C)$ is two-dimensional.
EDIT: Actually $H^1_{dR}(C) = 0$, so that this curve does not give an example of an affine singular variety whose algebraic de Rham cohomology differs from its singular cohomology. This is corroborated by a remark in the beginning of section 3.2 of Huber and Müller-Stach's Periods and Nori Motives (just control-F for "cusp"). Note that the remark is not in their first version, but it is in their sixth which can be found on Prof. Huber's website.
Thanks are due to Julian Rosen for both pointing out a mistake in the derivation defined in the original version as well as realizing the important and subtle point that $\Omega^2_C \neq 0$, being supported at the origin. My sincere apologies for posting prematurely to the (three or more) people who read this answer in the meantime. The derivation as defined originally was not a derivation, but can be replaced by another which serves the same purpose as the original:
$$ \frac{\mathbb C[x,y]}{(x^3-y^2)} \to \frac{\mathbb C[x]}{(x^3)} : x \mapsto x, y\mapsto 1 $$
This derivation still shows that $\eta$ and $x\eta$ are non-zero. The more important point is that the presence of $\Omega^2_C$ means that the forms $x\,dy$ and $x^2\,dy$ are not actually closed, so these forms do not contribute to cohomology as was incorrectly assumed before, because $\Omega^2_C$ is in fact two-dimensional and spanned by $d(x\,dy) = dx\wedge dy$ and $d(x^2\,dy) = 2x\,dx\wedge dy$. These 2-forms are zero everywhere but at the singularity. My assumption that they were closed was the mistake which led me to conclude the cohomology was two-dimensional. The computations and formulas in the post show that the other 1-forms are exact so that $H^1_{dR}(C) = 0$. At any rate this mistake is an instructive one I think, so I'll preserve the post as originally written. Again, the two mistakes of this post are in the definition of the derivation and the incorrect assumption that $x\,dy$ and $x^2\,dy$ are closed forms.

As Georges notes, the singular cohomology of its closed points is trivial.
Regardless of whether it's direct (it's not), we definitely have the sum
$$\Omega^1_C = \mathbb C[x]\,dx + \mathbb C[x]y\,dx + \mathbb C[x]\,dy + \mathbb C[x]y\,dy $$
and then because $3x^2\,dx = 2y\,dy$, the last summand is actually unnecessary. The first summand is obviously composed of exact forms, so the only forms in question are things like $x^ny\,dx$ and $x^n\,dy$ where $n\geq 0$. On the other hand,
$$ d\left( \frac{2x^{n+1}y}{2n+5} \right) = x^ny\,dx \quad \text{ for } n > 1 $$
$$d\left( \frac{3 x^ny}{2n+3} \right) = x^n \,dy \quad \text{ for } n > 2$$
For the lower $n$, one needs to divide by $x$'s and $y$'s which is only valid provided $\Omega^1_C$ is torsion-free at the origin (and later we'll see it isn't), so these need to be handled with care. This leaves $y\,dx$, $xy\,dx$, $x\,dy$, and $x^2\,dy$. As $d(xy) = x\,dy + y\,dx$, any $\mathbb C$-linear relation between $y\,dx$ and $x\,dy$ (except for multiples of $y\,dx = -x\,dy$) would show that they are both exact. Similarly, $d(x^2y) = 2xy\,dx + x^2\,dy$ so any $\mathbb C$-linear relation between $xy\,dx$ and $x^2\,dy$ (except for multiples of $2xy\,dx = -x^2\,dy$) would show that they are both exact.
The obvious place to look is the relation $3x^2\,dx = 2y\,dy$, which after multiplying with $y$ gives $$ x^2(3y\,dx - 2x\,dy) = 0 $$ as well as $$ x(3xy\,dx - 2x^2\,dy) = 0 $$ Hence the form $\eta = 3y\,dx - 2x\,dy$ is very relevant: if $\eta = 0$ then we get both linear relations from $\eta = x\eta = 0$ and then we would have shown that the first cohomology group is trivial.
More subtly, it turns out $\eta$ is not zero. What the first formula does show is that it is zero in any localisation of $\Omega^1_C$ away from $(0,0)$. This funny form also maps to zero in the cotangent space at zero, since $\eta \in (x,y)\Omega^1_C$. This implies that any first-order derivation cannot determine whether $\eta \neq 0$.
To see that $\eta \neq 0$ we will use the second-order derivation  (the correct derivation is above in the comment) $$ \frac{\mathbb C[x,y]}{(x^3-y^2)} \to \frac{\mathbb C[x,y]}{(x^2,y^2)} : x,y\mapsto 1, \, x^2,y^2 \mapsto 0$$ The universal property of $\Omega^1_C$ then promises that there's a unique map of $\frac{\mathbb C[x,y]}{(x^3-y^2)}$-modules $$ \Omega^1_C \xrightarrow{f} \frac{\mathbb C[x,y]}{(x^2,y^2)} : dx,dy\mapsto 1 $$ Since $f(\eta) = 3y-2x$ and $f(x\eta) = 3xy$ which are both non-zero, we've shown $\eta,x\eta \neq 0$. One way of thinking about $x\eta$ is that it is a Dirac delta supported at $(0,0)$, while $\eta$ is a (weak) derivative of a Dirac delta supported at $(0,0)$.
Now working modulo exact forms, we've shown that $[ydx] = -[xdy]$ and $[x^2\,dy]=-2[xy\,dx]$ in $H^1_{dR}(C)$. I claim these are non-zero and independent. 
Following up on David's comment we weight $\frac{\mathbb C[x,y]}{(x^3-y^2)}$ so that $\text{wt }x = 2$ and $\text{wt }y = 3$ and then weight $\Omega_{A/k}$ accordingly in order that $d$ preserves weight (The motivation for this is that we are measuring in units of $t$ under the rationalization map $t \mapsto (t^2,t^3)$).
Because $\eta$ (resp., $x\eta$) is not zero, the weight 5 piece (resp., weight 7) of $\Omega_{C}$ is two-dimensional, spanned by $y\,dx$ and $x\,dy$ (resp., $xy\,dx$ and $x^2\,dy$). On the other hand, the weight 5 piece (resp., weight 7) of $\frac{\mathbb C[x,y]}{(x^3-y^2)}$ is only one-dimensional, spanned by $xy$ (resp., $x^2y$). 
We conclude that $H^1_{dR}(C)$ is two-dimensional, supported in weights 5 and 7 and spanned by $[xdy]$ and $[x^2\,dy]$, respectively. See edit comment above.<|endoftext|>
TITLE: The Cayley Menger Theorem and integer matrices with row sum 2
QUESTION [5 upvotes]: I just filled a gap in my education by learning about the Cayley-Menger theorem, and the Cayley-Menger determinant:
If $P_0, \dots, P_n$ are $n+1$ point in $\mathbb{R}^n$, and $d_{i,j} = |P_i - P_j|$ is the Euclidean distance from $P_i$ to $P_j$, we first form the $n+1 \times n+1$ matrix of squares of the distances (say $B$) and then form an $n+2 \times n+2$ matrix $A$ which has $B$ in the lower right hand corner, and has all the elements in the first row and column $=1$ except for the one in the upper left hand corner, which is 0.
The Cayley-Menger theorem (which is an $n$ dimensional generalization of Heron's formula for the area of a triangle) says that if $V$ is the volume of simplex whose vertices are the $P_i$, then
$(-1)^{n-1} 2^n (n!)^2 V^2 = \det(A)$.
I was interested in the structure of $\det(A)$ as a polynomial.  There is a nice paper "The Cayley-Menger Determinant is Irreducible for $n \ge 3$" by Carlos d'Andrea and Martin Sombra (arXiV:math/0406359).  When I calculated the number of monomials in each of the polynomials obtained by evaluating $\det(A)$, I got the sequence (starting with $n=1$):
1,1,6,22,130,822, 6202, 52552
which is sequence A002137 in the OEIS: Number of n X n symmetric matrices with positive entries, trace 0 and all row sums 2.  There's no mention there of the Cayley-Menger determinant.
So the question I have, is there a one-to-one correspondence that one can find between the matrices and monomials in $\det(A)$?  Even nicer, would be to find the value for the coefficient of the monomial from the matrix.
Added Later: I should have looked at the reference in the OEIS: A. C. Aitken, On the number of distinct terms in the expansion of symmetric and skew determinants, Edinburgh Math. Notes, No. 34 (1944), 1-5.
In it he gives a recurrence for the number terms in a symmetric matrix with 0's on the diagonal, and produces the same sequence.  However, I still can't find a connection with the coefficients.

REPLY [9 votes]: Let us first consider an $n\times n$ matrix with zero diagonal and $n(n-1)$ indeterminates.
Every term (monomial) of the determinant corresponds to a permutation matrix $P$ with zero diagonal (i.e., an integer nonnegative matrix with trace 0 and row and column sums 1).
Let us now consider a symmetric $n\times n$ matrix with zero diagonal and $n(n-1)/2$ indeterminates $a_{ij}=a_{ji}$. Then every term corresponds to a permutation matrix $P$ as before, but the correspondence is not one-to-one, since symmetric elements are equal. We can get a one-to-one correspondence by associating the term to $P+P^T$ (the "symmetrized version" of $P$): an integer nonnegative symmetric matrix with trace 0 and row sums 2. The coefficient of the term is $\det(P+P^T)$.
Now, in the Cayley-Menger determinant, the entries of the first row and column are not indeterminates but ones. We have to argue that this does not cause a "loss of information". In fact, the matrix $P+P^T$ is the adjacency matrix of a 2-regular undirected graph $G$ on $n$ vertices, without loops but potentially with multiple edges. This graph is a straightforward representation of every monomial in the determinant. Setting all variables $a_{12},a_{13},\ldots,a_{1n}$ to $1$ corresponds to eliminating the edges incident to vertex 1, resulting in a graph $G'$ on $n-1$ vertices. However, since $G$ was 2-regular without loops, we can uniquely identify the missing edges: they are incident to the vertices of degree less than 2 in $G'$.
The coefficient of the monomial is still the determinant of $P+P^T$, which equals $\pm2^k$, where $k$ is the number of cycles of length at least 3 in the graph $G$. The sign is the sign of the (more precisely, of any) permutation $P$.<|endoftext|>
TITLE: Alexandrov angles in Riemannian manifolds
QUESTION [9 upvotes]: Dear all, I am teaching a course in Riemannian geometry, and I would like to prove some comparison theorems in the next lessons, building on the well-known theory of Jacobi fields, and of Rauch comparison Theorem for Jacobi fields. I would like to stress the fact that several arguments can work in the case of geodesic metric spaces, and I'd like to prove as soon as possible that the metric and differential notion of angle coincide on Riemannian manifolds. 
Let me briefly recall the well-known notion of Alexandrov angle. If $X$ is a geodesic metric space and $\gamma_1,\gamma_2$ are geodesics (parameterized by arc-length) exiting from a point $p$, then the Alexandrov angle between $\gamma_1$ and $\gamma_2$ is the quantity $$\angle_p (\gamma_1,\gamma_2)=\limsup_{t,t'\to 0} \overline{\angle}_p (\gamma_1(t),\gamma_2(t'))\ ,$$ where $\overline{\angle}_p (\gamma_1(t),\gamma_2(t))$ is the Euclidean angle (in the point corresponding to $p$) of the Euclidean comparison triangle for the triple $(p,\gamma_1(t),\gamma_2(t'))$.
It is well-known that, in the case when $X$ is a Riemannian manifold, then the above $\limsup$ is in fact a genuine limit, and the Alexandrov angle between two geodesics exiting from the same point coincides with the Riemannian angle between them.
Here is my question: Is there a direct proof of this fact? Here, by direct proof I mean a proof that does not use too much of the theory of CAT(k)-spaces. 
In the book by Bridson and Haefliger, the above statement is proved in Corollary II.1A.7, which in turn relies on Proposition II.1.7, where several tools form the preceding sections are used. On the contrary, if we want to prove the easier fact that, in a Riemannian manifold, the limit
$$\lim_{t\to 0} \overline{\angle}_p (\gamma_1(t),\gamma_2(t))$$ exists and is equal to the Riemannian angle, then an easy computation using normal coordinates seems to suffice. I was wondering if some local estimates (for example, the expansion of the metric in normal coordinates with the subsequent estimates on the deviation of the exponential from being an isometry) could work for computing the $\limsup$ in the definition of Alexandrov angle (the difficult case being when $t,t'$ convergence to $0$ at a very different speed).

REPLY [8 votes]: Your equality is two inequalities.
To show the upper bound you can use the triangle inequality --- come closer to $p$ along the geodesic and apply the local estimates.
(This is the "first variation inequality" it holds in any metric space where angles defined.)
The lower bound follows since the injectivity radius at $p$ is positive.
Indeed, if the angle is smaller, the geodesic $[\gamma_1(t)\gamma_2(\tau)]$ converges as $\tau\to0$ to an other geodesic distinct from $\gamma_1$.
This question is the baby case of so called "lemma about strong angle" in Alexandrov geometry
it also can be called "first variation formula".
(Hope it helps, sorry if I misinterpret your question.)<|endoftext|>
TITLE: Rank gradients and HNN-extensions
QUESTION [6 upvotes]: Let $\pi$ be a finitely presented group and let $\phi:\pi\to \Bbb{Z}$ be an epimorphism.
Given $n\in \Bbb{N}$ we denote by $\pi_n$ the kernel of $\pi\xrightarrow{\phi} \Bbb{Z}\to \Bbb{Z}/n$
and given a group $\Gamma$ we denote by $\operatorname{rank}(\Gamma)$ its rank, i.e. the size of a minimal generating set.
The rank gradient of $(\pi,\phi)$ is then defined as
$ \liminf \frac{\operatorname{rank}(\pi_n)}{n}.$
This definition, I think, basically goes back to Marc Lackenby.
If $\ker(\phi)$ is finitely generated, then
 $\mbox{rank}(\pi_n)\leq \mbox{rank}(\ker(\phi))+1,$
i.e. the rank gradient is zero.  Now suppose that $(\pi,\phi)$ is represented by an ascending HNN-extension, i.e. there exists an isomorphism $\pi\cong \langle A,t|tAt^{-1}=\varphi(A)\rangle$  where $\varphi$ is a monomorphism and such that $\phi$ is given by the obvious surjection on the HNN-extension onto $\Bbb{Z}$. In this case we also have
$\operatorname{rank}(\pi_n)\leq \mbox{rank}(A)+1,$
i.e.
the rank gradient is zero. The same evidently also holds for descending HNN-extensions.
My question is now, whether there exist such pairs $(\pi,\phi)$ where the ranks of the groups $\pi_n$ are not bounded but for which the rank gradient is zero.

REPLY [4 votes]: Here are 2 remarks.

Since you assume $\pi$ finitely presented, by Bieri-Strebel 78, $\phi$ is the structural epimorphism of an HNN extension over a f.g. subgroup. More precisely, we can write $\pi=\langle H,t\mid txt^{-1}=\kappa(x),x\in K\rangle$ for some injective homomorphism $\kappa$ of a finitely generated subgroup $K$ of $H$ into $H$, and $\phi:\pi\to\mathbf{Z}$ maps $t$ to 1 and $H$ to 0. A convenient way to view $\pi$ is as the semidirect product $\Lambda\rtimes\mathbf{Z}$, where $\Lambda$ is the iterated amalgamated sum $$\cdots H\ast_K H\ast_K H\cdots,$$ where each $K$ embeds into its left neighbor $H$ by the inclusion, and embeds into its right neighbor $H$ by $\kappa$. So a closely related question is to determine, in such a setting, whether the rank of $\Lambda_n$ grows linearly, where $\Lambda_n$ is the amalgamated product (with the same inclusions) $H\ast_K\dots \ast_K H$ of $n$ copies of $H$. Indeed, $\pi_n$ is generated by $\Lambda_n$ and $t$. In the ascending case you mention, $\Lambda_n$ is isomorphic to $H$ and hence has bounded rank, so $\pi_n$ has bounded rank as well.

I know that Lackenby and you are motivated by topology, but from a group theoretic point of view it is natural to ask the question in the case of finitely generated groups. (I first expected the (standard restricted) wreath product $\pi=F\wr\mathbf{Z}$, where $F$ is a nontrivial perfect group, to provide an example (here $\pi_n=F^n\wr\mathbf{Z}$), but it seems that for such groups $\text{rank}(\pi_n)=2$ although $\text{rank}(F^n)$ grows logarithmically: see Jacques Thévenaz, Maximal subgroups of direct products.)<|endoftext|>
TITLE: Drawings of complete graphs with $Z(n)$ crossings
QUESTION [13 upvotes]: Hill conjectured that the minimum number of crossings in a drawing of the complete graph $K_n$ in the plane is exactly 
$$Z(n) = \frac{1}{4} \bigg\lfloor\frac{n}{2}\bigg\rfloor \left\lfloor\frac{n-1}{2}\right\rfloor \left\lfloor\frac{n-2}{2}\right\rfloor\left\lfloor\frac{n-3}{2}\right\rfloor.$$
In the literature, two general constructions of drawings of $K_n$ with $Z(n)$ crossings appear: 
1) the cylindrical (or tin can) drawing, where vertices are placed on the boundaries of the bottom and the top circular face of a cylinder and edges are drawn as geodesics,
2) a $2$-page (or cycle) drawing where the vertices form a regular $n$-gon, with the diagonals that are "more horizontal than vertical" drawn inside the $n$-gon and the remaining diagonals drawn outside the $n$-gon. Recently Abrego et al. showed that all optimal $2$-page drawings of $K_n$ are basically the same (up to some boundary effects for odd $n$).
The question:

Are there other known classes of drawings of $K_n$ with $Z(n)$ crossings? I am especially interested in explicit constructions like the two above.

References:
B. M. Abrego, O. Aichholzer, S. Fernandez-Merchant, P. Ramos, and G. Salazar, The 2-page crossing number of $K_n$, 2012, arXiv:1206.5669
L. Beineke and R. Wilson, The early history of the brick factory problem, The Mathematical Intelligencer 32(2) (2010), 41--48
H. Harborth, Special numbers of crossings for complete graphs, Discrete Mathematics 244 (2002), 95--102
F. Harary and A. Hill, On the number of crossings in a complete graph, 
Proc. Edinburgh Math. Soc. (2) 13 (1963), 333--338

Edit:
Apparently there is a rather broad class of drawings with crossing number $Z(n)+O(n^3)$, which generalize the cylindrical drawings:
3) A spherical drawing is a drawing on the sphere where edges are drawn as shortest arcs.
Moon showed that a random spherical drawing of $K_n$ has expected crossing number $\frac{1}{64} n(n-1)(n-2)(n-3)$.
J. W. Moon, On the Distribution of Crossings in Random Complete Graphs, J. Soc. Indust. Appl. Math. 13 (1965), 506--510
I think the following construction must be known but I couldn't find any reference.
For $n$ even, if one places $n/2$ pairs of antipodal points on the sphere (so that no three pairs are on the same great circle), then the crossing number of the induced spherical drawing is $Z(n)+X(n)$ where the term $X(n)$ denotes the number of crossings of the $n/2$ arcs (half-circles) connecting the pairs of antipodal points (so $X(n) < n^2/8$). For the spherical analogue of cylindrical drawings, we have $X(n)=0$.
Question 2:


Is there some simple criterion for the positions of the pairs of antipodal vertices so that the half-circles can be drawn in a non-crossing way? Can one obtain, in this way, "antipodal" spherical drawings with $Z(n)$ crossings that are not cylindrical?


Question 3:


Dropping the condition that vertices should form antipodal pairs, are there spherical drawings with $Z(n)$ crossings that are not cylindrical?



Edit 2:
The antipodal spherical drawings can be generalized a bit:
3) An antipodal pseudospherical drawing: for even $n$, place $n/2$ pairs of vertices in the plane arbitrarily. Through every two pairs $(a_1, a_2)$ and $(b_1, b_2)$, draw a simple closed curve that visits the four points in the order $a_1,b_1,a_2,b_2$, and does not pass through other vertices. The curve represents four edges of $K_n$. Every two such curves cross precisely twice, either at vertices or at other points (the crossings of the emerging drawing of $K_n$). In particular, the curves form an arrangement of pseudocircles. Finally, for each pair $(a_1^i, a_2^i)$, select another pseudocircle $\rho_i$ passing through $a_1^i$ and $a_2^i$ and draw one of its segments $\gamma_i$ between $a_1^i$ and $a_2^i$, so that no two curves $\gamma_i$ and $\gamma_j$ cross.
From the three non-equivalent minimal drawings of $K_8$, the cylindrical and the $2$-page drawing are both antipodal pseudospherical drawings. The optimal $2$-page drawing of $K_{10}$ is not antipodal pseudospherical (it does not contain a perfect matching of edges $\gamma_i$: every such edge would have to have $6$ crossings).
Question 4:


For even $n\ge 10$, are there antipodal pseudospherical drawings of $K_n$ with $Z(n)$ crossings that are not cylindrical?

REPLY [5 votes]: Since I cannot add a comment,   I would like to let you know that another family of drawings there other known classes of drawings of $K_n$ with $Z(n)$ crossings was found.
http://www.csun.edu/~sf70713/publications/NewFamilyCCCG2014.pdf<|endoftext|>
TITLE: Why $\Omega X$ and $BG$ are adjoint functors?
QUESTION [8 upvotes]: This is definitely not a research level question. I believe this is "common sense" among homotopists, however after "extensive" googling for 2 days I could not find a proof of it online or in standard textbooks (Hatcher, Milnor-Stasheff, Husemoller). I asked on mathstackexchange but did not receive an answer. I also tried to use online resources like nlab. It seems to be this is related to the bar construction, but I do not think the proof should be that complicated. I read in lecture notes that this is a natural consequence of the long exact sequence of fibration $F\rightarrow E\rightarrow B$, and the adjointness follows from homotopy lifting property, but I do not know how to put everything together. Below is the original post on Math.SE website. I also want to know "what should I read" for this kind of question, like why Bott periodicity can be framed as $\Omega U\cong \mathbb{Z}\times BU$.
It is not clear to me why we have a bijection of the form $$Mor_{Top^{*}}(BG,X)\rightarrow Mor_{Mon}(G,\Omega X)$$where $Mon$ is the category of topological monoids and $Top^{*}$ the based topological spaces. It seems to be something  essentially trivial that the instructor did not bother to write down the proof in notes, but after thinking for 20minutes I still do not understand how to construct such a bijection to let $B$ and $\Omega$ be adjoint functors (or maybe I formulated it wrong somehow?). I think I need to this result to prove the well known result that $$\Omega BG\cong G, B\Omega X\cong X$$
Just to clarify definition, here $BG$ is the weakly contractible total space $EG$ quotient out by group action of $G$. $\Omega X$ is the group formed by mappings of the circle to $X$ with a fixed based point.
Sorry for the low level of this question.

REPLY [7 votes]: Are you willing to accept a proof of the adjunction in the homotopy category?
I think the more natural way to do that is to prove first that $\Omega B G \simeq G$ (homotopy equivalence) and $B \Omega X \simeq X$ (note that $X$ needs to be connected for this second statement to be true, so you have to restrict your category).  The first equivalence follows from writing out the long exact sequence in homotopy for the fiber sequence $G \rightarrow EG \rightarrow BG$. (You use Whitehead's theorem to pass from an isomorphism on all homotopy groups to a homotopy equivalence.)  The second equivalence follows from comparing the fiber sequence $\Omega X \rightarrow P X \rightarrow X$ to the fiber sequence $\Omega X \rightarrow E \Omega X \rightarrow B \Omega X$.
Then if you have a map $BG \rightarrow X$, you loop it and get (up to homotopy) a map $G \rightarrow \Omega X$, and if you have a topological monoid morphism $G \rightarrow \Omega X$, you apply the bar construction (which is functorial) to get (up to homotopy) a map $B G \rightarrow X$.  These are inverses on homotopy classes.<|endoftext|>
TITLE: What does the $q$-Catalan Numbers count?
QUESTION [5 upvotes]: I had completed a paper describing the $q$-Catalan numbers, which is the $q$-analog of the Catalan numbers.
The $n$-th Catalan numbers can be represented by:
$$C_n=\frac{1}{n+1}{2n \choose n}$$
and with the recurrence relation:
$$C_{n+1}=\sum^n_{i=0}C_i C_{n-i}\ \ \ \ \ \forall n\geq 0$$
Now, for the $q$-analog, I know the definition of that can be defined as:
$$\lim_{q\to 1}\frac{1-q^n}{1-q}=n$$
and we know that the definition of the $q$-analog, can be defined like this: 
$$[n]_q=\frac{1-q^n}{1-q}=1+q+q^2+q^3+\cdots+q^{n-1}$$
which this is the $q$-analog of $n$.
and that for that $q$-analog of ${2n\choose n}$:
$$C_n(q)=\frac{1}{[n+1]_q}\begin{bmatrix}2n\\ n\end{bmatrix}_q$$
So, everything up to this point I know what I'm doing, and I'm not sure if I did everything correct after this
So, in order to generate the $q$-Catalan Numbers, I will need to use the Lagrange inversion formula.
And, then I got something like this:
$$G(X)=\sum^\infty_{i=0}C_i x^i$$
where $G(x)$ is the generating function, and that
$$G(x)=G_q(x)=\sum^\infty_{i=0}C_n(q)x^n=\sum^\infty_{i=0}C_nx^n=1+x+x^2(1+q)+\cdots$$
Since I know that for Catalan Numbers, it's true:
$$G(x)=(G(x))^2+1$$
So, the $q$-analog will just be:
$$G_q(x)=G(x)G_q(x)+1$$
So the recurrence relation for the $q$-analog Catalan Numbers:
$$C_{n+1}(q)=\sum^n_{i=0}C_i C_{n-1}q^i$$
It just doesn't sound right here...
Also, I don't have a clue that what does the $q$-Catalan Numbers count, can anyone help me with that or give me like a clue?
Help appreciated!

REPLY [10 votes]: As Vasu commented already: there is not "the" q-analogue of the Catalan numbers. And indeed, you're mixing two different here.

Your first q-Catalan numbers defined by the $q$-binomials is MacMahon's q-Catalan numbers which is (and I don't actually know many others) the major index generating function on Dyck paths, where the descent set is given by the positions of the valleys.

Your second $q$-Catalan numbers given by the recurrence is, on the other hand, the area generating function on Dyck paths.


Both are deeply related in the context of the $q,t$-Catalan numbers appearing in the theory of symmetric function as a bigraded Hilber series of (the alternating part of) the space of diagonal coinvariants.
As Darij mentions, both (and as well the $q,t$-Catalan numbers and the space of diagonal coinvariants) can be found e.g. in Jim Haglund's book http://www.math.upenn.edu/~jhaglund/books/qtcat.pdf. You actually find quite a bit as well in our online project http://www.findstat.org/StatisticsDatabase/St000012 and http://www.findstat.org/StatisticsDatabase/St000027.<|endoftext|>
TITLE: Expected edit distance
QUESTION [26 upvotes]: The edit or Levenshtein distance between two strings is the minimum number of single symbol insertions, deletions and substitutions to transform one string into another.   For example $$\operatorname{E}(01010,00100)=2.$$
Let $E_n$ be a random variable giving the edit distance between two random binary strings of length $n$. 

What bounds can be found for
$$\lim_{n \to \infty} \frac{\mathbb{E}(E_n)}{n} = c\;?$$

According to my non-extensive simulations, $c \approx 0.288$.
There is related work by Luecker on the expected length of the longest common subsequence which has lower and upper bounds of $0.788071n$ and $0.826280n$. This was a line of work originally started by Chvatal and Sankoff. However the longest common subsequence length and $n$ minus the edit distance need not be similar.

REPLY [2 votes]: I did some simulations myself, (5000 of each string length 1-1000) and thought I would share them here. They are plotted against Carlo's conjectured bound of $1-1/\sqrt{2}$ (in blue).
To me, the conjecture seems quite likely, since the smallest average distance I get is 0.296 at $n=1000$.<|endoftext|>
TITLE: Volume of Gr(2,4)
QUESTION [7 upvotes]: Hello
I was wondering if anybody can direct me to a paper or a book regarding the volume of $Gr(2,4) $ or generic complex Grassmanian manifolds of order $k$. My own heuristic method seems not to work! It is based on the adaption of the same procedure one has to follow for finding the volume of complex projective spaces $\mathbb{C}P^n$ using Hopf fibration $\mathbb{C}P^n\cong S^{2n+1}/S^1$. Here the volume can be roughly given by dividing the volume of $2n+1-$sphere by volume of $S^1$. Therefore in analogy with this example, we can estimate the volume of $Gr(k,n)$ by dividing the volume of $U(n)$ by that of $U(n-k) \times U(k)$ which gives me $12\pi^4 r^{16}$ for $Gr(2,4)$ where $r$ is the radius of $S^1$ and I don't like it because $Gr(2,4) $ is $8$ dimensional! 
Thanks in Advance
AB

REPLY [7 votes]: The volume of a Grassmanian can be computed using Wirtinger's theorem: 

The volume of a $p$-dimensional complex submanifold $S$ of a complex Hermitian manifold $(X,\omega)$  is 
$$
\frac{1}{p\!}\int_S\omega^p.
$$

If $X=\mathbb{CP}^N$ the integral is equal to the degree of $S$ times the volume of $X$. Thus up to normalization factors, the volume of the Grassmanian $Gr(k,n)$ is its degree in the Plücker embedding $$Gr(k,n)\subset \mathbb{CP}^N, N=\binom{n}{k}-1.$$<|endoftext|>
TITLE: A spectral inequality for positive-definite matrices 
QUESTION [8 upvotes]: Question. Given a positive-definite $n \times n$ matrix $A = (a_{ij})$ with eigenvalues 
$$
\lambda_1 \leq \cdots \leq \lambda_n ,
$$
is there a sharp upper bound for the product $\lambda_2 \cdots \lambda_n$ in terms of the quantity
$$
\|A\|_\infty := \max_{1 \leq i, j \leq n} |a_{ij}| ?
$$
A classic inequality due to A. Hirsch states that the modulus of an eigenvalue of an $n \times n$ complex matrix $A$ is less than $n  \|A\|_\infty$, which implies that 
$$
|\lambda_2 \cdots \lambda_n| \leq n^{n-1}\|A\|_\infty^{n-1} .
$$
However, this seems like a rather rough estimate for positive-definite matrices. I'm interested in any estimate that is substantially better than this.
Motivation. Using Hirsch's inequality one can improve Lemma 4 in page 43 of Siegel's Lectures on Quadratic Forms to yield the following result:
Theorem.
If $A = (a_{ij})$ is a positive-definite $n \times n$ matrix, then for every $x \in \mathbb{R}^n$ we have that
$$
\frac{\det(A)}{n^{n-1}a_{11} a_{22} \cdots a_{nn}} \sum a_{ii} x_i^2 \leq \sum a_{ij} x_i x_j \leq n \sum a_{ii} x_i^2 .
$$  
The inequality on the left would be greatly improved if we had the sharp upper bound required in the question. This in turn would yield a better answer to this enclosure problem (see my answer to that question).
Addendum. If we apply the estimate in Suvrit's answer in the proof of the theorem above, the inequality is indeed improved to:
$$
\frac{\det(A)}{2^{n-1}a_{11} a_{22} \cdots a_{nn}} \sum a_{ii} x_i^2 \leq \sum a_{ij} x_i x_j \leq n \sum a_{ii} x_i^2 .
$$  
In fact, in the proof the estimate  for $\lambda_2 \cdots \lambda_n$ is applied to an auxiliary matrix $B = (b_{ij})$ whose diagonal entries are all $1$ and for which $|b_{ij}| < 1$ if $i \neq j$.   
In turn this yields the following improved bound for the enclosure problem:
Theorem. Let $E \subset \mathbb{R^n}$ be an $n$-dimensional ellipsoid centered at the origin and containing no other integer point. There exists a transformation $T \in GL(n,\mathbb{Z})$ such that $T(E)$ is contained in the ball of radius
$$
\left(\frac{3}{2}\right)^{(n-1)(n-2)/2} \frac{2^n}{\epsilon_n}\sqrt{2^{n-1}}
$$
centered at the origin. 
Here $\epsilon_n$ is the volume of the unit ball of dimension $n$. 
Does anyone know a better bound?

REPLY [11 votes]: $\newcommand{\trace}{\operatorname{trace}}$ 
The result below mentions a reasonably improved inequality.
Let $m = \frac{\trace(A)}{n}$, and $s^2= \frac{\trace(A^2)}{n}-m^2$. Then, Wolkowicz and Styan (Linear Algebra and its Applications, 29:471-508, 1980), show that 
\begin{equation*}
 \lambda_1 \ge \frac{\det(A)}{(m+s/\sqrt{n-1})^{n-1}}
\end{equation*}
Remark: As per the notation in the OP, $\lambda_1$ is the smallest eigenvalue---usually the literature uses $\lambda_1$ to be largest.
Thus, we obtain the upper bound
\begin{equation*}
 \lambda_2\lambda_3\cdots\lambda_n \le \left(m+ \frac{s}{\sqrt{n-1}}\right)^{n-1}.
\end{equation*}
This bound is tight. Consider for example,
If $A= \text{Diag}(1,2,2,\ldots,2)$, then the lhs is $2^{n-1}$, $m=2-1/n$, and $s^2 
= 1/n-1/n^2$, so that $s/\sqrt{n-1} = 1/n$. Thus, the bound on the rhs is tight. 
With $M := \max_{i,j}|a_{ij}|$, we see that $m \le M$ and $s^2 \le nM^2 - m^2$, which leads to an upper bound in terms of $M$ as desired
\begin{equation*}
 \lambda_2\lambda_3\cdots\lambda_n \le \left(M+ \sqrt{\frac{nM^2-m^2}{n-1}}\right)^{n-1} < \left(M + M\sqrt{\frac{n}{n-1}} \right)^{n-1},
\end{equation*}
which is better than the bound mentioned in the post (though we lost a bit by deleting $m^2$).<|endoftext|>
TITLE: Hahn's Embedding Theorem and the oldest open question in set theory
QUESTION [31 upvotes]: Hans Hahn is often credited with creating the modern theory of ordered algebraic systems with the publication of his  paper Über die nichtarchimedischen Grössensysteme (Sitzungsberichte der Kaiserlichen Akademie  der Wissenschaften, Wien, Mathematisch - Naturwissenschaftliche Klasse  116 (Abteilung  IIa), 1907, pp. 601-655). Among the results established therein is Hahn’s Embedding Theorem, which is generally regarded to be the deepest result in the theory of ordered abelian groups. The following are two of its familiar formulations:
(i) Every ordered abelian group is isomorphic to a subgroup of a Hahn Group.
(ii) Every ordered abelian group G is isomorphic to a subgroup G’ of a Hahn Group, the latter of which is an Archimedean extension of G’.
(For definitions and modern proofs, see: A. H. Clifford [1954], Note on Hahn’s theorem on ordered Abelian groups, Proceedings of the American Mathematical Society, vol. 5, pp. 860–863; Laszlo Fuchs [1963], Partially ordered algebraic systems, Pergamon Press.)
Hahn’s proofs (and all subsequent proofs) of (i) and (ii) make use of the Axiom of Choice or some ZF-equivalent thereof. Moreover, while writing before the complete formulation of ZF (Foundation and Replacement had yet to be included), Hahn further maintained that he believed his embedding theorem could not be established without the well-ordering theorem, which had been established by Zermelo using Choice (and was subsequently shown to be equivalent in ZF to Choice). To my knowledge, this essentially amounts to the earliest conjecture that an algebraic result is equivalent (in ZF) to an assertion equivalent to the Axiom of Choice. Surprisingly, both Hahn’s use of Choice and his conjecture are overlooked in the well-known histories of the Axiom of Choice, including the excellent one by Gregory Moore. Apparently without knowledge of Hahn’s conjecture, D. Gluschankof (implicitly) asked if (i) is equivalent to the Axiom of Choice in ZF in his paper The Hahn Representation Theorem for ℓ-Groups in ZFA, (The Journal of Symbolic Logic, Vol. 65, No. 2 (Jun., 2000), pp. 519-52). However, Gluschankof did not answer the question and, unfortunately, died shortly after raising it. R. Downey and R. Solomon (in their paper Reverse Mathematics, Archimedean Classes, and Hahn’s Theorem) establish a countable version of Hahn’s theorem without using Choice, but their technique does not extend to the general case.
This leads to my two questions:

Has anyone established or refuted Hahn’s Conjecture?
Assuming (as I suspect) the answer to 1 is “no”, is the status of Hahn’s Conjecture the longest standing open question in Set Theory?

Amendment (Response to request for references)
Asaf: There are numerous proofs of Hahn’s Embedding Theorem in the literature besides the especially simple one due to Clifford. One proof is on pp. 56-60 of Laszlo Fuchs’s Partially ordered algebraic systems [1963] Pergamon Press. On page 60 of the just-said work there are also references to several other proofs including those of Clifford, Banaschewski, Gravett, Ribenboim and Conrad. Another proof, closely related to the one in Fuchs (including all preliminaries) can be found in Chapter 1 of Norman Alling’s Foundations of Analysis over Surreal Number Fields, North-Holland, 1987. Another very nice treatment, including all preliminaries, can be found in Chapter 1 of H. Garth Dales and W. H. Woodin’s Super-Real Fields, Oxford, 1996.There is also an interesting proof in Jean Esterle's Remarques sur les théorèmes d'immersion de Hahn et Hausdorff et sur les corps de séries formelles, Quarterly Journal of Mathematics 51 (2000), pp. 2011-2019.                                
For a now slightly dated history of Hahn’s Theorem, see my:
Hahn’s Über die nichtarchimedischen Grössensysteme and the Origins of the Modern Theory of Magnitudes and Numbers to Measure Them, in From Dedekind to Gödel: Essays on the Development of the Foundations of Mathematics, edited by Jaakko Hintikka, Kluwer Academic Publishers, 1995, pp. 165-213. (A typed version of the paper can be downloaded from my website: http://www.ohio.edu/people/ehrlich/)
Finally, I note that the earliest, but largely forgotten, altogether modern proof of Hahn’s theorem may be found on pp. 194-207 of Felix Hausdorff’s, Grundzüge der Mengenlehre, Leipzig [1914]. It was the lack of familiarity with Hausdorff’s proof and the need for a concise modern proof that led to the plethora of proofs in the 1950s.

REPLY [9 votes]: Regarding 

In 1902 Beppo Levi introduced the Partition Principle stating that if S is a partition of A then |S|≤|A|. Although he coined this principle to argue against its use by Bernstein, the latter rejected the criticism and claimed that this is one of the more important principles of set theory. 

This is a trivial thm of ZFC, but definitely not provable in ZF+DC, in particular, in the case A = the reals and S = the Vitali partition R/Q, as Sierpinski established in 1920s that any injection R/Q\to R implies a non-measurable set of reals, whose existence is not provable in ZF+DC by Solovay. 
Regarding the oldest unsolved concrete problem in set theory, this is most likely one of Hausdorff's questions on pantachies, namely, 
is there a pantachy containing no $(\omega_1,\omega_1^\ast)$ gap 
(H, Untersuchungen uber Ordnungstypen, V, 1907, May 05)
I underline that this is a concrete problem of existence of a certain mathematical object, rather than an abstract question on interrelations of different forms of AC spread over the whole set universe.
See more on this problem in my survey "Gaps in partially ordered sets" (12.1) in Hausdorff's Gesammelte Werke Band 1A, Springer 2013, 367-405, with additional references to Goedel (who rediscovered the problem with full ignorance of the Hausdorff's original formulation), Solovay, Kanamori.<|endoftext|>
TITLE: Random rings linked into one component?
QUESTION [17 upvotes]: Let $S$ be a sphere of unit radius.
Let $C_n$ be a collection of unit-radius circles/rings whose centers
are (uniformly distributed)
random points in $S$, and which are oriented (tilted) randomly (again, uniformly).

Q1.
  As $n \to \infty$, does the probability that all the
  rings in $C_n$ are linked together in one component approach $1$?

By "linked together" I mean that if you pick up any one ring,
all the others are physically connected and would follow.
For example, below there are $n=5$ rings, four of which are
connected, but one (topmost) is not:

          



Q2.  Same as Q1, but with $S$ a sphere of some (perhaps large) radius
  $r > 1$.
Q3.  Same as Q1, except with $S$ an arbitrary convex
  body, e.g., a cube.

I feel the answer to Q1 should be Yes, but I am
less certain of Q3.
Exact computation of probabilities as a function of $n$ might be difficult,
but I am hoping there are relatively simple arguments to settle these
questions. Thanks for ideas!
Answered (1May13).
The combination of
Ori Gurel-Gurevich's
and
Benoît Kloeckner's postings constitute a rather complete answer,
establishing that the answer to all my questions is Yes, even
without the assumption that $S$ is convex.
Thanks for the interest!

REPLY [5 votes]: I think I can complement the answer of Ori Gurel-Gurevich to prove that indeed, when we deal with connected open sets (no need for convexity) the answer is positive.
1. There is a finite configuration of circles $C_1, \dots, C_N$ whose centers are in the domain $D$, such that any circle $C$ with center in the domain $D$ must be linked to at least one of the $C_i$.
One way to do this is to first take a very tight lattice $\Lambda$, and put horizontal circles $C_1,\dots, C_K$ with centers on $\Lambda\cap D$. Now, a circle with center in $D$ that is not linked to any of these $C_i$ ($i\leq K$) must be roughly horizontal.
Then, add circles $C_{K+1},\dots, C_N$ with center on the lattice but oriented along a given vertical plane. A circle not linked to any $C_i$ must be roughly horizontal and roughly vertical, thus don't exist.
2. The above construction is stable under small perturbation. This means that there are small open sets $U_1,\dots, U_N$ of the parameter space such that for all set of circles $C_1,\dots, C_N, C$ such that $C_i\in U_i$, $C$ must be linked to one of the $C_i$.
3. By adding more circles that link together the $C_i$ of 1., we could have assumed that the $C_i$ make a linked component (this is where we use the connectedness assumption on $D$, which in fact could be weakened). As in 2., this is stable under small perturbation, so in fact there are small open sets $U_1,\dots, U_N$ of the parameter space such that for all set of circles $C_1,\dots, C_N, C$ such that $C_i\in U_i$, the $N+1$ circles must be linked together.
4. Now, when $n\to \infty$ the probability that circles have been drawn in each of the $U_i$ increase to $1$ exponentially fast, so at some point our random configuration contains with high probability a set of circles that links all admissible  circles, including all the other randomly drawn ones.<|endoftext|>
TITLE: Satellite knot example
QUESTION [6 upvotes]: Can someone provide me with an example of a satellite knot with symmetry group which is neither cyclic nor dihedral?

REPLY [15 votes]: If by the symmetry group of a knot you mean the group of isometries of $S^3$ leaving the knot invariant, then this can only be cyclic or dihedral, apart from the special case of torus knots which can have an $O(2)$ group of symmetries.  By restricting symmetries to the knot itself one gets a homomorphism from the symmetry group of the knot to the symmetry group of a circle, which can only be cyclic or dihedral. The kernel of this restriction map must be zero by the Smith Conjecture which says that a nontrivial periodic homeomorphism of $S^3$ cannot fix a nontrivial knot pointwise. (The conjecture was proved by combining work of several people following Thurston's breakthrough work on hyperbolic structures on Haken manifolds.)
There might be other less rigid definitions of the symmetry group of a knot. For example one could take the mapping class group of the pair $(S^3,K)$ for a knot $K\subset S^3$.  For satellite knots this group will generally be infinite due to Dehn twists along incompressible tori in the knot complement.  One could define a finite "supersymmetry" group by taking the quotient of the mapping class group in which twists along tori are factored out. This can be larger than cyclic or dihedral since there could be for example rotational symmetries of solid tori containing the knot and bounded by incompressible tori.  Examples for this are easy to find, such as the Whitehead double of the figure eight knot.<|endoftext|>
TITLE: Why isn't $\langle x,y,z|xyzx^{-1}y^{-1}z^{-1}\rangle$ a hyperbolic surface group?
QUESTION [11 upvotes]: The group mentioned in the title, $\langle x,y,z|xyzx^{-1}y^{-1}z^{-1}=1\rangle$, is in between the torus fundamental group $\langle x,y|xyx^{-1}y^{-1}=1\rangle$ and the two-holed torus fundamental group $\langle x,y,z,w|xyzx^{-1}y^{-1}z^{-1}w^{-1}=1\rangle$. 
It is not the hexagonal presentation of the torus because the homology is not the same (gluing up a hexagon with that pattern gives you two vertices instead of one, giving you only two loops in the one-skeleton, so this is not the torus group).
Its Cayley graph can be realized as an infinite tiling of hexagons, each of which meet six to a vertex.
Such a tiling can be embedded in the hyperbolic plane, making the Cayley graph quasi-isometric to hyperbolic space, which means that the group is delta hyperbolic with a circle at infinity, implying that the group is Fuchsian, by the work of Gabai and others.
So it has a finite index surface subgroup. But this group is a subgroup of the RAAG with defining graph the diamond graph, I.e. $F_2\times F_2$. This can be seen by letting $a,b$ generate the first free group, $c,d$ generate the second subgroup, and letting $x=ab^{-1},y=bc^{-1}$, and $z=cd^{-1}$.
*Edit:*I meant to say that $a,c$ generate the first group and $b,d$ generate the second.\
So this implies that the diamond graph contains a hyperbolic subgroup. But in all the RAAG references, they say that RAAG's contain surface subgroups if they contain 5-cycles. So why does it seem as if a four-cycle (the diamond graph) contains a surface subgroup?

REPLY [8 votes]: I think Lee's and Steve's comments pretty much answer this question.  Let me try to summarize, and clear up a couple of misconceptions that seem to be lurking.  For convenience, I'll denote your group by $G$.
The map $G\to F_2\times F_2$.
Actually, I don't think the map $G\to F_2\times F_2$ that you describe is an injection.  The images of $x$ and $z$ commute, so the image also satisfies the relation $[x,z]=1$.  Your relator then becomes $[z^{-1}x,y]=1$.  It follows from these two relations that the image is isomorphic to $F_2\times\mathbb{Z}$.
Indeed, the theorem of Baumslag and Roseblade that I alluded to in my comments says more or less (there are some difficulties in getting the statement exactly right) that every fp subgroup of a free-cross-free group is virtually free-cross-free, so we could have guessed it would be of this form.
The group $G$.
If you have a presentation in which every generator (or its inverse) appears exactly twice then it is, indeed, natural to guess that it might be a surface group.  And it nearly is.  But there's one more thing you need to check: the link of the 1-vertex of the corresponding presentation complex.
This is called the Whitehead graph of the relations.  It's necessarily a union of cycles (in the case when each generator appears exactly twice), but the presentation complex is a surface exactly when the Whitehead graph consists of just one cycle.  Otherwise, the presentation complex is a surface with some points identified, and so the group is a free product of a surface group and a free group.
The Whitehead graph is easy to calculate; in this case, it turns out to consist of two cycles, and so we have two points identified, and $H_1$ tells us that the surface is in fact a torus.
Therefore, $G\cong\mathbb{Z}*\mathbb{Z}^2$, as Lee and Steve correctly said.  But I hope the above gives you some idea of how one might calculate this, rather than just pluck it from thin air.
The Convergence Group Theorem.
By the way, here's another way to see that something must have been wrong.  It follows from the Convergence Group Theorem of Tukia, Casson--Jungreis and Gabai, which you mention, that any torsion-free group which is virtually a surface group is in fact a surface group.  Therefore, if $G$ really were a subgroup of $F_2\times F_2$ and virtually a surface group, it would have had to have been a surface group.  Of course, there's only one candidate, which can be ruled out by looking at $H_1$.<|endoftext|>
TITLE: Can one make the category of pairs of topological spaces a model category?
QUESTION [5 upvotes]: Can you make the category whose objects are pairs of spaces $(X,A)$, and morphisms the obvious diagrams, into a model category?  Of course I want this to be done in a meaningful way, that is, agreeing with the adjoint functors $X\mapsto (X,\emptyset)$ and $(X,A)\mapsto X$?
There might be some intuitive reason that it is wrong to expect this, but I don't see it yet.  
Thanks!

REPLY [8 votes]: The answer has to be no. There's just no good homotopy theoretic way to talk about subspaces, because up to homotopy every map is an inclusion (via the mapping space construction). So even before you run into the completeness issue Karol raises you have a more fundamental issue of what the homotopy category would be. It seems to me that there's no good category theoretic/homotopy theoretic way to pick out a space and a subspace of it; the closest you can do is Tyler's comment and Karol's answer. Since neither of them mentioned this, I'll mention that it's also known as the Arrow Category $Arr(C)$ (in this case $Arr(Top)$). It's a well-studied object, but on the surface seems very different from what you were asking for. However, the lens of homotopy can't see the difference between an object in $Arr(Top)$ and an inclusion, so I guess that's the best you can do. The other obvious idea (taking the product model structure on $Top \times Top$ and then placing some restriction so the only pairs $(X,A)$ you get have $A\subset X$) fails for the same reason.<|endoftext|>
TITLE: Computing a large permanent
QUESTION [13 upvotes]: Is there a practical way to compute the permanent of a large ($91 \times 91$) $(0,1)$ matrix?
I have tried to use the matlab function written by Luke Winslow which works great for smaller matrices but it's not getting anywhere with this size.
UPDT: I am thinking of the incidence matrix of an order 9 projective plane. Does that help?

REPLY [8 votes]: The answer is unfortunately probably no, but there are a few things you could try.
There are algorithms that run in time polynomial in the value of the permanent, meaning that the permanent can be computed quickly if its value is small, but that is not going to help you for the incidence matrix of a 9x9 projective plane, whose permanent will be huge.  I think getting the exact value in your example will be hopeless unless you can exploit the geometric structure to dramatically reduce the size of the computation.
If you only want an approximation, then there's a little bit more hope.  Famously, Jerrum, Sinclair and Vigoda exhibited a fully polynomial randomized approximation scheme.  This was a fantastic theoretical achievement, but in practice the algorithm is not all that fast (having high exponents and multiplicative constants) and is not at all trivial to implement.
Your best bet may be to look at recent work on using belief propagation to approximate the permanent.  I am not up to speed on the latest developments but I would start by contacting the authors of "Approximating the Permanent with Fractional Belief Propagation" for suggestions.<|endoftext|>
TITLE: equivalence between katz and classical modular forms
QUESTION [5 upvotes]: $\newcommand{\CC}{\mathbb{C}}$
$\newcommand{\ZZ}{\mathbb{Z}}$
$\newcommand{\PP}{\mathbb{P}}$
$\newcommand{\QQ}{\mathbb{Q}}$
$\newcommand{\hH}{\mathcal{H}}$
$\newcommand{\eE}{\mathcal{E}}$
$\newcommand{\dD}{\mathcal{D}}$
$\newcommand{\aA}{\mathcal{A}}$
$\newcommand{\oO}{\mathcal{O}}$
$\newcommand{\Tate}{\text{Tate}}$
$\newcommand{\can}{\text{can}}$
$\newcommand {\spmatrix}[4]{\left[\begin{smallmatrix}#1&#2\\#3&#4\end{smallmatrix}\right]}$
$\newcommand{\Spec}{\text{Spec }}$
$\newcommand{\an}{\text{an}}$
EDIT: Most of the arguments I've used here are WRONG!
So, I've seen many definitions of modular forms, and I can't seem to find a detailed proof of their equivalence. I'll list the two main definitions I've seen, and try to prove/explain as much of their equivalence as I can, leaving the rest as questions. I hope some of you will find this interesting, and let me know if I'm thinking of something in the wrong way, or if I'm just flat out wrong.
I will highlight my questions or things I'm not clear about in bold.
For simplicity, we'll begin by restricting ourselves to the case of modular forms for $\Gamma(N)$, $N\ge 1$.
Because I'm having trouble writing matrices here, we'll always assume that $\gamma$ is a 2x2 matrix with entries $a,b,c,d$ (left to right, top to bottom).
$\textit{Definition 1 (classical)}$.
A modular form of weight $k$ for $\Gamma(N)$ is a function $f : \hH\rightarrow\CC$ with the following properties:
1(a) - $f$ is holomorphic
1(b) - $f(\gamma(z)) = (cz + d)^k f(z)$ for all $\gamma \in \Gamma(N)$
1(c) - $(cz+d)^kf(\gamma(z))$ has $q^{1/N}$ - expansion $(q = e^{2\pi iz})$ with no negative power terms.
$\qquad$
$\textit{Defintion 2 (Katz)}$
Let $R_0$ be a ring containing $1/N$ and a primitive $N$th root of unity. A modular form of weight $k$ for $\Gamma(N)$ defined over a ring $R_0$ is a function on triples $(E/R, \omega, \alpha_N)$, where $E/R$ is an elliptic curve over an $R_0$-algebra $R$, given by the structure map $p : E\rightarrow R$, $\omega$ is a basis of $\underline{\omega}_{E/R} := p_*\Omega^1_{E/R}$, and $\alpha_N : E[N]\stackrel{\sim}{\longrightarrow} (\ZZ/N\ZZ)_R^2$ is an isomorphism of group schemes, which takes values in $R$, and satisfies the following properties:
2(a) $f(E/R, \omega,\alpha_N)$ depends only on the $R$-isomorphism class of the triple $(E/R, \omega, \alpha_N)$.
2(a') For any $\phi : R\rightarrow R'$, we have $f(E'/R', \omega', \alpha_N') = \phi(f(E/R, \omega, \alpha_N))$, where $E', \omega', \alpha_N'$ denote $E, \omega, \alpha_N$ base changed to $R'$.
2(b) For any $\lambda\in R^*$, $f(E/R, \lambda\omega, \alpha_N) = \lambda^{-k}f(E/R,\omega, \alpha_N)$.
2(c) $f(\text{Tate}(q^n)/$$(\ZZ((q))\otimes_\ZZ R_0)$, $\omega_{\can}, \alpha_N)\in\ZZ[[q]\otimes_\ZZ R_0$ for all level structures $\alpha_N$ on $\Tate(q^n)$.
$\qquad$
This leads me to my first semi-question - $\underline{\omega}_{E/R}$ isn't always free right? So, I'm guessing this $f$ takes as input only triples where $E/R$ is an elliptic curve and $R$ is "small enough" that $\underline{\omega}_{E/R}$ is trivial over $R$?
It ought to be the case that a katz modular form defined over $\CC$, restricted to triples over $\CC$, is equivalent to a classical modular form. First suppose that $N \ge 3$, so that the modular curve $X(N)$ is a fine moduli scheme for a representable moduli problem. Then, we have the universal generalized elliptic curve $\eE := \eE_N\rightarrow X(N)$ with a canonical level $N$ structure $\aA_N$ such that any isomorphism class of generalized elliptic curve with level $N$ structure is a pullback of $(\eE_N,\aA_N)$. Over the affine non-cuspidal piece $Y(N)$, $\eE|_{Y(N)}$ is an elliptic curve, and over the cusps $\eE$ has fibers that are Neron $N$-gons.
In this case, let $U = \Spec R$ be a open set of $Y(N)$ such that $\underline{\omega}_{\eE/R}$ is free, with a basis $\omega_{\eE/R}$. Then, for any Katz modular form $f$ of weight $k$ and level $N$, $f(\eE/U,$ $\omega_{\eE/R}$ $, \aA_N)\in R$, which by definition consists of holomorphic functions on $U$ viewed as a subset of $Y(N)^\an$. Thus, for any map $p : \Spec\CC\rightarrow U$, letting $\tilde{p}$ denote the corresponding map of rings $\tilde{p} : R\rightarrow\CC$, note that this map is just "evaluate $f\in R$ at the point $p$ of $U$". Hence, we see that $f((p^*\eE)/\CC, p^*\omega_{\eE/R}, p^*\aA_N) =$ $\tilde{p}(f(\eE/U,$ $\omega_{\eE/R}$ $,\aA_N))$ is a holomorphic function "in $p$". Now, let $\dD\subset\hH$ be a connected fundamental domain for $\Gamma(N)$ (say, a union of $\text{SL}_2(\ZZ)$-translates of the standard fundamental domain for $\text{SL}_2(\ZZ)$). Then since the points of $\dD$ are in 1-1 correspondence with isomorphism classes of pairs $(E/\CC, \alpha_N)$, we've just shown that $f$ gives a holomorphic function on $U\cap\dD^\circ$ (the $^\circ$ denotes interior). Let $f'$ denote this function. Ie, for any $\tau\in U\cap\dD^\circ$, define $f'(\tau) = f(E_\tau/\CC, \omega_\tau, \alpha_{N,\tau})$, where $\omega_\tau$ is the pullback of the differential $\omega_{\eE/R}$ on $\eE$ via the map $\Spec\CC\mapsto\tau\in U$.
Now, using 2(a) and considering an open cover of $Y(N)$ that trivializes $\underline{\omega}_{\eE/R}$, one finds that $f'$ is a holomorphic function on $\dD^\circ$, Then, one can compute that condition 2(b) (weight $k$ homogeneity w.r.t. $\omega$) shows that $f'$ satisfies condition 1(b) (this is spelled out rather clearly in various sources). Now, some care will be needed to show that $f'$ extends to a holomorphic function on the entirety of $\hH$ (ie, want to prove that $f'$ is continuous on the boundaries of translates of $\dD$), though I expect this can be accomplished by considering different fundamental domains?
This leads to another question: What of the case $N < 3$, ie when no universal elliptic curve exists? For example, in the post:
Is there an elliptic surface over $Y(1)$?
Will Sawin gives a proof that there is no neighborhood of the points $j = 0$ or $j = 1728$ that has an elliptic curve above it with the property that the fiber above $j = a$ is an elliptic curve with $j$ invariant equal to $a$. Is there a nice way to see that a Katz modular form must be "continuous" at these points?
Lastly, we have the notion of holomorphicity at the cusps. My understanding of this is as follows. For a classical modular form $f'$, this is defined by looking at the expansion of $f'$ in terms of $q := q_N := q^{1/N} = e^{2\pi i z/N}$. We do this, because on the upper half plane, the map $e^{2\pi i z/N}$ gives a conformal equivalence from a vertical strip of width $N$ to the punctured unit disk. Now, passing to the quotient space $Y(N) = \Gamma(N)\backslash\hH$, we see that $q$ is invariant under the stabilizer of $\infty$ in $\Gamma(N)$, and hence is a holomorphic function in a neighborhood of $\infty\in X(N)$. Looking at fundamental domains, we see that in fact $q$ is a complex analytic local uniformizer at $\infty$, which does not extend to a global meromorphic function on all of $X(N)$. Now, let $t$ be an algebraic uniformizer at $\infty\in X(N)$ - ie, it's a globally meromorphic function with a simple pole at $\infty$, and is in the stalk $\oO_\infty$ of the algebraic structure sheaf of some algebraic model of $X(N)$. Then even though $q\notin\oO_\infty$, we have that $\widehat{\oO}_\infty\cong \CC[[t]] = \CC[[q]]$, where here we view $q$ as a formal power series in $t$, which can be accomplished since $t$ (thinking of it as a modular function, and hence having a fourier expansion), can be seen as a formal power series in $q$ of order 1.
Now, there's a natural map $\Spec\oO_\infty\rightarrow X(N)$, which extends to a map $i : \Spec\CC((t))\rightarrow\Spec\oO_\infty\rightarrow X(N)$. Intuitively speaking, the pullback of $\eE$ along $i$ should correspond to an infinitesimal family of elliptic curves with $j$-invariant and level $N$ structure parametrized by $t$. Now, identifying $\CC(t) = \CC(q)$ as per the previous paragraph, we can consider the pullback of $\eE$ along $i$ to be the same infinitesimal family, now reparametrized by $q$ (viewed as a power series in $t$). Complex analytically, $q$ actually gives a conformal equivalence of a neighborhood of $\infty$ with a open subset $V$ of the punctured unit disk. In this setting, noting that a modular form of level $N$ can be thought of as a function on lattices, and hence on elliptic curves with differential modulo isomorphism (in both cases with some level structure, see section 1.5 in Diamond/Shurman), it's not hard to see that viewing a modular form as a function on the "Tate family" over $V$, its $q$-expansion is actually just a function on $V$, which agrees with the notion of evaluating a Katz modular form at the Tate curve over $\CC((q))$. Hence, the "holomorphicity at $\infty$" conditions 1(c) and 2(c) are equivalent (may need to use the fact that congruence modular forms have bounded denominators).
The above two paragraphs show that the Tate curve is the pullback of $\eE$ by a rather special morphism $\Spec\CC((q))\rightarrow X(N)$. Namely, the morphism is obtained by mapping an algebraic uniformizer $t$ in $\oO_\infty$ to its power series in $q$ given by the realization of $t$ as a periodic meromorphic function on $\hH$.

REPLY [3 votes]: On $\omega_{E/R}$: Yes. I don't remember if this is always free. But the definition means exactly what it says $E/R$ is an elliptic curve, and $\omega$ is a basis for $\omega_{E/R}$. From this we can conclude that $\omega_{E/R}$ has a basis. I'd also like to point out that we can always make $\omega_{E/R}$ free by making $R$ larger - it's a line bundle, and its sixth power has a nowhere vanishing section, $\Delta$, so it's always trivialized on a sixfold cover.
On $f'$: Yes. The key point is that the definition of the function does not depend on the fundamental domain used, and every point is in the interior of some fundamental domain. (translate your favorite fundamental domain by an appropriate element of $SL_2(\mathbb R)$.) 
On continuity: Use the universal elliptic curve with full level $3$ or $4$ structure or something like that. This extends to $j=0$ and $j=1728$. You then get a function on this. The base curve will have a lot of automorphisms that change the level structure but fix the elliptic curve. Using $2$, we see that the function is invariant to these automorphisms, so it descends to the quotient, which gives the continuity (in fact analyticity) that you want.<|endoftext|>
TITLE: Counterintuitive consequences of the Axiom of Determinacy?
QUESTION [31 upvotes]: I just read Dr Strangechoice's explanation that if all subsets of the real numbers are Lebesgue measurable, then you can partition $2^\omega$ into more than $2^\omega$ many pairwise disjoint nonempty subsets.  Note that this paradoxical conclusion is a consequence of the Axiom of Determinacy.  
I've read of many counterintuitive consequences of the Axiom of Choice, but what other consequences come from the Axiom of Determinacy?  
(I am aware that AD is inconsistent with AC, and yes, AC being false is rather counterintuitive.)

REPLY [17 votes]: Graph combinatorics is very different under AD to what we are used to under AC. For example, under AD there is an acyclic graph on $\mathbb{R}$ with no 2-coloring (indeed, no coloring with countably many colors). Hall's marriage theorem is also false, even in a very trivial case: there is a 2-regular bipartite graph on $\mathbb{R}$ with no perfect matching.
These theorems both follow from the assertion that every set of reals has the Baire property, and the arguments can be found in Kechris-Solecki-Todorcevic's 1999 paper: Borel Chromatic Numbers.
I also have a recent paper on descriptive combinatorics that directly uses the axiom of determinacy, and not just measure or category arguments. For example, I show that under AD, for every n, there is an acyclic n-regular graph on $\mathbb{R}$ with chromatic number n+1, and an acyclic bipartite n-regular graph with no perfect matching. The games I use are based on partitions of free products of countable groups into two pieces.<|endoftext|>
TITLE: Defining Equations of a Flag Variety
QUESTION [5 upvotes]: I've been reading Fulton's Young Tableaux, and I'm trying to understand flag varieties. I want to understand the defining equations of a Flag Variety, but the coordinates in Fulton's Plucker relations, and in the equations defining the flag varieties are in the dual projective space. I tried to understand this proof, but I'm very confused.
Does anyone know another source, where they write down the flag's defining equations without using the dual space?
I've searched and searched, but I haven't been able to find anything. Everyone seems to mention it, but they always reference Fulton.
Thanks!

REPLY [9 votes]: Equations are for cutting schemes out of ambient schemes. Are you sure you want to cut your flag manifold out of projective space? There are easier places to find it.
To begin with, you could embed $Flags(n) \to \prod_{k=1}^{n-1} Gr_k(n)$, taking the flag to its list of subspaces. Then the equations are "for each $i < j$, the $i$-plane should be contained in the $j$-plane".
You could Plücker embed each Grassmannian $Gr_k(n)$ into projective space, or you could regard it as $GL(k) \backslash \backslash M_{k\times n}$, i.e. look at row-spans of $k\times n$ matrices. Then the equations above say that when you stack your $i\times n$ matrix and $j\times n$, the resulting $(i+j)\times n$ matrix should only have rank $j$, so all $(j+1)\times (j+1)$ determinants should vanish. There's some equations.
If you do Plücker embed, it means you only have the Plücker coordinates on those Grassmannians, and so you get Plücker relations between the Plücker coordinates of size $i$ and $j$. I find these much harder to remember than the determinants above.
Once you've Plücker embedded the Grassmannians, then you can Veronese them each by different amounts, then Segre the whole thing together, and you get all the projecively normal embeddings of the flag manifold. It's interesting to note that all the equations encountered along the way are linear or quadratic (a theorem of Ramanathan for general $G$).
Anyway one very good answer to your actual question is [Miller-Sturmfels], chapter 15 I think it is, as Victor Protsak suggested.<|endoftext|>
TITLE: How exactly does Schützenberger promotion relate to Striker-Williams promotion?
QUESTION [11 upvotes]: Schützenberger promotion, studied (for example) in Richard Stanley, Promotion and Evacuation, 2009, is a permutation of the set of all linear extensions of a finite poset. Since one can identify the linear extensions of a poset with saturated chains of order ideals in that poset, this allows one to also view Schützenberger promotion as a permutation of the set of the latter. The famous promotion of standard Young tableaux is a particular case of this.
Striker-Williams promotion, defined in Jessica Striker, Nathan Williams, Promotion and Rowmotion, arXiv:1108.1172v3, Definition 4.13, is a permutation of the set of all order ideals (not saturated chains of order ideals!) of a so-called "rc poset" (which is a poset with a map into $\mathbb Z^2$ satisfying certain conditions, best viewed as a way to draw its Hasse diagram on a grid; see below or §4.2 of Striker-Williams for an exact definition).
Apparently people are considering these two promotions to be closely related. However, the only direct relation I am aware of is Striker-Williams Theorem 4.12, which bijects Schützenberger promotion on standard tableaux on a two-rowed Young diagram with Striker-Williams promotion on a poset which looks like a triangle grid.
Questions:
1. Is this really the only relation? Is promotion of standard Young tableaux of a Young diagram with more than $2$ rows not a (known) case of Striker-Williams promotion?
2. I've seen some kind of promotion on semistandard Young tableaux being mentioned on the internet. Assuming it's not a typo, how is that defined?

Appendix:
Let me define the two notions involved for the sake of completeness. Probably the sources quoted give better definitions...
Definition of Schützenberger promotion: Let $P$ be a finite poset. Let $\mathcal L\left(P\right)$ denote the set of all linear extensions of $P$. We define a map $\partial : \mathcal L\left(P\right)\to \mathcal L\left(P\right)$ as follows:
Let $f \in \mathcal L\left(P\right)$ be a linear extension. We set $p=\left|P\right|$, and we view $f$ as a function $P\to\left\lbrace 1,2,...,p\right\rbrace$, i. e., as a labelling of the elements of $P$ by the numbers $1$, $2$, ..., $p$ (we get this labelling by labelling every element $v\in P$ with the number $\left| \left\lbrace w\in P \ \mid \ f\left(w\right)\leq f\left(v\right) \right\rbrace \right|$). Define a (dynamic) map $g:P\to\mathbb Z$ by $g = f$ (we will be modifying $g$, while $f$ remains static). If $p=0$, do nothing. Else, set $u$ to be the element of $P$ labelled $1$ (that is, the smallest element of $P$ with respect to $g$), and do the following loop:
While there exists an element of $P$ covering $u$:
let $v$ be the smallest (with respect to $g$) among the elements of $P$ covering $u$ (that is, the element $p$ of $P$ covering $u$ with smallest $g\left(p\right)$);
slid the label of $v$ down to $u$ (that is, set $g\left(u\right)$ to be $g\left(v\right)$, accepting that $g$ will temporarily fail to be injective);
set $u = v$.
Endwhile.
After the end of this loop, label $u$ with $p+1$ (that is, set $g\left(u\right) = p+1$), and then subtract $1$ from each label (i. e., replace $g$ by $g-\mathbf{1}$, where $\mathbf{1}$ is the constant function $P\to\mathbb Z,\ p\mapsto 1$).
The resulting $g$ is called the promotion of $f$, and denoted by $\partial f$. (It is more common to call it $f\partial$, so that $\partial$ is seen as a map acting from the right).
Definition of Striker-Williams promotion: Let $P$ be a finite poset. Let $J\left(P\right)$ denote the set of all order ideals of $P$. For every $q\in P$, define a map $t_p : J\left(P\right) \to J\left(P\right)$ as follows: Let $I \in J\left(P\right)$. If $I \bigtriangleup \left\lbrace p\right\rbrace$ (with $\bigtriangleup$ standing for "symmetric difference") is an order ideal of $P$, set $t_p\left(I\right) = I \bigtriangleup \left\lbrace p\right\rbrace$. Otherwise, set $t_p\left(I\right) = I$.
Let $\mathbb Z^2_{\operatorname*{ev}}$ denote the $\mathbb Z$-submodule of $\mathbb Z^2$ spanned by $\left(1,1\right)$ and $\left(2,0\right)$. In other words, let $\mathbb Z^2_{\operatorname*{ev}}$ be the set of all $\left(x,y\right)\in\mathbb Z^2$ for which $x+y$ is even.
Now, let $P$ be a finite rc-poset; this means a poset along with a map $\pi : P \to \mathbb Z^2_{\operatorname*{ev}}$ such that whenever an element $p_1$ of $P$ covers an element $p_2$ of $P$, we have $\pi\left(p_1\right)-\pi\left(p_2\right) \in \left\lbrace \left(-1,1\right), \left(1,1\right) \right\rbrace$. (See §4.2 of Striker-Williams for some good pictures of what this means.)
For every $p\in P$, let $\pi_1\left(p\right)$ denote the first coordinate of $\pi\left(P\right)$. Now, consider the composition of the maps $t_p$ in decreasing order of $\pi_1\left(p\right)$ (the relative order of the $t_p$ for distinct $p$ having the same $\pi_1\left(p\right)$ does not matter). This composition is Striker-Williams promotion.

REPLY [5 votes]: Regarding your first question, promotion of standard Young tableaux of rectangular shape (with any number of rows) can be viewed as as a special case of the piecewise-linear "lift" of Striker-Williams promotion. Specifically, there is an equivariant bijection between (on the one hand) the action of the Schutzenberger promotion operator on the set of semistandard Young tableaux of rectangular shape with $A$ rows and $B$ columns having entries between 1 and $n$, and (on the other hand) the action of the piecewise-linear 
promotion operator on the rational points in the (reverse) order polytope ${\cal O}(([A] \times [n-A])^{\rm op})$ with denominator dividing $B$. See http://jamespropp.org/gtt-promotion.txt for a more detailed and in-depth explanation.
Regarding your second question, my understanding is that there are two equivalent definitions of promotion for semistandard tableaux, one using jeu de taquin (which I don't really understand) and the other using Bender-Knuth involutions.  I could say more about the latter (and will if you like), but maybe that's enough for now.  One early reference for this is E. Gansner, On the equality of two plane partition correspondences, Discrete Math. 30 (1980), 121-132. (Can anyone suggest anything more recent?)<|endoftext|>
TITLE: Permutations of prescribed cycle types that multiply to the identity
QUESTION [8 upvotes]: Suppose that $\lambda_1,\lambda_2,\lambda_3$ are partititions of $n$.  When do there exist permutations $\sigma_1,\sigma_2,\sigma_3 \in S_n$ such that
(1) $\sigma_1\sigma_2\sigma_3$ is the identity;
(2) the $\sigma_i$ generate a transitive subgroup of $S_n$; and
(3) the cycle type of $\sigma_i$ is $\lambda_i$ for all $i$?
One can show that the total number of parts among the $\lambda_i$ must be an integer that is at most $n+2$ and that is congruent to $n$ modulo 2.  Is this condition sufficient?  If $\lambda_3 = (n)$ and the total number of parts of the $\lambda_i$ is exactly $n+2$, then the existence is known.
This problem is equivalent to the existence of an irreducible curve $X$ over the complex numbers and a nonconstant morphism $X \rightarrow \mathbb{P}^1$ whose branch locus is contained in $\{0,1,\infty\}$ (a Belyi map) and whose monodromy generators have cycle types $\lambda_1,\lambda_2,\lambda_3$.

REPLY [2 votes]: The answer is no.  No such triples exist for $n = 4$, $\lambda_1 = (3, 1)$, $\lambda_2 = \lambda_3 = (2, 2)$.  (Indeed, in $S_4$, the double transpositions and the identity form the Klein four subgroup.)
For other results on this problem, see for example http://arxiv.org/abs/math/0508434.<|endoftext|>
TITLE: Cofibrant replacements of a given object in a combinatorial model category
QUESTION [7 upvotes]: In a combinatorial model category, every $\lambda$-filtered colimit is a homotopy colimit for $\lambda$ regular big enough. So for $\lambda$ regular big enough, every $\lambda$-filtered colimit of a diagram of cofibrant replacements of a given object $X$ is a cofibrant replacement of $X$. Does the contrary hold, i.e. is the full subcategory of cofibrant replacements of a given object accessible ?
EDIT : the class of cofibrations is accessible so for $\lambda$ regular big enough, a $\lambda$-filtered colimit of cofibrations $\varnothing \to X_i$ is a cofibration ; the class of weak equivalences is accessible so for $\lambda$- regular big enough, a $\lambda$-filtered colimit of weak equivalences $X_i\to X$ is a weak equivalence ; so for $\lambda$ regular big enough, a $\lambda$-filtered colimit of cofibrant replacements of $X$ is a cofibrant replacement of $X$.

REPLY [3 votes]: (sorry I have troubles with comments, I post here even if it is not an answer) I have a new information. In On a fat small object argument, it is proved that in a λ-combinatorial model category, every cofibrant object is a λ-filtered colimit of λ-presentable cofibrant objects, which is close to what I wanted.<|endoftext|>
TITLE: Order of torsion group
QUESTION [13 upvotes]: What can one say about the order of the torsion group of an elliptic curve defined over the compositum of all quadratic extensions of $\mathbb{Q}$ ?

REPLY [6 votes]: Mazur's theorem is not needed here, and besides, it applies only to elliptic curves themselves defined over $\mathbb{Q}$ rather than the said compositum $K^{(2)}$ of all quadratic extensions of $\mathbb{Q}$ (which is how I understand the question).
A simple proof is available by purely local methods. First, if $F/\mathbb{Q}_p$ is a finite extension, then any abelian variety $A/F$ has $|A(F)_{\mathrm{tors}}| < \infty$. This follows from the theory of the formal group; for the elliptic curve case, see Chapter IV, Theorem 6.4 in Silverman's book (supplemented by Ch. VII Ex. 7.6 as noted by  René).
Given this, fix any prime. Since $\mathbb{Q}_p$ has just finitely many quadratic extensions, it is possible to embed $K^{(2)}$ in a finite extension $F/\mathbb{Q}_p$. Then the stated finiteness follows from the above quoted result.<|endoftext|>
TITLE: Steinberg reps of reductive groups over local fields vs finite fields
QUESTION [5 upvotes]: Let $G$ be a reductive group over a non-archimedean field $F$ with reisdue field $f$. 
Edit: The statements only make sense modulo tensoring by one-dimensional representations.

Are the unitary, square-integrable representation (modulo tensoring by one-dimensional  reps) of $G(F)$, which are not supercuspidal, in one-to-one correspondence with a certain subclass of representations of $G(f)$ (modulo tensoring by one-dimensional  reps)?

I am mostly interested in the case of $G=GL(n)$. The question has an easy answer if $n=2$.
For the case GL(2): The sq.int, non-supercuspidal reps are isomorphic to the Steinberg tensored by a one-dimensional representation $G(F)$. There is precisely one irreducible rep of $G(f)= G(o/p)$ contained in the Steinberg rep of $G(F)$, i.e., it is the Steinberg of $G(f)$.

More concretely, do they also in general admit a $\Gamma(p) = \{ \gamma \in G(o) : \gamma = 1 \bmod p \}$-invariant vector modulo tensoring by one-d. reps? Is their restriction to the Iwahori(= pullback of $B(f)$ to $G(o)$ for a fixed Borel subgroup) or its "Levi component" a one-dimensional representation?

REPLY [3 votes]: As a follow-up to Paul Broussous's answer, not every representation is depth-0, up to a central twist, but those that are are described by representations of finite groups associated to, but not necessarily of the same type as, $G$.  Specifically, they are quotients of parahoric subgroups, and the representations are cuspidal.  This is described in Proposition 6.8 of the second Moy–Prasad paper.<|endoftext|>
TITLE: Borel's Paris Lectures
QUESTION [9 upvotes]: I am trying to read Harish-Chandra's book on automorphic forms on Semisimple Lie groups, and he keeps referring to Borel's Paris lecture notes. Does anyone have an online version of these notes or know how I could get it? Ensembles fondamentaux pour les groupes arithmétiques et formes automorphes, Lectures at Institut H. Poincaré, Paris 1966 (Unpublished, notes by H. Jacquet, J.J. Sansuc and B. Schiffmann)
Thanks!

REPLY [13 votes]: The Paris lectures, along with others he gave later, were spliced together into a publication:  Introduction aux groupes arithmetiques (softcover, Hermann, Paris, 1969).   As his nominal assistant at IAS in 1968-69, I tried to help with the splicing process but didn't manage to clean up all the inconsistent notation and typos (especially on pages 90-94) which I later jotted down for myself.   There was some time pressure, not to mention his many other projects at the time, so the finished product is useful but imperfect.   By now it's also probably hard to locate outside some libraries.<|endoftext|>
TITLE: A question on the Mahler conjecture
QUESTION [7 upvotes]: In its asymmetric version, the Mahler conjecture states that if $K \subset \mathbb{R^n}$ is a convex body containing the origin as an interior point and 
$$
K^* := \{y \in \mathbb{R}^n : \langle y, x \rangle \leq 1 \mbox{ for all } x \in K \}
$$
is its polar body, then the product of volumes $|K| |K^*|$ is bounded below by 
$(n+1)^{n+1}/(n!)^2$. Equality is conjectured to hold only for simplexes. 
Is it known whether there is a unique minimum of $K \mapsto |K| |K^*|$ modulo linear equivalence? 
Is there some dimension dependent bound on the number of minima (modulo linear equivalence)?
For a given dimension is it least known that the number of minima (modulo linear equivalence) is finite?
This last would follow if it were known that the minima or local minima are isolated points in the space of linear equivalence classes of convex bodies containing the origin as an interior point. I guess that's only known for the explicit case of the simplex.
I know that determining whether the minima are polytopes is still open (and that in the symmetric case there are different linearly inequivalent classes of conjectured minima), but I don't remember having seen the problem of uniqueness discussed in the asymmetric case.

REPLY [5 votes]: No, it is not known that the minimum is unique, but it is believed to be.  In fact, this paper by Kim and Reisner proves that the simplex is (modulo linear equivalence) a strict local minimum; thus the whole conjecture would follow from uniqueness of local minima.<|endoftext|>
TITLE: Links which HOMFLY homology distinguish but the HOMFLY polynomial does not.
QUESTION [10 upvotes]: Does anyone know of a pair of different links which the HOMFLY polynomial does not distinguish, but HOMFLY homology does? Or does there exist such a pair of links? 

I'm assuming there does exist such a pair, but have never seen it. I've been looking for this for a few days and have had no luck finding or computing such an example. Any help would be greatly appreciated!

REPLY [11 votes]: Although $5_1$ and $10_{132}$ cannot be distinguished by Jones, Alexander and (uncolored) HOMFLY-PT polynomials, their HOMFLY homologies do tell them. (See the review by Gukov-Saberi.)
In addition, some mutant pairs can be distinguished by Khovanov homology. (See the paper by Wehrli.)<|endoftext|>
TITLE: Euler Sequence on Homogeneous Spaces
QUESTION [7 upvotes]: For the projective space, we have the very well known Euler sequence
$$0 \to \Omega_{P^n_k} \to \mathcal{O}_{P^n_k}(-1)^{n+1} \to \mathcal{O}_{P^n_k} \to 0.$$
Are there any generalizations to other rational homogeneous spaces?

REPLY [10 votes]: Here is how it works for the (complex) Grassmannian. I will leave you the pleasure to extend this point of view to others homogeneous spaces (for instance complete and incomplete flag manifolds).
Firs of all, let me give a slightly different point of view for the Euler exact sequence on the projective space. Let $V$ be a complex vector space of dimension $n+1$ and $P(V)$ its corresponding projective space of lines. Call $\underline V=P(V)\times V$ the trivial rank $n+1$ vector bundle with fiber $V$ on $P(V)$. Then you have the short exact sequence
\begin{equation}
0\to\mathcal O(-1)\to\underline V\to H\to 0,
\end{equation}
where $\mathcal O(-1)$ is the tautological line bundle on $P(V)$ and $H$ is the quotient rank $n$ vector bundle $\underline V/\mathcal O(-1)$.
Now, looking at the differential $d\pi_x\colon V\to T_{P(V),[x]}$ of the projection $\pi\colon V\setminus\{0\}\to P(V)$, it is straightforward to see that it defines a canonical isomorphism
$$
T_{P(V)}\simeq H\otimes\mathcal O(1).
$$
Tensoring the short exact sequence above by $\mathcal O(1)$ and using this canonical isomorphism you get the Euler exact sequence
$$
0\to \mathcal O\to\underline V\otimes\mathcal O(1)\to T_{P(V)}\to 0
$$
(to get your version just notice that $\underline V\otimes\mathcal O(1)\simeq\mathcal O(1)^{\oplus(n+1)}$, then take the dual exact sequence).
Now, let $V$ be of dimension $d$ and $G_r(V)$ be the Grassmannian of $r$-codimensional vector subspaces of $V$. Consider the tautological subbundle $S\subset \underline V=G_r(V)\times V$ and the associated short exact sequence
$$
0\to S\to\underline V\to Q\to 0,
$$
where $Q=\underline V/S$. Similarly as before, it is straightforwardly seen that there is a canonical isomorphism
$$
T_{G_r(V)}\simeq S^*\otimes Q\simeq\operatorname{Hom}(S,Q).
$$ 
Tensoring by $S^*$, you get your Euler sequence for the Grassmannian, that is
$$
0\to S^*\otimes S\to\underline V\otimes S^*\to T_{G_r(V)}\to 0.
$$ 
Of course, it reduces to the standard one whenever $r=d-1$, and playing with the short exact sequence you can give it your favorite shape (taking duals, tensoring with other vector bundles...).
Observe finally that 
$$
\det S^*\simeq\det Q
$$
is the (very) ample line bundle on $G_r(V)$ which gives the Plücker embedding into projective space and whose first Chern class generates $H^2(G_r(V),\mathbb Z)$.<|endoftext|>
TITLE: Number of Distinct Sums of Integers
QUESTION [8 upvotes]: Hello,
I'm new to MO. Please, bear with my inexperience. I try to be as precise as it is possible to me.
Given a multiset of integers $S$, I'm interested in the number of sub-multisets which sum up to distinct sums. I call this number $S_\sum$ and is defined as
$S_\sum = \left|\left\{ \sum_{i \in Q} i \quad \middle| \quad Q \subseteq S \right\}\right|$

EXAMPLE
$S = \{1,3,7,9\}$. Number of subsets is $2^4 = 16$, but the number of sub-multisets with distinct sums are $S_\sum = 15$, as the subsets $\{3,7\}$ and $\{1,9\}$ do add to the same number 10.

The upper bound is clearly the number of sub-multisets ($2^{|S|}$). The lower bound can be as small as $|S|+1$ (in the case of $S$ consisting of only equal numbers).
As I think, that it is difficult to come up with an exact solution, I'm looking for an estimation of the number in question if the set is chosen randomly.
The precise question:
For a randomly chosen multiset of integers $S$, how does $S_\sum$ relate in average to $|S|$ and $\sum_{i \in S} i$.

REPLY [4 votes]: Here is a possible formulation of your question: A multiset is a list $S$ of positive integers $x_1 \le x_2 \le \cdots \le x_k.$ The size is $k=|S|$ and the total is $t=\Sigma_{x \in S}x.$ Call $S$ a $(k,t)-$multiset (which could also be called an unordered partition of $t$ into $k$ positive parts.)  There are $2^k$ sums of some all or none of the $x_i$ and, as you say, the number of distinct sums could be as high as $2^k$ ( provided that $t \ge 2^{k}-1$) and as small as $k+1$ (provided that $t$ is a multiple of $k$). 

Since the number of $(k,t)-$multisets is finite, what can be said, for fixed $k,t$, about the average number of distinct sums for a "randomly" chosen $(k,t)-$multiset.

There are still choices of what is meant by random. We could write each possible multiset on a card and pick one at random: then $13,13,13,13$ would be as likely as $3,10,13,26$ for $k,t=4,52.$ Or, you could roll a fair $4$-sided die $52$ times, see how often each face comes up and just use the multiset of those 4 counts ( you would either need to start over in the unlikely event that some count is zero or else roll $48$ times and start each count at $1$.). In this second model $5,8,13,26$ is twenty four times more likely than $13,13,13,13$
Later: You have now clarified that you are particularly interested in $(k,2^k+r)$ designs where $r$ is "small". I don't know if you are thinking about $r=3$ or $r=3k$ or $r=k^3.$ Even for $r=3$ I don't think that explicitly generating all multisets would be feasible for $k=10$ (or something like that) I can see a possible opening for impressive probability and statistics arguments by those expert in the field. It am thinking about fixing $k$ and increasing $t$ so perhaps that is not so much your interest, however maybe these thoughts would be of interest: 

For $A \subseteq \{1,2,\cdots,k\}$ let $\Sigma_A=\Sigma_{i \in A}x_i.$ Perhaps you want to consider what the set $\mathcal{E}_S=\{(A,B) \mid  \Sigma_A=\Sigma_B\}$ could look like for a particular multiset $S$ of values. Certainly if $(A,B)$ is in $\mathcal{E}_S$ with $A,B$ disjoint so are  $(A \cup C,B \cup C)$  for any of the $2^{k-|A|-|B|}$ sets disjoint from both. That is a special case of $(A \cup A',B \cup B') \in \mathcal{E}_S$ when the unions are disjoint and $(A,B)$,$(A',B')$ both are.
For $t \lt \binom{k+1}{2}$ there are forced to be solutions of $x_i=x_j$ (with $i \ne j$ of course) each of which leads to $2^{k-2}$ other equal sums as just commented. For $t \lt 2^k-1$ we have $\mathcal{E}_S \ne \emptyset.$
How big must $t$ be relative to $k$ in order to have the $x_i$ distinct with probability greater than $ 1-1/k$ ( or some other $1 - \varepsilon$? ) Is it a sharp transition at some critical point?
The comment above about the isolation lemma seems very interesting. If I read it correctly, then if we take $t=mk$ then the least of the $x_i$ is unique of its value with probability exceeding $1-1/m.$ If we take that singleton set out of consideration then the next smallest is again unique of its weight with probability exceeding $1-1/m$ etc. So the expected number of distinct $x_i$ is at least $(1-1/m)k.$ 
How much larger (I'd guess quite a bit larger) does $t$ have to be in order to make it unlikely that there would be any cases of $x_i+x_j=x_k+x_{\ell}$? The same question if we ask that instead (or in addition) we are unlikely to see $x_i+x_j=x_k?.$ Again the isolation lemma says that for any given partition of the index set $\{1,\cdots,k\}$ or merely family $F$ of subsets of the index set, the $A \in F$ which minimizes $\Sigma_A$ is the only one of its weight with probability exceeding $1-1/m.$ (So it makes sense to restrict $F$ to at least have no member contain any other.)

Well maybe that is far enough with questions I can't answer.<|endoftext|>
TITLE: Dimension of incomplete matrix over finite fields.
QUESTION [7 upvotes]: Hi,
Suppose one has an incompletely specified $2^n \times 2^n$ matrix over some fixed finite field $\mathbb{F}_{p^k}$. In fact, one knows that the diagonal entries are zero and all other entries are non-zero. Graphically:
$$\left( \begin{array}{ccccc} 0 & ? & \cdots & ? & ? \\\ ? & 0 & \cdots & ? & ? \\\ ? & ? & \ddots & ? & ? \\\ ? & ? & \cdots & 0 & ? \\\ ?&?&\cdots&?&0  \end{array} \right)$$
where distinct $?$'s needn't be equal, but they must both be non-zero.

Is it true that, for every possible choice of non-zero entries in $\mathbb{F}_{p^k}$, the space spanned by the columns has dimension super-polynomial in $n$?

This holds over $\mathbb{F}_2$ where necessarily every $? =1$, the dimension being at least $2^n - 1$. I would also be very interested in any known methods for dealing with this type of problem.
Any help much appreciated.

Clarification: The finite field used should be fixed. So to answer the question negatively it suffices to find a sequence of filled-in $2^n \times 2^n$ matrices $(M_n)_{n \in \mathbb{N}}$ over some fixed finite field, whose ranks $rk(M_n)$ are bounded polynomially in $n$.

REPLY [10 votes]: It is true. Let $N=2^n$ be the dimension, and $q=p^k$ be the size of the field. Let $v_i$ be the $i$'th column vector. Define multiplication of vectors coordinate-wise, i.e., $uu'$ is vector whose $i$'th coordinate is $u_i u'_i$. Similarly, for a vector $u$ define $u^r$ be the result of raising all the elements of $u$ to $r$'th power. Note that the vectors $v_1^{q-1},\dotsc,v_N^{q-1}$ are linearly independent.
Suppose, now that the rank of the matrix is $r$, i.e, some $r$ column vectors span the column space. Without loss, every $v_i$ is a linear combination of $v_1,\dotsc,v_r$. Let $S$ be the set of all the products of $q-1$ vectors from $\{v_1,\dotsc,v_r\}$ (repetition are allowed).  If $$v_i=\sum_{j=1}^r \alpha_j v_j,$$ then $$v_i^{q-1}=\left(\sum_{j=1}^r \alpha_j v_j\right)^{q-1}=\sum_{j_1,\dotsc,j_{q-1}=1}^r (\alpha_{j_1}\dotsb\alpha_{j_{q-1}}) v_{j_1}\dotsb v_{j_{q-1}}$$ is a linear combination of vectors from $S$. Since $v_i^{q-1}$'s are linearly independent, we conclude that $|S|\geq N$. Since $|S|\leq r^{q-1}$, it follows that the rank is
$$r\geq N^{1/(q-1)}.$$
This argument is a variation on the proofs of Frankl–Wilson theorem, and of Ray-Chaudhuri–Wilson theorem.
Added on 25 Jan 2014: I just discovered that the same proof appeared as Theorem 4 in paper "Low rank co-diagonal matrices and Ramsey graphs" by Vince Grolmusz.<|endoftext|>
TITLE: Weakest choice principle required for Robertson-Seymour Graph Minor Theorem?
QUESTION [15 upvotes]: The main Robertson-Seymour Theorem states that finite graphs form a well-quasi-ordering under the graph minor relation.  In other words, in every infinite set of finite graphs, there exist two graphs $G$ and $H$ such that $G$ is a minor of $H$.
My question is:

What is a weakest consequence $C$ of the Axiom of Choice that is sufficient in ZF to imply the Graph Minor Theorem?

A little more precisely, what is a sentence $C$ of an appropriate logic (such as a form of second-order logic) such that:

$C$ is known to be a theorem of ZFC,
the Graph Minor Theorem is a theorem of ZF+$C$, and
there is no known sentence $D$ that is a theorem of ZF+$C$ with the Graph Minor Theorem being a theorem of ZF+$D$?

Ideally $C$ should be one of the already studied consequences of AC, and the quantification over "known" $D$ can also be restricted to such consequences.

Discussion
Another way to make my question precise would be to say that it is about the "first form" of the Graph Minor Theorem, rather than the "second form", in analogy with Ali Enayat writing about König's Lemma.  Forster and Truss showed that König's Lemma is equivalent to a rather weak choice principle:

Every countably infinite family of finite non-empty sets has a choice function.

and I am asking which choice principle is a weakest one that is known to be sufficient to prove the Graph Minor Theorem: some stronger form of choice seems to be needed than for König's Lemma.

T. E. Forster and J. K. Truss, Ramsey's theorem and König's Lemma, Archive for Mathematical Logic 46, 2007, 37–42.  doi:10.1007/s00153-006-0025-z

I am waiting for the following possibly relevant paper, although hopefully something has been written about this in the intervening 26 years:

H. Friedman, N. Robertson, and P. D. Seymour, The metamathematics of the graph minor theorem, in (S. G. Simpson, ed.) Logic and Combinatorics, Contemporary Mathematics 65, AMS, 1987, 229–261.

I am also aware of Timothy Chow's answer to a question of Ryan Williams which suggests rephrasing the Graph Minor Theorem as a statement of arithmetic (which therefore can be proved in ZF, with no notion of choice required).  However, I find it unsatisfactory to finesse away choice by only allowing graphs over vertices from $\mathbb{N}$, and encoding the graph minor relation as a particular relation over the natural numbers, about which we then ask the whether a particular arithmetical sentence is true.  To my mind, the key difficulty seems to be precisely that if we don't have canonization, then these questions require additional machinery.  (In fact, I would naively be tempted to use $C$ and $D$ as measures of just what we gain from canonization, in some sense.)
Harvey Friedman recently commented:

wqo statements, which are Pi-1-1. E.g., Kruskal's Theorem,
  graph minor theorem, Hilbert's basis theorem (appropriately
  formalized), etcetera. These are equivalent over RCA_0 to the well
  orderedness of familiar ordinal notations (in the case of GMT, the
  exact ordinal has not been determined, but has been determined for the
  important restriction to graphs of finite tree width).

Hence, I expect that we currently do not know $C$ such that ZF+$C$ can prove GMT, while ZF+GMT can also prove $C$.  But surely something is known about how "strong" a notion is needed?  For instance, which choice principle is Friedman referring to above in the bounded tree-width case?  This would suggest a candidate for $D$.
(See also a related question by Joe Shipman.)

REPLY [11 votes]: As far as I can see, the Robertson–Seymour theorem is provable in plain ZF.
First, the restricted version of the theorem for graphs whose vertices are natural numbers can be written as a $\Pi^1_1$ second-order arithmetical sentence, hence its provability in ZF follows from its provability in ZFC by the Schoenfield absoluteness theorem. (Note that the set of all such finite graphs is countable, hence well-orderable, thus all the common definitions of wqo coincide in this case without the need for any choice.)
Second, ZF proves that the general case is equivalent to the restricted case. I will take the definition of wqo requiring that every subset has a finite basis: this is the most natural formulation in set theory without choice, and it implies in ZF both the property given in the original question, and the property that for every sequence $\{x_i:i\in\mathbb N\}$, there exist $i< j$ with $x_i\le x_j$.
So, assume that $X$ is a set of finite graphs. Let $X'$ be the set of all graphs with integer vertices isomorphic to some graph in $X$. By the restricted version of the theorem, we can find a finite basis $B'\subseteq X'$, say $B'=\{G_i':i< n\}$. In ZF, we have choice for finite families, hence we can find $B:=\{G_i:i< n\}\subseteq X$ such that $G_i\simeq G_i'$ for each $i< n$. Then $B$ is a finite basis for $X$: if $G\in X$, there exists $G'\in X'$ isomorphic to $G$ (as the vertex set of $G$ is finite, it is in bijection with some $\{0,\dots,m-1\}\subseteq\mathbb N$, and we can lift the graph along this bijection). Since $B'$ is a basis of $X'$, some $G_i'\in B'$ is a minor of $G'$, hence $G_i\in B$ is a minor of $G$.<|endoftext|>
TITLE: Verifying the correctness of a Sudoku solution
QUESTION [43 upvotes]: A Sudoku is solved correctly, if all columns, all rows and all 9 subsquares are filled with the numbers 1 to 9 without repetition. Hence, in order to verify if a (correct) solution is correct, one has to check by definition 27 arrays of length 9. 
Q1: Are there verification strategies that reduce this number of checks ?
Q2: What is the minimal number of checks that verify the correctness of a (correct) solution ? 
   
   
   (Image sources from Wayback Machine: 
first
second)
The following simple observation yields an improved verification algorithm: At first enumerate rows, columns and subsquares as indicated in pic 2. Suppose the columns $c_1,c_2,c_3$ and the subsquares $s_1, s_4$ are correct (i.e. contain exactly the numbers 1 to 9). Then it's easy to see that $s_7$ is correct as well. This shows: 
(A1) If all columns, all rows and 4 subsquares are correct, then the solution is correct. 
Now suppose all columns and all rows up to $r_9$ and the subsquares $s_1,s_2,s_4,s_5$ are correct. By the consideration above, $s_7,s_8,s_9$ and $s_3,s_6$ are correct. Moreover, $r_9$ has to be correct, too. For, suppose a number, say 1, occurs twice in $r_9$. Since the subsquares are correct, the two 1's have  be in different subsquares, say $s_7,s_8$. Hence the 1's from rows $r_7, r_8$ both have to lie in $s_9$, i.e. $s_9$ isn't correct. This is the desired contradiction. 
Hence (A1) can be further improved to   
(A2) If all columns and all rows up to one and 4 subsquares are correct, then the solution is correct. 
This gives as upper bound for Q2 the need of checking 21 arrays of length 9. 
Q3: Can the handy algorithm (A2) be further improved ?

REPLY [19 votes]: $\DeclareMathOperator\span{span}$Here is an argument which works for general $n\times n$ Sudokus, $n\ge 2$, using some ideas from the other answers (namely, casting the problem in terms of linear algebra as in François Brunault’s answer, and the notion of alternating paths below is related to the even sets as in Tony Huynh’s answer, attributed to Zack Wolske).
I will denote the cells as $s_{ijkl}$ with $0\le i,j,k,l< n$, where $i$ identifies the band, $j$ the stack, $k$ the row within band $i$, and $l$ the column within stack $j$. Rows, columns, and blocks are denoted $r_{ik},c_{jl},b_{ij}$ accordingly. Let $X=\{r_{ik},c_{jl},b_{ij}:i,j,k,l< n\}$ be the set of all $3n^2$ checks. For $S\subseteq X$ and $x\in X$, I will again denote by $S\models x$ the consequence relation “every Sudoku grid satisfying all checks from $S$ also satisfies $x$”.
Let $V$ be the $\mathbb Q$-linear space with basis $X$, and $V_0$ be the span of the vectors $\sum_kr_{ik}-\sum_jb_{ij}$ for $i< n$, and $\sum_lc_{jl}-\sum_ib_{ij}$ for $j< n$.
Lemma 1: If $x\in\span(S\cup V_0)$, then $S\models x$.
Proof: A grid $G$ induces a linear mapping $\phi_G$ from $V$ into an $n^2$-dimensional such that for any $x'\in X$, the $i$th coordinate of $\phi_G(x')$ gives the number of occurrences of the number $i$ in $x'$. We have $\phi_G(V_0)=0$, and $G$ satisfies $x'$ iff $\phi_G(x')$ is the constant vector $\vec 1$. If $x=\sum_i\alpha_ix_i+y$, where $x_i\in S$ and $y\in V_0$, then $\phi_G(x)=\vec\alpha$ for $\alpha:=\sum_i\alpha_i$. The same holds for every grid $G'$ satisfying $S$; in particular, it holds for any valid grid, which has $\phi_{G'}(x)=\vec1$, hence $\alpha=1$. QED
We intend to prove that the converse holds as well, so assume that $x\notin\span(S\cup V_0)$. We may assume WLOG $x=r_{00}$ or $x=b_{00}$, and we may also assume that $r_{i0}\notin S$ whenever $r_{ik}\notin S$ for some $k$, and $c_{j0}\notin S$ whenever $c_{jl}\notin S$ for some $l$. By assumption, there exists a linear function $\psi\colon V\to\mathbb Q$ such that $\psi(S\cup V_0)=0$, and $\psi(x)\ne0$. The space of all linear functions on $V$ vanishing on $V_0$ has dimension $3n^2-2n$, and one checks easily that the following functions form its basis:

$\omega_{ik}$ for $0\le i< n$, $0< k< n$: $\omega_{ik}(r_{ik})=1$, $\omega_{ik}(r_{i0})=-1$.
$\eta_{jl}$ for $0\le j< n$, $0< l< n$: $\eta_{jl}(c_{jl})=1$, $\eta_{jl}(c_{j0})=-1$.
$\xi_{ij}$ for $i,j< n$: $\xi_{ij}(r_{i0})=\xi_{ij}(c_{j0})=\xi_{ij}(b_{ij})=1$.

(The functions are zero on basis elements not shown above.) We can thus write
$$\psi=\sum_{ik}u_{ik}\omega_{ik}+\sum_{jl}v_{jl}\eta_{jl}+\sum_{ij}z_{ij}\xi_{ij}.$$
If $r_{ik}\in S$, $k\ne0$, then $0=\psi(r_{ik})=u_{ik}$, and similarly $c_{jl}\in S$ for $l\ne0$ implies $v_{jl}=0$. Thus, the functions $\omega_{ik}$ and $\eta_{jl}$ that appear in $\psi$ with a nonzero coefficient individually vanish on $S$. The only case when they can be nonzero on $x$ is $\omega_{0k}$ if $x=r_{00}$ and $r_{00},r_{0k}\notin S$, but then taking any valid grid and swapping cells $s_{0000}$ and $s_{00k0}$ shows that $S\nvDash x$ and we are done. Thus we may assume that the first two sums in $\psi$ vanish on $S\cup\{x\}$, and therefore the third one vanishes on $S$ but not on $x$, i.e., WLOG
$$\psi=\sum_{ij}z_{ij}\xi_{ij}.$$
That $\psi$ vanishes on $S$ is then equivalent to the following conditions on the matrix $Z=(z_{ij})_{i,j< n}$:

$z_{ij}=0$ if $b_{ij}\in S$,
$\sum_jz_{ij}=0$ if $r_{i0}\in S$,
$\sum_iz_{ij}=0$ if $c_{j0}\in S$.

Let us say that an alternating path is a sequence $e=e_p,e_{p+1},\dots,e_q$ of pairs $e_m=(i_m,j_m)$, $0\le i_m,j_m< n$, such that

$i_m=i_{m+1}$ if $m$ is even, and $j_m=j_{m+1}$ if $m$ is odd,
the indices $i_p,i_{p+2},\dots$ are pairwise distinct, except that we may have $e_p=e_q$ if $q-p\ge4$ is even,
likewise for the $j$s.

If $m$ is even, the incoming line of $e_m$ is the column $c_{j_m0}$, and its outgoing line is the row $r_{i_m0}$. If $m$ is odd, we define it in the opposite way. An alternating path for $S$ is an alternating path $e$ such that $b_{i_mj_m}\notin S$ for every $m$, and either $e_p=e_q$ and $q-p\ge4$ is even ($e$ is an alternating cycle), or the incoming line of $e_p$ and the outgoing line of $e_q$ do not belong to $S$.
Every alternating path $e$ induces a matrix $Z_e$ which has $(-1)^m$ at position $e_m$ for $m=p,\dots,q$, and $0$ elsewhere. It is easy to see that if $e$ is an alternating path for $S$, then $Z_e$ satisfies conditions 1, 2, 3.
Lemma 2: The space of matrices $Z$ satisfying 1, 2, 3 is spanned by matrices induced by alternating paths for $S$.
Proof:
We may assume that $Z$ has integer entries, and we will proceed by induction on $\|Z\|:=\sum_{ij}|z_{ij}|$. If $Z\ne 0$, pick $e_0=(i_0,j_0)$ such that $z_{i_0j_0}>0$. If the outgoing line of $e_0$ is outside $S$, we put $q=0$, otherwise condition 2 guarantees that $z_{i_0,j_1}< 0$ for some $j_1$, and we put $i_1=i_0$, $e_1=(i_1,j_1)$. If the outgoing line of $e_1$ is outside $S$, we put $q=1$, otherwise we find $i_2$ such that $z_{i_2j_1}>0$ by condition 3, and put $j_2=j_1$. Continuing in this fashion, one of the following things will happen sooner or later:

The outgoing line of the last point $e_m$ constructed contains another point $e_{m'}$ (and therefore two such points, unless $m'=0$). In this case, we let $p$ be the maximal such $m'$, we put $q=m+1$, $e_q=e_p$ to make a cycle, and we drop the part of the path up to $e_{p-1}$.
The outgoing line of $e_m$ is outside $S$. We put $q=m$.

In the second case, we repeat the same construction going backwards from $e_0$. Again, either we find a cycle, or the construction stops with an $e_p$ whose incoming line is outside $S$. Either way, we obtain an alternating path for $S$ (condition 1 guarantees that $b_{i_mj_m}\notin S$ for every $m$). Moreover, the nonzero entries of $Z_e$ have the same sign as the corresponding entries of $Z$, thus $\|Z-Z_e\|<\|Z\|$. By the induction hypothesis, $Z-Z_e$, and therefore $Z$, is a linear combination of some $Z_e$s. QED
Now, Lemma 2 implies that we may assume that our $\psi$ comes from a matrix $Z=Z_e$ induced by an alternating path $e=e_p,\dots,e_q$. Assume that $G$ is a valid Sudoku grid that has $1$ in cells $s_{i_mj_m00}$ for $m$ even, and $2$ for $m$ odd. Let $G'$ be the grid obtained from $G$ by exchanging $1$ and $2$ in these positions. Then $G'$ violates the following checks:

$b_{i_mj_m}$ for each $m$.
If $e$ is not a cycle, the incoming line of $e_p$, and the outgoing line of $e_q$.

Since $e$ is an alternating path for $S$, none of these is in $S$. On the other hand, $\psi(x)\ne0$ implies that $x$ is among the violated checks, hence $S\nvDash x$.
It remains to show that such a valid grid $G$ exists. We can now forget about $S$, and then it is easy to see that every alternating path can be completed to a cycle, hence we may assume $e$ is a cycle. By applying Sudoku permutations and relabelling the sequence, we may assume $p=0$, $i_m=\lfloor m/2\rfloor$, $j_m=\lceil m/2\rceil$ except that $i_q=j_q=j_{q-1}=0$. We are thus looking for a solution of the following grid:
$$\begin{array}{|ccc|ccc|ccc|ccc|ccc|}
\hline
1&&&2&&&&&&&&&&&&\\
\strut&&&&&&&&&&&&&&&\\
\strut&&&&&&&&&&&&&&&\\
\hline
&&&1&&&2&&&&&&&&&\\
&&&&&&&&&&&&&&\cdots&\\
&&&&&&&&\ddots&&&&&&&\\
\hline
2&&&&&&&&&1&&&&&&\\
\strut&&&&&&&&&&&&&&&\\
\strut&&&&&&&&&&&&&&&\\
\hline
\strut&&&&&&&&&&&&&&&\\
\strut&&&&\vdots&&&&&&&&&&&\\
\strut&&&&&&&&&&&&&&&\\
\hline
\end{array}$$
where the upper part is a $q'\times q'$ subgrid, $q'=q/2$.
If $q'=n$, we can define the solution easily by putting $s_{ijkl}=(k+l,j-i+l)$, where we relabel the numbers $1,\dots,n^2$ by elements of $(\mathbb Z/n\mathbb Z)\times(\mathbb Z/n\mathbb Z)$, identifying $1$ with $(0,0)$ and $2$ with $(0,1)$. In the general case, we define $s_{ijkl}=(k+l+a_{ij}-b_{ij},l+a_{ij})$. It is easy to check that this is a valid Sudoku if the columns of the matrix $A=(a_{ij})$ and the rows of $B=(b_{ij})$ are permutations of $\mathbb Z/n\mathbb Z$. We obtain the wanted pattern if we let $a_{ij}=b_{ij}=j-i\bmod{q'}$ for $i,j< q'$, and extend this in an arbitrary way so that the columns of $A$ and the rows of $B$ are permutations.
This completes the proof that $x\notin\span(S\cup V_0)$ implies $S\nvDash x$. This shows that $\models$ is a linear matroid, and we get a description of maximal incomplete sets of checks by means of alternating paths.
We can also describe the minimal dependent sets. Put
$$D_{R,C}=\{r_{ik}:i\in R,k< n\}\cup\{c_{jl}:j\in C,l< n\}\cup\{b_{ij}:(i\in R\land j\notin C)\lor(i\notin R\land j\in C)\}$$
for $R,C\subseteq\{0,\dots,n-1\}$. If $R$ or $C$ is nonempty, so is $D_{R,C}$, and
$$\sum_{i\in R}\Bigl(\sum_kr_{ik}-\sum_jb_{ij}\Bigr)-\sum_{j\in C}\Bigl(\sum_lc_{jl}-\sum_ib_{ij}\Bigr)\in V_0$$
shows that $D_{R,C}$ is dependent. On the other hand, if $D$ is a dependent set, there is a linear combination
$$\sum_i\alpha_i\Bigl(\sum_kr_{ik}-\sum_jb_{ij}\Bigr)-\sum_j\beta_j\Bigl(\sum_lc_{jl}-\sum_ib_{ij}\Bigr)\ne0$$
where all basic vectors with nonzero coefficients come from $D$. If (WLOG) $\alpha:=\alpha_{i_0}\ne0$, put $R=\{i:\alpha_i=\alpha\}$ and $C=\{j:\beta_j=\alpha\}$. Then $R\ne\varnothing$, and $D_{R,C}\subseteq D$.
On the one hand, this implies that every minimal dependent set is of the form $D_{R,C}$. On the other hand, $D_{R,C}$ is minimal unless it properly contains some $D_{R',C'}$, and this can happen only if $R'\subsetneq R$ and $C=C'=\varnothing$ or vice versa. Thus $D_{R,C}$ is minimal iff $|R|+|C|=1$ or both $R,C$ are nonempty.
This also provides an axiomatization of $\models$ by rules of the form $D\smallsetminus\{x\}\models x$, where $x\in D=D_{R,C}$ is minimal. It is easy to see that if $R=\{i\}$ and $C\ne\varnothing$, the rules for $D_{R,C}$ can be derived from the rules for $D_{R,\varnothing}$ and $D_{\varnothing,\{j\}}$ for $j\in C$, hence we can omit these. (Note that the remaining sets $D_{R,C}$ are closed, hence the corresponding rules have to be included in every axiomatization of $\models$.)
To sum it up:
Theorem: Let $n\ge2$.

$S\models x$ if and only if $x\in\span(S\cup V_0)$. In particular, $\models$ is a linear matroid.
All minimal complete sets of checks have cardinality $3n^2-2n$. (One such set consists of all checks except for one row from each band, and one column from each stack.)
The closed sets of $\models$ are intersections of maximal closed sets, which are complements of Sudoku permutations of the sets

$\{b_{00},b_{01},b_{11},b_{12},\dots,b_{mm},b_{m0}\}$ for $0< m< n$
$\{c_{00},b_{00},b_{01},b_{11},b_{12},\dots,b_{mm},r_{m0}\}$ for $0\le m< n$
$\{c_{00},b_{00},b_{01},b_{11},b_{12},\dots,b_{m-1,m},c_{m1}\}$ for $0\le m< n$

The minimal dependent sets of $\models$ are the sets $D_{R,C}$, where $R,C\subseteq\{0,\dots,n-1\}$ are nonempty, or $|R|+|C|=1$.
$\models$ is the smallest consequence relation such that $D_{R,C}\smallsetminus\{x\}\models x$ whenever $x\in D_{R,C}$ and either $|R|,|C|\ge2$, or $|R|+|C|=1$.<|endoftext|>
TITLE: deRham cohomology of $S^n$ without Mayer-Vietoris
QUESTION [16 upvotes]: I've got a few days left at the end of a differential geometry class, and would like to compute the deRham cohomology of $S^n$.  We've just proved the Poincare lemma, so I know the cohomology of $R^n$, and homotopy invariance of cohomology is an easy consequence.
The way to compute $H^*(S^n)$ is with Mayer-Vietoris, for example as in Bott and Tu.  But I don't have time to fully develop the cohomological algebra for that, just to then apply it in only one case.  I'd rather get the computation for $S^n$ directly and use it for a couple of applications.
So, it seems to me that one could trace through the Mayer-Vietoris argument in this specific case. At least I'd like to show that a closed $k$-form on $S^n$ is exact when $0 < k < n$.
Here's how I think the argument could go:
Let $\omega$ be a $k$-form on $S^n$.
Let $U$ and $V$ be $S^n$ minus its north and south pole, respectively.  Then homotopy invariance and the Poincare lemma give $k-1$ forms $\alpha$ on $U$ and $\beta$ on $V$ with $d\alpha = \omega$ and $d\beta = \omega$.
Now use a partition of unity $f$ and set $\gamma = f_U \alpha + f_V \beta$, a $k-1$ form on $S^n$.  So $\omega - d\gamma$ is now supported on $U \cap V$.  Since $U\cap V$ is homotopic to $S^{n-1}$, induction gives that $\omega - d\gamma$ is exact, so $\omega = d\tau + d\gamma$.
The problem is that $\omega - d\gamma$ is exact when restricted to $U \cap V$, and I don't see how why $\tau$ would extend to a form on $S^n$.
Am I missing something?  Is there a better approach entirely?

REPLY [6 votes]: Suppose $1 < k < n$.
Let $\omega$ be a $k$-form on $S^n$.
Let $U$ and $V$ be $S^n$ minus its north and south pole, respectively.  Then homotopy invariance and the Poincare lemma give $k-1$ forms $\alpha$ on $U$ and $\beta$ on $V$ with $d\alpha = \omega$ and $d\beta = \omega$.
On $U \cap V$, $d(\alpha -\beta) = 0$.  Since $H^{k-1}(S^{n-1}) = 0$, there is a $k-2$ form $\tau$ on $U \cap V$ with
$d\tau = (\alpha - \beta) |_{U\cap V}$.
Now let $f_U, f_V$ be a partition of unity subordinate to $U,V$.  Then on $U\cap V$, $d(f_U\tau) + d(f_V\tau) = \alpha - \beta$.
Since $f_U\tau$ extends (by 0) to a form on $V$, $\beta + d(f_U \tau)$ is defined on $V$.
Since $f_V\tau$ extends to a form on $U$, $\alpha - d(f_V \tau)$ is defined on $U$.
Put $\sigma = \beta + d(f_U\tau) = \alpha - d(f_V\tau)$.  $\sigma$ is defined on all of $S^n$, and $d\sigma = \omega$, so that $\omega$ is exact.
In the $k=1$ case, $\alpha$ and $\beta$ differ by a constant, so they extend to all of $S^n$ and $\omega = d\alpha$.
When $k=n$, show that any $n$ form with integral 0 is exact, so $H^n(S^n)$ is generated 
by the volume form.  The only change from the $k < n$ case is that one needs to check that the 
integral of $\alpha-\beta$ is zero over $S^{n-1}$. Apply Stokes' theorem  to the lower hemisphere (for $\alpha$) and the upper hemisphere (for $\beta$), so that
$\int_{S^{n-1}} \alpha - \beta = \int_{S^n}\omega = 0$.<|endoftext|>
TITLE: Bounded Hamming distance
QUESTION [5 upvotes]: Definition 1. For each $n\in\mathbb{Z}^+$, the $n$-dimensional Hamming cube is the set of ordered $n$-tuples of $\lbrace 0,1\rbrace$, denoted by $\lbrace 0,1\rbrace ^n$.
Definition 2. The binary operation that turns $\lbrace 0,1\rbrace ^n$ into a group is $\oplus$ (XOR), which is bitwise addition reduced modulo $2$.
Definition 3. The sum of the digits of an element of $\lbrace 0,1\rbrace ^n$ is its Hamming weight. The Hamming weight of the XOR of two elements is the Hamming distance $h$ between them. In fact, $h$ is a metric on $\lbrace 0,1\rbrace ^n$.
Question. Given integers $0\le a\le b\le n$, what is known regarding maximal subsets $S$ of $\lbrace 0,1\rbrace^n$ such that: $\forall x,y\in S: a\le h(x,y)\le b$ ? Specifically, is anything known about bounds on cardinality, or exact cardinality, or even construction of these maximal sets?
I expect that the full answer is not known but I would be happy to read replies about, or receive direction toward resources for, special cases. For example, if $n$ is even and if $a=b=\frac{n}{2}$, then we turn to Hadamard matrices.

REPLY [6 votes]: My previous answer is already too long, and this is too much to include in a comment. But I found a paper that studies the problem you asked here:
R. M. Roth, G. Seroussi, Bounds for binary codes with narrow distance distributions, IEEE Transactions on Information Theory 58 (2007) 2760-2768
It seems a lot of interesting things are proved here. Among others, one of their propositions states that

Let $\mathcal{C}$ be an $(n,M,d)$ binary code of length $n$ and minimum distance $d$ with exactly $M$ codewords. Assume that the largest Hamming distance between two codewords $\boldsymbol{c}, \boldsymbol{c}' \in \mathcal{C}$ is $d_{\text{max}}$. Denote by $d_a$ and $d_g$ the arithmetic and geometric means respectively, of $d$ and $d_{\text{max}}$, i.e.,
$$d_a = \frac{d+d_{\text{max}}}{2} \quad \text{and} \quad d_g = \sqrt{dd_{\text{max}}}.$$
  Then
$$1-\frac{1}{M} \leq \begin{cases} \left(1-\frac{1}{n}\right)\left(\frac{d_a}{d_g}\right)^2 \quad &\text{if }\ 2d_a \not\in \{n+1, n+2\},\\\\ \frac{n}{d_g^2}\left(d_a- \frac{n+1}{4}\right) &\text{if }\ 2d_a \in \{n+1, n+2\}.\end{cases}$$

They also study equidistant codes, constant-weight codes, and self-complementary codes.
If you don't have access to the journal, a preprint is available on the first author's website:
http://www.cs.technion.ac.il/~ronny/PUB/2007/variance.pdf<|endoftext|>
TITLE: Vector fields on $(4n+1)$-spheres
QUESTION [26 upvotes]: If $n$ is odd then $S^{n-1}$ doesn't admit a nowhere-vanishing vector field, and if $n$ is even then there does exist one (Hairy Ball Theorem). We can then ask, on $S^{n-1}$, what is the maximum number $k(n)$ of linearly independent vector fields? Rewriting $n=2^{4a+b}(2s+1)$, Adams computes $k(n)=2^b+8a-1$.
In particular, on the $(4n+1)$-spheres, there is only one nowhere-vanishing vector field up to linear-independence, whereas in every other (odd) dimension there are more.
Example: on the circle $S^1$ there are the vector fields generated by (counter)clockwise rotation, but these are the same up to a scalar. This makes sense: I start flowing along this single dimension and then I have to continue flowing in that direction until I come back to my starting point.
I tried considering the difference between $S^3$ and $S^5$, which fiber over $\mathbb{C}P^1$ and $\mathbb{C}P^2$ respectively. A nowhere-vanishing vector field in both cases is given by taking the standard nowhere-vanishing vector field on the $S^1$-fiber. But for $S^3$ there are three linearly-independent fields (the $i,j,k$-directions when representing $S^3$ as the unit quaternions -- is there a way to see this using the fibration picture?), whereas for some reason $S^5$ can only admit the one.
What can be the differential/topological reasoning behind this? I.e. is there a down-to-earth way to deduce this result on $S^{4n+1}$, or for starters, $S^5$?
Could there possibly be an analogous index theorem going on here, in the same way that the Poincare-Hopf theorem provides us the Hairy Ball result?

REPLY [2 votes]: Thanks Misha for the reference! (Just rewriting it here to complete this thread).
 It seems that an off-the-cuff calculation (what I refer to as "down-to-earth") probably isn't going to suffice; there is some intricate stuff going on in the proofs involving the homotopy groups of the rotation groups of the spheres and orthogonal frames.
Homotopy Properties of the Real Orthogonal Groups (Whitehead), 1941, which focuses precisely on our case in question. The proof comes down to the obstruction of certain maps induced from rotation groups and their fixed subspaces on spheres.
Vector Fields on the n-Sphere (Steenrod, Whitehead) 1950, which studies particular fibrations of Stiefel manifolds over the spheres. It proves that for $n=2^k(2m+1)-1$ we have $k(n)<2^k$, which includes our case in question.<|endoftext|>
TITLE: Representing KO-theory using Clifford algebras
QUESTION [8 upvotes]: I'm trying to understand a statement Segal makes in this book:
Let $C_q$ be the real Clifford algebra associated to the standard negative definite form on $\mathbb{R^q}$ and let $\Phi_q(n)$ be the space of symmetric unitary (with respect to the automorphism of $C_q$ induced by $e_i \mapsto -e_i$) $n\times n$-matrices over $C_q$.
Segal then claims that $\Phi_q=\cup_n \Phi_q(n)$ represents $KO^q$ for $q\geq 1$ and that this is a reformulation of Bott periodicity.
Can someone indicate how this works or is there any good reference for this ?

REPLY [8 votes]: I learned about this in a course taught by Mark Hovey, but we didn't use a book and his lecture notes are not available online. As I recall, you first understand how the signature of a quadratic form classifies quadratic forms of given degree. Then you use this to classify Clifford algebras and notice the algebraic periodicity which will later be revealed to be Bott periodicity. If you want to be more concrete, let $q_{r,s}(x_1,\dots,x_n) = x_1^2 + \dots + x_r^2 - x_{r+1}^2 - \dots - x_{r+s}^2$ and then $C_{r,s}$ is the corresponding Clifford Algebra. You can reduce the classification to classifying $C_q = C_{q,0}$ by noting how $C_{r,s}$ splits up via tensor products. You seem to already understand this part of the process, so I won't say any more about it. The table classifying $C_q$ can be found here and it's 8 periodic.
Now we look at representations, and that's where your $\Phi_q$ comes in. Let $M_n$ be the Grothendieck group of representations of $C_n$ and note that $M_n$ is $\mathbb{Z}$ generated by some $\mathbb{R}^m$ if $n$ is not $3$ or $7$ mod $8$, and for all other $n$ we have $M_n\cong \mathbb{Z}\oplus \mathbb{Z}$ generated by $\mathbb{R}^m$ and $\mathbb{R}^m$. This is just observing which of the Clifford algebras $C_n$ in the table above split and $m$ is just the dimension of the corresponding vector space in the table considered as a vector space over $\mathbb{R}$. Now consider the standard inclusion $i:C_n\to C_{n+1}$ and note that you get $i^*:C_{n+1}-mod \to C_n-mod$ which induces $i^*:M_{n+1}\to M_n$.
The Atiyah-Bott-Shapiro Theorem says that there's a natural map $M_{n-1}/i^*M_n \to KO^{-n}(\ast) \cong \tilde{KO}(S^n)$
You can then compute all the quotients very simply, using the fact that you know the generators of $M_n$. For example, $M_0/i^*M_1$ is $\mathbb{Z}/2$ because $M_0$ is $\mathbb{Z}$ generated by $C_0 = \mathbb{R}$ and $M_1$ is $\mathbb{Z}$ generated by $C_1=\mathbb{C}$, and the map $i^*$ is just multiplication by 2 because it takes $\mathbb{C}\mapsto \mathbb{R}^2$. So the quotient is $\mathbb{Z} / 2\mathbb{Z}$. I recommend working out the rest of the $M_j/i^*M_{j+1}$ on your own. It's a good way to check your understanding of Clifford algebras and you'll see the Bott groups appear as if by magic.<|endoftext|>
TITLE: Are irrational multiples of central sets again central?
QUESTION [7 upvotes]: Let me begin by giving the relevant definitions. A set $A \subset \mathbb{N}$ is said to be central if and only if there exists a topological system $(X,T)$ (with $X$ a compact metric space, $T$ a continuous map on $X$) and a pair of points $x,y \in X$ with $y$ uniformly recurrent and proximal to $x$, such that for some neighbourhood $U$ of $y$ one has $A = \{n \in \mathbb{N} \ : \ T^n(x) \in U \}$. That $x$ and $y$ are supposed to be proximal means that there is a sequence $n_i \to \infty $ such that $d(T^{n_i}x,T^{n_i}y) \to 0$. That $y$ is uniformly recurrent means that for any neighbourhood $U$ of $y$, the set $\{n \in \mathbb{N} \ : \ T^n(y) \in U \}$ is syndetic (has bounded gaps).
Quite amazingly the above definition of centrality turns out to be equivalent to $A$ being a member of a minimal idempotent ultrafilter in $\beta \mathbb{N}$. A good reference for more details is this article by Vitaly Bergelson. Let me give some details. The ultrafilters $\beta \mathbb{N}$ are a compact left-topological semigroup (meaning, addition is continuous in the left argument; and of course some authors have the reverse convention). Any such group has a unique minimal two-sided ideal, call it $K$, and an ultrafilter is miniamal if and only if it belongs to $K$. An ultrafilter $p$ is idempotent if and only if $p+p = p$. One of course needs some work to get the whole theory up and running, but it can be done.
It is not hard to verify that if $A$ is central then so is $dA$ for a positive integer $d$. For this, replace $X$ by $X' := X_1 \cup \dots \cup X_2 \cup X_d$, with $X_i = X\times\{i\}$ a copy of $X$, and define $T'$ by $T'(x,i) = (x,i+1)$ for $i < d$, and $T'(x,d) = (T(x),1)$, and finally let $x',\ y' = (x,1),\ (y,1)$. Likewise, it can be shown that $A/d = \{n \ : \ nd \in A\}$ is central by considering the system $(X,T^d)$, and the same points.  Of course, there are a few things to check along the way that I skipped. Because centrality is a notion of largeness, in the sense that if $A$ is central and $A \subset B$ then $B$ is central, it is clear that also the set $\{ \lfloor \frac{n}{d} \rceil \ : \ n \in A \}$ is again central, because it contains $A/d$ ($\lfloor x\rceil$ stands for the closest integer).
Given the above, it appears to be reasonable that if $A$ is a central set, then central is also the set$\{ \lfloor \alpha n \rceil \ : \ n \in A \}$, where $\alpha \in \mathbb{R}^+$ is a real number. I think a possible way to attack the problem is by considering the "suspended" system $X' = X \times \mathbb{T}$ ($\mathbb{T} = \mathbb{R}/\mathbb{Z}$) with $T'(x,t) = (T^{\lfloor \alpha + t \rfloor}(x), \{ \alpha + t \} )$, $A' = A \times \mathbb{T}$, $x',\ y' = (x,0),\ (y,0)$. However, I can't quite seem to work out the details in a way that I would find convincing (and they say that oneself is the easiest to fool...).
Question: If $A$ is a central, then does $\{ \lfloor \alpha n \rceil \ : \ n \in A \}$ need to be central?

REPLY [5 votes]: The answer is yes, it was pointed out to me by Vitaly Bergelson that it follows from Theorem 6.1 (d) of a paper by Bergelson, Hindman and Kra (Trans. Amer. Math. Soc. 348 (1996), no. 3, 893–912).
The notation $g_{\alpha,\gamma}[A]$ is defined in the previous section and means
$$g_{\alpha,\gamma}[A]:=\{\lfloor \alpha n+\gamma\rfloor:n\in A\}$$
where $\lfloor\cdot\rfloor$ represents the floor function.
Note that $\lfloor x+1/2\rfloor=\lfloor x\rceil$<|endoftext|>
TITLE: Intersection of 2 visibility polygons
QUESTION [7 upvotes]: Let $P$ be a simple, closed and bounded polygon and $p_1,p_2 \in \mathrm{int}(P)$ be two points in its interior. Is it true that the intersection of the visibility polygons of $p_1$ and $p_2$ is connected?
The visibility polygon of a point $p\in P$ is the set of all $x \in P$ such that the line segment connecting $p$ and $x$ is contained in $P$.
It might be elementary, but we fail either finding a proof or a counterexample. Furthermore, in the literature I could only find details about the computational aspects of the visibility polygon, but not a single word about its properties. Any reference, example, sketch of proof etc. is welcomed!

REPLY [6 votes]: Yes. Here is an argument.
Let $p_1$ and $p_2$ both see $a$ and $b$.  I claim that every point along
the shortest path $\sigma$
connecting $a$ to $b$ inside $P$ is visible to both $p_1$ and $p_2$.
With this claim established, we know the intersection of
the visibility polygons is connected.
Start turning/sweeping the rays $p_1 a$ and $p_2 a$ counterclockwise 
along $\sigma$ toward $b$, simultaneously tracking the
same point $x$ along $\sigma$.  If both rays remain unobstructed throughout the
sweep, we are finished.
So suppose otherwise.  Then one or the other ray, say $p_2 a$, must encounter
an obstruction, a reflex vertex $v$ hitting $p_2 a$, blocking the visibility
to point $x \in \sigma$.
Then we have some exterior points of the polygon enclosed within the closed path
$p_2 x \cup \sigma(x,b) \cup b p_2$ of
points interior to $P$, contradicting the simplicity of the polygon,
i.e., the polygon has a hole:

REPLY [2 votes]: The answer is yes. I do not see a clean proof. Here is a proof by case-checking. Consider two points $q$ and $q'$ that are visible from both $p_1$ and $p_2$. There are several cases: 
1) Suppose four points $p_1$, $p_2$, $q$, $q'$ are in convex position, and points $p_1$, $p_2$ are opposite vertices of the convex hull. Then convex hull is completely inside the polygon, and in particular in the intersection of two visibility polygons. Thus, $q$ and $q'$ are in the same connected component. 
2) Four points $p_1$, $p_2$, $q$, $q'$ are in convex position, and points $p_1$, $p_2$ are adjacent vertices of the convex hull. Without loss, assume that the order is $p_1$, $p_2$, $q$, $q'$. Let $r$ be the intersection point of line segments $p_1q$ and $p_2q'$. Then the line segments $qr$ and $q'r$ are contained in the intersection of visibility regions.
3) Point $q$ is in the convex hull of $p_1$, $p_2$ and $q'$. Then the nonconvex $4$-gon $p_1qp_2q'$ is in the intersection of visibility polygons, and line segment $qq'$ is completely contained in it.
4) Point $p_1$ is in the convex hull of $p_2$, $q$ and $q'$. In that case the line segments $qp_1$ and $q'p_1$ are in both visibility polygons. Again, $q$ and $q'$ are connected.<|endoftext|>
TITLE: Algebraic Stratifications of $G$-varieties
QUESTION [11 upvotes]: My question is simple:
Given an algebraic group $G$ acting on a variety $X$ algebraically. If the orbits are of finite number then they form what is called an algebraic stratification of $X$.
Now my question is: Is this (the stratification by $G$-orbits) also a Whitney Stratification?
I think showing this by hand is quite painfull since the definition of a Whitney stratification is very unhandy.  
I would appreciate any reference.

REPLY [4 votes]: So Ulrich and Geordie were right, Tom Braden was the right person to ask and here is what he told me:
The answer is yes, in the case above $X$ is Whitney stratified.
The argument goes roughly as follows. In the paper [K] the following is shown. Let $X$ be an algebraic stratified variety and $ S$  a stratum.   The set $Sing(S )$ of points in $S $ which do not fulfil Whitney's b condition has the structure of a (semi-)variety of dimension strictly lower that $\dim S $. 
Now applied to our situation :
$Sing(S)$ is necessarly $G$-invariant, hence empty. This answers my question.  
[K] ``A Geometric Proof of Existence of Whitney Stratifications’’, Moscow Math. Journ., 5 (2005), no.1, 125—133<|endoftext|>
TITLE: Weyl group of the restriction of scalars of split reductive group
QUESTION [5 upvotes]: Let $G$ be a connected algebraic group defined over a field $E$ of characteristic $0$. Suppose $G$ reductive $E$-split and let $T \subset G$ a maximal (split) torus defined over $E$.
Set $G' = Res_{E/F} G$ where $E/F$ is a finite field extension, $T' = Res_{E/F} T$ and $S' \subset T'$ a maximal $F$-split torus. Then $T'$ is the centralizer of $S'$ in $G'$ and a maximal torus of $G'$.
Now, what is the relation between the Weyl group $W = N_G(T)/T$ of $(G,T)$ and the relative Weyl group $W' = N_{G'}(S') / T'$ of $(G',S')$ ?
Thanks

REPLY [6 votes]: While waiting for nosr to make the comments here into a full answer, I'll add some references to the literature.   The early sources are quite technical and don't provide much in the way of user-friendly examples, but they do show the origins of the ideas involved in Weil restriction:
A. Weil, Adeles and Algebraic Groups (Birkhauser, PM 23, 1982): this small monograph contains a verbatim copy of Weil's 1959-60 IAS lectures on Tamagawa numbers, along with some updates by T. Ono and a bibliography.  See section 1.3 for the starting formalism of Weil restriction of scalars.
A. Borel and J.-P. Serre, Theoremes de finitude en cohomologie galoisienne, Comment. Math. Helv. 39 (1964), 111-164: see section 2.8.  [This paper and the following one are available online, the second at numdam.org]
A. Borel and J. Tits, Groupes reductifs, Publ. Math. IHES 27 (1965): see especially section 6.
M. Demazure and P. Gabriel, Groupes algebriques, tome I, Masson and North-Holland, 1970: see I, section 4, 6.4 and 6.6.   (Their framework is more scheme-theoretic than the early ones.)
Much more recently, Weil restriction has played a large role in the book Pseudo-reductive Groups (Cambridge, 2010) by B. Conrad, O. Gabber, G. Prasad.    In their index, see the many specific topics listed under "Weil restriction".    Here I haven't tracked down an explicit comparison of the Weyl groups, but for instance Prop. A.5.15 comes close in its treatment of tori.
My only editorial comment would be that no one over the years seems to have provided an ideal intuitive discussion of what Weil restriction is all about and why it's important.   Some explicit examples could be very enlightening, including the three-dimensional simple group $\mathrm{SL}_2$.
ADDED: For a typical application of Weil restriction (in language close to Weil's) which brings out implicitly the connection between the maximal split tori (and Weyl groups) in the groups over the two fields involved, see 3.1.2 in the article by J. Tits, "Classification of algebraic semisimple groups", Proc. Symp. Pure Math. 9, Amer. Math. Soc., 1966 (proceedings of 1965 Boulder summer institute).   As in other sources, the emphasis is on comparing tori root systems and rather than directly on the Weyl groups.<|endoftext|>
TITLE: Example of codim 1 regular embedding that is not an effective Cartier divisor?
QUESTION [14 upvotes]: An effective Cartier divisor on a scheme $X$ (a closed subscheme of $X$ that is locally cut out by one equation that is not a zero-divisor) is always a regular embedding of codimension 1 (in the stalks, it is cut out by one equation that is not a zero-divisor).  The converse is true if $X$ is locally Noetherian, using Nakayama's lemma to "lift" from the stalk to an "honest neighborhood".  

Is there a (necessarily non-Noetherian) example of a codimension 1 regular embedding that is not an effective Cartier divisor?

(See section 8.4.7 of the March 23, 2013 version of the notes here for some discussion, if you care.)
I am asking this out of idle curiosity rather than any need for it, but I am genuinely curious.

REPLY [19 votes]: Let $(A,m,k)$ be a discrete valuation ring. Consider the subring $R\subset A^\mathbb{N}$ of converging sequences (for the discrete topology on $A$), i.e. ultimately constant sequences. We have a morphism $\varphi:R\to A$ sending any sequence to its limit. Put $\mathfrak{p}=\varphi^{-1}(m)$: this is the maximal ideal of sequences with noninvertible limit. 
Now it is easy to check that $\varphi$ induces an isomorphism $R_\mathfrak{p}\cong A$. In particular, $\mathfrak{p}$ becomes principal in $R_\mathfrak{p}$, and of course also in $R_\mathfrak{q}$ for all $\mathfrak{q}\neq\mathfrak{p}$ because then $\mathfrak{p}R_\mathfrak{q}=R_\mathfrak{q}$. So $V(\mathfrak{p})\subset\mathrm{Spec}\,(R)$ is a regular embedding of codimension 1.  
On the other hand, $\mathfrak{p}$ is not finitely generated. Indeed, if $J\subset\mathfrak{p}$ is a finitely generated ideal, there is an $N\in\mathbb{N}$ such that for each $u:\mathbb{N}\to A$ in $J$ and each $n>N$ we have $u(n)\in m$. Clearly there is no such $N$ for $\mathfrak{p}$.<|endoftext|>
TITLE: A question about the proof of Beilinson-Bernstein localisation
QUESTION [7 upvotes]: I'm trying to understand the proof of the Beilinson-Bernstein localisation theorem at the moment, but there's just one point where I'm having a mental block, and was wondering if anybody could clarify things for me.
Specifically, it's this (not quite general) theorem which I'm trying to prove:
Let $G$ be a semisimple algebraic group over $\mathbb{C}$, $B$ a Borel subgroup and $X=G/B$ the flag variety. Let $\mathfrak{g}$ be the Lie algebra of $G$, $\mathfrak{b}$ the Lie algebra of $B$, $\mathfrak{h}$ the Cartan subalgebra contained in $\mathfrak{b}$, $\mathfrak{b}=\mathfrak{h}\oplus\mathfrak{n}$ ($\mathfrak{n}$ nilpotent), $\mathfrak{g}=\mathfrak{n}^-\oplus\mathfrak{h}\oplus\mathfrak{n}$ (you get the picture). Writing $U(\mathfrak{g})$ for the universal enveloping algebra of $\mathfrak{g}$ we have $U(\mathfrak{g})=U(\mathfrak{h})\oplus(\mathfrak{n}^-U(\mathfrak{g})+U(\mathfrak{g})\mathfrak{n})$, by PBW.
Let $\lambda:B\to \mathbb{C}^\times$ be a character, and let $\mathcal{L}^\lambda$ denote the $G$-equivariant invertible sheaf on $X$ with fiber $\mathbb{C}^{-\lambda}$ at $eG$. That is, $B$ acts on the trivial $\mathbb{C}^{-\lambda}$-bundle $G\times\mathbb{C}^{-\lambda}$ by $b(g,m)=(gb^{-1},(-\lambda)(b)m)=(gb^{-1},\lambda(b^{-1})m)$, and $B\backslash G\times(\mathbb{C}^{-\lambda})$ is a $G$-equivariant $\mathbb{C}^{-\lambda}$-bundle on $X$; $\mathcal{L}^\lambda$ is then its sheaf of sections.
Since $\mathcal{L}^\lambda$ is $G$-equivariant we obtain a homomorphism $\alpha^\lambda:U(\mathfrak{g})\to\Gamma(X,\mathcal{D}_X^\lambda)$ where $\mathcal{D}_X^\lambda=\mathcal{L}^\lambda\otimes\mathcal{D}_X\otimes\mathcal{L}^{-\lambda}$ is the sheaf of differential operators on $\mathcal{L}^\lambda$. (tensor products taken over ${\mathcal{O}_X}$).
Then, what I would like to prove is that the restriction of $\alpha_\lambda$ to the centre $Z(\mathfrak{g})$ of $U(\mathfrak{g})$ factors through the character $\chi_\lambda$ (i.e. the map $Z(\mathfrak{g})\to U(\mathfrak{h})$ coming from the direct sum decomposition above, composed with the map $\lambda:U(\mathfrak{h})=\operatorname{Sym}(\mathfrak{h})\to\mathbb{C}$).
Of course, this isn't the whole theorem, but it's the only part I'm having trouble with.
I believe I am correct in thinking that $\mathcal{L}^\lambda$ is nothing more than the pushforward from $G$ of a certain subsheaf of $\mathcal{O}_G$, namely the one whose sections $f$ are those satisfying $f(gb)=\lambda(b)f(g)$ for $g\in G,b\in B$. So it should be enough to show that $Z(\mathfrak{g})$ acts on that in the right way. (For some reason, I find the action of $\mathfrak{g}$ on $\mathcal{O}_G$ much easier to think about than its action on $\mathcal{L}^\lambda$.)
But I am really stuck. I've looked in the book "D-modules, Perverse Sheaves and Representation Theory" by Hotta et al., where they seem to prove this on pages 278-279, but only found it confusing (and they gave the wrong definition of the Harish-Chandra homomorphism, which I found off-putting). I've also looked in Gaitsgory's notes (http://www.math.harvard.edu/~gaitsgde/267y/catO.pdf) where he seems to prove this (at least, the case $\lambda=0$) on pages 42-43, but that's also confusing. What's worse, is that apparently Gaitsgory's proof makes no use of the algebraic geometry of $\mathfrak{g}$, whereas Hotta et al. appear the Springer resolution of the nilpotent cone in an important way.
If anyone could enlighten me at all about this, I would be extremely thankful!

REPLY [3 votes]: This may not be the most efficient way to get to the result, but here's how I would think about it:
The universal enveloping algebra can be identified with right $G$-invariant differential operators on the group $G$, via the map sending an element of the Lie algebra to the corresponding left translation vector field (and vice versa with left and right switched).  In particular, the center of $U(\mathfrak{g})$ is given by bi-invariant differential operators on the group, in two different ways.
If I want to understand how elements of the center act on functions that satisfy $f(gb)=\lambda(b)f(g)$, then I should write them as $z=h(z) + n_1m_1+ \cdots+n_km_k$ where $m_i\in U(\mathfrak{n})\mathfrak{n}$ which is possible by the PBW theorem and the fact that central elements have weight 0 (here $h\colon Z(\mathfrak{g})\to U(\mathfrak{h})$ is the Harish-Chandra homomorphism).    Thus, $z\cdot f= d\lambda (h(z)) f$; here I'm using $d\lambda$ to distinguish between characters of the group and Lie algebra.  
Ok, I'm basically done, but I cheated a little here, since here I was using the map of the center to bi-invariant operators for the right action, and you really want the one that comes from the left action, which a priori might be different.  Let me be lazy, and note that we've now shown that the map of the center factors through some character, and that this is induced by some ring map $Z(\mathfrak{g}) \to U(\mathfrak{h})$ (maybe not the HC homomorphism).  Thus, it suffices to check it at a Zariski dense set of points; this follows from Borel-Weil, since we know how the center acts on the sections of $\mathcal{L}^\lambda$.<|endoftext|>
TITLE: Does every simplicial polytope have a topology-preserving contractible edge?
QUESTION [7 upvotes]: An edge of a triangulated manifold is said to be contractible if it may be contracted to a vertex without modifying the topological type of the underlying manifold. Otherwise, the edge is noncontractible. 
Not every edge of a triangulated manifold is contractible. For example, on a triangular bipyramid, the edges gluing the two pyramids together are noncontractible, while all other edges are contractible (yielding a simplex). On a $d$-simplex, no edge is contractible.
Dey, Edelsbrunner, Guha, and Nekyahev provide exact conditions for when an edge of a 2- or 3-manifold is contractible, noting that an edge $ab$ between vertices $a$ and $b$ is contractible iff  $link(ab) = link(a) \cap link(b)$.  I have found little in the literature about conditions for the contractibility of edges on higher-dimensional objects.
My questions concern the existence of contractible edges on polytopal $d$-spheres, ie, simplicial polytopes.  
1) Does every simplicial polytope other than the simplex have a contractible edge?
2) Does contraction of an edge of a simplicial polytope always yield another polytope?
3) Are these questions any easier to answer if we restrict the question to 4-polytopes?

REPLY [6 votes]: I brought this thread to the attention of Eran Nevo, who gave some more references, and pointed out that the answer to the converse of your question (2) is "no".
In his thesis, which I gave the arXiv link to above, and in the paper version of this "Higher minors and Van Kampen's obstruction", Nevo defines a strongly edge decomposable sphere to be a sphere which can be reduced to the simplex by edge contractions.  He shows that in any PL-manifold (in any dimension), an edge is contractible if and only if it satisfies the link condition.
Then Satoshi Murai in "Algebraic shifting of strongly edge decomposable spheres" shows that any squeezed sphere is strongly edge decomposable.  (Murai also has an earlier paper which is related, "Generic initial ideals and squeezed spheres".)  Since there are more squeezed $d$-spheres than polytopes for $d \geq 5$, there are non-polytopal spheres where an edge contraction leaves a polytopal sphere.  Though I didn't see $d=4$ settled explicitly one way or the other in the papers I looked at, I suspect that there are non-polytopal squeezed 4-spheres as well (and probably this is known).  Every squeezed $3$-sphere is polytopal.
Another paper which is somewhat relevant is Babson and Nevo "Lefschetz properties and basic constructions on simplicial spheres".
I also thought a bit about the forward direction of your question (2).  If $vw$ is a contractible edge in the boundary of the convex hull of $V$, it looks plausible to me that the edge contraction should correspond to passing to (boundary of) the convex hull of $V \setminus \lbrace v,w \rbrace \cup x$, where $x$ is a point on $vw$.  For as you 'slide' $v$ and $w$ together, the link condition seems to prevent new faces from being created.  (Obviously, there is a lot to be checked with this idea!)<|endoftext|>
TITLE: Plethysm of $\mathrm{QSym}$ into $\mathrm{QSym}$: can it be defined?
QUESTION [9 upvotes]: I will denote by $\Lambda$ the ring of symmetric functions, and by $\mathrm{QSym}$ the ring of quasisymmetric functions (both in infinitely many variables $x_1$, $x_2$, $x_3$, ..., both over $\mathbb Z$). See Victor Reiner, Hopf algebras in combinatorics, chapters 2 and 5, respectively, for the definitions.
It is known that one can define a plethysm $f\circ g \in \mathrm{QSym}$ for any $f\in\Lambda$ and any $g\in\mathrm{QSym}$. This is described, e. g., in Claudia Malvenuto, Christophe Reutenauer, Plethysm and conjugation of quasi-symmetric functions, Discrete Mathematics 193, 225-233 (1998). In a nutshell, if $g$ is a sum of monomials in the $x_1$, $x_2$, $x_3$, ..., one can construct $f\circ g$ by substituting these monomials as indeterminates into $f$. It takes some more work (and is less intuitive) to define $f\circ g$ when $g$ has negative coefficients, but the above should give some feeling for what $f\circ g$ is. Notice that $e_1\circ g = g$ (where $e_1$ is the $1$-st elementary symmetric function). For $g \in \Lambda$, the plethysm $f\circ g$ becomes the usual plethysm in $\Lambda$.
In section 3 of the preprint Michiel Hazewinkel, Explicit generators for the ring of quasisymmetric functions over the integers, it is claimed that this construction extends to all $f\in \mathrm{QSym}$, where the monomials are substituted into $f$ in lexicographic order. I cannot follow this claim, because it seems to me that the "addition formula"
(1) $f \circ \left(g+h\right) = \sum\limits_{(f)} \left(f_{(1)}\circ g\right) \left(f_{(2)}\circ h\right)$ (using Sweedler notation, where $\sum\limits_{(f)} f_{(1)} \otimes f_{(2)}$ is the first coproduct of $f$)
is no longer satisfied for general $f\in\mathrm{QSym}$, and the definition of $f\circ g$ outside the case of $g$ being a sum of monomials hinges on this formula (of course, there are better definitions in the $f \in \Lambda$ case which don't depend on this formula, but they don't look generalizable at all).
What I want to know, apart from whether or not my doubts on this definition are justified, is whether there is any reasonable definition of a plethysm of two elements of $\mathrm{QSym}$ known, or whether there are good reasons no such beast exists in nature.
[EDIT: At a second glance, if we take Hazewinkel literally, he isn't claiming this all; he is only defining $f\circ g$ for $f$ and $g$ being monomial quasisymmetric functions, which (I guess) he can do as he pleases. But I think he is trying to define $f\circ g$ for all $f\in\mathrm{QSym}$ and $g\in\mathrm{QSym}$. If I interpret his definition of $f\circ g$ as being only formulated for the monomial quasisymmetric functions, and then try to extend it using (1) to all $f\in\mathrm{QSym}$ and $g\in\mathrm{QSym}$, then I think I obtain a contradiction due to the non-cocommutativity of $\mathrm{QSym}$.]
[Let me remark that Hazewinkel's proof of the polynomial freeness of $\mathrm{QSym}$ does not depend on this kind of plethysm. He only ever uses it for $f\in\Lambda$. A clean version of his proof can be found in Chapter 6 of Michiel Hazewinkel, Nadiya Gubareni, V.V. Kirichenko, Algebras, Rings and Modules, Volume 3, and the only gap in it (the unproven footnote 13) can be filled in using Section 2 of David E. Radford, A natural ring basis for the shuffle algebra and an application to group schemes.]

REPLY [4 votes]: As per @darij's suggestion regarding my above comment: Loehr and Remmel's  A computational and combinatorial exposé of plethystic calculus is a good resource for a combinatorial point of view of plethysm; I think it clarifies what one should do when both functions are quasisymmetric.  Edit: Apparently the paper is Open Access.<|endoftext|>
TITLE: short character sums averaged on the character
QUESTION [7 upvotes]: Let $a$ be an integer, $p$ a prime (much) greater than $a$, and $\chi$ a Dirichlet character.
There is an abundant literature on the sums
$$S(\chi,a)=\sum_{i=1}^a \chi(i),$$
called short (or incomplete) character sums, and their average in various sense. Yet I have not found in this literature an answer to the following specific question (perhaps because of its very abundance):
Let us form the average of the modulus of $S(\chi,a)$ over $p$; that is, let's define:
$$AS(p,a) = \frac{1}{p-1} \sum_{\chi} |S(\chi,a)|,$$ where the sum is over the $p-1$ Dirichlet characters mod $p$. An easy upper bound for $AS(p,a)$, namely
$$AS(p,a) \leq \sqrt{a}$$
can be obtained by applying Cauchy-Schwarz. 

Is this upper bound essentially the best possible? Precisely, is it true or false that $AS(p,a)=o(\sqrt{a}),$ uniformly in $p >a$? (And what if one restricts to the domain where $p$ is not too large, say $p < a^K$ for some fixed constant $K$ -- so that the character sums are not too short?)

I have done some numerical experimenta, which are not really conclusive in what sense or another. I might very well have overlooked some paper in the literature treating this question. Thanks for any idea, guess, conjecture, proof or reference...

PS: here is the proof of the trivial upper bound: $AS(p,a) \leq \frac{1}{p-1} \sqrt{(p-1) \sum_{\chi} |S(\chi,a)|^2}$ by Cauchy-Schwarz, and $\sum_\chi |S(\chi,a)|^2$ can be expanded as $\sum_{\chi,i,j} \chi(i) \overline{\chi(j)}=\sum_{i,j} \sum_\chi \chi(ij^{-1})$ where $i,j$ run from $1$ to $a$ and
$ij^{-1}$ is computed modulo $p$; then the sum over $\chi$ is $0$ unless $i=j$ which
happens $a$ times, and it this case it is $p-1$, so $\sum_\chi |S(\chi,a)|^2=(p-1)a$, hence $AS(a,p)\leq \sqrt{a}$.

REPLY [6 votes]: In his beautiful talk at MSRI today (1 May 2017), Adam Harper announced the following theorem which settles this problem.  Let $p$ be a large prime, and let $x$ be less than $p$.  Put $L=\min (x,p/x)$.  Then 
$$ 
\frac{1}{p-1} \sum_{\chi \pmod p} \Big| \sum_{n\le x} \chi(n) \Big| 
\ll \frac{\sqrt{x}}{\min (({\log\log L})^{\frac 14}, ({\log \log \log p})^{\frac 14})}. 
$$ 
One might expect that the $(\log \log \log p)^{\frac 14}$ term above can be removed with more work.  In particular, Harper's result shows that when $x=o(p)$, the $L^1$ norm is $o(x)$.  MSRI should have the video of this lecture up in a short while.<|endoftext|>
TITLE: For a monoid with zero $M$, how many additive operations on $M$ can there be making $M$ a ring?
QUESTION [6 upvotes]: Let $(M,\times)$ be a monoid with zero. Let $\Sigma(M,\times)$ be the set of binary operations $+$ on $M$ such that $(M,+,\times)$ is a ring. Let $\sim$ be an equivalence relation on $\Sigma(M,\times)$ defined by $$+_1\sim+_2\iff(M,+_1,\times)\cong(M,+_2,\times).$$
Let's denote the quotient set $\Sigma(M,\times)/\sim$ by $\Sigma'(M,\times)$ and consider the number $$\mathrm{add}(M,\times)=|\Sigma'(M,\times)|.$$
I have several questions about the behavior of this function, none of which I know how to approach.
$(1)$ For an integer $n\geq 0$, is there always a monoid $(M,\times)$ such that $n=\mathrm{add}(M,\times)?$
$(2)$ The same question with the requirement that $M$ be finite.
$(3)$ The two previous questions are equivalent if $\mathrm{add}(M,\times)\geq\aleph_0$ for $M$ infinite. Is it true? It is false by Todd Trimble's answer. $(\mathbb Z,+)$ with a zero element adjoined is another example. 
$(4)$ Is there an upper bound to the values of $\mathrm{add}(M,\times)$ over all finite monoids with $0?$ What about all monoids with zero?

REPLY [12 votes]: I think the answer to (2) is yes (which also answers (1) and (4), of course).
First, note that $R=\mathbb{Z}/4\mathbb{Z}$ and $S=\mathbb{F}_2[x]/(x^2)$ have isomorphic multiplicative monoids (via the map $0\mapsto 0$, $1\mapsto 1$, $2\mapsto x$, $3\mapsto 1+x$).
Call this monoid $M$.
Then the direct product $M^n$ of $n$ copies of $M$ is the multiplicative monoid of
$T_k=R^k\times S^{n-k}$ for $k=0,\dots,n$, giving $n+1$ rings that are pairwise non-isomorphic, since they have different additive groups.
I claim that these are the only rings with $M^n$ as multiplicative monoid. Suppose $T$ is another. Then it follows just from the multiplicative structure that $T$ is commutative and has $n$ primitive idempotents $e_1,\dots,e_n$ such that, as a ring, $T\cong Te_1\times\dots\times Te_n$, where each $Te_i$ is a ring whose multiplicative monoid is $M$.
But it's quite easy to see that $R$ and $S$ are the only rings with multiplicative monoid $M$: the subring generated by $1$ can only be $\mathbb{Z}/4\mathbb{Z}$ or $\mathbb{Z}/2\mathbb{Z}$, so either the ring is isomorphic to $R$ or is a two-dimensional $\mathbb{F}_2$-algebra generated by an element $x$ with square zero, in which case it must be isomorphic to $S$.<|endoftext|>
TITLE: Are there five consecutive primes in arithmetic progression?
QUESTION [8 upvotes]: For example 
3 consecutive primes in arithmetic progression 
3,5,7  distance 2
151,157,163 distance 6

4 consecutive primes in arithmetic progression
251,257,263,269 distance 6
1741,1747,1753,1753 distance 6
76543,76561,76579,76597 distance 18

Are there five consecutive primes in arithmetic progression?

REPLY [12 votes]: Yes for the lengths you ask for this is known to exist. 
The minimal example is $9843019+ 30 n$ for $n=0,1,2,3,4$ (taken from the page at the end). 
A more common way to phrase this would be to ask about (five) consecutive primes in arithmetic progression. 
Indeed, it is conjectured that there are arbitrarily long (finite) arithmetic progressions of consecutive primes, however this is open. (Without the restriction of the primes being consecutive primes this is known by a well-known result of Green and Tao.)
The longest known arithmetic progression of consecutive primes has length ten. 
For further details one could start at http://en.wikipedia.org/wiki/Primes_in_arithmetic_progression (see the section towards the end on consecutive primes in AP) for record data related to this see http://users.cybercity.dk/~dsl522332/math/cpap.htm

REPLY [7 votes]: Yes.  In fact, there is a set of 10 consecutive primes that are in arithmetic progression.  I believe this is the longest known set.  See here.<|endoftext|>
TITLE: $n$-in-a-row game on $\mathbb{R}^2$
QUESTION [20 upvotes]: For integers $n$ such that $\:3< n\:$,$\:$ what is known about the following 2-player game:
Player_1 and Player_2 take turn choosing points on $\mathbb{R}^2$ that were not previously chosen, with Player_1 going first. $\:$ If there is ever a segment with $n$ points chosen by Player_1 and zero points chosen by Player_2, then Player_1 wins. $\:$ If infinitely many turns go by without that happening, then Player_2 wins.

(That includes, if Player_1 has a win, how long the win should take.)

REPLY [12 votes]: The points on the lines and the lines through the points...
The lines should be many with numerous joints,
the points should be few and should sit at right places.
What else? We can use high-dimensional spaces,
if we do not mind doubling enemy's strength.
It would not be bad if we bound the length
of games, tell who wins, and what way he should play.
The questions are many, but short is the day.
One cannot resolve all the problems at once 
by being just smart. Better leave them to Chance!
The law of large numbers is our best friend.
Who is unpredictable beats every trend.
Of course, he can lose, but the God in the sky
does know enough to condition, not try,
thus weaving each random irrational state
into the unerring decisions of Fate.


The first player wins playing more or less randomly after he prepares the playing field in the right way.
As it has already been said, we can play this game in $\mathbb R^d$. Since, when projecting a high dimensional configuration to the plane, we can avoid extra triple intersections but not extra double intersections of lines, we should allow the opponent to make two moves for each our move to get a stronger version of the game.
Fix $n$ (the number of points on the line we want). Choose small $\delta>0$ and large integer $d>0$ in this order. Consider the cubic lattice in $\mathbb R^d$ with $n$ points along each side ($n^d$ points total). Make $\delta n^d$ moves choosing a random lattice point at each move. If this point is already occupied, just put a point somewhere far away and forget of it. However, mark the occupied point as "intended". Let the opponent move twice after each of your moves. At the end of the game, just look at what you got. Pay attention only to the fully filled (with either actual, or intended points) lattice lines in coordinate directions. You should see the following.
1) The probability to fill each particular line is about $\delta^{n}$, so you should expect $L=n^{d-1}d\delta^{n}$ filled lines. Moreover, the probability to fill less than half of this amount is extremely small because filling two disjoint lines are negatively correlated events and intersecting pairs are few, so the expectation of the square is pretty close to the square of the expectation if $n,\delta$ are fixed and $d\to\infty$.
2) The probability that a given grid point has $k$ filled lines passing through it is at most $\delta\cdot \frac{[d\delta^{n-1}]^k}{k!}$, which is exponentially small in $d$ if $k>2ed\delta^{n-1}$ in the same regime  when $n,\delta$ are fixed and $d\to\infty$.
3) Each opponent's point that doesn't lie on the grid can interfere with at most one line.
4) Each opponent's point that lies on the grid can interfere with one of the filled lines only if it was intended during one of your moves. However, since during the whole game only $3\delta n^d$ points have been used, the chance to hit an already occupied point has never been higher than $3\delta$, so it is highly unlikely that the opponent could score much more than $6\delta^2 n^d$ such points.
Now it is time to count the blocked lines. First, $e^{-c(n,\delta)d}dn^d=e^{-c(n,\delta)d}\delta^{-n}nL$ come from "high efficiency" points (let the opponent grab them all, it is still nothing compared to $L$ if $d$ is large enough!). Second, the points put off the grid can block only $2\delta n^d=d^{-1}\delta^{-n+1}n L$ lines, which isn't much either. Last, the "efficient" but not "highly efficient" points on the grid block at most $6\delta^2 n^d\cdot 2ed\delta^{n-1}=2e\delta n L$ filled lines. Here we do not gain from $d$, but if we start with choosing $\delta$ so that $12e\delta n\ll 1$, we are still fine. Thus, normally at most $L/2$ lines will be blocked  and, with high probability, the outcome is that we have won the game by this moment.
Since the game is now of fixed length, one of the players must have a deterministic winning strategy. But the second player loses against the random strategy with positive probability no matter what he does, so it isn't he. The deterministic winning strategy for the first player can be defined explicitly in terms of conditional probabilities to win the random game from the current position (just move to the point that gives you the best chance to win the random game in the maximin sense (max over locations, min over all possible opponent's strategies) but the computation of those conditional probabilities is well beyond the human abilities.<|endoftext|>
TITLE: Basis theorem (due to Solovay?)
QUESTION [7 upvotes]: I'm finishing up my bibliography and I'm looking for a reference for the statement that, working in $L(\mathbb{R})$, the $\Delta^2_1$ sets form a basis for the $\Sigma^2_1$ predicates. I believe that it is due to Solovay.

REPLY [9 votes]: The result is indeed Solovay's Basis Theorem. 
It is a consequence of Moschovakis's Coding Lemma, and sometimes it is referred to as (a version of) the reflection theorem (for example, in section 8 of Steel's Handbook article). Perhaps the optimal reference (it is self-contained, and easily accessible) is Section 2.4 of 

Peter Koellner, and W. Hugh Woodin. Large cardinals from determinacy. In Handbook of set theory. Vols. 1, 2, 3, 1951–2119, Springer, Dordrecht, 2010. 

I do not think it appears explicitly in a paper by Solovay. It should date back to 1976 at the latest, which is when Kechris and Solovay observed that not all $\Pi^2_1$ sets can be uniformized if $V=L(\mathbb R)$ and choice fails (see section 30 in Kanamori's book).<|endoftext|>
TITLE: Reason for studying coherent sheaves on complex manifolds.
QUESTION [15 upvotes]: Hello everybody! I would be interested in knowing, what the reason is for investigating coherent sheaves on complex manifolds. By definition a sheaf $F$ on a complex manifold $X$ is coherent, when it is locally finitely presented (i.e. every point possesses a neighborhood $U$, such that $F \mid U$ is the image of $O_X^n\mid U$ under some surjective morphism) and locally of finite presentation (i.e. for an open súbset $U \subseteq X$ and a surjective morphism $f: O_U^n \rightarrow F \mid U$ of sheaves the kernel of $f$ is locally finitely presented).
My question now is: Why are such sheaves especially interesting? Why might somebody be interested in studying them or what is the intuition behind them?
Every help will be appreciated.

REPLY [10 votes]: Let me complete Georges' good answer by this. A natural class of
object to study on manifolds are vector bundles. There are many reasons to be interested in them, but to name one, there are the examples of the tangent and cotangent bundles,
of which the sections are vector fields and differential forms, two fundamental classes
of geometric object on the manifold. Now vector bundles are essentially the same thing
as locally free coherent sheaves (to a bundle one attach its sheaf which to an one set $U$ attached the section of the bundle on $U$). This is a reason to study at least certain
coherent sheaves, namely the locally free ones. But it turn out that the category of locally free coherent sheaves has an important short-coming: it is not abelian. Certain morphisms
have trivial kernel and cockerel, but are not isomorphisms, like the multiplication by
a function which vanishes at exactly one point from the structural sheaf to itself.
So one has to work with a larger category, which is abelian, namely the category of coherent sheaves.<|endoftext|>
TITLE: No Exceptional Eigenvalues of Weight 1/2 Maass Forms on $\Gamma_0(4)$?
QUESTION [5 upvotes]: Some colleagues and I were wondering if there is a citation out there which shows there are no exceptional eigenvalues, $\lambda$, of classical weight 1/2 Maass forms on $\Gamma_0(4)$, which is to say $\lambda=1/4+t^2>1/4$, so $t \in \mathbb{R}_{>0}$. 
A resource where somebody has computed the lowest eigenvalue for this case, like is done in LMFDB for the weight zero case, would be optimal. We're not looking to go through the guts of analogous methods to compute this for ourselves, if we can help it.

REPLY [4 votes]: A weight $1/2$ Hecke-Maass form of eigenvalue $(1-s^2)/4$ on $\Gamma_0(4)$ has a Shimura lift to a weight $0$ even Hecke-Maass form of eigenvalue $1/4-s^2$ on $\Gamma_0(1)$, see e.g. the proof of Theorem 1.5 in Baruch-Mao: A generalized Kohnen-Zagier formula for Maass forms (manuscript here). So the database for weight $0$ forms furnishes the required information for weight $1/2$ forms.<|endoftext|>
TITLE: Why is it hard to prove that the Euler Mascheroni constant is irrational?
QUESTION [45 upvotes]: Philosophically why should proving that $\gamma$ is irrational (let alone transcendental) be so much harder  than proving $\pi$ or $e$ are irrational?

REPLY [80 votes]: Philosophically, there is essentially only one way to prove that a number is irrational/transcendental, which is to use the fact that there is no integer between 0 and 1.  That is, one assumes that the number in question is rational/algebraic, and constructs some quantity that can be shown to be bounded away from 0, less than 1, and also an integer.  To get these estimates, one typically needs some rapidly converging series expansion that is closely related the number of interest. For example the reason that Fourier's proof that $e$ is irrational is so simple is that we have a ready-made rapidly converging series $\sum 1/n!$ for $e$ that allows us to  construct an integer between 0 and 1 from the assumption that $e$ is rational.  As for numbers like $\pi$ or $e^\pi$, they basically piggyback on the rapidly converging series $\exp z = \sum z^n/n!$ because $\exp(i\pi) = -1$.
In general, the less obvious it is how to relate your number to a suitable rapidly converging series, the harder it will be to prove irrationality/transcendence.  Apéry's dramatic success with $\zeta(3)$ was based on the highly non-obvious rapidly converging series representation
$$    \zeta(3) = \frac{5}{2} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3\binom{2n}{n}}. $$
But $\zeta(5)$ doesn't yield to exactly the same method because there is ample numerical evidence that the obvious analogous series that you would conjecture for it after seeing the above identity does not exist.  Later progress on $\zeta(2n+1)$, particularly by Rivoal and Zudilin, relies on taking various subtle linear combinations of them in order to eventually deduce the nonexistent integer between 0 and 1, and so are only able to prove that at least one of a set of zeta values is irrational.
So could there be another proof that Euler missed out there, based on an elementary but non-obvious identity for $\gamma$ that has the right properties for the usual machinery to grind through?
Maybe, but currently there is no method in sight for relating $\gamma$ in the right way to a suitable rapidly converging series.  If you like, that is the "philosophical" reason why we're stuck.
By the way, I'd highly recommend Making Transcendence Transparent by Burger and Tubbs if you want an accessible treatment of transcendental number theory.  They do an excellent job of showing how the same basic ideas underlie all the results in the area, while introducing the technical complications one at a time in digestible chunks.<|endoftext|>
TITLE: Normal forms for homogeneous cubic polynomials in $\mathbb{R}[x_1, x_2, x_3]$
QUESTION [5 upvotes]: Is there a standard normal form for homogeneous cubic polynomials in $\mathbb{R}[x_1, x_2, x_3]$?  Or, put another way, is there a nice way to describe the orbit space of the natural (diagonal) action of $GL(3, \mathbb{R})$ on $Sym^3(\mathbb{R}^3)$?
(This is related to my previous question Normal form for trace-free real cubic forms in 3 variables under SO(3)-action?, but now I'm interested in the full $GL(3)$ action and not just the $SO(3)$ action, and I also don't want to impose the trace-free condition.)

REPLY [10 votes]: Hi, Jeanne!  It may help to have some geometric explanation of the normal forms over $\mathbb{R}$.  The standard account is this:
If the projective cubic curve $F(x_1,x_2,x_3)=0$ is nonsingular (the 'generic' case), it has exactly three real flexes, they are distinct and lie on a line.  One can make a linear change to make them lie on the line $x_1{+}x_2{+}x_3=0$ and have this line intersect the curve at the three points where some $x_i=0$ (so that these are the three flexes).  From this, one sees that one can make a real linear change of variables so that 
$$
F = {x_1}^3 + {x_2}^3 + {x_3}^3 + 6\sigma\ x_1x_2x_3\ ,
$$
where $\sigma\not=-\tfrac12$ is a real number.   (When $\sigma=-\tfrac12$, the above cubic factors as a line $x_1{+}x_2{+}x_3=0$ and an irreducible quadratic form.)
Enumerating the singular cases over the reals gets a little messy, but the final result is that when the curve is irreducible and singular there is exactly one singular point, which is necessarily real, and it is either a hyperbolic node, elliptic node, or a cusp. These have the normal forms 
$$
F = {x_2}^2x_3  - \epsilon\ {x_1}^2 x_3 - {x_1}^3
$$
where $\epsilon$ is $1$, $-1$, or $0$, respectively.
If the curve is the union of a line and a smooth quadric, i.e., $F = LQ$, where $L$ is linear and $Q$ is nonsingular (possibly without real points), you can put the quadric in normal form, $Q = {x_1}^2+{x_2}^2\pm{x_3}^2$ and then use the stabilizer group of the quadric to normalize $L$.  In the $+$ case, you can always rotate, using $\mathrm{SO}(3)$, to make $L=x_1{+}x_2{+}x_3$ (say).  In the $-$ case, there are three cases, and you can rotate, using $\mathrm{SO}(2,1)$ to make $L$ be one of $x_1$, $x_1{+}x_3$, or $x_3$.
Finally, if the curve is the union of three (complex) lines, it depends on whether the lines are all distinct or not and whether they are concurrent or not.  In the distinct case, you get that $F$ lies on the orbit of either $F=x_1x_2x_3$ (all real, not concurrent), $x_1x_2(x_1{+}x_2)$ (all real, concurrent), $x_1({x_2}^2{+}{x_3}^2)$ (two complex conjugate, not concurrent), or $x_1({x_1}^2{+}{x_2}^2)$ (two complex, concurrent).  If they are not all distinct, you get either $x_1{x_2}^2$ (two distinct) or ${x_1}^3$ (all the same).
These are the normal forms for all the nonzero cubics in $3$ variables.
Remark:  You also asked about the 'orbit space' of $\mathrm{GL}(3,\mathbb{R})$ acting on $\mathrm{Sym}^3(\mathbb{R}^3)$.  The naïve quotient, endowed with the quotient topology, is not very nice because it is highly non-Hausdorf, as every open neighborhood of the fixed point $0$ meets every orbit.  Even when you remove $0$, so that the remainder can be thought of as $\mathrm{PGL}(3,\mathbb{R})\simeq \mathrm{SL}(3,\mathbb{R})$ acting on $\mathbb{P}\bigl(\mathrm{Sym}^3(\mathbb{R}^3)\bigr)\simeq \mathbb{RP}^9$, the orbit space with the quotient topology is still non-Hausdorf.  The whole point of developing Geometric Invariant Theory (aka GIT) is to figure out the right way to remove the bad points so that the remaining 'good' points (aka, the 'semi-stable' points) will make a nice quotient space when you divide out by the action of the group.  I don't know how good this theory is in the real case, though.  In any case, the nonsingular cubics are all 'good' points for the moduli space.<|endoftext|>
TITLE: Random graphs nonisomorphic to unit distance graphs
QUESTION [5 upvotes]: I've encountered an interesting problem but can solve it only partially:

Prove that random graph $G\sim G\left(n,\frac cn\right)$, $c=const$, almost surely is isomorphic to some unit distance graph on a plane if $c$ is sufficiently small and almost surely won't be ismorphic to any unit distance graph on a plane if $c$ is sufficiently large.

I can prove only the first part (for $c<1$, connected components of $G$ will almost surely contain no more than 1 cycle, and it's easy to show that such $G$ can be represented as a unit distance graph).
Some precisions: the model for random graphs is the one of Erdös and Renyi ($G$ has $n$ vertices and each edge is present with probability $c/n$), and ``almost surely'' means that the probability of the event goes to $1$ when $n\to\infty$.
A unit distance graph is a graph that can be represented by point in the plane, with two points joined by an edge if and only if their are at unit distance one from the other.

REPLY [6 votes]: The almost sure asymptotic chromatic number of $G$ goes to $\infty$ with $c$, see for example the precise result by Achlioptas and Naor in Annals of Math. 2005.
The chromatic number of a unit-distance graph (and in fact of the whole plane) is bounded above by $7$, see e.g. the math coloring book by Soifer (this is simple: one colors an hexagonal tiling of carefully chosen side length).
These two facts end the proof of your problem.<|endoftext|>
TITLE: What fields can be used for an inner product space?
QUESTION [9 upvotes]: Math people:
The title is the question: What fields can be used for an inner product space?
This question has been discussed in Math Stack Exchange with no definitive resolution.  A similar question appeared here, and an answer was accepted, but someone pointed out a serious problem with the answer.  
I am using the standard definition of inner product, which includes $\langle \mathbf{x}, \mathbf{x} \rangle > 0$ for all non zero vectors $\mathbf{x}$.  
It seems to me that any field of prime characteristic does not make sense, because it does not have an order relation that respects addition.  It also seems to me that the field $\mathbb{F}$ can be any ordered field or any subfield of $\mathbb{C}$ that is stable under complex conjugation (for any non-algebraists like me, the word "stable" seems to be standard here. Anyone else would use the word "closed").  I do not know if any other fields are possible.  Of course, an ordered field may or may not be a subfield of $\mathbb{R}$.
It seems to be rare for people, even mathematicians, to use any field other than $\mathbb{R}$ or $\mathbb{C}$ for an inner product space.
Can anyone clear this up?
EDIT: someone alerted me to a Wikipedia page that addresses this question (http://en.wikipedia.org/wiki/Inner_product_space#Definition).  Let me quote the relevant part : 
"There are various technical reasons why it is necessary to restrict the basefield to $\mathbb{R}$ and $\mathbb{C}$ in the definition. Briefly, the basefield has to contain an ordered subfield[citation needed] (in order for non-negativity to make sense) and therefore has to have characteristic equal to 0 (since any ordered field has to have such characteristic). This immediately excludes finite fields. The basefield has to have additional structure, such as a distinguished automorphism. More generally any quadratically closed subfield of $\mathbb{R}$ or $\mathbb{C}$ will suffice for this purpose, e.g., the algebraic numbers$\ldots$"
The Wikipedia article fails to explain why the basefield has to have additional structure.  They do not define a "distinguished automorphism" or provide a link to a definition.  I am not an algebraist.  I Googled the term and I could not find a definition of "distinguished automorphism".  I did find links to papers and books that probably do contain a definition.  The article states it is "necessary" to restrict the basefield to $\mathbb{R}$ and $\mathbb{C}$ in the definition but then contradicts itself by at least suggesting that the basefield can be  any quadratically closed subfield of $\mathbb{R}$ or $\mathbb{C}$.  
Stefan (STack Exchange FAN)

REPLY [8 votes]: Of course if you insist on condition $\langle \mathbf{x},\mathbf{x}\rangle > 0$, and not merely $\langle \mathbf{x},\mathbf{x}\rangle \ne 0$, then you must have an order.  
Let $F$ be a formally real field.  Then
$$
\langle \mathbf{x}, \mathbf{y}\rangle = \sum_{j=1}^n x_j y_j
$$
can be a reasonable inner product on $F^n$.  According to an ordering for $F$ (indeed, any ordering, since there may be more than one) we have $\langle \mathbf{x}, \mathbf{x}\rangle > 0$ if $\mathbf x \ne \mathbf 0$.  
Another part that you quote is what would be required for metric completeness.    Do you want that? If $F$ is a proper subfield of $\mathbb R$, then even the one-dimensional space is not complete.  
Something weaker than completeness will be enough to carry out the Gram-Schmidt process.  It requires only that square-roots of $\langle \mathbf{x}, \mathbf{x}\rangle$ exist.<|endoftext|>
TITLE: Best bounds toward Serre's uniformity conjecture
QUESTION [13 upvotes]: If $E$ is a non-CM elliptic curve over $Q$, then it is a famous theorem of Serre
that there is some integer $M(E)$ such that for any prime $\ell > M(E)$, the image of the
Galois representations $\rho_{E,l}: G_{\mathbb Q} \rightarrow Gl_2(\mathbb F_\ell)$
on the $\ell$-torsion points of $E$ is surjective. If we define $M(E)$ as the smallest integer
having this property, then Serre's bounded uniformity conjecture is that $M(E)$ is bounded above by an absolute constant (41 perhaps) when $E$ varies over all non-CM-elliptic curve over $\mathbb Q$.
Let $N(E)$ be the product of all primes where $E$ has bad reduction.
My question is:

What are the current best bounds (if any) of $M(E)$ in terms of $N(E)$, both unconditionally and under GRH (the second being the case that interests me most)?

I kind of get lost in the immense literature on the subject. As is well-known, there are four types of proper maximal subgroups of $Gl_2(\mathbb F_\ell)$, which are (1) Borel subgroups,
(2) (resp. (3)) Normalizer of split (resp. non-split) Cartan subgroups, (4) exceptional ones (icsoahedral, dodecahedral, etc), so if $\rho_{E,\ell}$ fails to be surjective, it lands in a
subgroup of one of those type, and we can define four integers $M^i(E)$ for $i=1,..,4$
,as the smallest integer such that for $\ell$ larger that $M^i(E)$, $\rho_{E,\ell}$ does not
fall into a subgroup of type (i). Please tell me if I am not correct (I am troubled by Lemma 17 page 197 of this paper of Serre)
, but it is known
that $M^4(E)$,$M^1(E)$,$M^2(E)$ are bounded by an absolute constant (independent of $E$)
due to results of Serre, Mazur, and Bilu-Parent (respectively and in chronological order,
the last one being very recent). So the only problem that remains for Serre's
uniformity conjecture would be to bound $M^3(E)$ uniformly, and for my question to bound it at least in terms of $N(E)$.

REPLY [7 votes]: So I think I can now answer my own question about the best known bound under GRH
for M(E) in terms of N(E). The accepted answer of Chris refers to a recent paper by Cojoacru which in its introduction reviews quickly the history and cites Serre's result (in his paper on Chebotarev) as the best. This result is (theorem 22, under GRH)
$$ M(E) \ll \log N(E) (\log \log 2N(E))^3.$$
This result is deduced from an other one:
Theorem 21': If $E$ and $E'$ are two non-isogenous elliptic curves over $\mathbb Q$, which can be deduce from each other by torsion by a quadratic character, then there exists, under GRH, $$p \ll (\log N(E,E'))^2 (\log \log 2 N(E,E'))^6$$ such that $a_p \neq a'_p$. Here $N(E,E')$ is the products of prime of ramifications of $E$ or $E'$, and $a_p=1+p-|E(\mathbb F_p)|$, $a'_p=1+p-|E(\mathbb F_p)|$ as usual.
Without the hypothesis that $E$ is obtained from $E'$ by torsion, Serre proves the same result with the exponent 6 replaced by 12 (Theorem 21).
Now I say that those results of Serre are not optimal anymore. More precisely, the $\log \log$ factors can be removed (under GRH) in Theorem 21, 21', and 22.
The proof uses the modularity of $E$, $E'$, which replaces those curves by modular forms. One then can use the Rankin-Selberg result, which gives Theorem 21 and 21' without $\log \log$ factors, cf. Iwaniec-Kowalski, Proposition 5.22 page 118. Theorem 22 without the \log\log factors follows by the same argument that Theorem 22 followed from theorem 21' with the \log \log factor. So in short, we have under GRH
$$M(E) \ll \log N(E).$$
I learnt about this when I submitted an article improving (among many other things fortunately) the exponent of $\log \log$ in Serre's theorem 22. The referee told me that the $\log \log$ factor was not needed. I guess that when no modularity is available, say for elliptic curve over number fields, where you can prove similar results with similar methods, then the best results are still Serre-like, that is with a $\log \log$-factor (which my method allows to reduce the exponent of, but not to completely remove).<|endoftext|>
TITLE: Proof by contradiction in a topos
QUESTION [5 upvotes]: In a topos which is not Boolean topos, can we use proof by contradiction?

REPLY [11 votes]: It depends on what examples you have in mind when you say "proof by contradiction". This topic has come up a number of times recently at MO, but I recommend to your attention the useful blog post by Andrej Bauer, which explains that there is a subtle distinction to be made between "proof of negation" and "proof by contradiction". 
If the proposition to be proved is already of the form $\neg p$, then it may help to recall that $\neg p$ is (by definition) the weakest assumption one could make such its conjunction with $p$ entails falsity (in symbols, $x \leq \neg p$ iff $x \wedge p \leq 0$). This is true in intuitionistic logic as well as in classical logic. So a proof of a negated proposition $\neg p$ would quite properly begin, "suppose $p$, then ... contradiction". Many people call this a proof by contradiction, because the structure of the argument-phrasing looks just like any old proof by contradiction. 
An example of this is Cantor's theorem (that there is no surjection from a set to its power set, or $\neg$ "there exists a surjection..."). This can be formulated in any topos and is true in any topos, Boolean or not. 
(If this helps, notice that in intuitionistic logic, we have that $\neg p$ is equivalent to $\neg \neg \neg p$: a negated proposition is always equivalent to its double negation.) 
But contrast this with for example the Hahn-Banach theorem: every locally convex topological vector space admits a continuous functional to the ground field. This proposition, which is not in negated form, is a prime example of something which has no constructive proof. A typical method of proof would be something like "by Zorn's lemma, there is a maximal closed subspace that admits such a continuous functional, and suppose this were not the whole space" and eventually derive a contradiction. This type of reasoning is not valid in a general topos. 
For another example, consider "$\sqrt{2}$ is irrational". This is a negative proposition: "$\neg (\exists p, q \in \mathbb{Z}_+ \; p^2 = 2 q^2)$". The usual arithmetic proofs are valid in any topos.

REPLY [3 votes]: No.
There is no need to say any more than that, since the answer is in the question, except that MathOverflow will not let me submit something with so few characters.<|endoftext|>
TITLE: Automorphisms of Plane Curves are linear
QUESTION [8 upvotes]: Hello guys,
I need to ask you for a proof or reference to the fact that: 
Every automorphism of any non-singular algebraic curve in $\mathbb{P}^2(\mathbb{C})$ of genus $g\geq 2$ is linear, i.e., can be viewed as an element of $\mbox{PSL}_3(\mathbb{C})$.
I saw the proof for higher dimensional hypersurfaces (Matsumura-Monsky) but they use the fact that the Picard group of the variety is torsion-free (and in this case is false, since $\mbox{Pic}^0(C)$ is an abelian variety of positive dimension).
Thanks!

REPLY [11 votes]: Let $C\subset\mathbb{P}^3$ be a smooth curve of degree $d$ and genus $g = \frac{(d-1)(d-2)}{2}\geq 4$. Let us consider the divisor $D = \mathcal{O}_C(1)$.

Consider the exact sequence 
$$0\mapsto \mathcal{I}_C\rightarrow \mathcal{O}_{\mathbb{P}^2}\rightarrow\mathcal{O}_C\mapsto 0,$$
that is 
$$0\mapsto \mathcal{O}_{\mathbb{P}^2}(-d)\rightarrow \mathcal{O}_{\mathbb{P}^2}\rightarrow\mathcal{O}_C\mapsto 0.$$
Twisting by $\mathcal{O}_{\mathbb{P}^2}(1)$ we get
$$0\mapsto \mathcal{O}_{\mathbb{P}^2}(-d+1)\rightarrow \mathcal{O}_{\mathbb{P}^2}(1)\rightarrow\mathcal{O}_C(1)\mapsto 0.$$
Finally, taking cohomology we have $H^{0}(C,\mathcal{O}_C(1))\cong H^{0}(\mathbb{P}^2,\mathcal{O}_{\mathbb{P}^2}(1))$. Therefore the linear system cut out on $C$ be the lines in $\mathbb{P}^2$ is complete, and $h^0(C,\mathcal{O}_C(1)) = 3$.
Let $A$ be another effective divisor on $C$ such that $deg(A) = d$ and $h^{0}(C,A)=3$. Then $deg(D-A) = 0$ and $h^{0}(C,D-A)\neq 0$. Therefore $D-A$ is linearly equivalent to $\mathcal{O}_C$, and $A$ is linearly equivalent to $D$.
We conclude that $D = \mathcal{O}_C(1)$ is the unique effective divisor of degree $d$ and with $h^{0}(C,D) = 3$.
Let $\phi:C\rightarrow C$ be an automorphism. Then $\phi^{*}\mathcal{O}_C(1)\cong \mathcal{O_C}(1)$. Therefore $\phi$ acts on the sections of $\mathcal{O}_{\mathbb{P}^2}(1)$. Since $\mathbb{P}^2 = \mathbb{P}(H^{0}(\mathbb{P}^2,\mathcal{O}_{\mathbb{P}^2}(1))) = \mathbb{P}(H^{0}(C,\mathcal{O}_{C}(1)))$ we conclude that $\phi$ in indeed the restriction to $C$ of a linear automorphism of $\mathbb{P}^2$.

Just a couple of remarks:

This is true for curves of degree $d=1,2$ as well. For a line it is trivial. If $C$ is a smooth conic an automorphism of $\mathbb{P}^2$ has to map five points of $C$ to other five points of $C$ in order to stabilize $C$. This group has dimension $dim(Aut(\mathbb{P}^2))-5 = 3$. On the other and $Aut(C) = PGL(2)$ has dimension $3$ as well. This is clearly false for smooth plane cubic. Indeed one needs nine points to stabilize a cubic. But $C\subseteq Aut(C)$.
If $X\subset\mathbb{P}^n$, with $n\geq 3$, is a smooth hypersurface of degree $d$ and $d\neq n+1$ (i.e. $\omega_X$ is not trivial) then any automorphism of $X$ is induced by an automorphism of $\mathbb{P}^n$. This is essentialy due to the fact that such a hypersurface is sub-canonical and the Picard group is torsion free. This fact is indeed true for any Calabi-Yau hypersurface with $(n,d)\neq (3,4)$.<|endoftext|>
TITLE: A simple and good reference about solitons
QUESTION [6 upvotes]: I want to study gradient Ricci solitons. Can anyone help me to find a simple and good reference about solitons and their applications?
Thanks

REPLY [2 votes]: As far as gradient Ricci solitons are concerned, this presentation, the first half of which gives an introduction to the subject, could be helpful.<|endoftext|>
TITLE: Is every submartingale a convex function of a martingale?
QUESTION [7 upvotes]: Is every submartingale a convex function of a martingale?

REPLY [7 votes]: Not every. Every convex function is either bounded below or monotonic. The first kind cannot produce any submartingale that is not bounded below. The second kind cannot produce any submartingale that, at some point in time, might be certain to increase. It is easy to produce a submartingale satisfying both conditions.
If it is bounded below, $|x|-B$ will do. In other words, given a positive submartingale, we can add signs to make it a martingale: at each step, choose the probability of switching signs to cancel out the growth in expectation.<|endoftext|>
TITLE: Does anyone know where I can get a copy of Gaunce Lewis's thesis?
QUESTION [5 upvotes]: Tracing through a trail of references I found myself needing something proven in Appendix A of "The Stable Category and Generalized Thom Spectra" by Gaunce Lewis (I believe this was his thesis at UChicago under Peter May, from 1978). My library doesn't seem to have a copy and I can't find anything online. Does anyone know where I could find this?
Before anyone suggests it, I'll remark that I can't contact Lewis himself because he passed away several years ago. I guess I'm hoping Peter May will come on here and know where a copy can be found.

REPLY [4 votes]: My apologies to everyone, and especially to Markus.  He got me to get this scanned, and I sent him a copy; I also promised to put it on my web page, but I only just now did so: http://www.math.uchicago.edu/~may/MISC/GaunceApp.pdf
Alex, that would be nice, and drop in and say hello some time.
Unfortunately, I do not actually have a copy of Gaunce's thesis, so what is posted is only Appendix A, ``Compactly generated spaces''.  Most, maybe all, of the rest evolved into much of Springer Lecture Notes Vol 1213, http://www.math.uchicago.edu/~may/BOOKS/equi.pdf
Gaunce's thesis was 1978, and SLN 1213 only appeared in 1986.  We did a ton of work in that time, and I doubt that there would be much besides Appendix A that is of current interest.
However, Appendix A is to my mind (and the minds of others) the nicest and most informative treatment of compactly generated spaces, and that part is not at all duplicated in SLN 1213.<|endoftext|>
TITLE: Conley Theorem (or fundamental theorem of dynamical systems)
QUESTION [9 upvotes]: Notations:  

$\mathcal{R}(f)$  denotes the chain recurrent set of $f$
$NW(f)$ denotes the non wandering set of $f$
$R(f)$  denotes the recurrent set of $f$  ($x: x\in \omega(x)$)

Given compact metric space X and homeomorphism  $f: X \rightarrow X$,
 $g: X \rightarrow \mathbb{R}$ is a complete Lyapunov function for $f$ if:
$\forall \ p \notin \mathcal{R}(f)$, $\ g(f(p)) < g(p)$
$\forall \ p, q \in \mathcal{R}(f)$, $\ g(p) = g(q)$ iff $ p \sim q$
$g(\mathcal{R}(f))$ is compact and nowhere dense in  $ \mathbb{R}$
Conley's Theorem (or fundamental theorem of dynamical systems):
Complete Lyapunov function exists for any homeomorphisms on compact metric spaces.
My Question:  I'd like to understand why this theorem is called fundemental theorem of dynamical systems, It seems to me it says that the whole interesting dynamic $ f $ is contained in the chain recurrent set of $f.$ this was what I was told, but it is not very clear to me.
An example to explain my question: In the case of all non-wandering of $ f $ we prove that given any neighborhood $U$ of $NW(f)$ there is a uniform time $T$ for which
$$
\text{Card} \{ k: f^k(x)\notin U \} \leq T
$$
In this sense it seems clear that the non wandering set contains much of the interesting dynamics of $ f $.  
However in the case of the chain recurrent set is not clear to me nothing that occurs in the previous example.
So like an explanation of why the whole interesting dynamic $ f $ is concentrated in the chain recurrent, and wanted to better understand why the theorem of Conley is called "fundamental".
Another question:
Would an example of a dynamic system where $NW(f)\setminus \overline{R(f)}$  is not empty, or at least the idea of what kind of dynamic phenomenon can produce these point.

REPLY [2 votes]: There is a classical example with $NW(f)\backslash \overline{R(f)}\neq\emptyset$, the so called Bowen's eye-like attractor (see the paper by Baladi, Bonatti and Bernard: Abnormal Escape Rates from Nonuniformly Hyperbolic Sets):

Every point on the dark curve is nonwandering, but only the two corners are recurrent (in fact these two are fixed).
Why is the theorem called Fundamental? One reason is that it is the 'correct' setting of $C^1$ stability conjecture (see the discuss here).
Theorem: Let $f$ be a diffeomorphism on a closed manifold $M$. Then the following are equivalent: 
1. the map $f$ is  structurally stable;
2. $\mathcal{R}(f)$ is hyperbolic;
3. $f$ is $\mathcal{R}$-stable.
This modern version is very succinctly comparing to the version using the nonwandering set, which involves with no-cycle condition and transversality condition.
Also I copied a few words from the paper mentioned in Barry's comment:
'The theorem is fundamental in the sense that it deals with the basic question of the field. It is also fundamental in that it encompasses such big ideas in such a small, concise statement.' 
Compared to Fundamental Theorem of Arithmetic and Fundamental Theorem of Algebra, Norton wrote: 
'the space on which the dynamics take place, can be decomposed uniquely into its basic dynamical parts: points whose dynamics can be described as exhibiting a particular type of recurrence, and points which proceed in a gradient-like fashion.'<|endoftext|>
TITLE: M-matrix plus S-matrix is P-matrix?
QUESTION [9 upvotes]: I am trying to prove that a mapping has a unique fixed-point by showing that its Jacobian is a P-matrix. In this particular case the Jacobian can be decomposed as the sum of two matrices and I would like to show the following result:

Let $M \in \mathbb R^{n \times n}$ be a M-matrix and $S \in \mathbb R^{n \times n}$ be a S-matrix (positive definite matrix) with non-negative entries. Is it the case that $M + S$ is a P-matrix?

The result holds trivially for $n=1,2$, but I was wondering if it holds for arbitrary $n$. I generated millions of random matrices with $n \le 10$ and the result seems to be true. I would truly appreciate any comments, references or help!
Some definitions
A matrix $M \in \mathbb R^{n \times n}$ is a M-matrix if (i) all off-diagonal entries are less than or equal to zero, and (ii) all principal minors of $M$ are positive. A matrix $P \in \mathbb R^{n \times n}$ is a P-matrix if all principal minors of $P$ are positive. Horn Johnson's Topics in Matrix Analysis (1991, pp. 112-125) discusses various equivalent definitions for these matrices.
Interestingly, all M-matrices are P-matrices and all S-matrices are P-matrices, but P-matrices are not close under addition. For example, $A = \begin{bmatrix} \frac 1 2 & -1 \\ 0 & \frac 1 2 \end{bmatrix}$ and $B = \begin{bmatrix} \frac 1 2 & 0 \\ -1 & \frac 1 2 \end{bmatrix}$ are both P-matrices but the sum $A+B$ is not. This example is taken from Parthasarathy's On Global Univalence Theorems (1983, p. 15).
Some Background
In many equilibrium models one proves the uniqueness of an equilibrium by showing that an equation $F(x) = 0$ has at most one solution. In many cases the Jacobian of $F$ can be decomposed as the sum of a positive diagonal matrix and M- or P-matrix, which is again a P-matrix. Then one invokes the univalence result of Gale and Nikaido's The Jacobian matrix and global univalence of mappings (1965) to show that the mapping admits at most one solution. The previous result would allow one to prove uniqueness in more general mappings and may be of interest to the community.

REPLY [6 votes]: This is not true in general.
$\left(\begin{array}{ccc} 4 & 0 & -16  \\ 2 & 4 & 0  \\ 0 & 2 & 4 \end{array}\right)= \left(\begin{array}{ccc} 1 & -2 & -16  \\ 0 & 1 & -2  \\ 0 & 0 & 1 \end{array}\right) +\left(\begin{array}{ccc} 3 & 2 & 0  \\ 2 & 3 & 2  \\ 0 & 2  & 3 \end{array}\right)$
A matrix with determinant $0$ is $M+S$. (Thanks to S. Sra for the correction. I multiplied by $4$ so everything's an integer now.)<|endoftext|>
TITLE: Does there exist a polar decomposition of matrices over finite fields? 
QUESTION [6 upvotes]: There exists a polar decomposition for matrices over the reals.
What I would like to know is if an analog has been shown for groups of matrices over finite fields. If not, it would be great to get some feedback as to how to proceed, or some reasons why it may not exist.
Specifically, I'm interested in decomposing invertible matrices over $\mathbb{F}_q$ as a product $x = s.o$ where $s$ is symmetric and $o$ is orthogonal.

REPLY [3 votes]: The answer is no for $\mathbb F_2$. A minimal invertible counterexample is $M = \left[\begin{matrix}1 & 1 & 0\\0 & 1 & 0\\0 & 1 & 1\end{matrix}\right]
$. This can be verified by observing that all orthogonal $3 \times 3$ matrices over $\mathbb F_2$ are permutation matrices. So if you multiply $M$ by each of the six permutation matrices, you never get a symmetric matrix.
The fact that $O_3(\mathbb F_2)$ only contains permutation matrices can be verified using the "inner-product preserving" property of orthogonal matrices.
Computational experiments show that a counterexample exists for $\mathbb F_3$ of size $3 \times 3$. For $\mathbb F_5$, there is a $2 \times 2$ counterexample.

Regarding alternatives to polar decomposition, in Gutin - Generalisations of singular value decomposition to dual-numbered matrices [1] I introduce an analogue of SVD for the ring of dual numbers equipped with the involution $a + b \varepsilon \mapsto a - b\varepsilon$. This analogue of SVD exists for every dual-numbered matrix, while the polar decomposition (its obvious generalisation) does not. What this means is that when generalising SVD it might be profitable to allow the matrix $S$ in $USV^*$ to not necessarily be Hermitian.
[1] - https://www.tandfonline.com/doi/full/10.1080/03081087.2021.1903830, but this has some formatting errors, so instead see https://arxiv.org/abs/2007.09693<|endoftext|>
TITLE: Quantum algorithms for dummies
QUESTION [13 upvotes]: I want to try my hand at designing quantum algorithms to solve certain problems.  I feel like I understand (for example) how Grover's algorithm and Shor's algorithm work, and I'm excited to apply the various "computing tricks" that aren't available in classical computing.  Unfortunately, my knowledge of such tricks is limited to their application in the few famous algorithms, and I'm not at all sure how to legitimately apply them in general.
Are there any resources available that will quickly bring me up to speed?

REPLY [4 votes]: I think this book is good for beginning:
"An introduction to quantum computing algorithms" by Pittenger A.O.
Also, the course by Prof. $Vazirani$ on $edx$ "(www.edx.org)" is good way for learning the necessary material in this way. 
Since, there are two main algorithms and other algorithms uses these basic algorithms, you can understand deeply these two quantum algorithms and modify your problem again to find an approach. These two basic and important algorithms are Grover's algorithm (for searching problem) and Shor's algorithm (for factorization). Also, Simons algorithm is good for studying. But, in general, you must learn Fourier sampling and superposition and the quantum gates and their works.<|endoftext|>
TITLE: The Kunen inconsistency and definable classes
QUESTION [6 upvotes]: There is a tension between (1) interpreting proper class talk in set theory as talk about first-order formulas and satisfaction; and (2) taking it to be an interesting and non-trivial result that there is no (non-trivial) elementary embedding from V into V and/or taking it to be an open question whether there can be such an e.e. in the absence of choice. Basically, there is a very simple proof that there can be no definable e.e. from V into V (see Suzuki (1999)). 
This tension was recently highlighted by Hamkins, Kirmayer, and Perlmutter (2012) (and pointed out here and here). There, the resolution was to give up on (1), since accepting it "does not convey the full power of the [Kunen's] theorem" (p. 1873). But this is perhaps the only place I've seen this issue addressed. For instance, Kanamori seems to hold both (1) and (2) in  The Higher Infinite: "By “class” in the ZFC context is meant definable class,... $x \in M$ is merely [a] facon de parler" (p. 33); and "[t]he following unresolved question [i.e. whether there could be an e.e. from V into V in the absence of choice] is therefore of foundational interest" (p. 324). 
My question is: how do other set theorists prefer to resolve this tension? 
Hamkins, J., Kirmayer, G., Perlmutter, N. (2012) ``Generalizations of the Kunen inconsistency". Annals of Pure and Applied Logic, 163, 1872–1890.
Suzuki, A. (1999) No elementary embedding from V into V is definable from parameters. Journal of Symbolic Logic 64, 1591-1594.

REPLY [10 votes]: My perspective on this issue is that there are a variety of ways to take the claim of the Kunen inconsistency, and we needn't pick a particular one as the only right one. Rather, we gain a fuller perspective of the result by understanding the full robust context including all of the interpretations.

Kunen proved his result in Kelly-Morse set theory, in large part in order that he could formalize what it means for a class function $j:V\to V$ to be (fully) elementary. In KM, we can prove that that there is a satisfaction class, a truth predicate for first-order truth, and with this class (which is definable) one can express the elementarity of $j$ as a single second-order assertion.
Meanwhile, using the observation (Gaifman) that any cofinal $\Sigma_1$-elementary embedding is $\Sigma_n$-elementary for any meta-theoretic natural number $n$, we can formalize the result in GBC as the claim that no class $j$ is a nontrivial cofinal $\Sigma_1$-elementary embedding. Thus, this kind of elementarity of $j$ becomes expressible as a first-order assertion about $j$.
We don't actually need full GBC, since for example global choice is not used, but only the usual AC for sets, and so this argument can be formalized in GB+AC. 
But actually, we don't need the full second-order part of GB, but only the ability to refer to the class $j$. So we can formalize the argument in $\text{ZFC}(j)$, the theory using ZFC where the axioms of replacement is allowed to use formulas in which the class $j$ appears. (But we only insist on elementarity of $j$ in the language without $j$.) This theory is used and suffices to show, for example, that the supremum of the critical sequence $\lambda=\sup_n\kappa_n$ exists. 
If one intends to rule out only definable class embeddings $j$, that is, ones which are classes in the ZFC sense of being first-order definable from set parameters, then as you mentioned, there is an easy argument ruling them out, and this argument does not use AC.  I do not know any set theorist, however, who takes this result as an answer to the question of whether one can prove the Kunen inconsistency in ZF. Rather, this example reveals the issues of formalization, and shows us that it may be important to take more care in our formal treatment of the result. 
Meanwhile, a purely first-order version of the Kunen inconsistency is formalizable in ZFC, with no talk of classes of any kind, as the claim that there is no nontrivial $j:V_{\lambda+2}\to V_{\lambda+2}$ for any $\lambda$. This version still uses AC, and it is open in ZF. It avoids the set/class issues underlying your question by noting that the Kunen inconsistency proof establishes more by restricting to $V_{\lambda+2}$. This set version of the result implies the full result in any set theory capable of showing that a purported class $j$ must have a closure point $\lambda$.
The wholeness axiom gets around the issue of the previous point by stating the theory ZFC + "$j:V\to V$ is nontrivial and elementary in the language with a function symbol for $j$. Elementarity is expressed by the scheme $\forall x[\varphi(x)\iff \varphi(j(x))]$. 
Various weakenings and strengthenings of the wholeness axiom are realized by making further claims about $j$, such as whether it has a critical point, whether it moves an ordinal, etc. Also, one can make claims about the extent to which $j$ may appear in the ZFC axioms. Officially, $j$ is allowed in the separation axiom but not in replacement, and so models of WA are not able to prove the supremum of the critical sequence exists. 
If one uses merely the model-theoretic concept of embedding, one would be considering $j:V\to V$ for which $x\in y\iff j(x)\in j(y)$. But now the point is that ZFC proves that there are nontrivial embeddings. For example, we can inductively define $j(y)=\{j(x)\mid x\in y\}\cup\{\{\emptyset,y\}\}$, and prove that this is a nontrivial embedding $j:V\to V$. (See my paper Every countable model of set theory embeds into its own constructible universe, to appear in the JML, for more information.) 

I prefer to understand the Kunen inconsistency in the rich context of all these results, rather than pick just one perspective and say that that perspective is the right one.<|endoftext|>
TITLE: elliptic curve with a degree 2 isogeny to itself?
QUESTION [5 upvotes]: I've come across the following question, which I think must be easy for experts: is there a complex elliptic curve $E$ with an isogeny of degree 2 to itself?
Of course one can  ask the same question for isogenies whose degree is not a square, or for higher dimensional abelian varieties etc.

REPLY [13 votes]: Expanding on Francois's answer, $E$ has an endomorphism of degree 2 if and only if its endomorphism ring $R=\operatorname{End}(E)$, which is an order in an imaginary quadratic field, has an element of norm 2. There are exactly three such orders, namely $\mathbb{Z}[i]$, $\mathbb{Z}[\sqrt{-2}]$, and $\mathbb{Z}[(1+\sqrt{-7})/2]$. So up to isomorphism over $\overline{\mathbb{Q}}$, there are exactly three elliptic curves with endomorphisms of degree 2. Equations for these curves and their degree 2 endomorphism are given in Advanced Topics in the Arithmetic of Elliptic Curves, Proposition II.2.3.1.
There are similarly only finitely many curves with a higher degree cyclic isogeny of fixed degree $d$. Using Velu's formulas, one could probably write them all down for small values of $d$.

REPLY [9 votes]: Yes, but the elliptic curve needs to have complex multiplication, since the multiplication-by-$n$ map has degree $n^2$. For an explicit example, you can take $E=\mathbf{C}/(\mathbf{Z}+i\mathbf{Z})$ with the isogeny being multiplication by $1+i$.<|endoftext|>
TITLE: Harish-Chandra modules of $\mathrm{PSL}_2(\mathbb{R})$
QUESTION [6 upvotes]: I am sorry to bother you with this question but I can't figure out this myself (and Mathematics Stack Exchange didn't help).
Is the category of Harish-Chandra Modules of $PSL_2(\mathbb{R})$ equivalent to the category of Harish-Chandra Modules of $SL_2(\mathbb{R})$ with even $K$-types?
The motivation of this question comes from the finite-dimensional case,   the finite-dimensional $PSL_2(\mathbb{C})$-representations being exactly the $SL_2(\mathbb{C})$-representations where $\{\pm I\}$ acts trivially.

REPLY [6 votes]: Yes, the central character of the even $K$-types is trivial, and of the odd ones is the sign character.
Also, this can easily be seen from the classification of irreducible representation on Hilbert spaces.
More generally, the category of rep on Hilbert spaces of a locally compact group $G$ with trivial central character coincides with that of the rep theory on Hilbert spaces of the locally compact group $G/Z$, here $Z$ being the center of $G$. So although, the categories are not semisimple, one can always decompose with respect to the center. 
Quote from Wikipedia: In 1973, Lepowsky showed that any irreducible (g,K)-module X is isomorphic to the Harish-Chandra module of an irreducible representation of G on a Hilbert space....(http://en.wikipedia.org/wiki/Harish-Chandra_module)<|endoftext|>
TITLE: Where do the product expansions of modular forms come from?
QUESTION [8 upvotes]: It is well-known that many modular forms can be expressed as infinite products. For instance, the most famous one is probably the expansion
$$\Delta(q) = q \prod_{n=1}^\infty (1-q^n)^{24}$$
for the discriminant cusp form of weight $12$ and level $1$. Another example is the cusp form of weight $2$ and level $11$
$$f(q) = q\prod_{n=1}^\infty(1-q^n)^2(1-q^{11n})^2,$$
which is attached to the elliptic curve $X_0(11) : y^2-y = x^3-x^2$. Such product expansions can be derived from the product expansion of the Dedekind $\eta$ function, by taking suitable combinations.
But why should such product expansions exist? Is there a reason to expect that they should exist, say, from the point of view of Galois representations?

REPLY [2 votes]: Many "natural" examples of automorphic infinite products (also known as Borcherds products) can be explained using the singular theta lift of Harvey-Moore and Borcherds.  These examples have the property that the exponents of the product are coefficients of a modular form.
For example, the 24 in the exponents of the product formula for $\Delta$ can be matched with the positive-power coefficients of $12 \theta(0;\tau) = 12 + 24q^{1/2} + 24 q^2 + 24 q^{9/2} + \cdots$ in the following way: Shimura gave a correspondence between forms of weight $k + 1/2$ on $\Gamma_0(4)$ (satisfying some conditions) and forms of weight $2k$ for $SL_2(\mathbb{Z})$ (for some character), taking $f(\tau) = \sum_n c(n)q^n$ to $-c(0)B_k/2k + \sum_n q^n \sum_{d|n} d^{k-1} c(n^2/d^2)$.  While this really works best for $k$ positive and even, you can remove the infinite constant term when $k=0$ to get something almost modular.  Applying this to $12 \theta$, you naturally get $\log (\Delta/q)$.
In other words, products like $\Delta$ arise by exponentiation of a Howe theta lift (of which Shimura's correspondence is a special case), although the lift may need to be regularized.  For example, the Koike-Norton-Zagier formula:
$$ j(\sigma) - j(\tau) = (p^{-1} - q^{-1}) \prod_{m,n>0} (1-p^m q^n)^{c(mn)}$$
arises as a lift of $j(\tau) - 744 = \sum_n c(n) q^n$ to $O(2,2)$, and the usual method of lifting involves a divergent integral of $(j-744)\theta$ over a fundamental domain of $SL_2(\mathbb{Z})$.
In general, I don't think these products are very naturally related to Galois representations.  It is easy to lift forms like $-12\theta$ to get products like $1/\Delta$ of negative weight, which are somewhat invisible to the Langlands program (as far as I know).  Instead, the products tend to show up naturally in subjects related to string theory, like the representation theory of infinite dimensional Lie algebras.  For example, the product expansion of $1/\Delta$ gives the partition function of free bosons propagating in 24-dimensional space, and the Koike-Norton-Zagier formula is the Weyl denominator formula for the Monster Lie algebra.  Borcherds has some expository overviews of this subject on his web page, e.g., number 28: "Automorphic forms and Lie algebras".<|endoftext|>
TITLE: Does the paper "On the cobordism ring $\Omega_*$ and a complex analogue II" exist?
QUESTION [13 upvotes]: I've been investigating the Milnor hypersurfaces, and every reference seems to point to the paper by Milnor, "On the cobordism ring $\Omega_*$ and a complex analogue II". Despite my best efforts, I cannot seem to find it. Was this paper ever published? If not, is there a draft that is available?

REPLY [23 votes]: Here's what John Milnor writes about this, in his collected works:

The projected Part II of this paper
  was never written. In fact I am
  chagrined to discover that I have
  never published any details about some
  of the announced results which were
  intended to appear in it. I was very grateful when Thom gave me permission to reprint his 1959 Bourbaki lecture, "Travaux de Milnor sur le cobordism", which gives a better account of this work than anything which I have published. 

(page 249 of Part III of the Collected Papers of John Milnor, AMS 2007).
The paper by René Thom which Milnor mentions can be accessed here.<|endoftext|>
TITLE: Seeking errata for Berger-Moerdijk Axiomatic Homotopy Theory for Operads
QUESTION [10 upvotes]: The paper I'm referring to can be found here. It came out in 2003. I've been told by professors before that I should be careful relying on things from this paper, because it contained errors.

Is there some place the errors in this paper are enumerated?

An error in a different paper of the authors (but related to this paper) was pointed out and corrected here. Is this the only known error in the paper?
My interest is in building model structures on categories of algebras over an operad. For a long time I thought there weren't any errors in that part of the paper. However, today I realized that in Remark 4.2 they claim that Prop. 4.1 implies every topological operad $P$ is admissible, i.e. $P$-alg inherits a model structure. This makes me suspicious. It would say for one thing that commutative monoids in Top form a model category, i.e. it would give us a model structure on a subcategory of Top where every object is a product of Eilenberg-Maclane spaces. Does that work?

REPLY [9 votes]: The only errors that I am aware of in this paper are the following:

A $G$ missing in the statement of Lemma 5.10 (pointed out by the authors after Lemma 2.5.3 in [1])
A small mistake in the proof of Proposition 5.1 (also pointed out by the authors and fixed in the Appendix of [2])

[1] I. Moerdijk and C. Berger, The Boardman-Vogt resolution of operads in monoidal model categories, Topology 45 (2006), 807-849.
[2] I. Moerdijk and C. Berger, On the derived category of an algebra over an operad, Georgian Math. J. 16 (2009), 13-28.<|endoftext|>
TITLE: Decidability of equality of expressions built using 1,+,-,*,/,^
QUESTION [27 upvotes]: Consider expressions built using number $1$, arithmetical operators $+, -, *, /$ and exponentiation ^ (in case of multiple values, the principal value is assumed, the same way as it implemented in Power function in Mathematica).
Is it a decidable problem to check if such an expression is zero?
If so, could you please point me to an algorithm that can solve this problem?
Update: I found a reference to Richardson's Theorem, that establishes undecidablity of equality in a wider set of expressions, in particular, including the logarithm and absolute value functions.

REPLY [8 votes]: The equational theory of $\langle {\bf N}, 0, 1, +, \times, \uparrow\rangle$ is decidable, but not finitely axiomatizable.<|endoftext|>
TITLE: Does an existence of large cardinals have implications in number theory or combinatorics?
QUESTION [44 upvotes]: Does an existence of large cardinals have implications in more down-to-earth fields like number theory, finite combinatorics, graph theory, Ramsey theory or computability theory? Are there any interesting theorems in these areas that can be proved based on assumptions of existence of certain large cardinals? Or they are too far to change anything here?
I know that existence of certain cardinals implies consistency of some axiomatic systems of the set theory, which in turn can be expressed as a statement about certain huge Diophantine equations having no solutions, but it looks unlikely that we could use these Diophantine equations for anything else.

REPLY [12 votes]: Applications of Friedman's Jump-free theorem (the proof of which requires large cardinals) to distance functions in lattice graphs is on his website:  
https://u.osu.edu/friedman.8/foundational-adventures/downloadable-manuscripts/
It is item 11: Applications of Large Cardinals to Graph Theory, October 23, 1997.
This remarkable paper by Friedman arose when a homework problem presented in an upper division undergraduate course at UCSD-CSE (solvable by a version of Ramsey's theorem) was extended slightly and no one could solve it (including the instructor and other faculty members). In January 1997, Friedman gave a lecture at UCSD where he presented his jump-free theorem.  It was just what we needed to solve the homework problem!
Another extension of this homework problem has been shown by Friedman (in the above paper) to itself require large cardinals and is equivalent to the jump-free theorem.
These problems can be viewed as a "toy versions" of the string theory landscape or multiverse concept.
A fanciful presentation for high school level students showing how these simply stated but hard problems arise is at
http://cseweb.ucsd.edu/~gill/MultUnivSite/<|endoftext|>
TITLE: Cuspidal automorphic representations as the space of $K$-finite vectors in a unitary cuspidal automorphic representation.
QUESTION [6 upvotes]: The definition I know of for a cuspidal automorphic representation of, say, $G=\mathrm{GL}_2$ over a number field $F$ (relative to a choice of compact open subgroup $K_f$ of $G(\mathbf{A}_F^\infty)$ and a maximal compact subgroup $K_\infty$ of $G(F\otimes_\mathbf{Q}\mathbf{R})$) is: an irreducible subquotient (equivalently subspace) of the space of cuspidal automorphic forms for $G$ over $F$. Since the full group $G(\mathbf{A}_F)$ does not preserve the property of $K=K_fK_\infty$-finiteness, "subrepresentation" here really means a subspace which is simultaneously a $G(\mathbf{A}_F^\infty)$-submodule and a $(\mathfrak{g}_\infty,K_\infty)$-submodule (and this can be phrased more concisely as a submodule for the appropriate global Hecke algebra). Now, I gather that there is a more ``analytic" theory, where one actually has a representation of $G(\mathbf{A}_F)$ on a Hilbert space. My first question is: 
Are these theories equivalent in some sense? 
I should make this more precise. I know that any cuspidal automorphic representation in the first sense is unitarizable. If I were to take the completion, would I get a unitary (admissible) Hilbert space representation of $G(\mathbf{A}^\infty)$ whose space of $K$-finite vectors was isomorphic to the representation with which I started?
I also know that two irreducible unitary admissible representations of, say, $\mathrm{GL}_2(\mathbf{R})$, are unitarily equivalent if and only if they are infinitesimally equivalent in the sense of having isomorphic associated $(\mathfrak{gl}_2(\mathbf{R}),\mathrm{O}_2(\mathbf{R}))$-modules (the spaces of $K$-finite vectors). My second question is:
Does this extend to unitary cuspidal automorphic representations (which I guess would be defined as irreducible admissible subrepresentations of the appropriately defined space of cusp forms in $L^2(G(F)\setminus G(\mathbf{A}_F),\omega)$, $\omega$ being the unitary central character)?
That is, are two unitary cuspidal automorphic representations isomorphic if and only if their spaces of $K$-finite vectors are isomorphic as modules for the Hecke algebras? And, out of curiosity:
If the answer to my second question is 'yes,' is the same true with $G$ an arbitrary connected reductive group of a number field?
The reason I ask is because, in some of the literature, it seems like both viewpoints are being used, sometimes at the same time, sometimes implicitly. So I would like to know if I can freely pass back and forth between them. 
Thanks!

REPLY [7 votes]: Yes. All these viewpoints are equivalent. We write "rep" for irreducible representation.
Edit on request:
An analytic cuspidal rep is subrepresentation $\pi$ of $L_0^2(G(F) \backslash G(A), \omega)$ for a central character $\omega$.
An algebraic cuspidal rep is an irreducible Harish-Chandra module and an irreducble $C_c^\infty(G(A_f))$-module. By Schur's lemma they automatically have a central character $\omega$. Moreover, you need appropiate growth conditions guaranteeing square-integrability. This guarantees also that the vectors are in the $L_0^2(G(F) \backslash G(A), \omega)$ vector space. Hence every algebraic cuspidal rep is a dense subspace of an analytic cuspidal rep. Let us now discuss the opposite inclusion.
Fact 1:  Every analytic cuspidal rep $\pi$ of $G(A)$ factors into a tensor product $\otimes_v \pi_v$ over the places $v$ of $F$ of unitary reps of $G(F_v)$. We write $\pi_\infty \otimes \pi_f$. See Flath's article in the Corvallis Proceedings Theorem 4.
Fact 2: Two distinct unitary reps have distinct $tr \; \pi_v : C_c^\infty(G(F_v)) \rightarrow \mathbb{C}$ functional, and according to Fact 1, this holds for $tr \; \pi : C_c^\infty(G(A)) \rightarrow \mathbb{C}$ and $\pi$ cuspidal analytic as well. See this related question: Character determines the representation?. Hence the $C_c^\infty(G(A))$-module structure determines uniquely the analytic cuspidal representation.
Fact 3: Every unitary representation of a real reductive Lie group has a unique Harish-Chandra Module associated to it. So the $C_c^\infty(G(A_f)) \otimes C_c^\infty(G(A_\infty))$-module structure is encoded uniquely in the Harish-Chandra module $\times C_c^\infty(G(A_f))$-module structure. This can be found on the related Wikipedia page, but you seem to believe this already.
So the missing point was fact 2, which allows an algebraic classification of unitary representations. This can be expressed in various languages, but I prefer the statement in terms of character distributions.<|endoftext|>
TITLE: Is there a Sudoku matroid?
QUESTION [16 upvotes]: This question is inspired from this one, where it is asked what is the minimum number of checks needed to verify that a Sudoku solution is correct.  Let 
$$
E=\{r_1, \dots, r_9\} \cup \{c_1, \dots, c_9\} \cup \{b_1, \dots, b_9\},
$$
where $r_i, c_i$, and $b_i$ are the set of rows, columns and boxes of the Sudoku.  We have an oracle, which given any $e \in E$, tells us if our Sudoku (which we cannot see) is correct on $e$.  The original question was to determine what is the minimum number of calls we need to make to the oracle to verify a correct Sudoku.  
For $S \subseteq E$, and $x \in E$, we say that $S$ implies $x$ if every Sudoku which is correct on all members of $S$, must also be correct on $x$.  Following Emil Jeřábek's notation, we write $S \models x$, if $S$ implies $x$. The original question asks for the smallest set $S$ such that $S \models x$ for all $x \in E$.  
The question here is: 

Is $\models$ a closure operator of a matroid with ground set $E$?

This is something I've been wondering myself, and I suspect the answer is yes.  This question was also explicitly asked by François Brunault, so I thought I'd publicize it independently.
I'll comment that the fact that $\models$ yields a matroid $M$ gives a very short proof of the original question (minus the verification that $M$ is a matroid of course).  
Proof. Phrased in the language of matroid theory, the original question asks what is the rank of $M$?  Now, let $S$ be a set of checks consisting of all boxes, 2 rows from each band, and 2 columns from each stack.  It is easy to see that the closure of $S$ in $M$ is all of $E$.  On the other hand, for each $e \in S$, we have $cl_M(S - e) \neq E$.  This is also easy; removing a box leaves a cell $x$ such that the row, column, and box containing $x$ are all unchecked, removing a row yields a band with two unchecked rows, and removing a column leaves a stack with two unchecked columns.  So, $S$ is an independent and spanning set of $M$, and hence a basis of $M$.  Thus, $r_M(E)=|S|=21$. 
Update. I wrote a blog post on this and some related questions on the Matroid Union Blog.

REPLY [10 votes]: By some sort of strange mathematical cosmic entanglement, it appears that  François Brunault answered his own question in the other thread while I was writing this question.  
The answer is indeed yes.  Feel free to click on the original question and vote up his answer!  Or vote this answer up (I've made it community wiki).<|endoftext|>
TITLE: Discrete Morse theory and chess
QUESTION [18 upvotes]: There are many mathematical objects that are similar to groups and Cayley graphs of groups but lack  homogeneity in some sense. Graphs of webpages with edges corresponding to links are one example. One long-studied example is chess. Some moves have inverses, and others do not. If we create a graph of all positions, it is certainly not homogeneous, and many methods developed for analyzing groups fail miserably (for instance, the graph is finite, so is delta-hyperbolic, but this gives no helpful information).
On the other hand, Morse theory is designed to help one study complex structures by picking out the points of greatest interest, ie the critical points. It has been used in several discrete settings such as video compression to help sort through mounds of intractable data.
My question is, has a version of discrete Morse theory been used to analyze chess? For instance, the critical points would perhaps correspond to positions with a local maximum or minimum number of moves; the absolute minima represent checkmates, the maxima represent positions with a lot of freedom.
Because chess is so well studied, I wonder if this hasn't been studied before. Does anyone know of a reference where Morse-like ideas were used to analyze chess?

REPLY [15 votes]: The quick answer to your question is no, discrete Morse theory has not been used to study chess moves yet (unless this has been done in some very obscure journal). I would like to highlight a few likely reasons for this current state of affairs.
If I understand your question correctly, you are interested in the following scheme:

build a directed graph $G$ whose vertices correspond to positions on the chessboard with an edge $a \to b$ if a valid move takes you from $a$ to $b$,
construct a discrete Morse function on $G$ whose discrete vector field is only allowed to pair those $a$ and $b$ which are related by bidirectional (i.e., invertible) arrows $a \leftrightarrow b$.

The basic problem with such an approach, simply put, is that $G$ is huge and might contain positions which have not been reached in the entire recorded history of the game. As such, there is no complete "input space" $G$ on which you could impose your discrete Morse function, even if the process of imposing such a function could be performed in a distributed way. By contrast, even the example you have given regarding video compression assumes that the entire uncompressed video is accessible in the first place.  
Note also that there is no known method for declaring that such-and-such cells will be critical when imposing a discrete Morse function. The discrete vector field must be acyclic, and depending on the topology of $G$ one might end up with many more critical cells than the "absolute maxima and minima" which were originally intended.
Finally, the following statement from your question is somewhat ambiguous:

Morse theory is designed to help one study complex structures by picking out the points of greatest interest, ie the critical points.

While it is certainly true that a discrete Morse function $f$ on a given cell complex $X$ allows one to construct a homotopy-equivalent Morse complex $M_f(X)$ built out of the critical cells of $f$, all the attaching map information is computed using the non-critical cells. Once you compute these attaching maps you can throw away those uncritical cells, but you must know all of them to begin with.
I think the idea is interesting, and may bear fruit if one restricts from the hopelessly unknown $G$ with "all possible positions" to the much more tractable $G' \subset G$ which is the subgraph spanned by all positions attained by grandmasters in recorded championship games.<|endoftext|>
TITLE: Sz.-Nagy dilation for uniformly convex Banach spaces
QUESTION [5 upvotes]: The Sz.-Nagy dilation theorem says that for a Hilbert space $H$ with nonexpansive operator $T$, there is a larger space $H'$ containing $H$ and a unitary operator $U$ on $H'$ such that for all $x \in H$ and all $n$, $T^n x  = P U^n x$ where $P$ is the projection from $H'$ to $H$.

Is there anything analogous to the Sz.-Nagy dilation theorem except for uniformly convex Banach spaces?


A few more details:
I have a result for linear isometries on uniformly convex Banach spaces with "power type" modulus of convexity, $\eta(\varepsilon) = C \varepsilon^p$. I want to extend the result to linear nonexpansive operators on the same class of spaces.  I need the larger space to be uniformly convex, ideally with the same power type.
It would be a bonus if this also holds for power-bounded operators $T$ (i.e. $\|T^n x\| \leq C\|x\|$).  I don't actually need the other map to be an isometry, just something that is power bounded from above and below (i.e. $c\|x\| \leq \|T^n x\| \leq C\|x\|$).

REPLY [6 votes]: The Akcoglu-Sucheston-Peller dilation theorem gives indeed a characterization of operators on a $L^p$-space with an isometric dilation on a $L^p$-space. These operators are the contractively regular operators. And it is well-known that we can find contractive operators without this property (some 2x2 matrix on $\ell^p_2$ with $p\not=$ 1,2,$\infty$)
Moreover, you cannot obtain your "bonus". Indeed, the existence of an isometric dilation fo $T$ imply that $T$ is a contraction. However, it seems to me that you need the notion of "loose dilation" with a power-bounded isomorphism. See the paper "Dilation of Ritt operators on Lp-spaces" on
https://sites.google.com/site/cedricarhancet/publications-1
With the methods of the paper, it is not very difficult to obtain a dilation result for $R$-Ritt operators (a nice class of power-bounded operators) on UMD spaces. I warn you that there exist power-bounded operators on a $L^p$-space without loose dilation on a $L^p$-space.
Finally, if you really need a isometry instead of a power-bounded isomorphism, it is possible with a ultraproduct argument.<|endoftext|>
TITLE: Weyl law for arithmetic Fuchsian groups known?
QUESTION [5 upvotes]: For congruence subgroups of $PSL(2,\mathbb{Z})$, the Weyl law for the eigenvalues of Maass cusp forms had been proven by Selberg. How is the status of such a Weyl law for eigenvalues of Maass cusp forms for arbitrary arithmetic nonuniform cofinite Fuchsian groups? Has it already been proven? Does one know at least that there are infinitely many linearly independent Maass cusp forms for these lattices?

REPLY [2 votes]: Consider the normalizer $\Gamma$ of $\Gamma_0(N)$ in $SL_2(\mathbb{Q})$ for $N$ squarefree, which is not a subgroup of $SL_2(\mathbb{Z})$, then you obtain the same Weyl law as classical. For my taste, these groups should also be called congruence subgroups, although they are not contained in $SL_2(\mathbb{Z})$. They are important if you want classically distinguish between nonisomorphic supercuspidal reps contained as factors of automorphic reps associated, which are associated to ramified quadratic extensions of $\mathbb{Q}_p$ at $p |N$.
I don't know a single example what happens if $\Gamma$ does not contain a congruence subgroup, but is arithmetic.
Btw, your question makes perfect sense, because the modified Rolecke-Selberg conjecture states (see the  introduction http://link.springer.com/content/pdf/10.1007%2FBF02572621.pdf) that apart from arithmetic lattices, there are at most finitely many discrete eigenvalues.
There is some computational evidence by Hejhal for this conjecture: 
Hejhal, Dennis A.(1-MN-SM)
On eigenvalues of the Laplacian for Hecke triangle groups.

Copy&Paste from MathSciNet: This paper is part of a series of articles in which computer experiments are performed to numerically compute the eigenvalues of the Laplacian. In this paper the Hecke triangle groups generated by z→−1/z, z→z+2cos(π/N) are considered. 
     The basic results are a computation of eigenvalues for the groups with N=3,4,6 (the congruence groups) with eigenvalue 14+R2 with R<25, and the conclusion that no even cusp forms exist when N=5, R<60 and N=7, R<40. The last result, which is similar to results of Winkler, gives evidence in support of the Phillips-Sarnak conjecture that one should have few if any cusp forms for nonarithmetic groups (except for the obvious ones caused by symmetries). 
     The procedure used is essentially the collocation method. One must also be careful in evaluating the K-Bessel functions that appear in the Fourier expansion of the forms.

What is sure is that if $\Gamma$ does not contain a congruence subgroup, then the Maass cusps forms on $\Gamma \backslash \mathbb{H}$ can not be lifted to a vector of an automorphic adelic space $SL_2(\mathbb{A})$ in any obviuous way, and one cannot expect them to have a nice $L$-functions. Also the contribution of the continuous spectrum is most likely not expressable in terms of classical Dirichlet L-functions. Recall that this is the main point to be understood for Weyl laws of non-uniform lattices.
I know that the Belyi's theorem generates an isomorphisms between nonsingular algebraic curves over $\mathbb{C}$ and $\Gamma \backslash \mathbb{H}$ compactified at the cusps, where $\Gamma$ is an arithmetic subgroup of $SL_2(\mathbb{Z})$, but noncongruence in general. This might be a source of such lattices and at least modular functions on them should exists (by Riemann Roch?).<|endoftext|>
TITLE: Composing two-term sums from the same primes
QUESTION [9 upvotes]: The following is an old result of Erdős and Turán (American Mathematical Monthly, 1934):
Given a set of $2^n + 1$ distinct positive integers, all of its two-term sums cannot be composed of the same $n$ primes. (An integer $m$ is said to be composed of primes $p_1, p_2, \dots$ is every prime factor of $m$ is one of those primes. And in two-term sums we exclude sums of two copies of the same integer.)
Let $f(n)$ be the minimum number such that given $f(n)$ distinct positive integers, all its two-term sums cannot be composed of the same $n$ primes. Erdős and Turán conjecture that the upper bound $f(n) \le 2^n + 1$ can be greatly improved. For example, they conjecture that $f(n) = O(n^{1+ \epsilon})$ for every fixed $\epsilon > 0$.
Does anyone know if any progress has been made on this problem since? In particular is there any polynomial upper bound known?
Also I would be interested to know about lower bounds on $f(n)$.

REPLY [9 votes]: This problem and generalizations of it are discussed in the following papers:

P. Erdős, A. Sárközy, C. Stewart, On prime factors of subset sums.  J. London Math. Soc. (2) 49 (1994), no. 2, 209–218.
C. Stewart, On prime factors of integers which are sums or shifted products. Anatomy of integers, 275–287,
CRM Proc. Lecture Notes, 46, Amer. Math. Soc., Providence, RI, 2008.
C. Stewart, R. Tijdeman, On prime factors of sums of integers. II. Diophantine analysis (Kensington, 1985), 83–98,
London Math. Soc. Lecture Note Ser., 109, Cambridge Univ. Press, Cambridge, 1986.

I am fairly sure that no improvement to the Erdos-Turan result is known. However, there is a very closely related problem where one considers two sets.
First note we can reformulate the Erdos-Turan result as the estimate
$$w(\prod_{a,b \in S} (a+b)) \gg \log(|S|)$$
where $w$ denotes the number of prime factors of an integer. Now a natural generalization of the above is the inequality
$$w(\prod_{a \in A, b \in B, |A|=|B|=k} (a+b)) \gg \log(k)$$
This was conjectured by Erdos and Turan and proved by Gyory, Stewart, and Tijdeman in Compositio 59 (1986). In this more general setting it was proved by Erdos, Stewart and Tijdeman in Compositio 66 (1988) that this estimate is nearly optimal. More specifically,
$$w(\prod_{a \in A, b \in B, |A|=|B|=k} (a+b)) \geq (1/8+\epsilon) \log(k)^2 \log\log(k)$$ can't hold for any $\epsilon >0$.<|endoftext|>
TITLE: Unbounded metrics on groups
QUESTION [6 upvotes]: If $G$ is an infinite group, is there necessarily an unbounded left-invariant metric on $G$?

REPLY [8 votes]: The answer is no. See Thm 1.2 of http://homepages.math.uic.edu/~rosendal/PapersWebsite/Property(OB)10.pdf. There is a property discussed in the intro of this paper which is equivalent to all left invariant metrics are bounded. It is known that certain large permutation groups have this property.<|endoftext|>
TITLE: Actions of Thompson group F
QUESTION [9 upvotes]: Does anybody know the actions of Thompson group F which are not conjugate to the standard one?
Motivation is to find actions such that the Schreier graph of the action does not contain a binary tree.
I've decided to ask a separate question on the motivation here

REPLY [5 votes]: $F$ is bi-orderable, so if you care about actions by homeomorphisms on the line, you can pick up a bi-ordering and produce a dynamical realization: a faithful almost free action ( see Proposition 3.4 and Example 3.5 in this paper of Navas http://arxiv.org/abs/0710.2466  for a "non-standard" action of $F$). For the discrete case, the proof of Proposition 1.8 in the same paper might be an inspiration... All the bi-orderings of $F$ were described by Navas and Rivas here http://arxiv.org/abs/0808.1688 . (As a side remark, in one dimensional dynamics one usually consider two actions equivalent if they are semi-conjugate, rather than conjugate) 
(this was meant to be just a comment, but don't have enough points...:)<|endoftext|>
TITLE: Classic applications of Baire category theorem
QUESTION [28 upvotes]: I've seen Baire category theorem used to prove existence of objects with certain properties. But it seems there is another class of interesting applications of Baire category theorem that I have yet to see.
First I found this MathOverflow problem:

Let $f$ be an infinitely differentiable function on $[0,1]$ and suppose that for each $x \in [0,1]$ there is an integer $n \in \mathbb{N}$ such that $f^{(n)}(x)=0$. Then $f$ coincide on $[0,1]$ with some polynomial.

I found another one from Ben Green's notes:

Suppose that $f:\mathbb{R}^+\to\mathbb{R}^+$ is a continuous function with the following property: for all $x\in\mathbb{R}^+$, the sequence $f(x),f(2x),f(3x),\ldots$ tends to $0$. Prove that $\lim_{t\to\infty}f(t)=0$.

Are there any other classic problems of this type?

REPLY [2 votes]: The famous Ornstein Isomorphism theorem in ergodic theory (any two Bernoulli shifts with the same entropy are isomorphic) has a simplified proof due to Burton and Rothstein, which uses Baire Category.<|endoftext|>
TITLE: Interpretation of multiplicity of a point 
QUESTION [12 upvotes]: Let $x$ be a (closed) point on an algebraic variety $X$ (of dimension $n$) defined over an algebraically closed field $k$. What is the multiplicity $mult_x(X)$, and how to compute it?
While having a hard time recently to compute the multiplicity of some surface singularities, I thought it might be useful to have a list of equivalent definitions. These are the ones I know of:
Notations: let $m_x$ be the maximal ideal of $x$ at $X$. For definitions 1 to 3 below assume (a neighborhood of $x$ in) $X$ is embedded in a projective space $\mathbb{P}^N(\mathbb{k})$. 
Geometric Definitions:

(Copying from Mumford, Algebraic Geometry I: This is valid only in the case $k = \mathbb{C}$.) For every linear subspace $L$ of dimension $N - n$ such that $x$ is a component of $L \cap X$, define a number $i(x;L \cap X)$ as follows: it is the unique number such that for every sufficiently small neighborhood $U$ of $x$ (in the classical topology), there is a neighborhood $U'$ of $L$ (in the space of $N-n$ dimensional linear subspaces of $\mathbb{P}^N(\mathbb{C})$) such that if $L' \in U'$ and $L'$ intersects $X$ transversally, then $i(x;L \cap X) = |L' \cap X \cap U|$. Then $mult_x(X)$ is the minimum of $i(x;L \cap X)$ as $L$ runs over all $N-n$ dimensional linear subspaces of $\mathbb{P}^N(\mathbb{C})$ for which $x$ is an isolated point of $L \cap X$.
(Corollary of (1), but holds also over positive characteristic - or at least so I think. For my problem this turned out to be the right definition to use) For every linear subspace $L$ of dimension $N - n$ such that $L \cap X$ is discrete, let $s(x;L \cap X)$ be the number of points of intersection (counted with intersection multiplicity) of $L \cap X$ other than $x$. Then $mult_x(X) = \min\lbrace  \deg(X) - s(x;L \cap X)\rbrace$ as $L$ runs over all $N-n$ dimensional linear subspaces of $\mathbb{P}^N(k)$ for which $L \cap X$ is discrete. 
(From Ramanujam's "On a geometric interpretation of multiplicity") Take a proper birational map $\phi: Y \to X$ such that the pull back of the maximal ideal of $P$ defines an effective Cartier divisor $D$ on $Y$. Then $mult_x(X) = (-1)^{n-1}D^n$.

Algebraic Definitions:

$mult_x(X)$ is $(n-1)!$ times the leading coefficient of the Hilbert–Samuel polynomial of $m_x$.
If $X$ is a hypersurface in a neighbourhood of $x$ defined by a single equation $f$, then $mult_x(X)$ is the integer $q$ such that $f \in m_x^q \setminus m_x^{q+1}$.

What other definitions are out there?

REPLY [3 votes]: Here is another equivalence: Let $\pi\colon Z\to \mathbb{P}^n$ be the blow-up of the point $x$, let $E\simeq \mathbb{P}^{n-1}$ be the exceptional divisor and let $\tilde{X}\subset X$ be the strict transform of $X$. Then, the intersection of $\tilde{X}$ with $E$ is a variety of degree $\mathrm{mult}_x(X)$ in $E\simeq \mathbb{P}^{n-1}$.
In the case where $X$ is a curve, you obtain finitely many points and the multiplicity is then just the number of points (that you need to count with multiplicity).<|endoftext|>
TITLE: Does every convex polyhedron have a combinatorially isomorphic counterpart whose faces all have rational areas?
QUESTION [25 upvotes]: Does every convex polyhedron have a combinatorially isomorphic counterpart whose faces all have rational areas?
Does every convex polyhedron have a combinatorially isomorphic counterpart whose edges all have rational lengths?
Does every convex polyhedron have a combinatorially isomorphic counterpart whose vertices all have rational $x,y,z$ coordinates?

Can multiple conditions above be combined?
Update: all polyhedra in question are in $\mathbb{R}^3$.

REPLY [3 votes]: The answer to the second question is yes for simplicial polyhedra, i.e., those with triangular faces. Indeed one may use Alexandrov's isometric embedding theorem to show that simplicial polyhedra with rational edge lengths are dense in the space of simplicial polyhedra, as discussed below.
Let $\ell_i$ be the edge lengths of a given simplicial convex polyhedron $P$ and $\ell_i'$ be rational numbers with $|\ell_i'-\ell_i|\leq\epsilon$. For each face $F_i$ of $P$ let $F_i'$ be the triangle with edge lengths $\ell_{i1}',\ell_{i2}',\ell_{i3}'$, where $\ell_{i1},\ell_{i2},\ell_{i3}$ are edge lengths of $F_i$. Assuming $\epsilon$ is small, the sum of the angles of $F_i'$ around each vertex will be less than $2\pi$ if $F_i'$ are glued together the same way that $F_i$ are. Thus, by Alexandrov's theorem, gluing $F_i'$ yields a convex polyhedron $P'$. By the uniqueness part of Alexandrov's theorem, $P'$ is close to $P$, after a rigid motion, and is therefore isometric to it since $P$ is simplicial. This in turn yields that the edges of $P'$ coincide with those of $F_i'$, because edges of $F_i'$ are the unique geodesics in $P'$ connecting their end points. So $P'$ has rational edge lengths as desired.<|endoftext|>
TITLE: If ZFC has a transitive model, does it have one of arbitrary size?
QUESTION [21 upvotes]: It is known that the consistency strength of $\sf ZFC+\rm Con(\sf ZFC)$ is greater than that of $\sf ZFC$ itself, but still weaker than asserting that $\sf ZFC$ has a transitive model. Let us denote the axiom "There is a transitive model of $T$" by $\rm St(\sf ZFC)$.
If $M$ is a transitive model of $\sf ZFC$ of size $\kappa$ then we can easily generate transitive models of any smaller [infinite] cardinal by using the downward Löwenheim–Skolem theorem and the Mostowski collapse. Note that we can use the latter because the former guarantees that the model uses the real $\in$ relation, so it is well-founded.
But the upward Löwenheim–Skolem makes use of compactness, which can easily generate models which are not well-founded, and therefore cannot be collapsed to a transitive model.
So while $\rm Con(\sf ZFC)$ can prove that there is a model of $\sf ZFC$ of any cardinality, can $\rm St(\sf ZFC)$ do the same, or do we have a refinement of the consistency axioms in the form of bounding the cardinality of transitive models?
It is tempting to take a countable transitive model and apply forcing and add more and more sets, but forcing doesn't add ordinals and the result is that we cannot increase the size of the model without bound.

REPLY [7 votes]: In a comment to François's answer, I point out that the least $\alpha$ such that $L_\alpha$ is a model of $\mathsf{ZFC}$ is countable. In what follows, "model" means "transitive model of $\mathsf{ZFC}$."
If $M$ is transitive, then $L^M=L_\beta\models\mathsf{ZFC}$ for $\beta=\mathsf{ORD}\cap M$, so the least height of a model is countable. Moreover, any model $M$ proves that there is a bijection between each level of its cumulative hierarchy, and one of its ordinals, so if $M$ has height $\kappa$, so is $\kappa$ its size. This proves that the existence of a model does not imply the existence of an uncountable one: If $M$ has least height, let $\alpha_0$ be least such that $L_{\alpha_0}$ is a model, and $\alpha_0$ is larger than the height of $M$ ($\alpha_0$ could be $\mathsf{ORD}$). We see that in $L_{\alpha_0}$ the height of $M$ is countable and there are no models of height larger than the height of $M$.
Asaf asked whether this generalizes, that is, whether for each $\kappa$ the existence of models of size $\kappa$ does not imply the existence of models of larger size. That this is indeed the case follows from extending the argument from the previous paragraph: Let $L_\alpha$ be a model of height (and size) $\kappa$, and let $\alpha_0$ be such that $\alpha<\alpha_0$ and $L_{\alpha_0}$ is a model. We may as well assume that $\alpha_0$ exists (that is, it is an ordinal), or else there is nothing to prove. Now let $X$ be an elementary substructure of $L_{\alpha_0}$ containing both $L_\alpha\cup\{L_\alpha\}$ (as a subset) and a bijection between $\alpha$ and $|\alpha|^{L_{\alpha_0}}$, and of size $\kappa$, which exists by a standard application of the Lowenheim-Skolem argument. The transitive collapse of $X$ is $L_\beta$ for some $\beta$, and has size $\kappa$. This means that if $\alpha_0$ is least (so the collapse of $X$ is again $L_{\alpha_0}$), then $L_{\alpha_0}$ is a model of $\mathsf{ZFC}$ plus the assertion that there are no set models of height (and therefore size) larger than $|\alpha|=\kappa$. 
Without choice, I do not know whether models of $\mathsf{ZF}$ of height $\kappa$ must have size $\kappa$.

[Edit: In response to the last paragraph above, Joel and Asaf pointed out some results. I am including them here, to increase visibility.]
In

Ali Enayat. Models of set theory with definable ordinals, Arch. Math. Logic 44 (3), (2005), 363–385. MR2140616 (2005m:03098), 

Ali discusses some results about transitive models of $\mathsf{ZF}$ that show that the situation is much more subtle than in the presence of choice. One of the most incredible results is due to Friedman, in

Harvey Friedman. Large models of countable height, Trans. Amer. Math. Soc. 201 (1975), 227–239. MR0416903 (54 #4966).

Let me quote from Harvey's paper:

The first examples of transitive models of $\mathsf{ZF}$ of power $\omega_1$ with countably many ordinals were constructed by Cohen. Later Easton, Solovay, and Sacks showed that every countable transitive model of $\mathsf{ZF}$ has an ordinal-preserving extension satisfying $\mathsf{ZF}$, of power  $2^{\omega}$.   We prove here that every countable transitive model $M$ of $\mathsf{ZF}$ has an ordinal preserving extension satisfying $\mathsf{ZF}$, of power $\beth_{M\cap\mathsf{ORD}}$. 

Harvey's argument uses forcing. For his first result, given $M$ a countable transitive model of $\mathsf{ZF}$, he says that $x\subset\omega^\omega$ is $M$-generic iff any finite sequence of distinct elements of $x$ is $M$-generic (for the product of the appropriate number of copies of Cohen forcing), $x$ is infinite, and dense. The models he builds are of the form $M(x)$ so, in particular, they are transitive. He shows that there are $M$-generics $x$ of size continuum with $M(x)$ a model of $\mathsf{ZF}$ (and $M(x)$ has the same height as $M$). He then builds on the machinery introduced here, and proves that, starting with an $M$-generic $x$, a family of sets $C_\alpha$, $\alpha\lt M\cap\mathsf{ORD}$, can be found with $|C_\alpha|\ge\beth_\alpha$, and such that $M[(C_\alpha)_\alpha]$, properly defined, is a model of $\mathsf{ZF}$ of the claimed size.   
Ali builds on this results to produce Paris models of $\mathsf{ZF}$, that is, models $M$ all of whose ordinals are first order definable in $M$. In prior work, he had shown that from the assumption that $L$ satisfies that there are uncountable transitive models of $\mathsf{ZFC}$, it follows that there are unboundedly many $\alpha<\omega_1^L$ such that $L_\alpha$ is Paris. He shows now that from the same assumption, we have that for every infinite $\kappa$ there are Paris models of $\mathsf{ZF}$ of size $\kappa$; this uses Harvey's result, since generic (or simply, ordinal preserving) extensions of $L_\alpha$ are Paris if $L_\alpha$ itself is Paris. It follows that there is a complete extension of $\mathsf{ZF}$ admiting in $L$ Paris models of size $\beth_\alpha$ for each countable $\alpha$. The theory, including the requirement that its models are Paris, can be described in $L_{\omega_1\omega}$. Since the Hanf number of this logic is $\beth_{\omega_1}$, the result follows.
This produces large transitive models indeed, in view of a result of Paris: If a completion $T$ of $\mathsf{ZF}$ has a well-founded model, then every Paris model of $T$ is well-founded.   

As pointed out by Mohammad here, Solovay's construction referenced in Friedman's answer (and a clear influence in Friedman's argument) can be found in 

Ulrich Felgner. Choice functions on sets and classes. In Sets and classes (on the work by Paul Bernays), pp. 217–255. Studies in Logic and the Foundations of Math., Vol. 84, North-Holland, Amsterdam, 1976. MR0424566 (54 #12525).<|endoftext|>
TITLE: zeta(3) in terms of derivatives of zeta at 1/2 and pi
QUESTION [19 upvotes]: Got numerical support that for odd $n$, $\zeta(n)$ might be
expressed in terms of the derivatives of $\zeta(\frac12)$.
Based on More Zeta Functions for the Riemann Zeros, Andre Voros, p.12, Table 3:
Conjecture:  For odd $n$, 
$$\zeta(n) = \left(\frac{2}{(n-1)!} (\log(|\zeta|)^{(n)} (\frac12) - 2^n \beta(n))\right)/(2^n-1)$$
$\beta(n)$ is Dirichlet beta function and it is a rational multiple of $\pi^n$
for odd $n$.
The derivative can be expressed in terms of $\zeta(\frac12),\zeta^{(k)}(\frac12)$
For $n=3$ get numerical support for:
$$\zeta(3) = (-\zeta'''(\frac12)/|\zeta(\frac12)| -3 \zeta''(\frac12) \zeta'(\frac12)/|\zeta(\frac12)|^2 -2 \zeta'(\frac12)^3/|\zeta(\frac12)|^3- \pi^3 / 4)/7 $$
The last equality holds with precision $10^4$ decimal digits.
One can eliminate the first derivative since there is closed form
for $\zeta'(\frac12)/\zeta(\frac12)$

Is this result true?

sage/mpmath code in case of typos of the latex.
#run in sage

import mpmath
from mpmath import mpf
mpmath.mp.pretty=True

def zeta3test():

    n=3
    mpmath.mp.dps=10^3
    zeta3=mpmath.zeta(3)

    Pi=mpmath.pi
    gamma=mpmath.euler
    z12=mpmath.zeta(1/mpmath.mpf(2))
    z1=mpmath.zeta(1/mpmath.mpf(2),derivative=1)
    z2=mpmath.zeta(1/mpmath.mpf(2),derivative=2)
    z3=mpmath.zeta(1/mpmath.mpf(2),derivative=3)

    # eliminate the first derivative

    #rh0=1/32*(72*Pi*mpmath.log(2)*mpmath.log(Pi)*z12+144*gamma*mpmath.log(2)*mpmath.log(Pi)*z12+72*mpmath.log(2)*z12*gamma*Pi+24*mpmath.log(Pi)*z12*gamma*Pi-144*z2*mpmath.log(2)-48*z2*mpmath.log(Pi)+216*mpmath.log(2)^3*z12+8*mpmath.log(Pi)^3*z12-48*z2*gamma-24*z2*Pi+8*z12*gamma^3+z12*Pi^3+72*mpmath.log(2)*z12*gamma^2+18*mpmath.log(2)*z12*Pi^2+24*mpmath.log(Pi)*z12*gamma^2+6*mpmath.log(Pi)*z12*Pi^2+216*mpmath.log(2)^2*mpmath.log(Pi)*z12+72*mpmath.log(2)*mpmath.log(Pi)^2*z12+216*gamma*mpmath.log(2)^2*z12+24*gamma*mpmath.log(Pi)^2*z12+108*Pi*mpmath.log(2)^2*z12+12*Pi*mpmath.log(Pi)^2*z12+32*z3+12*z12*gamma^2*Pi+6*z12*gamma*Pi^2)/z12
    z12a=mpmath.fabs(z12)
    rh1= -z3/z12a -3*z2*z1/z12a**2 -2* z1**3/z12a**3

    #print 'rh',mpmath.chop(rh0-rh1)
    #rhs= (mpmath.diff( lambda y:  mpmath.log(mpmath.fabs(mpmath.zeta(y))),1/2,n) - mpmath.pi^3 / 4 )/(7)
    rhs= (rh1 - mpmath.pi**3 / 4 )/(7)
    print mpmath.chop(zeta3-rhs)

def conjecture1(n):
    """

    voros, p. 12
    """
    assert n%2==1
    a1= mpmath.zeta(n)
    a2= (2/factorial(n-1) * mpmath.diff( lambda y:  mpmath.log(mpmath.fabs(mpmath.zeta(y))),1/2,n) - 2**(n) * dirichletbeta(n))/(2**n-1)
    print mpmath.chop(a1-a2)


def dirichletbeta(s):
    """
    dirichlet beta
    """
    return 4**(-s) * (mpmath.hurwitz(s,1/4)-mpmath.hurwitz(s,3/4))

REPLY [8 votes]: In fact for odd $n\ge3$ we have 
$$\Bigl.\frac{d^{n}}{ds^n}\log\zeta(s)\Bigr|_{s=\frac12}=
\frac{(n-1)!}{2}\Bigl(2^n L(n,\chi)+(2^n-1)\zeta(n)\Bigr)$$
The proof  (due to Voros) is the following:
It is well known that 
$$\frac{\zeta'(s)}{\zeta(s)}=-\frac{1}{s-1}+\sum_\rho\Bigl(\frac{1}{s-\rho}-\frac{1}{\rho}\Bigr)
+\sum_{n=1}^\infty \Bigl(\frac{1}{s+2n}-\frac{1}{2n}\Bigr)+B$$
Differentiating this we get
$$\frac{d^{n-1}}{ds^{n-1}}\frac{\zeta'(s)}{\zeta(s)}=\frac{(-1)^{n-2} (n-1)!}{(s-1)^{n}}-\sum_{\rho}
\frac{(n-1)!}{(\rho-s)^{n}}+\sum_{k=1}^\infty \frac{(-1)^{n-1} (n-1)!}{(s+2k)^{n}}.$$
Now take $s=\frac12$
We get 
$$\Bigl.\frac{d^{n}}{ds^n}\log\zeta(s)\Bigr|_{s=\frac12}=$$
$$(n-1)! 2^{n}-
\sum_{\rho}\frac{(n-1)!}{(\rho-\frac12)^{n}}+\sum_{k=1}^\infty \frac{(-1)^{n-1} (n-1)! 2^{n}}
{(4k+1)^{n}}.$$
When $n$ is odd the sum on the non trivial zeros is $0$ by symmetry. 
There is only the question of recognizing the sum.
Assuming $n$ even we have
$$\Bigl.\frac{d^{n}}{ds^n}\log\zeta(s)\Bigr|_{s=\frac12}\frac{1}{(n-1)!\,2^{n-1}}=$$
$$=2+2\sum_{k=1}^\infty \frac{1}{(4k+1)^{n}}$$
It is easily seen that 
$$2\sum_{k=1}^\infty \frac{1}{(4k+1)^{s}}=-2+L(s,\chi)+(1-2^{-s})\zeta(s)$$
from which the result follows.<|endoftext|>
TITLE: Showing a filter with a certain property on the power set of $\mathbb{Z}$ is a one point filter
QUESTION [5 upvotes]: Let $\mathcal{P}_0(X)$ the Power set of $X$ without the empty set and let $\dot{x}:=\{A\subseteq X: x \in A\}$ the one point filter generated by $x$. Furthermore let $$ \mathcal{A} := \{ f \in X^{\mathcal{P}_0(X)} : \ \forall A \in \mathcal{P}_0(X): f(A) \in A\} $$ be the set of the functions mapping subsets of the power set to elements of them. Let $\varphi$ be a filter on $\mathcal{P}_0(X)$ with $$\forall f \in \mathcal{A} \exists x_f \in X : f[\varphi]=\dot{x}_f.$$
Here $f[\varphi]$ stands for the filter generated from the filter basis:
$$  \{ \operatorname{image}f|_M \ : \ M \in \varphi\}$$
On the bottom of the post I included some notation, I hope this avoids notational issues.

At first we should show that $\varphi$ is already an ultrafilter.
Now we shall show that if $X=\mathbb{Z}$, then it follows that $\varphi$ is a one point filter.

For the first part my attempt was to use that a filter is a ultra filter when every set or its complement is in the filter. At first we look at the case where $x$ is independent of
$f$ and then (somehow) conclude that $\varphi$ must already be an ultrafilter.
If this is done we could look at the equivalence relation $\sim$ given by
\[ f \sim g \iff x_f=x_g \]
Now we use the equivalence classes given by this equivalence relation. If I am right, this reduce the problem again to the case where $x$ is independent of $f$.
For the second part my only idea was looking the the properties of $\mathbb{Z}$ which could help us (that the ultrafilter is already a one point filter). I think one of the properties which help is that $\mathbb{Z}$ is countable, but I don't see how it helps.

Let $S$ be a set (an arbitrary one), and $A,B\subseteq S$. We call $\varphi\subset \mathcal{P}(S)$ a filter when the following holds

$\varnothing\notin \varphi$
$A,B\in \varphi \implies A\cap B \in \varphi$
$A\subset B $ and $A\in \varphi \implies B\in \varphi$


An ultrafilter is a filter, such that there is no bigger filter, e.g. when $G$ is a filter and $F$ is an ultra filter and $G\supseteq F\implies G=F$. More convenient is the equivalence that for every subset $A$ of $S$ it holds that $A\in F \wedge S\setminus A \in F$.
Maybe this question is not appropiate to MathOverflow cause it is actually not a research question. I faced this problem on the group exercises of my topology class I am in (so nothing which should be handed in). If it is not welcome I will delete it  just leave a comment. I already posted that question on Math Stack Exchange

REPLY [7 votes]: For this problem, we shall use $\mathbb{N}$ instead of $\mathbb{Z}$ since $\mathbb{N}$ is easier to work with in this case. We shall say that $A$ has property $P$ almost everywhere (or for almost all $A$) abbreviated a.e. if $\{A\in P_{0}(X)|A\,\textrm{has property}\,P\}\in\varphi$.`
For $n>0$, let $f_{n}\in\mathcal{A}$ be the function where  if $A\in P_{0}(\mathbb{N})$ then $f_{n}(A)$ is the $n$-th element of $A$ whenever $|A|\geq n$ and $f_{n}(A)$ is the last element of $A$ whenever $|A|<n$. Let $y_{n}=x_{f_{n}}$. Then $f_{n}(A)=y_{n}$ for almost every $n$. If $i<j$, then clearly $y_{i}\leq y_{j}$. Furthermore, if $y_{n}=y_{n+1}$, then $f_{1}(A)=y_{1},...,f_{n+1}(A)=y_{n+1}$ for almost every $A\in P_{0}(\mathbb{N})$. However, if $f_{1}(A)=y_{1},...,f_{n+1}(A)=y_{n+1}$, then $A=\{y_{1},...,y_{n+1}\}$ making $\varphi$ a principal ultrafilter. We shall therefore assume that $y_{n}<y_{n+1}$ for all $n$.
We claim that $\varphi$ is an ultrafilter. Let $S=\{A\in P_{0}(X)|f_{1}(A)=y_{1},f_{2}(A)=y_{2}\}$. Then clearly $S\in\varphi$. Let $R\subseteq P_{0}(X)$. Let $h\in\mathcal{A}$ be a function where $h(A)=y_{1}$ for each $A\in R\cap S$ and $h(A)=y_{2}$ for each $A\in R^{c}\cap S$. Then the function $h$ is constant almost everywhere. It is clear that $h(A)=y_{1}$ a.e. or $h(A)=y_{2}$ a.e. If $h(A)=y_{1}$ almost everywhere, then $R\cap S\in\varphi$. If $h(A)=y_{2}$ a.e., then $R^{c}\cap S\in\varphi$. Therefore, we conclude that either $R\in\varphi$ or $R^{c}\in\varphi$. Therefore $\varphi$ is an ultrafilter.
We shall now prove that $\varphi$ is a principal ultrafilter. Assume for the sake of contradiction that $\varphi$ is a non-principal ultrafilter. Let $Y=\{y_{n}|n\in\mathbb{N}\}$. We claim that $Y\not\subseteq A$ for almost every $A$. To prove this assume that $Y\subseteq A$ for almost every $A$. Then the ultrafilter $\varphi$ is $\sigma$-complete. To see this, let $T=\{A\in P(X)|Y\subseteq A\}$, and let $P=\{R_{n}|n\in\mathbb{N}\}$ be a partition of $T$ into countably many pieces. Let $g\in\mathcal{A}$ be a function where if $A\in R_{n}$, then $g(A)=y_{n}$. Then the function $g$ is constant almost everywhere. In particular, $g(A)=y_{n}$ for almost every $A$. However, this implies that $R_{n}\in\varphi$ for some $n$ making the ultrafilter $\varphi$ $\sigma$-complete. On the other hand, it is well known that there are no non-principal $\sigma$-ultrafilters on $P_{0}(\mathbb{N})$ since $|P_{0}(\mathbb{N})|$ is far below the first measurable cardinal if one even exists. We conclude that $Y\not\subseteq A$ for almost every $A$.
Now let $t\in\mathcal{A}$ be a function where if $y_{1}\in A\subseteq\mathbb{N}$ and $Y\not\subseteq A$, then $t(A)=y_{n}$ where $y_{n+1}\not\in A$. Then $t(A)=y_{n}$ for almost all $A\in P_{0}(X)$. Then the function $t$ is constant almost everywhere. In particular, $t(A)=y_{n}$ for almost all $A\in P_{0}(X)$. However, this implies that $y_{n+1}\not\in A$ for almost all $n$. This contradicts the fact that $f_{n+1}(A)=y_{n+1}$ for almost every $A\in P_{0}(X)$. We therefore conclude that $\varphi$ can only be a principal ultrafilter.
$\textbf{Added 5/9/13}$
I mentioned in the comments how this problem can be phrased in terms of uniform spaces. I will now give a topological proof of the fact that every ultrafilter $\varphi$ on $P_{0}(\mathbb{N})$ where every $f\in\mathcal{A}$ is constant $\varphi$-a.e. is simply a principal ultrafilter. The topological proof of this fact that I am about to give is very similar to the purely combinatorial proof of this fact above. In order to translate this problem involving ultrafilters into a problem involving uniform spaces we will need the following result and definitions.
A uniform space $(X,\mathcal{U})$ is said to be non-Archimedean if $\mathcal{U}$ is generated by equivalence relations. For example, if $X$ is a compact space, then there is a unique uniformity $\mathcal{U}$ on $X$ compatible with the topology on $X$ and this uniformity is non-Archimedean if and only if $X$ is totally disconnected.
We define a partition space to be a pair $(X,F)$ where $X$ is a set and $F$ is a filter on the lattice of partitions of $X$. Clearly the partition spaces are in a one-to-one correspondence with the non-Archimedean uniform spaces since the partitions of a set are in a one-to-one correspondence with the filters on that set.
If $(X,F)$ is a partition space, then let $\mathfrak{B}^{*}(X,F)=\{\emptyset\}\cup\bigcup F$. Then $\mathfrak{B}^{*}(X,F)$ is a Boolean algebra. In particular, if $R\subseteq X$, then $R\in\mathfrak{B}^{*}(X,F)$ if and only if $\{R,R^{c}\}\setminus\{\emptyset\}\in F$ if and only if the characteristic function $\chi_{R}$ is uniformly continuous.
An ultrafilter $\mathcal{U}$ on $\mathfrak{B}^{*}(X,F)$ is said to be an $F$-ultrafilter if $\mathcal{U}\cap P\neq\emptyset$ for each $P\in F$.
$\mathbf{Theorem}$ Let $(X,F)$ be a separated partition space. Then the following are equivalent.

$(X,F)$ is complete (as a uniform space,i.e. Every Cauchy filter converges).
Every $F$-ultrafilter on the Boolean algebra $\mathfrak{B}^{*}(X,F)$ is of the form
$\{R\in\mathfrak{B}^{*}(X,F)|x\in R\}$ for some $x\in X$.

$\mathbf{Proof}$(Outline) 
$1\rightarrow 2$. If $(X,F)$ is complete and $\mathcal{U}$ is an $F$-ultrafilter, then $\mathcal{U}$ is a Cauchy filter. Therefore $\mathcal{U}$ converges to some point $x\in X$. However, $\{R\in\mathfrak{B}^{*}(X,F)|x\in R\}$ is the only $F$-ultrafilter that converges to the point $x\in X$.
$2\rightarrow 1$. Assume that every $F$-ultrafilter on $\mathfrak{B}^{*}(X,F)$ is of the form $\{R\in\mathfrak{B}^{*}(X,F)|x\in R\}$. Let $Z$ be a Cauchy filter on $X$. Then $Z\cap\mathfrak{B}^{*}(X,F)$ is an $F$-ultrafilter, so $Z\cap\mathfrak{B}^{*}(X,F)=\{R\in\mathfrak{B}^{*}(X,F)|x\in R\}$ for some $x\in X$. It turns out that the filter $Z$ converges to the point $x$. $\mathbf{QED}$
We shall give the space $\mathbb{N}$ the discrete uniformity.
Let $\mathcal{B}=\mathcal{A}\cup\{f:P_{0}(\mathbb{N})\rightarrow\mathbb{N}:|f[P_{0}(\mathbb{N})]|<\infty\}$. In other words, $\mathcal{B}$ is the union of the set $\mathcal{A}$ with the set of all functions $f:P_{0}(\mathbb{N})\rightarrow\mathbb{N}$ with bounded range. Now give $P_{0}(\mathbb{N})$ the coarsest uniformity such that every function $f\in\mathcal{B}$ is a uniformly continuous function. Let $(P_{0}(\mathbb{N}),F)$ be the corresponding partition space structure. Then $\mathfrak{B}^{*}(P_{0}(\mathbb{N}),F)=P(P_{0}(\mathbb{N}))$, and the $F$-ultrafilters are precisely the ultrafilters on the set $P_{0}(\mathbb{N})$ such that every function $f\in\mathcal{A}$ is constant almost everywhere. Therefore by the above theorem, every ultrafilter $\mathcal{U}$ on $P_{0}(\mathbb{N})$ such that every $f\in\mathcal{A}$ is constant $\mathcal{U}$-a.e. is principal if and only if the space $P_{0}(\mathbb{N})$ is a complete uniform space. We shall now prove that the uniform space $P_{0}(\mathbb{N})$ is complete.
Let $L:P_{0}(\mathbb{N})\rightarrow\mathbb{N}^{\mathcal{B}}$ be the function where
$L(A)=(f(A))_{f\in\mathcal{B}}$ for each $A\in P_{0}(\mathbb{N})$. Give $\mathbb{N}^{\mathcal{B}}$ the product uniformity where $\mathbb{N}$ still has the discrete uniformity. Then $\mathbb{N}^{\mathcal{B}}$ is a complete uniform space since the product of complete uniform spaces is always a complete uniform space. Therefore the space $P_{0}(\mathbb{N})$ is complete if and only if the image $L[P_{0}(\mathbb{N})]$ is a complete subspace of $\mathbb{N}^{\mathcal{B}}$. However, the space $L[P_{0}(\mathbb{N})]$ is complete if and only if $L[P_{0}(\mathbb{N})]$ is a closed subspace of $\mathbb{N}^{\mathcal{B}}$. We therefore only need to show that $L[P_{0}(\mathbb{N})]$ is closed in $\mathbb{N}^{\mathcal{B}}$.
Let $(x_{f})_{f\in\mathcal{B}}\in \overline{L[P_{0}(\mathbb{N})]}$. Then since every neighborhood of $(x_{f})_{f\in\mathcal{B}}$ intersects $L[P_{0}(\mathbb{N})]$, whenever $h_{1},...,h_{n}\in\mathcal{B}$ there is some $A\in P_{0}(\mathbb{N})$ where $x_{h_{1}}=h_{1}(A),...,x_{h_{n}}=h_{n}(A)$.
As before, let $f_{n}\in\mathcal{A}$ be the function where $f_{n}(A)$ is the $n$-th element of $A$ whenever $|A|\geq n$ and $f_{n}(A)$ is the last element of $A$ whenever $|A|<n$. Let $y_{n}=x_{f_{n}}$ for all $n$. Then for all $n$, there is some $A\in P_{0}(\mathbb{N})$ with $y_{i}=x_{f_{i}}=f_{i}(A)$ for $1\leq i\leq n$. In particular, since the sequence $(f_{i}(A))_{i\in\mathbb{N}}$ is increasing for all $A$, we conclude that the sequence $(y_{n})_{n\in\mathbb{N}}$ is increasing.
Assume that $x_{f_{n}}=y_{n}=y_{n+1}=x_{f_{n+1}}$ for some $n$. Then there is an $A\in P_{0}(\mathbb{N})$ with $f_{i}(A)=x_{f_{i}}=y_{i}$ for $1\leq i\leq n+1$. Furthermore, one can clearly see that the $A\in P_{0}(\mathbb{N})$ with $f_{i}(A)=x_{f_{i}}=y_{i}$ for $1\leq i\leq n+1$ is unique. In particular $A=\{y_{1},...,y_{n}\}$. Let $U$ be the neighborhood of $(x_{f})_{f\in\mathcal{B}}$ consisting of all systems $(z_{f})_{f\in\mathbb{B}}$ such that $z_{f_{1}}=x_{f_{1}},...,z_{f_{n+1}}=x_{f_{n+1}}$. Then $U$ is an open set, but $U$ intersects $L[P_{0}(\mathbb{N})]$ at only one point, namely the point $L(\{y_{1},...,y_{n}\})=(f(\{y_{1},...,y_{n}\}))_{f\in\mathcal{B}}$. Therefore since $U$ is an open neighborhood of $(x_{f})_{f\in\mathcal{B}}$, $U$ intersects $L[P_{0}(\mathbb{N})]$ at only the point $L(\{y_{1},...,y_{n}\})$, and $(x_{f})_{f\in\mathcal{B}}\in\overline{L[P_{0}(\mathbb{N})]}$, we conclude that 
$(x_{f})_{f\in\mathcal{B}}=L(\{y_{1},...,y_{n}\})$.
Now assume that $y_{n}<y_{n+1}$ for all natural numbers $n$. Let $Y=\{y_{n}|n\in\mathbb{N}\}$. We shall now show that $(x_{f})_{f\in\mathcal{B}}=L(Y)$.
We first claim that there is a neighborhood $U$ of $(x_{f})_{f\in\mathcal{B}}$ such that if $L(A)\in U$, then $Y\subseteq A$.
Let $h\in\mathcal{B}$ be the function where $h(A)=1$ if $Y\subseteq A$ and $h(A)=2$ if $Y\not\subseteq A$. Let $t$ be a function where if $Y\not\subseteq A$ and $y_{1}\in A$, then $t(A)=y_{n-1}$ where $n$ is the least natural number with $y_{n}\not\in A$. 
I claim that $x_{h}=1$. Suppose to the contrary that $x_{h}\neq 1$. Then there is some $A$ with $h(A)=x_{h},f_{1}(A)=x_{f_{1}}=y_{1},t(A)=x_{t}$.
Since $x_{h}\neq 1$, we have $h(A)=2$, so $Y\not\subseteq A$. Since $f_{1}(A)=y_{1}$, we have $y_{1}\in A$. Therefore by the definition of $t$, we have $x_{t}=t(A)=y_{n-1}$ where $n$ is the least natural number with $y_{n}\not\in A$.
Now, there is some $B\in P_{0}(\mathbb{N})$ with $h(B)=x_{h},t(B)=x_{t}=y_{n},f_{1}(B)=x_{f_{1}}=y_{1},f_{n+1}(B)=x_{f_{n+1}}=y_{n+1}$. Then since $f_{n+1}(B)=y_{n+1}$, we have $y_{n+1}\in B$. However, $f_{1}(B)=y_{1}$, so $y_{1}\in B$ and since $h(B)=x_{h}\neq 1$, we have $Y\not\subseteq B$. Therefore since $t(B)=y_{n}$, we have $y_{n+1}\not\in B$. This is a contradiction. Therefore, we conclude that $x_{h}=1$.
Let $g$ be the function where $g(A)$ is the least element of $A\setminus Y$ whenever $A\not\subseteq Y$ and $g(A)$ is the least element of $A$ whenever $A\subseteq Y$.
We claim that $x_{g}\in Y$. To prove this claim, suppose to the contrary that $x_{g}\not\in Y$. Then $x_{g}<y_{n}$ for some $n$. Thus, there is some $A\in P_{0}(\mathbb{N})$ with $g(A)=x_{g},f_{1}(A)=x_{f_{1}}=y_{1},...,f_{n}(A)=x_{f_{n}}=y_{n}$. Therefore the first $n$ elements of $A$ are $y_{1},...,y_{n}$. This contradicts the fact that $x_{g}\in A\setminus Y$ and $x_{g}<y_{n}$. We therefore conclude that $x_{g}\in Y$.
Now let $U\subseteq\mathbb{N}^{\mathcal{B}}$ be the set where $(z_{f})_{f\in\mathcal{B}}$ if and only if $z_{g}=x_{g}$ and $z_{h}=x_{h}$. Then $U$ is an open neighborhood of $(x_{f})_{f\in\mathcal{B}}$.
If $A\in P_{0}(X)$ and $(f(A))_{f\in\mathcal{B}}=L(A)\in U$, then $g(A)=x_{g}=h(A)=x_{h}$. Therefore $h(A)=x_{h}=1$ and $g(A)=x_{g}\in Y$. Since $h(A)=x_{h}=1$, we conclude that $Y\subseteq A$. However, since $g(A)=x_{g}\in Y$, we conclude that $A\subseteq Y$ as well, so $A=Y$. We conclude that $L(Y)$ is the only possible point in $L[P_{0}(\mathbb{N})]\cap U$. Therefore since $U$ is a neighborhood of $(x_{f})_{f\in\mathcal{B}}$ and
$(x_{f})_{f\in\mathcal{B}}\in\overline{L[P_{0}(\mathbb{N})]}$, we conclude that $(x_{f})_{f\in\mathcal{B}}=L(Y)$. Thus, the set $L[P_{0}(\mathbb{N})]$ is a closed set in $\mathbb{N}^{\mathcal{B}}$.
$\textbf{Added 5/10/13}$ 
I am now going to completely characterize the cardinalities $|X|$ such that if $\varphi$ is a filter where every function in $\mathcal{A}$ is constant almost everywhere then $\varphi$ is a principal ultrafilter. This characterization is a generalization of the proof that I gave at the top of this post. As one who is familiar with large cardinals might expect, the every such ultrafilter is a principal ultrafilter if and only if $|X|$ is non-measurable.
We say that a cardinal $\lambda$ is non-measurable if there is no non-principal $\sigma$-complete ultrafilter on $\lambda$. This is equivalent to saying that $\lambda$ is below the first measurable cardinal.
$\mathbf{Theorem}$ Let $X$ be a set.
I. Let $\varphi$ be a filter on $P_{0}(X)$ such that every function $f\in\mathcal{A}$ is constant $\varphi$-a.e. Then $\varphi$ is an $\sigma$-complete ultrafilter.
II. $|X|$ is non-measurable if and only if every filter $\varphi$ on $P_{0}(X)$ such that every function $f\in\mathcal{A}$ is constant almost everywhere is a principal ultrafilter.
$\mathbf{Proof}$.
I. For each $f\in\mathcal{A}$, let $x_{f}$ be the constant with $x_{f}=f$ $\varphi$-a.e.
Without loss of generality, assume that $X=\lambda$ for some cardinal $\lambda$. For all $n\in\omega$, let $f_{n}\in\mathcal{A}$ be the function where $f_{n}(A)$ is the $n$-th element of $A$ whenever $|A|\geq n$ and $f_{n}(A)$ is the last element of $A$ whenever $|A|<n$. Let $y_{n}=x_{f_{n}}$ for all $n$. Let $Y=\{y_{n}|n>0\}$. Clearly the sequence $(y_{n})_{n}$ is increasing.
If $y_{1}=y_{2}$, then $\varphi$ is a principal ultrafilter since if $f_{1}(A)=y_{1},f_{2}(A)=y_{2}$, then $A=\{y_{1}\}$, so since $f_{1}(A)=y_{1},f_{2}(A)=y_{2}$ $\varphi$-a.e., we have $\varphi$ be the principal ultrafilter generated by $A$. Now assume that $y_{1}<y_{2}$.`
Let $S=\{A\in P_{0}(\lambda)|f_{1}(A)=y_{1},f_{2}(A)=y_{2}\}$. Then clearly $S\in\varphi$. Let $R\subseteq P_{0}(X)$. Let $h\in\mathcal{A}$ be a function where $h(A)=y_{1}$ for $A\in R\cap S$ and $h(A)=y_{2}$ for $A\in R^{c}\cap S$. Then $h$ is constant almost everywhere and $h(A)\in\{y_{1},y_{2}\}$ for almost every $A\in P_{0}(\lambda)$ and $h(A)=y_{2}$. Therefore $h(A)=y_{1}$ for almost every $A\in P_{0}(\lambda)$ or $h(A)=y_{2}$ for almost every $A\in P_{0}(\lambda)$. If $h(A)=y_{1}$ for almost every $A\in P_{0}(\lambda)$, then $R\cap S\in\varphi$ and if $h(A)=y_{2}$ a.e., then $R^{c}\cap S\in\varphi$. Therefore either $R\in\varphi$ or $R^{c}\in\varphi$. Therefore $\varphi$ is an ultrafilter.
If $y_{n}=y_{n+1}$, and $f_{1}(A)=y_{1},...,f_{n+1}(A)=y_{n+1}$, then clearly $A=\{y_{1},...,y_{n}\}$, but since $f_{1}(A)=y_{1},...,f_{n+1}(A)=y_{n+1}$ for almost every $A$, we conclude that $\varphi$ is the principal ultrafilter generated by $A$.
Now assume that $y_{n}<y_{n+1}$ for all $n$. Let $Y=\{y_{n}|n\in\mathbb{N},n>0\}$. We claim that $Y\subseteq A$ for almost every $A\in P_{0}(\mathbb{N})$. Suppose to the contrary that $Y\not\subseteq A$ for almost every $A\in P_{0}(\mathbb{N})$. Let $t\in\mathcal{A}$ be a function such that if $Y\not\subseteq A$ and $y_{1}\in A$, then $t(A)=y_{n-1}$ where $n$ is the least natural number where $y_{n}\not\in A$. Since $Y\not\subseteq A$ and $y_{1}\in A$ for almost every $A\in P_{0}(\lambda)$, we have $t(A)\in Y$ for almost every $A\in Y$. Since the function $t$ is constant almost everywhere, we conclude that $x_{t}=y_{n}$ for some $n$. Therefore, $t(A)=y_{n},f_{1}(A)=y_{1},f_{n+1}(A)=y_{n+1}$ and $Y\not\subseteq A$ for almost every $A\in P_{0}(\lambda)$. However, this is a contradiction since for such $A$ we have that  $t(A)=y_{n},f_{1}(A)=y_{1},Y\not\subseteq A$ implies that $y_{n+1}\not\in A$, but clearly $y_{n+1}=f_{n+1}(A)\in A$. Therefore, we conclude that $Y\subseteq A$ for almost every $A\in P_{0}(\mathbb{N})$.
Since $Y$ is a countable set and $Y\subseteq A$ for almost every $A$, it is now fairly easy to show that the ultrafilter $\varphi$ is $\sigma$-complete. If $P=\{R_{n}|n\in\mathbb{N},n>0\}$ is a partition of $P_{0}(\lambda)$ into countably many pieces, then let $g$ be the function where if $Y\subseteq A$ and $A\in R_{n}$, then $g(A)=y_{n}$. Then $g$ is constant almost everywhere. Furthermore, since $Y\subseteq A$ for almost every $A$, we have $g(A)\in Y$ for almost every $A$. Therefore there is some $n$ where $g(A)=y_{n}$ for almost every $A$. Therefore since $g(A)=y_{n}$ and $Y\subseteq A$ for almost every $A$, and if $g(A)=y_{n}$ and $Y\subseteq A$, then $A\in R_{n}$. Therefore $A\in R_{n}$ for almost every $A$, so $R_{n}\in\varphi$. Since the ultrafilter $\varphi$ selects an element from every countable partition of $P_{0}(\lambda)$, then ultrafilter $\varphi$ is $\sigma$-complete.
II. $\rightarrow.$ If $X$ is of non-measurable cardinality, then $P_{0}(X)$ is of non-measurable cardinality as well. Furthermore, since every filter $\varphi$ such that every function $f\in\mathcal{A}$ is constant almost everywhere is a $\sigma$-complete ultrafilter, we have $\varphi$ be a principal ultrafilter.
$\leftarrow.$ If $X=\lambda$ is of measurable cardinality, then there is some measurable cardinal $\mu$ with $\mu\leq\lambda$. Since every measurable cardinal $\mu$ is $\mu$-supercompact, there is a normal ultrafilter $U$ over $[\mu]^{<\mu}=\{A\subseteq \mu:0<|A|<\mu\}$. Therefore if $\varphi$ is the ultrafilter over $P_{0}(\lambda)$ where $R\in\varphi$ if and only if $R\cap[\mu]^{<\mu}\in U$, then every $f\in\mathcal{A}$ is constant $\varphi$-a.e., but $\varphi$ is a non-principal ultrafilter. $\mathbf{QED}$<|endoftext|>
TITLE: spectral radius monotonicity
QUESTION [5 upvotes]: I encountered an inequality when reading a paper. Can someone help to show how to prove it?
Let be the spectral radius of matrix $A$ or $\rho(A)=\max\{|\lambda|, \lambda \text{ are eigenvalues of matrix }A\}$. For matrices $S$ and $T$ with positive spectral radii, and two arbitrary real positive numbers $a$ and $b$, such that $\rho(S) < a < b$ Is the following inequality true?
$$b\rho((bI-S)^{-1}T) \leq a\rho((aI-S)^{-1}T)$$
If the above is not true in general, will it be true if $S$ and $T$ are non-negative matrices?

REPLY [9 votes]: Not true in general, as noted by @SergeiIvanov, but true for (element-wise) nonnegative matrices.
Note that if $\rho(S) < b$, then $b(bI-S)^{-1}=(I-\frac{S}{b})^{-1}=\sum_{i=0}^\infty \frac{S^i}{b^i}$. In particular, thanks to this expansion, if $\rho(S)< a < b$, then $b(bI-S)^{-1}< a(aI-S)^{-1}$ in the componentwise ordering, and thus also $b(bI-S)^{-1}T \leq a(aI-S)^{-1}T$ for any nonnegative $T$. Now, it is a part of the Perron-Frobenius theorem that for any $A,B$ with $0 \leq A \leq B$ then $\rho(A) \leq  \rho(B)$, and that's all we need here.<|endoftext|>
TITLE: Does a connected manifold with vanishing Euler characteristic admit a nowhere-vanishing vector field?
QUESTION [9 upvotes]: A version of the "hairy ball" theorem, due probably to Chern, says that the Euler-characteristic of a closed (i.e. compact without boundary) manifold $M$ can be computed as follows.  Choose any vector field $\vec v \in \Gamma(\mathrm T M)$.  If $p$ is a zero of $\vec v$, then the matrix of first derivatives at $p$ makes sense as a linear map $\partial\vec v|_p : \mathrm T_p M \to \mathrm T_p M$.  By perturbing $\vec v$ slightly, assume that at every zero,  $\partial\vec v|_p$ is invertible.  Then $\chi(M) = \sum_{\vec v(p)=0} \operatorname{sign}\bigl( \det \bigl(\partial\vec v|_p\bigr)\bigr)$.
I am curious about the following potential converse: "If $M$ is closed and connected and $\chi(M) = 0$, then $M$ admits a nowhere-vanishing vector field."
Surely the above claim is false, or else I would have learned it by now, but I am not sufficiently creative to find a counterexample.  Moreover, I can easily see an outline of a proof in the affirmative, which I will post as an "answer" below, in the hopes that an error can be pointed out.  Thus my question:

What is an example of a compact, connected, boundary-free manifold with vanishing Euler characterstic that does not admit a nowhere-vanishing vector field?  (Or does no such example exist?)

REPLY [10 votes]: That a compact manifold M with vanishing Euler characteristic has a nonvanishing vector field was proved by Heinz Hopf,  Vektorfelder in Mannifaltigkeiten, Math. Annalen  95 (1925), 340-367.  A pretty convincing "intuitive proof" was outlined by Norman Steenrod in his book Fibre Bundles (Theorem 39.7), using a smooth triangulation and obstruction theory.  For a complete proof he refers to page 549 of the 1935 book Topologie by P. Alexandroff and H. Hopf<|endoftext|>
TITLE: When does an even-dimensional manifold fiber over an odd-dimensional manifold?
QUESTION [10 upvotes]: Are there simple necessary and sufficient conditions for an (oriented) even-dimensional compact smooth manifold to fiber over an (oriented) odd-dimensional manifold (with oriented fibers)?

For example, if $M \to N$ is a fiber bundle of compact manifolds with fiber $F$, then their Euler characteristics satisfy $\chi(M) = \chi(N)\chi(F)$.  But if $N$ and $F$ are odd-dimensional, $\chi(N) = \chi(F) = 0$, and so $\chi(M)$ must also vanish.  Is the converse true?  I.e. if $M$ has vanishing Euler class, does it fiber over an odd-dimensional manifold?  Or perhaps there's a hint that $\chi(M) = 0^2$, not just $0$, and so maybe some combination of Massey products also must vanish?

REPLY [5 votes]: No, I don't believe there's a simple solution. But
here's an approach to the problem which indicates how it can be fractured up.
Assume $M,N$ are closed and connected. 
If $f\: M \to N$ is homotopic to a smooth fiber bundle
with $M$ and $N$ compact, then the fibers are homotopy finite (i.e., they are homotopy equivalent to a finite complex). 
Conversely, it is a result first stated by Quinn (later proved by Gottlieb, and then differently by me) that if $f\: M^m \to N^n$ is such that its homotopy fiber $F$ (at some basepoint in $N$) is homotopy finite, then $F$ is a Poincare duality space of dimension $m-n$. Thus, $f$ gives rise to a fibered surgery problem. 
One can approach this problem in two steps:
Step 1: find a block bundle  $E \to N$ and a fiber homotopy equivalence $E\simeq M$.
This step can be attacked classical surgery techniques (here the dimension of the fiber should be $\ge 6$).  What one studies here is the map $\tilde S_N(M) \to \tilde S(M)$ from the fiberwise block structure space to the block structure space. 
Step 2: Study the map $S_N(M) \to \tilde S_N(M)$ from the fiberwise structure space to the fiberwise block structure space. This step involves higher algebraic $K$-theory a la Waldhausen. This step is only really understood in the "concordance stable range" which in this case requires $4n \le m$ (approximately).${}^\dagger$
The above is only meant to be an outline. I first learned about these ideas from the papers of Weiss and Williams, most notably:
Automorphisms of manifolds. Surveys on surgery theory, Vol. 2, 165–220, Ann. of Math. Stud., 149, Princeton Univ. Press, Princeton, NJ, 2001
An alternative approach which packages Step 1 and Step 2 into a single step
is in the third WW paper which can be obtained from Michael Weiss' website.
More recently, see the papers of Wolfgang Steimle, especially
Obstructions to stably fibering manifolds.
Geom. Topol. 16 (2012), no. 3, 1691–1724
${}^\dagger$ Added Later: According to Steimle, the "stable range" for the fibering problem is more complicated than what I wrote above. Rather than write it down, let me refer to his paper for the actual range.<|endoftext|>
TITLE: Modern Mathematical Achievements Accessible to Undergraduates
QUESTION [88 upvotes]: While there is tremendous progress happening in mathematics, most of it is just accessible to specialists. In many cases, the proofs of great results are both long and use difficult techniques. Even most research topologists would not be able to understand the proof of the virtually fibering conjecture or the Kervaire problem, to name just two recent breakthroughs in topology, without spending months on it. 
But there are some exceptions from this rule. As a topologist, I think here mostly about knot theory:

The Jones and HOMFLY polynomials. While the Jones polynomial was first discussed from a more complicated context, a rather simple combinatorial description was found. These polynomials help to distinguish many knots.
The recent proof by Pardon that knots can be arbitrarily distorted. This might be not as important as the Jones polynomial, but quite remarkably Pardon was still an undergrad then!

Or an example from number theory are the 15- and 290-theorems: If a positive definite integer valued qudadratic form represents the first 290 natural numbers, it represents every natural number. If the matrix associated to the quadratic form has integral entries, even the first 15 natural numbers are enough. [Due to Conway, Schneeberger, Bhargava and Hanke.] [Edit: As mentioned by Henry Cohn, only the 15-theorem has a proof accessible to undergrads.]
My question is now the following:

What other major achievements in mathematics of the last 30 years are there which are accessible to undergraduates (including the proofs)?

REPLY [3 votes]: How major was the discovery of the 15th tiling pentagon, https://en.wikipedia.org/wiki/Pentagonal_tiling#Mann.2FMcLoud.2FVon_Derau_.282015.29_Type_15? The history makes for a nice story.<|endoftext|>
TITLE: What arithmetic information is contained in the algebraic K-theory of the integers
QUESTION [54 upvotes]: I'm always looking for applications of homotopy theory to other fields, mostly as a way to make my talks more interesting or to motivate the field to non-specialists. It seems like most talks about Algebraic $K$-theory mention that we don't know $K(\mathbb{Z})$ and that somehow $K(\mathbb{Z})$ is worth computing because it contains lots of arithmetic information. I'd like to better understand what kinds of arithmetic information it contains. I've been unable to answer number theorists who've asked me this before. A related question is about what information is contained in $K(S)$ where $S$ is the sphere spectrum.
I am aware of Vandiver's Conjecture and that it is equivalent to the statement that $K_n(\mathbb{Z})=0$ whenever $4 | n$. I also know there's some connection between $K$-theory and Motivic Homotopy Theory, but I don't understand this very well (and I don't know if $K(\mathbb{Z})$ helps). It seems difficult to search for this topic on google. Hence my question:

Can you give me some examples of places where computations in $K(\mathbb{Z})$ or $K(S)$ would solve open problems in arithmetic or would recover known theorems with difficult proofs?

I'm hoping someone who has experience motivating this field to number theorists will come on and give his/her usual spiel. Here are some potential answers I might give a number theorist if I understood them better...The wikipedia page for Algebraic K-theory mentions non-commutative Iwasawa Theory, L-functions (and maybe even Birch-Swinnerton-Dyer?), and Bass's conjecture. I don't know anything about this, not even whether knowing $K(\mathbb{Z})$ would help. Quillen-Lichtenbaum seems related to $K(\mathbb{Z})$, but it seems it would tell us things about $K(\mathbb{Z})$ not the other way around. Milnor's Conjecture (or should we call it Voevodsky's Theorem?) is definitely an important application of $K$-theory, but it's the $K$-theory of field of characteristic $p$, not $K(\mathbb{Z})$.
There was a previous MO question about the big picture behind Algebraic K Theory but I couldn't see in those answers many applications to number theory. There's a survey written by Weibel on the history of the field, and that includes some problems it's solved (e.g. the congruence subgroup problem) but other than Quillen-Lichtenbaum I can't see anything which relies on $K(\mathbb{Z})$ as opposed to $K(R)$ for other rings. If $K(\mathbb{Z})$ could help compute $K(R)$ for general $R$ then that would be something I've love to hear about.

REPLY [66 votes]: $\newcommand\Z{\mathbf{Z}}$
$\newcommand\Q{\mathbf{Q}}$
I'm a number theorist who already thinks of the algebraic $K$-theory of $\Z$ as part of number theory anyway, but let me make some general remarks.
A narrow answer: Since (following work of Voevodsky, Rost, and many others) the $K$-groups of $\Z$ may be identified with Galois cohomology groups (with controlled ramification) of certain Tate twists $\Z_p(n)$, the answer is literally "the information contained in the $K$-groups is the same as the information contained in the appropriate Galois cohomology groups."
To make this more specific, one can look at the rank and the torsion part of these groups.  

The ranks (of the odd $K$-groups) are related to $H^1(\Q,\Q_p(n))$ (the Galois groups will be modified by local conditions which I will suppress), which is related to the  group of extensions of (the Galois modules) $\Q_p$ by $\Q_p(n)$. A formula of Tate computes the Euler characteristic of $\Q_p(n)$, but the cohomological dimension of $\Q$ is $2$, so there is also an $H^2$ term. The computation of the rational $K$-groups by Borel, together with the construction of surjective Chern classes by Soulé allows one to compute these groups explicitly for positive integers $n$. There is no other proof of this result, as far as I know (of course it is trivial in the case when $p$ is regular).
The (interesting) torsion classes in $K$-groups are directly related to the class groups of cyclotomic extensions. For example, let $\chi: \mathrm{Gal}(\overline{\Q}/\Q) \rightarrow \mathbf{F}^{\times}_p$ be the mod-$p$ cyclotomic character. Then one can ask whether there exist extensions of Galois modules:

$$0 \rightarrow \mathbf{F}_p(\chi^{2n}) \rightarrow V \rightarrow \mathbf{F}_p \rightarrow 0$$
which are unramified everywhere. Such classes (warning: possible sign error/indexing disaster alert) are the same as giving $p$-torsion classes in $K_{4n}(\Z)$. The non-existence of such classes for all $n$ and $p$ is Vandiver's conjecture. Now we see that:
The finiteness of $K$-groups implies that, for any fixed $n$, there are only finitely many $p$ such that an extension exists. An, for example, an explicit computation of $K_8(\Z)$ will determine explicitly all such primes (namely, the primes dividing the order of $K_8(\Z)$). As a number theorist, I think that Vandiver's conjecture is a little silly --- its natural generalization is false and there's no compelling reason for it to be true. The "true" statement which is always correct is that $K_{2n}(\mathcal{O}_F)$ is finite.
Regulators. Also important is that $K_*(\Z)$ admits natural maps to real vector spaces whose image is (in many cases) a lattice whose volume can be given in terms of zeta functions (Borel). So $K$-theory is directly related to problems concerning zeta values, which are surely of interest to number theorists. The natural generalization of this conjecture is one of the fundamental problems of number theory (and includes as special cases the Birch--Swinnerton-Dyer conjecture, etc.). There are also $p$-adic versions of these constructions which also immediately lead to open problems, even for $K_1$ (specifically, Leopoldt's conjecture and its generalizations.)
A broader answer: A lot of number theorists are interested in the Langlands programme, and in particular with automorphic representations for $\mathrm{GL}(n)/\Q$. There is a special subclass of such representations (regular, algebraic, and cuspidal) which on the one hand give rise to regular $n$-dimensional geometric Galois representations (which should be irreducible and motivic), and on the other hand correspond to rational cohomology classes in  the symmetric space for $\mathrm{GL}(n)/\Q$, which (as it is essentially a $K(\pi,1)$) is the same as the rational cohomology of congruence subgroups of $\mathrm{GL}_n(\Z)$. Recent experience suggests that in order to prove reciprocity conjectures it will also be necessary to understand the integral cohomology of these groups. Now the cohomology classes corresponding to these cuspidal forms are unstable classes, but one can imagine a square with four corners as follows:
 stable cohomology over $\mathbf{R}$: the trivial representation.
 unstable cohomology over $\mathbf{R}$: regular algebraic automorphic forms for $\mathrm{GL}(n)/\Q$.
 stable cohomology over $\mathbf{Z}$: algebraic $K$-theory.
 unstable cohomology over $\mathbf{Z}$: ?"torsion automorphic forms"?, or at the very least, something interesting and important but not well understood.
From this optic, algebraic $K$-theory of (say) rings of integers of number fields is very naturally part of the Langlands programme, broadly construed.
Final Remark:  algebraic K-theory is a (beautiful) language invented by Quillen to explain certain phenomena; I think it is a little dangerous to think of it as being an application of "homotopy theory". Progress in the problems above required harmonic analysis and representation theory (understanding automorphic forms), Galois cohomology, as well as homotopy theory and many other ingredients. Progress in open questions (such as Leopoldt's conjecture) will also presumably require completely new methods.<|endoftext|>
TITLE: Isoperimetric inequality on a Riemannian sphere
QUESTION [24 upvotes]: Consider a two-dimensional sphere with a Riemannian metric of total area $4\pi$. Does there exist a subset whose area equals $2\pi$ and whose boundary has length no greater than $2\pi$?
(To avoid technicalities, let's require that the boundary is a 1-dimensional smooth submanifold.)
If that fails, does there exists a set, say, with area between $\pi$ and $2\pi$ and length of the boundary no greater than that of the spherical cap of the same area? Or at least no greater than $2\pi$?
More generally, I am interested in any results saying that the isoperimetric profile of the round metric on the sphere is maximal in some sense (among all Riemannian metrics of the same area).
Notes.

The answer is affirmative for central symmetric metrics (i.e. if the metric admits an  $\mathbb{RP}^2$ quotient). This follows from Pu's isosystolic inequality: in $\mathbb{RP}^2$ with a metric of area $2\pi$ there exists a non-contractible loop of length at most $\pi$. The lift to the sphere is a loop of length at most $2\pi$ dividing the area in two equal parts.
One should not require that the set is bounded by a single loop. A counter-example is the surface of a small neighborhood of a tripod (formed by three long segments starting from one point) in $\mathbb R^3$. Here one can divide the area in half by two short loops, but one loop would be long. (However one can cut off 1/3 of the area by one short loop.)
In $S^3$ the similar assertion is false, moreover the minimal area of the boundary of a half-volume set can be arbitrary large.

REPLY [5 votes]: The result of F.Balacheff and S.Sabourau mentioned by Alvarez Paiva can be improved as follows:
For any Riemannian 2-sphere M there exists a tree T with vertices of degree at most 3 and continuous maps $f:M \rightarrow T$ and $g: T \rightarrow \mathbb{R}$, such that the following holds:

Preimage of an interior point $x \in T$ under $f$ is a simple closed curve; preimage of a vertex of degree 3 is a figure eight curve.
$length((f \circ g) ^{-1} (x)) < 26 \sqrt{Area(M)}$ for all $x \in \mathbb{R}$.

In particular, the sphere can be subdivided into 2 regions each of area $\geq \frac{Area(M)}{3}$ by a simple closed curve or a figure eight curve of length $< 26 \sqrt{Area(M)}$. It can be subdivided into 2 equal parts by a collection of curves of total length $< 26 \sqrt{Area(M)}$, but then there is no control over the number of curves.
The proof of this result is not written yet, but I plan to include it in my PhD thesis. The reason for constant 26 is the following: we obtain the above sweep-out by successively dividing regions of the sphere by short arcs of length $\leq 2 \sqrt{3} \sqrt{Area(A)} +\epsilon$. $A$ denotes the region and constant $2 \sqrt{3}$ comes from the paper of P.Papasoglu. Every subdivision produces new regions of area $\leq \frac{3}{4} Area(A)$. We proceed until the total area becomes small compared to injectivity radius of $M$. Then we use Besicovitch inequality to bound lengths of curves in the sweep-out of this region in terms of its boundary length. In the process of subdivision we accumulated $2 \sqrt{3} \sum_{i=0} ^{\infty} (\frac{\sqrt{3}}{2})^i \sqrt{Area(M)}+\epsilon< 26 \sqrt{Area(M)}$.
Although $26$ is better than $10^8$ in F.Balacheff and S.Sabourau's paper it seems far from the optimal and at the moment I do not see how these methods can be used to produce a much better bound. A somewhat different problem about the relationship between isoperimetric profile and the diameter is studied in Surfaces of small diameter with large width. It is shown that one can not bound the length of a cycle subdividing the surface into two parts of equal area in terms of diameter, but if it is allowed for one of the parts to be slightly smaller such a bound exists.<|endoftext|>
TITLE: Could we interpolate the compactness of compact operators?
QUESTION [5 upvotes]: Classical theorems of Marcinkiewicz and Riesz and their extensions to general Banach spaces by Calderón, Lions, Peetre, et al. allow us to interpolate the continuity of two operators, viz., the suitable "intermediate operators" of the two operators inherit the boundedness of the latter. Do these operators inherit other properties of the endpoint operators, such as compactness? A quick google search turns up a 2006 survey which states that the problem of interpolating compactness is open, and I would like to know what is known about this and other related problems.
Edit:The linked survey is from 2006, not 1991. Here's also a 2008 paper by the first author of the survey, which seems to indicate that the problem of interpolating compactness is still open. I'd still like to know if interpolating any other functional-analytic properties of operators has been considered (or is of interest, even).

REPLY [6 votes]: For $L^p$ spaces over countably generated $\sigma$-finite measure spaces (with $1 \leq p \leq \infty$), the answer is yes - compactness at one endpoint propagates to the midpoints. My reference for this is Davies' 'Heat Kernels and Spectral Theory', Theorem 1.6.1, but apparently this goes back to Persson 'Compact linear mappings between interpolation spaces' (1964) (I don't have access to the reference at the moment).
More precisely, when we have an operator $A$ which is compact on $L^{p_0}$ and $L^{p_1}$ ($1 \leq p_0 < p_1 \leq \infty$), we can extend $A$ to a compact operator on $L^p$ ($p_0 \leq p < p_1$). There's nothing special about compactness being on the lower endpoint, except that we would then have to take $p_1 < \infty$.
The key idea of the proof is that we can strongly approximate the identity operator on $L^q$ (for all $1 \leq q < \infty$) by a sequence $(P_{k})_{k=1}^\infty$ of finite rank projections $P$, of the form
$$ Pf = \sum_{r=1}^n \chi_{E_r} |E_r|^{-1} \int_{E_r} f(x) \; dx$$
where ${E_n}$ is a sequence of disjoint subsets of finite positive measure.
These projections are contractions on $L^q$ for all $1 \leq q \leq \infty$.
The rest of the proof is a nice interpolation argument.
Since $A$ is compact we have that $\lim_{k \to \infty} P_k A = A$ (as bounded operators on $L^{p_0}$), and since each $P_k$ is a contraction on $L^{p_1}$ we have
$$\lim_{k \to \infty}{||A - P_k A||}_{p_1} \leq 2||A||_{p_1}.$$
Interpolating between $p_0$ and $p_1$ for each $k$ then shows that $P_k A$ converges to $A$ as bounded operators on $L^p$.
I have no idea what the situation is for more general Banach spaces, and this is certainly the most simple setting to work in (and probably not what you were looking for), but I thought this argument was too nice not to post!<|endoftext|>
TITLE: First order decidability of rings vs Diophantine decidability
QUESTION [14 upvotes]: Are there known (preferably ``concrete'') examples of a ring $R$ (commutative, with 1) such that:  
$\bullet$ the first order theory of $R$ is undecidable, but
$\bullet$ the positive existential (= Diophantine) theory of $R$ is decidable?
The Diophantine theory consists of formulas of the form $\exists x S(x)$ where $x$ is an $n$-tuple of variables and  $S$ denotes a finite system of polynomial equations, with coefficients in (some subring of) $R$.

REPLY [11 votes]: Let $F=\mathbb{R}(t)$ be the field of rational functions in the variable $t$ with real coefficients. We regard $F$ as a structure of type $(+ \times -\,\, 0\,\, 1)$. Then

The (positive) existential theory of $F$ is effectively computable (e.c.)
The full first-order theory of $F$ is not e.c.

Proof of 1: Suppose that a system of polynomial equations has a solution $\bar{r}$ in $F$. Here $\bar{r}$ is a tuple of rational functions. Choose a real number $s$ that  is not a root of any of the denominators of the rational functions $r_i$ and substitute $s$ for $t$ in $\bar{r}$, to obtain a tuple of real numbers that satisfies the same system of equations. Conversely, a tuple of reals that satisfies a given system of equations is already a tuple of rational functions. It follows that the existential theory of $F$ is e.c. if and only if the existential theory of $\mathbb{R}$ is e.c. But the last statement is true, by a well-known theorem of Tarski.
Proof of 2: A proof of the undecidablity of the first-order theory of $F$ (actually $\mathbb{R}$ can be replaced by any archimedian formally real field) is the subject of a 1961 paper by Raphael Robinson here.  Especially, look at Section 3, "The Method of Julia Robinson." The argument shows (amazingly) that the natural numbers can be defined in $F$.<|endoftext|>
TITLE: Frobenius weights on etale cohomology and purity
QUESTION [7 upvotes]: Let $X_0$ be a smooth variety (for simplicity I'm willing to assume that X is a curve) over a finite field $k$, $X$ its geometric base change, and $\mathcal{F}$ an $l$-adic etale sheaf on $X$ with $\mathcal{F}(m)$ its Tate-twist. 

What is the relation between the weights of Frobenius on $H^r(X,\mathcal{F})$ and $H^r(X,\mathcal{F}(m))$? 

I remember seeing somewhere (in Milne?) that $H^r(X,\mathcal{F})\otimes \mathbb{Z}_l(m)\cong H^r(X,\mathcal{F}(m))$ under some restriction to $k$, but I can't find it now. Also, it strikes me as an odd result, considering that the analogue for Serre twists in the classical sheaf cohomology setting doesn't hold.
While searching for an answer in Milne, Etale Cohomology i read about the concept of Purity. Say, I'm willing to assume $\mathcal{F}$ is pure of some weight. 

Is there any relation between "Cohomological Purity" in the sense of Milne, namely that something happens in a certain codimension, and the concept of a sheaf being of "pure weight" $s$, i.e. having Frobenius eigenvalues with complex norm $|k|^{s/2}$?

REPLY [7 votes]: You remember correctly, that the effect of Tate twisting the sheaf $\mathcal F$ is just to multiply the Frobenius eigenvalues by a factor of $q$. This result is not surprising because the Tate twist $\mathcal F(m)$ has no relation to the Serre twist on a projective variety, they are just denoted the same way. 
The analogous operation to Tate twist in the classical cohomology of a complex variety is defined on the singular cohomology, not in coherent cohomology. There it has the effect of changing the natural $\mathbf Z$-structure on the complex cohomology by a factor $(2\pi i)$, shifting the weight filtration on the mixed Hodge structure by two steps and the Hodge filtration by one step.
Purity in $\ell$-adic cohomology has no direct relation to cohomological purity. But if you interpret a Frobenius eigenvalue of absolute value $q^{k/2}$ as something "k-dimensional", then both are statements that one thing or another is equidimensional, that it has pure dimension.<|endoftext|>
TITLE: Hardness of approximation of Dominating Set
QUESTION [6 upvotes]: It is stated throughout the computational complexity literature that the Dominating Set problem is NP-hard to approximate within a factor of $\Omega(\log n)$. To my knowledge, the first and only proof available (Lund and Yannakakis, 1994), relies on a well-known L-reduction from Set Cover to Dominating Set (also reported on Wikipedia), which implies that the two problems are equivalent in terms of approximation ratio. Because Set Cover is NP-hard to approximate within a factor of $\Omega(\log n)$, the same holds for Dominating Set.
I have reasons to believe that this may be an incorrect deduction.
Recall that, in Set Cover, the parameter $n$ is the size of the universe set. In contrast, the number of sets given as input, $m$, could be exponentially larger than $n$. Because the L-reduction from Set Cover to Dominating Set constructs a graph on $n+m$ vertices, this graph may have size exponential in $n$. Now, in Dominating Set, the "$n$" that is used in approximation bounds is in fact the number of vertices. It follows that, using this reduction, only a ratio of $O(\log \log n)$ can be deduced for Dominating Set, as opposed to $\Omega(\log n)$.
Can this proof be fixed in some easy way (or is my reasoning incorrect)?

REPLY [6 votes]: I had a private conversation with Dana Moshkovitz (whom I thank), who confirmed that, in Alon, Moshkovitz, and Safra (2006), the hard instances of Set Cover resulting from a rather involved gap-preserving reduction are all such that $m\leqslant {\rm poly}(n)$. Hence, after the L-reduction to Dominating Set, we have graphs on $|V|$ vertices, such that $|V|=n+m\leqslant a\cdot n^k+b$, for some constants $a>0$, $b\geqslant 0$, $k\geqslant 1$.  It follows that, since Set Cover is $NP$-hard to approximate within a factor of $\Omega(\log n)$, Dominating Set is $NP$-hard to approximate within a factor of $$\Omega(\log n) = \Omega\left(\log\left(\frac{|V|-b}{a}\right)^{\frac 1k}\right)= \Omega\left(\frac{\log(|V|-b)-\log a}{k}\right)=\Omega(\log |V|).$$ Determining optimal values for $a$, $b$, $k$ remains open.
It is my understanding that these matters have been overlooked by most authors, as no explicit mention to them is ever made, to the best of my knowledge. Usually it is just stated that Set Cover and Dominating Set are "equivalent" under L-reductions, hence the $c\cdot\log n$ hardness carries over to Dominating Set. Even the observation of the crucial inequality $m\leqslant {\rm poly}(n)$ resulting from the reductions to Set Cover has often been neglected by most authors (Feige being an exception), as well as the determination of an optimal constant factor for the hardness of approximation of Dominating Set.<|endoftext|>
TITLE: Elliptic curves over QQ with identical 13-isogeny
QUESTION [17 upvotes]: Dear MO Community, 
this is not a real maths question, but rather the hope that someone else has stored in his or her private archive some data I am interested in.

I'd like to know some pairs of non-isogenous elliptic curves over $\mathbf Q$ possessing the same cyclic isogeny of degree $13$, i.e. they both have an $13$-isogeny defined over $\mathbf Q$ and the kernels (over $\overline{\mathbf{Q}}$) of these isogenies are isomorphic as $Gal(\overline{\mathbf{Q}}/\mathbf{Q})$-modules.

A quick and naive search on my computer was without results so far.
Maybe someone knowns some examples of such pairs of elliptic curves and is willing to share them.
Thanks a lot.

REPLY [16 votes]: [Edited mostly to include the second example, corresponding to
$(t,X) = (3,-115/126)$]
Thanks to Jordan Ellenberg for

calling attention to this nice question on his blog.
I didn't remember an example in my "private archive", but the question
is close enough to some of my previous computations that I was able to
adapt those techniques here.  It turns out that there are infinitely many
such pairs (even up to quadratic twist); one example has both torsion
subgroups defined over the 7th cyclotomic field ${\bf Q}(\zeta_7)$:
the curve with coefficients $[0,-1,1,-2,-1]$, i.e.
$y^2 + y = x^3 - x^2 - 2x - 1$,
of conductor $147 = 3 \cdot 7^2$ and discriminant $-147$,
and the curve with coefficients
$$
[0,-1,1,-1424883795842044404862,-20702237422068075268318817670099],
$$
conductor $8480886141 = 3 \cdot 7^2 \cdot 13 \cdot 251 \cdot 17681$,
and discriminant $3 \cdot 7^2 13^{13} 251^{13} 17681$.
This felt familiar, and it turns out that I had already encountered
the quadratic twists of these curves by ${\bf Q}(\sqrt{-7})$
because one of them, also of conductor $3 \cdot 7^2$ but
discriminant $-3 \cdot 7^8$, is the Jacobian of the Shimura
modular curve computed in my paper

Elkies, N.D.: Shimura Curves for Level-3 Subgroups of the $(2,3,7)$
  Triangle Group, and Some Other Examples,
  Lecture Notes in Computer Science 4076
  (proceedings of ANTS-7, 2006; F.Hess, S.Pauli, and M.Pohst, ed.),
  302$-$316;
  arXiv:math/0409020.

(so it was already in my "public archive"...).  See page 11 of
the arXiv version:
Mark Watkins noted that this curve 147-B1(I) actually has 13-torsion
over the cubic field ${\bf Q}(\zeta_7^{\phantom1} + \zeta_7^{-1})$;
I then explained this observation from the Shimura-curve structure,
and noted (footnote 5) that the twist of $X_1(13)$ parametrizing
curves over ${\bf Q}$ with a $13$-torsion point over
${\bf Q}(\zeta_7^{\phantom1} + \zeta_7^{-1})$
has at least one more orbit of rational points, which yields
the curve of conductor $8480886141$.
As Jordan observes in his blog, and also in his comment here,
the question of finding pairs of curves with "the same" cyclic
$13$-isogeny is equivalent to finding rational points
(away from some degeneracy locus) on a certain surface $S$.
This surface turns out to be "honestly elliptic" of the simplest kind
(with $\chi=3$): the canonical class $K_S$ is positive but not ample,
with a two-dimensional space of sections that gives a map
$S \rightarrow {\bf P}^1$ whose fibers are curves of genus $1$.
This fibration has sections defined over ${\bf Q(i)}$ but not over ${\bf Q}$.
But many of the first few fibers have rational points small enough
to find by a brief computer search.  Any one such point yields
infinitely many rational points on its fiber, and thus infinitely
many pairs of $j$-invariants of elliptic curves with Galois-isomorphic
subgroups of order $13$.
The surface has a birational model
$
Y^2 = (X^2+4) A(X),
$
where $A(X)$ is the quadratic $A_2 X^2 + A_1 X + A_0$
whose coefficients $A_2,A_1,A_0$ are the following sextics in $t$:
$$
A_2(t) = t^6-4t^5+6t^4-2t^3+t^2-2t+1,
$$ $$
A_1(t) = -6t^5+26t^4-22t^3-4t^2+6t,
$$ $$
A_0(t) = 4t^6-8t^5+37t^4-74t^3+57t^2-16t+4.
$$
Thus we have for each $t$ a curve of genus $1$, though without
an obvious rational point (except for the degenerate $t=0,1,\infty$
where every $X$ makes $(X^2+4) (A_2(t) X^2 + A_1(t) X + A_0)$ a square
but the resulting elliptic curves $E,E'$ are isomorphic).
So I tried a few small values of $t$ with
Stahlke and Stoll's program ratpoints.
For $t=2$ the program reported an obstruction, and indeed
there's no $11$-adic solution.  Hence our elliptic fibration
has no section over ${\bf Q}$ (else we could specialize it at $t=2$),
though there are certainly sections over ${\bf Q}(i)$, namely $X=\pm 2i$
(and also the roots of $A(X)$).
Still we can look for rational points on individual fibers,
and we already succeed for $t=3$, finding a rational solution
at $X=-115/126$, and several solutions of larger height for
other small $t$.  An hour's exhaustive search up to height $50$ for $t_0$
and $500$ for $X$ finds three further solutions, including
$(t,X) = (33/17,0)$ which leads to the curves of conductor
$147$ and $8480886141$ exhibited above.  The solution
$(t,X) = (3,-115/126)$ corresponds to the curves
$$
[1, 1, 0, -2193228435814, -4048327365374399852],
$$
with conductor
$133333589432694 = 2 \cdot 3 \cdot 7 \cdot 181^2 \cdot 263 \cdot 607^2$,
and
$$
[1, 1, 0, -9358273692452696799, -11018986378569871927950945915],
$$
with conductor
$N = 18612166837338258 = 2 \cdot 3 \cdot 79 \cdot 181^2 \cdot 607^2 \cdot 3253$
(these curves were recovered from their $j$-invariants using J.Cremona's
conductor-minimizing Sage routine EllipticCurve_from_j);
both curves have $x$-coordinates in the same cubic field of discriminant
$181^2 607^2$, and $y$-coordinates in the quadratic extension
of that field by $\sqrt{-181 \cdot 607}$.
Details of the computation of the surface etc. coming soon
(but probably in a separate answer because this is already
quite long or a Mathoverflow answer...).<|endoftext|>
TITLE: integration by parts for the fractional Laplacian
QUESTION [7 upvotes]: Is there an integration by parts formula for fractional laplacians in $L^p(\mathbb{R}^N)$, something like 
    $$
    s\in(0,1),\qquad\int\limits_{\mathbb{R}^N}f[(-\Delta)^sg] =\int\limits_{\mathbb{R}^N}[(-\Delta)^{s}f]g 
    $$
or an intermediate formula involving "lower derivatives"?
Typically, I would like to know if
    $$
    \int\limits_{\mathbb{R}^N}f\cdot[(-\Delta)^sf] dx\geq 0
    $$
still holds true as for the usual Laplacian (say for well-behaved $f$)? Computing formally in the Fourier space with $\widehat{(-\Delta)^sf}(\xi)=|\xi|^{2s}\hat{f}(\xi)$ it seems obvious, but it is not clear to me from the Riesz potential representation of $(-\Delta)^{-s}f$. Also, what kind of regularity/decay at infinity do I need in order not to bother with boundary terms at infinity?

REPLY [8 votes]: You can integrate by parts:
$$
\int_{\mathbb{R}^d} (-\Delta)^s f(x) g(x)dx=\int_{\mathbb{R}^d} (-\Delta)^s g(x) f(x)dx. 
$$
Using Fourier and $L^2$ the equality is obvious. Let's do "by hand" in $d=1$ and $s=1/2$ (the other cases follow the same idea:
You have
$$
\int_{\mathbb{R}} (-\Delta)^{1/2} f(x) g(x)dx=\int_\mathbb{R} g(x)P.V.\int_\mathbb{R} \frac{f(x)-f(y)}{|x-y|^2}dydx$$
$$
=\int_\mathbb{R} P.V.\int_\mathbb{R} \frac{g(y)(f(y)-f(x))}{|x-y|^2}dydx=-\int_\mathbb{R} P.V.\int_\mathbb{R} \frac{g(y)(f(x)-f(y))}{|x-y|^2}dydx.
$$
From here
$$
\int_{\mathbb{R}} (-\Delta)^{1/2} f(x) g(x)dx=\frac{1}{2}\int_\mathbb{R} P.V.\int_\mathbb{R} (g(x)-g(y))\frac{f(x)-f(y)}{|x-y|^2}dydx
$$
$$
=\frac{1}{2}\int_\mathbb{R} (-\Delta)^{1/2}g(x)f(x)dx+\frac{1}{2}\int_{\mathbb{R}}P.V.\int_{\mathbb{R}} -f(y)\frac{g(x)-g(y)}{|x-y|^2}dydx
$$
$$
=\frac{1}{2}\int_\mathbb{R} (-\Delta)^{1/2}g(x)f(x)dx+\frac{1}{2}\int_{\mathbb{R}}P.V.\int_{\mathbb{R}} -f(y)\frac{g(x)-g(y)}{|x-y|^2}dydx.
$$
Now you can change variables again in the last integral and conclude the result.<|endoftext|>
TITLE: Classification of higher dimensional manifolds
QUESTION [9 upvotes]: It is known that a 2-connected closed smooth 6-manifold is homeomorphic to S^{6} 
or connected sum of (S^{3}xS^{3}). My question is whether we have a similar statement for (n-1)-connected closed smooth 2n-manifold (at least when n=4). If not, do we have a clear classification theorem?

REPLY [10 votes]: There is a series of papers from the mid-1960's by C.T.C. Wall on classification of highly connected smooth manifolds, starting with Wall, C. T. C., Classification of (n−1)-connected 2n-manifolds. Ann. of Math. vol. 75 1962 163–189. I think this paper answers your question.
The Math Review of this paper by Kervaire starts, "This paper is an application of almost everything known in differential topology to the problem of classifying (n−1)-connected differential 2n-manifolds under diffeomorphism." The classification of 2-connected 6-manifolds to which you refer is actually a later result in this series.
Wall addresses the implications of his results to PL classification; presumably, later developments about topological manifolds would give the classification up to homeomorphism.  There are probably more `modern' ways of formulating Wall's results, as the paper was written in the early days of surgery theory.  It would be a worthwhile exercise to compare the way in which classification results are done these days with the methods of the original paper.<|endoftext|>
TITLE: Infinitely many curves with isogenous Jacobians
QUESTION [18 upvotes]: Let $g\geq 4$. Are there infinitely many compact genus $g$ Riemann surfaces with (mutually) isogenous Jacobians?
Does the situation change in positive characteristic?

REPLY [4 votes]: I believe I have an example in genus 5:
Humbert curves (see either Varley's "Weddle's Surfaces, Humbert's Curves, and a Certain 4-Dimensional Abelian Variety", or exercise batch F in chapter 6 of ACGH) are in 1-1 correspondence with 5-tuples lines in the plane -- the plane being $I_C(2)$ -- up to projective transformations of the plane. These 5 lines are the image of the double cover from $W^1_4(C)$ to rank < 4 quadrics enveloping the canonical image of $C$. Each of the 5 lines is marked with 4 points - its intersection points with the other four lines, and is the 2-quotient (under the map $W^1_4\to I_C(2)$ ) of an elliptic curve which lies inside the Jacobian of $C$, where four points are the ramification points. The Jacobian of C is isogenous to the product of these five elliptic curves.
Giving the plane the projective coordinates $x,y,z$, we now consider the lines: $$x=0,\quad y=0,\quad x=z,\quad y=z,\quad x+y=az.$$
The $y$ (resp $x$) coordinate of the intersection points on the first (resp second) line are $0,1,a,\infty$; whereas the $x$ (resp $y$) coordinate of the intersection points on the third (resp fourth) lines are $0,1, 1-a, \infty$. Since $x\mapsto 1-x$ takes $0\to1,1\to0,\infty\to\infty, a\to1-a$, the four elliptic curves associated with these four lines are isomorphic. Finally, The elliptic curve associated with fifth line has an extra involution, (flipping $x,y$), so it is the same curve regardless of $a$. Hence if we pick any $a_1,a_2$ which give isogenous elliptic curves, we get isogenous Humbert curves.<|endoftext|>
TITLE: Inductive limit of C*-algebras 
QUESTION [6 upvotes]: It is known that an intertwining (or even approximately intertwining) diagram implies the isomorphism of the limit algebras. Under what conditions the converse holds?

REPLY [7 votes]: The converse holds if the inductive limits involve semiprojective building blocks.  That is: suppose
$$\varinjlim (A_i,\phi_i^{i'}) \cong \varinjlim (B_j,\psi_j^{j'})$$
(here my notation is $\phi_i^{i'}:A_i \to A_{i'}$, etc.), and each $A_i$ and $B_j$ are separable and semiprojective, then there exists subsequences $(A_{i_k}), (B_{j_k})$ and an approximate intertwining between these.<|endoftext|>
TITLE: Relating two characterizations of ${\mathfrak sl}_{n > 2}$ among simple Lie algebras
QUESTION [15 upvotes]: Let $\mathfrak g$ be a finite-dimensional simple Lie algebra over $\mathbb C$.
Theorem 1. The highest root is perpendicular to all but one simple root, except in the case ${\mathfrak g}={\mathfrak sl}_{n > 2}$, in which case it is perpendicular to all but two (the first and last).
Theorem 2. The space $Hom({\mathfrak g}\otimes {\mathfrak g},{\mathfrak g})$ is $1$-dimensional (spanned by the Lie bracket), except for $\mathfrak g = {\mathfrak sl}_{n > 2}$, in which case it is $2$-dimensional. (The symmetric product is $(A,B)\mapsto$ the traceless part of $AB+BA$.)
Bruce Fontaine showed me a proof of the second, using the first, via the geometric Satake correspondence. Much as I like that, I'm wondering if there is a more classical argument connecting these two facts.
(Feel free to retag if inspired.) EDIT: I forgot that ${\mathfrak sl}_2$ is the exception that proves the rule.

REPLY [5 votes]: As Evan points out, "modern" technology (including Littelmann paths and canonical bases) provides an improved way to think about tensor product decompositions for simple Lie algebras.   But your Theorems 1 and 2 could also be understood in purely classical terms, though I'm not sure how far anyone looked at these.   The assumption is that the classification of simple Lie algebras $\mathfrak{g}$ over $\mathbb{C}$ is in hand and we ignore rank 1.  Here $\mathfrak{sl}_{n+1}$ has Lie type $A_n$ with $n > 1$.  
Theorem 1 then is basically a case-by-case observation, using the known description of root systems as in Bourbaki (or for exceptional types more explicitly in Springer's table here).   
Then Theorem 2 is just counting the number of summands isomorphic to the adjoint module in the tensor product of this module with itself.  Here the module has highest weight equal to the highest root, which I'll call $\gamma$ (Bourbaki denotes it by $\widetilde{\alpha}$).   From the case-by-case study one knows (as indicated) that the multiplicity of the adjoint module here, or equivalently the dimension of the Hom space, is at least 2 for type $A_n$ and at least 1 for other types.   So the remaining problem is to make these bounds exact.  
It's a standard (but intricate) classical problem to work out such tensor product multiplicities for arbitrary finite dimensional highest weight representations.   However, the basic approach (going back to Brauer's Comptes rendus note in 1937 and further developed by Klimyk) typically involves a huge amount of cancellation along with a summation over the entire Weyl group $W$.  But the idea is quite simple: in our case, add to $\gamma$ the weights (= roots along with 0) of the second factor in the tensor product, each counted with its multiplicity: 1 for each root, $n$ for 0.  This gives the full list of irreducible summands with their multiplicities, but only if you transform each non-dominant weight in the list into the (shifted) dominant Weyl chamber via the dot-action of $W$ given by $w \cdot \mu = w(\mu+\rho)-\rho$.  When this unique weight is dominant in the strict sense, attach the sign of $w$ to the resulting multiplicity in the tensor product.
Brauer's method in fact implies here that we get at most $n$ occurrences of the adjoint   module as summands of the tensor product.  So the problem is to reduce this using Theorem 1.  This one does directly using the reflections $s_i$ corresponding to the simple roots along with standard root system information such as the fact that $s_i \rho = \rho - \alpha_i$ and that $s_i \gamma = \gamma$ whenever $\alpha_i$ is orthogonal to $\gamma$.   Thus $s_i \cdot (\gamma -\alpha_i) = \gamma$ (and the sign is $-1$) in the orthogonal situation.   There are more details to fill in, but the point is that it's all fairly straightforward and classical even though not transparent.
UPDATE: This question led me to consult a specialist (code name SK), who recalled a more general theorem but not its source.   I asked about that here.  Just now my
consultant has retrieved the original source in a 1996 paper by R.C. King and B.G. Wybourne here.  Like most of King's other work, the paper involves classical Lie theory of interest in mathematical physics.   The proof relies on techniques such as Schur functors and plethysm but not on Littelmann paths, etc.   (The article itself seems to be restricted to those with library subscriptions.)   
As I noted in my question,  Allen's theorem 1 translates into the more general hypothesis: the highest weight of the adjoint representation (i.e., the highest root) involves in each case just one or two fundamental weights, being orthogonal to the others.<|endoftext|>
TITLE: Which topological spaces are coset spaces of locally compact groups?
QUESTION [12 upvotes]: What is a topological characterization of the class of spaces that have the form $G/H$ for a locally compact, Hausdorff group $G$ and a closed subgroup $H$ ?
Such a space $X=G/H$ necessarily satisfies the following conditions:
1) $X$ is locally compact, Hausdorff.
2) $X$ is (topologically) homogeneous (i.e., for any points $x,y\in X$ there exists a homeomorphism $\phi$ of $X$ such that $\phi(x)=y$).
3) $X$ satisfies the Suslin condition (i.e., every collection of non-empty, disjoint, open subsets of $X$ is countable; this follows from the Haar-measure on $G$).
Are these conditions sufficient? Maybe this is easier if one restricts attention to the class of separable, metric spaces.
Note that homogeneity for $X$ means exactly that the homeomorphism-group Homeo(X) acts transitively on $X$. If $X$ is also locally compact, separable and metric, then Homeo(X) is a separable, complete metric space (in the compact-open topology), and it follows that $X$ is homeomorphic to $G/G_x$, where $G_x$ is the stabilizer subgroup (also called isotopy group). See for instance the following articles for these results:
Effros (1965) # Transformation groups and C*-algebras [Ann. Math. (2) 81]
Ungar (1975) # On all kinds of homogeneous spaces [Trans. AMS 212]
The question now is under which conditions there exists a locally compact subgroup of Homeo(X) which still acts transitively on $X$.
The motivation for this question is to clarify the notion of "homogeneous space". Sometimes in the literature, by a homogeneous space it is not understood a (topological) homogeneous space but a coset space $G/H$ where $G$ is usually even assumed to be a Lie group.

REPLY [2 votes]: This problem is considered in the recent papers by Hofmann, Kramer (http://arxiv.org/pdf/1301.5114.pdf) and Antonyan, Dobrowolski [Locally contractible coset spaces, Forum Mathematicum. 27:4 (2015), 2157–2175]. According to these papers, for a locally compact group $G$ and a closed subgroup $H$ in $G$ the homogeneous space $X=G/H$ is an Euclidean manifold if either $X$ is finite-dimensional and locally connected or $X$ contains a non-empty open set contractible in $X$. This result implies that the Hilbert and Menger cubes are not coset spaces of locally compact groups (in spite of the fact that they are topologically homogeneous).
Concerning non-metrizable compact coset spaces of locally compact groups, I think that all such spaces should be supercompact and Dugunji compact (at least this is true for compact topological groups, see http://arxiv.org/abs/1010.3329 and http://iopscience.iop.org/article/10.1070/SM1990v067n02ABEH002098/pdf).<|endoftext|>
TITLE: Enumerating/counting paths of a given length on a 2D lattice
QUESTION [8 upvotes]: All,
I'm wondering if anyone can point me to a reference on how to address the following problem. 
In my thesis work on lattice QCD many years ago I had to enumerate all possible paths of a given length on a
2D lattice, up to the symmetries of reflection and rotation (plus some other internal symmetries we won't 
concern ourselves with here). I've been wondering whether it would be
possible to compute the number of such paths without explicitly listing them all, for example using
generating-function techniques. The paths were used to generate coherent-states for fermionic operators. Each path 
had a quark at one end and an anti-quark at the other end.
To make this more concrete, I'll give some examples. Write the link in the plus and minus $x$
directions as 'x' and 'X', likewise use 'y' and 'Y' for the plus and minus $y$ directions, respectively. Note that these
'links' represent unitary operators so combinations like $xX$ and $Yy$ cancel, so are not counted. Then the first
few examples are:
Length 1: $x$
Length 2: $xx$, $xy$ 
(note here, for example, that the other paths $yy$, $YY$, $XX$, $xY$, $yX$, $yx$, 
          etc, are related to the two that I listed by symmetry, so are not counted).
Length 3: $xxx$, $xxy$, $xyx$, $xyX$
Length 4: $xxxx$, $xxxy$, $xxyx$, $xxyy$, $xxyX$, $xyxY$, $xyxy$, $xyyx$, $xyyX$, $xyXY$
and so on.
Let $A_n$ be the number of paths of length $n$. As $n$ gets large, it seems like there could be a simple asymptotic relation for $A_{n+1}/A_n$, since most paths would be extended by
adjoining the three possible directional links (that don't cancel) to the end of the path, so maybe the ratio would go to 3 (NOTE - which it seems to from Liviu's result)?
Again, this work was done years ago, but this problem has stuck in my mind. I never had any formal training in combinatorics or graphs,
but from what I've read this seems like it could be a tractable problem. For me, the issue of having to not count paths that are 
related by symmetries makes it quite difficult. 
Thanks for any information/references/thoughts on this! I plan to write some code to numerically check the behavior of $A_n$. My goal is to 
have someone point me in the right direction to get started on an 'analytic' solution or asymptotic estimate. By the way, the same 
problem arises in 3D, but I will stick to 2D first.
EDIT: Using Liviu's solution I found this sequence in OEIS, related to bending of a wire in 2D (the same problem). It is here: link text
Regards,
Tom

REPLY [5 votes]: I  am trying to make sense  of your problem.  From what I can gather    you are talking about  walks  with steps of size $1$ in each   of the  cardinal directions  East, West, North, South, (E,W,N,S).  To use your notation $E=x$, $W=X$, $N=y$, $S=Y$. You are not allowed to immediately backtrack, i.e., if  say you  take an $E$-step, then your  next step cannot be a $W$-step.   I will refer to such paths as admissible.   I think that you need to fix the starting point. Assume it is the origin.
The number of admissible paths of length $n$ starting at the origin is $4\cdot 3^{n-1}$.
Alas, there is a symmetry in the problem and this is where I am a bit  confused. Its looks to me  that the symmetry group  is the group $G$ of symmetries of the square with vertices $(\pm 1,0)$, $(0,\pm 1)$. (I could be wrong, but that is  what I  am getting from your description.) The center of this square is the origin, an the vectors  obtained    by joining the center with the vertices    are the unit vectors $\vec{E},\vec{W},\vec{N},\vec{S}$ pointing in the four cardinal directions.
The group has  eight elements  and it is   generated by   the reflection $R_x$ in the $x$-axis and the counterclockwise rotation $J$ by  ninety degrees.  The elements of this group are
$$  1, J, J^2, J^3, R_x, R_xJ, R_xJ^2, R_xJ^3. $$
Assuming that my guess is correct you need to count  admissible paths,    where two paths  that are related by one of the above eight symmetries are considered identical.  Denote by $Q_n$ the number of such  paths of length $n$.
Here you need to invoke   Burnside's theorem.   Here is what it says in this case.
For each $g\in G$  denote by $P_n(g)$ the number of admissible paths of length $n$ that admit $g$ as symmetry.   More precisely   a path
$$v_1\dotsc,v_n, \;\;v_1,\dotsc,v_n\in \lbrace E,W,N,S\rbrace $$
admits $g$ as symmetry if $g(v_1)\dotsc g(v_n)=v_1\dotsc v_n$.  Then Burnside's theorem states that
$$Q_n= \frac{1}{|G|}\sum_{g\in G} P_n(g). $$
There are only  three elements  $g\in G$ for which $P_n(g)\neq 0$. They are $1, R_x, R_y$, where $R_y$ denotes the reflection in  the $y$-axis. Putting all the above together  we deduce that
$$Q_n= \frac{1}{8}\Bigl( 4\cdot 3^{n-1}+ 2+2)=\frac{1}{2}\bigl(3^{n-1}+1\bigr). $$
Update.  I have worked out the details  taking into account the correct    symmetry group.
Here is what I found. If $n$ is odd,  $n=2m-1$, then
$$ Q_n=\frac{1}{4}\bigl(3^{n-1}+2\cdot 3^{m-1}+1\bigr). $$
If $n$ is even, $n=2m$, then
$$ Q_n = \frac{1}{4}\bigl( 3^{n-1}+4\cdot 3^{m-1}+1\bigr). $$
Here are  a few values.
$$Q_1=1, \;\; Q_2=2,\;\; Q_3=4,  \;\; Q_4= 10,\;\; Q_5=25,\;\; Q_6=70. $$
Note that I get  $Q_4=10\neq 8,9$.   In any case, the  details can be found here.   Maybe somebody can explain the discrepancy involving $Q_4$.<|endoftext|>
TITLE: Is it possible to reconstruct an order type from its initial segments?
QUESTION [17 upvotes]: Suppose

$T$ is a totally ordered set without a maximal element,
$\tau$ is the order type of $T$,
$S$ is the set of order types of all proper initial segments (downward closed subsets) of $T$. 

Is it always possible to unambiguously reconstruct $\tau$ from $S$?

REPLY [15 votes]: The answer is No.
Let $\eta$ be the order type of $\mathbb{Q}$, and $\omega_1$ - the order type of the set of countable ordinals. The order types $\eta$ and $\eta \cdot \omega_1$ are different (because they have different cardinality), but the set of order types of all proper initial segments of some instances of $\eta$ and $\eta \cdot \omega_1$ is the same. Actually, as proved by Joel David Hamkins, there are $2^{\aleph_1}$ distinct order types with this property: https://math.stackexchange.com/a/174404/19661

REPLY [14 votes]: No. Take $\omega_1$, with each element replaced by a copy of $\mathbb Q$. Then $S$ will contain a single order type.  (The rest is left as an exercise.)<|endoftext|>
TITLE: Can a harmonic number be a rational number for non-integer rational argument?
QUESTION [13 upvotes]: Define harmonic numbers for a complex argument $z$ as $H_z=\frac{\Gamma'(z+1)}{\Gamma(z+1)}-\Gamma'(1)$.
For $n\in\mathbb{N}$, $H_n$ are usual harmonic numbers $\sum^n_{k=1} k^{-1}$ . They are obviously rational and are known (Taeisinger 1915) to be non-integers for $n>1$.
Question: Is there a non-integer rational $q$ such that $H_q\in\mathbb{Q}$?

REPLY [11 votes]: The answer is "no". Your function $H_z$ which is the same as $\psi(z+1)+\gamma$, where $\psi$ is the digamma function, and $\gamma$ is the Euler-Mascheroni constant, takes transcendental values at non-integer rationals. This is a theorem of M. Ram Murty and N. Saradha, "Transcendental values of the digamma function".
Notice that at rational values the digamma function has an explicit evaluation given by Gauss's formula.<|endoftext|>
TITLE: Cohomology of twisted holomorphic forms on Fano threefolds
QUESTION [6 upvotes]: Given a Fano threefold $X$, its index $ind(X)$ is the largest integer $r$ such that there exists a divisor $H$ such that $rH \cong -K_X$.  Let $\mathcal{L}$ be the associated (ample) line bundle and define the twisted forms $\Omega^q(k) = \Omega^q \otimes \mathcal{L}^k$.
When do the twisted cohomology groups $H^p(X, \Omega^q(k))$ vanish?
The Kodaira-Nakano vanishing theorem states that the cohomology groups vanish for $p+q > 3.$  For simple examples such as $\mathbb{CP}^3$ and the quadric hypersurface, many more of the twisted cohomology groups vanish.  What results are known?  Can anything stronger be said if $X$ has a Kahler-Einstein metric?
Added:
There are few scattered results in the literature.  For example:
The only two threefolds with non-vanishing $H^0(X, \Omega^1_X (1))$ are the Mukai–Umemura threefold $V_{22}$ and $V_{18}$ [math.AG/0310390].

REPLY [5 votes]: The Hodge groups $H^1(X,\Omega^2_X)$ and $H^2(X,\Omega^1_X)$ are usually not zero.  These are the groups used to construct the Griffiths intermediate Jacobian of $X$.  For many types of Fano threefolds, e.g., smooth cubic hypersurfaces in $\mathbb{C}P^4$, the Torelli theorem holds -- one can uniquely reconstruct $X$ from its polarized intermediate Jacobian (or, equivalently, polarized weight 3 Hodge structure).
$\textbf{Update}$.  The OP asks for examples where $k$ is positive.  Let $X$ be a smooth, cubic hypersurface in $\mathbb{P}^4$.  I claim that $h^1(X,\Omega^2_X(1))$ is nonzero.  Let $S$ be a general hyperplane section of $X$ so that $\mathcal{O}_X(S)$ equals $\mathcal{O}_{\mathbb{P}^4}(1)|_X$.  Then I even claim that the induced map of cohomology groups,
$$
H^1(X,\Omega^2_X) \to H^1(X,\Omega^2_X(S)),
$$
is injective.  Using the long exact sequence of cohomology, to prove this, it suffices to prove that $h^0(S,\Omega^2_X(S)|_S)$ equals $0$.  
Let $C\cong \mathbb{P}^1$ be a smooth conic in the smooth cubic surface $S$.  Then $T_X|C$ is isomorphic to $T_C\oplus (N_{C/X})$, which is $\mathcal{O}_{\mathbb{P}^1}(2)\oplus ( \mathcal{O}_{\mathbb{P}^1}(1) )^{\oplus 2}$.  Thus $\Omega_X|_C$ is isomorphic to $\Omega_C\oplus T_C^{\dagger}$, i.e., $\mathcal{O}_{\mathbb{P}^1}(-2) \oplus (\mathcal{O}_{\mathbb{P}^1}(-1))^{\oplus 2}$, where $T_C^\dagger$ denotes the annihilator of $T_C$.  Taking the second exterior power, $\Omega^2_X|_C$ is isomorphic to $(\Omega_C\otimes T_C^\dagger)\oplus \bigwedge^2 T_C^\dagger$, i.e., $(\mathcal{O}_{\mathbb{P}^1}(-3))^{\oplus 2} \oplus \mathcal{O}_{\mathbb{P}^1}(-2)$.  Finally, $\Omega^2_X(+1)|_C$ is isomorphic to $(\mathcal{O}_{\mathbb{P}^1}(-1))^{\oplus 2} \oplus \mathcal{O}_{\mathbb{P}^1}$.  Therefore, $H^0(C,\Omega^2_X(S)|_C)$ is $1$-dimensional, with the unique global section (up to scaling) in the "direction of" $\bigwedge^2 T_C^\dagger$.  One way of thinking of this is, if you contract this $2$-form with a tangent vector in $T_C$ at a point $p$ of $C$, then you get the zero $1$-form.
However, for a general point $p$ of $S$, there are $27$ different conics passing through this point with $27$ different tangent spaces that span the entire tangent space of $S$ at $p$.  For a global section in $H^0(S,\Omega^2_X(S)|_S)$, its restriction to each of these $27$ conics $C$ containing $p$, the contraction of the $2$-form with $T_{C,p}$ gives zero, and the spaces $T_{C,p}$ span $T_{S,p}$.  Thus this $2$-form must contract to zero with every element in $T_{S,p}$.  Since $T_{S,p}$ is a codimension $1$ linear subspace of $T_{X,p}$, the only $2$-form that contracts to zero with all of $T_{S,p}$ is the zero $2$-form (there are nonzero $1$-forms that contract to zero with all of $T_{S,p}$).  Therefore, every global section of $\Omega^2_X(S)|_S$ must vanish at every sufficiently general point $p$ of $S$.  Since $\Omega^2_X(S)|_S$ is a locally free sheaf on an integral scheme $S$, this means that every global section is zero.
$\textbf{Second Update}$.  I want to add one word about why $N_{C/X}$ is isomorphic to $\mathcal{O}_{\mathbb{P}^1}(1)\oplus \mathcal{O}_{\mathbb{P}^1}(1)$, as opposed to $\mathcal{O}_{\mathbb{P}^1}(1)\oplus \mathcal{O}_{\mathbb{P}^1}(1)$.  This is equivalent to saying that the induced map $H^0(C,N_{C/X}) \to H^0(C,N_{C/X}|_{0,\infty})$ is surjective, i.e., the "evaluation map" from the parameter space of $2$-pointed conics in $X$ to $X\times X$ is submersive.  By Sard's theorem / generic smoothness, it suffices to prove the map is dominant (since we are in characteristic $0$).  Given general points $x_0$, $x_\infty$ of $X$, the line $L$ spanned by $x_0$ and $x_\infty$ intersects $X$ in a third point $x_1$, which is also a general point of $X$ (since $X$ is a cubic hypersurface and the points are general).  Thus there are $6$ lines $M$ in $X$ containing $x_1$.  For each such line, for the plan $\Pi = \text{span}(L,M)$, the intersection of $\Pi$ with $X$ equals $M\cup C$, where $C$ is a conic containing $x_0$ and $x_\infty$.  Also, in all likelihood, this computation of $H^1(X,\Omega^2_X(1))$ appears somewhere in the publications of Markushevich, Tikhomirov and Iliev, since this comes up in their analysis of the Abel-Jacobi maps associated to cubic threefolds.<|endoftext|>
TITLE: Computing the Grothendieck-Springer resolution for $G = SL_2$
QUESTION [5 upvotes]: Let $G = SL_2, \mathfrak{g} = \mathfrak{sl}_2$, $B$ the Borel subgroup, and $\mathfrak{u}$ the unipotent radical; so that $G/B = \mathbb{P}^1$; how does $\widetilde{\mathfrak{g}}$ decompose as a vector bundle over $\mathbb{P}^1$? Recall the definition:
$\widetilde{\mathfrak{g}} = $ {$(X, gB) \in \mathfrak{g}^* \times \mathbb{P}^1 | X|_{g \mathfrak{u}} = 0$}
This should be simple, but I'm having trouble.

REPLY [8 votes]: A paper of Bezrukavnikov (http://arxiv.org/abs/math/0604445) identifies this as $\mathcal{O}(-1) \oplus \mathcal{O}(-1)$ in Example 2.8.  What follows is an argument.
The identification of $G$-equivariant vector bundles on $G/B$ with $B$-representations gives us a short exact sequence of $B$-representations:
$$0 \rightarrow V_{-2} \rightarrow \mathfrak{b} \rightarrow V_0 \rightarrow 0$$
which gives us a short exact sequence in vector bundles:
$$0 \rightarrow \mathcal{O}(-2) \rightarrow \tilde{\mathfrak{sl}_2} \rightarrow \mathcal{O}(0) \rightarrow 0$$
There is no decomposition theorem of $SL_2$-equivariant vector bundles on $\mathbb{P}^1$, since this category is equivalent to $B$-representations and $B$ is solvable.  I claim that if we base change to the non-equivariant case, then the short exact sequence becomes:
$$0 \rightarrow \mathcal{O}(-2) \rightarrow \mathcal{O}(-1) \oplus \mathcal{O}(-1) \rightarrow \mathcal{O}(0) \rightarrow 0$$
Let $\mathcal{E}$ be the locally free sheaf on $\mathbb{P}^1$ associated to $\tilde{\mathfrak{sl}_2}$.  Take the long exact sequence of the short exact sequence of locally free sheaves and one obtains:
$$0 \rightarrow H^0(\mathbb{P}^1, \mathcal{E}) \rightarrow H^0(\mathbb{P}^1, \mathcal{O}(0)) \rightarrow H^1(\mathbb{P}^1, \mathcal{O}(-2)) \rightarrow H^1(\mathbb{P}^1, \mathcal{E}) \rightarrow 0.$$
I claim that the first map must be zero.  By Borel-Weil-Bott all cohomologies $H^i(\mathbb{P}^1, \mathcal{E})$ are sums of trivial representations (since $H^0(\mathbb{P}^1, \mathcal{O}(0) = H^1(\mathbb{P}^1, \mathcal{O}(-2))$ are trivial); in particular all global sections are $G$-invariant.  Above the Borel $$\left(\begin{array}{cc} * & * \\ 0 & * \end{array}\right)$$
take the point in the fiber
$$\left(\begin{array}{cc} a & b \\ 0 & -a \end{array}\right)$$
and use the $G$-action to try to move it around.  In particular, the action of
$$\left(\begin{array}{cc} 1 & t \\ 0 & 1 \end{array}\right)$$
fixes the Borel but moves the section to
$$\left(\begin{array}{cc} a & -2at + b \\ 0 & -a \end{array}\right)$$
so the only section which has a chance of invariant under $G$ has $a = 0$ on this fiber (we haven't even considered whether it extends globally).  But on this fiber it maps to zero to $\mathcal{O}(0)$ and so by equivariance, if it extends to a global section, this section also maps to zero.  So the first map in the short exact sequence is zero, thus the second map is an isomorphism, and the third map is zero.  So $\mathcal{E}$ has no cohomology, and by preservation of rank, it must be $\mathcal{O}(-1) \oplus \mathcal{O}(-1)$.
Note: this bundle is as a "twist" of the usual short exact sequence for the tautological bundle on $\mathbb{P}^1$:
$$0 \rightarrow \mathcal{O}(-1) \rightarrow \mathcal{O}(0) \oplus \mathcal{O}(0) \rightarrow \mathcal{O}(1) \rightarrow 0.$$<|endoftext|>
TITLE: Summing ratio of ratio of partial sums of binomial coefficients
QUESTION [5 upvotes]: I would like to approximate the following when $n \gg k$.
$\sum_{y = k + 1}^n \frac{\sum_{m = 0}^{k - 1} {y - 2 \choose m} (y - 1)}{\sum_{m = 0}^k {y - 1 \choose m}}.$
The formula can be re-written as
$\sum_{y = k + 1}^n \frac{(y - 1) + \sum_{m = 1}^{k - 1} {y - 1 \choose m + 1} (m + 1)}{\sum_{m = 0}^k {y - 1 \choose m}}.$
However since the partial sum of binomial coefficient does not closed form, I could not see any way to further simplify the formula. Any help is much appreciated. Thanks!

REPLY [3 votes]: $$ \sum_{m=0}^k \binom{z+1}{m} = \sum_{m=0}^k [  \binom{z}{m-1} + \binom{z}{m} ]
= \binom{z}{k} + 2 \sum_{m=0}^{k-1}\binom{z}{m},$$ when $z \geq k$.  So the sum
can be rewritten as
$$ \sum_{z=k}^{n-1} \frac{z}{2 + \binom{z-1}{k}/\sum_{m=0}^{k-1}\binom{z-1}{m}}. $$   Let's
call this summand $a_z$ .  Note that $a_k = k/2, a_{k+1} = (k+1)/(2+1/(2^k - 1))$, and for small
values of $j$, $a_{k+j} = (k+j)/(2 + \binom{k+j-1}{j-1}/(2^{k+j-1}-\sum_{m=0}^{j-1} \binom{k+j-1}{m}))$
.  Now as z increases, the denominator of $a_z$ eventually tends monotonically to $(z+k)/k$, which means
$a_z$ tends to $k$ from below.  Thus the entire sum has an upper bound (for $n \gg k$) of $(n-k)k$.
Further, for sufficiently large integers $l$, $a_{lk}$ is slightly larger than $lk/(l+1)$, so I imagine the actual sum differs from the upper bound by
a $O(n\log{k}) $ amount.
Gerhard "Ask Me About System Design" Paseman, 2013.05.08<|endoftext|>
TITLE: Evaluation of an $n$-dimensional integral
QUESTION [22 upvotes]: I asked the same question on math.se but got no answer there. Since it pertains to my current research, I decided to ask here:
Let $n\in 2\mathbb{N}$ be an even number. I want to evaluate
$$I_n
:=
\int_0^1\mathrm{d} u_1 \cdots \int_0^1 \mathrm{d} u_n \frac{\delta(1-u_1-\cdots-u_n)}{(u_1+u_2)(u_2+u_3)\cdots(u_{n-1}+u_n)(u_n+u_1)}.
$$
For small $n$, this is computable by simply parameterizing the $\delta$ function, and
$I_2 = 1$, $I_3 = \pi^2/4$, $I_4 = 2\pi^2/3$. The values of $I_5$ and $I_6$ are numerically $18.2642 \approx 3\pi^4/16$ and $51.9325\approx 8\pi^4/15$. I strongly suspect that
$$
I_{2n+2} \stackrel{?}{=} (2\pi)^{2n} \frac{(n!)^2}{(2n+1)!} =
\frac{(2\pi)^{2n}}{\binom{2n+1}{n}(n+1)}
= (2\pi)^{2n}\mathrm{B}(n+1,n+1),
$$
where $\mathrm{B}$ is the Beta function. Dividing by $(2\pi)^{2n}$, this is Sloane's A002457. For $I_6$, this conjecture is equivalent to
$$\int_0^1\mathrm{d}x \Bigl(\mathrm{Li}_2(\frac{x-1}{x})\Bigr)^2 \stackrel{?}{=} \frac{17}{180}\pi^4$$
(with $\mathrm{Li}_2$ the dilogarithm), which seems to be true numerically, but I could neither prove it nor find it in the literature.
As a last remark, it is possible to get rid of the $\delta$ function by using the identity
$$I_n = \int_{(0,\infty)^n}\mathrm{d}u \frac{f(\lvert u\rvert_1)}{(u_1+u_2)\cdots(u_n+u_1)} \Bigm/\int_0^\infty\mathrm{d}t \frac{f(t)}{t}$$
for any $f:(0,\infty)\to\mathbb{R}$ that makes both integrals finite. Using $f(t) = t 1_{[0,1]}(t)$ where $1_{[0,1]}$ is the characteristic function of the interval $[0,1]$, one can write $I_n$ as an integral over an $n$-dimensional simplex.

REPLY [5 votes]: Here is another proof, the main part of which was communicated to me by Dr. Peter Otte of Bochum University:
\begin{equation}
I_n := \int_{[0,1]^n}\mathrm{d}u\,\delta(1-\lvert u\rvert_1)
\frac{1}{\prod_{j=1}^n (u_j + u_{j+1})}
=
(2\pi)^{n-2}
\frac{[\Gamma(\frac{n}{2})]^2}{\Gamma(n)}.
\end{equation}
First, define
$$J_n(t) := \int_{[0,1]^n}\mathrm{d}u\,\delta(t-\lvert u\rvert_1)
           \frac{1}{\prod_{j=1}^{n-1}(u_j + u_{j+1})}.$$
for $t>0$. By scaling, $J_n(t) = J_n(1) =: J_n$ for all $t > 0$. Also,
    \begin{align}
      I_n &
      = \frac{1}{2}\int_{[0,1]^n}\mathrm{d}u\, \delta(1-\lvert u\rvert_1)
      \frac{2\lvert u\rvert_1}{\prod_{j=1}^n (u_j + u_{j+1})} \notag\\
      & =
      \frac{1}{2}\sum_{k=1}^n \int_{[0,1]^n}\mathrm{d}u\, \delta(1-\lvert u\rvert_1)
      \frac{u_k+u_{k+1}}{\prod_{j=1}^n (u_j + u_{j+1})} \notag\\
      & = \frac{n}{2} \int_{[0,1]^n}\mathrm{d}u\,
      \frac{\delta(1-\lvert u\rvert_1)}{(u_1+u_2)\dotsm(u_{n-1}+u_n)} =
      \frac{n}{2} J_n.
    \end{align}
Next, let $f\in L_1(0,\infty)$. Then
  \begin{equation}
    J_n
    =
    \int_{(0,\infty)^n}\mathrm{d}u\, \frac{f(\lvert u\rvert_1)}{\prod_{j=1}^{n-1}(u_j + u_{j+1})}
    \Bigm/\!
    \int_0^\infty\mathrm{d}t\, f(t).
  \end{equation}
  In particular,
  \begin{equation}
    J_n = \int_{(0,\infty)^n}\mathrm{d}u\, \frac{e^{-\lvert u\rvert_1}}{\prod_{j=1}^{n-1}(u_j + u_{j+1})},
  \end{equation}
We will need the Rosenblum-Rovnyak integral operator
$T: L_2(0,\infty)\to L_2(0,\infty)$,
see Rosenblum (1958) and Rovnyak (1970), defined via
\begin{equation}
  (Tf)(x) := \int_0^\infty \mathrm{d}y\, \frac{e^{-(x+y)/2}}{x+y} f(y)
  \quad (x\in(0,\infty)).
\end{equation}
for $f\in L_2(0,\infty)$. This is the special case $T = \mathcal{H}_0$ in
Rosenblum (1958), Formula (2.3).
The operator $T$ is unitary equivalent to the Hilbert matrix
$H:\ell_2(\mathbb{N})\to\ell_2(\mathbb{N})$,
\begin{equation}
  (H x)_j = \sum_{k=1}^\infty \frac{x_k}{j+k-1} \quad(j\in\mathbb{N}, x\in\ell_2(\mathbb{N}))
\end{equation}
and can be explicitly diagonalized: Following
Yafaev (2010), Sec. 4.2, we define the unitary operator
$U: L_2(0,\infty)\to L_2(0,\infty)$ via
\begin{equation}
  (Uf)(k) = \pi^{-1}\sqrt{k\sinh 2\pi k} \, \lvert \Gamma(1/2 - ik)\rvert
            \int_0^\infty\mathrm{d}x\, x^{-1} W_{0,ik}(x)f(x)
\end{equation}
for $f\in L_2(0,\infty)$ and $k\in(0,\infty)$, where the Whittaker
functions are given by
\begin{equation}
  W_{0,\nu}(x) = \sqrt{x/\pi} K_\nu(x/2) \quad (\nu, x\in(0,\infty)),
\end{equation}
with $K_\nu$ as the modified Bessel function of the second kind, see
DLMF.
In order to compute $J_n$, we will employ the following result due to
Rosenblum, see Yafaev, Prop. 4.1:
\begin{equation}
 (UTf)(k) = \frac{\pi}{\cosh(k\pi)}(Uf)(k)  \quad (k\in(0,\infty),
  f\in L_2(0,\infty).
\end{equation}
Proof of $I_n = (2\pi)^{n-2} \frac{[\Gamma(\frac{n}{2})]^2}{\Gamma(n)}$.
  Let $n\in\mathbb{N}_{\ge 2}$. From the definition of $T$ and the
  identity of $J_n$ above, we see that
  \begin{equation}
    J_n = \langle f_0, T^{n-1}f_0\rangle
  \end{equation}
  with $f_0(x) := e^{-x/2}$.
  From this and the identity of $UT$ above, we obtain
  \begin{equation}
    J_n =
    \langle Uf_0, UT^{n-1}f_0\rangle
    =
    \int_0^\infty\mathrm{d}k\, \lvert \hat{f}_0(k)\rvert^2 \Bigl(\frac{\pi}{\cosh(k\pi)}\Bigr)^{n-1},
  \end{equation}
  where $\hat{f}_0 := Uf_0$. In order to compute $\hat{f}_0$, we
  employ the classical formula
  \begin{equation}
    \lvert\Gamma(1/2 - ik)\rvert^2 = \frac{\pi}{\cosh(k\pi)}  \quad (k\in\mathbb{R}),
  \end{equation}
  which is a consequence of the reflection formula for the Gamma
  function, and
  \begin{equation}
    \int_0^\infty\mathrm{d}x\, x^{-1} W_{0,ik}(x)e^{-x/2} =
    \frac{\pi}{\cosh(k\pi)}
    \quad(k > 0),
  \end{equation}
  which follows from the special case $z=1/2$ and $\nu = \kappa = 0$ in
  DLMF. From the definition of $U$ above and the last two equations,
  we deduce
  \begin{equation}
    \lvert\hat{f}_0(k)\rvert^2 = 2\pi k\frac{\sinh(k\pi)}{\cosh(k\pi)^2}
    \quad (k > 0).
  \end{equation}
  This yields
  \begin{equation}
    J_n = 2\pi^{n-2}\int_0^\infty\mathrm{d}k\, k
    \frac{\sinh(k)}{\cosh(k)^{n+1}}
    =
    \frac{2\pi^{n-2}}{n}\int_0^\infty\mathrm{d}k\,\frac{1}{\cosh(k)^n}
  \end{equation}
  where we applied the substitution $\tilde{k} = k\pi$ and integrated
  by parts. This integral can be evaluated using the substitutions
  $y = \cosh(k)^{-1}$ and $x = y^2$, one after the other:
  \begin{align}
    J_n
    =
    \frac{2\pi^{n-2}}{n}
    \int_0^1\mathrm{d}y\, \frac{y^{n-1}}{\sqrt{1-y^2}}
    & =
    \frac{\pi^{n-2}}{n}
    \int_0^1\mathrm{d}x\, x^{n/2-1}(1-x)^{-1/2} \\
    & =
    \frac{\pi^{n-2}}{n} \mathrm{B}(n/2, 1/2),
  \end{align}
  since $k'(y) = - y^{-1}(1-y^2)^{-1/2}$.
  The claim then follows by expressing the Beta function via the Gamma
  function and then applying the classical duplication formula.<|endoftext|>
TITLE: Discrete disjoint covering of integer lattices
QUESTION [10 upvotes]: Is there a covering of $\mathbb{Z}^n$ by disjoint translates of the basis-and-origin minimal integer $n$-simplex? By haphazard I have such coverings for $\mathbb{Z}$, $\mathbb{Z}^2$ and $\mathbb{Z}^3$, where the wanted translations are lattices spanned by $\{2\}$, $\{(2,-1),(-1,2)\}$, and $\{(1,1,-1),(1,-1,1),(-1,1,1)\}$, but rhyme nor reason can I see in this sequence of families to extend.

REPLY [2 votes]: For fixed dimension $n$ there is an algorithm to find all such lattice
tilings. Namely, let $S_n$ be the set of all $n\times n$ matrices $A$
of determinant $n+1$ that are in Hermite normal form over
$\mathbb{Z}$. If the columns of $A$ are $v_1,\dots,v_n$, then there
are $n$ nonzero integer column vectors $u_1,\dots,u_n$ for which there
exist $0\leq a_i<1$ satisfying $\sum a_i v_i=u_i$. If the determinant 
of the matrix $M$ with columns $u_i$ is $\pm 1$, then the translates by
the lattice generated by $v_1,\dots,v_n$ of the origin and the vectors
$u_i$ gives a tiling of $\mathbb{Z}^n$. By a unimodular integral
change of basis we can convert the $u_i$'s to the unit coordinate
vectors.  This construction gives all the desired lattice tilings, and
is easy to implement algorithmically. For $n=4$ there are exactly two
Hermite normal forms such that $\det M=\pm 1$, namely,
  $$ \begin{bmatrix} 1 & 0 & 0 & 2\\\ 0 & 1 & 0 & 3\\\
          0 & 0 & 1 & 4\\\ 0 & 0 & 0 & 5\end{bmatrix},
  \qquad \begin{bmatrix} 1 & 0 & 0 & 1\\\ 0 & 1 & 0 & 0\\\
          0 & 0 & 1 & 1\\\ 0 & 0 & 0 & 5\end{bmatrix}. $$<|endoftext|>
TITLE: BRST cohomology definition
QUESTION [7 upvotes]: Is there written anywhere a full definition of BRST cohomology? All I have found so far is BRST cohomology in _______.
As far as I can see, BRST cohomology is the cohomology of a complex in which the differential has at least a piece that "looks like" the Chevalley Eilenberg differential.
That seems to include all known examples, but is hardly a precise definition.

REPLY [11 votes]: It is difficult to give a precise definition, because there are many cohomology theories which go by the name of BRST.  It might be helpful to give a couple of examples, not necessarily in chronological order.
In classical (i.e., non-quantum) physics, there is a notion of BRST cohomology which encodes symplectic reduction à la Marsden-Weinstein.  This was the subject of my answer to an earlier question.  It is the cohomology of a double complex and one of the differentials is the Chevalley-Eilenberg differential.  However, it does admit generalisations (e.g., to coisotropic reduction) where there is no Chevalley-Eilenberg differential, while still being referred to as BRST cohomology.
The original BRST cohomology arose in quantum field theory, where it arose as an "invariance" of the gauge-fixed Fadde'ev-Popov action for a gauge theory and plays an important role in proving the renormalisability of four-dimensional gauge theories.  The BRST differential again has a part which is the Chevalley-Eilenberg differential of the Lie algebra of the gauge group.  Again, there are generalisations (to theories with "open algebras") where there is still a BRST cohomology, but now there is no longer any Chevalley-Eilenberg differential.
In the context of two-dimensional conformal field theories (e.g., string theory), the BRST cohomology can be identified with a certain (relative) semi-infinite cohomology in the sense of Feigin.
In all cases precise definitions can be given, but they are different.<|endoftext|>
TITLE: examples of "exotic" moduli problems for elliptic curves?
QUESTION [10 upvotes]: Let $\textbf{Ell}$ be the category of elliptic curves over various base schemes, and where a morphism between $E\rightarrow S$ and $E'\rightarrow S'$ is a cartesian diagram with those two maps as columns.
A moduli problem for elliptic curves is then just a contravariant functor $\textbf{Ell}\rightarrow\textbf{Sets}$.
For example, we usual level $N$ moduli problem is the functor sending $E/S$ to the set of isomorphisms $E[N]\stackrel{\sim}{\longrightarrow}(\mathbb{Z}/N\mathbb{Z})_S^2$.
There are a ton of these functors, mostly coming from various cohomology theories, but the only such functors I can think of that land in the category of finite sets all have to do with torsion points on the elliptic curve.
Does anyone have any examples of a contravariant functor $F:\textbf{Ell}\rightarrow\textbf{Sets}$ such that for $E/S$ with $S$ connected, $F(E/S)$ is finite, and doesn't have to do with torsion data?
Ideally, the functor will actually land in the category of groups, be generically of some order $M\ge 3$, and always be of order $\le M$.

REPLY [11 votes]: Sure -- try the set of homomorphisms from $\pi_1^{\mathrm{et}}(E - O)$ to a fixed finite group G.  This is a "non-abelian level structure" of the sort considered by de Jong and Pikaart
http://arxiv.org/abs/alg-geom/9501003
Rachel Davis, a 2013 Wisconsin Ph.D. working with Nigel Boston, wrote her thesis about this kind of stuff in the case of elliptic curves.<|endoftext|>
TITLE: "Cohomology at the infinity": what does one call it
QUESTION [8 upvotes]: Suppose $X$ is a "good enough" Hausdorff topological space; we assume that $X$ is not compact. Now, for a natural number $k$ and an abelian group $G$, consider the group $\varinjlim_{\substack{U\subseteq X\\ \text{$X\setminus U$ is compact}}} H^k(U,G)$.  
Could you, please, give me a reference to a text where this object is defined? I would like to learn the standard term and notation for it.  
Thank you in advance,
Serge

REPLY [11 votes]: This is the cohomology of $X$ at $\infty$. You'll find a discussion of (singular) homology and cohomology at $\infty$ in Hughes & Ranicki, Ends of Complexes (CUP, 1996).<|endoftext|>
TITLE: Forcing mildly over a worldly cardinal.
QUESTION [9 upvotes]: A cardinal $\theta$ is worldly if $V_{\theta}$ is a model of ZFC.  We could force to collapse $\theta$ to a successor cardinal, for example, and destroy the worldliness of $\theta$, but is there a less catastrophic way to do so?  I'd like to know about a notion of forcing which makes $V_{\theta}$ no longer a model of ZFC in the extension, but in the mildest way possible.  Ideally, I'd like $\theta$ in the extension to be very similar to the original $\theta$.  In other words, and I'm pretty what I want is not possible, is there a way to force to change $V_{\theta}$ as much as possible while changing $\theta$ as little as possible? Is there any way to tease these two apart? I'm sorry that I'm not being very specific about what I'd like to preserve about $\theta$, but if anyone has any ideas based on what I described, I would appreciate your answer.

REPLY [10 votes]: I've got it! We can kill the worldliness of a singular worldly
cardinals as softly as we like.
Theorem. If $\theta$ is any singular worldly cardinal, then
for any natural number $n$ there is a forcing extension $V[G]$ in
which $\theta$ remains $\Sigma_n$ worldly, but not worldly,
meaning that $V_\theta^{V[G]}$ satisfies the $\Sigma_n$ fragment
of ZFC, but not ZFC itself.
Thus, such worldly cardinals can be killed as softly as desired.
Proof. First, we may assume without loss that the GCH holds, by
forcing it if necessary. Also, by forcing to collapse the
cofinality of $\theta$ to $\omega$, which is small forcing with
respect to $\theta$ and therefore preserves the worldliness of
$\theta$, we may assume that $\theta$ has cofinality $\omega$.
I claim that in $V$, we may find a set $A\subset\theta$ that is
$V_\theta$-generic for the class forcing
$\text{Add}(\text{Ord},1)$ to add a Cohen subset of the ordinals
over $V_\theta$. To see this, one simply finds ordinals $\theta_n$
with supremum $\theta$ such that $V_{\theta_n}\prec_{\Sigma_n}
V_\theta$, and then diagonalizes with respect to the
$\Sigma_n$-definable dense classes having parameters in
$V_{\theta_n}$ when extending $A$ up to $\theta_{n+1}$. Even
though the forcing is not even countably closed (since $\theta$
has cofinality $\omega$), nevertheless we can meet the dense class
before the next higher reflecting cardinal since we've limited the
complexity of the dense class. It follows that $\langle
V_\theta,A,{\in}\rangle$ satisfies $\text{ZFC}(A)$, the theory of
ZFC in which the class $A$ is allowed to appear as a predicate the
in the replacement scheme.
Now let $\mathbb{Q}$ be the class forcing over $V_\theta$ to code
$A$ into the GCH pattern. If $G\subset\mathbb{Q}$ is $V$-generic,
then it follows that $V_\theta^{V[G]}=V_\theta[G]$ is a model of
ZFC, and so $\theta$ is still worldly in $V[G]$.
But let me now modify the argument slightly, so as to preserve
only some amount of worldliness, while killing the rest. The idea is to find a set $A$ in $V$ that is $\Sigma_k$-generic
over $V_\theta$, but not fully generic for the definable dense
classes in the first step, where $k$ is much larger than $n$. We
can ensure that $\langle V_\theta,A,{\in}\rangle$ satisfies the
$\Sigma_k$ fragment of $\text{ZFC}(A)$, but not all of
$\text{ZFC}(A)$. This can be done by inserting coding information
to reveal an unbounded $\omega$-sequence when restricted to the
$\Sigma_{k+1}$ reflecting cardinals. In essence, one hides away
the cofinal $\omega$-sequence within the complex set of
$\Sigma_{k+1}$-reflecting cardinals. A very similar idea is used
in the the final section of our paper J. D. Hamkins, D. Linetsky,
J. Reitz, Pointwise definable
models of set theory.
The point now is that if $k$ is sufficiently larger than $n$, then
the $\Sigma_k$ genericity of $A$ will ensure that after one codes
$A$ into the GCH pattern of $V[G]$, one still gets that
$V_\theta^{V[G]}=V_\theta[G]$ will satisfy at least the $\Sigma_n$
fragment of ZFC. But it will not satisfy all of ZFC, because $A$
is definable in this model and $A$ reveals the unbounded
$\omega$-sequence of ordinals. So in $V[G]$, the ordinal $\theta$ is $\Sigma_n$-worldly, but not
worldly. QED
As observed earlier, we can extend this result to regular $\theta$ in the case that $\theta$ is measurable, simply by first performing Prikry forcing to singularie $\theta$ while preserving its worldliness, thereby reducing to the singular case above. 
Update. But in general, we cannot get the result for all regular worldly cardinals, because if the result holds for a regular worldly cardinal $\theta$, then in fact $\theta$ must be measurable in an inner model. To see this, suppose that $\theta$ is a regular worldly cardinal, which is another way of saying that $\theta$ is inaccessible, and suppose that the conclusion of the result is true for $\theta$. It follows that there is a forcing extension in which $\theta$ is a strong limit cardinal but not worldly, and so in particular $\theta$ is not inaccessible, and thus it is singular in $V[G]$. In other words, we have a forcing extension $V[G]$ in which $\theta$ is a singular cardinal. But this implies by a covering lemma argument with the Dodd-Jensen core model (recently explained to me by Gunter Fuchs) that $\theta$ is measurable in an inner model. So we cannot expect to kill inaccessibility softly down to worldly non-inaccessbility for all inaccessible cardinals.<|endoftext|>
TITLE: Modular reductions of simple characters
QUESTION [5 upvotes]: Given a (splitting) $p$-modular system $(K, \mathcal{O}, k)$ for a finite group $G$, any given simple character $\chi$ is afforded by some $KG$ module $V_\chi$, and there in general many non-isomorphic $\mathcal{O}G$ modules $B$ (free as an $\mathcal{O}$ module) such that $B\otimes_{\mathcal{O}} K\cong V_\chi$ (call $B$ an $\mathcal{O}$-form of $\chi$).  Further, the decomposition matrix tells us the composition factors of $B\otimes_\mathcal{O} k$.  My question is, what further restrictions are known on the isomorphism class of $B\otimes k$, given $\chi$? 
In particular, it is not hard to see that an $\mathcal{O}$-form $B$ can be chosen such that $B\otimes k$ is indecomposable.  Is the reduction of every $\mathcal{O}$-form of $\chi$ indecomposable?
Any advice would be greatly appreciated, everything I've read to date seems to skirt this question.

REPLY [3 votes]: Here is a concrete example that the reduction of such a form can be decomposable. Let 
$$G = \langle t, s \mid t^4 =s^2 = 1, t^s=t^{-1} \rangle$$ 
be the dihedral group of order $8$ and $\chi$ the irreducible character of degree $2$. Let $p=2$. Consider first the natural representation affording $\chi$, namely
$$ t \mapsto \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \quad
   s \mapsto \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}. $$
The reduction of $\mathcal{O}^2=M$ with respect to this representation is indecomposable, as is easily seen. 
Now assume that $1-i$ (where $i=\sqrt{-1}$) is a prime element of $\mathcal{O}$. Let $B\subseteq \mathcal{O}^2$ be the $\mathcal{O}$-submodule 
$$ B= \langle \begin{pmatrix} 1\\ i \end{pmatrix}, 
               \begin{pmatrix} 1 \\ 1 \end{pmatrix} \rangle, $$
which is the unique maximal $\mathcal{O}G$-submodule of $M$, by the way. The action of $t$ and $s$ on $B$ with respect to the new basis is given by the matrices
$$ t\mapsto \begin{pmatrix} -i & -1-i \\ 0 & i \end{pmatrix},\quad
    s \mapsto \begin{pmatrix} i & 1+i \\ 1-i & -i \end{pmatrix}. $$
Thus $t$ and $s$ act as identity on the reduction mod $2$, and thus $B\otimes_{\mathcal{O}}k \cong k^2$ is completely reducible.  
It follows that $B$ has $|k|+1$ maximal $\mathcal{O}G$-submodules. Each of these happens to have indecomposable reduction mod $2$ and a unique maximal submodule, namely $(1-i)B$.
When $\mathcal{O}$ is not ramified, however, then there are only two non-isomorphic forms and both have indecomposable reduction mod $2$.<|endoftext|>
TITLE: Diagram spectra and Algebraic Geometry
QUESTION [6 upvotes]: I was recently reading a paper titled "Model Categories of Diagram Spectra" and it was mentioned in the paper that the contents of the paper were also useful in algebraic geometry. I'm really interested in this subject and I'm vaguely aware that techniques from stable homotopy are finding ways into algebraic geometry. I'm relatively unfamiliar with Lurie's works on Derived Algebraic Geometry and stuff (which is where I'm guessing a lot of this stuff is used?) so I guess I'm wondering if anyone can suggest some beginner level literature on this subject. Also if anyone is willing to briefly sketch some of the ideas and applications of this too that would be awesome. I understand this question is kind of vague and open ended so maybe it should be a community wiki? Thanks!
Edit: to be more clear, I am particularly interested in a comment made in the paper on ph446 "examples of such symmetric monoidal categories arise in other fields, such as algebraic geometry." I would like to know what this is referring to.

REPLY [4 votes]: A nice paper to understand the connection between Spectra and Motivic Homotopy Theory is Mark Hovey's Spectra and symmetric spectra in general model categories. To form the classical category of spectra you start with Topological Spaces and an endofunctor $\Sigma$ which you wish to stabilize. The trick is to pass to sequences of spaces $(X_i)$ with structure maps $\Sigma X_i \to X_{i+1}$. To do symmetric spectra you build in group actions as you go. In this paper Hovey discussed how to do this in a more general model category $M$ and with a general endofunctor $G$. With enough hypotheses you can prove things like "the stable equivalences are the stable homotopy isomorphisms"
The work of Morel and Voevodsky was the starting place for motivic homotopy theory. They resolved the Milnor conjecture. Model categories came into the picture because they wanted a model structure on the category of schemes. This doesn't quite work, but if you fatten up the category by formally throwing in the missing colimits then it does work. The other trick is to use the Nisnevich site. The punchline is that you now have a category where both topological objects and algebraic objects live. We've done this in a way such that the topological spheres (well, technically simplicial) and the affine line $\mathbb{A}^1$ are spheres, i.e. monoidal units. If you want more details on this work I can add some but I'm going to assume you can go read about it on your own and I'm going to focus on where spectra come in instead.
In the introduction to the paper above, Hovey discusses the application to Morel-Voevodsky. There are several applications. For one, if you want to do spectra in the motivic setting you have to change what you mean by suspension because now there are different types of spheres, as mentioned above. So Hovey's work lets you do this, i.e. pass from unstable $\mathbb{A}^1$-homotopy theory to stable $\mathbb{A}^1$-homotopy theory. Jardine had proven stable equivalences are stable homotopy isomorphisms in the stable $\mathbb{A}^1$-homotopy theory using the Nisnevitch descent theorem. With Hovey's machine (and passage to a nicer model category which is equivalent to the Morel-Voevodsky one) gets this result in a cleaner way without relying on that theorem.
Motivic homotopy theory has become a thriving subject. Jardine has written about motivic symmetric spectra, Po Hu has written about motivic S-modules, and Ostvaer has written about motivic functors (which might be to your liking if you like the Diagram Spectra paper). Googling will give you these papers. I recommend checking out the presentations on Kyle Ormsby's website for an accessible introduction to the field.<|endoftext|>
TITLE: Axiomatic intersection theory
QUESTION [5 upvotes]: Is there an axiomatic intersection theory?
What I expect is something like:
An intersection theory is a functor from the category of schemes(or other spaces) to the category of algebras, with well-defined flat pullback and proper pushforward, and (maybe) projection formula or some other axioms.

REPLY [5 votes]: You might be interested in the bivariant theories of Fulton and MacPherson.<|endoftext|>
TITLE: Concrete examples of noncongruence, arithmetic subgroups of SL(2,R)
QUESTION [11 upvotes]: A  subgroup of $SL_2(\mathbb{R})$ is called arithmetic if it is commensurable with $SL_2(\mathbb{Z})$.
An arithmetic subgroup is called congruence if it contains a subgroup of type $\Gamma(N)$ for some $N\in \mathbb{N}$.

Question: What are concrete examples of subgroups of $SL_2(\mathbb{R})$, which are arithmetic, but not congruence?

I have heard the Belyi theorem produces some examples, but I have never seen a concrete one.

Further Question: Can such things exist in higher rank Lie groups (real rank $\geq 2$)?

REPLY [5 votes]: If you want to go really concrete, take any connected finite $3$-valent graph in the $2$-sphere with disks as components of the complement. 
This yields a conjugacy class of genus $0$ finite index torsion free subgroups of $\Gamma(1)=PSL(2,\mathbb{Z})$. In fact the index is $6n$ if the graph has $2n$ vertices (hence $3n$ edges and $n+2$ faces). ADDED: Indeed, since $\Gamma(1)\simeq Z/2*Z/3$, there is a transitive action of $\Gamma(1)$ on oriented edges of the graph, flipping orientation for $Z/2$, and rotating around source vertex for $Z/3$. The conjugacy class of stabilizers subgroups determines the graph.
But there are clearly infinitely many such graphs (up to oriented homeomorphism), whereas there are only finitely many congruence subgroups of genus $0$ in $\Gamma(1)$ (about $33$ conjugacy classes of torsion free ones, with maximum index $60$, according to this table).<|endoftext|>
TITLE: The classifying space of a gauge group
QUESTION [13 upvotes]: Let $G$ be a Lie group and $P \to M$ a principal $G$-bundle over a closed Riemann surface. The gauge group $\mathcal{G}$ is defined by
$$\mathcal{G}=\lbrace f : P \to G \mid f(p \cdot g) = g^{-1}f(p)g\ (\forall g \in G, p \in P) \rbrace.$$
I want to understand the following statement which I found in Atiyah-Bott (Roy. Soc. London Ser. A 308).

PROPOSITION 2.4. Let $BG$ be the classifying space for $G$. Then in homotopy theory
  $$B\mathcal{G} = \mathrm{Map}_P(M,BG).$$ Here the subscript $P$
  denotes the component of a map of $M$ into $BG$ which induces $P$.

This is proved by showing that $\pi:\mathrm{Map}_G(P,EG) \to \mathrm{Map}_P(M,BG)$ is a universal $\mathcal{G}$-bundle. 
My questions: 

Why is $\mathrm{Map}_G(P,EG)$ contractible? 
How do we see that $\pi:\mathrm{Map}_G(P,EG) \to \mathrm{Map}_P(M,BG)$ is locally trivial? Here for a $G$-equivariant map $f:P \to EG$, $\pi(f)$ is the map $P/G \to EG/G$ which is canonically induced by $f$. 

Notes:

It is easy to see that $\mathrm{Map}(P,EG)$ is contractible, because $EG$ is contractible. But, because of the $G$-actions, it does not seem that we can apply the same discussion to $\mathrm{Map}_G(P,EG)$.
For the second question, Atiyah and Bott say the following. But I can not understand it and I want to understand the details.

If $BG$ is paracompact and locally contractible, which is easily arranged, $\pi$ will be a locally trivial principal fibring, as follows easily from the homotopy properties of fibrings.

If there is a good exposition of this problem, please let me know.

REPLY [2 votes]: Actually, $EG$ has more structure that just being contractible. You can endow it with a continuous group multiplication (for the compactly generated topology on the product) and there is a contraction $F\colon [0,1]\times EG\to EG$ such that each $F_t$ is a group homomorphisms. Moreover, $G$ is a closed subgroup and we have $F_t(g)\cdot F_t(x)\cdot F_t(g)^{-1}=g\cdot F_t(x)\cdot g^{-1}$. Putting this together you see that you can apply this contraction to get a contraction of the space of equivariant maps $C(P,EG)^G$. 
However, this is limited to locally contractible $G$ and spelling out the details is a bit involved since you have to use different models of $EG$ that are not entirely compatible with each other. For instance, the claimed equalities can be checked when considering $EG$ as left continuous step functions from $[0,1]$ to $G$ (cf. Brown-Morris "Embeddings in contractible or compact objects", ), but the topology is the one from taking $EG=|\mathcal{E}G|$ for $\mathcal{E}G$ the nerve of the pair groupoid of $G$ (cf. Segal "Classifying spaces and spectral sequences" and "Cohomology of topological groups"). These between these models there exists a bijection that you can use to verify the assertions.<|endoftext|>
TITLE: Converse to Milnor's theorem on manifolds with nonnegative Ricci curvature
QUESTION [8 upvotes]: Disclaimer : I suspect the question I am about to ask is really hard, but I just want to know the status of such questions.
Thanks to Milnor, we know that the $\pi_1$ any compact manifold with nonnegative Ricci curvature has polynomial growth. 
I want to know if anything is known about the opposite direction : which manifolds whose $\pi_1$ has polynomial growth admit metric with nonnegative Ricci curvature ?
At least if you allow non compact manifolds, the answer cannot be 'all of them' (because of Whitehead's three manifold), and I feel like there are compact counter examples but I am not able to cook one.

REPLY [6 votes]: For a compact counterexample, take any nilmanifold $N/H$ modulo the action of a freely acting cocompact lattice $\Lambda$, assuming $N/H$ is not just Euclidean space and $\Lambda$ is not just virtually abelian. For instance, $N=N/H$ is the $3 \times 3$ real Heisenberg group and $\Lambda$ is the $3 \times 3$ integer Heisenberg group. The proof that this is a counterexample is to apply Wilking's theorem in Anton's answer, and the theorem that a finitely generated nilpotent group has polynomial growth.<|endoftext|>
TITLE: Is there an equivalent of Heisenberg's uncertainty principle in the decision sciences ?
QUESTION [15 upvotes]: From memories of a quantum mechanics class and Wikipedia:

In quantum mechanics, the uncertainty principle is any of a variety of mathematical inequalities asserting a fundamental limit to the precision with which certain pairs of physical properties of a particle known as complementary variables, such as position x and momentum p, can be known simultaneously. For instance, the more precisely the position of some particle is determined, the less precisely its momentum can be known, and vice versa.

In practical terms the product of the uncertainty in the position and the uncertainty in the momentum is at least of the order of magnitude of the Planck constant.
A reason why an analogous of this principle could be found in social sciences is that (in my understanding) its origins are purely mathematical. It is a Cauchy-Schwarz-like inequality applied to the scalar product Bra-ket under the dynamic of a particular pde. However, the nanoscopic nature of quantic measurements and the ~10^-34 Planck constant are so small that the uncertainty principle is irrelevant in physics of larger particles, not to mention the other sciences.
A common operation in decision sciences is < belief || outcome >  (or < price vector || vector of goods > at a larger scale) and the solution concepts in game theory, for example, allow tremblings as small as possible (epsilon > 0) with fixed outcomes! Therefore,
Is there (yet) an equivalent of Heisenberg's uncertainty principle in the game theory, decision theory etc. ?

REPLY [3 votes]: I believe there might be something similar to Heisenberg Uncertainty Principle in every decision model, particularly multicriteria decision making models.
If we consider a decision model as a set of criteria that rank a few alternatives differently according to some preference agregation function (that agregates criteria weights, for example additively in a linear model) an analogue to Heisenberg's principle precision idea is the discriminative power of the model. 
If you want to refine your appraisal of the alternatives you will consider a few more criteria you had not thought of before so that your model is as detailed and complete as possible, and considers some new possibly forgotten value measures. By doing this you increase the discriminative power on the "Criteria" side, but then you lose it on the "Alternatives" side. From the "Alternatives" side the ideal is to have just one criteria, according to which it is easy to see how they all rank against each other. If you start to have more criteria some alternatives will be best according to some criteria and some will be best according to another, so it is not so easy anymore to rank them. You then have to decide on criteria weights - which criteria are more important, and aggregate. With many criteria ultimately every alternative is best according to some of them and besides the subjectiveness of criteria weights determination that induces misjudgment, in some models these alternatives are considered equally competitive. For example, in DEA (Data Envelopment Analysis) criteria weights make no diference in this last cenario - if one alternative (or decision making unit) is best according to one criteria (input or output) it can’t be ranked below the others. 
It seems that by enlarging a decision making model with more criteria you increase discriminative power on one hand but lose it on the other. As if there was a minimum of "Uncertainty" attached to your decision making model, which can't discriminate beyond a certain limit. But this is just an idea. I can't quantify it.<|endoftext|>
TITLE: How long can it take to generate a $\sigma$-algebra?
QUESTION [7 upvotes]: I want to know if there is a $\sigma$-algebra such that for every countable ordinal $\alpha$ the $\sigma$-algebra can be generated in more than $\alpha$ steps but less than $\omega_{1}$ steps.
Given an algebra of sets $(X,\mathcal{A})$, let $\mathcal{A}_{0}=\mathcal{A}$, and for all ordinals $0<\alpha\leq\omega_{1}$, let $\mathcal{A}_{\alpha}$ be the algebra of sets generated by countable unions from the collection $\bigcup_{\beta<\alpha}\mathcal{A}_{\beta}$. Clearly $\mathcal{A}_{\omega_{1}}$ is the $\sigma$-algebra generated by $\mathcal{A}$. 

For each countable ordinal $\alpha$ does there exist an algebra of sets $(X,\mathcal{A})$ such that $\mathcal{A}_{\alpha}\neq\mathcal{A}_{\omega_{1}}$, but where $\mathcal{A}_{\beta}=\mathcal{A}_{\omega_{1}}$ for some countable ordinal $\beta$?
Does there exist a $\sigma$-algebra $(X,\mathcal{M})$ such that

If $(X,\mathcal{A})$ is an algebra of sets that generates $\mathcal{M}$, then 
$\mathcal{A}_{\alpha}=\mathcal{M}$ for some countable ordinal $\alpha$, and
for each countable ordinal $\alpha$ there is an algebra of sets $(X,\mathcal{A})$ that generates $\mathcal{M}$ but where $\mathcal{A}_{\alpha}\neq\mathcal{M}$?

REPLY [9 votes]: The answer to the first question is in the positive; thanks to a corollary drawn by Ken Kunen from a key theorem (about Boolean algebras) of Arnie Miller. More specifically, as shown in Theorem 9.2 of Arnie Miller's beautiful monograph, we have the following:

Theorem (Miller-Kunen) For every countable ordinal $\alpha$ there is a field $H$ of sets such that $o(H)=\alpha$.

In the above, $H$ is a field of sets means that $H$ is a family of subsets of some set $X$, and $H$ is closed under complements and finite unions; and $o(H)$ measures the length of the $\sigma$-algebra generated by $H$ [see p.10 of the aforementioned reference for the official definition].<|endoftext|>
TITLE: probability calculation
QUESTION [11 upvotes]: Given $m\cdot e$ balls, $b$ of which are black (suppose the rest are white balls). Randomly put the balls into $m$ baskets, with $e$ balls in each basket. What is the probability of the event that every basket has more white balls than black ones?

REPLY [3 votes]: Since the first bin contains $k$ balls with probability
$$\frac{{e\choose k}{me-e\choose b-k}}{{me\choose b}},$$
we get the recursion relation
$$p_e(m,b)=\sum_{k=0}^{\lfloor e/2\rfloor}\frac{{e\choose k}{me-e\choose b-k}}{{me\choose b}}p_e(m-1,b-k)$$
with $p_e(m,b)$ denoting the probability that each of the $m$ bins filled with $e$ balls contains at least as many
white than black balls where the total number of black balls is $b$ and with the bins 
filled randomly in an obvious sense. (If we want strict inequality,
we have to replace the upper summation-bound $\lfloor e/2\rfloor$ by
$\lfloor (e-1)/2\rfloor$.
Using the obvious initial condition $p_e(1,b)=1$ if $b\leq e/2$ (respectively 
$b$ strictly smaller than $e/2$ if we wish strict inequality) and $p_e(1,b)=0$ otherwise,
we can compute $p_e(m,b)$ by an algorithm needing roughly the computation of $2mb+m$
binomial coefficients
and having a memory requirement $b$ (by computing $p_e(a+1,0),\dots,p_e(a+1,b)$
using the values $p_e(a,0),\dots,p_e(a,b)$.<|endoftext|>
TITLE: Homotopy equivalences preserving structure
QUESTION [5 upvotes]: Suppose I have $X=X_1\cup X_2\cup…\cup X_n$ and $f:X \to Y$ where $Y$ has a similar decomposition.
Suppose I know that 
$f | X_{i_1}\cap…\cap X_{i_r} \to Y_{i_1}\cap…\cap Y_{i_r}$ is a homotopy equivalence for each nonempty intersection. Suppose that the spaces are nice enough that we have the homotopy extension property wherever we want it.
Then $f$ has a homotopy inverse preserving all of this structure and the homotopies themselves preserve the structure.
The isn't hard to prove by piecing together strong deformation retractions in mapping cylinders, but I'd rather just point to a reference. 
For anyone who hasn't seen this stuff before, the base case is that if $f:(X,A) \to (Y,B)$ is a map such that $f:X \to Y $ and $f|:A \to B$ are homotopy equivalences, then $f$ is a homotopy equivalence of pairs. The proof is to form the mapping cylinder, squash the mapping cylinder of f| to it's top, keeping the whole top fixed, and then squash the rest. Of course, all of the spaces should be nice enough for the homotopy extension theorem to be available.

REPLY [2 votes]: I think an answer is in 
tom Dieck, Tammo
Partitions of unity in homotopy theory. 
Composito Math. 23 (1971), 159–167. 
With regard to the result on pairs given by Steve, it could be  useful to note that the book Topology and Groupoids gives a result in 7.4.2(Addendum) which gives control over the homotopies involved. The utility of this is that it gives a key to one proof of a gluing theorem for homotopy equivalences, which was first given in the 1968 edition of this book and is applied by tom Dieck in his paper. The Addendum is  as  follows: 
We are dealing with the situation $f:(X,X^0) \to (Y,Y^0)$ where each pair has the HEP. 
Let $g^{0} :Y_{0} \to X_{0}$ be any homotopy inverse of $f^{0}$ and let 
$ H^0: f^0g^0 \simeq 1, K^0: g^0f^0 \simeq 1$ be homotopies. Then $g^0$ extends to a homotopy inverse $g$ of $f$ such that the homotopy $fg \simeq 1 $ extends $H^0$ while the homotopy $gf \simeq 1$ extends the sum 
$$
K^{0} + g^{0}H^{0}f^{0} - g^{0}f^{0}K^{0}
$$
of the homotopies
$$
g^{0}f^{0} = g^{0}f^{0}1_{X_{0}} \simeq g^{0}f^{0}g^{0}f^{0}
\simeq g^{0}1_{Y_{0}}f^{0} \simeq 1_{X_{0}}
$$
determined by $H^0,K^0$. 
I do not know of a counterexample to the idea of avoiding the kind of "conjugation" given above (though it has been given in the dual situation). The argument derives from the categorical result that if $a,b, c$ are morphisms in a category such that $ab, bc$ are defined and are isomorphisms, then $a,b,c$ are isomorphisms. 
Note that this Addendum easily gives a gluing theorem for $n$ subspaces with a common intersection. 
I'll add that the idea for this result came from generalising the proof that a homotopy equivalence of spaces (not necessarily base point preserving) induces an isomorphism of homotopy groups. 
My memory is that another paper relevant to the question, but to which I do not have easy access,  is 
Spanier, E. H.; Whitehead, J. H. C.
The theory of carriers and S-theory. Algebraic geometry and topology. A symposium in honor of S. Lefschetz, pp. 330–360. Princeton University Press, Princeton, N.J., 1957. 
particularly the work on carriers.<|endoftext|>
TITLE: bipartite graph coloring
QUESTION [5 upvotes]: Hi, I don't know much about graph theory so I would need to know if the following problem has a positive answer or a reference. There is a bipartite graph G with the two vertex sets V1, V2. Each vertex of V1 is connected to a least 2 vertices of V2 and with at most N vertices of V2. Each vertex of V2 is connected with some vertex of V1. Is there a function f(N) such that choosing f(N) colors to color the vertices of V2 I can ensure that each vertex of V1 is connected with at least two vertices of different colors?

REPLY [2 votes]: Answer is NO. Consider the following bipartite graph $G=(V,E)$.
$V(G)=\{ v_1, \ldots , v_n\}\cup \{u_{i,j}: 1 \leq i < j \leq n\}$,
$E(G)=\{v_i u_{i,j}:1 \leq i  \leq n  , \, \, 1 \leq  j \leq n\}$.
Let $V_1=\{u_{i,j}: 1 \leq i < j \leq n\}$ and $V_2=\{ v_1, \ldots , v_n\}$. Each vertex 
in $V_1$ is connected to exactly two vertices in $V_2$, so $N=2$, but we need at least $n$ colors to color the vertices of $V_2$, to ensure that each vertex of $V_1$ is connected with at least two vertices of different colors.<|endoftext|>
TITLE: Maximum number of Vertices of Hypercube covered by Ball of radius R
QUESTION [10 upvotes]: Let $R>0$ be given and let $H^n$ be the unit hypercube in $\mathbb{R}^n$. The problem I am facing is to find the maximum number of vertices of $H^n$ which can be covered by a closed $n$-dimensional ball of radius $R$. 
I would greatly appreciate any references to literature or ideas on how to approach the problem. I strongly suspect that a solution for any $R$ comes from placing the center of the ball at $(\frac{1}{2},\frac{1}{2},\ldots,\frac{1}{2},0,\ldots,0)$ with the number of nonzero coordinates being $\lfloor 4R^2 \rfloor=\lfloor D^2 \rfloor $, where $D$ is the diameter (for $D \leq n$).
Any references or ideas are welcome.
Thanks!

REPLY [2 votes]: Sometimes it is better to place the ball "in the corner". For example:
Let $n=8$ and $R=\sqrt{\frac{7}{8}}$. Then the ball with center $(\frac{1}{8}, \frac{1}{8}, \dots, \frac{1}{8})$ and radius $R$ covers exactly $9$ vertices (those with at most one coordinate equal to $1$), but a ball with radius $R$ cannot cover a $4$-dimensional unit hypercube, which requires radius at least $1$.
There is even no upper bound on the number of points a ball with radius $1$ can cover, provided the dimension is not bounded. Generalizing previous construction, a ball with center $(\frac{1}{n}, \frac{1}{n}, \dots, \frac{1}{n})$ and radius $\sqrt{\frac{n-1}{n}}$ covers $n+1$ vertices of the $n$-dimensional hypercube. As user21277 points out, for $n=16$ this is better than placing the ball in the center of a $4$-dimensional face.
In general, for $n>2i$, a ball with center $(\frac{i}{n}, \frac{i}{n}, \dots, \frac{i}{n})$ and radius $\sqrt{\frac{i(n-i)}{n}}$ covers ${n \choose i} + {n \choose i-1} + \cdots +{n \choose 0}$ vertices: those with at most $i$ coordinates equal to $1$. A ball with the same radius cannot cover all $2^{4i}$ vertices of a $4i$-dimensional unit hypercube.<|endoftext|>
TITLE: Etale Cohomology of Punctured Spectra of Local Rings
QUESTION [7 upvotes]: Let $R=\mathbb{C}[[x,y]]$ be a power series ring in two variables (or maybe more generally a strictly Henselian local ring) with maximal ideal $\mathfrak{m}$.  

What is $H^*_{et}(\operatorname{Spec}(R)\setminus\{\mathfrak{m}\}, \mathbb{G}_m)$?

My motivation is this:  I'm trying to understand the extent to which etale cohomology of $\mathbb{G}_m$ resembles the cohomology of $\mathcal{O}_X^*$ in the complex-analytic setting.  For example, one might expect that in the case above, $H^2$ ought to be $\mathbb{Z}$.
I'd be satisfied with an answer that computes $H^2$.

REPLY [12 votes]: With your particular choice of $R$, the $H^2$ is $0$. More generally, if $R$ is a strictly Henselian regular local ring of dimension $2$, then by the purity for the Brauer group (in this particular case it is known and due to, I believe, Grothendieck; for a proof see Grothendieck "Le groupe de Brauer II", Prop. 2.3) $H^2_{et}(R \setminus \{ \mathfrak{m} \}, \mathbf{G}_m) = H^2_{et}(R, \mathbf{G}_m) = 0$. The first equality is due to the purity because you're removing a closed subscheme of codimension $2$; the second equality is because $R$ is strictly Henselian and $\mathbf{G}_m$ is smooth: by Grothendieck "Le groupe de Brauer III" appendix, Thm. 11.7 2), the cohomology can thus be computed over the separably closed residue field, where it vanishes.
More generally assuming that $R$ is a strictly Henselian regular local ring of dimension $\ge 2$, the $H^2$ is always supposed to be $0$ by the aforementioned purity and the argument as above. This is open (as far as I know), although many cases are known, including $\dim R \le 3$. For a brief survey on precisely this question see Gabber "On purity for the Brauer group" in Oberwolfach report No. 37/2004.<|endoftext|>
TITLE: If a $d \log$ form is exact, is it zero?
QUESTION [5 upvotes]: Let $T = \mathrm{Spec}\  \mathbb{C}[x_1^{\pm 1}, x_2^{\pm 1}, \ldots, x_n^{\pm 1}]$  be an algebraic torus and $X$ a closed subvariety. Let $\eta$ be a differential form on $T$ of the form
$$\sum_I a_I\cdot \bigwedge_{i \in I} d \log x_i $$
where $I$ runs over subsets of $\{ 1,2, \ldots, n\}$ and $a_I$ are various constants. If the cohomology class of $\eta$ restricts to $0$ on $X$, is $\eta$ identically zero?
Cases where this is true: $X$ a subtorus (easy), $X$ a curve (easy), $X$ cut out by linear equations in the $x_i$ (Orlik-Solomon).

REPLY [4 votes]: I think this works. We prove a slightly stronger statement by induction on $\dim X$: Let $X$ be a quasi-projective variety over $\mathbb{C}$. Let $x_1$, $x_2$, ..., $x_n$ be units in $H^0(X, \mathcal{O})$. By "$d \log x_i$" I mean the class in $H^1(X)$ pulled back from the generator of $H^1(\mathbb{G}_m)$ by the map $x_i : X \to \mathbb{G}_m$. (I say this to address the possibility that $X$ isn't smooth, since I'm not sure whether Kahler differentials give cohomology classes in general.) Then, as before, I claim that:

If a polynomial $\eta$ in the $d \log x_i$ is exact, then it is $0$.

The base case, $\dim X=0$, is trivial.
Reduction We may (and do) assume that $X$ is smooth.
Proof Let $\tilde{X} \to X$ be a resolution of singularities. Since the class of $\eta$ is $0$ in $H^{\ast}(X)$, it pulls back to $0$ in $H^{\ast}(\tilde{X})$. So $\eta$ pulls back to $0$ on $\tilde{X}$, and is thus $0$. $\square$
Now, choose a simple normal crossing compactification $\bar{X}$ of $X$. Let $D_1$, $D_2$, ..., $D_m$ be the components of $D$. Our next goal is to prove
Claim For any component $D_1$ of $D$, the form $\eta$ has no pole on $D_1$.
Reduction We may assume that $x_2$, $x_3$, ...., $x_n$ don't have poles or zeroes along $D_1$, and $x_1$ vanishes on $D_1$.
Proof Let $d_i$ be the order to which $x_i$ vanishes on $D_1$. Let $d = GCD(d_1, d_2, \dots, d_m)$. Then by applying a monomial transformation to the $x$'s, we may assume that $x_1$ vanishes to order $d$ on $D_1$. $\square$
Let $Y = D_1 \setminus \bigcup_{i \geq 2} D_1 \cap D_i$ and let $Z$ be the union of $X$ and $Y$ inside $\bar{X}$. So $Y$ is a smooth hypersurface in $Z$. We claim that the $x_i$ extend to homolomorphic function on $Z$, and that $x_2$, ..., $x_n$ extend to nonvanishing functions. Proof: Suppose $x_i$ does not extend to $Z$. Then $x_i$ has a pole along some hypersurface in $Z$. By the previous Reduction, that hypersurface is not $Y$, so it must meet $X$. But $x_i$ is well defined on $X$. For $i \geq 2$, this argument also shows that $x_i^{-1}$ extends.
Proof of Claim Let $\omega = \mathrm{Res}_{Y} \eta$. The form $\omega$ is a $d \log$ form on $Y$, in the ring generated by $d \log x_2$, $d \log x_3$, ..., $d \log x_n$. The residue map is a well defined map $H^k(X) \to H^{k-1}(Y)$. More specifically, let $U$ be a tubular neighborhood of $Y$, so $\partial U \subset X$ is a circle bundle over $Y$. Then $\mathrm{Res}$ is the composition of restriction $H^k(X) \to H^k(\partial U)$ and the Gysin map $H^k(\partial U) \to H^{k-1}(Y)$. 
So $\omega$ is exact on $Y$. By induction, $\omega=0$. We have now established that the residue of $\eta$ along $Y$ is $0$. But $\eta$ clearly has at most a first order pole on $Y$, so it has no pole at all. $\square$.
Proof of theorem We have shown that $\eta$, as a form on $\bar{X}$, does not blow up on any of $D_i$. Since $\bar{X}$ is smooth, this shows that $\eta$ extends to $\bar{X}$. But, by Hodge theory, on a smooth projective variety, any exact holomorphic form is $0$.

Alternate proof As above, reduce to $X$ smooth; let $\overline{X}$ be a normal crossing compactification; let $D = \overline{X} \setminus X$. Let $\eta$ be a $k$-form. The condition that $\eta$ is generated by $d \log$ forms implies that $\eta$ is an element of $H^0(\overline{X}, \Omega^k(\log D))$. The condition that $\eta$ is exact says that the image of $\eta$ in $H^k(X, \mathbb{C})$ is $0$.
Conveniently, there is a spectral sequence $H^q(\overline{X}, \Omega^p(\log D)) \Rightarrow H^{p+q}(X, \mathbb{C})$ which degenerates at $E_1$. (I'm looking at Voisin's book, volume 1, Theorem 8.35, she cites Deligne Theorie de Hodge II.) So $H^0(\overline{X}, \Omega^k(\log D))$ injects into $H^k(X, \mathbb{C})$, which was the desired claim. 
Thomas Lam and I have finally started writing the paper where we needed this lemma; this is probably the proof we will use.<|endoftext|>
TITLE: Casselman-Shalika formula for split reductive groups
QUESTION [5 upvotes]: In the paper of Casselman and Shalika they give an explicit formula for the spherical Whittaker function of an unramified principal series.  Apparently, upon combining their formula with the Weyl character formula, one can get an evaluation of the Whittaker function in terms of the Satake parameter of the representation.  Can someone please explain how to do this, or give me a reference where it is done?

REPLY [2 votes]: I think something very close to what is asked in the original question is explained by Haines, Kottwitz and A. Prasad in their review article.<|endoftext|>
TITLE: Algebraic topology in low regularity
QUESTION [5 upvotes]: This question is triggered by a talk by Pierre Bousquet, who considered related questions (but not quite what I ask below).
Take a classical algebraic topological result, like the inexistence of retraction map $f:D^2\to \partial D^2$. Can we lower the regularity hypothesis (i.e., replace continuity with something weaker, or at least something not implying continuity) and still get a result?
Let me be more precise:

For which values of $p$ Does it exist a map $f:D^2\to \partial D^2$ in $W^{1,p}$
  such that the trace of $f$ on the boundary is the identity?

In the same spirit:

For which values of $s,p$ must each map $f:D^2 \to D^2$ in $W^{s,p}$
  have an almost fixed point in some sense (e.g. a sequence $x_n\to x$ such that $f(x_n)\to x$).

REPLY [3 votes]: This does not answer your specific questions, but degree theory has been extended to certain classes of non-continuous functions (including some Sobolev spaces) by Brezis and Nirenberg. See 
Brezis, H., & Nirenberg, L. (1995). Degree theory and BMO; part I: Compact manifolds without boundaries. Selecta Mathematica, New Series, 1(2), 197-263.
Brezis, H., & Nirenberg, L. (1996). Degree theory and BMO; part II: Compact manifolds with boundaries. Selecta Mathematica, New Series, 2(3), 309-368.<|endoftext|>
TITLE: Who constructed the projective plane of order $4$ from $K_6$?
QUESTION [12 upvotes]: I have been trying to hunt down the original reference for the construction of the projective plane of order $4$ from the complete graph on $6$ vertices.
The reference I have at hand are Cameron and van Lint (Chapter 6) and Beutelspacher.
Cameron and van Lint vaguely credit lecture notes by G. Higman, and I'm finding some of Beutelspacher's references hard to access. Does anyone know offhand where this first appeared? Any help will be greatly appreciated.

REPLY [6 votes]: All references converge on W.L. Edge, Some implications of the geometry of the 21-point plane, Math.Zeitschr. 87, 348--362 (1965) 
Available at http://www.maths.ed.ac.uk/cheltsov/edge2013/pdf/1965a.pdf
Section 6 lays out all the necessary ideas.<|endoftext|>
TITLE: On Geometric Langlands Correspondence
QUESTION [10 upvotes]: The Geometric Langlands correspondence introduced by Drinfeld and Laumon conjectures a 1 to 1 correspondence between 
(A) local systems on a projective smooth curve over a field 
and
(B) (Hecke eigen-)perverse sheaves on the algebraic stack of vector bundles of rank n on the curve (or more generally G-bundles). 
The conjecture seems in good way to be proven (see the recent preprint of Gaitsgory http://arxiv.org/abs/1302.2506). I have the following questions: Is it possible to replace curves by higher dimensional varieties? Is it possible to consider curves (or varieties) minus a finite set of points. Has it been studied? References? Can you formulate (approximatively) what would be the generalization of (B)?

REPLY [4 votes]: EDIT Few days ago a survey by A. Parshin appeared in arxiv.
I think it is the best place to look on the higher-dimensional Langlands.
From the abstract:

A brief survey is given of the classical Langlands correspondence
  between n-dimensional representations of Galois groups of local and
  global fields of dimension 1 and irreducible representations of the
  groups GL(n). A generalization of the Langlands program to fields of
  dimension 2 is considered and the corresponding version for
  1-dimensional representations is described. We formulate a conjecture
  on a direct image (=automorphic induction) of automorphic forms which
  links the Langlands correspondences in dimension 2 and 1. The direct
  image conjecture implies the classical Hasse-Weil conjecture on the
  analytical behaviour of the L-functions of curves defined over global
  fields of dimension 1.


Actually similar question has been asked. Higher-dimensional Langlands is something very intriguing. I did not follow recent advances, but let me mention some part which I know about.
Let us start with NON-geometric  local case. Langlands correspondence is roughly speaking "bijection" between representations of Galois group and representations of  GL(Local Field).
Abelian case (class field theory) is bijection between characters of Galois and characters Local field, if dualize we get Galois/[Galois,Galois] = (Local Field)^*.
One of the first questions to ask - whether it is possible to generalize local class field theory to higher dimensions ? 
The main idea by A. Parshin is that  in n-dimensions one should consider Milnor's (n)-th K-group of local field instead of (Local Field)^. In particular for n=1 K_1^Milnor(Field)=Field^. (By the way the definition of higher dimensional local field should also be given). 
Parshin also found higher analogs of various symbols and proved higher analogs of reciprocity laws. 
To the best of my knowledge there were no further developments in the field before Kapranov's paper  "Analogies between the Langlands correspondence and topological quantum field theory" . In that paper he gave certain vision what higher dimensional Langlands might be. 
His idea (quite amazing) that "representations" should be substituted by "k-representations" (i.e. representations in higher categories), (e.g. for surfaces we should consider 2-representations). 
So in n-th dimensions k-representations of dimension r of Galois group should correspond to (n-k)-representations of GL_r(n-Local Field)
In particular abelian version will correspond to Parshin's higher dimensional class field theory, since Milnor's K-groups correspond higher-representations.

Now about the geometric version. It is quite unclear for me.
In 1-dimension, we can think of flat connections as analogs of Galois representations.
It seems in higher dimensions we should consider gerbes with flat connections 
as analogs of higher representations of Galois group, whatever it means...
The geometric substitute for moduli space of vector bundles and Hecke-eigen sheaves is not clear for me. The reason is that in 1-dimension moduli space of vector bundles arise as standard coset  G_{out}\G(k((z))/G_{in}, however in higher dimensions I do not see how the group G( k((z,u)) ) may have finite-dimensional quotient. 
So I do not see something finite-dimensional where "n-Hecke eigensheaf" may live.
Well, it is probably just my own problem. I have speculated around these things in http://arxiv.org/abs/hep-th/0604128, but I am afraid it is very unclear...

Last year there appeared a paper Unramified two-dimensional Langlands correspondence,
which probably is the last mile achievement in the question. Unfortunately I have no time to follow these developments.<|endoftext|>
TITLE: the global m-th power reciprocity law and Quartic Reciprocity Law
QUESTION [6 upvotes]: I'm reading Cox "Primes of the form $x^2+ny^2$". And I read a chapter about the global m-th power reciprocity law. Now I'm not able to prove the quartic and cubic reciprocity laws. Where can i find the proofs of them using global m-th power reciprocity law. There is a link to Hasse's book "Bericht uber neuere Untersuchungen und Probleme as der Theorie def algebraischen Zahlkorper" in Cox's book. But I haven't found English translation of it. Any recommendations for such books or articles would be of great utility. 
@Dietrich Burde
I read the chapter about Cubic and Quartic Reciprocity laws a long time ago, and I knew these proofs, but I don't know they could be proved using $m$-th power reciprocity law. I didn't find this law in the book. Speaking about $m$-th power reciprocity law I mean this
$K$ is a number field containing a primitive $n$-th root of unity, and $\alpha, \beta \in \mathcal O_K$ are relatively prime to each other and to $n$. Then 
$$\biggl(\frac{\alpha}{\beta}\biggr)_{n} \biggl(\frac{\beta}{\alpha}\biggr)_{n}^{-1}=\prod_{\mathfrak p \mid n\infty}\biggl(\frac{\alpha,\beta}{\mathfrak p}\biggr)_{n},$$
where $\biggl(\frac{\alpha,\beta}{\mathfrak p}\biggr)_{n}$ is the $n$-th power Hilbert symbol. Bibliography seems to be quite usefull, i'm going to check it, thanks for it.

REPLY [4 votes]: A complete exposition of the derivation of Hilbert's product formula from Artin's reciprocity law, plus an application to cubic and quartic reciprocity, is indeed contained in the second volume of Hasse's report on class field theory. 
If you take Artin's reciprocity law for granted, the whole thing is little more than an exercise: by Artin, the Jacobi symbol $(\mu/\alpha)_\ell$ only depends on the residue class of $\alpha$ modulo the conductor of the extension 
$K(\sqrt[\ell]{\mu})$ of the field $K$ of $\ell$-th roots of unity, and from 
this observation the whole reciprocity law for $\ell = 3$ and $\ell = 4$ follows.
A slightly different approach would be using Furtwängler's trick, which is explained in the correspondence between Artin and Hasse
(the pdf of the German version is free; an English translation will appear at the end of 2013 from Springer).<|endoftext|>
TITLE: Fourier transform of a bounded function
QUESTION [10 upvotes]: This should really be well-known, but I was not able to find a definite answer to this question:
Is the Fourier transform of a bounded function always a borel measure (i.e. an order 0 distribution)?
In some sense, the distributional order corresponds to the order of a bounding polynomial and a bounded function can be bounded by an order zero polynomial. But I could not find any reference.
If the above statement is false, what is a counterexample?
If the statement is true: How does bounded variation correspond to all this? I.e. are there criterions for the Fourier transform to be of bounded variation?

REPLY [15 votes]: The answer is no : Fourier transform of the signum function is a constant times the distribution  $\lim_{\epsilon\to 0} 1_{|x|>\epsilon}/x$, of order $1$, also known as the principal value of $1/x$. Another keyword is Hilbert transform.<|endoftext|>
TITLE: Question about Shelah's version of "Shooting a club" found in PIF
QUESTION [6 upvotes]: Suppose $S \subset \omega_{1}$ is stationary co-stationary. Then there is a forcing notion $P_{S}$ which shoots a closed unbounded $C \subset S$ without collapsing cardinals (or
  changing cofinalities).

Let $P$ = $Levy$($\aleph_{0}$， $< \aleph_{1}$). So P is essentially adding $\aleph_{1}$ Cohen reals. If $G_{P} \subset P$ is (directed and) generic over V, then $G_{P}$ is also generic over $L$ (the constructible universe) as $P \in L$ and also over $L[S]$. By Theorem I 6.7 the forcing P satisfies the countable chain condition and $\aleph_{1}^{L[G_{P}]} = \aleph_{1}^{V} = \aleph_{1}^{V[G_{P}]}$ and $V$, $V[G_{P}]$ have the same cardinals and cofinalities. Let
$Q$ = {$C$ : $C$ is a closed bounded subset of $S$ which belongs to $L[S, G_{P}]$} and
    $C_{1} < C_{2}$ iff $C_{1} = C_{2} \cap (Max(C_{1}) + 1)$.
Clearly $Q$ is a forcing notion of power $\aleph_{1}$, so it cannot collapse cardinals or regularity of cardinals except possibly $\aleph_{1}$ (all finite subsets of $S$ belong to $Q$). So we shall prove that $P * \dot{Q}$ does not collapse $\aleph_{1}$. So let (in $V$) $N \prec (H(\lambda), \epsilon)$, $N$ countable, $(p, \dot{q}) \in P * \dot{Q} \in N$, $(p, \dot{q}) \in N$ , $\delta := N \cap \omega_{1} \in S$ and suppose $G_{P} \subset P$ is generic over $V$ and $p \in G_{P}$. Note that as $S$ is a stationary subset of $\omega_{1}$, there is such $N$ (in $V$). So it is enough to find $(p',\dot{q'}) \geq (p, \dot{q})$ which is $(N, P * \dot{Q})$-generic. As $P$ satisfies the countable chain condition, p is $(N, P)$-generic (by 2.9), hence $N[G_{P}] \cap V = N$. Clearly $\dot{Q}[G_{P}] \in L[S, G_{P}] \subset V[G_{P}]$, now $N[G_{P}]$ does not necessarily belong to $L[S, G_{P}]$, but $N[G_{P}] \cap L[S, G_{P}]$ is $N[G_{\gamma}] \cap L_{\delta}[S, G_{P}] = L_{\delta}[S, G_{P}] \in L[S, G_{P}]$ and is a countable set in $L[S, G_{P}]$.
In $L[S, G_{P}]$ we have an enumeration of $\dot{Q}[G_{P}] \cap N[G_{P}]$ (of length $\omega$), say $\langle q_{n} : n < \omega \rangle$ (but not of the set of dense subsets); in fact we have it even in $L[S, G_{P} \upharpoonright(\delta + 1)]$ (and as we use $Levy(\aleph_{0}, < \aleph_{1})$ not $Levy(\aleph_{0}, < \kappa)$ even in $L[S, G_{P} \upharpoonright \delta]$). Now in $L[S, G_{P}]$ there is a Cohen generic real over $V[G_{P} \upharpoonright (\delta + 1)]$ say $r^{*} \in$ $^{\omega}{\omega}$ and we use it to construct a sequence $\bar{C} = \langle C_{n} : n < \omega \rangle$ such that $C_{n} \in \dot{Q}[G_{P}]$ in $L[S, G_{P}]$ i.e. $\bar{C} \in L[S, G_{P}]$, e.g. we choose $C_{n}$ by induction on $n$; we let $C_{0} = \dot{q}[G_{P}]$ and we let $C_{n+1}$ be $q_{m(n)}$ where $m(n)$ is the first natural number $m$ such that: $m \geq r^{*}(n)$ and $\dot{Q}[G_{P}] \models$ "$C_{n} \leq q_{m}$". Let 
$G$ := {$q$ : $q \in \dot{Q}[G_{P}]$, and $q \in N[G_{P}]$ and for some $n$, $\dot{Q}[G_{P}] \models$ "$q \leq C_{n}$"}. 
Clearly $G \subset N[G_{P}] \cap \dot{Q}[G_{P}]$ is generic over $V[G_{P} \upharpoonright (\delta + l)]$. So $q^{\dagger} = (\bigcup_{n} C_{n}) \cup \{\delta\} \in \dot{Q}[G_{P}]$ is $(N[G_{P}], \dot{Q}[G_{P}])$-generic. Going to names we finish.


A few questions regarding the proof above (Theorem 4.4 of Shelah's PIF). Here '$p \geq q$' means '$p$ extends $q$'.
First let $A \in N[G_{P}]$ be a maximal antichain of $\dot{Q}[G_{P}]$.

How do we ensure that the $G$ constructed at the end of the proof intersect $N[G_{P}] \cap A$?
If we cannot ensure the above, how do we show that $q^{\dagger}$ is ($N[G_{P}]$, $\dot{Q}[G_{P}]$)-generic?

I think there is some crucial properties of Cohen reals that I am missing here, such as how Cohen-genericity translate to $\dot{Q}[G_{P}]$-genericity, if at all.
Much thanks in advance.

REPLY [3 votes]: Sorry this is sketchy, but I hope it helps!
A Cohen condition $p$ of length $n$ is going to determine $\langle C_m:m<n\rangle$.  We know there's going to be an $i<\omega$ such that $q_i$ extends both $C_{n-1}$ and a member of $N[G_P]\cap A$.  Extend $p$ to a Cohen condition $p'$ with $p'(n)=i$.  If $r^*$ is any Cohen generic extending $p'$, then running the construction with $r^*$ guarantees that $C_n$ extends a member of $N[G_P]\cap A$.  So a density argument tells us the construction is forced to work.
The point is that in Shelah's construction,  $m(n)$ is going to be equal to $r^*(n)$ very often because $r^*$ is Cohen.<|endoftext|>
TITLE: Does this property of a partially ordered set have a name?
QUESTION [9 upvotes]: What do you call a poset with this property? For any elements $a,b,c,d$ such that $\{a,b\}\le\{c,d\}$, there is an element x such that $\{a,b\}\le x\le\{c,d\}$. (Equivalently, for any finite sets $A\le B$, there is an element x such that $A\le x\le B$.)  
For example, any upper or lower semilattice has this property. Also, overlooking the fact that it's not a set, the class of all cardinal numbers has this property in ZF.

REPLY [8 votes]: In the world of partially ordered abelian groups, this is the interpolation property.
These groups are called partially ordered abelian groups with interpolation, or simply interpolation groups. Intuitively, I think about them as "almost as nice as lattice ordered abelian groups".
A simple example of a non-lattice ordered interpolation group is the set of all
polynomial functions $\mathbb R\to\mathbb R$.
Probably the most important subclass is the class of dimension groups. As proved by Effros, Handelman and Shen in 1980, dimension groups classify
the approximate finite dimensional $C^*$-algebras (via the $K_0$ functor).
I recommend this book by Goodearl -- very readable, he is an excellent writer.<|endoftext|>
TITLE: Least prime in an arithmetic progression and the Selberg sieve
QUESTION [16 upvotes]: My question concerns a technical step in the proof of Linnik's theorem on the least prime in an arithmetic progression, as presented in Chapter 18 of Iwaniec-Kowalski: Analytic number theory. 
The proof uses certain weights $\theta_b$ coming from the theory of the Selberg sieve. The sequence is supported on square-free numbers up to $y$ coprime with $q$ (the modulus of the arithmetic progression), and its main feature is, cf. (18.19)-(18.23) in the book,
$$ |\theta_b|\leq 1,\qquad \theta_1=1, $$
$$ 0<\sum_{b_1,b_2}\frac{\theta_{b_1}\theta_{b_2}}{[b_1,b_2]}\leq\frac{q}{\varphi(q)\log y}. $$
The corresponding Selberg upper sieve coefficients are defined by
$$ \sigma_m:=\sum_{[b_1,b_2]=m}\theta_{b_1}\theta_{b_2}, $$
so that with the notation
$$ \nu(n):=\sum_{b\mid n}\theta_b $$
we have for any integer $n\geq 1$
$$ \sum_{m\mid n}\sigma_m=\nu(n)^2\geq\sum_{m\mid n}\mu(m). $$
By (18.70) of the book we have, "applying a sieve of dimension 8 (see the Fundamental Lemma 6.3)",
$$ \sum_{n\leq x}\nu^2(n)\frac{\tau^3(n)}{n}\ll\left(\frac{\log x}{\log y}\right)^8. $$
Here $\tau(n)$ is the number of divisors of $n$.
Can anyone help me understand why this is true? The quoted lemma is about Brun's sieve (where $\sigma_m$ would be $\mu(m)$ restricted to certain integers), but even if I accept it for the Selberg sieve, I do not see the stated bound. Similarly, I do not understand why (18.75) in the book is true.
Added. Based on the expert response of Dimitris Koukoulopoulos I think that the proof of Linnik's theorem, as presented in Iwaniec-Kowalski's book, works fine if we replace the Selberg sieve $\sigma_m$ with an upper $\beta$-sieve $\beta_m$. More precisely, we redefine $\nu(n)$ so that 
$$\nu(n)^2=\sum_{m\mid n}\beta_m,$$
which makes sense as the right hand side is nonnegative.

REPLY [12 votes]: The proof given in Iwaniec-Kowalski is, as it stands, wrong. It can be easily fixed, as I explain below. 
In general, one can think of $\nu(n)^2$ as the characteristic function of integers with $P^-(n):=\min(p|n)\ge y$. So
$$
\sum_{n\le x} \frac{\nu(n)^2 f(n)}{n} \approx \sum_{ n\le x,\ P^-(n)\ge y } \frac{f(n)}{n} \asymp \prod_{y\le p\le x} \left(1+\frac{f(p)}{p}\right) 
$$
for any reasonable multiplicative function $f$ that is bounded on primes. However, an important restriction is that $f(p)<\kappa$ on average, where $\kappa$ is the sifting dimension. In this case the dimension is 1, whereas $f(p)=8$. So the sum in question, say $S$, does not satisfy a priori the claimed bound. In fact, in this case $S\gg(\log x/\log y)^8$:
If $P^+(n)=\max(p|n)$, then we have that
$$
S= \sum_{n\le x}\frac{\nu(n)^2\tau(n)^3}{n} 
   \asymp \sum_{P^+(n)\le x}\frac{\nu(n)^2\tau(n)^3}{n}  
$$
(this step is heuristic and employed for simplicity). If we let $\sigma_m=\sum_{[d_1,d_2]=m}\theta_{d_1}\theta_{d_2}$ and we open $\nu(m)^2$, we have that
$$
S \approx \sum_{P^+(m)\le x} \sigma_m \sum_{P^+(n)\le x,\ m|n} \frac{\tau(n)^3}{n}  
  = P(x) \sum_{P^+(m)\le x} \frac{\sigma_m g(m)}{m},
$$
where $g(m)$ is a multiplicative function with $g(p)=8+O(1/p)$ and $P(x)=\prod_{p\le x}(1+\tau(p)^3/p+\tau(p^2)^3/p^2+\cdots)\asymp(\log x)^8$. Hence
$$
S/P(x)\approx \sum_{P^+(d_1d_2)\le x} \frac{\theta_{d_1}\theta_{d_2}g([d_1,d_2])}{[d_1,d_2]}
= \sum_{P^+(d_1d_2)\le x} \frac{\theta_{d_1}\theta_{d_2}g(d_1)g(d_2)}{d_1d_2} \frac{(d_1,d_2)}{g((d_1,d_2))}.
$$
Writing $f(m)=\prod_{p|m}(1-g(p)/p)$ so that $m/g(m)=\sum_{n|m}f(n)n/g(n)$, we deduce that
$$
S/P(x)\approx \sum_{P^+(n)\le x}\frac{f(n)n}{g(n)} \left( \sum_{P^+(d)\le x,\ n|d} \frac{\theta_{d}g(d)}{d}\right)^2.
$$
When $y/2 \lt n\le y$, we have that
$$
\sum_{P^+(d)\le x,\ n|d} \frac{\theta_{d}g(d)}{d}
 = \frac{\theta_n g(n)}{n} = \frac{\mu(n)g(n)}{G} \sum_{k\le y,\ (k,q)=1,\ n|k} \frac{\mu^2(k)}{\phi(k)}
 = \frac{\mu(n)g(n)}{G} \cdot \frac{\textbf{1}_{(n,q)=1}}{\phi(n)}
$$
(note that there is an error in the definition of $\theta_b$ in Iwaniec-Kowalski, as one has to restrict them on those $b$ which are coprime to $q$. In fact, $\theta_b=(\mu(b)b/G) \sum_{k\le y,\ (k,q)=1,\ b|k}\mu^2(k)/\phi(k)$). So
$$
S \gtrsim \frac{P(x)}{G^2} \sum_{y/2 \lt n\le y,\ (n,q)=1} \frac{\mu^2(n) f(n)g(n)n}{\phi(n)^2}
 \gg_q (\log x)^8(\log y)^5,
$$
by the Selberg-Delange method. This is certainly bigger than what we would need.
In order to remedy this deficiency, one would have to choose $\nu(n)$ in another way, as weights from a sieve of dimension 8 or higher. The easiest choice to work with is, as GH also points out, is to define $\nu$ via the relation $\nu(n)^2=(1*\lambda)(n)$, where $(\lambda(d):d\le D)$ is a $\beta$ upper bound sieve of level $y$ and dimension 8, so that
$$
S = \sum_{n\le x} \frac{(1*\lambda)(n)\tau(n)^3}{n}.
$$
The point is that the sequence $(\lambda(d))_{d\le D}$ satisfies the Fundamental Lemma (Lemma 6.3 in Iwaniec-Kowalski) in the following strong sense:
(1) $\lambda(d)$ is supported on square-free integers with $P^+(d)\le y$
(2) whenever $\{a(d)\}_{d\ge1}$ is a sequence such that
$$
\bigg|\sum_{P^+(d)\le z}\frac{\mu(d)a(dm)}{d}\bigg|
\le Ag(m)\prod_{y_0\le p\le z}(1-g(p)/p)
\quad\text{when}\ P^-(m)>z,
$$ 
where $A\ge0$ and $y_0\ge1$ are some parameters and $g\ge0$ is multiplicative with 
$$
\prod_{w\le p\le w'} (1-g(p)/p)^{-1} \le \left(\frac{\log w'}{\log w}\right)^8\left(1+\frac{K}{\log w}\right)\qquad(w'\ge w\ge y_0),
$$
we have
$$
\sum_{d\le D} \frac{\lambda(d)a(d)}{d} = \sum_{P^+(d)\le y} \frac{\mu(d)a(d)}{d}
+ O_K\left( Ae^{-u}\prod_{y_0\le p\le y}\left(1-\frac{g(p)}{p}\right) \right),
$$
where $u=\log D/\log y$. 
Remark: if $a=g$, then the condition on $a$ holds trivially with $A=y_0=1$. Hence, the above statement is indeed a generalization of the usual Fundamental Lemma.
We have that
$$
S = \sum_{d\le D} \frac{\lambda(d)a(d)}{d},
$$
where 
$$
a(d) = \sum_{m\le x/d} \frac{\tau(dm)^3}{m}.
$$
If $P^-(m)>z$, then
\begin{align*}
\sum_{P^+(d)\le z} \frac{\mu(d)a(dm)}{d}
&= \sum_{P^+(d)\le z} \frac{\mu(d)}{d}\sum_{k\le x/(dm)}\frac{\tau(dkm)^3}{k} \\
&= \sum_{n\le x/m} \frac{\tau(nm)^3}{n} \sum_{dk=n,\,P^+(d)\le z}\mu(d).
\end{align*}
By M\"obius inversion, we then conclude that
\begin{align*}
\sum_{P^+(d)\le z} \frac{\mu(d)\tau(d)^3a(dm)}{d}
&= \sum_{n\le x/m,\,P^-(n)>z} \frac{\tau(nm)^3}{n} \\
&\ll \tau(m)^3 (\log x/\log z)^8 \\
&\asymp (\log x)^8\tau(m)^3\prod_{11\le p\le z}(1-\tau(p)^3/p),
\end{align*}
so that the second axiom holds for $a$ with $A\asymp(\log x)^8$, $y_0=11$ and $g=\tau^3$. We thus conclude that
\begin{align*}
S = \sum_{d\le D} \frac{\lambda(d)a(d)}{d} 
&= \sum_{P^+(d)\le y} \frac{\mu(d)a(d)}{d} 
+ O\left( e^{-\log x/\log y} \left(\frac{\log x}{\log y}\right)^8\right) \newline
&= \sum_{n\le x,\ P^-(n)>y} \frac{\tau(n)^3 }{n} 
+ O\left( e^{-\log x/\log y} \left(\frac{\log x}{\log y}\right)^8\right)  \newline
&\ll \left(\frac{\log x}{\log y}\right)^8.
\end{align*}
A more general remark: Selberg's sieve is not as flexible as the $\beta$-sieve as far as ``preliminary sieving'' is concerned because it carries inside it the sieve problem it is applied to, in contrast to the $\beta$-sieve weights that only depend on the sifting dimension via the $\beta$ parameter.
EDIT: the monotonicity principle II described in page 49 of "Opera de Cribro" by Friedlander-Iwaniec is a way to use Selberg's sieve weights for preliminary sieving. See also Proposition 7.2 in the same book.<|endoftext|>
TITLE: A Johnson-Lindenstrauss lemma for finite fields?
QUESTION [9 upvotes]: Given $m$ points in $\mathbb{R}^N$, the Johnson-Lindenstrauss lemma guarantees the existence of a linear operator $\mathbb{R}^N\rightarrow\mathbb{R}^n$ that nearly preserves pairwise distances between the points.  Here, we can take $n=\Omega(\log(m)/\varepsilon^2)$, where $\varepsilon$ is the level of distortion.
Is there a similar result for points in a vector space over a finite field, e.g. $\mathbb{F}_2^N$?  I assume a result of this form would be in terms of Hamming distance.

REPLY [2 votes]: The problem with Hamming distance is that it's bounded above by $N$, so if you have a subset of $\mathbb{F}_2^N$ with Hamming distances in that range, you're not going to be able to embed it in $\mathbb{F}_2^n$ for $n$ much smaller than $N$.
Perhaps more natural is to build in a scaling, so that you want to find an embedding $f$ of the subset $S$ of $\mathbb{R}^N$ into $\mathbb{R}^n$ in such a way that
$d(f(x),f(y)) \approx \frac{n}{N}\cdot d(x,y)$.
In other words, given two vectors $x$ and $y$, you want the proportion of coordinates in which they agree to be more or less left alone by the projection.  Then you could try a random projection as in Johnson-Lindenstrauss -- i.e. show that (if indeed this is true) a random choice of one of the $N \choose n$ coordinate projections gives you low distortion in this sense, when $n$ is not too horrifically small.<|endoftext|>
TITLE: Restrictions of null/meager ideal
QUESTION [12 upvotes]: Let I denote the null (resp. meager) ideal on reals. Is it consistent that for any pair of non null (resp. meager) sets A and B, there is a null (resp. meager) preserving bijection between A and B? In particular, is this true in the model obtained by adding $\omega_2$ Cohen (resp. random reals) over a model of CH?

REPLY [6 votes]: $\newcommand\continuum{\mathfrak{c}}$
Update. This proof strategy is hopeless, for the reasons explained in Miha's answer at mathoverflow.net/a/144538/1946, and the comments on it. But I'll leave it here for the record of this false attempt. 

Original answer: 
Your hypothesis follows from the continuum hypothesis, and more generally, from the
assertion that the additivity of the null ideal is the continuum, which is consistent with the failure of CH.
To see this, suppose that the additivity of the null ideal is the continuum, which means that the union of fewer than continuum many sets of measure
zero still always has measure zero, and in particular, every set of size
less than the continuum has measure zero. Fix any two
sets $A$ and $B$ that are not of measure zero. So they have size
continuum. Let us now build a bijection between them $\pi:A\to B$,
in such a way that $\pi$ takes every measure zero subset of $A$ to
a measure-zero subset of $B$ and conversely. We will construct
$\pi$ as the union of a increasing chain of partial functions
$\pi=\bigcup_{\alpha\lt\continuum}\pi_\alpha$, in a construction of length continuum, where
$\pi_\alpha:A_\alpha\to B_\alpha$ is a bijection of measure-zero
subsets of $A$ and $B$, respectively. Enumerate the Borel
measure-zero sets as $D_\alpha$ for $\alpha\lt\continuum$. Suppose
that $\pi_\alpha:A_\alpha\to B_\alpha$ is defined, and consider
the set $D_\alpha$. First, we may extend $A_\alpha$ to
$A_{\alpha+1}$ in such a way that $D_\alpha\cap A\subset A_{\alpha+1}$,
by also adding a measure zero part of $B$ to $B_\alpha$, forming
$B_{\alpha+1}$ and extending the bijection to
$\pi_{\alpha+1}:A_{\alpha+1}\to B_{\alpha+1}$. (I am using that
every non-measure-zero set contains a measure zero set of size
continuum, which I believe follows from our assumptions; please correct me if this is wrong. Edit: this is not correct, because of the possibility of Sierpinski and Luzin sets, which exist, as Miha pointed out, when the additivity numbers are large.)
Similarly, for the target side, we may extend in such a way also
that $D_\alpha\cap B\subset B_{\alpha+1}$. At limit stages of our
construction, we take the union of the bijections constructed so
far, and until the end of the constructin, this union domain and
target will still have measure zero by the assumption that the
additivity of the null ideal is the continuum.
The resulting map $\pi:A\to B$ is a bijection, since every point
will arise as a singleton in some $D_\alpha$, thereby getting
added to the domain and range at that stage. And every measure
zero subset of $A$ will be covered by some $A\cap D_\alpha$ for
some $\alpha$, which gets mapped to $B_{\alpha+1}$, which has
measure zero on the $B$ side. And similarly for the measure zero
subsets of $B$.
Thus, the construction resembles the familiar back-and-forth constructions, but at each stage, we have defined the bijection on only measure zero sets, which exhaust neither $A$ nor $B$. This gives time to diagonalize against the possible measure zero sets, since there are only continuum many Borel measure zero sets to consider. 
So the answer is yes, this situation is consistent.
Meanwhile, if the uniformity number (the smallest size of the non-measure zero set) is less than the continuum, then your hypothesis clearly fails, since there will be non-measure-zero sets of different cardinality, which can therefore not be bijective. So we seem to have trapped your statement between the assertions that the additivity number is big and the uniformity number is not small. 
A very similar argument seems to work also in the case of category rather than measure, under the assumption that the additivity of the meager ideal is the continuum. (And one similarly needs in this case that every non-meager set contains a meager set of size continuum.)<|endoftext|>
TITLE: Closed form for derivatives $\zeta^{(n)}(1/2)$
QUESTION [7 upvotes]: According to mathworld
 41,42. "Derivatives $\zeta^{(n)}(1/2)$ can also be given in closed form"
 with example for the first derivative.
What is the closed form? References?
The motivation is that this question
expresses $\zeta(3)$ in terms of $\zeta(1/2)$ and the first 3 derivatives,
so closed form possibly might result in closed form for zeta(3)
(unless the closed form is derived by the linked question).
Particaluraly intersted in the second and third derivatives.
On what the derivatives would depend helps too.

REPLY [3 votes]: Edit: My original answer was incorrect.
You can evaluate $\zeta'(\frac{1}{2})$ recursively in terms of $\zeta(\frac{1}{2})$ using the symmetric form of the functional equation:
$$ \zeta(s)\Gamma(\tfrac{s}{2}) \pi^{-s/2} = \zeta(1{-}s)\Gamma(\tfrac{1-s}{2}) \pi^{(s-1)/2}. $$
Differentiating both sides sides of the equation, plugging in $s=\frac{1}{2}$, and then solving for $\zeta'(\frac{1}{2})$, I get the value listed on the MathWorld website.
As Noam Elkies points out, taking higher derivatives, this process allows you to write $\zeta^{(2n+1)}(\frac{1}{2})$ in terms of the smaller even derivatives $\zeta(\frac{1}{2}),\zeta''(\frac{1}{2}), \zeta^{(4)}(\frac{1}{2}), \ldots, \zeta^{(2n)}(\frac{1}{2})$.<|endoftext|>
TITLE: Optimal inspection path on a sphere
QUESTION [15 upvotes]: Suppose you would like to "inspect" every point of a unit-radius
sphere $S \subset \mathbb{R}^3$ by walking along a path $\gamma$
on $S$, but you can only see a distance $d$ from where you stand.

Q1.
  For a given $d \in (0,\pi]$, what are the shortest
  such paths $\gamma(d)$?

Two other ways to define $\gamma(d)$:


The shortest path $\gamma$ such that every point on $S$
is no more than a distance $d$ from some point of $\gamma$.

The shortest path $\gamma$ such that the union of disks of radius $d$
centered on every point of $\gamma$ cover $S$.

For $d \in [\pi/2,\pi]$, $\gamma(d)$ is an arc of a great circle
of length $2\pi - 2d$.
So, for $d=\pi$, $\gamma$ is a single point;
for $d=3\pi/4$, $\gamma$ is an arc of length $\pi/2$;
for $d=\pi/2$, $\gamma$ is a semicircle.
As $d$ approaches zero, it might be that $\gamma$ is a type of
spiral, e.g.:

          


          
(Image from grabcad.com.)

What is quite unclear to me is when $d$ is less than but
close to $\pi/2$.
For example, suppose $d = \frac{5}{12}\pi = 75^\circ$.
The union of disks of this radius centered on the equatorial
semicircle that works for $\pi/2$ is shown in (a) below: a
strip connecting the north and south poles is uncovered
(blue circles are boundaries of disks of radius $d$ centered on $\gamma$):

   


One can construct a candidate $\gamma$ as in (b) above:
extend the semicircle at each end
so that all points on the equator are reached, and add short vertical sections,
red to reach the north pole, and green to reach the south pole.
It does not seem likely that such sharp turns yield the
shortest $\gamma$.
Perhaps this question can be answered even without precise understanding
of $\gamma(d)$ for all $d$:

Q2. Does $\gamma(d)$ vary continuously with respect to $d$?

I originally wanted to explore this for any (smooth, closed, compact) surface
in $\mathbb{R}^3$, but it already seems not so straightforward for the
sphere.

Q3. Has this question been studied before?  It feels classical.

The question could also be posed for $(d{-}1)$-dimensional spheres in $\mathbb{R}^d$,
again seeking an optimal inspection path for each $d$.
(My interest in $\mathbb{R}^3$ stems from two related concepts: Voronoi diagrams on a surface,
and the cut locus with respect to a path, although neither seems to help here.)
Thanks for any pointers or ideas!

(Related earlier MO questions:
Optimal paintbrush geodesics;
Shortest closed curve to inspect a sphere.)

REPLY [2 votes]: This new paper "develops efficient algorithms for (self-)repulsion of plane and space curves."
"Constraining curves to a surface yields Hilbert-like curves that are smooth and evenly-spaced."

Chris Yu, Henrik Schumacher, Keenan Crane.
"Repulsive Curves."
ACM Transactions on Graphics, 2021.
Author link.<|endoftext|>
TITLE: Second nonabelian group cohomology: cocycles vs. gerbes 
QUESTION [18 upvotes]: In 1965 Jean Giraud published two Comptes Rendus notes titled "Cohomologie non abélienne", and in 1971 he published a book with the same title.
In 1966 Tonny A. Springer's paper "Nonabelian $H^2$ in Galois cohomology" appeared, where he, in particular,
constructs nonabelian $H^2$ of a group in terms of group extensions and in terms of cocycles.
Springer writes that his definition for group cohomology "seems to be essentially equivalent to to that of Dedecker and Giraud".
Giraud in his book (page 452) writes that "la définition de $H^2$ en termes de gerbes ... redonne, dans ce cas, la théorie de Springer".
I do not understand the latter assertion.
Let $\Gamma$ be a group, and let $G$ be a group together with a "$\Gamma$-kernel": a homomorphism
$\kappa\colon \Gamma\to {\rm Out}(G)$, where ${\rm Out}(G):= {\rm Aut}(G)/{\rm Inn}(G)$ is the group of outer automorphisms of $G$.
Springer defines $H^2(\Gamma, G,\kappa)$  in terms of group extensions 
$$ 1\to G\to E\to \Gamma\to 1$$
inducing the "kernel" $\kappa$. He also describes $H^2(\Gamma, G,\kappa)$ in terms of 2-cocycles coming from the group extension.
(A 2-cocycle is a pair of maps $(f,g)$ of maps $f\colon \Gamma\to {\rm Aut}(G)$, $g\colon \Gamma\times \Gamma\to G$ satisfying certain conditions.)
Giraud defines $H^2$ in terms of gerbes (on the category of $\Gamma$-sets?).
Question: How can I get a gerbe $\mathcal{G}$ (i.e., a stack over the category of $\Gamma$-sets) from a group extension? In other words, for any $\Gamma$-set $S$,  I want to get a groupoid $\mathcal{G}_S$ defined in terms of the given group extension.
Conversely, I would like get a group extension from a gerbe.

REPLY [3 votes]: Nonabelian $H^2$ in Galois cohomology can be defined in terms of: (1) cocycles, (2) extensions, (3) gerbes. The relations between these three definitions are described in Section 2.2 of Le principe de Hasse pour les espaces homogènes : réduction au cas des stabilisateurs finis by Cyril Demarche and  Giancarlo Lucchini Arteche.<|endoftext|>
TITLE: What is a good reference (preferably thorough) for the Derived Category of a scheme/orbifold/stack?
QUESTION [7 upvotes]: I've sort of circled around the idea of derived categories a few times, read a few introductory papers ("Derived Categories for the working mathematician", e.g.), and feel now that this is something that I need to actually learn.
So what is a good reference textbook-type resource for this material? Expository papers are great to read, but they only go so far in terms of building up a working understanding of the material.
Ideally this reference would contain some information about derived categories of orbifolds and some discussion of the relation to moduli problems, cotangent complices, etc. Bonus points if it has many exercises to work through.

REPLY [4 votes]: I'll just mention some sources that are not already in the other question indicated by Franz's answer, in case you need to become more comfortable with the derived machinery itself. (The book by Huybrechts on Fourier-Mukai Transforms in the other answer will probably be the best, depending on your interests, and there is also "Fourier-Mukai and Nahm Transforms in Geometry and Mathematical Physics", by Bartocci, Bruzzo, and Ruipérez that contains additional information and good references).
So here are the other references that might help with derived categories themselves:
I think most of the texts or papers treating derived categories applied to algebraic geometry are a bit too terse when it comes to derived categories themselves. There is a nice book called "Interactions Between Homotopy Theory and Algebra", which is from a summer school that was held in Chicago, and in this book there are two nice lectures by Henning Krause on derivated categories. His second lecture actually consists of about 40 exercises, which might help you with understanding the machinery of derived categories without having to worry about how they work in algebraic geometry yet. These two chapters are also available on the arXiv (Lecture, Exercises), though I recommend the whole book too.
Yet another book is the edited volume "Handbook of Tilting Theory" (LMS 332), which is a series of longer (most introductory papers) on tilting theory, so you'll see plenty of applications to module categories and representation theory, but there is also a chapter on the Fourier-Mukai transform there. 
In the book "Derived Equivalences for Group Rings" (König, Zimmerman, et al.), there are several chapters that include introductions to aspects of derived categories including unbounded derived categories and many examples, which might also be useful.<|endoftext|>
TITLE: Embeddings of smooth projective surfaces
QUESTION [5 upvotes]: Let $X$ be a smooth projective surface not contained in $\mathbb{P}^3$. 
Is there any known condition on $X$ under which I can embed it into $\mathbb{P}^3$ such that the its image contains at most simple surface singularities. Any idea or references on this topic will be very helpful.

REPLY [9 votes]: Let $S' \subset \mathbb{P}^3$ be the birational projection of a smooth surface $S \subset \mathbb{P}^4$. The generic projection theorem of Gruson-Peskine (http://arxiv.org/abs/1010.2399v2) tells you that either $S'$ is smooth or has a curve of double points.
For instance, if $S$ is the Veronese surface in $\mathbb{P}^4$, then its projection in $\mathbb{P}^3$ (the Steiner surface) has a curve of double points.
So the answer to your question is "never", unless your surface embedds in $\mathbb{P}^3$ (which you don't want).
Edit : If by "embed into $\mathbb{P}^3$", you mean project down to $\mathbb{P}^3$ from another embedding, my answer is correct, but there is a much simpler proof : you can always embed your surface in $\mathbb{P}^5$ because the secant variety of a surface in any projective space has at most dimension $5$. Then a projection to $\mathbb{P}^3$ factors as a projection to $\mathbb{P}^4$ and a projection to $\mathbb{P}^3$. If you choose your projection generically from $\mathbb{P}^5$ to $\mathbb{P}^4$ then it has only simple singularities in $\mathbb{P}^4$. But a simple count of dimension shows that for any point $p \in \mathbb{P}^4$, there is a $1$-dimensional family of bisecants to $S$ passing through $p$. This gives you a curve of double points in $\mathbb{P}^3$.<|endoftext|>
TITLE: Compact open topology
QUESTION [24 upvotes]: What is the intuition behind using compact open topology for eg. in the case of Pontryagin dual ?

REPLY [26 votes]: I have a weakness for pictures. So here is one 
 (source)
The above shows a function $f: \mathbb  R \to \mathbb R$, a  compact set $C$, and an open set $U$. The condition $f(C) \subseteq U$ is that the graph of $f$ passes through the shaded part shown in the picture.

REPLY [7 votes]: The compact open topology is essential for getting compact sets in your function space–especially a version of the Arzelà-Ascoli theorem holds for spaces of continuous functions to uniform spaces equipped with the compact open topology ($C(X,Y)$ with the compact open topology where $X$ is an appropriate topological space and $Y$ is an appropriate uniform space).
In your case (Pontryagin duality) the uniform space $Y$ is just the circle group $\mathbb{T}$ and $X$ is your group $G$. Consider a compact neighbourhood $U$ of the neutral element $e\in G$, an open neighbourhood $I:=\exp((-2\pi i\epsilon, 2\pi i\epsilon))\subset\mathbb{T}$ of 1 where $0<\epsilon<1$. Now there is a maximal set of irreducible representations $S\subset \hat{G}\subset C(G,\mathbb{T})$ such that $\forall \pi\in S\ \pi(U)\subset I$. The definition of the compact open topology guarantees that $S$ is open (it is just the definition). But $S$ is also equicontinuous: For $n\in\mathbb{N}$ define $U_0:=U$,$U_{n+1}:=\{x\in U_n\mid x+x\in U_n\}$. Clearly for $\pi\in S$ we have $\pi(U_n)\subset \exp((-2\pi i \epsilon/2^n,2\pi i \epsilon/2^n))$. It suffices to prove equicontinuity at the point $e$ (shifting does not change the situation). Now we can apply Arzelà-Ascoli: $S$ is in fact precompact and we get a compact neighbourhood $\bar{S}$ of the neutral element of $\hat{G}$ thus $\hat{G}$ is locally compact.
There is also a categorical motivation for this topology: The category of topological spaces is not cartesian closed, thus in this category (and many usual related topologies) the adjunction mentioned by David White in the comment above (currying) does not exist—it is impossible to choose the right topology. However, in the category of compactly generated spaces it works: It is cartesian closed and the topology for function spaces is the compact open topology. Again, the compact open topology guarantees that there are enough compact sets such that the topology is actually compactly generated (I guess that this is also a consequence of a version of Arzelà-Ascoli using even continuity, which does not need a uniformity, instead of equicontinuity, but I am not sure).<|endoftext|>
TITLE: Variational Principle for the Entropy
QUESTION [7 upvotes]: Theorem:  Let be  $f$  a homeomorphism of a compact metric space $X$, then 
$$
h_{top}(f)=\sup_{\mu\in \mathcal{M}_{f}}~  h _\mu (f)
$$
Question: The above theorem is the famous variational principle for compact spaces, I'm looking for an example to see that the hypothesis $ f $ be a homeomorphism is really necessary.
Another known theorem is
Theorem: Expansive transformations of compact metric spaces have a measure with maximal entropy.
Question: This measure is unique?
Thank you in advance.

REPLY [2 votes]: The mapping $f$ does not really need to be a homeomorphism.  The variational principle of the entropy is valid for all continuous mappings.  (See e.g., the book of Peter Walters)
A simple (boring) example of an expansive dynamical system on a compact metric space having multiple measures with maximal entropy will be the direct sum of two copies of your favorite example that has a unique measure with maximal entropy.
Another (more interesting) example is when the system has zero topological entropy but is not uniquely ergodic.  In particular, there are minimal systems that are not uniquely ergodic and have zero topological entropy.  However, one might still consider such examples pathological.
Multiplicity of measures with maximal entropy could be interpreted as some kind of "phase transition".  If measures with maximal entropy model the equilibrium states of a system (in the sense of statistical mechanics), then multiplicity of maximal entropy measures would mean the system has multiple "macroscopically distinguishable" equilibrium states.
If you allow two-dimensional dynamics (two commuting maps rather than one), Burton and Steif found examples of two-dimensional subshifts of finite type that are strongly irreducible and have more than one measures of maximal entropy.   Häggström later showed that in fact essentially any statistical mechanics model on the lattice with finite-range interactions is "equivalent" to a strongly irreducible subshift of finite type, so that the shift-invariant Gibbs measures of the former are in one-to-one correspondence with the maximal entropy measures of the latter.<|endoftext|>
TITLE: The role of the Automatic Groups in the history of Geometric Group Theory
QUESTION [19 upvotes]: What is the role of the theory of Automatic Groups in the history of Geometric Group Theory?
Motivation:
When I read through the "Word Processing in Groups" I was amazed by the supreme beauty and elegance of the theory and of how robust it is. (That it started from conversations between (true artists) Cannon and Thurston made it a bit less shocking)
I'm aware of the early great results and that it was a big thing. I also read somewhere that it was one of the pillars of Geometric Group Theory in the 80s, but I also noticed that many (great, young) people in the field don't know about this theory. Is it less central? If yes, why? Many basic early questions are open...
Sometimes in such a fast moving field (ggt) a beginner loose perspective...

REPLY [11 votes]: I think there are a few different ways in which Automatic Groups affected the history of Geometric Group Theory.
One was mentioned by Derek Holt, which I will spin in a slightly different way: if you really want to know the group, and if it has an automatic structure, you had better know that structure. For an individual group, the way to find out is to plug its presentation into the programs that Derek mentions. For classes of groups things can be trickier, but the effort is rewarding and sometimes even quite beautiful, for instance the proofs on automaticity of braid groups and Euclidean groups in "Word processing in groups", and the proof by Niblo and Reeves on biautomaticity of cubulated groups; and I retain a lot of affection for my paper proving automaticity of mapping class groups (and no offense taken, Misha).
Another thread of influence is the theory of biautomatic groups. To me, this theory is not yet played out, although as others say perhaps the open questions are hard. However, there are various beautiful pieces of this theory which had some interesting effects, and I think there is still some gold to mine here. The theory starts with the papers of Gersten and Short. Some applications of that theory are: the proof by Bridson and Vogtmann that $Out(F_n)$ is not biautomatic when $n \ge 3$; and my proof that direct factors of biautomatic groups are biautomatic. I also like the beautiful papers of Neumann and Shapiro which describe completely all possible automatic and biautomatic structures on $Z^n$, and the paper of Neumann and Reeves and its followup by Neumann and Shapiro which describe how to construct biautomatic structures on central extensions. I like to think that one of the effects of the Gersten/Short theory is that it gives a hint to a "hierarchical" structure on a biautomatic group. Farb's thesis follows this idea up with his notion of relatively automatic and bi-automatic groups, and the thesis of my student Donovan Rebbechi pins some of this down by giving rigorous proofs of some statements in Farb's thesis regarding bi-automatic structures on relatively hyperbolic groups. The theory of biautomatic groups is definitely still alive; poking around on MathSciNet just this very moment I find a paper that slipped my notice and that I want to go read right now, by Bridson and Reeves studying the isomorphism problem for biautomatic groups.
A separate and very important thread of influence is how the theory of automatic groups stoked interest in certain classes of quasi-isometry invariants. Gromov had already proved that hyperbolicity of a group is equivalent to linearity of the Dehn function (the isoperimetric function). One of the big applications of an automatic structure is Thurston's theorem that an automatic group has a quadratic (or better) Dehn function, which combined with the theorem that every subquadratic Dehn function is actually linear proves that nonhyperbolic automatic groups have quadratic Dehn functions on the nose. Thurston's proof of the quadratic upper bound introduced the concept of a combing of a group, and this led to a whole industry of studying different classes of combings, and the upper bounds they impose on the Dehn function. I particularly like the proof by Hatcher and Vogtmann finding an exponential upper bound to the Dehn function of $Out(F_n)$, which proceeds by finding a quite broadly stretched (pun intended) combing for $Out(F_n)$. Finding bounds on Dehn functions, and pinning down exact Dehn functions can be far from obvious, e.g. Robert Young's proof that $SL(n,Z)$ has quadratic Dehn function for $n \ge 5$, and the proof by Handel and myself of the exponential lower bound for the Dehn function of $Out(F_n)$. The issue of lower bounds is not particularly connected to automatic groups in any mathematical sense, but the historical connection is what I am trying to emphasize. Combings and Dehn functions have continued to grow from these and various seeds, and although I don't want to overstate the particular influence of Thurston's theorem in this context, nonetheless it is (besides Gromov's) one of the earliest concrete computations of Dehn functions.<|endoftext|>
TITLE: A characterization of Hilbert spaces?
QUESTION [6 upvotes]: My question was prompted by an earlier MO by @Daniel:
    Duality map in strictly convex Banach spaces
I will even use his symbol   $\phi$   below.
Let   $B$   be an arbitrary Banach space. Let   $S := \{x\in B:\|x\|=1\}$   be its unit sphere. Let   $\Gamma := \{f\in B^*: \|f\|=1\}$    be the unit sphere in the dual space $B^*$.
QUESTION   Are the following two conditions on $B$ equivalent:

$B$   is isometric to a Hilbert space.
There exists an isometry   $\phi: \Gamma \rightarrow S$   such that   $\forall_{f\in\Gamma}\ f(\phi(f))=1$.

?
The finite-dimensional case is especially basic.
REMARK 0   Perhaps similar questions were asked in the past (on MO too?)--please, let me know.
REMARK 2 The case of   $\mathbb R^2$   and its two dual but isometric norms   $L_\infty\quad L_1$   is interesting. The general question related to the one above is to describe all Banach spaces which are isometric to their dual space. Is there any beside the Hilbert spaces and   $\mathbb R^2$   with the norm(s) just mentioned above?

REPLY [11 votes]: Yes it is true. Let me show that the existence of $\phi$ implies that the norm of $B^*$ is associated to an inner product. Then it follows easily that the spaces are Hilbert.
It suffices to verify that the norm is an inner product one on every two-dimensional subspace. (Indeed, this property is equivalent to the parallelogram law,  which involves only two-dimensional configurations.)
Let $V$ be a two-dimensional subspace of $B^*$ (equipped with the restriction of the norm of $B^*$) and $V^*$ the dual of $V$ (equipped with the norm dual to this restriction). There is a natural map $\pi:B\to V^*$ dual to the inclusion $V\to B^*$. Namely $\pi(x)(f)=f(x)$ for $x\in B$, $f\in V$. Note that $\pi$ does not increase the norm: $\|\pi(x)\|\le\|x\|$ for all $x\in B$.
Let $D$ be the unit ball of $V$ and $E\subset V$ the maximum-area ellipse contained in $D$ (i.e., the John ellipsoid of $D$). Let $\Sigma$ be the set of points where the boundaries of $D$ and $E$ meet. It is easy to see that $\Sigma$ contains at least two pairs of opposite points. The ellipse $E$ is a unit ball of  a Euclidean norm $\|\cdot\|_E$ on $V$. Since $E\subset D$, we have $\|f\|_E\ge\|f\|$ for all $f\in V$ and equality is attained only if $x$ is proportional to an element of $\Sigma$.
On $V^*$, there is a dual Euclidean norm, denoted by $\|\cdot\|^*_E$. There we have $\|y\|^*_E\le \|y\|$ for all $y\in V^*$. The norm $\|\cdot\|_E$ is associated to an inner product, which defines an isomorphism $I:V\to V^*$ preserving the Euclidean norm.
For every $f\in\Sigma$, we have $\pi(\phi(f))=I(f)$. Indeed, $\|\pi(\phi(f))\|\le\|\phi(f)\|=1$, hence $\|\pi(\phi(f))\|^*_E\le 1$. On the other hand, $\pi(\phi(f))(f)=f(\phi(f))=1$. Since $\|f\|_E=1$, this is possible only for $\pi(\phi(f))=I(f)$.
Now consider two linearly independent vectors $f,g\in\Sigma$ and look at the distance between $f$ and $-g$. Since $\phi$ is an isometry, we have 
$$
\|f+g\|=\|\phi(f)-\phi(-g)\| \ge \|\pi(\phi(f)-\phi(-g))\| .
$$
The r.h.s equals
$$
 \|\pi(\phi(f))-\pi(\phi(-g))\| = \|I(f)-I(-g)\|=\|I(f+g)\|
$$
since $\pi$ and $I$ are linear and $f,-g\in\Sigma$.
Thus $\|I(f+g)\|\le \|f+g\|$ and therefore
$$
\|I(f+g)\|_E^*\le\|I(f+g)\|\le \|f+g\| \le  \|f+g\|_E .
$$
But $I$ preserves the Euclidean norm, so the inequalities turn into equalities. In particular, $\|f+g\|=  \|f+g\|_E$, hence $f+g$ is proportional to an element of $\Sigma$.
Thus we have proved that, for every $f,g\in\Sigma$, the normalized bisector $\frac{f+g}{\|f+g\|}$ also belongs to $\Sigma$. Since $\Sigma$ is a closed subset of an ellipse and contains more than two points, it follows that $\Sigma$ is the whole ellipse. This means that $D=E$, so the norm on $V$ is Euclidean. Q.E.D.
Remark. The proof would be much easier (in fact, nearly trivial) if we assumed in advance that $\phi$ is a restriction of an isometry between  $B^*$ and $B$. Then it would be linear by Mazur-Ulam and one could just define the inner product of $f,g\in B^*$ by $2\langle f,g\rangle=g(\phi(f))+f(\phi(g))$.<|endoftext|>
TITLE: A sequence based on Catalan–Mihăilescu problem
QUESTION [12 upvotes]: It was conjectured by Catalan in 1844 that the only solutions of the equation $x^a-y^b=1$ over variables $a,b,x,y\in\mathbb{N^+}$ are trivial ones: $3^1-2^1=1$ and $3^2-2^3=1$. The conjecture was proved true by Preda Mihăilescu in 2002.
Let's consider the equation $3^a-2^b=n$  over variables $a,b\in\mathbb{N^+}$ with a parameter $n\in\mathbb{N^+}$. Let $f(n)$ be the number of solutions of this equation for a given $n$.

Is there a simple way to calculate $f(n)$ for a given $n$?
Can $f(n)$ take any values except $0$ and $1$ for $n>1$?
Is it possible that $f(n)=\infty$?
Are there arbitrary long runs of $0$'s and $1$'s in the sequence?
What is the asymptotic density of $0$'s in the sequence?

REPLY [7 votes]: Can $f(n)$ take any values except $0$ and $1$ for $n>1$?

Yes, $2^5 - 3^3 = 5 = 2^3 - 3^1$, but this is very exceptional!  

Is it possible that $f(n)=\infty$?

No. Indeed, $f(n)$ is $0$ or $1$ for $n\gt 13$  (and for the remaining ones all solutions are also known and I think there are never more than two, but deifinitely only very small, see link below). This was proved by Stroeker and Tijdeman (1982) however that it is only $0,1$ for large $c$ is a lot older (Herschfeld in the thirties).

Are there arbitrary long runs of $0$'s and $1$'s in the sequence?

For $0$ yes, for $1$ I am not sure at the moment but I doubt it (and it might be known, perhaps there is even a direct argument).

What is the asymptotic density of $0$'s in the sequence?

The density is $1$. This follows from the fact that the number of solutions $(x,y)$ of the diophantine inequality
$$ 0 \lt  2^x - 3^y \le c$$ is asymptotically $(\log c )^2/ (2 \log 2 \log 3)$.
So below $c$ the function $f$ can be (and is, due to above mentioned result) positive only about  $(\log c)^2$ times. (This is a special case of a result by Pillai.)
For further details the start of the paper of Waldschmidt "Perfect powers: Pillai's works and their devellopment" is a good starting point. Also you might look at http://oeis.org/A219551 which gives (something equivalent to) the exact values of $f(n)$ and some references (but note this is slightly different as absolute values are considered).

REPLY [5 votes]: The conjecture that there is at most a single solution to the equation $3^a-2^b=n$, provided $|n|>13$ (which implies that $f(n)$ is $0$ or $1$ for $n > 1$), dates back to Pillai and was proved by Stroeker and Tijdeman in 1982 (using lower bounds for linear forms in logarithms). One has (again, from linear forms in logarithms) inequalities of the shape
$$
\left| 3^a - 2^b \right| > 2^{0.9b},
$$
by way of example, valid for something like $b > 3$. This enables one to compute $f(n)$ efficiently and to show that the asymptotic density of zeros in the sequence of values of $f(n)$ is one. If you assume that $a$ and $b$ are positive, $3^a-2^b$ is odd and so there are never consecutive occurrences of $f(n)=1$.<|endoftext|>
TITLE: Show that this ratio of factorials is always an integer
QUESTION [37 upvotes]: show the formula always gives an integer
$$\frac{(2m)!(2n)!}{m!n!(m+n)!}$$
I don't remember where I read this problem, but it said this can be proved using a simple counting argument (like observing that $\frac{(3m)!}{m!m!m!}$ is just the number of ways of permuting m identical things of type 1, m of type-2 and m of type-3).

REPLY [2 votes]: I am unable to suggest a  $\textbf{combinatorial}$  interpretation of the fact that $~Q(m,n)=\dfrac{(2m)!(2n)!}{m!n!(m+n)!}~$ is an integer (and I do not pretend that what follows is original...) but the result itself is a easy consequence of the two simple propositions :
a) $~\forall x\in \mathbb{R}~~\forall y\in \mathbb{R}~~\lfloor x \rfloor+\lfloor x+y \rfloor+\lfloor y \rfloor~\leq~\lfloor 2x \rfloor+\lfloor 2y \rfloor,~$
b) if, for $~p~$ prime and $~m~$ integer $~\geq 1,~$ we name $~v_p(m)~$ the exponent of $~p~$ in the prime decomposition of $~m,~$ then $~v_p(m!)=\sum\limits_{k\geq 1}\Big\lfloor\dfrac{m}{p^k}\Big\rfloor.~$
Indeed, for all prime $~p~$ and all integers $~m,~n~$ $\geq 1,~$
$v_p(Q(m,n))=\sum\limits_{k\geq 1}\Big(\Big\lfloor\dfrac{2m}{p^k}\Big\rfloor+\Big\lfloor\dfrac{2n}{p^k}\Big\rfloor-\Big\lfloor\dfrac{m}{p^k}\Big\rfloor-\Big\lfloor\dfrac{n}{p^k}\Big\rfloor-\Big\lfloor\dfrac{m+n}{p^k}\Big\rfloor\Big)~\geq 0,$ what was to be shown.<|endoftext|>
TITLE: Is there an algebraic curve over Q which is not modular?
QUESTION [28 upvotes]: Every elliptic curve $E/\mathbf Q$ is modular, in the sense that there exists a nonconstant morphism $X_0(N) \to E$ for some $N$. 
It is tempting to extend this definition in a naïve way to an arbitrary projective curve over $\mathbf Q$; if $Y$ is such a curve, we might say that $Y$ is modular if there exists a nonconstant morphism $X_0(N) \to Y$ for some $N$. A necessary condition for $Y$ to be modular is that it should have at least one rational point, since $X_0(N)$ always has a rational point. For an elliptic curve, this condition is satisfied by definition. 
There are certainly a great deal of curves which are modular in this sense. But is there an example of a curve (with a rational point) which is not modular?

REPLY [21 votes]: It is conjectured that there are only finitely many curves over $\mathbf{Q}$ of given genus $g \geq 2$ which are covered by a modular curve, see Conjecture 1.1 in the following paper :
Baker, M. H. ;  González-Jiménez, E. ;  González, J. ;  Poonen, B. . Finiteness results for modular curves of genus at least 2. Amer. J. Math.  127  (2005),  no. 6, 1325--1387.
In fact, the authors prove a strong result towards this conjecture, namely that there are only finitely such curves which are new (in an appropriate sense).<|endoftext|>
TITLE: Proving that a generic variety with ample canonical bundle has no automorphisms
QUESTION [11 upvotes]: Let $X$ be a smooth projective connected variety over the complex numbers with ample canonical bundle. If $X$ is generic and $\dim X \leq1$, the automorphism group of $X$ is trivial, see for instance
Why is a general curve automorphism-free?
This question is about generalizing this to arbitrary dimension. Let me be more precise.
Suppose that $X$ is "generic". Is the automorphism group of $X$ trivial?
This is probably true, and there are three approaches to this sketched in the above MO question. The first two might not be feasible.

Use deformation theory, i.e., compute the tangent space at the moduli space, and use Lefschetz trace formula. Can somebody make this more precise in this case?
Count parameters using Riemann-Hurwitz. This is going to be problematic in the higher-dimensional case, even though there is a Riemann-Hurwitz formula, I am not sure the dimension of the moduli space is explicitly known (as opposed to the one-dimensional case where it equals $3g-3$).
Exhibit an $X$ as above with trivial automorphism group for any possible hilbert polynomial. In fact, the order of the automorphism group of $X$ is bounded (even explicitly) by a constant depending only on the Hilbert polynomial of $X$.

I think 3 is the most promising, but this would require me to come up with the following.
Let h be the hilbert polynomial of $X$. Then there exists a smooth projective connected variety $Y$ with ample canonical bundle and hilbert polynomial of the canonical bundle equal to $h$ such that Aut$(Y)$ is trivial.
So my problem is to do this for every occuring hilbert polynomial. Of course, writing down varieties $X$ as above with no automorphisms is not so difficult.

REPLY [12 votes]: As Francesco points out, the claim is  false when $\dim X>1$. The question is  discussed in 
[B. Fantechi, R. Pardini, Automorphisms and moduli spaces of varieties with ample canonical class via deformations of abelian covers, Comm.  Algebra  25 (1997), 1413-1441.  math.AG/9410006] (see in particular Thm. 6.6).<|endoftext|>
TITLE: Applications of visual calculus
QUESTION [16 upvotes]: Mamikon's visual calculus (see Mamikon, Tom Apostol, Wikipedia) is a very beautiful and surprisingly efficient tool.
The basis is

Mamikon's theorem. The area of a tangent sweep is equal to the area of its tangent cluster, regardless of the shape of the original curve.

For a nice picture, see this, and the following picture from Apostol's introduction:
    
 (source: Wayback Machine)
The above links provide interesting applications, like very easy ways to find the area of a cycloid and tractrix.

Question.

What are other new applications?
Are there new developments of visual calculus?
What are some similar visual results, which can simplify calculations, and can be included in visual calculus?

REPLY [3 votes]: Perhaps this previous MO question may help:
Taking “Zooming in on a point of a graph” seriously,
e.g., this answer link.<|endoftext|>
TITLE: Triangle area on surfaces of constant curvature
QUESTION [5 upvotes]: I am looking for an elementary derivation of the formula for the area of a geodesic triangle lying in a surface of constant curvature $\kappa$, depending on the angles and side length.
Of course, the formula can easily be derived from the Gauss–Bonnet formula to be
$$A = \frac{1}{\kappa}(\alpha + \beta + \gamma - \pi)$$
for $\kappa \neq 0$. However, I would like to have an elementary geometric proof.
Does anybody know a reference?

REPLY [7 votes]: All the "elementary derivations of that type" are cheating (it may look nice but it proves nothing). 
The only elementary way to introduce area is adding it as an axiom (which is already kind of cheating).
You have to say that there is a additive area-functional on the set of all polygons.
Then you probably want to prove that this functional is unique (or include it in the same axiom).
It remains to notice that your functional $A$ satisfies the same properties and nothing left to prove.<|endoftext|>
TITLE: Mapping class group of once-punctured torus
QUESTION [6 upvotes]: Let $T$ be the 2-dimensional torus and let $S$ be $T$ minus one point. Then Birman exact sequence of mapping class groups becomes an isomorphism
$$
\beta: Map(S)\to Map(T)=GL(2, {\mathbb Z}). 
$$
It is then essentially immediate that $\beta$ preserves Thurston's classification of elements of the mapping class group into three types: $\beta(f)$ is Anosov if and only if $f$ is pseudo-Anosov, etc. 
Question. Did anybody bother to record this elementary observation in the literature? 
I just need a reference, since anybody who knows anything about the mapping class group knows how to prove it (in several ways). (Please, do not write proofs, I know at least 4.)   
I was nearly sure that Farb and Margalit have it, but they do not. Same for Casson and Bleiler, same for Ivanov. Of course, maybe this is one of the cases when it is easier to write a proof then to find a reference.

REPLY [3 votes]: I think the correct answer is the book "Thurston's work on surfaces", or under its original French title "Travaux de Thurston sur les surfaces", by Fathi, Laudenbach, and Poenaru; english translation by Kim and Margalit.
The explicit statement you want is in the beginning of section 1.5.<|endoftext|>
TITLE: Collatz stopping-time and Poisson distribution, and connection to other problems?
QUESTION [9 upvotes]: I read many threads about Collatz here - so don't worry, this is no attempt to any proof, just asking about a curious fact:
This graph gives the stopping-time of Collatz sequences up to $n=10^8$

(source: http://en.wikipedia.org/wiki/File:CollatzStatistic100million.png ) and it's distribution looks very similar to a Poisson distribution.
Is there some known reason why the Collatz-sequence stopping time behaves like a poissonian distribution?
What are the connections to other mathematical problems: Does the Collatz-conjecture imply other conjectures, or do other conjectures imply the Collatz-conjecture?
Thank you!

REPLY [12 votes]: As to the observed distribution of total stopping times for integers $n \leq 10^8$,
I think heuristically this can be explained quite well by the obvious stochastic model
(multiply $n$ by $3/2$ or $1/2$, each with probability $1/2$, repeat this
until the number gets $\leq 1$ and count the number of steps this takes).
For literature on such stochastic considerations, see Lagarias'
annotated bibliography on the conjecture (http://arxiv.org/abs/math/0309224).
Proving anything is of course quite a different task!
The Collatz conjecture can be formulated in quite a number of equivalent
ways, see also Lagarias' bibliography. Though so far the conjecture is by far
not as well-embedded into known parts of mathematics as for example the Riemann
Hypothesis. Namely I am not aware of, say, important 'theorems' which are
proved up to the Collatz conjecture. Nevertheless, I think the fact that an
easy-to-state question like the Collatz conjecture seems so intractable
suggests that certain important things are not understood so far.
Hence I don't think the conjecture will seem as isolated as now for all
future -- but of course this is my personal view.<|endoftext|>
TITLE: Invariants of a $GL(3,\mathbb{R})$ action
QUESTION [6 upvotes]: I'm trying to understand the standard $GL(3,\mathbb{R})$ action on the 15-dimensional space of possible values for the derivative of the Riemann curvature tensor of a 3-dimensional manifold $M$ at a point, thought of as the codimension-3 subspace of the space
$ S^2(\Lambda^2 T^*M) \otimes T^*M = \{r_{ijkl,m} (dx^i \wedge dx^j) \circ (dx^k \wedge dx^l) \otimes dx^m \} $
defined by the 2nd Bianchi identities:
$r_{2323,1} + r_{2331,2} + r_{1223,3} = r_{3131,2} + r_{1231,3} + r_{2331,1} = r_{1212,3} + r_{1223,1} + r_{1231,2} = 0.$
Is anything known about normal forms and/or invariants for this action?

REPLY [5 votes]: There's a 'quasi-normal form' on a dense open set that can be described without too much trouble.  Here's one way to do it.
First, recognize that we are looking for a normal form for elements $Q\in W=\bigl[S^2(\Lambda^2(V^\ast))\otimes V^\ast \bigr]_0$ under the action of $G=\mathrm{GL}(V)$, where $V$ is a vector space of dimension $3$ and $W$ is the kernel of the $G$-module mapping $$
S^2(\Lambda^2(V^\ast))\otimes V^\ast\hookrightarrow 
\Lambda^2(V^\ast)\otimes \Lambda^2(V^\ast)\otimes V^\ast\to \Lambda^2(V^\ast)\otimes\Lambda^3(V^\ast)$$ 
defined to be the natural inclusion followed by fully skewsymmetrizing in the last two factors.  This $W$ is an irreducible $G$-module of dimension $15$.  
Now, it makes things a little easier to follow if one recognizes that there is a canonical isomorphism $\Lambda^2(V^\ast) = V\otimes \Lambda^3(V^\ast)$ (given by the obvious contraction), so that $W$ can also be understood as a subspace of $S^2(V)\otimes V^\ast\otimes \bigl(\Lambda^3(V^\ast)\bigr)^2$. In fact $W = W_0\otimes \bigl(\Lambda^3(V^\ast)\bigr)^2$, where $W_0\subset S^2(V)\otimes V^\ast$ is the $15$-dimensional subspace that is the kernel of the trace mapping $S^2(V)\otimes V^\ast\to V$.  So I am going to think of $Q$ as an element of $W_0\otimes \bigl(\Lambda^3(V^\ast)\bigr)^2$.  
The vector space $U = V^\ast\otimes \bigl(\Lambda^3(V^\ast)\bigr)^2$ has dimension $3$, and by pairing $Q$ with elements in $U^\ast$, one can generate a subspace $\delta Q\subset S^2(V)$ of dimension at most $3$; say that $Q$ is of full quadratic rank if $\delta Q$ has dimension $3$, i.e., $\delta Q$ lies in $\mathrm{Gr}_3\bigl(S^2(V)\bigr)$.  The set of $3$-dimensional subspaces of a $6$-dimensional vector space has dimension $9$, and there is another way to generate a $9$-parameter family $3$-dimensional subspaces of $S^2(V)$, namely, if $C\in S^3(V)$ is a nondegenerate cubic, thought of as a cubic polynomial function on $V^\ast$, then we can let $\partial C\subset S^2(V)$ denote the $3$-dimensional space of its partial derivatives. (In fact, 'nondegenerate' in this context means exactly that $\partial C$ has dimension $3$, the maximum dimension possible.)  
Now, these two methods of generating a $3$-dimensional subspace of $S^2(V)$ are related: For any $3$-dimensional subspace $P\subset S^2(V)$, there is always a cubic $C$ such that $\partial C\subset P$, and, most of the time, this cubic is unique up to multiples and satisfies $\partial C = P$.  One can see this as follows:  There is a natural exact sequence
$$
0\to S^3(V)\to S^2(V)\otimes V\to V\otimes \Lambda^2(V)\to \Lambda^3(V)\to 0.
$$
In particular, the image of the map $S^2(V)\otimes V\to V\otimes \Lambda^2(V)$ has dimension $8$.  If $P\subset S^2(V)$ has dimension $3$, then the restriction of this map to $P\otimes V\to V\otimes \Lambda^2(V)$ has rank at most $8$, so there must be at least a $1$-dimensional kernel, i.e., there must be a nonzero $C\in S^3(V)$ whose image under the above mapping lies in $P\otimes V$.  Say that a $3$-dimensional subspace $P\subset S^2(V)$ is uniquely partial if the kernel of $P\otimes V\to V\otimes \Lambda^2(V)$ has dimension $1$.  The set of uniquely partial $P$ is a dense open set in $\mathrm{Gr}_3(S^2(V))$.  In particular, for a dense open set of $Q\in W$, the subspace $\delta Q$ will be of dimension $3$ and uniquely partial.  Let us say that $Q$ is nonsingular if, in addition, the cubic $C\in S^2(V)$ (unique up to multiples) such that $\partial C = \delta Q$ is nonsingular.  The set of nonsingular $Q$ is a dense open set in $W$.  
Finally, put the nonsingular $C$ associated (uniquely up to multiples) to a given nonsingular $Q$ in normal form, i.e., take a basis $e_i\in V$ such that
$$
C = {e_1}^3+{e_2}^3+{e_3}^3 + 6\sigma\ e_1e_2e_3\ ,
$$
where $\sigma\not=-\tfrac12$ is a real number.  (In fact, one needs to disallow a few more values of $\sigma$ in order to make sure that $\partial C$ is uniquely partial, but I'll leave that to the reader.  For example, $\sigma=0$ is not allowed.)  The basis $e_i$ is uniquely determined by $C$ up to permutations.  (Of course, $C$ is only determined up to a multiple, so the basis $e_i$ is only defined by $Q$ up to permutation and simultaneous scaling by a real number.  This $S_3\times \mathbb{R}^*$-ambiguity is the reason that the normal form is only 'quasi-normal', as we will see.)  
Let $e^i$ be the dual basis.  Now, by definition, $Q$ lies in $\partial C\otimes V^\ast\otimes S^2(\Lambda^3(V^\ast))$, so there are numbers such that
$$
Q = \tfrac13 b_{ij} \frac{\partial C}{\partial e_i}\otimes e^j
\otimes (e^1\wedge e^2\wedge e^3)^{\otimes 2}
$$
The $9$ entries of $b = (b_{ij})$ are subject to three linear equations caused by the trace relation (i.e., the second Bianchi identity), and these relations are found to be $b_{ii}+\sigma(b_{jk}+b_{kj})=0$, where $(i,j,k)$ is any even permutation of $(1,2,3)$.  They must also satisfy $\det b\not=0$, since, otherwise, $Q$ would be degenerate.  
Obviously, these relations are invariant under the symmetric group and scaling.  One could normalize the scaling away (up to a $\pm1$) by requiring that the sum of the squares of the $b_{ij}$ be equal to $1$, and this would leave a finite group $S_3\times \lbrace\pm1\rbrace$ 
that preserves this quasi-normal form.  This brings the number of free parameters down to $6$, namely $\sigma$ and the $5$-sphere of the normalized $b_{ij}$ (which lie in a canonical $6$-dimensional subspace that depends on $\sigma$).  One could then use the finite group to normalize things further or use the invariant theory of this finite group to find invariant combinations of these quantities that will yield invariants of the original $\mathrm{GL}(V)$-action on the dense open set consisting of the nonsingular $Q$s.
With a little more work, one could actually get quasi-normal forms for all of the orbits in $W$, but that can get complicated.<|endoftext|>
TITLE: collapsing successor of singular
QUESTION [8 upvotes]: Let $\lambda$ be a singular cardinal.  Is it consistent that there is a forcing of size $\lambda^+$ that collapses $\lambda^+$ while preserving all cardinals below $\lambda$?
(Note that even without the size requirement this implies a failure of the Jensen covering property, so such a forcing does not necessarily exist.)

REPLY [5 votes]: The answer is no and it follows easily from the following theorem:

Theorem. Suppose $\kappa$ is a regular uncountable cardinal and $|P|\leq \kappa.$ Then $\Vdash_P cf(\kappa)=|\kappa|.$

For a proof of the theorem see
Singularizing forcing of "small" cardinality?<|endoftext|>
TITLE: Existence of dominating measure for weak*-compact set of measures
QUESTION [6 upvotes]: I have posted the following question also here a longer time ago, but due to no answers I thought it might fit better to MO.
Let $(\Omega,\mathcal F)$ be a measurable space and $\mathcal P$ a weak*-compact subset of the set of all probability measures on $\mathcal F$. Does there exist a probability measure $\mathbb Q$ such that every measure $\mathbb P\in\mathcal P$ is absolutely continuous to $\mathbb Q$, i.e. such that $\mathbb P$ dominates the mesures in $\mathcal P$?
The weak*-topology is the weakest topology such that all the linear functionals $L_Z:\mathbb P\mapsto \int_\Omega Zd\mathbb P$ for $Z:\Omega\rightarrow\mathbb R$ $\mathcal F$-measurable and bounded, are continuous.
Note that when the set $\mathcal P$ is countable one can easily find a dominating measure for $\mathcal P$.

REPLY [2 votes]: This more a comment than an answer but will be too long for that.  There are two basic approaches to the theory of finite measures---the topogical one and the one based on $\sigma$-algebras.  At the core of the first approach lies the duality between $C^b(S)$ (the bounded, continuous functions on a completely regular space $S$) and the space $M^t(S)$ of Radon measures thereon, for the second that between $L^\infty(\Omega)$ (the bounded measurable functions on a measure space $(\Omega,\cal A)$ and the space $\cal M(\cal A)$ of finite $\sigma$-additive real valued-measures.  All four of these spaces are Banach spaces in a natural way but these structures are not compatible with the above dualities.  This suggests that  one should try to provide the function spaces  with suitable locally convex topologies which are compatible.  In the case of $C^b(S)$ this means a suitable intrinsic locally convex topology which is complete and has the required dual space $M^t(S))$.
Due to the joint efforst of several mathematicians (in particular, R.C. Buck and the polish school, e.g., Orlicz and  Wiweger)  such a topology is known---it is the so-called strict topology  (strictly speaking, this is not always complete, but it is so for most of the standard classes of topological spaces).
There is such a topology in the second case but I have been unable to find it in the literature and this  is the reason that I am submitting this answer.  It has various descriptions (in itself, in my opinion, a hint that it is a useful and natural topology).  But for me the killer-diller fact is that it has the natural universal property for $\sigma$-additive measures with values in a Banach space  (in contrast to the topological case, where we can embed $S$ into the space $M^t(S)$ with corresponding universal property), we can here embed the $\sigma$-algebra $\cal A$ into $L^\infty(\Omega)$ in the natural way.
We close with a few remarks. 
$1$. We emphasise that it is important that the above topologies have intrinsic definitions (i.e., independent of the dual pairs).  The fact that they then turn out to be topologies which can be defined via the duality---typically the Mackey topology---is then a theorem to be proved.  This is, basically, the reason for
the relevance of all this to the OP which essentially asks for a characterisations of the weakly compact subsets of  the dual of $L^\infty$.
$2$.  Another reason for depositing this was to clarify the confusion about which is the relevant topology on the family of measures---it is clear from the formulation that the OP refers to the weak star topology of a subsset of the dual of $L^\infty$ with the topology referred to above.
$3$.  If one tries to weaken the topology of a Banach space in a non-trivial way (as we are doing here) then there are contraints on the kind of space one can get.  For example, one cannot get a barrelled space (closed graph theorem).  This means that the resulting
topology 
cannot be a member of one of the traditional classes of well-behaved LCS's (metrisable, inductive limits of Banach spaces, e.g.). Neither can they  be nuclear (except, of course, in trivial cases).  This, alas, seems to have been  a hindrance to their acceptance into the main body of functional analysis.  It is my belief that this is a great pity since they are precisely the tool required for extending and enriching the relationship between functional
analysis (duality theory) and measure theory (the Riesz representation theorem) which is one of the crown jewels of analysis.<|endoftext|>
TITLE: Another colored balls puzzle
QUESTION [11 upvotes]: This is a puzzle a colleague asked me recently.
Imagine you have $n$ balls in a bag that are colored from $1$ to $n$.  At each turn you take two balls at random out that have different colors and color one the color of the other. You then put them both back in the bag.  What is the expected number of turns before all the balls have the same color?
The pair of balls you choose is uniformly selected from the set of all pairs of different-colored balls.  You choose uniformly at random whether to paint the first the same as the second or vice versa.

REPLY [2 votes]: Let $B_i$ denote the number of balls of color $i$ ($i=1,\ldots,n$) after some turns, and define a quantity
$$
S=\sum_i B_i^2.
$$
Suppose that on the next turn, we draw two balls of colors $j\neq k$, and choose one at random to receive the color of the other. The expected value of $S$ after this turn is
$$
\frac{1}{2}\left(\sum_{i\neq j,k}B_i^2+(B_j+1)^2+(B_k-1)^2\right)+\frac{1}{2}\left(\sum_{i\neq j,k}B_i^2+(B_j-1)^2+(B_k+1)^2\right)=\sum_i B_i^2 +2.
$$
The expected value of $S$ increases by $2$ each turn. The initial value of $S$ is $n$, and when all balls are the same color the value of $S$ is $n^2$, so the expected number of turns is
$$
\frac{n^2-n}{2}.
$$<|endoftext|>
TITLE: Continuous dependence  of the expectation of a r.v. on the probability measure
QUESTION [7 upvotes]: $\newcommand{\bsV}{\boldsymbol{V}}$ $\newcommand{\bsE}{\boldsymbol{E}}$ $\newcommand{\bR}{\mathbb{R}}$ Suppose that $\bsV$ is an $N$-dimensional    real Euclidean space. Denote by $\newcommand{\eA}{\mathscr{A}}$ $\eA$ the space of symmetric positive semidefinite operators $A:\bsV\to \bsV$. To each  $A\in \eA$   we  can associate in a canonical fashion  a  centered Gaussian measure $\gamma_A$ on $\bsV$ which is concentrated on $(\ker A)^\perp$. For example, if $A$ is nondegenerate, then
$$ \gamma_A(dv)= \frac{1}{\sqrt{\det 2\pi A}} e^{-\frac{1}{2} (A^{-1}v,v)}dv, $$
while if $A=0$, then $\gamma_0$ is the Dirac measure concentrated at the origin. 
Fix a  locally Lipschitz   function $f:\bsV\to\bR$ which is positively homogeneous of degree $\alpha \geq 2$. For any $A\in \eA$    we denote by $\bsE_A(f)$ the expectation of $f$ with respect to the probability measure $\gamma_A$ on $\bsV$.   Consider the function
$$ \eA\ni A\mapsto \bsE_A(f)\in \bR. $$ 
This function is continuous  and positively homogeneous of degree $\frac{\alpha}{2}$, i.e.,
$$ \bsE_{tA}(f)=t^{\frac{\alpha}{2}} \bsE_A(f),\;\;\forall t>0,\;\;A\in\eA. $$
I am interested in  its  modulus of uniform continuity on the ball
$$\eA_1:=\bigl\lbrace A\in\eA;\;\;\Vert A\Vert\leq 1\bigr\rbrace, $$
I was able to prove that on this ball the above function is Holder continuous, with Holder exponent $\frac{1}{2N+3}$.  This suffices  for the applications I  have in mind, but I strongly suspect that it is far from optimal.  I believe that  the Holder exponent $\frac{1}{2}$  is   uniformly optimal in the following sense: there exist $C, r>0$ so that  for any $A, B\in\eA_1$ satisfying
$$\Vert A- B\Vert \leq r, $$
we have
$$\bigl\vert \bsE_A(f)-\bsE_B(f)\bigr\vert\leq C\Vert A-B\Vert^{\frac{1}{2}}. \tag{1} $$
Remark. To see that  the exponent $\frac{1}{2}$ is the best one can hope for consider the case $\bsV=\bR^2$, $f(x,y)=|xy|$ and  $\newcommand{\ve}{{\varepsilon}}$ and the Gaussian measures
$$ \gamma_{A_\ve}=\frac{1}{2\pi\ve} e^{-\frac{1}{2\ve^2}x^2-\frac{1}{2}y^2} |dxdy| $$
Then $\Vert A_\ve-A_0\Vert =\ve^2$,
$$\bsE_{A_0}(f)=0,\;\; \bsE_{A_\ve}(f)=\left(\int_{\bR}|x|e^{-\frac{1}{2}x^2} |dx|\right)^2 \ve. $$
My question is the following:  have you encountered  continuity results of this sort, and if so, can you  indicate some references that deal with this? Thanks!

REPLY [6 votes]: This problem reduces quickly to Holder continuity of the operator square root. That is, there exists a $C > 0$ such that
$$
\begin{align}
\lVert\sqrt{A}-\sqrt{B}\rVert\le C\lVert A-B\rVert^{1/2}&&{\rm(1)}
\end{align}
$$
for any positive semidefinite operators $A,B$.[1] 
Assuming (1), the proof of continuity with Holder exponent $1/2$ as mentioned in the question is quite direct. If $X$ is an $\mathbb{R}^N$-valued standard normal random variable, then $\sqrt{A}X$ has distribution $\gamma_A$. So, if $f$ has Lipschitz constant $K$ on the unit ball then, writing $\hat X=X/\lVert X\rVert$,
$$
\begin{align}
\left\lvert\mathbb{E}_A(f)-\mathbb{E}_B(f)\right\rvert
&=\left\lvert\mathbb{E}[f(\sqrt{A}X)-f(\sqrt{B}X)]\right\rvert\cr
&=\left\lvert\mathbb{E}[\lVert X\rVert^\alpha(f(\sqrt{A}\hat{X})-f(\sqrt{B}\hat{X}))]\right\rvert\cr
&\le K\mathbb{E}[\lVert X\rVert^\alpha]\lVert\sqrt{A}-\sqrt{B}\rVert\cr
&\le CK\mathbb{E}[\lVert X\rVert^\alpha]\lVert A-B\rVert^{1/2},
\end{align}
$$
as required.

[1] Holder continuity of the square root looks quite obvious, so I would expect it to be standard. I don't have a reference for this though, but I can give a proof now using the Taylor expansion of the square root (maybe there is a quicker proof). It is tempting to suggest that it holds with $C=1$ as in the 1-dimensional case, but this is possibly rather too strong when $A$ and $B$ do not commute.
Multiplying through by a scalar, we can assume that $\lVert A\rVert$ and $\lVert B\rVert$ are bounded by $1$. Then, write $A=1-X$, $B=1-Y$ for positive semidefinite operators $X,Y$ with $\lVert X\rVert,\lVert Y\rVert\le1$. By Taylor expansion,
$$
\sqrt{1-X}=1-\sum_{n=1}^\infty a_n X^n
$$
where
$$
\begin{align}
a_n&=\frac12\prod_{k=2}^n\left(1-\frac3{2k}\right)
\le\frac12\prod_{k=2}^n\exp\left(-\frac3{2k}\right)\cr
&\le\exp\left(-\frac32\log n\right)
=n^{-3/2}.
\end{align}
$$
Then,
$$
\sqrt{A}-\sqrt{B}=\sum_{n=1}^\infty a_n(Y^n-X^n).
$$
As $X,Y$ have norm bounded by 1, the term $Y^n-X^n$ has norm bounded by $n\lVert Y-X\rVert=n\lVert A-B\rVert$. As $X,Y$ are also positive semidefinite, $Y^n-X^n$ is bounded by 1 in norm. So,
$$
\begin{align}
\lVert\sqrt{A}-\sqrt{B}\rVert&\le\sum_{n=1}^\infty n^{-3/2}\min\left(n\lVert A-B\rVert,1\right)\cr
&=\lVert A-B\rVert\sum_{n\le\lVert A-B\rVert^{-1}}n^{-1/2}+\sum_{n > \lVert A-B\rVert^{-1}}n^{-3/2}.
\end{align}
$$
As $\sum_{n\le x}n^{-1/2}$ and $\sum_{n > x}n^{-3/2}$ are bounded by fixed multiples of $\sqrt x$ and $1/\sqrt x$ respectively (for $x\ge1$), both terms on the right hand side of the inequality above are bounded by a multiple of $\lVert A-B\rVert^{1/2}$. This gives (1) as required. Note that the constant $C$ does not depend on the dimension $N$, and (1) also holds in infinite dimensions.<|endoftext|>
TITLE: Another colored balls puzzle (part II)
QUESTION [8 upvotes]: The same colleague as in Another colored balls puzzle asked me the following variant which she called "part II".
Imagine you have $n$ balls in a bag that are colored from $1$ to $n$.  At each turn you take one ball uniformly at random from the bag and then a second uniformly at random from the remaining balls which have a different color to the first. You then color one ball the color of the other and put them both back in the bag. 
What is the expected number of turns before all the balls have the same color if:

You always paint the first ball the color of the second? Or...
You always paint the second ball the color of the first?

My intuition is that in the first case you are likely to be decreasing the number of balls with a  commonly occurring color  and in the second you are likely to be increasing the number of balls with a commonly occurring color. Hence the time to get all colors the same will be much less in the second case than the first.
Simulations.
For rule 1 I get the following approximate expected values for $n = 2 \dots 6$.
$1$, $4$, $10.33$, $22.48$, $45.23$.
For rule 2 I get the following approximate expected values for $n = 2 \dots 6$.
$1$, $2.500$, $4.416$, $6.698$. $9.310$.
To get some feeling for how the problem scales I also tried some other values up to $n=100$. For rule 1 and $n=100$ my simulation didn't get to all balls being equal even once in the time I gave it. However for $n=10$ I get about $618$ and for $n=20$ I get very roughly $520,000$.    For rule 2 I get a mean of roughly $865$ for $n=100$.  My blind guess is now that the mean for rule is 1 is at least exponential (in fact it looks a little like $2^{n-1}$) but that for rule two it is not quite  $\frac{e}{4} n^{1.55}$ but it may be something similar.

REPLY [2 votes]: For given $n$ it is possible to calculate the expected number of turns 
from an absorbing Markov chain on the integer partitions of $n$.
Each state of the Markov process corresponds to the ordered list of the number of equally colored balls.
The matrix of transition probabilities between the states is easily calculated according to rules 1 or 2, 
e.g.
Removing from position 1, adding to position 2 :  {3,1,1} -> {2,2,1}
    with probability  p = P(choosing position 1 out of all balls) * P(choosing position 2 out of remaining balls) 
                        =3/5 * 1/2 
Given the transition matrix with the unique absorbing state {n} and the initial state {1,1,1,...,1} it is possible to calculate 
the expected number of turns before absorption.
This is done in the  Mathematica program below. Here some results where the first number is n, the second the numerical result, and the third the exact fraction:
Rule 1
{{2, 1., 1}, 

{3, 4., 4}, 

{4, 10.3333, 31/3}, 

{5, 22.4852, 16729/744}, 

{6, 45.2173, 33913/750},

{7, 87.7733, 26707139046097/304274018880}, 

{8, 168.252, 129857255359868261/771803525388385},

{9, 322.292, 4555917617310039296830835441635/14135986803865219963776139264}, 

{10,620.346,822838635777324535445878391148603051494611/1326419190860455039655669536523314862820}}

Only the numerical values for n>10:
{{11, 1202.04}, {12, 2344.58}, {13, 4599.07}, {14, 9062.01}, 
{15, 17916.8}, {16, 35513.4},{17, 70522.1}, {18, 140231.},{19, 279122.}, {20, 555989.}}
Rule 2
{{2, 1., 1}, 

{3, 2.5, 5/2}, 

{4, 4.41667, 53/12}, 

{5, 6.6994, 2251/336}, 

{6, 9.31157, 866023/93005}, 

{7, 12.2249, 1009285097/82560060},

{8, 15.4166, 2246993235815929/145751872750176}, 

{9, 18.868, 2285085765293281062003190373/121108796080566797904702840}, 

{10, 22.5637, 618224636000595187350171250435705332105433100763641/27399140168645771065204844597274355963355735154297}}

Only the numerical values for n>10:
{{11, 26.4901}, {12, 30.6358}, {13, 34.9907}, {14, 39.546}, {15, 
  44.2936}, {16, 49.2266}, {17, 54.3383}, {18, 59.6231}, {19, 
  65.0754}, {20, 70.6904}}
Program:
(*
Defining Rule 1:
  input : lst = Partition of n , e.g. {5,3,2,1} with n=11
        k, m :  Positions to be changed ; remove ball from position k and add to position m
 output : list containing changed partition and probability for this change

 E.g. PartMove[{5, 3, 2, 1}, 1, 3] -> {{4, 3, 3, 1}, 5/33}
*)
PartMove[lst_, k_, m_] :=
 Module[{ll = lst, len = Plus @@ lst, prob},
  prob = (ll[[k]]/len*ll[[m]]/(len - ll[[k]]));
  ll[[k]] -= 1; ll[[m]] += 1;
  If[ll[[k]] == 0, ll = Drop[ll, {k}]]; {Reverse[Sort[ll]], prob}]

(*Alternatively
Defining Rule 2:
  input : lst = Partition of n , e.g. {5,3,2,1} with n=11
        k, m :  Positions to be changed ; add ball to position k and remove from  position m
 output : list containing changed partition and probability for this change

 E.g. PartMove[{5, 3, 2, 1}, 1, 3] ->   {{6, 3, 1, 1}, 5/33}
*)
 PartMove[lst_, k_, m_] :=
 Module[{ll = lst, len = Plus @@ lst, prob},
  prob = (ll[[k]]/len*ll[[m]]/(len - ll[[k]]));
  ll[[k]] += 1; ll[[m]] -= 1;
  If[ll[[m]] == 0, ll = Drop[ll, {m}]]; {Reverse[Sort[ll]], prob}]

  (*
Calculate all possible target partitions for a given partition with respective probabilities:
  input : lst = Partition of n , e.g. {5,3,2,1} with n=11

  output : list containing all possible changed partitions with their  probabilities
 Note: Depends on the chosen rule via PartMove

 E.g. for rule 1:
 Targets[{2, 1, 1}] -> {{{2, 2}, 1/6}, {{3, 1}, 1/3}, {{2, 1, 1}, 1/2}}
*)
  Targets[lst_] := Module[{len = Length[lst], pairs, res},
  If[len <= 1, {{lst, 1}},
   pairs = Select[Tuples[Range[len], 2], #[[1]] != #[[2]] &];
   res = Sort[PartMove[lst, #[[1]], #[[2]]] & /@ pairs];
    {#[[1, 1]], Plus @@ (#[[2]] & /@ #)} & /@ SplitBy[res, First]]]

(*  Define all possible states for chosen n (here n=4) *)

states = IntegerPartitions[4];

(* Define transition matrix PM for Markov chain *)
nn = Length[states];
Clear[PartIndex]; n = 1; (PartIndex[#] = n++) & /@ states;
PM = Table[0, {nn}, {nn}];
Do[(PM[[PartIndex[#[[1]]], k]] = #[[2]]) & /@
  Targets[states[[k]]], {k, 1, nn}]

(* Define submatrix Q for transient state changes *)
Q = (Drop[#, 1] & /@ Drop[PM, 1]);
(* Calculate fundamental matrix NPM of absorbing Markov chain*)
NPM = Inverse[ IdentityMatrix[Length[Q]] - Q];
(*Calculate expected number t of turns when starting from {1,1,..,1} = states[[-1]] *)
t = Plus @@ (#[[-1]] & /@ NPM);
{N[t], t} (*numerical and exact value of t*)<|endoftext|>
TITLE: Absolutely 2-summable operator on a Hilbert space
QUESTION [6 upvotes]: An bouneded linear operator $A \in L(X, Y)$ (here $X$, $Y$ are Banach spaces) is called absolutely $2$-summable if there exists a $C>0$ such that
$$ \left( \sum_{j=1}^N \| A x_j\|_X^2 \right)^{1/2} \leq C \cdot \sup \left\{ \left(\sum | \langle x_j, \omega \rangle | \right)^{1/2} \mid \omega \in X', \|\omega\|_{X'} = 1 \right\}$$
for any finite set of vectors $x_1, \dots x_N \in X$. The $2$-summable norm is the infimum of all such constants $C$.
Those operators are the natural analog to Hilbert-Schmidt-Operators: If $X$ and $Y$ are Hilbert spaces, than $A$ is absolutely $2$-summable if and only if $A$ is Hilbert-Schmidt, and the norms coincide.
Now the question is: If $X$ is a Hilbert space (but $Y$ isn't), can we restrict to elements $x_j$ of some orthonormal basis? That is, if we allow only elements of a given ONB to take for $x_1, \dots, x_N$, can the resulting $C$ be strictly smaller than the $2$-summable norm?

REPLY [3 votes]: I answered the OP's question in a comment, but the right question is whether you can estimate the $2$-summing norm of an operator $T$ from a Hilbert space $H$ by taking   $\sup (\sum \|Te_n\|^2)^{1/2}$, where the sup is over all ON bases $(e_n)$ for $H$.  In fact, if $H$ is infinite dimensional, this sup is equal to the $2$-summing norm (combine a dilation argument with the fact that $T$ is essentially zero on an infinite dimensional subspace).  More interesting is that the the sup is at least $2^{-1/2}$ times the $2$-summing norm of $T$. This is the gist of Tomczak's lemma (see Theorem 18.4 in her book "Banach-Mazur distances and finite-dimensional operator ideals").<|endoftext|>
TITLE: On the cardinality of perfect spaces with the countable chain condition
QUESTION [8 upvotes]: QUESTION: Does every regular perfect space with the countable chain condition have cardinality bounded above by the continuum? Is this at least true for perfectly normal ccc spaces?

Recall that a space is perfect if every closed set is a $G_\delta$ set (that is, a countable intersection of open sets). We say that a space has the countable chain condition if every family of pairwise disjoint non-empty open sets is countable. Perfectly normal just means perfect and normal.

If one strengthens the ccc to the condition every discrete set is countable then the answer is yes. This follows from Hajnal and Juhasz's result that every space where singletons are $G_\delta$ and discrete sets are countable has cardinality at most continuum. Regularity is not needed for this to be true ($T_1$ is enough).

(It's easy to prove that in a perfect space where closed discrete sets are countable, it's even true that every discrete set is countable. So from the above statement it follows that every Lindelof $T_1$ perfect space has cardinality at most the continuum, something that was proved by Alexandroff and Urysohn in their Memoire, at least for compact Hausdorff spaces, if I'm not mistaken).
On the other hand:

There are Hausdorff ccc perfect spaces of arbitrarily large cardinality.
For example, let $\kappa$ be any cardinal and $D_n=\{x \in 2^\kappa: |x^{-1}(1)|=n \}$. Set $X=\bigcup_{n<\omega} D_n$. Let $\tau$ be the refinement of the usual topology on $X$ obtained by making every $D_n$ closed discrete. In other words, a basic open set has the form $U \setminus \bigcup_{n\in F} D_n$ where $U$ is open in the topology on $X$ inherited from $2^\kappa$ and $F \subset \omega$ is finite. This space is ccc: indeed, let $\{U_\alpha: \alpha<\aleph_1\}$ be a pairwise disjoint family of open sets of cardinality $\aleph_1$. By the pigeonhole principle we can assume that for some fixed finite set $F$ we have $U_\alpha=V_\alpha \setminus \bigcup_{n \in F} D_n$, where $V_\alpha$ is open in the usual topology on $X$, for every $\alpha<\aleph_1$. So for all $\alpha, \beta <\aleph_1$ such that $\alpha \neq \beta$ we have that $V_\alpha \cap V_\beta \subset \bigcup_{n \in F} D_n$ which implies that $V_\alpha \cap V_\beta$ is empty, as $\bigcup_{n\in F} D_n$ is nowhere dense. But this is a contradiction since $X$ with the topology inherited from $2^\kappa$ is dense in the ccc space $2^\kappa$ and thus it is also ccc.
The space $(X, \tau)$ is more than perfect. As a matter of fact, let $G \subset X$ be any set and set $G_n=D_n \setminus G \cap D_n$. Note that $G_n$ is closed. Then $G=\bigcap_{n<\omega} (X \setminus G_n)$, which proves that $G$ is a $G_\delta$ set.

REPLY [5 votes]: The reason why my Hausdorff non-regular counterexample above is perfect is that it is a countable union of closed discrete sets. Now, Uspenskij constructed $\sigma$-closed discrete ccc regular spaces of arbitrarily large cardinality. So that answers my question in the negative. 
See this paper: http://dml.cz/dmlcz/106296<|endoftext|>
TITLE: Is every connected metrizable locally path connected space a length space?
QUESTION [15 upvotes]: Does every connected metrizable locally path connected topological space $X$ admit a compatible metric $d$ so that $(X,d)$ is a length space?
(Edit to correct definition: Recall that a metric space $(X,d)$ is a length space if for every $x$ and $y$ in $X$ and every $e>0$, there exists a rectifiable path from $x$ to $y$ whose length is less then $d(x,y)+e$.)
The answer is certainly yes for Peano continua, but this is not a trivial fact.
More generally the answer is apparently yes for such locally compact spaces, but local compactness is certainly not necessary: for example, familiar Hilbert space is a length space.
Do the above claims survive without local compactness?

REPLY [2 votes]: I think the answer is no.  There is a metrizable version of the Tangent Disc Topology, namely where instead of extending the Euclidean topology in the upper half-plane to all of $\mathbb{R}$ you extend it to a countable subset of $\mathbb{R}$.  This will be locally path-connected and connected (see Counterexamples in Topology by Steen/Seebach), but not locally compact, and as far as I can tell there will be no equivalent metric which is a length space.
To see this, pick just a single point $x \in \mathbb{R}$ and adjoin to the Euclidean topology on $\mathbb{H}^2$ neighborhoods of the form $D \cup \lbrace x \rbrace$, where $D$ is any open disc in $\mathbb{H}^2$ tangent to $x$, creating a topological space $(\mathbb{H}^2 \cup \lbrace x \rbrace, \tau)$.  Then if $C$ is the boundary of one of these tangent discs, points along $C$ do not converge to $x$ in $\tau$.  However, if $V$ is a vertical segment with endpoint $x$, then points along $V$ do converge to $x$, and clearly $V$ is a length-minimizing path.
If $c_n, v_n$ are points with the same $y$-coordinate along $C$ and $V$ respectively (assume all the $c_n$ are either to the left, or to the right, of $V$) with $v_n \rightarrow x$, then $|c_n - v_n| \rightarrow 0$.  Notice that the lengths of paths from $c_n$ to $x$ converges to zero (by triangle inequality and moving horizontally, then vertically), yet each $c_n$ lies outside the neighborhood $D \cup \lbrace x \rbrace$, impossible in a length space.
This last bit is the 'topological agreement' criterion, I'm not sure what it's normally called; see p. 27, no. (4) in A Course in Metric Geometry by Burago/Burago/Ivanov:

If $x \in X$ and $U$ is a nbhd of $x$, then $\inf \ell(x, c) > 0$ for
$c \in U^c$.

Especially there will be no equivalent metric that's a
length space, since an intrinsic length metric is topologically unique
and will have the same defect.
Hopefully this argument is right; even if there's a small error, I wanted to bump this question.  I'm not an expert in length spaces, perhaps replacing $x$ with $\mathbb{Q}$ is necessary, but those details seemed quite complicated.<|endoftext|>
TITLE: On the large cardinals foundations of categories
QUESTION [10 upvotes]: (This question was posted on math.SE over two weeks ago, but received no answer. I am therefore posting it here as well.)
It is well-known that there are difficulties in developing basic category theory within the confines of $\sf ZFC$. One can overcome these problems when talking about small categories, and perhaps at one or two levels when talking about larger categories (e.g. if the objects and the morphisms are all definable uniformly, I suppose it is possible to do prove some basic things).
But often we want to talk about larger and larger categories, and for that we need the ability to deal with higher and higher level of classes.
The easiest route to solve this is to use the Tarski-Grothendiek set theory, but that is equiconsistent with the existence of a proper class of inaccessible cardinals. While not a mind-boggling assumption, it is still quite a strong one. Even if people are only interested in one or two levels, often they just assume that some suitable number of inaccessible cardinals exist.
But I kept asking myself, what is wrong with just assuming that you have an $\omega$, or $\omega+1$ chain of models of $\sf ZFC$?
That is, a chain $\langle M_n\mid n\in\omega\rangle$ such that $(M_n,\in)\models\sf ZFC$ and $M_n\in M_{n+1}$. Perhaps we want another $M$ which contains all the models and the sequence as well. We can even assume they are countable if we really want to. This is a much weaker assumption in terms of additional axioms, and should be roughly equivalent to $\rm Con^\omega(\sf ZFC+St)$, where $\sf St$ is the axiom asserting a standard model exists.
So why are people jumping to large cardinals? I am certain that they are needed for some construction, but if we just want to talk about categories of categories and so on and so forth, why isn't the above assumption sufficient? Is it just because large cardinals, or rather universes, are easier to explain to the working mathematician? Or is there something we really can't do with this sort of chain of models and we can do with inaccessible cardinals?

REPLY [11 votes]: Allow me to make some comments as someone who converted to the universeful approach recently; but take it with a pinch of salt, as I have only been studying category theory for 2½ years.
I should briefly mention the trigger that led me to the pro-universe camp: about 6 months ago, I started learning about quasicategories and became convinced that the theory relied on far too much machinery to admit a workable elementary approach that would allow us to consider entities like the quasicategory of all Kan complexes within an NBG-like framework – which is the the raison d'être for quasicategories in the first place! So I decided that I would have to start taking universes seriously – and for this purpose the universes would not merely be for bookkeeping but for doing actual mathematics in. 
If all we wanted to do was to ensure that sets of large cardinality and high complexity exist, it would probably be enough to just have a chain $M_0 \in M_1 \in M_2 \in \cdots$ of (transitive) models of set theory (say, ZFC), but I find this very unsatisfactory. For instance, consider theorems that assert that some object with some universal property exist: there is no guarantee that a universal object in $M_0$ remains universal in $M_1$. Indeed, one such theorem says that, for every set $X$, there exists a set $\mathscr{P}(X)$ and a binary relation $[\in]_X \subseteq X \times \mathscr{P}(X)$ such that, for every binary relation $R \subseteq X \times Y$, there is a unique map $r : Y \to \mathscr{P}(X)$ such that $\langle x, y \rangle \in R$ if and only if $\langle x, r(x) \rangle \in [\in]_X$. Of course, it is well-known that powersets need not be preserved when passing from one model of set theory to another. But if something as trivial as powersets is not preserved, what hope is there of preserving more complicated universal objects like the free ind-completion of a category, or even the free monoid on one generator (i.e. $\mathbb{N}$!)? For a category theorist, to work in a setting where universal objects have to be qualified by a universe parameter is simply untenable. 
So, we have to find some kind of compromise: we need a chain of universes such that each universe is embedded in the next in as pleasant a way as possible, so that the mathematics in one universe agrees with the next as much as is feasible. The most ideal situation one could hope for goes something like this:

Let $\mathcal{C}$ be a category scheme, i.e. a definable function that assigns to each sufficiently nice universe $\mathbf{U}$ a category $\mathcal{C}(\mathbf{U})$ such that, for any universe $\mathbf{U}^+$ with a sufficiently nice embedding $\mathbf{U} \subseteq \mathbf{U}^+$, $\mathcal{C}(\mathbf{U})$ is a subcategory (in the strict sense) of $\mathcal{C}(\mathbf{U}^+)$.
We say that an inclusion $\mathbf{U} \subseteq \mathbf{U}^+$ is adequate for a category scheme $\mathcal{C}$ if the following conditions are satisfied:

$\mathcal{C}(\mathbf{U})$ is a full subcategory of $\mathcal{C}(\mathbf{U}^+)$: we do not get any new morphisms between objects in $\mathcal{C}(\mathbf{U})$ when passing to a larger universe.
The inclusion $\mathcal{C}(\mathbf{U}) \hookrightarrow \mathcal{C}(\mathbf{U}^+)$ preserves all limits and colimits that exist in $\mathcal{C}$ for $\mathbf{U}$-small diagrams: the most elementary kind of universal constructions are preserved when passing to a larger universe.
Moreover $\mathcal{C}(\mathbf{U})$ is closed in $\mathcal{C}(\mathbf{U}^+)$ under all limits and colimits for $\mathbf{U}$-small diagrams: so passing to a larger universe does not create new universal objects where none existed before.


So, when is $\mathbf{U} \subseteq \mathbf{U}^+$ adequate for the category scheme $\mathbf{Set}$, if $\mathbf{U} \in \mathbf{U}^+$? We take it as given that $\mathbf{U}$ and $\mathbf{U}^+$ are transitive models of ZFC. The first condition implies that this extension has no new subsets, and if there are no new subsets, there are no new functions either – so (1) holds if and only if the extension preserves powersets. By considering explicit constructions, one sees that equalisers are preserved, and as soon as we know (1), products are also preserved. We may then use the monadicity of $\mathscr{P} : \mathbf{Set}^\mathrm{op} \to \mathbf{Set}$ to deduce that colimits are preserved, so (1) implies (2). In particular, directed unions are preserved, so it follows that $\mathbf{U}$ embeds as an initial segment of the cumulative hierarchy of $\mathbf{U}^+$. What about (3)? Well, take a $\mathbf{U}$-set $I$ and a map $X : I \to \mathbf{U}$ in $\mathbf{U}^+$. By replacement in $\mathbf{U}^+$, we can form the disjoint union $\coprod_{i \in I} X(i)$, and if (3) holds, the cardinal of this set is in $\mathbf{U}$. Thus, $\mathbf{U}$ must actually be embedded as a Grothendieck universe in $\mathbf{U}^+$. Conversely, if $\mathbf{U}$ is embedded as a Grothendieck universe in $\mathbf{U}^+$, then (1), (2), and (3) are easily verified. 
But are Grothendieck universes enough? For instance, it would be good if the following were true:

Let $\kappa$ be the smallest (uncountable strongly) inaccessible cardinal, let $\mathbb{B}$ be a $\mathbf{V}_{\kappa}$-small category, and let $\mathcal{M}$ be the category scheme obtained by defining $\mathcal{M}(\mathbf{U})$ to be the free ind-completion of $\mathbb{B}$ relative to $\mathbf{U}$ for each Grothendieck universe $\mathbf{U}$. Suppose $\mathcal{M}(\mathbf{V}_{\kappa})$ admits a cofibrantly generated model structure. Then, for all Grothendieck universes $\mathbf{U}$, there is a (unique) cofibrantly generated model structure on $\mathcal{M} (\mathbf{U})$ extending the one on $\mathcal{M}(\mathbf{V}_{\kappa})$.

However, I do not know if this is true. What I do know is that embeddings of Grothendieck universes are adequate for $\mathcal{M}$; in fact, locally presentable categories and adjunctions between them are very well behaved under this kind of universe enlargement.<|endoftext|>
TITLE: Proof of the weak Goldbach Conjecture
QUESTION [23 upvotes]: What are the main ideas of Harald Helfgott's proof that all odd $n \geq 5$ is the sum of 3 primes?

REPLY [21 votes]: I think this blog post of Terry Tao, as well as the comments following it (including some from Helfgott) answer this question as completely as one could reasonably hope.
https://terrytao.wordpress.com/2012/05/20/heuristic-limitations-of-the-circle-method/<|endoftext|>
TITLE: When do two positive braids represent the same link?
QUESTION [10 upvotes]: Let $B_n$ be the braid group on $n$ strands, with the usual generators: $s_1, \ldots, s_{n-1}$ and their inverses, where $s_i$ is a positive half-twist interchanging the strands labelled $i$ and $i+1$.  Markov's theorem says that two elements of the braid group have closures giving the same knot in $S^3$ when they are related by a sequence of moves, each of which is either a conjugation $\beta \to s_i^{\pm} \beta s_i^{\mp}$ or the inclusion $B_n \to B_{n+1}$ taking $\beta \to \beta s_n^{\pm}$.
Recall a braid is positive if it can be written as $s_{i_1} \cdots s_{i_k}$.  Conjugation does not generally preserve positive braids, but it can: $s_1^{-1} (s_1 \beta) s_1$ is of course positive when $\beta$ is. 

If two positive braids represent the same knot, can they be related by a sequence of Markov moves in which only positive braids appear?

REPLY [4 votes]: The answer is "No". Consider two braids
$$
\beta_1 = s_1^3 s_2^3 s_3,\qquad \beta_2 = s_1 s_2^3 s_3^3
$$
They obviously represent the same knot. Moreover, they are conjugate. But
Claim 1: $\beta_2$ cannot be transformed to $\beta_1$ by successive conjugations by $s_i^{\pm 1}$ preserving positivity. I.e. you need to conjugate by longer braids. This I checked on a computer. Since there are only finitely many positive braids of given length this is a finite calculation.
Claim 2: $\beta_2$ cannot be transformed to $\beta_1$ by a sequence of Markov moves preserving positivity. This is conjectural. I checked it on a computer by going through all possible stabilizations with up to 10 strands.
On the other hand, if you allow conjugations by arbitrary short braids (=lifts of permutations) instead of only conjugations by Artin generators, the statement is true by the above cited paper "Fibered Transverse Knots and the Bennequin Bound" by John B. Etnyre, Jeremy Van Horn-Morris (https://arxiv.org/abs/0803.0758) and Lemma 6.7 from Kassel, Turaev - "Braid Groups" (thanks to Vera Vertesi for the latter reference).<|endoftext|>
TITLE: A generalisation of the theorem of Maschke
QUESTION [9 upvotes]: The theorem of Maschke tells us that every representation of a finite group is the direct sum of irreducible representation. More precisely:
Let $G$ be a finite group, $K$ a field whose characteristic does not divide the order of $G$, $V$ a $K$-vector space of finite dimension and $\rho:G\to {\rm GL}(V)$ a representation of $G$. If $W$ is a $G$-invariant vector space of $V$, then there exists a $G$-invariant complement $W'$ such that $V=W\oplus W'$. 
Now, I would like to prove or to find a counter-example to the following generalisation:
Let $G$ be a group  such that every element has finite order, $K$ a field of characteristic $0$, $V$ a $K$-vector space of finite dimension and $\rho:G\to {\rm GL}(V)$ a representation of $G$. If $W$ is a $G$-invariant vector space of $V$, then there exists a $G$-invariant complement $W'$ such that $V=W\oplus W'$.

REPLY [4 votes]: I. Schur explicitly proved that every periodic subgroup of ${\rm GL}(n,\mathbb{C})$ is completely reducible ( Corollary 36.3 in Curtis and Reiner's "Representation Theory of Finite Groups and Associative Algebras", Wiley and Sons, 1962), improving earlier results of Burnside,<|endoftext|>
TITLE: Matrix Inverse with Same Principal Minors
QUESTION [6 upvotes]: Given an invertible matrix $A \in \mathbb{R}^{n \times n}$, and index set $\langle n\rangle = \{ 1, \dots, n \}$, and the submatrix $A(\alpha)$ with the columns and rows of $A$ with indices $\alpha \subset \langle n \rangle$. 
Is there are characterisation of all matrices $A$ with
$\textrm{det } A(\alpha) = \textrm{det } A^{-1}(\alpha) \qquad \forall \alpha \subset \langle n \rangle  \qquad ?$

Example: Sufficient conditions are involutory ($A^{-1} = A$) or orthogonal ($A^{-1} = A^T$) matrices.

REPLY [2 votes]: Let $A=(a_{ij})$. Then we obtain a system of polynomial equations in the variables $a_{ij}$,
given by determinant conditions $\det A(\alpha)=\det A^{-1}(\alpha)$. If one determines a Groebner 
basis for such a system with a given $n$, we see that the dimension of the variety given by the 
polynomial equations is rather high.  We see that there are always many other solutions for $A$, which do not
satisfy $A=A^{-1}$ or $A^{-1}=A^T$. One reason is, that many determinants of the submatrices of size less
than $n$ can be chosen to be zero, for $A$ and $A^{-1}$.
I know this does not answer the question how to give a nice characterization for this variety other
than by determinants.
Therefore, let me just give a typical example for $n=3$, with $a,b\in \mathbb{R}$ and $b\neq 0$:
$$
A=\begin{pmatrix} 0 & 0 & -1/b \cr
a & 1 & 0 \cr
b & 0 & 0 \end{pmatrix}, \quad
A^{-1}=\begin{pmatrix} 0 & 0 & 1/b \cr
0 & 1 & -a/b \cr
-b & 0 & 0 \end{pmatrix}.
$$<|endoftext|>
TITLE: Reference for "lax monoidal functors" = "monoids under Day convolution" 
QUESTION [15 upvotes]: Suppose $A$ and $C$ two symmetric monoidal categories. Let's say that $A$ is small and $C$ is locally presentable, and let's assume also that the tensor product on $C$ preserves colimits separately in each variable. (What I'm about to say is true in greater generality, but let's just stick with these assumptions.)
Then the category $\mathrm{Fun}(A,C)$ of functors $A\to C$ can be endowed with the Day convolution product, which is a nice symmetric monoidal structure [LNM 137, pp. 1-38]. Here's the fact I'm interested in:
Proposition. There is a natural equivalence of categories between the category of lax (symmetric) monoidal functors and the category of (commutative) monoids in $\mathrm{Fun}(A,C)$ with respect to the Day convolution.
After a fair amount of searching, the earliest reference I've found for such a result is Mandell-May-Schwede-Shipley, Model categories of diagram spectra, Pr. 22.1.
But surely it goes much further back than that. In particular, I'd imagine that Brian Day himself would have known this in the 70s. It seems like the sort of thing that's actually a special case of something incredibly general that category theorists knew 40 years ago. So ...
Question. Where is the first place this result (or a generalization thereof) appears in the literature?

REPLY [11 votes]: This observation appears already in Day's thesis as Example 3.2.2. For some reason it is only stated for commutative monoids and symmetric (pro)monoidal functors and only as a correspondence of objects not an equivalence of categories, also no proof is given. (Day's thesis used to be available on Street's homepage but the link is dead now.)<|endoftext|>
TITLE: Fano plane drawings: embedding PG(2,2) into the real plane
QUESTION [16 upvotes]: By a drawing of the Fano plane I mean a system of seven simple curves and
seven points in the real plane such that

every point lies on exactly three curves, and every curve contains
exactly three points;
there is a unique curve through every pair of points, and every two
curves intersect in exactly one point;
the curves do not intersect except in the seven points under
consideration.

The familiar picture
 (source)
does not count as a drawing, since the last requirement is not satisfied:
there are two "illegal" intersections. In fact, this is easy to fix:
 (source)
However, this drawing is degenerate in the sense that two of the curves
just "touch" each other, without crossing, at some point. And here,
eventually, my question goes:

Is every drawing of the Fano plane degenerate?

(Although I can give a topological definition of degeneracy, it is a little technical and, may be, not the smartest possible one, so I prefer to suppress it here.)

REPLY [10 votes]: Here is a bit more symmetric version of the picture in the accepted answer:<|endoftext|>
TITLE: Possible ratios of Pythagorean fractions
QUESTION [7 upvotes]: A Pythagorean fraction is a number of the form $a/b$ or $b/a$ where $a$ and $b$ are the legs of a Pythagorean triple. Are there simple necessary and/or sufficient conditions for determining whether a rational number can be expressed as a ratio of two Pythagorean fractions? As an example, I would be interested to know if $4/9$ can be expressed as a ratio of two Pythagorean fractions. A brute force search of all primitive Pythagorean triples $(a,b,c)$ with $c < 6000$ found that $4/9$ could not be written as a ratio of two Pythagorean fractions when one of the fractions has a corresponding hypotenuse less than 6000.

REPLY [5 votes]: This is a sketch how to decide the question for $\frac{4}{9}$.
The question is if there are positive integers $a,b,d,e$ such that $\frac{4}{9}=\frac{a/b}{d/e}$ with $a^2+b^2$ and $d^2+e^2$ squares. Denoting $p:=\frac{9a}{b}=\frac{4d}{e}$, the question is if there is $p\in\mathbb{Q}^\times$ such that both $p^2+4^2$ and $p^2+9^2$ are rational squares. That is, the question is if the quadrics $x^2+4y^2=t^2$ and $x^2+9y^2=z^2$ over $\mathbb{Q}$ intersect in a point with $x,y\neq 0$. The intersection of the two quadrics is isomorphic to the elliptic curve 
$$ Y^2=8(X-1)(X+1)(9X-1) $$
according to a 1997 preprint by R.G.E Pinch: Square values of quadratic polynomials (which used to be here but is no longer available, unfortunately). If this elliptic curve has finitely many rational points (something that I cannot check at the moment for lack of time) then finding them explicitly will list all points $(x,y,t,z)$ lying on both quadrics, so the question if there is a point with $x,y\neq 0$ can be settled. Otherwise there surely will be a point with $x,y\neq 0$.
Added. It seems that François Brunault filled in the details (see his comments to this post), and $\frac{4}{9}$ is indeed not a ratio of two Pythagorean fractions.<|endoftext|>
TITLE: Lower bound on $L^2$ norm of mean curvature in general dimensions
QUESTION [7 upvotes]: Suppose $\Sigma\subset \mathbb{R}^{n+1}$ is a closed embedded hypersurface. We know that when $n=1$ 
$$
\int_{\Sigma} |H|^2 \geq \frac{4 \pi^2}{|\Sigma|}
$$
by Gauss-Bonnet and that this is saturated on the round circle -- here $|\Sigma|$ is the length of $\Sigma$ and $H$ the mean curvature.
Likewise, if $n=2$ we have
$$
\int_{\Sigma} |H|^2\geq 16\pi
$$
which is also saturated on the round sphere (of course as we now know this can be improved for positive genus surfaces).
I'm wondering to what extent one can get sharp lower bounds in dimensions $n+1>3$.  Specifically, something like
$$
\int_{\Sigma} |H|^2 \geq C_n |\Sigma|^{(n-2)/n}.
$$
Ideally, this bound would be sharp on the round sphere at least amongst convex competitors (I suspect otherwise it wouldn't be true).  
Any references would be appreciated.

REPLY [4 votes]: I have no idea about the general case but in the convex case the sphere is indeed optimal. Moreover the $L^1$ norm of $H$ attains its minimum at the sphere (among the convex surfaces with the same area). To deduce the result for the $L^2$ norm, just apply Cauchy-Schwarz,
Let $A$ be the convex body bounded by $\Sigma$ and $B$ the unit ball in $\mathbb R^{n+1}$. Then the area $|\Sigma|$ is proportional to $V_1(B,A)$ and the integral of the mean curvature is proportional to $V_2(B,A)$, where $V_k(B,A)$ is the mixed volume of $k$ copies of $B$ and $n+1-k$ copies of $A$. 
By the Alexandrov-Fenchel inequality, $\log V_k(A,B)$ is a concave function of $k$ ($k\in\{0,1,\dots,n+1\}$). This fact yields a lower bound for $V_2(B,A)$ in terms of $V_1(B,A)=C(n)|\Sigma|$ and $V_{n+1}(B,A)=V(B)$, a constant. Namely
$$
 C(n)\int_\Sigma H = V_2(B,A) \ge  V(B)^{1/n} V_1(B,A)^{(n-1)/n} = C_1(n)|\Sigma|^{(n-1)/n}
$$
If $A$ is a ball, the inequality turns to equality because so does the Alexandrov-Fenchel inequality.<|endoftext|>
TITLE: How to get 3-manifold, Knots from Number Fields
QUESTION [15 upvotes]: I'm reading a paper On the Torsion Jacquet-Langlands correspondence by Akshay Venkatesh and Frank Calegari.  
Truthfully speaking I have no idea what Jacquet-Landlands is.  I'm just trying to understand why there are knots in a paper on algebraic number theory and some of the players involved in that paper.
To keep matters simple, how do we pass between number fields and 3-manifolds and why is this beneficial?  Poking around this paper, I find a version of the congruence groups:
$$\Gamma_0(\mathfrak{n}) = \left\{ \left( \begin{array}{cc} a & b \\ c & d\end{array} \right) : \mathfrak{n}\big|\, c\right\} \subset PGL_2(\mathbf{O}_F)$$
where $\mathbf{O}_F$ is an order of a number field.  We get a 3-manifold by quotienting hyperbolic 3-space: $\mathbb{H}^3/\Gamma_0(\mathfrak{n})$.  Apparently, there's also another similar way to do it with quaternions.
They then proceed to look at look at some group-cohomology invariants of the group and then they use some spectral theory and the rest of paper mostly goes over my head. Well... we do get this:

However, along the way, we took many
  detours to explore related phenomena,
  some of which was inspired by the data
  computed for the first author by Nathan
  Dunfield. In view of the almost
  complete lack of rigorous
  understanding of torsion for
  (nonHermitian) locally symmetric
  spaces, we have included many of these
  results, even when what we can prove
  is rather modest.

The take-home message seems to be that we have constructed a large collection of infinite groups and actions on low-dimensional spaces of interest number theorists.  Arithmetic lattices look like they play an important role.
I guess I'm trying to understand better how this connection between knots and number fields works and how Calegari and Venkatesh are using it to get a handle of the many invariants (which I may save for later questions).

REPLY [26 votes]: I think that the connection between knots and number fields you may
be imagining from reading that paper is rather specious.
It's a theorem of Alan Reid that the only
arithmetic hyperbolic knot complement is the figure eight knot complement.
More generally, only finitely many Bianchi groups may admit covers
which are link complements; this follows from non-vanishing results
concerning
the interior cohomology of such manifolds (for large enough discriminant,
there are interior cohomology classes coming either from
quadratic base change if you are a number theorist or from totally geodesic surfaces fixed by complex conjugation if you are a topologist).
The Whitehead link does occur in the paper you mention, but only in the following context:
there are two specific arithmetic manifolds $W$ and $M$ discussed as examples which are obtained by Dehn surgery on the Whitehead link. By a theorem of Lickorish, all $3$-manifolds may be obtained by surgery on link complements, so this in itself has no particular relation to arithmetic. It is true that $W$ and
$M$ have fairly simple desciptions as Dehn fillings of a fairly simple link, and this
is somewhat related to arithmeticity --- in part because
the resulting manifolds have particularly
small volume.  In particular, the manifold $W$ coming from $(5,2)$, $(5,1)$ surgery
on the Whitehead link is the Weeks manifold,
which, by a theorem of Gabai, Meyerhoff, and Milley, is the
smallest volume orientable hyperbolic $3$-manifold. (In general, arithmetic manifolds
seem to be "over"-represented in the small volume hyperbolic manifolds.)
To summarize, the main topic of interest in the paper you mention is the study of arithmetic hyperbolic $3$-manifolds, with a particular interest in their integral cohomology
(and its surprising links to K-theory, Galois representations, and functoriality), but not really to the study of knots or links. If you want some background reading on these manifolds (written for topologists rather than number theorists), then Katie's suggestion (Machlachlan-Reid's book) is a good one.<|endoftext|>
TITLE: Did Smith correctly state the mass formula?
QUESTION [8 upvotes]: Did Smith correctly state the mass formula?
H.J.S. "normal form" Smith was the first, in 1867, to state the mass formula for integral quadratic forms in a genus of 4 or more variables. This was forgotten and the formula is usually attributed to Minkowski, who rediscovered it in 1885, and to Siegel, who corrected Minkowski in 1935. Conway and Sloane mention a number of erroneous sources after Siegel, but though they quote Smith worrying that he has made an error, they mention no errror, suggesting that he got it right. On the other hand, they say that in addition to the 1867 paper, Smith had an 1884 paper which won a prize from the French Academy jointly with Minkowski. So I expect that the judges compared the two results and would have noticed a discrepancy, suggesting that they were equally right or wrong.
I'm not sure what any of these papers actually state. The 1884 competition was about the specific case of the sum of five squares, so perhaps the entries imposed restrictions that saved their correctness and Minkowski's error was only in the 1885 extension? In particular, I think that he restricted to odd forms in 1884, but I forget my source for this. Conway and Sloane say that Smith's 1867 formula restricted to odd determinant.
More generally, what are good sources for the history of quadratic forms?

REPLY [3 votes]: Not entirely sure about Smith. As far as i know, Conway and Sloane's version is correct, I've used it, but they give no proof at all. This is one reason that Shimura got involved. He and his student Jonathan Hanke both published on proofs of the mass formula. Shimura wrote a book, I think that must be item 22 at http://www.ams.org/journals/bull/2006-43-03/S0273-0979-06-01107-4/
Let's see. All the difficulty lies in the 2-adic contribution. There have been attempts to make a canonical 2-adic representative for quadratic forms. See J. W. S. Cassels, Rational Quadratic Forms. On page 120 we read

We do not attempt to specify a unique
  canonical form [see Jones(1944), Pall
  (1945), or Watson (1976a)]: that is
  more a job for a parliamentary
  draftsman than for a mathematician. 

Note that C+S use Watson's version. 
There is a fair amount of stuff at http://zakuski.math.utsa.edu/~kap/forms.html and http://zakuski.math.utsa.edu/~kap/more_than_this.html which tends to the modern. 
Probably enough for now.<|endoftext|>
TITLE: A catalog of faithful representations of finite groups?
QUESTION [9 upvotes]: I want a reference that catalogs the smallest-dimensional faithful representation of every noteworthy finite group.  Specifically, I want representations on $\mathbb{R}^n$ and $\mathbb{C}^n$.  
Where can I find this reference?
UPDATE: Some have questioned my use of the word "noteworthy."  I really just want this information for as many groups as possible.  Also, the Atlas of Finite Group Representations might very well be an example of the desired reference, but which of the cataloged representations are actually faithful?  And are there other references available that give the information I want for other groups?

REPLY [2 votes]: This isn't quite what you want because of the irreducibility issue, but may be interesting nonetheless: https://people.maths.bris.ac.uk/~matyd/GroupNames/R.html<|endoftext|>
TITLE: Multivariate polynomial approximation of smooth functions
QUESTION [8 upvotes]: Let $f$ be a function defined on $[-1,1]^d$. Assume that all partial derivatives of $f$ up to order $r$ are continuous; and the $\infty$-norm of these partial derivatives are uniformly upper bounded by a constant. Let $p^*_n$ be the best degree $n$ approximation polynomial of $f$. That is,
$ p_n^* = \mathrm{argmin}_{p_n \in \mathcal{P}_n} \|f-p_n\|_{\infty},   $
where $\mathcal{P}_n$ is the set of all polynomials of degree n. I am interested in the rate of approximation.
For the single variable case, i.e., $d=1$, it is known that $\| f-p_n \|_{\infty} = \mathcal{O}(n^{-r})$. ($r$ is the order of smoothness defined above.)
My question is what is the rate of approximation when $d>1$.

REPLY [8 votes]: Short answer: The estimate is similar to that for functions in one variable.
Longer answer: The estimates of best approximation of a real-valued smooth function (by algebraic as well as by trigonometric polynomials) in terms of moduli of continuity of its derivatives are known as Jackson theorems. They were first proved for functions on  the unit interval by Dunham Jackson. For functions on the $n$-cube a result of this type was obtained by D. J. Newman and H. S. Shapiro [On approximation theory (Oberwolfach, 1963), pp. 208–219, Birkhäuser, Basel, 1964; MR0182828 (32 #310)]. I cannot give now an exact quotation, because I do not remember the result well and do not have an access to the paper. There were subsequent generalizations, many by M. Gansburg. One of the newer results is the following theorem from MR1920286 
Bagby, T.; Bos, L.; Levenberg, N.;
Multivariate simultaneous approximation. 
Constr. Approx. 18 (2002), no. 4, 569–577: 
If K is a connected, compact set in $\mathbb{R}^N$ such that every pair of distinct points $a,b$ of $K$ can be joined by a rectifiable arc in $K$ with length at most $s|a-b|$, where $s$ is a positive constant, and $f$ is a function of class $\mathcal{C}^r$ on an open neighborhood of $K$, then for each non-negative integer $n$, there exists a polynomial $p_n$ of degree at most $n$ defined on $\mathbb{R}^N$ such that for each multi-index $\alpha$ with $|\alpha|=\min\{r,n\}$,
$$\sup_K |D^{\alpha}(f-p_n)|\leq \frac{C}{n^{r-|\alpha|}}\sum_{|\gamma|\leq r}\sup_K|D^{\gamma}f|,$$
where $C$ is a positive constant depending only on $N$, $r$ and $K$.<|endoftext|>
TITLE: Examples of polynomial rings $A[x]$ with relatively large Krull dimension
QUESTION [14 upvotes]: If $A$ is a commutative ring we have the estimate 
$$
\dim (A)+1 \le \dim (A[x])\le 2\dim (A)+1
$$
for the Krull dimension, with $\dim (A)+1 = \dim (A[x])$ for Noetherian rings.
I am looking for nice examples of rings $A$ so that $A[x]$ has Krull dimension
$\dim (A)+2, \dim(A)+3,\ldots ,2\dim(A)+1$.

REPLY [8 votes]: Your question is essentially completely answered in the paper The dimension sequence of a commutative ring by Arnold and Gilmer (Amer. J. Math. 96 (1974), 385--408).
EDIT: The aforementioned paper by Arnold and Gilmer treats the case of an arbitrary finite number of variables. Since you are interested only in the case of one variable, the general but still rather concrete construction of rings with the desired properties given in Bourbaki's Algèbre commutative VIII.2 Exercice 7 should be sufficient.<|endoftext|>
TITLE: Prym varieties as Jacobian varieties
QUESTION [6 upvotes]: A generic abelian variety of dimension 2 or 3 is a jacobian of a curve.
Is there a canonical way to determine a curve whose jacobian is a prym variety of a unramified double cover of a curve of genus 3 or 4?

REPLY [12 votes]: Yes, assuming you mean a way to go from the double cover to the curve, (rather than how to go from the Prym variety to the curve, which is just a constructive Torelli argument).  This is based on the fact that curves of genus 3 and 4 are "trigonal", i.e. admit a degree 3 map to the projective line.  The corresponding "trigonal construction" is attributed to S. Recillas who apparently generalized work of Roth, and it may have antecedents in the work of Wirtinger.  An excellent reference is Beauville's paper: 
http://math.unice.fr/~beauvill/pubs/specprym.pdf
Basically, a double cover C-->D, where D is trigonal of genus g+1, induces a map of degree 8 on the corresponding third symmetric products of these curves.  By definition, that of D contains a projective line, whose preimage in the 3rd symmetric product of C, and modulo the induced involution, is a tetragonal curve E of genus g.  The Jacobian of E is isomorphic to the Prym variety of C-->D.
In fact the preimage curve in the symmetric cube of C, is a disjoint union of two copies of E, and both embed via the Abel map into copies of the Prym variety of C-->D, realized as the inverse image in Pic^3(C), of the g(1,3) on D. Then Beauville calculates the homology classes of those embedded curves and applies the criterion of Matsusaka.  Recillas himself gave an argument based on Hurwitz schemes.
The link with classical projective geometry is via the "complete quadrangle" or "complete quadrilateral".  (These objects had apparently been studied by Wirtinger and Roth.)  Note that 4 general points determine 3 pairs of distinct diagonals, hence three points of intersection of such pairs, together with a "double cover" of the triple consisting of the 6 diagonals.  On the canonical model of a tetragonal curve E there is a pencil of coplanar 4-tuples of points, hence also another trigonal curve D swept out by the associated pencil of coplanar triples of intersection points.  This curve D is presumably the Prym canonical model of the curve D associated to the double cover C-->D by the curve C of diagonals.
Here is a relevant classical reference:
 Roth P., U ̈ber Beziehungen zwischen algebraischen Gebilden vom Geschlecht drei und vier, Monatsh. fu ̈r Math. und Phys., 22, (1911), 64–88.
And some more recent ones:
http://www.math.lsa.umich.edu/~idolga/rat3.pdf
http://link.springer.com/article/10.1007/BF01766014#page-1<|endoftext|>
TITLE: Does there exist a space X whose suspension is homotopy equivalent to [0,1] rel ends but where X is not contractible?
QUESTION [6 upvotes]: As pointed out by David White in 
when mapping cone is contractible
there exists an acyclic CW-complex $X$ which is not contractible but whose suspension is contractible. Namely, let $a$ and $b$ be the two loops in $X=S^1\vee S^1$ and glue in two $2$-cells along the words $a^5b^{−3}$ and $b^3(ab)^{−2}$.
My question is: Is this possible if the suspension, $\Sigma X$, of $X$ is homotopy equivalent to the unit interval $I=[0,1]$ rel ends?
More precisely: if $f:\Sigma X \to I$ is a homotopy equivalence with inverse $g$ and homotopies $h_1:fg\simeq id_I$ and $h_2:gf\simeq id_{\Sigma X}$ such that $g(i) = [X\times i]$ for $i=0,1$, $h_1$ is a homotopy rel {$0,1$} and $h_2$ is a homotopy rel {$[X\times 0]$, $[X\times 1]$} then does $X$ have to be contractible?

REPLY [9 votes]: $\newcommand{\set}[1]{\lbrace #1 \rbrace}$I will assume that the notation $\Sigma X$ in the question denotes the unreduced suspension of the space $X$.
Quick answer: The notion of homotopy equivalence $\Sigma X\to I$ rel ends described in the question is actually equivalent to the contractibility of $\Sigma X$, since the inclusions of the "ends" $\set{0,1}$ into $\Sigma X$ and $I$ are both cofibrations. Thus the cited example of David White is also a counter-example for this question.
Slow answer: There is the canonical map $i:\set{0,1}\to \Sigma X$ given by $i(t)=[x,t]$ for any $t\in\set{0,1}$ and $x\in X$. Moreover, we have the inclusion $j:\set{0,1}\hookrightarrow I$. The question describes the notion of a homotopy equivalence under $\set{0,1}$ between $I$ and $\Sigma X$ (relative to the preceding maps $i$ and $j$), sometimes also called a cofibre homotopy equivalence. It is asked if the existence of such a cofibre homotopy equivalence implies that $X$ is contractible.
The answer is no, as follows from:

There exists a space $X$ which is not contractible, yet $\Sigma X$ is contractible. Any non-contractible, acyclic CW-complex can be used here, such as the example given by David White in the link provided in the question.
If $\Sigma X$ is contractible, then there is a cofibre homotopy equivalence $\Sigma X \to I$ of spaces under $\set{0,1}$.

To prove statement (2), it suffices to observe that the maps $i:\set{0,1}\to \Sigma X$ and $j:\set{0,1}\to I$ are both cofibrations: for example, the NDR condition is very simple to check in these cases (see, for example, section 6.4 of Peter May's book "A concise course in algebraic topology" for a description of NDR pairs). Then it is a well-known fact that any homotopy equivalence $f:\Sigma X\to I$ such that $f\circ i = j$ is actually a cofibre homotopy equivalence (see section 6.5 of the aforementioned book "A concise course in algebraic topology"). In conclusion, if $\Sigma X$ is contractible, then the projection map $f:\Sigma X \to I$ — defined by $f([x,t])=t$ for $x\in X$, $t\in I$ — is a homotopy equivalence and, by the above result, is therefore a cofibre homotopy equivalence of spaces under $\set{0,1}$.<|endoftext|>
TITLE: Grothendieck fibrations and classifying spaces
QUESTION [9 upvotes]: Suppose that $$F:\mathcal{D} \to \mathcal{C}$$ is a Grothendieck fibration of small categories, whose fibers are groupoids. Is there anything sensible which can be said about the induced morphism $$BF:B \mathcal{D} \to B\mathcal{C}$$ between classifying spaces? It seems that Quillen's theorems do not buy us much when the fibers are groupoids, since fibers are homotopy equivalent if and only if they are equivalent as groupoids. I also care about the special case of discrete fibrations (i.e. the Grothendieck construction of a presheaf of sets). Is there some way of expressing the homotopy type of $B \mathcal{D}$ in terms of the homotopy type of $B\mathcal{C}$ together with the data of the psuedofunctor $$\varphi:\mathcal{C}^{op} \to \mathcal{Gpd}$$ classified by the fibration $F,$ which is computationally tractable, at least for "nice enough" $\varphi$? I am aware of a spectral sequence involving the homology of a $\mathcal{C}$-module derived from $\varphi$ converging to the homology of $B\mathcal{D},$ but I would really like something computable from knowing only the homotopy type of $B\mathcal{C},$ not $\mathcal{C}$ itself.

REPLY [2 votes]: Since others seem to be interested in this question, p. 19 of this preprint http://arxiv.org/abs/1112.3996 appears to have nice spectral sequences for homology (resp.cohomology) of $B\mathcal{D}$ in terms of homology (resp. cohomology) of $B\mathcal{C}$ with coefficients twisted by a local system determined by $\varphi.$
Actually, I am a bit confused. Could anyone tell me why $H^{q}\left(BD_\left(\mspace{3mu} \cdot \mspace{3mu}\right),A\right)$ is a local system? It seems like I would need $\varphi$ to invert all morphisms.<|endoftext|>
TITLE: Cardinals without choice: interpolation (reference wanted)
QUESTION [9 upvotes]: Is there a published reference for this ZF theorem?
Let $m,n\in\mathbb{N}$. If $a_1,\dots,a_m$ and $b_1,\dots,b_n$ are cardinals such that $a_i\le b_j$ for all $i$ and $j$, then there is a cardinal $x$ such that $a_i\le x\le b_j$ for all $i$ and $j$.
It's enough if the proposition is stated for the case $m = n = 2$, as the rest follows by an easy induction.

REPLY [4 votes]: The comments by The User and Joel David Hamkins refer to a previous version of the answer which contained a mistake. The current version is completely disjoint of the previous one, and the comments no longer apply.
This appears in Tarski's book Cardinal Algebras as Theorem 2.28, called Interpolation Theorem, and the statement of the theorem is as follows:

If $n\leqq\infty,\ p\leqq\infty$, such that $a_i\leqq b_j$ for $i< n$ and $j < p$, then there is an element $c$ such that $a_i\leqq c\leqq b_j$ for every $i < n$ and $j < p$.

The theorem appears on page 27. The full citation is given below, one can also read about it on MathSciNet.

Tarski, Alfred. Cardinal Algebras. With an Appendix: Cardinal Products of Isomorphism Types, by Bjarni Jónsson and Alfred Tarski. Oxford University Press, New York, N. Y., 1949. xii+326 pp. 


One note about the use of $\infty$ here, Tarski assumes that there is an operator of summation of a countable sequence, and therefore the theorem holds in that case as well. This may require the axiom of choice (or rather a minor fragment thereof), but if we remove that case, then the proof goes through just fine.
The proof begins with the case of $n=p=2$, in which cardinals are manipulated by hand. It then proceeds to the case where $n$ is arbitrary and $p=2$, where a sequence is defined by induction and the existence of a supremum is guaranteed by a previous theorem. Here we use choice when $n=\infty$, but if $n$ is finite then the induction halts and the upper bound of the last step is our wanted $x$.
Next he proves for $n=2$ and $p$ arbitrary. For finite $p$ the proof is purely constructive, and I suspect it holds without the axiom of choice for $p=\infty$ as well (it relies on previous theorems which I haven't read thoroughly in search of choices).
Lastly he argues that the proof of the general case follows as the proof of the second case, with using the third case to reason instead of the first case. In either case, if we assume $p,n<\infty$ then there is no invocation of the axiom of choice.<|endoftext|>
TITLE: Can group cohomology be interpreted as an obstruction to lifts?
QUESTION [16 upvotes]: The standard way to view the first and second group cohomologies is this:
The Standard Story
Let $G$ be a group, and let $M$ be a commutative group with a $G$-action. Then the first cohomology has the following interpretation: $H^1(G,M)$ is bijective with sections (modulo conjugation by $M$) of the short exact sequence $$1\rightarrow M\rightarrow M\rtimes G\rightarrow G\rightarrow 1.$$
Furthermore, the group of $1$-cocycles, $Z^1(G,M)$ is bijective with the sections of this short exact sequence. (Not modulo anything.) In fact, this holds even if $M$ is non-abelian.
The second cohomology $H^2(G,M)$ is bijective with the set of isomorphism classes of group extensions
$$1\rightarrow M\rightarrow H\rightarrow G\rightarrow 1$$
for which there exists a (or equivalently for every) set theoretic section $s:G\rightarrow H$ such that $g\cdot m=s(g)m(s(g))^{-1}$. (Here $g\cdot m$ denotes the action of $g$ on $m$ coming from the $G$-module structure of $M$.)
Liftings
In a paper I have been reading, they have given an entirely different interpretation to the first cohomology. Namely:
Let $A$ and $B$ be groups, and let $C$ be a normal abelian subgroup of $B$. Let $\bar \phi:A\rightarrow B/C$ be a homomorphism. Assume $\phi$ has a lift $\alpha:A\rightarrow B$. Then $Z^1(A,C)$ is bijective with the set of lifts of $\phi$ to homomorphisms from $A$ to $B$.
The bijection goes like this: $\theta\in Z^1(A,C)$ goes to $\alpha\theta$.
My question is: can one give an interpretation in terms of lifts to the second cohomology, or to the group of $2$-cocycles?
More precisely:
Question
Let $A$ and $B$ be groups, and let $C$ be a normal abelian subgroup of $B$. Let $\bar \phi:A\rightarrow B/C$ be a homomorphism.
Is it true that there exists a lift of $\bar \phi$ to a homomorphism from $A$ to $B$ if and only if $H^2(A,C)$ is trivial? Or is there a $2$-cocycle one can define (how would one define it?) such that there exists a lift of $\bar \phi$ if and only if it is trivial in $H^2(A,C)$? Or perhaps the right group to look at is the group of $2$-cocycles $Z^2(A,C)$ rather than the cohomology group?
I don't know if such an interpretation exists, so this is just wishful thinking. Since this is the first time I've seen the interpretation of $Z^1(A,C)$ in terms of lifts, I was curious whether such an interpretation extends to the second cohomology.

REPLY [16 votes]: In your setup, the extension
$$
C \to B \to B/C
$$
represents a cohomology class $u\in H^2(B/C;C)$. The homomorphism $\phi: A\to B/C$ admits a lift $\bar\phi : A\to B$ if and only if $\phi^\ast(u)\in H^2(A;C)$ is zero.
To see this, note that the induced map on second cohomology is given by pullback of extensions, so $\phi^\ast(u)$ is represented by the bottom row in the diagram
$$
\begin{array}{ccccc}
C & \to & B & \to & B/C \newline
\| &  & \uparrow & & \uparrow \newline
C  & \to   & E  & \to & A.
\end{array}
$$
Now splittings of the bottom row correspond to liftings of $\phi$.
This gives plenty of examples where $\phi$ lifts but $H^2(A;C)\neq 0$. In fact, you could take any non-split extension $C\to B\to B/C$ such that $H^2(B;C)\neq 0$ and let $\phi: B\to B/C$ be the quotient map.<|endoftext|>
TITLE: Could the Jacobian conjecture be undecidable?
QUESTION [8 upvotes]: Most of us know the Jacobian conjecture.  Here's a version below for fixed positive integers $d$ and $n$:
$J(d,n)$:  If $f: C^n \rightarrow C^n$ is a polynomial map of degree $d$, and if the Jacobian determinant $\vert Jf \vert$ is nowhere vanishing (hence constant), then $f$ is injective (hence bijective).
We know that the sentence "For all $n$, $J(3,n)$" implies the sentence "For all $d,n$, $J(d,n)$."  In other words, the Jacobian conjecture has been reduced to degree $3$.
We also know that, for any fixed $n$, $J(3,n)$ is provably true or provably false.  This boils down to the completeness of the theory of algebraically closed fields of characteristic zero.  
But, do we know whether the sentence "For all $n$, $J(3,n)$" is provably true or false?  In other words, might the Jacobian conjecture be... (oh no).. undecidable?!
In other words, I could theoretically program my computer to set out to prove the Jacobian conjecture $J(3,1)$ (easy) and $J(3,2)$ then $J(3,3)$, etc.., and my theoretical computer would keep on going for epochs and epochs.  But would it ever halt?  Might this be undecidable?

REPLY [5 votes]: There's no way to rule out a priori that the Jacobian conjecture is undecidable (in your favorite axiomatic system).
As I pointed out in my answer to another MO question, a proof that some statement is decidable would automatically mean that its status would be resolved up to a finite computation.  You can be sure that if a conjecture as important as the Jacobian conjecture were reduced to a finite computation, then we would hear about it.  This observation yields an informal "proof" that (almost) all the major open problems you care to name are not known to be decidable.
See also this related MO question.<|endoftext|>
TITLE: "Inverse problem" for Brauer groups
QUESTION [8 upvotes]: This question is just a curiosity, but I'm really interested in the answer. It was originally posted on math.stackexchange (https://math.stackexchange.com/questions/368897/inverse-problem-for-brauer-groups), but hasn't received any responses despite some upvotes, so I'm posting it here.
Given a field $K$, we can form the set$^*$ $Br(K)$ consisting of equivalence classes of finite-dimensional central simple $K$-algebras which split over some Galois extension of $K$, modulo "are Morita-equivalent" (I hope I have that right, it's been a while). This set is actually a group, in a natural way: the tensor product over $K$ is well-defined on the equivalence classes, and has identity (the equivalence class of $K$ as an algebra over itself) and inverses (given by $R\mapsto R^{op}$). Actually $Br(K)$ turns out to be a second cohomology group, in a natural and useful way, but I don't really have a good understanding of that part.
My main question is, what groups are the Brauer group of some field? I know a couple trivial bits of the answer to this: $Br(K)$ is always abelian, and of cardinality at most $\aleph_0\times\vert K\vert$, and $Br(K)$ is always torsion. Within those constraints, I only know of one specific nontrivial Brauer group: $Br(\mathbb{R})=\mathbb{Z}/2\mathbb{Z}$ by Frobenius' Theorem on division algebras over $\mathbb{R}$. (I've seen the Brauer group of $\mathbb{Q}$ described by a short exact sequence, but I wasn't able to get an explicit description from that; is it known?) EDIT: As Emerton points out in a comment below, the Brauer group of $\mathbb{Q}$ (and much more) is known: it is $Br(\mathbb{Q})=\mathbb{Z}/2\mathbb{Z}\oplus\bigoplus_1^\infty\mathbb{Q}/\mathbb{Z}.$
My main question is: is there a known list of properties which are necessary and sufficient for a group to be $\cong Br(K)$ for some $K$?
There are many possible variations/elaborations of this question, which may not have deep significance but seem kind of interesting. For example, leaving the context of fields for a moment, there is an analogous notion of Brauer group for groups, and we can ask (although I'm not sure why we would ask): is there a group which is its own Brauer group? My second question is just: is there a good resource for this type of question, that is, for constructing Brauer groups of various objects to specification? I imagine the opposite direction (finding Brauer groups of fields we already care about) is much more useful, but I'm personally interested in this direction.
(As an aside, I'm not sure whether the "group theory" tag is appropriate here; if it is not, feel free to delete it, or let me know and I will delete it.)

$^*$ As a very minor aside, note that size issues don't arise here: since we specify "finite-dimensional," there are at most $\aleph_0\times\vert K\vert$ many such algebras up to isomorphism; using Scott's trick then lets us represent these equivalence classes in perfectly fine way.

REPLY [6 votes]: The Brauer group of a field $F$ decomposes as the direct sum of its $p$-primary components $\mathrm{Br}(F)_p$, with $p$ prime. A conjecture of Brumer and Rosen 
(Proc. AMS 19 (1968), 707-711) predicts that for every prime $p$, the component $\mathrm{Br}(F)_p$ is either 

trivial, 
contains a non-trivial divisible group, or 
$p=2$ and $\mathrm{Br}(F)_2$ is an elementary abelian 2-group. 

This was proved by Merkurjev (Brauer groups of fields, Comm. Algebra 11 (1983), 2611-2624) when $(F(\mu_p):F)\le2$, as well as by B. Kahn in his Thesis. 
Other related papers may be:
T. Wurfel, Ein Frieheitskriterium fur pro-$p$-Gruppen mit Anwendungen auf die Struktur der Brauer-Gruppe, Math. Z. 172 (1980), 81-88
I. Efrat, On fields with finite Brauer groups, Pacific J. Math. (177)(1997), 33-46.
I hope this helps.<|endoftext|>
TITLE: Reasons to prefer one large prime over another to approximate characteristic zero
QUESTION [65 upvotes]: Background:
In running algebraic geometry computations using software such as Macaulay2, it is often easier and faster to work over $\mathbb F_p = \mathbb Z / p\mathbb Z$ for a large prime $p$, rather than over $\mathbb Q$. (Note that working directly over $\mathbb C$ is not really possible for exact computations.)  Some basic model theory implies (more or less) that if you have a question that is capable of being answered by an algorithm, and the question has the same answer for $k$ as for $\bar{k}$, then its answer over $\mathbb Q$ will be the same as its answer over $\mathbb F_p$ for all but finitely many primes $p$.  The accepted wisdom is that with virtually no exceptions, if you want to answer an algebro-geometric question over $\mathbb Q$, you can get a reliable answer by picking a large prime such as $p=32003$ and doing your computations over $\mathbb F_p$.
I think people generally pick a prime near the top of the range their software can handle, which is relatively easy to remember; $32003$ and $31667$, for instance, both fit the bill when using Macaulay2.  However, I was wondering whether there are other mathematical characteristics of a prime that can affect how well it approximates characteristic zero. For instance, does some special pathology arise if $p$ is (or is not) a Mersenne prime, or if $(p-1)/2$ is (or is not) prime, or...?
Note that I only am asking about behavior that affects how well characteristic $p$ approximates characteristic $0$.  Questions that one would only ever ask in finite characteristic are not relevant.  Also irrelevant are questions that cannot be answered by an algorithm; for instance, "is $n\cdot1=0$ for some positive integer $n$" cannot be asked unless you include a bound on $n$.
The Question: Are there valid mathematical reasons for preferring one large prime over another to approximate characteristic $0$, other than the assumption that larger primes are generally better?

REPLY [6 votes]: I prefer primes like 1000003 and 1000000007 because it is easy to recognize small integers and rational numbers with small denominators in the output. For instance, modulo 1000003 we have
1/3=666669,
-1/3=333334,
1/5=600002.
I know that's not what you are asking from the mathematical point of view, but from practical point of view it is very helpful.
EDIT: To answer the OP question, my answer is no. There is no mathematical reason one prime can behave better than another one in an algebraic computation. Of course, larger primes are better because there are less coincidences when you ask if $x=0$. But that's it. I am happy to be proven wrong.<|endoftext|>
TITLE: Anomalies in the definition of Turaev's TQFT
QUESTION [11 upvotes]: In his book Quantum invariants of knots and 3-manifolds page 124, Turaev defined a TQFT $\tau$ axiomatically. 
For a cobordism $(M, \partial_{-}M, \partial_{+}M)$, a TQFT assignes a $k$-homomorhism $\tau(M)$ from the projective $k$-module $\tau(\partial_{-}M)$ to the projective $k$-module $\tau(\partial_{+}M)$. Here $k$ is a (ground) ring.
One of the axioms is a functoriality. Let $M_1$ and $M_2$ be a cobordisms and let $M$ be a cobordism obtained by gluing $M_1$ and $M_2$ along a homeomorphism $f$ from $\partial_+(M_1)$ to $\partial_{-}(M_2)$. Then the functoriality says that
$\tau(M)=k\tau(M_2)\circ f_{*} \circ \tau(M_1)$, where $\tau$ is a TQFT and $k$ is an invertible element called anomaly for the pair $(M_1, M_2,f)$. 
My question is whether this is associative or not. Namely, if we have two pairs $(M_1, M_2,f)$ and $(M_2, M_3,g)$, then we can calculate the anomaly in two way, gluing first along $f$ and then $g$, or gluing along $g$ first then $f$.
Are anomalies obtained in these ways same? If so, how do we prove it from the axioms of Turaev TQFT?
If we regard $(M_1 \cup_f M_2)\cup_g M_3$ and $M_1 \cup_f (M_2\cup_g M_3)$ are the same, then we should have the anomalies the same. Or if we think there is a homeomophism between these spaces, then the anomalies should be the same up to invertible factor. Right?

REPLY [10 votes]: I think in the section you refer to Turaev is just defining what one might call a "projective TQFT" or a "TQFT up to phase factors".  In later sections he discusses how to remove the anomaly by tweaking the cobordism category.
Turaev's book is probably not the best place to learn about Reshetikhin-Turaev TQFTs for a beginner.  You might want to supplement your reading with one or more of

The original Reshetikhin-Turaev papers
The book by Kauffman and Lins
The paper by Blanchet, Habeggar, Masbaum and Vogel
My TQFT notes from 1991<|endoftext|>
TITLE: Germs at infinity of sequence of integers
QUESTION [6 upvotes]: Consider the  $\mathbb Z$-module $\mathcal Z$ obtained as the set of sequences of integers $\mathbb Z ^ \mathbb N$ modulo the relation that two sequences are deemed equivalent when their difference is $0$ almost everywhere. This space is the space of germs at $+\infty$ of sequences of integers. I think this is a quite natural object to study, as for instance it appears naturally when you try to generalize the notion of fundamental group of the circle when dealing with paths with a non-compact source space: imagine the value of $z_n$ as being the "number of turns" done at step $n$, the classical case being recovered from the stationary sequences. In that sense it is a completion of $\mathbb Z$ with a rich structure in the non-finite part. Yet I have not been able to find any literature regarding this object. This may be so because either

nothing specific/interesting can be said about it, or

it is an obvious/trivial example of some algebraic concept I'm unaware of (I'm no algebraist myself).
So my question is: are the basic properties of this module known/interesting? For instance, is it a free module? Does anybody know a classical reference which may help me in studying this object?
Thanks in advance for your contributions.

REPLY [11 votes]: This abelian group, which can also be described as the quotient of the direct product $\prod_{\mathbb N}\mathbb Z$ by the direct sum $\sum_{\mathbb N}\mathbb Z$, is isomorphic to the direct sum of the following two pieces.  The first piece is a torsion-free, divisible abelian group (so you can view it as a vector space over $\mathbb Q$) of rank (i.e., dimension over $\mathbb Q$) equal to the cardinality of the continuum.  The second piece is the direct product, over all primes $p$, of the direct product groups $\prod_{\mathbb N}\mathbb Z_p$, where $\mathbb Z_p$ is the additive group of $p$-adic integers.  I believe this result is due to Balcerzyk.
Edit: I believe the Balcerzyk reference is "On factor groups of some subgroups of a complete direct sum of infinite cyclic groups" [Bull. Acad. Polon. Sci., Sér. Sci. Math., Astron., Phys. 7 (1959) 141-142.<|endoftext|>
TITLE: Waldhausen $K$-theory for $G$-spaces
QUESTION [11 upvotes]: I would guess that the following is true, and that somebody has worked it out, but I don't recall ever seeing it. Can anyone point me to any literature on it?
Let $G$ be a finite group. We know that there is a functor 'equivariant suspension spectrum' $X\mapsto \Sigma^\infty_G(X_+)$ from $G$-spaces to $G$-spectra. It seems that there should be another such functor $A_G$, 'equivariant Waldhausen $K$-theory spectrum'. The fixed point spectrum ought to split up according to conjugacy classes of subgroups of $G$ just as for the suspension spectrum:
$$
A_G(X)^G\sim \Pi_H\ A(X^H_{hW_GH})
$$
($X^H_{W_GH}$ is homotopy orbits for the action of $W_GH=(N_GH)/H$ on fixed points of $H$.)
On the one hand when $X$ is based and connected then there should be a description of $A_G(X)$ that makes it a special case of a $K$-theory of structured $G$-ring-spectra.
On the other hand there should be a splitting 
$$
A_G(X)\sim \Sigma^\infty_G(X_+)\times Wh^{diff}_G(X)
$$
where the other factor is related to equivariant pseudoisotopies and $h$-cobordisms.

REPLY [4 votes]: For the sake of people googling this, there's a pretty canonical reference avaliable now: Malkiewich and Merling.<|endoftext|>
TITLE: First order consequence of a combinatorial principle
QUESTION [5 upvotes]: (Base theory $RCA_0$)The principle says there exists a function g such that g dominates any X-recursive function for any X in the model. 
i.e. For any $f\le_T X$, $\exists b\in M$ such that $g(a)>f(a) \forall a>b$. Here the second order structure (countable) is $\langle M,S_M,+_M,\cdot_M,0_M,1_M \rangle$.
In the context of $\omega$ models, models satisfying such principle could be characterized by closure of high sets with respect to the existing sets in the model by Martin's characterization of high sets. I am curious about the first-order consequence of this principle?
EDIT: (some update) I was trying to separate the domination principle from $B\Sigma_2^0$. Starting from a countable first order model of arithmetic $M$ such that $\neg B\Sigma_2^0 + I\Sigma_1^0$ holds, I was trying to add into the second order part the dominating functions. Dominating functions exist by $B\Sigma_1^0$. However, I need to prove that after augmenting the model with a dominating function $f$, $M[f]$ preserves $I\Sigma_1^0$. I found some similar forcing arguments on trees (Lemma IX.2.4) in Simpson's book and the augmented set is one generic set. However, I am not sure how to impose the special requirement (i.e. dominating property) on the choice of the generic set.

REPLY [7 votes]: The simplest forcing to add a dominating function is Hechler forcing $\newcommand{\D}{\mathbb{D}}\D$. In set-theoretic circles, conditions in $\D$ are pairs $(s,f)$ where $s$ is a finite sequence of natural numbers and $\newcommand{\N}{\mathbb{N}}f:\N\to\N$, extension is defined by $(s,f) \leq_{\D} (t,g)$ if $t \supseteq s$, $g \geq f$, and $t(n) \geq f(n)$ for $|s| \leq n \lt |t|$. A $\D$-generic filter $G$ defines a function $g = \bigcup \lbrace s : (s,f) \in G\rbrace$ which eventually dominates every ground model function.
Since the statement you're trying to force is localized in the sense that you only want $g$ to dominate all total $X$-computable functions, you can get away with an index-based variant of Hechler forcing. In that case, conditions of $\D_X$ are pairs $(s,i)$ where $s$ is a (coded) finite sequence of natural numbers and $i$ is an index for a total $X$-computable function $\varphi_i^X$, extension is defined by $(s,i) \leq_{\D_X} (t,j)$ if $(s,\varphi_i^X) \leq_{\D} (t,\varphi_j^X)$ in the sense described above. A $\D_X$-generic filter defines a function $g$ as above which eventually dominates every total $X$-computable function.
Note that we cannot expect $\D_X$ conditions to form a set since "$\varphi_i^X$ is total" is a $\Pi^0_2(X)$-complete statement. This is not a major problem since generics are constructed externally and we understand what "$\varphi_i^X$ is total" means from outside the ground model. Note that if the set of conditions exists in the ground model, then $\D_X$ is just a variation on Cohen forcing. However, in general, the ground model will have a very different perception of $\D_X$ and the generic will be quite different from a plain Cohen generic set.
To see that $\D_X$ preserves $\Sigma^0_1$-induction, first show that if some extension $(t,j) \leq_{\D_X} (s,i)$ forces a $\Sigma^0_1$-statement (which may use a fixed ground model set parameter in addition to the generic function $g$) then there is another extension $(u,i) \geq (s,i)$ that also forces the same $\Sigma^0_1$-statement. It follows from this that if $A(x)$ is a $\Sigma^0_1$ statement in the forcing language, then the set $$\lbrace x \in \N : (s,i) \nVdash \lnot A(x)\rbrace$$ is actually $\Sigma^0_1$-definable over the ground model. By $\Sigma^0_1$-induction in the ground model, this set has a minimal element $x_0$ and there is an extension $(t,j) \geq (s,i)$ (even with $j = i$) such that $$(t,j) \Vdash A(x_0) \land (\forall x \lt x_0)\lnot A(x).$$ This shows that it is dense to either force $\forall x \lnot A(x)$ or to force that there is a minimal $x$ that satisfies $A(x)$. Therefore, forcing with $\D_X$ preserves $\Sigma_1$-induction.
The use of the indexed variant $\D_X$ instead of the full second-order forcing $\D$ is very useful here since $\D$ can be quite devastating to weak subsystems of second-order arithmetic. Indeed, if the ground model satisfies arithmetic comprehension, then every $\Pi^1_1$ statement becomes $\Sigma^0_2$ in the generic extension. So forcing with $\D$ will not preserve systems weaker than $\Pi^1_1$-CA0 containing ACA0. The index-based variant $\D_X$ is not so devastating since it is equivalent to Cohen forcing over any model of ACA0.<|endoftext|>
TITLE: Generalization of the Lefschetz fixed point theorem
QUESTION [9 upvotes]: I have encountered a certain generalization of the Lefschetz fixed point theorem as folklore, and I am hoping that someone out there knows its provenance or can otherwise refer me to a source where it is discussed or proved.  The statement of the theorem should look vaguely like this:
Let $X$ be a smooth, complete variety over $\mathbb{C}$ and $g$ an automorphism of $X$.  Suppose also that the fixed locus of $g$ is smooth and complete (but not necessarily discrete), and let $e(g)$ denote the Euler characteristic of this fixed locus.   Then
$$
e(g)=\sum _{i=0}^{\dim L} (-1)^i Tr (g | H^i(X)). 
$$
Any help or advice you can offer is much appreciated!

REPLY [9 votes]: One of the standard proofs of the Lefschetz formula proceeds by showing that the RHS of your equation computes the intersection number between the graph of $f$ and the diagonal inside $X \times X$. The usual case is when this intersection has expected dimension. The general case requires the excess intersection formula. Now, the normal bundle of the diagonal $\Delta$ in $X \times X$ is the tangent bundle of $\Delta = X$. After dividing by the normal bundle of $X^g$ in $X$, the excess intersection formula shows that the intersection number is given by the top Chern class of the tangent bundle of $X^g$. But this is just the topological Euler characteristic of $X^g$, $e(g)$. (This last fact is just the special case of your formula when $g = \mathrm{id}$!)<|endoftext|>
TITLE: Blue and red balls puzzle
QUESTION [7 upvotes]: I was sent this puzzle and wondered if it is known or if its origin is known? (I see colored ball puzzles are also in vogue.)
Consider a bag with $n$ red balls and $n$ blue balls. At each turn you take out a ball chosen uniformly at random. If the ball is red you put it back in the bag and take out a blue ball. If the ball is blue put it back in the bag and take out a red ball.
If you stop when all the balls are the same color, how many balls will be left?

REPLY [12 votes]: This appears to be a description of the "OK Corral process" as an urn problem. This stochastic process was apparently introduced by David Williams and Paul McIlroy in the 1998 article The OK Corral and the power of the law and was subsequently investigated by J.F.C. Kingman and S. E. Volkov. In the 1999 article Martingales in the OK Corral Kingman proved that the expected number of balls at the end of the process is asymptotic to $n^{\frac{3}{4}}$ times
$$2 \cdot 3^{-\frac{1}{4}}\pi^{-\frac{1}{2}} \Gamma\left(\frac{3}{4}\right) \simeq 1.0506511521875180068945465...$$ in the limit as $n \to \infty$. The numerical value of this constant appears to be in good agreement with Aaron's answer.
A subsequent article by Kingman and Volkov, Solution to the OK Corral model via decoupling of Friedman's urn, investigates the asymptotic distribution of the probability that a particular number of balls is left at the end of the process. The authors note in particular that this process is similar to running an urn model studied by B. Friedman in 1949 in reverse time. I might add that of the three articles it is this one in which the identity between the OK Corral model and the urn model is most immediately apparent.<|endoftext|>
TITLE: Proof that $L^2(0,T;X)^* = L^2(0,T;X^*)$
QUESTION [6 upvotes]: How is the proof that 
$$[L^2(0,T;X)]' = L^2(0,T;X')$$
looking like, where $X$ is a Hilbert space? I am asking for the proof that the dual space of $L^2(0,T;X)$ is the space $L^2(0,T;X^*)$. 
Is the proof much different to the $L^p(a,b)$ case?
I can't find any easy to understand proofs.

REPLY [6 votes]: To give you a reference: Diestel-Uhl, Vector measures, page 98, Chapter 4, Theorem 1:
$$L^p(\mu,X)^\ast = L^q(\mu,X^\ast)$$
if and only if $X^\ast$ has the Radon-Nikodym property with respect to $\mu$. 
Here $\mu$ is a finite measure and $p$ and $q$ as usual. Proof can be read there.
ADDED: It should be noted that $L^q(\mu,X^\ast) \subset L^p(\mu,X)^\ast $ always holds, without any condition on the Banach space $X$. The proof of this is quite the same as in the scalar case.<|endoftext|>
TITLE: Does the generalized $\Delta$-system lemma imply some weak version of the GCH?
QUESTION [17 upvotes]: Let $\Delta(\kappa, \mu)$ be the statement: "let $F$ be a family of cardinality $\kappa$ of sets of cardinality less than $\mu$. Then there is a family $G \subset F$ of cardinality $\kappa$ and a set $r$ such that $a \cap b=r$ for every $a,b \in G$".
We know that if $\kappa$ is a regular cardinal and $\lambda^{<\mu} < \kappa$, for every $\lambda < \kappa$ then $\Delta(\kappa, \mu)$ holds. My question is:

Does $\Delta(\kappa, \mu)$ for some regular uncountable $\kappa$ and some uncountable $\mu<\kappa$ imply some weak form of the GCH below $\kappa$?

REPLY [21 votes]: It is a very nice question! The answer is yes, natural instances of the $\Delta$ system property, which hold under GCH, are in fact equivalent to the GCH. 
Theorem. $\Delta(\omega_2,\omega_1)$ is equivalent to CH.
Proof: You've pointed out that CH implies the principle, since the
hypothesis you mention for this case amounts to
$\omega_1^{\lt\omega_1}<\omega_2$, which amounts to CH. So let us
consider what happens when CH fails. Let $T=2^{\lt\omega}$ be the
tree of all finite binary sequences, and label the nodes of $T$
with distinct natural numbers. Let $F$ be the subsets of $\omega$
arising as the sets of labels occuring on any of $\omega_2$ many
branches through $T$. Thus, $F$ has size $\omega_2$, and any two
elements of $F$ have finite intersection. I claim that this family
of sets can have no $\Delta$-system of size $\omega_2$, and
indeed, it can have no $\Delta$-system even with three elements.
If $r$ is the root of $a$, $b$ and $c$ in $F$, then $r=a\cap
b=a\cap c$, and so $a$ and $b$ branch out at the same node that
$a$ and $c$ do, in which case $b$ and $c$ must agree one step
longer, so $b\cap c\neq r$. QED
The same idea works for higher cardinals as follows: 
Theorem. For any infinite cardinal $\delta$, we have 
$\Delta(\delta^{++},\delta^+)$ is equivalent to
$2^\delta=\delta^+$.
Proof. If $2^\delta=\delta^+$, then your criterion, which amounts to 
$(\delta^+)^{\lt\delta^+}<\delta^{++}$, is fulfilled, and so the
$\Delta$ property holds. Conversely, consider the tree
$T=2^{\lt\delta}$, the binary sequences of length less than $\delta$. Let $F$ be a family of
$\delta^{++}$ many branches through $T$, regarding each branch $b$ as a subset of $T$, the set of its initial segments. Each such branch has size $\delta$, since the tree has height $\delta$. But for the same reason as before, there can be no $\Delta$ system with even three elements, since the tree is merely binary branching, and so three distinct branches cannot have a common root. This contradicts $\Delta(\delta^{++},\delta^+)$, as desired. QED
Corollary. The full GCH is equivalent to the assertion that $\Delta(\delta^{++},\delta^+)$ for every infinite cardinal $\delta$. 

Update. The same idea shows that the hypothesis you mention
is optimal: one can reverse the lemma from the conclusion to the hypothesis. 
Theorem. The following are equivalent, for regular $\kappa$
and $\mu\lt\kappa$:

$\Delta(\kappa,\mu)$
$\lambda^{\lt\mu}\lt\kappa$ for every $\lambda\lt\kappa$.

Proof. You mentioned that 2 implies 1, and this is how one usually
sees the $\Delta$ system lemma stated. For the converse, suppose
that $\lambda^{\lt\mu}\geq\kappa$ for some $\lambda\lt\kappa$.
Since $\kappa$ is regular and $\mu\lt\kappa$, this implies $\lambda^\eta\geq\kappa$
for some $\eta\lt\mu$. Let $T$ be the $\lambda$-branching tree
$\lambda^{\lt\eta}$, which has height $\eta$. Let $F$ be a family
of $\kappa$ many branches through this tree, where we think of a
branch as the set of nodes in the tree that lie on it, a maximal linearly
ordered subset of the tree $T$. Each such branch is a set of size $\eta$. I claim that this family has no
subfamily that is $\Delta$ system of size $\lambda^+$. The reason is
that because the tree is $\lambda$-branching, if we have $\lambda^+$
many branches with a common root, then at least two of them must
extend that root to the next level in the same way, a contradiction to it being a
root. Thus, the failure of 2 implies the failure of 1, as desired.
QED<|endoftext|>
TITLE: Are there proofs that you feel you did not "understand" for a long time?
QUESTION [75 upvotes]: Perhaps the "proofs" of ABC conjecture or newly released weak version of twin prime conjecture or alike readily come to your mind. These are not the proofs I am looking for. Indeed my question was inspired by some other posts seeking for a hint to understand a certain more or less well-establised proof, or some answers to those posts. I am interested in proofs at undergraduate levels. Since the question as asked in the title would be too personal, I suggest a longer and hopefully more positive version: 
Is there any proof that you feel you didn't understand fully until years later? 
The reason that I am interested in this question is that we are currently working on an assessment framework for assessing students' understanding of proof. Reading some previous posts on MO, It occurred to me that perhaps we are too naive in our approach just seeking for understanding logical structure, the key point and so on. It would be very informative if you kindly include in your answer the follow-up of the proof you mention.

REPLY [4 votes]: The recent answer by coudy reminds me that for a very long time I didn't really grok the proof of Minkowski's inequality from Hölder's inequality, even though it's fairly simple. It just looks like a clever trick, and not really the first thing one would think of. Put differently, one would like a more conceptual story lurking behind that proof. 
As I understand it now, Minkowski's inequality (or the triangle inequality for the $p$-norm) is really all about convexity of the unit ball. You can prove $L^p$ is locally convex with a little calculus, without Hölder's inequality. But more recently I'd been telling myself the following conceptual story: if the meaning of Hölder's inequality is that there is an isometric isomorphism $L^p \cong (L^q)^\ast$, induced by the standard pairing $\langle -, - \rangle: L^p \times L^q \to \mathbb{C}$, then this implies that the unit ball in $L^p$ is the intersection of sets 
$$H_g = \{f \in L^p: |\langle f, g\rangle| \leq 1\}$$ 
where $g$ ranges over the unit ball in $L^q$. But each $H_g$ is clearly convex, and therefore so is the intersection. The standard proof of Minkowski via Hölder can be seen as just a very tidied up rendition of this more conceptual explanation.<|endoftext|>
TITLE: Reference request: sheaves on closed sets
QUESTION [10 upvotes]: I am faced with a context in which the most natural notion of a sheaf $\mathcal F$ is as a functor on the category of compact subsets of a (locally compact Hausdorff) space $X$.  Specifically, I say a presheaf $\mathcal F$ is a sheaf iff it satisfies the three axioms:
$$\begin{align}
\mathcal F(\varnothing)&=0\\
0\to\mathcal F(K_1\cup K_2)\to\mathcal F(K_1)\oplus\mathcal F(K_2)\to\mathcal F(K_1\cap K_2)&\text{ is exact }\forall\text{ }K_1,K_2\subseteq X\\
\varinjlim_{\begin{smallmatrix}K\subseteq U\cr U\textrm{ open}\end{smallmatrix}}\mathcal F(\overline U)\to\mathcal F(K)&\text{ is an isomorphism }\forall\text{ }K\subseteq X
\end{align}$$
I would very much like to not have to rewrite the entire foundations of sheaf theory in this context!  I'm not sure how much the reader would appreciate such an exposition either.  Granted, I would only need to write the foundations I am using, but even this turns out to require many proofs which in retrospect are mostly trivial applications of compactness and direct limit arguments.
Is there a good reference for sheaves on the category of compact subsets somewhere in the literature?  I hope that there is a book which develops the relevant foundations (perhaps by comparing this notion to the more familiar notion of sheaves on open subsets) so I don't have to include too much baggage explaining the basics of such sheaves.
To be more specific about what I mean by "the basics of sheaf theory", I mean something like:

Sections are determined by their stalks
Functors $i_\ast$ and $j_!$ (for inclusions of closed and open subsets respectively)
A theory of Cech cohomology

I don't need anything about injective resolutions or making the category of sheaves into an abelian category.
EDIT: I withdraw this question because the category of sheaves I define above turns out to be equivalent to the (usual) category of sheaves via the functors:
$$\begin{align}
(\alpha_\ast\mathcal F)(K)&=\varinjlim_{\begin{smallmatrix}K\subseteq U\cr U\textrm{ open}\end{smallmatrix}}\mathcal F(U)\\
(\alpha^\ast\mathcal F)(U)&=\varprojlim_{\begin{smallmatrix}K\subseteq U\cr K\textrm{ compact}\end{smallmatrix}}\mathcal F(K)
\end{align}$$

REPLY [6 votes]: Let's label the three conditions you wrote as 1'), 2'), and 3)'. Combine 1') and 2') by saying for any finite cover by compact subsets... (I hope you follow what I mean). Also, in 3)', you better mean that $\bar U$ is compact, otherwise it doesn't make sense.
So now we're reduced to 1) and 2). Also, let's not talk about sheaves of abelian groups, but sheaves of sets. Sheaves of abelian groups are just abelian group objects in the category of sheaves of sets, so there is no harm to start with the latter. Now, if you want to say you are doing "sheaf theory" you better actually have a Grothendieck topology. Here is what you can do in your situation:
Given your space $X$, define the category $K\left(X\right)$ as the poset of of compact subsets of $X$ and their inclusions. You can define a Grothendieck pre-topology, by saying a covering family is a finite family of jointly surjective inclusions. Being a sheaf for this topology is equivalent to condition 1) (i.e. 1') and 2)'). Let us call this topology $J$.
Consider the category $O_c\left(X\right)$ of open subsets of $X$ which have compact closure. Since $X$ is locally compact, these form a basis for $X$. There is  a Grothendieck pre-topology on this category which is the usual one (restricted to this subcategory), except we only allow finite covering families. Sheaves for the associated Grothendieck topology will in general not be sheaves on $X$ in the classical sense, unless $X$ is compact. However, given a sheaf $F$ on $K(X)$, we can define $F(U)$ for a $U$ in $O(X)_c$ by $\varinjlim F(K)$ running over all compacts containing $U,$ but note, this is the same as $F(\bar U),$ since the poset of compact subsets containing $U$ has $\bar U$ as a terminal object. Note, we may also just simply remark that there is a functor $$cl:O_c(X) \to K(X)$$ induced by taking closures and what we have done is defined $cl^*F$. Then, 1) implies that $cl^*F$ is a sheaf for finite open covers as well, since $X$ is locally compact. It would seem that this implies 2) follows automatically, however I haven't checked carefully. 
EDIT: This fails in general! Condition 2) is equivalent to for all $F$ sheaves on $K(X)$, $cl_!cl^*F \cong F,$ i.e. the co-unit of the adjunction $$cl_! \dashv cl^\star$$  needs to be an iso on $Sh(K(X))$, which is if and only if $cl^{\star}$ is full and faithful when restricted to $Sh(K(X)).$ Since $cl$ is itself full and faithful, it implies that $cl_*$ is, and hence we get that $F$ must lie in the image of the pullback topos $$Sh(O_c(X)) \times_{Set^{K(X)^{op}}} Sh(K(X))$$ in $Sh(K(X))$. Concretely, this means there is a finer Grothendieck topology $J'$ on $K(X)$ which does the trick. It is possible to write down explicitly, what the covers are, but I will not attempt to do so here. To get an idea of how to do this though, look at how I construct the "compactly generated Grothendieck topology" here: http://arxiv.org/abs/0907.3925
*The following should work with either $J$ or $J'$ * 
If all this is right, then all you are talking about is a sheaf on $K(X)$ with the Grothendieck topology I have mentioned. Given such a sheaf $F,$ if you restrict to a compact $C$, $F|C$ defines an ordinary sheaf on $C,$ by extending it to opens the way I described. Since every point $x$ in $X$ has a compact neighborhood, it should follow that sections are determined by their stalks, by reducing to the case of ordinary sheaves on compact spaces. 
Now, if $i$ is a closed inclusion, $i:V \to X,$ then $i$ induces a functor $$i^{-1}:K(X) \to K(V),$$ by intersecting with $X.$ This functor preserves covers, so the functor $$i_*:Sh(K(V)) \to Sh(K(X)),$$ defined by $$i_*F(C)=F(C \cap V)$$ for all compact $C$ in $X,$ is well defined. It has a left adjoint $i^{\star},$ which is simply given by procompositing with $i.$
Note: The following probably fails now, with the new topology:
I am not sure how to get $j_!$ for open inclusions, since for open covers this is usually induced by the inclusion functor $O(U) \to O(X),$ but we have no inclusion functor $K(U) \to K(X).$ BUT, we do have an inclusion functor $$K(V) \to K(X);$$ it is right adjoint to the functor $i^{-1}.$ It also preserves covers, so we get an induced functor $$i_!:Sh(K(V)) \to Sh(K(X))$$ by left Kan extension. It is in fact left adjoint to $i^*$ so, from a closed inclusion we get both $i_*$ and $i_!.$
Cech cohomology is automatic, since it makes sense in any topos. This is always true<|endoftext|>
TITLE: Plancherel formula for non-second-countable (non-unimodular) groups
QUESTION [6 upvotes]: The Plancherel formula for unimodular, second-countable, type 1 groups can be found in A Course in Abstract Harmonic Analysis by Gerald Folland (theorem 7.44) or here. It states that we can get a square-integrable function on the group from its Fourier transform by taking traces and integrating with respect to the Plancherel measure (alternatively: we can decompose the regular representation into irreducible representations using the Plancherel measure).
In Representation Theory and Noncommutative Harmonic Analysis by Alexander Kirillov the same theorem is stated but without the restriction “second-countable” (theorem 6.12). It is just a survey book—there is neither a proof nor an explicit reference.
First question: Do you know a reference for this theorem?
Regarding non-unimodular groups: Duflo and Moore proved that the Plancherel formula still works for non-unimodular groups, but you have to introduce some additional unbounded, positive, “semiinvariant” operators to scale the stuff correctly (see this paper). They require the group to be of type 1 (of course) and second countable.
Second question: Is it known whether this works for non-second-countable groups? Is there any point where second-countability is thought to be crucial?
Third question: Kirillov also mentions generalisations to non-type-1 groups. Then it is not enough to consider irreducible representations, but according to him there is a similar statement. Do you know what theorem he means and do you know any reference? These are his words:

Another generalization is possible for groups which are not of type I. In this case, the integral on the right hand side of the formula is computed over the larger space $\tilde{G}$ and the ordinary trace is replaced by the trace in the sense of the corresponding factor.

REPLY [3 votes]: This is an incomplete answer, and the OP seems to have left MathOverflow several years ago, but since the only answer that has been posted thus far gives an incorrect "answer" to the first question, which has not been stricken out in anyway, I would like to post a reference which does answer the first question of the OP --- and perhaps reviving this question will prompt other users who know the literature better than me to point out other sources!
Regarding Q1: contrary to what the earlier answer says (at time of writing), 18.8.1 in Dixmier's book does not address cases where $G$ is not second-countable. (There seems to have been confusion arising from the usage, standard at the time, of "separable" to mean "second countable", as discussed in comments to that earlier answer.) However, in the notes to Chapter 18 we find (English translation):

18.9.2. The Plancherel formula can be generalised to non-separable postliminal unimodular locally compact groups. [452]

where [452] is the paper

J. Dixmier, Traces sur les ${\rm C}^\ast$-algèbres. Ann. Inst. Fourier 13 (1963) 219–262. NUMDAM link

The precise statement can be found as Théorème 3 (Section 16).
As already mentioned, "non-separable" here means "not 2nd countable"; postliminal (postliminaire) means GCR in the sense of Kaplansky. Note that for ${\rm C}^*$-algebras, the equivalence of GCR with Type I (Glimm's theorem) relies on a separability assumption for at least one direction.
A quick look at Dixmier's paper shows that his method is to consider the Hilbert algebra $L^1(G)\cap L^2(G)$ and the induced (Plancherel) weight on the von Neumann algebra generated by this Hilbert algebra; this weight is tracial and its restriction to the group ${\rm C}^*$-algebra $A$ is a densely-defined lower-semicontinuous trace on $A$; he then appeals to earlier results in his paper concerning such traces, where the GCR assumption on $A$ is used to bypass the disintegration-of-von-Neumann-algebras approach of Segal and others (which did require $L^2(G)$ to be separable).
Regarding Q2: The reliance on traces may explain, to some extent, why this approach cannot help with non-unimodular groups. I admit I've not looked up how one bypasses non-unimodularity for 2nd-countable Type I groups; it might be relevant that ${\rm C}^*$-algebras which do not support any densely-defined faithful tracial state can have WOT closures that support faithful normal semifinite traces in the sense of von Neumann algebras. The real $ax+b$ group is an example of such a phenomenon, and the Plancherel theorem for that group can actually be found in the Folland book which the OP mentions.
Regarding Q3: I suspect Kirillov is alluding to decompositions of the left regular representation of $G$ as a direct integral of factor representations, which can be seen as a weakened subtitute for a decomposition as a direct integral of irreducible representations. I think some Lie group cases that are not Type I were worked out by various authors in the 1970s/1980s but I am not familiar with the literature. Folland's book (the one mentioned in the original question) has some commentary on this topic but no proofs.

Coda: I am a bit surprised to see that Dixmier's paper has relatively few citations, even though it seems to have been written at the same time as several other "foundational" papers of his. Perhaps this reflects the fact that only the second-countable case (a.k.a. separable Type I ${\rm C}^*$-algebras) made it into his book, and it is the book which has become a standard reference rather than the series of papers that went into his book.
To be honest: the lack of citations, even in papers which make explicit reference to Plancherel-type theorems for non-2nd-countable groups that are "close to abelian or compact", does make me slightly concerned that there may be problems with Théorème 3 in Dixmier's paper, perhaps known to experts but not recorded in the literature. Of course I hope that this is not the case!<|endoftext|>
TITLE: Vector bundles on Stein manifolds
QUESTION [10 upvotes]: This might be standard if true (if so, I shall be grateful if provided with a reference). Given a smooth map from a Stein manifold $X$ to $\operatorname{Gr}(k,n)$ (the Grassmannian of $k$ planes in $\mathbb{C}^n$), is there a holomorphic map in its homotopy class?

REPLY [11 votes]: As has been established in the comments, the answer to your question is yes. It is a special case of a general result known as the Oka principle which has been strengthened over time. The key is that $\operatorname{Gr}(k, n)$ is an Oka manifold as are all homogeneous spaces for complex Lie groups.

Theorem: Let $X$ be a Stein manifold and $Y$ an Oka manifold. The inclusion
$$\mathcal{O}(X, Y) \hookrightarrow \mathcal{C}(X, Y)$$
is a weak homotopy equivalence. In particular, every continuous map is homotopic to a holomorphic map.

Here $\mathcal{O}(X, Y)$ denotes the collection of holomorphic maps $X \to Y$, $\mathcal{C}(X, Y)$ denotes the collection of continuous maps $X \to Y$, and both spaces are equipped with the compact-open topology.
There are more general versions of the result stated above. In particular, there is an analagous result for holomorphic and continuous sections of stratified holomorphic fiber bundles over Stein manifolds with Oka fibers - the above statement corresponds to the trivial bundle $X\times Y \to X$.
If you are interested in learning more about this story and its history, have a look at Stein Manifolds and Holomorphic Mappings (The Homotopy Principle in Complex Analysis) by Forstnerič, in particular Chapter $5$.

Allow me to explain somewhat carefully (mostly for my own benefit) why the claim "every continuous map is homotopic to a holomorphic map" follows from the first part of the theorem.
A topological space $X$ is compactly generated if it is Hausdorff and $A \subseteq X$ is closed if and only if $A\cap K$ is closed in $K$ for every compact $K \subseteq X$.
Such spaces are ubiquitous. Every locally compact Hausdorff space is compactly generated, in particular topological manifolds. In addition, metric spaces and CW complexes are also compactly generated.
In his paper A Convenient Category of Topological Spaces, Steenrod showed that the category of compactly generated spaces has the properties one would desire. In particular, he proved the following (Theorem $5.4$).

Theorem: Let $W$, $X$, and $Y$ be compactly generated spaces. The map 
\begin{align*}
\mu : \mathcal{C}(X\times W, Y) &\to \mathcal{C}(W, \mathcal{C}(X, Y))\\
f &\mapsto \mu f
\end{align*}
where $(\mu f(w))(x) = f(x, w)$ is a homeomorphism.

Applying the above to $W = [0, 1]$, we see that 
$$\mathcal{C}([0, 1], \mathcal{C}(X, Y)) = \mathcal{C}(X\times [0,1], Y).$$
So a path between $f$ and $g$ in $\mathcal{C}(X, Y)$ gives rise to a homotopy between $f$ and $g$, and vice versa. Therefore, we see that
$$\pi_0(\mathcal{C}(X, Y)) = [X, Y]$$
where the right hand side denotes the collection of homotopy classes of continuous maps $X \to Y$. 
The inclusion $\mathcal{O}(X, Y) \hookrightarrow \mathcal{C}(X, Y)$ induces an isomorphism $\pi_0(\mathcal{O}(X, Y)) \cong \pi_0(\mathcal{C}(X, Y))$. That is, each path-connected component of $\mathcal{C}(X, Y)$ contains a unique path-connected component of $\mathcal{O}(X, Y)$. So for every $f \in \mathcal{C}(X, Y)$, there is $g \in \mathcal{O}(X, Y)$ in the same path-connected component as $f$, and hence homotopic to $f$. Therefore, every continuous map $X \to Y$ is homotopic to a holomorphic one. 
Moreover, if $f$ and $g$ are holomorphic maps which are homotopic, $f$ and $g$ belong to the same path-connected component of $\mathcal{C}(X, Y)$, and hence the same path-connected component of $\mathcal{O}(X, Y)$. Therefore $f$ and $g$ are homotopic through holomorphic maps. That is, there is a homotopy $H : X\times [0, 1] \to Y$ between $f$ and $g$ such that $H(\cdot, t) : X \to Y$ is holomorphic for all $t\in [0, 1]$.

Given the title of your question, it seems you had the Oka-Grauert Theorem in mind. Namely, the fact that on a Stein manifold, the classification of topological complex vector bundles coincides with the classification of holomorphic vector bundles. A proof of this theorem using classifying maps to Grassmannians and the Oka Principle is also discussed in Chapter $5$, see pages $190 - 191$.<|endoftext|>
TITLE: Is there an analogue of spin/oscillator representation for the general linear Lie algebra?
QUESTION [11 upvotes]: (Work over complex numbers)
Let $V$ be an orthogonal space. Let $Pin(V)$ be the double cover of the orthogonal group $O(V)$. Then $Pin(V)$ has a basic spin representation which we can think of as the exterior algebra on a maximal isotropic subspace $V'$ in $V$.
Analogously, let $W$ be a symplectic vector space. Let $sp(W)$ be the symplectic Lie algebra. Then this has a basic "oscillator" representation which we can think of as the symmetric algebra on a maximal isotropic subspace $W'$ in $W$. 
Both constructions are completely analogous and proceed by embedding the Pin group in a Clifford algebra for $V$ in the first case and by embedding $sp(W)$ in the Weyl algebra for $W$ in the second case. These algebras are fundamental objects associated to an orthogonal/symplectic form.
My question is if there is an analogue of the above picture for the general linear group? 
A possibly related point: over the real numbers, the symplectic group has a nontrivial double cover, the metaplectic group, and the oscillator representation can be integrated to this group. Again over the real numbers, the general linear group $GL(n, {\bf R})$ has a nontrivial double cover $\widetilde{GL(n, {\bf R})}$ (what is its name and where can I find basic information about it?). Specific questions:

What is an analogue of the spin/oscillator representation for $gl(n)$?
What is the analogue of the Clifford/Weyl algebra for $gl(n)$? Is it some algebra we build from ${\bf C}^n \oplus ({\bf C}^n)^*$?
Is there a natural "smallest" faithful representation of $\widetilde{GL(n, {\bf R})}$? Does it coincide with the answer to 1.?

REPLY [2 votes]: The Clifford / Weyl algebra duality stems from the symmetric / antisymmetric duality. For me the most natural way to generalize this to $GL(V)$ is kind of vacuous, i.e. the full tensor algebra $\bigotimes V$. Perhaps for $SL(V)$ we could take some $n$-ary operation and quotient out some relation cooked up from the volume form...
There is a defined and studied notion of oscillatory representation for $A$-series algebras but the real form is different, namely $\mathfrak{su}(p,q)$. See "On the Segal-Shale-Weil representations and harmonic polynomials" by Kashiwara and Vergne. This generalization considers the oscillatory representation of $Sp$ as a prominent example of a highest weight unitarizable module. In this framework the spin representation disappears, however the kernel of the Dirac operator in Lorentzian signature is an example of such a representation. All these representations are quite special and much is known about them. For example they all can be realized as kernels of systems of invariant differential operators (see "Differential Operators and Highest Weight Representations" by Davidson, Enright and Stanke) and if I remember correctly,  they are in some sense attached to minimal nilpotent orbits. I guess the reference for this last fact could be "Annihilators and associated varieties of unitary highest weight modules" by Joseph. Finally, unitarizable highest weight modules are examples of so called minimal representations which also exists for other real forms and perhaps some of these minimal representations could serve as generalization of the oscillatory representation to the $GL$ case.<|endoftext|>
TITLE: What is the 31st homotopy group of the 2-sphere?
QUESTION [21 upvotes]: What is $\pi_{31}(S^2)$, the 31st homotopy group of the 2-sphere ?

This question has a physics motivation:

There are relations between (2nd and 3rd) Hopf fibrations and (2- and 3-) qubits (quantum bits) entanglement; see Pinilla and Luthra - Hopf Fibration and Quantum Entanglement in Qubit Systems.

Maybe there are relations between classification of qubits entanglements and sphere homotopy groups, and  we are interested in the classification of 4-qubits entanglements.



I tried fo find the solution on the net, with help of math fans, but without success.
Wikipedia gives only to the 22nd group homotopy of the 2-sphere.
This article of John Baez gives interesting references, like Allen Hatcher, Stable homotopy groups of spheres  or a link with braids (Berrick, Cohen, Wong, and Wu - Configurations, braids, and homotopy groups). One speaks of a book of Kochman Stanley O.: Stable Homotopy Groups of Spheres: A Computer-Assisted Approach.

But I am totally unable to find the answer.
A subsidiary question would be: Until what rank do we know these high homotopy group of the 2-sphere?

REPLY [35 votes]: My apologies for updating this very old question.
As already mentioned, the 31st homotopy group of $S^2$ is the same as the 31st homotopy group of $S^3$. Serre's mod-C theory shows that this is a finite abelian group, and moreover that for any prime $p$ the first $p$-primary torsion occurs in $\pi_{2p} S^3$. This means that we don't have to check any primes $p$ for which $2p > 31$, which leaves us with the primes $p=3,5,7,11,13$.
The main tool for calculating these is the EHP spectral sequence. For $S^3$ the EHP sequence is pretty simple: there is a long exact sequence
$$
\dots \to \pi_{n-1}(S^{2p-1}) \to \pi_{n}(S^3) \to \pi_{n}(S^{2p+1}) \to \pi_{n-2}(S^{2p-1}) \to \dots
$$
that goes through $n > 3$.
The next point is the stable range. At odd primes and for odd spheres, the stable range is larger than that given by the Freudenthal theorem: if $n$ is odd, the stabilization map $\pi_{n+k}(S^n) \to \pi_k^S(S^0)$ is an epimorphism if $k=(n+1)(p-1) - 2$ and an isomorphism if $k < (n+1)(p-1)-2$. This means that all the groups relevant to computing $\pi_{31}(S^3)$ appear in the stable ranges of $\pi_*(S^{2p \pm 1})$ for $p \geq 5$, and those stable ranges are moreover very sparse (they only include the "image of J", which is well-known). If I have calculated correctly, $\pi_{31}(S^3)$ includes no $p$-torsion for $p \geq 5$.
That leaves only $p=3$, where we have to switch to a different algorithm ("I cannot do this, so I better look up what Toda calculated"). Toda showed that the 3-primary part of $\pi_{31}(S^3)$ is $\Bbb Z/3$. I found this in his 2003 text, "Unstable 3-primary Homotopy Groups of Spheres", but that was simply because I had it handier.
Combining this with the information from Mark Grant's answer gives $\Bbb Z/3 \oplus \Bbb Z/2 \oplus \Bbb Z/2 \oplus \Bbb Z/2 \oplus \Bbb Z/2$.<|endoftext|>
TITLE: Reference request: affine transforms + circle inversion?
QUESTION [5 upvotes]: This problem cropped up in the context of scale-insensitive methods for generating random variables.
Let $X=R^n \cup \{\infty\}$.  Suppose we consider a set of transforms $\cal{T}$ from $X\rightarrow X$.  We construct them by concatenating functions chosen from the following set:

Invertible linear transform ($x \mapsto Ax$, for $A\in GL_n(R)$)
Translation ($x \mapsto x+b$)
Circle inversion ($x \mapsto x/|x|^2$; 0 and $\infty$ swap)

The set $\mathcal{T}$ is very similar to the Mobius transformations, which are built from:

Rotation and scaling ($x \mapsto sAx$, for $A\in SO_n(R)$ and $s$ a positive scalar)
Translation ($x \mapsto x+b$)
Circle inversion and reflection ($x \mapsto Mx/|x|^2$, where $M$ reflects through the first coordinate; 0 and $\infty$ swap)

I would like to know if $\cal{T}$ has a standard name, and if any of the properties of the Mobius transformations generalize to $\cal{T}$.  For instance, Mobius transformations in $R^2$ preserve generalized circles; are generalized ellipsoids in $R^n$ preserved by $\mathcal{T}$?  Is there a property analogous to the cross-ratio?  Any references would be greatly appreciated.

REPLY [6 votes]: $\mathcal{T}$ is not a Lie group when $n>1$.  
Actually, the OP did not say whether he wanted $\mathcal{T}$ to be all possible sequences of compositions of these generating sets, but, if he did, then it is clear that $\mathcal{T}$ is not a Lie group, in the sense that it is not defined as the set of solutions of some system of PDE for transformations of $\mathbb{R}^n$.  For one thing, the group that they generate would properly contain the conformal group $\mathrm{O}(n{+}1,1)$ acting on $S^n$, which is known to be a maximal Lie group, i.e., there is no group (in Lie's sense) between the conformal group and the full diffeomorphism group.  (NB: The group of analytic diffeomorphisms of $S^n$ is not a subgroup of the full diffeomorphims in Lie's sense because it is not defined as the set of solutions of some system of PDE.)
In particular, no group $G$ that contains $\mathcal{T}$ can preserve any geometric structures of the kind the OP mentions because this would define a PDE that $G$ satisfies.
(By the way, note that $\mathcal{T}$, as the OP defined it, does not consist of smooth transformations of $S^n$ only when $n>1$, since the non-conformal affine transformations do not extend smoothly to $\infty$ except when $n=1$.)<|endoftext|>
TITLE: Are residually finite, perfect groups residually alternating?
QUESTION [7 upvotes]: Dear all,
I am interested in residually finite, perfect groups. Are all of them known to be residually alternating? If not, how could one construct a counterexample?
A group $G$ is residually alternating if for every $g \in G$ there exists a finite alternating quotient $G/N$ such that $g \notin N$.
By a result by Katz and Magnus free groups are known to be residually alternating. This has recently been extended by Henry Wilton.
Elisabeth

REPLY [13 votes]: I think the question needs to be made a little more precise. One can take a finite simple group $G$ which is not any alternating group. Then  $G$ is perfect, residually finite (!) and is not residually alternating. 
If we ask for finitely generated infinite perfect groups which are not residually alternating, then things are a little more involved: you can take $G=SL_n({\mathbb Z})$ for $n\geq 3$. Then (by the congruence subgroup property for $SL_n({\mathbb Z})$) the only simple quotients are $SL_n({\mathbb Z}/p{\mathbb Z})$ for some prime $p$. These are not alternating groups if $n$ is large enough. [$SL_n({\mathbb Z})$ is perfect, and residually finite, as can be easily seen].

REPLY [7 votes]: I guess you are looking for finitely generated counterexamples. Perfect crystallographic space groups provide one source of examples. These are virtually abelian groups that are extensions of a finitely generated abelian group by a finite perfect group. They are residually finite, and not residually alternating - they are not even residually finite simple.
As a specific example, let $X = A_5$ given by the presentation $\langle a,b \mid a^2=b^3=(ab)^5=1 \rangle$, and let $a$ and $b$ act on ${\mathbb Z}^4$ as in the deleted permutation module. For example, their actions could be given by the integral matrices
$a \to \left(\begin{array}{rrrr}0&0&0&1\\0&1&0&0\\-1&-1&-1&-1\\1&0&0&0\end{array}\right),\ \ \ b \to \left(\begin{array}{rrrr}0&0&1&0\\1&0&0&0\\0&1&0&0\\0&0&0&1\end{array}\right).$
Now let $G$ be the semidirect product of ${\mathbb Z}^4$ with $X$ using this action. Then $G$ itself is not perfect, but its commutator subgroup $G'$ has index 5 in $G$ and is perfect. There are examples like this for all finite perfect groups $X$.

REPLY [5 votes]: If you allow the group to be finite, any non-alternating finite simple group
is a counterexample. Otherwise you can still obtain counterexamples from
wreath products of such groups with the infinite cyclic group.
To give a specific counterexample: let 
$$
  G := (\mathbb{Z},+) \wr {\rm PSL}(2,7) \ = \ (\mathbb{Z},+)^8 \rtimes {\rm PSL}(2,7),
$$
where ${\rm PSL}(2,7) \cong \langle (3,7,5)(4,8,6), (1,2,6)(3,4,8) \rangle$
acts on $(\mathbb{Z},+)^8$ by permuting the factors. Then $G'$ is perfect,
it is residually finite as $(\mathbb{Z},+)$ is so, and it does not admit a surjection
to a nontrivial alternating group.
The example can be constructed in GAP as follows:
gap> LoadPackage("rcwa");
gap> G := WreathProduct(CyclicGroup(IsRcwaGroupOverZ,infinity),PSL(2,7));;
gap> StructureDescription(G);
"Z wr PSL(3,2)"
gap> IsPerfect(G); # not yet (as Derek remarked) ...
false
gap> G := DerivedSubgroup(G);; # ... but now we have our example.
gap> IsPerfect(G);
true
gap> StructureDescription(G);
"(Z x Z x Z x Z x Z x Z x Z) . PSL(3,2)"<|endoftext|>
TITLE: Good effective versions of theorems of Artin and Brauer
QUESTION [7 upvotes]: The theorem of Artin and Brauer of the title are the famous theorem in the theory of representation of finite groups. 
For example, Artin's theorem is the statement that for every character $\chi$ of a finite group $G$, there are a sequence of cyclic subgroups $H_1,\dots,H_r$ (possibly with repetition), one-dimensional characters $\chi_i$ of $H_i$ for $i=1,\dots,r$, signs $\epsilon_i = \pm 1$ for $i=1,\dots,r$ 
and an integer $d \geq 1$, such that $$(1)\ \ \ \ \   \chi = \frac{1}{d} \sum_{i=1}^r \varepsilon_i \  Ind_{H_i}^G \chi_i.$$
Brauer's theorem states similarly that if we weaken the assumption that the $H_i$ are cyclic,
assuming just that they are elementary, then such a writing (1) exists with $d=1$.
I'd like to know if there is a version of these theorems with an explicit control of the complexity of the writing (1) in term of $\chi(1)$ and perhaps of $|G|$. More specifically, if all the $\epsilon_i$ were $+1$, then one one would have $\frac{1}{d} \sum_i [G:H_i] = \chi(1)$ . In general of course, the $\epsilon_i$ can be $+1$ or $-1$, and $\frac{1}{d} \sum_{i=1}^r [G:H_i]$ will be larger that $\chi(1)$ but

Do you know a version a version of Artin or Braueur with an explicit bound on $\frac{1}{d} \sum_{i=1}^r [G:H_i]$ in terms of $\chi(1)$ and $|G|$, or a place where can I find some ? 

In the case of the Brauer's theorem, one has $d=1$, so the question is simply to get a bound on $\sum_{i=1}^r [G:H_i]$. That would also directly give a bound on $r$, the number of subgroups $H_i$ involved, and I would be also interested in a version with such a bound on $r$. In the case of Artin's theorem, there is the further and orthogonal question of finding a  bound on $d$, but that is not what primarily interests me here.
It seems pretty clear to me that the usual proofs (e.g. In Serre's book on representation of finite groups) of Artin and Braueur are effective, so give
an upper bound as asked but a huge one. I am looking for something better, or even the best 
possible bound if it is known.
ADDED: the completely explicit form of Artin's theorem mentioned in Denis's answer is the following:
$$ \chi = \sum_C \alpha_C [G:C]^{-1} Ind_C^G 1,$$
the sum being on cyclic subgroup $C$ of $G$, and 
$$\alpha_C = \sum_{C \subset B} \mu([B:C]) \chi(b)$$
the sum being now and cyclic subgroups $B$ containing $C$ and $b$ being any
generator of $B$. 
This is a nice-looking formula, but the complexity of this formula, in the sense of my question, is big. Indeed, this complexity $\gamma$ is the sum over $C$ (cyclic subgroup of $C$)
of $[G:C]$ times $[G:C]^{-1} |\alpha_C \chi(b)|$, that is this complexity is
$$ \gamma = \sum_C |\alpha_C \chi(b)| \leq \dim \rho \sum_C |\alpha_C|.$$
Now a trivial lower bound for $\sum_C |\alpha_C|$
in the case $G=(\mathbb Z /p\mathbb Z)^n$ is $(p^n-1)/(p-1)$, since this already the value of $\alpha_{\{1\}}$, i.e. the number of subgroup of order $p$ of $G$. This shows that the complexity of this formula is not an order
of magnitude better than $|G| \dim \rho$. For my application this is grossly insufficient. I would expect something in $\dim \rho \log|G|$.

REPLY [5 votes]: Work of Snaith, and of Robert Boltje, on Explicit Brauer induction should be helpful here.
Their results are essentially equivalent, but Boltje shows that there is a unique explicit Brauer induction formula which commutes with restriction, while Snaith obtains a unique explicit form of Brauer's induction theorem which commutes with induction. These explicit forms are different in general. Snaith's work is more topological, which Boltje's work is more algebraic, working with a nice generalization of the Burnside ring (where monomial representations replace permutation representations). However, the formulae are indeed complicated, and may not meet the criteria you impose.
Later edit: Note also that in Artin's induction theorem, each maximal cyclic subgroup of $G$ must occur as one of the $H_{i}$, and J.A. Green showed in around 2005 that similarly in Brauer's theorem, you need to use a conjugate of every maximal Brauer elementary subgroup.
This seems to indicate that the sums you are considering are unavoidably large.<|endoftext|>
TITLE: When is a matrix similar to a non-negative matrix?
QUESTION [13 upvotes]: Consider a real square matrix $A$ of size $n\times n$. Under which conditions on $A$ does there exist a row-stochastic matrix $U$ (non-negative, rowsums = 1), such that $A'=U^{-1}AU$ is a non-negative matrix? In other words, does there exist a row-stochastic matrix $U$ such that the linear system $AU = UA'$ will have a solution with non-negative $A'$? Geometrically, we are asking: When are the columns of the product $AU$ inside the cone generated by the columns of $U$. 
If $U$ exists, how can you determine it? If there is no analytic solution, I'd also be happy with a numerical procedure that converges to $U$. 
Consider the following examples for $n=2$:
a) the trivial case: Let $A$ be already non-negative, then $U$ can be the identity and we are done. If all entries of $A$ are strictly positive we have an infinite number of possible matrices $U$. The cone spanned by the columns of $U$ only needs to contain the columns of $A$.
b) only one solution: consider $A=\left( \begin{array}{cc}
 -1/2 & 1\\\ 
1/4 & 1/2\end{array}\right)$ which can be transformed with $U=\left( \begin{array}{cc}
 1 & 0\\\ 
1/2 & 1/2\end{array}\right)$ into $U^{-1}AU = A'=\left( \begin{array}{cc}
 0 & 1/2\\\ 
1 & 0\end{array}\right)$
c) no solution: consider $A=\left( \begin{array}{cc}
 -1/2 & 1\\\ 
1/4 & 1/8\end{array}\right)$ for which one cannot find an appropriate $U$. 
Thanks for any suggestions!

REPLY [4 votes]: If the column of $AU$ belongs (strictly) inside the space generated by the columns of $U$ then you have a dominant eigenvector inside $U$ (ie an eigenvector whose eigenvalue is real positive and of maximum modulus). It is Perron-Frobenius theorem. 
EDIT: Now given a matrix $A$, a necessary condition is to check that it admits a real positive dominant eigenvalue whose eigenvector is positive. Another necessary condition is that the trace of $A$ is positive (see the comment of @JReichardt below).
In your second example the positive eigenvector whose eigenvalue is positive ($ 0.40\ldots$) is not dominating as the other eigenvalue is $-0.77\ldots$.
V.<|endoftext|>
TITLE: Does the signature admit a homotopy coherent refinement?
QUESTION [14 upvotes]: Cobordism genera can often be refined to $E_\infty$-orientations in the sense of Ando-Blumberg-Gepner-Hopkins-Rezk:
1) the mod 2 Euler characteristic $MO\to H\mathbb{F}_2$;
2) the $\widehat A$-genus $MSpin\to KO$ (Ando-Hopkins-Rezk, Joachim);
3) the Todd genus $MSpin^c\to K$ (Joachim);
4) the Witten genus $MString\to tmf$ (Ando-Hopkins-Rezk).
Now, the signature (and Arf invariant, etc.) lifts to an $\mathbb{L}$-theory orientation for $PL$-bundles (e.g. Ranicki), and $MPL$ and $\mathbb{L}$ are both $E_\infty$-ring spectra.
My question is the following: is it known whether this orientation has an $E_\infty$ refinement? If the answer is yes, I would appreciate a reference.

REPLY [7 votes]: [Since my comment above appears to have been helpful, I am repeating it here.]
I must admit I am unfamiliar with L-theory. Nevertheless, I came across a recent article on the arXiv which is related: Commutativity properties of Quinn spectra by Gerd Laures and James McClure. It states in remark 1.4 that the Sullivan–Ranicki orientation from $MSTop$ to L-theory is a ring map of symmetric ring spectra, which I assume to mean a map of associative or $A_\infty$ monoids. Immediately before that remark, it is also stated that the authors are unaware of any previous result on the multiplicativity of the symmetric signature.<|endoftext|>
TITLE: Hilbert metric and cross-ratio of points on simplices
QUESTION [7 upvotes]: Background and motivation:
Consider the cone $C\subset \mathbb{R}^d$ of vectors with non-negative components, and let $\Delta\subset C$ be the simplex of probability vectors (those for which $\sum v_i = 1$).  The cone (and hence the simplex) can be equipped with the Hilbert metric, which has applications to Perron-Frobenius theory, among other things.
In these applications, the following step is important:  given a $d\times d$ stochastic matrix $A$ with strictly positive entries, one has $A\Delta\subset \Delta$, and one wishes to estimate the diameter of $A\Delta$ in the Hilbert metric.  Using some explicit formulas for the Hilbert metric, it can be shown that
$$
(1)\qquad\qquad \mathrm{diam}(A\Delta) = \sup_{v,w\in\Delta} d(Av,Aw) = \max_{i,j} d(Ae_i,Ae_j),
$$
where $e_i$ are the standard basis vectors.  Geometrically, this can be stated as follows: the diameter of $A\Delta$ is achieved by considering only its extreme points.
The proof of (1) that I know relies on some matrix computations and doesn't feel particularly geometrically informative.  It uses the characterisation of the Hilbert metric in terms of a partial order -- there is also a characterisation of the Hilbert metric in terms of a cross-ratio.  In the present case it boils down to fixing two points $x,y\in C$, letting $w,z$ be the points at which the line through $x,y$ intersects $\partial C$, and setting $d(x,y)$ to be the log of the cross-ratio of $w,x,y,z$.
I wonder if there is a more geometric proof of (1) using this description of $d$.  This motivates the following question, which can be stated without reference to the Hilbert metric but would (1).
Question: (can be read independently of the above)
Let $\Delta,\Delta'$ be simplices of the same dimension and suppose that $\Delta'\subset \Delta$.  Let $\ell$ be a line that intersects $\Delta'$ in an interval.  Let $x,y\in \Delta'$ be the endpoints of this interval, and let $w,z$ be the points where $\ell$ intersects $\partial \Delta$.  Let $\Theta(\ell)$ be the cross-ratio of the points $w,x,y,z$.
Compactness of $\Delta'$ implies that there exists $\ell$ maximising $\Theta$.  (We assume that $\Delta'\cap\partial\Delta=0$ so that $\Theta<\infty$.)  It can be shown that the supremum is attained when $\ell$ is one of the edges of $\Delta'$, but the proof I know is non-geometric.  Is there a geometric proof of this fact?

REPLY [4 votes]: The following is taken from On convex projective manifolds and cusps Adv. Math. 277 (2015), 181–251.
If Ω is properly convex, a function $f:\Omega\to{\mathbb R}$ satisfies the maximum principle if for every compact subset $K\subset\Omega$ the restriction $f|K$ attains its maximum at an extreme point of $K$. 
Corollary 1.9 (Maximum principle). If $C$ is a closed convex set in a properly convex domain $\Omega$, then the distance of a point in $\Omega$ from $C$ satisfies the maximum principle.
The proof is based on (1.8) which uses a symmetry argument.<|endoftext|>
TITLE: Effect of abc conjecture on Fermat's Last Theorem
QUESTION [14 upvotes]: A website ( http://www.math.unicaen.fr/~nitaj/abc.html#Consequences ) says that the $abc$ conjecture implies that there are only finitely many solutions to the equation $x^n+y^n=z^n$ with $\gcd(x,y,z)=1$ and $n\ge 4$. This one I have proven.
Lang's Algebra (p. 196) says that the $abc$ conjecture implies that for all $n$ sufficiently large, there are no solutions to the equation $x^n+y^n=z^n$ with $x,y,z\ne 0$. This one I have not proven (is it true?).
What are the strongest known assertions about Fermat's Last Theorem that follows from the $abc$ conjecture, and how are they proven? I searched the web but people tend to just say vague things like "implies asymptotic version of FLT" and such.

REPLY [23 votes]: There is a slight ambiguity what "the ABC conjecture" is as there are some variation. However, the most common and what you likely mean is this fomulation (or something equialent to it): 
For every $\epsilon >0$, there is a $C_{\epsilon}$ such that: if $a+b=c$, with positive coprime intergers, then 
$$c < C_{\epsilon} \ \text{rad}(abc)^{1 +\epsilon}.$$ 
Now, for the equation $x^n + y^n = z^n$ this means that 
$$
z^n < C_{\epsilon}\ \text{rad}(x^ny^nz^n)^{1 +\epsilon}.
$$
Yet  $\text{rad}(x^ny^nz^n) = \text{rad}(xyz)$ so one actually has
$$
z^n < C_{\epsilon} \ \text{rad}(xyz)^{1 +\epsilon}.
$$
and since $z$ is the largest and since $\text{rad}(a) \le a$ this further means
$$
z^n < C_{\epsilon}  z^{3 + 3\epsilon}.
$$
Now, take $\epsilon =1/4$, say. Then on the one hand this cannot hold for any $z>1$ for $n$ sufficiently large (so no solution for large $n$, what you ask) and also not for any $n \ge 4$ fixed for $z$ sufficiently large (so only finitely many for fixed $n$) or also only finitely many couples $(z,n)$ that fulfill this (for $n \ge 4$).
[Added:] Since the question was changed to remove the gcd condition in Langs's version, I add for completeness, that each solution (for given $n$) implies the existence of a coprime one (for this $n$) so since above establishes there are no coprime solutions for some $n$ then there are none at all. [End Added] 
However, to make these things effective/explicit one would need to know something about  $C_{\epsilon}$ (in dependence of $\epsilon$).   
Regarding "strongest possible": I think this is about what can be said, from the conjecture in the way I stated it. If one assumes stronger conjectures where one would have an explicit dependence of $C_{\epsilon}$ on $\epsilon$ then one could give explicit bounds. But (I think) the argument essentially always passes throught the last displayed equation and checking what this yields.<|endoftext|>
TITLE: Why are Schur multipliers of finite simple groups so small?
QUESTION [36 upvotes]: Given a finite simple group $G$, we can consider the quasisimple extensions $\tilde G$ of $G$, that is to say central extensions which remain perfect.  Some basic group cohomology (based on the standard trick of averaging a cocycle to try to make it into a coboundary) shows that up to isomorphism, there are only finitely many such quasisimple extensions, and they are all quotients of a maximal quasisimple extension, which is known as the universal cover of $G$, and is an extension of $G$ by a finite abelian group known as the Schur multiplier $H^2(G,{\bf C}^\times)$ of $G$ (or maybe it would be slightly more accurate to say that it is the Pontryagian dual of the Schur multiplier, although up to isomorphism the two groups coincide).
On going through the list of finite simple groups it is striking to me how small the Schur multipliers are for all of them; with the exception of the projective special linear groups $A_{n-1}(q)=PSL_n({\bf F}_q)$ and the projective special unitary groups ${}^2 A_{n-1}(q^2) = PSU_n({\bf F}_q)$, all other finite simple groups have Schur multiplier of order no larger than 12, and even the projective special linear and special unitary groups of rank $n-1$ do not have Schur multiplier of size larger than $n$ (other than a finite number of small exceptional cases, but even there the largest Schur multiplier size is 48).  In particular, in all cases the Schur multiplier is much smaller than the order of the group itself (indeed it is always of order $O(\sqrt{\frac{\log|G|}{\log\log|G|}})$).  For comparison, the standard proof of the finiteness of the Schur multiplier (based on showing that every $C^\times$-valued cocycle on $G$ is cohomologous to $|G|^{th}$ roots of unity) only gives the terrible upper bound of $|G|^{|G|}$ for the order of the multiplier.
In the case of finite simple groups of Lie type, one can think of the Schur multiplier as analogous to the notion of a fundamental group of a simple Lie group, which is similarly small (being the quotient of the weight lattice by the root lattice, it is no larger than $4$ in all cases except for the projective special linear group $PSL_n$, where it is of order $n$ at most).  But this doesn't explain why the Schur multipliers for the alternating and sporadic groups are also so small.  Intuitively, this is asserting that it is very difficult to make a non-trivial central extension of a finite simple group.  Is there any known explanation (either heuristic, rigorous, or semi-rigorous) that helps explain why Schur multipliers of finite simple groups are small?  For instance, are there results limiting the size of various group cohomology objects that would support (or at least be very consistent with) the smallness of Schur multipliers?
Ideally I would like an explanation that does not presuppose the classification of finite simple groups.

REPLY [31 votes]: The Schur multiplier $H^2(G;{\mathbb C}^\times) \cong H^3(G;{\mathbb Z})$ of a finite group is a product of its $p$-primary parts
$$H^3(G;{\mathbb Z}) = \oplus_{ p | |G|} H^3(G;{\mathbb Z}_{(p)})$$
as is seen using the transfer. The $p$-primary part $H^3(G;{\mathbb Z}_{(p)})$ depends only of the $p$-local structure in $G$ i.e., the Sylow $p$-subgroup $S$ and information about how the subgroups of $S$ become conjugate or "fused" in $G$. (This data is also called the $p$-fusion system of $G$.)
More precisely, the Cartan-Eilenberg stable elements formula says that 
$$H^3(G;{\mathbb Z}_{(p)}) = \{ x \in H^3(S;{\mathbb Z}_{(p)})^{N_G(S)/C_G(S)} |res^S_V(x) \in H^3(V;{\mathbb Z}_{(p)})^{N_G(V)/C_G(V)}, V < S\}$$
One in fact only needs to check restriction to certain V above. E.g., if S is abelian the formula can be simplified to $H^3(G;{\mathbb Z}_{(p)}) = H^3(S;{\mathbb Z}_{(p)})^{N_G(S)/C_G(S)}$ by an old theorem of Swan. (The superscript means taking invariants.) See e.g. section 10 of my paper linked HERE for some references.
Note that the fact that one only need primes p where G has non-cyclic Sylow $p$-subgroup follows from this formula, since $H^3(C_n;{\mathbb Z}_{(p)}) = 0$.
However, as Geoff Robinson remarks, the group $H^3(S;{\mathbb Z}_{(p)})$ can itself get fairly large as the $p$-rank of $S$ grows. However, $p$-fusion tends to save the day. The heuristics is:
Simple groups have, by virtue of simplicity, complicated $p$-fusion, which by the above formula tends to make $H^3(G;{\mathbb Z}_{(p)})$ small.
i.e., it becomes harder and harder to become invariant (or "stable") in the stable elements formula the more $p$-fusion there is. E.g., consider $M_{22} < M_{23}$ of index 23: $M_{22}$ has Schur multiplier of order 12 (one of the large ones!). However, the additional 2- and 3-fusion in $M_{23}$ makes its Schur multiplier trivial. Likewise $A_6$ has Schur multiplier of order 6, as Geoff alluded to, but the extra 3-fusion in $S_6$ cuts it down to order 2.
OK, as Geoff and others remarked, it is probably going to be hard to get sharp estimates without the classification of finite simple groups. But $p$-fusion may give an idea why its not so crazy to expect that they are "fairly small" compared to what one would expect from just looking at $|G|$...<|endoftext|>
TITLE: Is the site of (smooth) manifolds hypercomplete?
QUESTION [12 upvotes]: By site of manifolds Man, I mean the category of manifolds (maybe submanifolds to obtain a small category) with continuous maps between them. A Grothendieck topology is given by open covers. Actually, I am more interested in the corresponding smooth site but the question may be posed for both.
Daniel Dugger states this (implicitely) in "Sheaves and Homotopy Theory" (http://math.mit.edu/~dspivak/files/cech.pdf) by saying that Cech weak equivalences of presheaves on manifolds can be detected stalkwise. However, this paper is unfinished and the proof is missing. In the paper "Universal Homotopy Theories" he uses the hypercompletion of presheaves on manifolds instead.
In "Differential Cohomology in a cohesive (infinitiy)-topos" (http://ncatlab.org/schreiber/files/cohesivedocumentv032.pdf) Urs Schreiber proves that the subsite consisting of the manifolds R^n is cohesive which implies that it is hypercomplete. However, the proof cannot be generalized since there are non-contractible manifolds.
Since hypercompleteness is a local criterion it suffices to check that the subsites Man|X (overcategory) are hypercomplete. In an attempt to prove this, I found a criterion in HTT saying that an (infinity)-topos which is locally of homotopy dimension <=n is hypercomplete (7.2.1.12) and I hoped to show this for simplicial presheaves on a subsite Man|X (every manifold in such a subsite has the same dimension) using that the (representables of the) contractible open sets generate this (infinity)-topos under colimits. But I failed to identify the corresponding overcategory since this should be the (infinity)-overcategory.

REPLY [11 votes]: I think your idea to reduce the question to small slice topoi works perfectly. I will use it to show that every sheaf on $Man$ (either the continuous or the smooth version) is the limit of its Postnikov tower. This implies that the topos is hypercomplete since truncated objects are hypercomplete.
Let $F$ be a sheaf. Since limits are computed objectwise, we must show that for every $M\in Man$, $F(M)$ is the limit of $\{(a\tau_{\leq n} F)(M)\}$, where $a$ means sheafification and $\tau_{\leq n}$ objectwise truncation. Given $M\in Man$, let $Open(M)$ denote the small site of $M$, and let $i: Open(M) \to Man$ be the forgetful functor. The key is that the restriction functor $i^\ast$ (on presheaves) preserves sheaves and commutes with sheafification; I guess this is intuitively obvious, but as David points out in the comments to the OP we must be careful with intuition here, so I will explain in more details below. Assuming this, we get
$$ (a\tau_{\leq n}F)(M)=(a\tau_{\leq n} i^\ast F)(M).$$
We know that $Sh(M)$ is locally of homotopy dimension $\leq dim(M)$ since objects of $Open(M)$ have covering dimension $\leq dim(M)$, so that $i^\ast (F)$ is the limit of its Postnikov tower in $Sh(M)$. Evaluating at $M$ gives what we wanted.
This argument even shows that "Postnikov towers are convergent" in the topos of sheaves on $Man$, in the sense of HTT (which is stronger than just saying that sheaves are limits of their Postnikov towers, and much stronger than hypercompleteness).
Back to the claim that $i^\ast$ preserves sheaves and commutes with sheafification. Descent is equivalent to Cech descent for open covers so it's clear that $i^\ast$ preserves sheaves. To prove that it commutes with sheafification it suffices to show that $i^\ast(S)\subset T$ for some classes of maps $S$ and $T$ such that sheaves on $Man$ (resp. on $Open(M)$) are the $S$-local (resp. $T$-local) presheaves. We can use for $S$ and $T$ the classes of maps of the form $F\to \tilde F$ where $F$ is a simplicial presheaf and $\tilde F$ is its degreewise sheafification (see Dugger-Hollander-Isaksen, Theorem A.6). That $i^*(S)\subset T$ now follows from the fact that $i^\ast$ commutes with sheafification for sheaves of sets.<|endoftext|>
TITLE: Sequences equidistributed modulo 1
QUESTION [5 upvotes]: Let $\alpha$ be any positive irrational and $\beta$ be any positive real. We have the following results.
H. Weyl (1909): The fractional part of the sequence $\alpha n$ is equidistributed modulo 1.
I. Vinogradov (1935): The fractional part of the sequence $\alpha p_n$ is equidistributed modulo 1 where $p_n$ is the $n$-th prime.
E. Hlawka (1975): The fractional part of the sequence $\beta \gamma_n$ is equidistributed modulo 1 where $\gamma_n$ is the imaginary part of the $n$-th zero the Riemann zeta function.
The common thing in each of the above three celebrated results is that the sequences are of the form $as_n$ where $a$ is a positive real and $s_n$ has the property that the sequence
$$
\frac{s_1}{s_n}, \frac{s_2}{s_n}, \ldots , \frac{s_{n-1}}{s_n}
$$
approaches equidistribution modulo 1 as $n \to \infty$.
Question: I would like a nontrivial counterexample of a positive real $a$ and a sequence $s_n$ such that the fractional part of the sequence $as_n$ is equidistributed modulo 1 but the sequence of the ratios $s_i/s_n$ do not approach equidistribution modulo 1 as $n \to \infty$.  
Edit: I am explaining what I mean by nontrivial because the example given by Noam indicates that it is necessary to explain it explicitly. The examples of Weyl, Vinogradov and Hlawka are nontrivial because there is no assumption on the normality constant. If we take the constant to be normal, we indirectly already assume what we want to prove and so we can construct many artificial examples.

REPLY [3 votes]: Consider all possible pairs $(a,s_n)$ of a positive irrational $a$ and an integer sequence $s_n$  with exponential growth. It seems likely (maybe even easy to show) that it is almost always true that the sequence of fractional parts $[r_n]=\{s_na\}|_{n \in \mathbf{N}}$ is equally distributed in $[0,1).$ For example for any fixed sequence $s_n$ (or even all the sequences $b^n$) the set of appropriate $a$ will have density $1$.
You've already rejected this, but take $s_n=10^{n^{2}}.$ then given a desired sequence $r_n^{'}$ in $[0,1)$ we can easily describe an explicit $a$ such that $|\{s_na\}-r_n^{'}| \lt 10^{1-2n}$ (use the first $2j-1$ digits of $r_j'$ followed by the first $2j+1$ of  $r^{'}_{j+1}$ etc. for $j \ge 1$.)
You will probably be as displeased with this, but I'll mention it anyway: Pick any positive irrational $a$ you wish and any desired sequence $r_n^{'}$ in $[0,1).$ Then we can also pick an integer sequence $s_n$ one term at a time so that for each $n$, $|\{s_na\}-r_n^{'}| \lt 10^{-n}$ and $\frac{s_1}{s_n} \lt \frac{s_2}{s_n} \lt \ldots \lt \frac{s_{n-1}}{s_n} \lt10^{-n}$: Suppose $s_{k-1}$ has been determined and consider in order $s_k=10^ks_{k-1},1+10^ks_{k-1},2+10^ks_{k-1},\cdots  .$ From the theorem of Weyl, you are sure to eventually arrive at a choice   with $\{s_ka\}$ differing from $r^{'}_k$ by no more than $10^{-k}.$<|endoftext|>
TITLE: The relations between the Perelman's entropy functional and notions of entropy from statistical mechanics
QUESTION [9 upvotes]: I am looking for the relations and analogies between the Perelman's entropy functional,$\mathcal{W}(g,f,\tau)=\int_M [\tau(|\nabla f|^2+R)+f-n] (4\pi\tau)^{-\frac{n}{2}}e^{-f}dV$, and notions of entropy from statistical mechanics. Would you please explain it in details?

REPLY [10 votes]: For metrics on $S^{2}$ with positive curvature, Hamilton introduced the
entropy $N\left(  g\right)  =-\int\ln(R\operatorname{Area})Rd\mu.$ If the
initial metric has $R>0,$ he proved that this is nondecreasing under the Ricci
flow on surfaces; note that $Rd\mu$ satisfies $(\frac{\partial}{\partial t}-\Delta
)(Rd\mu)=0.$ Let $T$ be the singular time; then
$$
\frac{d}{dt}N\left(  g\left(  t\right)  \right)  =2\int\left\vert
\operatorname{Ric}+\nabla^{2}f-\frac{1}{2\tau}g\right\vert ^{2}d\mu+4\int%
\frac{\left\vert \operatorname{div}(\operatorname{Ric}+\nabla^{2}f-\frac
{1}{2\tau}g)\right\vert ^{2}}{R}d\mu,
$$
where $\tau=T-t$ and $\Delta f=r-R.$ ($f$ satisfies $\frac{\partial
f}{\partial t}=\Delta f+rf$; since $n=2$, $\operatorname{Ric}=\frac{1}{2}Rg$)
Perelman's entropy has the main term: $\int fe^{-f}d\mu,$ which is the
classical entropy with $u=e^{-f}$ as Deane Yang wrote. (Besides Section 5 of
Perelman, further discussion of entropy appeared later in some of Lei Ni's
papers as well as elsewhere.) Even though this term is lower order (in terms
of derivatives), geometrically it is the most significant as can be seen by
taking the test function to be the characteristic function of a ball
(multiplied by a constant for it to satisfy the constraint); technically, one
chooses a cutoff function. Thus Perelman proved finite time no local
collapsing below any given scale only assuming a local upper bound for $R,$
since the local lower Ricci curvature bound (control of volume growth is
needed to handle the cutoff function) can be removed by passing to the
appropriate smaller scale.
Heuristically (ignoring the cutoff issue), since the constraint is $\int(4\pi\tau)^{-n/2}e^{-f}d\mu=1,$ if
we take $\tau=r^{2}$ and $e^{-f}=c\chi_{B_{r}},$ then $c\approx\frac{r^{n}%
}{\operatorname{Vol}B_r}.$ So, if the time and scale are bounded from above, by Perelman's monotonicity, we have $$-C\leq\mathcal{W}(g,f,r^{2})\lessapprox r^{2}%
\max_{B_{r}}R+\ln\frac{\operatorname{Vol}B_r}{r^{n}},$$ yielding the volume ratio lower bound.
Added December 12, 2013:
For all of the following, see Perelman, Ni, Topping, etal. Let $\mathcal{N}
=\int fe^{-f}d\mu$ be the classical entropy. Then, under $\frac{\partial
}{\partial t}g=-2(\operatorname{Ric}+\nabla^{2}f)$ and $\frac{\partial
f}{\partial t}=-\Delta f-R$, we have $-\frac{d\mathcal{N}}{dt}=\mathcal{F}
(g,f)\doteqdot\int(R+\left\vert \nabla f\right\vert ^{2})e^{-f}d\mu$
(Perelman's energy). If the solutions are on $[0,T)$, then $\mathcal{F}
(t)\leq\frac{n}{2\left(  T-t\right)  }\int e^{-f}d\mu$, which implies that
$\frac{d}{dt}(\mathcal{N}-(\frac{n}{2}\int e^{-f}d\mu)\log(T-t))\geq0$. Let
$\mathcal{W}(g,f,\tau)=(4\pi\tau)^{-n/2}\left(  \tau\mathcal{F}+\mathcal{N}
\right)  -n\int(4\pi\tau)^{-n/2}e^{-f}d\mu$ (Perelman's entropy). Under
$\frac{\partial}{\partial t}g=-2\operatorname{Ric}$ and $\frac{\partial
f}{\partial t}=-\Delta f+|\nabla f|^{2}-R+\frac{n}{2\tau}$, we have have
$\frac{d\mathcal{F}}{dt}=2\int|\operatorname{Ric}+\nabla^{2}f|^{2}e^{-f}
d\mu-\frac{n}{2\tau}\mathcal{F}$ and $\frac{d\mathcal{N}}{dt}=-\mathcal{F}
+\frac{n}{2\tau}\int e^{-f}d\mu-\frac{n}{2\tau}\mathcal{N}$. So, by coupling
with $\frac{d\tau}{dt}=-1$, we obtain (Perelman's entropy formula)
\begin{align*}
(4\pi\tau)^{n/2}\frac{d\mathcal{W}}{dt}  & =\frac{n}{2\tau}\left(
\tau\mathcal{F}+\mathcal{N}\right)  -\mathcal{F}+\tau\frac{d\mathcal{F}}
{dt}+\frac{d\mathcal{N}}{dt}\\
& =2\tau\int|\operatorname{Ric}+\nabla^{2}f-\frac{1}{2\tau}g|^{2}e^{-f}d\mu.
\end{align*}<|endoftext|>
TITLE: Differentiable manifolds by Serge Lang question
QUESTION [8 upvotes]: I have started reading "Introduction to differentiable manifolds" by Serge Lang. In this book, Lang takes a different approach, by immediately introducing manifolds on arbitrary Banach spaces. His approach uses little to no multilinear algebra and he states the following in the foreword: "The orgy of multilinear algebra in standard treatises arises from unnecessary double dualization and an abusive use of the tensor product." What exactly does he mean by this? Is there something inelegant about the traditional treatment of finite dimensional manifolds and differential forms on them?

REPLY [8 votes]: I'm not sure what Lang had in mind with "unnecessary double dualization", but here's an example that occurred to me in ancient times when I was trying to understand differential geometry better.  Some (many? most?) people define tangent vectors to a manifold to be certain derivations on the smooth functions, and they define the cotangent space to be the dual of the tangent space.  So a cotangent vector is a function taking as arguments tangent vectors, which are themselves functions taking as arguments smooth functions.  In this sense, a cotangent vector is a doubly-dualized function.  But one can avoid these dualizations by defining a cotangent vector at a point $p$ to be an element of $m/(m^2)$ where $m$ is the maximal ideal in the ring of germs of smooth functions at $p$.  In other words, $m$ consists of the smooth germs that vanish at $p$ and $m^2$ consists of those that vanish to second order, so the quotient is "first-order data about a germ at $p$, omitting the value (zero-order data) at $p$."  That picture captures pretty well my intuition of what a cotangent vector should be.  (I think that the $m/(m^2)$ definition is used more in algebraic geometry than in differential geometry.)<|endoftext|>
TITLE: Questions about the proof of Stickelberger's theorem on discriminants
QUESTION [11 upvotes]: I was going through the proof of Stickelberger's theorem about discriminants in the book 'Algebraic Number Theory' by Richard A. Mollin, and I am having some problems in understanding the proof. I will state the theorem and the proof, and I will be highly grateful if anyone can answer my questions. I have also asked this question in MSE but have not got any answers.
$\textbf{Theorem :}$ If $K$ is an algebraic number field then $\Delta_K$, the discriminant of $K$, satisfies $$\Delta_K\equiv 0,1\pmod{4}.$$
$\textbf{Proof :}$ Let $\lbrace a_1,\ldots ,a_n\rbrace\subseteq\mathfrak{O}_K$ be an integral basis for $K$ and $\sigma_1,\ldots\sigma_n :K\to \mathbb{C}$ be all the embeddings of $K$. Then we have by definition, $$\sqrt {\Delta_K}=\det([\sigma_i(a_j)])$$ and this can be written as $$\sqrt{\Delta_K}=\sum_{\pi\in A_n}\prod_{i=1}^n\sigma_i\left(a_{\pi (i)}\right)-\sum_{\pi\not\in A_n}\prod_{i=1}^n\sigma_i\left(a_{\pi (i)}\right):=P-N.$$
Now for each embedding $\sigma_i$ we have, $$\sigma_i(P+N)=P+N,\hspace{5mm}  \sigma_i(PN)=PN$$ and hence $P+N$, $PN\in\mathbb{Q}$.
Hence we have $P+N$, $PN\in\mathbb{Z}$, because $P$ and $N$ are both algebraic integers. Now using the identity $$(P-N)^2=(P+N)^2-4PN$$ it follows that $\Delta_K\equiv0,1\pmod{4}.$
$\underline{\textbf{My questions}}:$ 

 How can we apply $\sigma_i$ to $P+N$ and $PN$, I mean how does it follow that $P+N$, $PN\in K$ ?
 Why is $\sigma_i(P+N)=P+N$ and $\sigma_i(PN)=PN$ ?
 From the above how does it follow that $P+N$, $PN\in\mathbb{Q}$ ?

REPLY [5 votes]: This is just a clarification of Dietrich Burde's remarks taken from Schur's original proof. Let $K$ be generated by a root $\beta$ of a monic polynomial $f \in {\mathbb Z}[x]$. Then each $\alpha_i$ is a rational polynomial in $\beta$, and each $\sigma_j$ permutes the roots of $f$; thus we may apply the $\sigma_j$ to $P$ and $N$. Since $P+N$ and $PN$ are symmetric functions of these roots (even permutations fix $P$ and $N$, odd permutations switch them), these expressions are rational algebraic integers, i.e., ordinary integers.<|endoftext|>
TITLE: What is the ring structure of the complex topological K-theory of a non-singular complex quadric?
QUESTION [5 upvotes]: I would like to know the ring structure of $K(Q_n)$ explicitly where $Q_n \subset \mathbb{P}^{n+1}$ is the non-singular $n$-dimensional complex quadric and $K(Q_n) = K^0(Q_n)$ is the complex topological $K$-theory of $Q_n$ (with analytic topology). What is it? Can anyone provide a reference?

Ideally, I would like an expression for $K(Q_n)$ with generators for which I know how to compute the Chern character. Also, as it happens, I am most interested in the case where $n$ is odd.
To my great surprise, I have not been able to find this in the literature (with one qualification, see below). I am certain that such a basic calculation must be well-known to experts but I have been unable to find it. Hence this question.
While I have been aware of the basics of $K$-theory for years, this is the first time I have really had to work with it so I am very inexpert. In the last week I grokked relevant-seeming chunks of the books of Karoubi, Atiyah, Hatcher but I'm still quite green.
Motivation
My interest in this ring arose in the course of a problem I have been thinking about but I am hoping that the fact that $K(Q_n)$ is such a basic object will be sufficient motivation to justify this question.
What I do know

Since the quadric has a cell decomposition with only even-dimensional cells, $K^1$ vanishes and $K = K^0$ is free Abelian with rank equal to the number of cells (and generators corresponding to the cells). For $n$ even this rank is $n+2$ (because there is an extra cell in middle dimension) for $n$ odd, it is $n+1$.
The ordinary cohomology $H^* = H^{\rm even}$ is of course free Abelian of the same rank (Lefshetz tells us the restriction map from $H^*(\mathbb{P}^{n+1}, \mathbb{Z})$ is an isomorphism in all dimensions below $2n+2$ except middle dimension for $n$ even). The Chern character thus embeds $K$ as a maximal-rank lattice inside $H^*(Q_n, \mathbb{Q})$. However it is not the same as the lattice $H^*(Q_n, \mathbb{Z})$.
Since we know $H^*(Q_n, \mathbb{Q})$ as a ring, it might be satisfactory to know the images of the Chern character on a set of generators of $K(Q_n)$.
The cases $n=1, 2, 4$ are easy since $Q_n$ is respectively $S^2$, $S^2\times S^2$, $G(2, 4)$ (the complex Grassmannian).
$Q_n$ is diffeomorphic to the real oriented Grassmannian $\tilde G(2, n+2)$ and so is a homogeneous space $SO(n+2)/SO(2)\times SO(n)$. There are tools for calculating $K$-theory for homogeneous spaces pioneered by Atiyah and Hirzebruch (I believe). Subsequently Hodgkin introduced a spectral sequence which seems to allow relatively straightforward (if lengthy) calculation in many cases, including $Q_n$.
I managed to find a paper where the above technique is apparently used to calculate $K(\tilde G(k, n))$ for general $k, n$: Sankaran, Zvengrowski "K-theory of Oriented Grassmann Manifolds", Math. Slovaca 47(3). It looks right though I would probably be tempted to work from first principles myself than to specialize their results to my $k=2$ case.

Bottom line
Surely I am missing the obvious here? I find it astonishing that I should need to use the methods of Atiyah-Hirzebruch-Hodgkin for such a simple space. Perhaps if I thought more carefully about $\mathbb{P}^{n+1}/Q_n$ or $Q_{n+1}/Q_n$ (bearing in mind natural cell decompositions) then I could use the exact sequences either for the pairs $Q_n \subset \mathbb{P}^{n+1}$ or $Q_{n} \subset Q_{n+1}$ to work this out?

I am tempted to believe the reason I cannot find this in the literature is that it is so trivial. What am I missing?

REPLY [3 votes]: I guess I might as well answer my own question as it might help somebody in the future. I ended up working this out using the methods of Hodgkin. (The Atiyah-Hirzebruch SS leaves one with a series of extension problems and so only gives the Abelian group structure.)
In fact I have written up a careful proof of this in the appendix to this paper: http://arxiv.org/abs/1308.0949 so I will just give the statement here and refer to the paper for the proof (especially as the proof, though fairly routine, is quite long).
Hodgkin's results allow one to compute the K-theory of homogeneous spaces $G/H$ when $\pi_1(G)$ is torsion-free. For this reason we represent the quadric $Q$ as:
$$Q = \frac{Spin(n+2)}{Spin^c(n)}$$
where $Spin^c(n)$ is the double cover of $SO(2)\times SO(n) \subset SO(n+2)$ in $Spin(n+2)$.
Given this, representations of $Spin^c(n)$ give vector bundles on $Q$ and hence classes in $K(Q)$. There is one representation (and hence class in $K(Q)$) which I wish to highlight. To define it, we consider the double cover $Spin(2)\times Spin(n)$ of $Spin^c(n)$. Now if $RSO(2) \simeq \mathbb{Z}[t, t^{-1}]$ is the representation ring of $SO(2)$ then $RSpin(2) \simeq \mathbb{Z}[t^{1/2}, t^{-1/2}]$. Also, for $n$ odd, $Spin(n)$ has the unique spin representation $\delta$ (of dimension $2^{(n-1)/2}$). Neither $t^{-1/2}$ nor $\delta$ descends to a representation of $Spin^c(n)$ but their product does. We let $X$ be the bundle on $Q$ associated to the representation $t^{-1/2}\delta$ of $Spin^c(n)$. With this defined, we can state:
Proposition
Let $Q \subset \mathbb{P}^{n+1}$ be an $n$-dimensional non-singular quadric ($n \ge 3$, odd). Let $L = \mathcal{O}(1)-1 \in K(Q)$ and let $X$ be the bundle defined above, then:

$1, L, L^2, \cdots, L^{n-1}, X$ are a $\mathbb{Z}$-basis for the torsion-free ring $K(Q)$
$L^{n+1} = 0$
$LX = 2^{(n+1)/2} - 2X$
$2^{(n+1)/2}X = 2^n - 2^{n-1}L + \cdots + 2L^{n-1} - L^n$ (this is equivalent to the previous bullet but shows why we need $X$ instead of $L^n$)

There is also a slightly-complicated formula for $X^2$ which I will suppress and a similar statement for $n$ even except that now there are two bundles $X^+$, $X^-$.<|endoftext|>
TITLE: On finite groups with same complex-valued character table
QUESTION [7 upvotes]: What are the necessary and sufficient conditions for two finite groups $G$ and $H$
to have same complex-valued character table?
Is there any criterion for which one could know about the character table similarity of two finite groups without direct computations of each table?
Obviously two isomorphic groups have same character table, but according to the case of $Q_8$ and $D_8$, I'm searching for a weaker criterion.

REPLY [13 votes]: Finite groups have the same complex character tables if and only if their group algebras 
are isomorphic as quasi-Hopf algebras (if and only if the group algebras
are twisted forms of each other as Drinfel'd quasi-bialgebras, if and only if there is non-associative bi-Galois algebra over these groups).
For details see arXiv:math/0001119.<|endoftext|>
TITLE: Algorithm to find exponential map of differential operators acting on function
QUESTION [8 upvotes]: I am trying to write a computer program which computes the action of the exponential of a differential operator on a function, for any given differential operator.
Examples:
$\exp(\varepsilon \partial_x) f(x) = f(x + \varepsilon)$
$\exp(\varepsilon x \partial_x) f(x) = f(x \exp(\varepsilon) )$
$\exp(\varepsilon x^2 \partial_x) f(x) = f\left( \frac{x}{1-\varepsilon x}\right)$
$\exp\Big(\varepsilon (x \partial_y + y \partial_x) \Big) f(x, y) = f\Big(\cosh(\varepsilon) x + \sinh(\varepsilon) y, \sinh(\varepsilon) x + \cosh(\varepsilon) y\Big) $
All these equalities are varifiable by Taylor expanding the variable $\varepsilon$ around zero.
My ideas:

Taylor expand the exponential of differential and try to guess the function yielding the same expansion (think this requires going through differential equations, which my program cannot do).
Recognize Lie algebras equivalent to the differential form, contruct a Lie algebra matrix and exponentiate.

I'm stuck because I do not know precisely how to proceed, except for particular cases in which the Lie algebra identification is obvious.
I also tried to look for papers both on Google and Google Scholar to get something out of it, but I didn't manage to find an explanation for such an algorithm.

Do you know such an algorithm to find the action of such an exponential map on a function?
Do you have any useful references for this?
Is the Lie-algebra/Lie-group approach correct and valid for all types of differential operators?

EDIT:
I identified the case of linear operators, such as
$ \exp\Big( \epsilon ( \sum_{i,j} a_{ij} x_i \partial_{x_j} ) \Big ) f(x_1, \ldots, x_N) $
which would be easily solved by constructing a matrix 
$M_{ij} = a_{ij}$
and then exponentiating it, using well known algorithms for matrix exponentiation, then make the matrix act on the function parameters $x_1, \ldots, x_N$ and substitute the resulting vector as new parameters.
e.g.:
the hyperbolic rotation mentioned earlier:
$\exp\Big(\varepsilon (x \partial_y + y \partial_x) \Big) f(x, y) = f\Big(\cosh(\varepsilon) x + \sinh(\varepsilon) y, \sinh(\varepsilon) x + \cosh(\varepsilon) y\Big) $
has matrix:
$ \begin{pmatrix} 0 & \varepsilon \end{pmatrix}$
$\begin{pmatrix} \varepsilon & 0 \end{pmatrix} $
which results by exponentiation in:
$ \begin{pmatrix} \cosh(\varepsilon) & \sinh(\varepsilon) \end{pmatrix} $
$ \begin{pmatrix} \sinh(\varepsilon) & \cosh(\varepsilon) \end{pmatrix} $
The problem concerns general differential operators, such as
$\exp\Big(\varepsilon x^n \partial_{x}\Big) f(x) $
or maybe even multivariable non-homogeneous differential operators, such as:
$\exp\Big(\varepsilon ( x^2 y^3 \partial_x + x y^5 \partial_y ) \Big) f(x, y) $
How do I find a non-infinite formula for the action on $f(x, y)$?
4) Is the matrix algebra approach a good way?
5) Is an analysis of the structure of the Taylor expansion a good way?
SOLUTION
Given a differential operator $D$, the exponential action $\exp(t \, D) f(x_1,\ldots)$ is given by the partial differential equation:
$\partial_t g(t, x_1, \ldots, x_n) = D [g(t, x_1, \ldots, x_n)] $
$ g(0, x_1,\ldots,x_n) = f(x_1, \ldots, x_n)$
Then $g(1, x_1, \ldots, x_n)$ is the result of the exponential action.

REPLY [3 votes]: A partial answer: What you call the "exponential function" is the so-called flow semigroup, see Engel-Nagel, Section II.3.28.
Another reference on the Lie derivative is  the monograph by Chicone and Swanson.<|endoftext|>
TITLE: von Staudt-Clausen for other special values
QUESTION [6 upvotes]: The von Staudt-Clausen theorem expresses that the Bernoulli numbers' denominators have a very special form (see the wikipedia page on the theorem for more details).
What interests me is that those Bernoulli numbers have a link with the Riemann $\zeta$ function (also discussed in this other question) :
$$\forall n>1, \zeta(1-n)=\frac{-B_n}n$$
(and through the functional equation, likewise to the right -- with an explicit factor with $\pi$ powers)
I interpret this as an example of an arithmetical function with an arithmeticity result for some special values, where a theorem gives results on denominators.
Here comes the question now the big picture is in place : there are also arithmeticity results for special values of $L$ functions of Dirichlet characters. Are there results similar to the von Staudt-Clausen theorem in this case? And for other $L$-functions for which arithmeticity results do exist (say, elliptic modular forms)?
Notice that I checked the proof of the theorem for the Bernoulli numbers, and the one I found (in Hardy&Wright's "Introduction to the theory of numbers") uses the definition of the sequence as coefficients of a series expansion... something which isn't applicable as far as I know in the setting(s) of my question.

REPLY [3 votes]: For an elliptic curve over $\mathbb{Q}$, the analogous question is what is the denominator of the algebraic part of $L(E,\chi,1)$ as $\chi$ varies in the Dirichlet characters. 
First if $\chi$ is trivial, then the precise question is the denominator of $L(E,1)/\Omega^+$ where $\Omega^+$ is the least positive period of the Néron differential. If it is non-zero, then the only odd primes $p$ that can divide this denominator are those for which the Galois module $E[p]$ is reducible. But even those need not arise depending on which elliptic curve in the isogeny class is taken. BSD would tell you what the denominator is; some divisor of the torsion order.
Now for other twists the same is true, only those odd prime can divide the denominator (and very often the values are integers in fact). Since these values are integral sums of modular symbols it all boils down to study the denominator of modular symbols.
For more general modular forms one has to be careful when making the question precise. It depends a lot on what period is chosen and there may not always be a canonical best choice.<|endoftext|>
TITLE: What is an interpretation of the relation in the cohomology of the pure braid groups?
QUESTION [10 upvotes]: In 1968, Arnol'd proved that the integral cohomology of the pure braid group $P_n$ is isomorphic to the exterior algebra generated by the collection of degree-one classes $\omega_{i,j}\ (1 \le i < j \le n)$, subject to the following relation:
$$
\omega_{k,l} \omega_{l,m} + \omega_{l,m}\omega_{m,k} + \omega_{m,k}\omega_{k,l} = 0.
$$
The classes $\omega_{k,l}$ are realizable as differential forms on the pure configuration space of $n$ points in $\mathbb C$ as follows:
$$
\omega_{k,l} = \frac{1}{2 \pi i} \frac{dz_k - dz_l}{z_k - z_l},
$$
from which it can be seen that the classes $\omega_{k,l}$ are computing the winding number of the $k^{th}$ point around the $l^{th}$. In his paper (which can be found here), Arnol'd merely remarks that the above relation can be seen to hold for the forms $\omega_{k,l}$ by direct computation.
My question is, does the above relation have an interpretation in the context of winding numbers? How would I know to find this relation if I were trying to compute the cohomology of the pure braid group on a desert island?

REPLY [11 votes]: You know some kind of relation has to hold, since the configuration space of three points in the plane $C_3 \mathbb R^2$ has the homotopy-type of $(S^1 \vee S^1) \times S^1$, so $H^2$ only has rank $2$.  The homotopy-equivalence comes from noticing the Faddell-Neuwirth fibration $C_1 (\mathbb R^2 \setminus \{0,1\}) \to C_3 \mathbb R^2 \to C_2 \mathbb R^2$ is trivial.  But why a relation of that specific type?  You can follow this line of reasoning to its conclusion and derive the relation from a close inspection of this model, say, using cellular cohomology. Because of the action of the symmetric group, there's basically no other relation possible (modulo a small sign issue). 
Another way to go about it would be to think of cohomology as dual to homology, but for that you need a compact manifold.  So you could compactify the manifold in the Fulton-Macpherson manner.  In this model, $\omega_{i,j}$ is dual to the subspace of the compactified configuration space where $j$ is directly above $i$.  Cup product corresponds to intersection product, so your relation boils down to saying that the homology class where $m$ is directly over $l$ and $l$ is directly over $k$ (or any cyclic permutation of that) is a boundary. You can write that as a boundary of a class -- the idea is to swing the bottom point of the configuration around to be the top point.  
Those are two things that come to mind.  I suspect someone like Dev Sinha or Fred Cohen have cuter ways of thinking of this.<|endoftext|>
TITLE: Proving a variety is not unirational
QUESTION [7 upvotes]: It is known that if a variety is unirational then it is rationally connected. However, there are no known examples of rationally connected varieties which are not unirational.  In these notes, at the top of page 6, it is said that the problem is that showing a variety is not unirational is difficult, and in fact "there are no ways in practice" to do so. 
Since these notes were published in 2001, and since a quick search didn't turn up anything, my question is: has there been any recent progress in this area? That is, are there any new methods for proving a given variety is not unirational? Or are there any recent cases in which a class of varieties is proven to be not unirational?

REPLY [4 votes]: Just so there is an answer: the problem of equivalence / non-equivalence of rational connectedness and unirationality is still open.<|endoftext|>
TITLE: Why are affine Lie algebras called affine?
QUESTION [19 upvotes]: Hi. I was wondering if someone could explain why we call affine Lie algebras affine. Thanks!
Oliver

REPLY [22 votes]: It's not easy to separate out the purely mathematical from the historical question here: What is the mathematical justification for use of the label "affine" and how did this label get attached to certain Lie algebras?   The history in this case would be challenging to sort out, partly because some of the people involved are likely to remember it differently.  Here is my own approach to answering both questions, which is not guaranteed to be exact.  
The study of what are now usually called Kac-Moody algebras began in the mid-1960s with the simultaneous and independent thesis work by Kac (Moscow) and Moody (Toronto).   Though their motivations differed, both of them arrived at a construction of (typically infinite dimensional) Lie algebras using generators and relations analogous to those found earlier in the study of finite dimensional simple Lie algebras over $\mathbb{C}$.   The starting point is a generalized version of the classical Cartan matrix, leading to versions of root systems and Weyl groups as well.   
But it took a while for terminology to settle down.  For intance, Moody preferred at first the term "Euclidean Lie algebra" for what later became known as affine.   This is probably related to a traditional geometric trichotomy: spherical, euclidean, hyperbolic.  (Such terminology is not unreasonable in the study of generalized Cartan matrices: see for instance Cor. 15.11 in Carter's 2005 textbook Lie Algebras of Finite and Affine Type).  However, by the 1970s the GCM's and the Lie algebras themselves were being classified as finite, affine, or indefinite type.   Those of finite type are the classical ones, while the affine Lie algebras involve affine root systems and affine Weyl groups (as pointed out in comments here).   
At first the work of Kac and Moody seemed to me to be a creative but standard sort of generalization popular in dissertations.  But the striking 1972 paper by I.G. Macdonald on affine root systems and the Dedekind $\eta$-function, followed soon by Kac's explanation in terms of the "Weyl-Kac" character formula for an affine Lie algebra, gave the entire subject a much higher profile.  (Connections with mathematical physics provided separate impetus, especially for the study of affine Lie algebras.)   
To extract the term "affine" from all this history is not entirely straightforward.   Certainly the affine Weyl groups were in use decades earlier in connection with compact Lie groups; these were also beginning to surface in the 1960s in the work of Iwahori-Matsumoto on BN-pairs and Bruhat decomposition in Chevalley groups over local fields.    Generalizations of classical root systems were also in the air.   But on balance it seems to me that the classification of generalized Cartan matrices using the above trichotomy provided the immediate rationale for the term "affine Lie algebra", with affine roots and affine Weyl groups not far behind.<|endoftext|>
TITLE: What are the main structure theorems on finitely generated commutative monoids?
QUESTION [38 upvotes]: I should read J. C. Rosales and P. A. García-Sánchez's book Finitely Generated Commutative Monoids and L. Redei's book The Theory of Finitely Generated Commutative Semigroups.  I haven't.   But here's what I've heard so far:

If a commutative monoid is finitely generated it is finitely presented.

Finitely generated commutative monoids have decidable word problems, the isomorphism problem for them is decidable, and indeed the first-order theory of finitely generated commutative monoids is decidable.

If a finitely generated commutative monoid is cancellative ($a + b = a' + b \Rightarrow a = a'$) then it embeds in a finitely generated abelian group.

If a finitely generated commutative monoid is cancellative and torsion-free (for any natural number $n \ge 1,$ $n a = n b \Rightarrow a = b$) then it embeds in a finitely generated free abelian group.  (This follows easily from the previous claim.)

If a commutative monoid is a submonoid of $(\mathbb{N},+,0)$ it is called a numerical monoid and of course it is cancellative.
A lot is known about numerical monoids, though I don't believe they have been "classified" in any useful sense.


If we drop the property of being cancellative we get an enormous wilderness of finitely generated commutative monoids, so there shouldn't be any simple 'classification theorem'.  But there still might be interesting structure theorems which help us understand this wilderness, just as there are for (say) compact topological abelian groups.  What are they?

REPLY [6 votes]: [In a comment to the OP, I asked whether a statement along the lines of Corollary 1 below would count as an "interesting structure theorem", and this post expands on John Baez's yes to that question.]
Let $H$ be a multiplicatively written monoid; unless stated otherwise, $H$ need not be commutative, cancellative, or whatever. An atom of $H$ is a non-unit $a \in H$ such that $a \ne xy$ for all non-units $x, y \in H$. We write $H^\times$ and $\mathscr A(H)$, resp., for the group of units and the set of atoms of $H$.
Denote by $\mathscr F(X)$ the free monoid on a basis $X$; we'll refer to the elements of $\mathscr F(X)$ as $X$-words. Given a non-unit $x \in H$, we define $\mathcal Z_H(x)$ as the set of all $\mathscr A(H)$-words $\mathfrak a$ such that $x = \pi_H(\mathfrak a)$, where $\pi_H$ is the canonical (monoid) epimorphism $\mathscr F(H) \to H$ (see the section "Pills of factorization theory" in this answer if something in the terminology or in the notation doesn't look familiar). The $\mathscr A(H)$-words in $\mathcal Z_H(x)$ are the atomic factorizations of $x$ (relative to $H$).
We write $\preceq_H$ for the "shuffling preorder" induced on $\mathscr F(\mathscr A(H))$ by the divisibility preorder${}^{(1)}$ $\mid_H$ on $H$; roughly speaking, this means that an $\mathscr A(H)$-word $\mathfrak a$ is $\preceq_H$-smaller than an $\mathscr A(H)$-word $\mathfrak b$ if and only if $\mathfrak a$ is a subword of a permutation of $\mathfrak b$ modulo $\mid_H$-equivalence${}^{(2)}$ on the letters.
We call $H$ atomic (resp., an FF-monoid) if $\mathcal Z_H(x)$ is non-empty (resp., finite and non-empty modulo $\preceq_H$-equivalence) for every non-unit $x \in H$; and an FmF-monoid if, for each non-unit $x \in H$, the set of $\preceq_H$-minimal $\mathscr A(H)$-words in $\mathcal Z_H(x)$ is finite and non-empty modulo $\preceq_H$-equivalence.

Examples. (i) If $H$ is the multiplicative monoid of the ring of integers, then $\mathscr A(H)$ is the set of (positive or negative) primes, and the $H$-words $(-2) \ast 3 \ast (-5)$ and $(-3) \ast 5 \ast 2$ are $\preceq_H$-equivalent atomic factorizations of $-30$.
(ii) If $p \in \mathbf N^+$ is a prime, $n$ is an integer $\ge 2$, and $H$ is the multiplicative monoid of the ring of integers modulo $p^n$, then the $H$-word $\bar{p}^{\ast (n+1)}$ is an atomic factorization of $\bar{0}$ but is not $\preceq_H$-minimal (here, $\bar{x}$ is the residue class mod $p^n$ of an integer $x$).

We say that $H$ is finitely generated modulo units (or shortly, an f.g.u. monoid) if there is a finite subset $A \subseteq H$ such that $H \setminus H^\times$ is contained in the subsemigroup generated by the set $H^\times A H^\times$; and is locally finitely generated modulo units (or shortly, an l.f.g.u. monoid) if, for each $a \in H$, the smallest divisor-closed submonoid${}^{(3)}$ of $H$ containing $a$ is an f.g.u. monoid.

Examples. (iii) The monoid $H$ considered in Example (i) is not atomic: $0$ is a non-unit of $H$ with no atomic factorizations. Note, though, that $0$ is an irreducible of $H$ (see the definition before Theorem 2 below), and $H$ has the weaker property that every non-unit is a product of irreducibles.
(iv) The monoid $H$ considered in Example (ii) is FmF, but not FF; in particular, if $k$ an integer $\ge n$ and $u_1, \ldots, u_k \in H$ are units, then the $H$-word $\bar{p}u_1 \ast \cdots \bar{p}u_k$ is an atomic factorization of $\bar{0}$. (Note that two $\mathscr A(H)$-words are $\preceq_H$-equivalent only if they have the same number of letters.)

With these premises in place, we have:

Theorem 1. Every acyclic${}^{(4)}$, l.f.g.u. monoid is FmF.

Most notably, this implies the following:

Corollary 1. If $H$ is a unit-cancellative${}^{(5)}$, commutative monoid and the quotient monoid $H/H^\times$ is finitely generated, then $H$ is FF.

The corollary is essentially due to A. Geroldinger and F. Halter-Koch, cf. Proposition 2.7.8.4 in their monograph:

Non-Unique Factorizations. Algebraic, Combinatorial and Analytic Theory, Pure Appl. Math. 278, Chapman & Hall/CRC, Boca Raton (FL), 2006.

These results are basically a generalization of the existence part of the fundamental theorem of arithmetic and, at least to some extent, bring about the same kind of "structural content" (though atomic factorizations are, in general, far from being unique in any sensible way). In fact, they are part of a bigger picture; and though the OP asked only about finitely generated (commutative) monoids, it is perhaps worth seeing what can be done in a greater generality.
In particular, taking an irreducible of $H$ to be a non-unit $a \in H$ such that $a \ne xy$ for all non-units $x, y \in H$ with $HxH \ne HaH \ne HyH$ leads to:

Theorem 2. If $H$ is a Dedekind-finite${}^{(6)}$ monoid satisfying the ACCP${}^{(7)}$, then every non-unit of $H$ factors as a product of irreducibles.

To my knowledge, a proof of Theorem 2 for commutative rings was first published by D.D. Anderson and S. Valdes-Leon in

Factorization in Commutative Rings with Zero Divisors, Rocky Mountain J. Math. 26 (1996), No. 2, 439-480;

their proof (see loc. cit., Theorem 3.2) carries over verbatim to commutative monoids (and is, in some sense, straightforward from the definitions). In a different direction, we have:

Theorem 3. The conditions below are equivalent:

$H$ is acyclic and satisfies the ACCP.
$H$ is unit-cancellative and satisfies the ACCPR and the ACCPL${}^{(8)}$.

Moreover, each of these conditions implies that $H$ is atomic.

Note that every atom is irreducible, and the converse is true if $H$ is acyclic. Insofar as I'm aware, the terms "atom" and "atomic" were first coined by P.M. Cohn, whom I believe deserves credit also for the statement and the proof of the following (see Question 395980 for discussion on this "controversial point"):

Corollary 2. If $H$ is cancellative and satisfies the ACCPR and the ACCPL, then it is atomic.

There is more to the story. For further details and generalizations, I can only address the interested reader to an article of mine on monoids and preorders (here) and to a couple of related papers with Austin A. Antoniou (here) and Laura Cossu (here).
Notes.
(1) $a \mid_H b$, for some $a, b \in H$, if and only if $b \in HaH$.
(2) Given a preorder $\preceq$ on a set $X$, we say that two elements $x, y \in H$ are $\preceq$-equivalent if $x \preceq y \preceq x$.
(3) A submonoid $K$ of $H$ is divisor-closed if $x \mid_H y$ for some $x \in H$ and $y \in K$ implies that $x \in K$ (among other things, this guarantees that $H$ and $K$ have the same units).
(4) $H$ is acyclic if $xyz \ne y$ for all $x, y, z \in H$ such that either $x$ or $z$ is not a unit.
(5) $H$ is unit-cancellative if $xy \ne x \ne yx$ for all $x, y \in H$ such that $y$ is a non-unit. Obviously, every cancellative monoid is unit-cancellative; moreover, every acyclic monoid is unit-cancellative and the two notions coincide on the level of commutative monoids. On the other hand, Adian's embedding theorem (that is, the special case of Guba's embedding theorem where the left and right graphs of a finite semigroup presentation are cycle-free) implies at once the existence of a finitely generated, cancellative monoid with trivial group of units that is not acyclic.
(6) $H$ is Dedekind-finite if every left- or right-invertible element is a unit (or equivalently, if $xy = 1_H$ for some $x, y \in H$ implies $yx = 1_H$). It is fairly clear that every unit-cancellative or commutative monoid is Dedekind-finite.
(7) $H$ satisfies the ACCP if there is no (infinite) sequence $a_1, a_2, \ldots$ of elements of $H$ such that $Ha_nH \subsetneq Ha_{n+1}H$ for each $n \in \mathbf N^+$ (a set of the form $HaH$ with $a \in H$ is called a principal ideal of $H$).
(8) $H$ satisfies the ACCPR if there is no (infinite) sequence $a_1, a_2, \ldots$ of elements of $H$ such that $a_nH \subsetneq a_{n+1}H$ for each $n \in \mathbf N^+$ (a set of the form $aH$ with $a \in H$ is called a principal right ideal of $H$); and it satisfies the ACCPL if the opposite monoid of $H$ satisfies the ACCPR.<|endoftext|>
TITLE: Is this combination of generalized polygamma and dilogarithm actually zero? $\Im\;\psi^{(-2)}(1+i)+\frac1{4\pi}\text{Li}_2(e^{-2\pi})-\log\sqrt{2\pi}+\frac{5\pi}{24}+\frac12$
QUESTION [20 upvotes]: I encountered this quantity in my calculations and tried to simplify it. Approximate numeric calculations suggested it could be zero (more precisely, it is certainly less than $10^{-4\times10^3}$ in absolute value).
$\hspace{1in}$$\Im\;$$\psi^{(-2)}$$(1+\;$$i$$)+\frac1{4\pi}$$\text{Li}_2$$(e^{-2\pi})-\log\sqrt{2\pi}+\frac{5\pi}{24}+\frac12\stackrel{?}{=}0$.
But I am stuck finding the proof. Could you please help me? I am also curious if this formula could be generalized for other arguments of $\psi^{(-2)}(z)$, and what is the value of the real part $\Re\;\psi^{(-2)}(1+i)$ in terms of simpler functions.

The polygamma function of the negative order  $-2$ can be defined as:
$$\psi^{(-2)}(z)=\int_0^z\log\Gamma(x)\mathrm dx.$$

REPLY [4 votes]: The following identities hold for all real $x>0$.  Your equality is the second identity at $x=1$.
$$
-x + \frac{\pi}{12}(6x^2-1) + x\log{x} - x \log(2\pi) + 2\Im{\psi^{(-2)}(ix)} + \frac{1}{2\pi}\text{Li}_2(e^{-2\pi x}) = 0,\\
x + \frac{\pi}{12}(6x^2-1) - x\log{x} - x \log(2\pi) + 2\Im{\psi^{(-2)}(1+ix)} + \frac{1}{2\pi}\text{Li}_2(e^{-2\pi x}) = 0.
$$
My proofs are series manipulations, which I'll now sketch. To prove the first identity, consider the product expansion for the gamma function:
$$
\Gamma(z) = \frac{e^{-\gamma z}}{z}\prod_{n=1}^\infty\left(1+\frac{z}{n}\right)^{-1}e^{z/n}.
$$
Take the logarithm of both sides; the RHS is now an infinite sum.  Expand the $\log\left(1+\frac{z}{n}\right)$ term as power series about $z=0$, and switch the order of summation.  Now we have:
$$
\log{\Gamma(z)} = -\gamma z - \log{z} + \sum_{k=2}^\infty \frac{(-1)^k}{k}\zeta(k)z^{k}
$$
The polygamma $\psi^{(-2)}(z)$ (by which I mean the Mathematica function $\texttt{PolyGamma[-2,z]}$) is the antiderivative of $\log{\Gamma(z)}$ with $\psi^{(-2)}(0)=0$.  Integrating term-by-term, plugging in $z=ix$, and taking just the imaginary terms, we obtain
$$
\Im{\psi^{(-2)}(ix)} = -x\log{x} + x + \sum_{\ell=1}^\infty \frac{(-1)^\ell}{2\ell(2\ell+1)}\zeta(2\ell) x^{2\ell+1}.
$$
To obtain the appropriate series expansion for $\text{Li}_2(e^{-2\pi x})$, start with its second derivative
$$
\frac{d^2}{dx^2} \frac{1}{2\pi}\text{Li}_2(e^{-2\pi x}) = \frac{2\pi}{e^{2\pi x}-1} = x^{-1} + \sum_{k=1}^\infty \frac{B_k}{k!}(2\pi)^k x^{k-1}.
$$
Here, $B_k$ is the $k^{\rm th}$ Bernoulli number.  After integrating twice, the series we get for $\text{Li}_2(e^{-2\pi x})$ is
$$
\frac{1}{2\pi}\text{Li}_2(e^{-2\pi x}) = \frac{\pi}{12} + x\log{x} + (\log(2\pi)-1)x + \sum_{k=1}^\infty \frac{B_k(2\pi)^k}{k(k+1)k!} x^{k+1}.
$$
The $\frac{\pi}{12}$ term comes from the fact that $\text{Li}_2(1) = \zeta(2) = \frac{\pi^2}{6}$.  Using the relation between Bernoulli numbers and even zeta values, this series may be rewritten
$$
\frac{1}{2\pi}\text{Li}_2(e^{-2\pi x}) = \frac{\pi}{12} + x\log{x} + (\log(2\pi)-1)x - \frac{\pi}{2}x^2 - \sum_{\ell=1}^\infty \frac{(-1)^\ell}{\ell(2\ell+1)} \zeta(2\ell) x^{2\ell+1}.
$$
To obtain my first identity, just add the series for $2\Im{\psi^{(-2)}(ix)}$ and $\frac{1}{2\pi}\text{Li}_2(e^{-2\pi x})$; everything cancels beyond the quadratic term.  The second identity---and identities for every $m + ix$, $m \in \mathbb{Z}$---follow from the first via the formula
$$
\psi^{(-2)}(z+1) = \psi^{(-2)}(z) + \frac{1}{2}\log(2\pi) + z\log{z} - z,
$$
which is itself a corollary of the functional equation $\Gamma(z+1)=z\Gamma(z)$.
Taking more antiderivatives will give us identities involving $\Re\psi^{(-3)},\Im\psi^{(-4)},$ etc., but this isn't particularly interesting now that we know where they come from.  All of joro's identities are now accounted for, I believe.
The real part $\Re\psi^{(-2)}(ix)$ or $\Re\psi^{(-2)}(1+ix)$ picks out the odd, rather than even, zeta values in my series expansion.  Because these numbers, $\gamma,\zeta(3), \zeta(5), \zeta(7), \ldots$, remain so mysterious to number theorists, I expect that $\Re\psi^{(-2)}(1+i)=1.13063...$ does not have an closed-form expression in terms of "simpler" elementary/special function values, appropriately defined. I also expect that a proof of any such statement, even irrationality, is beyond the scope of current knowledge---but I would be glad to be corrected on either point.<|endoftext|>
TITLE: Why do rigid spaces have "not enough points"?
QUESTION [17 upvotes]: In Brian Conrad's notes 
here for the 2007 Arizona winter school, bottom of p18, he says that there is an affinoid rigid-analytic space and a sheaf of abelian groups on it equipped with a non-zero section such that all stalks vanish (at all the "usual" points corresponding to maximal ideals in the affinoid algebra). He uses this to motivate Berkovich spaces etc, and explains why the existence of such a section does not contradict anything (the resulting open cover on which the section vanishes might not be an admissible cover) but does not give an explicit example of such an affinoid/sheaf/section. 
What is an explicit example?

REPLY [10 votes]: If you are familiar with Berkovich spaces, you can do the following construction. Let $X$ be an affinoid space of positive dimension and pick a point $x$ in $X$ that is not a rigid point. Consider the inclusion map $i\colon x \to X$. Then the sheaf $F = i_*\mathbb{Z}$ does the job. Since the space $X$ is Hausdorff, the point $x$ is closed and the sheaf $F$ has no section on the open set $X\setminus \{x\}$.
Let me try to rewrite this example purely in terms of rigid geometry in a simple case: $X$ is the closed unit disc over an algebraically closed field $k$ and $x$ is the point at its boundary (the Gauss point). The previous sheaf may then be described as follows: for an affinoid domain $V$ of $X$, we have $F(V)=0$ if every connected component of $V$ is contained in an open unit disc (with any center) and $F(V)=\mathbb{Z}$ otherwise.<|endoftext|>
TITLE: Is the equivalence between a $\Sigma^0_1$ and a $\Pi^0_1$ formula defining the same recursive set provable in a sufficiently strong arithmetic ?
QUESTION [7 upvotes]: Let $A$ be a recursive set. $A$ is recursively enumerable, so $A$ may be defined by a $\Sigma^0_1$ formula, i.e. by $\exists \overrightarrow{a} \phi (\overrightarrow{a}, n)$, where $\phi$ contains no quantifiers (Matiyasevich's theorem). The complement of $A$ is r.e. so $A$  may be defined by a $\Pi^0_1$ formula $\forall \overrightarrow{a} \psi (\overrightarrow{a}, n)$. Is there a result guaranteeing that the equivalence (true in $\mathbb{N}$)
$\forall n (\exists \overrightarrow{a} \phi (\overrightarrow{a}, n) \leftrightarrow \forall \overrightarrow{a} \psi (\overrightarrow{a}, n))$ is provable, say, in Peano's arithmetic ? 

The reason I ask is the following. Simpson (in Subsystems of second-order Arithmetic), following Friedman, defines a subsystem of second-order arithmetic, $\textbf{RCA}_0$, with, as restricted comprehension axiom scheme, the universal closure of
$\forall n ( \phi(n) \leftrightarrow \psi(n) ) \rightarrow \exists X \forall n (n\in X \leftrightarrow \phi (n) )$
for $\phi$ any $\Sigma^0_1$ formula and $\psi$ any $\Pi^0_1$ formula. So any set this axiom guarantees the existence of will be recursive (or recursive in some other sets if $\phi$ happens to contain free set variables). My problem is that Simpson seems to assume the converse (see eg p. 64) : that this axiom scheme postulates the existence of all recursive sets. But, as far as I can tell, this requires that the equivalence above be provable in his system for appropriate $\phi$ and $\psi$, and I don't see how to prove this.

REPLY [8 votes]: No, in general, a true $\Delta^0_1$ assertion may not necessarily be provably $\Delta^0_1$ in a given theory. For example, assume $\text{Con}(\text{PA})$ is true, and consider the formula $\phi(a)$ asserting that $a=a$ and the formula $\psi(a)$ asserting that "$a$ is not the code of a proof of a contradiction in $\text{PA}$," which is expressible as saying that $a$ does not solve a certain specific diophantine equation. 
Since we assumed there is no such proof, we have that $\exists a\ \psi(a)$ is equivalent to $\forall a\ \psi(a)$, since these are both true sentences. But there can be no proof of this equivalence in $\text{PA}$, if it is consistent, since $\text{PA}$ proves the former sentence, but if it were to prove the latter sentence, it would be proving its own consistency.<|endoftext|>
TITLE: Homological characterization of smooth maps
QUESTION [6 upvotes]: Let $A \to B$ be a finitely generated homomorphism between two commutative noetherian rings.
As far as I understand, in various generalizations of this situation, such a map is called smooth if $B$ is a perfect object in $D(B\otimes_A B)$. (See for example Definition 2.2 of http://arxiv.org/pdf/1006.4721v2.pdf).
Is this true for commutative noetherian rings?
That is, is a finite type map $A\to B$ is smooth if and only if $B$ has a finite projective dimension over $B\otimes_A B$? Note that one direction (that smoothness implies finite projective dimension) is true because of EGA IV Proposition 17.12.4 which says that the kernel of $B\otimes_A B\to B$ is locally generated by a regular sequence. But does the converse holds?
Edit: the answer below shows that surjections, for example, are a counterexample. What if in addition we assume flatness of the map $A\to B$, so that $B\otimes^L_A B \cong B\otimes_A B$?

REPLY [6 votes]: Yes to the second question. More generally, if $f:A \to B$ is a flat homomorphism of noetherian commutative rings such that the flat dimension of $B$ over $B\otimes_A B$ is finite, then $f$ is regular. See Rodicio: Smooth algebras and vanishing of Hochschild homology, Comm. math. Helv. 65 (1990) 474-477. When $f$ is of finite type then you get your answer, since in this case regularity and smoothness are the same thing.<|endoftext|>
TITLE: How can I randomly draw an ensemble of unit vectors that sum to zero?
QUESTION [19 upvotes]: Inspired by this question, I would like to determine the probability that a random knot of 6 unit sticks is a trefoil.  This naturally leads to the following question:
Is there a way to sample uniformly from the set of ensembles of $n$ unit vectors $\{ v_i \}_{i=1}^n$ in $\mathbb{R}^d$ that sum to zero?  I would like some sort of analytic expression for the distribution (something I might be able to prove a theorem with), but also an algorithmic process to implement in code.
UPDATE:  It looks like what I want is basically the Hausdorff measure of an algebraic variety.  Can I use the construction of this measure produce an analytic expression for the distribution?

REPLY [9 votes]: Cantarella and Shonkwiler have now posted a paper on the equilateral random polygon case. Their paper seems to cover a lot of what you're interested in, including a new Markov chain algorithm as well as a new lower bound of 3/4 for the probability that a random equilateral hexagon is un-knotted.
That bound is really a huge underestimate, though -- they report that their Markov chain gives about a probability of $(1.3\pm0.2)\times10^{-4}$ for knottedness.

Older post
Cantarella, Deguchi and Shonkwiler have recently begun investigating various ensembles of random polygons which, while not of the particular form you specified, may still be of interest.  
I can't summarize their approach better than the abstract of their paper:

We build a new probability measure on closed space and plane polygons. The key construction is a map, given by Knutson and Hausmann using the Hopf map on quaternions, from the complex Stiefel manifold of 2-frames in n-space to the space of closed n-gons in 3-space of total length 2. Our probability measure on polygon space is defined by pushing forward Haar measure on the Stiefel manifold by this map. A similar construction yields a probability measure on plane polygons which comes from a real Stiefel manifold.
  The edgelengths of polygons sampled according to our measures obey beta distributions. This makes our polygon measures different from those usually studied, which have Gaussian or fixed edgelengths. One advantage of our measures is that we can explicitly compute expectations and moments for chordlengths and radii of gyration. Another is that direct sampling according to our measures is fast (linear in the number of edges) and easy to code.
Some of our methods will be of independent interest in studying other probability measures on polygon spaces. We define an edge set ensemble (ESE) to be the set of polygons created by rearranging a given set of n edges. A key theorem gives a formula for the average over an ESE of the squared lengths of chords skipping k vertices in terms of k, n, and the edgelengths of the ensemble. This allows one to easily compute expected values of squared chordlengths and radii of gyration for any probability measure on polygon space invariant under rearrangements of edges. 

In a follow-up paper Cantarella, Grosberg, Kusner and Shonkwiler compute the total expected curvature for these random $n$-gons and thus extract bounds on knotting probabilities of hexagons and heptagons.

It turns out there is a relatively large literature on "random equilateral polygons", which is the model you are interested in. I don't think there is any known analytic formula for the probability - you would be looking for the measure of some rather complicated semi-algebraic sets corresponding to the components of the configuration space which correspond to knotted polygons. Thus people have primarily been studying properties of such random knots with Monte Carlo sampling.
Ken Millett is one of the experts on random knots and would be a great person to chat with about these questions. Here's one of his papers titled "The Generation of Random Equilateral Polygons"; which contains a discussion of methods used in the literature following variations of the idea that Joseph O'Rourke described above.
Of course, the hexagonal case has been studied numerically but I couldn't find a precise estimate in my searches. An early paper by Ken Millett which shows it as a data point on a curve of knotting probability versus number of edges can be found here.<|endoftext|>
TITLE: Octonions and the Fano plane.
QUESTION [20 upvotes]: Does the Fano plane mnemonic for octonion multiplication have any deeper meaning?
http://upload.wikimedia.org/wikipedia/commons/2/2d/FanoPlane.svg
The symmetry group of the Fano plane is PSL(2,7), the second-smallest nonabelian simple group. It is also the smallest Hurwitz group, and the group of automorphisms of the Klein quartic.
http://en.wikipedia.org/wiki/PSL(2,7)
I guess I'm wondering if Hurwitz' classification of normed division algebras and Hurwitz' theorem on automorphisms of Riemann surfaces are directly related in some way.
http://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(composition_algebras)
http://en.wikipedia.org/wiki/Hurwitz%27s_automorphisms_theorem
EDIT: Apparently the Fano plane mnemonic is originally due to Freudenthal (1951). Unfortunately I don't read German, nor do I have access to Freudnthal's book to scan for the word "Fano".
http://www.ams.org/mathscinet-getitem?mr=MR13:433a
EDIT: John Baez suggests an idea here (from 2001), but I can't find any follow-up discussion.
http://math.ucr.edu/home/baez/octonions/node4.html

REPLY [5 votes]: Yes. 
As referenced above, G=PSL(2,7) has 168 elements and is the automorphism group of the Klein quartic as well as the symmetry group of the Fano plane.
There are 30 possible ways of arranging 7 objects into 7 triads such that each pair of objects belongs to exactly 1 triad. There are 16 possible 7-bit sign reversals used to generate the 30*16=480 valid Octonions (of which Baez' shows one).  All 480 permutations of Fano plane and cubic are shown here (15MB pdf).
There are 168 permutations of "twisted Octonions" for each of the 30 sets of triads (168*30=5040=7!).
The reference of Baez' to the twisted Octonions are treated in more detail by Chesley here. They have been integrated in my work with Mathematica code based on his c code. It was used to produce the WikiPedia referenced Fano plane and cubic in the posted question. I have also validated the 7! twisted Octonion permutations with it.
The Baez Fano cubic is shown in a pic here (sorry, I can't post pics in MathOverFlow yet). It has yellow nodes highlighting the Real, Complex, Quaternion and Octonion plane. It is one of the few Octonions that have the sign mask (per Chesley) of 00, meaning it is one of the canonical 30 sets of 7 triads. 
My term "flipped" (in the pic referenced above and the .pdf files) refers to the need to rearrange the Fano mnemonic nodes from the algorithms default "flattened" sequence in order to apply the same directed arrow logic as the "non-flipped" Fano planes and cubics. The flattening procedure is simply taking the first occurrence of each number of 1-7. 
The Octonions are known to be associated with the 240 vertices of E8. It is the relationship between the flipped and non-flipped Octonions that provides the 2 to 1 mapping to E8. These are shown integrated in my (as yet) very speculative work on a ToE, shown in a comprehensive (40MB pdf) here.<|endoftext|>
TITLE: Lawvere's fixed point theorem and the Recursion Theorem
QUESTION [18 upvotes]: Building off of Qiaochu's comment on my answer to a previous mathoverflow question, I would like to know: can the Recursion Theorem, $$\forall e\exists k[\Phi_e\text{ is total }\implies \Phi_{\Phi_e(k)}\cong\Phi_k],$$ be gotten as a corollary of Lawvere's Fixed Point Theorem: $$ \text{If $\mathcal{C}$ is Cartesian closed and }f: A\rightarrow B^A\text{ is an epimorphism, then every $g: B\rightarrow B$ has a fixed point.}$$
I tried to work this out myself, but I couldn't seem to come up with the right category to live in; I feel like the proof should be fairly simple, and quite illuminating. Alternatively, if it can't be done (at least in a reasonable way), I'd like to know what the obstacle is.
(I apologize if this question is too low-level for MO; my own background in category theorem is somewhat limited, so I don't know how whether this is actually a research-level question. As partial justification, I would like to point out that from a computability theory perspective, this doesn't seem entirely trivial.)

REPLY [17 votes]: There is indeed a very close connection between Lawvere's fixed-point theorem and Recursion theorem, but one has to look at it the right way. Namely, it all becomes clear once we do it in the effective topos.
Let us start by recalling Lawvere's theorem. (I use $X \to Y$ and $Y^X$ as synonyms for the set of all functions from $X$ to $Y$.)

Theorem (Lawvere): If $e : A \to B^A$ is onto then every $f : B \to B$ has a fixed point.

Proof. There is $x \in A$ such that $e(x)(y) = f(e(y)(y))$ for all $y \in A$, because $e$ is onto and $x \mapsto f(e(y)(y))$ is a map from $A$ to $B$. Then $e(x)(x) = f(e(x)(x))$ so $e(x)(x)$ is a fixed point of $f$. QED.
Now here is Recursion theorem written so that it is most similar to Lawvere's theorem. I explain below why this is the Recursion theorem.

Recursion theorem: Suppose countable choice holds and $e : \mathbb{N} \to B^{\mathbb{N}}$ is onto. Then every total relation $R \subseteq B \times B$ has a fixed point.

We say that $x \in B$ is the fixed point of $R$ if $R(x,x)$. Note that total relations can also be viewed as multivalued maps, so this is a fixed point theorem which generalizes the instance of Lawvere's fixed point theorem in which $A = \mathbb{N}$.
Proof.
Because $R$ is total, for every $n \in \mathbb{N}$ there is $y \in B$ such that $R(e(n)(n), y)$. Therefore, by countable choice, there is a map $c : \mathbb{N} \to B$ such that $R(e(n)(n), c(n))$ for all $n \in \mathbb{N}$. As $e$ is onto there exists $k \in \mathbb{N}$ such that $e(k) = c$. But then $e(k)(k)$ is a fixed point of $R$ because $e(k)(k) = c(k)$. QED.
Of course, you are asking yourself what the theorem has to do with Recursion theorem from computability theory. Note that the proof is intuitionistic and uses countable choice, therefore it is valid in the effective topos. To get the connection with the classical recursion theorem, we need to understand what the object of partial computable maps looks like in the effective topos. In fact, it is just the function space $\mathbb{N} \to \mathbb{N}_\bot$ where I do not really want to get into the internal definition of $\mathbb{N} _{\bot}$, let me describe it as a numbered set instead: the underlying set of $\mathbb{N} _\bot$ is $\mathbb{N} \cup \lbrace \bot \rbrace$. A number $r$ realizes $\bot \in \mathbb{N} _\bot$ if the $r$-th Turing machine diverges on input $0$, and it realizes $n \in \mathbb{N} _\bot$ if the $r$-th Turing machine halts and outputs $n$ on input $0$.
Another way to explain the object of partial computable maps $\mathbb{N} \to \mathbb{N} _\bot$ in the effective topos is that this is the object of those partial maps whose domain is a countable subset of $\mathbb{N}$ (which of course is just the internal version of the classic theorem that partial computable maps have c.e. sets as their domains).
Anyhow, $\mathbb{N} \to \mathbb{N}_\bot$ is countable in the effective topos. This can be proved from the axioms of synthetic computability, but a shortcut is just to observe that there is an effective enumeration of partial computable maps, which realizes an enumeration $\varphi : \mathbb{N} \to (\mathbb{N} \to \mathbb{N} _\bot)$ in the effective topos.. But then, since by $\lambda$-calculus $$(\mathbb{N} \to (\mathbb{N} \to \mathbb{N} _\bot)) \cong (\mathbb{N} \times \mathbb{N} \to \mathbb{N} _\bot) \cong \mathbb{N} \to \mathbb{N} _\bot$$
we see that we may apply Recursion theorem to $\mathbb{N} \to \mathbb{N} _\bot$. So, given any $f : \mathbb{N} \to \mathbb{N}$, consider the total relation $R$ defined on $\mathbb{N} \to \mathbb{N} _\bot$ by
$$R(u,v) \iff \exists k \in \mathbb{N} . u = \varphi_k \land v = \varphi_{f(k)}.$$
There is a fixed point $u$ and so by definition of $R$ there is $k$ such that $u = \varphi_k$ and $u = \varphi_{f(k)}$. And we have the usual recursion theorem as a consequence.
Let's do another one, just to convince you this is the recursion theorem. There is an enumeration $W$ of all countable subsets of $\mathbb{N}$ (yes, there are countably many countable subsets of $\mathbb{N}$ in the effective topos, and that is a way cool axiom if you like to smoke weird stuff). A typical exercise in recursion theorem asks for $n$ such that $W_n = \lbrace{ n \rbrace}$. Because the countable subsets of $\mathbb{N}$ satisfy the condition of recursion theorem, we get such a set simply by considering the total relation $R$ defined by
$$R(S,T) \iff \exists m \in \mathbb{N} . S = W_m \land T = \lbrace m\rbrace.$$
Indeed, a fixed point is a countable set $S$ such that for some $m$ we have $S = W_m$ and $S = \lbrace m \rbrace$.
I could go on, but I am in fact preparing a paper about this which should appear on arXiv in a couple of days. See also my materials on synthetic computability (older material has suboptimal proofs of recursion theorem).<|endoftext|>
TITLE: Question about tetrahedron decomposition
QUESTION [9 upvotes]: Are there tetrahedra which can be subdivided into three non-overlapping parts similar to the original? I believe this would require splitting one face into three parts. I know some types of tetrahedra for which this decomposition is impossible. In 2d, for right triangles you get a decomposition into two similar parts by dropping a perpendicular from the right angle to the hypotenuse, and I would be surprised if there were other 2d solutions.

REPLY [7 votes]: In the case where the three parts are each congruent to one another, the answer to your question is no: there is no such decomposition of a tetrahedron.
The terminology needed to find such an answer in the literature is "reptile" or "$k$-reptile simplices."
Citation for proof:
Safernová, Z.: Perfect tilings of simplices. Bc. degree thesis. Charles University, Prague (2008).
Unfortunately (for many) this thesis is written in Czech.
Fortunately, though, there is a more general paper on this topic, entitled "On the Nonexistence of $k$-reptile Tetrahedra." In particular, see Theorem 1.1 (p. 600, pdf 2/11) for the citation above; alternatively, see page 2 of the arxiv version here.
The citation for this latter paper is:
Matoušek, J., & Safernová, Z. (2011). On the Nonexistence of k-reptile Tetrahedra. Discrete & Computational Geometry, 46(3), 599-609.
If you relax the condition and require the simplices be similar to one another but not necessarily congruent, then the term "irreptile" is sometimes used (at least in the $2D$ case). Sadly, I do not know of any work on $k$-irreptile tetrahedra.<|endoftext|>
TITLE: Connection between subnet and superfilter
QUESTION [5 upvotes]: Let's define a net and subnet in this way:

A net is any function of the form $n:(P,\le)\to X$ where $(P,\le)$ is a (preordered) directed set.
A net $m:(P',\le)\to X$ is a subnet of the net $n:(P,\le)\to X$ iff there is a function:
$$\theta:(P',\le)\to (P,\le)$$
which is increasing:
$$x'\le y' \to \theta(x')\le\theta (y')$$
and cofinal:
$$(\forall p\in P)(\exists p'\in P')(p\le\theta(p'))$$
and $m=n\circ\theta$.
The filter assigned to a net $n:(P\le)\to X$ is the filter generated by
$$\lbrace \lbrace n(x)\mid x\ge p  \rbrace \mid p\in P\rbrace$$
on X. We denote this filter by $\mathcal F_n$,

My question is:
Are there nets $m:(P',\le)\to X$ and $n:(P,\le)\to X$ with
$$\mathcal F_n\subseteq \mathcal F_m$$
such that $m$ is not a subnet of $n$?

REPLY [6 votes]: Yes.  Let Z be the integers, let X consist of a single point and let m : {0} -> X and n : Z -> X be constant functions. Then n and m give the same filter but m cannot be
a subnet of n since no single integer in Z is cofinal.<|endoftext|>
TITLE: Philosophy behind Yitang Zhang's work on the Twin Primes Conjecture
QUESTION [100 upvotes]: Yitang Zhang recently published a new attack on the Twin Primes Conjecture. Quoting Andre Granville :

“The big experts in the field had
  already tried to make this approach
  work,” Granville said. “He’s not a
  known expert, but he succeeded where
  all the experts had failed.”

What new approach did Yitang Zhang try & what did the experts miss in the first place?

REPLY [3 votes]: At a fairly non-detailed level, the Simons Foundation has a nice article on the subject, which I understand has also been picked up by Wired:
https://www.simonsfoundation.org/features/science-news/unheralded-mathematician-bridges-the-prime-gap/
To summarize very briefly, he is building on work by Goldston, Pintz, and Yıldırım.  The article describes his innovation as "... to use not the GPY sieve but a modified version of it, in which the sieve filters not by every number, but only by numbers that have no large prime factors."
A lot of smart people were working very hard on this problem, and surely the overview article makes it sound a lot easier than it actually was!  But the article does make one feel one understands a little of the overview or "philosophy" of the proof.<|endoftext|>
TITLE: Algebraic closure of a polynomial ring
QUESTION [6 upvotes]: What could be conditions on $k\in\mathbb{C}[x,y,z]$ that would ensure that any polynomial $f\in\mathbb{C}[x,y,z]$ that is algebraically dependent of $k$ is indeed a polynomial in $k$, ie $f\in\mathbb{C}[k]$. References?

REPLY [11 votes]: $\def\CC{\mathbb{C}}$A necessary and sufficient condition is that $k$ cannot be written as $h(\ell(x,y,z))$ for $h \in \CC[t]$ of degree $>1$ and $\ell \in \CC[x,y,z]$. Clearly, this is a necessary condition since, if $k = h(\ell)$, then $k$ and $\ell$ are integrally dependent. We now prove sufficiency.
Let $A = \CC[k]$ and let $B$ be the integral closure of $A$ in $\CC[x,y,z]$. I claim that $B \cong \mathbb{C}[\ell]$ for some $\ell \in \mathbb{C}[x,y,z]$. First of all, $B$ is finite over $A$ and so finitely generated and one-dimensional. Also, it is normal. So $\mathrm{Spec}(B)$ is a smooth curve of some genus and some number of punctures. Since $B \subset \CC(x,y,z)$, the curve is unirational which, for curves, is the same thing as rational. So $\mathrm{Spec}(B)$ is genus zero. Also, $B \subset \CC[x,y,z]$ so $B$ has no units other than $\CC^{\times}$ and we see that $\mathrm{Spec}(B)$ has only one puncture. In short, $\mathrm{Spec}(B) \cong \mathbb{A}^1$ and $B \cong \CC[\ell]$.
In particular, we have $k=h(\ell)$ for some $h \in \CC[t]$. If $h$ has degree $\geq 2$, as noted above, than $k$ and $\ell$ are algebraically dependent and $\ell \not \in \CC[k]$.
Conversely, suppose that $h$ has degree $1$, in which case $B = A$. I claim that, if $f$ and $k$ are algebraically dependent, then $f \in \CC[k]$. Proof: If $f$ and $k$ are algebraically dependent then $p(k) \cdot f$ is integral over $\CC[k]$ for some polynomial $p$. So $p(k) f \in B = \CC[k]$ and we deduce that $f \in \mathrm{Frac} \ \CC[k]$. If $f$ is in $\mathrm{Frac}\ \CC[k]$ but not $\CC[k]$, then the ring $\CC[k,f]$ will contain a unit not in $\mathbb{C}^{\times}$, contradicting that $k$ and $f$ are both in $\CC[x,y,z]$.<|endoftext|>
TITLE: The first eigenvalue of the Schrödinger operator is simple.
QUESTION [6 upvotes]: Hello,
let $(M,g)$ be a compact and connected Riemannian manifold (possibly with $\partial M\neq \emptyset$). We consider the Friedrichs extension of $L=-\Delta +V: C^{\infty}(M,\mathbb{R})\subset L^2(M)\rightarrow L^2(M)$ with bounded potential $V:M\rightarrow \mathbb{R}$ (and Dirichlet/Neumann boundary conditions for nonempty boundary). Then the spectrum of L is a discrete sequence $\lambda_0 < \lambda_1 \leq \lambda_2\leq ...$
I'm trying to understand why the first eigenvalue is always simple. I've found the same question here:
First eigenvalue of Schrödinger operator is simple
The answer given there is very helpful, but I still couldn't understand the situation on the whole.
First, the answer remarks the operator $L$ satisfies the maximum principle, i.e. for $f\geq 0$ and $f\neq 0$ there is a unique solution $u$, s.t. $Lu=f$. The solution $u$ is positive. 
My questions are, whether this is true for all $f\in L^2_{+}:=\lbrace f\in L^2(M) : f\geq 0 \text{ }a.e.\rbrace$ and do I need M to be connecet? Where can I find a proof? 
Second, the given answer suggest to use the Krein-Rutman thm. for $L^{-1}$. But the Krein-Rutman version I've found, doesn't state that the largest eigenvalue of $L^{-1}$ must be simple:
Krein-Rutman-thm:
"Let $X$ be a Banach space, $K\subset X$ a total cone and $T\in L(X)$ compact positive with $r(T)>0$. Then $r(T)$ is an eigenvalue with a positive eigenvector." (r(T) is the spectral radius of T)
In our situation we can put $X=L^2(M)$, $K=L^2_{+}$ (?) and $T=L^{-1}$.
How to get the simplicity out of the above Krein-Rutman-thm? 
I will remark something else, maybe it will help you to answer my questions:
On the other hand, I've found a stronger result which provides that the eigenvalue $r(T)$ is simple. But then the cone $K$ must have non empty interior and T must be strongly positive. The problem is that $L^2_{+}$ has empty interior. Therefore I can't use this result for the choice of cone I've made above.
I hope you can help me.
Regards

REPLY [3 votes]: Roughly, the trick is not to view $L$ as an operator on $L^2$, but on $C^0$
I will use the following version of Krein-Rutmann which is proven in "Du, Yihong: Order Structure and Topological Methods in Nonlinear Partial Differential Equations, Vol. 1: Maximum Principles and Applications.":
Let $X$ be a Banach space, $C \subset X$ a solid cone (i.e. a cone with nonempty interior) and $T : X \longrightarrow X$ a compact linear operator which is strongly positive, i.e. $Tu \in C$ if $u \in C$. Then the spectral radius $r(T)$ fulfills $r(T) > 0$ and is a simple eigenvalue admitting an eigenvector $\psi \in \mathrm{int} C$ and there is no other eigenvalue that admits an eigenvector in $C$. Furthermore, all other eigenvalues $\lambda$ fulfill $|\lambda| < r(T)$.
By partial integration, you easily show that the operator is bounded from below, so $L + \alpha$ is strongly positive from $C^2$ to $C^0$ for some $\alpha$ big enough. It is also well-known that the inverse $T := (L + \alpha)^{-1}$ exists and is a compact operator on $C^0$. By the strong maximum principle, $Lu>0$ implies $u>0$, so $T$ is a strongly positive operator.
For the strong maximum principle, you can consult Evans: Partial Differential Equations, for example. To get the statement on a manifold instead of an area in $\mathbb{R}^n$, use a partition of unity.
Now it is easy to show that the set of positive functions is actually a solid cone in $C^0$ (even though it has empty interior in $L^2$), so we can apply the Krein-Rutmann theorem.
By elliptic regularity, every $L^2$ eigenfunction is $C^\infty$ and because the manifold is compact, is bounded, hence in $C^0$. Conversely, every $C^0$ Eigenfunction is in $L^2$, again because the domain is bounded. Hence the eigenvectors and eigenfunctions of $L$ are the same, whether viewed as operator on $L^2$ or on $C^0$<|endoftext|>
TITLE: Yitang Zhang's preprint on Landau-Siegel zeros
QUESTION [45 upvotes]: The recent sensational news on bounded gaps between primes made me wonder: what is the status of Yitang Zhang's earlier arXiv preprint on Landau-Siegel zeros? If this result is correct, then (in my opinion) it is even bigger news for analytic number theory. Has anyone checked this paper carefully?

REPLY [11 votes]: I started a careful study of the paper but stopped after already stumbling over Lemma 2.3. Let me cite from an email in Jan 2008

I can't follow the proof of Lemma 2.3, which is a key
  in proving Lemmas 2.4-2.6 and therefore also Prop. 2.7
  and therefore also the Theorem:
  As far as I understand, the paper estimates (see last line on
  page 8) the Supremum (over the s in Omega_1) of the
  left sum via standard integral-estimation by L^2 times
  the supremum (over the w in R_1) of the respected sum.
  The last equals the sum at a special w in R_1
  (Maximum-principle of continuous functions), but this
  w is (highly) dependent on the psi. Thats why I dont
  understand how one can then use the great sieve as in
  page 9 top in order to estimate the initial sum at the
  left of 2.11, because the great sieve applies only
  when the coefficients are (of course) independent of
  psi.
  Contrary, if the way of proving Lemma 2.3 was
  actually as sketched above, then I dont see it
  necessary to go the extra way over the integral, but
  one could estimate immediately. That's why I think I
  may have missed a point.
  ...
  (I don't think
  that a variation of the statement could help, since
  Lemmas 2.4-2.6 are using very precisely the full
  statement of Lemma 2.3, the same with Prop. 2.7).

Yitang Zhang should at least provide a respective comment at his arxiv-article where/how incomplete the paper is. Frankly put, not being fully transparent about the state of this work and thus have other people spend their time on it (not knowing where it is lacking) is absolutely ridiculous.<|endoftext|>
TITLE: Surfaces ruled over elliptic curves
QUESTION [6 upvotes]: Ground field $\Bbb{C}$. Algebraic category. Elliptic surfaces are those surfaces endowed with a morphism onto some smooth curve, with generic fiber an elliptic curve.
Suppose $E$ is an elliptic curve and consider the ruled surface
$$ S=\frac{E\times\Bbb{P}^1}{G} $$
where $G$ is a group of translations of $E$, acting on $\Bbb{P}^1$. Then $S$ is an elliptic surface, for the projection on the second factor induces a morphism $S\rightarrow\Bbb{P}^1$ whose fibers are elliptic curves ($F=E/G$, in fact).
Now, let $p\colon S\rightarrow C$ be an elliptic surface and suppose $S$ is ruled over an elliptic curve $E$. 
Is $S$ isomorphic to the above example? Any hint for attacking this? Thank you.

REPLY [5 votes]: One more attempt. 
All the fibers of $p\colon S\to C$ dominate $E$, so   they are all isogenous to $E$ and  by Kodaira's classification of elliptic fibers they all have smooth support. Hence $p$ is a so-called "quasi-bundle'' and by a result of Serrano [F.Serrano, 
Isotrivial fibred surfaces
Ann. Mat. Pura Appl. (4) 171 (1996), 63–81]  is of the form $(F\times  B)/G$, where:

$F$ is the elliptic fiber
$B$ is a smooth curve
$G$ is a finite group that acts faithfully on $F$,  on $B$ and diagonally on the product $F\times B$
the action of $G$ on $F\times B$ is free. 

The surface $S$ has  exactly two morphims to a curve, $p_1\colon S\to F/G$ and $p_2\colon S\to B/G$. The general fiber of $p_1$ is $B$ and the general fiber of $p_2$ is $F$.
Since $S$ is ruled over an elliptic curve $E$,  we have $B=\mathbb P^1$ and $F/G=E$. So $G$ acts on $F$ by translation and $S$  is as required.
A final remark: since $G$ acts  faithfully also on $\mathbb P^1$, the possibilities for $G$  are $G=\mathbb Z_2\times \mathbb Z_2$ and $G=\mathbb Z_m$, $m\ge 2$.<|endoftext|>
TITLE: Permutations of $(Z/pZ)^*$
QUESTION [22 upvotes]: Let $p$ be a prime integer, and let $(\mathbb Z/p\mathbb Z)^*$ be the set of non-zero elements of $\mathbb Z/p \mathbb Z$.
Denote by $S((\mathbb Z/p \mathbb Z)^*)$ the group of permutations of $(\mathbb Z/p \mathbb Z)^*$.
Say that a map $a:(\mathbb Z/p \mathbb Z)^*\to S((\mathbb Z/p \mathbb Z)^*)$ satisfies condition (A) if, for any two distinct elements $i,j\in  (\mathbb Z/p \mathbb Z)^*$, $a(i)-a(j)\in S((\mathbb Z/p \mathbb Z)^*)$. 
For example, let $a(i)(k) = ik.$ This satisfies condition (A). The same is true if we permute the functions $a'(i) = a(c(i))$, or relabel the objects $a''(i)(k) = i \cdot b(k)$, or both. Are these modifications of $a(i)(k) = ik$ the only ways to get a map satisfying condition (A)?

If $a$ satisfies (A), are there $b,c\in S((\mathbb Z/p \mathbb Z)^*)$ such that, for all $i\in (\mathbb Z/p \mathbb Z)^*$ and all $k\in (\mathbb Z/p \mathbb Z)^*$, $a(i)(k)=c(i)\cdot b(k)$, where the dot is multiplication in $\mathbb Z/p \mathbb Z$?

Note: it would probably be sufficient to prove that, if $a$ satisfies (A), then, for all $i,j,l\in (\mathbb Z/p \mathbb Z)^*$, $a(i)a(l)^{-1}a(j)=a(j)a(l)^{-1}a(i)$. Or in simpler terms, if $a(1)$ is the identity (one can reduces to this case) then the $a(i)$ commute.
edit I've corrected the question -- and the paragraph before it -- thanks to comments by François Brunault and Victor Protsak, who noted that the original formulation was incorrect due to an irrelevant $b^{-1}$.

REPLY [2 votes]: A similar concept is an orthomorphism of a group $G$. This is an automorphism $\theta: G \rightarrow G$ with the property that $g^{-1}\theta(g)$ is a bijection (equivalently an automorphism). Two orthomorphisms $\theta$, $\eta$ are orthogonal if $\theta^{-1} \eta$ is an orthomorphism.
A set of $k$ orthogonal orthomorphisms correspond to a set of $k$ mutually orthogonal latin squares with specified symmetries. In particular, the examples you give above are the prototypical examples of orthogonal orthomorphisms, and they give a set of $p-1$ MOLS of order $p$. From these one can easily construct the (desarguesian) projective plane of order $p$.
It seems to me that your question relaxes the condition that the orthomorphisms be automorphisms of $G$: you simply want functions. The relation to mutually orthogonal latin squares should still hold however. So you are essentially looking for a non-desarguesian projective plane of order $p$. As far as I know this problem is open, though none are known to exist. (And people have looked.)
Non-desguesian projective planes exist at prime power orders - so I guess that there will be inequivalent sets of functions with the properties you desire there.<|endoftext|>
TITLE: How to cite a sequence from The On-Line Encyclopedia of Integer Sequences (OEIS)?
QUESTION [16 upvotes]: In my paper I want to provide a reference for a sequence (in this case - A001970) from The On-Line Encyclopedia of Integer Sequences (OEIS).
However, I couldn't find an official bib entry for it (there is an unofficial OEIS2BibTeX). Even the respective FAQ/Wiki entry is a bit vague on the issue, providing only a suggestion:

A text reference might say:
The On-Line Encyclopedia of Integer Sequences, published electronically at http://oeis.org, 2010, Sequence A000108
or, if it is clear who "discovered" the sequence, something like
J. H. Conway, Sequence A007970 in The On-Line Encyclopedia of Integer Sequences (2010), published electronically at http://oeis.org.

Moreover, in most of papers I saw people add N. [J. A.] Sloane but drop the year, e.g.:

N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, Sequence A000009.

So, is there a "good practice" for citing OEIS sequences?

REPLY [3 votes]: For those who, like me, just want to copy and paste BibTeX for OEIS which approximates the general suggested reference style, here you go:
@misc{oeis,
    Author = {{OEIS Foundation Inc.}},
    Note = {Published electronically at \url{http://oeis.org}},
    Title = {The {O}n-{L}ine {E}ncyclopedia of {I}nteger {S}equences},
    Year = YYYY}

Of course, fill in YYYY with the current year.
Usage:
\cite[A008277]{oeis}<|endoftext|>
TITLE: Borel constructions, equivariant cohomology, and homotopy quotients of monoid actions.
QUESTION [5 upvotes]: Let $M$ be a (discrete) monoid acting on a space $X.$ We may take the quotient of $X,$ by this action, $X/M$ that is the coequalizer of the action map $M \times X \to X$ against the projection map. $X$ with its action can also be encoded as a functor $$F_X:M \to Top$$ with $F_X(*)=X,$ and $X/M$ is simply the colimit of this functor. One may also take the homotopy colimit of $F_X$ and call this the homotopy quotient of $X$ by its action, $X//_hM$. (Of course, an internal version of this can be used for non-discrete topological monoids, however the discrete case is what I care about right now) By a theorem of Thomason, we have that the homotopy quotient of the $1$-point space by its trivial $M$-action, $pt//_hM$ i.e. the homotopy colimit of the constant functor $$M \to Top$$ with value $pt,$ is $BM$- the classifying space of $M.$ When $M=G$ is in fact a group, this fact is well known and can be proven by replacing $pt$ with $EG$ with its free principle $G$-action and taking the ordinary quotient to obtain $BG$. Moreover, when  $M=G$ is a group, $X//_hG$ can be modelled by a Borel construction.
Question: Is there a model for $X//_hM$ by a Borel construction for monoids? If so, how close to the story for groups is this?
Question: Has the case of homotopy quotients by monoid actions been well studied? (Can someone point to some references if so?) Has, e.g., equivariant cohomology been developed to calculate cohomology of $X//_hM$ in terms of the action $F_X$?

REPLY [5 votes]: This probably depends on your definition of homotopy colimit, but it you mean ``the geometric realisation of the simplicial replacement" then it seems to me that $X /\ \!/_h M$ is homeomorphic to $X \times_M EM$, where $EM$ is the simplicial model (i.e. the nerve of $M \wr M$). The main difference from the case of groups is that $\pi : EM \to BM := *\times_M EM$ is not anything like a bundle.
Homotopy quotients by monoid actions play a central role in "group-completion", for which I would recommend D. McDuff, G. Segal: Homology Fibrations and the "Group-Completion" Theorem, as well as G. Segal: Classifying spaces and spectral sequences.
For example, in the situation you describe there is a spectral sequence
$$Tor_{k[M]}^s(k, H_t(X;k)) \Rightarrow H_{s+t}(X /\ \!/_hM;k)$$
given by filtering $EM$ by skeleta.<|endoftext|>
TITLE: Objects which can't be defined without making choices but which end up independent of the choice
QUESTION [35 upvotes]: It happens a lot of times that when one defines a new object (ring, module, space, group, algebra, morphism, whatever) out of given data, one first chooses some additional structure. And sometimes (often?) the constructed object a posteriori happens to be independent of the choice.
Examples which come to mind are the following:  

The trace of an endomorphism, defined by choosing a basis.
Derived functors like $Ext$, $Tor$, defined by choosing projective/injective resolutions.

Now I have the feeling, that "natural" objects that are not dependent on choices should exist independently of these choices. So they should still exist in a universe where there's no way to choose and so I would expect that existence to be provable there. So I would expect there to be some way to avoid choosing anything in the first place in order to define such objects.
For instance, one can define the trace as the composite map:
 $End_k(V)\cong V^*\otimes V \to k$ where $V^ *\otimes V\to k$ is evaluation and $V^ *\otimes V \to End_k(V)$ is given by $f\otimes v\mapsto (w\mapsto f(w)v)$.
So my questions are the following:
 Do you know examples of things which are natural in some sense but which can't be defined without choosing something first?
 Are $Ext$, $Tor$, etc. examples? i.e. is there a way to define derived functors without choosing resolutions?
 If things like in 1. exist, is there some way to make the statement precise that they can't be defined without choosing anything? Can such results be proven?
 Assuming again things like in 1. exist. Where exactly does the above informal "philosophical" argument fail? What is the deeper reason for the existence (or nonexistence) of such objects?

REPLY [3 votes]: In topology, many maps are only characerized by homotopy properties. Proving that a certain homotopy class exists often requires the construction of an actual element in it, and this element is typically not uniquely determined. Aa advanced jargon phrase is that a ''construction depends on a contractible space of choices''. Here are two examples.
If $P \to X$ is a $G$-principal bundle, a ''classifying map'' $f: X \to BG$ is characterized by the property that $f^{\ast} EG \cong BG$, and this is unique up to homotopy. What one does is to prove that the space of $G$-equivariant maps $g:P \to EG$ is contractible (if $X$ is paracompact). Any such $g$ descends to a classifying map $X \to BG$ (and any classfying map is covered by such a $g$). Therefore, $f$ depends on a contractible space of choices and is therefore unique up to homotopy in the strongest possible sense. The concrete construction depends on data such as local trivializations of the bundle and partitions of unity and is therefore not unique or ''natural'' or ''canonical''.
The other example I want to mention is the Pontrjagin-Thom isomorphism, both of whose direction depend on choices. It states a bijection of the set of bordism classes of framed $n$-manifolds in $R^{m+n}$ and homotopy classes of maps $S^{m+n} \to S^m$.
Only after descending to discrete invariants (homotopy and bordism groups), the construction is well-defined. Passing from a bordism class to a homotopy class requires choosing an embedding of the manifold into a euclidean space and the choice of a tubular neighborhood and the choice of a concrete framing. In the reverse direction, you need a representative of the homotopy class by a smooth map and a regular value.<|endoftext|>
TITLE: How many perfect matchings in a regular bipartite graph?
QUESTION [5 upvotes]: We have a $d$-regular bipartite graph $G = (X,Y,E)$ with $|X| = |Y| = n$ and $|E| = nd$.

What is an upper bound on the number of perfect matchings of $G$?

REPLY [11 votes]: To elaborate on Tony's answer, for a graph $G$ with an even number of vertices and degree sequence $d_1,\dots,d_n$ the number of matchings in $G$ is at most $\prod_{i=1}^n (d_i!)^{\frac{1}{2d_i}}$, with equality achieved only at graphs which are a disjoint union of complete bipartite graphs. The proof is a consequence of the Bregman inequality for permanents. See "The maximum number of perfect matchings in graphs
with a given degree sequence", by Alon and Friedland
Notice that there is a conjecture that something similar holds for the number of matchings of size $t$ where $t$ is not necessarily half the number of vertices. It is conjectured that this is also maximized at disjoint union of complete bipartite graphs, but is open in general (even when we restrict to only bipartite graphs).<|endoftext|>
TITLE: Quotients in Sums of Rings
QUESTION [8 upvotes]: Suppose we are given a commutative ring $R$ with a unit. Suppose that $R$ is the direct product of two rings $R\cong R_1\times R_2$. It's straightforward to show that any ideal $I\subset R$ maps to an ideal $I_1\times I_2\subset R_1\times R_2$ by the above isomorphism. It is, however, not straightforward at all to give a proper description of $I_1$ and $I_2$. 
To be precise, my problem is the following, arising from the book "The Connective K-Theory of Finite Groups" by Bruner and Greenlees.
Let $C_n$ be the cyclic group with $n$ elements; then in odd degrees $2i-1$ we have 
$ku_{2i-1}(BC_n)\cong \left(\mathbb{Z}[\alpha]/(1+\alpha+\ldots+\alpha^{n-1})\right)/(1-\alpha)^i$
I am especially interested in the case where $n=p^k$ for $p$ an odd prime and $k\geq2$. By the Chinese remainder theorem we have an isomorphism 
$$\mathbb{Z}[\alpha]/(1+\alpha+\ldots+\alpha^{p^k-1})\cong\prod\limits_{\begin{array}{c}d\mid p^k\\ d>1\end{array}}\mathbb{Z}[\alpha]/\Phi_d(\alpha)$$
where $\Phi_d(\alpha)$ is the $d^{th}$ cyclotomic polynomial.
If I don't make any silly mistakes the isomorphism should be given by mapping the residue class of $\alpha$ to the tuple which has the residue class of $\alpha$ at every single entry. However, if I take $p=2$ and $k=2$ the first component of $ku_{4j+2s+1}(BC_4)$ would result to be $\mathbb{Z}/p^j$ instead of $\mathbb{Z}/p^{j+1}$.
It would be great if anyone could give me hint or a reference how to solve this issue.

REPLY [6 votes]: Pardon me for not editing my previous answer, but this is really a different answer, not an improvement on my previous answer, which addresses different aspects of the (neighborhood of) the question.
The fact that $Z[x]/(fg) \longrightarrow Z[x]/(f) \times Z[x]/(g)$ is not iso here makes the attempt to replace $Z[x]/(1+x+\cdots+x^{n-1})$ by $\prod_d Z[x]/(\Phi_d(x))$ fail!
Take the simplest case, n=4. Let $R = Z[x]/(1+x+x^2+x^3)$, $R_1=Z[x]/(1+x)$ and $R_2 = Z[x]/(1+x^2)$.  The map $1-x : R \longrightarrow R$ and the map  $1-x : R_1 \times R_2 \longrightarrow R_1 \times R_2$ do not have the same cokernel!  The cokernel of the former is Z/4, but the coker of the latter is $Z/2 \times Z/2$ !  The cokernel of $R \longrightarrow R_1 \times R_2$ is $Z/2$, but the map induced on this coker by 1-x is trivial.  The snake lemma's ker-coker sequence is amusing here, and if I knew how to make mathJax display commutative diagrams, I'd have used fewer words to say all this.
This shows that Smith Normal Form is not invariant under monomorphisms!  Lovely example.<|endoftext|>
TITLE: Are small knots generic?
QUESTION [10 upvotes]: A knot in S^3 is small if its complement does not contain a closed incompressible surface. Is it a generic property for knots, meaning that among all knots with less than $n$ crossings, the proportion of small knots goes to 1 when $n$ goes to infinity?

REPLY [3 votes]: Following up on the first comment by Misha: Your question is very sensitive to the model you choose.  I vaguely remember a talk by Ken Millet (http://math.ucsb.edu/~millett/) in which he gave a natural model of knot generation where the generic knot seemed to be a connect sum of $O(n)$ copies of the trefoil.   If you tweaked the model, then the generic knot was the unknot.  
And to reply to another comment above: one could condition on the knot being hyperbolic.  Then the model is more interesting to analyze.  However such a model is "unusable in practice" -- you can't actually generate knots this way because the waiting time is too long.<|endoftext|>
TITLE: Chern Character Isomorphism for non-finite CW complexes, resp. for non-CW complexes
QUESTION [6 upvotes]: This is a question I asked at Math.SE but got no answers: https://math.stackexchange.com/q/397164/7110/
Atiyah and Hirzebruch showed in their paper "Vector bundles and homogeneous spaces" that $\mathrm{K}^\ast(X) \otimes \mathbb{Q} \cong \mathrm{H}^\ast(X; \mathbb{Q})$, where $\mathrm{H}^\ast$ denotes singular cohomology, if $X$ is a finite CW complex.
Somewhere on the Internet I saw the statement $\mathrm{K}^\ast(X) \otimes \mathbb{Q} \cong \check{\mathrm{H}} {}^\ast(X; \mathbb{Q})$, where $\check{\mathrm{H}} {}^\ast$ denotes Cech cohomology, if $X$ is a compact Hausdorff space.

First, I'm looking for a reference for this fact $\mathrm{K}^\ast(X) \otimes \mathbb{Q} \cong \check{\mathrm{H}} {}^\ast(X; \mathbb{Q})$.
Second, can the statements be extended to non-compact spaces, i.e., do we have something like $\mathrm{K}^\ast_{\text{cpt}}(X) \otimes \mathbb{Q} \cong \mathrm{H}_{\text{cpt}}^\ast(X; \mathbb{Q})$ for (locally finite) CW complexes? Or something analogous for the Cech cohomology and locally compact Hausdorff spaces?

REPLY [2 votes]: I have found a reference:

"M. Karoubi, Les isomorphismes de Chern et de Thom-Gysin en K-theorie, Seminaire Henri Cartan, vol. 16, no. 2, Expose no. 16, 1963-1964".

There the isomorphism $\mathrm{K}^0(X) \otimes \mathbb{Q} \cong \mathrm{\check{H}}^{ev}(X; \mathbb{Q})$ is shown for every compact Hausdorff space $X$. But the Theorem 2 from there should also be applicable to $\mathrm{K}^\ast(X) \otimes \mathbb{Q}$ and $\check{\mathrm{H}}^\ast(X; \mathbb{Q})$.
For locally compact Hausdorff spaces $X$ one can indeed just consider the one-point compactification $X^+$ and apply the above isomorphism to the pair $(X^+, \infty)$.<|endoftext|>
TITLE: How much of character theory can be done without Schur's lemma or the Artin-Wedderburn theorem?
QUESTION [42 upvotes]: This is a somewhat imprecise question, as I am not sure how exactly how to formalise how to do mathematics "without" a certain key tool, but hopefully the intent of the question will still be clear.
Let $G$ be a finite group.  Traditionally, a character $\chi: G \to {\bf C}$ on $G$ is defined as being the trace of a finite-dimensional unitary representation $\rho: G \to U(V)$ of $G$, and then representation-theoretic tools (including Schur's lemma) can then be used to derive the basic results of character theory, including the following assertions:

The irreducible characters form an orthonormal basis of the space $L^2(G)^G$ of class functions, and all other characters are natural number combinations of the irreducible characters.
The space of characters form a semiring with identity; in particular, for three irreducible characters $\chi_1,\chi_2,\chi_3$, the structure constants $\langle \chi_1 \chi_2, \chi_3 \rangle$ are natural numbers.
For any character $\chi$, $\chi(1)$ is a positive integer.  
For any character $\chi$, $\chi(g^{-1})=\overline{\chi(g)}$ for all $g$.
For any irreducible character $\chi$, the convolution operation $f \mapsto f * \chi(1) \chi$ is a minimal idempotent in $L^2(G)$, and one has the Fourier inversion formula $f = \sum_\chi f * \chi(1) \chi$ for all class functions $f$.  Furthermore, the image $I_\chi$ of the convolution operation $f \mapsto f * \chi(1) \chi$ in $L^2(G)$ (or ${\bf C} G$) is an irreducible $G \times G$ representation.
If $\chi$ is an irreducible character of $G$, and $\eta$ is an irreducible character of a subgroup $H$ of $G$, then the structure constant $\langle \chi, \operatorname{Ind}^G_H \eta \rangle_{L^2(G)^G} = \langle \operatorname{Res}^H_G \chi, \eta \rangle_{L^2(H)^H}$ is a natural number.

Note that representation theory does not make an explicit appearance in the above character theory facts (induction and restriction of characters can be done at a purely character-theoretic level without explicit reference to representations), other than as one of the conclusions to Fact 5.
Note also that one can define characters directly, without mention of representations.  The space $L^2(G)^G$ of class functions is a finite-dimensional commutative algebra under the convolution operation, and one can then locate the minimal idempotents $f$ of this algebra; $f(1)$ will then be positive, so one can define an "irreducible" character $\chi$ associated to this idempotent by the formula $f = \chi(1) \chi$ with $\chi(1) := f(1)^{1/2}$ being the positive square root of $f(1)$.  One can then define an arbitrary character to be a natural number combination of the irreducible characters.  This definition of character automatically gives Facts 1, 4, and 5 above (the $G \times G$-irreducibility of $I_\chi$ can be obtained by computing that the dimension of $\operatorname{Hom}^{G \times G}(I_\chi,I_\chi)$ is one), and if one is allowed to use Schur's lemma (or the Artin-Wedderburn theorem), one can also show that this definition is equivalent to the usual representation-theoretic definition of a character which then gives the remaining Facts 2, 3, and 6.
My (imprecise) question is whether one can still recover Facts 2, 3, and 6 from this non-representation-theoretic definition of a character if one is "not allowed" to use Schur's lemma or the Artin-Wedderburn theorem.  Now, I do not know how to rigorously formalise the concept of not being allowed to use a particular mathematical result; my first attempt was to phrase the problem with the complex numbers replaced by the field of definition of the characters $\chi$ (or the cyclotomic field of order $|G|$), so that the underlying field is not algebraically closed and so Schur's lemma or Artin-Wedderburn do not directly apply.  But this is hardly any constraint at all since one can immediately pass to the algebraic closure of these fields in order to bring Schur or Artin-Wedderburn back into play.  Another option is to take a constructivist (or maybe reverse mathematics) point of view, and only permit one's mathematical reasoning to work with representations as long as they can be constructed from characters using explicit, "functorial", representation-theoretic constructions (e.g. tensor sum, tensor product, orthogonal complement, isotypic component, Schur functors, induction, or restriction) from concrete representations (e.g. trivial representation, regular representation, or quasiregular representation), but prohibit any argument that requires one to make "arbitrary" or "non-functorial" choices on representations (and in particular, to split an isotypic representation or module into irreducibles).  However, I do not know how to formalise this sort of mathematical reasoning (perhaps one has to introduce a suitable topos?).  It has also been suggested to me (by Allen Knutson) that perhaps the correct setting for this framework is that of quantum groups over roots of unity rather than classical groups, but I am not familiar enough with quantum groups to formalise this suggestion.
Leaving aside the question of how to properly formalise the question, one can make some intriguing partial progress towards Facts 2, 3, 6 without invoking Schur or Artin-Wedderburn.  The isotypic representation $I_\chi$ has character $\chi(1) \chi$ by construction, so in particular this shows that $\chi(1)^2$ is a positive integer which is in some sense the "square" of Fact 3.  By considering the $\chi_3\otimes \overline{\chi_3}$-isotypic component of $I_{\chi_1} \otimes I_{\chi_2}$, viewed as $G \times G$ representations, one can similarly show that the square $|\langle \chi_1 \chi_2, \chi_3 \rangle|^2$ of the structure constant $\langle \chi_1 \chi_2, \chi_3 \rangle$ is a natural number, and by viewing these representations instead as $G$-representations one also gets that $\chi_1(1)\chi_2(1) \chi_3(1) \langle \chi_1 \chi_2, \chi_3 \rangle$ is a natural number.  These two facts don't quite establish Fact 2, but they do at least show that Fact 3 implies Fact 2.  Finally, for Fact 6, the Frobenius reciprocity  $\langle \chi, \operatorname{Ind}^G_H \eta \rangle_{L^2(G)^G} = \langle \operatorname{Res}^H_G \chi, \eta \rangle_{L^2(H)^H}$ is an easy algebraic identity if one defines induction and restriction in character-theoretic terms, and the same sort of arguments as before show that $|\langle \chi, \operatorname{Ind}^G_H \eta \rangle_{L^2(G)^G}|^2$ and $\chi(1) \eta(1) \langle \chi, \operatorname{Ind}^G_H \eta \rangle_{L^2(G)^G}$ are natural numbers, so again Fact 3 will imply Fact 6.  (Conversely, it is not difficult to deduce Fact 3 from either Fact 2 or Fact 6.)
So it all seems to boil down to Fact 3, or equivalently that the dimension of any minimal ideal of $L^2(G)$ (or ${\mathbf C} G$, if you prefer) is a perfect square.  This is immediate from the Artin-Wedderburn theorem, and also follows easily from Schur's lemma (applied to an irreducible representation of this ideal) but I have been unable to demonstrate this fact without such a representation-theoretic (or module-theoretic) tool.  
Characters do give a partial substitute for Schur's lemma, namely that the dimension of the space $\operatorname{Hom}^G(V,W)$ of $G$-morphisms between two representations $V,W$ is equal to $\langle \chi_V, \chi_W \rangle_{L^2(G)^G}$, where $\chi_V, \chi_W$ are the characters associated to $V, W$.  This gives Schur's lemma when the characters $\chi_V, \chi_W$ are irreducible in the sense defined above (defined through minimal idempotents and square roots rather than through irreducible representations).  At the level of $G \times G$-representations, the isotypic components $I_\chi$ are irreducible (in both the character-theoretic and representation-theoretic senses) and so one has a satisfactory theory at this level (which is what is giving the "square" of Facts 2, 3, and 6) but at the level of $G$-representations there are these annoying multiplicities of $\chi(1)$ which do not seem to be removable without the ability to reduce to subrepresentations for which Schur's lemma applies.
Somehow the enemy is a sort of phantom scenario in which a group-like object $G$ has an irreducible (but somehow not directly observable) "ghost representation" of an irrational dimension such as (say) $\sqrt{24}$, creating an isotypic component $I_\chi$ whose dimension ($24$, in this case) is not a perfect square.  This is clearly an absurd situation, but one which appears consistent with the weakened version of representation theory discussed above, in which Schur's lemma, the Artin-Wedderburn theorem, or "non-constructive" representations are not available.  But I do not know if this is a genuine limitation to this sort of theory (e.g. can one cook up a quantum group with such irrational representations?), or whether I am simply missing some clever argument.
(My motivation for this question, by the way, is to explore substitutes for character-theoretic or representation-theoretic methods in finite group theory, for instance to find alternate proofs of Frobenius's theorem on Frobenius groups, which currently relies crucially on Fact 6.)

REPLY [23 votes]: I spent a long time writing an answer to this question, but MO did not believe I was a human being ( I did mis-spell one of the test words, but everyone deserves a second chance, I think ), so it seems to have disappeared. I am not sure I have the energy to do it again right now, but here (in precis, though not precise) are three points I thought worth making/suggesting:

The group determinant has been mentioned: in a 1991 Proc AMS paper, Formanek and Sibley proved that the group determinant determines the group. Perhaps you could use the analogue of the group determinant to tease out the properties that an algebraic structure $G$, whose "formal character theory" satisfies the properties you can get without Schur and Wedderburn, would have. Such a "group determinant" would not a priori have the property that the multiplicity of an irreducible factor equals its degree.
It is possible to get quite far into the structure of the group algebra just using the symmetric algebra structure of the complex group algebra of $G$ induced by the linear form $t$ with $t(\sum_{g \in G} a_{g}g ) = a_{1}.$ Since $t$ vanishes on nilpotent elements, it follows that no non-zero right ideal of $\mathbb{C}G$ consists of nilpotent elements and that for each minimal (two-sided) ideal $A$ of $\mathbb{C}G$, $Z(A)$ is $1$-dimensional.
You might argue that this is using representation theory, since it is not a priori immediately obvious that $t$ vanishes on nilpotent elements until one notes that (up to the multiple $|G|$ ) $t$ is the trace afforded by the regular representation.
Frobenius's theorem, and other normal complement theorems of a similar nature suggest 
that there might be an analogue (under certain hypotheses) of the transfer homomorphism
but when the target group is not necessarily Abelian. That theorem can be proved in the case that $H$ is solvable by using the usual transfer homomorphism, and what the theorem says in the end (in the general case)  is that the identity homomorphism from $H$ to itself extends to a homomorphism from $G$ onto $H.$ If one tries a transfer-type proof, it looks as though it almost would work, except that the order of products matters when the target group is non-Abelian. Nevertheless, the Theorem does in the end say that the homomorphism you would like to define by "transfer" is well-defined after all.<|endoftext|>
TITLE: Smith Normal Form of powers of a matrix
QUESTION [17 upvotes]: What invariants of a matrix determine the Smith Normal Form (SNF) of all the powers of a matrix? 
The question makes sense over any PID $R$.   If we let $M = M_n(R)$ and $G=Gl_n(R)$, then SNF is a parameterization
of the double coset space $G\backslash M/G$ and I am asking about the image of a sequence $(y^i : i \geq 1)$ under the projection $M \longrightarrow G \backslash M/G$.  Clearly this factors through the quotient of $M$ under conjugation by $G$.
It is easy to produce examples to show that the characteristic polynomial is insufficient, even for 2x2 matrices.
The reason I care is that, in computing local cohomology groups for graded one dimensional rings, one often comes across a ring $S$, free as a module over a PID $R$, and an element $y \in S$ for which one wants to know all the quotients $S/(y^i)$ explicitly as an $R$-module.  
If we let $A_i$ be the cokernel of $y^i$, then we have short exact sequences 
$ 0 \longrightarrow A_i \longrightarrow A_{i+j} \longrightarrow A_j \longrightarrow 0$ relating these quotients (at least when $det(y) \neq 0$).  
I conjecture that for any $y \in M$ there exist an integer $d > 0$ and a diagonal matrix $D$ such that $SNF(y^{i+d}) = D*SNF(y^i).$
The work on the possible values of SNF(AB), given SNF(A) and SNF(B), masterfully recounted in Fulton's "Eigenvalues, invariant factors, highest weights, and Schubert calculus",
Bull. AMS 37 (2000), no. 3, 209–249, is probably relevant, though $SNF(A^i)$ is in some sense the worst case, since $SNF(AB)$ is most constrained when $det(A)$ and $det(B)$ share few factors.  If I understood that work, perhaps I would know that the answer is already known.
I should note that nothing crucial depends on this question:  I am simply curious.

REPLY [4 votes]: EDIT: My argument about indecomposability is complete nonsense. Indecomposable modules of this type need not be monogenic.
We can split the question into a question local at each prime of $R$, because one can recover a finitely-generated module from its localizations. So we can work over a complete DVR. Over a complete DVR, we can sometimes decompose a matrix, up to conjugacy, into two blocks separated by $0$s, so we can reduce to the indecomposable case.<|endoftext|>
TITLE: Constructing Polynomial Count Varieties
QUESTION [7 upvotes]: I have some naive questions about polynomial-count affine varieties over $\mathbb{C}$:

Are all reductive algebraic groups strongly polynomial-count?
Are products of strongly polynomial-count varieties also strongly polynomial-count?  What about (disjoint) unions?
If X is strongly polynomial-count variety, and F is a finite group acting on X, is X//F also polynomial-count?  More generally, is the property of strongly polynomial-count invariant under étale equivalence.  
If G is reductive algebraic group acting on a variety X, and the orbit-type stratification of X consists of strongly polynomial count quasi-projective subvarieties, then is X//G also strongly polynomial-count?   
Are there general conditions on a variety X and algebraic group G for X//G to be strongly polynomial count?

Basically, I would like to know if there are operations that allows one to cook up polynomial count varieties from other polynomial count varieties.
See the Appendix here for the definition of polynomial count:  here
EDIT:  Just to be completely honest, when I posted this, I did have a sense for some of the questions, but I wanted to learn more about a concept I am just now learning to work with, and felt that they all fit a general theme so just included them all without detailing what I thought I knew and didn't.

REPLY [7 votes]: A combination of easy and hard questions here. The easy ones:
(1) No. For example, the group scheme $\{ (x,y) : x^2+y^2=1 \}$, with multiplication $(x_1, y_1) (x_2, y_2) = (x_1 x_2 - y_1 y_2, x_1 y_2 + x_2 y_1)$ has $q - (-1)^{(q-1)/2}$ points over a field with $q$ elements. Or, similarly, the group scheme $x^3=1$ has $1$ or $3$ points depending on whether $q$ is $1$ or $2$ mod $3$. However, connected split reductive groups are polynomial point count. Moreover, we can make an arbitrary reductive group be polynomial strongly point count by changing the base ring $R$: For example, switching from $\mathbb{Z}$ to $\mathbb{Z}[i]$ in the first example or from $\mathbb{Z}$ to $\mathbb{Z}[(-1+\sqrt{-3})/2]$ in the second.
(2) Yes and yes, this is immediate from the definition.
(3) Polynomial point count is not invariant under etale equivalence, because it is easy to build examples where $X$ is polynomial point count and $X/G$ is not. For example, take $X$ to be the elliptic curve $y^2=x^3-x$ with the points $(x,y) = (-1,0)$, $(0,0)$ and $(0,1)$ deleted. $X$ is not polynomial point count. Let $G = \mathbb{Z}/2$ act on $X$ by $(x,y) \mapsto (x,-y)$. Then the quotient is the affine line with $3$ points deleted, with point count $q-3$.
It is harder to find an example the other way, with $X$ polynomial point count and $X/G$ not polynomial point count. Here is one. Let $E \subset \mathbb{P}^2$ be a smooth cubic curve. Take two copies of $\mathbb{P}^2$, each with $E$ embedded in them, and glue them along $E$. Then take the disjoint union of that with another copy of $E$. Call this $\bar{X}$. So $\bar{X}$ has $2(q^2+q+1) - \#E(\mathbb{F}_q) + \#E(\mathbb{F}_q)= 2(q^2+q+1)$ points over $\mathbb{F}_q$.  Let $X$ be $\bar{X}$ with four points deleted from the copy of $E$ which is disjoint from the projective planes. So $X$ has point count $2(q^2+q+1)-4$, another polynomial. Let $\mathbb{Z}/2$ act trivially on the $2$-dimensional component of $X$ and act by a fixed point free involution on $E$ with the $4$ points deleted. So the quotient of the one dimension component is $\mathbb{P}^1$ minus $4$ points, and the point count of the whole quotient is $2 (q^2+q+1) - \#E(\mathbb{F}_q) + (q-4)$.
This mostly demonstrates that polynomial point count is a weird notion; a better condition is that all the eigenvalues of Frobenius on compactly supported cohomology be powers of $q$. I believe that this does pass to quotients as I think (but I am not an expert) that $H^{\ast}(X/G, A) = H^{\ast}(X,A)^G$ whenever $|G|$ is invertible in the coefficient ring $A$.
(4) seems harder to me.<|endoftext|>
TITLE: Is deciding whether a Turing machine *provably* runs forever equivalent to the halting problem?
QUESTION [25 upvotes]: Assume for this question that ZF set theory is sound.
Now consider the language "PROVELOOP," which consists of all descriptions of Turing machines M, for which there exists a ZF proof that M runs forever on a blank input.
It's clear that PROVELOOP is recursively-enumerable, and hence reducible to the halting problem.  I can also prove that PROVELOOP is undecidable (details below).  But I can't see how to prove that PROVELOOP is Turing-equivalent to the halting problem!  (This is contrast to, say, the set of all descriptions of Turing machines that provably halt, which is just the same thing as the set of all descriptions of Turing machines that do halt!)
I'm guessing that there's a reduction from HALT that I haven't thought of, though it would be exciting if PROVELOOP were to have intermediate degree like the Friedberg-Muchnik languages.  In any case, whatever the answer, I assume it must be known!  Hence this question.

Proof that PROVELOOP is undecidable.  Consider the following problem, which I'll call "Consistent Guessing" (CG).  You're given as input a description of a Turing machine M.  If M accepts given a blank input, then you need to accept, while if M rejects you need to reject.  If M runs forever, then you can either accept or reject, but in either case you must halt.
By adapting the undecidability proof for HALT, we can easily show that CG is undecidable also.  Namely, suppose P solves CG.  Let Q take as input a Turing machine description $\langle M \rangle$, and solve CG for the machine $M(\langle M \rangle)$ by calling P as a subroutine.  Then $Q(\langle Q \rangle)$ (i.e., Q run on its own description) must halt, accept if it rejects, and reject if it accepts.
Let's now prove that CG is Turing-reducible to PROVELOOP.  Given a description of a Turing machine M for which we want to solve CG, simply create a new Turing machine M', which does the same thing as M except that if M accepts, then M' goes into an infinite loop instead.  Then if M accepts, then M' loops, and moreover there's a ZF proof that M' loops.  On the other hand, if M rejects, then M' also rejects, and there's no ZF proof that M' loops (by the assumption that ZF is sound).  If M loops, then there might or might not be a ZF proof that M' loops -- but in any case, by calling PROVELOOP on M', we separate the case that M accepts from the case that M rejects, and therefore solve CG on M.  So $CG \le_{T} PROVELOOP$, and PROVELOOP is undecidable as well.
One more note.  In the comments of this blog post, Andy Drucker supplied a proof that CG is not equivalent to HALT, but rather has Friedberg-Muchnik-like intermediate status.  So, the situation is
$0 \lt_{T} CG \le_{T} PROVELOOP \le_{T} HALT$
with at least one of the last two inequalities strict.  Again, I'm sure this is all implicit in some computability paper from the 1960s or something, but I wouldn't know where to find it.

REPLY [5 votes]: This is a complement to other answers, giving some "higher level" reasons why $PROVELOOP$ must be complete.
For a recursively enumerable set $A$, the following are equivalent:

$A$ is Turing complete (i.e. $A \equiv_T 0'$).
$A$ computes a fixed-point free (FPF) function (i.e. a total function $f$ such that $\varphi_e \neq \varphi_{f(e)}$ for all indices $e$).
$A$ computes a diaginally non-recursive (DNR) function (i.e. a total function $g$ such that  if $\varphi_e(e)$ halts then $\varphi_e(e) \neq g(e)$).

The equivalence of (1) and (2) is the Arslanov Completeness Criterion and the equivalence with (3) was observed by Jockusch [Degrees of functions with no fixed points, in Logic, methodology and philosophy of science, VIII (Moscow, 1987), 191–201]. Conditions (2) and (3) are very useful tests for completeness of recursively enumerable sets.
A variation of Scott's $CG$ argument shows that $PROVELOOP$ computes a separating function for the disjoint recursively enumerable sets $$A_0 = \lbrace e : \varphi_e(e){\downarrow} = 0\rbrace, \qquad A_1 = \lbrace e : \varphi_e(e){\downarrow} = 1\rbrace.$$ That is $PROVELOOP$ computes a $\lbrace0,1\rbrace$-valued function $h$ such that $h(e) = 0$ if $e \in A_0$ and $h(e)= 1$ if $e \in A_1$. Then $g(e) = 1-h(e)$ is easily seen to be DNR. Since $g \leq_T h \leq_T PROVELOOP$ and $PROVELOOP$ is recursively enumerable, it follows that $PROVELOOP$ is complete.<|endoftext|>
TITLE: In what rigorous sense are Sperner's Lemma and the Brouwer Fixed Point Theorem equivalent?
QUESTION [33 upvotes]: I understand that one can give a proof of each of these propositions assuming the truth of the other.  But this seems a bit squishy to me, since there is a trivial sense in which any two true theorems are equivalent (to any proof of Theorem A, prepend "Assume Theorem B", and vice versa; the objection "But the proof of Theorem A doesn't really use the assumption that Theorem B holds" seems more psychological than mathematical).
One might try to formalize the notion of equivalence by considering the lengths of proofs, saying "There is a derivation of Theorem A from Theorem B that is significantly shorter than any proof of Theorem A from scratch, and vice versa", but this too is squishy, in two distinct ways: the length of a proof depends on the formalization procedure one chooses, and "significantly shorter" is vague. Moreover, it's hard to imagine how one could work with this notion of equivalence, since the totality of all short proofs is going to be hard to get a handle on, for the usual reasons.
Can one find some sort of mathematical context (a topos, perhaps?)  in which there is a rigorously defined (and not vacuously true) meaning of the equivalence between Sperner and Brouwer?
(For a recent article that discusses this equivalence and gives pointers to relevant literature, see "A Borsuk-Ulam Equivalent that Directly Implies Sperner's Lemma" by Nyman and Su in the April 2013 issue of the American Mathematical Monthly, a version of which is available online at http://willamette.edu/~knyman/papers/Fan_Sperner.pdf .)
See also the related thread Sperner's lemma and Tucker's lemma.

REPLY [7 votes]: Here is another perspective from theoretical computer science. Sperner's theorem represents the complexity class PPAD; This complexity class (described by Christos Papadimitriou) is represented also by finding approximate fix points and several important applications of the fixed-point theorem are known to be PPAD-complete, most famously computing Nash equilibrium of (even 2-players) games. Here is a nice introduction to PPAD by Paul Goldberg, that I found in the post "Brower, Sperner and PPAD$  in this course site of Noam Nisan.<|endoftext|>
TITLE: Dirac measures dense in space of measures?
QUESTION [12 upvotes]: Let $I$ be a compact interval and $\mathcal{M}(I)$ the space of (signed) Borel measures. We equip it with the weak topology, i.e. a sequence $\mu_n$ converges to zero if and only if
$$ \left|\int_I f(x) \mathrm{d}\mu_n(x)\right| \longrightarrow 0$$
for all $f \in C(I)$.
Now the question is the following: Let $V \subset \mathcal{M}(I)$ be the vectorspace of all finite linear combinations of Dirac measures supported at different points in $I$. Is $V$ dense in $\mathcal{M}(I)$?
For example if $I = [0,1]$, the sequence
$$ \mu_n =  \frac{1}{N}\sum_{j=1}^N \delta_{j/N},$$
$\delta_{j/N}$ being the Dirac measure supported at $j/N$, weak*-converges to the Lebesgue measure as $\mu_n$ is just the approximation by Riemann sums. Hence one can easily get all measures that are absolutely continuous w.r.t. the Lebesgue measure. 
However, there are more measures (singular measures) that are neither point measures nor Lebesgue measures and I don't have an idea how to reach those.

REPLY [8 votes]: This question has been answered but it is part of a larger picture which might be of interest to the OP.  If $S$ is a set, then   $\Lambda(S)$, the set of formal linear combinations of elements of $S$, is a vector space, the free vector space over $S$.  If $S$ has a (topological or analytic) structure, then this can be used to provide the vector space with a suitable locally convex topology.  Thus if $S$ is a compact interval, or indeed a compact space $K$, then we give $\Lambda(K)$
the topology defined by all seminorms whose restrictions to $K$ are continuous.  This space will not usually be complete so we complete it to get $\tilde \Lambda(K)$.  Then
1) $\Lambda(K)$ is dense in the latter;
2) this space has the universal property that each continuous mapping from $K$ into a Banach space has a unique lifting to a continuous linear mapping on $\tilde \Lambda(K)$;
3)  the dual is $C(K)$;
4)  it can be naturally identified with space of Radon measures on $K$ provided with a topology which is not identical with the weak $\ast$ topology but has the same convergent sequences and the same dual but with the, in our opinion significant, advantage that it is complete.
An important bonus is that this construction has a vast array of interesting variants.  Thus one can consider bounded continuous functions on a completely regular space, smooth functions on a manifold (compact or not, with or without boundary), holomorphic functions on complex manifolds, bounded, uniformly continuous functions on uniform spaces, to get spaces of bounded Radon measures, distributions, analytic functionals, uniform measures to name but a few of the interesting possibilities.<|endoftext|>
TITLE: Why don't more mathematicians improve Wikipedia articles?
QUESTION [61 upvotes]: Wikipedia is a widely used resource for mathematics. For example, there are hundreds of mathematics articles that average over 1000 page views per day. Here is a list of the 500 most popular math articles. The number of regular Wikipedia readers is increasing, while the number of editors is decreasing (graphs), which is causing growing concern within the Wikipedia community. 
WikiProject Mathematics is relatively active (compared to other WikiProjects, but not compared to MathOverflow!), and there is always the need for more experts who really understand the material. An editor continually faces the tension between (1) providing a lot of advanced material and (2) explaining things well, which generates many productive discussions about how mathematics articles should be written, and which topics should have their own article. 
Regardless of the long term concerns raised about whether Wikipedia is capable of being a resource for advanced mathematics (see this closed question), the fact is, people are attempting to learn from Wikipedia's mathematics articles right now. So improvements made to articles today will benefit the readers of tomorrow. 

Wikipedia is a very satisfying venue for summarizing topics you know well, and explaining things to other people, due to its large readership. Based on the number of mathematicians at MathOverflow, who are willing to spend time (for free!) answering questions and clarifying subjects for other people, it seems like there is a lot of untapped volunteer potential here. So in the interests of exposing the possible obstacles to joining Wikipedia, I would like to know:

Why don't mathematicians spend more time improving Wikipedia articles?

Recent efforts intended to attract new participants and keep existing ones include the friendly atmosphere of the Teahouse, as well as WikiProject Editor Retention. 
In case anyone is interested, my Wikipedia username is User:Mark L MacDonald (which is my real name).

REPLY [16 votes]: Here's the math WikiProject's "Current activity" page, which lists each day's new articles, articles for which deletion is proposed, articles needing expert attention, articles needing references, and articles needing various other sorts of work:  http://en.wikipedia.org/wiki/Wikipedia:WikiProject_Mathematics/Current_activity  PS added in August 2016: The "current activity" page hasn't worked for more than a year.  But there is this page that includes new articles, which often need work: https://en.wikipedia.org/wiki/User:Mathbot/Changes_to_mathlists
Here's the page listing "requested articles"---mathematics articles that do not yet exist: http://en.wikipedia.org/wiki/Wikipedia:Requested_articles/Mathematics
(Inexperienced or non-logged-in users are no longer allowed to create NEW articles, but they can create drafts in the user space, which can be moved by experienced users into the article space.)
DO NOT begin a Wikipedia article by writing "Let $\lbrace T_n\rbrace_{n=1}^\infty$ be a sequence of bounded linear operators on a Banach space $B$."  That fails to even hint to the lay reader that the article is not about theology, music, chemistry, banking, politics, etc.
The title of the article should be a singular noun phrase except when there's some special reason to use the plural.  One such reason is that the article is about a set of things (in something more like the layman's understanding of "set" than that of the mathematician); for example "Maxwell's equations".
One might begin by writing "In mathematics, an oriented matroid is . . ."  Or "In algebra,..." or "In number theory,..." or "In geometry,..." or "In calculus,..."  But DON'T start by saying "In functional analysis,...", or "In category theory," or "In topology,....".  Again with those, the lay reader is given no reason to suspect that mathematics is what it is about.  There's no need for that sort of context-setting phrase if the title of the article is "Mathematical induction" or "Algebraic equation" or "Arithmetic of $p$-adic numbers".
Usually, the word or phrase that is the title of the article should appear in or near the first sentence set in bold, not necessarily verbatim identical to the title (e.g. it may be plural where the title is singular).
Do not begin a biographical article by writing "John Schmon was born in Putford in 1801.  His father was a polecat farmer and his mother was a quantum software designer.  He attended this school and that school.  His older brother died when he was 10....."  The reader should be told right at the beginning what John Schmon was notable for, thus: "John Schmon (1801–1998) was a Nevian mathematician who discovered this theorem and that theorem and made fundamental contributions to the understanding of whatever."
Use lower case initial letters in article titles and section headings except where there is a reason to use a capital (e.g. a person's name).  The first letter of a section heading is capital except in rare instances where there's a reason to use lower case.  Thus a section may be called
Applications of the theorem to population genetics
but should NOT be called
Applications of the Theorem To Population Genetics
If you create a new article, it's a good idea to create redirect pages from alternative names, commonplace misspelling and miscapitalizations, commonplace misnomers, etc.  Thus "Schmon's Theorem" (with a capital "T") might redirect to "Schmon's theorem" (the proper title).
One of the easiest mistakes to make in creating a new article is to leave it as an "orphan", i.e. an article to which no other articles link.  If you go to the "toolbox" menu and choose "what links here" you can see which other articles link to it.  In the relatively early days of Wikipedia (in 2002 or 2003), I created a new article titled "exponential growth".  I then used Google to find more than 150 occurrences of "exponential growth" or "grows exponentially" or "growing exponentially" in other Wikipedia articles, and linked them to the new article.  Much more recently, in 2011, I found that 60 Wikipedia articles mentioned Gerald J. Toomer, in most cases by citing one of his works, but there was no article about him.  I created the links to the new article about him BEFORE creating the article about him.  These are "red links": links to an article that does not exist, and are red in color, whereas links to existing articles are blue.  One should create appropriate red links in anticipation of the future existence of an article even when one intends never to create it oneself.  A red link invites others to click on it and then create the new article.

I've put whatever came to mind instantly into these comments.  I could say a lot more if I took more time.
Most of what appears above is codified in Wikipedia's style manuals and guidance pages.<|endoftext|>
TITLE: Motivation for Frankl's conjecture?
QUESTION [8 upvotes]: Frankl's conjecture, open since 1979, says that if $F$ is a union-closed family of subsets of $X$, then there is some $x \in X$ such that $x$ appears in at least half the sets in $F$.
What was the motivation for this conjecture? Is it a generalization of some simpler known fact?

REPLY [16 votes]: Frankl originally stated the dual of the problem as written here, i.e., in terms of intersections instead of unions.  This seems to have been in the 1979 edition of the Handbook of Combinatorics [edit: No such edition exists, and I'm not sure of the original source.  See comments below], which isn't that easy to find, but the current edition is up on Google Books, and he states it in this form at the beginning of his (updated) chapter on Extremal Set Systems as Conjecture 2.1:  For an intersection-closed family of subsets of $[n]$, there is an element that is contained in at most half the subsets.
I always assumed that the conjecture was following in the spirit of Erdos-Ko-Rado type theorems.  Let me expand below:
The Erdos-Ko-Rado Theorem bounds the size of a family of small sets $\mathcal{F}$ with pairwise nonempty intersection (see erdos-ko-rado for details).  An extension of this due to Hilton-Milner says that a maximum-size family of small subsets of $[n]$ with pairwise nonempty intersection has an element contained in all of the subsets. 
In the intersection-closed family $\mathcal{G}$ taken by considering all intersections of sets from $\mathcal{F}$ (satisfying the Erdos-Ko-Rado/Hilton-Milner conditions), this says that there is an element contained in all subsets from $\mathcal{G}$.  If one makes it this far, it's reasonable to ask how few a number of subsets an element may be contained in, which is exactly what Frankl's Conjecture concerns.
UPDATE MAY 2016:  I've had cause to look into the provenance more.  Apparently Frankl asked the question in '79, and talked about it with several people, but never published it.  The first appearance in print arises from a problem session at the 1984 conference on Graphs and Order in Banff, where Duffus presented several forms of the problem.  The conference proceedings are in the volume "Graphs and order".  From the summary there, Duffus appears to have already known several partial results on the conjecture that were only published much later.
See also the recent survey article of Bruhn and Schaudt.<|endoftext|>
TITLE: Sheaves on Contractible Analytic Spaces
QUESTION [13 upvotes]: Let $(X,\mathcal{O}_X)$ be a contractible complex analytic space. Suppose that $\mathcal{F}$ is a coherent sheaf of $\mathcal{O}_X$-modules. Can we invoke the fact that $X$ is contractible to conclude, in some cases, that $\mathcal{F}$ is isomorphic to $\mathcal{O}_X^{\oplus n}$ for some $n$? If you like, you may take $X$ to be the analytic space associated to a complex affine variety.
I ask because contractibility is often a useful condition when attempting to prove a fibre bundle is trivial.

REPLY [30 votes]: The so-called Oka-Grauert principle states that for any Stein space $X$ the holomorphic and the topological classification of complex vector bundles on $X$ coincide. 
The original reference is 
Hans Grauert, Analytische Faserungen über holomorph-vollständigen Räumen, Math. Ann. 135 (1958), 263-273.
As a consequence, every locally free, coherent sheaf $\mathscr{F}$ defined on a contractible subvariety $X$ of $\mathbb{C}^n$ is free.
Of course, if $\mathscr{F}$ is not locally free this is no longer true. For instance, take a closed analytic subvariety $Z \subset X$; then the ideal sheaf $\mathscr{I}_Z \subset \mathscr{O}_X$ is coherent but not free.<|endoftext|>
TITLE: Sperner's lemma and Tucker's lemma
QUESTION [9 upvotes]: In their article "A Borsuk-Ulam Equivalent that Directly Implies Sperner's Lemma" (American Mathematical Monthly, April 2013), Nyman and Su write "[W]e are unaware of a direct proof that Tucker's lemma implies Sperner's lemma".
Could there be a mathematical obstruction to finding a derivation of Sperner's lemma from Tucker's lemma?  E.g., could there be a mathematical context (perhaps some fragment of ZF as a background theory), and in that context two propositions S and T that are recognizable as versions of Sperner's lemma and Tucker's lemma, such that T is true but S is false?  Or a computational context in which finding "Sperner's maguffin" (a fully labeled n-simplex) is demonstrably harder than finding "Tucker's maguffin" (a complementary edge)?
See the related thread In what rigorous sense are Sperner's Lemma and the Brouwer Fixed Point Theorem equivalent?.

REPLY [8 votes]: Hello, I'm one of the authors of this paper: arXiv:1305.6158 (PDF)
The way I see it, "Sperner's Lemma", as usually stated, is actually the "wrong" version of the theorem.  The versions in my paper, which use cubic and octahedral labels, are in some sense, the more "natural" statements, at least as far as analogy with Tucker goes.  As we show, they are implied by Tucker's lemma quite naturally via geometric embeddings.
Topologically, the various Sperner-like theorems are obviously the same.  But it seems that you need a topological theorem (e.g. Brouwer) to establish their equivalence.  For the same reason, it seems like you need a topological theorem to prove Sperner from Tucker.  At the very least, the geometric approach doesn't make any sense, because the antipodality condition doesn't have a good analogy.
As for your question about mathematical contexts where T is True and S is false, I can't think of any context where that would really make sense.  Maybe T can be provable and S can be unprovable, e.g. if you restricted yourself to a context where Brouwer, etc. can't be proven (maybe by accepting only axioms that yield constructive proofs).<|endoftext|>
TITLE: Langlands product
QUESTION [7 upvotes]: In his 'Märchen' Langlands considers for a local field $F$ a certain abelian category $\Pi(F)$ whose objects are given by isomorphisms classes of irreducible admissible representations of $GL_n(F)$, where $n \in \mathbf{N}$ runs over all natural numbers. For $[\pi],[\pi']$ represented by cuspidal reps $\pi,\pi'$ of $GL_n(F)$ and $GL_{n'}(F)$ he introduces the sum $[\pi] \boxplus [\pi]$ as the class of the unique irreducible quotient of the parabolic induction of $\pi\otimes \pi'$ where $GL_n(F)\times GL_{n'}(F)$ is viewed as a Levi component of the obvious standard parabolic of $GL_{n+n'}(F)$. Since he wants to recognize $\Pi(F)$ as the category of representations of some proalgebraic group via the Tannaka formalism he poses the problem of defining a tensor product $[\pi]\boxtimes [\pi']$ as a class of a representation of $GL_{n\cdot n'}(F)$. Moreover $\boxplus$, $\boxtimes$ and $\oplus$, $\otimes$ should correspond to each other under the Langlands correspondence. Of course since it has now been proven one could define $\boxtimes$ via the Local Langlands correspondence.
My question is: Is there any known elementary construction of $[\pi]\boxtimes [\pi']$ for any example with $n,n' > 1$ ?

REPLY [7 votes]: Basically, no.  Marc Palm's answer addresses L-functions, but that is a long long way from determining the irrep -- you'd need L-functions and epsilon-factors of twists, plus an impressive ability to translate such information into a construction of a smooth irrep (if you want more than existence).
To me, your question is the most important outstanding problem in the local Langlands program.  I anticipate that a deeper understanding of types (redoing Bushnell-Kutzko in such a way that the type of a putative Langlands tensor product is clear) will be needed. Whatever it is, it won't be "elementary".<|endoftext|>
TITLE: Equivariant $K$-theory, singular vectors, and flag manifolds
QUESTION [5 upvotes]: For a homogeneous space $M = G/B$, with $G$ a (complex) semi-simple Lie group, it is very well-known that equivariant vector bundles $E$ over $M$ correspond to representations $(V_{\lambda},\lambda)$ of $B$, and the differential operators on $E$ are closely linked to the representation theory of $G$.
For the special case of a flag manifold, which is to say, when $B$ is a Borel subgroup of $G$,  differential operators from $E$ to itself correspond to homomorphisms of the Verma module $U({\frak g})\otimes_{U({\frak b})} V_{\lambda}$. These homomorphisms are in turn classified by the so-called singular vectors of $V_{\lambda}$, which is to say the vectors killed by the action of the positive niradical. Moreover again, these singular vectors correspond to solutions of certain hyper-geometric functions.
What I would like to know is how all this relates to equivariant K-theory. Is there some characterization of the singular vectors correspond to a Fredholm operator. Also, can the defining equivalence relation of the equivariant K-theory group $K^0$ be nicely reformulated in terms of representation theory and singular vectors?

REPLY [3 votes]: The ideas were first developed by Givental and Lee in the context of quantum equivariant K-theory https://arxiv.org/abs/math/0108105, where they defined quantum K-theory as a certain lift of quantum cohomology. Then their results were generalized from flag varieties to Nakajima quiver varieties (at least of type A).
So to answer the first part of your question the (difference or q-) hypergemetric functions are (up to a certain normalization) singular vectors that you have mentioned. Based on what we know from Givental, Lee and the follow-up fork is that those functions are nothing but K-theoretic J-functions of the corresponding variety. One can also check explicitly that they reproduce quantum K-ring relations in the semiclassical limit.
I am gonna give an example for $T^*\mathbb{P}^1$ (which you can find in 3.5.3 of https://arxiv.org/pdf/1412.6081.pdf). So the corresponding singular vector is a hypergeometric function of type $_2\phi_1$ and it satisfies a difference equation (which is an integral of motion of some integrable system). The normalization factor consisting of the ratio of theta functions (formula (3.42)) has a meaning of its own and is called $\textit{elliptic envelope}$ by Aganagic and Okounkov https://arxiv.org/abs/1604.00423.<|endoftext|>
TITLE: Order type of the smallest set containing the identity function and closed under exponentiation
QUESTION [24 upvotes]: Let $E$ be the smallest set of functions $\mathbb{N}^+\to\mathbb{N}^+$ containing the identity function $n \mapsto n$ and closed under exponentiation  $(f,g) \mapsto \left(n \mapsto f(n)^{g(n)}\right)$, i.e. $E=\{n \mapsto n, n \mapsto n^n, n \mapsto n^{n^n}, n \mapsto (n^n)^n, n \mapsto (n^n)^{n^n},\ \dots\}$. Let $E$ be ordered by eventual domination. 
Is $E$ well-ordered? What is the least ordinal that cannot be embedded in $E$?

REPLY [18 votes]: As Joel showed, the set $E$ is well-ordered with order type no more than the Cantor ordinal $\epsilon_0$. In fact, its order type is exactly $\epsilon_0$. This can be proved by constructing the order isomorphism between $\epsilon_0$ and $E$.
First, note that if $F,G\in E$ are of the form $F(n)=n^{n^{f(n)}}$, $G(n)=n^{n^{g(n)}}$ then $F^G$ is of also of the form $n\mapsto n^{n^{h(n)}}$, where $h(n) = f(n)+n^{g(n)}$. So, let me define $E^\prime$ to be the smallest subset of $\mathbb{N}^{\mathbb{N}}$ containing the zero function $n\mapsto0$ and such that for any pair $f,g\in E^\prime$ then the function $n\mapsto f(n)+n^{g(n)}$ is in $E^\prime$. Then the map taking $f\in E^\prime$ to $n\mapsto n^{n^{f(n)}}$ is an order isomorphism from $E^\prime$ to $E$.
I'll now define a map $\theta\colon\epsilon_0\to E^\prime$ and show that it is an order isomorphism. By Cantor normal form any ordinal $\alpha < \epsilon_0$ can be written uniquely as $\alpha=\omega^{\beta_1}+\cdots+\omega^{\beta_k}$ for ordinals $\beta_1\ge\cdots\ge\beta_k$ less than $\alpha$. Write,
$$
\theta(\alpha)(n)=n^{\theta(\beta_1)(n)}+\cdots+n^{\theta(\beta_k)(n)}.
$$
This defines $\theta(\alpha)$ in terms of its values on smaller ordinals. Note that if $k=0$ then $\theta(\alpha)=0$ is in $E^\prime$. On the other hand, if $k\ge1$, then $\alpha = \omega^{\beta_1}+\gamma$ for ordinals $\beta_1,\gamma < \alpha$ and,
$$
\theta(\alpha)(n)=n^{\theta(\beta_1)(n)}+\theta(\gamma)(n).
$$
So $\theta(\alpha)$ is in $E^\prime$ whenever $\theta(\beta_1)$ and $\theta(\gamma)$ are. Transfinite induction then defines $\theta\colon\epsilon_0\to E^\prime$.
To show that $\theta$ is onto, it just needs to be shown that for any two ordinals $\alpha,\gamma < \epsilon_0$ then $n\mapsto\theta(\alpha)(n)+n^{\theta(\gamma)(n)}$ is also in the image of $\theta$. Write $\alpha$ in Cantor normal form as above, and let $\tilde\beta_1\ge\cdots\ge\tilde\beta_{k+1}$ be the ordinals $\beta_1,\ldots,\beta_k,\gamma$ arranged into decreasing order. Setting $\tilde\alpha=\omega^{\tilde\beta_1}+\cdots+\omega^{\tilde\beta_{k+1}}$,
$$
\theta(\tilde\alpha)(n)=n^{\theta(\tilde\beta_1)(n)}+\cdots+n^{\theta(\tilde\beta_{k+1})(n)}
=\theta(\alpha)(n)+n^{\theta(\gamma)(n)}.
$$
So, $\theta$ is a surjective map from $\epsilon_0$ to $E^\prime$.
It just remains to be shown that $\theta$ is (strictly) order preserving. I'll show that if $\alpha > \gamma$ are ordinals then $\theta(\alpha)(n) > \theta(\gamma)(n)$ for large $n$. By induction, it can be assumed that this is true whenever $\alpha,\gamma$ are replaced by smaller values. Again, using Cantor normal form,
$$
\alpha=\omega^{\beta_1}+\cdots+\omega^{\beta_k}, \gamma=\omega^{\tilde\beta_1}+\cdots+\omega^{\tilde\beta_j}
$$
where $\beta_1\ge\cdots\ge\beta_k$ and $\tilde\beta_1\ge\cdots\ge\tilde\beta_j$. Letting $i$ be the smallest number such that one of $\beta_i\not=\tilde\beta_j$, $i > j$ or $i > k$ holds then, as $\alpha > \gamma$, we have $i\le k$ and $\beta_i > \tilde\beta_r$ for all $r=i,\ldots,j$. Then,
$$
\theta(\alpha)(n)-\theta(\gamma)(n)\ge n^{\theta(\beta_i)(n)}-n^{\theta(\tilde\beta_i)(n)}-\cdots-n^{\theta(\tilde\beta_j)(n)}.
$$
If $j < i$ then the right hand side is just $n^{\theta(\beta_i)(n)}$. On the other hand, if $j\ge i$ then, using the induction hypothesis, $n^{\theta(\beta_i)(n)}/n^{\theta(\tilde\beta_r)(n)}\to\infty$ for $r\ge i$, so the right hand side tends to infinity. In either case, $\theta(\alpha)(n) > \theta(\gamma)(n)$ for large $n$.<|endoftext|>
TITLE: Growth of Thompson's group $F$
QUESTION [7 upvotes]: EDIT(August 2013): I accepted Mark's answer as being the state of art- there are two relevant references, one in the answer and one in the comments. The minimal growth rate of $F$ remains unknown with no conjectural answer. END OF EDIT
EDIT: Mark Sapir pointed a reference (in the comments) giving a  lower bound  of $2^{1/4}$ for the minimal rate. Is this the state of art? The third question remains unanswered. If the answer is NO then the lower bound jumps suddenly to $\frac{\sqrt{5}+3}{2}$ by known results. END OF EDIT
What is it known about the minimal growth rate of the Thompson's group $F$? Is there an easy lower bound? Is there a generating set growing slower than the standard one?

REPLY [11 votes]: These questions have been studied (perhaps except the third one). See Section 5.8.7 in my book  and the references there.<|endoftext|>
TITLE: Integer lattice points on a hypersphere
QUESTION [16 upvotes]: Is the following statement true?

For every integer $n\ge2$ and every integer $k\ge0$ there exists a hypersphere in $\mathbb{R}^n$  (circle, sphere etc) containing exactly $k$ integer lattice points on its surface.

REPLY [17 votes]: There are several related and very interesting problems and theorems:  

Schinzel's theorem - solves the problem in $\mathbb{R}^2$ using so-called Schinzel circles. It seems intuitively clear that it generalizes to higher dimensions by slightly adjusting radius of a hypersphere so that it contains exactly the same lattice points as its section in lower dimension, but of course, a rigorous proof is needed. Indeed, there is:
Kulikowski's theorem - gives explicit construction in $\mathbb{R}^3$ and generalizes to all higher dimensions:

W. Sierpiński, "Elementary Theory of Numbers: 2nd English Edition", page 386, the last paragraph:

T. Kulikowski [1] has proved that for any natural number n there 
  exists a sphere (in the three-dimensional space), on the boundary of 
  which there are precisely n points whose coordinates are integers. He 
  generalized this theorem for spheres in spaces of an arbitrary $\ge 3$ 
  dimension.

And similar problems related to interior points:

Steinhaus' theorem
Browkin's theorem<|endoftext|>
TITLE: How closed-form conjectures are made?
QUESTION [59 upvotes]: Recently I posted a conjecture at Math.SE:
$$\int_0^\infty\ln\frac{J_\mu(x)^2+Y_\mu(x)^2}{J_\nu(x)^2+Y_\nu(x)^2}\mathrm dx\stackrel{?}{=}\frac{\pi}{2}(\mu^2-\nu^2),$$
where $J_\mu(x)$ and $Y_\mu(x)$ are the Bessel function of the first and second kind.
It is supported by numerical calculation with hundreds of digits of precision for many different values of $\mu, \nu$. The question is open for several days with +500 bounty on it and is not resolved yet. But my question here is not about if this conjecture true or false.
Obviously, several possible closed forms matched my numeric calculations, for example:
$$\left(\frac{\pi}{2}+7^{-7^{7^{7^{7^{7^{\sqrt{5}+\sin \mu\nu}}}}}}\right)(\mu^2-\nu^2).$$
But for some reason that I cannot clearly explain (or even understand) I selected the simpler one, and I am strongly inclined to search for its proof rather than a disproof. I believe most people would feel and behave exactly the same way. 

(1) Are there any mathematical or philosophical reasons supporting this position?

Why when we calculate some sum or integral (which do not contain explicit tiny quantities like $10^{-10^{10^{.^{.^{10}}}}}$) with thousands of digits of precision and it matches some simple closed-form expression, we inclined to believe this is the exact equality rather than an accidental very close value? 

(2) Are there known cases when such intuition turned out to be wrong?

And one more question:

(3) Do you believe there can be exact closed forms for some infinite sums or integrals, that cannot be proved in $ZFC$ or any its reasonable extension (like adding some large cardinal axioms) - so to speak, equalities that hold without any reason.

REPLY [5 votes]: There is an aspect that has not been mentioned so far, namely that of usability. I do not believe that there is a context in which the OP's additional term of $7^{-7^{7^{7^{7^{7^{\sqrt{5}+\sin \mu\nu}}}}}}$ could actually have been used in a further calculation - that is, in its exact form, not just for an estimate.
Indeed, when there was suspicion that the value of $\frac\pi2 (\mu^2 - \nu^2)$ were no correct, the next best conjecture for the value of the integral would probably be something like $(\frac\pi2 + \epsilon) (\mu^2 - \nu^2)$ with suitable bounds on $\epsilon$: Many conjectures of this form have been made in other contexts.
Could it be that usability is actually the main criterium that is employed when making conjections? I believe that this is entirely possible.<|endoftext|>
TITLE: Homotopy left-exactness of a left derived functor
QUESTION [8 upvotes]: Let
$$
F: \mathcal{C} \leftrightarrows \mathcal{D} :G
$$
be a Quillen adjunction between model categories. Consider the corresponding adjunction of total derived functors
$$
\mathbb{L}F: \mathrm{Ho}(\mathcal{C}) \leftrightarrows \mathrm{Ho}(\mathcal{D}) :\mathbb{R}G
$$
It is then a well-known fact that $\mathbb{L}F$ preserves homotopy colimits. Suppose $F$ preserves finite limits. Under what conditions does $\mathbb{L}F$ preserves finite homotopy limits?
Motivation: in the case I am interested in $\mathcal{C}$ and $\mathcal{D}$ are categories of simplicial presheaves on some sites, endowed with either the local injective or local projective model structures. The functors $F$ and $G$ arise from a functor between the underlying sites. Homotopy left-exactness of $\mathbb{L}F$ is what I need to claim that $\mathbb{L}F \dashv \mathbb{R}G$ induces a geometric morphism of the $\infty$-topoi presented by $\mathcal{C}$ and $\mathcal{D}$.

REPLY [10 votes]: I do not know the answer for a general Quillen adjunction, but I will attempt to give a complete answer in the case you're interested in, when the adjunction $(F,G)$ is of the form $(f_!,f^\ast)$ for $f\colon C\to D$ a continuous functor between sites:
Claim. The functor $\mathbb{L}f_!$ preserves finite homotopy limits iff $f$ is a morphism of site, i.e., iff $f$ induces a geometric morphism between $1$-topoi.
(So for instance if finite limits exist in $C$ and $D$ and $f$ preserves them, then $\mathbb{L}f_!$ preserves finite homotopy limits.)
The proof of the claim will use higher topos theory. The underlying $(\infty,1)$-adjunction to $(F,G)$ is the adjunction 
$$ f_!: Shv_\infty^\wedge(C)\to Shv_\infty^\wedge(D): f^\ast$$
between the $(\infty,1)$-topoi of hypercomplete sheaves, so we must figure out when this $f_!$ preserves finite limits. First of all it is clear that if $f_!$ preserves finite limits, then $f$ is a morphism of sites: the restriction of $f_!$ to $0$-truncated objects still preserves finite limits.
The other implication is less trivial: assume that $f$ is a morphism of sites. Let $Shv_\infty^\sim(C)$ denote the $1$-localic $(\infty,1)$-topos whose $0$-truncated objects are sheaves of sets on $C$ (see HTT Definition 6.4.5.8): its hypercompletion is the topos $Shv_\infty^\wedge(C)$ we're interested in (concretely, $Shv_\infty^\sim(C)\simeq Shv_\infty(C')$ where $C'$ is a site with finite limits equivalent to $C$). By definition of $1$-localic topoi, the functor $f^\ast\colon Shv_\infty^\sim(D)\to Shv_\infty^\sim(C)$ has a left exact left adjoint since its restriction to $0$-truncated objects has (by assumption). The functor $f_!\colon Shv_\infty^\wedge(C)\to Shv_\infty^\wedge (D)$ is the composition of the functors
$$ Shv_\infty^\wedge(C) \hookrightarrow Shv_\infty^\sim(C) \stackrel{f_!}{\to} Shv_\infty^\sim(D) \to Shv_\infty^\wedge(D)$$
all of which preserve finite limits.
ETA:
This leaves open the question of whether $f_!\colon Shv_\infty(C)\to Shv_\infty(D)$ preserves finite limits for $f$ a morphism of site. It's true if $C$ and $D$ have finite limits, by the above. My hunch is that it's not true in general, i.e., that the appropriate higher notion of morphism of sites is stronger than the usual one.
In case $f:C\to D$ is cocontinuous (but not necessarily continuous), then the functor $f_\ast\colon PSh_\infty(C)\to PSh_\infty(D)$ preserves sheaves (this follows from the corresponding fact for sheaves of sets because sieves are $0$-truncated). The induced adjunction $(f^\ast,f_\ast)$ between sheaves is always a geometric morphism because $f^\ast$ is the composition of three functors that preserve finite limits, as above. In particular, $f_\ast$ preserves hypercomplete objects and you get an induced geometric morphism between hypercompletions.<|endoftext|>
TITLE: How many triangulations of the genus $g$ surface on $n$ vertices?
QUESTION [16 upvotes]: By "a triangulation of $X$", I mean a simplicial complex whose geometric realization is homeomorphic to $X$. Tutte showed that the number of combinatorially distinct triangulations $t(n)$ of the $2$-dimensional sphere on $n+3$ vertices is asymptotically given by 
$$t(n) \approx \frac{1}{16} \sqrt{\frac{3}{2 \pi}}n^{-5/2} \left( \frac{256}{27} \right)^n.$$
What is known about the number of combinatorially distinct triangulations $t(n,g)$ of the genus $g$ surface? 
(1) It seems likely to me that for fixed $g$, the growth is still roughly exponential, and maybe even 
$$ t(n,g) = \exp \left( c_1 n + c_2 \log n + c_3 + o(1) \right)$$ for some constants $c_1, c_2, c_3 \in \mathbb{R}$ which only depend on $g$. For example in the case $g=0$, Tutte's result gives that $c_1 = 256 / 27$, $c_2= -5 / 2$, and $c_3 = \log \left( \frac{1}{16} \sqrt{\frac{3}{2 \pi}} \right)$. Is the growth of $T(n,g)$ always exponential in $n$, and if so can we at least compute the base of the exponent $c_1$ for higher $g$?
Updated: The rooted version of this question is answered in the paper: Z.-C. Gao. The number of rooted triangular maps on a surface. Journal of Combinatorial Theory, Series B, 52(2):236 – 249, 1991. For fixed genus $g$, Gao establishes strong results along the lines of the above.
(2) At the other extreme, is anything known about the rate of growth of $t(n,g)$ as $n \to \infty$ if $g \approx cn^2$ for some constant $0 < c< 1/12$?
(3) Finally,  let $T(n)$ be the total number of combinatorial types of triangulated surfaces on $n$ vertices. What is the rate of growth of $T(n)$ as $n \to \infty$? 
Updated: What I would really like to know is if we can not establish the rate of growth of $T(n)$, is there at least an upper bound of order $$T(n) = n^{o(n)}?$$
Or let $\tilde{T}(n)$ denote the number of labelled triangulated surfaces on $n$ vertices. Is it true that
$$\tilde{T}(n) = n^{n+o(n)}?$$
Since $\tilde{T}(n) \le n! T(n)$, the second inequality would follow from the first.
Gao's results give that if we assume that the genus $g$ is bounded, then we get the stronger bound
$$T(n) = n^{O(n /\log n),}$$
but I still don't know how to handle $g$ growing with $n$.

REPLY [10 votes]: I don't think a nice asymptotic formula like the one you've mentioned from Tutte's work is available for higher $g$ to the best of my understanding; it is entirely possible that someone who regularly deals with such things will come along and show us otherwise.
Meanwhile, the state-of-the-art on counting triangulations of genus $g$ surfaces is the lovely paper "The KP hierarchy, branched covers, and triangulations" of Goulden and Jackson, available here: see Theorem 5.4. There are two obstacles in terms of going from this Theorem to a Tutte-type asymptotic formula:

The Goulden-Jackson formula is presented in terms of the face count rather than the vertex count, and
As is typical in the field, the formula relies on a recursively defined function $f(n,g)$ where (again) $n$ counts the faces rather than the vertices.

I think 1. is surmountable, but 2. might make things difficult: I have no idea how fast $f(n,g)$ grows, but I am sure various simplifications are possible if you set $g = cn$ in their formula.<|endoftext|>
TITLE: Do operations generate well-ordered sets only?
QUESTION [10 upvotes]: I've read   @TauMu's question   about the set of functions   $\mathbb N\rightarrow\mathbb N$   generated from the identity map by repeatedly applying exponentiation of two already accepted functions. @TauMu asks if such a set is well-ordered with respect to eventual domination.
It seems to me that it would be hard, perhaps impossible, to obtain a set which is not well-ordered regardless of the choice of a binary operation   $\mathbb N\times\mathbb N\rightarrow\mathbb N$.   Here is a patient formulation of my more general question:
Given an operation   $\tau : \mathbb N^2\rightarrow\mathbb N$,   let   $\bigcirc^{\tau}\subseteq\mathbb N^{\mathbb N}$   be the smallest set such that it has the identity map   $I_{\mathbb N}$   as its element,   $I_\mathbb N\in\bigcirc^\tau$,   and   $h:=\tau\circ(f\triangle g)\in\bigcirc^{\tau}$   for every   $f\ g\in\bigcirc^{\tau}$.
REMARK (an explanation of the notation above)
$$\forall_{n\in\mathbb N}\quad h(n) := \tau(f(n)\ g(n))$$
Finally,
QUESTION (edited twice after the 1st and 2nd comment by @Joseph Van Name)   Does there exists an operation   $\tau:\mathbb N^2\rightarrow \mathbb N$   dominating the identity in each variable (see below), and such that   $\bigcirc^{\tau}$   contains an infinite strictly decreasing sequence with respect to the relation of eventual domination?
By   $\tau$   dominating the identity in each variable I mean that:
$$\forall_{k\ n\in\mathbb N}\quad \tau(k\ \ n)\ \ \ge\ \ \max(k\ \ \ n)$$

REPLY [4 votes]: I think the following example shows that the extra condition (nondecreasing in all arguments) proposed in Emil Jeřábek´s answer is necessary.
Consider the operation $$\tau(m,n)=2 \max(m,n)^n -\min(m,n).$$
It is clear that $\tau(m,n) \geq \max(m,n)$. Now let $g_0=I_\mathbb{N}$, $g_{i+1}=\tau\circ(g_i \triangle g_0)$, $h=\tau\circ(g_1 \triangle g_1)$ and $f_i=\tau\circ(g_i \triangle h)$. Then $\{f_i: i \in \mathbb{N} \}$ is a strictly decreasing sequence with respect to eventual domination.<|endoftext|>
TITLE: Effective Chebotarev without Artin's conjecture
QUESTION [14 upvotes]: $\DeclareMathOperator\Frob{Frob}$Iwaniec and Kowalski, in their famous book Analytic Number Theory states a strong form
of the effective Chebotarev density theorem page 143, and prove it assuming both GRH for Artin's $L$-function and Artin's conjecture
(that the Artin $L$-function have no pole except maybe at $s=1$). But they also add, with no argument nor reference,
that the same theorem is true assuming only GRH, not the Artin's conjecture. I'd like to know how this can be proved.
Let me explicit my question for a reader that would not want to open Iwaniec and Kowalski's book. Let $L/\mathbb Q$ be a finite extension
Galois group $G$, and let $\rho$ be an irreducible complex representation of $G$, that one can as well suppose non-trivial (as Artin's conjecture is known for the trivial representation). For $x>0$ a real, let $\psi (\rho,x) = \sum_{p^n < x} \operatorname{tr}\rho(\Frob_p^n) \log p$, where the sum is on all prime power of the form $p^n$, where the prime $p$ is unramified for $\rho$. Then the crucial point in proving Iwaniec-Kowalski's effective Chebotarev is the following estimate
$$(1) \quad \psi(\rho,x) = O(x^{1/2} \log x \log x^d q(\rho)),$$
where $q(\rho)$ is the Artin's conductor of $\rho$ and the implied constant is absolute. Estimate (1) is proved under GRH and Artin in Iwaniec-Kowalski (Theorem 5.15)

How to prove (1) under GRH alone, as Iwaniec and Kowalski suggest is possible ?

I have already given some thoughts to the question, but I am not able to solve it. Surely one of the ideas involved should be the following:
even assuming only GRH for the Artin's $L$-function, not Artin's conjecture, we know that $L(\rho,s)$ has no pole except
maybe on the critical line,  for by the theorem of Brauer, $L(\rho,s)$ is a quotient of products of Hecke $L$-functions, and those functions have no poles on the critical strip except perhaps at $s=1$ by a deep result of Hecke, and no zeros either, under GRH, except maybe on the critical line. Moreover, one also knows using the same argument of Brauer that $L(\rho,s)$ is
meromorphic function of order $1$, that is quotient of two entire functions of order $1$, and therefore that $L(\rho,s)$ has a nice Weierstrass product formula (like (5.23)).
Then the natural way to go is to try to follow the proof given by Iwaniec and Kowalski
under GRH and Artin, using the above remarks to avoid using Artin. I see several issues with that method, the most important being the following:
a crucial step in the method is the estimate of the number of zeros $N(T)$ (with their positive multiplicity) of $L(\rho,s)$ on the critical segment between $s = 1/2 -iT $ and $s=1/2 + iT$ - see Theorem 5.8.
This estimate is obtained by integrating $L'/L$ on a suitable rectangle intersecting the critical line on that segment. Yet in the presence of poles
this method will not count the number $N(T)$ of zeros, but the difference $N(T)-P(T)$ where $P(T)$ is the number of poles (with their multiplicity) on the same segment. So even a good estimate for $L'/L$ hence of $N(T)-P(T)$ will not prevent $N(T)$ and $P(T)$ to be arbitrary large. Then, when one computes $\psi(\rho,x)$ by  some explicit formula (such as (5.53)), both zeros and poles contribute and if these are too many, that is if $N(T)+P(T)$ is too large, the precise estimate (1) will be completely ruined.

Edit after Sausage and Frank's comments and Denis's answer : I will answer Frank's question and this will allow me to explain why Sausage and Denis's suggestion are not sufficient (or so I think -- perhaps I am missing something). The estimate (1) of Lagarias-Odlyzko leads easily by linear combination and then applying Cauchy-Schwarz and the fact the sum of the square of the dimension of irreducible representation of $G$ is $|G|$ to the following form of Chebotarev, for $C$ a subset of $G$ invariant by conjugation, and $M$ the product of primes
ramified in $L$ :
$$(2) \quad \psi(C,x) = \frac{|C|}{|G|}Li(x) + O \left( \sqrt{x} \sqrt{|C|} \log x (\log x + \log M + \log |G|)\right).$$
Here $\psi(C,s) = \sum_{p^n < x, \Frob_p^n \in C} \log p$.
Formula (2) is stated in a slightly weaker form page 144 of Iwaniec-Kowalski, weaker just because they use a substandard majoration for discriminant. Formula (2) was also stated and proved earlier, by Murty, Murty, and Saradha under GRH and Artin: see Modular Forms and the Chebotarev Density Theorem,
American Journal of Mathematics, Vol. 110, No. 2 (Apr., 1988), pp. 253-281. Curiously, this paper is not mentioned in Iwaniec-Kowalski.
Now let us compare (2) with the form of effective Chebotarev proved by Lagarias and Odlyzko, and slightly improved soon after by Serre (because Lagarias and Oslyzko use the same substandard lower bound on the discriminant than later Iwaniec and Kowalski, as Serre use better bounds).
$$ (3)  \quad \psi(C,x) = \frac{|C|}{|G|}\mathrm{Li}(x) + O \left( \sqrt{x} |C| \log x (\log x + \log M + \log |G|)\right)$$
It is clear how (2) is better than (3): we gain a factor $\sqrt{|C|}$.
How comes Lagarias and Odlyzko and Serre didn't see it? Well, essentially this is because
they use, for counting the zero of $L(\rho,s)$, the zero on $\zeta_L(s)$. The problem is that
while $L(\rho,s)$ is an $L$-function of degree $\dim \rho$ (and the various $L(\rho,s)$ therefore have in quadratic average a degree equal to $\sqrt{|G|}$), the function $\zeta_L$
is an $L$-function of degree $|G|$. So to get (1) or (2), one needs to get an optimal bound for each $L(\rho,s)$ individually, as a function of degree $\dim \rho$, and the argument
suggested by Sausage can not be sufficient, nor can the one given by Denis, since it is
already used by Lagarias and Odlyzko.
So the question remains :

How to prove (1), or (2), without assuming Artin ?
Or is it possible that there is a mistake in Iwaniec-Kowalski and proving (2) without Artin is not possible ?

REPLY [14 votes]: Since no one answered my question, I have asked the author of the book. Emmanuel Kowalski
told me that this remark they make (namely that the form (1) or (2) they give of Chebotarev can be proved using GRH alone, without Artin) is mistaken. In the state of our knowledge,
Artin is necessary to get such a precise form.<|endoftext|>
TITLE: An operation on binary strings
QUESTION [8 upvotes]: Consider the “product” $\gamma = \alpha \times \beta$ of two binary strings $\alpha$, $\beta$ $\in \lbrace 0,1\rbrace^+$ which one gets by replacing every 1 in $\beta$ by $\alpha$ and each 0 in $\beta$ by $\overline{\alpha}$, with $\overline{\alpha}$ being the negation of $\alpha$, which one gets by replacing every 1 in $\alpha$ by 0 and vice versa.
Formally:
$$(\alpha \times \beta)[k] = \begin{cases} 1 & \text{if}\ \ \alpha[k\ \text{mod}\ a] = \beta[k\ \text{div}\ a] \\\ 0 & \text{otherwise} \end{cases}$$
for $a=|\alpha|, b=|\beta|, k = 0,\dots,ab-1$. 
Maybe it comes as a surprise - at least for me it did - and it's a little bit cumbersome to prove, that the operation $\times$ - even though it is not commutative - is associative, i.e. $\alpha \times (\beta \times \gamma) = (\alpha \times \beta) \times \gamma$.

Is there an elegant argument to see that $\times$ is associative? (I
  had to go through a couple of case discriminations and some
  equivalencies of modulo arithmetic like $(k\ \text{mod}\ ab)\
 \text{div}\ a = (k\  \text{div}\  a)\ \text{mod}\ b$ to get to the
  result.)

Since $1$ is a neutral element ($\alpha \times 1 = 1 \times \alpha = \alpha$), the tuple $(\lbrace 0,1\rbrace^+,\times,1)$ is a monoid.
In this monoid it seems that each $\sigma$ has a unique “factorization” into “primes” (upto associativity). If  the length of $\sigma$ is prime, $\sigma$ itself is necessarily “prime”. If $\sigma$ has length $2^n$ it can have up to $n$ prime factors, e.g. $10010110 = 10 \times 10 \times 10$. But it can also be prime, e.g. $1110$.

In which contexts and under which name has this monoid been
  investigated?

REPLY [7 votes]: I think it's less confusing if you swap the roles of 0 and 1, as then the basic operation you're using to generate the entries of $\alpha\times\beta$ is addition mod 2.
Then, if you write $\alpha_i$ for the $i$th entry of a string $\alpha$, associativity follows because the entries of $\alpha\times\beta\times\gamma$ are just the $\alpha_i+\beta_j+\gamma_k$ (mod 2), listed in lexicographic order of $(k,j,i)$.

REPLY [2 votes]: The monoid has a central subgroup $\{0,1\}$ (the group of units). One needs to take that into account when talking about primes. There is the induced congruence on the monoid $u\sim v$ iff $u=v$ or $u=0\cdot v$. The factor-monoid is free. That can be proved using Levi's theorem (Theorem 1.8.4 in my book ): 

Suppose that $S$  is a cancellative semigroup without an identity element, in which every 
  element is a product of indecomposable  elements, and  for every four elements $a,u,v,c$ from 
  $S$ the equality $au = vc$, implies either $u=c$ or $u=bc$ or $c=bu$ for some $b$. Then  $S$
  is a free semigroup and the set of indecomposable elements of $S$ is its free generating set.

 Correction. The factor-monoid is not free. Indeed, $11\cdot 111=111111=111\cdot 11$. It would be interesting to find a presentation of this monoid. I think that the defining relations should be related to periodicity and a presentation should not be too complicated.<|endoftext|>
TITLE: Natural Isomorphism of $S(V[1])$ and $(\bigwedge V)[n]$
QUESTION [6 upvotes]: Let $V:=\oplus_{j\in\mathbb{Z}}V_j$ be a graded $\mathbb{F}$-vector space over
the field $\mathbb{F}$. The graded tensor product of graded vector spaces is given
by 
$V \otimes W:= \oplus_{j\in \mathbb{Z}}\oplus_{p+q=j}V_p\otimes V_q$
and for the graded vector space $\mathbb{F}[j]$, which is $\mathbb{F}$ in degree
$j$ and te zero vector space $\{0\}$ otherwise, the shift $V[j]$ is given by
$V[j]:=\mathbb{F}[j]\otimes V$
We then define a monoidal structure on the category of graded vector space,
(more or less) given by the rule on homogeneous elements 
$v\otimes w= (-1)^{deg(v)deg(w)}w\otimes v$
Then there is the decalage isomorphism
$
dec: V_1[1]\otimes \cdots \otimes V_n[1] \to (V_1 \otimes \cdots \otimes V_n)[n]
$
given by $dec(v_1[1]\otimes \cdots \otimes v_n[1])= 
(-1)^{\sum_{j=1}^n(n-j)deg(v_j)}(v_1\otimes \cdots \otimes v_n)[n]$.
Now in work on graded (stuff), it is frequently said, that this isomorphism
defines a natural isomorphism of the symmetric graded tensor-algebra of $V[1]$
and the antisymmetric graded tensor algebra, that is
$S(V[1])\simeq (\bigwedge V)[n]$
*The question is: How does the decalage induces such an algebra isomorphism? Or
What is the natural isomorphism? *
If $dec$ itself would be the isomorphism, then
$dec(v[1] \vee w[1])= (dec(v_1)\wedge dec(w))[2]$ should hold, but this isn't true in general.

REPLY [2 votes]: Let $V$ be a $\mathbb Z$ graded vector space over a field $\mathbb k$ of characteristic $0$ (for simplicity). The suspension $V[1]$ of $V$ is the graded vector space
$V[1]:= V\otimes_{\mathbb k} \mathbb k[1],$ with $\mathbb k[1]$ concentrated in degree $-1$, and $\mathbb k_{-1}=\mathbb k$. With $s$ we denote the "suspension" morphism $s:V\rightarrow V[1]$, $x\mapsto sx:=x$ of degree $-1$; in other words, $|sx|=|x|-1$, where $|\cdot|$ denotes the degree of any homogeneous element in $V$ (or any other graded vector space). In general, we write $s^n: V\rightarrow V[n]$, for any integer $n$. We introduce the graded symmetric resp. antisymmetric algebras over $V$:
$$S(V ) = T (V )/ \langle x \otimes y − (−1)^{|x||y|}
y \otimes x \rangle,~~ \text{resp.}~~\Lambda (V ) = T (V )/ \langle x \otimes y + (−1)^{|x||y|} y \otimes x \rangle, $$
For any $n\geq 0$ the decalage is a canonical isomorphism of graded vector spaces (not of algebras)
$$\Phi_n: S_n(V[1] )\rightarrow \Lambda(V)[n], $$
where $$\Phi_n(sx_1,\cdots, sx_n):= (-1)^{\sum_{i=1}^n(n-i)|sx_i|}s^{n}(x_1\wedge\cdots\wedge x_n). $$
Why that sign? The sign follows from the Koszul rule, once one removes the suspensions $s$ from the string $sx_1,\cdots, sx_n$ one by one applying $n$-times the desuspension morphism $s^{-1}$.<|endoftext|>
TITLE: A family of words counted by the Catalan numbers
QUESTION [37 upvotes]: In recent work with Michael Albert and Nik Ruškuc, a family of words has arisen which is counted by the Catalan numbers. I've looked at Richard Stanley's Catalan exercises in EC2 and his Catalan addendum, but I don't see anything that looks to be clearly equivalent, and a bijection to Dyck paths isn't jumping out at me. So I have two questions:


Has anyone seen these words, or some equivalent objects, before?
Do you see a nice bijection between these words and any family of "classic" Catalan objects such as Dyck paths or noncrossing partitions?


Let $w$ be a word of length $n$ over the natural numbers (including $0$). Then $w$ lies in our family if it satisfies two rules:

For all $k\le n-1$, $w_{k+1}\ge w_k-1$.
If $w$ contains an $i\ge 1$, then the first $i$ lies between two $i-1$s.

(The word "between" does not imply contiguity, so rule 2 means that when we read $w$ from left to right, we should see an $i-1$ before we see the first $i$, and then at some point after that we should see another $i-1$.)
The number of words of length $n$ that satisfy these conditions is equal to the $n-1$st Catalan number.
For example, the only word of length $2$ that satisfies these rules is $00$, for length $3$ there are two such words, $000$ and $010$, for length $4$ there are five,
$$
0000, 0010, 0100, 0101, 0110,
$$
and for length $5$ there are $14$,
$$
00000, 00010, 00100, 00101, 00110, 01000, 01001,
$$
$$
01010, 01011, 01021, 01100, 01101, 01110, 01210.
$$

REPLY [3 votes]: This answer summarizes partial progress which was interesting enough to lead to its own question.
Consider sequences which only obey your first condition. Call these $L$-words. ($L$ stands for Lukasiwiecz, because they are more or less the partial sums of the sequences studied by Lukaswiecz.)
There are Catalan many $L$-words. Given an $L$-word, let $p$ be the number of pairs $(i,i+1)$ for which your second rule is violated. So we want to count pairs with $p=0$. Also, let $q+1$ be the number of occurrences of $0$ in the $L$-word. Here is a table:
$$\begin{matrix}
L-\mbox{word} & p & q \\
000 & 0 & 2 \\
010 & 0 & 1 \\
001 & 1 & 1 \\
011 & 1 & 0 \\
012 & 2 & 0 \\
\end{matrix}$$
It appears that the number of $L$-words of length $n$ with given values of $p$ and $q$ only depends on $(p+q, n)$! See my question for more.<|endoftext|>
TITLE: Characterization of amenable actions
QUESTION [7 upvotes]: Let $(X,\mu)$ be a $G$-space, i.e. a measure space with a Borel quasi-invariant $G$-action. Say that $X$ is amenable (equivalently, that the action is amenable) if there is a $G$-fixed point in every affine space over $X$ with an $\alpha$-twisted action, where $\alpha$ is a corresponding cocycle.
If $X$ is an amenable $G$-space, it follows more or less from definition that for every compact metric space $Y$ there is an  $G$-equivariant measurable map $\varphi : X \to M(Y)$, where $M(Y)$ is the collection of probability measures on $Y$.
My question is - is this property equivalent to the above definition of amenable action or strictly weaker?

REPLY [3 votes]: We have resolved this question recently in Proposition 3.16 of
https://arxiv.org/pdf/2003.03469.pdf
We prove, among other things, that an action of an exact locally compact group G on a von Neumann algebra M is amenable (in the sense of Claire Anantharaman-Delaroche, i.e., there is a $G$-equivariant projection $M\bar\otimes L^\infty(G)\to M$) if and only if there is a unital G-equivariant ucp map $L^\infty(G)\to Z(M)$, the center of $M$, or equivalently, a unital $G$-equivariant ucp map $C_{ub}(G)\to Z(M)$, where $C_{ub}(G)=UC_b(G)$ denotes the algebra of bounded uniformly continuous functions $G\to \mathbb{C}$ as in Jesse Peterson's answer.
The existence of $C_{ub}(G)\to Z(M)$ is equivalent to the existence of a $G$-equivariant projection $Z(M)\otimes C_{ub}(G)\to Z(M)$; one can also replace $C_{ub}(G)$ by $L^\infty(G)$ here. But $\otimes$ here denotes the $C^*$-tensor product, and it cannot be replaced by the von Neumann tensor product $\bar\otimes$ in general, even if $G$ is discrete.
Indeed, if $G$ is not exact, then $A=\ell^\infty(G)$ is a commutative $G$-$C^*$-algebra with respect to the translation $G$-action. And this action (or equivalently the $G$-action on its spectrum $\beta G$) is amenable if and only if $G$ is exact. This means that the $G$-action on the bidual von Neumann algebra $A^{**}$ is not von Neumann amenable (in the sense of Delaroche).<|endoftext|>
TITLE: Field generated by the Fourier coefficients of a modular form
QUESTION [6 upvotes]: Let $f = \sum_n a_n q^n$ be a cuspidal newform of weight $k$ on $\Gamma_0(N)$ for some $N$.  Let $K_f$ be the number field generated by the $a_q$ as $q$ runs over all primes.
My question: if we consider the field generated by all but one of these $a_q$, can this field be smaller than $K_f$?  
(edited based on Sawin's answer to only look at Fourier coefficients at primes)

REPLY [12 votes]: No, strong multiplicity one says that all-but-finitely-many of the $a_p$'s determine all the others.
Edit in response to comment/query, and further in response to subsequent comments: ... and, once all the other coefficients are determined by strong multiplicity one (for newforms), invoke Shimura's results (arguably going back to Fricke-Klein in principle) that automorphisms of the complex numbers over $\mathbb Q$ (not merely Galois automorphisms on an algebraic closure of $\mathbb Q$) can be applied to the Fourier coefficients of a holomorphic modular form, producing another. In the case at hand, applying such an automorphism over the field generated by all the given Fourier coefficients, and subtracting, would either give a modular form with only finitely-many non-zero $a_p$'s (impossible), or $0$ (giving the desired result).
[Thanks to MF1 for corrections.]<|endoftext|>
TITLE: Can the fact that the square of an integer is a natural number be categorified?
QUESTION [29 upvotes]: If $a$ and $b$ are natural numbers, then $a-b$ is an integer and so the square $(a-b)^2$ is a natural number.  In particular
$$ (a-b)^2 \geq 0.   \qquad (1)$$
Combining this fact with the identity
$$ ab + ba + (a-b)^2 = a^2 + b^2 \qquad (2)$$
we obtain the inequality
$$ ab + ba \leq a^2 + b^2 \qquad (3)$$
which can be viewed as a special case of either the arithmetic mean-geometric mean inequality or the Cauchy-Schwarz inequality.
Of course this argument can be generalised; for instance, if $v, w$ are elements of a (real or complex) inner product space, then
$$ \langle v-w, v -w \rangle \geq 0 \qquad (1')$$
and by combining this with the identity
$$ \langle v, w \rangle + \langle w, v \rangle + \langle v-w, v-w \rangle = \langle v,v \rangle + \langle w, w \rangle \qquad (2')$$
we conclude the inequality
$$ \langle v, w \rangle + \langle w, v \rangle \leq \langle v, v \rangle + \langle w, w \rangle \qquad (3')$$
which is closely related to the Cauchy-Schwarz inequality (indeed one can amplify (3') to the Cauchy-Schwarz inequality, as discussed in this blog post of mine).
The inequalities (3) or (3') can be interpreted in various categories.  For instance, (3) in the category of finite sets becomes

Theorem 1.  Let $A, B$ be finite sets.  Then there exists an injection from $(A \times B) \uplus (B \times A)$ to $(A \times A) \uplus (B \times B)$.

while (3) in the category of finite-dimensional vector spaces becomes

Theorem 2.  Let $V, W$ be finite-dimensional vector spaces.  Then there exists an injective linear map from $(V \otimes W) \oplus (W \otimes V)$ to $(V \otimes V) \oplus (W \otimes W)$.

In the category of unitary representations of a compact (or finite) group $G$, a little character theory (or complete reducibility together with Schur's lemma) allows one to similarly interpret (3') in this category:

Theorem 3.  Let $V, W$ be finite-dimensional unitary representations of a compact group $G$.  Then there exists an injective $G$-equivariant linear map from $(V \otimes W) \oplus (W \otimes V)$ to $(V \otimes V) \oplus (W \otimes W)$.
Theorem 3.  Let $V, W$ be finite-dimensional unitary representations of a compact group $G$.  Then there exists an injective linear map from $\operatorname{Hom}_G(V,W) \times \operatorname{Hom}_G(W,V)$ to $\operatorname{Hom}_G(V,V) \times \operatorname{Hom}_G(W,W)$.

One can presumably state similar theorems for modules of semisimple algebras over an algebraically closed field, or for various types of vector bundles (or finite covers) over a fixed base space, although I won't attempt to do so here.
Anyway, these results suggest that there may be a way to categorify the inequalities (1), (3), (1'), (3') and/or the identities (2), (2'), for instance by finding a bijective proof (or perhaps an "injective proof") of Theorem 1.  However, despite the simplicity of these inequalities and identities this seems to be a surprisingly difficult task.  In the ordered case in which one knows that $a=b+k$ or $b=a+k$ for some natural number $k$, one can interpret $(a-b)^2$ as $k^2$, so it is not difficult to categorify Theorem 1 if one possesses an injection from $A$ to $B$ or vice versa, and similarly for Theorem 2 and Theorem 3.  From this and the trichotomy of order one can finish off Theorem 1 or Theorem 2, though this is an argument which requires one to make a number of arbitrary choices (as the injections here are not canonical) and so one might consider this to be an incomplete categorification.  And in any event, this trick does not seem to recover the full strength of Theorem 3, since there need not be an injective $G$-equivariant map from $V$ to $W$ or vice versa.  So I'm wondering if there is another way to categorify these results?  For the vector space results (Theorem 2 and Theorem 3) it seems natural to try to use K-theory somehow to achieve this goal (since K-theory already has a formalism for taking formal differences of vector spaces), but I don't know enough K-theory to take this idea further.
[Closely related questions to these were discussed some years ago at the n-category cafe here and here regarding the categorification of the Cauchy-Schwarz inequality, but the results of the discussion were inconclusive, although David Speyer did show that there was an obstruction to categorifying Theorem 3 in the case $G = Z/2Z$ in that the injection could not be natural.]

REPLY [18 votes]: Probably the most natural thing to ask for Theorem 1 is as follows.  Let $\mathcal{A}$ be the category whose objects are pairs of finite sets, and whose morphisms are pairs of bijections.   Let $\mathcal{B}$ be the category of finite sets and functions.  (Allowing more morphisms in $\mathcal{A}$ or fewer morphisms in $\mathcal{B}$ would make the problem harder.)  We have functors $F,G:\mathcal{A}\to\mathcal{B}$ given by $F(A,B)=(A\times B)\uplus(B\times A)$ and $G(A,B)=(A\times A)\uplus(B\times B)$.  We then ask whether there is a natural injective map $j:F\to G$.  I think the answer is negative. Indeed, the subset $j^{-1}(A\times A)\subseteq F(A,B)$ would have to be preserved by the action of $\text{Aut}(A)\times\text{Aut}(B)$ on $F(A,B)$, so it would have to be $(A\times B)$ or $(B\times A)$, wlog the former.  Now $j$ gives an $\text{Aut}(A)\times\text{Aut}(B)$-equivariant map from $A\times B$ to $A\times A$, which consists of two equivariant maps $p,q:A\times B\to A$.  For fixed $b$ we have an $\text{Aut}(A)$-equivariant map $p(-,b):A\to A$, and (provided that $|A|>2$) the only possibility is $p(a,b)=a$.  Similarly $q(a,b)=a$, and we see that $j$ cannot be injective. 
[Added later]
This whole story may be related to Thomason's paper "Beware the phony multiplication on Quillen's $\mathcal{A}^{-1}\mathcal{A}$".  Here $\mathcal{A}$ is a symmetric bimonoidal category (like the category of finite-dimensional vector spaces, under $\oplus$ and $\otimes$).  From this Quillen constructed a new category $\mathcal{A}^{-1}\mathcal{A}$.  The objects are pairs $(A,B)$, where $A$ and $B$ are objects of $\mathcal{A}$.  I won't spell out the morphisms except to say that $(A,B)$ is supposed to represent the "formal difference" $A-B$, and everything is guided by that.  It is reasonable to hope that there should be a tensor product on $\mathcal{A}^{-1}\mathcal{A}$ compatible with the original tensor product on $\mathcal{A}$.  However, Thomason showed that several purported constructions of such a tensor product contain subtle errors, and that a wide class of approaches are doomed to fail.<|endoftext|>
TITLE: Can one bound the Quadratic Points on Curves?
QUESTION [5 upvotes]: Let $C$ be a nonsingular projective curve defined over $\mathbb{Q}$, which does not admit a map of degree 1 or 2 to $\mathbb{P}^1$ or to an elliptic curve. It is then a consequence of Corollary 3 of [1] that $C$ possesses only finitely many quadratic points; i.e., the set 
$$\Gamma_C := \left\{p \in C : [\mathbb{Q}(p) : \mathbb{Q}] = 2\right\}$$
is finite. In particular, there exists a bound $D_C$ such that, if $D$ is a squarefree integer satisfying $|D| > D_C$, then $C(\mathbb{Q}(\sqrt{D})) = C(\mathbb{Q})$. 
It is natural to ask if there is an algorithm to effectively compute $D_C$, given a model of $C$. Going through the proof of the Harris-Silverman result does not suggest any such algorithm to me (unless the rank of the Jacobian of $C$ is 0).

Given a model of $C$, is there an algorithm to effectively compute $D_C$?

I would also like to ask if $D_C$ is known when $C$ is the modular curve $X_0(N)$; there are only 55 values of $N$ for which $X_0(N)$ does not have finitely many quadratic points (see Theorem 4.9 in [2] for this list). Let $S$ be this set of 55 integers. 

Given $N \notin S$, is there an algorithm to effectively compute $D_{X_0(N)}$?

[1]: J. Harris, J.H. Silverman. Bielliptic Curves and Symmetric Products. Proc. Amer. Math. Soc. 112 (1991), 347-356
[2]: F. Bars. On Quadratic Points of Classical Modular Curves. Manuscript, 2012, available from the author's website.

REPLY [4 votes]: I was going to point out that the specific result you cite for quadratic points is actually a theorem of Joe Harris and mine (Proc. Amer. Math. Soc. 112 (1991), 347-356), but I see that stankewicz has already given the reference, so I won't repeat the link. What Abramovich and Harris did later (and this is much harder) is generalize the result to points of degree $d$ for all $d\ge2$, although the obvious generalization turns out to be false, one can't merely assume that there are no maps of degree at most $d$ to an elliptic curve or to $\mathbb P^1$. Anyway, as far as I know, all finiteness results of this sort rely on a theorem of Faltings' (generalizing Vojta's proof of the Mordell conjecture) that is highly noneffective. However, for what it's worth, it is possible to give an effective upper bound for $\#D_C$. More generally, one can give effective constants $K_1$ and $K_2$ such that $\#D_C$ has at most $K_1$ points whose height is greater than $K_2$. Unfortunately, we can't get our hands on those putative $K_1$ points of large height, although one suspects that there aren't any such points.<|endoftext|>
TITLE: Eigenvalues of permutations of a real matrix: can they all be real?
QUESTION [22 upvotes]: For a matrix $M\in GL(n,\mathbb R)$, consider the $n!$ matrices obtained by permutations of the rows (say) of $M$ and define the total spectrum $TS(M)$ as the union of all their spectra (counting repeated values separately).  

Can all these $n!\cdot n$ eigenvalues be real? 

Denote by $c(M)$ the number of pairs of non-real eigenvalues in $TS(M)$.
For a matrix of rank 1, its TS is trivially real. But trying a continuity argument in a neighborhood of such a matrix will fail miserably, e.g. if $J=J_n$ denotes the all-1-matrix and $I=I_n$ the unit matrix, it is easy to show that $c(J+\epsilon I)=c(I)$ for all $\epsilon\in\mathbb R$ (corresponding permutations of both matrices have the same eigenvalues), but $c(I)$ is far from $c(J)=0$, e.g. $c(I_5)=118$.
Examples for $c(M)=0$:   
For $n=3$, take $M=\pmatrix{ 4&3&0\cr2&1&-2\cr0&0&1}$.   
For $n=4$, take $M=\pmatrix{ 
83& 81& 64& 58\cr 79& 67& 65& 63\cr 74& 71& 58& 53\cr 67& 53& 79& 80}$. 
(The image of the distribution of $c(M)$ in this related thread suggests that the probability for a random $4\times4$ matrix to have $c(M)=0$ must be extremely small, maybe $10^{-20}$ at best.)
For $n=5$, so far I have been only able to get $c(M)$ as low as $11$; one such matrix is $$M=\pmatrix{ 
9885& 9887& 9887& 9765& 9894\cr
 9887& 9888& 9883& 9887& 9891\cr
 9887& 9883& 10013& 9765& 9755\cr
 9752& 9762& 10141& \color{red}{7013}& 9789\cr
 9772& 10149& 9922& 9654& \color{red}{-47650}}.$$    Note that an environment of $M$ contained in $c^{-1}(11)$ cannot be very ‘big’: change e.g. $M_{1,1}$ by only $\pm.005$ and already $c(M)$ will go up! (Of course my search wasn't for integer matrices, rather once I’d found a real $M$ with $c(M)$ that small, I have tweaked it to obtain a matrix with not-too-big integer entries.)
There should be $M\in GL(5,\mathbb R)$ with $c(M)$ smaller than that, and I'd even conjecture with $c(M)=0$. But given that the average of $c(M)$ for random $5\times5$ matrices appears to be about $175$, finding those is just way beyond my computer’s capacities, and so is the $n\ge 6$ case. Human intelligence is needed.
UPDATE: Here is a different $M\in GL(4,\mathbb R)$ which should be one of the smallest integer ones with $c(M)=0$: $$ M=\pmatrix{7& 5& 5& 6\cr 5& 3& 7& 2\cr 5& 7& 2& 9\cr 6& 2& 9& 0}$$ It has full rank but, like $J$, is not in the interior of $c^{-1}(0)$, due to the fact that several eigenvalues are repeated in the TS, e.g. the EVs $\pm1$ occur 10 times each and the EV $20$ occurs 12 times.  
And finally I have found $M\in GL(5,\mathbb R)$ with $c(M)=0$!! 
$$M=\pmatrix{\color{red}{4188} &\color{red}{4588}&4948&4925&4919\cr
4948&4979&5001&5008&4990\cr
4988&4989&4989&4998&5065\cr
5077&5032&5005&5015&4948\cr
4966&4923&5096&4948&\color{red}{-24543}}$$
UPDATE 2: Even nicer but very very tight: $$
M=\pmatrix{0&0&1 \cr0&1&3 \cr1&3&2} \qquad
M=\pmatrix{0&0&0&1\cr0&0&1&4\cr0&1&5&8\cr1&4&8&2} \qquad M=\pmatrix{0&0&0&0&1\cr0&0&0&1&2\cr0&0&1&144&18\cr0&1&144&5839&409\cr1&2&18&409&3}$$ 
The existence of $M$'s with such special shapes for $n=3,4,5$ is of course a huge heuristic argument in favor of a positive answer to the initial question.
To find those, I actually minimized instead of $c(M)$  the continuous function $\sum\limits_{z\in TS(M)}\arctan\left|\frac{\Im(z)}{\Re(z)}\right|$. Note that each complex root contributes at worst $\pi/2$ to this sum. 
UPDATE 3: I couldn't resist to try $n=6$, even though each TS takes my poor computer already about 3 sec. My best is $c(M)=9$, which seems not too bad compared with $c(I_6)=948$.
Here goes:
$$M=\pmatrix{0& 0& 0& 0& 0& 6\cr 0& 0& 0& 0& 2& 9204\cr 0& 0& 0& -1& -145& -265335\cr 0& 0& -1& 20
54947& 30426445& 5683742\cr 0& 2& -127& 30426614& 368233489& 735312954\cr 6& 9195& -
265314& 5683632& 735312686& 47613387}$$

REPLY [2 votes]: Here is finally a solution for the $n=7$ case! Putting $T=2$, we have $c(M)=0$, where
$$M=\pmatrix
{1&T^{17}&T^{{32}}&T^{45}&T^{56}&T^{65}&T^{72}
\\0&1&\color{red}2\cdot T^{17}& T^{32}&T^{45}&T^{56} &T^{65}
\\0&0&1&\color{red}{\frac98}\cdot T^{17}& \color{red}{\frac{171}{64}} \cdot T^{32}&T^{45}&T^{56}
\\0&0&0&1&\color{red}{\frac98}\cdot T^{17} &T^{32}&T^{45}
\\0&0&0&0&1&\color{red}2\cdot T^{17}&T^{{32}}
\\0&0&0&0&0&1&T^{17}
\\0&0&0&0&0&0&1}
$$
Similarly to what I had said in the comments of S. Carnahan's answer, this is again based on a triangular matrix $A$ with constant diagonals, given by $A_{ij}=T^{k(18-k)}$ where $k=j-i$ for $j\geqslant i$. For this kind of matrices, the TS seems to have very few complex roots (why?), e.g. $m(A)=42$ which is quite a bit less than $m(I_7)=8796$, and with the above adjustments the complex roots of $A$ can be eliminated.
Note that like in the solution for $n=6$ given by S. Carnahan, the first diagonal is not concave.<|endoftext|>
TITLE: Why did Bourbaki's Élements omit the theory of categories?
QUESTION [28 upvotes]: QUESTION
They had plenty of time to adopt the theory of categories. They had Eilenberg, then Cartan, then Grothendieck. Did they feel that they have established their approach already, that it's too late to go back and start anew?
I have my very-very general answer: World is Chaos, Mathematics is a Jungle, Bourbaki was a nice fluke, but no fluke can last forever, no fluke can overtake Chaos and Jungle. I'd still like to have a much more complete picture.
Appendix: CHRONOLOGY

1934:   Bourbaki's birth (approximate date);
1942-45:   Samuel Eilenberg & Saunders Mac Lane - functor, natural transformation, $K(\pi,n)$;
1946 & 1952: S.Eilenberg & Norman E. Steenrod publish "Axiomatic..." & "Foundations...";
1956: Henri Cartan & S.Eilenberg publish "Homological Algebra";
1957: Alexander Grothendieck publishes his "Tohoku paper", abelian category.


(Please, feel free to add the relevant most important dates to the list above).

REPLY [14 votes]: It might be interesting to look at the Appendix to Exposé I of SGA 4. In a footnote this is described as follows:
Nous reproduisons ici, avec son accord, des papiers secrets de N. BOURBAKI.
While the appendix treats mainly the theory of universes, it makes use of the language of categories. Moreover, some internal references hint to the existence of some more "papiers secrets" containing a draft of a chapter about categories.<|endoftext|>
TITLE: Anything special (historical?) about surface $x\cdot y\cdot z\ +\ x+y+z=0$?
QUESTION [22 upvotes]: QUESTION
I wanted to introduce and develop the complex logarithm from scratch. As the result I've arrived a couple of months ago at the following identity after which the road to complex logarithm is wide open:
$$\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}\quad +\quad \frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\qquad =\qquad 0$$
This holds over any field of course, and it is a parametrization of the following surface, call it   $L$:
$$x\cdot y\cdot z\ +\ x+y+z\ \ =\ \ 0$$
Could you provide any references and information about this surface and the above formula. A knowledgeable friend of mine is sceptical about a geometric interest of this surface   $L$.   I still believe that in some ways   $L$   must be interesting when it rests at the foundation of the complex logarithm.

(Please, feel free to remove/add tags).

A connection
(I mean a connection between surface   $L$   and the complex logarithmic function. I'll write below just a little bit less pedantically than in a textbook for students).
Let's restrict ourselves now to the field of complex numbers   $\mathbb C$.   Let   $a\ b\ c\in\mathbb C^*$,   where   $\mathbb C^* := \mathbb C\setminus \{0\}$.   Then we may consider a pretty much canonical piecewise linear loop   $\gamma := \overline{abca}$.   When   $0\notin\triangle(abc)$   (a closed solid triangle is meant) then we want to show that
$$\int_{\gamma} F = 0 $$
where   $\forall_{z\in\mathbb C^*}\ F(z):=\frac 1z$.   At first my goal is more modest. I want to show that when the diameter of the triangle is much smaller than   $\max(|a|\ |b|\ |c|)$ (so that it already follows that   $0$   does not belong to the triangle) then a crude approximation of the integral is very small. How small? Regular simplicial subdivisions of the triangle lead to about   $n^2$   triangles of diameter about   $\frac 1n$ (everything up to a multiplicative constant). Our integral above is a sum of about   $n^2$   integrals over perimeters of all these small triangles (because the terms which come from the inside of the original triangle will cleanly cancel out--cleanly, I promise). Thus I want the crude approximations of the integrals over the perimeters of the small triangles to converge to   $0$   faster than   $\frac 1{n^2}$.   Then the above integral indeed will be equal to   $0$.
Let
$$A :=\frac{b+c}2\qquad B:=\frac{a+c}2\qquad C:=\frac{a+b}2$$
Then a crude approximation of the above integral over   $\gamma$   can be defined as
$$\Lambda\ :=\ \frac{b-a}C + \frac{c-b}A + \frac{a-c}B$$
Due to the identity above we get:
$$\Lambda\ =\ \frac 14\cdot\frac{a-b}C\cdot\frac{b-c}A\cdot\frac{c-a}B$$
A similar formula holds for each small triangle of the consecutive simplicial subdivision. When the original triangle   $\triangle(abc)$   is disjoint (outside) a disk of radius   $r>0$,   around   $0$,   then all respective values   $A'\ B' C'$,   corresponding to the small triangles, have modules greater than   $r$. Thus the sum of the crude approximations will be of the magnitude about   $(r\cdot n)^{-3}$.   Since the number of summands is of the order   $n^2$,   the whole sum will be arbitrarily close to   $0$.
The internal terms of the sum of the crude approximations cancel out (cleanly :-) because we have selected the mid-points of the edges of the triangles. Thus the whole sum of the crude approximations of the integrals for the triangles of a subdivision approximates arbitrarily well the original integral over   $\gamma$.
(Now one can study integrals of   $F(z):=\frac 1z$   over homotopic paths, etc).

REPLY [13 votes]: Just wanted to see what it looks like...



Now added origin and axes $\pm 1$ in each coordinate.<|endoftext|>
TITLE: Amalgamation of two ccc algebras may collapse the continuum
QUESTION [5 upvotes]: The claim that appears in the title of this question is mentioned in the paper "On Shelah's amalgamation" by Judah and Roslanowski. I'd really like to see a proof of this fact, but unfortunately I couldn't find any reference for this claim.
So my question is: Are there relatively simple examples of amalgamation of ccc algebras that collapse the continuum? If there aren't any simple examples, then a reference for a harder example would be welcome as well.

REPLY [6 votes]: Suppose CH holds and there is a Souslin  tree $T$.  (For example, suppose $V=L$.)  Then $T$ itself can be regarded as a forcing notion; it has ccc and it adjoins a generic path through $T$.  There is also a ccc forcing notion that specializes $T$; a condition here is a finite partial function from $T$ into $\omega$ such that any two comparable nodes in the domain of the condition are assigned distinct values; the union of the generic filter is a function from all of $T$ into $\omega$ that never takes the same value at two comparable nodes of $T$.  If you force with both of these, then the specializing function restricted to the generic path gives a one-to-one map of $\omega_1$ into $\omega$.  So you've collapsed $\aleph_1$; since I assumed CH, this means you've collapsed the cardinal of the continuum.<|endoftext|>
TITLE: Asymptotic bounds on $\pi^{-1}(x)$ (inverse prime counting function)
QUESTION [8 upvotes]: What are the current best asymptotic bounds on $\pi^{-1}(x)$, where $\pi(x)$ denotes the prime counting function (number of primes at most $x$)?
In other words, I am curious about the state of the art for estimating the $n^{th}$ prime. From the prime number theorem, it seems clear that $\pi^{-1}(x)=\Theta(x \log x)$. Can someone point me in the direction of literature that answers the question? Please excuse my inability to find such literature myself...

REPLY [4 votes]: Since "the state of the art" slightly changed since 2013, I am adding this new result on lower and upper bounds on $p_n$. (This is complementary to the answers given earlier.)
Christian Axler proved in https://arxiv.org/abs/1706.03651 that:
(1) For every $n\ge46254381$,
$${\small
p_n < n \Big(\log n + \log\log n - 1 + {\log\log n - 2\over\log n} - {(\log\log n)^2 - 6\log\log n + 10.667 \over 2\log^2 n} \Big).
}
$$
(2) For every $n\ge2$,
$${\small
p_n > n \Big(\log n + \log\log n - 1 + {\log\log n - 2\over\log n} - {(\log\log n)^2 - 6\log\log n + 11.508 \over 2\log^2 n} \Big).
}
$$<|endoftext|>
TITLE: Strong Whitney embedding theorem for non-compact manifolds
QUESTION [34 upvotes]: $\newcommand{\RR}{\mathbb{R}}$The present question arises from some confusion on my part regarding the precise statement of the strong Whitney embedding theorem for non-compact manifolds.
The strong Whitney embedding theorem is usually stated as follows.

Theorem: If $M$ is a smooth $n$-dimensional manifold, then $M$ admits a smooth embedding into $\RR^{2n}$.

In fact, the theorem is stated in essentially this form in Whitney's original article "The self-intersections of a smooth $n$-manifold in $2n$-space". For definiteness, I will assume that all manifolds are Hausdorff, second countable, and smooth.

Question 1: Can we always take the embedding in the above theorem to be closed? If so, is there a reference for such a statement of the theorem?

It seems that Whitney's original proof produces an embedding whose image is not closed when $M$ is open. In fact, immediately after the construction, Whitney explicitly poses the following problem: "Does there exist an imbedding, for $M$ open, with no limit set?"
Having thought about the matter for a short while, I am inclined to believe that Whitney's trick (introduced in the aforementioned article by Whitney) allows the cancellation of infinitely many double points in a manner that preserves closed immersions. Is this correct? Or is my argument getting trapped in some pitfall?
My second question concerns possible dimensional restrictions in the above embedding theorem, stemming from the failure of Whitney's trick for $n=2$.

Question 2: Does every $2$-dimensional manifold embed in $\RR^4$? If so, can we also take the embedding to be closed in this case?

Here is the suggested proof in Whitney's article: "For $n=2$, we imbed the sphere, projective plane, or Klein bottle in $E^4$, and add the necessary number of handles to obtain the given manifold." I can see that this procedure should work for compact surfaces, but I am unable to carry it out in the non-compact case.
Finally, I would also be interested to hear about more recent, good references concerning Whitney's strong embedding theorem.

REPLY [26 votes]: Regarding question 1, yes you can always ensure the image is closed.  You prove the strong Whitney by perturbing a generic map $M \to \mathbb R^{2m}$ to an immersion, and then doing a local double-point creation/destruction technique called the Whitney trick.  So instead of using any smooth map $M \to \mathbb R^{2m}$, start with a proper map -- one where the pre-image of compact sets is compact.   You can then inductively perturb the map on an exhausting collection of compact submanifolds of $M$, making the map into an immersion that is also proper.  
Regarding question 2, generally speaking if a manifold is not compact the embedding problem is easier, not harder.  Think of how your manifold is built via handle attachments.  You can construct the embedding in $\mathbb R^4$ quite directly.  Think of $\mathbb R^4$ with its standard height function $x \longmapsto |x|^2$, and assume the Morse function on $M$ is proper and takes values in $\{ x \in \mathbb R : x > 0 \}$. Then I claim you can embed $M$ in $\mathbb R^4$ so that the Morse function is the restriction of the standard Morse function.  The idea is every $0$-handle corresponds to creating an split unknot component in the level-sets, etc.  
edit: The level sets of the standard morse function on $\mathbb R^4$ consists of spheres of various radius.  So when you pass through a critical point (as the radius increases) either you are creating an split unknot component, doing a connect-sum operation between components (or the reverse, or a self-connect-sum), or you are deleting a split unknot component.  By a split unknot component, I'm referring to the situation where you have a link in the $3$-sphere.  A component is split if there is an embedded 2-sphere that contains only that component, and no other components of the link.   So a split unknot component means that component bounds an embedded disc that's disjoint from the other components. 
Regarding your last question, the Whitney embedding theorem isn't written up in many places since all the key ideas appear in the proof of the h-cobordism theorem.  So Milnor's notes are an archetypal source.  But Adachi's Embeddings and Immersions in the Translations of the AMS series is one of the few places where it occurs in its original context.  You can find the book on Ranicki's webpage.<|endoftext|>
TITLE: A curious sequence of rationals: finite or infinite?
QUESTION [19 upvotes]: Consider the following function repeatedly applied to a rational
$r = a/b$ in lowest terms:
$f(a/b) = (a b) / (a + b - 1)$.
So, $f(2/3) = 6/4 = 3/2$. $f(3/2) = 6/4 = 3/2$.
I am wondering if it is possible to predict when the sequence is finite,
and when infinite.  For example, $f(k/11)$ seems infinite for
$k=2,\ldots,10$, but, e.g., $f(4/13)=f(13/4)$ is finite.
Several more examples are shown below.

$$ \frac{1}{k} \to \frac{k}{k} {=} 1 $$
$$ \frac{3}{7} \to \frac{21}{9} {=} \frac{7}{3} \to \frac{21}{9} {=} \frac{7}{3} $$
$$ \frac{5}{12} \to \frac{60}{16} {=} \frac{15}{4} \to \frac{60}{18} {=} \frac{10}{3}
\to \frac{30}{12} {=} \frac{5}{2} \to
\frac{10}{6} {=} \frac{5}{3} \to
\frac{15}{7} \to \frac{105}{21} {=} \frac{5}{1} \to \frac{5}{5} = 1 $$
$$ \frac{5}{23} \to \frac{115}{27} \to \frac{3105}{141} {=} \frac{1035}{47}
\to \frac{48645}{1081} = \frac{45}{1} \to 1 $$
$$ \frac{4}{11} \to \frac{44}{14} {=} \frac{22}{7} \to \frac{154}{28} {=} \frac{11}{2} \to
\frac{22}{12} {=} \frac{11}{6}
\to \frac{66}{16} {=} \frac{33}{8} \to \frac{264}{40} {=} \frac{33}{5}
\to \frac{165}{37} \to \cdots \to \infty ?$$

I am hoping this does not run into Collatz-like difficulties,
but it does seem straightforward to analyze...
If anyone recognizes it, or sees a way to tame it even partially, I would be interested to learn.  Thanks!

REPLY [6 votes]: Note:  I've appended some additional material that extends the analysis of, in effect, the pre-images of the pre-images of the fixed point $n/m=1$ of the function $f$, to look at general pre-images and pre-images of other fixed points (i.e., $n/m = (m^2-m+1)/m$ for $m>1$.  The new stuff follows a boldface "Added later."  End of note
The families Karl Fabian found can be subsumed under a single rule:  

$f^{(2)}(a/b)=1$ if and only if $${a\over b}={n+d\over n+d'}$$ where
  $n>0$,
  $dd'=n(n-1)$, and $(n+d,n+d')=1$.

It's not hard to show that if $a/b$ has the given form, then $f(a/b)=n$, and it's easy to show that $f(n)=f(n/1)=1$.  (The assumption that $n+d$ and $n+d'$ are relatively prime is crucial.)  The tricky part is showing that nothing else gets to $1$ in two steps.
It's clear that $f(a/b)=1$ if and only if $a+b-1=ab$, which can be rewritten as $(a-1)(b-1)=0$, so the only numbers that get to $1$ in one step are integers $n$ and their reciprocals $1/n$.  It's also clear that $f(a/b)\ge1$ for all $a/b$.  Therefore, the numbers that get to $1$ in two steps are those that get to some integer $n$ in one step.  Let's see how that can happen.
If $ab/(a+b-1)=n$, then we must have $ab=nk$ and $a+b-1=k$ for some integer $k$.  Writing $b=nk/a$, we wind up with $a^2-(k+1)a+nk=0$, so
$$a={k+1\pm\sqrt{(k+1)^2-4nk}\over2}.$$
For $a$ to be an integer, we must have a square inside the square root:
$$(k+1)^2-4nk = m^2,$$
which can be rewritten as
$$(k+1-2n)^2-m^2 = 4n(n-1),$$
or
$$(k+1-2n+m)(k+1-2n-m)=4n(n-1).$$
The two terms on the left hand side have the same parity (they differ by $2m$), hence they must both be even, i.e., $k+1-2n+m=2d$ and $k+1-2n-m=2d'$ where $dd'=n(n-1)$.  From this we get $k+1-2n=d+d'$ and $m=d-d'$, so
$$a={2c+d+d'\pm(d-d')\over2}.$$
Choosing the positive sign gives $a=n+d$.  (Choosing the negative sign gives $b=n+d'$.  If you like, you can assume $a\ge b$ and $d\ge d'$.)
To do just one example, let $n=16$.  The choices for $dd'$ are $240\cdot1$, $80\cdot3$, $48\cdot5$, and $16\cdot15$, leading to the four possibilites for $a/b$ (with $a>b$) for which $f(a/b)=16$:
$${a\over b} = {256\over 17}, {96\over19}, {64\over21}, {32\over31}.$$
Note how a factorization like $dd'=6\cdot40$ fails:
$$f((16+6)/(16+40)) = f(22/56)=f(11/28)= {11\cdot28\over38}={154\over19}.$$
Added later:  If I've done everything correctly, a similar analysis gives the following nice result about pre-images:

Let $n/m$ be a fraction with $n\ge m$ and
  $(n,m)=1$.  Then $f(a/b)=n/m$ if and
  only if
$${a\over b} = {n+d\over n+d'}$$
where $dd'=n(n-m)$ and $(n+d,n+d')=m$.

Let's see how this applies to the other fixed points for $f$, namely when $n=m^2-m+1$, for the first few values of $m$.
Skipping the case $m=1$ (which you can check gives the result noted earlier), let $n/m = 3/2$, so that we need $dd'=3$ and $(3+d,3+d')=2$.  The only factors are $3$ and $1$, which indeed satisfy $(6,4)=2$, but this only gives $a/b = 6/4=3/2$.  In other words, the fixed point $3/2$ has no pre-image other than itself.
For $n/m=7/3$, we have $n(n-m)=28$, for which the possible factorizations are $28\cdot1$, $14\cdot2$, and $7\cdot4$.  But the only one of these for which $(7+d,7+d')=3$ is $14\cdot2$, which gives $a/b = 21/9 = 7/3$, so $7/3$ also has no pre-image other than itself.
You can check that the same thing happens for $n/m = 13/4$:  The only factorization $dd'$ of $13(13-4)$ for which $(13+d,13+d')=4$ is $39\cdot3$, giving $a/b=52/16 = 13/4$.
But $n/m = 21/5$, finally, is interesting:  There are two factorizations that work, namely $84\cdot4$ and $24\cdot14$.  The first, as before, gives the fixed point again, $$a/b = (21+84)/(21+4) = 105/25 = 21/5.$$  But the other one gives $$a/b = (21+24)/(21+14) = 45/35 = 9/7.$$  Note, however, that $9/7$ has no pre-image:  The factorizations of $9(9-7)=18$ are $18\cdot1$, $9\cdot2$, and $6\cdot3$, none of which produce anything divisible by $7$, much less a pair $(9+d,9+d')$ with $7$ as a common divisor.
In summary (for now), the only fraction with infinitely many pre-images is $n/m=1$ (since $n(n-m)=0$ has infinitely many divisor pairs!); the size of the pre-image set for all other fractions is bounded by the number of divisor pairs of $n(n-m)$.  For some of the fixed points of $f$, the pre-image set is just the fixed point itself, while for others (e.g. $n/m = 21/5$), the pre-image set contains additional points.  There may be some simple criterion that identifies the fixed points that have no additional pre-images, but I don't offhand see one, possibly because I haven't thought hard enough about it (but maybe because I'm just blind to the obvious).<|endoftext|>
TITLE: Spinoffs of analytic number theory
QUESTION [11 upvotes]: What are some techniques and theorems of analytic number theory that have proved useful outside of number theory?

REPLY [2 votes]: The circle method was an inspiration behind the solution of the toroidal semiqueens problem by Eberhard, Manners and Mrazović ("Additive triples of bijections, or the toroidal semiqueens problem", JEMS 21 (2019), no. 2, 441–463).
The problem is equivalent to counting permutations $\pi_1,\pi_2$ of $\mathbb{Z}/n\mathbb{Z}$ whose pointwise sum is also a permutation. The proof adapts the circle method to the group $(\mathbb{Z}/n\mathbb{Z})^n$.<|endoftext|>
TITLE: Varieties which become isomorphic to algebraic groups over an algebraic closure
QUESTION [9 upvotes]: My question is as follows:

Let $k$ be a field of characteristic zero and let $\overline{k}$ be an algebraic closure. Let $V$ be an algebraic variety over $k$ and let $\overline{V}=V \times_k \overline{k}$. Suppose that $\overline{V}$ admits the structure of an algebraic group. Then is $V$ itself a principal homogeneous space for some algebraic group?

Note that the answer to my question is yes when $V$ is projective; in this case it is well-known that such a $V$ is a principal homogeneous space for its Albanese variety.
So I am really interested in the case where $V$ is affine. Here I am not aware of an analogue of the Albanese variety for linear algebraic groups.
Even if the answer is no in general, I would still be interested in some positive results for special cases, e.g. for reductive groups or semisimple groups.

REPLY [7 votes]: Here is a counterexample with $V$ finite.
Suppose that $k'$ is a separable extension of degree $5$ of $k$, and set $V = \mathop{\rm Spec} k \sqcup \mathop{\rm Spec} k'$. Suppose that $V$ is a homogenous space; then it is a group scheme, since $V(k) \neq \emptyset$.
If $\overline k$ is the separable closure of $k$, then $V_{\overline k}$ is the disjoint union of six copies of $\mathop{\rm Spec} \overline k$, and these copies form a group of order $6$. The Galois group of $k$ acts on $V_{\overline k}$ by permuting the non-identity component transitively. But a group of order $6$ contains elements of order $2$ and $3$, so this is impossible.
I am convinced that one can get geometrically connected examples using forms of $\mathbb G_{\rm a}^n$.<|endoftext|>
TITLE: Derivative of a determinant of a matrix field
QUESTION [8 upvotes]: Let $A(x_1,...,x_n)$ be an  $n\times n$ matrix field over $R^n$. 
I am interested in the partial derivative determinant of $A$ in respect to $x_i$. In can be shown that:
$\frac{\partial{\det(A)}}{\partial{x_i}} = \det(A)\cdot\sum_{a=1}^{n}{\sum_{b=1}^{n}{ A^{-1}_{a,b} \cdot \frac{ \partial{A_{b,a}} }{ \partial{x_i} }}}$
I was able to prove this using induction and careful, boring calculations, but I was wondering if there was any intuition behind this formula?

REPLY [6 votes]: Another proof, using the characteristic polynomial
$$
\det(A+tI) = t^n+t^{n-1}\text{Tr}(A) + t^{n-2}c_2(A) + \dots+ t c_{n-1}(A) + \det(A)
$$
where $c_i(A) = \text{Tr}(\Lambda^i A)$ is the $i$-th characteristic coefficient.
Namely, assume that $A$ is invertible. Then
$$
\det(A+tX) = t^n\det(t^{-1}A+X) = t^n\det(A(A^{-1}X + t^{-1}I)) = t^n\det(A)\det(A^{-1}X+t^{-1}I)
$$
$$
 = t^n \det(A) \Big( t^{-n}+t^{1-n}\text{Tr}(A^{-1}X) + t^{2-n}c_2(A^{-1}X) + \dots+ t^{-1} c_{n-1}(A^{-1}X) +    \det(A^{-1}X)\Big)
$$
$$
 = \det(A)\Big(1+t\text{Tr}(A^{-1}X) + O(t^2)\Big)
$$
Thus
$$
d\det(A)X = \partial_t\big|_0 \det(A+tX) = \partial_t\big|_0  \det(A)\Big(1+t\text{Tr}(A^{-1}X) + O(t^2)\Big)
$$
$$ 
 = \det(A)\text{Tr}(A^{-1}X) = \text{Tr}(\text{Adj}(A)X)
$$
Where $\text{Adj}(A)$ is the adjugate of $A$ which satifies Cramer's rule $\text{Adj}(A).A=A.\text{Adj}(A)=\det(A).I$.
Since invertible matrices are dense, the formula follows. 
This poof can be simplified a little: Prove it for $A=I$ first, and then use that $\det:GL(n)\to (\mathbb R\setminus 0,\cdot)$ is a group homomorphism.
Finally note, that Jacobi's formula
$$ 
\det(e^A) = e^{\text{Tr}(A)}
$$ 
is a consequence, by inserting $tA$ instead of $A$ and differentiating.
Both sides are 1-parameter subgroups and satisfy the same ODE with the same initial condition. 
EDIT: Misprints corrected; Thanks to Peter Kravchuk for pointing them out.<|endoftext|>
TITLE: What is the least ordinal than cannot be embedded in $\mathbb{R}^\mathbb{R}$?
QUESTION [12 upvotes]: Let $\mathbb{R}^\mathbb{R}$ be the set of functions $\mathbb{R}\to\mathbb{R}$ patially ordered by eventual domination. Obviously, every ordinal below $\omega_1$ can be embedded in $\mathbb{R}^\mathbb{R}$ using only constant functions.
What is the least ordinal than cannot be embedded in $\mathbb{R}^\mathbb{R}$?

REPLY [14 votes]: Let me get things started with some simple observations.
Note that given any countable sequence of functions $f_n$, we can
by diagonalization construct a function eventually dominating all
of them, $f(x)=\max_{n\leq x}f_n(x)$. It follows that we may by
transfinite recursion construct an embedding of $\omega_1$ into
your order: at successor stages, add one to the previous function;
at limit stages, use the diagonalization just described.
So actually, since $\mathbb{R}$ is order-isomorphic to bounded
intervals of itself, we can therefore also embed $\omega_1$ into
the order many times, on top of one another. So this gives
strictly larger ordinals mapping in.
More generally, the bounding number $\mathfrak{b}$ is the size
of the smallest unbounded family of functions, and any family of
size less than $\mathfrak{b}$ will be bounded above. Thus, the
recursive construction actually shows that we can find an
embedding of $\mathfrak{b}$ into $\mathbb{N}^{\mathbb{N}}$ under
eventual domination. Thus, we also get strictly larger ordinals
than $\mathfrak{b}$ embedding in, by using the bounded-interval trick again.
There are diverse independence results concerning the exact
value of $\mathfrak{b}$. Under CH, it is the same as the
continuum, of course, but when CH fails, it can be far larger than
$\omega_1$.
Using Péter's idea, once we have a map from $\mathfrak{b}$ into the order, then we may conclude that the class of ordinals that map into the order is closed under sums of length $\mathfrak{b}$. Thus, any ordinal up to $\mathfrak{b}^+$ is is order-embeddable into $\mathbb{R}^\mathbb{R}$ under eventual domination. So $\mathfrak{b}^+$ is a lower bound for your desired ordinal.
I guess the same idea shows that whenever an ordinal $\kappa$ embeds in, then the class of ordinals will be closed under sums of length $\kappa$, and so all ordinals up to $\kappa^+$ will also map in. Thus, the smallest ordinal not embedding in must be a cardinal, and furthermore, it must be a regular cardinal for the same reason.
Update.  It is relatively consistent that the answer is $\mathfrak{c}^+$, even when the continuum $\mathfrak{c}$ is very large, and much larger than $\mathfrak{b}$. The reason is that by forcing, we can undertake a very long forcing iteration of length $\kappa$ to add a dominating real at each stage, and thereby get a model with continuum $\kappa$, such that $\kappa$ embeds into the order (and so the smallest ordinal not embedding into the order is $\kappa^+$). Now, the point is that with further ccc forcing, we can make $\mathfrak{b}$ small or whatever we like, but meanwhile, we still have our old functions showing that $\kappa$ maps into the order.

REPLY [6 votes]: One can define an eventually increasing sequence $\{f_\alpha:\alpha<\omega_1\}$ by transfinite recursion on $\alpha$. By inserting sequences in the intervals $(f_\alpha(x),f_{\alpha+1}(x))$ we can see that $\omega_1$-sums (and of course, $\omega$-sums) do not lead out of the representable ordinals. This gives that all ordinals $<\omega_2$ are representable and this is clearly sharp if CH holds, by Will's above answer. 
If CH fails, larger ordinals can be represented, if, e.g. Martin's Axiom holds.<|endoftext|>
TITLE: The "right" $C^*$ algebraic proof of Bott Periodicity
QUESTION [10 upvotes]: In learning about the K-theory of $C^*$-algebras, I have encountered the following 3 proofs of Bott periodicity:
$\bullet$ An argument based on Moyal quantization found in "Elements of Noncommutative Geometry."
$\bullet$ An argument based on Toeplitz algebras in Murphy's book.
$\bullet$ A seemingly brute force (but elementary) argument in Rordam, Larsen and Laustsen's book where they prove a variety of density results about projections in matrix algebras.
Which of these proofs is the most "geometrical" in the sense that it has a nice geometric interpretation when we restrict our attention to commutative $C^*$-algebras? As a student interested in noncommutative topology and geometry, if I wanted to study one of these proofs in gory detail, which should it be? 
If there are other proofs that are more geometrical or more essential for understanding the phenomenon of Bott periodicity, I would be happy to hear about those too.

REPLY [9 votes]: I opine that everyones favorite should be the proof using Toeplitz operators (as described in Higson-Roe); essentially due to Atiyah (in ''Bott periodicity and the index of elliptic operators''). In the commutative case, it is the cleanest and most memorable proof. It gives an explicit homotopy inverse to the Bott map as a map $\Omega U \to Z \times BU$ spaces). It is fairly elementary. It is directly linked to the Toeplitz index theorem (that relates the most elementary topological invariant, the winding number, to an index). The Toeplitz index theorem is, moreover, one of the building blocks of the Atiyah-Singer index theorem. The proof can be generalized to the Real case. It sets the stage for Cuntz' proof of periodicity in $KK$-theory.
There is one drawback that I am willing to take serious: there is, as far as I can see, no straightforward generalization to the Clifford-linear case.

REPLY [4 votes]: Probably the argument that you're looking for is based on the "Dirac / Dual Dirac" method.  The idea is to exploit the product structure in KK-theory as much as possible - this makes the proof functorial in practically every possible way, and it suggests lots of generalizations (for instance, a variation on the argument can prove the Baum-Connes and analytic Novikov conjectures in a lot of cases).  Here's the structure of the argument.
First, construct a "Dirac class" $\delta$ in the K-homology group $K^1(S\mathbb{C})$ and a "dual Dirac class" $\beta$ in the K-theory group $K_1(S\mathbb{C})$ ($S$ means suspension).  The Dirac class is the K-homology class of the Dirac operator on $S^1$, or alternatively the K-homology class of the Toeplitz extension.  The dual Dirac class is also known as the Bott class (because of it's role in this proof); it can be viewed as the K-theory class of the Clifford multiplication operator.  
Second, prove that the map $K_0(\mathbb{C}) \to K_1(S\mathbb{C})$ given by multiplication by $\beta$ and the map $K_1(S\mathbb{C}) \to K_0(\mathbb{C})$ given by multiplication by $\delta$ are inverses and hence both are isomorphisms.  This is a direct calculation, and it's really the meat of the proof.  It can be reduced to the Toeplitz index theorem, for instance.
Third, simply observe that the map $K_0(A) \to K_1(SA)$ given by multiplication by $\beta$ is an isomorphism by naturality properties of K-theory products.
This is all worked out in a number of places; the expository paper "Group C*-algebras and K-theory" by Guentner and Higson or the textbook "Analytic K-homology" by Higson and Roe both provide this argument in some form, though both references demand some effort to get to the finish line.  The first reference might be my favorite; as an added advantage, you'll learn the basics of E-theory along the way.<|endoftext|>
TITLE: Dijkgraaf-Witten TQFT vs. Representation Theory?
QUESTION [12 upvotes]: From what I had read, group characters can be "glued" together in a topological fashion and there is something to this effect in the paper by Dijkgraaf and Witten. TQFT seems to be a topological generalization of representation theory.
Dijkgraaf and Witten write down a few interesting formulas in Section 6:

$\displaystyle Z(S^3) = \frac{1}{|G|} $
$\displaystyle Z(S^2 \times S^1) = 1$
$\displaystyle Z(\mathbb{R}P^3) = \frac{1}{2}(1 + (-1)) = 0$
$\displaystyle Z( M ) = \frac{1}{|G|} \sum_{\gamma \\, : \\, \pi_1(M)\to G} e^{2\pi i S}$
$\displaystyle Z(S^1 \times S^1 \times S^1) =  \frac{1}{|G|}\sum_{[g,h]=[h,k]=[k,g]=1} W(g,h,k) = \sum_{g \in C} r(N_g; c_g)$, the sum over certain conjugacy classes.
$\displaystyle Z(S^3/\mathbb{Z}_n) = \langle \varnothing |(TST)^n|\varnothing \rangle = \left\{ \begin{array}{ll}\frac{1}{2}(1 + (-1)^{n/2}) & n \text{ even } \\\\
\frac{1}{2} & n \text{ odd }  \end{array} \right.$

A few years later, there seem to have been written Freed-Hopkins-Lurie-Teleman seem mainly concerned with the categorical structure of this story.  
About a year ago, I wrote a short group theory note generalizing the 5/8 bound using a tiny bit of TQFT (itself based on an MO question).  
I wonder, what these TQFT's can say about surfaces (or higher dim topological spaces) or groups aand their representations ?

REPLY [16 votes]: Strictly speaking, this answer is not about the 3d TQFT which you mention in your question, but rather a 2d version of Dijkgraaf-Witten theory (described in Section 2 of Freed-Hopkins-Lurie-Teleman). 
To every finite group $G$, there is a 2d TQFT $Z_G$ which assigns to a closed orientable surface $\Sigma$ the following sum over isomorphism classes of $G$ local systems $P \to \Sigma$:
$$
Z_G(\Sigma) = \sum 1/|Aut(P)| = |Hom(\pi_1(\Sigma),G)|/|G|.
$$
In the framework of TQFT as a symmetric monoidal functor $Bord \to Vect$, this assigns to a circle the space of class functions on $G$ (which is a commutative Frobenius algebra under convolution). We can extend further and define $Z_G$ on a point to be the category of representations of $G$ (or alternatively, the group algebra of $G$, depending on your set-up).
Analysing this TQFT on surfaces allows you to recover interesting group-theoretic identities. For example, by cutting up the surface $\Sigma$ into pairs of pants, recovers the following formula (probably first due to Frobenius):
$$
Z_G(\Sigma) = \sum_{V\in \widehat{G}} \left(\frac{\dim V}{|G|}\right)^{\chi(\Sigma)}.
$$
There are similar formulas involving the other entries in the character table for $G$, by considering surfaces with boundary (which can be thought of as counting $G$ local systems on a closed surface with singularities). 
These formulas were used by Hausel and Rodriguez-Villegas to compute data about the Hodge numbers of character varieties in their paper Mixed Hodge Polynomials of Character Varieties. 
The recent work of Ben-Zvi and Nadler Character Theory of a Complex Group is in some sense a categorified analogue of this TQFT, but where the finite group is replaced by a complex reductive group (as explained in the introduction). In ongoing work of myself with the authors, we are trying to understand what this structure says about the cohomology of character varieties.
At the risk of over advertising my own work, let me also mention this paper: Spin Hurwitz Numbers and TQFT, which describes an analogue of the Dijkgraaf-Witten TQFT for surfaces with spin structure.
There are probably many other references which I will try to add later...<|endoftext|>
TITLE: A double grading of catalan numbers
QUESTION [32 upvotes]: This is something I found in trying to work on Vince Vatter's excellent question. I have no solution, but a much more precise conjecture.
Recall that a rooted planar tree is a rooted tree where, for every vertex, the children of that vertex come equipped with an order. I will think of the vertices of tree as members of an asexually reproducing species, and therefore use language like "sibling", "cousin", "child", "parent", "generation" etc. When I speak of generations, I am counting from the oldest, so the root is the fist generation; the children of the root are the next generation, and so forth.
For any vertex, I will think of the ordering of its children as the birth order of the children.
Define a vertex $v$ to be "crucial" if $v$ is the youngest member of its generation, all the other members of that generation are childless, but $v$ has children. For example, in the tree
$$\begin{matrix}
 & & a & & \\
 & & \downarrow & & \\
 & & b & & \\
 & \swarrow & \downarrow & \searrow \\
c &  & d & & e \\
& & \downarrow & & \downarrow \\
& & f & & g \\
& & \downarrow & \searrow & \\
& & h & & i \\
& & & & \downarrow \\
& & & & j \\
\end{matrix}$$
the crucial elements are $a$, $b$, $f$ and $i$. So the root is always crucial.
Let $c(p,q,n)$ be the number of planar trees on $n$ vertices with $p$ crucial elements and where the root has $q$ children. For example, of the $5$ planar trees on $4$ elements, there is one tree each with $(p,q)$ equal to $(1,2)$, $(2,1)$, $(3,1)$, $(2,2)$ and $(1,3)$.
The number of planar trees, $\sum_{p,q} c(p,q,n)$, is the Catalan number $C_{n-1}$ (item $e$ on Stanley's famous list). Also, $\sum_{p} c(p,1,n)$ is $C_{n-2}$ -- this counts planar trees where the root only has one child, so just delete the root to get planar trees on $n-1$ vertices.
Planar trees on $n$ vertices are in bijection with sequences $a_1$, $a_2$, \dots, $a_{n-1}$ of integers such that $a_1 =0$ and $0 \leq a_{i+1} \leq a_i+1$ (item $u$ on Stanley's list). The bijection is to consider $j$ to be in generation $a_j$, with the parent of $j$ being the largest $k$ such that $a_k = a_j-1$ and $k \lt j$. Then put a root at the top in generation $-1$.
The lists of integers in Vatter's question are a subset of lists above. Specifically, they are the ones where the root is the only crucial vertex. (Doug Zare's answer was very helpful in making me realize this description.) So our goal is to show that $\sum_q c(1,q,n)$ is Catalan. After a lot of computation, I came to the following conjecture:

The integer $c(p,q,n)$ only depends on $p+q$ and $n$. The array of integers $c(p+q, n)$ is the Catalan triangle.

I'm bad at indexing, so I'll give the values for planar trees on $10$ vertices. In total, there are $4862$ such trees in total. There are $429$ of them with $(p,q) = (1,2)$ and the same number with $(p,q) = (2,1)$. In total, the number of trees with each value of $(p,q)$ is:
$$\left(
\begin{array}{ccccccccc}
 0 & 429 & 429 & 297 & 165 & 75 & 27 & 7 & 1 \\
 429 & 429 & 297 & 165 & 75 & 27 & 7 & 1 & 0 \\
 429 & 297 & 165 & 75 & 27 & 7 & 1 & 0 & 0 \\
 297 & 165 & 75 & 27 & 7 & 1 & 0 & 0 & 0 \\
 165 & 75 & 27 & 7 & 1 & 0 & 0 & 0 & 0 \\
 75 & 27 & 7 & 1 & 0 & 0 & 0 & 0 & 0 \\
 27 & 7 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
 7 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
\end{array}
\right)$$
Note that the numbers 1, 7, 27, 75, 165, 297, 429, 429 appear as a row of the Catalan triangle. In particular, proving this proves Vatter's conjecture, since the sum $\sum_q c(1,q,n)$ would then be the sum of a row of the Catalan triangle, and hence a Catalan number.
Federico Ardila found a matroid whose Tutte polynomial $T_n(x,y)$ obeyed $T_n(1,1) = C_n$ and $T_n(x,y) = T_n(y,x)$. I have checked numerically that $T_n(x,y)$ appears to be $\sum c(p,q,n) x^p y^q$. Federico gives an explicit generating function for his polynomial (Theorem 3.6). From this formula, one can check that the coefficient of $x^p y^q$ depends only on $p+q$. I will add that this is a very unusual property for a Tutte polynomial to have, and I would be interested to know any general property of a matroid which implies it.
Federico gives a combinatorial interpretation for the coefficient of $x^p y^q$ in $T_n(x,y)$ in terms of Dyck paths. For him, $q$ is the number of times the Dyck path returns to $0$, which is easily related to the number of children of the root. However, his $p$ is the number of up steps before the first down step. I do not see why this should be the number of crucial vertices.
What's going on here?

REPLY [3 votes]: I wanted to add a simple combinatorial proof that the coefficient of $x^p y^q$ is dependent only on $p+q$. Given a Dyck path of length $2n$, let $q$ be the number of returns to $0$ and let the initial segment be $0 1 2 3 \cdots (p-1) p (p-1)$. I will introduce an operation which sends a path of type $(p,q)$, with $q \geq 2$, to one of type $(p+1, q-1)$. 
So, take a Dyck path with $q \geq 2$. Write it as a matching parentheses sequence; we can decompose it as
$$( D_1 ) ( D_2 ) ( D_3 )  \cdots ( D_q )$$
where each $D_i$ is, itself, a matching parentheses sequence. We map it to
$$((D_1)D_2) (D_3) \cdots (D_q)$$
It is easy to see that this map is invertible for $p \geq 2$.<|endoftext|>
TITLE: A technical question related to Zhang's result of bounded prime gaps
QUESTION [13 upvotes]: Here is a link on the internet: https://www.dropbox.com/s/su3uak2a057yrqv/YitangZhang.pdf
Can someone teach me how to use trivial estimation to reach (6.1) on page 24? Namely, how to impose $(d,P_0)<D_1$ and replace $\theta$ by $\Lambda$ with acceptable O-term?
Thanks a lot!

REPLY [10 votes]: Let me remind some of notations for readers:
$$\mathcal{L}=\log x,$$
$$k_0=3.5\times 10^6, \varpi=\frac{1}{1168},$$
$$D_0=\exp(\mathcal{L}^{\frac{1}{k_0}}),  P_0=\prod_{p < D_0}p,$$
$$D_1=x^{\varpi},P=\prod_{p < D_1}p,$$
$$D=x^{\frac{1}{4}+\varpi}, D_2 = x^{\frac{1}{2}-\epsilon}$$
Here, the effect of replacing $\theta$ by $\Lambda$ produces an error of 
$$O(x^{\frac{1}{2}}\mathcal{L}^B),$$
for some positive $B$. So, this change is negligible compared to $O(x\mathcal{L}^{-A})$.
For imposing $(d,P_0)< D_1$, we claim that the following sum is also negligible compared to $O(x\mathcal{L}^{-A})$:
$$
\sum_{\substack{{D_2 < d {<} D^2} \\\ {d|P} \\\ {(d,P_0)\geq D_1}}} \sum_{c\in C_i(d)} |\Delta(\Lambda,d,c)|
$$
Note that the trivial bound for $|\Delta(\Lambda,d,c)|$ is:
$$
|\Delta(\Lambda,d,c)|=O(\frac{x\mathcal{L}}{d})$$
Since we have $(d,P_0)\geq D_1$, the number $w(d)$ of distinct prime divisors of $d$, must satisfy 
$$
w(d )\geq \mathcal{L}^{\varpi(1-\frac{1}{k_0})}.$$
The sum is bounded by:
$$
x\mathcal{L} \sum_{\substack{{ d {<} D^2} \\\ {w(d )\geq \mathcal{L}^{\varpi(1-\frac{1}{k_0})} } }} \frac{\tau_{k_0}(d)}{d }$$
Standard argument now applies, and the sum above is bounded by:
$$
\frac{1}{2^{\mathcal{L}^{\varpi(1-\frac{1}{k_0})}}}\sum_{d < D^2} \frac{2^{w(d)}\tau_{k_0}(d)}{d}=O(\frac{\mathcal{L}^B  }{2^{\mathcal{L}^{\varpi(1-\frac{1}{k_0})}}}),$$
for some positive constant $B$. 
Combining all together, we obtain an upper bound:
$$
\sum_{\substack{{D_2 < d {<} D^2} \\\ {d|P} \\\ {(d,P_0)\geq D_1}}} \sum_{c\in C_i(d)} |\Delta(\Lambda,d,c)| =O(\frac{x\mathcal{L}^{B+1}}{2^{\mathcal{L}^{\varpi(1-\frac{1}{k_0})}} })$$
Hence, our claim follows.<|endoftext|>
TITLE: Construction of exotic spheres that do not bound parallelizable manifolds
QUESTION [8 upvotes]: There are at least two ways to construct homotopy spheres that bound parallelizable manifolds, namely Milnor's plumbing construction and Brieskorn's method of singularities, and each of these methods has the virtue that it produces every element of $bP_{n+1}$ for every $n$.


What is the state of the art in constructing elements of $\Theta_n\setminus bP_{n+1}$?

REPLY [3 votes]: I don't really know about the state of the art, but I've come across a couple of examples in the literature at least. In the article:
Frank, David L., An invariant for almost-closed manifolds, Bull. Amer. Math. Soc. 74 (1968) 562–567, MR0222906
the unique exotic $8$-sphere and an order-$3$ element of $\Theta_{10} \cong \mathbb{Z}/6$ are shown to be in the image of Milnor's plumbing construction; see Examples 1 and 2 on page 565. Since they're exotic (i.e. not the standard sphere) and even-dimensional, they do not bound any parallelisable manifold (since $bP_{2n+1} = \{S^{2n}\}$). Also, in the article:
Sperança, L. D., Pulling back the Gromoll-Meyer construction and models of exotic spheres, Proc. Amer. Math. Soc. 144 (2016), no. 7, 3181–3196, MR3487247
there is an explicit description of clutching diffeomorphisms that realise these two examples as twisted spheres; see Theorem 4.6 (plus the description in the middle of page 3187 of "reentrance").<|endoftext|>
TITLE: Universal covering of compact surfaces
QUESTION [8 upvotes]: Is there any elementary (i.e. without using analytical methods like the theory of Riemann surfaces or more elaborate results from differential geometry) way to show that the universal covering of the compact oriented surface of genus $g>0$ is homeomorphic to $\mathbb R^2$?

REPLY [11 votes]: You can build a certain covering space of the surface $S$ rather explicitly as a nested union of closed discs $D_1 \subset D_2 \subset D_3 \subset \cdots$, each contained in the interior of the next, from which it follows that the union is $\mathbb{R}^2$ and, being simply connected, is therefore the universal covering space. To construct these discs, start by representing your surface in the usual fashion as the quotient of a $4g$-gon, with oriented 1-cells assigned labels, and with side pairings respecting orientations and labellings. The disc $D_1$ is a single copy of the $4g$-gon $Q$ with appropriate side labels, and the map $D_1 \to S$ is the one given by the quotient map. Then $D_2$ is constructed by attaching additional copies of $Q$ to the periphery of $D_2$, exactly as they ought to be: one copy of $Q$ attached to each edge of $D_1$, and additional copies of $Q$ attached to each vertex so as to fill out $4g$ copies of $Q$ attached around each vertex of $D_1$. Et cetera.<|endoftext|>
TITLE: global sections of structure sheaf on the toric Calabi-Yau
QUESTION [5 upvotes]: Let P be a lattice polytope and lying in $ N \times {1} \subset N \times \mathbb{R}$. Let $\sigma$ be the cone over this polytope and $X_\sigma$ be the corresponding toric variety, which is an affine,Gorenstein, toric variety. 
Is there a simple algorithm for computing the ring of global sections of the structure sheaf on X_{\sigma}? I would like the presentation in terms of generators and relations if possible. If this is not possible in general, I would be happy if this is possible under some nice, fairly general situation. Thank you for your help. 
I suppose a different version of this question is: Are there computer programs which are freely available and which can make these computations quickly?

REPLY [2 votes]: Probably this answer comes too late, but anyway:
What you want to do can be divided into two tasks:

Find a Hilbert basis of (the dual of) your cone.
Compute the toric ideal for the Hilbert basis.

I.e. let $M=\mathbb Z^n$ be the dual lattice of $N\times\mathbb Z$. You start with your cone $\sigma$, dualize it and obtain $\sigma^\vee$. Now you compute a minimal generating set $H:=\{h_1,\ldots,h_m\}\subseteq M$ of the semigroup $\sigma^\vee\cap M$. Then you get a surjection
$$
k[x_1,\ldots, x_m]\to k[\sigma^\vee\cap M],\ x_i\mapsto x^{h_i}
$$
The kernel of this morphism is the ideal $I$ you want. If you denote $H$ as a matrix, it can be described as
$$
I=( x^u - x^v\ | u,v\in\mathbb Z^m_{\ge 0},\ H\cdot(u-v)=0).
$$
Thus you can obtain generators of $I$ by just taking all vectors of $\ker H$, splitting them into positive and negative part, and then take the corresponding binomials.
This you probably knew already, but now for the software:

I know two (three) frameworks for computing Hilbert bases: 4ti2 and Normaliz. You can enter your cone via rays or give the facet inequalities, i.e. the rays of $\sigma$. Thus, dualization can be done on the fly. Macaulay2 is also able to compute Hilbert bases via the Polyhedra package, but the first two are probably faster.
To compute the toric ideal you can use the groebner method of 4ti2 (see Usage->groebner on the 4ti2 page). Singular provides some algorithms for computing toric ideals on its own, via the toric.lib package. If performance is an issue, 4ti2 is probably faster.

Both Macaulay2 and Singular provide packages interfacing 4ti2, which makes dealing with the algebraic aspect easier.<|endoftext|>
TITLE: Transitive subgroup of $S_p$ containing a $p$-cycle and a double transposition
QUESTION [5 upvotes]: Let $p$ be a prime other than 5 or 7.  Are $A_p$ and $S_p$ the only subgroups of $S_p$ that contains a $p$-cycle and a double transposition?
As for $p = 5$, the dihedral group $D_{10}$ contains a 5-cycle and a double transposition.  For $p = 7$, the group $PSL_3(\mathbb{F}_2)$ (acting on the projective plane of order 2) contains a 7-cycle and a double transposition.

REPLY [6 votes]: There is a way to prove this via analogy with a proof approach which works for transpositions and 3-cycles.  
For transpositions and 3-cycles on 4 or more points, the same proof (more or less) works easily to show that a primitive permutation group on $n$ points containing a transposition (resp., a 3-cycle) must be $S_{n}$ (resp., or $A_{n}$):
Define a relation on the indices by $i \sim j$ means "there is a transposition (resp., 3-cycle) in $G$ sending $i$ to $j$".
This relation is reflexive: Since $G$ is primitive, $G$ is transitive. $n \geq 4$ means that transpositions and 3-cycles have fixed points. The transitivity of $G$ allows a fixed point of a transposition or 3-cycle to be moved (by conjugation) to any desired index.
This relation is symmetric: If $\sigma$ is a transposition or a 3-cycle sending $i$ to $j$, then $\sigma^{-1}$ is, respectively, a transposition or 3-cycle sending $j$ to $i$.
This relation is transitive: This one takes more work. The cases where any two of $i$, $j$ and $k$ are equal are trivial, so assume they are distinct. Suppose $\alpha$ sends $i$ to $j$ and $\beta$ sends $j$ to $k$. If $\alpha$ and $\beta$ are both transpositions, then $< \alpha , \beta>$ acts as $S_{3}$ on {$i, j, k $}. Since this also contains the transposition $ (i,k) $, we are done. If $\alpha$ and $\beta$ are both 3-cycles, then $ < \alpha, \beta> $ acts as $A_{3}$, $A_{4}$ or $A_{5}$ on the union of the sets of indices moved by $\alpha$ and $\beta$. This alternating group contains all 3-cycles moving these points, so it contains the one sending $i$ to $k$.
So that relation is an equivalence relation. Moreover, from its definition, it is clear that it is a $G$-invariant equivalence relation on the indices. Then, since $G$ contains a transposition (resp., a 3-cycle), this equivalence relation has fewer than $n$ equivalence classes. Since $G$ is primitive, this means that all indices are equivalent.
In the case of transpositions, generating the whole $S_{n}$ is easy: since any two indices are switched by a transposition, $G$ contains all transpositions and thus $G = S_{n}$.
In the case of 3-cycles, it is slightly trickier: start with the 3-cycle $(1,2,a)$ which is in $G$. This generates $A_{3}$ on these 3 points. Now extend this subgroup by adjoining a 3-cycle which sends $1$ to an unmoved index. This results, as noted before, in $A_{4}$ or $A_{5}$. Again we extend this generating set by adjoining a 3-cycle which sends a moved index to an unmoved one, until all indices have been moved.
The claim is that the result of extending the group by adjoining the next 3-cycle is a new alternating group on all the moved points. If the next 3-cycle has only one unmoved point, this is immediate from the orbit-stabilizer theorem. If the next 3-cycle has 2 unmoved points, then the resulting group is transitive on $r \geq 6$ points and has a point stabilizer of order at least $\frac{ (r-2)!}{2}$. This means the extended group has order at least $\frac{ r (r-2)!}{2}$, so its index in $A_{r}$ is at most $r-1$. Since $r \geq 6$, $A_{r}$ has no subgroups of index less than $r-1$ except for itself. So the extended group is an alternating group, and $G$ has enough 3-cycles to generate $A_{n}$. Thus $G = A_{n}$ or $S_{n}$.  
Now, just as moving from transpositions to 3-cycles introduced new hoops to jump through, moving from 3-cycles to double transpositions makes carrying out this idea even trickier. We now assume, as the problem specifies, that $n \geq 9$.
We define the same relation and want to prove it's an equivalence relation. Reflexivity only requires $n \geq 5$, so we're good. Symmetry is immediate.
Transitivity is more work: Suppose $\alpha$ is a double transposition which sends $i$ to $j$, and $\beta$ is a double transposition which sends $j$ to $k$. As before, the problem is trivial if any of $i$, $j$ and $k$ coincide. So assume $i$, $j$ and $k$ are distinct. The conjugate $\beta \alpha \beta$ is also a double transposition and it sends $i$ to $k$ unless $\beta$ also moves $i$. So write $\alpha = (i,j)(k,l)$ and $\beta = (j,k)(i,m)$. Here $i$, $j$, $k$ and $l$ are distinct by assumption, and $m$ is not $i$, $j$ or $k$. If $m = l$, then $\alpha \beta$ sends $i$ to $k$ ( and, in fact, $< \alpha, \beta >$ acts as $V$ on {$i ,j , k, l$} ). If $m$ is a fifth index, then $< \alpha, \beta >$ acts as the dihedral group of order 10 on {$i, j, k, l, m$} and this group has just enough double transpositions to send any index to any other. So transitivity is done.
As before, this relation is an equivalence relation. Since $G$ by assumption contains double transpositions, this equivalence relation, as before, makes all indices equivalent and there is a double transposition in $G$ switching any two desired indices.
The trickiest part of this proof is building up the set of double transpositions to generate $A_{n}$.
We start with the subgroup generated by a single double transposition, $(1,2)(i,j)$. This group has 2 orbits for its action on its set of moved points. The number of orbits for the action of the group we build on the set of moved points never needs to exceed 2, since we may adjoin our double transposition to switch two points in different nontrivial orbits and get at most one other orbit from the other transposition. In fact, continuing in this way as long as we have two orbits of moved points gives us, at each step, a group acting transitively on its set of moved points or a group acting with two orbits on its moved points, one of size 2 and the other (the large orbit) on which the group acts faithfully.
If the group we build is ever transitive on its set of moved points and there are still unmoved points, adjoin a double transposition which switches a moved point with an unmoved point. (This also keeps the number of orbits less than or equal to 2, with a large orbit and a 2-point orbit when there are 2 orbits on the moved points.)
Before going further, it is good to describe how to handle the extreme cases, where the group acts as $ (S_{r-2} \times S_{2}) \cap A_{r}$ or $A_{r}$ on its $r \geq 4$ moved points. If we have $A_{r}$, then if there are no unmoved points $r = n$ and $G$ contains $A_{n}$ as desired. If $r < n$, then adjoin a double transposition which switches a moved point with an unmoved point. If the two points of the other transposition are already moved, then the resulting group can only be $A_{r+1}$. If the two points of the other transposition are both unmoved, we likewise must have $ (S_{r+1} \times S_{2}) \cap A_{r+3}$. If one is moved and the other is unmoved, then the resulting group is transitive on $r+2$ points, consisting entirely of even permutations, and containing $A_{r}$ in a point stabilizer. Since $r \geq 4$, $r+2 \geq 6$ and, following the reasoning used before with 3-cycles, the group obtained must be $A_{r+2}$.
If the group acts as $ (S_{r-2} \times S_{2}) \cap A_{r}$ on $r > 4$ points (the inequality is strict to exclude the position we start from, with a lone double transposition), then adjoin a double transposition which switches a point in the large orbit and a point in the 2-point orbit. If the two points of the other transposition are both unmoved, we have $ (S_{r} \times S_{2}) \cap A_{r+2}$ as our new group. If both points of the other transposition are moved, then the maximality of $ (S_{r-2} \times S_{2}) \cap A_{r}$ in $A_{r}$ (here I am using the assumption that $r > 4$) implies that the new group must be $A_{r}$. If one point of the other transposition is moved and one is unmoved, then considering which 3-cycles are in the resulting group (remember that $r > 4$ so $ (S_{r-2} \times S_{2}) \cap A_{r}$ has 3-cycles in it) shows it must be $A_{r+1}$.
So if, at any point in this process, we obtain a group that acts as $A_{r}$ or $ (S_{r-2} \times S_{2}) \cap A_{r}$ on its $r > 4$ moved points, then the adjunction process will ultimately give us $A_{n}$ on our $n$ points when we have adjoined enough double transpositions. However, unlike the situation with 3-cycles, we don't settle into this situation so easily.
Applying the procedure to the subgroup generated by our first double transposition, $(1,2)(i,j)$, we adjoin a double transposition switching $2$ and $i$. The other transposition of this adjoined element switches two moved points, two unmoved points, or one moved point and one unmoved point. If it switches two moved points, it is $(2,i)(1,j)$ and the result is $V$ on {$1,2,i,j$}. If it switches one moved point and one unmoved point, the result is $D_{10}$ on 5 points. If it switches two unmoved points, the result is $D_{8}$ on the 4 points of the large orbit. The basic idea of the rest of the proof is this: continue all the cases of this adjunction procedure until they all lead to dead ends (via $ (S_{r-2} \times S_{2}) \cap A_{r}$ or $A_{r}$ moving more than 4 points, or what immediately follows this). (It is in working out these cases that the assumption that $n \geq 9$ winds up being fully used.)
One issue that description doesn't capture, however, is the fact that $V$ leads to another infinite family of groups: $V$ can be thought of as $ (S_{2} \wr S_{2}) \cap A_{4}$. $ (S_{2} \wr S_{t}) \cap A_{2t} $ is transitive on $2t$ points for any $t \geq 2$, and we adjoin a double transposition which switches a moved point (which we will call $1$) with an unmoved point (which we will call $2t+1$). However, the other transposition might be $(2,2t+2)$, in which case we have extended $ (S_{2} \wr S_{t}) \cap A_{2t}$ to $ (S_{2} \wr S_{t+1}) \cap A_{2t+2}$. To handle this, prove that $t > 4$ implies that that group is a maximal subgroup of $A_{2t}$. This will work (though it's written in a way that shows I haven't polished it):  
Suppose $K$ is strictly intermediate between $ (S_{2} \wr S_{t}) \cap A_{2t}$ and $A_{2t}$.  

Show that $K$ is doubly transitive. It's already given that $K$ has elements which send two indices in different pairs (thinking of $ (S_{2} \wr S_{t}) \cap A_{2t}$ as the stabilizer of a pairing in $A_{2t}$) to two indices in different pairs, and two paired indices to two paired indices. We only need to show that $K$ can send two paired indices to two indices in different pairs and vice versa. This is done by conjugating an element of $K$ outside $ (S_{2} \wr S_{t}) \cap A_{2t}$ appropriately.  
A 2-point stabilizer in $K$ contains $ (S_{2} \wr S_{t-1}) \cap A_{2t-2}$, so $K$ is, in fact, triply transitive.  
A 3-point stabilizer in $K$ contains $ (S_{2} \wr S_{t-2}) \cap A_{2t-4}$, so either it is transitive or it has a fixed point and an orbit of size $2t-4$. It can't be transitive because then $K$ would be quadruply transitive and contain double transpositions, so $K$ would contain all double transpositions, contradicting $K < A_{2t}$. So it has a fixed point and an orbit of size $2t-4$.  
This means that $K$ preserves a $S(3,4,2t)$ Steiner system. To extend a set of 3 points to a set of 4, take the (pointwise) 3-point stabilizer, and adjoin the extra fixed point one gets outside the orbit of size $2t-4$.  
We want to identify the points of this $S(3,4,2t)$ Steiner system with the elements of an affine space over $ \mathbb{Z} / (2) $. Equivalently, we want to identify an arbitrary chosen point with the zero vector and identify the points of the remaining $ S(2, 3, 2t-1) $ Steiner system with nonzero vectors in an $ \mathbb{Z} / (2) $ vector space.  
We define addition in our $S(3,4,2t)$ Steiner system as follows: $0+v = v+0 = v$ for any $v$, $v+v = 0$ for any $v$, and, if $v$ and $w$ are different nonzero vectors, then $v+w$ is the third point of the Steiner system block determined by $v$ and $w$.  
Commutativity of this addition law is easy to check. Every vector it its own additive inverse, so that's not an issue.  
It remains to check associativity. It is easy to check that $ (a+b)+c = a+(b+c) $ when any one of $a$, $b$ or $c$ is zero. Therefore assume that they are all nonzero. It is likewise easy to prove associativity when any two of them are equal. Therefore assume they are distinct. It is easy to prove associativity when $a+b = c$ (that is, they form a block of the $S(2,3,2t-1)$ Steiner system). So assume that $a+b \neq c$.  
How many fixed points does the pointwise stabilizer of $a$, $b$ and $c$ have? When $a$ and $b$ are fixed, $a+b$ also appears as a fixed point. Then, when $a$, $b$ and $c$ are fixed, we have $a$, $b$, $c$, $a+b$, $a+c$ and $b+c$ as fixed points. These must all be distinct (the cancellation laws are easy to verify for this addition, and the assumptions on $a$, $b$ and $c$ establish the rest of the distinctness), so we have at least 6 fixed points. However, $ (S_{2} \wr S_{t-2}) \cap A_{2t-4}$ is, when $t > 4$, non-regularly transitive on $2t-4$ points, so that a point stabilizer in it fixes only 2 points and the pointwise stabilizer of $a$, $b$ and $c$ (in the stabilizer of 0 in $K$) has 5 fixed points (and so is unable to accommodate our 6 fixed points). This is enough to contradict the existence of the $S(3,4,2t)$ Steiner system with these extra properties when $t > 4$, and so is enough to contradict the existence of $K$ when $t > 4$.  

That gives a way of doing it by bare hands. Enjoy the tour of the small permutation groups you get out of this!<|endoftext|>
TITLE: A conjecture on intersection of some intervals.
QUESTION [9 upvotes]: It was proved here that if $a\in \mathbb{N}_{\geq3}$ then
$$\bigcap_{i = 1}^{a} \bigcup_{j = 0}^{i-1} \left[\frac{1+aj}{i},\frac{a(j+1)-1}{i}\right] = \varnothing \tag{1}$$
It may be conjectured that forcing $i\ne b$, where $1\leq b< a$, renders $(1)$ untrue, that is, the result is not an empty interval.
We look at the diagram here from the link above for $a=5$ to get a better picture:
  
Red is the interval, Yellow are the gaps between the intervals that cause the intersection to be a null set, white gaps do not effect the intersection. The conjecture here says that if we were to remove any one of the top $4$ strips, then there will form a region of intersection. 
I tried analyzing the gaps but everything seems to meet up at a dead end.
What tools may one employ to handle such problems?

REPLY [6 votes]: As I commented, the original result can be restated as that for every $x \in [0,1]$ (or $x\in \mathbb R$) there is some $1\le i\le a$ so that $ix$ is within $1/a$ of an integer (and the proof is the pigeonhole principle -- two of the fractional parts of $0,x,2x,...,ax$ must be within $1/a$ of each other, so their difference is within $1/a$ of an integer). Your conjecture is that for any $1\le b \lt a$ there is some $x$ so that the only one of $x,2x,...,ax$ within $1/a$ of an integer is $bx$. For example, for $a=5$ and $b=2$, then if we take $x \in [\frac{44}{100},\frac{45}{100}] \cup [\frac{55}{100},\frac{56}{100}]$ then the only one of the first $5$ multiples within $1/5$ of an integer is the second.
If $2b \gt a$ then $x=1/b$ works. Since $b\lt a$, $i/b$ is within $1/a$ of an integer only when $i$ is a multiple of $b$.
For $2b \le a$ we can modify this to $x=1/b + 1/(2ab) = \frac{2a+1}{2ab}$. This is designed so that $bx = \frac{2ab + b}{2ab} = 1 + \frac{1}{2a}$ is within $1/a$ of $1$, but $2bx = 2 + \frac{1}{a}$ just barely misses being within $1/a$ of $2$. Larger multiples of $bx$ are also too large, $bix - i = \frac{i}{2a} \ge \frac{1}{a},$ while $(bi-1)x$ is too small to be within $1/a$ of $i$ when $bi-1 \le a$. $i - \frac{(bi-1)(2a+1)}{2ab} = \frac{2a - (bi-1)}{2ab} \ge \frac{a}{2ab} = \frac{1}{2b} \ge \frac{1}{a}.$<|endoftext|>
TITLE: Example for non equivalent rational full CFTs with same modular invariant (partition function)
QUESTION [10 upvotes]: I am looking for a counter example which shows, that a full rational 2D CFT (with respect to a given chiral subtheory) is not characterized by its modular invariant partition function. People tell me such an example exists, but noone could point out a concrete example or reference to me.

In the framework of Fuchs, Runkel, Schweigert etc. full CFTs are classified by Morita equivalence classes of Frobenius algebras in a modular tensor category $\mathcal C$ (the representation category of the chiral CFT), so my question more concretely:

I am looking for two special symmetric haploid Frobenius algebra objects $A,A'$ in a modular tensor category $\mathcal C$, which are not Morita equivalent but whose full centres $Z(A),Z(A')$ are equivalent as objects (they cannot be equivalent as algebra objects due to a result of Kong and Runkel), i.e. have the same modular invariant partition function $Z(A)_{ij}=Z(A')_{ij}$.


I am also fine with an example in any other formulation of CFT any reference in physics literature, non-degenerate braided subfactors or similar...

REPLY [6 votes]: There is an SU(3) example here in arXiv:math/0008056 pages 21 and 22 for $SU(3)_9$ in $E_6$ and also in $E_6 \times  Z_3$ - or see also pages 11 and 12 of arXiv:1002.2348,  page 6 of arXiv:0906.4252<|endoftext|>
TITLE: How many subsets of $\mathbb{R}$ are order isomorphic to $\mathbb{Q}$?
QUESTION [13 upvotes]: How many subsets of $\mathbb{R}$ are order isomorphic to $\mathbb{Q}$?  
How many subsets of the long line $\omega_1\times[0,1)$ are order isomorphic to $\mathbb{Q}$?  

I can see that results in both cases are between $\mathfrak{c}$ and $\mathfrak{2^c}$.

REPLY [9 votes]: There are $2^{\aleph_0}$ subsets of $\Bbb Q$ which are order isomorphic to $\Bbb Q$.
To see this, note that $\Bbb{Q\setminus N}$ is order isomorphic to $\Bbb Q$, and consider for every $A\subseteq\Bbb N$ the set $\Bbb{Q\setminus N}\cup A$.
Since there are no more than $2^{\aleph_0}$ countable subsets to $\Bbb R$ the answer has to be $2^{\aleph_0}$.

REPLY [2 votes]: There are continuum many countable subsets of the continuum (because $\mathfrak{c}^{\aleph_0}=2^{\aleph_0}$). Thus the answer is $\mathfrak{c}$. See this question.<|endoftext|>
TITLE: Is it possible to construct an infinite subset of $\Bbb R$ that is not order isomorphic to any proper subset of itself?
QUESTION [23 upvotes]: Is it possible to construct an infinite subset of $\Bbb R$ that is not order isomorphic to any proper subset of itself?

REPLY [18 votes]: The answer is yes in ZFC. We can construct a dense infinite set
$A\subset\mathbb{R}$ such that the only order-preserving map
$f:A\to A$ is the identity. In particular, $A$ is not
order-isomorphic with any proper subset of itself.
To see this, note first that any order-preserving map
$f:B\to\mathbb{R}$ defined on a dense set $B\subset\mathbb{R}$ can
be extended to a total order-preserving map $\bar
f:\mathbb{R}\to\mathbb{R}$ defined on the closure of $B$, by
defining $\bar f(x)=\sup_{y\leq x, y\in B}f(y)$. Further, note
that any such monotone map will have at most countably many points
of discontinuity, since every discontinuity will be a jump
discontinuity. Thus, there are precisely continuum many such
order-preserving functions $\mathbb{R}\to\mathbb{R}$, since any
one of them is determined by countably much information about
their values on a countable dense set and the information about
what their values are on the countably many points of
discontinuity.
We may therefore enumerate all order-preserving functions
$f_\alpha:\mathbb{R}\to\mathbb{R}$ in a sequence of length
continuum, $\alpha\lt\mathfrak{c}$.
Let's now build the set $A$ by a transfinite process, making
promises at each stage about some reals being definitely in $A$
and other promises about keeping some reals out of $A$, in such a
way that we kill off $f_\alpha$ at stage $\alpha$ as a possible
order-preserving map from $A$ to $A$. We may begin at stage $0$ by
placing all the rational numbers into $A$, so that it will
definitely be dense. Suppose we have carried out our process up to
stage $\alpha$, and $f_\alpha$ is the next non-identity
order-preserving map $\mathbb{R}\to\mathbb{R}$ presented for our
consideration. Since $f_\alpha$ is order-preserving and not the
identity, it must be that there is an interval $(a,b)$ with
$(f(a),f(b))$ disjoint from $(a,b)$. Since we've made fewer than
continuum many promises so far, there must be an $x\in (a,b)$ such
that we've made no promises about $x$ or $f_\alpha(x)$. In this
case, we place $x$ into $A$ and promise to keep $f_\alpha(x)$ out
of $A$. This will prevent $f_\alpha$ from being an
order-isomorphism of $A$ to a proper subset of $A$.
The end result is that $A$ is dense, but is strongly rigid in the
sense that there is no non-identity order-preserving map from $A$
to $A$. In particular, $A$ is not order-isomorphic with any proper
subset of itself.<|endoftext|>
TITLE: Determinant of non-symmetric sum of matrices
QUESTION [5 upvotes]: Given three real, symmetric matrices $A\succ0$ and $B$, $C⪰ 0$.
How can it be shown that:
$$\det(A^2+AB+AC) \leq \det(A^2 +BA +AC+BC) ? \qquad (\star)$$
Where $A^2$ is symmetric and positive definite. Eigenvalues of $BA$, $AC$, and $BC$ are all $> 0$, but symmetry is lost.
Thank you!

REPLY [6 votes]: Because the original question has changed so much, I am writing a new answer.
The key point to recognize is that you are trying to prove a submodularity property. Indeed, we see that we may equivalently prove
\begin{equation*}
  \log\det(A) + \log\det(A+B+C) \le \log\det(A+B) +\log\det(A+C).
\end{equation*}
One common way to verify submodularity is to prove the diminishing marginals property: in our case, it amounts to showing that for a fixed $B \succeq 0$, the function
\begin{equation*}
 f(A) := \log\det(A+B)-\log\det(A)
\end{equation*}
is monotonically decreasing (since we are dealing with hermitian positive definite matrices, this means $f(A) \le f(C)$ if $C \succeq A$ in the semidefinite order).
To verify this, simply check if $\nabla f(A) \preceq 0$ for all $A$. But this is easy since 
\begin{equation*}
 \nabla f(A) = (A+B)^{-1} - A^{-1} \preceq 0,
\end{equation*}
where the latter inequality follows as the map $X \mapsto X^{-1}$ is well-known to be (operator) decreasing.<|endoftext|>
TITLE: Simple abelian varieties over non algebraically closed fields.
QUESTION [5 upvotes]: I was wondering what people would normally mean by a simple abelian variety $A$ where $A$ is defined over a field $k$ that is not algebraically closed.
The definition I found in for example on the bottom of page 29 in Langs book Abelian Varieties states:

An abelian variety $A$ is simple if $0$ and $A$ are the only abelian subvarieties of A.

But this definition is not entirely unambiguous for me. And other sources that I read only define being simple over algebraically closed fields.
My main question is:
Would $A$ still be called simple if $A$ contains a nontrivial subvariety $B$ that is defined over $\bar k$ but such that $0$ and $A$ are the only subvarieties that are defined over $k$?
The reason I am asking is because I want to apply the results of http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.172.6978 to the weil restriction of an elliptic curve $A=Res_{K/\mathbb Q}E$. One of the hypothesis of that article is that $A$ must be simple. But depending on the definition of simple it might be so that $Res_{K/\mathbb Q}E$ is never simple as variety over $\mathbb Q$

REPLY [8 votes]: You're right: "simple abelian variety $A/K$" is ambiguous when $K$ is not algebraically closed.  One should say geometrically (or absolutely) simple or K-simple to emphasize which is meant.
If you do encounter "simple abelian variety $A/K$" in the literature, then in my experience it usually means "geometrically simple".  Somehow the typical perspective is that the geometric properties of the abelian variety are considered and described first, and the arithmetic properties only later.  A big example of this is that $\operatorname{End} A$ almost always means $\operatorname{End} A_{/\overline{K}}$, and then you can endow it with the structure of an $\operatorname{Aut}(\overline{K}/K)$-module if you like.  In my mind the notable exception is Honda-Tate theory, where the isogenies, the Weil polynomials, the endomorphism algebras and so forth really do all take place over a fixed finite field $\mathbb{F}_q$, and extending the ground field can change everything.
Having said all that I did glance through Waldschmidt's paper, and my impression is that he means $\mathbb{Q}$-simple abelian variety, so yes, including certain Weil restrictions.  I think that if you read the paper more carefully, this will become even more apparent: the simplicity assumption must come up in the proof, after all!  Further, he mentions several times that he is restricting to this special case for simplicity (!!), and makes references to a more general result holding for any commutative group $G_{/\mathbb{Q}}$.  So at some point you should try to track that down: maybe any abelian variety will do...<|endoftext|>
TITLE: First-order axiomatization of free groups
QUESTION [9 upvotes]: Is there a way to axiomatize [non-abelian] free groups in first-order logic using the language of groups (which contains the binary operation symbol $\cdot$, and the constant symbol $e$)?
Is there one particular axiom, or even a schema, from which we can prove that $G$ is a free group? (Regardless to the cardinality of a generating set.)
I should clarify that I'm not interested in augmented languages where we allow additional constant symbols for the generating set (in which case we can just write a schema stating when the various strings are equal).

REPLY [4 votes]: More can be said than non-first-order axiomatizability. Since the free group $  \mathbb{Z}^{(\omega)}$ is an $L_{\infty, \omega}$-elementary substructure of the non-$\aleph_{2}$-free group $\mathbb{Z}^{\omega}$, there is no axiomatization of the class of free groups (or of $\aleph_{2}$-free groups) in $L_{\infty, \omega}$. 
If large cardinals exist, the outcome is very different. Mekler proved that if there is a strongly compact cardinal $\kappa$, then the class of free abelian groups is definable in $L_{\infty, \infty}$ (and in $L_{\kappa, \kappa}$ in fact). Furthermore, if the class of free abelian groups is definable in $L_{\infty, \infty}$, then there is an inner model with a measurable cardinal.<|endoftext|>
TITLE: Bruhat-Schwartz functions and derivatives in p-adic numbers
QUESTION [5 upvotes]: First of all, I am not an expert in neither classical, nor $p$-adic functional analysis, but anyway, I stumbled over the following lately:
Let $\varphi:\mathbb{Q}_p\rightarrow\mathbb{C}$. 
Canonically, I would define a derivative of such a function in the following way:
$\varphi'(x):=\lim\limits_{h\rightarrow 0} \frac{\varphi(x+h)-\varphi(x)}{|h|_p}, $
provided of course that the limit exists. Is this the usual definition of the derivative?
In real functional analysis, there is the notion of Schwartz functions, which are rapidly decreasing functions supported on $\mathbb{R}^n$, i.e. functions, where for multiindices $\alpha,\beta$
$\sup_{x\in\mathbb{R}^n}|x^\alpha D^\beta (f)(x)|<\infty.$
To my knowledge, the term Bruhat-Schwartz function generalizes this to the case where the domain is a locally compact abelian group, e.g. the field of $p$-adic numbers. Thus there should be an analogue notion where one considers $p$-adic derivatives.
Now I found in some sources, that a Bruhat-Schwartz function on $\mathbb{Q}_p^n$ is a locally constant function with compact support. Unfortunately, I don't see at all how you get from the one definition to the other.

REPLY [2 votes]: Your formula for the derivative isn't even correct in the most classical setting : why do you divide by the norm of the increment instead of the increment itself?
Schwartz-Bruhat functions for totally disconnected spaces are generally defined to be the locally constant ones ; at least that's what one takes when considering distributions in a $p$-adic setting.<|endoftext|>
TITLE: Dynamics in one matrix variable
QUESTION [12 upvotes]: Are dynamical systems
$$X \mapsto F(X)$$
studied where $X \in \mathrm{M}_n$, $\mathrm{M}_n:=\mathrm{Mat}(n,\mathbb{C})$ or $\mathrm{Mat}(n,\mathbb{R})$, and $F$ is a (properly defined noncommutative) polynomial function?
For example, I was wondering if anybody has studied "Julia sets" and "Mandelbrot set" for mappings of the form
$$X \mapsto X^2 + C$$
with $C\in\mathrm{M}_n$.
One could presumably ask the same questions in the case of "one Clifford variable", i.e. for $X\in$ the Clifford algebra $\mathcal{C}\ell(V,Q)$ where $V$ is a (real or complex) finite dimentional vector space and $Q$ a quadratic form on $V$. In this case, for example, $X^2+C$ would reduce to $Q(X)\cdot 1+C$ for all $X \in V$.

REPLY [3 votes]: To answer your first, but not the rest of the questions, I thought about fractional linear matrix transformations (or at least their denominators), that is, densely defined maps on $n \times n$ matrices, of the form $X \mapsto (I + AXB)^{-1}$ (Fixed points of two-sided fractional matrix transformations, Fixed point theory and applications, 2007, doi:10.1155/2007/41930), but only with respect to fixed points. At least for the purpose of describing the fixed points, these can be rewritten in the form of matrix Riccati equations, and are closely connected to transformations of the type $X \mapsto X^2 + C$. But this probably isn't what you're looking for.<|endoftext|>
TITLE: Needle probing for a convex body
QUESTION [11 upvotes]: Suppose there is an unknown closed convex body $K$ of
volume vol$(K) = V$ inside the
unit cube $[-\frac{1}{2}, \frac{1}{2}]^d$ in $\mathbb{R}^d$.
You are permitted to probe with a (one-dimensional)
ray $r$, which detects whether
$r$ includes some point of $K$, or instead if $r \cap K = \varnothing$.
My question is:

Q1.
  What is the fewest number of such
  needle probe rays that guarantee
  hitting $K$, as a function of its volume $V$?

I am especially interested in $\mathbb{R}^3$.
For example, for $V \ge \frac{1}{2}$,
one ray through the origin parallel to an axis suffices.
Here such a ray touches the boundary of a body $K$ with
vol$(K)=\frac{1}{2}$:

          


To be specific:

Q2. How many needle probes suffice for $V \ge \frac{1}{4}$ in $\mathbb{R}^3$?

Four parallel rays in a grid pattern do not suffice, but nine do suffice (I believe):

    



Q3. In $\mathbb{R}^3$, is it sometimes more efficient to use
  nonparallel rays?

This feels like a classic problem, but I am not finding literature.
Thanks for pointers or ideas!
Update 1 (28May13). Here is Yoav Kallus' example showing that dropping the middle of the $9$ points of
the grid permits a triangle with area larger than $\frac{1}{4}$ to avoid detection:

    


Update 2 (30May13). Q2 and Q3 are now answered by Benjamin Dickman and Douglas Zare: for $V=\frac{1}{4}$, three needles suffice, and more are needed for parallel needles. Q1, in its full generality, is difficult, and so I've added the "open problem" tag.

REPLY [9 votes]: If $V \gt 1/2$ then there must be some pair of points $(x,y,z)$ and $(-x,-y,-z)$ contained in the region, and the convex hull of these two points contains the origin. This generalizes to show that $3$ needles connecting the centers of opposite faces (the axes) suffice for $V\gt 1/4$: 
Consider the orbits of points under the volume-preserving action of changing signs. Generic points $(x,y,z)$ have orbits of size $8$: $(\pm x, \pm y, \pm z)$. If $V \gt 1/4$ then there must be some orbit so that $3/8$ of the points are contained in the region. Given any $3$ of these points, $2$ must have different signs in at least $2$ coordinates since if the first is $(x,y,z)$ and the second is $(x,y,-z)$, then the third point can't be only a single sign change away from both. If the signs differ in $2$ coordinates, suppose without loss of generality these are $(x,y,z)$ and $(x,-y,-z)$. The convex hull of these points contains the midpoint $(x,0,0)$ which is on the $x$-axis. If the points differ in $3$ signs, then their midpoint is $(0,0,0)$ on all axes. 
There is a limit to how much this generalizes, but perhaps the full action of the cubic symmetries will restrict the volumes of regions which avoid the diagonals, too.<|endoftext|>
TITLE: Supremum of measure of sets of measure less or equal to 1/2.
QUESTION [8 upvotes]: Let $(X,d)$ be a metric space equipped with a probability measure $\mu$ (defined on the Borel $\sigma$-algebra on the topology induced by the metric $d$). I am interested in the different values that the following can take
$\sup\lbrace\mu(A):A\subset X\text{ is measurable and }\mu(A)\leq1/2\rbrace~~~~~~~~~~(\ast)$.
If there exists a measurable set $A\subset X$ such that $\mu(A)=1/2$, then it clearly is the case that $(\ast)$ is $1/2$.
If there is no set of measure $1/2$, then it is possible for $(\ast)$ to be different than $1/2$ (Dirac probability measure comes to mind). What I am wondering is if it could still be $1/2$.
In other words, I'm wondering if we could prove the existence (ideally with an example) or the nonexistence of a probability measure on a metric space $(X,d)$ such that for every measurable $A\subset X$, $\mu(A)\neq1/2$ and such that there is a sequence $\lbrace A_n\rbrace$ where $\mu(A_n)\to1/2$.
A few obvious remarks, if we were to find a probability measure as described in the above paragraph, the sequence $A_n$ cannot be nested, otherwise continuity from above or continuity from below would imply that the union or the intersection of the $A_n$ has a measure of $1/2$. Furthermore, the $A_n$ cannot be pairwise disjoint, otherwise $\mu$ cannot be a probability measure: for all $\epsilon>0$, infinitely many of the $A_n$ are contained in $(1/2-\epsilon,1/2+\epsilon)$, which implies by countable additivity that $\mu(X)\geq\infty$.

REPLY [6 votes]: Short version: the set $\{\mu(B):B\in\mathcal{B}\}$ is a closed set for any  probability space $(X,\mathcal{B},\mu)$. For atomic spaces this follows from an elementary topological argument, and for non-atomic spaces it is a closed interval by a  classical (and easy) result of Sierpinski.
Longer version with details:
Let $A_n=A'_n\cup A''_n$ where $A'_n$ is the atomic part of $A_n$ and $A''_n$ is the rest. By passing to a subsequence we may assume that $\mu(A'_n)$ converges to some $\alpha\in [0,1/2]$. 
I claim that there is an atomic set $B'$ such that $\mu(B')=\alpha$. Indeed, let $\{ x_k\}$ be the collection of the $\mu$-measures of all atoms of all $A'_n$. 
The set $X=\{ \sum_{k\in I} x_k: I\subset \mathbb{N}\}$ can be checked to be closed, since it is the continuous image of the $\{0,1\}^\mathbb{N}$ under the map $I\to \sum_{k\in I} x_k$, where we identify sequences $\omega$ in $\{0,1\}^\mathbb{N}$ with the set $I=\{ j: \omega_j=1\}$. Since $\alpha$ is in the closure of $X$ (as $\mu(A'_n)\to\alpha$), $\alpha\in X$, as claimed.
Now the non-atomic part of the space has mass at least $1/2-\alpha$ (since it has mass at least $\mu(A''_n)$ for each $n$). But it is well known that if $\nu$ is a non-atomic measure of mass $c$, then for any $c'\in [0,c]$ there is a measurable set of $\nu$-mass $c'$ (see for example wikipedia). Applying this to $\nu$=restriction of $\mu$ to non-atomic part and $c'=1/2-\alpha$, we get a measurable non-atomic set $B''$ of $\mu$-measure $1/2-\alpha$.
Hence $B=B'\cup B''$ satifies $\mu(B)=1/2$.<|endoftext|>
TITLE: Mathematical model for Hanoi Towers
QUESTION [13 upvotes]: The strategy for the Hanoi Tower puzzle is quite simple. It is based on parity only. In an $n$-pieces puzzle, $2^n-1$ moves are sufficient to carry the whole pile from one pole to another one. My question is

Is there any mathematical (algebraic, logical, ...) structure associated with the Hanoi Tower puzzle? Does the optimal strategy translate into some mathematical result about this structure?

REPLY [2 votes]: The fundamental group of the Menger Cube is an uncountable locally free and residually free group that contains the fundamental groups of all one-dimensional separable metric spaces as subgroups. It is known to frighten children and mathematicians alike. One can work within this group using a reduced path calculus; however, there is also an interesting word calculus (due to Hanspeter Fischer and Andreas Zastrow) that can be framed in terms of a variation of the Towers of Hanoi game. This is possible because the Menger cube may be realized as the inverse limit of finite groups each of which is the state-graph $X_n$ of a variation of the classical game using $n+1$ two-sided disks.
Hanspeter even created an ipad app that allows the user to move around within the approximating graphs $X_2$ (image below) while seeing the game state at the same time. It had dramatic whooshing and ringing sounds so naturally I spent too much time playing around with it.

Hanspeter Fischer, Andreas Zastrow, Word calculus in the fundamental group of the Menger curve, Fundamenta Mathematicae 235 (2016) 199-226. https://arxiv.org/abs/1310.7968
From the introduction of the cited paper regarding the details of the variation:

In our version of the Towers of Hanoi, the placement of the disks is restricted to within the well-known unique shortest solution of
  the classical puzzle, while we allow for backtracking within this solution and for the
  turning over of any disk that is in transition. We color the disks white on one side
  and black on the other. Then the state graph of this new “puzzle” is isomorphic
  to $X_n$, with edges corresponding to situations where all disks are on the board
  and vertices marking the moments when disks are in transition. The exponents of
  the edge labels ($x^{\pm 1}$ or $y^{\pm 1}$) indicate progress (“$+1$”) or regress (“$−1$”) in solving the classical puzzle (we add a game reset move when the classical puzzle is solved) and their base letters indicate whether the two disks to be lifted at the respective vertices of this edge are of matching (“$x$”) or mismatching (“$y$”) color. Hence, each edge-path through $X_n$ corresponds to a specific evolution of this game, as recorded by an observer of the movements of the $n + 1$ disks.
Our word calculus can be modeled by aligning an entire sequence of
  such puzzles with incrementally more disks into an inverse system,
  whose bonding functions between individual games simply consist of
  ignoring the smallest disk. Subsequently, every combinatorial notion
  featured in the description of the generalized Cayley graph has a
  mechanical interpretation in terms of this sequence of puzzles<|endoftext|>
TITLE: Has the Fundamental Theorem of Algebra been proved using just fixed point theory?
QUESTION [15 upvotes]: Question:

Is there already in the literature a proof of the fundamental theorem of algebra as a consequence of Brouwer's fixed point theorem? 

N.B. The original post contained superfluous information, but it did generate one answer with a source that claims such a proof is impossible, and another answer with a source that claims to carry out precisely such a proof. Clearly these cannot both be correct.

REPLY [4 votes]: Yes you can prove FTA using BFPT.  See also recent paper by Daniel Reem relating to Open Mapping Theorem.
I often looked at the 1949 note of B.H. Arnold (a professor of mine in bygone times) but missed the flaw.  Maybe someone could post the essence of the retraction by him and Ivan Niven.  Thank you.
You have proved FTA when you prove that any irreducible *real polynomial has degree 1 or two.
The existence of an *irreducible one of higher degree leads to a real vector space of that same dimension on which polynomial product induces a ring structure, in fact, integral domain.
Consider the squaring map on this real algebra, and scale it to map the whole thing to a Cartesian sphere of dimension one less.  Scaling the domain vector variable also, we get a squaring mapping from unit sphere to unit sphere, in fact from RP(n-1) to the sphere. The mapping is topologically continuous (on Hausdorff spaces) due to bilinearity of the original product operation.
The integrity condition (entire domain) shows that this mapping is injective. Everything in sight is compact Hausdorff, so such a 1-1 mapping induces a homeomorphism to the image.
Throw in "connectedness" and the (Brouwer) Invariance of Domain Theorem shows that in fact the image of RP(n-1) must be the whole sphere.  The fact that these two spaces are thus homeomorphic leads to the conclusion that n<=2 .
To show that RP is not topologically the same as S(n-1), construct a non-trivial loop in RP by projecting a longitude from North to South pole on its universal covering sphere.
This loop could not be homotopic to the constant loop (at the chosen base point) since the homotopy would lift to the UC keeping end-points invariant.
What does this have to do with BFPT?  As Dr. Terence Tao points out in his blog, the latter theorem is necessary in some form to prove Invariance of Domain.<|endoftext|>
TITLE: Group actions with finite stabilizers and compact quotients 
QUESTION [5 upvotes]: Let $G$ be a discrete group that acts on a contractible finite dimensional $G$-complex $X$ with the following properties: 

$X/G$ is compact (i.e. the action is cocompact) 
Each stabilizer $G_\sigma$ admits a cocompact action on a contractible 
finite dimensional $G_\sigma$-complex with finite stabilizers 

Question: Is there a finite dimensional contractible $G$-complex $Y$ 
with finite stabilizers such that $Y/G$ is compact ? 
The question can be stated more conceptually by help of the following definition: Let $\mathscr{F}$ be the class of all finite groups and define the class $K_i\mathscr{F},\;i\ge 0$ inductively by 

$K_0\mathscr{F} := \mathscr{F}$ 
$K_i\mathscr{F}$ includes all groups $G$ that admit a finite dimensional contractible $G$-complex $X$ such that (1) $X/G$ is compact and (2) the stabilizers are in $K_{i-1}\mathscr{F}$. 

Then the question is equivalent to 
Question: Is $K_1\mathscr{F} \subsetneqq K_2\mathscr{F}$ ? 
Remark: If condition (1) is dropped, we get the classical Kropholler classes $H_i\mathscr{F}$. There (among many other results) $H_1\mathscr{F} \subsetneqq H_2\mathscr{F}$ is known: For example, the free abelian group of countably infinite rank belongs to $H_2\mathscr{F}\setminus H_1\mathscr{F}$.

REPLY [2 votes]: It follows from the proof of prop. 5.1 in 
'Lück, W. and Weiermann, M.,On the classifying space of the family of virtually cyclic subgroups,Pure and Applied Mathematics Quarterly, Vol. 8(2) (2012), 497-555 (it's on the arxiv)'
that if $X$ is a cocompact $G$-CW-complex such that all its stabilizers have cocompact classifying space for proper actions, then $X \times E_{\mathcal{F}}G$ is $G$-homotopy equivalent to a cocompact $G$-CW-complex. Here,  $E_{\mathcal{F}}G$ is any model for the classifying space for proper actions of $G$.
Hence, the answer to your question is yes, if you assume additionally that all stabilizers have cocompact classifying space for proper actions.
Without this assumption, I think this is an open problem. 
On the other hand, every group in $K_2\mathcal{F}$ is of type $FP_{\infty}$. Therefore, it has a finite dimensional model for proper actions by a result of Kropholler and Mislin. Hence,  $K_2\mathcal{F} \subseteq H_1\mathcal{F}$.<|endoftext|>
TITLE: References on Taylor series expansion of Riemann xi function
QUESTION [9 upvotes]: I am looking for the references on Taylor series expansion of Riemann xi function at $\frac{1}{2}$.
$$ \xi (s)=\sum_0^{\infty}a_{2n}(s-\frac{1}{2})^{2n}$$
where
$$a_{2n}=4\int_1^{\infty}\frac{d[x^{3/2}\psi'(x)]}{dx}\frac{(\frac{1}{2}ln(x))^{2n}}{(2n)!}x^{-1/4}dx$$
and
$$\psi(x)=\sum_{m=1}^{\infty}e^{-m^2\pi x}=\frac{1}{2}[\theta_3(0,e^{-\pi x})-1]$$
Specifically I would like to know how fast $a_{2n}$ goes to zero.
Has anyone proved that 
$$a_0>a_2>a_4>...>a_{2n}>...>a_{\infty}=0$$ 
Thanks a lot!

REPLY [3 votes]: An accurate asymptotic estimate for the Taylor coefficients was obtained via the saddle point method in
Griffin+Ono+Rolen+Zagier, Jensen polynomials for the Riemann zeta function and other sequences - arXiv 1902.07321 www.pnas.org/cgi/doi/10.1073/pnas.1902572116
A short verification of the asymptotic formula was posted to the Pari/GP users mailing list, with references
Learning with GP: Griffin,Ono,Rolen,Zagier asymptotic formula for xi(s)
http://pari.math.u-bordeaux.fr/archives/pari-users-1908/msg00022.html
J. Gélinas<|endoftext|>
TITLE: Biggest ball included in an intersection of balls
QUESTION [5 upvotes]: I would like to prove that  for any family of balls $\{B(c_i,r_i)\}_i \subset \mathbb{R}^d$ such that $\{c_1, \dots, c_n\} \subset \bigcap_i B(c_i,r_i) $ and $\forall i, r_i \geq 1$, there exists a ball of radius $1-\frac{\theta_d}{2}$ included in the intersection $\bigcap_i B(c_i,r_i)$
where $\theta_d/2 = \frac{1}{2}\sqrt{\frac{2d}{d+1}}$ denotes the ratio between the diameter and the radius of the smallest enclosing ball of a regular simplex. 
Intuitively, it seems that the way of making the smallest intersection is to assign all points $c_i$ to the vertices of a regular simplex of diameter $1$ and all $r_i$ to $1$. By doing so, one can check that the ball of radius $1-\frac{\theta_d}{2}$ centered at $x$ the barycenter of $\{c_1,\dots,c_n\}$ is included in the intersection of balls $ \bigcap_i B(c_i,r_i)$.
Indeed, in the case of a simplex, the radius of the biggest ball centered at $x$ and included in the intersection of balls is $1-\text{Radius}(\sigma) = 1 - \frac{\theta_d}{2}$ (hence the constant is tight in this case).
I am having difficulties to prove that this case is indeed the worst case. 
I was just able to prove that the result holds when all balls have the same radius.
Does this result seems familiar to someone? I would really appreciate any comment, idea or reference.
ps : The topology tag is here for several reason. One of them is that the biggest radius of the ball included in the intersection corresponds to the weak feature size of the complement of the intersection. Another one is that this result is linked to a collapsibily result.

REPLY [5 votes]: OK, let's try.
First, $n=d+1$ (Helly's theorem).
Second, if the balls of radii $r_i-\theta_d$ do not have a common point, there is $\theta<\theta_d$ such that the balls of radii $r_i-\theta$ have exactly one common point, say, the origin. Also, we can ignore the balls such that $0$ is not on their boundary. Thus, we get $m\le d+1$ vectors $v_i$ of length $r_i-\theta$ with $|v_i-v_j|\le \min(r_i,r_j)$. Moreover, the vectors $v_i$ cannot lie in one half-space (otherwise there would be more common points), so $\sum_i a_iv_i=0$ for some $a_i\ge 0$, not all of them $0$. 
Now just note that as soon as the angle between $v_i$ and $v_j$ is obtuse, decreasing each length keeping $|v_i-v_j|$ constant will increase the angle and that in the case of the equilateral triangle, setting $r_i=1$ will also increase the angle. Thus, if $e_i$ are unit vectors in the directions of $v_i$, we have $\langle e_i,e_j\rangle> -1/d$ and we still have $\sum b_i e_i=0$ with $b_j\ge 0$. However, the matrix with $1$'s on the diagonal and off-diagonal terms satisfying $0>A_{ij}>-1/d$ is positive definite. We thus run into a contradiction.
I hope this holds water. Check everything and let me know if there are any gaps. :)<|endoftext|>
TITLE: Biholomophic non-Algebraically Isomorphic Varieties
QUESTION [25 upvotes]: Recently, when writing a review for MathSciNet, the following question arose: 
Is it true that two smooth complex varieties that are biholomorphic are algebraically isomorphic?  The converse is true just because polynomials are holomorphic.
I was mainly interested in the affine case since that was the context of the review I was writing.
I knew about exotic affine spaces, so two smooth complex affine varieties that are real analytically isomorphic do not have to be algebraically isomorphic.  But real analytic maps between complex manifolds need not be holomorphic in general (just think of complex conjugation on $\mathbb{C}$).  So that is not enough.
I posted the question on my Facebook page, and it got answered:

I was reminded that Serre's GAGA Theorem implies that it is true for projective varieties.  But there are  quasiprojective counterexamples provided on MO.  See the answer of Georges Elencwajg given here.
Then it was pointed out that the answer in the link above is a manifold which is both affine and non-affine.  So what about two affine varieties?
I then found a reference for such an example.  Two exotic affine spaces that are biholomorphic yet not algebraically isomorphic.  See here.  This is a very nice result.  Since it is in dimension 3, it pretty much shows that at the very first point where something can go wrong, it does.  A similar result for exotic affine spheres (apparently due to appear in J of Alg. Geo.) is here.

So I finally come to my questions.

Question 1:  Is the example in the
  reference I found the first
  example of two affine varieties
  that are biholomorphic but not
  algebraically isomorphic?
Question 2:  Is there some general
  reason to believe that all exotic
  affine spaces should be biholomorphic?

Comment:  It seems to me within context to ask similar questions about objects, not only maps between objects.  Here is some of what I have learned in that regard: all smooth complex varieties are complex manifolds since they are covered by smooth affine open sets with polynomial (hence holomorphic) transition charts.  Conversely, a closed analytic subspace of projective space is algebraic by Chow's Theorem.  But there are compact complex manifolds that are not algebraic (see here), which also shows that a closed analytic subspace of affine space need not be algebraic.

REPLY [8 votes]: Question 1: The first example published seems to be the following (Corollary 4 in "Embeddings of Danielewski surfaces", G. Freudenburg and L. Moser-Jauslin, Math.Z. 2003 )
For any $a\in \mathbb{C}^*$, the surfaces in $\mathbb{C}^3$ given by
$x^2z-y^2-a$ and $x^2z-(1+x)y^2-a$ 
are algebraically not isomorphic, but holomorphically isomorphic.
Question 2: The answer should be no because of the strong analytic cancellation theorem of Zaidenberg (see http://arxiv.org/pdf/alg-geom/9506005v1.pdf page 5). It says that if $X,X'$ are contractible non-biholomorphic surfaces of general-type, then $X\times \mathbb{A}^1$ and $X'\times\mathbb{A}^1$ are not biholomorphic but are both exotic affine spaces.<|endoftext|>
TITLE: Maximal norm-1 projection
QUESTION [5 upvotes]: Suppose I have a real unitary matrix $U$ and a unit vector $\mathbf{x}, \|\mathbf{x}\|_2 = 1$. What is the solution to the following problem?
$$
\widehat{\mathbf{x}} = \arg\max_{\mathbf{x}, ~\|\mathbf{x}\|_2 = 1}\|U\mathbf{x}\|_1.
$$

REPLY [2 votes]: The following may be of interest: K. Drakakis, "On the calculation of the l2 -> l1
induced matrix norm", Int. J. Algebra, 2009 (3), No 5, 231-240 (pdf).<|endoftext|>
TITLE: An "inchworm-like" random walk on an integer interval
QUESTION [12 upvotes]: Imagine I place $k$ stones on an infinite one-dimensional integer interval $Z$ s.t. no stone is more than some distance $d$ from any other stone.  For example, if $d=1$ and $k = 5$, we might place the five stones are positions $(-2,-1,0,1,2)$.  
I then proceed to do the following:  
For each of some arbitrary number of discrete time steps, I select one of the $k$ stones with uniform probability.  I then lift the stone up and uniformly select an unoccupied site to place it back down which satisfies the dual criterion of: (a) being at most a distance $d$ from both of the stone's nearest neighbors, and (b) not disturbing the initial ordering of the stones along the integer interval (note that the stone's original position can be reselected).  If $d=1$, no stone can be moved from its original position and the center of mass of the stones/system, $C_m$, will be immobile.  However, if $d=2$, starting from the state $(-2,-1,0,1,2)$, the center of mass will move to either the right or left with some per step probability of $\frac{1}{k}$ (until we leave this initial system state).
My question is - provided some number of stones $k$, and some "leashing distance" $d$, how can we characterize the dynamics of the random walk taken by the above system's center of mass, $C_m$?  What is the average step time and size, and, in terms of Euclidean distance, what mean square displacement can we expect after some number of steps, $T$?
Update (due to a suggestion by Douglas Zare) - The $k$ stones must be kept in order, and the distance between a stone and its two nearest-neighbors along the interval (or one nearest-neighbor if the stone occupies the left-most or right-most position of the chain) can be at most $d$.

REPLY [7 votes]: This is a particular case of random walks with internal degrees of freedom (aka semi-Markov random walks or covering Markov chains). In the case when the translation group is just $\mathbb Z$ (like in this question) this is a Markov chain on $\mathbb Z \times X$ with translation invariant transition probabilities. In the "inchworm" situation one can take for $X$ the (finite) set of all configurations in which the leftmost stone is at the point $0\in \mathbb Z$. Then states of the worm are described by the pairs $(n,\phi)\in \mathbb Z\times X$, where $n$ is the position of the leftmost stone, and $\phi$ is the shifted configuration. The center of masses will then be within uniformly bounded distance from $n$. It is obvious that in these coordinates the transition probabilities of the chain describing worm's motion are translation invariant, so that this is a random walk with internal degrees of freedom on $\mathbb Z$. 
The paper "Random walks with internal degrees of freedom. I. Local limit theorems" of Krámli and Szász (MR0699788) describes the asymptotic behaviour of such chains in terms of the properties of the quotient chain on $X$. In particular, there is an explicit formula for the non-zero variance of the displacement  in terms of the stationary distribution on $X$ (which in our case should also be quite explicit - it's a nice exercise).<|endoftext|>
TITLE: Homomorphisms from powers of Z to Z
QUESTION [10 upvotes]: I believe it is known that if I is a set of non-measurable cardinality, then any homomorphism $Z^I\to Z$ factors through a finite power.  Here $Z$ is the group of integers.  Can anyone give a reference for this?

REPLY [12 votes]: This runs under the name Łoś-Eda Theorem. A reference is the book Paul C. Eklof, Alan H. Mekler, Almost Free Modules (2002): 
Call a set $I$ $\omega$-measurable if its cardinality is greater or equal to the first measurable cardinal. This is equivalent to $I$ being uncountable and supporting a non-principal countably complete ultrafilter. 
First note that $\mathbb{Z}$ is slender (Cor. III.2.4). Then, by Cor. III.3.6 (and the discussion before Lemma III.3.5), if $I$ is not $\omega$-measurable, the natural map 
$$\phi: \bigoplus_{i \in I}\operatorname{Hom}(\mathbb{Z},\mathbb{Z}) \to \operatorname{Hom}(\prod_{i \in I}\mathbb{Z},\mathbb{Z}),\; (g_i)_i \mapsto \big((m_i)_i \mapsto \sum_i g_i(m_i)\big)$$
is an isomorphism. 
Remarks: 1) If $I$ is $\omega$-measurable, not all homomorphisms $\prod_I \mathbb{Z} \to \mathbb{Z}$ factor through a finite subset of $I$. For, let $D$ be a non-principal countably complete ultrafilter on $I$ and let $K_D = \lbrace x \in \prod_I \mathbb{Z} \mid I \setminus \sup(x) \in D\rbrace$. Then it's not hard to show that the composition $\prod_I \mathbb{Z} \twoheadrightarrow  \prod_I \mathbb{Z}/K_D \cong \mathbb{Z}$ doesn't factor through a finite subset of $I$ (the latter isomorphism uses II.3.3). 
2) Irrespective whether $I$ is $\omega$-measurable or not, there is a canonical isomorphism 
$$\operatorname{Hom}(\prod_{i \in I}\mathbb{Z},\mathbb{Z}) \cong \bigoplus_D \operatorname{Hom}(\mathbb{Z},\mathbb{Z})$$
where $D$ runs through all countably complete ultrafilters on $I$ (Cor. III.3.7).<|endoftext|>
TITLE: complete metric space
QUESTION [7 upvotes]: Hallo, I have the following question:
Let $(X,d)$ be a complete metric space. Is then $(X,\operatorname{dist})$ also complete? Here by $\operatorname{dist}$ I mean the metric induced by $d$ by: $\operatorname{dist}(x,y)=\inf(L(\gamma))$, where the infimum is taken over all paths joining $x,y$ and $L(\gamma)=\sup(\sum_{i=1}^{n-1}d(\gamma(t_{i}),\gamma(t_{i+1})))$, where the supremum is taken over all partitions $P=(0=t_{1}< ... < t_{n}=1)$ of $[0,1]$, where $\gamma$ is defined. Intuitively it should be but I dont know how to prove it. If its true, how can one prove it?
cheers
denis

REPLY [7 votes]: The intrinsic metric induced by a complete metric space is always a complete extended metric. To prove this, we shall write $d^{\sharp}$ for the extended metric $dist(x,y)=\inf(L(\gamma))$. 
Also, I want to remark that $d^{\sharp}$ is in general not a metric since we may have $d^{\sharp}(x,y)=\infty$ even when $X$ is path-connected. For instance, if we take some fractal $X$ such the boundary of the Koch snowflake, then $dist(x,y)=\infty$ for every pair of points $x,y\in X$. 
However, $d^{\sharp}$ is always an extended metric (By an extended metric I mean that $d^{\sharp}$ is a metric in every way except for the fact that we may have $d(x,y)=\infty$). Notions such as Cauchy sequences and completeness still hold if we use extended metrics instead of metrics. Furthermore, if we let $d'(x,y)=Min(d^{\sharp}(x,y),1)$, then $d'$ is a metric with the same uniform structure as $d^{\sharp}$(and hence the same Cauchy sequences, topology,...).
We shall now prove that $d^{\sharp}$ is complete using standard arguments.
To prove that $d^{\sharp}$ is complete assume that $(x_{n})_n$ is a Cauchy sequence with respect to the extended metric $d^{\sharp}$. Then since $d(x,y)\leq d^{\sharp}(x,y)$ for each $x,y\in X$, the sequence $(x_{n})_{n}$ is Cauchy with respect to the metric $d$. Since $(X,d)$ is a complete metric space the sequence $(x_{n})_{n}$ converges to some point $x\in X$ with respect the metric $d$. Since $(x_{n})_{n}$ is Cauchy with respect to $d^{\sharp}$, we may take a subsequence $(y_{n})_{n}$ of $(x_{n})_{n}$ such that $d^{\sharp}(y_{n},y_{n+1})<\frac{1}{2^{n}}$ for all $n$. In this case, there is some path $\gamma_{n}:[0,1]\rightarrow X$ such that $\gamma_{n}(0)=y_{n},\gamma_{n+1}(1)=y_{n+1}$ and
$L(\gamma_{n})<\frac{1}{2^{n}}$. If we let $\gamma:[0,\infty]\rightarrow X$ be the function where $\gamma(r+n)=\gamma_{n}(r)$ for $n\geq 0$ and $r\in[0,1]$ and $\gamma(\infty)=x$, then $\gamma$ is a continuous function. However, the restricted function $\gamma|_{[k,\infty]}$ is a path from the point $y_{k}$ to $x$ with $L(\gamma|_{[k,\infty]})<\sum_{n=k}^{\infty}\frac{1}{2^{k}}=\frac{1}{2^{k-1}}$. Therefore $d^{\sharp}(y_{k},x)<\frac{1}{2^{k-1}}$, so the sequence $(y_{n})_{n}$ also converges to $x$ with respect to the generalized metric $d^{\sharp}$.<|endoftext|>
TITLE: Normality of Chaitin's constant
QUESTION [9 upvotes]: Can anyone provide an overview of the proof that Chaitin's constant is normal, or better yet, the guiding intuition?
Even if we replace the existential quantifiers in the assertion of non-normality by explicit functions of the universally quantified variables, I don't see how an oracle in possession of those functions could solve the halting problem, yielding the desired contradiction.
(In fact, I would've guessed that the normality of Chaitin's constant was in the class of undecidable things!)
Putting the question more starkly: if 90% of the bits were 0's, how would knowing this give you a way to solve the halting problem?

REPLY [7 votes]: Consider a program $P_n$ which first unpacks $\Omega_n$, the first $n$ digits of $\Omega$, then runs all finite programs tallying $2^{-k}$ each time a program of length $k$ halts. $P_n$ continues until it gets within $2^{-n}$ of $\Omega_n$, and then halts. If you run all finite programs forever, then you see all contributions to $\Omega$, so you get arbitrarily close to $\Omega$, and $P_n$ must halt at some point. $P_n$ runs a copy of itself, but it doesn't observe itself halt. So, the contributions of programs which halt up to this point (at least $\Omega_n - 2^{-n} \ge \Omega - 2^{-n+1}$), plus $2^{-|P_n|}$ for $P_n$, must be less than $\Omega$. This means $|P_n|\ge n$.
Let $c_1(n)$ be the length of the part of $P_n$ which says to run all finite programs, tallying their contributions to $\Omega$, and halt if it gets within $2^{-n}$. $c_1(n)$ is $O(\log n)$. For any $n$, there can't be a way unpack $\Omega_n$ with fewer than $n-c_1(n)$ bits. If $\Omega$ were not normal, then for infinitely many $n$, $\Omega_n$ could be compressed saving at least $c_2 n + c_3$ bits. If $90\%$ of the digits of $\Omega$ were $0$s then it would take fewer than $n/2 + c_4$ bits to encode $\Omega_n$. So, if $\Omega$ is not normal, then some $P_n$ could have fewer than $n$ bits, a contradiction.<|endoftext|>
TITLE: Ideal Membership without Certificate?
QUESTION [12 upvotes]: I have a homogeneous ideal $I=\langle f_1,\ldots,f_r\rangle$ of the polynomial ring $\mathbb C[X_1,\ldots,X_n]=:R$ where each of the $f_i$ is actually over $\mathbb Z$. My computations are usually performed over $\mathbb Q$. As an additional note, the $f_i$ are all homogeneous of the same degree. I also have a monomial $\mu = X^\alpha$, where $\alpha\in\mathbb N^n$ is some multi-index. I don't know if it matters that $\mu$ is a monomial.
Several computer algebra systems (magma, sage, macaulay2) can tell me that $\mu\in I$, but only in SAGE I could find a method to compute $g_1,\ldots,g_r\in R$ with $\sum_{i=1}^r g_i f_i = \mu$. Unfortunately, this method seems to crash the SAGE kernel, because after a couple of minutes the execution simply stops with no result, and SAGE has forgotten everything from the session. I have two questions:

How do Computer Algebra systems check ideal membership without actually creating a certificate? In other words, how do they know $\mu\in I$ without having the $g_i$ that prove it?
I would really like to know those $g_i$. Given that SAGE just flat-out crashes on me, do you know any other options I could try? There might be routines in other computer algebra systems that I am simply unaware of and which I could try.

REPLY [2 votes]: My recollection of the Gröbner engine of Macaulay 2 is that it's rather black-boxed. Surely others here know better than I how to access its parts directly.
The first step is to extend to a Gröbner basis $G$. But without reimplementing the Buchberger algorithm I don't know how to express the new generators in terms of the old.
Then, you can $\rm scan()$ through ${\rm flatten\ entries\ generators}\ G$, replacing p by p%G, and if they're different print out (G, (p-p%G)/G). Continue cycling through until $p=0$.<|endoftext|>
TITLE: The status of 'the consistency of NF relative to ZF'
QUESTION [11 upvotes]: One of the responses to my Mathoverflow question No. 122658 hinted that a proof (or the outline of a
proof) of the consistency of NF relative to ZF was on the horizon and was to be presented at a meeting
in Cambridge scheduled for April of this year. Does anyone have additional information about this?

REPLY [5 votes]: Saying that Randall Holmes has now posted his proof of the consistency of NF is misleading. He has posted five different proofs at his homepage http://math.boisestate.edu/~holmes/, and explains

The difficulty is not that I do not know how the proof goes...this I am fairly certain about; what is unclear is how to lay it out so that another human being can understand it. Different approaches suggest themselves, and I have tried several.

The oldest of these proofs has been published before July 2014, see http://www.cs.nyu.edu/pipermail/fom/2014-July/018031.html.
All five versions contain a section Conclusions to be drawn about NF including the following paragraph

It seems clear that this argument, suitably refined, shows that the consistency strength of NF is exactly the minimum possible on previous information, that of TST + Infinity, or Mac Lane set theory (Zermelo set theory with comprehension restricted to bounded formulas). Actually showing that the consistency strength is the very lowest possible might be technically tricky, of course. I have not been concerned to do this here. It is clear from what is done here that NF is much weaker than ZFC.


The Conclusions section of Gabbay's paper includes the following paragraph

Given our proof, we can examine it to see how much set-theoretic strength it really uses, and thus see relative to what system we have proved NF consistent. We have not used the Axioms of Choice or Replacement in the proofs of this paper: we have proved NF consistent relative to Zermelo set theory (Z).

Note that consistency relative to Zermelo set theory is not the minimum possible, which would be Mac Lane set theory (Zermelo set theory with comprehension restricted to bounded formulas). It remains unclear whether Gabbay just didn't care about this detail, or whether his proof really uses comprehension for unbounded formulas in an essential way.
Andres Caicedo indicates that Gabbay withdrew his paper a while ago. However, http://www.gabbay.org.uk/papers.html#submitted still links to a version of the proof from May 2015. I guess that Gabbay's proof simply should not appear in a journal before one of Randall Holmes proofs, because those are older and have been checked much more thoroughly by many more mathematicians.

This question raises the general issue what must be done before a long standing open problem can be declared as solved. As long as the author of the proof is not yet satisfied with the presentation of his proof and still makes steady progress towards a better presentation, either waiting patiently for a final version, or reading and trying to understand a preliminary version seem to be reasonable options.<|endoftext|>
TITLE: Conjugacy classes of the Ree groups
QUESTION [5 upvotes]: Dear all,
Consider the finite simple Ree groups $G=^2\hspace{-1mm}G_2(3^{2n+1})$ where $n$ is a positive integer. I would like to know the orders of conjugacy class representatives of $G$ and from there to know the number of conjugacy classes of $3$-singular elements of $G$ (an element is 3-singular if its order is divisible by 3). This is probably well-known but I couldn't find it somewhere in the literature.
Thank you in advance!

REPLY [7 votes]: I think there is enough information in Chapter III of Ward's paper to determine the classes of 3-singular elements, and the paragraph numbers below refer to that).
Let $P \in {\rm Syl}_3(G)$. Then $|P|=q^3$, with upper and lower central series of $P$  both $1 < C < P_1 < P$ (using Ward's notation), where $|C|=q$, $P_1$ is elementary abelian of order $q^2$, and $P/P_1$ is elementary abelian.
The normalizer of $P$ is a semidirect product of $P$ with a cyclic group of order $q-1$, which acts fixed-point-freely on $C$ and on $P/P_1$, but acts as a cycle of order $(q-1)/2$ on the middle layer $P_1/P$. The element of order 2 in this cyclic group has centralizer $C_2 \times L_2(q)$.
The $q-1$ elements in $C \setminus \{1\}$ form a single class. The $q^2-q$ elements of $P_1 \setminus C$ (which have order 3) split into two classes (paragraph 7). These elements centralize an element of order 2, and there are also two classes of elements of order 6. Finally, the $q^3-q^2$ elements in $P \setminus P_1$ split into three classes of elements of order 9 (paragraphs 9,10,11).<|endoftext|>
TITLE: Temperley-Lieb algebras for other Weyl groups?
QUESTION [10 upvotes]: The Temperley-Lieb algebra has the same generators as the $S_n$ group algebra, and the same commuting relations, but its other relations are different. A nice diagrammatic interpretation can be seen in the Wikipedia article.

Is there a standard analogue for other Weyl groups? Perhaps some specialization of the Hecke algebra?

There is a standard basis for the Temperley-Lieb algebra, which can be indexed by 321-avoiding permutations: pick a reduced word for $v$ in $S_n$'s generators, and interpret the generators inside T-L instead. This association is well-defined, because the 321-avoiders are also the "fully commutative" elements, meaning that each has only one reduced word up to commutation relations.
This subset of $S_n$  is also the set of "lambda-cominuscule" elements (meaning, the $T$-action on the Bruhat cell $BvB/B$ includes the dilation action), and is an ideal in left/right weak Bruhat order. In Weyl groups other than $S_n$ the lambda-cominuscule condition is strictly stronger than "fully commutative".

Assuming the first question, do the lambda-cominuscules, or some other subset of W, similarly give a basis?

Of course references would be most appreciated.

REPLY [8 votes]: First I don't think you have stated the defining relations correctly. [AK: you were quite right, and I've edited the question.]
It sounds as though you should start with the work of Richard Green.
The following papers seem relevant.
http://arxiv.org/abs/q-alg/9712018
http://arxiv.org/abs/math/0102003
http://arxiv.org/abs/math/0108076<|endoftext|>
TITLE: Are roots of transcendental elements transcendental?
QUESTION [6 upvotes]: This looks extremely easy, but then again it's late at night...
Let $k$ be a commutative ring with unity. An element $a$ of a $k$-algebra $A$ is said to be transcendental over $k$ if and only if every polynomial $P\in k\left[X\right]$ (with $X$ being an indeterminate) such that $P\left(a\right)=0$ must satisfy $P=0$.
Let $n$ be a positive integer. Let $A$ be a $k$-algebra, and $t$ be an element of $A$ such that $t^n$ is transcendental over $k$. Does this yield that $t$ is transcendental over $k$ ?
There is a rather standard approach to a problem like this which works if $k$ is reduced (namely, assume that $t$ is not transcendental, take a nonzero polynomial $P$ annihilated by $t$, and consider the product $\prod\limits_\omega P\left(\omega X^{1/n}\right)$, where $\omega$ runs over a full multiset of $n$-th roots of unity adjoined to $k$; this product can be seen to lie in $k\left[x\right]$ and annihilate $t^n$; this all requires a lot more work to put on a solid footing). There are even easier proofs around when $k$ is an integral domain or a field (indeed, in this case, if $t$ is not transcendental over $k$, then $t$ is algebraic over $k$, so that, by a known theorem, $t^n$ is algebraic over $k$ as well, hence not transcendental). I am wondering if there is a counterexample in the general case or I am just blind...

REPLY [6 votes]: $\def\F{{\mathbb F}}$
This is just a proof of Qiaochu's example.
Let  $k=\F[Y,Z]/(Y^2,YZ,Z^2)$, $a=\overline{Y}$, $b=\overline{Z}$. In the ring $k[X]$ we have $I=(aX^3-b)=\{aP(X)X^3-bP(X): P(X)\in \F[X]\}$, since $a(aX^3-b)=b(aX^3-b)=0$. So each element of this ideal contains both even and odd powers of $X$. Thus in $k[X]/I$ the element $t=\overline{X}$ is algebraic, while $t^2$ is not.<|endoftext|>
TITLE: Can all $\aleph_2$-dense subsets of $\mathbb{R}$ be isomorphic?
QUESTION [17 upvotes]: Let $\kappa$ be an infinite cardinal. For a subset $A \subseteq \mathbb{R}$, we say that $A$ is $\kappa$-dense if $|A \cap (a, b)| = \kappa$ for every interval $(a, b)$. By Cantor, any two $\aleph_0$-dense sets are order-isomorphic. 
If the CH holds, then there are many non-isomorphic $\aleph_1$-dense sets. However, Baumgartner showed that it is consistent with $2^{\aleph_0}=\aleph_2$ that all $\aleph_1$-dense sets are isomorphic, giving a generalization of Cantor's result. My question is, can this be pushed further? Is it consistent with $2^{\aleph_0} \geq \aleph_3$ that all $\aleph_2$-dense sets of reals are isomorphic? And what about for larger cardinals still? Can we have the result for $\aleph_1$-dense sets and $\aleph_2$-dense sets simultaneously? 
Baumgartner himself asked if all $\aleph_2$-dense sets of reals could be isomorphic in his original paper, which I found linked here on MO (all thanks to François Dorais). But I have been unable to find out if his question has subsequently been answered. If anyone knows the answer, or can provide a reference, I would be grateful.

REPLY [8 votes]: In section 5 (Concluding remarks and recent developments) of the paper Baumgartner's isomorphism problem for $\aleph_2$-dense suborders of $\mathbb{R}$, by Moore and Todorcevic, the following is stated:

He (Itay Neeman) moreover tentatively announced that, in spite of the gap in the
  present paper, he was able to show that if there is a weakly compact
  cardinal, then there is a forcing extension in which $BA_{\aleph_2}$ and $MA_{\aleph_2}$
  both hold.

If the claim is true, it means that the problem is solved now!!!
Also in the notes Reflection of clubs, and forcing
principles at $\aleph_2$ by Itay Neeman, the last theorem stated the consistency of $BA_{\aleph_2}.$<|endoftext|>
TITLE: Proving that every term of the sequence is an integer
QUESTION [16 upvotes]: Let $m,n$ be nonnegative integers.
The sequence $\{a_{m,n}\}$ satisfies the following three conditions.

For any $m$, $a_{m,0}=a_{m,1}=1$
For any $n$, $a_{0,n}=1$
For any $m\ge0, n\ge1$, $a_{m+1,n+1}a_{m,n-1}=a_{m+1,n-1}a_{m,n+1}+a_{m+1,n}a_{m,n}$

Prove that $a_{m,n}$ is an integer for any $m\ge0, n\ge0$.
I don't have any good idea. I need your help.

REPLY [8 votes]: This looks like it might be a "rank two elliptic divisibility sequence", i.e., let $E$ be an elliptic curve, let $P$ and $Q$ be independent points in $E(\mathbb{Q})$, and write
$$
  x(mP+nQ) = A(m,n)/D(m,n)^2.
$$
Then the doubly-indexed sequence $D(m,n)$ is a rank two elliptic divisibility sequence, or as they were named by Kate Stange, an elliptic divisibility net. Kate made a detailed study of these sequences in her thesis (Algebra and Number Theory, 5-2 (2011), 197-229; it's also available on the ArXiv at http://arxiv.org/abs/0710.1316), including a description of the recurrence relation(s) that they satisfy. 
Anyway, I don't know for sure if your rank two sequence fits into Kate's elliptic net framework, but it's quite possible that it does.
(I'm actually cheating a little bit here, one really needs to use division polynomials instead of writing $x(mP+nQ)$ as a fraction in case there's cancellation between the numerator and the denominator. But this will only affect primes of bad reduction.)<|endoftext|>
TITLE: A question on semi-stratifiable space
QUESTION [5 upvotes]: This question is also posted here.
A space $X$ is callled semi-stratifiable space if it has a $g$-function such that: for any point $x$ of $X$ and a sequence $\{x_n\}$ of $X$ if $x \in g(n,x_n)$, then $x_n \to x$.
Note that every Moore space is semi-stratifiable. We know the cardinality of a star countable Moore space is not greater than $\mathfrak c$.
A topological space $X$ is said to be star countable if whenever $\mathscr{U}$ is an open cover of $X$, there is a countable subspace $K$ of $X$ such that $X = \operatorname{St}(K,\mathscr{U})$.

Is there a star countable semi-stratifiable space $X$ with $|X|> \mathfrak c$?

Thanks for your help.

REPLY [2 votes]: As a counterexample to this question we can consider the Katetov extension $\kappa\omega$ of the  discrete space of all finite ordinals $\omega$. 
By definition, $\kappa\omega$ is the space of all ultrafilters on $\omega$ with the topology in which a neighborhood base of an ultrafilter $\mathcal U$ consists of the sets $\{\mathcal U\}\cup U$ where $U\in\mathcal U$. Here we identify $\omega$ with the set of principal ultrafilters on $\omega$. So, $\kappa\omega=\omega\cup\omega^*$ where $\omega^*$ is the set of free ultrafilters on $\omega$. The space $\kappa\omega$ is separable and hence star-countable. On the other hand, $\kappa\omega$ has cardinality $2^{\mathfrak c}>\mathfrak c$. Also the space $\kappa\omega$ is semi-stratifiable. This is witnessed by the function $g$ defined by $g(n,\mathcal U)=\{\mathcal U\}$ if the ultrafilter $\mathcal U$ is principal and $g(n,\mathcal U)=\{\mathcal U\}\cup(\omega\setminus n)$ if $\mathcal U$ is free. 
Since the subspace $\omega^*$ of free ultrafilters is discrete and uncountable, the separable space $\kappa\omega$ has uncountable network weight, so is not $\omega$-monolithic. This answers question 
Is every semi-stratifiable space $\omega$-monolithic?<|endoftext|>
TITLE: Examples of (Phi,Gamma)-modules
QUESTION [5 upvotes]: What is the (Phi,Gamma)-module of an elliptic curve over Z_p, expressed by a direct construction ?

REPLY [12 votes]: (By an elliptic curve over $\mathbb Z_p$, I assume you mean an elliptic scheme over spec $\mathbb Z_p$, or which amount to the same, an elliptic curve over $\mathbb Q_p$ which has good reduction.) The Galois representation on $V_p(E)$ is then crystalline, so its $(\phi,\Gamma)$-module $D$ will be triangulate -- in other words, an extension of
two $(\phi,\Gamma)$-modules of dimension $1$, a sub-object $D_1$ and a quotient $D_2=D/D_1$.
As you probably know (since otherwise you would have begun by this case), a 
$(\phi,\Gamma)$-module of dimension $1$ is described by a continuous character $\delta: \mathbb Q_p^\ast \rightarrow \mathbb Q_p^\ast$. More precisely, it is always isomorphic
to the Robba ring as a module over the Robba ring, with actions that are twisted from the original one, which  by $\delta_{|\mathbb Z_p^\ast}$ for the $\Gamma$-action and by $\delta(p)$ for the $\phi$-action.
Now let us describe the characters $\delta_1$ and $\delta_2$ corresponding to $D_1$ and $D_2$.
Let $\alpha$ and $\beta$ be the two roots of the polynomial $X^2-a_pX + p =0$, ordered so that $v_p(\alpha) < v_p (\beta)$. Here $a_p$ is as usual $E(\mathbb F_p) -1 - p$, and $v_p$ is the $p$-valuation. So let $\delta_1$ be the character which sends $z \in \mathbb Z^\ast_p$ to $1$ and $p$ to $\alpha$ (so the character $z \mapsto \alpha^{v_p(z)}$ in short),
and let $\delta_2$ be the character which sends $z \in \mathbb Z^\ast_p$ to $1$ and $p$ to $\beta$.
This describes $D_1$ and $D_2$. Am I done? not quite. I need to say which extension
it is. But I have to run, so I will finish later... (later...) So if you go look to the paper
of Colmez on triangulline representations, you will see that there is a result computing the Ext group $Ext^1(D_1,D_2)$ in the category of $(\phi,Gamma)$-modules, and that in our case
the dimension is $1$. So the extension $D$ has two possibility: either it is trivial, or it is non-trivial (and then the proof of that Colmez's theorem give you and explicit description of what it is). 
When your elliptic curve $E$ is super singular, $D$ is always the non-trivial extension.
When it is ordinary, it is more complicated. Let $\rho_{E,p}$ be the Galois representation
of $G_{\mathbb Q_p}$ on the Tate module of $E$. It is always reducible, but it may be decomposable or not (with a conjecture saying that it is decomposable if and only if E has CM). Well the extension $D$ is split if and only if $\rho_{E,p}$ is decomposable.
Okay, this describes explicitly $D$. I practice, it is not fundamental to know when $D$ is a split extension or not, the fact that it is an extension of the very concrete 
$D_1$ and $D_2$ is enough. If you want to understand all this better, I think that COlmez' paper on triangulline representation is the best place to start.<|endoftext|>
TITLE: Can one recover a metric from geodesics?
QUESTION [30 upvotes]: Assume there are two Riemannian metrics on a manifold ( open or closed) with the same set of all geodesics.  Are they proportional by a constant? If not in general, what are the affirmative results in this direction?

REPLY [2 votes]: For the sake of completeness, here is the local story, assuming that geodesics are understood as maps from intervals to the manifold and "the same" is understood literally (rather than "up to reparameterization"). Equivalently, two metrics have the same exponential map. 
Results along these lines are well-know but hard to find (with usual references pointing to old papers by Eisenhart which I find unreadable), while this should be standard textbook material...
I will say that two Riemannian metrics with the same Levi-Civita connection form an LC pair. I will say that an LC pair is trivial if the metrics are homothetic to each other,  i.e. are scalar multiples of each other. 
The first thing to observe is that two Riemannian metrics on the given manifold have the same exponential map if and only if they form an LC pair. One direction is immediate, as geodesics on a Riemannian manifold are curves defined by the differential equation $\nabla_{c'} c'=0$. The opposite direction is harder, it is proven (for instance) in: 
M.Spivak, "Comprehensive Introduction to Differential Geometry" (Publish Or Perish, 2000), volume 2, chapter 6, Appendix 1.
Theorem. Suppose that $(M,g_1)$ is a Riemannian manifold such that $M$ admits another Riemannian metric $g_2$ 
such that $g_1, g_2$ form a nontrivial LC pair. Then $(M,g_1)$ is locally isometric to a product of Riemannian manifolds. 
Proof. I will need several ingredients. Recall that the holonomy group $Hol_{p,g}$ of the Riemannian metric $g$ (rel. basepoint $p\in M$)  depends only on the Riemannian affine connection $\nabla$, $Hol_{p,g}=Hol_{p,\nabla}$. Our discussion is local, it suffices to restrict to loops contained in a totally convex neighborhood of $p$. Since $\nabla$ is a Riemannian connection (for a metric $g$) then $G=Hol_{p,\nabla}$ will preserve the quadratic form $g_p$ on $T_pM$. Thus, $G$ is a relatively compact subgroup of $O(n)=Aut(T_pM, g_p)$. This group is essentially independent of the basepoint: Holonomy groups at different base-points $p_1, p_2$ in $M$ are "conjugate" via the parallel transport along a path from $p_1$ to $p_2$.  
Next, comes a fact from elementary representation theory. Suppose that $V$ is a finite-dimensional real vector space and $G< GL(V)$ is a (relatively) compact subgroup whose action on $V$ is irreducible, i.e. $V$ does not admit a nontrivial $G$-invariant direct sum decomposition $V=V_1\oplus V_2$. Then any two $G$-invariant quadratic forms $q_1, q_2$ on $V$, are scalar multiples of each other. 
Applying this result to the holonomy groups of Riemannian metrics $g_1, g_2$ on a connected manifold $M$ we obtain:
If $g_1, g_2$ have the same Levi-Civita connection $\nabla$ then either the holonomy groups $Hol_{p,\nabla}$ are reducible for some (equivalently, every) $p\in M$ or the metrics $g_1, g_2$ are  conformal to each other: 
$$
g_2= e^{2f}g_1, f\in C^\infty(M). 
$$ 
Another ingredient that we need is deRham's theorem: 
If $g$ is a Riemannian metric on $M$ whose holonomy is reducible, then $M$ locally splits as a Riemannian direct product.
See e.g. Theorem 3.1 on p. 228 in Petersen's book "Riemannian Geometry." 
Applying this to an LC pair $g_1, g_2$, we conclude:
Either (1) $(M,g_1)$ locally splits as a Riemannian direct product or (2) the metrics $g_1, g_2$ are conformal to each other. 
Consider the case (2):
$$
g_2= e^{2f}g_1, f\in C^\infty(M). 
$$
I will prove that the function $f$ is constant, i.e. $g_1, g_2$ form a trivial LC pair. Let $grad(f)$ denote the gradient field of $f$ with respect to $g_1$. Then, since $g_1, g_2$ form an LC pair, one obtains:
$$
X(f)Y + Y(f)X - g_1(X,Y) grad(f)=0, 
$$
for any two vector fields $X, Y$ on $M$. I claim that $grad(f)$ is identically zero on $M$, i.e. $f$ is constant. 
Indeed, consider vector fields $X=Y= grad(f)$. Then, the equation becomes 
$$
2 ||grad(f)||^2 grad(f) - ||grad(f)||^2 grad(f) = ||grad(f)||^2 grad(f)=0,
$$
i.e. $grad(f)$ is identically zero. Thus, the metric $g_2$ is a constant multiple of $g_1$. qed 
Clearly, the "converse" to this theorem holds as well assuming that $(M,g_1)$ has a global nontrivial deRham decomposition. 
Applying the theorem inductively, one obtains the following corollary:
Corollary. Let $(M,g)$ be a Riemannian manifold which has local deRham decomposition 
$$
M=M_0\times M_1\times ... \times M_m, g= g_0 \oplus g_1\oplus ... \oplus g_m
$$
where $g_0=\delta_{ij}$ is the standard flat metric on ${\mathbb R}^k$ (and the rest of the factors are non-flat and do not split any further). Suppose that $g, h$ is an LC pair on $M$. Then (locally) $h$ has the form
$$
h= h_0 \oplus a_1 g_1\oplus ... \oplus a_m g_m
$$
where $h_0$ is a constant metric tensor on ${\mathbb R}^k$ and $a_1,...,a_m$ are certain constants.<|endoftext|>
TITLE: etale cohomology of an abelian variety and its dual
QUESTION [8 upvotes]: Let $A$ an abelian variety over a field $k$ and $A^{*}$ the dual abelian variety.
How can we relate the étale cohomology of $A$ with etale cohomology of $A^{*}$?

REPLY [8 votes]: Let me add something to what Timo Keller says rightly. What he said can be reformulated
as $H^i_{et}(A,{\bf Z_\ell})$ and $H^i_{et}(A^*,{\bf Z_\ell})(1)$ are canonically dual
of each other. This should be understand that there are dual not only as $\bf Z_\ell$-modules,
but also as $Gal(\bar k / k)$-representation, and here the (1) takes its significance: it is a twist by the $\ell$-adic cyclotomic character.
Now, more can be said. Every abelian variety $A$ over $k$ as a polarization $p$ over $k$, which is in particular a symmetric isogeny $p: A \rightarrow A^\ast$. Now any isogeny induces isomorphism between the $V_\ell$'s (that is $T_\ell \otimes \bf Q_\ell$), hence taking again
the wedge power, we get for all $i$ an isomorphism
$$H^i_{et}(A,{\bf Z_\ell}) \otimes {\bf Q_\ell} \rightarrow H^i_{et}(A^*,\bf Z_\ell) \otimes \bf Q_\ell.$$ This isomorphism preserves other structures (the $Gal(\bar k/k)$-action), but is not canonical, since it depends on the choice of $p$. But this is certainly a relation between the stale cohomology of $A$ and $A^\ast$ that is worth mentioning, and that is 
logically independent of the one given by Timo.<|endoftext|>
TITLE: What is the largest number of k-element subsets of a given n-element set S such that…
QUESTION [5 upvotes]: Given a set S of n elements. What is the largest number of k-element subsets of S such that every pair of these subsets has at most one common element?

REPLY [8 votes]: You're asking what the number of blocks of a maximum packing is.
An ordered pair $(S, \mathcal{B})$ of a finite set $S$ of cardinality $\vert S \vert = v$ and a finite set $\mathcal{B}$ of $k$-subsets of $S$ is a packing of order $v$ and block size $k$ if any pair of elements of $S$ appears in at most one element of $\mathcal{B}$. The elements of $S$ are the points, and those of $\mathcal{B}$ are the blocks.
A maximum packing of order $v$ and block size $k$ is a packing with the most blocks among all packings of block size $k$ on $v$ elements. What you're asking is what the cardinality $\vert \mathcal{B} \vert$ of the block set of a maximum packing.
A corollary of the Johnson bound (which is proved by S.M. Johnson in A new upper bound for error-correcting codes, IRE Trans. IT-8 (1962) 203–207) states that a packing of order $v$ and block size $k$ has at most
$$\left\lfloor\frac{n}{k}\middle\lfloor\frac{n-1}{k-1}\middle\rfloor\right\rfloor$$
blocks. As Butch mentioned, a Steiner $2$-design attains the upper bound with equality. The existence of such perfectly packed designs is solved in the affirmative in the asymptotic sense:

There exists a constant $v_k$ that depends only on $k$ such that for all $v > v_k$ such that $\frac{{{n}\choose{2}}}{{{k}\choose{2}}}$ is an integer, there exists a Steiner $2$-design of order $v$ and block size $k$.

The above result is a corollary of Wilson's Fundamental Existence Theorem. So the problem is when a maximum packing is not a Steiner $2$-design because $\frac{{{n}\choose{2}}}{{{k}\choose{2}}}$ isn't an integer (and, of course, finitely many cases of small order $v$ for each $k$). For small $k$, the necessary and sufficient conditions on $v$ are known, e.g., for $k=3$, you can get a Steiner $2$-design whenever the integer condition is met. A comprehensive list of results like this can be found in the Handbook of Combinatorial Designs. A quick summary is that for $k \leq 5$, the integer condition is also sufficient. But for larger $k$, there are definite exceptions for the existence of Steiner $2$-designs even if the order $v$ meets the necessary condition.
So, the tougher problem is the packing case where a maximum one does miss some pairs. Some cases are easy to solve. For instance, take a Steiner $2$-design $(S, \mathcal{B})$ of order $v$ of block size $3$. In this case, we have $v \equiv 1, 3 \pmod{6}$ because of the integer condition. You delete one point, say $\infty$, from the point set $S$ and throw away all $2$-subsets resulted in from those blocks containing $\infty$. Then you get a packing on  the point set $S\setminus\{\infty\}$ with triples $\mathcal{B}'$ in which exactly $\frac{v-1}{2}$ pairs don't appear anywhere. You can show that it is a maximum packing. So the case $v \equiv 0, 2 \pmod{6}$ is also completely solved. The remaining case $v \equiv 5 \pmod{6}$ is not as simple. But as Butch mentioned, we have the answer: it's one step behind the Johnson bound.
The best results for the general case is given in the paper I linked to in this answer to a recent MO question here:
Uniform 4-hypergraph avoiding 2-cycles
In fact, the question there is, if you translate it into the language of design theory, asking if a maximum packing can get $c\cdot v^2$ blocks for some positive constant $c$ when $v$ approaches infinity.
The result given in the paper is that for sufficiently large orders, you can always attain the Johnson bound minus some constant. The history of this problem and its generalizations are also given in the paper.<|endoftext|>
TITLE: An elementary expression for $_3F_2(1,1,9/4;2,2;-1)$
QUESTION [14 upvotes]: Consider the following series:
$$S=\sum_{n=1}^\infty\frac{(-1)^n\ \Gamma\left(\frac{5}{4}+n\right)}{n^2\ \Gamma(n)}.$$
It can be expressed in terms of a hypergeometric function:
$$S=-\frac{5}{16}\Gamma\left(\frac{1}{4}\right)\ { _3F_2}\left(1,1,\frac{9}{4};2,2;-1\right).$$
I tried to find an expression of $S$ using elementary functions and ended up with this conjecture:
$$S\stackrel{?}{=}\frac{\Gamma\left(\frac{1}{4}\right)}{16}\Bigg(8\sqrt[4]{8}-16-\ln\frac{24\sqrt{69708+49291\sqrt{2}}+3168\sqrt{2}+4481}{4096}\\\\+\arctan\frac{24\sqrt{49291\sqrt{2}-13260}}{6913}\Bigg).$$My derivation of this formula is quite long and uses a non-rigorous, heuristic approach involving heavy use of techniques like guessing sequence formulas using Mathematica command FindSequenceFunction and OEIS superseeker, and recognizing approximate numeric quantities using RootApproximant, TranscendentalRecognize, Inverse Symbolic Calculator and Mathematica command
WolframAlpha[ToString[value], IncludePods -> "PossibleClosedForm", TimeConstraint -> ∞]

The formula holds with at least $1800$ decimal digits of precision.
Now I am looking for a rigorous way to prove this formula. Can you suggest any ideas how to do that?

REPLY [22 votes]: Use formula 16.5.2 from DLMF:
$$_3F_2\left(1,1,\frac94;2,2;-1\right)=\int_0^1{_2F_1}\left(1,\frac94;2;-t\right)dt=\frac45\int_0^1\frac{1-(1+t)^{-5/4}}tdt=\\\\\frac{2}{5}\left(8-4\sqrt[4]{8}+\pi-6\ln2+4 \ln
   (1+\sqrt[4]{2})-4 \arctan\sqrt[4]{2}+2\ln(1+\sqrt2)\right).$$
Some elementary transformations show that this result is equivalent to your conjecture.<|endoftext|>
TITLE: Wrong-way Frobenius reciprocity for finite groups representations
QUESTION [12 upvotes]: This is a typical lazy mathematician question, so do not hesitate to close it and recommend me to do my homeworks... 
Let $H$ be a subgroup of a finite group $G$, and let $Res_H^G$ and $Ind_H^G$ the restriction and induction functors between the categories of linear representations of $G$ and $H$, respectively. I'm familiar with Frobenius reciprocity stating that $Ind_H^G$ is the right adjoint to $Res_H^G$. Moreover, by choosing the traditional model for the induced representation, namely,
$Ind_H^G(U)=  \{f:G\to U \text{ such that } f(h g)=h\cdot f(g)\text{ for any } h\in H\}$
I also know how to write explicitly the natural isomorphism
$
Hom_{Rep(H)}(Res_H^G(W),U) \stackrel{\sim}{\to} Hom_{Rep(G)}(W,Ind_H^G(U)).
$
Now, I've just learnt that actually $(Res_H^G,Ind_H^G)$ is an ambidextrous adjunction, i.e. that $Ind_H^G$ is also the left adjoint to $Res_H^G$, but I'm not familiar with this fact: how is the isomorphism
$
Hom_{Rep(H)}(U,Res_H^G(W)) \stackrel{\sim}{\to} Hom_{Rep(G)}(Ind_H^G(U),W)
$
explicitly defined?

REPLY [18 votes]: The point is that there are two ways to describe the restriction functor. The first way is as $\text{Hom}_{\mathbb{C}[G]}(\mathbb{C}[G], -)$, thinking of $\mathbb{C}[G]$ as a $(\mathbb{C}[G], \mathbb{C}[H])$-bimodule, which by the tensor-hom adjunction means that it has left adjoint $\mathbb{C}[G] \otimes_{\mathbb{C}[H]} (-)$. The second way is as $\mathbb{C}[G] \otimes_{\mathbb{C}[G]} (-)$, thinking of $\mathbb{C}[G]$ as a $(\mathbb{C}[H], \mathbb{C}[G]$)-bimodule, which by a second application of the tensor-hom adjunction means means that it has right adjoint $\text{Hom}_{\mathbb{C}[H]}(\mathbb{C}[G], -)$. 
It remains to write down a natural isomorphism between the left and right adjoints. Very explicitly, the obvious candidate is to try something like
$$\mathbb{C}[G] \otimes_{\mathbb{C}[H]} V \ni \sum g \otimes v_g \mapsto (g \mapsto v_g) \in \text{Hom}_{\mathbb{C}[H]}(\mathbb{C}[G], V)$$
where $V$ is a left $\mathbb{C}[H]$-module and $v_g \in V$. But we need to check that this is well-defined. On the LHS we have $gh \otimes v_{gh} = g \otimes h v_{gh} = g \otimes v_g$ for any $h \in H$, hence $h^{-1} v_g = v_{gh}$, whereas on the RHS we need functions such that if $g \mapsto v_g$ then $hg \mapsto hv_g = v_{hg}$ for any $h \in H$, hence $h^{-1} v_g = v_{h^{-1} g}$. These don't quite match up, so our map is slightly wrong. It should be
$$\mathbb{C}[G] \otimes_{\mathbb{C}[H]} V \ni \sum g \otimes v_g \mapsto (g \mapsto v_{g^{-1}}) \in \text{Hom}_{\mathbb{C}[H]}(\mathbb{C}[G], V)$$
and then everything matches up; the inverse map sends a function $(g \mapsto v_g)$ to $\sum g^{-1} \otimes v_g$.

Alternatively, instead of working with representations we can work with unitary representations on Hilbert spaces (in the finite-dimensional case this recovers the same theory). Here for every pair of unitary representations $V, W$ we have a natural identification $\text{Hom}(V, W) \cong \text{Hom}(W, V)$ given by taking adjoints, so categories of unitary representations are dagger categories. Any adjunction between dagger categories is automatically ambidextrous. 
In general, if $f : R \to S$ is a homomorphism of rings, we get a restriction functor $S\text{-Mod} \to R\text{-Mod}$ which can be described in either of the ways above and hence which has both a left and a right adjoint, which one might call "induction" and "coinduction" or something like that. These are left and right Kan extension along $f$ (thinking of $R, S$ as $\text{Ab}$-enriched categories with one object).<|endoftext|>
TITLE: constructive Serre classes
QUESTION [9 upvotes]: A Serre class (of abelian groups) is a class of abelian groups closed under subgroups, quotients, and extensions.  For instance, finitely generated groups and finite groups are both Serre classes.
However, in constructive mathematics, these are no longer examples, at least not with the usual definition of "finite" (= in bijection with $\{0,1,\dots,n\}$ for some $n\in\mathbb{N}$).  In particular, finite sets are not closed under subsets and quotients, so there is no reason that finite groups should be either.
There are other weaker constructive notions of "finite", some of which are described here: subfinite, finitely indexed, subfinitely indexed.  It seems that subfinitely indexed sets — the subquotients of finite sets — are closed under subsets, quotients, and finite products, so that the subfinitely indexed groups should be a Serre class even constructively.  Classically, of course, all subfinitely indexed sets are finite.
My original question was:

Is there a Serre class of abelian groups in constructive mathematics which reduces classically to the finitely generated ones?

As pointed out by Ingo, this has a trivial and uninteresting answer; the real question is

Is there a description of the Serre envelope of the class of finitely generated abelian groups which is more explicit than the trivial inductive one?

REPLY [4 votes]: Any subclass $\mathcal{C}$ of an abelian category determines a smallest Serre class containing it, by iteratively adding (the zero object and) the object $Y$ for any exact sequence $X \to Y \to Z$ where $X$ and $Z$ are objects of the previous stage. Note the missing zeros at the ends of the sequence; alternatively, one can iteratively add the zero object, subobjects, quotients, and extensions (objects $Y$ such that there is a short exact sequence $0 \to X \to Y \to Z \to 0$ with $X$ and $Z$ previously added).
Anyway, this construction can in particular be performed with the class $\mathcal{C}$ of finitely generated abelian groups. Classically, its closure will coincide with $\mathcal{C}$, as $\mathcal{C}$ is already a Serre class.
However, this construction only answers the question in a technical sense, since it describes the sought-after class in rather abstract terms. The construction thus only shows that the question could profitably be reformulated in order to better capture its spirit: "Is there a description of the Serre envelope of the class of finitely generated abelian groups which is more explicit than the trivial inductive one?"<|endoftext|>
TITLE: C* Algebras, Foliations and Dynamical Systems
QUESTION [9 upvotes]: I am a Ph.D student involved in topics like integrability of foliations arising from center stable bundles of partially hyperbolic dynamical systems. These are generally only continuous bundles so one either has to use the dynamics of the map, find some meaningful approximations or try to generalize the theorems of smooth integrability and foliation theory to continuous case (it is more doable in the case of foliation theory). 
In studying these topics I have come to see that there are alot of interactions with geometry of foliations which I became interested with. Where smoothness fails and you can not use Frobenius theorem, you can make some intersting connections to foliation theory (for instance Novikov theorem) to see whether a system is integrable or not.
Following this I have been hearing the "new applications of C* algebras to dynamical systems" by finding "C* algebra of a dynamical system". and I have also seen in Connes book that C* algebras give a new way of studying foliation spaces. So I became interested in these, although I have only recently met them and did not have much chance to read about them. 
My first question would, what direction (books, questions ideas) would you propose to someone who wants to learn this/these field. My second question is, what is vaguely the use of C* algebras about foliations and dynamical systems. What kind of their properties is studied through their use? Finally does C* algebras study the case of continuous foliatons :-)
Thanks

REPLY [10 votes]: A very good place to start is Connes' book "Noncommutative Geometry", available for free on his website.  It's a huge book, but it's possible to skip around quite a bit to get what you need.
To begin, I'll remark that the foliations which are accessible to C*-algebraic techniques are generally smooth in the horizontal direction and integrable.  By now it may be possible to relax these assumptions; I'm not really an expert.  But I think the techniques are mainly useful for handling cases where the transversal behavior is very bad, e.g. the irrational rotation folation on the torus.
To associate a C*-algebra to a foliated manifold, one can use the foliation groupoid construction.  The objects in this groupoid are just the points of the manifold, and there is a morphism between two points if and only if they lie on the same leaf.  If the leaves are smooth then one can define a convolution product on the space of smooth compactly supported functions on the foliation groupoid, and this can be completed to form a (possibly noncommutative) C*-algebra.  If the foliation comes from a group action (e.g. the irrational rotation action on the torus) then this generalizes the "crossed product" construction in the theory of C* dynamical systems.
With the C*-algebra of a foliated manifold in hand, the idea is to relate invariants of the C*-algebra (e.g. K-theory, cyclic homology) to the geometry of the foliation.  Many of the best results that I know about are organized around index theory leafwise elliptic operators.  For instance, Connes used these ideas together with the Lichnerowicz vanishing theorem to produce nontrivial topological obstructions to the existence of leafwise positive scalar curvature metrics.  That said, I'm not sure how much contact there is with problems that are of interest to experts in dynamics and foliation theory.<|endoftext|>
TITLE: Boys and Girls Revisited
QUESTION [21 upvotes]: Consider a country with $n$ families, each of which continues having children until they have a boy and then stop. In the end, there are $G$ girls and $B=n$ boys.
Douglas Zare's highly upvoted answer to this question 
computes the expected fraction of girls in a population formed of complete families and explains why we shouldn't expect it to equal $1/2$.  My current question concerns a different statistic, namely the probability that there are more boys than girls (after all families have finished reproducing).  This probability turns out to be exactly $1/2$, and I'm looking for an intuitive explanation of why.  
Indeed, for fixed $n$, it's not hard to see that
$$Prob(G=k)=\binom{n+k-1}{k}\cdot {1\over 2^{n+k}}$$
(The binomial coefficient is the number of ways to assign $k$ indistinguishable girls to $n$ distinguishable families.) 
Therefore 
$$Prob(G < B)=\sum_{k=0}^{n-1}\binom{n+k-1}{k}\cdot {1\over 2^{n+k}}$$
It's not hard to check that this sum is exactly equal to $1/2$ (and therefore, in particular, independent of $n$).  
That is, regardless of the number of families, we always have the surprisingly (to me) simple formula
$$Prob(G < B )=1/2$$
(Note that this implies $Prob(G>B)$ is strictly less than $1/2$ --- because there's always some probability weight on the event $G=B$ --- though an application of Stirling's formula shows that $Prob(G>B)$ converges to $1/2$ as $n$ gets large.)
My question is:

Is there some simple intuitive reason I should have expected this result?

REPLY [14 votes]: This is just a variant presentation of fedja's truly wonderful solution.  It took me a while to catch on to the idea there, so I'm offering this in case it helps clarify it for anyone else.  All credit for the insight, though, properly belongs to fedja.
For the purpose of determing the probability that the $n$ families wind up with more boys than girls altogether, it's convenient to imagine births take place in the following fanciful way.  There is a "magic fertility wand" which is passed from family to family and which causes the family that possesses it to produce one child per day until they have a boy, at which point they pass the wand along to the next family.  If and when every family has their boy, the wand is given to a pair of rabbits, who use it to produce a male or female offspring each day without cease.  All births take place at a hospital/veterinary clinic.
On the $(2n-1)$st day, the hospital reports whether they've delivered a preponderance of males or females.  Obviously it's a 50:50 chance for either result.  If it's more males, there are at least $n$ of them, which means there are exactly $n$ boys and possibly some male rabbits, but certainly no more than $n-1$ girls -- in any event, the families are done reproducing and there are more boys than girls.  If, on the other hand, the hospital reports more female births, then there have been fewer than $n$ males born, which means the wand hasn't yet reached the rabbits, so all the births have been boys and girls, with at least $n$ of them girls, so the boys will never outnumber the girls.<|endoftext|>
TITLE:  A Canonical Form Theorem for $n$-forms? 
QUESTION [12 upvotes]: I have been working with what I call ``measured-manifolds'', i.e., an $n$-dimensional 
smooth manifold $M$ together with a non-vanishing $n$-form $\omega$ on $M$. 
At a certain point I realized that a lot of things would be greatly simplified if I 
knew that at each point $p$ of $M$ I could find a  "canonical coordinate system"
$(x_1,\ldots,x_n)$, meaning that in these coordinates 
$\omega = dx_1\wedge\ldots,\wedge\  dx_n$.  I couldn't recall ever seeing 
such a result, and I spent about a day looking in the obvious textbooks, 
searching with Google, and asking friends if they had heard of such 
an existence theorem. Then suddenly I realized that it was obvious !
If $(y_1,\ldots,y_n)$ is any coordinate system centered at $p$ and 
$\omega = \rho(y_1,\ldots , y_n) dy_1\wedge \ldots\wedge dy_n$ then define
$x_1(y_1,\ldots , y_n) := \int_0^{y_1} \rho(t,y_2,\ldots,y_n) dt$ and let 
$x_i := y_i$ for $i = 2,3,\ldots,n$. Then 
$dx_1 = \rho(y) dy_1 + \sum_{j=2}^n a_j(y) dy_j$ 
so $dx_1 \wedge(dy_2 \wedge \ldots \wedge dy_n) = \omega$
and $(x_1,\ldots,x_n)$ are indeed canonical coordinates.
Now I am not so dumb as to imagine I am the first person to observe this 
triviality, and I would even be willing to bet that E. Cartan knew it way 
back when, and that a few zillion others have noticed it and used it 
since.  So my question is, where can I find the best reference for the 
existence of canonical coordinates for $n$-forms.

REPLY [7 votes]: This probably will not actually answer your question about earliest references, but it's too long for a comment. 
You are right that this 'normal form' was known to É. Cartan, and I'll be that it was known to Lie as well earlier than that.  However, I am not sure that you'll be able to find it explicitly stated as a result in Cartan's works.  More likely, he would have regarded it as obvious, and for the reason that you gave:  The proof is one line.
The reason I have for claiming this is probably overkill, but here goes:  Cartan classified the 'infinite transitive primitive Lie groups' in a series of early papers starting around 1904.  ('Infinite' means 'not finite dimensional, 'transitive' has its usual meaning, 'primitive' means 'leaves no nontrivial foliation invariant', and 'Lie group' means 'a transformation (pseudo-)group defined as the solutions of some system of PDE'.)  His list (all of which were known to Lie) goes like:

All transformations (the PDE system is empty)
Transformations that preserve a volume form up to a constant multiple (Jacobian determinant is constant)
Transformations that preserve a volume form (Jacobian determinant is 1)
Transformations that preserve a symplectic form on a domain, up to a constant multiple. (Only happens in even dimensions.  THe PDE system is the obvious one.)
Transformations that preserve a symplectic form on the domain. (Only happens in even dimensions.  THe PDE system is the obvious one.)
Transformations that preserve a contact structure.  (Only happens in odd dimensions.  The PDE system is the obvious one.)

Cartan implicitly assumes that there is (locally) only one kind of volume form when stating this result for Items 2 and 3, just as he implicitly assumes Darboux' Theorem in the statement of Items 4 and 5.  From this, I conclude that he knew then that there was, up to diffeomorphism, only one kind of volume form (i.e., form of top degree).  However, I don't remember him actually stating this 'normal form' for volume forms explicitly (and I'm traveling right now and don't have access to the works of Cartan, so I can't check); it's more likely that he just assumed that 'everybody knows this'.
Remark:  By the way, if you look at the above list, you probably will object that it doesn't have some other obvious items, such as the biholomorphic transformations in a given (even) dimension and the transformations that preserve a volume form up to a sign (i.e., the transformations for which the absolute value of the Jacobian determinant is 1 , etc.  Cartan's result needs some interpretation.  While he didn't say this explicitly, he generally assumed in those papers that it didn't matter whether one was working with real or complex variables and he didn't bother to distinguish groups that had the same Lie algebra of 'infinitesimal transformations'.  When Singer and Sternberg revisited this work in the 1960s, they did kind of clean this up and extend the list to distinguish real and complex variables, so that the list gets a bit longer, but only by the obvious additions that this entails.<|endoftext|>
TITLE: Asymptotic independence in a multinomial setting.
QUESTION [5 upvotes]: Let $(X_1,\ldots,X_r)$ be a multinomial vector with parameters $n$ and $1/r$, i.e., we throw $n$ balls into $r$ bins, with a uniform probability for each ball to land in each of the bins.  As is well known, $X_i$ and $X_j$ are dependent, with $\mathrm{cov}(X_i, X_j) = -n/r^2$.
Now let $r = r(n) \rightarrow \infty$, so we have a sequence of multinomial vectors (of increasing length), the nth vector being $(X_{1,n},\ldots,X_{r(n),n})$.  For a fixed index $k$, I know how to prove that if $r(n)/n \rightarrow 0$ as $n \rightarrow \infty$, then the normalized binomial RVs $[X_{k,n} - n/r(n)]/\sqrt{n(1/r(n))(1 - 1/r(n))}$ converge in distribution to a $N(0,1)$ RV (using the Lindeberg-Feller Theorem for triangular arrays).
However, I want to show that the resulting limiting RVs, for different $k$'s, are independent of each other.  The right way to think about it, I believe, is to extend each of the multinomial vectors with infinitely many zeros (say) to the right, i.e., to define $X_{i,n} = 0$ for $i > r(n)$; we end up with a a sequence of well-defined processes $X_1, X_2,\ldots$, where $X_n = \{X_{1,n}, X_{2,n}, \ldots\}$.  To show independence between the elements of the limiting process, it is enough to show that for each fixed $k$, the joint distribution of the first $k$ elements of the processes (properly normalized) converge in distribution to $k$ independent standard normal RVs.
But how do I go about this?  Simply proving that the limiting covariance is zero is not enough, I think, since zero covariance implies independence only in a multivariate normal setting, and I haven't proved that the limiting RVs have jointly a multivariate normal distribution.  The problem seems to me too elementary to be new, so I will be grateful for any references for an existing proof, or for ideas how to proceed.  Thanks.

REPLY [5 votes]: The simplest way is to couple your process with something obvious. It can be done in many ways. I would do the following.
Consider $r$ bins represented as disjoint unit intervals on $\mathbb R$. Run the Poisson process of intensity $\lambda=n/r$. Look at what we get in the bins. If the total number of balls $N$ is wrong, throw out or add $|N-n|$ balls at random from the whole configuration. You'll end up with the same distribution as if you threw in exactly $n$ balls in the original way. Now, before the correction procedure, the numbers of balls in bins are independent and have expectations and variances $n/r$ each. Also, since the total variance of $N$ is $n$, with high probability you'll need to use only $C\sqrt n$ balls during the correction. The claim now is that as long as the number of balls in the bin is not greater than $Cn/r$ (which has probability at least $(1-C^{-1}$), the correction can change the number of balls in the bin by only $C'r^{-1}\sqrt n$ on average, which is much less than the scaling size $\sqrt{n/r}$ if $r$ is large. Thus, the correction is invisible after scaling for each fixed finite set of bins and you can merely ignore it when passing to the limit.<|endoftext|>
TITLE: Unpublished works of Woodin on SCH and Radin forcing
QUESTION [13 upvotes]: There are many unpublished results of Hugh Woodin on ''singular cardinals hypothesis'' and '' Radin forcing''. Some of his results are published later by others, but it seems that there are still many unpublished results.
Does any one know some of them to state here. 
Is there any way to get a copy of Woodin's unpublished results?

REPLY [12 votes]: (Below, I refer to the Handbook. This is the Handbook of Set Theory, Foreman, Kanamori, eds., Springer, 2010.)
The bulk of these results appears in notes by James Cummings. You may want to ask him about them. Originally these notes were intended for a book on Radin forcing to be coauthored by him and Hugh, but significant portions of it appear in his Handbook article. (The book exists in draft form.)
There are a few additional facts about Radin forcing that did not make it into the article, but James has write ups of them, and of course the theory has expanded since. 
For $\mathsf{SCH}$ specifically, some details appear in Gitik's paper The negation of the singular cardinal hypothesis from $o(\kappa)=\kappa^{++}$, APAL 43 (1989), 209-234. The state of the art in this regard is described in Gitik's Handbook article, his papers, and those of his co-authors, particularly Carmy Merimovich. 
Some applications of Radin/Prikry-like forcing in the context of determinacy are not in any of the above. You can see some in the Koellner-Woodin Handbook article, and yet others in the proofs of the derived model theorem, most of which can be seen (perhaps in preliminary form) here.
The one thing I do not think is in either place is a discussion of $T$-degrees, or constructibility degrees. Hugh discusses some of this (briefly) in The cardinals below $|[\omega_1]^{<\omega_1}|$, APAL 140 (2006), 161–232, but there is a bit more than this. 
Richard Ketchersid wrote up the basics in a nice article, More structural consequences of $\mathsf{AD}$, in Set Theory and Its Applications, Contemporary Mathematics, vol. 533, Amer. Math. Soc., Providence, RI, 2011, pp. 71-106. To define these degrees, assume the axiom of determinacy, so we have Martin's cone measure on Turing-invariant sets of reals. Given sets of ordinals $T$ and $S$, define $S\lt T$ iff for almost every $r$ (in the sense of Martin's measure)
 $$ L[T,x]\cap\mathbb R\setminus L[S,x]\ne\emptyset. $$
It turns out that for any two sets of ordinals $S,T$, precisely one of the following holds:

$S\lt T$, in which case for almost every $x$, the reals of $L[S,x]$ are in $L[T,x]$.
$T\lt S$, in which case for almost every $x$, $\mathbb R\cap L[T,x]\subset L[S,x]$.
For almost every $x$, $\mathbb R\cap L[T,x]=\mathbb R\cap L[S,x]$.

If option 3 holds, we say that $S$ and $T$ have the same degree. The relation $\lt$ induces a well-ordering of degrees. This is established, together with the basic properties of these degrees, via Prikry-like arguments.
It is possible that there are yet additional results not in any of the above. In that case, I doubt there are formal written notes, but there may be accounts in notes from seminar talks.<|endoftext|>
TITLE: A function from partitions to natural numbers - is it familiar?
QUESTION [17 upvotes]: I've come across a function from the set of integer partitions to the natural numbers which I don't recognise but which probably ought to be familiar; it arises in the homogeneous Garnir relations for graded Specht modules (see Kleshchev, Mathas & Ram: "Universal graded Specht modules", Proc. LMS 105). I hope someone will recognise it and point me to a useful reference.
The function (let's call it $f$) is defined recursively; to begin with, $f(\varnothing)=1$.  Now suppose we have a non-empty partition $\lambda$, and take a box $(i,j)$ in the Young diagram for which $i+j$ is maximised.  Let $\mu$ be the partition you get by removing the first $i$ rows from the Young diagram and let $\nu$ be the partition you get by removing the first $j$ columns from the Young diagram.  Now recursively define
$f(\lambda) = \binom{i+j}if(\mu)f(\nu)$.
For example, take $\lambda=(5,3,3,1)$. Let $(i,j)=(3,3)$, so that $\mu=(1)$ and $\nu=(2)$. One can calculate $f(\mu)=2$ and $f(\nu)=3$, so that $f(\lambda)=120$.
It's not hard to show that $f$ is well-defined (i.e. doesn't depend on the choice of $(i,j)$).  I have two other definitions of $f$ (also recursive) which I'll omit for now (it's non-trivial to show that they are equivalent).
Please let me know if you've seen this function (or a similar-looking one) before.  Since it's a function from partitions to positive integers, it ought to count tableaux of some kind.

REPLY [3 votes]: Okay, I agree that this answer comes actually a little late...
If you type some values of your function into FindStat, you will see David Speyer's answer automatically generated, since it is obtained as a natural statistic obtained after applying a natural combinatorial map.
To have some values, for size up to $4$ it is given by
$$
[1] => 2,
[2] => 3,
[1,1] => 3,
[3] => 4,
[2,1] => 6,
[1,1,1] => 4,
$$
$$
[4] => 5,
[3,1] => 8,
[2,2] => 6,
[2,1,1] => 8,
[1,1,1,1] => 5.
$$<|endoftext|>
TITLE: Fundamental group of a topological pullback
QUESTION [26 upvotes]: This should be such an elementary problem in algebraic topology that I'm almost too embarrassed to ask, but here goes.
Let $f: X\to Z$ be a surjective fibration, and let $g: Y\to Z$ be any map. Assume all spaces are path-connected, and base points $x,y,z$ chosen so that $f(x)=g(y)=z$. Form the pullback in the topological category,
$$
\begin{array}{ccc}
E & \to & X \newline
\downarrow & & \downarrow \newline
Y & \to & Z.
\end{array}
$$ 
Note that $E$ need not be path-connected.

Is it possible to express $\pi_1(E,e)$ for a given choice of base point $e\in E$, in terms of $f_\sharp: \pi_1(X,x)\to \pi_1(Z,z)$ and $g_\sharp: \pi_1(Y,y)\to \pi_1(Z,z)$?

REPLY [10 votes]: Expressed in terms of the homotopy pullback $N(f,g)$ of a pair of based maps $f\colon X\longrightarrow A$ and $g\colon Y\longrightarrow A$, the long exact sequence is Corollary 2.2.3 of May and Ponto ``More concise algebraic topology''.  The result of which it is a 
corollary, Proposition 2.2.2, describes the pointed set of based maps $[Z,N(f,g)]$
for any based space $Z$. The dual result for homotopy pushouts is Proposition 2.1.2.  The corollary is used heavily in the study of fracture theorems for localization and completion later on in the book.<|endoftext|>
TITLE: Homotopy pullbacks and pushouts of spectra
QUESTION [7 upvotes]: Is it true that homotopy pullbacks and homotopy pushouts coincide in the category of spectra? I had a feeling that this is the case, but don't know where to find a proof or how to prove it.
Thanks!

REPLY [2 votes]: Although the answer is sketched in the comments, I wanted to remark that this statement is proved carefully by Cary Malkiewich as Proposition 6.2.11 in Parameterized Spectra, A Low Tech Approach. He credits Model Categories of Diagram Spectra as the first place this was proven, but I couldn't find the result there.<|endoftext|>
TITLE: zeros of a homogeneous polynomial
QUESTION [10 upvotes]: Hi All,
Let $F$ be a finite field, $\lambda\in F$, and $$p_\lambda (x,y,z)=\left|\begin{array}{ccc}x & y & z \\ y & z & x +\lambda z \\ z& x+\lambda z & y+\lambda x+\lambda ^2z \end{array}\right|.$$
Can one choose $\lambda$ so that $(0,0,0)$ is the only zero in $F^3$ for $p_\lambda (x,y,z)$?
Also, I'm wondering if the matrix that defines $p_\lambda$ looks familiar to anyone. Thanks!

REPLY [9 votes]: The existence of $\lambda$ has by now been proved by A.Kumar's answer
and the ensuing comments, but this still doesn't explain the behavior
of the determinant $p_\lambda(x,y,z)$: usually the zero-locus of
$3 \times 3$ determinant of linear forms is an irreducible curve,
not the union of three lines.  A.Kumar's answer does strongly
hint at the reason, though.  Here's an explanation, as well as
a formula for the number of $\lambda \bmod p$ for which
$p_\lambda$ has no nontrivial zeros.
Let $k$ be any field, $R$ the ring $k[T] \left/ (T^3 - \lambda T^2 - 1) \right.$,
and $r \in R$ the element $y+xT+zT^2$.  Then multiplication by $r$
is a $k$-linear transformation of $R$ whose matrix with respect to
the basis $(T,1,T^2)$ is none other than
$$
\left(
\begin{array}{ccc}
x & y & z \\
y & z & x + \lambda z \\
z & x + \lambda z & y + \lambda x + \lambda ^2 z
\end{array}
\right)\phantom0.
$$
Thus $p_\lambda(x,y,z)$ is nonzero iff $r$ is invertible in $R$,
and this is true for all nonzero $r \in R$ iff $R$ is a field
iff the cubic $T^3 - \lambda T^2 - 1$ is irreducible.
Taking $T = 1/t$ then recovers A.Kumar's polynomial $t^3 + \lambda t + 1$.
Now as F.Brunault observes, if $k$ is finite then
the existence of a $\lambda$ that makes $t^3 + \lambda t + 1$
irreducible is an elementary counting argument: in degree $3$,
irreduicble still means no rational root; if $t^3 + \lambda t + 1 = 0$
then $t \neq 0$ and $\lambda = -(t^2 + t^{-1})$, so each nonzero $t$
arises exactly once as a root, and $t=0$ does not arise at all.
Since there are $\left|k\right|$ choices of $\lambda$, at least one
of them must make $t^3 + \lambda t + 1$ irreducible.
To enumerate such $\lambda$ we need only count the polynomials
$t^3 + \lambda t + 1$ that split completely.  These are parametrized
by solutions of $t_1 + t_2 + t_3 = 0$, $t_1 t_2 t_3 = -1$ up to permutation.
These equations give an elliptic curve isogenous with the Fermat cubic
(if $a^3+b^3+c^3 = 0$ then
$$
(t_1^{\phantom.}, t_2^{\phantom.}, t_3^{\phantom.}) = \frac{-1}{abc} (a^3, b^3, c^3)
$$
satisfies $t_1 + t_2 + t_3 = 0$ and $t_1 t_2 t_3 = -1$; this gives
the isogeny in one direction).  The number of rational points on
this curve is known, and eventually one finds (if I did this right)
that the count is $(p+1)/3$ if $p \equiv -1 \bmod 3$, and
$(p+1-a)/3$ if $p \equiv +1 \bmod 3$, where in the latter case
$a \equiv -1 \bmod 3$ is determined uniquely by $4p = a^2 + 27 b^2$.
In particular, for large $p$ the count is $p/3 + O(p^{1/2})$,
as predicted by Čebotarev (since $1/3$ is the proportion of
$3$-cycles in the symmetric group $S_3$).<|endoftext|>
TITLE: What is the discriminant divisor of a surface fibered over a curve?
QUESTION [8 upvotes]: Let $\pi:X\rightarrow C$ be a flat and proper morphism over $\mathbb{C}$ where
$X$ is a smooth projective surface and $C$ is a smooth projective curve. Assume that all the fibers of $\pi$, except finitely many, are smooth projective genus $g$ curves. 
Q1:Then what is the exact definition of the discriminant divisor of $\pi$ and more importantly, what is its geometrical content? 
My first guess is that it should probably keep track of critical points of $f$, i.e., points $x\in X$ such that $d\pi_x$ has not maximal rank, but then, it is not a divisor
(codimension one cycle) on $X$...
Q2: Do we have a general definition of the "discriminant divisor" associated to a 
sufficiently nice map between two smooth complex (varieties) manifolds?

REPLY [2 votes]: I actually spent some time thinking about what this should be, not sure if it gives the right point of view: Let $f: X \to S$ be a flat, projective, local complete intersection, with $S$ normal, of (constant) relative dimension $n$. Denote by $p: H \to s$ the $f$-singular locus over $s = f(H)$. Then I would define $\mathcal D_{X/S} = p_* c_{n+1}^H(\Omega_{X/S})$ in the appropriate Chow group. Here $c_{n+1}^H(\Omega_{X/S})$ denotes a/the localized Chern class, which is a refinement of the usual Chern class. One could probably weaken the definition to assume that $H$ is only proper over $s$. 
This has the virtue of coinciding with the discriminant divisor in the cases where I know there is a notion of discriminant, possibly up to some silly factor which can potentially be baked into the definition of the discriminant in that context, see Proposition 1.1 and Theorem 1.2 in http://www.math.chalmers.se/~dener/Multiplicity-of-discriminants-revision-Oct-7-2013.pdf (which has since appeared in Crelle).<|endoftext|>
TITLE: Conway's game of life for random initial position
QUESTION [47 upvotes]: What is the behavior of Conway's game of life when the initial position is random? -- We can ask this question on an infinite grid or on an $n$ by $n$ table (planar or on a torus). Specifically suppose that to start with every cell is alive with probability $p$ and these probabilities are statistically independent. This question was motivated by a recent talk by Béla Bollobás on bootstrap percolation.
Many thanks for all the answers. A related question that I thought about is what is the situation for "noisy" versions of Conway's game of life? For example if in each round a live cell dies with probability $t$ and a dead cell gets life with probability $s$ and both $t$ and $s$ are small numbers and all these probabilities are independent.
Another example is to consider the following probabilistic variant of the rule of the game itself ($t$ is a small real number):
Any live cell with fewer than two live neighbours dies with probability $1−t$.
Any live cell with two or three live neighbours lives with probability $1−t$ on to the next generation.
Any live cell with more than three live neighbours dies with probability $1−t$.
Any dead cell with exactly three live neighbours becomes a live cell with probability $1−t$.
Following some comments below I asked about the computational power of such a noisy version over here.
Update: Related question Is there any superstable configuration in the game of life?

REPLY [4 votes]: About the random version, a relevant reference is (IIRC):
Peter Gacs, Reliable computation with cellular automata. Journal of Computer System Science, 32(1):15–78, February 1986. Conference version at STOC’ 83.
(As well as subsequent works.) Essentially given a Turing-complete automaton - not sure about the terminology here - it can be made robust to noise by adding error-correction mechanisms. It is not about GoL, and tastes a bit more like low-temperature Glauber dynamics mixed with directed percolation, but the ideas might be relevant here.<|endoftext|>
TITLE: If $\kappa \rightarrow (\alpha)^r_2$ holds for every $r\in \omega$, then is $\kappa$ an $\alpha$-Erdős cardinal?
QUESTION [5 upvotes]: If $\kappa \rightarrow (\alpha)^r_2$ holds for every $r\in \omega$, then is $\kappa$ an $\alpha$-Erdős cardinal? (or rather, does $\kappa \rightarrow (\alpha)^{<\omega}_2$ hold?)
$\kappa \rightarrow (\alpha)^r_2$ means that $\forall f: [\kappa]^r \rightarrow 2$, $\exists A\in[\kappa]^\alpha$ such that $|f"[A]^r|=1$.
$\kappa \rightarrow (\alpha)^{<\omega}_2$ is $\forall f: [\kappa]^{<\omega} \rightarrow 2$, $\exists A\in[\kappa]^\alpha$ such that $\forall l \in \omega$, $|f"[A]^l|=1$.
Clearly if $\kappa \rightarrow (\alpha)^r_2$ holds for every $r\in \omega$, then splitting $f:[\kappa]^{<\omega}\rightarrow 2$ into functions $f_r:[\kappa]^r\rightarrow 2$ and applying the partition relations, we get that for each $l\in \omega$ we can find an $A_l\in[\kappa]^\alpha$ where $|f"[A_l]^l|=1$. But we can't simply take the intersection of these $A_l$ as this may not have size $\alpha$. (It could easily be empty!)
We can also find the $A_l$ such that $A_i\supseteq A_{i+1}$ for all $i$, by restricting the $f_r:[A_{r-1}]^r\rightarrow 2$ and finding homogenous sets for these. But again, the intersection of all of these may not have size $\alpha$.
In particular, does this hold when $\alpha=\kappa$?
What about if $\kappa \rightarrow (\alpha)^r_\gamma$ holds for every $r\in \omega$? Then does $\kappa \rightarrow (\alpha)^{<\omega}_\gamma$ hold?
If this is not the case, then is there a name for these $\kappa$ where $\kappa \rightarrow (\alpha)^r_2$ holds for every $r\in \omega$?
EDIT: In view of the answers; $\kappa$ is weakly compact implies $\kappa\rightarrow (\kappa)^r_\gamma$ for any $r<\omega$, and $\gamma<\kappa$. But does $\forall \alpha<\kappa$, $\kappa \rightarrow (\alpha)^2_2$ imply $\forall \alpha<\kappa,r<\omega, \gamma<\kappa, \kappa\rightarrow (\alpha)^r_\gamma$?

REPLY [8 votes]: The two properties are not the same.
If $\kappa$ is weakly compact, then $\kappa\rightarrow(\kappa)^r_\lambda$ for any $r<\omega$ and $\lambda<\kappa$.  These cardinals are compatible with $V=L$.
A cardinal $\kappa$ for which $\kappa\rightarrow(\kappa)^{<\omega}_2$ is called a Ramsey cardinal, and these cardinals imply that $0^\sharp$ exists.

REPLY [6 votes]: Unfortunately $\kappa\rightarrow(\alpha)^r_2$ for every $r$ is much weaker than $\kappa\rightarrow(\alpha)^{<\omega}_2$. In fact, the least $\kappa$ for which $\kappa\rightarrow(\alpha)^r_2$ holds is $\exp_{r-1}(|\alpha|)^+$ (the general Erdos-Rado theorem), and so the least $\kappa$ for which this holds for all $r$, is $\exp_\omega(|\alpha|)$, a singular cardinal. Here $\exp_r(\tau)$ is the $r$-fold iterated power of $\tau$. 
If $\alpha$ is limit, then the least $\kappa$ for which $\kappa\rightarrow(\alpha)^{<\omega}_2$ holds, is strongly inaccessible.<|endoftext|>
TITLE: Can we invert barycentric subdivision?
QUESTION [8 upvotes]: With apologies to fellow algebraic topologists, I confess that I have no idea how to answer this innocent-looking question:

(1) Let's say we know that a finite simplicial complex $S$ is the barycentric subdivision of some other simplicial complex $K$. Can we find $K$ (up to simplicial isomorphism) from knowledge of $S$ alone? 

Nor can I find answers to the related question:

(2) Given a simplicial complex $S$, can we decide if an "inverse-subdivision" $K$ exists? That is, can we decide if $\text{Sd}K = S$ has a solution $K$?

Ideally, one would like an affirmative answer to:

(3) Are there efficient algorithms which decide (2) and solve (1), given $S$?

Efficient in this case would mean polynomial time in the number of simplices in $S$ (this might be super-exponential in the number of simplices in $K$, if $K$ exists).

Background:
Let $K$ be an abstract simplicial complex. Its barycentric subdivision $\text{Sd}K$ is defined to be the simplicial complex whose vertices are the simplices $\sigma \in K$, and each $d$-simplex is a sequence $\sigma_0 < \sigma_1 < \cdots < \sigma_d$ of strict face relations in $K$. 
It is well-known that the geometric realizations of $\text{Sd}K$ and $K$ are homeomorphic, and barycentric subdivisions are used with remarkable frequency in basic algebraic topology (e.g., in proving the equivalence of simplicial and singular homology, or that simplicial homology satisfies the excision axiom, see Hatcher Ch. 2.1).

Update:
From the comments, there doesn't even appear to be a consensus on whether non-isomorphic simplicial complexes can have isomorphic barycentric subdivisions. Maybe someone wants to take a stab at this foundational question as well.

REPLY [3 votes]: [As requested by Vidit Nanda, I am reposting a slightly edited version of my comment above as an answer. Nevertheless, I hope someone will eventually give a satisfactory answer to this question.]
The following article is relevant for this question: "The insufficiency of barycentric subdivision" by Ross L. Finney (published in Michigan Mathematical Journal, Volume 12, Issue 3 (1965), pages 263-272).
Let $K$ and $L$ be locally finite simplicial complexes. The cited article claims that $K$ and $L$ are isomorphic if their barycentric subdivisions are isomorphic. It also states that any isomorphism between the barycentric subdivisions of $K$ and $L$ is induced by an isomorphism $K\to L$, as long as $K$ is connected and verifies the following two conditions:

$K$ is not isomorphic to a simplex or to the boundary of a simplex, and
the geometric realization of $K$ is not a $1$-manifold without boundary (i.e. not homeomorphic to $S^1$ or to $\mathbb{R}$).

Perhaps the analysis leading to this latter result may help with problem 3 in the question.<|endoftext|>
TITLE: Cohomological dimension of knit products
QUESTION [10 upvotes]: Let $G$ be a group with complementary subgroups $A$ and $B$ (meaning $A\cap B=1$ and $AB=G$).
If $A$ and $B$ are both normal in $G$, then $G\cong A\times B$ is a direct product. If $A$ is normal, then $G\cong A\rtimes_{\varphi} B$ is a semi-direct product. More generally, $G\cong A\bowtie B$ is a knit product, or Zappa-Szép product, or general product of $A$ and $B$.
My question concerns the cohomological dimension of such knit products. To ask a specific question:

Does anyone know of any examples where $\operatorname{cd}(A\bowtie B)<\operatorname{cd}(A\times B)$? 

(Edit: Other than free products $A\ast B$.)
More generally I would be interested in any references concerning the cohomology of knit products. Is there any weaponry akin to the Lyndon-Hochschild-Serre spectral sequence?

REPLY [3 votes]: First of all, the free product of nontrivial groups is never a knit product of these groups. 
Secondly, a similar question for short exact sequences of groups was already asked in this MO post, with examples similar to Dieter's. 
The third remark is that, in view of the examples, the natural inequality to expect is subadditivity:
$$
cd(A\bowtie B) \le cd(A) + cd(B). 
$$ 
I do not know how to prove this inequality in general (apart from semidirect products). Something might follow from this paper where the related concept of "factorable group" is considered and the author did some homological computations (I did not look at the details though). 
Suppose, however, that instead of cohomological dimension $cd(G)$ of groups, one considers the asymptotic dimension $asdim(G)$ (I assume from now on, that both groups $A$ and $B$ are finitely generated). Definitions and basic results can be found, for instance here. It then follows from the "Hurewicz theorem" for asymptotic dimension, using the product map 
$$
A\times B \to A\bowtie B
$$
that
$$
asdim(A\bowtie B) \le asdim(A) + asdim(B).
$$<|endoftext|>
TITLE: famous papers/results by non professional mathematicians
QUESTION [8 upvotes]: Possible Duplicate:
What recent discoveries have amateur mathematicians made? 

Dear overflowers
Out of curiosity: do you know any famous papers and/or results by non professional mathematicians? (I realize that 'non professional mathematicians' is quite vague, so let's also say 'amateur'?)
Thank you for your answers!
Edit: I meant of course math papers. Also, I am interested in somewhat recent examples, say the last 2 centuries.

REPLY [4 votes]: Robert Ammann (from Wikipedia):

Robert Ammann [...] was an amateur mathematician who made several significant and groundbreaking contributions to the theory of quasicrystals and aperiodic tilings.
Ammann attended Brandeis University, but generally did not go to classes, and left after three years. He worked as a programmer for Honeywell. After ten years, his position was eliminated as part of a routine cutback, and Ammann ended up working as a mail sorter for a post office.
He discovered several new aperiodic tilings, each among the simplest known examples of aperiodic sets of tiles. He also showed how to generate tilings using lines in the plane as guides for lines marked on the tiles, now called "Ammann bars".
Ammann's discoveries came to notice only after Penrose had published his own discovery and gained priority.

He published one paper with Grünbaum and Shephard.

REPLY [3 votes]: Kurt Heegner is a nice example. Sadly he died before the mathematical community realized that 
his proof of the class number 1 problem was essentially correct.<|endoftext|>
TITLE: How much one can earn on a white noise ? 
QUESTION [7 upvotes]: Consider the simplfied math. model for asset price (it is nevertheless quite practical for specific situations see "PS" part below) assume price "p(n)" at moment "n" is equal to N(0,1) - i.i.d - independent Gaussians.
Assume we can possess not more than one  asset in any moment. I.e. if buy it we can keep it or sell it, but cannot buy another until we sold the previous one.
Our profit is accumulation of differences between the prices we bought  and sold.
Informal Question What are the "best" trading strategies ?
Mathematically rigorous  Question 1 What is the strategy which will maximize expectation value of the profit for trading time n=1...N ?
Conjecture (YES/NO) Is it true that the answer to the question above is given by the following simple strategy  if the price is greater than zero  - sell, if less than zero buy.
See MatLab code for details. 
Variations on the question
Profit here is random variable, and what means "best" random variable is ambiguous. The simplest version is to  take expectation as in question above, but more profound measures of quality can be something like E(profit)/std(profit), I mean taking "risk" into account.

Here is MatLab code for simple strategy - buy if price < threshold1, sell if price > threshold2.
Results of simulation suggests that best choice is threshold1=threshold2=0.
It motivates the conjecture above.
enter code herefunction     profit = TradeStrategy2(threshold1, threshold2 )
len = 1e6;
p =  randn(1,len);

flagBought = 0; profit = 0;
for k=1:len
    if (p(k) < threshold1 ) && (flagBought == 0)
        pSave = p(k);
        flagBought = 1;
    elseif (p(k) > threshold2 ) && (flagBought == 1)
        profit  = profit  + (p(k)-pSave) ;
        flagBought = 0;
    end;
end;
fprintf(1,'Strategy 2, threshold1 =, %f, threshold2 =, %f profit/len = , %f, \n ', threshold1, threshold2, profit/len );

profit = profit/len;

end

Simulation results:
Strategy 2, threshold1 =, -0.000000, threshold2 =, 0.000000 profit/len = , 0.398951, 
Strategy 2, threshold1 =, -0.100000, threshold2 =, 0.100000 profit/len = , 0.395956, 
Strategy 2, threshold1 =, -0.100000, threshold2 =, 1.000000 profit/len = , 0.281029, 
Strategy 2, threshold1 =, -1.000000, threshold2 =, 0.100000 profit/len = , 0.281722, 
Strategy 2, threshold1 =, -1.000000, threshold2 =, 1.000000 profit/len = , 0.242517, 
Strategy 2, threshold1 =, -3.000000, threshold2 =, 3.000000 profit/len = , 0.004250, 

Similar questions can be asked for arbitrary random process.
I guess, that the general answer should also be known,
however I have not yet received answer at 
https://quant.stackexchange.com/questions/8110/mathematical-theories-of-sub-optimal-trading-strategies-under-idealized-assu

PS
Motivation for considering such simplified model of the price is the following.
Consider the "trend" part of some real price, the simplest model for the trend is linear trend: p(t) = At + B + noise. The most simple situation is to take A=0.
So we almost come to our model. Please notice that choice of "B" does not affect the optimal stategy, so we can assume it to be zero.
However, physiologically,  it is better to assume it to be B = million or billion - very big number. It will explain the second assumption of the model - you cannot buy two assets at one time - that it is just because your resources are limited - you have 1 million (or billion), but not two.

REPLY [5 votes]: As I already replied to your QSE question, almost any kind of such problem can be solved by means of the dynamic programming discovered by Bellman and deepened by such people as Bertsekas and Shreve to name a few. Btw, the joint work of the latter two, "Stochastic Optimal Control" is one of the best in the field. So are the books written by Shreve on the stochastic analysis and stochastic finance.
No matter which evolution (in discrete-time) of the assets/group of assets you assume, as long as it can be restated in the Markovian form (perhaps, giving up the number/finetness of dimensions), you can apply stochastic optimal control principles as per the reference above. the framework is rather broad - namely you can deal with Borel spaces which is often enough in practice. Moreover, even optimal stopping problems or double optimal stopping problems like the one you are interested in, can be restated as classical additive cost problems if you use enough of auxiliary variables. 
Even writing the dynamic programming recursions/fixpoint equations often gives some intuition for the shape of the optimal strategy. In general, of course, its worth computing the strategy and the value of the optimal control problem - and there is a whole range of approximate dynamic programming methods to you service.

I believe, Waldemar has given a nice answer to your question which didn't require any involved techniques. Let me show though how DP would work.
We have the following Markovian model for the capital of the trader
$$
  \mathbf x_{k+1} = \mathbf x_k + \mathbf u_k(\mathbf z_{k+1} - \mathbf y_k)
$$
$$
  \mathbf y_{k+1} = \mathbf z_{k+1}
$$
where $\mathbf z_k\sim N(0,1)$ is a price, $\mathbf u_k$ is your control variable: it's $1$ when you hold the asset and $0$ otherwise. Also, $\mathbf y_k$ is an auxiliary variable that we use to account for the correlation in price changes. 
Let us fix a time horizon $N>0$, so that the final reward is $r_N(x,y) = x$. We don't have any running cost, so $r(x,y) = 0$. Let $J^*_n(x,y)$ denote the optimal value on the horizon $[n;N]$, so that
$$
  J^*_N(x,y) := c_N(x,y) = x
$$
and we have recursions
$$
  J^*_n(x,y) = \max_u \int_{\Bbb R^2} J^*_{n+1}(x',y')Q(\mathrm dx'\times \mathrm dy'|x,y,u)
$$
where $Q$ is the transition kernel derived from the update equations above. 
Now, by simplifying the dynamics above you get iterations
$$
  J^*_n(x,y) = \max\left(\Bbb E {J^*}_{n+1}(x,Y),\Bbb E {J^*}_{n+1}(X + x-y,Y)\right)
$$
where $\Bbb E$ is taken w.r.t. to iid $X,Y\in N(0,1)$. Doing computations by hand, you'll get for $n = 0,1,2,\dots,N-1$ that
$$
  J^*_n(x,y) = x + (N-n-1)\gamma + y^-
$$
where $y^- := \max(0,-y)$ and $\gamma := \Bbb E Y^-$. Whenever in your iterations you have to find the maximum, you get the optimal decision for $u$. In such a case you just have to decide whether $x$ (corresponding to $u = 0$) or $x-y$ (to $u=1$) is bigger, so that the optimal policy is buy/hold when the price is negative, and sell/don't buy if the price is positive which agrees with outcomes in other answers. The total expected reward is hence
$$
  J^*_0(0,0) = (N-1)\gamma.
$$

REPLY [5 votes]: I'm not sure if I understand the model correctly as it seems to me that the answer here is very natural if you reinterpret your model in terms of expected "price increments" ("price changes"). 
The model assumes that the price level - not change - is given by N(0,1). Thus every time the price is positive the expected increment is negative. If the price level is negative the expected increment is positive. Therefore, the optimal strategy can be stated as: buy if the expected increment is positive and sell if it is negative. Of course it is exactly the conjectured optimal strategy.<|endoftext|>
TITLE: Why is Faltings' "almost purity theorem" a purity theorem?
QUESTION [36 upvotes]: My understanding of purity theorems is that they come in several flavors:
1) Those of the form "this Galois representation is pure, i.e. the eigenvalues of $Frob_p$ are algebraic numbers all of whose absolute values have size $p^{w/2}$". I don't think that this is the kind of purity I'm interested in.
2a) Purity in algebraic geometry 1: on a smooth algebraic variety the ramification locus of a morphism is a pure codimension 1 subvariety (Zariski, Nagata etc).
2b) Purity in algebraic geometry 2: absolute cohomological purity. Basically -- if $Y$ is a pure codimension $d$ subscheme of $X$ then the local cohomology groups $H^i_Y(X,\mathbf{Z}/n\mathbf{Z})$ should vanish away from $i=2d$ (under various hypotheses, e.g. $X$ locally Noetherian, $n$ invertible everywhere etc) and should be $\mathbf{Z}/n\mathbf{Z}$ etale locally if $i=2d$ (SGA5, Gabber etc).

Faltings proved an "almost purity theorem" and I think that I'm supposed to be regarding it as some sort of analog of a purity theorem above. Faltings' work occurs in the context of "almost mathematics", where one is working over the integers $R$ in a certain type of (non-discrete) valuation ring $K$, so $R$ has a maximal ideal $m$ and the idea is that instead of working in the category of $R$-modules, one works in the category of $R$-modules up to $m$-torsion (some localised category); this is the category of "almost $R$-modules". 
Faltings' almost purity theorem (or perhaps some beefed-up version due to Scholze) says something like this:
Theorem: If $K$ is a perfectoid field, $A$ is a perfectoid $K$-algebra, and $B/A$ is finite etale, then $B$ is also perfectoid and $B^o/A^o$ is almost finite etale.
Whatever does this have to do with the purity theorems mentioned at the beginning of this post?

REPLY [42 votes]: This is really just an elaboration of Emerton's comment: You should read Mark Kisins' review of Faltings's paper "Almost etale extensions".
But I wanted to elaborate: Faltings regards the almost purity theorem as an analogue of Zariski-Nagata purity. In Faltings's original setup, it was formulated as follows. Consider the rings
$$ R_m = \mathbb{Z}_p[p^{1/p^m},T_1^{\pm 1/p^m},...,T_n^{\pm 1/p^m}]
$$
Each of them is smooth over $\mathbb{Z}_p[p^{1/p^m}]$ (in particular regular), and the transition maps are finite and etale after inverting $p$. Let $S_0$ be a finite normal $R_0$-algebra, which is etale after inverting $p$, and let $S_m$ be the normalization of $S_0\otimes_{R_0} R_m$. The idea is that there is ramification of $S_0$ in the special fibre, and you want to get rid of it, by adjoining the chosen tower of highly ramified rings $R_m/R_0$. It is not hard to see that this actually works almost at the generic point of the special fibre: At the generic point, the local ring is a discrete valuation ring, and the statement is that the discriminant of the extension of discrete valuation rings becomes arbitrarily small as $m\to\infty$ (this boils down to some more or less classical ramification theory).
Now, assume that you were lucky, and for some $m$, the ramification at the generic point of the special fibre is not just very close to zero, but actually zero on the nose. Zariski-Nagata purity tells you that the ramification locus of $S_m$ over $R_m$ has to be pure of codimension $1$, but you know that there is no ramification at the codimension $1$ points (which are either in characteristic $0$, or equal to the generic point of the special fibre) -- thus, there is no ramification at all, and $S_m$ over $R_m$ is etale.
The almost purity theorem says that this result extends to the almost world: In the limit as $m\to \infty$, $S_\infty$ over $R_\infty$ is almost etale (in the technical sense defined in almost ring theory).<|endoftext|>
TITLE: How fast does Ricci flow converge on the three-sphere?
QUESTION [10 upvotes]: Suppose I have a metric $g_0$ on the $\mathbb S^3$, and let $g_t$ be the solution to Ricci flow (with surgery) with initial metric $g_0$.  What are some general results which give upper bounds on the extinction time of this flow?
Really, I want to flow for some (hopefully short) time $t$ so that all the pieces of the manifold at time $t$ are almost isometric to the standard metric on $\mathbb S^3$ (up to scaling).  I believe it must be well-known that each piece is very close to the standard $\mathbb S^3$ as it becomes extinct, and it is easy to see that the standard metric of radius $r>0$ becomes extinct in time $t\propto r^3$.  Thus hopefully it suffices to get a bound on the extinction time.
I know of the papers by Colding--Minicozzi and Perelman.  I am hoping their bounds can be improved, since the don't seem good enough for what I want to do.  Basically, I hope to avoid knowing things about the "width" of nontrivial homotopy classes in $\mathbb S^3$ (which is what both Colding--Minicozzi and Perelman use).  Is it realistic to expect these can be replaced with quantities like the volume, injectivity radius, curvature, etc. of the original metric?

REPLY [11 votes]: I'm not sure how to answer your question - I believe the Perelman or Colding-Minicozzi widths are the only known way to estimate extinction time in general. 
At the time of extinction of a component, the sphere might not be round. Consider a dumbbell rotationally-symmetric metric on $S^3$ which develops a neck singularity at finite time. As one increases the width of the neck until it becomes convex, there must be a time in between when the pinch and the extinction occur simultaneously. The singularity at this time might be like a peanut, with neck remaining until the singularity, so it does not approach a round sphere. At the tips of the peanuts, the metric should be approaching a type II singularity, which has rescaled limit a Bryant soliton. However, I think it is believed that this sort of singularity is non-generic. 
A special case in which one may estimate the extinction time is when the metric has positive scalar curvature. If the minimum scalar curvature is $R_{min}(0)$ at time $0$, then the solution must go extinct at time $3/(2R_{min}(0))$ by the maximum principle for the evolution of the scalar curvature (see e.g. Prop. 2.1). A similar estimate holds for $\lambda(g_0)$, the minimal eigenvalue of the operators $-4\Delta+R$. Since $\lambda(g_0)\geq R_{min}(0)$, this might give an extinction estimate when $R_{min}$ does not. 
During the Ricci flow-with-surgery when $\lambda(g_t) <0$, one has that the (scale-invariant) quantity $Vol(g_t)(-\frac16\lambda(g_t))^{3/2}$ is decreasing with respect to time. One also knows that $\lambda(g_t)$ must approach $0$ at some point for Ricci flow-with-surgery on $S^3$. However, I don't know of any way to show how fast this quantity approaches zero without invoking the width, since one must have some sort of topological input.<|endoftext|>
TITLE: Geometry defined by foliation.
QUESTION [9 upvotes]: In $\mathbb R^3$ there are 3 natural foliations given by the lines parallel to each axis, which intersect transversally. Let $M^n$ a manifold with $n$ foliations by lines or circles that intersect transversally, is there a way to define a metric such that this fibrations are geodesics?

REPLY [10 votes]: I'm editing my original answer to add information about the $n=2$ and  $n=3$ cases.
The first question to answer is whether such a metric exists locally, and the answer to this is yes for $n=2$, probably (see below) for $n=3$, and no for $n>3$ (in general; the exact answer will depend on the foliations you specify).
In the case $n=2$, the (local) answer is obviously yes, since, locally, any pair of transverse curve foliations is locally equivalent to the standard coordinate foliations, and such metrics obviously exist to make them geodesic. There will be many such local metrics (depending on an arbitrary function of $2$ variables) for a given pair of foliations, but there might be some interesting global obstructions.  Here is how one can see the local problem:  Locally, any pair of transverse curve foliations on a surface can be described in some coordinate system $(x,y)$ as the curves parallel to the $x$-axis and the $y$-axis, i.e., as either the curves satisfying $dy=0$ or as the curves satisfying $dx=0$.  In this coordinate system, a metric is described by the coefficients $ds^2 = E\ dx^2+2F\ dxdy + G\ dy^2$, where $E$, $G$, and $EG-F^2$ are all positive.  Set $E = e^2$ and $G = g^2$ for some positive functions $e$ and $g$.  The condition that $dy=0$ describe a family of $ds^2$-geodesics is equivalent to the condition that $d\xi = 0$ where $\xi = e\ dx +(F/e)\ dy$, while the condition that $dx=0$ describe a family of $ds^2$-geodesics is equivalent to the condition that $d\eta = 0$ where $\eta = (F/g)\ dx +g\ dy$.  Thus, one can choose $F=F(x,y)$ arbitrarily, and then the equation $d\xi=0$ becomes a (nonlinear) first-order scalar equation for $e$ (which can easily be solved by the method of characteristics), while the equation $d\eta=0$ becomes a (nonlinear) first-order scalar equation for $g$ (which can also easily be solved by the method of characteristics).  In fact, $e$ will be uniquely specified by its restriction to a single curve in the family $dy=0$, while $g$ will be uniquely specified by its restriction to a single curve in the family $dx=0$.  One just has to choose $e$ and $g$ so that the inequality $eg>|F|$ hold everywhere in the domain.  This shows the nature of the local solutions, and it may be useful in understanding possible global obstructions to existence on a surface endowed with two transverse curve foliations.
In the case $n>3$, the generic $n$-tuple of transverse curve foliations will not support a metric that all of the leaves (i.e., curves) be geodesic.  The reason is this:  To specify a curve foliation is to specify a vector field up to multiples, so this requires a choice of $n{-}1$ functions of $n$ variables.  To specify $n$ such foliations that are transverse requires a choice of $n(n{-}1)$ functions of $n$ variables.  Of course, two of these might be locally equivalent, and the group of diffeomorphisms (which depends, locally, on $n$ functions of $n$ variables) acts, so that, modulo diffeomorphisms, these structures depend on choosing $n(n{-}2)$ functions of $n$ variables.  However, metrics in dimension $n$ depend on $\tfrac12 n(n{+}1)$ functions of $n$ variables, and so, modulo diffeomorphism, they depend on $\tfrac12 n(n{-}1)$ functions of $n$ variables.  Since $n(n{-}2)>\tfrac12 n(n{-}1)$ when $n>3$, it follows that there must be an $n$-tuple of curve foliations that is not, even locally, geodesic for any metric.
There remains the (local) question for $n=3$.  The problem is formally determined, being 6 first order equations for 6 unknowns, and one would expect that this would imply that there is such a metric, at least locally in the real-analytic case.  This does indeed turn out to be the case.  A calculation shows that this first order system can be placed in Cauchy form, and therefore one can get local existence and uniqueness of such metrics for a triple of independent curve foliations that are analytic in some local coordinate system.  However, the characteristic variety (which is of degree $6$) turns out, in the generic case, to be the union of three lines and a plane cubic curve.  Generically, the cubic curve meets each of the three lines (which are in general position) in three distinct points.  Thus, the characteristic variety is singular in the generic case.  Consequently, I don't imagine that there are sufficiently strong hyperbolic existence theorems to prove the existence of solutions for smooth data.<|endoftext|>
TITLE: Arithmetic products of Cantor sets.
QUESTION [7 upvotes]: Let $A,B\subseteq \mathbb{R}$ be two Cantor sets. What is known about the arithmetic product
$AB=\lbrace ab|a\in A, b\in B\rbrace$? In particular, what is known in the case that the sets are self-similar?

REPLY [2 votes]: I) If $K$ and $K'$ are arbitrary Cantor sets with
dim$_HK=\overline{\dim}_BK$ and $HD (K)+HD(K')<1$,   then  $\dfrac{K}{ K'}$ and  $K\cdot K'$ have zero Lebesgue measure.
II) If $\tau(K)\cdot\tau(K')>1$, then  $\dfrac{K}{ K'}$ and  $K\cdot K'$ contain an interval.
III) Let  $C_\alpha$ and $C_\beta$ be two middle Cantor sets with  $\dfrac{\log \alpha}{\log \beta}=\dfrac{n_0}{m_0},~(m_0,~n_0)=1$ and $\dfrac{1}{\gamma}<\tau(C_\alpha)\cdot\tau(C_\beta)\leq1$, where  $\gamma:=\alpha^{-\dfrac{1}{n_0}}$. 
Then $\dfrac{C_\alpha}{ C_\beta}$ and $C_\alpha \cdot  C_\beta$  contain  an interval.
See the proofs in section 3 of my article on arXiv.<|endoftext|>
TITLE: Realization problem for Betti numbers
QUESTION [15 upvotes]: In Analysis Situs, Poincaré studies the following question: 
which sequences of integers $b_0,\ldots,b_n$ are the Betti numbers of an orientable compact manifold of dimension $n$?. 
He knows that necessary conditions are $b_k=b_{n-k}$ and if $n=4k+2$, $b_{2k+1}$ is even. 
Then he computes the homology of a product of spheres, reducing the problem to finding a manifold of dimension $4k$ with $b_0=b_{2k}=b_{4k}=1$ and the other are 0. He proposes the symmetric product of two spheres $S^{2k}$, missing that they are singular for $k>1$.
For $n=1$ and $2$, the projective plane and the quaternionic plane answer the question and I was very surprised to learn that the Hirzebruch signature theorem implies that a manifold of dimension 12 with $b_4=0$ has signature divisible by 62 (see page 8 of this paper). Hence, the smallest odd value of $b_6$ of a 12-dimensional manifold with vanishing other Betti numbers is at least 63. 
Here is the question: do we know the smallest odd value of $b_6$? More generally, do we know other obstructions for realizing an arbitrary sequence of Betti numbers?

REPLY [7 votes]: Suppose we are given non-negative integers $b_0, b_1, \dots, b_n$ with $b_k = b_{n-k}$. Is there a closed orientable manifold $M$ with $b_i(M) = b_i$? First we need $b_0 = b_n \geq 1$. It is enough to answer the question with $b_0 = b_n = 1$ as we can then just take the disjoint union with the appropriate number of spheres. So from now on, $b_0 = b_n = 1$.
If $n = 2m + 1$, then there is a closed smooth orientable manifold $M$ with $b_i(M) = b_i$, namely 
$$M = b_1(S^1\times S^{n-1})\ \#\ \dots\ \#\  b_m(S^m\times S^{m+1}).$$
If $n = 2m$ and $b_m$ is even, then there is a closed smooth orientable manifold $M$ with $b_i(M) = b_i$, namely 
$$M = b_1(S^1\times S^{n-1})\ \#\ \dots\ \#\ b_{m-1}(S^{m-1}\times S^{m+1})\ \#\ \tfrac{b_m}{2}(S^m\times S^m).$$
Suppose now that $n = 2m$ and $b_m$ is odd. If $m$ is odd, then there is no manifold realising these Betti numbers: the intersection form on the middle dimension is a non-degenerate skew-symmetric form, so the middle Betti number is necessarily even. If $m$ is even, things are more complicated. First of all, such examples can exist, e.g. $\mathbb{CP}^2$. However, the above approach of taking connected sums of products of spheres can't work because the resulting manifolds are all nullcobordant, but a manifold with such Betti numbers is not: $b_m(M)$ and $\sigma(M)$ have the same parity and $\sigma(M)$ is a cobordism invariant.
Note that a closed smooth orientable $2m$-dimensional manifold with Betti numbers $b_0 = b_m = b_n = 1$ and all others zero has rational cohomology ring $\mathbb{Q}[\alpha]/(\alpha^3)$ where $\deg\alpha = m$. For this reason, call such a manifold a rational projective plane and denote it by $\mathbb{QP}^2$. If such a manifold exists in dimension $n$, then if $b_m$ is odd, there is a closed smooth orientable manifold $M$ with $b_i(M) = b_i$, namely
$$M = b_1(S^1\times S^{n-1})\ \#\ \dots\ \#\ b_{m-1}(S^{m-1}\times S^{m+1})\ \#\ \tfrac{b_m-1}{2}(S^m\times S^m)\ \# \mathbb{QP}^2.$$
So the natural question to ask is: 

For which $n$ does there exist a rational projective plane? 

Well, we have $\mathbb{CP}^2$, $\mathbb{HP}^2$ and $\mathbb{OP}^2$ which are actually simply connected and have integral cohomology ring $\mathbb{Z}[\alpha]/(\alpha^3)$; by the solution of the Hopf invariant one problem, these are the only such manifolds up to homotopy equivalence. 
In Smooth manifolds with prescribed rational cohomology ring, Fowler and Su show (Theorem A) that for $n \geq 8$, a rational projective plane can only exist in dimensions of the form $n = 8(2^a + 2^b)$ for some non-negative integers $a$ and $b$.1 In Rational analogs of projective planes, Su shows (Theorem 1.1) the existence of a (simply-connected) rational projective plane in dimension $32$. Later, Kennard and Su also proved the existence of (simply-connected) rational projective planes in dimensions $128$ and $256$, see On dimensions supporting a rational projective plane, Theorem 1.1. The paper also contains non-existence results for simply-connected rational projective planes in certain dimensions.2
It may seem that the lack of a projective plane in a given dimension prevents the realisability of any collection of Betti numbers with middle Betti number odd, but that is not true. For example, when $m$ is even, $\mathbb{CP}^m$ has $b_m = 1$ regardless of whether or not a rational projective plane exists (thanks to Will Sawin for pointing this out). That is, the question of realisability of Betti numbers does not reduce to the existence of rational projective planes, it is much more complicated.
Finally, it is worth noting that if $n = 2m = 4l$ and $b_{4j} \geq 1$ for $j = 1, \dots, l - 1$ and $b_m$ is odd, then there is a closed smooth orientable manifold $M$ with $b_i(M) = b_i$, namely
$$M = b_1'(S^1\times S^{n-1})\ \#\ \dots \# \ b_{m-1}'(S^{m-1}\times S^{m+1})\ \#\ \tfrac{b_m'}{2}(S^m\times S^m)\ \#\ \mathbb{HP}^l$$
where $b_{4j}' = b_{4j} - 1$ for $j = 1, \dots, l - 1$ and $b_i' = b_i$ otherwise.

1 The definition of rational projective plane in that paper includes the hypothesis of being simply connected. However, the proof of Theorem A does not use this property, so the result holds for the definition used above.
2 I don't know the proof of these results, so I'm not sure if the simply connected hypothesis can be removed.<|endoftext|>
TITLE: Is the homeomorphism class of a connected open set of C determined by its fundamental group? 
QUESTION [6 upvotes]: Let $U,U'\subseteq\mathbf{C}$ be two connected open sets such that $\pi_1(U)\simeq\pi_1(U')$.
Q: Does this imply that $U$ is homeomorphic to $U'$?
In the case where the $\pi_1$'s are trivial then the answer is yes, this is a consequence of 
Riemann's mapping theorem. May be one should try to prove it when the $\pi_1$'s are free
groups on $n$ generators...

REPLY [10 votes]: The answer is indeed no as David Cohen has pointed out, and more generally the answer is determined by the complements of the sets U and U'.
The complete solution is effectively due to R.L. Moore (1925), the key fact being every nondegenerate monotone upper semicontinuous decomposition of the 2-sphere yields a 2-sphere.
Thus to decide if two open connected planar sets U and U' are homeomorphic, let C and C' denote the respective complements of U and U' in the extended plane. 
Now take the respective topological quotients K and K' of C and C', collapsing each component of C or C' to a point. Then U and U' are homeomorphic iff K and K' are homeomorphic.<|endoftext|>
TITLE: How does  Yitang Zhang use Cauchy's inequality and Theorem 2 to obtain the  error term coming from the $S_2$ sum
QUESTION [22 upvotes]: I have been reading Yitang Zhang's paper now for one and a half weeks and also volunteered to give a popular talk on the paper next week at Stockholm University. 
Today I  found a detail in the proof that seems outright wrong, and I am starting to worry that it is a serious problem (in particular since I promised to give a talk about the result next week).
Can anyone explain the last three lines on page 22, or if there indeed is a mistake on those lines give an alternative argument for obtaining the error term that we want to have?   Zhang has the following sum
$$ \mathcal E_i= \sum_{ d < D^2, d |  \mathcal P } \tau_3(d) \rho_2(d) \sum_{ c \in \mathcal C_i ( d ) } | \Delta( \theta,d,c) | .   $$
Then he says: By Cauchy's inequality and Theorem 2 we have 
$$ \mathcal E_i \ll x \mathcal L^{-A}. $$
Here  $\mathcal L$ denotes  $\log x$.  I do not see this. The reason is that Theorem 2 which is given in the following way: For $ 1 \leq i \leq k_0 $ we have
$$ \sum_{ d < D^2 , d | \mathcal P }  \sum_{ c \in \mathcal C_i(d) } | \Delta (\theta,d,c)|\ll x \mathcal L^{-A}, $$
is given in $L^1$-norm, Cauchy's inequality would need something in $L^2$-norm. The natural inequality to use would be  $$  \| f g \|_1  \leq \| f \| _\infty \| g \| _1, $$ where the first function would be the divisor function $\tau_3(d) \rho_2(d)$ and the second would be the sum in $c$. Thus it seems that Theorem 2 in its current form should not really give anything better than
$$  \mathcal  E_i  \ll \left( \max_{ d < D^2, d | \mathcal P} \tau_3(d) \rho_2(d) \right)   \sum_{ d < D^2, d | \mathcal P } \sum_{ c \in \mathcal C_i(d) } | \Delta( \theta,d,c ) | \ll  \left( \max_{ d < D^2, d | \mathcal P } \tau_3(d) \rho_2(d) \right) x \mathcal L^{-B}  $$
for any $B>0$. Here $ D^2 $ is a little more than $\sqrt x$ (to be precise $ x^{ 1 / 2  + 1 / 584 } $), and $\mathcal P$ is the product of all primes up to a small power of $x$.
Now the sup norm for the divisor function $\tau_3(n)$ certainly grows faster than any power of $\log x$ (even on square free numbers, e.g on primorials), even if it on the average grows like a power of $\log x$). The function $\rho_2(d)$ is defined on page 7 is multiplicative, has support on square free numbers and defined to be $v_p-1$ on the primes, where $v_p$ are the number of residue classes of $\mathcal H$ mod p. With the exception of finitely many $p$ this will be $k_0-1 =3.5\cdot 10^6-1$, i.e really large and contribute much more than the divisor function. Thus this should not get what we need, i.e. a bound of the form $x \mathcal L^{-B}$.
If my concerns are correct, I guess that you can start looking closely at the proof of theorem 2 and see if the same proof holds with the divisor functions thrown in.
If I have made som simple error in the above reasoning  I would appreciate your help to understand it, since I would like to understand at least all details on how Theorem 1 implies Theorem 2 before my talk next week (Of course as much as possible of the proof of Theorem 2 also).
Reference: Yitang Zhang: Bounded gaps between primes
           http://annals.math.princeton.edu/articles/7954

REPLY [3 votes]: The incorporation of the factor of divisor function, $\tau_?(d)$, in the Bombieri-Vinogradov Theorem is familiar feature in sieve application. For example, this occurs in GPY and Graham's paper "Small gaps between product of two primes" page 748.<|endoftext|>
TITLE: Applications of Chevalley Restriction Theorem
QUESTION [10 upvotes]: Let $G$ be a simple linear algebraic group (over $\mathbb{C}$, say) and $\mathfrak{g}$ be its Lie algebra, $\mathfrak{t}\subset \mathfrak{g}$ the Lie algebra of a maximal torus in $G$ and $W$ the corresponding Weyl group. The Chevalley Restriction Theorem (CRT) states that there is an isomorphism
$$ \mathbb{C}[\mathfrak{g}]^{G}\simeq \mathbb{C}[\mathfrak{t}]^{W}$$
obtained by restriction. This is the Lie-theoretic statement: $G$-invariant functions on $\mathfrak{g}$ are trace functions (ie, sums of 'eigenvalues'). Admittedly, this is a result that I feel I understand quite well - and that intuitively makes sense - but I don't know too many applications of it.
Question: what are some (nontrivial) applications of the above isomorphism? (ie, what cool things can be proved using the CRT?)
For example, CRT can be used to show that the defining ideal of the nilpotent cone $N\subset \mathfrak{g}$ is $\mathbb{C}[g]^{G}_{+}$, the ideal generated by the $G$-invariant functions without constant term (this is in Chriss & Ginzburg, Ch. 3; although there are proofs of this fact bypassing CRT), and this can be used to show that the algebra defining the intersection $N\times_{\mathfrak{g}} \mathfrak{t}$ is the coinvariant algebra $\mathbb{C}[\mathfrak{t}]/\mathbb{C}[\mathfrak{t}]^{W}_{+}$. 
EDIT 1: As mentioned in Konstantin Ardakov's answer, the CRT can be applied to prove the Harish-Chandra description of the centre of the universal enveloping algebra. I was preferably looking for straight applications of the CRT as opposed to generalisations, but I welcome all comments/suggestions that people may have.

REPLY [7 votes]: Let $G$ be a split reductive group over a finite field $k$ (of sufficiently large characteristic), $X$ a smooth geometrically connected projective curve over $k$ and $\mathcal{M}$ the algebraic stack that classifies Hitchin pairs on $X$ (see reference below, but these are pairs $(E,\varphi)$ where $E$ is a $G$-bundle on $X$ and $\varphi$ is a section of a certain bundle derived from $E$). 
The Chevalley restriction map for $G$ suitably twisted over $X$ is essentially the the Hitchin fibration $f:\mathcal{M}\to A$ where $A$ is an affine space; for nice enough $a\in A$ the fibers $f^{-1}(a)$ can be written as a nice sum of orbital integrals related to the ones appearing in various trace formulae for groups over number fields, and this is an essential tool in Ngô's proof of the fundamental lemma.
Here, twisted over $X$ means twisted by a torsor that gives the data for a group scheme on $X$ locally isomorphic to $X\times_k G$ in the étale topology.
See for instance Ngô's paper "Fibration de Hitchin et Endoscopie".<|endoftext|>
TITLE: Algebraic K-theory and Homotopy Sheaves
QUESTION [18 upvotes]: Recently, when I was reading the definition of higher algebraic K-theory, I tried to give myself some motivation by looking at derived algebraic geometry. The constructions for algebraic K-theory provide us with a $\mathrm{K}$-theory space. Certain constructions also give us explicit deloopings, and if the ring is commutative we even get an $E_\infty$ structure on the resulting spectrum. Now, all of a sudden, we have a sheaf of $E_\infty$-rings over the Zariski site of a ring $R$: namely, for the affine open $\mathrm{Spec}(R_f)$ of $\mathrm{Spec}(R)$, we assign the $\mathrm{K}$-theory spectrum $\mathcal{K}(R_f)$. We can, of course, do this over the etale site as well.
My questions are:

Does this assignment satisfy Zariski, etale descent? 
If not, what type of modifications can we do for it to satisfy the above-mentioned descents?

REPLY [24 votes]: The question already has good answers but I think there is still more to be said.
References
As already mentioned, algebraic K-theory satisfies Zariski descent.  For regular noetherian schemes this is due to

Kenneth S. Brown, Stephen M. Gersten, Algebraic K-theory as generalized sheaf cohomology, Higher K-Theories, Lecture Notes in Mathematics Volume 341, 1973, pp 266-292.

It was generalized to finite dimensional noetherian schemes by

R. Thomason, Higher algebraic K-theory of schemes and of derived categories, The Grothendieck Festschrift, vol. III, Progress in Mathematics, vol. 88, Birkhäuser, Basel, 1990, pp. 247–435.

However, there is a stronger statement: it satisfies descent with respect to the Nisnevich topology (which lies between Zariski and etale).  This is due to

Yevsey A. Nisnevich, The completely decomposed topology on schemes and associated descent spectral sequences in algebraic K-theory, Algebraic K-theory: connections with geometry and topology, 1989, pp 241-341.

and was generalized to finite dimensional noetherian schemes in the same paper of Thomason.
In the following paper the above results are extended to finite dimensional quasi-compact quasi-separated schemes.

Andreas Rosenschon, P.A. Ostvær, Descent for K-theories, Journal of Pure and Applied Algebra 206, 2006, pp 141–152.

Sketch of a proof
I learned from Peter Scholze that in modern language a proof can be given, after identifying Quillen K-theory with the K-theory of the stable infinity-category of perfect complexes, by using the following characterization of extendibility of perfect complexes: a perfect complex on an open subscheme $U \subset X$ can be extended up to quasi-isomorphism to a perfect complex on $X$, if and only if its class in $K_0(U)$ lies in the image of $K_0(X)$.  This kind of thing is discussed in

Bhargav Bhatt, Algebraization and Tannaka duality, 2014, arXiv.

This gives a homotopy fibre sequence of connective spectra $K(X \text{ on } Z) \to K(X) \to K(U)$ where $Z \subset X$ is the closed complement of an open subscheme $U \subset X$.  I am not sure in exactly what generality this proof works.  The significance of working with infinity-categories is that the presheaf of infinity-categories $U \mapsto \mathrm{Perf}(U)$ satisfies descent, unlike its triangulated shadow.
Etale descent
Finally, let me mention that the question of etale descent is closely related to the Lichtenbaum-Quillen conjecture.  This is now a theorem of Rost and Voevodsky and it implies that K-theory does satisfy etale descent in sufficient large degrees.  The theorem of Trobaugh that Steven Landsburg mentioned, about etale descent for mod-$\ell$ Bousfield-localized K-theory, is in

R. Thomason, Algebraic K-theory and étale cohomology, Ann. Sci. Ecole Norm. Sup. 18 (4), 1985, pp. 437–552.

and is also generalized in the paper of Rosenschon and Ostvær.
[Edit]
Sketch of a proof, part 2
Let me give more details on the proof, now that I understand the details a little better.  The main ingredients are

Nisnevich descent for perfect complexes, as a prestack of stable infinity-categories $X \mapsto Perf(X)$.
Compact generation of $D(X)$ by the perfect complexes, and also for the version "with support" on a closed subscheme $Z \subset X$, i.e. for the full subcategory $D_Z(X)$ of complexes that vanish on $X-Z$.
The Thomason-Neeman localization theorem: an exact sequence of stable infinity-categories induces a fibre sequence of K-theory spectra.

Then the proof is as follows.  Let $j : U \hookrightarrow X$ be an open immersion and $p : Y \to X$ an etale morphism defining an elementary Nisnevich square
$$\require{AMScd}
\begin{CD}
W @>>> Y \\
@VVV @VV{p}V \\
U @>>{j}> X
\end{CD}$$
We want to show that the square
$$\require{AMScd}
\begin{CD}
K(X) @>{j^*}>> K(U) \\
@V{p^*}VV @VVV \\
K(Y) @>>> K(W)
\end{CD}$$
is a homotopy cartesian square of connective spectra.  (In the noetherian case, Morel-Voevodsky showed that Nisnevich descent is equivalent to this Brown-Gersten-style excision property; in the non-noetherian case, this should really be taken as the "correct" definition of the Nisnevich topology.)
For this it is sufficient to show that there is an equivalence on homotopy fibres.
2) and 3) imply that the homotopy fibres are given by $K_{X-U}(X)$ and $K_{Y-W}(Y)$, respectively.
Then the equivalence follows from the fact that there is already an equivalence
$$ Perf_{X-U}(X) \stackrel{\sim}{\longrightarrow} Perf_{Y-W}(Y) $$
at the level of perfect complexes, by 1).
This proof works for quasi-compact quasi-separated schemes.  By the way, the same proof works for (qcqs) derived schemes.  By far the most non-trivial part of the proof is 2), which was established in [B. Toen, Derived Azumaya algebras and generators for twisted derived categories, arXiv:1002.2599].  A proof of 1), attributed to Drinfeld, is given in [D. Gaitsgory, Notes on geometric Langlands: Quasi-coherent sheaves on stacks, pdf].  In fact, this probably even gives a proof for the spectral schemes of Lurie, modulo a key point in the proof of 2) which I do not know how to do in the setting of $E_\infty$-ring spectra (but this is probably just my ignorance).<|endoftext|>
TITLE: Counting conjugacy classes in simple groups of Lie type
QUESTION [11 upvotes]: Finite groups of Lie type include those obtained as rational points of a connected simple algebrraic group over a finite field $k = \mathbb{F}_q$ of characteristic $p$: these are split or quasi-split.   There are also several families obtained less directly from algebraic groups: the Suzuki groups $^2\!B_2(q)$ and the Ree groups $^2\!G_2(q),\: ^2\!F_4(q)$, where $q = p^{2n+1}$ with $n \geq 1$ is an odd power of (respectively) $2,3,2$.  [Caution: Sometimes $q$ is written here as $q^2$ to emphasize the similarity of group orders to those of untwisted groups.]
Counting the total number of conjugacy classes in such a group is a natural problem, relative to the determination of ordinary characters.  Case-by-case study has been done for many of the families, especially the exceptional types; but it's unclear how much can be expressed uniformly.   (Older results are surveyed in Chapter 8 of my 1995 AMS book on conjugacy classes.)    It helps to assume the ambient algebraic group is simply connected, in which case Steinberg proved for split and quasi-split types that the number of semisimple classes (whose elements have order prime to $p$) is $p^r$ with $r$ the Lie rank.   For Suzuki or Ree groups, $r$ is the rank of the BN-pair: 1, 1, 2.
For Suzuki groups the total number of classes is $q + 3$ (Deriziotis), while for Ree groups of type $G_2$ the number is $q+8$ (Ward).  The latter groups came up recently here.

How many conjugacy classes do the Ree groups of type $F_4$ have?

This should be computable from known data, though not easily.  Judging from all special cases I'm aware of, the answer should be given by a polynomial $q^2 + aq + b$, where $a, b \in \mathbb{Z}$ are independent of $q$.  [Note however that for most other families of groups of Lie type, there are extra complications related to isogeny type, bad primes, and such.]
The underlying question, of course, is whether one can predict a priori what the polynomials will look like for all groups (coming from simply connected algebraic groups) starting with the highest degree term $q^r$.

REPLY [8 votes]: According to

Fleischmann, Peter; Janiszczak, Ingo.
  On the computation of conjugacy classes of Chevalley groups.
  Appl. Alg. in Eng., Comm. and Comp. 1996, 7(3), 221--234

the class number of ${\rm F}_4(q)$ is

$q^4 + 2q^3 + 6q^2 + 10q + 19$ if $q = 2^n$,
$q^4 + 2q^3 + 7q^2 + 15q + 30$ if $q = 3^n$, and
$q^4 + 2q^3 + 7q^2 + 15q + 31$ if $q = p^n$ where $p > 3$

(see page 233).
According to Frank Lübeck's database on finite groups of Lie type,
the class number of $^2{\rm F}_4(q^2)$ is $q^4+4q^2+17$.
In that database you also find class numbers for many other types of finite
groups of Lie type.<|endoftext|>
TITLE: Measures of entangledness of an open curve
QUESTION [8 upvotes]: Let $\gamma$ be a simple (non-self-intersecting) open curve in $\mathbb{R}^3$.
I am seeking a measure of its degree of "entangledness," some measure that accords
with the intuition one senses with a tangled fishing line.
One measure is to connect the two ends of $\gamma$ and use a measure of its degree of
knottedness, e.g., its unknotting number, or, perhaps, its writhing number.
But it would seem these depend on how the ends are connected, rather than on $\gamma$ alone.
Have other natural measures been proposed? I'd appreciate pointers. Thanks!
       
A year later, I remain interested, especially in some type of energy
measure along the lines attempted by Qfwfq. It would be especially pleasing
to have a measure that somehow measures the effort it would take to straighten
a tangle.

REPLY [3 votes]: Perhaps, one could define the "entangledness" of an open curve (of unit lenghth and parametrized in arclength) as the minimum 
$$\mathrm{min}_{H}\;E(H), $$
over all possible smooth simple "strightening" homotopies $H:[0,1]\times [0,1]\to\mathbb{R}^3$ with $H(0,s)=\gamma(s)$ and $H(1,s)=s\cdot \gamma\;'(0)+\gamma(0)$ $\forall s\in[0,1]$, of 
the total work (energy) $E(H)$ needed to strighten up the curve, supposing the curve has an evenly distributed unit mass:
$$E(H)=\frac{1}{2}\int_0^1\int_0^1|\frac{\partial H}{\partial t}(t,s)|^2\;dt\;ds$$
I'm not aware if this idea has been made effective somewhere in the literature...<|endoftext|>
TITLE: When is a 0-1 matrix a one-intersection incidence matrix?
QUESTION [7 upvotes]: The following problem is what motivated my previous MO question.
It is easily seen that for any given 0-1 matrix $M$, one can always find
a set $\mathcal P$ of points, and a set $\mathcal C$ of simple curves in
the plane, so that their incidence matrix is exactly the matrix $M$.
Suppose, however, that any pair of curves from $\mathcal C$ is now
allowed to intersect in at most one point (be it a point of $\mathcal P$
or any other point), and let's say that the matrix $M$ is
realizable if such $\mathcal P$ and $\mathcal C$ can be found.
Clearly, a necessary condition for this is that the scalar product of any
two rows of $M$ be at most $1$, but this condition is insufficient: say,
for $q$ large enough, by the Szemeredi-Trotter theorem, the point-line
incidence matrix of the finite projective plane $PG(2,q)$ has two many
incidences to be realizable. What are other reasonable necessary /
sufficient conditions for $M$ to be realizable? What are "small" examples
of non-realizable 0-1 matrices?

Added June 03, 2013
Here is a very specific question along these lines. Everything I presently can say on the problem above is essentially symmetric in $\mathcal P$ and $\mathcal C$. This suggests that, perhaps, there is some duality between points and curves involved. Accordingly, I wonder whether, by any chance, it can be true that $M$ is realizable if and only if its transpose $M^t$ is realizable?

REPLY [4 votes]: (I suspect a lot of this is known to Seva, but it's probably helpful to have it written down.) 
One lower-bounding argument: Anything that's representable by lines and points is representable by curves and points. When checking if this is so, we can delete any row with just two $1$s, as you can draw a line between any two points. (this generalizes Seva's point in the comments.) Similarly, there is no need for lines to be parallel, so we can delete a column with just two $1$s. Thus we can assume that every row and column has at least three $1$s, which gives a minimum of $7$ vertices: One vertex must be on three edges, each of which contains two other vertices. This is attained uniquely by the Fano plane.
To get an upper-bounding argument, we just need an effective Szemeredi-Trotter theorem. We can do this just by looking at the proof, e.g. on Wikipedia. We construct a graph whose number of edges $e$ is just the number of $1$ entries of the matrix, minus the number of rows. Using the explicit bound $e \leq 4 n$ or $e\leq \left(64 m^2/n^2\right)^{1/3}$. This gives an explicit inequality.
For the projective plane over $\mathbb F_q$, we have $n= q^2+q+1$, $m=q^2+q+1$, $e = q (q^2+q+1)$. Thus it can only be representible for $q \leq 4$. So the projective plane over $\mathbb F_5$ is an example. I guess this isn't very reasonably small.
I think a simple Euler characteristic bound might work better. If there are $P$ points and $C$ curves, $I$ incidences, and $c$ bonus crossings where two curves cross away from a points, we get a graph in the plane with $P+c$ vertices and $I-C +2c$ edges, so it has $2+I+c-P-C$ faces. With the inequality $3F \leq 2E$, since two curves cannot intersect in more than two points, so all faces are triangles are larger, we get the inequality:
\[ 6+ 3I + 3c -3P - 3C \leq 2I - 2C+4c\] 
\[ I \leq 3P + C + c-6\]
For the projective plane, $c=0$ because each pair of edges already intersect at a point. $I=(q+1)(q^2+q+1)$, $P=q^2+q+1$, $C=q^2+q+1$, so we get
\[(q+1)(q^2+q+1) \leq 4 (q^2+q+1) - 6\]
which improves it to the projective plane over $\mathbb F_3$.
The main explanation for the gap between the lower bounding technique and the lower bounding technique is that the first one always makes curves going off to infinity. If the curves went off to infinity, the Fano plane would no longer be possible.<|endoftext|>
TITLE: are these functors exact?
QUESTION [5 upvotes]: Let $X$ be a smooth projective algebraic variety over a field of characteristic zero and let $D$ be a simple normal crossing divisor on $D$. Put 
$j: U \hookrightarrow X$
for the inclusion of the complement $U=X-D$ on $X$. 
Are the functors $j_\ast$ and $j_!$ exact?
I think the answer is yes. 
I would be very grateful if someone could provide me with a proof.

REPLY [5 votes]: I assume you're asking about quasicoherent sheaves.
Since the inclusion of $U$ into $X$ is an open immersion, $j_!$ is exact.  You can find this in Tag 03DJ.
Since $j$ is an affine morphism, $j_*$ is exact (EGA 2 Corollary 5.2.2).  $j$ is affine, because the affine property of morphisms is étale local on the target, and the inclusion of the complement of a principal closed subscheme (such as the locus defined by $x_1 x_2 \cdots x_r = 0$ in affine space) is affine.  See e.g., Tag 07ZT<|endoftext|>
TITLE: Goldbach's conjecture and Euler's idoneal numbers
QUESTION [5 upvotes]: Recently, I stumbled upon an interesting statement regarding Quadratic Forms. It is quite well-known and, as I will describe briefly, equivalent to Goldbach's conjecture.
Let $p,q$ be odd primes and consider $Q(x,y) = x^2 - y^2$. A straightforward argument shows that there are only two non-negative integer solutions to $Q(x,y) = pq$. Namely, when $q \leq p$,
$\displaystyle (x,y) = \left(\frac{p+q}{2},\frac{p-q}{2}\right)$ and $\displaystyle (x,y) = \left(\frac{pq+1}{2},\frac{pq-1}{2}\right)$.
Now, suppose we go in the other direction. Given an integer $x = n \geq 2$, can we always find some integer $0 \leq y \leq n - 2$ and primes $p,q$ such that $Q(n,y) = pq$? Clearly, if this was the case, then $n^2 - y^2 = pq$ and it follows that $n = p + y$ and $n = q - y$. Hence, $2n = p + q$.
At this point, I would like to point out that I understand that this is no easier to resolve. However, it let me think that there might be an even deeper connection (which may or may not be true) between Quadratic Forms and Goldbach's conjecture. 
After some thought, I began looking at Euler's Idoneal numbers. These are positive integers, $D$, such that if an integer is representable as $x^2 \pm Dy^2$ in only one way with $x^2$ coprime to $Dy^2$, then it is guaranteed to be a prime, a power of a prime or twice one of these. Obviously, $D = 1$ is somewhat related to what I said above, but this time the focus is on the possibility of taking prime values.
Next, I thought to consider the quadratic form
$Q(a,b,c,d) = a^2 + D_1 b^2 + c^2 + D_2 d^2$
and integral representations for even integers. Trivially, every integer is represented by this in the case $D_1 = D_2 = 1$, according to Lagrange, but this is of no help. What I wondered is whether I could choose $D_1$ and $D_2$, idoneal numbers, so that there was only one positive integer solution to
$Q(a,b,c,d) = n$
which would imply that $a^2 + D_1 b^2$ and $c^2 + D_2 d^2$ are representable in only one way and, hence, (possibly) prime numbers. Of course, I mean one way up to $Q(a,b,c,d) = Q(c,b,a,d)$ (and possibly more cases depending on $D_1 = D_2$) and there is no guarantee that $(a^2, D_1 b^2) = (c^2, D_2 d^2) = 1$ must be the case.
In some cases, this is quite easy. For example,
$54 = a^2 + 12b^2 + c^2 + 40d^2$
has the only solution $(a,b,c,d) = (1, 1, 1, 1)$ in positive integers. It is easily seen that this example is useless in general: the coefficient $40$ is quite large compared to $54$ together with the fact that $6, 14, 42$ and $54$ are not sums of two squares; this forces no other solutions. I don't know whether there can be many numbers for which the above works.
I realise that this approach is very weak. In fact, I read a few pages off notes by Pete Clark which suggested that the number of solutions to an equation $q(x_1, x_2, ..., x_n) = N$, for a positive-definite integral quadratic form $q$, is asymptotically $N^{n/2-1}$. I interpret this as saying that, in general, the above approach fails: there is only a certain number of those quadratic forms, since there are only 65 idoneal numbers known.
My questions are the following.

Is it possible that Euler's idoneal numbers, together with some more results (say Gauss' three-square theorem), could imply Goldbach's conjecture for many (some class of) numbers?



Is there a criterion, or something of that nature, which allows one to identify integers that are (essentially) uniquely represented by a given quadratic form?


I will appreciate any references on any related topics. Thank you!

I apologise if something in my post doesn't make sense. I am yet to begin my undergraduate degree in Maths next academic year, but I plan to complete an essay project on a similar topic this summer.

REPLY [4 votes]: There should be only finitely many such $n$. The reason is that there should be too many pairs of primes $(p_i,q_i)$ such that $p_i=a_i^2+b_i^2 D_1$ and $q_i=c_i ^2+d_i ^2 D_2$, and $n=p_i+q_i$ and if we have just two different such representations then we do not get the uniqueness that you want.
If a prime $ p= a ^ 2 + b ^ 2 D $ (where $D$ is such an idoneal number (Cox calls these numbers Eulers convenient numbers, See  Cox, Primes of the form $p=x^2+n y^2$, Corollary 2.27 and page 59-60, which I believe might be one of the best references for these questions)), then in particular any prime $q$ such that $q \equiv p \pmod {4D}$ should also be representable in the same form, i.e. $ q = \alpha ^ 2 + \beta^2  D $.
Then my assertion follows from  a variant of Goldbach conjecture for arithmetic progressions, such that given $a \pmod m$ and $b \pmod m$ , $(m,a)=(m,b)=1$, then there should exists an $n_0$ such that for each even integer $n \geq n_0$ where $n \equiv a+b \pmod m$  there should exist at least two distinct representations $ n = p _ 1 + q _ 1 =   p _2  + q_2 $  where $p_i,q_i$ are prime numbers, $ p_i \equiv a \pmod m $ and $ q_i \equiv b \pmod m $. Although not proven since it is more general than Goldbach's conjecture, it should hold (and also we get an asymptotic for the expected number of representations) by probabalistic/heuristic reasoning (Cramer model...)).<|endoftext|>
TITLE: Tight vs. overtwisted contact structure
QUESTION [7 upvotes]: I know the familiar differences between tight and overtwisted contact structures. For example,  each homotopy class of plane-bundles on a three-manifold has an overtwisted representative but tight contact structures are less common. Also a tight contact structure gives a genus bound on an embedded closed surface and therefore can tell things about the topology of the ambient manifold. My question is what were the motivations behind distinguishing between the two originally? Is there possibly an intuition which makes the two fundamentally different and then proving the above differences comes as a consequence or it was these differences which led to distinguishing between the two? In other words I am trying to see the difference between the two without appealing to the statements like the above.

REPLY [4 votes]: As Douglas Zare said above, the discovery of the dichotomy between tight and overtwisted contact structures is certainly inspired by the existence of taut and non-taut foliations in dimension three (or probably more correctly between Reebless foliations and foliations with a Reeb component).
It is not very hard to imagine that if you knew the turbulization procedure for foliations (that is, if you have a loop that is transverse to the foliation, then you can modify the foliation in a neighborhood of this loop to create a Reeb component), then you would think that the construction is quite similar to a Lutz twist (creating a loop-full of overtwisted disks close to a transverse loop in a contact manifold).
The properties of overtwisted contact structures and foliations with a Reeb component are relatively similar, for example, every 3-manifold admits them, but it's not true that every 3-manifold admits a tight contact structure or a Reebless foliation, by performing a Lutz twist or a turbulisation you can modify a contact structure into one that is overtwisted and a foliation into one that has a Reeb component, but it is usually very hard two undo these constructions.  As you said in your post the behavior of submanifolds is much more restricted if the contact structure is not overtwisted/if the foliation is Reebless, etc.
The main difference between the theory of foliations and contact structures is probably that in the latter case you have Gray stability which tells you that contact structures are stable under continuous deformations.
Nonetheless note that by a recent result
Eynard-Bontemps, Hélène, On the connectedness of the space of codimension one foliations on a closed 3-manifold, Invent. Math. 204, No. 2, 605-670 (2016). ZBL1345.57033.
you can always find a path through foliations to connect any two foliations that are homotopic as plane fields.
By contrast the paper
Bowden, Jonathan, Contact structures, deformations and taut foliations, Geom. Topol. 20, No. 2, 697-746 (2016). ZBL1338.53053.
shows that if you suppose that the two foliations are taut, it might be necessary to move through non-taut foliations to find the connecting path.
Of course it is true that the last two results I cite are very recent, but comparing them to
Eliashberg, Y., Classification of overtwisted contact structures on 3-manifolds, Invent. Math. 98, No.3, 623-637 (1989). ZBL0684.57012.
should show that there really are many formal links between both theories.<|endoftext|>
TITLE: Construction of the spectral sequence of Katz/Oda
QUESTION [5 upvotes]: In their famous paper "On the differentiation of De Rham cohomology classes..." Katz and Oda construct the spectral sequence for de Rham cohomology for the situation of a smooth morphism
$\pi: X \rightarrow S$
of smooth $k$-schemes ($k$ a field), where $S$ is assumed affine.
There is a step which is not clear to me: in Lemma 8 of this paper they say that the equality
$(*) \qquad \mathbb R^0\Gamma_X =\Gamma_S\circ \mathbb R^0\pi_*$
yields a spectral sequence of composition
$(**) \qquad E_2^{a,b}=R^a \Gamma_S\circ \mathbb R^b\pi_* \Rightarrow \mathbb R^{a+b}\Gamma_X$.
They seem to consider $(*)$ as an equality of functors on the category of complexes of abelian sheaves on $X$ - Arguments involving quasi-coherence and affineness of $S$ only appear in the next step.
My question is a very simple one which I nevertheless can't figure out:
Why do they have $(**)$?
It is a spectral sequence of composition, hence only exists when one knows that $\mathbb R^0 \pi_* $ sends injective objects in the category of complexes of sheaves on $X$ to $\Gamma_S$-acyclic ones. I don't see why this is true: imagine a two-term complex $I^0\rightarrow I^1$ of injective abelian sheaves on $X$, then $\mathbb R^0 \pi_*$ of it is $ker(\pi_*I^0 \rightarrow \pi_*I^1)$. This has no reason to be acyclic for $\Gamma_S$, hasn't it?
A similar problem arises a bit later in Lemma 9 when they consider the equality
$(+) \qquad \mathbb R^0\Gamma_S = H^0\circ \Gamma_S$
which I see as equality of functors on the category of complexes of abelian sheaves on $S$.
Can anybody give a hint how to cleanly resolve these two problems?

REPLY [7 votes]: Let me try to say the same thing as Will but in a different way. Let $f \colon A \to B$ be left exact between abelian categories and assume there are enough injectives. Then one can distinguish between derived functors 
$$\newcommand{\R}{\mathrm R} \R^i f \colon A \to B,$$
as well as hyper-derived functors 
$$\newcommand{\RR}{\mathbb R} \RR^i f \colon \mathrm{Kom}^+(A) \to B,$$
and the total derived functor
$$ \newcommand{\RRR}{\mathbf R} \RRR f \colon D^+(A) \to D^+(B).$$
Now suppose we also have $g \colon B \to C$ with the same hypotheses and that $f$ maps injective objects of $A$ to $g$-acyclic objects of $B$. Then there is a natural isomorphism 
$$ \RRR(g \circ f) \cong \RRR g \circ \RRR f.$$
This isomorphism specializes not only to the usual spectral sequence 
$$ E_2^{p,q} = \R^p g( \R^q f(X)) \implies \R^{p+q}(g \circ f)(X),$$
for $X$ any object of $A$, but also to the spectral sequence
$$ E_2^{p,q} = \R^p g( \RR^q f(X^\bullet)) \implies \RR^{p+q}(g \circ f)(X^\bullet),$$
for $X^\bullet$ any object of $\mathrm{Kom}^+(A)$, in a similar way. In particular one needs no extra hypotheses to get the latter spectral sequence; even though the domain of $\RR^i f$ is $\mathrm{Kom}^+(A)$, one should not assume that complexes of injectives are sent to acyclics.<|endoftext|>
TITLE: Finite simple groups: smallest nonsplit automorphic extensions
QUESTION [5 upvotes]: Let $S$ be a finite nonabelian simple group such that the exact sequence 
$$1 \to S \to {\rm Aut}(S) \to {\rm Out}(S) \to 1$$
is nonsplit, where $S$ is identified with ${\rm Inn}(S)$. Then there is always a (not necessarily unique) minimal subgroup $A$ in ${\rm Out}(S)$ with respect to the condition  that 
$$1 \to S \to S.A \to A \to 1$$
is nonsplit. Where can I find a classification of all such pairs $(S,A)$? In particular, is it always true that $|A|=2$? 
I checked it with the ATLAS that  $S$ cannot be sporadic and, if it is alternating, then $S\cong A_6$ and $S.A\cong M_{10}$. 
Update: (inspired by Derek Holt's example below)
Can there be nonisomorphic minimal nonsplit extensions $S.A_1$  and $S.A_2$ for a given simple group $S$ with $A_1\cong A_2$?

REPLY [4 votes]: It is not always true that $|A|=2$. The outer automorphism of $A_6 \cong {\rm PSL}_2(9)$ induced in $M_{10}$ is the product of a field automorphism and a diagonal automorphism of ${\rm PSL}_2(9)$ and I believe that it is true in general that the product of a field and a diagonal automorphism of the same order of a finite simple group of Lie type gives rise to a nonsplit extension. I checked this by computer in the case of $S = {\rm PSL}_3(64)$, where we get a nonsplit extension $S.A$ with $|A|=3$.
I am guessing that these are essentially the only instances of nonsplit extensions of finite simple groups by their automorphism groups, but I am not certain, and I am afraid that I not aware of any definitive results of this type. It is an interesting question.<|endoftext|>
TITLE: One can earn nothing on the Brownian motion, true ? 
QUESTION [5 upvotes]: Consider any discrete time stochastic process $p(n)$ (price) with independent increments $\xi_k$ and $E(\xi_k)=0$. E.g. Brownian motion (i.e. $\xi_k = N(0,1)$). 
Consider some "trading strategy" which means - we can "buy" at some moments $t_k$, keep for some time, and then sell. The profit is difference between the price we sell and we buy. 
Assume we can buy only one asset, not more. It is slightly informal definition, but hope it is clear. 
Question/Conjecture Is it true that E(profit) = 0 for any trading  strategy, under assumption above that price is described by the independent increment process ?  (May be I need also assume that distributions are symmetric and/or identical ).
Any information on specific processes like $\xi_k= \pm 1$, or $\xi_k = N(0,1)$ is highly welcome.
Motivation:  since we cannot predict the price, we should not be able to earn something,
so $E(profit)<=0$, however if there exist a strategy, such that $E(profit) < 0$ (strictly less than zero) , we can try to consider "inverse" (not sure it is well-defined) strategy and get  $E > 0$. So it seems the only choice is to have $E(profit) = 0$.
Details: I assume that we are trading for some time n=0...N, and at last moment "N" we MUST sell on the price p(N) (if we have an asset). 
Example: Consider the simple case $\xi_k = +1$ or $-1$ with probabilities $1/2$.
Consider the trading only for n=0,1,2.
I assume that for n=0, p(0) = 0, and we buy asset for this price.
It seems the conjecture is true in this case. It seems, there are just 4 strategies,
and for all of them E(profit) = 0.
Look:
Strategy 1. "As soon price>0, sell it, and then do nothing" - it seems good strategy and we should have positive profit - but no way - the trouble is if the price goes down p(1)=-1, p(2) = -2 , we did not sell for n=1, so we must sell at n=2, so we get big loss = -2.
It compensates possible profits.
Strategy 2. "Keep untill the end" - obviously E(profit) = 0.
Strategy 3. "Sell at n=1",  - obviously E(profit) = 0.
Strategy 4. " if p(1) =-1, then sell, otherwise keep until the end".
It seems it is crazy strategy we sell when price lowered down - so we for sure have a loss,
but it is compensated by the fact that if p(1) =1, p(2) = 2 we will get profit = 2. 
(It is somewhat opposite strategy to strategy 1).

Combinatorial question By the way how many trading strategies are there for such binomial distribution $\xi_k= \pm 1$ ? 

PS
The question seems to me well-defined mathematical question, if something is unclear, please tell me I'll explain. 
Please avoid discussion whether it is realistic model for price or not, or something like, that. Please treat it as a mathematical problem. 
If one proposes a trading strategy with $E(profit) \ne 0$ it can be easily checked by MatLab or Excel or whatever.

REPLY [3 votes]: This is a specific case of Doob's optional sampling theorem<|endoftext|>
TITLE: Are there general position results in singular algebraic sets?
QUESTION [5 upvotes]: Let $X$ be a real algebraic set, and let $Y \subset X$ be its singular set.  In this question I'll focus on the analytic topology, so we can just imagine that $X$ is the zero set, in $\mathbb{R}^n$, of a certain polynomial $f(x_1, \ldots, x_n)$ with real coefficients.  It's a classical result that $X$ can be triangulated with $Y$ as a subcomplex.  Here is my question:

If $Y$ has codimension $d$ in $X$ (i.e. $X$ has dimension $n$ as a simplicial complex and $Y$ has dimension $n-d$) and $f: M^m\to X$ is a continuous map from a closed manifold of dimension $m\leq  d-1$, is $f$ necessarily homotopic, inside $X$, to a map $g: M^m\to X\setminus Y$?

I rather doubt that this is true in full generality.  I'd be quite interested in counterexamples (the simpler the better!) or any partial results of this nature.  
Note that if $X$ were a smooth manifold and $Y$ were a submanifold, a positive answer to the question is one of the standard consequences of Thom's transversality theorem.
The question could be asked more generally for simplicial complexes $Y\subset X$.  In this generality it's clearly false: let $X = Z \vee Z$, where $\pi_1 Z \neq 0$, and let $Y$ be the wedge point.  Then $\pi_1 X = \pi_1 Z * \pi_1 Z$ and if $\gamma\in \pi_1 Z$ is non-trivial, every loop representing $\gamma*\gamma$ must pass through the wedge point $Y$.  Here's a second question: is there an example of this form in which $X$ is actually a real algebraic set?
Finally, I'll mention that Lemma 2.5 of this paper gives a result somewhat along the lines I'm looking for, but just deals with loops in simplicial complexes (under somewhat strong hypotheses).

REPLY [3 votes]: What about $X$ given by the equation 
$$ (x_1^2 + \ldots + x_n^2)(y_1^2 + \ldots + y_n^2) = 0? $$
This is a wedge of $\mathbb{R}^n$ with itself, and $Y = \{0\}$. Then $X-Y$ is disconnected, so the claim is false with $M = S^0$. 
If you are not happy with $M$ being disconnected, you can take $M=S^1$ and use the above construction of $X\vee X$ where $X$ is your favorite variety with $\pi_1 X \neq 0$. As you said, there is a loop which will always pass through the origin.
Now I anticipate your question: what about $X$ irreducible in the Zariski topology? Then I don't know and I'm looking forward to see such examples in this thread.<|endoftext|>
TITLE: Lazarsfeld-Mukai bundles are stable on a K3 surface of picard number 1
QUESTION [6 upvotes]: Let $S$ be a K3 surface with $Pic(S) = \mathbb{Z}.[C]$, where $C$ is a smooth curve. Let $A$ be a complete, base point free $g_d^r$ on $C$. How to show that the Lazarsfeld-Mukai bundle $E_{C,A}$ associated to $(C,A)$ is $C$-stable on $S$?  (The bundle $E_{C,A}$ is as defined in Lazarsfeld's paper 'Brill-Noether-Petri without degenerations').

REPLY [3 votes]: You can argue as follows. Put $E:=E_{C,A}$. You need only to know : $c_1(E)$ is the positive generator of $\mathrm{Pic}(S)$,  there is a homomorphism $\mathscr{O}_S^r\rightarrow E$ which is generically surjective, and $H^0(S,E^*)=0$.
If $E$ is not stable,  it admits a torsion-free quotient $\mathscr{F}$ with $c_1(\mathscr{F})\leq 0$.  The bidual $F$ of $\mathscr{F}$ is a vector bundle with the same $c_1$, and a generically surjective homomorphism $\mathscr{O}_S^r\rightarrow F$. Let $s:=\rm{rk}( F)$; 
 we get a generically surjective homomorphism $\wedge^s\mathscr{O}_S^r\rightarrow \det(F)$. Therefore $\det(F)=\mathscr{O}_S$, and there is a direct factor $\mathscr{O}_S^s$ of $\mathscr{O}_S^r$ such that the induced homomorphism $\mathscr{O}_S^s\rightarrow F$ has nonzero determinant, hence is an isomorphism. But this contradicts 
$H^0(S,E^*)=0$.<|endoftext|>
TITLE: Deformations of p-divisible groups
QUESTION [9 upvotes]: Given a p-divisible group over $\mathbb{F}_p$, Grothendieck-Messing theory tells us that deforming the group to $\mathbb{Z}_p$ is the same as finding an admissible filtration of the Dieudonne-module evaluated on $\mathbb{Z}_p$. (I think the way this is done is that given the group on $\mathbb{F}_p$, we pick some lift to $\mathbb{Z}_p$ and "define" the Dieudonne-module to be the Lie algebra of the universal vector extension; this is independent of the lift. But the filtration one gets from the universal extension does depend on the lift, and in fact, determines it.) 
I've been told that this sometimes applies in greater generality, i.e. deforming the group to an object not in the Crystalline site of $\mathbb{F}_p$. In particular, the space that represents the deformation functor. And this can be done because there is an integrable connection. 
So my questions are:

Where does this connection come from? Is it just the fact that the Dieudonne module is a crystal on the Crystalline site and every crystal is equipped with an integrable connection?
(This is the main one) How does having the connection save the day?

REPLY [13 votes]: The theory can be summarized as follows: First, to every $p$-divisible group $G$ over a scheme $S$ in characteristic $p$, you can attach a Dieudonne crystal $\mathbb{D}(G)$ over $S$. What this means is that, for any $S$-scheme $U$, and any divided power thickening of $U$--that is, a closed immersion of $\mathbb{Z}_p$-schemes $U\hookrightarrow T$ such that $U$ is cut out in $T$ by a nilpotent ideal equipped with divided powers (note that this is usually an additional structure)--we have a vector bundle $\mathbb{D}(G)\vert_T$ over $T$. Of course, you need all these vector bundles to patch together in a nice way. 
It in fact has the structure of an $F$-crystal, but we don't need that here. There are many constructions of this crystal: First, using Grothendieck's idea of universal vector extensions, Messing (and then Mazur-Messing) built a crystal, which, however, is only defined on a smaller site, the so-called nilpotent crystalline site. The general construction is due to Berthelot-Breen-Messing.
When $S$ is smooth and admits a smooth lift $\widetilde{S}$ over $\mathbb{Z}_p$, giving such a crystal is equivalent to giving a vector bundle over $\widetilde{S}$ equipped with a (topologically quasi-nilpotent integrable) connection. The point is that the connection tells you how to differentiate sections of the vector bundle along vector fields, and hence lets you use Taylor series to identify the evaluations of the vector bundle along 'infinitesimally close' points of $\widetilde{S}$. To make sense of such a series, you need divided powers, and the 'topologically quasi-nilpotent' condition ensures that this series is always truncated at a finite level. In general, one has to work locally, and with divided power envelopes, but there exists a similar such description. See Theorem 6.6 of Berthelot-Ogus, 'Notes on crystalline cohomology'.
Once you have the crystal $\mathbb{D}(G)$, Messing showed that you can use it linearize the deformation theory of $p$-divisible groups. Namely, the restriction of $\mathbb{D}(G)$ to the Zariski site of $S$ has a natural (Hodge) filtration, given, as you point out, by the Lie algebra of the universal vector extension of $G$. Then, for any nilpotent divided power thickening (we need not just the ideal sheaf, but also the divided powers on it to be nilpotent) $U\hookrightarrow T$ in the crystalline site of $S$ over $\mathbb{Z}_p$, lifting $G\vert U$ over $T$ is equivalent to deforming the Hodge filtration on $\mathbb{D}(G)\vert_{U\hookrightarrow U}$ to a direct summand of $\mathbb{D}(G)\vert_{U\hookrightarrow T}$.
This specializes to the case you refer to in your question, by taking $S$ to be $\text{Spec }\mathbb{F}_p$ and $T$ to be $\text{Spec }\mathbb{Z}/p^n\mathbb{Z}$ (it is known (see 2.4.4 of de Jong's 'Crystalline Dieudonne theory...') that giving a $p$-divisible group over $\mathbb{Z}_p$ is equivalent to giving a compatible system of $p$-divisible groups over $\mathbb{Z}/p^n\mathbb{Z}$ as $n$ varies). The divided powers are the canonical ones: $p\mapsto\frac{p^n}{n!}$. Unfortunately, there is a hitch when $p=2$: these divided powers are not nilpotent, so Grothendieck-Messing theory is very delicate then. In fact, the statement as you have it is no longer true in this situation (one has to restrict to connected $p$-divisible groups).<|endoftext|>
TITLE: Fractional chromatic number, find reference to a particular alternate definition for
QUESTION [8 upvotes]: I'm searching for a reference to a particular alternate definition of the fractional chromatic number of graphs.  
Let me review the most common definition and basic properties first.  
Let $ G $ be a (finite simple) graph.  For any natural number $ n $, we define the multi-chromatic number $ \chi_n(G) $ as the least $ m $ such that you can have a multi-coloring function $ c $ mapping the vertices of $ G $ to $ n $ element subsets of $ [m] $ such that the color of any two adjacent vertices have no intersection.  We then define the fractional chromatic number of the graph as $ \chi^*(G) = \inf_n \chi_n(G)/n $.  
It can be proved that this infimum is equal to the limit and equal to $ \chi_k(G)/k $ for some number $ k $, and so it is always rational as well.  The fractional chromatic number of a graph is bounded from above by its chromatic number $ \chi(G) $ and from below by its clique number $ \omega(G) $.
Now the alternate definition I'm asking for.  

I define the multi-chromatic number as $ \chi_n(G) = \chi(GK_n) $.  Here $ GK_n $ is the graph you get by replacing each vertex of $ G $ by an $ n $-clique and each edge by $ n^2 $ edges linking each vertex from one corresponding clique to the other.  I then define the fractional chromatic number the same as above.

(I use the notation $ GK_n $ because the graph is the lexicographical product of $ G $ and $ K_n $.)
It is easy to prove that this definition gives the same value for the multi-chromatic number as the usual definition.  Indeed, if you have a coloring of $ GK_n $ with $ m $ colors, you get a natural $ n $-multi-coloring of $ G $ with $ m $ colors by defining the color of each vertex to the set of colors of the corresponding vertices in $ GK_n $, and this construction can be reversed as well.
While it's easy to see that the definition is equivalent, I'd like a reference from an existing book or article that gives this definition.  This would show that people other than me find this definition natural.  
I know there are other alternate definitions of the fractional chromatic number, namely there is one that gives it as the solution of a particular linear program, one using Kneser graphs, or one that uses graph powers.  I'm not interested in those for this question.
I have searched in Pavol Hell, Jaroslav Nesetril, Graphs and Homomorphisms, and in Edward R. Scheinerman, Daniel H. Ullman, Fractional Graph Theory.  Both of these talk about the fractional chromatic number in detail, but they don't give the definition I need.

REPLY [2 votes]: Wilfried Imrich, Sandi Klavžar, Product Graphs, Structure and Recognition gives this as theorem 8.38 on page 268 as well.  I should have looked in this book earlier, but now Dan Stahlke's answer made me check it.<|endoftext|>
TITLE: Congruences between CM and non-CM modular forms
QUESTION [6 upvotes]: Let $g$ be a cuspidal modular eigenform of weight 2 and level $N$ that has CM, so comes from a Groessencharacter of an imaginary quadratic field $K$. Let $p$ be a prime not dividing $N$.
(1) Is it possible that $g$ can be congruent mod $p$ to an eigenform that does not have CM by $K$?
I suspect that this may be possible [EDIT: it definitely is possible], so I'd also be interested in the following more specific question:
(2) Suppose $g$ is new of level $N$ and "$p$-isolated" (not congruent modulo $p$ to any other eigenform of that weight and level). Let $\ell$ be a prime not dividing $N$ that is 1 mod p and split in $K$, and let $g'$ be either of the two eigenforms at level $N\ell$ corresponding to $g$. Can there exist eigenforms of level $N\ell$, congruent to $g'$ mod $p$, that do not have CM by $K$?

REPLY [6 votes]: $\newcommand\rhobar{\overline{\rho}}$
$\newcommand\Q{\mathbf{Q}}$
$\newcommand\GL{\mathrm{GL}}$
$\newcommand\F{\mathbf{F}}$
$\newcommand\Frob{\mathrm{Frob}}$
If $\rhobar: G_{\Q} \rightarrow \GL_2(\F) = \GL(V)$ is a (projectively) dihedral representation, then congruences between $\overline{\rho}$ and non-projectively dihedral forms are captured by $H^1_{\Sigma}(\mathbf{Q},W)$, where the adjoint representation of $V$ decomposes as $1 \oplus \chi \oplus W$, and the subscript $\Sigma$ denotes that we are considering the appropriate Selmer group. One can see this by noting that projectively dihedral deformations correspond to deforming the induced character, which (with fixed determinant) comes down to $H^1_{\Sigma}(\Q,\chi)$. The group $H^1_{\Sigma}(\Q,W)$ can certainly be non-zero, even in the minimal case. In your second problem, the assumption amounts to asking (in particular) that $H^1_{\emptyset}(\Q,W) = 0$. By the Greeberg-Wiles Euler characteristic formula, this certainly implies that the dual Selmer group also vanishes. If you now allow ramification at an auxiliary prime $\ell$, then the condition on the dual Selmer group becomes more restrictive, and hence it is still zero. Thus, by the Greeberg-Wiles formula again, we deduce that (with $\Sigma$ indicating no condition at $\ell$):
$$|H^1_{\Sigma}(\Q,W)| = \frac{|H^1(\Q_{\ell},W)|}{|H^0(\Q_{\ell},W)|} = |H^0(\Q_{\ell},W(1))|,$$
The second equality coming from local duality ($W$ is self dual, so its Cartier dual is $W(1)$). So you get new deformations whenever the group on the right is non-trivial. If $\rhobar(\Frob_{\ell})$ has eigenvalues $\alpha_v$ and $\beta_v$ then the eigenvalues of $W \oplus \chi$ are $\alpha_v/\beta_v$, $\beta_v/\alpha_v$, and $1$. So:

If $\chi(\Frob_{\ell}) = 1$, you get new deformations if the ratios of the eigenvalues in some order are $\ell$,
If $\chi(\Frob_{\ell}) = -1$, then $\rho(\Frob_{\ell})$ has projective order two, and so the eigenvalues of $W \oplus \chi$ are $-1$, $-1$, and $1$, and hence the eigenvalues of $W$ are $1$ and $-1$. Hence you get deformations if $\ell \equiv \pm 1 \mod p$.

In either of these cases, the existence of deformations gives rise to modular forms, assuming (for example) we are in a context in which $R_{\Sigma} = \mathbf{T}_{\Sigma}$. Those lifts can not all be CM, since otherwise one would not see any cohomology coming from $W$. 
If you want to insist, for example, that the newforms have level structure $\Gamma(N) \cap \Gamma_0(\ell)$ (rather than $\Gamma_1(\ell)$ or higher powers of the conductor at $\ell$) then you could (for example) insist that $\chi(\Frob_{\ell}) = +1$, that $\ell \not\equiv 1 \mod p$, and the ratio of eigenvalues is $\ell$. Then the only possible local deformations are of Steinberg type, and they exist by 1.
Edit: I just saw that you want to insist that $\ell \equiv 1 \mod p$. Assuming that you also want $\Gamma_0(N \ell)$-level structure, I think one might be in good shape if $\rhobar(\Frob_{\ell})$ is (for example) trivial. Note that primes which split completely have relatively small density, so one might not find them by accident. Just to please Noam, here's an example:
$$E:= [0,0,0,-1,0] \ \text{of conductor $32$ with CM}$$
$$A:= [0,0,0,-332,-2752] \ \text{of conductor $32 \cdot 61$ with no CM}$$
$$D:= [0,0,0,-3256,-67984] \ \text{of conductor $32 \cdot 373$ with no CM},$$
and then $E[3] = A[3] = D[3]$.
Extra: One can also go the high powered route: Khare-Wintenberger lifting theorems allow one to find global lifts (in the potentially diagonalizable situation in which one is in here) as long as one has local lifts. So a lift which is Steinberg at $\ell$ up to twist at level $\Gamma_0(N \ell)$ will exist (assuming $\rhobar$ satisfies the Taylor-Wiles conditions) as long as $\alpha_{\ell}/\beta_{\ell}$ is $\ell$ or $\ell^{-1} \mod p$. Since the lift will be Steinberg at $\ell$, it won't be CM.<|endoftext|>
TITLE: $(1+\epsilon)$-injective Banach spaces, complex scalars
QUESTION [16 upvotes]: It is well known that a real Banach space which is $(1+\epsilon)$-injective for every $\epsilon >0$ is already 1-injective (Lindenstrauss Memoirs AMS, 1964, download here).  Using common terminology, If $E$ is a $\mathcal{P}_{1+\epsilon}$-space for every $\epsilon >0$ then $E$ is a $\mathcal{P}_1$-space.
The proof of Lindenstrauss seems valid only for real scalars.  Has a proof of the corresponding statement for complex scalars appeared in the literature?
This result is easier if $E$ is a dual space, and the proof in Semadeni's book seems to work for complex scalars.
[Edit 7/1/2013] After 1 month and 138 views, no answer is posted.  Two experts (not on MO) have told me they did not know of a reference.  This is likely not in the literature, which is somewhat surprising to me.
With editorial license, I am changing the question: Give a proof of the statement for complex scalars.
[Edit 8/20/2013] Per a reader's suggestion, be warned that the proof I offered below turned out to be incorrect, as mentioned at the end.  Maybe the idea can be rescued.

REPLY [9 votes]: Edit: I apologize for not catching the mistake sooner, but at least this can serve as an example of how science evolves.  If you don't make any mistakes, you aren't thinking.
I will have to uncheck this answer since the question is still open.

The statement for complex scalars is true.  In the mid-to-late 60's, not long after Lindenstrauss' memoir was published, the theory of the "injective hull of a Banach space" was completely worked out.  It is pretty well covered in Section 11 of Lacey, "The Isometric Theory of Classical Banach Spaces".  There we see that every Banach space $X$ is isometrically embedded with an "essential embedding" into a unique $C(K)$ space where $K$ is compact and extremally disconnected (sometimes called "Stonean").  For our theorem, it therefore suffices to prove that if $X$ is a $P_{1+\epsilon}$ space for every $\epsilon > 0$ then $X$ has no proper essential extension.
We use the criterion for essential extensions given by Lacey, p. 89: if $X\subset Y$ then $Y$ is an essential extension of $X$ if and only if the only seminorm on $Y$ which is dominated by the norm on $Y$ and equal to the norm on $X$ is the norm on $Y$ itself.
For a $P_{1+\epsilon}$ subspace $X\subset Y$, we define a seminorm $\rho$ on $Y$ by $$\rho(y) = \inf\{\|P(y)\|: P \: \text{is a projection of}\:  Y \: \text{onto}\: X\}.$$
The proof of the triangle inequality $\rho(y_1 + y_2)\le \rho(y_1) + \rho(y_2)$ uses the following lemma:  Given $y_1,\: y_2 \in Y$ which are linearly independent (mod $X$) and projections $P_1,\: P_2$ from $Y$ onto $X$, there exists a projection $P:Y \twoheadrightarrow X$ with $P(y_i) = P_i(y_i),\: i=1,2$. (Here we use the fact that $X$ has the extension property: bounded linear operators into $X$ can be extended.)  Since $X$ is $P_{1+\epsilon}$ for all $\epsilon > 0$, we have $\rho(y) \le \|y\|$ on $Y$ and $\rho(x) = \|x\|$ for $x\in X$.  But if $Y$ is a proper extension of $X$, $\rho$ can not be equal to the norm on $Y$.  Thus $Y$ is not an essential extension by the criterion mentioned. QED
Notice that this proof is valid for real and complex scalars.
[Edit 7/6/2013]:  Oops! I just reread this and I realize that the function $\rho$ as defined above does NOT in general satisfy the triangle inequality.  I believe the lemma, as stated, is true, but that is not sufficient to prove the triangle inequality.  I can provide a counterexample if requested, but I must withdraw the claim.  I still have hopes that a proof showing that $X$ has no proper essential extension can be found.  In particular, as follows from the above remark about the injective hull, it suffices to show that $X$ is not a proper subspace by an essential embedding into any $C(K)$ with $K$ extremally disconnected.  This is equivalent to showing that, for any isometric embedding of $X$ onto a proper subspace of such a $C(K)$, $K$ will have a proper closed subspace which is norming for $X$ (see Lacey).<|endoftext|>
TITLE: Hyperelliptic modular curves in characteristic p
QUESTION [7 upvotes]: Ogg characterized the finitely many N such that $X_0(N)_{\mathbb{Q}}$ is hyperelliptic, and Poonen proved in "Gonality of modular curves in characteristic p" that for large enough N, $X_0(N)_{\mathbb{F}_p}$ is not hyperelliptic.
Question: Are there any N such that $X_0(N)_{\mathbb{Q}}$ is not hyperelliptic but for some p not dividing N $X_0(N)_{\mathbb{F}_p}$ is hyperelliptic? 
I'm also interested in this question for other modular curves of the form $X_H$, where H is a congruence subgroup.

REPLY [7 votes]: A curve $C$ is hyperelliptic if and only if the canonical map $C \to \mathbb P^*(\Omega^1(C))$, which sends a point $p$ to the codimension 1 subspace $V_p \subset\Omega^1(C)$ of all one forms vanishing at $p$, is a two to one map. In the hyperelliptic case its image will be a $\mathbb P^1$, and the degree two map will be the quotient by the hyperelliptic involution. The map $\mathbb P^1 \to \mathbb P^*(\Omega^1(C))$ will basically be a Veronese embedding and hence its image will lie on a lot of quadrics. One can use this to search for possible candidates. I found none for $\Gamma_0(N)$ with $N \leq 151$. I think my search is provably complete, showing that for $N \leq 151$ and all $p$ the only hyperelliptic $X_0(N)_{\mathbb F_p}$ are the ones who are already hyperelliptic over $\mathbb Q$. Together with the general bound $${\rm gon_{\mathbb F_p^2}} X_0(N) \geq [PSL_2(\mathbb Z) : \Gamma_0(N)]\frac {p-1} {12(p^2+1)}$$ which is mentioned in Poonen's paper this implies that $X_0(N)_{\mathbb F_p}$ will never be hyperelliptic modulo $2,3$ or $5$ unless of course $X_0(N)_{\mathbb Q}$ is.
Now for $N \geq 151$ the genus of $X_0(N)$ already starts to become reasonably big. So I would not expect there to be any examples with $p>5$ and $N>151$ either. But I don't see how to prove it yet since Poonen's bound depends heavily on $p$.
Using this idea for $X_H$ with other congruence subgroups $H$ I found an example though. There might be more but this is just the first one I found. Anyway here is the explicit example:
Consider the modular curve $X_1(37)/\langle 4\rangle$, it is a double cover of $X_0(37)$ and its genus is $4$. Its global one forms are generated by the modular forms with q-expansion:
\begin{gather}
x_0 := q - 2q^{5} + 2q^{6} - 3q^{7} + 2q^{9} + O(q^{10})\\
x_1 := q^{2} - q^{5} - q^{6} + O(q^{10})\\
x_2 := q^{3} - 2q^{7} + O(q^{10})\\
x_3 := q^{4} - q^{5} + q^{6} - 2q^{7} + 2q^{9} + O(q^{10}).\\
\end{gather}
These satisfy the relations:
\begin{gather}
- x_{1}^{2} + x_{0} x_{2} - 2 x_{2} x_{3} = 0 \\
x_{1}^{2} x_{2} - 2 x_{1} x_{2}^{2} + 2 x_{2}^{3} -  x_{0} x_{1} x_{3} +
x_{1}^{2} x_{3} + x_{1} x_{2} x_{3} - 4 x_{2}^{2} x_{3} -  x_{0}
x_{3}^{2} + x_{1} x_{3}^{2} + 4 x_{2} x_{3}^{2} + 2 x_{3}^{3} = 0, \\
\end{gather}
which describe the canonical model $X_1(37)/\langle 4\rangle $ as the intersection of a smooth quadric and cubic in $\mathbb P^3$.
Modulo $2$ however $x_0,x_1,x_2,x_3$ satisfy two additional relations:
$$
x_{1} x_{2} + x_{2}^{2} + x_{0} x_{3} + x_{2} x_{3} =0 \\
x_{2}^{2} + x_{1} x_{3} + x_{2} x_{3} = 0,\\
$$
which shows that in this case the image of the canonical embedding is a $\mathbb P^1$. 2 is the only prime for which it happens for $X_1(37)/\langle 4 \rangle$.<|endoftext|>
TITLE: Is there a nice characterisation of topoi with nice meta-logical properties?
QUESTION [8 upvotes]: First-order order classical logic with standard semantics has a proof theory: it is complete, sound and effective. 
In higher order logic with standard semantics  one cannot obtain a proof theory - it cannot be simulateously complete, sound and effective.
Now, the internal language of a topos is higher order typed intuitionistic logic. Presumably like higher order classical logic it won't have nice meta-logical properties either.
Is there a nice characterisation for toposes with these nice properties for its internal language?

REPLY [18 votes]: The undesirable properties of higher-order logic are created by an insufficient notion of model. That is, we cannot have all three, soundness, completeness and effectivness (decidability of proof checking), if we insist that formulas be interpreted in the "standard" set-theoretic way. Henkin semantics does not suffer from this defficiency.
What this says is not that something is wrong with higher-order logic, but that something is wrong with those who refuse to look at semantic models, even when they are right in front of their faces, because these models are "unintended", "philosophically unacceptable", "not what mathematicians think", etc. This phenomenon of refusing to accept new interpretations of old theories is quite persistent, and always very harmful. Didn't someone stall progress in noneuclidean geometry because it was unthinkable that there would be strange new models? Aren't imaginary numbers so called because they were unthinkable and did not "really exist"? Doesn't higher-order classical logic suffer because it is being denied its natural notion of models on the grounds that they are non-standard?
The natural notion of model for intuitionistic higher-order logic (IHOL) is that of a topos. With respect to topos semantics, it is a standard result that IHOL enjoys soundness, completeness and effectivness.
We may specialize this standard fact to classical higher-order logic (CHOL). The result then is that, with respect to Boolean topos semantics, CHOL enjoys soundness, completeness and effectivness. From here on, we may prove various technical theorems which allow us to cut down on the class of Boolean toposes which is stil sufficient for completeness. And then it is not much of a surprise that we cannot cut down just to a single topos known as classical sets, which is called "Paradise" by its prisioners.<|endoftext|>
TITLE: Tightening Zhang's bound
QUESTION [18 upvotes]: Inspired by a blogpost by Scott Morrison and ongoing discussion there I decided to create this community wiki to track progress on the original bound of Yitan Zhang.
The original bound was $70,000,000$. The accepted answer should contain latest known improvement.

As of 4.6. 2013 there is a polymath project devoted to improving this bound. The proposal can be found at http://polymathprojects.org/2013/06/04/polymath-proposal-bounded-gaps-between-primes/
Links to various references: http://michaelnielsen.org/polymath1/index.php?title=Bounded_gaps_between_primes
A good place to start is to read notes by Terence Tao and his blog post on the topic.

REPLY [10 votes]: Original approach
A set of integers $H$ is called admissible if it avoids at least one residue class modulo $p$ for each prime $p$. In other words
$$
\forall p \in \mathcal{P} :\text{cardinality of} \, \lbrace x \bmod p \, | \, x \in H \rbrace \leq p-1.
$$
Let $Q(k_0)$ denote the assertion that for any admissible set $H$ of cardinality $k_0$ there are infinitely many translates $n+H$ that contain at least two primes. The bound on the gap is then $\mathrm{diam}\, H$.
Zhang deduces his bound from the following result:
T1: $Q(3,500,000)$ is true
In Zhang's paper the length $k_0$ is determined by the following inequality (1) that has to hold for some natural number $l_0$
$$
(1+4\varpi) (1-\kappa_2) > \left(1 + \frac{1}{2l_0+1}\right) \left(1 + \frac{2l_0+1}{k_0}\right) (1+\kappa_1),
$$
where
$$
\kappa_1 = \delta_1 \left( 1 + \delta_2^2 + k_0 \log\Bigl(1+\frac{1}{4\varpi} \Bigr) \right) \binom{k_0+2l_0}{k_0}
$$
$$
\kappa_2 = \delta_1 (1+4\varpi) \left(1 +\delta_2^2 + k_0 \log\Bigl(1+\frac{1}{4\varpi} \Bigr) \right) \binom{k_0+2l_0+1}{k_0-1}
$$
$$
\varpi = 1/1168
$$
and
$$
\delta_1 = (1+1/4\varpi)^{-k_0}
$$
$$
\delta_2 = \sum_{j=0}^{1/4\varpi} \frac{k_0\log(1+\frac{1}{4\varpi}))^j}{j!}.
$$
The admissible set that Zhang uses is $H = \{ p_{k_0+1}, \ldots, p_{2k_0}\}.$
Current record
Terence Tao & Scott Morrison: 4,802,222 
Terence Tao established another inequality on $k_0$ that manages to remove most of inefficiency of Zhang estimate.
$$
1+4\varpi > \left(1 + \frac{1}{2l_0+1}\right) \left(1 + \frac{2l_0+1}{k_0}\right) (1+\kappa)
$$
where
$$
\kappa := \sum_{1 \leq n < 2 + \frac{1}{2\varpi}} \Bigl(1 - \frac{2n \varpi}{1 + 4\varpi}\Bigr)^{k_0/2 + l_0} \prod_{j=1}^{n} \left(1 + 3k_0 \log\Bigl(1+\frac{1}{j}\Bigr)\right).
$$
Moreover $l_0$ is allowed to be real number. Scott Morrison then found that for $l_0 = 291.7$ one gets $k_0 = 341,640$ which is the best possible $k_0$ for given $\varpi = 1/1168$.
Paper by Richards suggest to take as admissible set $H_m = \{ \pm 1, \pm p_{m+1}, \ldots, \pm p_{m+k_0/2+1} \}$ for $m$ large enough. This leads to bound 
$$
2p_{m+\lceil k_0/2 \rceil + 1} \quad \text{for }  k_0  \text{ even}
$$
and 
$$
p_{m+\lfloor{k_0/2}\rfloor-1} + p_{m+\lfloor{(k_0+1)/2}\rfloor-1} \text{ for } k_0 \text{ odd.}
$$
For given $k_0=341,640$ program written by Scott Morrison found that $m=5553$ gives the smallest bound of $4,802,222$.<|endoftext|>
TITLE: Bounds for $\sum_{k=1}^\infty\frac{(-1)^{k+1}x^k}{(k-1)!\zeta(2k)}$
QUESTION [5 upvotes]: Let $$ f(x) = \sum_{k=1}^\infty\frac{(-1)^{k+1}x^k}{(k-1)!\zeta(2k)}$$

Are there lower bounds, upper bounds or (unlikely) simpler closed form
  for $f(x)$?

The bounds for Bernoulli numbers I found give the trivial bounds.
Wolfram Alpha times out.
mpmath's nsum returns small negative values.

REPLY [15 votes]: The Wikipedia page
on the Riesz function which Barry Cipra cited has a link to a
paper
by Wilf showing that there are infinitely many real $x$ where
the function vanishes, though rather sparse: the number of such $x \geq X$
is asymptotically proportional to $\log X$.  The function certainly
does not approach zero as $x \rightarrow \infty$;
even the estimate $f(x) = o(x^{1/4+\epsilon})$
would be equivalent to the Riemann hypothesis, as Riesz noted
in his original 1916 paper in Acta Math.
Numerically, the first few zero crossings beyond $1.15$ are at about
$19326.551$, $22521.798$, $51868.607$ if I did this right.
Here's a plot for $x \in [5\cdot 10^3, 10^5]$, comparing $f(x)$ 
with the main term
$$
{\rm Re}\left(
  \frac {\Gamma\left(1 - \frac{\rho_1}{2}\right)}{\zeta'(\rho_1)}
  x^{\rho_1/2}
\right)
$$
in the asymptotic expansion of $f(x)$ for large $x$.
    
 (source)
Here $\rho_1 = \frac12 + 14.1347\ldots i$ is the first
complex zero of the Riemann $\zeta$ function.
The factor $x^{\rho_1/2}$ oscillates with amplitude $x^{1/4}$.
The coefficient is small (absolute value about $7.775 \cdot 10^{-5}$)
because of the complex Gamma factor; further complex zeros provide
additional terms, but the $\rho_2$ term is already smaller by a factor of
almost $300$, and further terms should be smaller yet, assuming
no $\zeta$ zeros get very close to each other (let alone collide
or worse).  The discrepancy between the two plots is accounted for almost entirely by
the first trivial zero of $\zeta$ at $-2$, which gives a term
$1/\left(2\zeta'(-2)x\right)$ that is asymptotically negligible but makes
$f(x)$ negative for $1.15 < x < 5000$.  It takes a long time for
the growth of $x^{1/4}$ to overcome the small Gamma factor:
$\left|\phantom.f(x)\right|$ first exceeds $1$ around $x = 10^{19}$.
The asymptotic expansion comes from the contour-integral formula
$$
f(x) = \frac1{2\pi i}
 \int_{c-i\infty}^{c+i\infty} \frac{\Gamma(s+1)}{\zeta(-2s)} x^{-s} ds
$$
with $-1 < c < -1/2$.  NB the denominator really is $\zeta(-2s)$,
not $\zeta(+2s)$; we encounter zeros of $\zeta$ by moving the contour
to the right.  The Wikipedia page gives this formula but without
the factor of $(2\pi i)^{-1}$ $-$ maybe somebody reading this can fix
that error.
To compute $f(x)$ numerically, we can't quite use the formula
$$
f(x) = \sum_{n=1}^\infty \mu(n) \frac{x}{n^2} e^{-x/n^2}
$$
because it doesn't converge fast enough to locate the large zeros.
Instead, by applying the same trick to the sum over (say) $k \geq 4$
instead of $k \geq 1$ we get an expansion such as
$$
f(x) = \frac{6x}{\pi^2} - \frac{90x^2}{\pi^4} + \frac{945x^3}{2\pi^6}
+ \sum_{n=1}^\infty \mu(n) \frac{x}{n^2}
\left( \exp\left(\frac{-x}{n^2}\right)
   - 1 + \frac{x}{n^2} - \frac{x^2}{2n^4}
\right)
$$
which was used to compute $f(x)$ to enough accuracy to draw the
plot shown above.  For large $x$ this technique still requires
$x^{1/2 + \epsilon}$ terms.   It may be barely feasible to compute $f$
this way for $x \approx 10^{19}$ large enough to see $\left|\phantom.f(x)\right|>1$;
but the asymptotic expansion using the first handful of zeta zeros
should be much better for this purpose.<|endoftext|>
TITLE: Where to look for corrections of papers?
QUESTION [20 upvotes]: When I start reading a paper, is there some easy way to find a list of corrections for that paper?
For example, it happens occasionally that some result of a paper turns out to be wrong, or at least that the proof is incorrect and not salvageable, and I might not myself easily be able to spot the error (it will after all have been missed by both the original author(s) as well as the reviewers of the paper). And in such a situation, I would hate to end up using such a result in a crucial way only to later discover that it was not correct.
Another thing is that I might have been working on something similar which turns out to be a special case of such a result. In that case, if I am not aware that the proof (or possibly even the stronger result) is incorrect, I might stop working on that particular problem, even if I felt close to a solution, believing that it had already been settled.
In these cases, it would be nice to have some centralized list, where I could find the paper and see if it has had any corrections.
(I had a hard time figuring out what to tag this question, so please feel free to change the tags).

REPLY [3 votes]: I've found more errors in published papers when I went through their proofs, than as published corrections.<|endoftext|>
TITLE: The Erdős-Turán conjecture or the Erdős' conjecture?
QUESTION [30 upvotes]: This has been bothering me for a while, and I can't seem to find any definitive answer.  The following conjecture is well known in additive combinatorics:  

Conjecture: If $A\subset \mathbb{N}$ and $$\sum_{a\in A} \frac{1}{a}$$ diverges, then $A$ contains arbitrarily long arithmetic progressions.

This is often called either "The Erdős-Turán conjecture" or "Erdős's conjecture on arithmetic progressions."  

I want to make sure I quote things correctly in my writing, and my main question is, who should this conjecture be attributed to? 

I am asking because I find a lot of conflicting pieces of information regarding how it should be referred to.  For example, the Wikipedia page and the two Wolfram pages, page A and page B, are in disagreement.  Wikipedia seems to be adamant that calling it the Erdős-Turán conjecture is incorrect, however I am skeptical, as it does not seem to be so clear.  Also, both of the choices are used in many papers in the literature.
I went back and read the original 1936 paper of Erdős and Turán, "On some sequences of integers."  (Math Sci Net MR1574918).  In that paper, they conjecture that $r(N)=o(N)$, where $r(N)$ is a maximal subset of $\{1,\dots,N\}$ without 3 term arithmetic progressions.  They also mention the (incorrect) conjectures of Szekeres on $k$ term progressions, but other than that, they don't seem to mention any conjectures regarding longer progressions at all. 
As far as I can tell, the above conjecture first appears in the 1972 paper of Erdős, "Résultats et problèmes en théorie des nombres."  (Math Sci Net MR0396376)  This suggests that the conjecture should be named after Erdős, however, I believe that there must some reason why this conjecture has often been attributed to Turán as well.  I can't imagine that it would be done for no reason, and it is entirely possible that the reason why credit is given to both Erdős and Turán was not written down, or at least, is not easy to find.
I am hoping someone can shed some light on this, and provide a clearer answer for what to call the above conjecture.  
Thanks for your help,

REPLY [21 votes]: One of the best places to track these things down is The mathematical coloring book, by Alexander Soifer, Springer 2009. Chapter 35 is on "Monochromatic arithmetic progressions", and section 35.4, "Paul Erdős’s Favorite Conjecture", is on the problem you ask about. 
As far as I can tell, the question is sometimes called the Erdős-Turán conjecture for two reasons: First, it extends their older conjecture (now Szemerédi's theorem). Second, it was first popularized in connection to Turán, see below.
Soifer identifies a problem paper in the premier issue of Combinatorica as the first place in print were the conjecture appears with the attached value of $\$3000$:  On the combinatorial problems which I would most like to see solved, Combinatorica, 1 (1981), 25–42, submitted Sep. 15, 1979.
Soifer adds that apparently Erdős first offered this amount in a 1976 talk, "To the memory of my lifelong friend and collaborator Paul Turán", in a conference the University of Manitoba, see
Problems in number theory and Combinatorics, in Proceedings of the Sixth Manitoba Conference on Numerical Mathematics (Univ. Manitoba, Winnipeg, Man., 1976), Congress. Numer. XVIII, 35–58, Utilitas Math., Winnipeg, Man., 1977. (Erdős calls it "A very attractive conjecture of mine.") 
In Problems and results on combinatorial number theory, II, J. Indian Math. Soc. (N.S.), 40(1–4) (1976), 285–298, he stated the conjecture with an attached prize of $\$2500$. In page 288 of this paper, we read (mildly edited for typos):

$2.$ An old conjecture in number theory states that for every $k$ there are $k$ primes in arithmetic progression. This conjecture would immediately follow if we could prove that for every $k$ and $n\gt n_1(k)$, $r_k (n) \lt \pi (n)$. 

Here, as usual, $r_k(n)$ is the size of the largest subset of $\{1,2,\dots,n\}$ without $k$-term progressions.

This method of proof seems very attractive, it tries to prove a difficult property of the primes by just using the facts that the primes are numerous. In some cases I have been successful with this simple minded approach. In this connection I recently formulated the following conjecture for the proof or disproof of which I offer $2500$ dollars:
Let $a_l \lt a_2 \lt\dots$ be an infinite sequence of integers satisfying $\sum\frac1{a_i}=\infty$. 
  Then for every $k$ there are $k$ $a$'s in an arithmetic progression. 
(The truth of this conjecture would imply that for every $k$ there are $k$ primes in arithmetic progression). One can put this problem in a finite form as follows: Put $$g_k (n) = \max\sum_{a_i>n}\frac1{a_i},\quad G (k) = \lim_{n=\infty} g_k (n),$$ where the maximum is extended over all sequences $\{a_i\}$ not exceeding $n$ which do not contain an arithmetic progression of $k$ terms. The $2500$ dollar conjecture is equivalent to $G(k) \lt\infty$ for every $k$ (it is not quite trivial to show that $G (k) = \infty$ implies the falseness of the conjecture).

By the way, the prize currently attached to the conjecture is actually $\$5000$. Soifer indicates that the update occurred in the paper Some of my favorite problems and results, in The Mathematics of Paul Erdős, vol. I, Graham, R. L., and Nesetril, J., (eds), 47–67, Springer, Berlin, 1997. (The paper appeared posthumously, and was written in 1996.)

I offer $5000 for a proof (or disproof) of this [problem]. Neither Szemerédi nor Furstenberg’s methods are able to settle this but perhaps the next century will see its resolution. 

Soifer then tries to identify when the conjecture was first stated. 

I searched for evidence in the ocean of his writings, and found three indicators. First, in a paper submitted on September 7, 1982 to Mathematical Chronicle (now called New Zealand 
  Journal of Mathematics) that appeared the following year [E83.03], Paul writes: "This I conjectured more than forty years ago."
In the same year, 1982, Paul spoke at the Conference on Topics in Analytic Number Theory in Austin, Texas. I read in the proceedings (published in 1985 [E85.34], 
  p. 60): "I conjectured more than 40 years ago that if $a_1 \lt a_2\lt \dots$ is a sequence of integers for which $\sum_{i=1}^\infty \frac1{a_i}=\infty$ then the $a$’s contain arbitrarily long arithmetic progressions." 
Thus, both of these publications indicate that the conjecture was posed before 1942. On the other hand, in the 1986 Jinan, China, Conference proceedings (published in 1989 [E89.35]), Paul writes (p. 142): "About 30 years ago I conjectured that if $\sum_{n=1}^\infty 1/{a_n}=\infty$, then the $a$'s contain arbitrarily long arithmetic progressions." This would date the birth of the conjecture to about 1956.

The cited papers are:

[E83.03]: Combinatorial problems in geometry, Math. Chronicle 12 (1983), 35–54. 
[E85.34]: Some solved and unsolved problems of mine in number theory, in Topics in 
Analytic Number Theory (Austin, Tex., 1982), Univ. Texas Press, Austin, TX, 1985, 59–75. 
[E89.35]: Some problems and results on combinatorial number theory, in Graph theory 
and its applications: East and West (Jinan, 1986), Ann. New York Acad. Sci. 576, 132–145, New York Acad. Sci., New York, 1989. 

For the finitude of $G_k$, see for example ArxiV:math/0703467.
By the way, the conjecture in the 1972 paper you identify seems to be restricted to $3$-term progressions. Even this version remains open (and carries an attached prize of $\$250$.)<|endoftext|>
TITLE: Integer solution to special system of linear equations
QUESTION [8 upvotes]: This problem appear in my research, but I am unable to solve it.
There should be an easy argument, but I have not yet found it.
Informal version
An integer $k\geq 2$ is fixed.
We are given a matrix (left in image), 
where each column contains a block with entries in $1,\dotsc,k-1.$
The matrix has the special property that each row sums to a multiple of $k.$
Can we always find some blocks that when we remove $k$ from their (nonzero) entries,
all row sums become zero?
Here is the picture to illustrate the problem for an example with $k=5$,
where the given matrix is the left one, and the corrected one is on the right.

Formal version of problem
It may be formulated as follows:
We are given an integer $k\geq 2.$ (One may assume it is prime, if it helps).
We fill in columns in a matrix $M$ such that the following holds:
Each column $j$ is of the form $(0,0,\dotsc,0,c_j,c_j,\dotsc,c_j,0,\dotsc,0)$ where $0\lt c_j \lt k.$
The sum of each row $j$ in $A$ is of the form $kb_j$ with $b_j$ integer.
Let $A$ be the matrix with 1 at the non-zero positions of $M.$
Then we may write the last requirement as $Ac=kb$ where $c$ is the vector of column values,
and $kb$ is the sum of the entries in respectively row.
Now, my question is, can we always find a vector $x\in \{0,1\}^s$ (entries either 0 or 1) such that $A(c-kx)=0$?
(This corresponds to selecting a subset of columns in the matrix $M$
such that when $k$ is subtracted from the non-zero elements in them,
all row sums becomes zero.
I believe the problem can be reduced to the linear system $Ax=b$
where $A$ and $b$ are as above, and we still seek a vector solution with 0,1 entries.
Is there some theory that may tell if this is solvable?
The case $k=2$ is the same as the following setup, if it helps with the image:
a finite set of intervals covers [0,1], such that each point is covered an even number of times. Then, we may always color the intervals red and black, such that each point is covered by the same number of black intervals, as red intervals.
I have an inductive proof of this special case, but it does not translate to larger $k.$

REPLY [8 votes]: The problem is indeed equivalent to solving $Ax=b$ with $x\in\{0,1\}^n$ (for $A$ a matrix of size $m\times n$). To see that this always has a solution, consider the optimization problem
$$\text{maximize } \mathbf{1}^TAx\quad \text{s.t. } 
\begin{pmatrix}
-I_n\\\\ I_n\\\\ A
\end{pmatrix}x\leqslant 
\begin{pmatrix}
0\\\\ 1\\\\ b
\end{pmatrix}.
$$
The constraint matrix is totally unimodular, hence there is an integer optimal solution, i.e. an optimal solution $x^*\in\{0,1\}^n$. 
Clearly, $\mathbf{1}^Tb$ is an upper bound for the optimal objective value, which is achieved precisely by the feasible solutions $x$ with  $Ax=b$. Now $\tilde x=c/k$ is a fractional solution with objective value $\mathbf{1}^Tb$. This implies $\mathbf{1}^T A x^* =\mathbf{1}^Tb$, thus $Ax^*=b$.<|endoftext|>
TITLE: What is characteristic function of maximum of i.i.d. random variables?
QUESTION [7 upvotes]: Is is possible to get characteristic function of maximum of i.i.d. random variable sequence? Such as $X_1, X_2$ are two i.i.d random variables, then what is characteristic function of $X=\max(X_1,X_2)$?

REPLY [6 votes]: Assume that the random variables $X$ and $Y$ are defined on the probability space $(\Omega,\mathcal F,\mu)$. Let $\Delta:=\{(x,y)\in\Bbb R^2,x\lt y\}$. We have by independence
$$
E\left[e^{it\max(X,Y)}\right]=\int_{\Bbb R^2}e^{it\max(x,y)}\mathrm d\mu_X\otimes\mu_Y(x,y).
$$
Splitting over $\Delta$ and its complement, and denoting $F$ the common cumulative distribution function of $X$ and $Y$, we thus get 
$$E\left[e^{it\max(X,Y)}\right]=2E\left[F(X)e^{itX}\right]-\int_{\Bbb R}\mu(X=x)e^{itx}\mathrm d\mu_X(x).$$
Some remarks:

This gives an explicit formula in terms of the common distribution function.
If $\mu(X=x)=0$ for all $x$ (for example when $X$ has a density), then the formula is simpler.
This can be extended to $\max(X_1,\dots,X_d)$.
We get an analogous formula for $\min$ instead of $\max$.<|endoftext|>
TITLE: What is the largest possible operator norm of a sparse (0,1)-matrix?
QUESTION [5 upvotes]: Inspired by this question, I was wondering about the following problem:
Consider all $n\times n$ $(0,1)$-matrices with $k$ ones.  Which of these matrices has the largest operator norm?  And how does this largest possible operator norm behave with $n$ and $k$?
If any progress has already been made on this problem, I assume it's in the guise of directed graphs.  For the case of symmetric matrices with zero diagonal, which graphs give the largest possible largest eigenvalue?  $(k/n)$-regular graphs?

REPLY [4 votes]: If $k=ab$ with $a\leq n$ and $b\leq n$, then an $a \times b$ rectangle of $1$s surrounded by $0$s hits the optimal matrix norm of $\sqrt{k}$.  Otherwise, $\sqrt{k}$ is not achievable, but it's clear that one can get very close to $\sqrt{k}$ with very close to a rectangle.
A $k/n$-regular graph has a largest eigenvalue of $k/n$ which is not that great. A better thing to do is a complete graph on about $\sqrt{k}$ vertices plus $n-\sqrt{k}$ isolated vertices, which gives a largest eigenvalue of about $\sqrt{k}$.<|endoftext|>
TITLE: Why does the generalised Galvin-Prikry Theorem only hold at Ramsey cardinals?
QUESTION [13 upvotes]: The Galvin-Prikry theorem says that Borel sets are Ramsey. This means that for every Borel set $S\subseteq[\omega]^\omega$, there is an $A\in[\omega]^\omega$ such that either $[A]^\omega \subseteq S$ or $[A]^\omega \cap S=\emptyset$.
My question is, for which $\kappa>\omega$ does the analogous statement hold (call it the $\kappa$-Galvin-Prikry theorem):
For every open set $S\subseteq [\kappa]^\omega$, there is an $A\in[\kappa]^\kappa$ such that  $[A]^\omega\subseteq S$ or $[A]^\omega \cap S=\emptyset$.
With the open sets given by the usual topology on $[\kappa]^\omega$, with basic open sets $[\eta,\omega]=\{s\in[\omega]^\omega:\eta \prec s\}$, where $\prec$ denotes initial segment.
At $\omega$, the Galvin-Prikry theorem is equivalent to Nash-Williams' "every block contains a barrier". (See page 644 in http://www.cs.elte.hu/~kope/bqoint.pdf ). Ie, for every Block $B\subseteq[\omega]^{<\omega}$, there is an $A\in[\omega]^\omega$ such that $B\cap[A]^{<\omega}$ is a barrier.
The $\kappa$-version is equivalent to "every $\kappa$-block contains a $\kappa$-barrier", with the definitions given in Shelah's paper:

Better quasi-orders for uncountable cardinals
Israel Journal of Mathematics September 1982, Volume 42,
Issue 3, pp 177-226
http://link.springer.com/article/10.1007%2FBF02802723

The important implication here can be seen because the $\kappa$-Galvin-Prikry theorem is equivalent to $\kappa\stackrel{\mbox{open}}{\longrightarrow}(\kappa)^\omega$ which means $\forall f:[\kappa]^\omega\rightarrow 2$ with $f$ continuous, $\exists A\in[\kappa]^\kappa$ such that $|f"[A]^\omega|=1$. Which in turn is equivalent to $\forall f:B\rightarrow 2$ with $B\subset[\kappa]^{<\omega}$ a $\kappa$-block, $\exists A\in[\kappa]^\kappa$ such that $|f"[A]^{<\omega}\cap B|=1$. From this "every $\kappa$-block contains a $\kappa$-barrier" quickly follows.
On page 2 of the quoted paper, Shelah says that for "every $\kappa$-block contains a $\kappa$-barrier" to hold, $\kappa$ must be Ramsey. But he gives no proof.
If we assume that $\kappa$ is Ramsey, then it isn't too difficult to show that every $\kappa$-block contains a $\kappa$-barrier.
So, by Shelah, $\kappa$ is Ramsey iff the $\kappa$-Galvin-Prikry theorem holds. But I haven't managed to prove what Shelah omitted.
So why does the $\kappa$-Galvin-Prikry theorem imply that $\kappa$ is Ramsey?
Or the same for any of the other equivalent versions I mentioned.

REPLY [5 votes]: The Galvin-Prikry theorem for $\kappa > \omega$ was given by E.M. Kleinberg and R.A. Shore in their paper "On Large Cardinals and Partition Relations".
Let $\kappa > \omega$. If we denote by $\kappa \overset{\circ}{\rightarrow} (\kappa)^{\omega}_{\lambda}$ the assertion of $\kappa \rightarrow (\kappa)^{\omega}_{\lambda}$ restricted to $\kappa-$Borel functions, we have:
Theorem [Kleinberg-Shore, 1971]. Let $\kappa > \omega$. Then, $\kappa$ is Ramsey iff $\kappa \overset{\circ}{\rightarrow} (\kappa)^{\omega}_{\lambda}$ for each $\lambda < \kappa$.<|endoftext|>
TITLE: General bound for the number of subgroups of a finite group
QUESTION [14 upvotes]: I am interested in the following:
Let $G$ be a finite group of order $n$. Is there an explicit function $f$ such that 
$|s(G)| \leq f(n)$ for all $G$ and for all natural numbers $n$, where $s(G)$ denotes
the set of subgroups of $G$?

REPLY [3 votes]: In a similar vein to Geoff Robinson's answer, observe that any proper subgroup of a group of order $n$ can be generated by at most $\Omega(n) - 1$ elements, where $\Omega$ counts the number of prime factors (with multiplicity) of $n$. This is tight, with equality in the case of a product of prime cyclic groups.
This gives an upper bound of:
$$ \binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{\Omega(n) - 1} + 1$$
subgroups, where the final $+ 1$ includes $G$ itself. Now we can invoke Michael Lugo's strict upper bound from https://mathoverflow.net/a/17236/39521 to obtain a closed-form upper bound:
$$ |s(G)| \leq \left\lceil \binom{n}{\Omega(n) - 1} \dfrac{n - \Omega(n) + 2}{n - 2 \Omega(n) + 2} \right\rceil \leq \dfrac{2 \times n^{\Omega(n) - 1}}{(\Omega(n) - 1)!}$$
Note that this is tight infinitely often: in the case where $n = p$ is prime (so $G = C_p$), we get:
$$ \dfrac{2 \times p^0}{0!} = 2 $$
where the two subgroups are the trivial group and $G$ itself.<|endoftext|>
TITLE: What can be done with computability logic that previous logic systems can't?
QUESTION [6 upvotes]: I've been reading a lot about computability logic lately and I'm superficially aware that it unifies classical, intuitionistic and linear logics.
What I'm seeking to know is:
Can computability logic be used as a foundation for mathematics (or rather arithmetic) in the same sense classical logic has been proposed before by Russell, Whitehead and other authors? Is it subject to the same restrictions (like Gödel theorems) like the other logics (considering that it conservatively extends classical logic)?
I hope I've been clear enough.
References:
http://www.cis.upenn.edu/~giorgi/cl.html
http://www.csc.villanova.edu/~japaridz/ICL.pdf

REPLY [2 votes]: I would say yes, it can. However, in the research I found that you should take 2 things into account.
First, don't take the Turing-Halting problem as basis. Because, I didn't find a way to directly reason about halting problems. For a proof, you need to hop between the proof-steps. This does not seem to be possible with the halting problem. However, you can take as basis the inequality of 2 computable predicates. If you have p1 and p2 and you write p1 >= p2 when for all x p2(x) => p1(x). Then some reasoning becomes possible.
Second, instead of deterministic functions, you should use non-deterministic (but still computable) functions. Again, there is no practical known way to transform deterministic functions directly into other deterministic functions. However, if you allow non-deterministic functions (as intermediate), then these become possible.
In this way you get a rather simple logic based on computation. Still, if you want to use as the basis of mathematics, you get problems when going to uncountable sets. You are hitting the failure of Hilbert's program, or you should find a way to repair that.
I hope this is somewhat in the direction of the answer you were looking for.<|endoftext|>
TITLE: Is there any superstable configuration in the game of life?
QUESTION [35 upvotes]: This question spins off of Gil Kalai's recent question on Conway's game of life for a random initial configuration.  
There are numerous configurations in the game of life that are known to be stable---such as blocks, beehives, blinkers and toads---in the sense that if they appear on an otherwise empty board or on part of the board that remains otherwise empty, then they will persevere (or at least reappear on some period) into the indefinite future. All of the common examples of such configurations, however, seem to disintegrate when placed into a hostile environment; when they are hit by a glider or other spaceship, for example, these common stable configurations can be completely ruined.
My question is whether there is any superstable configuration, which can survive even in any hostile environment.
Question 1. Is there any superstable configuration in the game of life? 
Specifically, let us define that a finite configuration is superstable, if it can survive in any environment, no matter how hostile, meaning that if it should ever appear on the board, then it will definitely reappear later in exactly that same position, regardless of what else is happening on the board. Perhaps the position is somehow isolated, absorbing whatever is happening around it; or perhaps it is a strong source of some kind, spewing out gliders or other objects, regardless of what else is around it; or perhaps it is some core surrounded by encircling vacuum-cleaners, traveling patterns that sweep up whatever might interfere. 
This question is related to Gil Kalai's, in that if there are such superstable configurations, then we will expect that the infinite random position will have them with some (albeit very small) density, which will enable us to prove lower bounds on the density of the expected living infinite random position.
One can also imagine a glider version of superstability, where the pattern survives, but with some nonzero displacement:
Question 2. Is there any superstable glider? 
That is, is there a finite pattern that, regardless of the environment in which it is placed, will repeat itself at some future time with some displacement? A strong form of such a superstable glider would ask also that it be a vacuum cleaner, meaning that it glides around in any given environment while leaving only empty cells in its wake. 
Question 2b. Is there a superstable glider vacuum?
I can imagine a small glider that erases everything in its path; or perhaps there is a kind of moving wall, which steadily pushes against whatever it faces, leaving emptiness behind. If there were such a superstable glider vacuum that also moved in a definite direction, then of course there could be no superstable stationary position, since otherwise we could vacuum it up. 
Another alternative would seem to be that every finite configuration in the game of life is destructible, in the sense that one can design for it an especially hostile environment, leading to eventual death.
Question 3. Is every finite configuration destructible? 
In other words, can every finite configuration in the game of life be extended to a larger configuration whose development leads in finite time to a position with no living cells? A weaker version of this would ask merely that the configuration be extended to a configuration such that eventually, the original configuration does not recur on any subportion of the board.

REPLY [3 votes]: I think it could be of interest to mention that something meeting the definition of a finite superstable configuration exists if we add arbitrarily little noise to the original rules of the GoL (Conway's game of life for random initial position). The configuration is trivial – completely empty finite region. Via an argument in the spirit of the second Borel–Cantelli lemma, we see that “if it should ever appear on the board, then it will definitely reappear later in exactly that same position, regardless of what else is happening on the board”. Sorry if this is off-topic.<|endoftext|>
TITLE: The infinite X in Conway's game of life
QUESTION [13 upvotes]: In Conway's game of life, take the initial position to be two infinite diagonal lines of live cells, with a single cell in common. Does this thing converge to a stable configuration? I.e., is the state of each cell (or finite region) eventually periodic?

REPLY [21 votes]: What I get from an X of size $11121\times11121$ at just around the point where information travels to the tips.
Even from Xs ten times as long, there is Methuselah-like ebbing and flowing of debris near the center amid a pool of still lifes and blinkers still thousands of generations on.

Just going from experience working on the Busy Beaver of 5, I would imagine this question might be enormously difficult to settle, owing to the globally fractal and locally random nature of the picture.<|endoftext|>
TITLE: Solving a quadratic equation for an hermitian matrix
QUESTION [5 upvotes]: I am looking for a procedure to find solution(s) for a square matrix equation
$H^T H = S$
where $H = H^\dagger$ is a hermitian ($n\times n$) matrix and $S$ is a given symmetric complex matrix. Due to hermiticity of $H$, $S$ should satisfy $n$ conditions $(\operatorname{im}(\operatorname{tr}(S^i))=0,\quad i = 1,...,n)$.
I am interested in simple solutions for small $n=3$ matrices. For $n=2$, this can be solved by an explicit parametrization of $H$ which leads to a quadratic equation giving four solutions. Since the problem is similar to taking a square root of a matrix, presumably there are $2^n$ solutions for this problem, too.
note: This question is perhaps similar to this one. Here, the equation is simpler but applies to complex, not real matrices.

REPLY [3 votes]: The first thing to keep in mind is that having $\mathrm{Im}\bigl(\mathrm{tr}(S^i)\bigr)=0$ for $i=1,\ldots,n$ is not sufficient to guarantee that $S = H^TH$ has a solution with $H$ Hermitian, even when $n=2$.  For example, consider
$$
S = \begin{pmatrix}1&i\\ i&-1\end{pmatrix},
$$
which has $\mathrm{tr}(S) = \mathrm{tr}(S^2) = 0$.  It is easy to show, however, that $S$ is not of the form $H^TH$ for any $2$-by-$2$ Hermitian matrix $H$.  Another example satisfying the condition $\mathrm{Im}\bigl(\mathrm{tr}(S)\bigr)=\mathrm{Im}\bigl(\mathrm{tr}(S^2)\bigr)=0$ for which it is easy to show there is no solution is 
$$
S = \begin{pmatrix}-1&0\\ 0&a\end{pmatrix},
$$
where $a$ is any real number other than $-1$.  
On the other hand $H^TH=-I_2$ has a $1$-parameter family of solutions
$$
H = \begin{pmatrix}s&ic\\ -ic&s\end{pmatrix},
$$
where $c$ and $s$ are any real numbers satisfying $c^2-s^2=1$, while $H^TH=I_2$ has two $1$-parameter families of solutions, the first being
$$
H = \begin{pmatrix}s&c\\ c&-s\end{pmatrix}
$$
where $c^2+s^2=1$ and the second being
$$
H = \begin{pmatrix}c&is\\ -is&c\end{pmatrix},
$$
where $c$ and $s$ are any real numbers satisfying $c^2-s^2=1$.
These examples show that, even when $n=2$, neither invertibility nor having an orthogonal basis of eigenvectors guarantees solvability while, sometimes, there are an infinite number of solutions.
The second thing to keep in mind is that there is a natural $\mathrm{O}(n,\mathbb{C})$-equivariance built in to the problem, one that is important in understanding the equation that needs to be solved: Let $S_n(\mathbb{C})$ denote the vector space of symmetric $n$-by-$n$ matrices with complex entries (which has real dimension $n^2{+}n$), and let $H_n$ denote the (real) vector space of Hermitian symmetric $n$-by-$n$ matrices (which has real dimension $n^2$).  The OP's question then deals with understanding the preimages of elements of $S_n(\mathbb{C})$ with respect to the quadratic mapping $\sigma:H_n\to S_n(\mathbb{C})$ defined by 
$$
\sigma(H) = H^TH.
$$
Now, $\sigma$ has the $\mathrm{O}(n,\mathbb{C})$-equivariance
$$
\sigma(A\cdot H) = A\cdot \sigma(H)
$$
where, for any $A\in\mathrm{O}(n,\mathbb{C})$, i.e., $A$ satisfying $A^TA = I_n$, one has
$$
A\cdot H = \bar A\ H\ A^T\quad\text{for}\ H\in H_n
\qquad\text{and}\qquad
A\cdot S = A\ S\ A^T\quad\text{for}\ S\in S_n(\mathbb{C}).
$$
Thus, the problem can be naturally posed as a question about $\mathrm{O}(n,\mathbb{C})$-orbits in $S_n(\mathbb{C})$.  While the 'generic' orbit is easy to understand, and the structure of the 'non-generic' orbits is known, the latter can be somewhat complicated.  The 'generic' orbits are described as follows:  Say that $S\in S_n(\mathbb{C})$ is generic if $S$ has $n$ distinct eigenvalues, i.e., if there exists an $S$-eigenbasis $(v_1,\ldots, v_n)$ of $\mathbb{C}^n$ with $Sv_i = \lambda_i v_i$ where the $\lambda_i\in \mathbb{C}$ satisfy $\lambda_i\not=\lambda_j$ when $1\le i<j\le n$.  In this case, as usual, one has ${v_i}^Tv_j = 0$ for $i\not=j$ while ${v_i}^Tv_i\not=0$. One can, by scaling, assume that ${v_i}^Tv_i = 1$ for all $i$, and this determines the $S$-eigenbasis $(v_1,\ldots, v_n)$ up to the individual signs of the $v_i$.  In other words, when $S$ is generic, one can write $S = V^T\Lambda V$ where $V^TV=I_n$ while $\Lambda = \mathrm{diag}(\lambda_1,\ldots,\lambda_n)$ and, once the eigenvalues are fixed in some order, $V$ is unique up to premultiplication by an element of the form $E = \mathrm{diag}(\epsilon_1,\ldots,\epsilon_n)$ where ${\epsilon_i}^2=1$.  (Such elements $E$ form a finite subgroup of $\mathrm{O}(n,\mathbb{C})$ of order $2^n$.)
Now, for dimension reasons, $\sigma$ obviously cannot be surjective; in fact, the image of $\sigma$ must have real codimension at least $n$ in $S_n(\mathbb{C})$.  Indeed, as the OP points out, the image does satisfy $n$ real equations, i.e., $\mathrm{Im}\bigl(\mathrm{tr}(S^i)\bigr)=0$ for $i=1,\ldots,n$.  This follows because $\bar S = HH^T$, and so one has
$$
\mathrm{tr}(S^k)=\mathrm{tr}\bigl((H^TH)^k\bigr) = \mathrm{tr}\bigl(H^T(HH^T)^{k-1}H\bigr)
=\mathrm{tr}\bigl((HH^T)^{k-1}HH^T\bigr) = \mathrm{tr}({\bar S}^k)
=\overline{\mathrm{tr}(S^k)}.
$$ 
However, as the above examples with $n=2$ show, these $n$ equations (which merely establish that the characteristic polynomial of $S$ has real coefficients) are not sufficient to characterize the image of $\sigma$.
Now, suppose that $S$ is generic, with distinct eigenvalues $\lambda_1,\ldots,\lambda_n$ and write $S = V^T\Lambda V$ as above. (One finds $\Lambda$ and $V$ by the usual process of taking the $\lambda_i$ to be the roots of the characteristic polynomial of $S$ and then solving the linear equations $(S-\lambda_iI)v_i=0$ and normalizing the resulting $v_i$). If $H\in H_n$ satisfies $H^TH=S$, then by acting on the pair $(S,H)$ by $A = V\in\mathrm{O}(n,\mathbb{C})$, one is reduced to the case $(V{\cdot}S, V{\cdot}H) = (\Lambda, V{\cdot}H)$.
Thus, it suffices, in the generic case, to describe the solutions to $H^TH = \Lambda$ for $H\in H_n$, where $\Lambda=\mathrm{diag}(\lambda_1,\ldots,\lambda_n)$ with the $\lambda_i$ distinct.
To do this, write $H = a + i b$ where $a=a^T$ and $b = -b^T$ are real.  Then 
$$
\Lambda = H^TH = (a-ib)(a+ib) = a^2 + b^2 + i(ab-ba).
$$
Thus, both $a^2{+}b^2$ and $ab{-}ba$ must be diagonal.  Pursuing this, one sees that the assumption that the $\lambda_i$ are distinct implies that $a$ and $b$ can be simultaneously block-diagonalized, where the blocks have size either $1$ (for each real eigenvalue) or size $2$ (for each conjugate pair of non-real eigenvalues).  In particular, the real eigenvalues must all be nonnegative (at most one of which can be $0$), and the non-real eigenvalues correspond to a $2$-by-$2$ submatrix of the form
$$
S' = \begin{pmatrix}\lambda^2&0\\0&{\bar\lambda}^2\end{pmatrix}
$$
where $\lambda^2\not={\bar\lambda}^2$.  There are two solutions $H'$ for this, namely
$$
H' = \pm\begin{pmatrix}0&\bar\lambda\\ \lambda&0\end{pmatrix}.
$$
Assembling these blocks, one gets the complete solutions $S$, which are $2^k$ in number, where $k$ is the number of real eigenvalues of $S$ plus one-half the number of non-real eigenvalues of $S$ minus the number of $0$ eigenvalues of $S$ (if any).
As the above discussion when $n=2$ shows, the nongeneric cases are more complicated.  When $n=2$, if $S$ has a real double eigenvalue $\lambda$, then it is of the form
$$
S = \begin{pmatrix}\lambda + p & ip\\ ip & \lambda - p\end{pmatrix}
$$ 
for some $p\in\mathbb{C}$.  If $\lambda\le0$, then there is no solution unless $p=0$, in which case, the solutions lie in a $1$-parameter family
$$
H = \begin{pmatrix}c & is\\ -is & c\end{pmatrix}
$$
where $c$ and $s$ are real numbers satisfying $c^2-s^2 = \lambda$.  If $\lambda>0$ and $p=0$, then there is, in addition to the above solution, another $1$-parameter family
$$
H = \begin{pmatrix}c & s\\ s & -c\end{pmatrix}
$$
where $c$ and $s$ are real numbers satisfying $c^2+s^2 = \lambda$.  If $\lambda>0$ and $p$ is not zero, then there are exactly $2$ solutions:  Let $r>0$ satisfy $r^2 = |p|^2/(4\lambda)>0$, and set $u-iv = p/(2r)$, then $\lambda = u^2+v^2$, and the two solutions are
$$
H = \pm\begin{pmatrix} r+u & v+ir\\ v-ir & r-u\end{pmatrix}.
$$
There is a similar analysis of the nongeneric cases for higher $n$ (particularly in the case $n=3$, which the OP wanted to understand), and I will leave this to the interested reader.<|endoftext|>
TITLE: similarity transformation into symmetric matrices
QUESTION [6 upvotes]: I want to determine some structures of matrices that can be transformed into a symmetric matrices using similarity transformation, i.e.,
$B=T^{-1}AT$
where $T$ is the similarity transformation matrix. 
Here, $A$ is a non-negative matrix and the diagonal elements of $A$ is supposed to be the same $(diag(A)=[a, a, \ldots , a])$.
Any ideas?

REPLY [2 votes]: An addition to Dietrich Burde's answer: Since symmetric matrices are diagonizable, they are semisimple (each invariant subspace has an invariant complement). Thus $B$ has to be semisimple ($\iff$ the minimal polynomial is square free).<|endoftext|>
TITLE: Rank of a 0-1-matrix
QUESTION [5 upvotes]: Suppose $K$ is a field of characteristic $0$. Let $M \in K^{n \times m}$ be a matrix such that every entry of $M$ is either $0$ or $1$. About this matrix, I know further that each sum over a column and a row is a constant number, i.e.
$$\sum_{i=1}^n M_{ij} = c_{\text{column}} ~~ \forall j=1,...,m$$
and
$$\sum_{j=1}^m M_{ij} = c_{\text{row}} ~~ \forall i=1,...,n$$
for some $c_{\text{column}}, c_{\text{row}}$ both lying in $\{1,2,3,...\}$.
Question: Are there criteria that enables us to say something about the rank of $M$? I.e. are there theorems of the form
'' If some condition is satisfied, then
$$\text{rank}(M) \geq f(n,m,c_{\text{column}}, c_{\text{row}})$$
for some (hopefully not too complicated) function $f$''?
I have found a paper of Odlyzko from '79 in which he shows that a $0$-$1$-matrix with constant row-sums is of full rank if the number of distinct row vectors exceeds a certain number. Unfortunately, in my case I do not have sufficiently many row-vectors but I have some additional information, for example, I know that the column-sum is also constant.
Regards,
FW

REPLY [2 votes]: Even in the special case $K = \mathbb R$, $n=m$ and $c_{\rm row} = c_{\rm column}$, this is not an easy problem!
A good place to start looking would likely be the following paper.

On the minimum rank of regular classes of matrices of zeros and ones
  by Brualdi, Manber and Ross
  http://www.sciencedirect.com/science/article/pii/0097316586901135

You may also want to see Section 3.10 of Brualdi's 2006 "Combinatorial Matrix Classes" book.  I suspect you will find it contains just about all of what is known.<|endoftext|>
TITLE: Fundamental group of an hyperbolic $4$-manifold
QUESTION [9 upvotes]: Good afternoon everyone, 
I have a very general question about hyperbolic manifolds and their fundamental groups in high dimension (at least $4$). If the theory of surfaces and $3$-manifolds provide a lot of constructive examples of compact hyperbolic manifolds, it is not that obvious that in higher dimensions, such objects exists.
A way to construct a $4$ dimensional hyperbolic manifold is to find a cocompact lattice in $\text{O}(n,1)$. 
Let $\varphi = x_1^2 + ... + x_n^2 - \sqrt{p} x_{n+1}^2$ , and $\Gamma = \text{Aut}(\varphi) \cap \text{Gl}(n+1, \mathbb{Z}[\sqrt{p} ])$. A theorem by Borel ensures us that $\Gamma$ is a cocompact lattice in $\text{Aut}(\varphi) \simeq \text{O}(n,1) $. By Selberg's lemma, we can find a finite index subgroup of $\gamma$ which is torsion free, and this rises to the construction of a compact hyperbolic $4$-manifold. 
My question is : can one tell me the algebraic structure of $\Gamma = \text{Aut}(\varphi) \cap \text{Gl}(n+1, \mathbb{Z}[\sqrt{p} ])$ ? Is there a known method to compute it ? Is there a general method for all quadratic forms with coefficient in a number field ?
Thank you very much for your attention !
Selim

REPLY [14 votes]: For general $p$, the only known method is to construct a Dirichlet fundamental domain and read off the group presentation from it. The procedure for computation of a fundamental domain is called "Jorgenesen's algorithm: List elements of $\Gamma$ (using the embedding to $GL(2(n+1),Z)$ via restriction of scalars). For each $\gamma\in \Gamma$ construct the bisector of $o, \gamma(o)$ in the unit ball $H^4$. On each step, check if conditions of Poincare's fundamental domain theorem hold. If it they do, the process stops and you got your fundamental domain $D$. 
Now, generators of $\Gamma$ are face-pairing transformations for $D$. The relators correspond to cycles of 2-dimensional faces of $D$. 
For small $p$'s one can do better: If $p=1$ (and $n=4$), the group is a reflection group (whose fundamental domain is a simplex), so you get an explicit presentation. 
You can find a brief discussion of a similar procedure in the complex-hyperbolic setting in this paper: D. Cartwright, T. Steger, Enumeration of the 50 fake projective planes, C. R. Acad. Sci. Paris, Ser. I 348 (2010), 11-13. 
To indicate limits of our present knowledge of higher-dimensional hyperbolic manifolds/orbifolds of finite volume: There is not a single known lattice in $O(n,1)$, $n\ge 1000$, for which we know the abelianization. All what we know that for each $i$, every such arithmetic lattice (note that $n> 7$) contains a congruence subgroup $\Gamma_i$ so that abelianization of $\Gamma_i$ has rank $\ge i$.<|endoftext|>
TITLE: Tutte polynomials, graph complements and degree sequences
QUESTION [13 upvotes]: Harary and Akiyama asked whether there exists a non self-complementary (SC) graph $G$ having the same chromatic polynomial as its complement. 
It was later shown that there indeed exist such graphs and it was conjectured that all such graphs have the same degree sequence as their complement.
As it turns out this conjecture is false and for every $n \geq 9$ congruent to $0,1$ modulo 4 one can construct a graph having the same chromatic polynomial as its complement but different degree sequence. 
Extending this problem, it is possible to show that for any $n \geq 8$ congruent to $0,1$ modulo 4 there exist a non SC graph having the same Tutte polynomial as its complement. 
What I currently cannot solve is the following

Question 1. Is there a graph $G$ having different degree sequence from
  $\overline{G}$ but the same Tutte
  polynomial?

 

 Question 2. Is there a non SC graph having the same Tutte polynomial as
   its complement and order congruent to
   $0$ modulo $4?$ 

What I basically tried was generating all non SC graphs with $n(n-1)/4$ edges and checked the respective Tutte polynomials. As it turns out such a graph has to have order $n \geq 12$ and there are way too many graphs to be checked by a standard computer in this manner. Hence I need a more clever approach for searching. A nice thing would be if I could quickly filter through the output of nauty, removing graphs with the same degree sequence as their complement. I am afraid Sage is too slow for that.
I recall someone on MO (G. Royle?) saying that the Tutte polynomial is not ''good'' at distinguishing degree related invariants hence I suspect that the answer to Question 1 is positive.
Edit. The paper solving the original question of Akiyama and Harary is this one. While the thing for the Tutte polynomial can be found in this preprint. I would like to point out that it is just a preprint hence it may contain errors.

REPLY [5 votes]: If you allow $n=8$, the answer to question 1 2 is yes.
The graph on $8$ vertices with edges:
[(0, 3), (0, 5), (0, 6), (0, 7), (1, 4), (1, 6), (1, 7), (2, 4), (2, 6), (3, 5), (3, 7), (4, 5), (4, 7), (6, 7)]

is a solution.
Found $8$ solutions for $n=8$.<|endoftext|>
TITLE: Please recommend some literature on the systematical theory of the elliptic systems!
QUESTION [5 upvotes]: Now I'm interested in the theory of elliptic systems, for example, both the linear and nonlinear case, the exsitence and regularity results, and is there a Fredholm alternative result for the linear or nonlinear elliptic system. Further more, may I say that we can parallelly extened the $L^2$-theory of the second-order scalar elliptic equations to the corresponding $\mathbb{L}^2$-theory of the second-order elliptic systems? Any answer and reference will be appreciated!

REPLY [3 votes]: All you mentioned can be extended to elliptic systems. The only things you should be careful about are the maximum principles and DeGiorgi-Nash-Moser type regularity results. This means for linear systems the theory is very satisfactory. You can start with strongly elliptic systems and the variational approach to them, because the theory is much simpler and it covers a lot of important systems already. A good place to start would be McLean's book, Wloka's book, and perhaps Nirenberg's article. Wloka's book also contains a comprehensive introduction to the general elliptic theory, in the style of Agmon-Douglis-Nirenberg.<|endoftext|>
TITLE: Does Zhang's theorem generalize to $3$ or more primes in an interval of fixed length?
QUESTION [78 upvotes]: Let $p_n$ be the $n$-th prime number, as usual:
$p_1 = 2$, $p_2 = 3$, $p_3 = 5$, $p_4 = 7$, etc.
For $k=1,2,3,\ldots$, define
$$
g_k = \liminf_{n \rightarrow \infty} (p_{n+k} - p_n).
$$
Thus the twin prime conjecture asserts $g_1 = 2$.
Zhang's theorem (= weak twin prime conjecture) asserts $g_1 < \infty$.
The prime $m$-tuple conjecture asserts
$g_2 = 6$ (infinitely many prime triplets), 
$g_3 = 8$ (infinitely many prime quadruplets), "etcetera" (with $m=k+1$).
Can Zhang's method be adapted or extended to prove $g_k < \infty$
for any (all) $k>1$?

Added a day later: Thanks for all the informative comments and answers!
To summarize and update (I hope I'm attributing correctly):
0) [Eric Naslund] The question was already raised in the
Goldston-Pintz-Yıldırım
 paper.  See Question 3 on page 3:

Assuming the Elliott-Halberstam conjecture, can it be proved that
  there are three or more primes in admissible $k$-tuples with large enough $k$?
  Even under the strongest assumptions, our method fails to prove
  anything about more than two primes in a given tuple.

1) [several respondents]
As things stand now, it does not seem that Zhang's technique or any
other known method can prove finiteness of $g_k$ for $k > 1$.
The main novelty of Zhang's proof is a cleverly weakened estimate
a la Elliott-Halberstam, which is well short of "the strongest assumptions"
mentioned by G-P-Y.
2) [GH] For $k>1$, the state of the art remains for now as it was pre-Zhang,
giving nontrivial bounds not on $g_k$ but on
$$
\Delta_k := \liminf_{n \rightarrow \infty} \frac{p_{n+k} - p_n}{\log n}.
$$
The Prime Number Theorem (or even Čebyšev's technique) trivially yields
$\Delta_k \leq k$ for all $k$; anything less than that is nontrivial.
Bombieri and Davenport obtained $\Delta_k \leq k - \frac12$;
the current record
is $\Delta_k \leq e^{-\gamma} (k^{1/2}-1)^2$.
This is positive for $k>1$ (though quite small for $k=2$ and $k=3$,
at about $0.1$ and $0.3$), and for $k \rightarrow \infty$ is
asymptotic to $e^{-\gamma} k$ with $e^{-\gamma} \approx 0.56146$.
3) [Nick Gill, David Roberts] Some other relevant links:
Terry Tao's June 3 exposition of Zhang's result
and the work leading up to it;

The "Secret Blogging Seminar" entry and thread that has already
brought the bound on $g_1$ from Zhang's original $7 \cdot 10^7$
down to below $5 \cdot 10^6$;
A PolyMath page
that's keeping track of these improvements with links to the
original arguments, supporting computer code, etc.;
A Polymath proposal
that includes the sub-project of achieving further such improvements.
4) [Johan Andersson] A warning: phrases such as
"large prime tuples in a given [length] interval"
(from the Polymath proposal) refer not to configurations
that we can prove arise in the primes but to admissible
configurations, i.e. patterns of integers that could all be prime
(and should all be prime infinitely often, according to the
generalized prime $m$-tuple [a.k.a. weak Hardy-Littlewood] conjecture,
which we don't seem to be close to proving yet).  Despite appearances,
such phrasings do not bear on a proof of $g_k < \infty$ for $k>1$,
at least not yet.

REPLY [23 votes]: This has now achieved by James Maynard "Small gaps between primes."<|endoftext|>
TITLE: How to find a topic to do research with as a Post-Doc?
QUESTION [6 upvotes]: I will soon finish my PhD in arithmetic geometry. My advisor told me that I will have to find my next research topic on my own. How do I do that? (Except for "continue where the PhD thesis ends") Can you give me hints/advice?

REPLY [8 votes]: You should read literature in the area of your previous research, in search of interesting unsolved
problems. Also attend conferences and talk to specialists in your field. When I was on this stage
of my career, I found surveys with lists of unsolved problems in my field, and tried to solve them.
Now I make such lists myself, to help young researchers, and I suppose they exist in every field
of mathematics. You have to read a lot to become an expert in your area.<|endoftext|>
TITLE: Intuition for coordinate patches on Proj of a graded ring.
QUESTION [9 upvotes]: Hi Mathoverflow. This question is about building intuition for the Proj construction. When I first started learning about schemes, I found the construction of the structure sheaves on Spec and Proj very confusing. However, after enough time had passed I began to understand the construction for Spec: It is sort of an algebraic partition of unity argument. The construction of the structure sheaf on Proj is still very mysterious to me. For completeness it goes something like this:


Let $G$ be a graded ring. Take $ f \in G $ homogeneous of degree $ d \geq 1 $. We have a homeomorphism $D_+(f) \cong{\rm spec} \, G_{(f)} $ which sends $D_+(fg)$ to $D(g^{\deg f} / f^{\deg g})$. It first expands the prime to $G_f$ and then contracts the result to $G_{(f)}$. Motivated by this, we can define $$ \mathscr{O}_{{\rm Proj}G}(D_+(f)) = G_{(f)}$$
    and prove that the map $G_{(f)} \to G_{(fg)}$ defined by $ a / f^n \mapsto a g^n / (fg)^n$ is localization at $ g^{\deg f} / f^{\deg g}$.


I can fill in the details but they are messy and I have very little idea what the details actually mean. Thinking about $ \mathbb{P}^n $ as a variety, I understand why $D_+(X_0)$ should be $ {\rm spec} \mathbb{C} [X_1/X_0,X_2/X_0, \dots, X_n / X_0] $ but I don't really understand how this intuition translates into the commutative algebra which is boxed above. 
Also, for homogeneous elements of degree higher than $1$ I have no idea what is going on. I understand that geometrically the veronese map should be involved but I don't understand how that intuition translates into the messy proof which I am able to write down. 


 Question : Is anyone able to explain the idea behind this construction? Note that this explanation could lie in either the realm of algebraic geometry or commutative algebra.


It is very possible that there is not a nice way to think about this construction which would make me sad because it is so fundamental. I hope that there are some good responses which help me fix my ignorance!
EDIT: I just went through the proof again and wrote up the details.
Let $ G$ be a graded ring. The first order of business is to explain why we have a homeomorphism $ D_+(f) \cong {\rm Spec} G_{(f)} $ when $ f \in G $ is homogeneous with degree $d \geq 1 $. Fix $ \mathfrak{p} \in D_+(f)$. Then $ \mathfrak{p}$ is the generic point of some irreducible closed subset $Z$ in $ {\rm Spec} G$. Since $ f \not\in \mathfrak{p}$ it follows that $ G_f \mathfrak{p}$ is the generic point of the irreducible closed subset $ Z \cap {\rm Spec} G_f$ in $ {\rm spec} G_f$. Assume that $g \in G $ is homogeneous. Then $ g $ vanishes at $ \mathfrak{p} $ iff $ g / 1 $ vanishes at $ G_f \mathfrak{p}$ iff $g^{\deg f} / f^{\deg g}$ vanishes at $ G_f \mathfrak{p}$ iff $g^{\deg f} / f^{\deg g} \in G_f \mathfrak{p} \cap G_{(f)} $. This proves that the map $D_+(f) \to {\rm Spec} G_{(f)}$ is injective. We need to prove that the map is surjective. Let $ \mathfrak{q}$ be a prime ideal in $G_{(f)}$. Motivated by the above argument, we define
$$ \mathfrak{p}_n = \{ g \in G_n : g^{\deg f} / f^{\deg g} \in \mathfrak{q} \} $$
Then $ \mathfrak{p} = \oplus_{n \geq 0} \mathfrak{p}_n \in D_+(f)$ maps to $ \mathfrak{q} $. Since $ g $ vanishes at $ \mathfrak{p} $ iff  $g^{\deg f} / f^{\deg g} \in G_f \mathfrak{p} \cap G_{(f)} $ the map is a homeomorphism which sends $D_+(fg)$ to $D(g^{\deg f} / f^{\deg g})$. All that remains is to prove that the map $ G_{(f)} \to G_{(fg)}$  defined by $ a / f^n \mapsto a g^n / (fg)^n$ is the localization map. From the affine case we know that $ G_{f} \to G_{fg}$ is the localization map at $ g / 1 $, therefore the only non units which $ G_{(f)} \to G_{(fg)}$ sends to units are powers of $ g^{\deg f} / f^{\deg g}$. I suspect that this last bit can be made rigorous.

REPLY [4 votes]: You in fact already found out the most natural way to think about the Proj construction, namely in terms of the projective space. Consider $X := \mathbb{P}^n$ with homogeneous coordinates $[x_0: \cdots : x_n]$. Then what are the basic open sets in $\mathbb{P}^n$? These are simply the complement of zero sets $V(f)$ of homogeneous polynomials $f \in S:= k[x_0, \ldots, x_n]$. I suggest that you try to see for yourself that $U_f := X \setminus V(f)$  is indeed isomorphic to $Spec~ S_{(f)}$. (Hint: at first check this for linear polynomials - which follows almost by definition. Then for an arbitrary $f$, write $U_f$ as the union of $U_f \setminus V(x_i)$ and figure out what is the coordinate ring of $U_f \setminus V(x_i)$. $U_f \setminus V(x_i)$ is also the complement in $U_{x_i}$ of the zero set of an element in the coordinate ring of $U_{x_i}$ - what is this element?)
For me at least it was illuminating to get the hands dirty: start with the definition of projective space as the space of lines in $k^{n+1}$ (with the natural topology given by identification of the lines not passing through a hyperplane with $k^n$) - and check explicitly what the notions in the definition of the Proj construction mean. 
To get flavours of more general situations, I would again go through examples. E.g. for the scenarios that $S_0$ may not be a field, consider $X := \mathbb{C}^2 \times \mathbb{P}^2(\mathbb{C})$ and see why this is isomorphic to $Proj~ A[z_0, z_1, z_2]$, where $A := \mathbb{C}[x,y]$. For a bit more involved example, consider the subvariety $V$ of $X$ defined by the zero set of polynomials $f \in A$ and a homogeneous polynomial $g \in \mathbb{C}[z_0, z_1, z_2]$, and check that $V$ is isomorphic to the subscheme of $Proj ~ A[z_0, z_1, z_2]$ defined by the ideal generated by $f$ and $g$. Can you see how to interpret the zero degree elements of the graded ring?<|endoftext|>
TITLE: Is this similar to a known combinatorial identity?
QUESTION [18 upvotes]: (Apologies if this is  too obscure.)
In joint work with Izzet Coskun we came across the following kind of combinatorial identity, but we weren't able to prove it, or to identify what kind of identity it is. (We looked in some references, but to the outsider it can be difficult to distinguish one insanely complicated sum of binomial coeffficients from another...)
The identity looks like the following. Say $n$ is a fixed natural number, and $i \leq n$ is an even natural number. (There is an analogous formula when $i$ is odd.) Then
$\begin{align}
\sum_{m=2}^{\frac{i}{2}} (x-1) \left(x-2\right)^{2m-3} \ \sum_{j=0}^{i-2m} \binom{n-1-j}{i-2m-j} 2^{j+2} \sum_{l=0}^j (-1)^l \binom{n+1}{l} \, x^{i-2m-l} & \\\\
+ \sum_{j=0}^{i-2} \binom{n-1-j}{i-2-j} 2^{j+1} \sum_{l=0}^j (-1)^l \binom{n+1}{l} \, x^{i-2-l} &\\\\ 
+ \sum_{j=0}^{i-2} \binom{n-1-j}{i-2-j} 2^{j-1} \sum_{l=0}^j (-1)^l \binom{n+1}{l} (i-2-l) \, x^{i-l-1} & \\\\
- \sum_{j=0}^{i} \binom{n-1-j}{i-j} 2^{j-1} \sum_{l=0}^j (-1)^l \binom{n+1}{l} (i-l) \, x^{i-l-1} & \\\\
\qquad =C_i \left(x-2 \right)^{i-1} 
\end{align}
$
where $C_i$ is some constant that can be read off fairly easily by looking at the coefficient of $x^{i-1}$ on the left-hand side. (There are various rewritings one could do to the left-hand side, but it's not clear how much this helps.)
Given that there are various irregularities in the last three terms on the left-hand side, it seems unlikely that the whole thing can be reduced to a simple form. But it would nevertheless be very useful to know if this looks similar to any known combinatorial identities.
Maybe a simpler warmup question would be: how to see that the sum of the last three terms on the left-hand side is divisible by $x-2$? That might help get going with an inductive argument. 
Any ideas would be much appreciated!

REPLY [3 votes]: (For completeness, let me write as an answer the comment I made on Zack Wolske's answer.)
It turns out that this identity is tractable using the Wilf-Zeilberger machinery, as suggested by David Lehavi in comments to the question. Christoph Koutschan was able to prove the identity
using the Mathematica package $\mathtt{HolonomicFunctions}$ which he developed in his thesis:
Advanced Applications of the Holonomic Systems Approach (PhD Thesis). RISC-Linz, Johannes Kepler University, September 2009. 
Unfortunately I don't understand enough about the method to describe it accurately here, but very roughly, the idea is to prove identities of this kind by calculating the ideal of recurrences and differential operators that each one satisfies, and then checking that a certain (explicitly computable!) number of initial values agree.  Follow the above link for a real explanation!<|endoftext|>
TITLE: Modular symbols and degeneracy maps
QUESTION [7 upvotes]: Consider the space of (cohomological) modular symbols of level $N$,
$$ \operatorname{Symb}_{\Gamma_0(N)}(\mathbb{Z}) = \operatorname{Hom}_{\mathbb{Z}[\Gamma_0(N)]}(\mathrm{Div}^0(\mathbb{P}^1_\mathbb{Q}), \mathbb{Z}) \cong H^1_c(Y_0(N)(\mathbb{C}), \mathbb{Z}).$$
If I have a prime $q \nmid N$, then there's two injective maps $i_1, i_2: \operatorname{Symb}_{\Gamma_0(N)}(\mathbb{Z}) \hookrightarrow \operatorname{Symb}_{\Gamma_0(Nq)}(\mathbb{Z})$: one is just the natural inclusion, and the other is given by acting by $\begin{pmatrix} q & 0 \\\ 0 & 1 \end{pmatrix}$.
Say we choose a modular form $f$ of weight 2 and level $\Gamma_0(N)$ (with coefficients in $\mathbb{Z}$, for simplicity). Let $\operatorname{Symb}_{\Gamma_0(N)}(\mathbb{Z})[f]$ denote the rank 2 submodule where the Hecke operators act as they do on $f$. Similarly, let $\operatorname{Symb}_{\Gamma_0(Nq)}(\mathbb{Z})[f]$ be the subspace at level $Nq$ where the Hecke operators away from $q$ act as they do on $f$.
Do the images of $\operatorname{Symb}_{\Gamma_0(N)}(\mathbb{Z})[f]$ under $i_1$ and $i_2$ generate $\operatorname{Symb}_{\Gamma_0(Nq)}(\mathbb{Z})[f]$? This is certainly true after tensoring with $\mathbb{Q}$, but I'm worried that the sum of the images might not be saturated.

REPLY [12 votes]: Please allow me to replace $\mathbb Z$ by $\mathbb Z_\ell$ in your question, as the problem of being saturated or not is local.
Assuming $f$ is a cusp form, the answer is yes if and only if your modular form f is not Eisentein mod $\ell$ (that is, if the Galois representation $\rho_f$ is still irreducible 
modulo $\ell$). This is (in the difficult sense) the famous Ihara's Lemma. Indeed,
this lemma states that the map 
$H^1(X_0(N),\mathbb Z/\ell) \times H^1(X_0(N),\mathbb Z/\ell) \rightarrow H^1(X_0(Nq),\mathbb Z/\ell)$ is injective, cf Ribet's article at the ICM 1982/1983, page 10.
Under the hypotheses $f$ not Eisenstein, one has $H^1(X_0(N),\mathbb Z/\ell)[f] = H^1_c(Y_0(N),\mathbb Z/\ell)[f]$ and the same result with $N$ replaced by $Nq$, and this implies what you want.
If $f$ is Eisenstein, this is always true.<|endoftext|>
TITLE: Letter from Grothendieck to Tate on "crystals"
QUESTION [19 upvotes]: I have downloaded from  this link  a quite poor quality scan of the letter dating May 1966 that Grothendieck sent to Tate mentioning his ideas about generalizing Monsky-Washnitzer cohomology. I am trying to put it in LaTeX, at least for my own reference. So, first of all, do you know if it ever appeared in print somewhere or if the typing has already been done by someone? And, secondly, does anyone has a better-quality file? The big problem with this is that many top and bottom lines are missing, sometimes making it impossible to reconstruct the original sentence.

REPLY [22 votes]: In the light of past events ("Les Archives Grothendieck"), we now have an "ameliorated" version of the letter:
Cote n° 6. Lettre Tate (mai 66) (Cristaux) : lettre (1966), tapuscrit, notes manuscrites. 1966 - [à partir de 1970].

[scan], and project of transcription [pdf]

See [here] for more.<|endoftext|>
TITLE: A Proof that the Natural Numbers Form a pca?
QUESTION [5 upvotes]: It is known that the set of natural numbers with the operation ab = {a}b, where {a} represents the index of a recursive function, forms a partial combinatory algebra (pca). All the references I have so far consulted mention the set of natural numbers as an example of a pca with no proof. If possible, I would appreciate if someone could refer me to such proof.

REPLY [10 votes]: For a formalization of Turing machines you could look at

Andrea Asperti and Wilmer Ricciotti: Formalizing Turing Machines. Lecture Notes in Computer Science Volume 7456, 2012, pp 1-25.

The book

Martin Davis. Computability and Unsolvability. Courier Dover Publications, 1982 (3rd reprint).

contains an amazing amount of detail about Turing machines.
Given these references, let us take as granted the details of the numbering $\varphi$ of partial computable maps, the s-m-n and u-t-m theorems from computability theory, and one bit of knowledge: if we fix one argument of a partial computable map of two arguments we obtain a partial computable map again.
With s-m-n and u-t-m theorems in hand, it is not hard to show that Kleene's first algebra is a partial combinatory algebra. We just need to construct the $\mathsf{K}$ and $\mathsf{S}$ combinators.
To get the $\mathsf{K}$ combinator, consider the partial computable map
$$f(x, y) = x.$$
By the s-m-n theorem there exists $p$ such that $\varphi_{s(p, x)}(y) \simeq f(x,y)$ for all $x, y \in \mathbb{N}$. Here $s$ is the total computable map appearing in the s-m-n theorem and $\simeq$ is Kleene equality "if one side is defined then so it the other and they are equal". Now the map $x \mapsto s(p, x)$ is total and computable, hence there is $\mathsf{K}$ such that $\varphi_{\mathsf{K}}(x) = s(p, x)$ for all $x$. We now have (I write $a \cdot b$ for Kleene application, i.e., $a \cdot b = \lbrace a \rbrace b = \varphi_a(b)$):
$$(\mathsf{K} \cdot x) \cdot y = \varphi_{\mathsf{K}}(x) \cdot y = s(p,x) \cdot y = \varphi_{s(p,x)}(y) = f(x,y) = x.$$
For the $\mathsf{S}$ combinator, let $u$ be the universal computable map from the u-t-m theorem. It has the property that, for all $t$ and $m$,
$$u(t, m) \simeq \varphi_t(m).$$
In other words, $u(t,m) \simeq t \cdot m$ so the u-t-m theorem is just saying that Kleene application is computable.
Consider the map
$$g(x,y,z) = u(u(x, z), u(y,z)).$$
We have
$$g(x,y,z) \simeq u(x,z) \cdot u(y,z) \simeq (x \cdot z) \cdot (y \cdot z).$$
We just need a code for $g$, which we get by appying the s-m-n theorem twice.
By the s-m-n theorem there is $p$ such that $\varphi_{s^{(2)}(p, x, y)}(z) \simeq g(x,y,z)$ for all $x,y,z$. Let $r$ be such that $\varphi_r(x,y) = s^{(2)}(p, x, y)$ for all $x, y$. Note that $\varphi_r$ is total because the map $s^{(2)}$ from the s-m-n theorem is total. By the s-m-n theorem there is $q$ such that $\varphi_{s(q,x)}(y) = \varphi_r(x,y)$ for all $x,y$. Let $\mathsf{S}$ be such that $\varphi_\mathsf{S}(x) = s(q,x)$ for all $x$. Now finally compute
$$((\mathsf{S} \cdot x) \cdot y) \cdot z \simeq
(s(q,x) \cdot y) \cdot z \simeq
\varphi_{s(q,x)}(y) \cdot z \simeq
\varphi_r(x,y) \cdot z \simeq$$
$$
s^{(2)}(p,x,y) \cdot z \simeq
\varphi_{s^{(2)}(p,x,y)}(z) \simeq
g(x,y,z) = (x \cdot z) \cdot (y \cdot z).$$
Also, since $\mathsf{S} \cdot x \cdot y = s^{(2)}(p,x,y)$ we see that $\mathsf{S} \cdot x \cdot y$ is defined for all $x, y$, as required.<|endoftext|>
TITLE: Defining holomorphic functions in terms of Banach algebras, and similarly for C*-algebras
QUESTION [8 upvotes]: Let $C$ be the category of commutative Banach algebras and let $U : C \to \text{Set}$ be the usual forgetful functor. The holomorphic functional calculus guarantees that every holomorphic function $f : \mathbb{C}^n \to \mathbb{C}$ defines a natural transformation $U^n \to U$ (in a way which is compatible with composition, etc.). Does every natural transformation $U^n \to U$ have this form?
Similarly, let $C$ be the category of commutative C*-algebras and let $U : C \to \text{Set}$ be the usual forgetful functor. The continuous functional calculus guarantees that every continuous function $f : \mathbb{C}^n \to \mathbb{C}$ defines a natural transformation $U^n \to U$. Does every natural transformation $U^n \to U$ have this form? 
In both cases the first thing to try would be to check if $U$ is representable, and in both cases I think it can't be.

REPLY [5 votes]: I'll just consider a natural map $\phi:U\to U$; the multivariate case should be similar.
Let $D_n$ be the closed disk of radius $n$ centred at the origin, and let $A_n$ be the Banach algebra of functions that are continuous on $D_n$ and holomorphic on the interior.  Let $\iota_n\in A_n$ be the identity map.  By naturality for the restriction maps $A_{n+1}\to A_n$, the functions $\phi(\iota_n)\in A_n$ will fit together to give an entire function $f$.  Now consider an arbitrary Banach algebra $B$ and an element $b\in B$.  Choose $n>\|b\|$, so functional calculus gives a homomorphism $\beta:A_n\to B$ with $\beta(\iota_n)=b$.  (Here we can use the simplest version of functional calculus using convergent power series; we do not need need anything about the spectrum of $b$, or $B$-valued contour integrals.)  Naturality of $\phi$ with respect to $\beta$ tells us that $\phi(b)=f(b)$, as required.
I think that $A_n$ represents the functor $U_n(B)=\{b\in B:\|b\|\leq n\}$ (or should that be $\rho(b)\leq n$?), and $U$ is the colimit of these. 
UPDATE:
I didn't read Simon Henry's comment properly before; now I see that the argument above is more or less what he proposed.<|endoftext|>
TITLE: Why is $(\mathbb{Z}/3\mathbb{Z})^3$ not a class group of an imaginary quadratic number field ?
QUESTION [22 upvotes]: In general, it seems not known which finite abelian groups are class groups of quadratic number fields.
For imaginary quadratic number fileds, I  read that $(\mathbb{Z}/3\mathbb{Z})^3$ is the smallest 
abelian group which does not occur. It does occur as the class group of $\mathbb{Q}(\sqrt{188184253})$.
What other groups are known not to be the class groups of imaginary quadratic number fields ? 
How does one prove this  ? 
What is the situation for class groups of real quadratic number fields ?

REPLY [3 votes]: Hello MO, the system will not let me make a comment only an answer, but I just wanted to point out that, contrary to what "v08ltu" commented, the case of $(Z/3)^k$ is not clearly easier for the methods of Arno and Watkins, but in fact in one aspect is close to the worst case (aside from genera at 2).
The point is, you can have heaps of $(a,b,c)$ forms with $a\sim d^{1/4}$, the multiplicative structure is all gone. In the notation of these papers, the $A(s)=\sum_a 1/a^s$ is only bounded by $|A(1/2+it)|\ge 1-(h-1)/d^{1/8}$, and you want to keep this away from zero (the trick of Watkins is to consider smaller $a$ multiplicatively, and larger ones additively in a variant of $A$, but here all is additive). It seems that Watkins has just enough to get $h=81$ to work, as $d\sim 80^8\sim 2^{52}$, which was his sieveing limit. I sincerely doubt that $(Z/3)^5$ is feasible the same way, as the bound jumps by $3^8$. Watkins barely mentions this at the end of his work, saying that workload is at least fourth-power growth in sieving limit from $h/d^{1/4}$ "but also the lower bounds on $|A(s)|$ become worse", and I think they particularly jam you up here, being 8th power not 4th.
On the other hand, once you actually get to the sieving problem, even with a higher limit, it is then easier as you can demand that all small primes (up to $d^{1/4}$ which is at least almost 10000) are inert, whereas Arno and Watkins split up into a myriad of cases to handle alternative behavior vis-a-vis accounting of smallish split primes and their effect. Sieving up to $728^8\sim 10^{23}$ could be feasible, as Sorenson has done $10^{25}$ for pseudosquares (using doubly-focussed methods).<|endoftext|>
TITLE: Bass' stable range of $\mathbf Z[X]$
QUESTION [31 upvotes]: Let $n$ be a positive integer and $A$ be a commutative ring. The ring $A$ is said to be of Bass stable range $\mathrm{sr}(A)\leq n$ if for $a, a_1, \dots, a_n \in A$ one has the following implication:
$$1 \in \langle a, a_1, \dots, a_{n}\rangle \implies \exists \  x_1, \dots, x_n \in A, 1 \in \langle a_1+x_1a,\  \dots, a_n + x_n a\rangle.$$
(Above, $\langle \cdot \rangle$ denotes the ideal generated by the elements inside).
Of course, one says that $A$ is of stable range $n$ if $\mathrm{sr}(A) \leq n$ and $\mathrm{sr}(A) \not \leq n-1$.
Bass proved that if $A$ is noetherian of Krull dimension $d$ then $\mathrm{sr}(A)\leq d+1$.
Examples are known; for example Vaserstein proved that $\mathrm{sr}(k[x_1,\dots,x_n]) = n+1$ when $k$ is a subfield of the real numbers. 
My question is : is the stable range of the ring of integer polynomials $\mathbf Z[X]$ known?
What I wrote before shows that $\mathrm{sr}(\mathbf Z[X]) \leq 3$ and it seems very likely that $\mathrm{sr}( \mathbf Z[X])=3$.
A refinement of my previous question is : could you provide an explicit unimodular triplet of polynomials $(P_1,P_2,P_3) \in \mathbf Z[X]$ showing that $\mathrm{sr}(\mathbf Z[X]) \not \leq 2$?

REPLY [10 votes]: The goal of my answer is to elaborate on steps (1) and (2) of the accepted answer of Steven Landsburg. In particular, I would like to make clear how a "small enough ideal" looks like and how to produce a non-reducible row like the one in Oblomov's comment. This consists essentially in revisiting the proof of [2, Proposition 1.9], with some K-theoretical shortcuts. 


Addendum. I realized that the last paragraph of [2, Proof of Proposition 1.9] is flawed, so that the unimodular row $(21 + 4x, 12, x^2 + 20)$ is not guaranteed to be non-stable in $\mathbb{Z}[X]^3$. Indeed, it is assumed there that $SK_1(\mathcal{O}, 2\mathcal{O}) = \mathbb{Z}/2\mathbb{Z}$ for $\mathcal{O} = \mathbf{Z} + \mathbf{Z} \sqrt{-5}$,  whereas $SK_1(\mathcal{O}, 2\mathcal{O})= 1$ by the Bass-Milnor-Serre Theorem (In other words, $f = 2$ does not have the property $(*)$ defined at the top of page 191). This glitch in the final step of the proof is harmless in the sense that changing sligthly the conductor $f$ yields a valid, but different, non-reducible unimodular row. Setting for instance $f = 4$, i.e., $B = \mathbf{Z} + 4 \mathbf{Z} \sqrt{-5} \subset \mathcal{O}$, and considering a suitable quadratic residue symbol, will do the job. My answer below is yet another way to address this problem.


The Bass-Milnor-Serre theorem [3, Theorem 11.33] is key to understand step (1):
let $F$ be a totally imaginary number field and $m$ is the number of elements of finite order in $F^{\ast}$. For each ideal $I$ in the ring of integers $R$ of $F$, the relative special Whitehead group $SK_1(R, I)$ is a cyclic group of finite order $r$. For each prime $p$, $\text{ord}_p(r)$ is the nearest integer in the interval 
$\lbrack 0, \text{ord}_p(m) \rbrack$ to 
$$\min_{\mathfrak{p}} \left\lfloor \frac{\nu_{\mathfrak{p}}(I)}{\nu_{\mathfrak{p}}(pR)} - \frac{1}{p - 1} \right\rfloor$$ 
where $\lfloor x \rfloor$
 denotes the greatest integer $\le x$ and $\mathfrak{p}$ ranges over the prime ideals of $R$ containing $p$.
Independently of the above theorem, it follows from [1, Corollary 5.2] that $SK_1(R, I)$ surjects onto $SK_1(R, J)$ if $I \subset J$ and $R/I$ is semi-local. For $R$ and $I$ as in the theorem, the group $SK_1(R, I)$ increases in size as $I$ decreases with respect to inclusion. This is reflected in the above formula and it is immediate to see that for every decreasing sequence of ideals $I$, the corresponding sequence $(SK_1(R, I))_I$ stabilizes. Moreover, we can find $I$ such that $SK_1(R, I)$ has the maximal cardinality $m$ and $I$ is then small enough in the sense that any ideal contained in $I$ yields the same maximal $SK_1$ group.
We are now (almost) in position to provide an order $A$ in some $R$ as above such that $\mathbf{Z}[X]$ surjects onto $A$ and $SK_1(A)$ is not trivial. Indeed, consider $R = \mathbf{Z}[i]$, the Gaussian integers, set $A = \mathbb{Z} + 4i \mathbb{Z}$ and $I = 4 \mathbf{Z}[i]$ (this $I$ is sufficiently small for our purpose, but doesn't maximize $SK_1$, unlike $8 \mathbf{Z}[i]$). We will establish that $SK_1(A)$ surjects onto $SK_1(R, I) \simeq \mathbf{Z}/2\mathbf{Z}$, the latter isomorphism being given by Bass-Milnor-Serre's theorem. To do so, let us prove the following lemma:

Let $S$ be a one-dimensional Noetherian domain and let $J$ be an ideal of $S$ such that $S/J \simeq \mathbf{Z}/n\mathbf{Z}$ for some $n \ge 0$. Then $SK_1(S) = SK_1(S, J)$.

Proof.
In the exact sequence [3, Theorem 13.20 and Example 13.22]
$$K_2(S/J) \rightarrow SK_1(S, J) \rightarrow SK_1(S) \rightarrow SK_1(S/J)$$
the last term, namely $SK_1(S/J)$, is trivial since $S/J$ is Artinian and hence of stable range $1$. In addition, the image of $K_2(S/J)$ in $SK_1(S, J)$ is also trivial. Indeed $K_2(S/J)$ is generated by the Steinberg symbol $\{-1 + J, -1 + J\}_{S/J}$ because $S/J\simeq \mathbf{Z}/n \mathbf{Z}$ [3, Exercises 13A.10 and 15C.10]. As $\{-1, -1\}_S$ is a lift of the previous symbol in $K_2(S)$ our claim follows and the proof is now complete. 
By the previous lemma we have $SK_1(A) = SK_1(A, I)$. Since the inclusion $A \subset R$ induces an epimorphism from $SK_1(A, I)$ onto $SK_1(R, I)$, we deduce that $SK_1(A) \neq 1$.
It is actually possible to find a non-trivial element in $SK_1(A)$ by considering the power residue symbol $\binom{12}{1 + 4i}_2 = -1$ [2, Section 11C]. This yields an non-trivial Mennicke symbol $[12, 1 + 4i]$ and eventually the non-reducible row $(12, 1 + X, X^2 + 16)$ using the ring epimorphism induced by $X \mapsto 4i$.
Erratum. I claimed erroneously in former versions of this answer that the same technique easily applies to $\mathbb{Z}[X^{\pm 1}]$. My conclusions were wrong and I fail to answer the following


Question. 
    Is the stable rank of $\mathbb{Z}[X^{\pm 1}]$ equal to $3$?



[1] H. Bass, "K-theory and stable algebra", 1964.
[2] F. Grunewald et al., "On the groups $SL_2(\mathbf{Z}[x])$ and $SL_2(k[x,y])$", 1994.
[3] B. Magurn, "An algebraic introduction to K-theory", 2002.<|endoftext|>
TITLE: Heuristic for Montgomery's conjecture
QUESTION [6 upvotes]: This is my third question on this site regarding Montgomery's conjecture -- and I apologize
if this is too much -- but I am still not understanding well why this conjecture is believed to be true. 
The conjecture I am talking about is as follows (I am giving the slightly corrected version of Freidlander-Granville). 
et $q>1$ be an integer, $a$ an integer coprime to $q$, $\psi(q,a,x) = \sum_{p^\alpha < x, p^\alpha \equiv a \pmod{q}} \Lambda(n)$. Then:
Conjecture for $x>q$, one has $\psi(x,q) = \frac{x}{\phi(q)} + O(x^{1/2+\epsilon} q^{-1/2})$, with an implied constant depending only on $\epsilon$.  
In his answer to my preceding question, Matt Young gives the following heuristic:
For $\chi$ a non-principal Dirichlet character of $(\mathbb Z/q\mathbb Z)^\ast$, one has under GRH
$\psi(\chi,x) = O(x^{1/2+\epsilon})$. Now $\psi(q,a,x)$ is the arithmetic average of the (approximately $q$) terms $\chi^{-1}(a) \psi(\chi,x)$, and if those (weighted) terms are in random position one should expect their sum to have a norm roughly $q^{1/2}$ the norm of the individual term (by Einstein's Brownian motion theorem if you like), giving the conjecture.
This heuristic helped me a lot then but now I'd like to go further. 

Is there a good reason to believe that the $\chi^{-1}(a) \psi(\chi,x)$ are kind of randomly distributed, for every $a$ relatively prime to $q$? Could this expectation have some link,
  through the explicit formulas, with the conjectures on the position of zeros on the critical line of Dirichlet L-functions (assuming GRH)?

REPLY [8 votes]: This is not exactly what you asked, but I believe that there is a simpler heuristic for Montgomery's conjecture which does not involve characters.
Let $A$ be the arithmetic progression $a\pmod q$, where $a$ and $q$ are coprime to each other. We consider a sequence of Bernoulli random variables $(X_n)_{n\in A}$ such that
$$\textbf{Prob}(X_n=1)=\frac{q}{\phi(q)\log n} \quad\text{and}\quad \textbf{Prob}(X_n=0)=1-\frac{q}{\phi(q)\log n}.$$
The event $X_n=1$ models the event that $n$ is prime. (There are about $\le x/q$ elements in $A$ up to $x$ and about $x/(\phi(q)\log x)$ of them are primes.) Let $\Psi(x)=\sum_{n\le x} X_n\log n$. This is a random model for $\psi(x)$ (or rather $\theta(x)=\sum_{p\le x}\log p$, but this is merely a technicality). Note that
$$
\mathbb{E}[\Psi(x)] = \frac{q}{\phi(q)}\sum_{n\le x,\,n\equiv a\pmod q}1 
=  \frac{q}{\phi(q)}(x/q+O(1))=\frac{x}{\phi(q)} + O(q/\phi(q))
$$
and
$$
\text{Var}[\Psi(x)] = \frac{q}{\phi(q)} \sum_{n\le x,\, n\equiv a\pmod q}\log n 
\sim \frac{x\log x}{\phi(q)}.
$$
The Central Limit Theorem implies that the random variable $(\Psi(x) - x/\phi(q))/\sqrt{x\log x/\phi(q)}$ tends to the standard normal distribution. In  particular, almost surely as $x\to\infty$, we have that
$$
\left|\Psi(x) - \frac{x}{\phi(q)} \right| \ll x^\epsilon \sqrt{\frac{x\log x}{\phi(q)}}.
$$
(In fact, one should expect this to become accurate as soon as $x/q$ becomes a bit large, that is to say there are enough summands in $\Psi(x)$.) So we deduced Montgomery's conjecture for $\Psi(x)$, the random model of $\psi(x)$.<|endoftext|>
TITLE: Are there small examples of non-pivotal finite tensor categories?
QUESTION [16 upvotes]: I'm looking for small concrete examples of non-pivotal finite tensor categories to do some calculations with.
Here a finite tensor category is, according to Etingof-Ostrik, a rigid monoidal category whose underlying category is equivalent to the category of finite dimensional modules over a finite dimensional algebra.  A tensor category is pivotal if there's an isomorphism of tensor functors between the identity functor and the double dual functor.
The example I was able to find in the literature is the category of representations of a 72-dimensional Hopf algebra in Remark 2.11 of Andruskiewitsch-Angiono-Iglesias-Torrecillas-Vay.  (In fact, they give three such examples.)  But I was hoping for a smaller example.
Bonus question: What I would really like is an example of a category where not only is the double dual functor nontrivial, but there's no invertible object X such that the double dual is isomorphic as a tensor functor to conjugation by X.  (Note that the 72 dimensional Hopf algebras mentioned above do not give counterexamples, because their duals are pivotal.)
Easier question: I'd also love to hear about any other examples of non-pivotal finite tensor categories beyond the three from AAITV, even if they're not smaller.

REPLY [3 votes]: Monoidal categories $\mathcal{C}$ which admit a monoidal natural isomorphism $\Phi_X:X^{**} \rightarrow \beta\otimes X\otimes \beta^*$ are called $\textit{quasi-pivotal}$ . For $\mathcal{C}=$Rep$(H)$ with $H$ a finite dimensional Hopf algebra, quasi-pivotal structures on $\mathcal{C}$ are in bijection with pairs $(l,b)$ where $l$, $b$ are group-like elements in $H$, $H^*$ respectively satisfying $$S^2(h)=b(h_3)\;b^{-1}(h_1)\; l\;h_2 \; l^{-1} \;\;\;\; \text{for all}\;\; h\in  H.$$
Such a pair $(l,b)$ is called a $\textit{pair in involution}$. The paper Generalized Taft algebras and pairs in involution, constructs a family of Hopf algebras that do not admit a pair in involution, thereby providing an answer to the bonus question.

"Book Hopf algebras" provide an example of Hopf algebras that admit a quasi-pivotal structure but don't always admit a pivotal structure. See this paper for further details.

Furthermore, by a result of Shimizu, a quasi-pivotal structure on $\mathcal{C}$ yields a pivotal structure on $\mathcal{Z}(\mathcal{C})$. Thus, the Drinfeld doubles $D(H)$ of the Hopf algebras admitting a pair in involution (discussed above) provide examples of pivotal Hopf algebras.<|endoftext|>
TITLE: Hodge decomposition in Minkowski space
QUESTION [16 upvotes]: This question is motivated by the physical description of magnetic monopoles. I will give the motivation, but you can also jump to the last section.
Let us recall Maxwell’s equations: Given a semi-riemannian 4-manifold and a 3-form $j$. We describe the field-strength differential form $F$ as a solution of the equations
$\mathrm{d}F=0$
$\mathrm{d}\star F=j$ (where $\star$ denotes the Hodge star).
If the second de-Rham-cohomology vanishes (for example in Minkowski space), $F$ is exact and we can write it as $F=\mathrm{d}A$, where $A$ denotes a 1-form.
Now let us consider monopoles: We use two 3-forms $j_m$ (magnetic current) and $j_e$ (electric current) and consider the equations
$\mathrm{d}F=j_m$
$\mathrm{d}\star F=j_e$.
Essentially, it is described in this paper, but the author Frédéric Moulin (a physicist) uses coordinates. Now he assumes that (in Minkowski space) $F$ can be decomposed using two potentials — into an exact (in the image of the derivative) and a coexact (in the image of the coderivative) form: $F=\mathrm{d}A-\star\mathrm{d}C$. Is there a mathematical justification for this assumption (maybe it is just very pragmatic)?
The actual question
Given a 2-form $F$ on 4-dimensional Minkowski space (more generally: semi-riemannian manifolds)—are there any known conditions such that $F$ decomposes into an exact and a coexact form: $F=\mathrm{d}A+\star\mathrm{d}C$)?
For compact riemannian manifolds there is the well-known Hodge decomposition: There is always a decomposition into an exact, a coexact and a harmonic form. In the non-compact case you might be able to get rid of the harmonic form by only considering “rapidly decaying” forms (Wikipedia suggests that, but I do not have a good reference, in euclidean space there is the Helmholtz decomposition, and non-trivial (smooth) harmonic 1-forms do not vanish at infinity).
That is why I also ask: Are there “rapidly decaying” harmonic 2-forms in Minkowski space? Any references where I could see what is known about harmonic forms and Hodge theory in the semi-riemannian case are also welcome.

REPLY [11 votes]: Willie's answer is of course correct. But the part that I think is most interesting in the physical context is only briefly mentioned in point 3. of the last paragraph. Let me expand on that.
The relevant condition that make everything work nicely is that the background Lorentzian spacetime be globally hyperbolic. There are two important spaces of forms, $\Omega^p_0$ and $\Omega^p_{SC}$, $p$-forms with compact and spacelike compact supports, respectively. A set $X$ is spacelike compact if it is contained in the causal influence set of a compact set $Y$, $X\subseteq J(Y)$.
On Lorentzian manifolds, the D'Alembertian operator $\square = d\delta + \delta d$ is the analog of the Laplacian operator $\Delta$ in Riemannian signature. Note that I'm using the notation $\delta = (-1)^p{\star d \star}$. It is not elliptic, but it is hyperbolic with very nice properties. Instead of the nice analytical properties of the elliptic Laplacian, we have the following exact sequence (the image of each map is equal to the kernel of the next one):
$$ 0 \to \Omega^p_0 \stackrel{\square}{\to} \Omega^p_0 \stackrel{G}{\to} \Omega^p_{SC} \stackrel{\square}{\to} \Omega^p_{SC} \to 0 , $$
where $G=G^+-G^-$ is the causal Green function, defined as the difference of the retarded ($G^+$) and advanced ($G^-$) Green functions. The retarded and advanced Green functions of the D'Alembertian always exist on Globally hyperbolic manifolds (see, for instance, here or here).
A version of the result that you want is this: if $F\in \Omega^2_{SC}$, $d F = 0$ and $\delta F = 0$, then there exist two forms $A\in \Omega^1_{SC}$ and $B\in \Omega^3_{SC}$, with $\delta A = 0$ and $d B = 0$, such that $F = d A + \delta B$. Moreover, there exist $\alpha \in \Omega^1_0$ and $\beta \in \Omega^3_0$ such that $A=G\alpha$ and $B=G\beta$. This result can be found, for instance, as Prop.2.2 here. I think you can straight forwardly adapt the proof to the case with non-vanishing currents $j_e$ or $j_m$.<|endoftext|>
TITLE: Is there an infinite depth irreducible finite index maximal subfactor (other than Temperley Lieb) ?
QUESTION [5 upvotes]: A subfactor $N \subset M$ is maximal if it admits no non-trivial intermediate subfactors $N \subset P \subset M$.   
Is there an infinite depth irreducible finite index maximal subfactor (other than Temperley-Lieb $A_{\infty} $) ?

REPLY [6 votes]: The infinite depth subfactor coming from SU(3) at any index above 9 gives an example.  Here the Q-system is $V_{(1,0)} \otimes V_{(0,1)} \cong V_{(1,1)} \oplus V_{(0,0)}$ so the only possible sub-objects are the whole thing or the trivial, so it's certainly maximal.<|endoftext|>
TITLE: Torelli type theorem for sextic threefolds
QUESTION [9 upvotes]: I have a question to you:
In the work of Rapoport and Deligne they classify smooth complete intersections with Hodge level 1, which are the following:

Complete intersection of two quadrics in $\mathbb{P}^{2n+3}$.
Complete intersection of three quadrics in $\mathbb{P}^{2n+4}$.
Cubic hypersurface in $\mathbb{P}^4$.
Complete intersection of a cubic and a quadric in $\mathbb{P}^5$.
Cubic hypersurface in $\mathbb{P}^6$.
Quartic hypersurface in $\mathbb{P}^4$

For the first three cases there are Torelli type theorems for their Intermediate Jacobians.
My question is: Do you know any reference or idea for a proof in the case of the complete intersection of a cubic and a quadric in $\mathbb{P}^5$ (which is a Fano solid of index 1, whose Intermediate Jacobian is a p.p.a.v. of dimension 20), or something similar?
Thanks a lot!

REPLY [2 votes]: For the cases which you are interested in there appears to be only partial results available in the literature. Namely, as special cases of more general results.
For quartic threefolds and cubic fivefolds one knows that a generic Torelli theorem holds. This is a special case of a generic Torelli theorem for certain hypersurfaces, due to Donagi [1].
I don't know of an analogue of Donagi's work for complete intersections, but here Flenner [2] has shown an infinitesimal Torelli theorem in certain cases. This in particular applies to the case of an intersection of a quadric and cubic.
Also, I believe that the Torelli theorem for intersections of two quadrics of odd dimension is actually due to Donagi [3], not Reid (at least I can't find this statement in Miles Reid's PhD thesis).
Good luck with this problem if you are working on it! If you make any progress, or know more than I do, please do contact me as I would be most interested.
References:
[1] - Donagi. Generic Torelli for projective hypersurfaces. Compositio Math. 50 (1983), no. 2-3, 325–353
[2] - Flenner. The infinitesimal Torelli problem for zero sets of sections of vector bundles. Math. Z. 193 (1986), no. 2, 307–322. 
[3] - Donagi. Group law on the intersection of two quadrics.
Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4) 7 (1980), no. 2, 217–239.<|endoftext|>
TITLE: How to make Schubert calculus into a Hopf (actually a PCH) algbera?
QUESTION [5 upvotes]: The parallels between the formulas in Schubert calculus and in the theory of the representations of symmetric groups (par Geissinger-Zelevinsky) are so apparent (e.g. Giambelli formula), that one must wonder how to directly define the co-multiplication on Schubert (co)cells.

REPLY [2 votes]: If you're talking about Grassmannians, then you can use the direct sum maps
$Grass(k,n) \times Grass(k',n') \to Grass(k+k', n+n')$
to get the comultiplication. This induces a map on cohomology the other way. Then take the limit as $n,n' \to \infty$ and then take the limit $k,k' \to \infty$. This fixes two problems: 1) the Grassmannians aren't the same in the finite case, and 2) the values of $k,n$, etc. give truncations of the ring of symmetric functions, so you need to remove that restriction.
According to Symmetric polynoms are Hopf algebra ? What for one needs co-product ? the bialgebra is enough to get the whole Hopf structure.
Positivity (to get the PSH algebra structure) follows from geometric considerations (i.e., all structure coefficients are intersection numbers).<|endoftext|>
TITLE: One question on cup product and torsion elements
QUESTION [10 upvotes]: As we know, by universal coefficient theorem, $H^{1}(X,\mathbb{Z})$ is torsion-free. My question is: 
for cup product $H^{1}(X,\mathbb{Z})\otimes H^{1}(X,\mathbb{Z})\rightarrow H^{2}(X,\mathbb{Z})$
could $a\cup b$ be a torsion element in $H^{2}(X,\mathbb{Z})$.

REPLY [3 votes]: You can also make an example involving a closed surface. Let $X$ be the connected sum of a torus $T$ and a projective plane, and let $f:X\to T$ be nontrivial on $H^2$. Two elements of $H^1(T)$ whose cup product is nonzero mod $2$ will pull back by $f$ to two elements of $H^1(X)$ whose cup product generates $H^2(X;\mathbb Z/2)=\mathbb Z/2$.<|endoftext|>
TITLE: Bound on the order of a finite group generated by elements $a$ and $b$ of order 2 and $n \geq 3$ such that the sum of the images of $a$, $b$ and $b^{-1}$ under any ordinary representation has only rational eigenvalues
QUESTION [5 upvotes]: Assume that $G = \langle a, b \rangle$ is a finite non-abelian group which is generated by an
involution $a$ and an element $b$ of order $n$ ($n\geq 3$) such that for every (complex) representation
$\varphi$ of $G$ the matrix $\varphi(a) + \varphi(b) + \varphi(b^{-1})$ has only rational eigenvalues.

Question: Is there an upper bound on the order of $G$?

REPLY [2 votes]: If you are looking for a more abstract proof based on representation and group theory, I suggest you to work on a more general question as follows:
Assume that $G=\langle S\rangle$  is a finite  group which is generated by $S=S^{-1}$ ($1\not\in S$) such that for every complex representation $\phi$  of G  the matrix $\sum_{s\in S} \phi(s)$  has only rational eigenvalues. Then is $|G|$  bounded above by a function of $|S|$?
The answer to the above question is positive and a crude known bound is
$\frac{|S|(|S| − 1)^{2|S|} − 2}{|S| − 2}$.
Note that, what you need by the representation theoretic assumption is that the  eigenvalues of the linear transformation $T=\sum_{s\in S}s$ on the vector space $\mathbb{C}(G)$ are all rational (and so integer). Note that $T$ is an element of the group ring $\mathbb{C}(G)$.
You may find some related graph theoretic results in the following paper:
Alireza Abdollahi and E. Vatandoost, Which Cayley graphs are integral?, The Electronic Journal of Combinatorics 16 (2009), #R122.
Sorry for the self-promotion!<|endoftext|>
TITLE: What analysis should I know for studying Arakelov Theory?
QUESTION [9 upvotes]: Hi!
I have a fairly good background in Algebraic Geometry (say at the level of Hartshorne's book and some Intersection Theory from Fulton) and since I think working over $\text{Spec } \mathbb{Z}$ is fun, I would like to learn some Arakelov Theory.
My background in differential geometry and analysis is not that good, though - I know basic definitions in both fields and have taken some courses, but I have forgot a lot, and what more, I seem to need complex differential geometry, which I have never studied. From what I understand, residuce currents is important in Arakelov Theory. 
So my question is:
What books, what articles should I read to get a good analytical / complex differential geometric background (covering for example, residue currents) sufficient to study Arakelov Theory?

REPLY [2 votes]: I think the amount of analysis you need to know is fairly modest. If you know distribution theory and some introductory material on pseudo-differential operators (say first half of Shubin), you should be right to go. In particular there is no "hard analysis" being involved in the literature I went through. While "soft analysis" is by no means easy, it has an algebraic flavor somewhat similar to algebraic geometry. 
If you are reading papers, it might make more sense to learn material (like Sobolev embedding) and prove things yourself on the spot than reading through big books to have a decent understanding of the subject. For index theory: I think Faltings-Zhang's book has a nice section on the application of local index theorem to Arakelov theory. Since you asked the question as an undergraduate I imagine you might be very familiar with the background already. Good luck!<|endoftext|>
TITLE: Fredholm alternative result for general elliptic system?
QUESTION [7 upvotes]: Now I have known that Fredholm alternative result is valid for the strong elliptic system. But I'm not sure that is it still valid for the general elliptic system, in which the second-order heading coefficient matrix $A(x)$ is only positive definite, rather than strong elliptic type $$A(x)\xi\cdot\xi\geq\lambda|\xi|^2,\quad\forall\xi\in\mathbb{R}^{mn},\quad a.e. x\in\Omega$$
Any answer and reference will be appreciated!

REPLY [2 votes]: To complement Andrew's answer, Fredholm alternative holds for properly elliptic systems (with complementing boundary conditions). In 3 or more dimensions, any elliptic system in the sense of Douglis-Nirenberg is properly elliptic. Note that this is much more general than strongly elliptic systems. Strongly elliptic systems (with appropriate boundary conditions) have index 0, while a general elliptic system can have a nonzero index. Please see Agmon-Douglis-Nirenberg '64.<|endoftext|>
TITLE: Class forcing: Pelletier vs Friedman
QUESTION [8 upvotes]: [Apologies in advance for a fluffy question]
I'm reading this old paper by Pelletier, where he gives a Boolean-valued model version of class forcing, assuming that the Boolean algebra in question can be written as an $Ord$-indexed increasing union of CBAs $B_\alpha$. The model is given by induction, as usual, with a twist:

$^RV^B_0 = \emptyset$
$^RV^B_{\alpha+1} = \lbrace u \in B_{\alpha+1}^{dom(u)} |\ dom(u) \subseteq {}^RV^B_\alpha\rbrace$
$^RV^B_\lambda = \bigcup_{\alpha \lt \lambda} {}^RV^B_\alpha$ for $\lambda$ limit.
$^RV^B = \bigcup_{\alpha \in Ord} {}^RV^B_\alpha$

So in a sense, our hierarchy is restricted as to what can appear at the $\alpha$th stage. Given a certain condition on $B$ called ARP (I can supply details if desired), he shows that $^RV$ is a model of ZFC-Powerset. Another condition on $B$ gives us Powerset, and all this works in the usual Easton situation (ARP seems to be related to Easton support, but I may be wrong).
On the other hand, Friedman (although I'm reading his book), given a proper class $P$ of conditions defines names and interpretations as per usual for an ordinary forcing with no restrictions to get $M[G]$ (here $M$ satisfies a relativised $V=L$). He then shows that with a pretameness condition on P - that every set-indexed collection of dense (definable-with-parameters) classes in $P$, at the cost of passing to a $q\leq p$, there are subsets of each class whose down-closure is dense below $q$ - we have that $M[G]$ is a model of ZF(C)-Powerset. Then with an additional tameness condition on $P$, which refers to the  cumulative hierarchy of the base model $M$, he shows $M[G]$ is a model of ZF(C).
What I want to know is how these two approaches can get away with introducing the stratification in the two different parts of constructing the new, class-forced model. Pelletier uses ARP to show that $|| - ||$ is well-defined (so not having to quantify over classes) using his stratification of $B$, then the other condition to show that powerclasses are sets. Friedman uses pretameness to show $\Vdash$ is definable and then proves the relevant axioms hold, then uses tameness, and its condition using $V_\alpha^M$ to get powersets.
Perhaps it is just the different approaches of BVM and forcing using conditions, but I'm trying to take a third approach, and seeing where the stratification restriction is used seems to be crucial for what I need.
EDIT:  Then again, Jech just says to form the BVM $V^{B_\alpha}$ for each $\alpha$ and then take the union of them all, so this stratifies in yet another way.

REPLY [4 votes]: I think the last paragraph of Joel Hamkins's answer provides a useful way to think about these sorts of forcing.  Imagine that you have an inaccessible $\kappa$ (or, perhaps better, a $\kappa$ such that $V_\kappa$ is an elementary submodel of the universe) and that you're forcing with a poset $\mathbb P\subseteq V_\kappa$.  You get a forcing extension of the universe (since $\mathbb P$ is a mere set in this picture), but your objective is to get a ZFC model by restricting this model to ranks $<\kappa$.  Several issues arise.  You need to decide what "restricting to ranks $<\kappa$" means.  It could mean taking $V_\kappa$ in the forcing extension.  It could mean taking those elements of the forcing extension that have names (in the ground model) of ranks $<\kappa$.  I've seen both versions, and I don't think they're equivalent.
If you want to use Boolean-valued models, then the complete Boolean algebra $\mathbb B$ associated to $\mathbb P$ is one level higher in the cumulative hierarchy than $\mathbb P$, so you need to reconsider anything involving ranks of names.
These issues are "formal" in the sense that they concern how you want to set up the machinery.  But there are also some substantive issues that you must face no matter how you set up the machinery.  For example, as Joel mentioned, you'd better make sure you haven't added $\kappa$ new reals, as they would ruin ZFC restricted to ranks $<\kappa$.  (Whether they ruin the power set axiom or the replacement scheme seems to depend on how you choose to "restrict to ranks $<\kappa$".)<|endoftext|>
TITLE: Semidirect product decomposition of the Borromean rings group
QUESTION [15 upvotes]: Let $X=S^3\setminus B$ be the link complement of the Borromean rings.
    (source)
Then $G=\pi_1(X)$ has a presentation of the form
$$
G = \langle \; a,b,c \mid [a,[b^{-1},c]],\; [b,[c^{-1},a]], \; [c,[a^{-1},b]]\;\rangle,
$$
where any one of the relations is redundant.
Removing any component of the link results in a trivial link of two components. The effect on the fundamental group is to send one of the generators, say $c$, to $1$, resulting in a homomorphism $G\to F_2$ to the free group on $2$ generators. There results a split extension
$$
1\to K\to G\to F_2\to 1.
$$

Is the kernel $K$ a free group?

REPLY [5 votes]: I believe that $K$ is an infinitely generated free group.  First we find an infinite presentation $K = \langle X \mid R \rangle$, and then we argue that we can pick an infinite subset of $X$ that freely generates.
Presentation: Note that we can obtain the Borromean rings from the handlebody of genus two by drilling out the commutator curve and attaching a two-handle to the boundary along the usual separating curve.  Since the attaching curve for the two-handle is trivial in the handlebody, we can take the universal cover of the handlebody and then attach infinitely many two-handles, and drill infinitely many lines.  After drawing a few pictures, we can organize all of this information as follows. 
Let $D^2$ be the Poincare disk model for the hyperbolic plane.  Pick hyperbolic isometries $a$ and $b$ that have perpendicular axes and equal (very long) translation lengths.  Thus $F_2 = \langle a,b \rangle$ is a free group of rank two.  Let $T$ be the regular four-valent tree isometrically embedded in $H^2$, arising as the Cayley graph of $F_2$.  Let $v_0$ be the vertex of $T$ corresponding to the identity of $F_2$.  The universal cover of the handlebody above is obtained as a three-dimensional neighborhood of $T$, and $F_2$ is the deck group. 
We define $X = \{ x_q \}$ where $q$ ranges over the connected components of $H^2 - T$.  (These correspond to the drilled lines.)  This gives the generating set for $K$ subject to the relations $R = \{ x_p X_q x_r X_s \}$ where the regions $p,q,r,s$ are arranged counter-clockwise about a vertex of $T$. (These correspond to the attached two-handles.  Because we started with the commutator curve, the two-handles are attached to the neighborhood of $T$ via a "baseball curve" around each vertex.)  Here we use the convention that $X_q = (x_q)^{-1}$.  (Also, $p$ is always to the "north-east" of $v$ - we use translates of the picture about $v_0$ to set conventions.)  For future use, let $w_v$ be the relation coming from vertex $v$. 
Subset that freely generates:  We recursively color the regions of $H^2 - T$ as follows.  (1) All regions start white. (2) In the first step, we pick two regions $p, q$ adjacent to $v_0$.  Color both of $p$ and $q$ black.  (3) In general, pick a vertex $v$ having exactly two white regions $p, q$ adjacent to $v$.  Note that $p$ and $q$ must be adjacent to each other.  Color $p$ black and color $q$ grey. 
The set of black regions gives a subset $X' \subset X$ of the generators.  It is an exercise to prove inductively that $X'$ generates.  Next, for a contradiction, suppose that $w$ is a word in the $X'$ that is trivial in $K$.  Thus we can write $w$ as a product of conjugates of relations from $R$.  Recall that relations are in bijection with the vertices of $T$.  Pick one relation $w_v$ appearing in the product so that $v$ is as far as possible from $v_0$.  Of the four regions adjacent to $v$, there are two regions $p, q$ that are combinatorially further than the others from $v_0$.  Say $p$ and black and $q$ is grey.  Then $x_q$ is not freely cancelled in the product of conjugates, a contradiction. QED<|endoftext|>
TITLE: Principal series of finite group of Lie type
QUESTION [7 upvotes]: I have a naive question on complex representations of finite groups of Lie type.
Let $\bf G$ be a reductive group (say connected, with connected center, for safety)
defined over a finite field $\mathbb F_q$, and let $G={\bf G}(\mathbb F_q)$. Assume first that $G$ is split over $\mathbb F_q$, so there is a maximal torus $\bf T$ in $\bf G$ defined and split over $\mathbb F_q$. Let $g \in T = {\bf T}(\mathbb F_q)$ be a regular element.

Then is it true that if $\pi$ is an irreducible complex representation of $G$, of character $\chi = tr\  \pi$, then $\chi(g) \neq 0$ implies $\pi$ is a principal series ?

By a principal series, one means of course a sub representation of $Ind_B^G \psi$, where $B$ is a Borel containing $T$ and $\psi: T \rightarrow \mathbb C^\ast$ a one-dimensional character. 
A glance at the table of characters found for example in Fulton-Harris shows that this is true
for $GL_2$, so my intuition tells me that it should be true, and probably easy, in general,
but the argument escapes me and I haven't seen this statement in the literature (e.g. in
Curtis' survey or Digne and Michel's book). 
Also if the answer is yes, then can one remove the assumption that $G$ is split (that is
we work with a general $\bf G$ as above, and take for $\bf T$ a quasi-split group, and ask the same question) ?

REPLY [5 votes]: So, I think the answer to your question is yes. This may not be the slickest proof but I think it works. Firstly let $\mathrm{pr}_G$ be the projection map from the space of all class functions to the subspace of uniform functions, i.e. the subspace spanned by Deligne-Lusztig virtual characters, (see Digne and Michel - 12.11 and 12.12). Now for a pair $(\mathbf{S},\theta)$ consisting of an $F$-stable maximal torus and character $\theta$ of $\mathbf{S}^F$ let $c_{\mathbf{S},\theta} \in \mathbb{C}$ be such that
$$\mathrm{pr}_G(\chi) = \sum_{}c_{\mathbf{S},\theta}R_{\mathbf{S}}^{\mathbf{G}}(\theta)$$
where this sum is taken up to conjugacy. The characteristic function of a semisimple element is a uniform class function, (see DM - Proposition 12.20), in particular we have $(\mathrm{pr}_G(\chi))(g) = \mathrm{pr}_G(\chi(g))$. Hence if $\chi(g)$ is non-zero this implies that
$$\mathrm{pr}_G(\chi)(g) = \sum_{}c_{\mathbf{S},\theta}R_{\mathbf{S}}^{\mathbf{G}}(\theta)(g) \neq 0$$
However $R_{\mathbf{S}}^{\mathbf{G}}(\theta)(g) \neq 0$ implies that the conjugacy class of $g$ intersected with $\mathbf{S}^F$ is non-empty (see Carter - Proposition 7.5.3). As your element is regular this implies that $\mathbf{S}$ is rationally conjugate to your chosen torus $\mathbf{T}$. Now $c_{\mathbf{S},\theta} \neq 0$ implies that $\chi$ occurs in $R_{\mathbf{S}}^{\mathbf{G}}(\theta)$ with non-zero multiplicity, hence in $R_{\mathbf{T}}^{\mathbf{G}}(\theta')$ with non-zero multiplicity where $(\mathbf{T},\theta')$ is any pair rationally conjugate to $(\mathbf{S},\theta)$. However this says that $\chi$ is in the principal series as $R_{\mathbf{T}}^{\mathbf{G}}(\theta')$ is simply the Harish-Chandra induction of $\theta'$.
Note that this argument does not require that $F$ is split.<|endoftext|>
TITLE: Are there Carlitz analogues of quadratic residues and reciprocity?
QUESTION [5 upvotes]: Let $q$ be a prime power. I will use the notations of Keith Conrad's Carlitz extensions paper (but I'll work over $\mathbb{F}_q$ rather than $\mathbb{F}_p$).
The most general question I'm asking here is whether there is anything resembling the classical theory of quadratic residues (including results like quadratic reciprocity, supplementary laws, Gauss sums, Gauss' lemma etc.) in the context of Carlitz polynomials -- i. e., over $\mathbb{F}_q\left[T\right]$ instead of $\mathbb Z$, with "squaring" replaced by application of $\left[N\right]$ for some fixed polynomial $N\in\mathbb{F}_q\left[T\right]$. (This is in line with the usual idea that $\mathbb{F}_q\left[T\right]$ is an analogue of $\mathbb Z$, and applying a Carlitz polynomial in the former ring is like taking a power in the latter.)
Of course, unlike $\mathbb Z$, where every prime apart from $2$ is $\equiv 1 \mod 2$, there is no irreducible $N\in\mathbb{F}_q\left[T\right]$ such that all but finitely many monic irreducibles in $\mathbb{F}_q\left[T\right]$ are $\equiv 1\mod N$. So there is no obvious analogue of squaring that would mirror the "the product of two nonsquare residues is a square residue" property. But there is still an irreducible in $\mathbb{F}_q\left[T\right]$ which, in some sense, is simpler than the others: namely, $T$. Whenever $\pi\in\mathbb{F}_q\left[T\right]$ is an irreducible such that $\pi\equiv 1\mod T$, the elements $u$ of $\mathbb{F}_q\left[T\right] / \pi$ which can be written as $\left[T\right]\left(v\right)=v^q+Tv$ for $v\in\mathbb{F}_q\left[T\right] / \pi$ form an $\mathbb{F}_q\left[T\right]$-submodule of the Carlitz module $C\left(\mathbb{F}_q\left[T\right] / \pi\right)$. They can be viewed as the elements annihilated by $\left[\dfrac{\pi-1}{T}\right]$.
My more concrete question is in how far they share properties with quadratic residues in $\mathbb Z$. I don't dare formulate any conjectures (lacking computational data and number-theoretical intuition), but one could ask how the property of a monic irreducible $\phi$ to be a "$T$-quadratic residue" modulo another $\psi$ correlates with the same in the other direction.
(Fun fact: From the cyclicity of the Carlitz module $C\left(\mathbb{F}_q\left[T\right] / \pi\right)$, it follows immediately that if $\pi$ is a monic irreducible in $\mathbb{F}_q\left[T\right]$ satisfying $\pi\equiv 1\mod T$, then $-T$ is a $p-1$-th power in $\mathbb{F}_q\left[T\right]$. Of course, this also follows from Hilbert's theorem 90 or cyclicity of the group $\left(\mathbb{F}_q\left[T\right] / \pi\right)^\times$.)
If $T$ is a bad (because not invariant under $\mathbb{F}_q$-automorphisms or for whatever other reason) analogue of $2$, we might consider $T^q - T$ instead -- it doesn't appear like irreducibility is important here...

REPLY [14 votes]: In the context of Carlitz polynomials, the analogue is the classical quadratic reciprocity law in ${\mathbf F}_q[T]$ (for odd $q$, of course). For a monic irreducible $\pi$ in ${\mathbf F}_q[T]$, and any $A$ in ${\mathbf F}_q[T]$, let $(\frac{A}{\pi})$ be $1$ if $A \equiv \Box \bmod \pi$ and $A \not\equiv 0 \bmod \pi$, $-1$ if $A \not\equiv \Box \bmod \pi$, and $0$ if $A \equiv 0 \bmod \pi$. Then the main law of quadratic reciprocity in ${\mathbf F}_q[T]$ is that for distinct monic irreducible $\pi$ and $\widetilde{\pi}$, 
$$
\left(\frac{\widetilde{\pi}}{\pi}\right) = (-1)^{({\rm N}\pi-1)/2 \cdot ({\rm N}\widetilde{\pi}-1)/2}\left(\frac{\pi}{\widetilde{\pi}}\right),
$$
where ${\rm N}(f) := |{\mathbf F}_q[T]/(f)| = q^{\deg f}$. This is due to Dedekind, and can be found in Mike Rosen's Number Theory in Function Fields with a proof that is much simpler than quadratic reciprocity in $\mathbf Z$ and doesn't mention Carlitz polynomials at all. But it can be explained using Carlitz extensions of ${\mathbf F}_q(T)$ in exactly the same way quadratic reciprocity in $\mathbf Z$ can be explained using cyclotomic extensions of ${\mathbf Q}$.
Let's recall first what is done in the case of the integers. For an odd prime $p$ the cyclotomic extension ${\mathbf Q}(\zeta_p)$ has a cyclic Galois group over $\mathbf Q$ of order $p-1$ and thus it contains exactly one quadratic extension of $\mathbf Q$ (corresponding to the subgroup of squares in the Galois group). This quadratic extension is ${\mathbf Q}(\sqrt{p^*})$, where $p^* = (-1)^{(p-1)/2}p$.  For any odd prime $q$ other than $p$, we will compute the Frobenius element $\text{Frob}_q({\mathbf Q}(\sqrt{p^*})/{\mathbf Q})$ in two ways. We will interpret $\text{Gal}({\mathbf Q}(\sqrt{p^*})/{\mathbf Q})$ as $\{\pm 1\}$, which can be done in a unique way (all groups of order 2 are uniquely isomorphic to each other), so the Frobenius element is 1 or $-1$.
(1) By knowledge of how primes split in quadratic extensions of ${\mathbf Q}$, $q$ splits in ${\mathbf Q}(\sqrt{p^*})$ iff $X^2 - p^*$ splits mod $q$, so this Frobenius element is $(\frac{p^*}{q})$.
(2) From the standard isomorphism of $\text{Gal}({\mathbf Q}(\zeta_p)/{\mathbf Q})$ with $({\mathbf Z}/(p))^\times$, $\text{Frob}_q({\mathbf Q}(\zeta_p)/{\mathbf Q})$ is $q \bmod p$. The restriction of $\text{Frob}_q({\mathbf Q}(\zeta_p)/{\mathbf Q})$ to ${\mathbf Q}(\sqrt{p^*})$ is $\text{Frob}_q({\mathbf Q}(\sqrt{p^*})/{\mathbf Q})$, and the image of $q \bmod p$ under the natural restriction map $\text{Gal}({\mathbf Q}(\zeta_p)/{\mathbf Q}) \rightarrow \text{Gal}({\mathbf Q}(\sqrt{p^*})/{\mathbf Q})$ is $(\frac{q}{p})$. Therefore $\text{Frob}_q({\mathbf Q}(\sqrt{p^*})/{\mathbf Q}) = (\frac{q}{p})$.
By (1) and (2), $(\frac{p^*}{q}) = (\frac{q}{p})$, and this is the main law of quadratic reciprocity after we recall that $p^* = (-1)^{(p-1)/2}p$ and use the supplementary law $(\frac{-1}{q})  = (-1)^{(q-1)/2}$.
Now let's turn to the Carlitz setting. Pick a monic irreducible $\pi$ in ${\mathbf F}_q[T]$, where $q$ is odd.  Let $\Lambda_\pi$ be the set of roots of the Carlitz polynomial $[\pi](X)$. Set $K = {\mathbf F}_q(T)$, so $K(\Lambda_\pi)/K$ is a Galois extension with Galois group isomorphic to $({\mathbf F}_q[T]/\pi)^\times$ by using the action of Carlitz polynomials on $\Lambda_\pi$. Any monic irreducible $\widetilde \pi$ other than $\pi$ is unramified in $K(\Lambda_\pi)$, and $\text{Frob}_{\widetilde{\pi}}(K(\Lambda_\pi)/K) = \widetilde{\pi} \bmod \pi$.`
Letting $d = \deg \pi$, the group $({\mathbf F}_q[T]/\pi)^\times$ is cyclic of odd order $q^d-1$, so it has a unique subgroup of index 2 (the subgroup of squares). Therefore $K(\Lambda_\pi)$ has a unique quadratic subextension of $K$ inside it. It turns out to be  $K(\sqrt{\pi^*})$, where $\pi^* = (-1)^{({\rm N}\pi-1)/2}\pi$.`  (EDIT: This is explained below.)
If you run through the above proof of quadratic reciprocity in $\mathbf Z$ using the analogous constructions in the Carlitz setting that I describe above, then you'll obtain $$\left(\frac{\pi^*}{\widetilde{\pi}}\right) = \left(\frac{\widetilde{\pi}}{\pi}\right)$$ in exactly the same way that one obtains $(\frac{p^*}{q}) = (\frac{q}{p})$, and the main law of quadratic reciprocity in ${\mathbf F}_q[T]$ is an unraveling of this equation once you recall the definition of $\pi^*$ and use the supplementary law $(\frac{-1}{\widetilde{\pi}}) = (-1)^{({\rm N}\widetilde{\pi}-1)/2}$.
EDIT: I wrote in passing above that $K(\Lambda_\pi)$ contains a square root of $\pi^* := (-1)^{({\rm N}\pi-1)/2}\pi$. This is analogous to ${\mathbf Q}(\zeta_p)$ containing a square root of $p^* = (-1)^{(p-1)/2}p$, but explaining the containment is one place where simple analogies between ${\mathbf Q}$ and ${\mathbf F}_q(T)$ can break down.  Classically there are several ways of showing $\sqrt{p^*}$ lies in ${\mathbf Q}(\zeta_p)$.
(1) Ramification. The unique quadratic field in ${\mathbf Q}(\zeta_p)$ ramifies only at $p$, and there turns out to be just one quadratic extension of ${\mathbf Q}$ ramified only at $p$.
(2) Gauss sums.  Define $G = \sum_{a \bmod p} (\frac{a}{p})\zeta_p^a$, which by construction lies in ${\mathbf Q}(\zeta_p)$. Show $G^2 = p^*$.
(3) Calculating a norm in a second way. Setting $X = 1$ in the identity $X^{p-1}+\cdots+X+1 = \prod_{i=1}^{p-1}(X-\zeta_p^i)$ gives us $p = \prod_{i=1}^{p-1} (1 - \zeta_p^i)$. This says $p = {\rm N}_{{\mathbf Q}(\zeta_p)/{\mathbf Q}}(1- \zeta_p)$. The product of the terms at $i$ and $p-i$ in the product is $-1$ up to a square factor (because $\zeta_p$ is a square of some $p$th root of unity). Therefore $p$ is $(-1)^{(p-1)/2}$ times a square in ${\mathbf Q}(\zeta_p)$, so $(-1)^{(p-1)/2}p$ is a square in the $p$th cyclotomic field.
If we try to adapt these methods to find the unique quadratic extension of $K = {\mathbf F}_q(T)$ inside $K(\Lambda_\pi)$, the first two do not work directly.
(1) There is not just one quadratic extension of $K$ ramified only at $\pi$ among the "finite" places (those places other than the place associated to $1/T$). One choice is $K(\sqrt{\pi})$ and another is $K(\sqrt{c\pi})$ where $c$ is a nonsquare in ${\mathbf F}_q^\times$. 
(2) If we define $G = \sum_{A \bmod \pi} (\frac{A}{\pi})[A](\lambda)$ for any fixed choice of nonzero $\lambda$ in $\Lambda_\pi$, then $G = 0$ if ${\rm N}(\pi) > 3$ (not if ${\rm N}(\pi) = 3$). To prove $G=0$ we exploit additivity of Carlitz polynomials: the coefficients $(\frac{A}{\pi})$ are $\pm 1$, so 
$$
\sum_{A \bmod \pi} \left(\frac{A}{\pi}\right)[A](\lambda) = 
\left[\sum_{A \bmod \pi}\left(\frac{A}{\pi}\right)A\right](\lambda), 
$$
and the polynomial inside the brackets on the right only matters mod $\pi$ since its Carlitz action is being applied to $\lambda$, a root of $[\pi](X)$.  Therefore the vanishing of $G$ is the same as the vanishing of $\sum_{A \bmod \pi} (\frac{A}{\pi})A \bmod \pi$, and Darij Grinberg gives a proof of that in his comments to this answer.  (When ${\rm N}(\pi) = 3$ the sum $G$ is $2\lambda \not= 0$.)
The idea of the third method does carry over to the Carlitz setting.  
(3) Starting from the factorization $[\pi](X)/X = \prod_{A \not\equiv 0 \bmod \pi} (X - [A](\lambda))$, set $X = 0$ (not $X = 1$: the roots of $[\pi](X)$ are more analogous to $1 - \zeta_p^i$ instead of to the $p$th roots of unity themselves) and get $\pi = \prod_{A \not\equiv 0 \bmod \pi} [A](\lambda)$. This says $\pi = {\rm N}_{K(\Lambda_\pi)/K}(\lambda)$.` The product of the terms at $A$ and $-A$ is $[A](\lambda)[-A](\lambda) = -[A](\lambda)^2$ because $[-A](X) = [-1]([A](X)) = -[A](X)$.  Therefore up to a square factor in $K(\Lambda_\pi)$, $\pi$ equals $(-1)^{({\rm N}\pi- 1)/2}$, so $(-1)^{({\rm N}\pi- 1)/2}\pi$ is a square in $K(\Lambda_\pi)$.<|endoftext|>
TITLE: The Jones-Sato-Wada-Wiens polynomial for prime numbers and differential calculus?
QUESTION [5 upvotes]: After works of Davis, Matijasevic, Putman and Robinson between 1960 and 1970, we know that every recursively enumerable set of numbers can be represented by a polynomial.
In particular, it's the case for the set of prime numbers.
In 1976, Jones-Sata-Wada-Wiens published (see here) the following polynomial $P$ of degree $25$ in $26$ variables $a$, $b$, $c$,..., $z$, admitting the property that $P(\mathbb{N}^{26})\cap \mathbb{N} = \mathbb{P}$, the set of prime numbers !
\begin{align}
P(a,b,...,z)=&\phantom-\,(k+2)\Big(1\\
       &-(wz+h+j-q)^2\\ 
       &-\left[(gk+2g+k+1)(h+j)+h-z\right]^2\\  
       &-[2n+p+q+z-e]^2\\ 
       &-\left[16(k+1)^{3}(k+2)(n+1)^2+1-f^2\right]^2\\ 
       &-\left[e^{3}(e+2)(a+1)^2+1-o^2\right]^2\\ 
       &-\left[(a^2-1)y^2+1-x^2\right]^2\\ 
       &-\left[16r^2y^{4}\left(a^2-1\right)+1-u^2\right]^2\\ 
       &-\left[\left(\left(a+u^2\left(u^2-a\right)\right)^2-1\right)(n+4dy)^2+1-(x+cu)^2\right]^2\\  
       &-\left[n+l+v-y\right]^2-\left[\left(a^2-1\right)l^2+1-m^2\right]^2\\ 
       &-\left[ai+k+1-l-i\right]^2\\ 
       &-\left[p+l(a-n-1)+b\left(2an+2a-n^2-2n-2\right)-m\right]^2\\ 
       &-\left[q+y(a-p-1)+s\left(2ap+2a-p^2-2p-2\right)-x\right]^2\\ 
       &-\left[z+pl(a-p)+t\left(2ap-p^2-1\right)-pm\right]^2\\
       &\Big)
\end{align}
In "The Book of Prime Number Records" Paulo Ribenboim reports:
"It should be noted that this polynomial also takes on negative values, and that a prime number may appear repeatedly as a value of the polynomial."
I'm agree that this polynomial encodes an algorithm, but it's also a concrete polynomial!
We certainly can't generate large primes with a direct use of it in a reasonable time, but if we analyse it with the tools of differential calculus, maybe we can find some generic extremal points admitting neighborhood areas of positive range, and then generate easily computable sequence of prime numbers, as for the Catalan-Mersenne conjecture:
$2$, $2^{2}-1$, $2^{2^{2}-1}-1$, $2^{2^{2^{2}-1}-1}-1$, $2^{2^{2^{2^{2}-1}-1}-1}-1$...  a sequence of prime numbers ?
I don't know if it's possible, but it's worth a try... now there exists very powerfull supercomputers:
$10^{6}$ times more powerfull than in 1989 when Paulo write its book...
In my opinion this formula can't be useful too for a global understanding of the prime numbers, in particular, we certainly can't use it to upgrade our statistic understanding of primes and prove the Riemann hypothesis. My point is closer to Green-Tao theorem than RH.

Main question: Does differential calculus can be helpful to get results on primes by analyzing $ P $?

For those who already know: What's the minimal size of a solution for $P$ to get a fixed prime ?

REPLY [15 votes]: Added to address additional/clarifies questions. 
a. As mentioned below there are various  'formulas' that will yield arbitrarily large primes, however, they are not efficient in a certain way. Personally, I doubt one can get one (or at least a 'better' one) from considering this polynomial. Now, perhaps, I am wrong. I would be curious to learn if this were the case. 
b. For Green-Tao and related results the following as some 'warning': The sixty-first Putnam competition (2000) had the following question (paraphrasing): the values the polynomial $Q=X^2+Y^2$ takes on $\mathbb{Z}^2$, contains infinitely many triples of consecutive integers. This is not hard to show. Yet, then in the book "The William Lowell Putnam Mathematical Competition 1985-2000: Problems, Solutions, and Commentary" by Kedlaya, Poonen, Vakil there is mentioned the related question (p.279), (then) stated as open problem (again paraphrasing): 

For $Q=X^2+Y^2$ the set $Q(\mathbb{Z}^2)$ contains arbitrarily long arithmetic progressions.  

So if one cannot do this (directly) for this polynomial I am rather doubtful one can expect much for the one given here. One should note that, now, this problem is solved, but only (at least this is the 'official' solution on Kedlaya's webpage) via the result of Green and Tao [the set contains all primes congruent $1$ mod $4$, this is a set of positive relative density in the primes and thus it contains arbitrarily long arithmetic progressions].   

As remarked in comments it is always hard to say that something is definitely not useful in particular if the use refers to a vague notion as understanding the primes better. 
Yet, I will try to demonstrate by an analogy, somewhat close to the problem at hand, at least in my opinion, what is about the intuition. 
Recall that Lagrange's Four Squares Theorem asserts that every non-negative number is the sum of four squares of integers. Or in other words, for $L= X_1^2 + X_2^2 + X_3^2 + X_4^2$ one has $L(\mathbb{Z}^4)=\mathbb{N}$ (where we choose the convention that $\mathbb{N}$ includes $0$).
Now, this is on the one hand an interesting result in its own right, and on the other hand the fact that one can characterize the non-negative integers among all integers in such a way/by such a formula is sometimes also used. In fact, it is used  in considerations close to the one at hand see its mention in the context of  Hilbert's tenth problem.
So this is fine and interesting. What however does not seem like a very feasible idea is to try to understand the non-negative integers better by analysing the polynomial $X_1^2 + X_2^2 + X_3^2 + X_4^2$.
And, to some extent the situation for the primes and this polynomial seems comparable. Yes, one has this characterization of primes, this is very interesting but not so interesting to understand the primes (in the sense, say, some analytic number theorist would like to understand them, frequency, gaps, etc.): the description is not very convenient to work with, the primes do not come out in a systematic way, there are various polynomials having the same property (so why this and not another), the fact that such a polynomial exists is nothing very specific to the primes but also true for all recursively enumerable sets, and so on.
There are also various other prime generating formulas that have some interest, yet also they are not used typically to study primes.
A reason why the idea might seem tempting could be that one somehow thinks that polynomials are relatively easy to understand and handle. However, a key point of all these investigations was, very informally, to show that polynomials are in fact not simple, but can be used to "encode" complex things and thus there can be no algorithm to solve general Diophantine equations.<|endoftext|>
TITLE: Chern class of a logarithmic connection
QUESTION [8 upvotes]: Let $X$ be a smooth complex projective algebraic variety and $E$ a line bundle on $X$. It is a classical result that if $E$ carries an integrable connection, then the first Chern class $c_1(E)$ vanishes. I am interesting in the following variant: let $D$ be a normal crossing divisor and $\nabla: E \to E \otimes \Omega^1_X(\log D)$ an integrable connection with logarithmic singularities along $D$. Denote 
$$
\mathrm{Res}_{D_i} \nabla
$$ the residue of $\nabla$ at an irreducible component $D_i$ of $D$. As $E$ has rank one, it can be identified with a complex number. 
Proposition. One has $c_1(E)=-\sum_i \mathrm{Res}_{D_i} \nabla \cdot [D_i]$
Could anybody explain me how to prove such a result?

REPLY [4 votes]: There exist a paper due to Makoto Ohtsuki about a Residue Formula for Chern Classes Associated with Logarithmic Connections:
http://projecteuclid.org/download/pdf_1/euclid.tjm/1270215030<|endoftext|>
TITLE: Motive of CM elliptic curve and modular forms
QUESTION [11 upvotes]: I am trying to get some insight into the Deligne/Scholl construction of the motive of a modular form. First of all I would like to understand the case of weight two, especially when there is complex multiplication. Unfortunately in his paper "Motives for modular forms" Scholl says
"We do not treat here the case $k = 0$, which corresponds to cusp forms of weight 2; the associated motives are then given by the Jacobians of modular curves, and are well understood."
Question 0: Could anyone explain this or give a readable reference?
More precisely, I am interested in the following question. If $E$ is an elliptic curve over a number field $K$, one can look at the Chow motive with rational coefficients
$$
h^1(E)
$$ cut off from the total motive $(E, \nabla, 0)$ by $\nabla$ minus the two projectors associated to some rational point (over an extension of $K$ if necessary). In general, this motive is indecomposable.
However, when $E$ has complex multiplication by a quadratic imaginary field $F=\mathbb{Q}(\sqrt{-d})$, one gets a decomposition of $h^1(E)_F$ into two rank 1 motives with coefficients in $F$.
On the other hand, one can look to the weight two modular form $f$ attached to $E$ (modularity in the CM case, which is much easier!). Thanks to the construction quoted by Scholl one gets a rank two motive with coefficients in the field generated by the Fourier coefficients of $f$.
Question 1: (General elliptic curve) What is the relation between these two motives? One would like to say there are the same, but the coefficient field are different,  aren't?
Question 2 (CM case): Does the geometric construction using Jacobians of modular forms also give the decomposition into two pieces of rank one? If so, how?

REPLY [4 votes]: Since this question has come alive again, let me point out that the Hecke operators cannot give a splitting of $h^1(E)$ into two pieces over $F$, since the Hecke correspondences on a modular curve are always defined over $\mathbf{Q}$.
(If you allow some extra stuff like the Atkin--Lehner operator on $X_1(N)$, then you can get some slightly larger fields showing up, but they will still be totally real, and thus not capable of "seeing" the splitting of $h^1(E)$.) (EDIT: Actually I am not sure if this is true, sorry!)
If you take a higher-weight CM cuspform $f$ of weight $> 2$, associated to some Groessencharacter $\Psi$, then the situation is even worse: you can define a motive associated to $f$ using Kuga--Sato varieties, and a motive associated to $\Psi$ using the product of $k-1$ copies of an elliptic curve with CM by $K$. These motives really should be the same, because their $\ell$-adic Galois representations are the same for every $\ell$, but there is (as far as I know) no natural way of writing down a correspondence that gives an isomorphism between them in the category of Chow motives.<|endoftext|>
TITLE: Normal bundle of an exceptional divisor
QUESTION [5 upvotes]: Let's take $\phi:\mathcal{X}\rightarrow B$ a family of deformations of complex surfaces above the one dimensional disk $B=D_r(0)\subset \mathbb{C}$. Suppose that $X:=\phi^{-1}(0)$ has one ordinary double point $p$, while all other fibers are smooth. So we can take $x,y,z$ complex coordinates around $p$ such that $\phi(x,y,z)=x^2+y^2+z^2$.
Now take the function $m:\widetilde{B}:=D_{\frac{1}{2}}(0)\rightarrow B$, $m(t)=t^2$ and consider the deformation $\widetilde{\mathcal{X}}=\mathcal{X}\times_B \widetilde{B}$. If $\widetilde{p}\in \widetilde{\mathcal{X}}$ is the only point mapping to $p$ then in local coordinates the germ of $\widetilde{\mathcal{X}}$ in $\widetilde{p}$ is isomorphic to the germ in the ordigin of $V(x^2+y^2+z^2-t^2)\subset \mathbb{C}^4$.
I consider the blow up of $\widetilde{\mathcal{X}}$ in $\widetilde{p}$: its exceptional divisor $E$ is a quadric in $\mathbb{P}^3$ and so is isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$ by Segre embedding. Finally it shoul come out that $N_{E/\widetilde{\mathcal{X}}}\simeq\mathcal{O}_{\mathbb{P}^1}(-1)\boxtimes\mathcal{O}_{\mathbb{P}^1}(-1)$ and $\mathcal{O}_{\mathbb{P}^1}(-1)\boxtimes\mathcal{O}_{\mathbb{P}^1}(-1)=p_1^*\mathcal{O}_{\mathbb{P}^1}(-1)\otimes p_2^*\mathcal{O}_{\mathbb{P}^1}(-1)$ where $p_1$ and $p_2$ are the two projections from $\mathbb{P}^1\times\mathbb{P}^1$)
My questions are:
1) I have read that for example for the blow up of the origin in $\mathbb{A}^2$, the normal bundle of the exceptional divisor $\mathbb{P}^1$ is $\mathcal{O}_{\mathbb{P}^1}(-1)$, but why is that?
2) In the specific case of my construction in which way can i prove that $N_{E/\widetilde{\mathcal{X}}}$ is actually the box product?

REPLY [6 votes]: For the first question, and what Anton mentions, that the normal bundle of the exceptional divisor of the blow up of a smooth subvariety in a smooth variety is $\mathscr O(-1)$ see Thm 8.24 in [Hartshorne]. This takes care of your first question. 
However, you have to be careful, because 

this fails if the ambient scheme is singular, as in your case, where $\widetilde{\mathcal{X}}$ is singular, and
you have to keep in mind that the $(-1)$ here means a specific embedding, so it might not translate to "the natural" $(-1)$ of any given scheme, say $\mathbb P^n$ (this actually does not happen in your case, but  only by luck).

To see that the normal bundle you are looking for (I assume, because what you wrote makes no sense) is the box product do this:
First let's give names to our players: Let $A$ denote the blow up of $\mathbb A^4$ at the origin, $P\simeq \mathbb P^3$ the exceptional divisor, $B\subset A$ the blow up of $\widetilde{\mathcal{X}}$ at the same point, and $E$ the exceptional divisor of that. Presumably you want $N_{E/B}$.
Observe that $A$ is smooth and hence all divisors are Cartier. Then consider the restrictions of $\mathscr O_A(P)$ to both $P$ and $B$ and then the restrictions of these to $E$:
$$
\mathscr O_A(P)|_P \simeq N_{P/A}\simeq \mathscr O_{\mathbb P^3}(-1)
$$
$$
\left(\mathscr O_A(P)|_B\right)|_E \simeq \mathscr O_B(E)|_E \simeq N_{E/B}
$$
and hence
$$
N_{E/B}\simeq \mathscr O_A(P)|_E \simeq \mathscr O_{\mathbb P^3}(-1)|_E\simeq \mathscr O_{\mathbb P^1}(-1)\boxtimes \mathscr O_{\mathbb P^1}(-1).
$$<|endoftext|>
TITLE: Duality between K-theory and K-homology in the non-spin^c case.
QUESTION [13 upvotes]: I posted this question on Math.SE (https://math.stackexchange.com/questions/409444/), but got no answer. So I repost it here.
Let M be a closed manifold. Then there is a cap product $K^\ast(M) \times K_\ast(M) \to K_\ast(M)$ between the K-theory of M and its K-homology. For a definition of it one could see my prior question on Math.SE about it: https://math.stackexchange.com/questions/402170/ - take there A = C(M).
Now if M is spin$^c$, it has a fundamental class $[M] \in K_\ast(M)$ and it is well-known that the cap product with $[M]$ induces the Poincare duality $K^\ast(M) \stackrel{\cong}\to K_\ast(M)$. (See also my other question about it: Duality between K-theory and K-homology in the non-compact, spin$^c$ case.)
In the book "Spin Geometry" by Lawson, Michelsohn it is shown (on page 257) that every class in $K_{cpt}(TM) \cong K_0(M)$ is the difference of two Atiyah-Singer operators with coefficients (if M is spin and even dimensional), i.e., this translates to saying that the Poincare duality map is surjective onto $K_0(M)$ in this case.
Then it is written: "For non-spin manifolds, one can argue similarly by using the signature operator with coefficients". This means that there is some class $[D] \in K_\ast(M)$ (the class of the signature operator) such that the cap product with [D] is onto on $K_0(M)$.
Now the question is, if this statement generalizes:

Is it crucial that it is the signature operator? Could we also take the Euler characteristic operator?
Is there always (i.e., in the non-spin$^c$ case and not only for even dimensional manifolds) a class $[D] \in K_\ast(M)$ such that the cap product with $[D]$ is onto on $K_\ast(M)$ (and not only on $K_0(M)$)?
Maybe even if M is not orientable? Or even when M is not a manifold?
Are there other sufficient conditions besides spin$^c$ such that such a map is injective?

If one of the statements above is true, it would be nice to have some references. Thanks!

REPLY [8 votes]: This is nicely understood by duality (in the sense of dualizable objects in monoidal categories) in the monoidal category KK. The relevant results are collected and referenced on the nLab at Poincaré duality algebra:
For $X$ a closed manifold equipped with a twist $\chi$ (the class of a circle 2-bundle, realized as a $U(1)$bundle gerbe etc., let $C_\chi(X)$ be the corresponding twisted groupoid convolution C*-algebra, well defined up to Morita equivalence and hence KK-equivalence. This is the $C^\ast$-algebra such that
$$
  KK_\bullet(\mathbb{C}, C_\chi(X)) \simeq K^{\bullet+\chi}(X)
$$
is the $\chi$-twisted K-theory of $X$. 
Then here is the statement: The dual object of $C_\chi(X)$ in $(KK, \otimes)$ is
$$
  \left(
     C_\chi\left(X\right)
  \right)^\vee
  \simeq
  C_{-\chi - W_3(T X)}(X)
  \,,
$$
where $W_3(T Q)$ is the third integral Stiefel-Whitney class of the tangent bundle of $X$.
This formula implies everything in this business. 
For instance it implies that $C_\chi(X)$ is self-dual, and hence (since duality in KK is duality between K-homology and K-cohomology) that $X$ exhibits K-Poincaré duality, precisely if $\chi = 0$ and $W_3(T X) = 0$. The latter is precisely the condition of spin^c structure.
But you get much more by playing with the formula, and this is I think what you are after. For instance given a map of closed manifolds $i \colon Q \to X$ with $X$ carrying a twist $\chi$ and $Q$ carrying the twist $i^\ast \chi$, we get the corresponding pullback map of twisted convolution algebras
$$
  i^\ast \colon C_\chi(X) \to C_{i^\ast \chi}(Q)
  \,.
$$
Now since both objects have dualty, we can form the corresponding dual morphism by the general formula in monoidal categories:
$$
  i_! := (i^\ast)^\vee : (C_{i^\ast \chi}(Q))^\vee \to (C_\chi(X))^\vee
  \,,
$$
hence by the above
$$
  i_! : C_{-i^\ast \chi - W_3(T Q)}(Q)  \to C_{-\chi-W_3(T X)}(X)
  \,.
$$
By just relabelling the original twist for transparency as $\chi \mapsto - \chi - W_3(T X)$ this is equivalently a morphism
$$
  i_! : C_{i^\ast \chi + W_3(N_i Q)}(Q)  \to C_{\chi}(X)
  \,.
$$
(Here $N_i Q$ denotes the normal bundle of $Q$ in $X$ via $i$.)
By postcomposition in KK and using the general relation that $KK(\mathbb{C}, A) \simeq K(A)$ this yields a homomorphism
$$
  i_! : K^{\bullet + i^\ast \chi + W_3(N_i Q)} \to K^{\bullet + \chi}(X)
$$
from the $(i^\ast \chi + W_3(N_i Q))$-twisted K-theory of $Q$ to the $\chi$-twisted K-thory of $X$.
This is the general twisted Umkehr map in twisted K-theory. Notice again that it is simply the dual morphism of $i^\ast$ in $(KK,\otimes)$.
If now the normal bundle has $Spin^c$-structure then that term drops out, and so on. But in general it is like this.
If here we think of $X$ as the target spacetime for a type II superstring, of $\chi$ as the instanton sector of the B-field, and of $Q \hookrightarrow X$ as the worldvolume of a D-brane, then the $(i^\ast \chi + W_3(N_i Q))$-twisted K-cocycles on $Q$ are the (underlying instanton sectors of) the Chan-Paton gauge fields subject to the 
Freed-Witten-Kapustin anomaly cancellation condition and their image under the above twisted Umkehr map is the D_brane charge of the D-brane with that Chan-Paton bundle. 
All this just means really: the dual map of $i^\ast$ picks up twists as above. See at Poincaré duality algebra for citations. All these results go through also for $G$-equivariant twisted K-theory.<|endoftext|>
TITLE: Derived categories of toric varieties and convex geometry
QUESTION [9 upvotes]: Toric varieties and convex polyhedra are intimately connected. Some of this can be found in standard text books (the connection between divisors and mixed volumes seems to be a popular example).
One of the most important objects that are associated to an algebraic variety is its derived category. So I'm wondering: are there any constructions or properties in convex geometry that are reflected in the derived category of associated toric varieties?

REPLY [6 votes]: That's a great question! 
In the last few years there is plenty of research on the subject - and a few (interesting) open questions. 
Before Kawamta's result - note that toric manifolds satisfy $rk(K(X))= \chi(X)$ (this is a result on the fan), so conceptually they are a good candidates to have a full exceptional collection (as Kawamata's result shows), see Remark 2.3 in Uhera's work on derived categories of toric threefolds.
Kawamata's result (mentioned in the answer above) is very deep but the exceptional collections he constructs can be quite hard to track (the construction relies on various ideas from toric Mori theory). Pay attention that Kawamata's construction of e.c is not necessarily strong - which is a condition one typically wants an exceptional collection to satisfy. 
If you will look at the first examples of (full strongly) exceptional collections discovered by Beilinson for $X=\mathbb{P}^n$ (which is, of course, toric) you will see that they are given by rather natural elements $\mathcal{E}_1 = \left \{ \mathcal{O},\mathcal{O}(1),...,\mathcal{O}(n) \right \} $ and $\mathcal{E}_2 = \left \{ \mathcal{O},\Omega^1(1),... , \Omega^n(n) \right \}$. In fact $\mathcal{E}_1$ is given by line bundles, that is elements of $Pic(X)$. 
This led A. King to ask - which toric manifolds $X$ admit exceptional collections of line bundles in $Pic(X)$. For a while it was strongly believed that any toric manifold admits such a collection - but this was proved to be wrong by L. Hille and M. Perling. 
In particular, this question (which toric admit e.c of line bundles) turned out to be quite an elusive question regarding toric manifolds. For instance - there is no clear algorithm that determines on the basis of the toric fan $\Sigma$ whether the corresponding manifold has such a collection.
On the other hand - there's plenty of research and results which shows that exceptional collections of line bundles for toric manifolds $X$ are a rather interesting class of exceptional collections - for instance just to mention a few - A. Bondal, Costa and Miro-Roig, H. Uehara, Borisov and Hua, Dey, Lason and Michalek, Bernardi and Tirabassi...
Personally, I am interested in connections between such collections - and ideas from mirror symmetry - they turn to be related to solutions of toric Landau-Ginzburg system (which are given in terms of the fan) - if your'e interested - there's some details here.<|endoftext|>
TITLE: the graded pieces of the gamma-filtration of Quillen K-theory and Chow groups of a regular scheme
QUESTION [7 upvotes]: Let $X$ be a regular scheme and consider Grothendieck's $\gamma$-filtration $F^nK(X)$ on $K(X)$.  For the graded pieces, one has $Gr^0K(X) = CH^0(X)$ and $Gr^1K(X) = \mathrm{Pic}(X) = CH^1(X)$.  Does this continue to hold, i.e., do we have $Gr^pK(X) = CH^p(X)$?
I found that for $X/k$ smooth quasi-projective, $CH^q(X,p) \otimes \mathbf{Q} = K_p(X)^{(q)} \otimes \mathbf{Q}$, so this holds after rationalising.

REPLY [5 votes]: This does not hold in general.
An explicit counterexample is given in:
Karpenko, Nikita A. Codimension 2 cycles on Severi-Brauer varieties. K-Theory 13 (1998), no. 4, 305–330. (Reviewer: Jean-Pierre Tignol) 16K20 (14C15 19E15). (Available on the author's webpage https://sites.ualberta.ca/~karpenko/publ/ch2.pdf).
Fix a field $k$. Let $p$ be an odd prime. Let $A$ be a central division $k$-algebra of degree $p^2$, exponent $p$, and assume $A$ decomposes into a product $D_1\otimes D_2$ of two smaller algebras (both necessarily of degree $p$). Let $X$ be the Severi-Brauer variety associated to $A$. By Proposition 4.7 of the article above, the quotient $\text{Gr}^2K(X)$ contains torsion but, by Proposition 5.3 of the article above, the group $\text{CH}^2(X)$ is torsion free.
Similarly, one can show if $A$ is a division algebra of index $8$ and exponent $2$ decomposing into a product of smaller algebras then the same conclusion holds. A more general statement is true for the prime $2$ but it depends on the indices of the tensor powers of $A$.<|endoftext|>
TITLE: A Question Regarding the Relation Between 0-sharp and Koepke's Bounded Truth Predicate.
QUESTION [7 upvotes]: In Jech's SET THEORY (a very early edition to which I have access), it is shown that the existence of 0-sharp implies the existence of a truth definition for the constructible universe L.  Does the converse hold?  I ask this question because in Koepke's paper "Turing Computations on Ordinals", he defines an ordinal computable "bounded truth predicate" (at least as I understand it, and I possibly don't understand it correctly) for each stage L_i of the constructible universe L.  Are these two notions (0-sharp and Koepke's bounded truth predicate) at all related, and if so, in what way?  At first glance (to me, at least) they seem to be because in Jech's text, the satisfaction relation is defined for each stage L_i of the hierarchy for L, just as the satisfaction relation seems so defined for Koepke's bounded truth predicate.

REPLY [6 votes]: The answer is no, one can have models of ZFC set theory with a definable truth predicate for first-order truth in $L$, but without having $0^\sharp$. 
One way to build such a model is like this. In Kelly-Morse KM set theory, you can prove the existence of a truth predicate for first-order truth for the whole universe $V$, and then by forcing you can code this class into the GCH pattern, for example, in order to make it definable. The result is a forcing extension $V[G]$ which has a first-order definable class predicate for first-order truth in the ground model $V$. From this, one can easily define a truth predicate for first-order truth in $L$. 
But meanwhile, KM is weaker than $0^\sharp$ in consistency strength, and so we can find such a model $V$ and hence also $V[G]$ without $0^\sharp$. The theory KM is weaker than $0^\sharp$ because its consistency follows from the existence of a single inaccessible cardinal: if $\kappa$ is inaccessible, then $V_\kappa$ is a model of KM when equipped with its full second-order part $V_{\kappa+1}$. In contrast, $0^\sharp$ implies the consistency of a proper class of inaccessible cardinals (and more), since under $0^\sharp$ the Silver indiscernibles are all inaccesible in $L$.
This kind of example shows another direct way to build the desired model. Start with $\kappa$ inaccessible. So $V_\kappa$ is a model of ZFC, and remains a model of ZFC(S) even when we add the satisfaction class S for first-order truth in $V_\kappa$. Now we may force to make S definable in a forcing extension $V_\kappa[G]$. In $V_\kappa[G]$, we have a definable class for first-order truth in $V_\kappa$, from which we can define satisfaction in its $L$. But we needn't have $0^\sharp$, since in fact we could have started with $V=L$.<|endoftext|>
TITLE: Computing a cobordism group of manifolds endowed with a real vector bundle with constraints on the Stiefel-Whitney classes
QUESTION [10 upvotes]: I am interested in computing the cobordism group of oriented manifolds $M$ of dimension 7 endowed with real vector bundles $N$ of rank 5 with the following conditions on the Siefel-Whitney classes:
$ w_1(N) = 0 \;, \quad w_2(N) = w_2(TM) \;, \quad  w_5(N) = 0 \quad (1)$
The first equation says that $N$ is an orientable bundle. The second says that $TM \oplus N$ is spin. The third is equivalent to the vanishing of the Euler class of $N$. 
So to paraphrase the above, I want to know if given an oriented 7-manifold $M$ endowed with a bundle $N$ satisfying (1), it is always possible to find an oriented 8-manifold $M'$ bounded by $M$ together with a bundle $N'$ satisfying (1) and extending $N$ to $M'$.
If we ask the same question disregarding the bundle $N$, the obstruction is given by the stable homotopy group $\pi_7(MSO)$, which can be shown to vanish using the results of C.T.C. Wall. So given a 7-dimensional oriented manifold, one can always find an 8-dimensional manifold bounded by the latter.
Now taking into account $N$ and ignoring the last two conditions in (1), I believe that the obstruction to finding a bordism is given by $[S, \Sigma^{-7}MSO \wedge \Sigma^5 BSO]$, where $S$ is the sphere spectrum, $\wedge$ the smash product and $[.,.]$ denotes the homotopy classes of maps. Already at this level I am not sure how to compute this group. 
Edit: As $N$ has to be an actual bundle and not only a stable one, the bordism group should be in this case $[S, \Sigma^{-7}MSO \wedge BSO(5)]$.
Finally, one has to take into account the last two constraints of (1). To this end, I imagine that one should use the fact that there is a map from  $\Sigma^{-7}MSO \wedge \Sigma^5 BSO$ into $\Sigma^2H\mathbb{Z}_2 \wedge \Sigma^5H\mathbb{Z}_2$, determined by $(w_2(N) - w_2(TM), w_5(N))$ and that the relevant spectrum is in some appropriate sense the kernel of this map. 
Any hint about how to compute this cobordism group, or reference to similar computations in the literature would be greatly appreciated.
For people curious about it, the motivation to compute this group comes from the physics of the M5-brane. The worldvolume of the M5-brane is oriented and 6-dimensional. It is embedded in an 11-dimensional manifold $X$ which is spin, with normal bundle $N$. The spin condition on $X$ and the orientability of $M$ account first two conditions in (1). To compute the global gravitational anomalies of the effective field theory on the worldvolume, one has to consider mapping tori of the worldvolume, endowed with a vector bundle $N$ satisfying (1). And it turns out that the best way to express the anomaly is in terms of an 8-dimensional manifold bounded by the mapping torus. This is why knowing if such bounded manifolds exist is crucial.

REPLY [7 votes]: This is similar to András's answer, but a bit different...
Start with $BSO(5)$ and kill $w_5$, i.e take the fiber $F$ of
$w_5:BSO(5)\to K(Z_2,5)$. Now consider the composition
$\xi: BSpin \times F \to BSO\times BSO(5)\to BO\times BO \to BO$,
where the first two arrows are products of the obvious map and 
the last one corresponds to addition of vector bundles.
Finally take the Thom spectrum of the pullback of the universal 
bundle over BO along $\xi$.
The reason is that $N$ is classified by a map to $F$, and the 
difference $\nu-N$ is classified by a map to $BSpin$.
(I assume you want the spin structures on your manifolds and bordisms too.
Spin structures on $\nu-N$ are in bijection with spin structures on $TM+N$.)
Then the Thom spectrum is $MSpin\wedge \Sigma^{-5}TF$, where $TF$ is the
Thom space for $F\to BSO(5)$, so the bordism group is $\Omega_{12}^{Spin}(TF)$. One can try to compute this via the Atiyah-Hirzebruch spectral sequence,
or also by using the Adams spectral sequence.<|endoftext|>
TITLE: What is the source of this famous Grothendieck quote?
QUESTION [44 upvotes]: I've seen the following quote many times on the internet, and have used it myself. It is usually attributed to Grothendieck.

It is better to have a good category with bad objects than a bad category with good objects.

Question: Does anyone know the source of this quote, or at least when it first appears, or when it was first attributed to Grothendieck?
I wish I were able to properly cite this popular and insightful quote.

Added: On a related topic, another quote I wish I could cite properly, usually attributed to Manin, reads:

Proofs are more important than theorems, definitions are more important than proofs.

Does anyone know a source for this quote?

REPLY [14 votes]: I would be surprised if the purported Grothendieck quote is really his.  He does not lean to the short and sweet. It sounds more like an adaptation of another thing Deligne says in "Quelques idées maitresses de l'oeuvre de Grothendieck" (p. 13): "if the decision to let every commutative ring define a scheme gives standing to bizarre schemes, allowing it gives a category of schemes with nice properties."<|endoftext|>
TITLE: Projections in a W*-algebra as a continuous lattice?
QUESTION [5 upvotes]: A continuous lattice is a complete lattice $L$ in which every element $y$ is equal to $\bigvee${$x \in L \mid x \ll y$} where $x \ll y$ ("x approximates y" or "x is way below y") if for any directed set $D \subseteq L$, $y \leq \bigvee D$ implies that there is a $d \in D$ such that $x \leq d$.
It is known that the set of projections in an arbitrary W*-algebra is a complete orthomodular lattice. I would like to know for which kind of W*-algebras this lattice is also continuous.

REPLY [2 votes]: The already given answer and comments have essentially solved the problem; however I
will show the way the question can be studied from the point of view of
"classical" lattice theory (say, the times of von Neumann and
Birkhoff, then Halperin, Kaplansky, F. and M. Maeda)
For complete lattices, it is well known that algebraic implies
Scott-continuous implies meet-continuous.
{Remember the
definitions: a complete lattice is a partial order where every subset
has a supremum (equivalently, every subset has a infimum). It is
meet-continuous when increasing joins distribute over meet
(dually for join continuity); it is Scott continuous when every element
$a$ is the join of elements $b$ "way below" $a$ (meaning: every
increasing join that is at least $a$ has a member that is at least $b$);
it is algebraic when meet-continuous and each $a$ is the join of
join-inaccessible elements $b$ (elements that are "way below"
themselves).} 
The term "continuous" tout-court might seem suitable for
Scott-continuity, since the latter coincides with embeddability (with
preservation of all meets and increasing sups) in a power of the real
interval $[0,1]$ (the "archetipe" of the continuum), in the same ways
as algebraicity uses the two-element chain $\{0,1\}$ (the "archetipe"
of discreteness) instead of $[0,1]$. However, this also means that such
a concept of continuity is a generalization of (and not a opposition to)
discreteness.                                                                                             
For classical objects of lattice theory (like complemented modular
lattices or orthomodular lattices) which are related to (discretely or
continuously valued) dimension functions, the term "continuity" is
traditionally associated with meet-continuity and its dual (von
Neumann's [meet and/or join] "continuous geometries").                     
{For the concept of semiorthogonality (used below) and related
ones, see S. Maeda papers, especially the last paper (1961, freely
available inside projecteuclid) about dimension lattices.}
The point (to be proved below): for such classical (complete, relatively
semi-orthocomplemented) lattices, Scott-continuity coincides with
algebraicity and even "compactly atomistic" (hence usually even more:
direct product of "finite discrete factors": the discrete subcase of
continuous geometries).                                                                                                        
The lemma: in a complete (relatively) semiorthocomplemented lattice,
"$b$ way below $a$" implies "$b$ join inaccessible" (hence compact
in the meet-continuous case).                                                                                                  
Proof: suppose $b$ inceasing join of the $b_i$. Fix a
semi-orthocomplement $c$ of $b$ in $a$; then $b_i\oplus c$ is a
increasing family, with join $a$ (it contains $c$ and the join $b$ of
the $b_i$; conversely, each $b_i\oplus c$ is contained in $b\oplus                                                             
c=a$). By definition of "$b$ way below $a$",  $\exists i:b_i\oplus                                                           
c\geq b$. Adding $c$, $b_i\oplus c\geq b\oplus c$. Now, write
$b=b_i\oplus c_i$; so: $b_i\oplus c\geq b_i\oplus c_i\oplus c$; hence:
$c_i=0$, $b_i=b$.                                                                                                              
From the lemma it follows that a Scott-continuous relatively
semiorthocomplemented lattice is algebraic (it is meet-continuous, hence
"$b$ way below $a$" implies "$b$ compact" and so each element is
join of compact elements); then it is also atomistic (i.e. sectionally
semicomplemented and atomic: use relatively complemented [whci follows
from semi-ortho-complemented] and weakly atomic [which follows from
algebraic]: every interval $[a,b]$ contains a covering $\{a',b'\}$ and
the realtive orthocompement $c'$ of $a'$ in $b'$ is an atom in $[a.b]$:
if not, $c'$ properly splits in $x\oplus y$ and so between $a'$ and $b'$
one has $a'\oplus x$).                                                                            
Once one has a compactly atomistic lattice, a weak (semimodularity)
condition (like the covering property, which is satisfied by
meet-continuous geometries and projection ortholattices of
AW$^*$-algebras and their Jordan analogues) is well known to imply a
decomposition into a direct product of subdirectly irriducible factors
of the same kind (see "matroid lattices" in F. Maeda - S. Maeda book,
theory of symmetric lattices).                                                                                                 
In the semimodular and orthomodular case (i.e. dimension ortholattices),
the irreducible factors are exactly the finite-dimensional
orthocomplemented projective geometries; conclusion: Scott-continuous
semimodular otholattices are exactly the direct products of
(irreducible) orthocomplemented finite-dimensional projective
geometries. [The W$^*$ or AW$^*$ algebras that give such ortholattices
are exactly the direct products, in the category of such algebras, of
matrix $*$-algebras over the real, complex or quaternional $*$-field.]                                                         
{The direct product as (algebras or) rings (with involution) of
matrix $*$-algebras with scalar entries is a $*$-regular ring hence it
is not even a C$^*$-algebra (unless there are only finitely many
nontrivial factors); the direct product in the suitable category is
smaller (one takes only the bounded elements of the $*$-regular ring,
see Berberian's book "Baer$^*$-rings").}
In the complemented modular case (i.e. meet-continuous geometries), the
irreducible factors are exactly the (possibly infinite dimensional)
projective geometries (associated to vector spaces over a sfield, except
possibly nonarguesian planes, and lines and points). Conclusion: the
Scott-continuous complemented modular lattices are exactly the (possibly
reducible) projective geometries; subcase: the Scott-continuous
continuous geometries are exactly the direct products of
finite-dimensional projective geometries.                                                                                      
Note: this generalizes the characteriazion of Scott-continuous boolean
algebras in the "compendium of continuous lattices".<|endoftext|>
TITLE: the following inequality is true，but I can't prove it
QUESTION [36 upvotes]: The inequality is
\begin{equation*} 
\sum_{k=1}^{2d}\left(1-\frac{1}{2d+2-k}\right)\frac{d^k}{k!}>e^d\left(1-\frac{1}{d}\right)
\end{equation*}
for all integer $d\geq 1$. I use computer to verify it for $d\leq 50$, and find it is true, but I can't prove it.  Thanks for your answer.

REPLY [47 votes]: [Edited mostly to fix a typo noted by David Speyer]
The following analysis simplifies and completes the
"routine but somewhat unpleasant" task of
recovering the actual inequality from the asymptotic analysis.
The idea is that once we've obtained the asymptotic expansion
$$
\sum_{k=1}^{2d} \left( 1 - \frac1{2d+2-k} \right) \frac{d^k}{k!}
\sim e^d \left( 1 - \frac1{d} + \frac1{d^2} - \frac2{d^3} \cdots \right)
$$
by expanding $1 - 1/(2d+2-k)$ in a power series about $k=d$,
we should be able to replace $1 - 1/(2d+2-k)$ by something smaller
that can be summed exactly and is close enough that the result
is within a small enough multiple of $e^d$ to maintain the
desired inequality.
Because it takes about $2m$ terms of the power series in $k$
to get within $O(1/d^m)$, I had to match the power series to within $O(k-d)^6$.
Let
$$
A_6(k) =
\frac{d+1}{d+2}
- \frac{k-d}{(d+2)^2}
- \frac{(k-d)^2}{(d+2)^3}
- \frac{(k-d)^3}{(d+2)^4}
- \frac{(k-d)^4}{(d+2)^5}
- \frac{(k-d)^5}{(d+2)^6}
- \frac{(k-d)^6}{2(d+2)^6},
$$
so the final term has denominator $2(d+2)^6$ instead of $(d+2)^7$.
Then
$$
1 - \frac1{2d+2-k}
 = A_6(k) + \frac{(k-d)^6(2d-k)}{2(d+2)^6(2d+2-k)} \geq A_6(k)
$$
for all $k \leq 2d$.  Hence
$$
\sum_{k=1}^{2d} \left( 1 - \frac1{2d+2-k} \right) \frac{d^k}{k!} >
\sum_{k=1}^{2d} A_6(k) \frac{d^k}{k!}.
$$
On the other hand, since $A_6(k)$ is a polynomial in $k$,
the power series $\sum_{k=0}^\infty A_6(k) d^k/k!$ is elementary
(see my earlier answer for the explanation; David Speyer implicitly
used this too in the calculation "with the aid of Mathematica").
I find
$$
\sum_{k=0}^\infty A_6(k) \frac{d^k}{k!}
= \frac{2d^6 + 22d^5 + 98d^4 + 102d^3 + 229d^2 + 193d + 64}{2(d+2)^6} e^d
$$ $$
= \left(
 1 - \frac1d + \frac{2d^5 + 5d^4 + 69d^3 + 289d^2 + 320d + 128}{2d(d+2)^6}
\right) \cdot e^d > \left(1 - \frac1d\right) e^d.
$$
We're not quite finished, because we need a lower bound on
$\sum_{k=1}^{2d} A_6(k) d^k/k!$, not $\sum_{k=0}^\infty$.
However, once $d$ is at all large the terms with $k=0$ and $k>2d$
are negligible compared with our lower bound
$$
\sum_{k=0}^\infty A_6(k) d^k/k! - \left(1 - \frac1d\right) e^d
\geq \frac{2d^5 + 5d^4 + 69d^3 + 289d^2 + 320d + 128}{2d(d+2)^6} e^d >
\frac{d^4}{(d+2)^6} e^d.
$$
Indeed the $k=0$ term is less than $1$, and for $k>2d$ we have
$A_6(k) < A_6(2d) = 1/2$ while $d^k/k!$ is exponentially smaller than $e^d$:
$$
\sum_{k=2d+1}^\infty \frac{d^k}{k!} <
 2^{-2d} \sum_{k=2d+1}^\infty \frac{(2d)^k}{k!} <
 2^{-2d} \sum_{k=0}^\infty \frac{(2d)^k}{k!} = (e/2)^{2d}.
$$
So we're done once
$$
1 + \frac12 \left(\frac{e}{2}\right)^{2d} < \frac{d^4}{(d+2)^6} e^d,
$$
which happens once $d \geq 14$.  Since the desired inequality
has already been verified numerically up to $d=50$, we're done. QED<|endoftext|>
TITLE: A category with weak equivalences which is not a model category
QUESTION [9 upvotes]: I'm only considering complete and cocomplete categories. A pair $(\mathfrak{X} , \mathfrak{W}) $ is, by definition, a category with weak equivalences if  $ \mathfrak{X} $ is a category and $ \mathfrak{W} $ is a subcategory satisfying the $2$ out of $3$ axiom.
I was wondering if there are nice examples of categories with weak equivalences for which there is no model structure. 
Thank you

REPLY [7 votes]: Here is another (more combinatorial) answer.  Let $\kappa$ be any (nice) cardinal, and let $\mathcal{A}$ be the union of the full subcategory of $\mathbf{Set}$ consisting of all sets of cardinality less than $\kappa$ and the subcategory of all isomorphisms in $\mathbf{Set}$.  $\mathcal{A}$ is closed under retracts and 2-of-3, but it is NOT the subcategory of weak equivalences of a model structure on $\mathbf{Set}$.  We can see this just by listing all of the (nine) model structures on $\mathbf{Set}$, and noting that the subcategories of weak equivalences are

all morphisms
all isomorphisms
all morphisms between nonempty sets (plus the identity on $\emptyset$)

(See Tom Goodwillie's answer to https://mathoverflow.net/questions/29653.) In particular, model categories can determine whether you're empty or nonempty, but they can't differentiate between different set sizes.<|endoftext|>
TITLE: Negative sectional curvature and constant curvature
QUESTION [11 upvotes]: Good morning everyone,
I was wondering about the difference between manifolds carrying a Riemannian metric with negative sectional curvature and hyperbolic manifolds. I was told once "there are very few properties in constant negative sectional curvature that cannot be extended to the negative sectional curvature case".
Can one list the topological properties a manifold with negative (but variable) sectional curvature must have?

REPLY [18 votes]: Sullivan proved that every closed hyperbolic manifold has a stably parallelizable finite cover. This is not true for say complex hyperbolic manifolds (of real dimension $>2$). See Farrell's "Lectures on Surgical Methods in Rigidity". 
Real Pontryagin classes of complete hyperbolic (or more generally conformally flat) manifolds vanish. 
Orientable closed hyperbolic manifolds have even Euler characteristic, while there is a complex hyperbolic surface constructed by Mumford of Euler characteristic 3. (Proof: every orientable closed manifold of dimension not divisible by 4 has even Euler characteristic, and in dimensions divisible by 4 
the Euler characteristic is the Betti number in the middle dimension, which vanishes for hyperbolic manifolds because their signature iz zero by the fact 2 above).<|endoftext|>
TITLE: Belyi's theorem for function fields
QUESTION [13 upvotes]: Belyi's theorem states that every smooth projective algebraic curve $C$ defined over $\bar{\mathbb{Q}}$
admits a map $C\to\mathbb{P}^1$ ramified only over $0,1,\infty$.
Is there an analogue of this theorem with $\mathbb{Q}$ replaced by a global function field (i.e. finite extension $\mathbb{F}_q(t)$)?
I am especially interested in the existence of a tamely ramified map.

REPLY [8 votes]: In a comment on the accepted answer, the OP asks about tame ramification.  Saidi shows in Theorem 5.6 here that a smooth projective curve $C$ over a field $k$ of characteristic $p>2$ is defined over $\overline{\mathbb{F}_p}$ if and only if $C$ admits a map $C\to \mathbb{P}^1$ with only tame ramification over $\{0,1,\infty\}$.
The proof is rather easy.  I'll first sketch the argument that a curve defined over $\overline{F_p}$ admits a map as claimed. A result of Fulton shows that any curve admits a map $g: C\to \mathbb{P}^1$ with ramification indices at most two, hence a tamely ramified map if the characteristic is different from $2$.  Let $q$ be such that the ramification values are defined over $\mathbb{F}_q$.  Then composing $g$ with the map $$z\mapsto z^{q-1}$$ gives a map $C\to \mathbb{P}^1$ with the desired property.
To see the other direction, one may use Riemann existence or just observe that maps with the desired property don't deform and the moduli space of such maps is locally finite type, hence any $k$-point comes from an $\overline{\mathbb{F}_p}$-point.<|endoftext|>
TITLE: Bijection between Littlewood-Richardson tableaux and ordinary skew tableaux?
QUESTION [5 upvotes]: This question related to this,
but since that question does not ask the same thing, the answer to that question do not fully answer this.
Question: Is there a bijection between the set of Littlewood-Richardson tableaux of 
skew shape $\tau/\sigma$ and weight $\rho$ and 
the set of ordinary skew tableaux of shape $\lambda/\mu$ and weight $\nu$
for some $(\lambda,\mu,\nu)$?
In the question above, Richard Stanley gives a nice bijection 
in the case when $\tau/\sigma$ and $\rho$ have a quite special form,
(the LR-tableaux are essentially of the form of a disjoint union of
an ordinary tableau, and horizontal strips), and these can be set in bijection with
a set of non-skew tableaux with fixed shape and weight.
So, if we also allow to map to skew tableaux, can we 
extend this to cover all LR-tableaux, in some way?
For example, if we have that LR-tableaux $(\tau/\sigma,\rho)$
can bijectively be mapped to semi-standard (non-skew) tableaux with weight $\tau-\sigma$ and shape $\rho$, but this only works under the condition that $\tau_{i+1}=\sigma_{i}$ for all $i\geq 0.$ By allowing to map to skew tableaux as well, can we overcome this restriction?

REPLY [2 votes]: There has not been any answer to this, so I'll give a partial answer for the moment:
There is a bijection described by Littlewood and Richardson in the following sense:
$$K_{\lambda/\mu,\pi} \leftrightarrow \bigsqcup_\nu c_{\lambda/\mu,\nu} \times K_{\nu,\pi}$$
Here, I abuse notation and let $K$ and $c$ denote the sets of ordinary skew tableaux and Littlewood-Richardson tableaux.
This is the closest to a bijection that I can get as for now.<|endoftext|>
TITLE: Coding a model of $0^\sharp$ from a $\Pi^1_1$ Gale-Stewart game
QUESTION [12 upvotes]: As a preface to this question, this is my first time asking on Math overflow, and this seemed like the sort of question that would be acceptable here. However, I apologize if it is not.
A method for coding models from a Gale-Stewart game proceeds as follows: Our aim is to obtain a model of the form $L_\gamma[a]$ such that $L_\gamma[a]\models$ "$a$ is a large cardinal property" where $a$ can be coded by a sequence of ordinals, $\mathcal{S}$. Since $a$ can be coded by a sequence of ordinals, we see that it can be coded by a real $x\in{}^\omega\omega$ by by coding each well-order in the sequence, of by coding a definition of the sequence. With that in mind, a rough set up of the game has a legal play $z\in{}^\omega\omega$ that codes both $\gamma$ and $\mathcal{S}$. One of the players wins when the model $L_\gamma[a]$ produced is the one we were looking for. This technique is due to Martin and Solovay.
One can show that $\Pi^1_1$-determinacy is equivalent to the existence of $0^\sharp$ (left to right is due to Harrington, and right to left is due to Martin), but every proof of "$\Pi^1_1$-determinacy implies that $0^\sharp$ exists" that I know of goes through the following lemma:

$\mathbf{Lemma}$: If there is a real $x$ such that every $x$-admissible ordinal is a cardinal in $L$, then $0^\sharp$ exists.

The question then arises as to whether or not it is possible to get $0^\sharp$ from $\Pi_1^1$-determinacy more directly. In particular, my question is as follows:

Is it possible to construct a $\Pi^1_1$ game that produces a model satisfying the statement "$0^\sharp$ exists"? 

Edit:
I've realized that this is actually a rather difficult question to answer as stated. I asked it because I could not have been the first person to think of this, and was curious if anyone else had made progress. However, the question was ill-formed in some sense, but here is related question that is perhaps more answerable:

As far as I know, there is no proof of "$0^\sharp$ exists from $\Pi^1_1$-determinacy that does not utilize the above lemma. Is this simply because no one has worked on this since, or because there is something about $\Pi^1_1$ sets that make this difficult?

From what I understand, $0^\sharp$ is a well-studied object insofar as much is known about it. There is also a fair bit of work that has been done on determinacy, but I have not seen as much in conjunction with $\Pi^1_1$ determinacy. Most of what I have seen is in regards to OD-Determinacy, Projective Determinacy, and Determinacy in $L(\mathbb{R})$. Now, if one were interested in trying to prove $0^\sharp$ exists from $\Pi^1_1$-determinacy, it seems that a good approach would be to try and produce a model that contains $0^\sharp$ from a $\Pi^1_1$ game. The next question is whether or not the above technique of Martin and Solovay would be the best approach. This technique was used with stronger determinacy hypotheses, so I wonder if the low complexity of $\Pi^1_1$ sets would make it the case that a different approach would be better.
A good starting place may be the known proofs of "$0^\sharp$ exists from $\Pi^1_1$-determinacy. We define a set $A\subset{}^\omega\omega$ as follows: $a\in A$ if, and only if there is a binary relation $R$ on $\omega$ which is recursive in $a$ such that $(\omega, R)$ is isomorphic to an end extension of $(L_{\omega_1(a)},\in)$ where $\omega_1(a)$ denotes the first $a$-admissible ordinal. We then prove that $A$ is $\Sigma^1_1$, closed under Turing equivalence, and non-empty. This is the game that Harrington used to prove that $0^\sharp$ exists. However in an exercise in Martin's notes on determinacy, he defines a different game in the hint.
We play a game $G$ on $^{<\omega}\omega$. For each play of $G$, $I$'s part codes a relation $R$ on $\omega$ and let $II$'s part of the play code a relation $E$ on $\omega$. If $R$ is not a well-ordering of $\omega$, $I$ loses. If $R$ is a wellordering of $\omega$, let $\beta$ be its order type. Then, $II$ wins if and only if $(\omega;E)$ is a model of extensionality, and there is a $g:L_\beta\rightarrow\omega$ that embeds $(L_\beta;\in)$ into $(\omega;E)$ as an initial segment. We then show that the set we are playing over is $\Pi^1_1$, and that $I$ has no winning strategy in $G$.
This seems a bit closer to the idea presented in the first real paragraph of this question. In particular, perhaps one could construct a $\Pi^1_1$ game that codes a nontrivial elementary embedding $j:L_\alpha\prec L_\beta$ for limit ordinals $\alpha$ and $\beta$ using a similar idea. This is not exactly what we are looking for, but it's getting closer.

REPLY [16 votes]: This problem is very much open. 
Cheng Yong calls Harrington's $\star$ the assumption that there is a real $x$ such that all $x$-admissible ordinals are $L$-cardinals. From the work of Yong we know that Second- and even Third-order arithmetic do not suffice to prove that Harrington's $\star$ implies the existence of $0^\sharp$. Whether this was possible was originally asked by Woodin, who proved that Second order arithmetic suffices to show that $\mathrm{Det}(\Sigma^1_1)$ implies Harrington's $\star$.
This appears in Yong's dissertation, Analysis of Martin-Harrington theorem "$\mathrm{Det}(\Sigma^1_1)\leftrightarrow 0^\sharp$ exists" in higher order arithmetic, National University of Singapore, 2012. 
Here, 

Second order arithmetic, $Z_2$, is formalized as: $(\mathsf{ZFC}-$ Power set axiom$)+$ All sets are countable. 
Third order arithmetic, $Z_3$, is $(\mathsf{ZFC}-$ Power set axiom$)+\mathcal P(\omega)$ exists $+$ All sets have size at most continuum. 
Fourth order arithmetic, which Yong shows proves that Harrington's $\star$ implies the existence of $0^\sharp$, is $(\mathsf{ZFC}-$ Power set axiom$)+\mathcal P^2(\omega)$ exists $+$ All sets have size at most $2^{2^{\aleph_0}}$.

In joint work by Ralf Schindler and Yong, the situation has been further clarified: $Z_2+$ Harrington's $\star$ is equiconsistent with $\mathsf{ZFC}$, and $Z_3+$ Harrington's $\star$ is equiconsistent with $\mathsf{ZFC}+$ There is a remarkable cardinal. See here. 
As question 8.0.14 in his thesis, Yong asks whether $Z_2$ proves Harrington theorem. This is tricky, since $Z_2$ proves that $\mathrm{Det}(\mathbf\Sigma^1_1)$ is equivalent with "For all reals $x$, $x^\sharp$ exists." The reason why the lightface version is open, Yong admits, is that the only route we have from $\mathrm{Det}(\Sigma^1_1)$ to the existence of $0^\sharp$ is via Harrington's $\star$, so the question is effectively asking for a different argument. There are several sufficiently different proofs of Harrington's theorem. For example, there is the argument in  Recursive Aspects of Descriptive Set Theory by Mansfield and Weitkamp, via non-$\omega$-models of $\mathsf{KP}$. There is Harrington's original proof, using Steel's forcing. There is Sami's proof, from Analytic determinacy and $0^\sharp$: A forcing-free proof of Harrington's theorem. All use Harrington's $\star$ in an essential fashion. 
The question of whether a different route is possible has been studied, of course, but unsuccessfully.<|endoftext|>
TITLE: Why is TopGrp the category of topological groups and continous homomorphisms protomodular?
QUESTION [8 upvotes]: Why is TopGrp the category of topological groups and continous homomorphisms protomodular? I know it is, and I have several indirect proofs, but am not able to prove this directly by showing that the split short five lemma holds. Please help!!! Thank you!

REPLY [6 votes]: Given the fact that Borceux and Clementino (et al. from references) are reasonably explicit and constructive in their proofs, a "direct proof" should be obtainable in any case just by systematically unpacking all the lemmas they use. This is an instance of a general metamathematical method called "beta-reducing" a proof or computation, akin to cut elimination in proof theory. Let's take a look. 
The proof of theorem 50 in Borceux-Clementino (specialized to the theory of groups) explains that the short split five lemma is a statement expressible in the language of finite limits, so that by a Yoneda argument, it should hold in $\mathbf{Grp}(\mathrm{Top})$, given that the short split five lemma holds in $\mathbf{Grp}$. We can unpack this Yoneda lemma argument to give a direct proof. 
Thus, suppose given a commutative diagram in the category of topological groups 

\begin{array}{ccccc}
 \ker(q) & \xrightarrow{i'} & F & \xrightarrow{q} & B \\
 \wr \downarrow & & \downarrow \pi & & \downarrow 1_B\\
 \ker(p) &\xrightarrow{i}&E&\xrightarrow{p}& B &
\end{array}

where $p$ and $q$ are assumed to be split epic. We want to show $\pi$ is an isomorphism of topological groups. Let $U(E)$ be the underlying topological space of $E$. Then $\hom(U(E), -)$ sends this diagram of topological groups to a diagram of ordinary groups. Since this representable preserves kernels, split epics, etc., we infer from the split five lemma in $\mathbf{Grp}$ that $\hom(U(E), \pi)$ is a group isomorphism. In particular, there is a continuous map $s: E \to F$ which is sent to $1_E$ by $\hom(U(E), \pi)$; in other words, such that $\pi \circ s = 1_E$. We argue similarly that $\hom(U(F), \pi)$ is a group isomorphism, so that there exists a unique continuous map in $\hom_{\mathrm{Top}}(F, F)$ which maps to $\pi$ under $\hom(U(F), \pi)$. Since both $1_F$ and $s \circ \pi$ are such maps, we also have $s \circ \pi = 1_F$. Finally, since the forgetful functor $\mathbf{Grp}(\mathrm{Top}) \to \mathrm{Top}$ reflects isomorphisms, we have that $\pi$ is an isomorphism in $\mathbf{Grp}(\mathrm{Top})$. 
But we might as well go whole hog and make it even more direct, by following the diagram chase implicit in the preceding paragraph and constructing the inverse of $\pi$ explicitly. (I'll use additive notation here, even though we're in the context of groups and not abelian groups.) Thus, let $j$ be a section of the split epi $q$. We have $p \pi (j p) = q j p = 1_B p = p$, so $p(1_E - \pi j p) = 0$. It follows that $1_E - \pi j p$ factors through the kernel $i: \ker(p) \to E$; write $1_E - \pi p j = i g$ for some (unique) $g: E \to \ker(p)$. Let $\phi: \ker(q) \to \ker(p)$ be the isomorphism on display above. I claim that the continuous map 
$$s = i'\phi^{-1} g + j p: E \to F$$ 
is inverse to $\pi: F \to E$. Indeed, in one direction, we have 
$$\pi(i'\phi^{-1} g + j p) = \pi i' \phi^{-1} g + \pi j p = i \phi \phi^{-1} g + \pi j p = i g + \pi j p = (1_E - \pi j p) + \pi j p = 1_E.$$ 
In the other direction, to prove $s \pi = 1_F$, we first note that 
$$\pi (1_F - s \pi) = \pi - \pi s \pi = \pi - 1_E \pi = 0$$ 
so that in particular, $0 = p \pi (1_F - s \pi) = q (1_F - s \pi)$. Therefore $1_F - s\pi$ factors through the kernel of $q$: we have $1_F - s\pi = i'h$ for some unique $h: F \to \ker(q)$. From the equation displayed above, we thus have $0 = \pi i' h = i \phi h$. Since $i \phi$ is monic, this implies $h = 0$. Therefore $1_F - s\pi = i' h = 0$, and this completes the proof.<|endoftext|>
TITLE: Do we have a "topological assembly map" in the Baum-Connes conjecture?
QUESTION [8 upvotes]: In the equivariant Atiyah-Singer index theorem, when $G$ is a compact group acting on a manifold $M$ and $R(G)$ is the representation ring of $G$. We have the analytic index
$$
\text{a-ind}: K^*_G(TM)\rightarrow R(G)
$$
given by considering the kernel and cokernel of a elliptic operator as representations of $G$.
We also have the topological index 
$$
\text{a-ind}: K^*_G(TM)\rightarrow R(G)
$$
given by Bott periodicity or the Chern character and Todd class, etc.
Now if the group $G$ is not compact, we cannot apply the above method in the naive way. However, in the content of noncommutative geometry we have the assembly map
$$
\mu: K^G_*(M)\rightarrow K _ * (C^* _ r (G) ) 
$$
which is a generalization of the analytic index map and the Baum-Connes conjecture claims that if we take $M$ to be the unversal proper $G$ space $\mathcal{E}G$, then the assemble map
$$
\mu: K^G_*(\mathcal{E}G)\rightarrow K _ * (C^* _ r (G) ) 
$$
is an isomorphism. For more details we can refer to Alain Valatte's book Introduction to the Baum-Connes Conjecture.
My question is not about the Baum-Connes conjecture. but the following: Since the assembly map is an analytic index, does it make sense to define a topological index in this setting?
I've heard that the word "assembly map" originally came from surgery theory in topology but I'm not sure whether or not that gives us the desired definition.

REPLY [6 votes]: There is an interpretation of the Baum-Connes assembly map as the index of a $C^*_r(G)$-linear operator in case $G$ is discrete and torsion-free. The group $K_*(BG)$ can be represented by triples $(M, E, f)$, where $M$ is a closed spin$^c$-manifold, $E$ is a complex vector bundle and $g \colon M \to BG$ is a continuous map with respect to an equivalence relation given in Section 5 here. 
Let $\mathcal{V} = \widetilde{M} \times_{\lambda} C^*_r(\pi_1(M))$ be the bundle of right Hilbert $C^*_r(\pi_1(M))$-modules over $M$, where $\lambda \colon \pi_1(M) \to Aut(C^*_r(\pi_1(M)))$ (module automorphisms) is the left multiplication with elements of $\pi_1(M)$. Let $D$ be the Dirac operator on $M$. We can modify $D$ to act on smooth sections of the bundle $S \otimes E \otimes \mathcal{V}$ over $M$, where $S$ is the spinor bundle. Mishchenko and Fomenko have developed a theory of how to associate and index to these $A$-linear operators for some $C^*$-algebra $A$. This can be found here. The map $[M,E,f] \mapsto [ind(D^{E \otimes \mathcal{V}}_+)]$ is the assembly map.
They also prove an index theorem for these operators, which involves a topological index on the right hand side. For this you first need a Chern character, which can be obtained for nice enough $C^*$-algebras from the Künneth theorem for $K$-theory, i.e. we have isomorphisms
$$
K_0(C(M,A)) \otimes \mathbb{R} = K_0(C(M) \otimes A) \otimes \mathbb{R} \cong \left(K_0(C(M)) \otimes K_0(A) \oplus K_1(C(M)) \otimes K_1(A) \right) \otimes \mathbb{R} 
$$
which yields a map $K_0(C(M,A)) \to K_0(C(M)) \otimes K_0(A) \otimes \mathbb{R}$ by projection to one summand and finally a Chern character
$$
ch \colon K_0(C(M,A)) \to H^{\rm even}(M; K_0(A) \otimes \mathbb{R})
$$
by applying the usual Chern character to $K_0(C(M))$. The group $K_0(C(M,A))$ is the natural home for (formal differences of) isomorphism classes of Hilbert $A$-module bundles over $M$. In particular $E \otimes \mathcal{V}$ represents an element in $K_0(C(M,C^*_r(\pi_1(M)))$. A corollary of the Mishchenko-Fomenko index theorem is now that 
$$
{\rm ind}(D^{E \otimes \mathcal{V}}_+) = \int_M \hat{A}(M)\cdot ch(E \otimes \mathcal{V})
$$
and there you have your topological index (if I have done everything correct). I glossed over some things here, e.g. the grading on $S$. I also recommend the notes by Higson about the Baum Connes conjecture. (I recommend reading everything by Higson. He does a marvelous job at explaining things!)<|endoftext|>
TITLE: Visual pictures of rotation and torsion
QUESTION [7 upvotes]: In vector analysis / differential geometry we have rotation and torsion. The formalisms are certainly well known. But how could I best explain the geometric pictures and their difference (as a coach of an exercises group of bachelors who just started this topic)?

REPLY [7 votes]: Here is a beautiful animation created by Urs Hartl at Universität Münster:
   
Urs created this in Maple. The worksheet (.mw), which allows user-input 
of curves, is here.<|endoftext|>
TITLE: Homeomorphisms and disjoint unions
QUESTION [32 upvotes]: Let $X$ and $Y$ be compact subsets of $\mathbb{R}^n$.  Assume that $X \sqcup X \cong Y \sqcup Y$ (here $X \sqcup X$ is the disjoint union of two copies of $X$, considered as a topological space, and similarly for $Y \sqcup Y$).  Then I'm pretty sure that we must have $X \cong Y$.  This clearly holds if $X$ and $Y$ are connected, but I can't seem to prove it in general.  Can anyone help me?

REPLY [35 votes]: The result you want is false.  Counterexamples are given in
Yamamoto, Shuji and Yamashita, Atsushi,
A counterexample related to topological sums. 
Proc. Amer. Math. Soc. 134 (2006), no. 12, 3715–3719.
These counterexamples are compact subsets of $\mathbb{R}^4$.<|endoftext|>
TITLE: What is about nonassociative geometry?
QUESTION [30 upvotes]: At the end of a conference given by Alain Connes in 2000 (here is a video in French), a member of the audience asked a question. I transcribed and translated it for you below:  
Audience: You showed that noncommutative geometry greatly simplifies the physics in a very elegant way, by actually losing a property that seems a bit simple. Similary, if we work with algebras, complex numbers, quaternions and octonions..., are there people investigating some nonassociative geometries?
Alain Connes: Ok, I can explain you about this right now...
The video cuts off at that time, even before Alain finishes his explanation, so I post this question here :  

What is about nonassociative geometry
  ?

Sir Michael Atiyah (here at 58'): Connes' theory is very beautiful, but it only deals with associative algebras, so it can't deals with the octonions, so that's why I think this theory is not quite finished.

REPLY [14 votes]: There is progress in this direction by mathematicians here: Jordan operator algebra 
(see also this Physics post : Non-associative operators in Physics)  
Warning:  the Jordan operator algebras exchange the associativity by the commutativity, in fact their product $\circ$, given by $a_1 \circ a_2 = \frac{1}{2}(a_1*a_2+a_2*a_1)$, is commutative nonassociative, whereas $*$ is noncommutative associative.  So it's an advanced but it's not really satisfying.
A satisfying advanced would be with nonassociative and noncommutative algebras.<|endoftext|>
TITLE: What is the genus of the limit of a family of singular curves?
QUESTION [6 upvotes]: Let $\mathcal{X}$ be a flat family of (proper) algebraic curves. If generic fibers in $\mathcal{X}$ are non-singular of genus $g$, then the geometric genus (i.e. genus of the desingularizations) of special fibers must be $\leq g$. Does this inequality of the geometric genus remain valid also in the case that generic fibers of $\mathcal{X}$ are singular?
More specifically, if generic fibers of $\mathcal{X}$ are singular rational curves, then can the special fiber (assume it is reduced and irreducible) of $\mathcal{X}$ be non-rational?

REPLY [12 votes]: If the base of the family is a quasi projective variety, by taking sections and base change you can reduced to the case where $\mathcal X$ is a surface fibered over a smooth curve $B$.
Then one can normalize $\mathcal X$ and then solve the remaining singularities. 
In this way one gets a new  suface $\mathcal X'$ fibered over the same base $B$. 
The general fiber $F$ of $\mathcal X'$ is the normalization the general fiber of $\mathcal X$ and the special fiber of $\mathcal X'$ contains a component $D$ that maps  onto the original special fiber $X_0$ (I'm assuming, as in the question, that $X_0$ is reduced and irreducible).  Now one can use the adjunction formula on the smooth surface $\mathcal X'$ and 
Zariski's lemma (Barth-Peters Van de Ven , p.90 of the old edition), to show that $p_a(D)\le p_a(F)=g(F)$.<|endoftext|>
TITLE: Generalized basis
QUESTION [5 upvotes]: In quantum mechanics, people introduce the notion of "continuous basis" (I actually don't know the mathematical denomination of it). It is not a Schauder basis. I would like to know what could be a good definition of it, and what are the possible difficulties of defining it.
We have a Hilbert space $\mathcal{H}$, and a family of "vectors", $ \left\lbrace|x \rangle \right\rbrace_{x\in \mathbb{R}} $ such that any vector $|\psi \rangle $ in $\mathcal{H}$ can be written
" $|\psi \rangle = \int_{\mathbb{R}} \psi(x)|x \rangle $ "
and
$\int_{\mathbb{R}} |x \rangle \langle x | dx = Id$

We don't say where those $|x \rangle $ live, and in general they are not in $\mathcal{H}$. Usual example from physics $\mathcal{H}=L^2(\mathbb{R})$, and where we think of $|x \rangle$ as the delta distribution.
It looks a little bit like the spectral theorem that allows to write self adjoint operators as integral with a projector valued measure on the spectrum. I also came across that word, "direct integral" that may have a link.
we usually also take "orthonormal basis", i.e 
$\langle x |y \rangle = \delta(x-y)$ (what is the meaning of seing it as a distribution in 2 variables)

Of course the decomposition of a vector has to be unique. In the example $\mathcal{H}=L^2(\mathbb{R})$ the coefficients is itself a function $\psi\in L^2(\mathbb{R})$, then unique in the sense "equal almost everywhere" 
Among all the questions such presentation raises:

is the hilbert space structure playing any role? example in defining $\langle x |$ as the "dual basis". Because since those vectors are not even in the hilbert space, their scalar product is not defined.
is there any difficulty to define an abstract vector space generated by the family $ \left\lbrace|x \rangle \right\rbrace_{x\in \mathbb{R}} $. (Then of course there is still the problem of why a vector from our starting hibert space is equal to some construction in another abstract space.)
If this whole business is actually well defined, we can see fourier transform as a particular case of writing a vector in the following generalized basis (also from physics)

$|p\rangle:= \left\lbrace x\rightarrow \frac{1}{\sqrt{2\pi}} e^{ipx}\right\rbrace$
and the theorems of Fourier transform being an isometry and being invertible would just say that $|p\rangle $ is an "orthonormale" basis.

REPLY [7 votes]: You may want to have a look at Gelfand triples. In the example of the impulse operator and the Fourier transform you consider the triple consisting of the Schwartz space, the Hilbert space $L^2(\mathbb{R}^n)$ and the dual of the Schwartz space. The momentum operator is defined on the Schwartz space. The Fourier transform is well-defined on this triple: All the three spaces get mapped to themselves. The waves $e^{ipx}$ are elements of the dual of the Schwartz space and are a complete set of generalised eigen values of the momentum operator. 
Let me sketch the general situation: You have a Hilbert space $H$ and a nuclear topological vector space $S$ linearly embedded into the Hilbert space such that the scalar product is continuous with respect to the topology of $S$. We consider self-adjoint operators $T$ which are defined on $S$. Then generalised eigenvectors as elements of the dual space of $S$ can be defined in the obvious way. Let $\sigma$ be the spectrum of $T$. We can decompose the Hilbert space as a direct integral using a measure $\mu$
$H=\int^\oplus H(\lambda)\mathrm{d}\mu(\lambda)$
such that $T$ acts as a multiplication by $\lambda$ on each space $H(\lambda)$. There exists a unitary operator $U$ mapping $H$ to some space $L^2(X)$ where $T$ acts as a multiplication by a $\sigma$-valued function $a$. It can be proven that for each $x$ in $X$ there exists a $\phi_x$ in the dual space of $S$ such that for every $f\in S$ the functions $Uf\colon X\to\mathbb{C}$ and $x\mapsto\phi_x(f)$ are equal almost everywhere. The functional $\phi_x$ is a generalised eigenvector corresponding to the eigenvalue $a(x)$, that means, given a test function $f\in S$:
$\phi_x(Tf)=a(x)\phi_x(f)$
The space $H(\lambda)$ corresponds, informally spoken, to a space of functions defined on $a^{-1}(\left\{\lambda\right\})$. Thus, by taking linear combinatons, for every element $x\in H(\lambda)$ there is a generalised eigenvector $\phi$ for the eigenvalue $\lambda$. If you choose an orthonormal basis of each space $H(\lambda)$ the corresponding generalised eigenvectors are “complete” in a sense related to the direct integral decomposition.
In the book Introduction to Axiomatic Quantum Field Theory by Bogolubov, Logunov and Todorov you can read a short description, but they do not prove it. They refer to Generalized Functions, Vol. 4: Applications of Harmonic Analysis by Gelfand and Vilenkin.

REPLY [3 votes]: I believe the situation you asked about is addressed by the notion of "rigged Hilbert space"; see for example http://en.wikipedia.org/wiki/Rigged_Hilbert_space .  (If I remember correctly, I came across this notion in the book of Bogolyubov and Shirkov on quantum field theory.)<|endoftext|>
TITLE: The image of a measurable set under a measurable function.
QUESTION [8 upvotes]: Let $f:X \rightarrow (Y, \mathcal{Y})$ be an abstract function, with $\mathcal{Y}$ a $\sigma$-algebra on $Y$. Endow $X$ with $f^{-1}(\mathcal{Y})$. Is then $f(X)$ a measurable set in $Y$? If not, are there simple conditions on $f$ making $f(X)$ measurable? If $\mathcal{Y}$ were a $\sigma$-ring, would this modify anything?
More concisely (and generally): when is the image of a measurable set under a measurable function a measurable set?

REPLY [11 votes]: There are actually positive results if you change the context a little bit.
Suppose that $X$ is a separable complete metric space, i.e., a Polish space, 
and assume that $Y$ is something like $\mathbb R^n$, a Polish space that carries
a measure that interacts nicely with the topology like the Lebesgue measure.
Now, if $f:X\to Y$ is Borel measurable, then for every Borel set $B\subseteq X$
the image $f[B]$ is not necessarily Borel in $Y$, but it is Lebesgue measurable in $Y$.<|endoftext|>
TITLE: The number of lattice paths below y=n/m x for gcd(m,n) = 1
QUESTION [6 upvotes]: The motivation of my question is the recent preprent of Armstrong, Rhoades and Williams http://arxiv.org/abs/1305.7286 on rational Catalan combinatorics.
An important starting point of this paper is the fact that the number of lattice paths with steps $(1,0)$ and $(0,1)$ from the origin to $(m,n)$, where $gcd(m,n)=1$ weakly below the diagonal $y=\frac{n}{m} x$ equals
$$
\frac{1}{m+n}\binom{m+n}{m}.
$$
This can be proved (as done by Bizley, http://www.jstor.org/discover/10.2307/41139633?uid=3737528&uid=2&uid=4&sid=21102085756863) by considering all cyclic permutations of each path from the origin to $(m,n)$ and demonstrating that among these there is exactly one path below the diagonal.
I would like to know of any other proof of this fact, for example using generatingfunctionology.  
One approach I looked at (from the paper by Gessel and Ree http://people.brandeis.edu/~gessel/homepage/papers/faber.pdf) extends the recurrence $B(m,n)=B(m,n-1)+B(m-1,n)$ to all of $\mathbb N^2$ and then uses suitable initial values $B(m,0)$ and $B(0,n)$ to obtain generating functions for paths below $y=\frac{1}{m} x$.  But it seems that the initial values for the general case are unpleasant.

REPLY [7 votes]: You can use the approach of my paper A factorization for formal Laurent series, 
Journal of Combinatorial Theory, Series A
28 (1980) 321-337. Although this problem is not considered in that paper, Theorem 4.1 and its proof give the following result: Let $m$ and $n$ be relative prime positive integers and let $p(k)$ be the number of paths from $(0,0)$ to $(km,kn)$ weakly below the diagonal $y=\frac{n}{m}x$. Expand 
$$\log\left(\frac{1}{1 -t^{n}x-t^{-m}y}\right)$$
as a power series in $x$ and $y$ and let $f$ be the constant term in $t$. Explicitly, we have
$$f= \sum_{i=1}^\infty \frac{1}{(m+n)i} \binom{(m+n)i}{mi} x^{mi}y^{ni}.$$
Then 
$$\sum_{k=0}^\infty p(k) x^{mk}y^{nk} = e^f.$$
This is equivalent to Grossman's formula, which was proved by Bizley. In particular, $p(1) = \binom{m+n}{m}/(m+n)$.
(Of course in the final result we could replace $x^my^n$ with a single variable.)
The method of this paper can also be used to count paths that stay strictly below (or above) the line $y=\frac{n}{m}x$.<|endoftext|>
TITLE: Explanations for mathematicians, about the falsifiability (or not) of string theory
QUESTION [30 upvotes]: Like many other mathematicians, I think string theory very attractive. This theory has wonderfully influenced many new topics in mathematics (I myself have worked on one of them), but it's not the issue here (see for example there for this point).  
Unfortunately, it is well known that string theory is not appreciated by some of the physics community, because of its alleged non-falsifiability. Nevertheless, here and there, I hear some say that there are some possible indirect experiments...  
So, to do justice to string theory, I ask those who can to explain here (for mathematicians) one reason why this theory is non-falsifiable or one experiment that would test it.

REPLY [6 votes]: I was encouraged to post this answer from Physics Stack Exchange to the nearly equivalent question "What experiment would disprove string theory?" here as well.
One can disprove string theory by many observations that will almost certain not occur, for example:

By detecting Lorentz violation at high energies: string theory predicts that the Lorentz symmetry is exact at any energy scale; recent experiments by the Fermi satellite and others have showed that the Lorentz symmetry works even at the Planck scale with a precision much better than 100% and the accuracy may improve in the near future; for example, if an experiment ever claimed that a particle is moving faster than light, string theory predicts that an error will be found in that experiment
By detecting a violation of the equivalence principle; it's been tested with the relative accuracy of $10^{-16}$ and it's unlikely that a violation will occur; string theory predicts that the law is exact
By detecting a mathematical inconsistency in our world, for example that $2+2$ can be equal both to $4$ as well as $5$; such an observation would make the existing alternatives of string theory conceivable alternatives because all of them are mathematically inconsistent as theories of gravity; clearly, nothing of the sort will occur; also, one could find out a previously unknown mathematical inconsistency of string theory - even this seems extremely unlikely after the neverending successful tests
By experimentally proving that the information is lost in the black holes, or anything else that contradicts general properties of quantum gravity as predicted by string theory, e.g. that the high center-of-mass-energy regime is dominated by black hole production and/or that the black holes have the right entropy; string theory implies that the information is preserved in any processes in the asymptotical Minkowski space, including the Hawking radiation, and confirms the Hawking-Bekenstein claims as the right semiclassical approximation; obviously, you also disprove string theory by proving that gravitons don't exist; if you could prove that gravity is an entropic force, it would therefore rule out string theory as well
By experimentally proving that the world doesn't contain gravity, fermions, or isn't described by quantum field theories at low energies; or that the general postulates of quantum mechanics don't work; string theory predicts that these approximations work and the postulates of quantum mechanics are exactly valid while the alternatives of string theory predict that nothing like the Standard Model etc. is possible
By experimentally showing that the real world contradicts some of the general features predicted by all string vacua which are not satisfied by the "Swampland" QFTs as explained by Cumrun Vafa; if we lived in the swampland, our world couldn't be described by anything inside the landscape of string theory; the generic predictions of string theory probably include the fact that gravity is the weakest force, moduli spaces have finite volume, and similar predictions that seem to be satisfied so far
By mapping the whole landscape, calculating the accurate predictions of each vacuum for the particle physics (masses, couplings, mixings), and by showing that none of them is compatible with the experimentally measured parameters of particle physics within the known error margins; this route to disprove string theory is hard but possible in principle, too (although the full mathematical machinery to calculate the properties of any vacuum at any accuracy isn't quite available today, even in principle)
By analyzing physics experimentally up to the Planck scale and showing that our world contains neither supersymmetry nor extra dimensions at any scale. If you check that there is no SUSY up to a certain higher scale, you will increase the probability that string theory is not relevant for our Universe but it won't be a full proof
A convincing observation of varying fundamental constants such as the fine-structure constant would disprove string theory unless some other unlikely predictions of some string models that allow such a variability would be observed at the same time

The reason why it's hard if not impossible to disprove string theory in practice is that string theory - as a qualitative framework that must replace quantum field theory if one wants to include both successes of QFT as well as GR - has already been established. There's nothing wrong with it; the fact that a theory is hard to exclude in practice is just another way of saying that it is already shown to be "probably true" according to the observations that have shaped our expectations of future observations. Science requires that hypotheses have to be disprovable in principle, and the list above surely shows that string theory is. The "criticism" is usually directed against string theory but not quantum field theory; but this is a reflection of a deep misunderstanding of what string theory predicts; or a deep misunderstanding of the processes of the scientific method; or both.
In science, one can only exclude a theory that contradicts the observations. However, the landscape of string theory predicts the same set of possible observations at low energies as quantum field theories. At long distances, string theory and QFT as the frameworks are indistinguishable; they just have different methods to parameterize the detailed possibilities. In QFT, one chooses the particle content and determines the continuous values of the couplings and masses; in string theory, one only chooses some discrete information about the topology of the compact manifold and the discrete fluxes and branes. Although the number of discrete possibilities is large, all the continuous numbers follow from these discrete choices, at any accuracy. 
So the validity of QFT and string theory is equivalent from the viewpoint of doable experiments at low energies. The difference is that QFT can't include consistent gravity, in a quantum framework, while string theory also automatically predicts a consistent quantum gravity. That's an advantage of string theory, not a disadvantage. There is no known disadvantage of string theory relatively to QFT. For this reason, it is at least as established as QFT. It can't realistically go away.
In particular, it's been showed in the AdS/CFT correspondence that string theory is automatically the full framework describing the dynamics of theories such as gauge theories; it's equivalent to their behavior in the limit when the number of colors is large, and in related limits. This proof can't be "unproved" again: string theory has attached itself to the gauge theories as the more complete description. The latter, older theory - gauge theory - has been experimentally established, so string theory can never be removed from physics anymore. It's a part of physics to stay with us much like QCD or anything else in physics. The question is only what is the right vacuum or background to describe the world around us. Of course, this remains a question with a lot of unknowns. But that doesn't mean that everything, including the need for string theory, remains unknown.
What could happen - although it is extremely, extremely unlikely - is that a consistent, non-stringy competitor to string theory that is also able to predict the same features of the Universe as string theory can emerges in the future. (I am carefully watching all new ideas.) If this competitor began to look even more consistent with the observed details of the Universe, it could supersede or even replace string theory. It seems almost obvious that there exists no "competing" theory because the landscape of possible unifying theories has been pretty much mapped, it is very diverse, and whenever all consistency conditions are carefully imposed, one finds out that he returns back to the full-fledged string/M-theory in one of its diverse descriptions.
Even in the absence of string theory, it could hypothetically happen that new experiments will discover new phenomena that are impossible - at least unnatural - according to string theory. Obviously, people would have to find a proper description of these phenomena. For example, if there were preons inside electrons, they would need some explanation. They seem incompatible with the string model building as we know it today.
But even if such a new surprising observation were made, a significant fraction of the theorists would obviously try to find an explanation within the framework of string theory, and that's obviously the right strategy. Others could try to find an explanation elsewhere. But neverending attempts to "get rid of string theory" are almost as unreasonable as attempts to "get rid of relativity" or "get rid of quantum mechanics" or "get rid of mathematics" within physics. You simply can't do it because those things have already been showed to work at some level. Physics hasn't yet reached the very final end point - the complete understanding of everything - but that doesn't mean that it's plausible that physics may easily return to the pre-string, pre-quantum, pre-relativistic, or pre-mathematical era again. It almost certainly won't.<|endoftext|>
TITLE: Can the unsolvability of quintics be seen in the geometry of the icosahedron?
QUESTION [36 upvotes]: Q1. Is it possible to somehow "see" the unsolvability of quintic polynomials
in the $A_5$ symmetries of the icosahedron (or dodecahedron)?
Perhaps this is too vague a question.
Q2. Are there consequences of the unsolvability of $A_5$ that can
be viewed in terms of the geometry of the icosahedron?
I am seeking some connection for pedagogical reasons,
from the geometry to the group theory.
One could argue from orbits of vertices of the icosahedron under its various symmetries
that $A_5$ has no (nontrivial) normal subgroups, but it is a long way
from there to the unsolvability of $A_5$ for students who have not yet
studied group theory.
Thanks for any connections you can suggest!
Update 1. Answered by Barry Cipra and Geral Edgar:

      

No longer $2.25, but still inexpensive through used-booksellers.
Update 2. And here is Jerry Shurman's book (PDF download), as cited by Barry and Sam Hopkins:

REPLY [27 votes]: This is an old question, but I wanted to write a proof that $A_5$ is simple via symmetries of the icosahedron, using as little group theory as possible. I don't think that it can lead to a proof of the unsolvability of the quintic without the usual group theory (permutations, normal subgroups, quotients, solvable groups), and, of course, field theory.
Theorem: The group of rotations of the icosahedron is of order 60, simple, and isomorphic to the alternating group $A_5$. Therefore, $A_5$ is simple.
Proof: Consider a regular icosahedron, the convex hull of the 12 points $(0,\pm 1,\pm\phi)$, $(\pm 1,\pm\phi,0)$ and $(\pm\phi,0,\pm 1)$ in $\mathbb R^3$, where $\phi=\frac{1+\sqrt 5}2$ is the golden ratio. It has 12 vertices, 30 edges and 20 faces.
Let $G$ be its group of rotations. It has order $2*30=60$, because if $e$ is an oriented edge, then for any oriented edge $e'$, there is exactly one rotation that carries $e$ to $e'$. Any nontrivial rotation $g$ is of one of three types, according to whether its axis contains a vertex, an interior point of an edge (which must be the midpoint) or an interior point of a face (which must be the center). Its order is 5, 2 or 3, respectively.
To rotate a rotation: Notice that conjugation in $SO(\mathbb R^3)$ is particularly transparent: if $g$ is the rotation of angle $\omega$ about an axis $r$, then any conjugate $hgh^{-1}$ is the rotation of the same angle $\omega$ about the axis $h(r)$.
Proof that $G$ is simple: A normal subgroup $N$ that contains a nontrivial element $g$ must also contain the cyclic group of powers of $g$, and their conjugates, which are all the elements of the same type of $g$. To show that $N=G$, it's enough to see that $G$ can be generated with all the elements of any one of the three types.

With vertex rotations: Any vertex $v$ can be moved to a neighbor position $v'$ by rotating about the third vertex $v''$ of any of the two faces $f$ that contain $v$ and $v'$. So you can move any vertex to any place. After that you can rotate about that final place to put the icosahedron in any position.
With face rotations: If an edge $e$ joins $v$ with $v'$, then you can move $v$ to $v'$ using a rotation about any of the two faces $f$ and $f'$ that meet at $e$. If you move forth using $f$ and back using $f'$, then you return $v$ to its original position, so the composite is a rotation about vertex $v$. Again, we see that we can move and rotate vertices, so we generate $G$.
With edge rotations: Similar to the other cases.

 
The compound of 5 octahedra: Consider the construction of three orthogonal golden rectangles inscribed in the icosahedron. The six short sides are edges of the icosahedron, and we can paint them with the same color. Observe that, whenever an edge $e$ is chosen, there is a unique way to build the rectangles in such a way that the edge $e$ is one of the short sides, so in this way the 30 edges can be painted with 5 different colors. The midpoints of the edges of each color are the vertices of an octahedron, so we have formed a compound of 5 octahedra.
Proof that $G=A_5$:  Each rotation permutes the colors, so we have a morphism $G\to S_5$. The vertex rotations have order 5, so they must give 5-cycles, which are even permutations. Similarly, the face rotations give 3-cycles, which are also even. Finally, consider a half-turn rotation $g$ about the midpoint of an edge $e$ of color $E$. Let $a,b,c,d$ be the consecutive sides of the quadrilateral obtained by joining the two faces that share the edge $e$. The edges $a,b,c,d,e$ have different colors $A,B,C,D,E$. It's now easy to see that $g$ interchanges $A$ with $C$ and $B$ with $D$, leaving $E$ fixed. This permutation is also even. Since any nontrivial rotation of the icosahaedron gives a nontrivial even permutation, we have an injective morphism from the group $G$, of order $60$, to the group $A_5$, of the same order. It must be an isomorphism. $\square$
Exercise: Notice that, at any vertex, the five colors meet, giving a cyclical order of the five colors. Show that the 12 cyclical orders found at the vertices are different.
Regarding the question Q2, there is an interesting (though not quite elementary) application: The fact that the icosahedron group $G$ is simple is a step in the construction of Poincaré's homology sphere, a 3-manifold $M$ that has the same homology of the sphere $S^3$, but is not homeomorphic to $S^3$, because its fundamental group is nontrivial. A construction of this kind allowed Poincaré to reject a preliminary wrong version of his famous conjecture, which was originally stated in terms of homology (a 3-manifold with the homology of the sphere is homeomorphic to a sphere, as happens in dimension 2).
A construction of the Poincaré sphere: Consider the 2-to-1 universal cover $\pi:S^3\to SO(\mathbb R^3)$, where $S^3=S_{\mathbb H}$ is the sphere of unit quaternions. (Explicitely, if $q$ is a unit quaterion, then $\pi(q)$ acts on $\mathbb R^3=\{v=x\mathrm i+y\mathrm j+z\mathrm k:\ x,y,z\in\mathbb R\}$ by conjugation, sending $v\mapsto qvq^{-1}$.) Its kernel is $\{\pm 1\}$. Define the binary icosahedral group $2G:=\pi^{-1}(G)$. It's nonabelian and of order 120. Let $M=SO(\mathbb R^3)/G=S^3/2G$ be the Poincaré homology sphere.
Fundamental group and homology of the Poincaré sphere: Notice that, since the quotient map $q:S^3\to M$ is a universal cover, we have $\Pi_1(M)=2G$. Since the Poincaré sphere is an orientable compact 3-manifold with one connected component, its bottom and top homology groups $H_0(M)$ and $H_3(M)$ are isomorphic to $\mathbb Z$. It remains to be proven that $H_1(M)$ and $H_2(M)$ are trivial. By Poincaré duality, it is enough to prove that $H_1(M)=0$, since $H_2(M)\cong H^1(M)\simeq H_1(M)$.
The homology group $H_1(M)$ is obtained by abelianization of the fundamental group $\Pi_1(M)=2G$, so we must prove that the commutator subgroup $[2G,2G]$ equals $2G$. Observe that $\pi([2G,2G])=[G,G]=G$, where the first equality holds because $\pi$ maps $2G$ onto $G$, and the second one holds because $G$ is nonabelian and simple, so the nontrivial normal subgroup $[G,G]$ must be equal to the whole group. So $\pi$ restricts to a surjective morphism $[2G,2G]\to G$ and there are two possibilities:

The morphism is 2-to-1, so $[2G,2G]=2G$ and we are done.
The morphism is an isomorphism.

To rule out the second possibility, we prove that $-1$, which is in the kernel of $\pi$, is an element of $[2G,2G]$. We do so by expressing it as a product of commutators of elements of $2G$. For that, observe that $i\in 2G$ because the rotation $\pi(i)$, which maps $i\mapsto i$, $j\mapsto -j$ and $k\mapsto -k$, preserves the set of vertices of the icosahedron. In the same way we see that $j\in 2G$, and then we can compute $[i,j]=iji^{-1}j^{-1}=ij(-i)(-j)=ijij=kk=-1.$
I think that there is also a more conceptual proof that the restriction $\pi|_{[2G,2G]}^G:[2G,2G]\to G$ is not 1-to-1 along these lines: If it was 1-to-1, then $\pi$ would admit a section along $G$. Since the discrete subgroup $G$ is ``pretty dense'' in $SO(\mathbb R^3)$, this section can (hopefully) be extended to a section defined on the whole rotation group $SO(\mathbb R^3)$, which is absurd.
Off-topic comment about the density of $G$ in $SO(\mathbb R^3)$: I think that any rotation $h$ is of the form $h=gk$, with $g\in G$ and $k$ a rotation of angle at most $\frac{2\pi}6$. The situation I have in mind is the following. Let $h$ be a rotation of angle $\frac{2\pi}6$ about the direction $(1,1,1)$. It sends the convex hull of
$\{(0,\pm 1,\pm\phi), (\pm 1,\pm\phi,0), (\pm\phi,0,\pm 1)\}$ to the convex hull of 
$\{(0,\pm\phi,\pm 1), (\pm\phi,\pm 1,0), (\pm 1,0,\pm\phi)\}$.
The picture shows the superposition of the two icosahedra, known as compound of two icosahedra. The rotation $h$ can be decomposed as $g=hk$ in 8 ways. More visually, there are 8 rotations of angle $\frac{2\pi}6$ that move the red icosahedron to the yellow icosahedron.
 
Since the angle-of-rotation distance in $SO(\mathbb R^3)$ is just the double of the geodesic distance in $S^3$ (which is the angle of vectors), to verify what I claim we just have to prove that $2G$ is $\frac{2\pi}{12}$-dense in $S^3$. This, if true, can be checked numerically with error of less than 1 degree. But it would be more interesting to have a picture of the Voronoi-cell decomposition of $S^3$ with respect to the discrete set $2G$. I wonder if this leads to the other usual presentation of the Poincaré sphere (as a quotient of the dodecahedron, identifying opposite faces by a $\frac{2\pi}{10}$ right-handed twist).<|endoftext|>
TITLE: Are sums of sequences decidable?
QUESTION [25 upvotes]: Suppose that $f,g$ are rational functions with integer coefficients such that $\sum_{n=0}^{\infty}f(n)$ and $\sum_{n=0}^{\infty}g(n)$ both converge. Is it decidable whether
$\sum_{n=0}^{\infty}f(n)=\sum_{n=0}^{\infty}g(n)$? If this problem is decidable, then what is the computational complexity of this problem? If this problem is undecidable, then what is the turing degree of this problem? I am also interested in answers concerning generalizations of this problem such as when we take double sums or when we include factorial functions and exponential functions as well. This question does not have much to do with my current research. Instead, I am mainly interested in this question out of curiosity.

REPLY [14 votes]: It seems pretty clear that this is an open problem, so I will do the job of (trying to) put this question out of its misery by summarizing  a few of the remarks in the comments. This answer is community wiki.
Replacing $f(n)$ by $f(n)-g(n)$, one may as well ask whether the sum is zero or not. 
 Variations: Suppose one makes the restriction that $f(n)$ is an even function, or alternatively (and almost equivalently) consider summations from $-\infty$ to $\infty$. Then the degree of $f(n)$ must be at most $-2$, and one has (taking $R$ through a sequence of half integers and $C_R$ the corresponding circle centered at zero):
$$0 = \lim_{R \rightarrow \infty} \frac{1}{2 \pi i} \oint_{C_R} \frac{\pi \cos \pi z}{\sin \pi z} f(z) dz =
\sum_{-\infty}^{\infty} f(n)  + \sum_{f(\alpha) = \infty}  \pi \cot(\pi \alpha) \mathrm{Res}(f)(\alpha)$$ 
Literally speaking, this is only correct if $f(z)$ has simple poles, but the general case is no more difficult. It follows that determining whether the sum is equal to zero or not is the same as verifying an identity in
$$\overline{\mathbf{Q}}(e^{\pi \beta_1}, \ldots, e^{\pi \beta_n}),$$
for a fixed set of algebraic numbers $\beta_i$. By Schanuel's conjecture, the only such relationships are the "obvious" ones, so in this case one conjecturally has an algorithm.
Periods: Passing now to the general (no longer even) case but specializing in a different direction: for certain special $f(n)$, the sum in question is a period (e.g. $\zeta(n)$ for $n \ge 2$ an integer). Kontsevich and Zagier raise the question whether determining the equality of two periods is decidable or not; the expectation is that it should be, but there are no real ideas in this direction. There certainly are no known general algorithms for proving equalities. 
General Remarks: By expanding into partial fractions, one has the general formula:
$$\sum f(n) = \sum_{n=0}^{\infty} \sum_{r,s} \frac{(-1)^r \cdot r! \cdot A_{r,s}}{(n + \alpha_s)^{r+1}}
 = \sum_{r,s} A_{r,s} \cdot \psi^{(r)}(\alpha_i),$$ 
Where $\psi^{(r)}(z)$ is the Polygamma function, so $\psi^{(0)}(z) = \psi(z)$ is the logarithmic derivative of the Gamma function. Note that this only makes sense when one has convergence, which happens if and only if $\sum A_{0,s} = 0$ (this also means one can replace $\psi(z)$ by $\psi(z) + \gamma$ if you wish). Explicitly, for those who worry about such things, the $r=0$ term can be written (since $\sum A_{0,s} = 0$) as:
$$\sum_{n=0}^{\infty} \sum_{s} \frac{A_{0,s}}{(n + \alpha_s)}
= \sum_{n=0}^{\infty} \sum_{s} \frac{A_{0,s}}{(n + \alpha_s)} - \frac{A_{0,s}}{n+1}
= \sum_{s} A_{0,s} (\psi(\alpha_s) - \psi(1)) = \sum_s A_{0,s} \psi(\alpha_s)$$
The functions $\psi^{(r)}(z)$ can also be thought of as  Hurwitz zeta functions. The most general hope would be that the only relationships between values of polygamma functions at algebraic arguments are those which are occurring for "motivic" reasons. There's not much to motivate this beyond the related explanations (which are already conjectures) in the easier cases discussed above.<|endoftext|>
TITLE: Action of an isomorphism in cohomology as the intersection with the class of the graph
QUESTION [6 upvotes]: Let $X$ and $Y$ be two complex manifolds of dimension 2 and let $\varphi:X\rightarrow Y$ be an isomorphism. 
I have read that the action of $\varphi^*:H^2(Y,\mathbb{Z})\rightarrow H^2(X,\mathbb{Z})$ can be seen as $\omega\mapsto {p_1}_*([\Gamma]\cdot p_2^*(\omega))$ for all $\omega\in H^2(Y,\mathbb{Z})$. Here $p_1$ and $p_2$ are the two projections from $X\times Y$, $\Gamma$ is the graph of $\varphi$ and i interpret ${p_1}_*([\Gamma]\cdot p_2^*(\omega))$ as $\mathscr{P}({p_1}_*(\mathscr{P}^{-1}(\mathscr{P}(\Gamma)\wedge p_2^*(\omega))))$, where $\mathscr{P}$ is the Poincarè duality.
I think i can see why this is working, but do you know a formal proof for that or do you know where i can find it?
Thank you

REPLY [5 votes]: I believe this is a special case of a more general fact; I am not sure of all the signs off the top of my head, but here  is the idea. 
If $M$ and $N$ are orientable $d$-manifolds, the Künneth theorem gives
$$H^d(M \times N; \mathbb{Q}) \cong \bigoplus_k H^k(M; \mathbb{Q}) \otimes H^{d-k}(N; \mathbb{Q}).$$
To the second factor we first apply Poincare duality $H^{d-k}(N; \mathbb{Q}) \cong H_k(N; \mathbb{Q})$ and then the usual duality $H_k(N;\mathbb{Q})\cong H^k(N;\mathbb{Q})^\ast$ to rewrite this as
$$H^d(M \times N; \mathbb{Q}) \cong \bigoplus_k \text{Hom}\big( H^k(N; \mathbb{Q}), H^k(M; \mathbb{Q}) \big).$$
For any continuous map $f\colon M \to N$, the graph $\Gamma_f$ of the map $f$ is a $d$-dimensional submanifold of $M \times N$, so we can consider its class
$$[\Gamma_f] \in H^d(M \times N; \mathbb{Q}).$$
Claim: under the above isomorphism, we have
$$[\Gamma_f] = \bigoplus_k \big[\ f_{(k)}^*\colon H^k(N; \mathbb{Q}) \to H^k(M; \mathbb{Q}) \big]$$
where $f_{(k)}^*$ denotes the map induced on $k$-th cohomology by $f$.
There might be some signs missing here, but other than that I believe this is correct. (I don't know a reference for this, so if anyone does have a reference, I'd be grateful.) In particular, it doesn't seem that you need to assume that $\phi$ is an isomorphism, only continuous, or that $M$ and $N$ are surfaces.
As an aside, if you apply this twice with $M=N$, once to $\text{id}\colon M\to M$ and once to $f\colon M\to M$, you should be close to proving the Lefschetz fixed point theorem.

Edit: as David Speyer points out, this aside requires knowing that intersection of submanifolds is Poincare dual to the cup product. A reference for this can be found in the textbook "Differential forms in algebraic topology" by Bott and Tu, in 6.31 (on page 69): "under Poincare duality the transversal intersection of closed oriented submanifolds corresponds to the wedge product of forms". (From one perspective, this is the most important thing to know about cup product!)
Also, Mark Grant has kindly provided a reference for the fact I claimed above: this is proved in his answer to this Math Overflow question from 2010, by reducing the claim to the case of the diagonal embedding (i.e. when $f=\text{id}$).<|endoftext|>
TITLE: Levi decomposition in disconnected linear algebraic group (characteristic 0)?
QUESTION [10 upvotes]: For algebraic groups or Lie groups, the subject of Levi decompositions tends to be surrounded by some mystery in the literature (and in an older question raised here).   While I postpone further my intention to post a careful note on my homepage someday, I'm curious about the status of disconnected algebraic groups in this discussion.  Take the ground field $K$ to be algebraically closed of characteristic 0 throughout.

If $G$ is a disconnected linear algebraic group over $K$, of positive dimension, does it always have a Levi decomposition; if so, what can be said about conjugacy of Levi subgroups?

First I have to outline briefly the history of the concept of Levi decomposition.  This was first carried out for a finite dimensional Lie algebra $\mathfrak{g}$ and makes sense over an arbitrary field of characteristic 0:   Levi's theorem says that the solvable radical of $\mathfrak{g}$ has as a semidirect complement some semisimple subalgebra, while Mal'cev showed the conjugacy of any two such "Levi subalgebras" under special automorphisms of $\mathfrak{g}$ (which can be defined algebraically).   This is treated efficiently in modern language by Bourbaki in Chapter I, 6.8 of their treatise  Lie Groups and Lie Algebras. 
To carry this over to connected Lie groups (real or complex), one then has to use the nice correspondence between Lie subgroups and Lie subalgebras.   (But this is developed using exponentials.)    In the 1950s Chevalley experimented with linear algebraic groups in characteristic 0, using a sort of formal exponentiation process to imitate the Lie group framework.
Borel works some of this out in somewhat more modern language in II.7.9 of his Benjamin notes, reproduced in his GTM 126 second edition.   For the Lie algebra $\mathfrak{g}$ of a connected algebraic group $G$, the basic observation is that any Lie subalgebra equal to its derived algebra is algebraic, i.e., is the Lie algebra of some connected closed subgroup of $G$.   (And this subgroup is unique, since we are in characteristic 0.)   In particular, a semisimple Levi subalgebra of $\mathfrak{g}$ is the Lie algebra of a connected semisimple subgroup $H$ of $G$ whose intersection with the solvable radical is finite.
(In some sources the product of subgroups here is supposed to be semidirect, but this is artificial for cases like the general linear group.)   Using Jordan decomposition, this becomes the more useful decomposition of $G$ into a semidirect product of a connected reductive subgroup and the unipotent radical.
The conjugacy theorem for $\mathfrak{g}$ then carries over to the group in a natural way.    Unfortunately, there seems to be no complete account of this in a textbook, though Demazure-Gabriel translate some of it into scheme language in II, Section 6.  In any event, my question above doesn't seem to be answered directly, since the characteristic 0 correspondence between closed subgroups and their Lie algebras requires connected subgroups.
Footnote:  In prime charactertistic things tend to fall apart in general, as indicated in my 1967 note here.  But a recent paper by George McNinch offers numerous useful refinements here.

REPLY [3 votes]: See G. P. Hochschild: Basic Theory of Algebraic Groups and Lie Algebras (Graduate Texts in Mathematics)-Springer (1981) VIII, Theorem 4.3<|endoftext|>
TITLE: How to define the canonical sheaf on singular varieties
QUESTION [12 upvotes]: Let $X$ be a variety which might be singular, how to defined the canonical sheaf $K_X$ on $X$? 
When $X$ is a proper, irreducible variety over $\mathbb{C}$, Ueno defined $K_X$ as the pushforward of the canonical sheaf of its nonsigular model (see Chapter2 in his book " Classification Theory of Algebraic Varieties and Compact Complex Spaces"). However, I was wondering if there is a well-defined canonical sheaf on singular variety which is not proper (over $\mathbb{C}$ is OK for me). I will be very appreciated if someone can point out referenced of this kind.

REPLY [21 votes]: For $X$ normal, saying that canonical divisor is the pushforward of the canonical divisor of a resolution of singularities is totally fine (it even works in characteristic $p > 0$ if you happen to have a resolution).  In particular, if $\pi : Y \to X$ is a resolution of singularities, then $\pi_* K_Y$ is $K_X$ (here we define $\pi_* K_Y$ by simply throwing away any components of $K_Y$ that get contracted to non-divisorial varieties).  
However:  The pushforward of the canonical sheaf $\omega_Y$ is not the canonical sheaf $\omega_X$ in general.  Indeed, $\pi_* \omega_Y = \omega_X$ is very close to requiring that $X$ has rational singularities.  (Actually one definition of rational singularities, typically attributed to Kempf, is that $X$ is Cohen-Macaulay and $\pi_* \omega_Y = \omega_X$.
A good exercise is to show that if $X = \text{Spec} k[x,y,z]/(x^3+y^3+z^3)$ and $\pi : Y \to X$ is the blowup of the cone point, then $\pi_* \omega_Y = \mathfrak{m} \cdot \omega_X$ where $\mathfrak{m}$ is the maximal ideal of the origin.  (Hint:  use the adjunction formula and the formula for the canonical divisor when blowing up a point on $\mathbb{A}^3$).
The canonical sheaf
Ok, so what is the right definition of the sheaf $\omega_X$ in general?
Well, for a projective variety of dimension $d$, $i : X \hookrightarrow  \mathbb{P}^N$, define
$$i_* \omega_X := \text{Ext}^{N-d}\big(i_* O_X, O_{\mathbb{P}^N}(-N-1)\big).$$  Since $i$ is a closed embedding, this uniquely determines $\omega_X$.
For $X$ quasi-projective, you can define this by localizing.  There are generalizations which apply to other schemes of finite type over $k$ defined using the $f^!$ functor for $f : X \to k$ the structural map, but I won't get into that here.
The canonical sheaf is S2
It turns out that for any variety, $\omega_X$ is an S2 sheaf, this means it satisfies Hartog's phenomena (search math overflow).  In particular, the sheaf is determined by its codimension-1 behavior.  There's an easy way to see this, it turns out that if $h : X \to Z = \mathbb{P^d}$ is a generic projection to a hyperplane of the same dimension, then 
$$h_* \omega_X = \text{Hom}(h_* O_X, O_Z(-d-1)).$$
Now, it easily follows that this sheaf is S2 since it is reflexive on $Z$ and reflexive sheaves on $Z$ are always S2 (see for example Hartshorne's Generalized divisors on Gorenstein schemes).
Why does this matter?  Well it means that if $U$ is the regular locus of $X$, and furthermore $X \setminus U$ has codimension 2 on $X$ (which happens for example if $X$ is normal), then if $j : U \hookrightarrow X$ is the inclusion, then $j_* \omega_U = \omega_X$ since both sheaves are S2 and they agree outside a codimension-2 set.
Back to divisors
This also explains our first statement about divisors.  Indeed, any divisor, like $\pi_* K_Y$ is determined outside a codimension 2 set, it is determined on $U$ in fact.  And so if $X$ is normal, $\pi : Y \to X$ is an isomorphism outside of a codimension-2 set of $X$, and so the canonical divisor on that set works fine as a canonical divisor everywhere.  In particular, it can be computed on $Y$ as claimed.
For non-normal $Y$, something can be done, but the formula isn't quite so simple (you also have to describe by what exactly you mean by a divisor on a non-normal variety).<|endoftext|>
TITLE: If a unitsquare is partitioned into 101 triangles, is the area of one at least 1%?
QUESTION [46 upvotes]: Update: The answer to the title question is no, as pointed out by Tapio and Willie. I would be more interested in lower bounds.
Monsky's famous theorem with amazingly tricky proof says that if we dissect a square into an odd number of triangles, they cannot have the same area. However, this proof does not give any quantitative bounds, while from the compactness it follows that there should be some bound, e.g. on the size of the largest triangle that is well separated from the reciprocal of the number of triangles. I don't think the original proof gives any bounds, and I am not aware of any (substantially) different proofs.
So if a unitsquare is partitioned into n (odd) triangles, then what is the smallest $f$ such that the largest will have area at least $f(n)$? Monsky says that $f(n)>1/n$, but how much bigger? If we partition to $n-1$ (even) equal area triangles and then cut one of them, then this gives $f(n)\le 1/(n-1)$, which is not sharp as shown by Tapio and Willie. But can anyone give a better lower bound than $1/n$?

REPLY [32 votes]: As suggested in a comment above, I had asked a version of this question years ago. It makes sense to look at upper and lower bounds for the quantity $f(n)-\frac1n$. It is easily seen that 
$$ f(n) - \frac1n \le \frac1{n^2-n} $$
and indeed has shown that 
$$ f(n) - \frac1n = O\big(\frac1{n^3}\big), $$
see Bernd Schulze's paper "On the area discrepancy of triangulations of squares and trapezoids", The Electronic Journal of Combinatorics 18(1) (2011), #P137, 16 pp.
Computational experiments from an unpublished Diploma thesis (Katja Mansow, Ungerade Dreieckszerlegungen, TU Berlin 2003, 49 pages, in German) suggest that the true order of this difference is, however, singly-exponentially small. From gap-theorems in semi-algebraic geometry there is a doubly-exponential lower bound (details to be worked out).<|endoftext|>
TITLE: Continuity of the spectrum with respect to the metric
QUESTION [6 upvotes]: The following question is quite natural, but I am not aware of a reference dealing with it: let $M$ be a compact (smooth) manifold (posssibly with boundary) and $E$ a vector bundle on $M$ with an Hermitian or Euclidean metric, and let $g_u,\: u\in[0,1]$ be a smooth family of Riemannian metrics on $M$. For each $p=0,\ldots,{\rm dim}\: M$ we get a map ${\rm sp}(\Delta^p):\: [0,1] \to {\mathbb R}^{\mathbb N}$ which associates tu $u$ the ordered spectrum (with multiplicities) of the Laplace operator on $p$-forms on $(M,g_u)$ with coefficients in $E$. The question is: is this map continuous (with respect to the product topology on ${\mathbb R}^{\mathbb N}$)?
I believe that this is true, and that you can prove it using the min-max characterization of the spectrum, but as it seems a rather basic question I would think that somebody already answered some version of it.

REPLY [8 votes]: However, for continuity of individual eigenvalues (with appropriate provisos in the presence of multiplicity), the min-max characterization works fine. More refined statements (e.g. about the behavior of the spectrum for smooth or analytic one-parameter families of metrics) are contained in Kato's classic book `Perturbation theory for Linear Operators'.<|endoftext|>
TITLE: Reference request: Affine Grassmannian and G-bundles 
QUESTION [8 upvotes]: Let $G$ be an affine algebraic group over an algebraically closed field $k$ of zero characteristic. The set of cosets $X_G=G(k((t))/G(k[[t]])$ is called the Affine Grassmannian of $G$ and can be given the structure of an ind-$k$-variety, so that for a closed embedding of groups $H\to G$, we get a natural morphism $X_H\to X_G$, which is a closed embedding if $G/H$ is affine. I am interested in a (as detailed as possible) reference for this, but from a specific perspective as will be explain in the following.
There are several possible (equivalent) constructions. A direct approach, in the language of ind-varieties, can be found in Kumar's "Infinite grassmannians and Moduli spaces of G-bundles" for example (for a reductive $G$, which is fine for me), which describes the ind-variety structure explicitly. Apart of using some representation theory, that I don't know well enough, it also lacks an explicit universal property which makes it difficult to operate with and in particular to construct morphsims to and from it.
A more abstract approach  is to describe a functor $\operatorname{Gr}_G:kAlg\to Set$ for which $X_G$ is the set of $k$-points. Here is were the $G$-bundles (torsors) appear. There are the "global" and "local" approaches, in which roughly, $\operatorname{Gr}_G (A)$ classifies $A$-families of $G$-Bundles on a curve or a formal disc resp. together with a trivializaion away from a point. Now that one has a functor, it is possible to show that it is an ind-scheme. It is this approach that I would like to have a reference for.
I would like to mention, that this approach is outlined in Gaitsgory's seminar 
notes, and a more competent student would probably be able to fill in the details by herself, but unfortunately I find it difficult, so I was hopping there might be a more thorough treatment available.

REPLY [2 votes]: The following papers might also be helpful:
G. Faltings, Algebraic loop groups and moduli spaces of bundles, J. Eur. Math. Soc. 5 (2003), 41-68.
G. Pappas, M. Rapoport, Twisted Loop groups and their affine flag varieties, Adv. Math. 219 (2008), no. 1, 118-198
G. Pappas, X. Zhu,  Local models of Shimura varieties and a conjecture of Kottwitz, http://arxiv.org/abs/1110.5588v4, to appear in Invent. math.<|endoftext|>
TITLE: What´s essential to learn about complex spaces and several complex variables for an algebraic geometer?
QUESTION [14 upvotes]: Hi, I don´t know if this question is suitable for this site. The field of several complex variables is too broad, so I would like to know what´s essential to learn about complex spaces and several complex variables for an algebraic geometer? Any references?
Thanks in advance.

REPLY [15 votes]: If you want a specific goal to understand, one that has connections to most of modern complex geometry and provides motivation for many of the techniques used there, I suggest the link between ampleness in algebraic geometry and positivity in complex geometry. The first applications of these ideas were Kodaira's embedding theorem, which shows that for line bundles positivity is equivalent to ampleness, and the various associated vanishing theorems. Both are excellent starting points and provide managable goals for study. 
They are all covered in the books Francesco mentioned, perhaps most thouroughly in Griffiths-Harris. I would also recommend Demailly's Complex analytic and differential geometry, whose Chapters 5-7 go into great detail on these matters (without superfluous discussion, that you may want to get elsewhere, like in Griffiths-Harris, Huybrechts or Voisin's books or Lazarsfeld's "Positivity" series). Once you feel comfortable with this you can look at Demailly's Analytic methods in algebraic geometry, which gives a very modern snapshot of the complex-geometric notions of algebraic geometry and their applications. (Beware, that text is hardcore.)
The basic yoga of positivity in complex geometry is that ampleness of a line bundle $L$ is equivalent to the positivity of the curvature form of a smooth hermitian metric on $L$. This allows us to treat global algebraic questions involving ampleness and cohomology by looking at pointwise estimates of positive differential forms on our manifold. Once there, all of the machinery of Riemannian and complex geometry is available and hard global questions get converted into extremely computationally messy problems of linear algebra. For certain things, like cohomology of adjoint bundles $K_X \otimes L$, these methods work very well, for others they work less well or not at all, but it's always good to have another tool with which to attack problems.
As I said, historically the first major applications of these methods were the Kodaira-Nakano vanishing theorems and the Kodaira embedding theorem. A more recent, quite impressive, application is Siu's proof of the invariance of plurigenera of algebraic manifolds of general type. His proof makes great use of the modern versions of positivity and currently there is no algebraic proof of the same result. You can find it in his preprints on the arXiv or at the end of Demailly's "Analytic methods" manuscript, but again, this is hard.<|endoftext|>
TITLE: Best upper bound on the number of divisors of $n$ that are larger than $N$.
QUESTION [5 upvotes]: I am looking for the best upper bound on $$\sum_{\substack{d | n\\ d \geq N}} 1.$$ 
I know that
$$
d(n) = \sum_{\substack{d | n}} 1 \leq e^{O(\frac{\log n}{\log \log n})}.
$$
For my application, I need something like 
$$\sum_{\substack{d | n\\ d \geq N}} 1 \leq \frac{o(n^{\epsilon})}{\log N} \quad \forall \epsilon > 0.
$$ 
A reference where the bound can be found or a simple proof would be appreciated.
Thanks.
EDIT: Johan Andersson has pointed out that the third display follows from the second. (Thanks.) I am still interested to learn what the best known bound is.

REPLY [2 votes]: This is not an answer to what the best known upper bound is, but rather a comment that the (known) average distribution of divisors indicates you might not expect to do any better than the bounds on the divisor function itself.  Tennenbaum's "Introduction to Analytic and Probabilistic Number Theory", $\S$ 6.2 p. 207 says

For each integer $n$, let us define a
  random variable $D_n$ taking the
  values $\log d/\log n$, as $d$ runs
  through the the set of the $\tau(n)$
  divisors of $n$, with uniform
  probability $1/\tau(n)$.  The
  distribution function $F_n$ of $D_n$
  is then defined by $$
> F_n(u):=\text{Prob}(D_n\le
> u)=\frac{1}{\tau(n)}\sum_{d|n,d\le
> n^u}1\quad (0\le u\le 1). $$ It is
  clear that the sequence
  $\{F_n\}_{n=1}^\infty$ does not
  converge pointwise on $[0,1]$. 
  However, we shall see the sequence of
  Cesaro means $$
> G_N(u):=\frac{1}{N}\sum_{n\le N}F_n(u)
> $$ is uniformly convergent on $[0,1]$.
  Remarkably, the limit is the
  distribution function of a probability
  law well known to specialists: the
  arcsine law, with density
  $1/(\pi\sqrt{u(1-u)})$.  Large and
  small values have high probability: if
  $D$ is a random variable with this
  distribution law, we have $$
> \text{Prob}(D<0.01\text{ or
> }D>0.99)\approx0.128 $$ This indicates
  that, on average, an integer has many
  small (and correspondingly many large)
  divisors.

Update:  One thing we learn is that the relevant parameter is not $N$, but $u:=\log_n(N)$.<|endoftext|>
TITLE: Extensions of Galois representations
QUESTION [8 upvotes]: Let $G=Gal(\bar{\mathbb Q}/{\mathbb Q})$ be the absolute Galois group of the rationals. Fix two continuous group homomorphisms $\alpha,\beta: G\to {\mathbb Q}_l^\times$, where $l$ is a prime and ${\mathbb Q}_l$ the field of $l$-adic numbers.
Let $H(\alpha,\beta)$ be the ${\mathbb Q}_l$ vector space of all continuous maps $\chi:G\to{\mathbb Q}_l$ such that 
$$
\omega=\left(\begin{array}{cc}\alpha&\chi\\\  \ &\beta\end{array}\right)
$$
is a group homomorphism.
My question is: is the space $H(\alpha,\beta)$ one-dimensional? 
I can prove this if $\alpha=\beta$ and for the proof one can assume that $\beta=1$. In this case the defining relation for $H(\alpha,\beta)$ is
$\chi(xy)=\alpha(x)\chi(y)+\chi(x)$.

REPLY [15 votes]: You're asking if $Ext^1_G(\alpha,\beta)$ is one-dimensional. The short answer is no,
yet there are many cases where the answer is yes. Actually, the dimension of this group is not known in all cases, and when it is known, in general it is by a deep theorem,
not a trivial computation.
To be more precise, we have $Ext^1_G(\alpha,\beta) = Ext^1_G(1,\alpha^{-1} \beta) = H^1(G,\alpha^{-1} \beta)$ (continuous group cohomology). Setting $\chi=\alpha^{-1} \beta$
one is reduced to compute $H^1(G,\chi)$.  
The answer will depend mainly on the sign of $\chi$, that is $\chi(c)$ where $c$ is a complex conjugacy in $G$. We say that $\chi$ is even if $\chi(c)=1$, odd otherwise.
Also, let us assume for simplicity (1) that there is no prime $p$
where you character $\chi$ restricted to $Gal(\overline{ \mathbb Q_p} /\mathbb Q_p)$ is equal
to the cyclotomic or to the trivial character. 

Then
  the generic answer is $H^1(G,\chi)$ has dimension $1$ if $\chi$ is odd, $0$ if $\chi$ is even. But there are exceptions. It is always true, and easy to prove, that the dimension of $H^1(G,\chi)$ is at least its expected value. 

More precisely:
(a) Fixing a prime number $l$ which is regular in the sense of Kummer, then it is always true
that $H^1(G,\chi)$ is of dimension 1 or 0 according to the parity. This follows from Mazur-Wiles's proof of the main conjecture of Iwasawa, and the fact that the $l$-adic zeta function has no zero when $l$ is regular.
(a') For a fixed $l$, which is non-regular, then the same result will be true except 
for a finite, non-zero, number of exceptions, where $dim \ H^1(G,\chi)$ will be higher.
(same argument, using the fact that the $l$-adic zeta function has finitely many zero in any case, being an Iwasawa function, but has at least one zero, since $l$ is not regular).
(b) When $\chi = \chi_{cycl}^n \epsilon$, where $n$ is an integer, $\epsilon$ a finite order character, then it is expected that $H^1(G,\chi)$ has dimension $0$ or $1$ according to the parity. This is even a theorem except if $n$ is negative and $\chi$ even (and hypothesis (1)  above is satisfied, that is $n \neq 0,1$). This is a deep result of  Soulé combined with some lemmas of Bloch-Kato.
As far as I know, these are the only known results. 
In the cases we have left aside (excluded by our hypothesis (1)), the answer may be more complicated. For instance $H^1(G,\chi_{cycl})$ is of infinite dimension, as is easy to see using the Kümmer's exact sequence. 
For a more thorough discussion, please see my lecture notes on the Bloch-Kato conjecture for the Hawaii 2009's conference, available
on my web page.<|endoftext|>
TITLE: Freeness of a Z[x]-module
QUESTION [22 upvotes]: Definition: Call a mapping $f: \mathbb{Z} \rightarrow \mathbb{Z}$
a generalized polynomial if for any distinct integers $m$ and $n$
we have $(m - n)|(f(m)-f(n))$.
It is easy to check that polynomial functions $f \in \mathbb{Z}[x]$ are
generalized polynomials, that not all generalized polynomials are polynomials
and that the generalized polynomials form a ring under pointwise addition and
multiplication. Call the latter ring $R$.

Question: Is $R$, viewed as a $\mathbb{Z}[x]$-module, free? 
  And if yes, how does a basis look like?

REPLY [3 votes]: I posted a bunch of comments here, but I thought it would be better to write an answer, instead.
First, I've written up a lot about these functions, and functions with the same condition on  sets $S\subseteq \mathbb Z$. 
If you define $$q_k(x) =  \frac{\operatorname{lcm}(1,2,\dots,k)}{k!}\left(x+ \left\lfloor {\frac{k-1}{2}} \right\rfloor\right)_k$$ then these are rational polynomials in $R$ and they have the property that, for all $n\in\mathbb Z$, $q_k(n)\neq 0$ for only finitely many $k$. So given any infinite sequence $(a_k)_{k\in\mathbb N}$ of integers, we get a unique $f\in R$ defined by:
$$f(n)=\sum_{k} a_kq_k(n)$$
A little harder to prove, but not really hard, is that every $f\in R$ is expressed in this form. Essentially, you find $a_k$ by induction.
This gives an explicit abelian group isomorphism between $R$ and $\mathbb Z^{\mathbb N}\cong\mathbb Z^{\mathbb Z}$.
The proof is essentially using Chinese remainder there inductively to determine $a_k$, to match the values of $f(0),f(1),f(-1),f(2),f(-2),\dots.$ This is why you get the term $\operatorname{lcm}(1,2,\dots,k)$.
This also helps show that $R\cap \mathbb Q[x]$ has the $q_k$ as $\mathbb Z$-generators. Given a polynomial $q(x)\in Q[x]\cap R$ is of degree $d$, if $q(n)=\sum_{k\in N} a_k q_k(n)$ then $p(n)=\sum_{k=0}^{d} a_kq_k(n)$ is another polynomial of degree $d$ which agrees at $d+1$ values, so must agree everywhere.
Side note: $R$ is actually an integral domain, not just a ring. If $f\in R$, then $f$ can be zero only for finitely many values, so $fg=0$ implies $f=0$ or $g=0$. 
Using $R'=\left\{f:\mathbb N\to \mathbb Z\mid \forall n,m: n-m\mid f(n)-f(m)\right\}$, there is a natural ring homomorphism $R\to R'$ be sending $f\in R$ to $f_{\mid\mathbb N}\in R'$. This is $1-1$ since if $f_{\mid\mathbb N} =0$, then $f$ has infinitely many zeros. The map is not onto - it is possible to define $f\in R'$ such that you can't even extend $f$ to $-1$ and keep the divisibliity property.<|endoftext|>
TITLE: Which categories are the categories of models of a Lawvere theory? 
QUESTION [13 upvotes]: Background: a Lawvere theory $T$ is a category with finite products such that each object is a power of a fixed object $x$. Given a Lawvere theory $T$, the category $\text{Mod}_T$ of models of $T$ is the category of product-preserving functors $T \to \text{Set}$. Any such category is locally finitely presentable (if I'm not mistaken). Gabriel-Ulmer duality says, among other things, that locally finitely presentable categories are categories of models of generalizations of Lawvere theories. Do we have something like this result for categories of models of Lawvere theories (in the sense that we can identify some categorical properties that are satisfied precisely by such categories)?

REPLY [14 votes]: Adámek and Rosický [On sifted colimits and generalized varieties] have shown that a category $\mathcal{C}$ is equivalent to the category of models for a (finitary) Lawvere theory in $\mathbf{Set}$ if and only if it satisfies these conditions:

$\mathcal{C}$ is locally small and cocomplete.
There exists a small family $\mathcal{G}$ of objects in $\mathcal{C}$ such that, for every object $G$ in $\mathcal{G}$, the hom-functor $\mathcal{C}(G, -) : \mathcal{C} \to \mathbf{Set}$ preserves sifted colimits, and for every object $A$ in $\mathcal{C}$, there exists a small sifted diagram of objects in $\mathcal{G}$ whose colimit in $\mathcal{C}$ is $A$.

These conditions are easily seen to be analogues of the conditions for $\mathcal{C}$ to be locally finitely presentable, except for replacing "filtered" (or "directed") everywhere by "sifted". 
Incidentally, any such $\mathcal{C}$ is locally finitely presentable, and we may also replace "cocomplete" in condition (1) with "complete", just as for locally finitely presentable categories.<|endoftext|>
TITLE: Torsion and parallel transport
QUESTION [21 upvotes]: There's a close relationship between curvature and the holonomy group; the holonomy theorem of Ambrose and Singer, for example. It seems to me that there should be an analogous result for torsion. I believe that torsion measures the extent to which certain geodesic parallelograms don't close. Is there a theorem that makes this more precise?

REPLY [27 votes]: Here is another way to think of the relation between torsion and parallel transport, one that some may find more congenial than many of the other interpretations that have been proposed:
Start with a manifold $M$, and a connection $\nabla$ on $TM$.  Consider the bundle $\hat TM=\mathbb{R}\oplus TM$ (where, by $\mathbb{R}$, I really mean the trivial bundle $M\times\mathbb{R}$).  Define a connection $\hat\nabla$ on $\hat TM$ by the rule
$$
\hat\nabla_X\ \begin{pmatrix}a \\\\ Y\end{pmatrix} 
= \begin{pmatrix}da(X) \\\\ aX + \nabla_XY\end{pmatrix}
$$
for any function $a$ on $M$ and any vector fields $X$ and $Y$ on $M$.
I like to think of $\hat TM$ as a sort of 'extended' tangent bundle to $M$.  Of course, it canonically contains $TM= 0\oplus TM\subset \mathbb{R}\oplus TM$ as a subbundle, and one sees that, under this identification of $\begin{pmatrix}0\\ Y\end{pmatrix}$ with $Y$, one has $\hat\nabla_X Y = \nabla_XY$.  Also, by its definition, $\hat\nabla$-parallel translation in $\hat TM$ along a curve in $M$ will keep the $\mathbb{R}$-component of a section constant, so that the 'affine hyperplane subbundles' $a=const$ in $\hat TM$ are preserved under $\hat\nabla$-parallel translation.
Now, the curvature endomorphism of $\hat\nabla$, namely
$$
R^{\hat\nabla}(X,Y) = \hat\nabla_X\hat\nabla_Y-\hat\nabla_Y\hat\nabla_X - \hat\nabla_{[X,Y]},
$$
is just
$$
R^{\hat\nabla}(X,Y)\begin{pmatrix}a \\\\ Z\end{pmatrix}
= \begin{pmatrix}0&0\\\\T^{\nabla}(X,Y) & R^{\nabla}(X,Y)\end{pmatrix}
\begin{pmatrix}a \\\\ Z\end{pmatrix}.
$$
Thus, the condition that $T^\nabla(X,Y)=0$ is the condition that $\hat\nabla$-parallel translation around closed loops preserve the splitting $\hat TM = \mathbb{R}\oplus TM$, at least to lowest (i.e., second) order.  In particular, if one thinks of the hyperplane subbundle $a=1$ as a sort of 'shadow copy' of the actual tangent plane, the torsion measures how its 'zero', i.e., $\begin{pmatrix}1\\0\end{pmatrix}$, moves in that 'shadow  tangent plane' when one uses $\hat\nabla$ to parallel translate around in a loop.
In particular, note that the curvature of $\hat\nabla$ incorporates both the torsion and curvature of $\nabla$, and the vanishing of the torsion of $\nabla$ says precisely that $\hat\nabla$-parallel translation preserves, to lowest order, the origins in the 'shadow copies' of the tangent plane.
NB:  It is a notational accident that, in this interpretation, $T(X,Y)$ represents 'infinitesimal translation' while $R(X,Y)$ represents 'infinitesimal rotation'.
Remark 1:  The addition of a line bundle to the tangent bundle may seem somewhat artificial.  However, there is a way to present this construction more naturally in a dual formulation:  First, note that the vector bundle $J^1(M,\mathbb{R})$ of $1$-jets of smooth functions on $M$ has a natural identification with the vector bundle $\mathbb{R}\times T^*\!M$ by identifying the element $j^1_xf$ with $(f(x), \mathrm{d}f_x)\in \mathbb{R}\times T^*\!M$.
Second, note that a connection $\nabla$ on $TM$ naturally extends to a connection on $T^*\!M$, uniquely defined by the conditions that
$X\bigl(\eta(Y)\bigr) = (\nabla_X\eta)(Y) + \eta(\nabla_XY)$
for all $1$-forms $\eta$ and vector fields $X,Y$ on $M$.  Using this identification of $J^1(M,\mathbb{R})$ with $\mathbb{R}\times T^*\!M$, we can define a connection $\hat\nabla$ on $J^1(M,\mathbb{R})$ by
$$
\hat\nabla_X(a,\alpha) = \bigl(Xa + \alpha(X),\ \nabla_X\alpha\bigr)
$$
for all functions $a$, $1$-forms $\alpha$, and vector fields $X$ on $M$.
Then one easily computes that
$$
R^{\hat\nabla}(X,Y)(a,\alpha) = \bigl(\alpha\bigl(T^\nabla(X,Y)\bigr),\ R^{\nabla}(X,Y)(\alpha)\bigr),
$$
which is the dual of the above formula on the 'extended tangent bundle'.
Remark 2:  Finally, going back to the OP's original request for how to think about torsion in terms of failure of 'quadrilaterals' to close, I checked on the coordinate expressions and looked at the Taylor series.  This is, of course, very classical; it's in Schouten's work, but it might be worth making explicit in the following way:  Let $\nabla$ be a connection on $TM$, fix a point $p$ and $p$-centered coordinates $x=(x^i)$.  Define the connection coefficients $\Gamma^i_{jk}$ by the usual rule
$$
\nabla_{\partial_i}\partial_j = \Gamma^k_{ij}\ \partial_k\ ,
$$
where $\partial_i$ are the dual vector fields, i.e., $dx^i(\partial_j) = \delta^i_j$.
Choose $v,w\in T_pM$ be tangent vectors and write $v=v^i\ \partial_i(p)$ and $w=w^i\ \partial_i(p)$.  Let $a(t)=\exp_p(tv)$ be the $\nabla$-geodesic starting at $p$ with initial velocity $v$, let $b(t)\in T_{a(t)}M$ be the $\nabla$-parallel translate along $a$ of $w$, and set $c(s,t) = \exp_{a(t)}(sb(t))$.  Then the functions $c^i= x^i(c(s,t))$ have Taylor expansions in $s$ and $t$ of the form
$$
c^i(s,t) = tv^i+sw^i - \tfrac12\Gamma^i_{jk}(0)\bigl(t^2v^jv^k+2st\ v^jw^k+s^2w^jw^k\bigr) + R^i_3(s,t).
$$
Thus, if switching $(t,v)$ and $(s,w)$ in this construction is to yield the same result up to second order in $s$ and $t$ for all $v$ and $w$, one must have $\Gamma^i_{jk}(0) = \Gamma^i_{kj}(0)$.  In particular, if all the 'attempted parallelograms' close to second order at all points, the torsion of $\nabla$ must vanish.<|endoftext|>
TITLE: Do there exist "expanding" $1$-skeletons of simple $4$-polytopes?
QUESTION [7 upvotes]: Let $\{ G_n \}_{n \ge 1}$ be a sequence of graphs such that the number of vertices of $G_n$ tends to $\infty$ as $n \to \infty$. We say that $\{ G_n \}_{n \ge 1}$ is an expander family if
$\lambda_2( G_n)$ is bounded away from $0$ as $n \to \infty$. Here $\lambda_2(H)$ is the smallest positive eigenvalue of the normalized graph Laplacian of $H$.
If every $G_n$ is the $1$-skeleton of a simple $3$-polytope, you can not have an expander family. This follows from the planar separator theorem.
What if every $G_n$ is the $1$-skeleton of a simple $4$-polytope? If you relax "simple" then there are obvious examples — in particular, cyclic polytopes in dimension $4$ have complete graphs for their $1$-skeletons, so they are expanders.
If it is hard to come up with an example for simple $4$-polytope, does it make it easier if we allow the dual graph of an arbitrary simplicial $3$-sphere?

REPLY [7 votes]: As far as I know there are no examples of graphs of simple 4-polytopes (or simple $d$-polytopes, for fixed $d$) that are expanders, and not even such examples for the dual graphs for triangulations of spheres of fixed dimensions.  Some information on expansion properties of graphs of polytopes can be found in my chapter polytope skeletons and paths of the Handbook of Discrete and Computational Geometry (Goodman and O'Rourke, eds.). 
There is a conjecture that graphs of polytopes are not very good expanders: Let $d$ be fixed. The conjecture asserts that a graph of every simple $d$-polytope with $n$ vertices 
can be separated into two parts, each having at least $n/3$ vertices by removing $O(n^{1-1/(d-1)})$ vertices.
This is known to be false for dual graphs to triangulations. It is known that there are dual graphs to triangulations of $S^3$ which cannot be separated even by $O(n / \log n)$ vertices. This is a construction from G. L. Miller, S.-H. Teng, W. Thurston, and S. A. Vavasis. "Separators for sphere packings and nearest neighbor graphs." J. ACM, 44(1):1-29, 1997. (PDF download) (I would be happy to see a nice exposition of this construction.)
On the positive side it is known that dual to neighborly $d$ polytopes with $n$ facets are $\epsilon$-expanders for $\epsilon =n^{-4}$. This is far from the good expansion in the question but good enough to give a polynomial upper bound in $n$ for the diameter. Unfortunately general simple $d$ polytopes can be fairly poor expanders as seen by gluing two large simplicial polytopes along a facet and taking the dual.<|endoftext|>
TITLE: Consecutive Integer Squared Square
QUESTION [6 upvotes]: Is it possible to construct a squared square out of consecutive integer squares? 
Be it 1,2,3,...n or k,k+1,k+2,...n.

REPLY [3 votes]: If a solution was possible to tile a 70x70 square with squares $$1^2+2^2+\cdots+24^2$$ then a perfect squared square of order 24 would exist.  Order 24 has been completely enumerated.  The compound perfect squared squares (CPSSs - compound; subrectangles allowed) of order 24 were generated by Duijvestijn, Federico and Leeuw in 1979 and 1982.  Only 1 CPSS of order 24 was found, T.H. Willcocks 175x175, discovered in 1948.
The 26 simple perfect squared squares (SPSSs - simple; no subrectangles allowed) of order 24 were enumerated by Duijvestijn in 1991, the smallest SPSS of that order being 120x120.  These results have been confirmed a number of times by other researchers using different approaches and software.
The sum of squares formula gives;$$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$
$$For\ n = 32,\ \sum_{i=0}^n i^2 = 11440, \sqrt{11440}= 106.95794..$$
$$For\ n = 33,\ \sum_{i=0}^n i^2 = 12529, \sqrt{12529}= 111.93301..$$
If perfect squared squares existed with a side less than 110 in length then they would need to be of order 32 or less (if the 33 squares from 1 to 33^2 will not fit in a square of side 110, neither will a larger set of 33 squares, nor will any order higher than 33).  All orders of perfect squares squares up to and including order 32, both simple and compound, have been enumerated, (order 33 has also been done this year by Jim Williams) see www.squaring.net.  The smallest perfect squared squares found up to order 33 were the 3 SPSSs with sides of 110.  There are 2 in order 22, and 1 in order 23.  So these are the smallest possible perfect squared squares .  Note that the 112 side SPSS found by Duijvestijn in 1978 is the lowest order (21) perfect squared square, but not the smallest.
Gambini achieved the same result, that 110 is the smallest perfect squared square in his 1999 thesis, using a packing program.  I do not know of any consecutive square tiling, unless more than 1 of each square is allowed.
http://www.mathpages.com/home/kmath147.htm Sum of Consecutive Nth Powers Equals an Nth Power, has a generalisation to cubes.<|endoftext|>
TITLE: Quantum 9j symbols?
QUESTION [5 upvotes]: A formula for (SU2) quantum 6j symbols exists. A formula expressing ordinary (q=1)
9j symbols in terms of 6j symbols is long known. Unfortunately, combining both (I tried it myself) got tricky - the associated graph K3,3 is nonplanar, at least one knot-type
crossing is needed and first of all, this ruins the symmetry.
Can I find the quantum analogon of the standard sum over the product of three 6j 
symbols in the literature (or can someone post it here)?

REPLY [4 votes]: to save you a trip to the library, here's the relevant paragraph from V.G. Turaev, Quantum Invariants of Knots and 3-Manifolds, page 343.<|endoftext|>
TITLE: congruence on words: having the same (scattered) subwords of length at most n
QUESTION [21 upvotes]: For a fixed finite alphabet $A=\{a,b,...\}$, write $x \sim_n y$ if the two words $x$ and $y$ have the same (scattered) subwords of length at most $n$. The relation $\sim_n$ is a congruence of finite index and [SS83] asks what is the number of congruence classes. Has there been any progress on this question? I cannot find any recent paper mentioning it.
Write $k=\mid A\mid$ for the cardinal of $A$. Since two different words of length at most $n$ cannot be congruent, there must be at least $\mid A^{\leq n}\mid=k^n+k^{n-1}+\cdots+1=\frac{k^{n+1}-1}{k-1}$ congruence classes. And since each class is characterized by a subset of $A^{\leq n}$, there are less than $2^{\mid A^{\leq n}\mid}\leq 2^{k^{n+1}}$ congruence classes. 
For $n=1$, $\sim_1$ means "same set of occurring letters" and obviously there are $2^k$ congruence classes.
But observe that, for $n>1$, not all subsets of $A^{\leq n}$ are realizable sets of subwords. E.g., if $x$ has $aa$ and $bb$ as subwords of length $n=2$, it must also have $ab$ or $ba$ (at least one of them).
There is a very large gap between the obvious lower and upper bounds given above. Can we narrow it? The question I am interested in is for fixed $k$ and as a function of $n$, is the number of classes simply exponential or doubly exponential? (or something else?)
Added June 17th: 
Write $C_{n,k}$ for the number of classes. I did some computations for $k=3$, i.e., when $A=\{a,b,c\}$ has three letters: 
\[ 
C_{0,3}=1,    \quad 
C_{1,3}=8,    \quad 
C_{2,3}=152,  \quad
C_{3,3}=5312, \quad
C_{4,3}=334202.
\]
This leaves me perplexed. Much bigger than $k^n$ but much smaller than $2^{k^n}$.
Added June 26th: 
For $k=2$, $C_{n,2}$ can be bounded by $2^{2n^2+1}$, hence is "simply" exponential. Indeed, when $A=\{a,b\}$, a shortest witness for a congruence class does not have $n+1$ consecutive $a$'s (or $b$'s) and does not alternate more than $2n$ times between $a$'s and $b$'s. Hence each congruence class has a witness of length $\leq 2n^2$.
References:
[SS83] J. Sakarovitch and I. Simon's. "Subwords", chapter 6 in M. Lothaire's  Combinatorics on words, 1983.
Acknowledgments:
Jean-Éric Pin pointed me to the [SS83] ref for the open question.

REPLY [4 votes]: I could generalize my earlier proof for the $k=2$ case and show that, for any fixed $k$, $C_{n,k}$ is in $2^{O(n^k)}$ hence simply exponential.
When some given $n$ is understood, we say that $x\in A^*$ is minimal
if $x$ has minimal length inside its $\sim_n$-class. Since $\sim_n$ is a
congruence, all factors of a minimal $x$ are themselves minimal.
For a fixed $k$-letter alphabet $A$, write $l_{n,k}$ for the
length of the longest minimal word wrt $\sim_n$. Thus
$C_{n,k}\leq k^{l_{n,k}+1}$ since every congruence class has a
minimal representative.
A word $x\in A^*$ is rich if each letter of $A$
occurs at least once in $x$, otherwise $x$ is poor. A minimal poor word has length $\leq l_{n,k-1}$ since at least one letter is not used. 
We decompose a word $x$ under the form $x=r_1 \cdots r_m\cdot x'$
where $r_1$ is the shortest rich prefix of $x$, $r_2$ is the
shortest rich prefix of the rest, etc., until there only
remains a poor suffix $x'$. 
E.g., assuming $k=3$, $x=bbaaabbccccaabbbaa$
is written $x=bbaaabbc\cdot cccaab \cdot bbaa$.
Also, if $x$ is poor then
$m=0$ and $x'=x$.
Assume now that $x$ is minimal. Then $\mid{x'}\mid\leq l_{n,k-1}$
since $x'$ is poor. Also, for any $i=1,\ldots,m$,
$\mid{r_i}\mid\leq 1+l_{n,k-1}$, since $r_i$ minus its last
letter is not yet rich. If now $m\geq n$ then
$x$ already has all possible subwords (so
$\mid{x}\mid=kn$ if $x$ is minimal). Otherwise $m<n$ and
$\mid{x}\mid\leq n\cdot l_{n,k-1}+n-1$. We deduce
$l_{n,k}\leq \max(kn,n\cdot l_{n,k-1}+n-1)$.
Since $l_{n,1}=n$ we see that $l_{n,k}<
n^k+n^{k-1}+\cdots+n+1$. Hence for fixed $k$, $C_{n,k}$ is in
$2^{O(n^k)}$.<|endoftext|>
TITLE: When is Ext*(M,N) finitely generated as a Ext*(M,M) module?
QUESTION [12 upvotes]: Let A be a finite dimensional algebra over a field k and M,N a finitely generated A-module.
Im searching for examples where the module $ Ext^{o} (M,N)  $ is a finitely generated $ Ext^{o}(M,M) $ -module(via yonedaproduct) for every finitely generated A-modules M,N.
This is the case for example when A is a cocommutative Hopfalgebra by a result of Friedlander and Suslin and its a conjecture that this also holds for a general Hopfalgebra.
see for example: http://www.math.uiuc.edu/K-theory/0085/paper.pdf

REPLY [2 votes]: This is true whenever condition (Fg) from the theory of support varieties holds.
A finite dimensional algebra $A$ with Jacobson radical $J$ is said to satisfy condition (Fg) if the Hochschild cohomology ring ${\rm{HH}}^{\ast}(A)$ of $A$ is Noetherian and ${\rm{Ext}}^{\ast}_A(A/J,A/J)$ is finitely generated as an ${\rm{HH}}^{\ast}(A)$-module. The ring ${\rm{HH}}^{\ast}(A)$ acts on ${\rm{Ext}}^{\ast}_A(M,N)$ through the graded center of ${\rm{Ext}}^{\ast}_A(M,M)$ as explained in section 3 of [Solberg, Support varieties for modules and complexes]. Condition (Fg) implies that ${\rm{Ext}}^{\ast}_A(M,N)$ is finitely generated as an ${\rm{HH}}^{\ast}(A)$-module [Propositions 5.5 and 5.7 of that paper]. This in turn implies that ${\rm{Ext}}^{\ast}_A (M,N)$ is finitely generated as an ${\rm{Ext}}^{\ast}_A(M,M)$-module.
See the introduction to arXiv:1003.2867 for classes of algebras that satisfy condition (Fg). In particular (Fg) holds when $A$ is a representation finite self-injective algebra over an algebraically closed field.<|endoftext|>
TITLE: Are Hyperbolic Groups Residually Amenable
QUESTION [10 upvotes]: It is a well-known conjecture (or maybe just a question) that all hyperbolic groups are residually finite. What happens if we weaken the conclusion; in particular
Are all hyperbolic groups residually amenable?
What is known in this direction?

REPLY [10 votes]: Proposition 7 of this paper establishes that every hyperbolic group is residually amenable iff every hyperbolic group is residually finite.<|endoftext|>
TITLE: How can I tell if a group is linear?
QUESTION [17 upvotes]: The basic question is in the title, but I am interested in both necessary and sufficient conditions. 
I know the Tits' alternative and Malcev's result that finitely generated linear groups are residually finite, but I don't know any purely group-theoretical sufficient conditions. 
Though it's not purely group-theoretical, what if the group acts simplicially on a (finite) simplicial complex? Does this imply linearity? Does it imply linearity over $\mathbb{C}$?

REPLY [4 votes]: One more necessary condition. Let $T$ be a matrix from $SL_n(\mathbb{C})$. Then the set of all matrices $B$ such that $\lim_{n\to \infty} T^{n} BT^{-n} = 1$ is a nilpotent subgroup (an exercise, first noticed by Margulis, I think, a proof can be found here.). This implies, for example,  that the group $\langle a,b,t \mid tat^{-1}=a^2, tbt^{-1}=b^2\rangle$ is not linear (this group is residually finite).<|endoftext|>
TITLE: Is the algebraic Grothendieck group of a weighted projective space finitely generated ?
QUESTION [6 upvotes]: This is to be confronted with Joseph Gubeladze' paper : "Toric varieties with huge Grothendieck group" !

REPLY [3 votes]: In our paper with N. Pavic we have proved that over an algebraically closed field of char. 0, if the weights $a_0, \dots, a_n$ are coprime, so that singularities of  $\mathbb{P}(a_0, \dots, a_n)$ are isolated, then $K_0(\mathbb{P}(a_0, \dots, a_n)) = \mathbb{Z}^{n+1}$, see Application 3.2 in 
https://arxiv.org/pdf/1809.10919.pdf
Our proofs rely on comparison of $K_0(X) = K_0(Perf(X))$ with $G_0(X) = K_0(D^b(X))$ using $K$-groups of Orlov's singularity category and completion arguments. The same methods apply to compute $K_0(X)$ for any quasi-projective variety $X$ with isolated quotient singularities.<|endoftext|>
TITLE: Does an origin-centered ellipse in the plane intersect each $L^p$-circle at most 8 times?
QUESTION [6 upvotes]: The question is in the title: Let $E$ be an origin-centered ellipse in ${\mathbb R}^2$ and let $S$ be an "$L^p$-circle":  $S = \{(x,y) : |x|^p + |y|^p = \text{const}\}$, where $1 \leq p \leq \infty$.  Is it true that $E$ and $S$ have at most $8$ points of intersection?
(Of course, you should ignore the case that $p = 2$ and $E$ and $S$ coincide.)
This is easy to show if $E$ is a circle, but it's not completely obvious (to me) when $E$ is a more general ellipse, though it seems true from drawing pictures.
Motivation: I was trying to fill in (and maybe simplify) the details of "Remark 2" of the paper "On the best constant in the Khintchine-Kahane inequality" by Latala and Oleszkiewicz (http://www.mimuw.edu.pl/~rlatala/papers/chinsm.pdf).  I posted a slight variant of the question at MSE (https://math.stackexchange.com/questions/408680/how-many-points-of-intersection-between-an-ellipse-and-an-l-p-circle) which would have helped with the simplification, but unfortunately the variant is false.  Still, the above question seems of moderate interest, and no one at MSE answered it.

REPLY [7 votes]: $\def\sign{\mathop{\rm sign}}$First of all, it is enough to prove the statement when $p=u/v$ is rational, $u$ is even and $v$ is odd (such numbers are dense on the real line). We need this to simplify the last argument.
Let the equation of the ellipse be $f(x,y)=ax^2+2bxy+cy^2=1$; one may assume that $b\neq 0$ (either by a small variation argument, or by considering it directly --- this case is easy). Let us bound the number of local extrema of $f(x,y)$ on $|x|^p+|y|^p=1$; if there are at most 8 of them, then we are done: between every two, there is at most one intersection point. Since $b\neq 0$, the points where one of the coordinates vanish are not extremal (the ellipse passing through such point has a tangent not parallel to the coordinate axes).
The extremal points satisfy the system of Lagrange equations
$$
  2(ax+by)=p\lambda|x|^{p-1}\sign x, \quad
  2(bx+cy)=p\lambda|y|^{p-1}\sign y;
$$
obviously, $\lambda\neq 0$, so this system yields 
$$
  \frac{at+b}{bt+c}=|t|^{p-1}\sign t,
$$
where $t=x/y$. So, it suffices to show that there are at most four such values of $t$ (each corresponds to two symmetrical extremal points). Notice that $|t|^{p-1}\sign t=t^{(u-v)/v}$ by our assumptions on $u$ and $v$. Now, substituting $t=s^v$ we obtain
$$
  as^v+b=bs^u+cs^{u-v}.
$$
The last polynomial equation has at most four roots by an easy application of Descartes' rule of signs.
ADDENDUM. By the way, you have mentioned on MathSE that in your case the minor axis is in the positive quadrant, and you were interested in the intersection points lying in this quadrant as well. In this case we may assume that $a,b,c>0$, and we are interested in positive values of $t$. The rule of signs shows that there is at most one $t$ whenever $u-v>v$, that is --- if $p>2$. Otherwise it claims that there is at most one $t$ in the adjacent quadrant.
ADDENDUM2. Here is the explanation about the limiting case. Assume that you have more than 8 common pointsof an ellipse and an $\ell_p$-circle. Blowing up or away your ellipse a bit, you may reach the situation when there are more than 8 points of transversal intersection. Then you mark a point on the ellipse between every such neighboring points (or simply mark 10 of them); these marked points are alternately inside and outside the $\ell_p$-circle. Finally, if you change $p$ a bit, all these 10 points will still be alternating, guaranteeing 10 intersection points.
The same method works for $b=0$.
Finally, if it is easier for you, you may pass to rational $p$ of a desired form AFTER writing down the (almost) polynomial in $t$ --- in this case the statement is more visible...
And, even more finally, you may just repeat the proof of the Descartes' rule for an `almost polynomial' --- all you (perhaps) need is that all the powers differ by at least one, which can be reached by an appropriate substitution.
ADDENDUM3. Just to mention. The same method allows to prove that the affine images of $\ell_p$- and $\ell_q$-circles also have at most 8 common points, if they have the same center.<|endoftext|>
TITLE: Haar measure on $O(n)$ reduced to simpler probability space
QUESTION [5 upvotes]: The background of this question is how a random variable $X$ on the orthogonal group $O(n)$ whose distribution is the normalized Haar measure $\mu$, i.e., $\mu( O(n) ) = 1$, can be realized on a computer that has access to a number of "simpler" random variables, like:

a perfect coin
the uniform probability measure on $[0,1]$ or $S^1$
Gaussian variables on $\mathbb R$ with arbitrary standard deviation.

An algorithm needs to use the random variable $X$, but I am not aware of an explicit formula that would allow for such a reduction.

REPLY [10 votes]: Starting from a real $N\times N$ matrix and essentially performing a $QR$ decomposition, then if the initial real matrix elements are independent identically distributed Gaussian, then the matrices $Q$ will be Haar-distributed on $O(N)$. Please see the full construction in the following article by: Francesco Mezzadri<|endoftext|>
TITLE: Weights on equivariant cohomology?
QUESTION [6 upvotes]: Let $X$ be a quasi-projective variety over the complex numbers, equipped with an action of a linear algebraic group $G$. 
Is there a natural mixed Hodge structure on its equivariant cohomology?
Is it pure if $X$ is smooth projective? 
What if we ask the analogous question for $l$-adic equivariant cohomology for varieties over finite fields?
More generally is there a formalism of "mixed equivariant derived categories" such that the six functors have the expected effects on weights?

REPLY [4 votes]: In the $\ell$-adic setting a reference would be Laszlo-Olsson's papers on six operations for Artin stacks:
http://arxiv.org/abs/math/0512097
http://arxiv.org/abs/math/0603680
http://arxiv.org/abs/math/0606175
Also see:
http://arxiv.org/abs/1211.5948
Regarding mixed equivariant derived categories in the Hodge sense, as far as I know there isn't any canonical reference in the literature for this. However, I claim that the Bernstein-Lunts approach (for $X/G$, $G$ linear algebraic) works just fine for mixed Hodge modules (or even $\ell$-adic coefficients, but I am more comfortable with the mixed Hodge setting so not quite as sure about $\ell$-adic). The point is that Bernstein-Lunts mainly just use the six operations in their approach. So everything goes through formally in the same way for mixed Hodge modules. The only place where you might worry is something funny happening with weights. However, there is no problem if all your approximation spaces are algebraic (hence the linear algebraic requirement) and all pushforwards are along proper (algebraic) maps and pullbacks are along smooth (in the algebraic sense) maps.
I remember going through things carefully for $B\backslash G/B$ a couple of years ago. Many of these checks are also done in O. Schnurer's thesis. A condensed/article version of the latter can be found here:
http://arxiv.org/abs/0809.4785
I would also suggest just asking Wolfgang.
I do not know what the state of the art in defining mixed motivic sheaves at the moment is.
Some comments not directly related to the question (but I am reading between the lines and assuming this is where you are coming from Jan): As far as graded representation theory type applications go, the desire for a mixed equivariant derived category usually manifests itself in trying to prove splitting of some sequences, purity/formality type results, and/or to get a grading on Ext-spaces. In each of these situations one can avoid having to invoke a high powered theory of mixed equivariant categories by just working on an approximation space and the mixed (non-equivariant) derived category on this space (we are just unwinding the Bernstein-Lunts approach/Borel construction). I would claim that often even working with a mixed (non-equivariant) derived category can be avoided, since usually the only way one has of getting a handle on Ext-spaces is by interpreting them as cohomology of some space, or a convolution algebra formalism, or using a suitable fibre functor (I am thinking along the lines of Soergel bimodules, Geometric Satake, etc.). So as long as you know there is a functorial mixed structure on cohomology groups you can get away without explicitly invoking mixed categories. Having said that, there is one caveat: coming from the graded representation theory perspective, mixed sheaves offer the opportunity to impose gradings in a functorial way (note: Ext-groups in the mixed derived category are ordinary vector spaces, it is Ext-groups in the non-mixed derived category that inherit mixed structures via realization).  I don't know how one can (in general) cheat to get these mixed strucures on Ext-spaces that behave functorially. In special situations you can get by via devices like Soergel bimoules, working with $\mathbb{C}^*$-coherent sheaves on cotangent bundles, etc. But these are replacements that require a lot of extra work (often worth it for characteristic $p$ applications). In characteristic $0$ the conceptual approach still would be to go through mixed categories. Of course, getting gradings this way is not at all a triviality: a grading corresponds to having split Tate structures which can be quite difficult to prove (eg. I don't know how see directly that Ext between Vermas, in just the ordinary $G/B$ case is split Tate). In representation theory type settings, there is the (somewhat mysterious) phenomenon that intersection cohomology has a basis given by algebraic cycles (I am thinking Geometric Satake, fibres of the Bott-Samelson resolution etc.). As Wolfgang would (probably) say, if we had a fully functional theory of motivic sheaves all the pain would go away!
A comment on Dan Peterson's answer:  I am not sure about a 6 functor formalism for simplicial varieties. There is a brief discussion in Bernstein-Lunts about some of the difficulties in defining six functors and proving their properties in this setting (I don't remember the precise section).<|endoftext|>
TITLE: Hecke equidistribution
QUESTION [21 upvotes]: For a prime $p\equiv 1\pmod{4}$, we can write $p=a^2+b^2=N(a+bi)$. Therefore
$$
a+bi=p^{1/2}e^{i\varphi}
$$
where $\varphi\in [0,2\pi]$. I know that Hecke proved that $\varphi$ is equidistributed. I am looking for a reference for this nice result. I would be thankful if one can give me a reference.

REPLY [16 votes]: A very down-to-earth treatment of this result of Hecke is in Chapter 5 of the nice book Geometric and Analytic Number Theory, by Hlawka, Schoißengeier, and Taschner. By down-to-earth, I mean that they deal directly with this specific case of Hecke's result, and that they prove it using very little -- the method is a modification of the Korevaar--Newman--Zagier approach to the prime number theorem, and so doesn't need any quantitative zero-free region (just a statement that there are no zeros of the appropriate objects on the line $\Re(s)=1$).<|endoftext|>
TITLE: Random RSK and Plancherel Measure
QUESTION [9 upvotes]: Let $(X_1,X_2,\ldots)$ be a sequence of i.i.d. random variables. It is known that if these random variables are distributed uniformly on the unit interval, then applying the RSK algorithm to this sequence (and looking at the recording tableau) gives the (infinite) Plancheral measure on Young Tableaux. Restricting the above sequence to length $n$ to give the Plancheral measure on partitions of $n$. The proof is straightforward as the sequence of $X_i$ induce uniformly random permutations and RSK follows through. 

Here's my question: what is known in
  the case of other distributions for
  the $X_i$? I suppose one can
  equivalently say what is known for non
  uniformly random permutations but I'd
  like to stick to the former.

In particular, is anything known about the resulting limit shape of the tableau like the result of Logan-Shepp-Vershik-Kerov? I did some simulations with other distributions and it seems the limit shape is the same! Here's a picture of the usual Plancheral limit shape:
(source)

REPLY [7 votes]: Another large family of distributions that contains both $U[0,1]$ inputs (leading to the usual Plancherel measure) and $U\{1,2,\ldots,d\}$ (as in Ryan O'Donnell's answer) as special cases is the following. Let $(\alpha, \beta, \gamma)$ be an element of the Thoma simplex, i.e., a triple where
$\gamma\in [0,1]$ and $\alpha$ and $\beta$ are vectors
$$ \alpha = (\alpha_1, \alpha_2, \ldots) \ \ \textrm{ with }\alpha_1\ge\alpha_2\ge \ldots \ge 0,
$$
$$ \beta = (\beta_1, \beta_2, \ldots) \ \ \textrm{ with }\beta_1\ge\beta_2\ge \ldots \ge 0,
$$
and such that 
$$
\sum_n \alpha_n + \sum_n \beta_n + \gamma = 1.
$$
We can associate with $(\alpha, \beta, \gamma)$ a probability distribution of a random variable $X$ that is supported on $(0,1) \cup \{1,2,3,\ldots\} \cup \{-1,-2,-3,\ldots\}$, defined by
$$ \mathbb{P}(X \in (a,b)) = \gamma (b-a) \qquad (0<a<b<1), 
$$
$$ \mathbb{P}(X = n) = \alpha_n \qquad (n=1,2,\ldots), 
$$
$$ \mathbb{P}(X = -n) = \beta_n \qquad (n=1,2,\ldots). 
$$
If we now take a sequence of i.i.d. random variables $X_1,X_2,\ldots$ with the same distribution as $X$ and apply the RSK algorithm to them, the resulting random infinite Young tableau has a very natural probability distribution studied by Vershik and Kerov, and supposedly related to the representation theory of the infinite symmetric group (a subject I know nothing about). 
Note, however, that requires a modified version of RSK in which the rules for the insertion tableau are different: the rows and the columns of the insertion tableau are weakly increasing, but it is not allowed to have the same negative integer twice in the same row and it is not allowed to have the same positive integer twice in the same column.
This result was proved by Sergei V. Kerov and Anatol M. Vershik (SIAM J. Algebraic Discrete Methods, 7(1):116–124, 1986; available here)
Some additional consequences are discussed in the paper "Robinson-Schensted-Knuth algorithm, jeu de taquin, and Kerov-Vershik measures on infinite tableaux" by Piotr Śniady (SIAM J. Discrete Math. 28 (2014), 598–630; available at these links: journal version, arXiv version.)<|endoftext|>
TITLE: Who is the commutator sheaf?
QUESTION [7 upvotes]: Let $G$ be a reductive algebraic group (say $GL_n$) and $[\cdot,\cdot]: G \times G \to G$ the commutator map taking $(g,h) \mapsto ghg^{-1} h^{-1}$.  Note that $[\cdot,\cdot]$ and therefore $R[\cdot,\cdot]_* \mathbb{Q}_{G \times G}$ is $G$-conjugation equivariant.  So I want to think of it as being like a class function.


Is there a description of $R[\cdot,\cdot]_* \mathbb{Q}_{G \times G}$ in terms of character sheaves?

REPLY [4 votes]: I don't really have an answer to this, but I have a number of comments that became too long.
First let me give some more context, to explain where I'm coming from: The quotient of $G\times G$ by the diagonal conjugation action parameterizes local systems on a punctured torus, and the commutator map $R$ describes the restriction of a local system to a circle around the puncture. Thus the complex $R_\ast \mathbb Q_{G\times G}$ encodes the cohomology of twisted moduli stacks of local systems on the torus (for example, the costalk at the identity element of $G$ is given by Borel-Moore chains on the moduli of local systems on a torus). More generally, one can consider local systems on any punctured surface together with the restriction to the monodromy around the puncture and obtain a similar complex.
As Will Sawin hinted at, the finite group analogue of this problem has a nice solution, sometimes referred to as the Frobenius formula. For example,  Hausel and Rodriguez-Villegas use this formula in for the finite group $G(\mathbb F_q)$ to obtain information about the cohomology (and mixed Hodge structure) of the moduli of $G$ local systems. This can be related to your question via the "trace of Frobenius" operation. 
I interpret your question as essentially asking for the categorified analogue of the Frobenius formula.
To see the kind of thing that could happen, it might be helpful to go through a simple 
Example: $G=T=\mathbb C^\times$.
In this case, the commutator map $R$ is just the constant map at the identity, and thus the sheaf $R_\ast \mathbb Q_{T\times T}$ is supported at the identity of $T$. Character sheaves on $T$ are local systems. Thus you are essentially asking: "How to express the skyscraper sheaf at the identity in terms of local systems". Just as for finite abelian groups, the delta function at 1 is the sum of all characters (Plancherel formula), one can think of the skyscraper sheaf as an "integral" of local systems of all monodromies (this can be made precise via the Mellin transform for $D$-modules). One thing to note about this example is that the sheaf $R_\ast \mathbb Q$ is continuous in the central character parameter - it has no hope of being a direct sum of character sheaves (or extension etc.).
The answer to this question for general reductive groups, is the subject of ongoing work of myself with David Ben-Zvi and David Nadler. The idea is as follows: Pick a $\lambda \in \mathfrak h ^\ast$ (where $\mathfrak h$ is the Lie algebra of the torus of $G$). One can look at the projection $(R_\ast \mathbb Q)^\lambda$ of the sheaf $R_\ast \mathbb Q$ into the category of character sheaves (meaning the triangulated category generated by character sheaves) with central character $\lambda$. This complex has an interpretation in terms of the Character Theory topological field theory $\chi ^\lambda$ of Ben-Zvi and Nadler. More precisely, it is the object associated with the punctured torus, considered as a cobordism from $\emptyset$ to $S^1$. One consequence of this is that $(R_\ast \mathbb Q)^\lambda$ can be computed in terms of the combinatorics of the Hecke Category/Character Sheaves (it is not clear to me how to do this at the moment though). Sorry, I can't give a reference...<|endoftext|>
TITLE: Is it possible to prove in ZF that a non-trivial compact connected Hausdorff space is uncountable?
QUESTION [13 upvotes]: Let $X$ be a compact, connected Hausdorff space with at least two points.
In $\mathrm{ZF}+\mathrm{AC}_\omega(\mathbb R)$, any countable compact Hausdorff space is metrizable, and from this it can be shown that $X$ is uncountable.
In $\mathrm{ZF}$, however, that result does not hold. Does anyone know if it's still possible to prove that it is uncountable?

REPLY [3 votes]: Choice is not needed.

EDIT 1: Thanks to @dfeuer for pointing out my original argument required Dependent Choice, and for helping me arrive at a choice-less proof.
EDITS 2&3: I discovered a new flaw in the `proof', namely that $x^*$ could be an element of $C_n$ (this should to be avoided to guarantee empty intersection of $C_n$'s). Will work to fix. Unfortunately in this "Leap Frog" argument it seems difficult to fix $U$ at stage $n$ without invoking DC. One solution would be  to define  set $K$ which cuts $C_{n-1}$ between  $x^*$ and $x^{**}$.  Then the $U$'s could be neighborhoods of $K$.  I don't know if there's an explicit way to define a $K$.

I will say $X$ is countable if there is an injection $f:X\to \omega$, where $\omega$ is the set of natural numbers. Uncountable means not countable.
By a continuum I shall mean a connected compact Hausdorff space.

Theorem (ZF). Every non-degenerate continuum is uncountable.
Proof. Let $X$ be a non-degenerate continuum.
For a contradiction suppose $X$ is countable.  Apparently $X$ must be infinite, and so we may enumerate $X=\{x_i:i<\omega\}$  where the $x_i$'s are distinct. 
Let $C_0=X$.  
Suppose $n\geq 1$ and non-degenerate continua $C_0\supseteq C_1\supseteq ... \supseteq C_{n-1}$ have been defined.
Let  $x^*$ be the element of $C_{n-1}$ with least subscript. 
Let $x^{**}$ be the element of $C_{n-1}$ with least subscript greater than $x^*$'s. 
Let $\mathcal U_n=\{U\subseteq X:U \text{ is open, }x^*\in U\text{, and }x^{**}\notin \overline U\}.$ Since $X$ is Hausdorff, $\mathcal U_n\neq\varnothing$. Let $$\mathcal C_n=\{C\subseteq C_{n-1}:x^{**}\in C,\;C\text{ is connected, and }(\exists U\in \mathcal U_n)(C\cap U=\varnothing)\}.$$Let $C_n=\overline{\bigcup \mathcal C_n}$. Then $C_n$ is a continuum, and is non-degenerate because some elements of $\mathcal C_n$ are non-degenerate.  This is true because compactness and normality of $X$ implies the quasi-component of $x^{**}$ in $C_{n-1}\setminus U$, $U\in \mathcal U_n$, is connected, and this quasi-component must meet $\partial U$ in order for $X$ to be connected. (Choice is not needed to prove normality, nor is it needed to prove these quasi-components are connected.)
Continuing in this manner, we construct a nested sequence $(C_n)$ of non-empty compact sets.  Their intersection must be non-empty.  But on the other hand we ensured each point of $X$ is eventually not in $C_n$. Contradiction. $\blacksquare$

Related: In 2013 Horst Herrlich and  Kyriakos Keremedi proved that "Connected separable metric spaces need not have continuum size in ZF". 
My follow-up question: Is every separable metric continuum equinumerable with the reals, in ZF?<|endoftext|>
TITLE: What is Kirillov's method good for?
QUESTION [19 upvotes]: I am planing to study Kirillov's orbit method. I have seen Kirillov's method in several branch of mathematics, for instance, functional analysis, geometry, .... Why is this theory important for mathematics and mathematical physics? Is there any property which connects this theory to several areas of math?
More precisely, I want to know about applications of metaplectic quantization on coadjoint orbit .

REPLY [3 votes]: Let $\hat{G}$ be the set of irreducible representations of a group $G$. Kirillov's method provides a geometric method for understanding $\hat{G}$ as the orbits of $G$ in the dual of the Lie algebra. It is also a means to introducing non-commutative harmonic analysis to engineering.
Tony Dooley from the university of Bath in the UK gave a talk on it and the Kirillov character formula as part of a seminar at my university so it may be productive to check out his homepage.<|endoftext|>
TITLE: Visualizing functions with a number of independent variables
QUESTION [7 upvotes]: I need to graph real valued functions (for exposition and analysis).
The issue is: there are more independent variables so that the conventional graphing methods can't be used, and furthermore I don't want to slice the functions.

These functions are like $s=f_1(x,y,z,t)$ and $s=f_2(x,y,z,t,k)$
I also have vector functions of the same type: v = g1 (x,y,z,t) and v = g2 (x,y,z,t,k)

The motivation is to see the functions intuitively at one go and maybe compare them. I know that there is a limitation of physical dimensions. Domain coloring has its own limits in this regard. My question is:

Do we have a visualization methodology for such requirements? 

It will be a benefit if one could refer to a software tool.
Any thoughts/suggestions are welcome.

REPLY [2 votes]: A tool for visualizing functions that is sometimes as powerful as graphs is using mapping diagrams. 
The simple idea is that each real variable is represented on an independent parallel axis in 3 space. One can place a point in space not on any axis to represent the function and arrows from points on the domain axes to the function point and arrows to points on the target axes corresponding to the values of the function for the domain points.
See Alfred Inselberg. Parallel Coordinates: Visual Multidimensional Geometry and Its Applications  (Springer, Oct 8, 2009) for more on multi-dimensionsal connections and http://users.humboldt.edu/flashman/MD/section-1.1VF.html for an (draft) introduction to visualizing functions of one variable with mapping diagrams.<|endoftext|>
TITLE: What are some examples of mathematicians who had an unconventional education?
QUESTION [19 upvotes]: Possible Duplicate:
Famous mathematicians with background in arts/humanities/law etc 

What are some examples of mathematicians who had an unconventional education and yet, went on to make an impact on mathematics? Here is an example: Edward Witten. He does not have a formal undergraduate degree in mathematics or physics but won the Fields medal.
More precisely, I am looking for examples of mathematicians whose undergraduate mathematical education was in a different field or was hindered by circumstances like war and poverty or did not have a formal degree as was the case of Srinivasa Ramanujan.

REPLY [2 votes]: Anastacio da Cunha is a rather unknown eighteen century portuguese mathematician who published a significant encyclopedia on elements of calculus, algebra and geometry. His innovative contributions were mainly on Calculus. He was unusually rigorous for his time (http://www-history.mcs.st-andrews.ac.uk/Biographies/Cunha.html)
This mathematician did not learn much about mathematics and physics in school. In these subjects he was an auto-didact (http://www-history.mcs.st-andrews.ac.uk/Biographies/Cunha.html)
Anastácio da Cunha survived the terrible 1755 Lisbon earthquake. He was in the army for 10 years. One day he was arrested and imprisioned by the inquisition during three years for his heretical views. Apparently his health did not recover from that period and he died some years after (http://www-history.mcs.st-andrews.ac.uk/Biographies/Cunha.html).
Maybe one can say that he had not only an unconventional education but also an unconventional life.<|endoftext|>
TITLE: Distortion of tree embedding in Alexandrov spaces
QUESTION [8 upvotes]: It is a well-known theorem first proved by Bourgain that any map $\varphi:T_n\to H$ from the binary tree of height $n$ to a Hilbert space has distortion at least $C \sqrt{\ln n}$ where $C$ is a universal constant. Distortion is the least $D$ such that for some $s$ and all $x,y\in T_n$, we have
$$ s d(x,y) \leq d(\varphi(x),\varphi(y)) \leq sD d(x,y)$$
(This also generalizes in some way to many more Banach spaces).
Some of the proofs use the linear structure of $H$, but others are purely metric. I therefore wonder:

Has this distortion estimate been generalized to spaces of non-negative curvature (Alexandrov spaces)?

REPLY [3 votes]: See the direct proof of markov convexity in proposition 2.1 of the following paper
Mendel, Manor; Naor, Assaf Markov convexity and local rigidity of distorted metrics. J. Eur. Math. Soc. (JEMS) 15 (2013), no. 1, 287–337.  
the only thing that is used there is lemma 2.3 of the above references, which holds with p=2 for alexandrov spaces of nonnegative curvature. Therefore the answer to your question is known to be positive. The observation that the proof of the above reference works in this setting has been shown in talks that I heard over the past few years, and also stated explicitly in the following paper
Sean Li, Coarse differentiation and quantitative nonembeddability for Carnot groups<|endoftext|>
TITLE: Subgroups of algebraic groups
QUESTION [5 upvotes]: Is anyone aware of a result (or a counterexample) along the following lines: let $G$ be an algebraic group over $\mathbf Z$.  Let $H$ be a finite group such that $H$ occurs as a subgroup of $G(\bar{\mathbf F}_p)$ for all but finitely many primes $p$.  Then $H$ also occurs as a subgroup of $G(\mathbf C)$.
I only really need this for $H$ solvable and $G = \mathrm{GSp}(4)$ if that helps at all.

REPLY [10 votes]: The functor of injective homomorphisms from $H$ to $G$ is represented by a scheme of finite type over $\mathbb Z$ (a locally closed subscheme of the product $G^H$). If this has points over $\overline{\mathbb F}_p$ for infinitely many $p$, then it must dominate $\mathop{\rm Spec}\mathbb Z$, so it has points over $\mathbb C$.<|endoftext|>
TITLE: The Nth number with M prime factors
QUESTION [5 upvotes]: Hi.
Suppose we arrange all natural numbers in a matrix P defined as follows:
P[I][J] = The Jth number with I prime factors. So P looks something like:
1
2 ,    3 ,    5 ,    7 ,   11 ,   13 ,   17 ,   19 ,   23 ,   29 ,   31 ,   37 ,   41 ,   43 ,   47 , ...
4 ,    6 ,    9 ,   10 ,   14 ,   15 ,   21 ,   22 ,   25 ,   26 ,   33 ,   34 ,   35 ,   38 ,   39 , ...
8 ,   12 ,   18 ,   20 ,   27 ,   28 ,   30 ,   42 ,   44 ,   45 ,   50 ,   52 ,   63 ,   66 ,   68 , ...
16 ,   24 ,   36 ,   40 ,   54 ,   56 ,   60 ,   81 ,   84 ,   88 ,   90 ,  100 ,  104 ,  126 ,  132 , ...
32 ,   48 ,   72 ,   80 ,  108 ,  112 ,  120 ,  162 ,  168 ,  176 ,  180 ,  200 ,  208 ,  243 ,  252 , ...
64 ,   96 ,  144 ,  160 ,  216 ,  224 ,  240 ,  324 ,  336 ,  352 ,  360 ,  400 ,  416 ,  486 ,  504 , ...
I noticed that P[i][j] = P[i-1][j]*2 if and only if j < O(1.666^i).
Examples:
i =  2 AND j <   2
i =  3 AND j <   4
i =  4 AND j <   7
i =  5 AND j <  13
i =  6 AND j <  22
i =  7 AND j <  38
i =  8 AND j <  63
i =  9 AND j < 102
i = 10 AND j < 168
i = 11 AND j < 268
i = 12 AND j < 426
I suppose that there is a more accurate approximation of the condition above.
What work has been previously done on the relation between "The Nth number with M prime factors" and "The Nth number with M-1 prime factors"?
Thanks

REPLY [2 votes]: Each sequence of "$n$-almost-primes" opens with a string of even numbers that are twice the entries that begin the previous sequence.  The first odd $n$-almost-prime is $3^n$.  Thus the sequence the OP has observed, $2,4,7,13,22,38,63,\ldots$, is basically the sequence A078843 $(3,5,8,14,23,39,64,\ldots)$ in the OEIS.  There's a recursive formula given there, attributed to Max Alekseyev:
$$a(n) = a(n-1) + \text{appi}3(n-k,[3^n/2^k]),$$
 where $k = \text{ceil}(nc)$ with $c = \log(5/3)/\log(5/2) = 0.55749295\ldots$ and $\text{appi}3(k,n)$ is the number of $k$-almost-primes not divisible by $3$ and not exceeding $n$.  This might supply the "more accurate approximation" the OP referred to.  The ratio of consecutive terms in A078843 seems to be tending to around $1.53$-something, far less than the OP's $1.666$, but all that could be quite illusory.
The link from the OEIS entry to the MathWorld entry on "Almost Primes" may also be of interest.<|endoftext|>
TITLE: Does every equivalence class of Hecke characters contain a distinguished element?
QUESTION [5 upvotes]: Let $k$ be a number field and let $I_k$ denote the idele group of $k$. Let 
$$|\cdot|: (x_v) \mapsto \prod_{v \in \Omega_k} |x_v|,$$
denote the adelic norm map.
If $I_k^1$ denotes the kernel of this map, then we have a short exact sequence
$$1 \to I^1_k \to I_k \to \mathbb{R}_{>0} \to 1. \qquad (*)$$
Next, recall that a Hecke character for $k$ is simply a continuous character
$$\chi:I_k \to S^1 \subset \mathbb{C}^*,$$
which is trivial on $k^* \subset I_k$. 
We say that two Hecke characters are equivalent if their restrictions to $I^1_k$ are equal. It follows easily from the sequence $(*)$ that every Hecke character equivalent to a fixed Hecke charater $\chi$ has the form $\chi|\cdot|^{it}$, for some $t \in \mathbb{R}$.


Does every equivalence class of Hecke characters contain a distinguished element?


I won't deny that this question is slightly vague; what I want is something like a canonically defined Hecke character in each equivalence class.
If my calculations are correct, then if an equivalence class of Hecke characters contains a character $\chi$ of finite order, then $\chi$ is the unique character of finite order in its class. I certainly count such a character as being distinguished. The problem is therefore with Hecke characters of infinite order, which I have to say I don't understand that well. Perhaps there is a character in the class whose L-function has certain special properties? Many of the references I have come across about Hecke characters choose a splitting of the exact sequence $(*)$ in order to decompose Hecke characters. I certainly don't view this as canonical as there is no canonical choice of splitting.

REPLY [6 votes]: Dear Daniel, the answer is yes. 
An easy key lemma:

Lemma: Let $\alpha: \mathbb R^\ast_+ \rightarrow \mathbb C^\ast$ be a continuous character,
and $n \geq 1$ an integer. Then there exists one and only one character 
$\beta: \mathbb R^\ast_+ \rightarrow \mathbb C^\ast$ such that $\beta^n = \alpha$.
Proof: Using the isom $\mathbb R^\ast_+ \simeq \mathbb R$, and the polar decomposition,
we are reduce to prove that if $\alpha: \mathbb R \rightarrow \mathbb R$ (resp. $\alpha: \mathbb R \rightarrow \mathbb R / \mathbb Z$) then there exists a unique $\beta$ of the same type such that $\alpha=n \beta$. But any character $\alpha$ as above is of the form $\alpha(x)=ax$ (resp. $\alpha(x)= ax \pmod{\mathbb Z}$) for a unique $a \in \mathbb R$.
It is thus clear that taking $\beta(x) = (a/n) x$ works and is the unique possible choice for $\beta$. QED

Now consider the composed map: $i: \mathbb R^\ast_+ \hookrightarrow I_k^\infty \hookrightarrow I_k \rightarrow I_k/k^\ast$, where the first map is the diagonal embedding of $\mathbb R^\ast$
in each of the component at infinity of $I_k$ (the product of which I call $I_k^\infty$).
This map is clearly injective. Now call a Hecke character $\chi: I_k/k^\ast \rightarrow \mathbb C^\ast$ good if it is trivial on the image of $i$.
Prop: for any Hecke character $\chi$, there are exactly one good character in its equivalence class. 
Proof: Note that $|i(x)|=x^n$ if $n=[k:\mathbb Q]$ (recall that the $|\ |$ on the component $\mathbb C$ is the square of the complex modulus).
If $\chi$ and $\chi'$ are good characters in the same class, then $\psi = \chi (\chi')^{-1}$ is trivial on $I_k^1$ and on $i(\mathbb R^\ast)$, hence on $I_k$ since any element $x$ on $I_k$ can be written as $(x/ i(y)) i(y)$, with $y = |x|^{1/n} \in \mathbb R^\ast_+$,
hence is in $I_k^1 i(\mathbb R_+^\ast)$. Hence the uniqueness. 
For the existence, by the lemma there exists $\beta: \mathbb R_\ast^+ \rightarrow \mathbb C^\ast$ such that $\beta^n = \chi \circ i$, and consider $\chi' = \chi\ \  \beta^{-1}(|\bullet|)$.<|endoftext|>
TITLE: Counter example of upper semicontinuity of global fiber dimension on the source
QUESTION [6 upvotes]: We know that if $f : X\to Y$ is a morphism between two affine varieties over an algebraically closed field $k$, then the function that assigns to each point of $X$ the dimension of the fiber it belongs to is upper semicontinuous on $X$.
Does anyone know of a simple counterexample when $X$ is not irreducible (but remains an algebraic set over $k$, i.e a finitely generated $k$-algebra) to the global statement?
Edit: to avoid ambiguity I am looking for a counterexample in case $X$ is not irreducible when the dimension of the fibers is measured globally, i.e. $n\geq 0$, the set of $x\in X$ such that $\dim(f^{-1}(f(x) ) ) \geq n$ is closed in $X$.
Edit2: in his comments @dorebell linked an answer here https://mathoverflow.net/a/184925/3333 where a counterexample to the upper semicontinuity of global dimension on the source is given with $X$ and $Y$ affine and irreducible (it works even if the counterexample is explained looking at the dimension of fibers from the target)

REPLY [4 votes]: For what its worth, I can give you a non-Noetherian example, even with both schemes affine, irreducible (and of finite Krull dimension).
Set $R = k[x,y,x/y, x/y^2, x/y^3, ...]$ and $S = k[y]$.  We have the obvious map $S \hookrightarrow R$ which induces
$$
X = \text{Spec }R \to Y = \text{Spec }S.
$$
Now, away from the origin of $S$, $y$ is invertible and $R[y^{-1}] = k[x,y,y^{-1}]$ has all fibers with dimension $1$.  On the other hand, once we set $y = 0$ in $R$, we notice that $x = (x/y) y$ is a multiple, as is $(x/y) = (x/y^2) y$, and so is $(x/y^n)$ for all $n$.  This is already a maximal ideal, so the fiber over $y = 0$ is $0$-dimensional.<|endoftext|>
TITLE: At which level is it currently possible to write formal proofs?
QUESTION [21 upvotes]: I am wondering whether I should try to have some fun using proof systems. I have never used such a system, but I have some experiences in logic and programming. My question is: At which level of abstraction is it currently possible to perform formal proofs using proof assistants with reasonable effort?
Of course many branches of contemporary research rely heavily on large and complex theories which fill books but have not been formalised at all. On the other hand there are theorems which are thousands of years old (irrationality of $\sqrt{2}$) whose proofs can be found in every example-page on the websites of the proof assistants. I would not find it very compelling to perform elementary proofs in say arithmetics, elementary number theory or euclidean geometry, I would prefer a context where there is already some more abstraction available. There must be some level in-between were it is still possible to perform proofs without having to formalise whole theories. How can we describe this level?
Let me give an illustrative example: We could try to prove existence and uniqueness of the Haar measure. It is an old theorem from 1940 and it only uses very established and common theories. The proof is technical and you would definitely have to do some hard work to get all inequalities right in a fully formalised version, but we might think—it is probably naive—that it would be a manageable project. However, if it turns out that there are not yet any definitions and theorems from measure theory in the proof assistant, if you would have to formalise some stuff about Banach spaces and some concrete function spaces first, and if even the available material for general topology is very limited, it might turn out to be a project taking years.
So, can anybody tell me what is already available and usable for reusage in proof assistants? Are there basic frameworks for say general topology and for dealing with the most common algebraic structures available which are usable and well-structured and not totally messed up and which are still maintained? With which kinds of structures can you actually work (measure spaces? categories? manifolds? function spaces?), this includes, of course, the most fundamental lemmata/theorems (since the community of formal proofs is not that big, of course I do not expect any of such collections to be as exhaustive as some informal, large books). Are there large differences between the major proof assistants (Isabelle, Coq, Mizar…) with respect to such support?
(I am not asking for the state of current big projects to formalise a theorem, for example by Hales, but I want to know where the software is “ready to use”)

REPLY [12 votes]: One can possibly make some headway by switching from a traditional formulation of a field of mathematics to a "synthetic" formulation, where the objects under study are not explicitly built out of smaller pieces, but are universally characterized axiomatically. The historical prototype of such an approach is "synthetic differential geometry", where instead of defining what a smooth manifold is in the traditional way (which would take a lot of code), one imposes an axiom that guarantees that all objects/types under study behave like differentiable spaces with smooth functions between them.
Such synthetic formulations of traditional mathematics naturally lend themselves to axiomatization in proof management systems, specifically those based on type theory. Indeed, since intuitionistic type theory is precisely the internal logic of locally cartesian closed categories such as the smooth toposes of synthetic differential geometry, these are by design the systems that formalize synthetically formulated theories.
This gets even more pronounced as one passes from ordinary type theory to homotopy type theory, since here basic axioms are already so much more expressive. For instance where in ordinary type theory one formulates group theory in the traditional "piecewise" form, in homotopy type theory one finds that group theory is already built in, right out of the box: in an $\infty$-topos $\mathbf{H}$ group objects are equivalent to pointed connected objects $\mathbf{B}G$, and the slice $\infty$-topos $Act(G) \simeq \mathbf{H}_{/\mathbf{B}G}$ over such is the $\infty$-category of infinity-group actions. In thomotopy type theory this means that a dependent type over a pointed connected type $\mathbf{B}G$ is a group representation (even an $\infty$-group representation up to coherent homotopy), dependent sum now is forming  the induced representation, dependent product the co-induced representation. This yields a synthetic formulation of group theory and representation theory in homotopy type theory without adding a single extra axiom.
This option hasn't been explored much yet, as far as I can see, for doing formalized proofs. But I think it would be possible and worthwhile to do so.
One is therefore naturally led to wonder if one can usefully combine the synthetic description of differential geometry and that of higher gauge theory, aka higher group representation theory to find a synthetic formulation of modern higher differential geometry, that would naturally express modern concepts such as differential cohomology, D-module theory, étale stacks and the like. I have been exploring this a little under the name differential cohesive homotopy type theory where all this naturally exists, synthetically. Myself, I am not coding computer managed proofs myself, but for many of the statements that one proves in differential cohesive infinity-toposes it is or would be fairly straightforward to do so. Once on the n-Category Café we went through some basic exercises in this context and for instance proved in Coq from the axioms the long exact sequences that characterize differential cohomology, see here. 
For more details, with Mike Shulman we have written an introduction to the synthetic axiomatization of higher differential geometry and higher gauge theory in homotopy type theory:

Urs Schreiber, Mike Shulman, Quantum gauge field theory in cohesive homotopy type theory

Two weeks ago at the meeting of the Canadian Mathematical Society I advertized this approach further, see

Urs Schreiber, Synthetic Quantum Field Theory<|endoftext|>
TITLE: Quasicrystals and the Riemann Hypothesis
QUESTION [90 upvotes]: Let $0 < k_1 < k_2 < k_3 < \cdots $ be all  the zeros of the Riemann zeta function on the critical line:
$$  \zeta(\frac{1}{2} + i k_j) = 0 $$
Let $f$ be the Fourier transform of the sum of Dirac deltas supported at these points.  In other words:
$$    f(x) = \sum_{j = 1}^\infty e^{ik_j x}  $$
This is not a function, but it's a tempered distribution.  Matt McIrvin made a graph of it:
(source)
McIrvin seems to get an infinite linear combination of Dirac deltas supported at logarithms of powers of prime numbers.  But Freeman Dyson seems to claim that this is only known to be true assuming the Riemann Hypothesis.  So, my question is: 
1) What do people know about $f$ without assuming the Riemann Hypothesis?
2) What do people know about $f$ assuming the Riemann Hypothesis? 
3) Can we prove that $f$ is a linear combination of Dirac deltas supported at prime powers, if we assume the Riemann Hypothesis? 
4) Is there some property of $f$, like being a linear combination of Dirac deltas supported at prime powers, that's known to imply the Riemann Hypothesis?
Part of why I'm confused is that J. Main, V. A. Mandelshtam, G. Wunner and H. S. Taylor
have a paper containing some equations (equations 8 and 9) that seem to imply
$$  \sum_{j = 1}^\infty \delta(k - k_j) = - \frac{1}{\pi} \sum_{p} \sum_{m = 1}^\infty \frac{\ln p}{p^{m/2}} e^{i k \ln{p^m}} $$
where the sum over $p$ is a sum over primes.  This seems to be just what we need to show $f$ is an infinite linear combination of Dirac deltas supported at logarithms of prime powers!
I may be getting some signs and factors of $2 \pi$ wrong, but I don't think that's the main problem: I think the problem is whether a formula resembling the above one is known to be true, or whether it's only known given the Riemann Hypothesis.
Here's what Freeman Dyson said about this in 2009.  Unfortunately he omits some details I'm dying to know:

The proof of the Riemann Hypothesis is a worthy goal, and it is not for us to ask whether we can reach it. I will give you some hints describing how it might be achieved. Here I will be giving voice to the mathematician that I was fifty years ago before I became a physicist. I will talk first about the Riemann Hypothesis and then about quasi-crystals.
There were until recently two supreme unsolved problems in the world of pure mathematics, the proof of Fermat's Last Theorem and the proof of the Riemann Hypothesis. Twelve years ago, my Princeton colleague Andrew Wiles polished off Fermat's Last Theorem, and only the Riemann Hypothesis remains. Wiles' proof of the Fermat Theorem was not just a technical stunt. It required the discovery and exploration of a new field of mathematical ideas, far wider and more consequential than the Fermat Theorem itself. It is likely that any proof of the Riemann Hypothesis will likewise lead to a deeper understanding of many diverse areas of mathematics and perhaps of physics too. Riemann's zeta-function, and other zeta-functions similar to it, appear ubiquitously in number theory, in the theory of dynamical systems, in geometry, in function theory, and in physics. The zeta-function stands at a junction where paths lead in many directions. A proof of the hypothesis will illuminate all the connections. Like every serious student of pure mathematics, when I was young I had dreams of proving the Riemann Hypothesis. I had some vague ideas that I thought might lead to a proof. In recent years, after the discovery of quasi-crystals, my ideas became a little less vague. I offer them here for the consideration of any young mathematician who has ambitions to win a Fields Medal.
Quasi-crystals can exist in spaces of one, two, or three dimensions. From the point of view of physics, the three-dimensional quasi-crystals are the most interesting, since they inhabit our three-dimensional world and can be studied experimentally. From the point of view of a mathematician, one-dimensional quasi-crystals are much more interesting than two-dimensional or three-dimensional quasi-crystals because they exist in far greater variety. The mathematical definition of a quasi-crystal is as follows. A quasi-crystal is a distribution of discrete point masses whose Fourier transform is a distribution of discrete point frequencies. Or to say it more briefly, a quasi-crystal is a pure point distribution that has a pure point spectrum. This definition includes as a special case the ordinary crystals, which are periodic distributions with periodic spectra.
Excluding the ordinary crystals, quasi-crystals in three dimensions come in very limited variety, all of them associated with the icosahedral group. The two-dimensional quasicrystals are more numerous, roughly one distinct type associated with each regular polygon in a plane. The two-dimensional quasi-crystal with pentagonal symmetry is the famous Penrose tiling of the plane. Finally, the one-dimensional quasi-crystals have a far richer structure since they are not tied to any rotational symmetries. So far as I know, no complete enumeration of one-dimensional quasi-crystals exists. It is known that a unique quasi-crystal exists corresponding to every Pisot–Vijayaraghavan number or PV number. A PV number is a real algebraic integer, a root of a polynomial equation with integer coefficients, such that all the other roots have absolute value less than one${}^1$. The set of all PV numbers is infinite and has a remarkable topological structure. The set of all one-dimensional quasi-crystals has a structure at least as rich as the set of all PV numbers and probably much richer. We do not know for sure, but it is likely that a huge universe of one-dimensional quasi-crystals not associated with PV numbers is waiting to be discovered.
Here comes the connection of the one-dimensional quasi-crystals with the Riemann hypothesis. If the Riemann hypothesis is true, then the zeros of the zeta-function form a one-dimensional quasi-crystal according to the definition. They constitute a distribution of point masses on a straight line, and their Fourier transform is likewise a distribution of point masses, one at each of the logarithms of ordinary prime numbers and prime-power numbers. My friend Andrew Odlyzko has published a beautiful computer calculation of the Fourier transform of the zeta-function zeros${}^2$. The calculation shows precisely the expected structure of the Fourier transform, with a sharp discontinuity at every logarithm of a prime or prime-power number and nowhere else.
My suggestion is the following. Let us pretend that we do not know that the Riemann Hypothesis is true. Let us tackle the problem from the other end. Let us try to obtain a complete enumeration and classification of one-dimensional quasicrystals. That is to say, we enumerate and classify all point distributions that have a discrete point spectrum...We shall then find the well-known quasi-crystals associated with PV numbers, and also a whole universe of other quasicrystals, known and unknown. Among the multitude of other quasi-crystals we search for one corresponding to the Riemann zeta-function and one corresponding to each of the other zeta-functions that resemble the Riemann zeta-function. Suppose that we find one of the quasi-crystals in our enumeration with properties that identify it with the zeros of the Riemann zeta-function. Then we have proved the Riemann Hypothesis and we can wait for the telephone call announcing the award of the Fields Medal.
These are of course idle dreams. The problem of classifying one-dimensional quasi-crystals is horrendously difficult, probably at least as difficult as the problems that Andrew Wiles took seven years to explore. But if we take a Baconian point of view, the history of mathematics is a history of horrendously difficult problems being solved by young people too ignorant to know that they were impossible. The classification of quasi-crystals is a worthy goal, and might even turn out to be achievable.
[1] M.J. Bertin et al., Pisot and Salem Numbers, Birkhäuser, Boston, 1992.
[2] A.M. Odlyzko, Primes, quantum chaos and computers, in Number Theory: Proceedings of a Symposium, 4 May 1989, Washington, DC, USA (National Research Council, 1990), pp. 35–46.

REPLY [4 votes]: Thank you for posting this. I have some doubts about the worth of my own answer but I think the plots look nice so I will post them anyways.
A lazy but true formula for the Riemann zeta function when $s>0$ is:
$$\zeta(s)=\lim\limits_{k\to \infty}\left(\sum _{n=1}^k \frac{1}{n^s}+\frac{k^{1-s}}{s-1}-\frac{k^{-s}}{2}\right), \tag{1}$$
according to Mathematica:
Clear[s, n, k];
Limit[Sum[1/n^(s), {n, 1, k}] + k^(1 - (s))/((s) - 1) - (k^(-(s)))/2, 
 k -> Infinity, Assumptions -> s > 0] 

A plot of the above formula on the critical line:
Clear[t, m, s, M, r, b, M, v, d, x, k, q, z, nn, H, T, TT, g, t];
(*b=N[Sum[Sum[(BernoulliB[2*r]/((2*r)!))*(-d k^(1-2 r) E^(I x) \
Abs[StirlingS1[2*r-1,m]] Gamma[1+m,s Log[d*k]] \
Log[d*k]^(-1-m)),{m,1,2*r-1}],{r,1,q-1}],30];*)
b = 4;
M = 7;
v = 2;
k = 20;
q = 8;
c = 1;
dd = 2;
nn = 1;

z[t_] = Sum[1/n^(1/2 + I*t), {n, 1, k}] + 
   k^(1 - (1/2 + I*t))/((1/2 + I*t) - 1) - (k^(-(1/2 + I*t)))/2;
z1 = ListLinePlot[Table[N[Re[z[t]]], {t, 0, 60, 1/10}], 
   DataRange -> {0, 60}, ImageSize -> Large, 
   PlotStyle -> {Thickness[0.004]}];
z2 = Table[
   Graphics[{Arrowheads[0.025], 
     Arrow[{{Im[ZetaZero[n]], -1/2}, {Im[ZetaZero[n]], 0}}]}], {n, 1, 
    13}];
z3 = Table[
   Graphics[
    Style[Rotate[Text[N[Im[ZetaZero[n]]], {Im[ZetaZero[n]], -1}], 
      90 Degree], Medium]], {n, 1, 13}];
Show[z1, z2, z3]


which is the familiar looking zeta function.
The Fourier transform of the real part of $(1)$ is as in the question:
$$l(x)=\int_{t_1}^{t_2} \Re\left(\zeta \left(\frac{1}{2}+i t\right)\right)\ e^{i t x}\,dt = \sum _{n=1}^k \left(\frac{i d^{\frac{1}{2}-i t} n^{-\frac{1}{2}-i t} e^{i t x}}{2 (\log (d)+\log (n)-x)}-\frac{i d^{\frac{1}{2}+i t} n^{-\frac{1}{2}+i t} e^{i t x}}{2 (\log (d)+\log (n)+x)}\right)+\frac{i d^{\frac{1}{2}+i t} k^{-\frac{1}{2}+i t} e^{i t N[x]}}{4 (\log (d)+\log (k)+x)}-\frac{i d^{\frac{1}{2}-i t} k^{-\frac{1}{2}-i t} e^{i t x}}{4 (\log (d)+\log (k)-x)}-\frac{1}{2} i e^{x/2} \text{Ei}\left(\frac{1}{2} i (i+2 t) (x-\log (d k))\right)+\frac{1}{2} i e^{-\frac{x}{2}} \text{Ei}\left(\frac{1}{2} (2 i t+1) (x+\log (d k))\right)
,\tag{2}$$
when $d=1$.
A plot of the Fourier transform $l(x)$ when setting $t_1=-100$ and $t_2=+100$:
h[t_] = Sum[(I d^(1/2 - I t) E^(I t x) n^(-(1/2) - I t))/(
     2 (-x + Log[d] + Log[n])) - (
     I d^(1/2 + I t) E^(I t x) n^(-(1/2) + I t))/(
     2 (x + Log[d] + Log[n])), {n, 1, k}] + -(1/2) I E^(x/2)
     ExpIntegralEi[1/2 I (I + 2 t) (x - Log[d k])] + 
   1/2 I E^(-x/2) ExpIntegralEi[1/2 (1 + 2 I t) (x + Log[d k])] - (
   I d^(1/2 - I t) E^(I t x) k^(-(1/2) - I t))/(
   4 (-x + Log[d] + Log[k])) + (
   I d^(1/2 + I t) E^(I t N[x]) k^(-(1/2) + I t))/(
   4 (x + Log[d] + Log[k]));

t1 = -100;
t2 = +100;
L = Table[N[(h[t2] - h[t1])], {x, 1/200, dd, 1/200}];
l1 = ListLinePlot[
   Re[Sum[Sum[L*MoebiusMu[d], {d, Divisors[z]}]/z^c, {z, 1, nn}]], 
   DataRange -> {1/200, dd}, PlotRange -> {-80, 80}, 
   ImageSize -> Large, PlotStyle -> {Thickness[0.004]}];
l2 = Table[
   Graphics[{Arrowheads[0.025], 
     Arrow[{{Log[n], -35}, {Log[n], 0}}]}], {n, 1, 7}];
l3 = Table[
   Graphics[
    Style[Rotate[Text[Log[n], {Log[n], -52}], 90 Degree], Large]], {n,
     1, 7}];
Show[l1, l2, l3]


Notice that no scaling/shrinking/stretching of the x-axis is needed because the factor $2\pi$ has been left out as in $f(x)$. The frequency spikes are exactly at logarithms of $n$.
Now consider the relation:
$$\boxed{\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}} \tag{3}$$
confirmed here.
and plot:
$$m(t)=\sum _{z=1}^{\infty} \zeta \left(\frac{1}{2}+i t\right)\left(\sum\limits_{d \mid n} \frac{\mu (d) }{z^c d^{i t+\frac{1}{2}-1}}\right) \tag{4}$$
nn = k;
m[t_] = Sum[
  Sum[(Sum[1/n^(1/2 + I*t), {n, 1, k}] + 
       k^(1 - (1/2 + I*t))/((1/2 + I*t) - 1) - (k^(-(1/2 + I*t)))/2)*
     MoebiusMu[d]/N[d]^(1/2 + I*t - 1), {d, Divisors[z]}]/z^c, {z, 1, 
   nn}]; m1 = 
 ListLinePlot[Table[N[Re[m[t]]], {t, 1/10, 60, 1/10}], 
  DataRange -> {0, 60}, PlotRange -> {-3, 7 + 1/2}, 
  ImageSize -> Large, PlotStyle -> {Thickness[0.004]}];
m2 = Table[
   Graphics[{Arrowheads[0.025], 
     Arrow[{{Im[ZetaZero[n]], -1}, {Im[ZetaZero[n]], 0}}]}], {n, 1, 
    13}];
m3 = Table[
   Graphics[
    Style[Rotate[Text[N[Im[ZetaZero[n]]], {Im[ZetaZero[n]], -2}], 
      90 Degree], Medium]], {n, 1, 13}];
Show[m1, m2, m3]


The formula $(3)$ and $(4)$, for the von Mangoldt function is integrable and it is the right hand side of $(2)$:
A plot of this formula is:
nn = k;
t1 = -100;
t2 = 100;
T = Table[N[(h[t2] - h[t1])], {x, 1/200, dd, 1/200}];
f1 = ListLinePlot[
   Re[Sum[Sum[T*MoebiusMu[d], {d, Divisors[z]}]/z^c, {z, 1, nn}]], 
   DataRange -> {1/200, dd}, PlotRange -> {-80, 80}, 
   ImageSize -> Large, PlotStyle -> {Thickness[0.004]}];
f2 = Table[
   Graphics[{Arrowheads[0.025], 
     Arrow[{{Log[n], -35}, {Log[n], 0}}]}], {n, 1, 7}];
f3 = Table[
   Graphics[
    Style[Rotate[Text[Log[n], {Log[n], -52}], 90 Degree], Large]], {n,
     1, 7}];
Show[f1, f2, f3]


We notice that the heights of the spikes at logarithms are proportional to the first few terms of the von Mangoldt function:
$${\infty ,\log (2),\log (3),\log (2),\log (5),0,\log (7)\,...}$$
A Fourier transform that is closer to the picture in the question can be computed by setting the upper and lower integration limits $t_1=-500$ and $t_2=+500$:
nn = k;
h[t_] = Sum[(I d^(1/2 - I t) E^(I t x) n^(-(1/2) - I t))/(
     2 (-x + Log[d] + Log[n])) - (
     I d^(1/2 + I t) E^(I t x) n^(-(1/2) + I t))/(
     2 (x + Log[d] + Log[n])), {n, 1, k}] + -(1/2) I E^(x/2)
     ExpIntegralEi[1/2 I (I + 2 t) (x - Log[d k])] + 
   1/2 I E^(-x/2) ExpIntegralEi[1/2 (1 + 2 I t) (x + Log[d k])] - (
   I d^(1/2 - I t) E^(I t x) k^(-(1/2) - I t))/(
   4 (-x + Log[d] + Log[k])) + (
   I d^(1/2 + I t) E^(I t N[x]) k^(-(1/2) + I t))/(
   4 (x + Log[d] + Log[k]));
t1 = -500;
t2 = 500;
T = Table[N[(h[t2] - h[t1])], {x, 1/1000, dd, 1/1000}];
f1 = ListLinePlot[
   Re[Sum[Sum[T*MoebiusMu[d], {d, Divisors[z]}]/z^c, {z, 1, nn}]], 
   DataRange -> {1/200, dd}, PlotRange -> {-500, 500}, 
   ImageSize -> Large, PlotStyle -> {Thickness[0.003]}];
f2 = Table[
   Graphics[{Arrowheads[0.025], 
     Arrow[{{Log[n], -250}, {Log[n], -100}}]}], {n, 1, 7}];
f3 = Table[
   Graphics[
    Style[Rotate[Text[Log[n], {Log[n], -400}], 90 Degree], 
     Large]], {n, 1, 7}];
Show[f1, f2, f3]



Answer:
The integrable/Fourier transformable function that gives the plot of $$f(x) = \sum_{j = 1}^\infty e^{ik_j x}$$ is:
$$f(x)=\int_{t_1=-\infty}^{t_2=+\infty} \Re\left(\sum _{z=1}^{\infty} \zeta \left(\frac{1}{2}+i t\right)\left(\sum\limits_{d \mid z} \frac{\mu (d) }{z^c d^{i t+\frac{1}{2}-1}}\right)\right)\ e^{i t x}\,dt = \sum _{z=1}^{\infty}\sum\limits_{d \mid z} \frac{\mu (d) }{z^c}\left(\sum _{n=1}^k \left(\frac{i d^{\frac{1}{2}-i t} n^{-\frac{1}{2}-i t} e^{i t x}}{2 (\log (d)+\log (n)-x)}-\frac{i d^{\frac{1}{2}+i t} n^{-\frac{1}{2}+i t} e^{i t x}}{2 (\log (d)+\log (n)+x)}\right)+\frac{i d^{\frac{1}{2}+i t} k^{-\frac{1}{2}+i t} e^{i t N[x]}}{4 (\log (d)+\log (k)+x)}-\frac{i d^{\frac{1}{2}-i t} k^{-\frac{1}{2}-i t} e^{i t x}}{4 (\log (d)+\log (k)-x)}-\frac{1}{2} i e^{x/2} \text{Ei}\left(\frac{1}{2} i (i+2 t) (x-\log (d k))\right)+\frac{1}{2} i e^{-\frac{x}{2}} \text{Ei}\left(\frac{1}{2} (2 i t+1) (x+\log (d k))\right)\right)
,\tag{5}$$
Instead of depending on the zeta zeros this formulation of $f(x)$ depends on the Möbius/Mertens function.
The Fourier transform of the term with the Bernoulli numbers can be found here, but it requires integration by parts twice and then expanding the product in the expression inside the remaining integral from the second integration by parts. The convergence is slow for the Fourier transform when including the Bernoulli numbers in the Euler Maclaurin formula.
Disclaimer, if anything is wrong in the latex expressions, trust the Mathematica code. Also the blocks of code must be run/executed in the order they have been placed in the answer. The whole block of Mathematica code<|endoftext|>
TITLE: What would remain of current mathematics without axiom of power set?
QUESTION [29 upvotes]: The power set of every infinite set is uncountable. An infinite set (as an element of the power set) cannot be defined by writing the infinite sequence of its elements but only by a finite formula. By lexical ordering of finite formulas we see that the set of finite formulas is countable. So it is impossible to define all elements of the uncountable power set. The power set axiom seems doubtful. Therefore my question.

REPLY [31 votes]: Several standard theories intensely studied by set theorists do not have the power set axiom. 
One of these is the theory ZFC without the power set axiom, usually denoted $\text{ZFC}^-$. One should take care with the proper axiomatization of this theory, as we discuss in What is the theory ZFC without the powerset?, V. Gitman, J.D. Hamkins, T. Johnstone; the main point being that one should use collection+separation and not just replacement, since these are no longer equivalent without the power set axiom. 
Part of the attraction of $\text{ZFC}^-$, which is much stronger than the theory KP discussed below, but still lacks power set, is an abundance of natural models, such as the following:

HC, the universe of hereditarily countable sets. This is the land of the countable, where everything is countable. The sets in HC are precisely those sets that are countable and have only countable members and members-of-members and so on. Quite a bit of mathematics can be fruitfully undertaken in HC. 
More generally, $H_{\kappa^+}$, the universe of sets of hereditarily size at most $\kappa$. This universe satisfies $\text{ZFC}^-$, but can have some power sets, namely, as long as the power set has size at most $\kappa$. But meanwhile, there is a largest cardinal in this univese, $\kappa$ itself, and the powerset of $\kappa$ does not exist. 
More generally, $H_\delta$ for any regular cardinal $\delta$. When $\delta$ is an inaccessible cardinal, this is the same as $V_\delta$, the rank initial segment of the universe in the von Neumann hierarchy, and in this case it is a model of ZFC and a Grothendieck universe. 

These models and other models of $\text{ZFC}^-$ are used in arguments throughout set theory, from iterated ultrapowers in large cardinals to their use in forcing axioms and elsewhere. 
Another commonly studied theory without the power set axiom is Kripke-Platek set theory KP, which is a very weak set theory at the heart of the subject known as admissible set theory, in which an enormous amount of classical mathematics can be undertaken. There are numerous natural models of KP, such as:

The hyperarithmetic universe $L_{\omega_1^{CK}}$, of sets that are coded by well-founded hyperarithemtic relations on the natural numbers. This is the smallest admissible set, the smallest transitive model of KP. One interesting thing about this world is that every ordinal is not only hyperarithmetic, but actually computable. 
There are many other admissible ordinals $\alpha$, ordinals for which $L_\alpha\models$KP. 
One can relativize the admissibility concept to oracles $x$, forming $\omega_1^x$, the least admissible ordinal in $x$, so that $L_{\omega_1^x}[x]$ is the smallest model of KP containing $x$. 
The universes $L_\lambda$ and $L_\zeta$ arising in the theory of infinite time Turing machines, where $L_\lambda$ is the collection of sets coded by a well-founded infinite-time writable relation on $\omega$, and $L_\zeta$ are the sets coded by a well-founded infinite-time eventually writable relation. These universes both satisfy natural strengthenings of KP, but not the power set axiom.

And there are numerous other set theories without the power set axioms, including various strengthenings of KP that still lack the power set axiom and have natural models that are used for various purposes. 
All these models are intensely studied, and set theorists pay detailed attention to what is or is not possible to achieve in the models, depending on how strong it is. The crux of many arguments is whether the given model is strong enough to undertake a given set-theoretic construction or not. For example, one will often pay attention to the details of a mathematical construction to find out if it can be performed using only $\Sigma_1$-collection instead of, say, $\Sigma_2$-collection, in order to know whether or not it can be performed inside one of these models. 
Let me add that although set theorists are giving enormous attention to these set theories without the power axiom, the reason isn't usually because of doubt about the truth of the power set axiom, but rather it is just that they want to undertake certain constructions inside these natural models, and so they need to know whether these models are strong enough to undertake that construction or not. 
So one can be interested in set theory without the power set axiom without having doubt about that axiom. We study set theories without power set, while retaining it in our main background theory, because we want to know what is possible to achieve without power sets in those models. 
Lastly, concerning your remarks about definability, I refer you as I mentioned in the comments to an answer I wrote to a similar proposal, which I believe show that naive treatment of the concept of definability is ultimately flawed.<|endoftext|>
TITLE: A procedure to determine if an automorphism of a closed 2-manifold extends to an automorphism of a handlebody
QUESTION [8 upvotes]: In a paper of Casson and Gordon's "A loop theorem for duality spaces and fibred ribbon knots. Invent. Math. 74 (1983), no. 1, 119–137" they give a necessary criterion for a fibred knot to be a ribbon knot.   
The criterion is that the monodromy of the bundle has to extend to an automorphism of some handlebody (technicality, the fibre has a circle boundary so one caps-off the surface with a disc to get a closed surface). The handlebody is not known from simply the knot in advance, as it comes from a generalized Dehn lemma that they prove.   
My question:

Q: Given an outer automorphism of the fundamental group of a closed orientable surface, is there an efficient procedure to determine if the automorphism extends to some handlebody that has the surface as boundary? 

By the regular 3-manifold Dehn lemma, an automorphism of the surface extends to some handlebody if and only if it preserves the class of curves that bound discs in the handlebody -- which can be phrased entirely in terms of the homomorphism $\pi_1 \Sigma \to \pi_1 H$ where $\Sigma$ is the surface is the boundary of the handlebody $\Sigma = \partial H$.  So this is easy to check. 
My question has the do with the case where you want to know if such a handlebody exists.   Presumably there is a computable procedure for this, but it does not seem immediate to me.

REPLY [7 votes]: This problem is solved in the paper "Algorithmic compression of surface automorphisms" by Casson and Long.  They remark that their motivation is (partly) the paper of Casson and Gordon.<|endoftext|>
TITLE: Nearly all math classes are lecture+problem set based; this seems particularly true at the graduate level.  What are some concrete examples of techniques other than the "standard math class" used at the *Graduate* level?
QUESTION [30 upvotes]: In the fall, I am teaching one undergraduate and one graduate course, and in planning these courses I have been thinking about alternatives to the "standard math class".  I have found it much easier to find different models for undergraduate classes than for the graduate classes.  
There are many blogs and other online resources focusing in large part on math pedagogy, often taking "nonstandard" approaches to the math course (Moore method, inquiry based learning, flipped or inverted classroom, Eric Mazur's "Peer instruction",...).  For instance, Robert Talbert's blog discusses the practicalities of the "flipped" classroom, and here Dave Richeson discusses teaching undergraduate topology using the modified Moore method.  Note that although Moore originally used his method in a graduate topology course, most discussions of it and its adaptations are at the undergraduate level.
Of course, the basic principles of the methods would still apply at the graduate level, but I still find myself wanting to see specific examples of different approaches people have used to teach a graduate course. 

Are there similar discussions of teaching a nonstandard math course at the graduate level? Or at least links to syllabi, course webpages, etc. for such courses?

Further background:
First, it is of course hard to pin down exactly what is meant by "standard math class"; for the purposes of this question, let us highlight the following:

In class meetings, the vast majority of time is spent with the professor lecturing.
Nearly all the assessment is done via problem sets and quizzes/exams.
The textbook runs in parallel to the class lectures.  Although lip-service my be made to reading it, there is little structural dependence on it, except perhaps as a source of problems: lectures do not depend on students having read the book, and problems are not asked on material not presented in lecture

Second, though descriptions of the form "I once took a course where..." could be interesting, It would be particularly valuable to have links to course webpages with syllabuses/etc. describing in more practical detail what happened, or discussions from the actual professors doing the teaching about what they did, their reasoning behind it, how it went it practice, etc., as opposed to just a collection of anecdotes.
Finally, though what I have discussed above tend to be rather drastic changes from the "standard math class", of course things run on a continuum, and smaller deviations are possible. For example, I have taken graduate courses where student presentations of selected material played a role, and where a part of the grade was given to an expository paper.  But in the courses I took these were usually rather minor deviations from the standard structure.  Examples of similar smaller variations would be valuable if they were particularly well documented, or had explanations of why these variations were particularly important and not just window dressing.

REPLY [4 votes]: A good strategy for a class which I endured for a couple classes as an undergraduate and to a lesser extent as a graduate student for a couple classes is when the students make presentations for at least a substantial portion of the total number of class sessions, and make most of the students grade based on presentations.
If the students make presentations for class, then there are several obvious advantages. First of all, if the students make presentations, then they be better prepared for teaching and making presentations at seminar and conferences. Being able to talk in front of people with out becoming paralyzed is a very valuable skill to have. The students may even grow comfortable enough with making presentations, that they may even become comfortable enough to tell a couple appropriate jokes during the presentations during class. Also, when making a presentation on a certain topic or result, the student should have a deeper understanding of that particular topic. A couple years ago, I had to make a presentation on a certain paper, and I can still recall from memory several details about that paper that I made in that presentation. Furthermore, if the students give presentations for class, then class can still be held when the professor is away at a conference of something. This surely beats cancelling class altogether. I once made a presentation outside (yes outside) in front of two other students while the professor was away at a conference :).
Of course, for this strategy to work, the class cannot contain too many students, and the students have to be motivated and able enough to give high quality presentations.<|endoftext|>
TITLE: Analysis of the boundary of the Mandelbrot set
QUESTION [13 upvotes]: Motivation: The Mandlebrot set is a simply connected set with an infinitely complex boundary, but CAN one move from interior to the exterior of this topological space by just crossing over a finite set of points?
Application: I am modeling the leakage of emission of EM waves, through a fractal forest where dispersion occurs every time an edge of the forest is crossed
Let $\mathbb{M}\subset \mathbb{C}$ be the Mandelbrot Set. Let $\partial \mathbb{M}\subset \mathbb{M}$ be its boundary. Consider two points $\mathbb{z_1}\in \mathbb {M}$ and $\mathbb{z_2}\in\mathbb{C}\notin \mathbb{M}$
My question is: 

does there exist a path $\mathbb{P}$ $ \subset \mathbb{C}$ with initial point $\mathbb{z_1}$ and terminal point $\mathbb{z_2}$? 
Is $\mathbb{P}\bigcap\partial\mathbb{M}$ a finite set?

REPLY [22 votes]: Your notation is unusual, and I am not sure whether I entirely understand it.
I shall take your question to mean: Can every point of the Mandelbrot $M$ set be connected to $\infty$ by a path that intersects the boundary $\partial M$ in only finitely many points?
This is connected to a very famous conjecture, namely that The Mandelbrot set is locally connected. (See The deep significance of the question of the Mandelbrot set's local connectedness?.)
Indeed, if the Mandelbrot set is locally connected, then - in particular - every point of the boundary of M is accessible from $\infty$. Note that "finitely many points" can be replaced by "one point". (Recall that the Mandelbrot set is full, and hence every interior component is simply connected.)
If the Mandelbrot set is not locally connected, then it follows e.g. from the theory of "fibers", as formulated by Dierk Schleicher, that there exists a point of the boundary that is not accessible from the complement, and hence the answer to your question would be negative.
(Caution. This statement is not true for general compact sets: a compact and full connected set can have every point accessible, but not be locally connected. The statement above uses the specific structure we know about the Mandelbrot set.)
If you start asking for curves with specific geometric properties, the question will become more subtle. For example, if you ask for smooth curves, the answer is 'no' in general (as the Mandelbrot set spirals at many points).<|endoftext|>
TITLE: Reference for a theorem on crossing changes of links
QUESTION [5 upvotes]: I've recently stumbled upon a paper of Scharlemann on crossing changes: 
link text
In particular I am interested in understanding Theorem 2.2 (page 6):
"Theorem: If links A and B
are related by a crossing change, and both are composite,
then the crossing change takes place within a proper summand."
Where can I find a proof of this result?
Also, is it possible to find an equivalent statement on diagrams?
Thanks

REPLY [2 votes]: The reference is given in the paper you cite:
M. Eudave-Mun ̃oz, Primeness and sums of tangles, Trans. Am. Math. Soc. 306, 773-790 (1988)
The arguments are purely combinatorial, but there should be a simplification using sutured manifold theory along the lines of Scharlemann and Thompson's paper "Unknotting number, genus, and companion tori" by Scharlemann and Thompson MR0929535
Off the top of my head, I don't see a way to convert either type of argument into a something about diagrams.<|endoftext|>
TITLE: Symplectic blow-up
QUESTION [5 upvotes]: Blow-ups of points can also be performed in the symplectic category; for a given point $p\in (X,\omega)$ we choose a Darboux chart around $p$ and then use the symplectic cut corresponding to the standard hamiltonian S^1-action (Diagonal action) to define the blow-up.
It should be possible to define blow-ups of general symplectic submanifolds (which is J-holomrphic with respect to some fixed compatible almost complex structure) in a similar way.
Given $X\subset Y$ (a symplectic and J-holomorphic submanifold) we need to put a symplectic structure on a neighborhood of $X$ in $N_X^Y$ (normal bundle) which by symplectic neighborhood theorem is symplectomorphic to a neighborhood of $X \subset Y$ and such that the action of $S^1$ (multiplication by $e^{i\theta}$) is Hamiltonian.
I can't find a reference in the literature where the details are written? Is it any where out there?

REPLY [3 votes]: I think "Birational equivalence in the symplectic category" - Invent.Math (by V.Guillemin and S.Sternberg) covers all details that you want. (Section 8).<|endoftext|>
TITLE: Fundamental groups of normal complex quasi-projective varieties
QUESTION [8 upvotes]: I would like to know if there is an explicit example of a finitely presented group that can not be realised as the (topological) fundamental group of a normal complex quasi-projective variety?

REPLY [6 votes]: More obstructions for this realizability problem can be found in the paper On the fundamental groups of normal varieties by Donu Arapura, Alexandru Dimca, and Richard Hain (http://arxiv.org/abs/1412.1483).
For an explicit example, take the right-angled Artin group corresponding to a path on 4 vertices,
$$
G=\langle a,b,c,d \mid [a,b]=[b,c]=[c,d]=1 \rangle.
$$
It is known from http://arxiv.org/abs/0902.1250 that this group (or, for that matter,  any RAAG whose associated graph is not a complete multipartite graph) is not the fundamental group of a smooth complex quasi-projective variety. In the paper quoted above, the authors show that such a RAAG cannot be realized as the fundamental group of a normal complex quasi-projective variety.<|endoftext|>
TITLE: When is the reduced subscheme of a Cohen-Macaulay scheme also Cohen-Macaulay?
QUESTION [13 upvotes]: Let $X$ be a Cohen-Macaulay scheme (I will be interested in the case when this is ${\rm Spec}(A/I)$ where $A$ is a polynomial ring over a field and $I$ is a homogeneous ideal). I would like to know some conditions under which the reduced subscheme $X_{\rm red}$ is also Cohen-Macaulay.
For a more specific question, assume further that $X$ is cut out by a single equation in $Y$, which is also Cohen-Macaulay (and is of the form ${\rm Spec}(A/J)$ with $J$ homogeneous).

REPLY [10 votes]: Very interesting question! I do not have a full answer, but here are a few cases when one can say something not completely trivial. I will assume as in your last paragraph that $X = {\rm Spec (S/(f))}$ and $Y={\rm Spec(S)}$. Furthermore, let's assume $Y$ is normal. 
Then $X_{\rm red}$ would be Spec of $S/I$, where $I$ is the intersection of the minimal primes $P_1,\dots,P_n$ of $f$. The main points now are:
1)The isomorphism class $[I]$ is an element in the class group $Cl(S)$, equal to $[P_1]+\dots+ [P_n]$ in that group.
2)$X_{\rm red}$ would be Cohen-Macaulay  if and only if $I$ is CM as an $S$-module, as can be seen from the exact sequence $0\to I\to S\to S/I\to 0$.
Thus, knowing the Cohen-Macaulay elements in the class group $Cl(S)$ would be helpful.
Let me discuss some concrete examples. First, if $S$ is an UFD (e.g if it is the cone of some smooth projective complete intersection of dimension at least $3$), then $X_{\rm red}$ will always be CM. That is because $Cl(S)$ is trivial, so $I$ is isomorphic to $S$. 
Now, let's say $S= \mathbb C[x,y,u,v]/(xy-uv)$. The class group is isomorphic to $\mathbb Z$, generated by the class of $P=(x,u)$ with $Q=(x,v)=-P$. The class $n[P]$ is isomophic to $P^n$ if n positive and $Q^n$ if $n$ negative, so has $|n|+1$ generators. The only CM elements must be isomorphic to $S$, $P$ or $Q$. In other words, $I$ is CM if and only if it has at most $2$ generators. Even just knowing the number of generators of the minimal primes $P_1, \dots, P_n$ can help you decide. For example, if there are $3$ primes with $2, 3, 7$ generators than they can not add up to $0$ or $+-[P]$, thus  $I$ will not be CM. 
The CM elements of the class groups have been studied by people such as Bruns and Gubeladze over toric rings. For example it is known, see the intro here that all torsion classes over such rings are CM. It provides another sufficient criterion: if $S$ is a toric ring, and the multiplicity of $f$ at each $P_i$ is constant, then $X_{\rm red}$ is CM. 
In many cases (toric, determinantal) one can prove that there only finitely many such CM isoclasses. (Shameless plug: K. Kurano and I have had a few results on  this problem for more general settings, it is being written).<|endoftext|>
TITLE: Uncertainty principle on finite groups
QUESTION [12 upvotes]: For a finite group $G$ with normal subgroup $H$, the induced representation $\text{Ind}_H^G(1)$ decomposes as a sum of irreducibles with the multiplicities equal to the dimensions, because it is is the pullback of the right regular representation of $G/H$.  For a subgroup which is not normal, this need not be true.  For example, take $G=S_3$ and $H$ a subgroup of order $2$.  The index $3$ can only be written as a sum of squares as $1^2+1^2+1^2$, but $S_3$ only has two distinct $1$ dimensional representations.

Is the converse true?  That is, if $\text{Ind}_H^G(1)$ decomposes into irreducible representations with multiplicity equal to the dimension, is $H$ necessarily normal in $G$?  If not, is there some other characterization of subgroups which have this property?

The question arises in generalizing the 'Uncertainty Principle' for finite abelian groups
$$
|G|\le \sharp \text{supp}(f)\cdot \sharp\text{supp}(\hat f),
$$
to finite groups in general, and when the corresponding inequality is tight.

REPLY [10 votes]: Yes. If an irreducible representation $U$ occurs in the induced representation with multiplicity $\dim(U)$, then it follows by Frobenius Reciprocity that the trivial representation occurs with multiplicity $\dim(U)$ in the restriction of $U$ to $H$, and so $H$ acts trivially on $U$. So your condition implies that $H$ acts trivially on the induced representation, which is only the case if $H$ is normal in $G$.

REPLY [7 votes]: Yes, it is. If for any irreducible $\rho$, $\langle\text{Ind}_H^G(1),\rho\rangle$ is either 0 or dim $\rho$, then $H$ is the intersections of $\ker \rho$ for those $\rho$ for which this inner product is not 0, 
and intersections of kernels of irreducible characters are precisely the normal subgroups of $G$.
Indeed, by Frobenius recoprocity,
$$
\langle \text{Ind}_H^G(1),\rho\rangle = \langle 1,\text{Res}_H\rho\rangle =\dim\rho\Longrightarrow H\leq \ker\rho.
$$
This gives you the inclusion $H\leq \bigcap \ker\rho,$ the intersection running over those $\rho$ with non-trivial inner product with $\text{Ind}_H^G1$.
To get equality, notice that the hypothesis on the inner products implies that the dimensions of $\text{Ind}_H^G1$ and of $\text{Ind}_K^G 1$, where $K=\bigcap \ker\rho$, are the same.<|endoftext|>
TITLE: Calculus book in the spirit of the 18th century
QUESTION [10 upvotes]: I have been reading a lovely 18th century book by John Rowe called An Introduction to the Doctrine of Fluxions. It presents calculus completely motivated by geometric questions - finding tangents to curves; finding and maximising lengths, areas, and volumes; finding radii of curvature and evolutes, etc. etc. There are many more of these books, by L'Hospital, Hayes, Emerson, MacLaurin, Simpson, etc., always illustrated with copious figures. Are there any modern books on calculus written in this spirit?

REPLY [8 votes]: Here is an outstanding modern example:
http://www.pdmi.ras.ru/~olegviro/Shchepin/index.html
On Euler's footsteps, by Evgeny Shchepin.
On a more advanced level,
 MR1656255 Stalker, John Complex analysis, Fundamentals of the classical theory of functions. Birkhäuser Boston, Inc., Boston, MA, 1998.
It is an interesting, unusual book, though I strongly disagree with many statements in it.<|endoftext|>
TITLE: Is Lemma A.1.5.7 in Higher Topos Theory correct?
QUESTION [24 upvotes]: Hello to everyone,
I am studying the properties of combinatorial model categories, following the exposition given by Jacob Lurie in Higher Topos Theory ([HTT] from now on), in section A.2.6.
At some point, he needs to show that in a presentable category $\mathcal C$ and a large enough set $S$ of morphisms in $\mathcal C$, the class generated by $S$ under transfinite pushouts is the same as the class generated by $S$ under retracts and transfinite pushouts; that is: we don't need retracts. This is accomplished in Proposition A.1.5.12.
In the proof of Proposition A.1.5.12, he needs to replace a sequence of morphisms with a tree, satisfying some additional condition; the existence of such a replacement would be Lemma A.1.5.7, but I have problems in understanding why the proof should be correct.
In particular, with the notations used there, I could consider for every $\beta \in A$ the subset $B := \{\alpha \in A \mid \alpha \preceq \beta\}$; this would be $\preceq$-downward closed by construction; since it has a final object, we obtain
$$
Y_B^\prime := \varinjlim_{\alpha \in B} Y_\alpha^\prime \simeq Y_\beta^\prime
$$
On the other side, condition (1) implies that $B$ has a final object also when thought as subset of $(A,\le)$; it follows that
$$
Y_B := \varinjlim_{\alpha \in B} Y_\alpha \simeq Y_\beta
$$
i.e. $Y_\beta \simeq Y_\beta^\prime$, so that the diagram shouldn't be changed. But then, I don't see how to prove that $\{Y_\alpha\}_{\alpha \in A'}$ is a $S$-tree (Definition A.1.5.1 in [HTT]).
Therefore, my questions are:

do you agree with me that the result is seemingly false or can you explain me how the proof is supposed to work?
do you think that the Proposition A.1.5.12 is correct?
do you have any other reference for a proposition which is similar to Proposition A.2.6.8 (which is used in the proof of the Smith's characterization of combinatorial model structures)?

Edit. I found a related question here. Even though it doesn't answer my question, it fixes the notations I am using, hence I am signaling it for your convenience.

REPLY [40 votes]: Looks like a typo. Condition $(4)$ should say that $B$ is downward closed under $\leq$, not under $\preceq$ (otherwise, $Y_B$ is not defined).<|endoftext|>
TITLE: Are Turaev--Viro invariants secretly a discretized path integral?
QUESTION [12 upvotes]: Turaev--Viro http://www.ams.org/mathscinet-getitem?mr=1191386 defined an invariant of three-manifolds $M$ denoted $TV(M)$, which was subsequently shown by Kevin Walker to coincide with $\left|WRT(M)\right|^2$ where $WRT(M)$ is the Witten--Reshetikhin--Turaev invariant of $M$.  Both invariants depend on a choice of quantum group $U_q(\mathfrak g)$ and a root of unity $q$.
Witten's "physics definition" of $WRT(M)$ http://www.ams.org/mathscinet-getitem?mr=990772 is via an "integral over all $\mathfrak g$-connections on $M$".  I have read in many places that the Turaev--Viro model is essentially a discretization of the path integral (which only converges if we use a quantum group; the nonconvergent analogue for a classical group is the "Ponzano--Regge model").

How can one see that the Turaev--Viro model as a discretization of the path integral?

The definition of $TV(M)$ is as a "state sum" over all assignments of representations of $U_q(\mathfrak g)$ to each of the edges of (a fixed triangulation of) $M$.  On the other hand, one would naively expect that a discretization of the path integral would be a "state sum" over all assignments of "elements" of $U_q(\mathfrak g)$ to each of the edges of $M$ (this being the most obvious way to think about a discretized connection on $M$).
How does the connection between these two state sums go?

REPLY [6 votes]: Perhaps I asked this question a bit too quickly.  An explanation exactly as I was looking for appears at the beginning of Section III of this paper: http://arxiv.org/abs/hep-th/0401076<|endoftext|>
TITLE: Meaning of historical fluxion notation
QUESTION [6 upvotes]: I've noticed that in 18th century English books on calculus writers would say that 'the fluxion of $ax$ is $a\dot{x}$' and 'the fluxion of $x^n$ is $n x^{n-1} \dot{x}$'. What does this extra '$\dot{x}$' at the end of the formulas for fluxions (derivatives) signify?
P.S. Apologies if historical questions are not allowed here.

REPLY [6 votes]: In the physics applications that Newton was interested in, his functions were mostly functions of time. Since he was usually differentiating with respect to time, it was OK that his dot notation, unlike Leibniz's notation, didn't indicate what he was differentiating with respect to. Newton used the symbol $o$ to indicate a fixed infinitesimal interval of time. So in Newton's notation $\dot{x}$ would be Leibniz's $dx/dt$, while $\dot{x}o$ would be $dx$ --- an infinitesimal difference, called a "moment."
Newton also used a notational shortcut that caused confusion. He used a convention that when it was clear from context that he was talking about a moment, then $\dot{x}$ would implicitly mean $\dot{x}o$, i.e., the $o$ could be left out because it was too cumbersome to write it all the time. Because of this, English mathematicians began to muddy the notational waters by not distinguishing the two notions. See Boyer, p. 201, and p. 114 for the definition of "moment."
So when someone using Newton's notation says that the fluxion of $x^n$ is $nx^{n-1}\dot{x}$, what they mean could be two different but equivalent things in Leibniz notation. Either:
(1) $d(x^n)=nx^{n-1}dx$ (where $dx$ is the same as $\dot{x}o$, and the $o$ is implicit); or
(2) $d(x^n)/dt=nx^{n-1}dx/dt$
In 17th- and 18th-century mathematics, the difference between (1) and (2) is purely a matter of dividing both sides of the equation by $dt$. (There was no notion, as in modern NSA, that a derivative is the standard part of the ratio of infinitesimals, which is not quite the same thing as the ratio of infinitesimals.)
Boyer, The History of the Calculus and its Conceptual Development.  https://archive.org/details/TheHistoryOfTheCalculusAndItsConceptualDevelopment<|endoftext|>
TITLE: Does a proof of Selberg's 3.2 inequality exist?
QUESTION [28 upvotes]: A well-known inequality of Montgomery and Vaughan (generalizing a result of Hilbert) states that
$$ \left |\sum_{r \neq s} \frac{w_{r} \overline{w_{s}}  }{\lambda_r - \lambda_s} \right| \leq \pi \delta^{-1} \sum_{r} |w_{r}|^2$$
where $\{\lambda_{r}\}$ are an increasing sequence of $\delta$-separated real numbers ($|\lambda_{r+1} - \lambda_{r}| \geq \delta$) and $\{w_{r}\}$ are complex numbers. The constant $\pi$ is known to be sharp. There is a further `weighted' generalization of this inequality (also due to Montgomery and Vaughan) that states 
$$ \left |\sum_{r \neq s} \frac{w_{r} \overline{w_{s}}  }{\lambda_r - \lambda_s} \right| \leq \frac{3}{2} \pi  \sum_{r}  \delta_{r}^{-1}|w_{r}|^2$$
where $\delta_{r}>0$ is a real number such that $|\lambda_{r}-\lambda_{s}| \geq \delta_{r}$ for any $s \neq r$ and the rest of the notation is the same as above. 
Here the constant $\frac{3}{2}\pi$ is not optimal, indeed it is conjectured that the $\frac{3}{2} \pi$ can be replaced by $\pi$. Proving this, however, has remained open for over 40 years. Refining the constant would have a number of applications to sieve theory (indeed this inequality even has a minor role in the ongoing Polymath project to refine Zhang's prime gap theorem). This article of Montgomery is a good place to read about the role of the inequality in number theory (as of 1978, at least).
There have been a number of refinements to the constant over the years. In his 1978 survey article Montgomery states that Selberg has an unpublished proof that shows $\frac{3}{2} \pi \approx 4.71$ can be replaced by $3.2$. Curiously, in 1984 E. Preissmann published a (18 page!) proof showing that the inequality holds with the constant $\frac{4}{3}\pi \approx 4.18$ (which is inferior to that claimed by Selberg). In addition, I have read that there is a proof of the inequality with constant $\sqrt{22} \approx 4.69$ given by Jörg Brüdern (in Einführung in die analytische Zahlentheorie, Springer Verlag 1995), which would be yet inferior to Preissmann's result. This leads me to ask:


Does there exist a copy of Selberg's proof? 


Of course, I'd be interested to know of any results related to the problem  beyond those listed above.

REPLY [3 votes]: I think it is highly unlikely that such a proof exists. This is quite far from definitive, but that's usually the case with this kind of nonexistance answers. My reasoning is:

He was a very prolific writer (after WWII, anyway). His collected papers add up to more than 900 pages. As original a mathematician as Selberg was, I can't imagine him dismissing this result as "not important enought to publish".
The survey is from 1978, so Selberg (professor at Princeton until 1987) had more than enough time to publish such a relevant proof.
Finally, as Lucia mentions on the comments, a lot of work has gone into cataloguing and publishing his research since he retired. The first edition of the Springer volumes dates from 1989 and 1991, the newest from 2014. Springer is known to sometimes include things like unpublished papers to letters is his Collected Papers series, so if the proof was ever written down, it should have surfaced by now.<|endoftext|>
TITLE: Existence of Geodesics in continuous metrics
QUESTION [5 upvotes]: I learned that if we are given a $C^0$ Riemannian metric on a smooth manifold $M$, geodesics (i.e. length minimizing curves) are absolutely continuous, and if the metrics is $C^{0,\alpha}$, then the geodesics are even $C^{1, \beta}$ where $\beta = \alpha/(2-\alpha)$.
However, I did not find any results on the existence of geodesics. A priori, it could be possible that no geodesics exist at all, or am I wrong? 
Are there results that there exist "enough" geodesics in some sense, like when two points are close enough, or for almost any two points (suppose maybe that $M$ is compact)?
A problem in considering the geodesics equation seems to be that it involves the Christoffel symbols which are not continuous. If we have a solution to the geodesic equation, is it length minimizing for two points close enough?
It is known that given a continuous metric $g$, there exists a sequence of smooth Riemannian metrics that converges to $g_n$ such that the corresponding distance functions converge to the one associated to $g$. What happens with the geodesics (i.e. solutions to the geodesics equation) in this case? Do they converge in some sense?

REPLY [3 votes]: Ok, thank you Misha for the comments, let me try to fill out the hints you gave myself. I try to prove the following: 
Let $g_n$ be a sequence of complete smooth metrics that converge in $C^0$ against the continuous metric $g$. Let $p$, $q \in M$ such that they are not cutpoints of each other w.r.t. any $g_n$. Let $\gamma_n$ be a $g_n$-geodesic connecting $p$ and $q$, then $\gamma_n$ converges to a $g$-geodesic in $C^0$.
By theorem 4.5 in the paper quoted above, for each $\eta>0$, there exists a number $N \in \mathbb{N}$ such that
$$ (1-\eta)d_n(p, q) \leq d(p, q) \leq (1+\eta) d_n(p, q) ~~~~ \forall n\geq N$$
Hence for $\delta = \varepsilon/(1+\eta)$, $|s-t| < \delta$ implies
$$ \varepsilon > (1+\eta)|s-t| = (1+\eta)d_n(\gamma_n(s), \gamma_n(t)) \geq d(\gamma_n(s), \gamma_n(t))$$
Hence the set of $\gamma_n$ with $n \geq N$ is equicontinuous and contained in some compact set by the same reasining as in the proof of the quoted thm. 4.5. Hence it subconverges in $C^0$ to a path $\gamma$. By the assumption that $p$ and $q$ are not cutpoints for any $g_n$, $\gamma_n$ is uniquely determined, so $(\gamma_n)$ converges itself (they fulfill a differential equation depending smoothly on the metric and hence are $C^0$ close if the metrics are $C^0$ close).
I do not yet see why $\gamma$ is a $g$-geodesic though.<|endoftext|>
TITLE: Fourier series representing a continuous function?
QUESTION [6 upvotes]: This is maybe not really research level, but I have not found anything in the literature, and asking on math.stackexchange wasn't successful either.
Fourier series define an isometry $L^2(\mathbb{Z}) \rightarrow L^2(S^1), (a_k)_{k \in \mathbb{Z}} \mapsto \sum a_k z^k \colon S^1 \rightarrow \mathbb{C}$ (all Hilbert spaces are complex). It is known (and easy) that, when we start with a sequence $a_k$ which is actually in $L^1(\mathbb{Z})$, we end up with a continuous function on $S^1$. My question is: What more can we say about sequences $a_k$ whose associated Fourier series is a continuous function on $S^1$? In particular, is there an easily checkable if and only if-criterion? I would guess that not, but I have never seen such a question discussed in the literature.

REPLY [9 votes]: This is really a glorified comment, but the formatting in answer boxes is easier (and I think the information deserves greater visibility). Apologies to the OP if he already knew this.
From Katznelson's introduction to Harmonic Analysis (the Dover version, 2nd corrected edition). Everything takes place on the circle ${\bf T}$.
Chapter IV, Theorem 2.4.

(a) There exists a continuous function $f$ such that for all $p<2$, $\sum \vert\widehat{f}(n)\vert^p = \infty$.
(b) [paraphrased by YC] There exists a sequence $(a_n)_{n\in{\bf Z}}$ which belongs to $\ell^p$ for every $p>2$, but which is not the Fourier series of any finite Borel measure on ${\bf T}$.

Quite a lot has been done, in the classical days of the subject, on how smoothness/Hölder conditions on a function are reflected in the rate at which its Fourier series decays: see Chapter I, Section 4 of the same book.
On a positive note, if the Fourier series is lacunary, then it is the Fourier series of a continuous function iff it belongs to $\ell^1$. (Chapter V, paragraph 1.4; this is sometimes stated as "lacunary sets are Sidon sets")
The circumstantial evidence, as I understand it, is that there is no way to characterize the Fourier series of continuous functions by means of a naive "sequence space" condition on the sequence. However, as Mike Jury has pointed out, one can use Fejer's theorem to get some kind of algorithm-flavoured criterion.<|endoftext|>
TITLE: Upper limit on the central binomial coefficient
QUESTION [13 upvotes]: What is the tightest upper bound we can establish on the central binomial coefficients $ 2n \choose n$ ?
I just tried to proceed a bit, like this:
$ n! > n^{\frac{n}{2}} $
for all $ n>2 $. Thus,
$ \binom{2n}{n} = \frac{ (n+1) \ldots (2n) }{n!} < \frac{\left(\frac{\sum_{k=1}^n (n+k) }{n}\right)^n }{n^{n/2}} = \frac{ \left( \frac{ n^2 + \frac{n(n+1)}{2} }{n} \right) ^n}{n^{n/2}} = \left( \frac{3n+1}{2\sqrt{n}} \right)^n $
But, I was searching for more tighter bounds using elementary mathematics only (not using Stirling's approximation etc.).

REPLY [5 votes]: Since it gives tighter bounds, I will reproduce my answer from MathSE.

For $n\ge0$, we have (by cross-multiplication)
$$
\begin{align}
\left(\frac{n+\frac12}{n+1}\right)^2
&=\frac{n^2+n+\frac14}{n^2+2n+1}\\
&\le\frac{n+\frac13}{n+\frac43}\tag{1}
\end{align}
$$
Therefore,
$$
\begin{align}
\frac{\binom{2n+2}{n+1}}{\binom{2n}{n}}
&=4\frac{n+\frac12}{n+1}\\
&\le4\sqrt{\frac{n+\frac13}{n+\frac43}}\tag{2}
\end{align}
$$
Inequality $(2)$ implies that
$$
\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}\frac{\sqrt{n+\frac13}}{4^n}\text{ is decreasing}}}\tag{3}
$$
For $n\ge0$, we have (by cross-multiplication)
$$
\begin{align}
\left(\frac{n+\frac12}{n+1}\right)^2
&=\frac{n^2+n+\frac14}{n^2+2n+1}\\
&\ge\frac{n+\frac14}{n+\frac54}\tag{4}
\end{align}
$$
Therefore,
$$
\begin{align}
\frac{\binom{2n+2}{n+1}}{\binom{2n}{n}}
&=4\frac{n+\frac12}{n+1}\\
&\ge4\sqrt{\frac{n+\frac14}{n+\frac54}}\tag{5}
\end{align}
$$
Inequality $(5)$ implies that
$$
\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}\frac{\sqrt{n+\frac14}}{4^n}\text{ is increasing}}}\tag{6}
$$
Note that the formula in $(3)$, which is decreasing, is bigger than the formula in $(6)$, which is increasing. Their ratio tends to $1$; therefore, they tend to a common limit, $L$.

Theorem $1$ from this answer says
$$
\lim_{n\to\infty}\frac{\sqrt{\pi n}}{4^n}\binom{2n}{n}=1\tag{7}
$$
which means that
$$
\begin{align}
L
&=\lim_{n\to\infty}\frac{\sqrt{n}}{4^n}\binom{2n}{n}\\
&=\frac1{\sqrt\pi}\tag8
\end{align}
$$
Combining $(3)$, $(6)$, and $(8)$, we get
$$
\boxed{\bbox[5pt]{\displaystyle\frac{4^n}{\sqrt{\pi\!\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\!\left(n+\frac14\right)}}}}\tag9
$$<|endoftext|>
TITLE: What fraction of n-point sets in the unit ball have diameter smaller than 1?
QUESTION [13 upvotes]: This question is inspired by a recent talk by Matt Kahle on random geometric complexes. 
Some simple notation: let $\mathcal{B} \subset \mathbb{R}^d$ be the unit ball in $d$-dimensional Euclidean space with the usual norm $\|\cdot\|$ and let $\mathcal{B}^k$ denote the $k$-fold product of this ball with itself for any positive integer $k$. For each $n \in \mathbb{N}$, define $K_n \subset \mathcal{B}^n$ as follows:
$$ K_n = \lbrace(p_1,\ldots,p_n) \in \mathcal{B}^n \text{ so that } \|p_i - p_j\| < 1 \text{ for all } 1 \leq i,j \leq n\rbrace.$$
Thus, $K_n$ consists of those $n$-tuples of points in the unit ball whose diameter is bounded above by $1$. I would like to know what fraction of $\mathcal{B}^n$ lies in $K_n$ asymptotically as $n$ increases. More precisely, let $\mu$ be the usual $d$-dimensional Lebesgue measure, and define 
$$\chi(n) = \frac{\mu(K_n)}{\mu(\mathcal{B})^n}.$$

How does $\chi(n)$ behave for fixed dimension $d$ as $n \to \infty$?

While an answer to this specific question would be great, I would be much more interested in some insight regarding how one should tackle such problems. The Calculus approach of setting up some integral fails horribly -- at least, I should confess that I tried $n = 2 = d$ and got hopelessly stuck with what appears to be an elliptic integral of the murderous kind -- so I expect that there is no closed formula for $\chi(n)$. But surely someone has worked out asymptotic envelopes for such a basic quantity!

REPLY [15 votes]: I'm going to show that, for any $x>2^{-d}$, this is $O(x^{n})$. By Ricardo's lower bound, this is tight.
Given a set of $n$ points of diameter at most $1$, take all the points you get from those points by rounding the coordinates up or down to multiples of $\epsilon$. This new set of points will contain the old set of points in its convex hull, and since the new set of points will be of a distance no more than $\sqrt{d}\epsilon$ from the old set of points, will have a diameter at most $1+2\sqrt{d}\epsilon$. Thus each configuration of points of diameter at most $1$ in a sphere of radius $1$ is contained in the convex hull of some configuration of $\epsilon$-lattice points of diameter at most $1+2\sqrt{d}\epsilon$ in a sphere of radius $1+\sqrt{d}\epsilon$. Let $N$ be the (finite) number of such configurations of lattice points. Then since the sphere has the greatest volume of any convex body with a given diameter, the convex hulls each take up at most $(1+2\sqrt{d} \epsilon)^d/2^d$ of the volume of the sphere, so landing in any hull has probability
\[ N \left(\frac{ 1 + 2 \sqrt{d} \epsilon}{2}\right)^{dn} \]
If we let $\epsilon$ go to $0$ we have the desired result.
To be more specific, we can set $\epsilon = n^{-1/(d+1)}$, and get the upper bound
\[\ln \left(2^{nd} \chi(n)\right) = O\left( n^{d/(d+1)} \right) \]
using the trivial upper bound $N < 2^{O(\epsilon^{-d} )}$<|endoftext|>
TITLE: Equivariant Cohomology of a Complex Projective Variety
QUESTION [5 upvotes]: Suppose that I have a complex projective variety $X$ endowed with an algebraic action of a complex torus $T$. Suppose also that the set $X^T$ of fixed points is finite. I would like to relate the equivariant cohomologies $H_{T}(X;\mathbb{C})$ and $H_{T}(X^T;\mathbb{C})$ using the existing localization theory. Yet, I am finding it difficult to verify some of the hypotheses conventionally imposed in the main theorems (ex. that $H_T(X;\mathbb{C})$ is a free module over $H_{T}(*;\mathbb{C})$) for the examples I am considering. What are some of the most general conclusions that follow in the situation described above, and what sorts of things should I investigate in order to relate the equivariant cohomologies?

REPLY [5 votes]: Usually one uses a Białynicki-Birula‎ decomposition derived from a general circle in $T$ to find cycles that give a basis of this free module. But this decomposition is less useful when the space is singular.
For a basic example, let $X$ be the triangle $\{ [x,y,z] : xyz = 0 \} \subset {\mathbb P}^2$. This space has $H^1$ and $H^1_T$, but its fixed points do not.<|endoftext|>
TITLE: Axioms for mutual information
QUESTION [8 upvotes]: I am interesting in axiomatic justifications for concepts in information theory. I have found many axiomatizations for Shannon's entropy and for the Kullback-Leibler divergence, as well as their variations (for some examples, see this survey or these books: 1, 2). However, I have not found any results directly for the mutual information.
Are there any results that show that some intuitive properties uniquely characterize the mutual information of a joint distribution of two random variables?

REPLY [3 votes]: My paper https://arxiv.org/pdf/2108.12647.pdf provides an answer to this question. In particular, here is a quote from the introduction:
Our main result is Theorem~6.6, where we prove that the mutual information $\mathbb{I}(\mathcal{X},\mathcal{Y})$ of an ordered pair of random variables is the unique function (up to an arbitrary multiplicative factor) on pairs of random variables satisfying the following axioms:<|endoftext|>
TITLE: Is there an online encyclopedia of Diophantine equations (OEDE)?
QUESTION [26 upvotes]: Hello all!
I'm just wondering if there is an online encyclopedia of Diophantine equations (OEDE), analogous to the OEIS for sequences.
While trying to solve one Diophantine equation, I reduced the solution to that of a very similar Diophantine equation with smaller exponents (i.e., almost a descent, but not back to the exact same equation). What I want to do is type in the equation
$$a^2 + b^p = c^2$$
and find references, papers, solutions, etc. I know I can Google it, but there appears to be no standard format for equations, so the results are suspect, and what hits I do get are extremely time-consuming to obtain and filter through.
Thanks for any pointers!
Kieren.

REPLY [10 votes]: As mentioned in the comments, Piezas' site here has families of infinite solutions for many equations, and some complete characterizations.
There is a thread with a similar question here at math.stackexchange which may be helpful, if you remove the "online" component of your question. The answers in the thread state:
Diophantine Equations by Mordell
The Algorithmic Resolution of Diophantine Equations by Smart
Algorithms For Diophantine Equations by de Weger
History of the Theory of Numbers, Volume 2 by Dickson
EDIT: Another of the replies said Number Theory, Volume 1 and 2 by Cohen, but that he had not read either. Since someone just added it to the comments in this thread, I'm now including it here for completeness.
I would add that An Introduction to Diophantine Equations by Andreescu, designed for math olympiad training, has many equations and their solutions.
One of the answers in the math.stackexchange thread says "I imagine a complete catalog and bibliography today, bang up to date but in the same detail as Dickson, would require dozens of volumes and would be more a Wikipedia style project than anything one could reasonably contemplate publishing in book form."
This resonates with your online idea, so perhaps someone should initiate it. Certainly it would be very useful, especially to prevent wasting research time on rediscoveries.<|endoftext|>
TITLE: Finite generation and Henselization
QUESTION [9 upvotes]: Now I understand the answers.
I am trying to understand Henselian Weierstrass Theorem in Hironaka's Idealistic exponents of singularity, page 76 - 77. A glance at the paper. 
At some point there is a ring $R$, which is Noetherian, Henselian, and local, with maximal ideal $m$. A module $A$ such that 

$\frac{A}{mA}$ is finitely generated over $\frac{R}{m}$.
$A$ is finitely generated over the Henselization $S$ of $R[z]$ with respect to $(m,z)$.

And then he concludes, 'thanks to Hensel's lemma', that 
(3). $A$ is finitely generated over $R$.
I don't understand how that works. It should be something standard/simple since it is stated just like that, but I don't see it. Could you explain how the argument goes?
PS: If you are looking at the paper, the notation used is: 

$R$ is what in the paper is called $R_{j+1}$, 
$A$ is what in the paper is $G_j$, 
$m$ is $M_{j+1}$, 
$z$ is $z_{j+1}$, 
and $S$ is what in the paper is called $R_{j}$.

REPLY [4 votes]: (Note: There is no new content in this answer; it is simply an explanation of parts of user's excellent response. I'm writing it as an answer, because that seems less perverse than splitting it across many comments.)
The key result that is being used here is the following consequence of Zariski's Main Theorem + Hensel's lemma: Suppose that $f:A\to B$ is a map of finite type (between noetherian rings), with $A$ local henselian. Suppose that $\mathfrak{n}\subset B$ is  a prime above $\mathfrak{m}$ such that the localization $B_{\mathfrak{n}}$ is quasi-finite over $A$ at the closed point (i.e. $B_{\mathfrak{n}}/\mathfrak{m}B_{\mathfrak{n}}$ is finite over the field $A/\mathfrak{m}$). Then $B_{\mathfrak{n}}$ is finite over $A$, and is in particular itself henselian local.
Here, Zariski's Main Theorem is (or can be) used in the following incarnation (see Raynaud, 'Anneaux locaux henseliens', p. 42): Let $f:A\to B$ be as in our hypotheses in the first paragraph (but don't assume that $A$ is henselian). Let $B'$ be the integral closure of $A$ in $B$ (note that $B'$ is automatically finite over $A$); then there exists $f\in B'$, $f\notin \mathfrak{n}$ such that $B'_f=B_f$. In other words, $\text{Spec }B\to\text{Spec }B'$ is an open immersion in a neighborhood of $\mathfrak{n}$. Therefore, there exists a prime $\mathfrak{n}'\subset B'$ such that $B'_{\mathfrak{n}'}=B_{\mathfrak{n}}$. So, replacing $B,\mathfrak{n}$ by $B',\mathfrak{n}'$, we can assume that $B$ is finite over $A$.
If we now assume that $A$ is in addition henselian, then Hensel's lemma implies that $B$ is a product $B=\prod_{i=1}^rB_{\mathfrak{n}_i}$, where $\mathfrak{n}_i$ ranges over all the primes of $B$ that lie over $\mathfrak{m}$; see for example p. 2 of Raynaud's book. This immediately implies that $B_{\mathfrak{n}}$ is finite over $A$.
Let's see why this implies Hironaka's claim. As in user's answer, let $S/I$ be the quotient of $S$ that acts faithfully on the $R$-module $A$. As observed already in the answer, the map $R\to S/I$ is quasi-finite at the closed point; also, $S/I$ is the henselization of the localization of an $R$-algebra $B$ of finite type at a prime $\mathfrak{n}\subset B$ (this is the point where you use the fact that $I$ is finitely generated; $B$ is what user denotes as $S'/I'$). Now, we can apply the result from the first paragraph to conclude that $B_{\mathfrak{n}}$ is finite over $R$ and is therefore itself henselian. This implies that $S/I=B_{\mathfrak{n}}$, which shows that $S/I$ (and hence $A$) is finite over $R$.<|endoftext|>
TITLE: Intersection of localization with finitely generated subalgebra of fraction field
QUESTION [10 upvotes]: Let $R$ be a (commutative) noetherian integral domain. Let $I$ be a prime ideal of $R$. Let $S$ be a finitely generated $R$-subalgebra of $\mathrm{Frac}(R)$.

Is $S \cap R_I$ necessarily finitely generated as an $R$-algebra?
If not, is the image of $S \cap R_I$ in $R_I/IR_I = \mathrm{Frac}(R/I)$ finitely generated as an $R/I$-algebra?
If not, what if $R$ is not just noetherian but excellent (e.g., a finitely generated algebra over a field)?

REPLY [2 votes]: The answer to all three questions is "no".
This answer uses the same strategy as my earlier answer, but is a lot simpler in details because it puts most of the complexity into the ring $S$ rather than the ideal $I$. Let $k$ be an infinite field.
Let $p_1$, $p_2$, ..., $p_9$ be $9$ generic points in $\mathbb{P}_k^2$; let $x$, $y$ and $z$ be the homogenous linear coordinates on $\mathbb{P}^2$. Let $A = k[w_1, w_2, \ldots, w_9,x,y,z]$.
We will consider $A$ as a ring graded by $\mathbb{Z}^{10}$, where we separately keep track of the degrees in the $w_j$'s but lump the $x$, $y$ and $z$ degrees together.
So an element of $A$ is homogenous if it is of the form $\prod_i w_i^{r_i} \cdot f(x,y,z)$ with $f$ a homogenous polynomial of some degree, which we will denote $d$.
All our rings will be graded subrings of $A$.
Let $N_9$ be the graded subring of $A$ where $\prod w_i^{r_i} \cdot f(x,y,z)$ is in $N_9$ if, for each $i$, the polynomial $f(x,y,z)$ vanishes to order $\geq d-r_i$ at $p_i$. Let $N_8$ be the analogous subring of $k[w_1, \ldots, w_8, x,y,z]$, using the points $p_1$ through $p_8$. Nagata showed that $N_8$ is finitely generated over $k$, and $N_9$ is not.
Let $S = N_8[w_1]$. Choose coordinates on $\mathbb{P}^2$ so that $x$ and $y$ vanish at $p_9$ and $z$ does not. Set $\tilde{x} = w_1 w_2 \cdots w_8 x$, $\tilde{y} = w_1 w_2 \cdots w_8 y$ and $\tilde{z} = w_1 w_2 \cdots w_8 w_9 z$. Set $R = k[w_1, w_2, \ldots, w_9, \tilde{x}, \tilde{y}, \tilde{z}]$. So $R \subset S$; as $S$ is a finitely generated $k$ algebra, it is a finitely generated $R$ algebra.
Set $I = w_9 R$. We claim that
$$S \cap R_I = N_9. \quad (\ast)$$
Furthermore, we claim that the kernel of $N_9 \to R_I/I R_I$ is $w_9 N_9$. So the image is $N_9/(w_9 N_9)$, which we will show below is also not finitely generated.
Since $S$ is a graded subring of $A$, as is $R$, and $I$ is a graded ideal, the intersection $S \cap R_I$ is a graded subring.
Let $g=\prod w_i^{r_i} f(x,y,z)$ be a homogenous element, with $\deg f =d$. 
First, suppose that $g \in N_9$. Then clearly $g \in S$. Also, since $f$ vanishes to order $\geq d-r_9$ at $p_9$, we can write $w_9^{r_9} f(x,y,z)$ as a polynomial in $x$, $y$ and $z w_9$. Note that $x$, $y$ and $z$ are $(w_1 \cdots w_8)^{-1} \tilde{x}$, $(w_1 \cdots w_8)^{-1} \tilde{y}$ and $(w_1 \cdots w_8)^{-1} \tilde{z}$ so they are in $R_I$. Therefore, $w_9^{r_9} f(x,y,z)$ is in $R_I$. Since $w_1$ through $w_8$ are also in $R_I$, this shows that $g$ is in $R_I$. So $N_9 \subseteq S \cap R_I$.
Now, suppose that $g$ is in $S \cap R_I$. Since $g \in S$, we see that $f$ vanishes to order $\geq d-r_i$ at $p_i$ for $1 \leq i \leq 8$. 
Since $g$ in $R_I$, we can write $g=a/b$ with $a \in R_I$ and $b \in R_I - I$. Since $R$ is a UFD, we can assume that $a$ and $b$ are relatively prime in 
$R = k[w_1, w_2, \ldots, w_9, \tilde{x}, \tilde{y}, \tilde{z}]$. If $b$ has factors other than the $w_j$'s, then $a/b$ won't be in $R$, sos we may assume that $b$ is of the form $\prod_{j \leq 8} w_j^{q_j}$. 
We can then write $g = \prod_{j \leq 8} w_j^{s_j} \cdot w_9^{s_9} h(\tilde{x}, \tilde{y}, \tilde{z}) = \prod_{j \leq 8} w_j^{s_j+d} \cdot w_9^{s_9} h(x,y,w_9 z)$ for some constants $s_j$.
If we write $h(x,y,w_9 z) = w_9^r f(x,y,z)$ then $f$ vanishes to order $d-r$ at $p_9$ and $d-r \geq d-(r+s_9)=d-r_9$. 
So we have shown that $g$ is in $N_9$. We have now established $(\ast)$.
Similar arguments show that $g \in S \cap I R_I$ if and only if $f$ vanishes to order $> d-r_9$ at $p_9$, if and only if $w_9$ divides $g$ in $N_9$. 
I'll omit the details (but please challenge me if you think I'm wrong!).
So the image of $N_9$ in $R_I / I R_I$ is $N_9 / w_9 N_9$. We now recall the following lemma:
Lemma: Let $N$ be a positively graded $k$-algebra whose degree $0$ part is $k$. Let $J$ be a finitely generated graded ideal of $N$. Then $N/J$ is finitely generated if and only if $N$ is.
Proof: Clearly, if $N$ is finitely generated, then any quotient of $N$ is. Conversely, suppose that $a_1$, $a_2$, ..., $a_m$ is a list of generators for $J$ as an ideal, and $\bar{b}_1$, 
..., $\bar{b}_n$ is a list of generators for the $k$-algebra $N/J$. Let $b_i$ be a lift of $\bar{b}_i$ to $N$.
We may assume that the $a_i$ and $b_i$ are homogenous and of positive degree. We claim that $N$ is generated by the $a$'s and $b$'s. 
We show, by induction on $N_d$, that $N_d$ is contained in the $k$-subalgebra of $N$ generated by the $a$'s and $b$'s. 
Let $g \in N_d$. Then there is a polynomial $p$ so that $g \equiv p(a_1, \ldots, a_m) \bmod J$. Thus, $g = p(a_1, \ldots, a_m) + \sum b_i h_i$ for some $h_i \in N$. The $h_i$ are of lower degree, so inductively, the $h_i$ are in the subalgebra generated by the $a$'s and $b$'s, and thus $g$ is in this algebra. $\square$
Since $N_9$ is not finitely generated, this lemma shows that $N_9 / w_9 N_9$ isn't either.<|endoftext|>
TITLE: Pell Numbers and the Primes
QUESTION [6 upvotes]: I have observed that the Pell Numbers: a(0) = 1, a(1) = 2; for n > 1, a(n) = 2*a(n-1) + a(n-2)
and the related integer sequence:  a(0) = 2, a(1) = 0; for n > 1, a(n) = 2*a(n-1) + a(n-2)
together share an interesting property.
For any index number 'n', if n is a prime number, then it divides exactly into a(n) in just one, and only one, of the two sequences.  If n does not divide into a(n) in either sequence, then n is non-prime.
The sequences, and the primes that they generate, begin thus:
1  2  5  12  29  70…    3  5  11  13  19  29 …
2  0  2  4   10  24 …   2  7  17  23  31  41 …
I would like to find further information on this, such as who first discovered it, and whether it has been proved.
The sequence 2, 0, 2, 4, 10, 24, ...  is not listed in OEIS, so this may not be very well known.

REPLY [3 votes]: Your claim is true. Note that we can solve the recurrences to give
$$a_p=\frac{1}{\sqrt{2}}\left((1+\sqrt{2})^{p+1}-(1-\sqrt{2})^{p+1}\right)$$
for the first sequence. So for a prime $p$, we have that $p$ divides $a_p$ iff $p$ divides $2+2^{\frac{p+1}{2}}$, that is iff $2$ is a quadratic nonresidue modulo $p$, or in other words $p$ is $\pm 3 \pmod{8}$. (I used the fact that $\binom{p+1}{k}\equiv 0\pmod{p}$ for $2\le k\le p-1$.)
Similar analysis shows that for the second sequence we have $$a_p=\sqrt{2}\left((1+\sqrt{2})^{p-1}-(1-\sqrt{2})^{p-1}\right),$$
and so $a_p\equiv 2^{\frac{p+1}{2}}-2\pmod{p}$. So in the second sequence $p$ divides the $p$th term iff $2$ is a quadratic residue mod $p$, or in other words $p$ is $\pm 1\pmod{8}$.<|endoftext|>
TITLE: Elementary tools for proving congruences of modular forms
QUESTION [7 upvotes]: My impression is that the specialists in the field use geometric modular forms when proving congruences of modular forms. While this is probably the right way, I don't think I will be able to get a working knowledge of this point of view fast enough, as this is for an undergraduate summer research project.
Do you know of any references that of anything that involves congruences of modular forms being proved by elementary means? Mostly, I'm looking for examples of methods and tools rather than a specific theorem. 
More specifically, I'm interested in congruences between modular forms of different weight and the same level. The only fact I know in this case is that the Eisenstein series $E_{p-1}\equiv 1\pmod{p}$, and multiplying by this Eistenstein series give you equivalences between modular forms of different weight. Also (though much less trivial), the converse is also true. However, with only this fact, it seems the only hope of proving anything is to come up with very explicit formulas for what is happening.
(I know there is a short paper by Serre that determines the structure of the ring of modular forms, under the full SL_2(Z), reduced mod p http://math.bu.edu/people/potthars/writings/serre-1.pdf. However, this does not generalize to modular forms of a given level, since it uses the structure of the ring of modular forms under the full modular group.
A couple of other papers I've found are "Congruences between systems of eigenvalues of modular forms" and "A study of the local components of the Hecke algebra mod l" by Jochnowitz, but I've only just started reading.
This was originally posted on stackexchange: https://math.stackexchange.com/questions/420509/elementary-tools-for-proving-congruences-of-modular-forms.)

REPLY [3 votes]: To understand congruences between modular forms, one needs first to understand the abstract notion of congruences between elements of modules with (Hecke) operators acting on them.
For this, Ghate's note, as recommended by François, is a veri good introduction.
But then I assume you want to go further, and understand how people prove, or use, congruences between modular forms. The problem with the traditional definition of modular forms as holomorphic functions on the upper half plane is that obviously, this is an analytic, not algebraic definition, and that therefore some (hard) work
is needed to reveal the arithmetic nature of modular forms, in particular to study congruences
between them. This hard work generally involve some serious algebraic geometry, such as 
defining, constructing and studying moduli space of elliptic curves with various structures
over an arithmetic basis, etc. This is likely to be overwhelming for an undegraduate student.
Yet ti understand seriously the aspect you are mentioning (congruences between modular forms
of same level but various weights) 
An alternative is to start with a different kind of object, somehow related to modular forms,
but with a more direct connection to arithmetic. I am thinking of either "modular symbols"
or "modular forms over quaternion algebra".
Let me just mention the second here. Let $D$ be a quaternion algebra over $\mathbb Q$
which is ramified at infinity (that is $D \otimes \mathbb R = \mathbb H$) and $G=D^\ast$
the algebraic group over $\mathbb Q$ of its invertble element. Define a "level" K as a
compact open subgroup of $G(\mathbb A_f)$, and a "modular form of weight 2 and level K"
as simply a function form $G(\mathbb A_f)$ to $\mathbb C$ which is left-invariant by $K$ and right invariant of $G(\mathbb Q)$. You get this way a notion which has a lot of analogy with modular forms: they form a finite-dimensional vector space, which has a natural
structure over $\mathbb Z$ (just take the same functions with values in $\mathbb Z$),
a natural action of Hecke operators, etc. Actually this is just more than a simple analogy: a deep theorem of Jacquet-Langlands tells you that the space of "modular forms of weight 2 for $D$" defined as above is actually a big subspace of the space of traditional modular forms
of weight 2, in a way which respects the Hecke opertaors. But you don't need to understand fully this theorem, less alone its difficult proof, to use it as a motivation to study 
modular forms over $D$, and their congruences.
Now it turns that certain theorem about congruences between modular forms are much simpler
to prove (but still non-trivial) for their $D$-counterpart. One example of this is the 
famous "Ribet level-raising theorem" which is quite difficult to prove for classical modular forms but is relatively simple and very beautiful to prove for modular forms for $D$.
Understanding this theorem may seem a worthwhile goal for a summer project of a very good and very motivated undergrad. One reference is section 1 of Taylor's early and very deep paper "on Galois rep. associated with Hilbert modular forms" at inventiones. Unfortunately it is
not extremely reader-friendly, as the main ideas (basically some computations on the Bruhat-Tits tree) are not explicitly explained.<|endoftext|>
TITLE: Potentially good, semi-stable reduction => good reduction ?
QUESTION [8 upvotes]: Does a smooth proper variety having semi-stable reduction as well as potentially good reduction have good reduction ?
Note that over a $p$-adic field, this is true for the Galois representations in the $p$-adic étale cohomology of $X$.
(With a bit more details: fix a field $K$ complete for a discrete valuation, with ring of integers $\mathcal{O}_K$, and a smooth proper $K$-scheme $X$. We say that $X$ has good reduction if it is the generic fibre of a smooth proper $\mathcal{O}_K$-scheme $\mathcal{X}$. We say that $X$ has semi-stable reduction if it is the generic fibre of a flat proper $\mathcal{O}_K$-scheme $\mathcal{X}$ such that étale-locally on $\mathcal{X}$ there exists a smooth morphism $\mathcal{X}\to \text{Spec}(\mathcal{O}_K[T_1,\dots,T_r]/(T_1\dots T_r-\pi))$ where $\pi$ is a prime element of $\mathcal{O}_K$. We say that $X$ has potentially one of the two properties above if it has it after a finite extension $K'/K$.)

REPLY [9 votes]: No. Take $K=\mathbb Q_3$. Consider the projective genus $0$ curve $x^2+y^2+3z^2$. This has bad reduction, since it has no rational points. It is semistable, since after adjoining $i$ it has exactly that form. It has potentially good reduction, since all genus $0$ curves do. 
So you at least need to say it has good reduction etale locally.<|endoftext|>
TITLE: A strange matrix equality
QUESTION [10 upvotes]: Let $A$ and $B$ be $n\times n$ real matrices.
When $n=2$, we have the equality
$$A\Big(\mbox{Trace}(B)A-\mbox{Trace}(A)B\Big) B=B\Big(\mbox{Trace}(B)A-\mbox{Trace}(A)B\Big) A.$$

Can we give an interpretation to this equality?
Are there similar equalities when $ n = 3,4,...$?

REPLY [13 votes]: Let us rewrite it using the commutators $[P,Q]=PQ-QP$, as follows:
 $$
tr(B)[A^2,B]=tr(A)[A,B^2].
 $$
Now, for matrices $X$ of size~$2$, we have $X^2=tr(X)X-det(X)I$ (a particular case of Cayley--Hamilton), so
 $$
tr(B)[tr(A)A-det(A)I,B]=tr(A)tr(B)[A,B]=tr(A)[A,tr(B)B-det(B)I],
 $$
since $I$ commutes with everything. 
This might be the most economic proof of your identity; moreover, it is known (Procesi, Razmyslov) that every identity with traces for $n\times n$-matrices does follow from the Cayley--Hamilton identity, so generalisations of your identity should be obtained in a similar way.<|endoftext|>
TITLE: inner product of two gaussian random vectors?
QUESTION [7 upvotes]: Suppose that $x, y\sim N(0,I_n)$ are independent. Consider the inner product $\langle x, y\rangle$. Intuitively, $y$ behaves like a random vector of length $\sqrt n$, so $\langle x, y\rangle$ is close to $\langle x, z\rangle$, where $z$ is uniform on $\sqrt{n}S^{n-1}$. Now, $\langle x,z\rangle \sim N(0,\|z\|^2)$, that is, $\langle x,z\rangle \sim N(0, n)$. Therefore $\langle x,y\rangle$ is close to an $N(0,n)$ variable. 
I want to show that the density function of $\langle x,y\rangle$ is close to the density function of an $N(0,n)$ variable, without resorting to the explicit formula of the density function of $\langle x,y\rangle$ (which can be obtained by characteristic function/Fourier transform and involves Bessel function). 
Basically I'd like to formalize the intuition above. Write $\langle x,y\rangle = Z + W$, where $Z\sim N(0,n)$ and $W$ is small most of the time but not independent with $Z$. This gives that the cumulative density function of $\langle x,y\rangle$ and that of $Z$ are close, but this is not enough to argue that the p.d.f's are close. Is it possible to argue that following this outline?

REPLY [8 votes]: $S_n:=\langle x,y\rangle =\sum_{i=1}^nx_iy_i$ is the $n$th partial sum of a sequence of standardized and i.i.d. random variables, and the characteristic function of each summand, namely the function $t$ $\mapsto$
$$
  \int \mathrm{e}^{-t^2x^2/2} \frac{\mathrm{e}^{-x^2/2}}{\sqrt{2\pi}}\,\mathrm{d}x
 = \frac{1}{\sqrt{1+t^2}}
$$
is square integrable. So a standard form of the CLT for densities, see Feller II (1971, p. 516, Theorem 2 and the remark following it), shows the uniform convergence of the density of $S_n/\sqrt{n}$ to the standard normal density.<|endoftext|>
TITLE: A sum-product estimate in Z/p^2Z
QUESTION [8 upvotes]: We are interested in a sum-product type estimate. Let $p$ be an odd prime, and let $A$ be the order $p-1$ subgroup of $(\mathbb{Z}/p^2\mathbb{Z})^\times$. That is, let $A = \langle g^p \rangle$, where $(\mathbb{Z}/p^2\mathbb{Z})^\times = \langle g \rangle$. We seek an estimate of the size of the sum set
$$|A + A|.$$
The standard references, e.g. Additive Combinatorics by Tao & Vu, do not discuss this problem in the setting of $\mathbb{Z}/p^2\mathbb{Z}$, but instead focus on analogous results in finite fields. Similar estimates address situations where $|A + A|$ is small (e.g., on the order of $K|A|$), whereas in our situation, the product set $|A\cdot A|$ is very small, and we wish to show that $|A+A|$ must be large. Do results exist that address this problem more directly?

REPLY [2 votes]: Heath-Brown and Konyagin stop just short of proving that $|A+A|\gg |A|^{3/2}$ in their paper on Gauss sums and Heilbronn's sum.
(Here $A$ is the same subgroup as above; since $A\cup\{0\}=\{0^p,\ldots,(p-1)^p\}$, we may write Heilbronn's sum as $H_p(a)=\sum_{n=0}^{p-1}e(an^p/p^2)=1+\sum_{x\in A}e(ax/p^2)$.)
By Cauchy-Schwarz, we have $|A\pm A|\geq |A|^4/E_+(A)$, where $E_+(A)$ is the cardinality of the set of "additive quadruples" $\{(a,b,c,d)\in A^4\colon a+b=c+d\}$.
Using Stepanov's method, Heath-Brown and Konyagin show that $E_+(A)\ll (p-1)^2+(p-1)p^{3/2}$. (This follows from their lemma 4 and the argument at the bottom of page 7.)
Combining the previous two results yields $|A\pm A|\gg |A|^{3/2}$, where the hidden constant is $1+o(1)$. In [http://arxiv.org/abs/math/0304217], Konyagin proves the same bound for multiplicative subgroups of $\mathbb{F}_p$ (see lemma 5).
Shkredov gives the improved bound $E_+(A)\ll |A|^{42/17}(\log|A|)^{10/17}$ in [http://arxiv.org/abs/1208.6124v1], and a further improvement $E_+(A)\ll |A|^{22/9}(\log|A|)^{2/3}$ in [S]. Thus up to logs, our additive energy argument gives $|A\pm A|\gg |A|^{3/2+1/18}$.
You can do better using an argument of Vyugin and Shkredov [VS] from [http://arxiv.org/abs/1102.1172]. Theorem 5.5 from [VS] shows that $|R\pm R|\gg |R|^{5/3}(\log|R|)^{-1/2}$, where $R$ is a multiplicative subgroup of $\mathbb{F}_p$ of size at most $p^{1/2}$. The proof relies on the following results from [VS]: Lemma 2.3 and Corollary 2.7, which are true for any finite abelian group, and Corollary 5.1 and Lemma 5.4, whose analogs in [S] are Proposition 8 and Corollary 9. Thus we have $|A\pm A|\gg |A|^{5/3}(\log|A|)^{-1/2}$.
Finally, you can apply Proposition 8 from [S] to get a lower bound on |A+A+A|.

Proposition 8 ([S]) Let $A$ be the multiplicative subgroup of $\mathbb{Z}_{p^2}$ of order $p-1$. Suppose that $Q_1,Q_2,Q_3$ are $A$-invariant subsets of $\mathbb{Z}_{p^2}$ and $|Q_1||Q_2||Q_3|\ll p^5$. Then


$\sum_{z\in Q_1}Q_2\ast (-Q_3)(z)\ll p^{-1/3}(|Q_1||Q_2||Q_3|)^{2/3}.$



(Note that Shkredov uses $\circ$ to denote convolution with $-B$; we're also using the convention that $B(x)$ is the indicator function of $B$.)
If $Q_1=Q_2-Q_3$, then the left hand side of the above equation is equal to $|Q_2||Q_3|$. Choosing $Q_1=A+A+A$, $Q_2=A$ and $Q_3=-(A+A)$, we get that


$|A+A+A|\gg p^{1/2}|A|^{1/2}|A+A|^{1/2}\gg p^{11/6}/(\log|A|)^{1/4}$.<|endoftext|>
TITLE: Does the gluing procedure in Robert Wald’s book *General Relativity* yield a Hausdorff spacetime?
QUESTION [11 upvotes]: Before I state my problem, let me provide some definitions pertaining to the Cauchy Problem in General Relativity.

Definition 1: A triplet $ (\Sigma,h,k) $ is called an initial data set if $ (\Sigma,h) $ is a Riemannian $ 3 $-manifold, $ k $ is a symmetric $ 2 $-form on $ \Sigma $, and both $ h $ and $ k $ satisfy the so-called Einstein constraint equations for vacuum (we do not need to know what these equations are). $ \quad \spadesuit $
Definition 2: Let $ (\Sigma,h,k) $ be an initial data set. A triplet $ (M,g,i) $ is called a vacuum development of $ (\Sigma,h,k) $ if

$ (M,g) $ is a spacetime manifold (i.e., a Lorentzian $ 4 $-manifold) that satisfies the vacuum Einstein Field Equations;
$ i $ is a smooth embedding of $ \Sigma $ into $ M $ such that $ {i^{*}}(g) = h $ and $ {i^{*}}(\Pi) = k $, where $ \Pi $ is the second fundamental form (i.e., the extrinsic curvature) of $ \Sigma $ in $ M $; and
$ i[\Sigma] $ is a smooth Cauchy surface in $ M $. $ \quad \spadesuit $


The Cauchy Problem in General Relativity is to prove the existence of a vacuum development for every initial data set. The proof proceeds by working locally on small patches of the initial data set and ends by performing some gluing to obtain a global solution.
It is a well-known result from the theory of hyperbolic PDE’s that if $ U $ is a local coordinate patch in $ \Sigma $ (i.e., $ U $ is homeomorphic to an open subset of $ \mathbb{R}^{3} $), then we can find a vacuum development $ (M_{U},g_{U},i_{U}) $ of the initial data set $ (U,h|_{U},k|_{U}) $ such that $ M_{U} $ is an open neighborhood of $ U \times \lbrace 0 \rbrace $ in $ U \times \mathbb{R} $ and $ {i_{U}}[U] = U \times \lbrace 0 \rbrace $.
Knowing now that local vacuum developments exist, we have to glue them together in order to obtain a global-in-space vacuum development of $ (\Sigma,h,k) $. Once the Hausdorff property is established, paracompactness is for free by a 1968 result of Robert Geroch, which states: If a smooth $ 4 $-manifold is Hausdorff and admits a smooth Lorentz metric, then it is automatically paracompact.
In Robert Wald’s book General Relativity, the following method for gluing is proposed:

By the paracompactness of $ \Sigma $ as a $ 3 $-manifold, we can find a locally finite open cover $ \mathcal{U} $ of $ \Sigma $ consisting of coordinate patches. As mentioned above, for each $ U \in \mathcal{U} $, we have a vacuum development $ (M_{U},g_{U},i_{U}) $ of $ (U,h|_{U},k|_{U}) $.
For each $ p \in \Sigma $, define $ \mathcal{U}_{p} := \lbrace U \in \mathcal{U} ~|~ p \in U \rbrace $. We have from (1) that $ \mathcal{U}_{p} $ is a finite collection for each $ p \in \Sigma $.
For each $ p \in \Sigma $, let $ W_{p} \subseteq \Sigma $ be a neighborhood of $ p $ such that $ W_{p} $ is contained in all the open sets belonging to $ \mathcal{U}_{p} $, i.e., $ W_{p} \subseteq \bigcap \mathcal{U}_{p} $. Using a result known as ‘local geometric uniqueness’, we can then find a vacuum development $ (M_{p},g_{p},i_{p}) $ of $ (W_{p},h|_{W_{p}},k|_{W_{p}}) $ such that there exists an isometric embedding $ j_{p,U}: (M_{p},g_{p}) \to (M_{U},g_{U}) $ for each $ U \in \mathcal{U}_{p} $.
Using the embeddings $ j_{p,U} $ to make identifications, we can glue the $ (M_{p},g_{p}) $’s together in order to obtain a vacuum development of $ (\Sigma,h,k) $. Although Wald does not clearly describe this step, this is how I interpret it. For each $ U \in \mathcal{U} $, consider the union $ \tilde{M}_{U} := \displaystyle \bigcup_{p \in U} {j_{p,U}}[M_{p}] \subseteq M_{U} $. We wish to glue the $ (\tilde{M}_{U},g_{U}|_{\tilde{M}_{U}}) $’s together by identifying points as follows. Given $ U_{1},U_{2} \in \mathcal{U} $, $ x_{1} \in \tilde{M}_{U_{1}} $ and $ x_{2} \in \tilde{M}_{U_{2}} $, we say that $ x_{1} \sim x_{2} $ if and only if there exist a $ p \in U_{1} \cap U_{2} $ and a $ q \in M_{p} $ such that $ {j_{p,U_{1}}}(q) = x_{1} $ and $ {j_{p,U_{2}}}(q) = x_{2} $. Finally, define $ \displaystyle M := \bigsqcup_{U \in \mathcal{U}} \tilde{M}_{U} \bigg/ \sim $.


Problem: How do we show that $ M $ is Hausdorff?
This has eluded my best efforts for the past month, so I would appreciate it if anyone on MathOverflow could offer any insight. Thank you very much!

REPLY [6 votes]: Your construction won't work as stated. The problem is that $M_p$ can be too small. Let me give a slightly silly example of why it doesn't work. 
Cover $\mathbb{R}^3$ with the two local coordinate charts $U = \lbrace x_1 < 1\rbrace$ and $V = \lbrace x_1 > -1\rbrace$, and prescribe on it trivial initial data. An admissible "local solution" has $M_U$ and $M_V$ being the following subsets of the Minkowski space $\mathbb{R}^{1,3}$:
$$ \begin{align}
M_U &= \lbrace x_1 < 1 - |t|\rbrace \newline
M_V &= \lbrace x_1 > -1 + |t| \rbrace \end{align} $$
Let $W_p = \lbrace |x_1| < 1\rbrace$ for every $p \in U\cap V$. And let 
$$ M_p = M_U \cap M_V \cap \lbrace |t| < \frac12\rbrace $$
So far all the objects are admissible ones that could, in principle, be the ones you picked out from steps 1 through 3. 
By your step 4, you have two distinct points corresponding to what would've been the point $t = \frac12, x_1 = x_2 = x_3 = 0$, since the ones in $M_U$ and $M_V$ are not identified since that point technically lies outside of $M_p$ for every $p$. Those two points are not separated by neighborhoods in the space-time you constructed, which violates Hausdorff axiom.

If I am not mistaken, this would be the primary obstruction. Possible resolutions are to either make $M_p$ larger or allow us to make $M_U$ and $M_V$ smaller. (With the caveat that I haven't written down a full argument saying that this will work.) In either case you need to get yourself down to something analogous to Ringström's case where the images of gluing maps $j_{p,U}$ cover the neighborhood of intersection. 

(Edit 19.06.2013) Ah, I found a reference where the argument is given in slightly more detail. 

Choquet-Bruhat and York, "The Cauchy Problem". MathSciNet relay link

It depends on a slightly stronger local uniqueness theorem than you supposed. (See Hawking-Ellis, Large Scale Structure... section 7.5.) And some details are still missing. But it appears to be more easily filled-in. And considering the footnote on page 106, they certainly did worry about Hausdorff-ness in their construction.<|endoftext|>
TITLE: Dehn filling  of hyperbolic 3-manifolds and Gromov volume
QUESTION [6 upvotes]: Let $N$ be a hyperbolic 3-manifold with finite (Gromov) volume and with finitely many tori cusps boundary. If we do Dehn fillings along some cusps of $N$, we obtain a new manifold, denoted by $N_1$. Suppose $N_1$ is irreducible, my question is:
(1) is it possible that $N_1$ isn't hyperbolic but contains a hyperbolic JSJ piece?
(2) if (1) is possible, then whether the Gromov norm of $N_1$ is strictly less than the Gromov norm of $N$? 

Remark: By Thurton's book (1978) Proposition 6.5.2, the Gromov norm of $N_1$ is no larger than the Gromov norm of $N$.
Thank you.

REPLY [7 votes]: In Theorem 6.5.6 in Thurston's notes, he indicates that volume decreases strictly under hyperbolic Dehn filling, referring to Theorem 6.4 (volume rigidity, see the appendix to Dunfield's paper for more details http://arxiv.org/abs/math/9802022). I think his proof could be extended to the case of non-hyperbolic Dehn fillings, combining the arguments of 6.4 and 6.5.5, but this hasn't been written down anywhere. 
Alternatively, in my thesis I gave an argument that volume decreases strictly. First, perform an orbifold Dehn filling along the slope to get a hyperbolic Dehn filling, in which volume (and therefore Thurston norm) strictly decreases. Then map this orbifold filling to the Dehn filling along the slope, so that the Gromov norm strictly decreases.<|endoftext|>
TITLE: List of open problems of formal languages
QUESTION [5 upvotes]: As we know, there are some open problems of formal languages. I am wondering if there is a somehow complete list of open problem of formal languages. If there isn't such a list, can we make it one as answer?

REPLY [9 votes]: Here is a link which describes some of the open problems:

https://cs.uwaterloo.ca/~shallit/Talks/open10r.pdf<|endoftext|>
TITLE: Quintic polynomial solution by Jacobi Theta function.
QUESTION [14 upvotes]: Does someone have a good and rigorous reference for the solution of quintic ploynomial equation with Jacobi Theta function, in English?
Mathworld and Wikipedia don't give a good English reference, at least from what I skimmed over.

REPLY [13 votes]: There's also a very nice, modern exposition by Mark L. Green:  On the analytic solution of the equation of fifth degree, Compositio Mathematica 37 (1978), 233–241.  A pdf of the paper is available online:  http://www.numdam.org/item?id=CM_1978__37_3_233_0<|endoftext|>
TITLE: The proportion between permutations and derangements.
QUESTION [5 upvotes]: Denote the number of derangements by $D_N$. It's known that $D_N/N! \rightarrow 1/e$. Therefore $N!/e$ is an approximation for $D_N$.
I'm trying to bound the difference between this approximation and $D_N$. Namely, to bound from above $|N!/e - D_N|$ as a function of $N$.
I believe it will help me to know from which value of $N_0$, it holds that for every $N > N_0$: $|N!/e - D_N| < 2^{-k}$ for some $k$.
Any help will be appreciated!

REPLY [16 votes]: I might as well turn my comment into an answer.  From the Taylor expansion of the exponential function applied to $e^{-1}$, we have:
$$\frac{N!}{e} = \left( \sum_{k=0}^N (-1)^k \frac{N!}{k!} \right) + \left( \sum_{j=N+1}^\infty (-1)^j \frac{N!}{j!} \right)$$
where the first sum is an integer, and the second sum has absolute value between $\frac{1}{N+2}$ and $\frac{1}{N+1}$.
From the Wikipedia article on derangements, we see that the first sum is in fact equal to the number of derangements.  This can be deduced by an inclusion-exclusion argument.  Your error term $\left| \frac{N!}{e} - D_N \right|$ is therefore between $\frac{1}{N+2}$ and $\frac{1}{N+1}$.  For your last question, if you want $\left| \frac{N!}{e} - D_N \right| < 2^{-k}$ for all $N\gt N_0$, it is necessary and sufficient that $N_0 > 2^k-2$.<|endoftext|>
TITLE: Book on the Three body Problem
QUESTION [12 upvotes]: Hi all, I am looking for a good book about the famous (infamous perhaps?) three body problem -  both theoretical and numerical hardless and accomplishments.
can you help? Thanks

REPLY [5 votes]: If you want a more popular book, then
Florin Diacu & Philip Holmes: Celestial Encounters:
The Origins of Chaos and Stability
is a must.<|endoftext|>
TITLE: Is there a stronger (but widely believed) version of the Chowla conjecture?
QUESTION [7 upvotes]: In Terry Tao's notes on the Chowla conjecture, the said conjecture states that for any fixed integer $m>0$ and nonzero $(a_1,\ldots,a_m)\in\{0,1\}^m$, 
$$
\lim_{x\rightarrow\infty}\frac{1}{x}\sum_{n\leq x}\mu(n+1)^{a_1}\cdots\mu(n+m)^{a_m}=0,
$$
where $\mu$ denotes the Möbius function.  I would like information about the rate of this conjectured convergence.  Are there any widely believed conjectures along these lines?

REPLY [11 votes]: I imagine the "correct" conjecture is the Möbius $s$-tuples conjecture in the form that if $s \in \mathbb{N}$, $\alpha_1, \ldots, \alpha_s \in \mathbb{N}$ with at least one $\alpha_i$ odd, and $d_1, \ldots, d_s \in \mathbb{Z}$ distinct, then for all $\varepsilon > 0$ we have that
$$\sum_{n \leq x}{\mu(n + d_1)^{\alpha_1} \mu(n + d_2)^{\alpha_2} \cdots \mu(n + d_s)^{\alpha_s}} \ll_{\varepsilon} x^{1/2 + \varepsilon}$$
uniformly for all $|d_i| \leq x$. However, I don't imagine there is any good evidence for this other than the fact that the Riemann hypothesis implies the case $s = 1$. I don't think this problem has really been studied enough for there to be a widely-believed generalisation of Chowla's conjecture.
A form of this conjecture is assumed by Ng in this paper on the distribution of the Möbius function in short intervals; he assumes that this conjecture holds with "for all $\varepsilon > 0$" replaced by "for some $0 < \beta_0 < 1/2$".<|endoftext|>
TITLE: Non-enumerative proof that there are many simple permutations?
QUESTION [9 upvotes]: Terence Tao asked for a non-enumerative proof that a positive proportion of permutations are derangements and got a great answer. Inspired by this, I'd like to ask about another family of permutations.
An interval in the permutation $\pi$ (thought of in one-line notation) is a contiguous sequence of entries which is also contiguous in value. For example, the entries $43$ form an interval in the permutation $514362$. Every permutation of length $n$ has trivial intervals of lengths $0$, $1$, and $n$, and permutations with only these trivial intervals are (sometimes) called simple.
Like fixed points, it can be shown that the number of nontrivial intervals in a random permutation is Poisson distributed (see this paper by Corteel, Louchard, and Pemantle), although unlike fixed points, the expected number of nontrivial intervals in a random permutation is $2$. It follows that (in the limit) the proportion of simple permutations is $1/e^2$.
Is there a "non-enumerative" proof that a positive proportion of permutations are simple?

REPLY [6 votes]: I think the main idea of Brendan McKay's answer to Terence Tao's question still works. It's a bit more complicated (and less non-enumerative), but still "robust" in the sense of Tao's question. 
First one should somehow "estimate away" those permutations that have an interval of length 3 or more. The proportion of such permutations is only $O(1/n)$. 
Then, since the average number of intervals of length 2 is 2, it suffices to show that (i) there can't be many more permutations with 1 interval than with none, and (ii) there can't be many more permutations with 2 intervals than with 1. 
There is a variety of ways to define a "switching operation" that provides (up to a $O(1/n)$ error) the necessary comparisons. For instance, consider the operation of taking a symbol that belongs to an interval of length 2 and switching it with the first symbol. The inverse operation is to look at the first symbol (say $i$), and switch it with a neighbor of $i-1$ or $i+1$.
For a permutation with $k$ intervals (of length 2 and disjoint), there are (in general) $2k$ ways of performing the "interval destroying" operation, and, regardless of $k$, 4 ways of performing the "interval creating" inverse. Taking $k=1$ this shows that asymptotically there are only twice as many permutations with one interval as with none, and with $k=2$ it shows that there are roughly as many permutations with two intervals as with one.<|endoftext|>
TITLE: Perverse sheaves for easy stratifications
QUESTION [7 upvotes]: Let $X$ be a complex variety equipped with a stratification. Let us assume, that all strata are contractible and in addition, that all strata closures are smooth.
Is there an "easy" quiver description of the category of perverse sheaves in this case?
For example does this category only depend on the poset of strata?
This sounds too optimistic, so what additional information is needed? 
For example if we are also given the conormal bundles of the strata closures, can one reconstruct the category from this data?

REPLY [2 votes]: Since your question is somewhat open-ended ("what additional information is needed?"), it appears difficult to provide a complete answer. However, at least one flavor of answer may be found in David Treumann's thesis:

D Treumann, Exit paths and constructible stacks, Compositio Mathematica, Vol 145, No. 06, pp 1504- 1532 (2009)

The main result 2-categorifies a theorem of MacPherson, which states that the category of perverse sheaves over a (topologically) stratified space $X$ is equivalent to the category of set-valued functors from the exit path category of  $X$. 
The exit path category of $X$ has all points of $X$ as objects, and morphisms are (homotopy classes of) continuous paths which we allow to ascend up to higher strata, but never descend to lower ones. Asking that all strata be contractible does not, in general, allow one to pass from the exit path category to the poset of strata without losing structure. On the other hand, if your stratification of $X$ arises from a regular CW decomposition, then the associated exit path category retracts onto the poset of cells.
Stringing all of this together, one set of sufficient conditions which guarantee that the category of perverse sheaves is recoverable from the poset of strata, is that the stratification comes from a regular CW decomposition of $X$. 
(I've also described a purely combinatorial 2-categorified version of the dual entrance path category for regular CW complexes in Sec 4.1 of this preprint).<|endoftext|>
TITLE: when is an algebra map conjugate to a star algebra map
QUESTION [5 upvotes]: Take a (unital) algebra map $f:A\to B$ between two unital C* algebras - not necessarily star preserving. Under what circumstances is there a $b\in B$ so that $g(a)=b\ f(a)\ b^{-1}$ is a star algebra map? If not, is there another method of modifying $f$ to get a star algebra map?

REPLY [8 votes]: I will assume "algebra map" means "homomorphism." Certainly a necessary condition is that $f$ is bounded. When $B=B(H)$ this question (is every bounded homomorphism similar to a *-homomorphism?) is known as the "Kadison Similarity Problem" and is still open. The answer is known to be affirmative when $A$ is a nuclear C*-algebra, and in a few other cases. I am not an expert in this area, but it is my impression that the answer is believed to be "no" in general. This survey by Ozawa gives a quick introduction and some pointers to the literature (especially the work of Pisier).<|endoftext|>
TITLE: Group scheme over a DVR whose special fibre is the image of points under reduction mod p
QUESTION [6 upvotes]: Let $R$ be a complete discrete valuation ring with maximal ideal
$\mathfrak{p}$ and algebraically closed residue field $k$. Denote
the field of fractions of $R$ by $F$. Let $G$ be an affine flat
group scheme of finite type over $R$, and assume that the generic
fibre $G_{F}$ is smooth. Since $G$ is not necessarily smooth over
$R$ the reduction mod $\mathfrak{p}$ map on the points
\[
\rho:G(R)\longrightarrow G(k)
\]
is not necessarily surjective. Let $H_{k}$ denote the image of $\rho$
provided with its canonical affine reduced structure.
Does there exist a group scheme $G'$ of finite type over $R$ such
that $G'_{F}\cong G_{F}$ (i.e., the generic fibres coincide) and
$G'_{k}\cong H_{k}$ (i.e., the special fibre of $G'$ is $H_{k}$)? 
An affirmative answer would give some information about the image
$H_{k}$, for example, we would have $\dim H_{k}\geq\dim G_{F}$ by
Chevalley's upper semicontinuity theorem. Therefore any example where
$\dim H_{k}<\dim G_{F}$ would provide a negative answer.
Added: 
A way to see that $H_{k}$ is closed in $G(k)$ is by
using a theorem of Greenberg (in "Rational points in Henselian discrete
valuation rings") which implies that there exist two integers $r\geq s\geq1$
such that the image of any $x\in G(R/\mathfrak{p}^{r})$ in $G(R/\mathfrak{p}^{s})$
has a lift to $G(R)$. This implies that $\mathrm{Im}(G(R)\rightarrow G(R/\mathfrak{p}^{s}))=\mathrm{Im}(G(R/\mathfrak{p}^{r})\rightarrow G(R/\mathfrak{p}^{s}))$.
Using the Greenberg functor, we can consider each $G(R/\mathfrak{p}^{i})$
as (the $k$-points of) a linear algebraic group over $k$ (equipped
with its reduced structure in case $G$ is not smooth over $R$),
and by another theorem of Greenberg ($\S\,5$, Prop. 2 and Coroll. 5
in "Schemata over local rings I") each map $G(R/\mathfrak{p}^{i})\rightarrow G(R/\mathfrak{p}^{j})$
with $i\geq j\geq1$ induces an algebraic homomorphism. Thus $X:=\mathrm{Im}(G(R)\rightarrow G(R/\mathfrak{p}^{s}))$
is a closed subgroup of $G(R/\mathfrak{p}^{s})$ and hence $H_{k}=\mathrm{Im}(X\rightarrow G(k))$
is a closed subgroup of $G(k)$.

REPLY [5 votes]: Note that the question does not make sense until we know that the image of $\rho$ is closed, and such closedness is not obvious since not even construcibility is apparent at the outset (e.g., Chevalley's theorem on constructibility is not relevant here, as $\rho$ is in no sense an algebraic morphism).  Also, it isn't said if $G'$ is supposed to be separated, perhaps even affine, but presumably you want $G'$ to be (quasi-)affine? 
The only way I know to see the closedness of the image of $\rho$ is by a procedure that answers the entire question. Namely, consider the canonically associated group smoothening $f:\widehat{G} \rightarrow G$ in the sense of section 7.1 of the book "Neron Models".  Since $G$ is a flat affine $R$-group of finite type, and the group smoothening is an $R$-homomorphism built from a "dilatation" process that entails forming a certain relatively affine open chart in a blow-up, it is immediate from the construction that the smooth group $\widehat{G}$ is affine, and by design $\widehat{G}(R) = G(R)$ and $f_F$ is an isomorphism.  Consequently, since $\widehat{G}(R) \rightarrow \widehat{G}(k)$ is surjective by smoothness of $\widehat{G}$, it follows that $\rho(G(R))$ is the image on $k$-points of the special fiber $k$-homomorphism $f_k:\widehat{G}_k \rightarrow G_k$. Now we can apply our experience with morphisms of algebraic groups over fields to know that $\rho(G(R))$ is closed, since (i) it is just the set of $k$-points of the image of $f_k$ since $k$ is algebraically closed and (ii) the image of $f_k$ is closed (as for any homomorphism between finite type group schemes over a field).
This argument gives more, namely that $\dim H_k \le \dim \widehat{G}_k = \dim \widehat{G}_F = \dim G_F$, whereas you noted that if the $G'$ you seek is to exist then $\dim H_k \ge \dim G_F$, so we would be forced to have $\dim H_k = \dim G_F$, but $\dim G_F = \dim G_k$ by $R$-flatness of $G$, so in such cases $H_k$ would have to contain the identity component $G_k^0$ and then $f_k$ has finite kernel and it has image containing $G_k^0$.  So we see that in such cases $H_k$ is a union of some of the connected components of $G_k$, and hence $G'$ can be taken to be the open subscheme of $G$ given by removing the components of $G_k$ not in the image of $f_k$.
The upshot is that the answer is affirmative if and only if $\rho(G(R))$ contains $G_k^0(k)$, and that this is also equivalent to the condition that the special fiber of the group smoothening morphism has finite kernel. (It also makes me wonder if perhaps you would prefer simply to work with $\widehat{G}$ in general and forget about $G'$.)   
--
We can say more, by relating the group smoothening to the normalization of $G$. Note that the normalization map $\widetilde{G} \rightarrow G$ is finite since $R$ is excellent, so $\widetilde{G}$ is finite type over $R$ and visibly $R$-flat with the same generic fiber as $G$ over $F$.
Suppose we're in the case with an affirmative answer, so $f_k$ is quasi-finite, and hence $f$ is quasi-finite (as $f_F$ is an isomorphism). Since $f$ is an isomorphism between (possibly disconnected) generic fibers and $\widehat{G}$ is smooth (hence normal), $f$ uniquely factors through the normalization, say via $h:\widehat{G} \rightarrow \widetilde{G}$. But $f$ is quasi-finite, so $h$ is also quasi-finite. Yet $h_F$ is an isomorphism, so by Zariski's Main Theorem and normality considerations we see that $h$ is an open immersion. The quasi-finite $f_k$ must be finite (as for any quasi-finite homomorphism between finite type group schemes over a field), so $h_k$ is finite too.  Yet $h_k$ is an open immersion (since $h$ is), so $h_k$ identifies $\widehat{G}_k$ with a union of connected components of $\widetilde{G}_k$ (with their natural open subscheme structure).  Since $\widetilde{G}$ is $R$-flat (of finite type), so smooth points in its special fiber are precisely where the Zariski-open $R$-smooth locus of $\widetilde{G}$ meets $\widetilde{G}_k$, and all smooth $k$-points of $\widetilde{G}_k$ lift to $\widetilde{G}(R) = G(R) = \widehat{G}(R)$ (first equality by finiteness of normalization), we conclude that $\widehat{G}$ is exactly the smooth locus in $\widetilde{G}$ in these cases.
But in fact we need much less about the normalization to deduce the same conclusion:  suppose that the $R$-flat $\widetilde{G}$ merely has non-empty smooth locus in its special fiber, or in other words that the Zariski-open $R$-smooth locus has non-empty special fiber.  That provides a dense open locus of smooth points in some irreducible component $X$ of the special fiber of $\widetilde{G}$, and those all lift to $\widetilde{G}(R)=G(R)$ yet clearly $X$ maps finitely onto an irreducible component $Y$ of $G_k$ (as $G_k$ and $\widetilde{G}_k$ are equidimensional of the same dimension, due to $R$-flatness and equidimensionality of their common generic fiber $G_F$), so a dense open locus in $Y(k)$ lifts to $G(R)$, forcing $\rho(G(R))$ to contain a dense open locus in some irreducible component (= connected component) of $G_k$.  Hence, that would force $\dim H_k = \dim G_k$.
To summarize, the following conditions are equivalent:  $G'$ exists, $H_k$ contains $G_k^0(k)$, $f_k$ has finite kernel, $\widetilde{G}_k$ has non-empty smooth locus, $\widehat{G}$ is an open subscheme of $\widetilde{G}$.  In such cases, $\widehat{G}_k \rightarrow G_k$ is a finite homomorphism with image $H_k$, and $\widehat{G} \rightarrow G$ has open image that we can take as $G'$.<|endoftext|>
TITLE: Cauchy-Schwarz inequality for bilinear forms valued in an abstract vector space
QUESTION [7 upvotes]: I previously posted this question on Math.SE but didn't receive an answer.  It is perhaps a little vague; part of what I want to know is what question I should ask.
First, consider the following form of the Cauchy-Schwarz inequality: let $V$ be a real vector space, and suppose $(\cdot, \cdot) : V \times V \to \mathbb{R}$ is a symmetric bilinear form which is positive semidefinite, that is, $(x,x) \ge 0$ for all $x$.  Then for any $x,y \in V$ we have $|(x,y)|^2 \le (x,x) (y,y)$.   
I'd like to know what happens if we replace $\mathbb{R}$ by some other space $W$.  Suppose at first that $W$ is a real vector space, equipped with a partial order $\le$ that makes it an an ordered vector space, as well as a multiplication operation $\cdot$ that makes it an algebra.  Then it makes sense to speak of a positive semidefinite symmetric bilinear form $(\cdot, \cdot) : V \times V \to W$, and ask whether it satisfies the Cauchy-Schwarz inequality $(v,w)\cdot(v,w) \le (v,v) \cdot (w,w)$.

Under what conditions on $W$ does this "generalized Cauchy-Schwarz inequality" hold?

At a minimum I expect we will need some more structure on $W$; in particular I assume we would like the multiplication and the partial ordering in $W$ to interact in some reasonable way, so that for instance $w\cdot w \ge 0$ for all $w \in W$.  Are there other properties that $W$ should have?
There are lots of proofs of the classical Cauchy-Schwarz inequality; presumably one should try to find one of them which generalizes.  But I couldn't immediately see how to do this.

My motivating example is the quadratic variation form from probability. For instance, we could take $V$ to be the vector space of continuous $L^2$ martingales on some filtered probability space over some time interval $[0,T]$, and $W$ to be the vector space of continuous adapted processes of bounded variation, mod indistinguishability, with pointwise multiplication and the partial order $X \le Y$ iff $X_t \le Y_t$ for all $t$ almost surely. Then the quadratic variation $\langle M,N \rangle$ is a symmetric positive semidefinite bilinear form from $V \times V$ to $W$.  
In this case I can prove the Cauchy-Schwarz inequality pointwise: fix $M,N \in V$.  For almost every $\omega$, for all $t \in [0,T]$ and all $q \in \mathbb{Q}$ I can say
$$q^2 \langle M,M \rangle_t(\omega) \pm 2 \langle M,N \rangle_t(\omega) + \frac{1}{q^2} \langle N,N \rangle_t(\omega) = \langle q M \pm \frac{1}{q} N \rangle_t(\omega) \ge 0$$
and then letting $q$ be a rational very close to $\sqrt{\langle N,N \rangle_t(\omega) / \langle M,M \rangle_t(\omega)}$ shows that $$|\langle M,N \rangle_t(\omega)| \le \sqrt{\langle M,M \rangle_t(\omega) \langle N,N \rangle_t(\omega)}$$
which is what we want.  But I have used in an essential way the fact that $W$ is a function space, and it would be nice to see if this can be avoided.

REPLY [3 votes]: For a commutative $C^*$-algebra your generalized Cauchy–Schwarz inequality holds, almost:

Kaplansky, 1953 Let $\mathscr{A}$ be a commutative $C^*$-algebra with unit.  Let $\mathscr{X}$ be a right $\mathscr{A}$-module. Let  $\left<\,\cdot\,,\,\cdot\,\right>$ be a $\mathscr{A}$-valued inner product on $\mathscr{X}$. Then for all $x,y\in\mathscr{X}$,
  $$ \left< x,y\right>\left<y,x\right>\ \leq\ \left<x,x\right>\,\left<y,y\right>. $$

"Almost" because by "$\mathscr{A}$-valued inner product on $\mathscr{X}$" we mean a
 sesquilinear map $\left<\,\cdot\,,\,\cdot\,\right>\colon \mathscr{X}\times\mathscr{X}\to\mathscr{A}$ with for all $x,y\in\mathscr{X}$ and $a\in \mathscr{A}$,

$\left<x,x\right>\geq 0$;
$\left<x,x\right>=0$ iff $x=0$;
$\left<x,y\right>^* = \left<y,x\right>$;
$\left<x,ya\right>=\left<x,y\right>a$.

Note that such an inner product is not just linear in the second coordinate, but also an $\mathscr{A}$-module map.
There is a modified version for (not necessarily commutative) $C^*$-algebras, see Proposition 2.3 of [2]:

Pasckke, 1973 Let $\mathscr{A}$ be a $C^*$-algebra.  Then for any $\scr{A}$-valued inner product $\left<\,\cdot\,,\,\cdot\,\right>$  on a right $\mathscr{A}$-module  $\mathscr{X}$, and $x,y\in\mathscr{X}$,
  $$ \left< x,y\right>\left<y,x\right>\ \leq\ 
\left<x,x\right>\,\left\|\left<y,y\right>\right\|. $$

And there is also a version for completely positive maps, see Exercise 3.2 and 3.4 of [3]:

For a $2$-positive map $f\colon \mathscr{A}\to\mathscr{B}$ between $C^*$-algebras, we have, for all $a,b\in\mathscr{A}$,
  $$ f(a^*b) \,f(b^*a)\ \leq\ f(a^*a)\, \left\|f(b^*b)\right\|.$$


[1] I. Kaplansky, Modules over operator algebras, American Journal of Mathematics, 1953
[2] W.L. Paschke, Inner product modules over $B^*$-algebras, Transactions of the American Mathematical Society, 1973
[3] V. Paulsen, Completely bounded maps and operator algebras, Cambridge University Press, 2002<|endoftext|>
TITLE: Reference for Ring Structure on Group Cohomology
QUESTION [9 upvotes]: As a graded $\mathbb{Z}$-module, the structure of the group cohomology $H^{*}(\mathbb{Z}/n\mathbb{Z};\mathbb{Z})$ is extremely well-known. Yet, I am having difficulty finding a reference concerning its cup product structure. I assume this is also well-known, but I would appreciate a reference containing a precise statement of the $\mathbb{Z}$-algebra structure.

REPLY [4 votes]: Well, it's kind of a homework exercise. In particular, in Ken Brown's "Cohomology of Groups" it's Exercise V.3.2 which recommends using diagonal approximation. Here is my solution for completeness, using only the basic machinery:
Let $G=\langle t\rangle$ be a finite cyclic group of order $n$ and let $F$ be the periodic resolution $\cdots\rightarrow\mathbb{Z}G\stackrel{t-1}{\rightarrow}\mathbb{Z}G\stackrel{N}{\rightarrow}\mathbb{Z}G\stackrel{t-1}{\rightarrow}\mathbb{Z}G\stackrel{\varepsilon}{\rightarrow}\mathbb{Z}\rightarrow 0$, where $N=\sum_{i=0}^{n-1} t^i$ is the norm element.  Let $\Delta:F\rightarrow F\otimes F$ be the diagonal approximation map whose $(p,q)$-component $\Delta_{pq}:F_{p+q}\rightarrow F_p\otimes F_q$ is given by  $\Delta_{pq}(1) =$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;1\otimes 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ for $p$ even
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;1\otimes t\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ for $p$ odd and $q$ even
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum_{0\le i < j\le n-1}t^i\otimes t^j\;\;\;\;$ for $p$ and $q$ odd
(Check that this is a diagonal approximation!)
Consider the cohomology groups $H^{2r}(G,M)\cong M^G/NM$ and $H^{2r+1}(G,M')\cong Ker(N:M'\rightarrow M')/IM'$ where $I=\langle t - 1\rangle$ is the augmentation ideal of $G$.
The cup product in $H^*(G,M\otimes M')$ is given by $u\smallsmile v=(u\times v)\circ\Delta$ with $\langle u\times v,x\otimes x'\rangle=(-1)^{deg(v)\cdot deg(x)}\langle u,x\rangle\otimes\langle v,x'\rangle$.
Choose representatives $\langle u,x\rangle=m\in M$ of $H^i(G,M)$ with $m\in M^G$ for $i$ even and $m\in Ker(N:M\rightarrow M)$ for $i$ odd, and choose representatives $\langle v,x'\rangle=m'\in M'$ of $H^j(G,M')$ with $m'\in M'^G$ for $j$ even and $m'\in Ker(N:M'\rightarrow M')$ for $j$ odd.
If $i$ is even then $(-1)^{deg(v)\cdot deg(x)}=(-1)^{deg(v)\cdot i}=1$, and if $j$ is even then $(-1)^{deg(v)\cdot deg(x)}=(-1)^{-j\cdot deg(x)}=1$, and if both $i$ and $j$ are odd then $(-1)^{deg(v)\cdot deg(x)}=(-1)^{-j\cdot i}=-1$.
Thus the cup product element of $H^{i+j}(G,M\otimes M')$ is represented by $m\otimes m'$ for $i$ or $j$ even and is represented by $-\sum_{0\le p < q\le n-1} t^pm\otimes t^qm'$ when $i$ and $j$ are both odd.<|endoftext|>
TITLE: embeddings of graphs into surfaces
QUESTION [8 upvotes]: There is a vast literature on embeddings of graphs into surfaces. 
I am interested in embeddings of graphs that 
belong to the given homotopy class. Here is the precise formulation. 
I have two finite graphs $\Gamma, \Gamma'$ and a homotopy equivalence $f: \Gamma\to \Gamma'$. 
I know that there exists an embedding $\iota: \Gamma'\to S$, where $S$ is a fixed closed oriented surface. I am interested in finding an embedding $j: \Gamma \to S$ so that $j$ is homotopic to $\iota\circ f$. This is, of course, impossible without further restrictions on topology of $\Gamma$, since one can take, for instance, $\Gamma=K_5$, $\Gamma'$ to be the rose with $6$ petals and $S=S^2$. 
Edit: Therefore, let us assume, say, that
$$
(*) \quad  \chi(\Gamma) \ge \chi(S)-1
$$ 
and that $\iota_*: \pi_1(\Gamma')\to \pi_1(S)$ is surjective. 
Question 1. Does the inequality (*) together with surjectivity assumption above imply that  $\iota\circ f$ is homotopic to an embedding for every $f$ and every embedding $\iota: \Gamma'\to S$?
In fact, the inequality (*) seems to be way too generous. Assuming that $\Gamma=K_n$ and taking into account the formula for the genus of $K_n$, one arrives to:
Question 2. Suppose that, $\chi(S)<0$ and replace (*) with 
$$
\chi(\Gamma)> 3  \chi(S).  
$$
Does it follow that  $\iota\circ f$ is homotopic to an embedding?

REPLY [9 votes]: The answer in general is no. One obstruction is that a 4-valent vertex has three local resolutions as a pair of trivalent vertices, and only two of them may be realized in a given surface.
Edit I have simplified the counterexample below.
More precisely, let $\Gamma'$ be a graph that contains two simple closed curves $\alpha$ and $\beta$ intersecting in a single point $p$, which is a 4-valent vertex of $\Gamma'$. Embed $\Gamma'$ in a surface $S$ so that $\alpha$ and $\beta$ are two simple closed curves intersecting transversely at $p$. Among the three resolutions of $p$, one produces a graph $\Gamma$ homotopic to $\Gamma'$ where $\alpha$ and $\beta$ are disjoint. If $\Gamma$ were mapped injectively into $S$ with the same homotopy type as $\Gamma'$, the curves $\alpha$ and $\beta$ would have disjoint representatives. But two simple closed curves intersecting transversely in a point cannot have homotopic disjoint representatives.
As a simple example, one may take $\Gamma' = \alpha\cup\beta$ and embed it in the torus $T$ in the usual way as a spine.<|endoftext|>
TITLE: Can symplectic blow up increase symplectic capacities?
QUESTION [5 upvotes]: Let $N$ be a symplectic submanifold of $M$. Symplectic blow up of $M$ along $N$ is an operation replacing a tubular neighborhood of $N$ with the projectivization of that neighborhood. So it decreases the volume. I have a question on the change of symplectic capacities.
A symplectic capacity $c$ is a function from the set of symplectic manifolds to $[0, \infty]$ satisfying

$c(M_1) \leq c(M_2)$ if we can embed $M_1$ into $M_2$ symplectically,
$c(M, k\omega) = |k| c(M, \omega)$ for $k \neq 0$, and
$c(B^{2n}(r)) = c (B^2(r) \times \mathbb{R}^{2n-2}) = \pi r^2$, where $B^{2n}(r)$ is a $2n$-dimensional ball of radius $r$.

Symplectic capacities may not change after symplectic blow ups. But it seems to me that it is impossible that symplectic blow ups increase symplectic capacities. I couldn't prove this. Can symplectic blow up increase symplectic capacities?

REPLY [2 votes]: The answer is yes.
Let $c$ be the Gromov width except we put $c(M)=\infty$ if $M$ admits an embedding of $B^{2n}(r)$ with $0$ blown up for some $r$.
Using that Gromov width is a capacity it is easy to check 1-3 above, and blowing up $B^{2n}(1)$ at 0 changes this capacity from 1 to $\infty$.<|endoftext|>
TITLE: Stability of Pu's isosystolic inequality
QUESTION [5 upvotes]: The volume of a Riemannian metric on the projective plane is $2\pi$ and length of every non-contractible loop is greater than $\pi - \epsilon$ for some small, positive number $\epsilon$. Is this metric close to the canonical metric?
The question is somewhat vague on purpose. I'm mostly interested in the best constant 
for a bilipschitz equivalence in terms of $\epsilon$, but I also wonder whether for some sufficiently small $\epsilon$ one can conclude that the curvature is close to 1. 
Stability of inequalities is a well-trodden research topic in convex geometry and I was wondering what was known about this in systolic geometry.

REPLY [5 votes]: There is no Lipschitz or even Gromov-Hausdorff stability - just consider a round metric with long hairy tails of small area.
One can hope for stability with respect to intrinsic flat distance in the sense of Sormani-Wenger or some similar metric. This distance is basically Federer's flat distance between isometric images is $L^\infty$ (just like the Gromov-Hausdorff distance is the Hausdorff distance in $L^\infty$). The stability in this sense probably amounts to uniqueness of the equality case in the class of integral current spaces arising as limits of projective planes.<|endoftext|>
TITLE: Vanishing cohomology of de-Rham Witt complex
QUESTION [6 upvotes]: Let $X$ be a smooth scheme over $\mathbb{F}_{p}$ for a prime number $p$. As far as I understand,
there is a surjective morphism from 
$\Omega^\bullet_{W\mathcal{O}_X} \to W \Omega_{X}^\bullet$ which induces an isomorphism 
$\Omega^\bullet_{W\mathcal{O}_X}/(T+Fil^n \Omega^\bullet_{W\mathcal{O}_X}) \to W_{n}\Omega_X^\bullet$ where $T$ is the graded differential ideal consisting of $p-$torsion elements of $\Omega^\bullet_{W\mathcal{O}_X} $ and $Fil^n$ is the kernel of the natural projection map from $\Omega^\bullet_{W \mathcal{O}_X} \to \Omega^\bullet_{W_n\mathcal{O}_X} $. A reference for this is "Complexe de de-Rham Witt et Cohomologie Cristalline" by Luc Illusie. 
The question is: Suppose there exists an integer $N$ such that $H^i(W_{n}\Omega_{X}^\bullet) $ vanish for all $i>N$. Is there any condition on $X$ or $n$ that we can impose so that $H^i(\Omega^\bullet_{W\mathcal{O}_X})$ vanish as well for $i>N$? In other words when does $H^i(T+Fil^n \Omega^\bullet_{W\mathcal{O}_X})$ vanish? Any idea/reference in this direction will be very helpful.

REPLY [5 votes]: If $A$ is a finitely generated ring then so is each $W_n(A)$, and in fact one can bound the number of generators needed by a function of only $p$, $n$, and the number $d$ of generators of $A$. Write $N_p(d,n)$ for such a function. (See below.) Then $\Omega^i_{W_n(A)}$ vanishes for $i>N_p(d,n)$. In particular, this is true when $A$ is the polynomial ring in $d$ variables. Now if $X$ is a smooth scheme over dimension $d$, then it locally admits an etale map to affine $d$-space. Since $W_n$ preserves etaleness of maps, $\Omega^i_{W_n(X)}$ vanishes for $i>N_p(d,n)$. Since $W_n\Omega^i_X$ is a quotient of $\Omega^i_{W_n(X)}$, it also vanishes for $i>N_p(d,n)$. It follows that $H^i(W_n\Omega_X^\bullet)$ and $H^i(\Omega_{W_n(X)}^\bullet)$ vanish for $i>d+N_p(d,n)$. So in particular, your assumption always holds.
Now you ask about the vanishing of $H^i(\Omega^\bullet_{W\mathcal{O}_X})$. This probably involves some nasty details about inverse limits that I'd rather not think about. (For instance, it's not even clear to me what exactly you mean by ${W\mathcal{O}_X}$. Are you actually taking the limit over $n$ in some category or are you treating it as a pro-object? Etc.) But I hope what I've said helps.
[On $N_p(d,n)$: if I'm not mistaken, it can be taken to be $d+(1+p^d+\cdots+p^{nd})$. To show this, you can show that $W_n(A)$ is generated, as a module over the subring generated by Teichmueller lifts of the $d$ generators of $A$, by lifts of module generators of $W_{n-1}(A)$ and by the $V^n$ of Teichmueller lifts of monomials in the generators of $A$, where all exponents in the monomials are at most $p^n$ (thus making $p^{nd}$ additional generators). I hope I got that right. Also my index $n$ is normalized so that $W_0(A)=A$. So it is what would be $n-1$ for most people.]<|endoftext|>
TITLE: Are there only finitely varieties of general type dominated by a given variety in the following sense
QUESTION [5 upvotes]: Let $X$ be a smooth projective complex algebraic variety. I believe the answer to the following question should be well-known. It is an instance of the Iitaka conjecture.
There are only finitely many isomorphism classes of smooth projective varieties of general type $Y$ such that there exists a surjective proper birational morphism from $X$ to $Y$.

REPLY [8 votes]: The answer is yes. 
You can look at this paper:
Guerra,  Pirola, 
 On the finiteness theorem for rational maps on a variety of general type, 
Collect. Math. 60 (2009), no. 3, 261–276. 
It contains an overview of the topic, and besides it gives some effective bounds for the number of such maps.<|endoftext|>
TITLE: Is there an accepted definition of $(\infty,\infty)$ category?
QUESTION [79 upvotes]: For probably twenty years, category theorists have known of some objects in the Platonic universe called "(weak) $\infty$-categories", in which there are $k$-morphisms for all $k\in \mathbb N$, with some weak forms of composition and associativity and ....  It was recognized a bit later that it is worth recording two numbers for each category: an $(m,n)$-category for $m,n \in \mathbb N\cup\lbrace \infty\rbrace$ with $n\leq m$ has morphisms up to dimension $m$ (and above that only equalities), but everything above dimension $n$ is invertible (in some weak sense).  Thus an $(\infty,0)$-category is a "higher groupoid", which Grothendieck's homotopy hypothesis says should be the same as a homotopy type.  (Many people take the homotopy hypothesis to be a definition; others prefer to take it as a check, so that a definition is good if the homotopy hypothesis is a theorem.)  Just to confuse the language, an $(\infty,1)$-category is what many people now call by the name "$\infty$-category".  In any case, for quite a while these objects were expected to exist but definitions were not available (or perhaps too available, but without enough known to decide which was the correct one).
More than a decade ago, Leinster provided A survey of definitions of n-category (Theory and Applications of Categories, Vol. 10, 2002, No. 1, pp 1-70.).  Much more recently, Bergner and Rezk have begun a Comparison of models for (∞,n)-categories (to appear in Geometry and Topology).  These works tend to focus on $(\infty,n)$-categories with $n<\infty$ (although some of Leinster's definitions do include the $n=\infty$ case).
My impression, almost certainly very biased by the people that I happen to hang out with, is that by now it is known that "$(\infty,0)$-category" might as well mean "Kan simplicial set" and "$(\infty,1)$-category" might as well mean "simplicial set in which all inner horns are fillable".  By "might as well mean", I mean that there are many reasonable definitions, but all are known to be Quillen-equivalent.  For $n< \infty$, "$(\infty,n)$-category" can mean "iterated complete Segal space", for example.  It is not clear how to take $n=\infty$ in this approach.
A very tempting definition of "$(\infty,n)$-category", which I believe is discussed in Lurie's (to my knowledge not yet fully rigorous) paper On the Classication of Topological Field Theories, is that an $(\infty,n)$-category should be an $(\infty,1)$-category enriched in $(\infty,n-1)$-categories.  This is based on the observation that the coherence axioms of being an enriched category only require invertible morphisms above dimension $1$.  Thus as soon as you've settled on a notion of "$(\infty,1)$-category" (this seems to have been settled) and a notion of "enriched" (settled?), you get a notion of "$(\infty,n)$-category".  This approach is developed in Bergner–Rezk, for example.
In any case, it is less clear to me how to use these inductive definitions to get to $(\infty,\infty)$.  Let $\mathrm{Cat}_n$ denote the $(\infty,n+1)$-category of $(\infty,n)$-categories.  Then there are clearly inclusion functors $\mathrm{Cat}_n \to \mathrm{Cat}_{n+1}$, but the colimit of this sequence does not deserve to be called $\mathrm{Cat}_\infty$, since any particular object in it has morphisms only up to some dimension.  I have seen it proposed that one should instead take the (inverse) limit of the "truncation" functors $\mathrm{Cat}_{n+1} \to \mathrm{Cat}_n$; I am unsure exactly how to define these, and even less sure why this deserves the name $\mathrm{Cat}_\infty$.
Thus my question:

Is there by now an accepted / consensus notion of $(\infty,\infty)$-category?  For instance, if I have decided on my favorite meaning of $(\infty,1)$-category, is there some agreed-upon procedure that produces a notion of $(\infty,\infty)$-category?  Or are there known comparison results that can push all the way to $n=\infty$?

A closely related MathOverflow question was asked a little less than a year ago (or via Wayback Machine here), but received no answers.
Also, I am certainly unaware of many papers in the literature, and likely I have misrepresented even those papers I mentioned above.  My apologies to all parties for doing so.  It is precisely because of these gaps in my knowledge of the literature that I pose this question.

REPLY [8 votes]: This is more of a general comment on the direction of multiple category theory rather than an answer to the question but I hope it will be useful in directing attention to the origins of some of these questions in homotopy theory and for suggesting lines, or a line,  for further development. In this I will possibly still continue Out of Line! 
I looked at the Berner-Rezk paper referred to in the above, and found the following diagram for compositions in $2$-categories. 
 (source)
Compare this with the simplicity of describing the following in a double category. 
 (source)
One writes this as an array $(a_{ij})$ with composition $[a_{ij}]$, and this notation is easily extended to $n$ dimensions without using operads, though these might be useful in the cubical context.  Note also that an inverse in a specific direction is easy to envisage. 
This is why I have found it very difficult to conjecture and prove results in the globular setting, while the cubical setting has allowed a new approach to basic algebraic topology using filtered spaces and homotopy classes of mappings. 
My writing to Alexander Grothendieck in 1982 did stimulate his writing of "Pursuing Stacks" in 1983, but he noted that the model of $\infty$-groupoids did not model all homotopy types, and proceeded to look for weak categorical models, which are clearly important. But when in 1985 (or 6)  I told him of Loday's work that (strict) $n$-fold groupoids model homotopy $n$-types, he exclaimed: "That is absolutely beautiful!"  This work relates to classical parts of homotopy theory, such as $n$-ads, and the Blakers-Massey and Barratt-Whitehead theorems, and gives some new explicit calculations of  homotopy $n$-types in a nonabelian context. 
Another aspect of Loday's work is his notion of "resolution" of a space: the first stage resolves in one direction, the second in a different direction, and so on, ending up with a resolution of spaces by $n$-cubes of fibrations, which has been clearly explained by Richard Steiner, J. London Math. Soc. (2)  {34} (1986) 169--176. 
What I am arguing for is an eclectic approach, in which the advantages and difficulties of various approaches are considered. 
Update: I'll add a reference to J.C. Morton, "cubical $n$-categories and finite-limits theories", arxiv: 1001.2628, and some further references there.  Another relevant paper is arXiv:1204.5101 [pdf, ps, other]
n-Fold groupoids, n-types and n-track categories, 
David Blanc, Simona Paoli

September 1, 2014    Here is a link to a presentation I gave at the IHP, Paris, June 5, 2014 on Intuitions for cubical methods in nonabelian algebraic topology) (Wayback Machine). 
December 26, 2015 I would also like to add this presentation in June 2015 at Aveiro on 
A philosophy of modelling and computing homotopy types (Wayback Machine), 
which illustrates  the aim of doing specific nonabelian colimit computations in homotopy theory. 
October 26, 2016  A writeup of part of the above presentation is available on the arXiv as Modelling and Computing Homotopy Types:I.<|endoftext|>
TITLE: Numbers with known finite irrationality measure greater than 2
QUESTION [22 upvotes]: For a real number $\alpha$, let the irrationality measure $\mu(\alpha) \in \mathbb{R}\cup \{\infty\}$ be defined as the supremum of all real numbers $\mu$ such that 
$$ \left| \alpha-\frac{p}{q}\right| \le \frac{1}{q^\mu} $$ has infinitely many solutions $\frac{p}{q} \in \mathbb{Q}$, where $q \ge 1$.
It's known that if $\alpha$ is rational, then $\mu(\alpha)=1$ and if $\alpha$ is algebraic and irrational, then $\mu(\alpha)=2$ by Roth's Theorem. The set of all $\alpha$ such that $\mu(\alpha) >2$ has Lebesgue measure 0 by Khinchin's Theorem.
One can explicitly write down real numbers $\alpha$ such that $\mu(\alpha) = \infty$ (e.g. Liouville's constant) and transcendental real numbers $\alpha$ such that $\mu(\alpha)=2$ (e.g. $\alpha =e$, see also this thread). It's also possible to find upper bounds on $\mu(\alpha)$ for some real numbers such as $\pi$. But I don't know of a single example of a real number $\alpha$ whose irrationality measure is known, finite and greater than 2.
Question: Is there an example of a real number whose irrationality measure $\mu(\alpha)$ is known exactly and satisfies $2<\mu(\alpha)<\infty$?

REPLY [6 votes]: The irrationality measure of the Champernowne constant $C_b$ in base $b>2$ is exactly $b$.<|endoftext|>
TITLE: Banach Algebra Counterexample
QUESTION [7 upvotes]: Can someone give me an example of a Banach Algebra which does not have an isometric representation in a Hilbert Space ?
(if possible, can you add a proof or a reference ? )
Thank you very much !

Note added by YC: this question has also been asked on MSE where someone has given a much better, elementary example to the OP's question. (However, my example also works for the question of topologically isomorphic representations, not just the isometric ones.)

REPLY [6 votes]: Any Banach algebra which is not Arens regular cannot be embedded as a closed subalgebra of B(H), even if you allow for isomorphic embeddings that have closed range yet are not isometric.
If you are only interested in Banach $\ast$-algebras and isometric $\ast$-homomorphic embeddings, then it is easier to find examples, as Owen Sizemore has indicated.

[Not directly relevant, but perhaps of background interest to the OP]
By the way, although the question asks about isometric embeddings, there are interesting and slightly unexpected examples of Banach algebras $A$ for which there is an injective homomorphism $A\to B(\ell^2)$ that has closed range, showing that non-selfadjoint operator algebra theory has to be a lot wilder than the self-adjoint case. Examples include: $\ell^p$ for $1\leq p <\infty$ with pointwise product; and the algebras $C^k([0,1]^m)$ of $k$-times continuously differentiable functions on the $m$-cube. You can even get radical commutative Banach algebras embedded into $B(\ell^2)$ in this way, see

MR0410386 (53 #14136) P. G. Dixon, Radical Q-algebras.
  Glasgow Math. J. 17 (1976), no. 2, 119--126.<|endoftext|>
TITLE: How do we express measurable spaces using type theory?
QUESTION [18 upvotes]: A measurable space $(X,\mathcal X)$ consists of a set $X$ equipped with a $\sigma$-algebra of subsets $\mathcal X$. I would like to write computer programs involving measurable spaces, but to the best of my knowledge, no language (including Haskell) has the ability to deal with completely unstructured sets. On the other hand, type theory is an alternate foundation for mathematics which is well-grounded computationally. 
Given a measurable space $(X,\mathcal X)$, is there an equivalent presentation using Martin-Löf dependent type theory?
With no constraints on the measurable space, the answer is probably no; a counterexample would be appreciated. What if $\mathcal X$ is a countably generated $\sigma$-algebra?

REPLY [9 votes]: I am not exactly sure what you are looking for.
There is this work in inductively generated sigma-algebras in type theory:
   http://link.springer.com/article/10.1007%2Fs001530100123
More generally, I would be inclined to an algebraic approach to measure theory. 
For instance, using Riesz spaces (of real valued measurable functions modulo zero measure) 
http://logicandanalysis.org/index.php/jla/article/view/14/12
or the related pointfree (localic) approach
http://www.cs.bham.ac.uk/~sjv/Riesz.pdf
http://homepages.inf.ed.ac.uk/als/Research/Sources/mrs.pdf
There is also these formalizations in type  theory:
https://www.lri.fr/~paulin/ALEA/ (randomized algorithms)
http://www.cs.unibo.it/~sacerdot/PAPERS/jfr.pdf (Lebesgue’s Dominated Convergence Theorem in Matita).
I'd be happy to elaborate, but would need some more information about what you are searching about.<|endoftext|>
TITLE: Distortion of malnormal subgroup of hyperbolic groups
QUESTION [14 upvotes]: Let $G$ be a countable, Gromov-hyperbolic group. 
We say that $H$ is hyperbolically embedded in $G$ if $G$ is relatively hyperbolic to {$H$} (in the strong sense). This definition is due to Osin.
A theorem of Bowditch says that infinite, finitely generated, almost-malnormal and quasi-convex subgroups of $G$ are hyperbolically embedded in $G$. Later Osin has proved that the conditions are necessary (even in the wider context of relatively hyperbolic groups). 
Quasi-convex subgroups are not necessary malnormal but they have always finite height by a result of Gitik, Mitra, Rips and Sageev. The height of $H\subset G$ is defined to be the maximal $n$ such that there exist $g_1,\ldots,g_n\in G$ with $g_1Hg_1^{-1}\cap\ldots\cap g_nHg_n^{-1}$ infinite (but all the $g_iHg_i^{-1}$ different). 
I would like to know how distorted a malnormal subgroup can be in $G$. 
Is there some class of groups for which malnormal implies quasi-convex? Examples?
What about the relatively hyperbolic case?

REPLY [7 votes]: There's a weak sort of quasiconvexity proved by Kapovich:
if one has an acylindrical graph of hyperbolic groups, with
finitely generated edge groups which embed q.i. into the 
vertex groups, then the edge groups embed q.i. into the graph
of groups. Acylindrical here is weaker than malnormality:
it says that the stabilizers of the action on the tree
have bounded length, whereas malnormality of the edge
groups would correspond to stabilizers having length 1. 
Incidentally, if one had a finitely-generated malnormal
hyperbolic subgroup of a hyperbolic group which was not quasiconvex,
then the double of the group along the subgroup is finitely
presented and does not contain any Baumslag-Solitar
subgroups. However, this group is
not hyperbolic. This would be quite interesting, as I 
believe there are no known examples of such 
groups which are of type F (acting properly cocompactly on a 
contractible locally compact complex). As Henry points out
in the comments, there are finitely presented examples of Brady
which are not hyperbolic, but which are also not type F (they
occur as normal subgroups of a hyperbolic group, so are
far from being malnormal subgroups).<|endoftext|>
TITLE: Orbit structure of linear representations of complex Lie groups
QUESTION [5 upvotes]: Let $G$ be a semisimple complex Lie group (or perhaps a reductive algebraic group over $\mathbb{C}$) and $V$ an irreducible finite-dimensional representation of $G$, determined by its highest weight.  Is there a complete description of the orbits of $V$, or $\mathbb{P}(V)$, under $G$?
In the case of the adjoint representation of $SL_n$ or $GL_n$, I expect to recover the Jordan canonical form of a matrix, or something of the sort.

REPLY [9 votes]: As Allen has indicated, there is no hope to solve this question even for $G=GL_n.$ However, there are some representations that are mild enough to be analogous to the vector (or standard) representation of $GL_n$ on $\mathbb{C}^n$  (all non-zero vectors belong to the same orbit) or to the adjoint representation of $GL_n$, where the classification problem is equivalent to determining the conjugacy classes of matrices. The following classes of "well-behaved" representations, listed in the order of increasing complexity, have been extensively studied: multiplicity-free actions, prehomogeneous vector spaces, Kostant--Rallis actions. In particular, $\mathbb{C}^n, S^{2}\mathbb{C}^n$ and $\Lambda{^2}\mathbb{C}^n$ are multiplicity-free and $M_{n,n}$ is a Kostant--Rallis action. For all these classes, the invariant algebra is either trivial (scalars) or a free commutative algebra. Unfortunately, most representations fall outside of these classes, even if you restrict attention to the irreducible polynomial representations of $GL_n.$<|endoftext|>
TITLE: What new primitive recursive functions are needed to reconcile Turing time complexity with Gödel time complexity?
QUESTION [6 upvotes]: Let me begin with an example.
Consider the computable function $f(x) = 2x$.  A Turing machine can implement this function in $O(|n|)$ steps: simply walk to the end of the input string, write a $0$, and return.  However, the "best" definition of $f(x) = 2x$ built from Gödel's primitive recursive functions will use primitive recursion to take $x$ successors of $x$ and return the result.  If we intuitively hold that each "call" to the primitive successor function requires one algorithmic step, then the function $f(x) = 2x$ requires $O(2^{|n|})$ steps.
From this example, we can see that the intuitive notion of time complexity associated with Turing machine computation ("Turing complexity") and the intuitive notion of time complexity associated with Gödel's recursive functions ("Gödel complexity") do not coincide.  My goal is to tweak the definition of the recursive functions to make it so these concepts do coincide.
One attempt is to add $f(x) = 2x$ to the list of primitive recursive functions.  But there are still problems: for example, the function $f(x) = x - 1$ is still $O(|n|)$ on a Turing machine, but $O(2^{|n|})$ when expressed as the combination of primitive recursive functions. 
This is my question:

Problem
Find a set of new primitive recursive functions that can be added to the original three primitive recursive functions such that the best-case Turing complexity of any computable function is always equal to the best-case Gödel complexity of that function.

REPLY [3 votes]: Unless I misunderstand something:

The best-case representation for $g(x) = x+1$ as a primitive recursive function takes time $1$, if we agree that each "call" to the primitive successor function requires one algorithmic step. This is trivial: computing $g(x)$ always requires exactly one invocation of the primitive successor function.
Common Turing machine models (such as those that require $O(|x|)$ steps to compute $x-1$, as in the question) will not be able to compute $g(x) = x+1$ in a bounded amount of time. 

The same holds for computing $f(x) = x-1$. The usual primitive recursive definition of this function uses no successor operations at all, and runs in constant time if we define the projection functions to use short-circuit evaluation by not evaluating the arguments that they don't return.  The definition is given by recursion as $f(0) = 0$ and $f(x+1) = \pi^2_1(f(x),x)$.
So the goal of the question cannot be achieved, under the Turing machine conventions in the question, but not for the reason suggested. 
This is a key point about what happens if we try to capture the complexity of a primitive recursive function by counting function invocations. The successor operation can add 1 to any number in one "step", regardless of the size of the number; the primitive recursion combinator can, in one "step", determine whether a given number $z$ is 0, and also compute $z-1$ if $z$ is positive.<|endoftext|>
TITLE: Where did Sophus Lie write the group commutator for two one parameter groups
QUESTION [16 upvotes]: If $X,Y$ are vector fields and $\def\Fl{\operatorname{Fl}}\Fl^X_t$ and $\Fl^Y_t$ their local flows, let $[\Fl^X_t,\Fl^Y_t]:= \Fl^Y_{-t}\Fl^X_{-t}\Fl^Y_t\Fl^X_t$ denote the group commutator of the flows. Then
$$
\partial_t|_0 [\Fl^X_{t},\Fl^Y_{t}] =0, \qquad
\frac12 \partial_t^2|_0 [\Fl^X_{t},\Fl^Y_{t}] = [X,Y].
$$
See

Markus Mauhart, Peter W. Michor: Commutators of flows and fields. Archivum Mathematicum (Brno) 28,3-4 (1992), 228–236, (pdf)

for an extension of this to formal bracket expressions of arbitrary length.
Where did Sophus Lie write this? And where did he compute something like
$$
\lim_n \frac1n [\Fl^X_{1/n},\Fl^Y_{1/n}]^n\quad ?
$$

REPLY [16 votes]: Looking around on the internet, I found an English translation of Lie's 1891 paper Die Grundlagen für die Theorie der unendlichen kontinuierlichen Transformationsgruppen. I.  (I.e., The foundations of the theory of infinite continuous transformation groups - I) at the location 
http://neo-classical-physics.info/uploads/3/0/6/5/3065888/lie-_infinite_continuous_groups_-_i.pdf
In this English translation (I haven't gone to look for the original German), Lie gives the formula for the commutator of the flows of two vector fields in the form that you require.  Look at Paragraphs 35-37, noting especially the displayed equation (45) and the first displayed equation in Paragraph 36, which is exactly the first formula you asked about.
It's quite possible that these formulae appeared even earlier in Lie's papers.  He mentions, in the introduction, a paper of 1883 that might well also have these formulae. I'll check when I get the chance.
Concerning the Trotter formula (your second formula), I have no idea.<|endoftext|>
TITLE: what is the cyclic cover trick?
QUESTION [16 upvotes]: What do people mean by the "cyclic cover trick"? I have found this expression a couple of times with no complete explanation, both talking about curves and surfaces...

REPLY [26 votes]: The "cyclic cover trick" can refer to more than one thing.  One example is as follows.  Let $L$ be an invertible sheaf on a smooth projective scheme such that some power $L^{\otimes d}$ has a global section $s$ whose zero scheme $D$ is a smooth Cartier divisor (all of these smoothness conditions are not strictly necessary).  Let $\nu:Y\to X$ be the associated degree $d$ branched cover branched over $D$ and whose restriction over $X\setminus D$ is the $\mathbb{\mu}_d$-torsor corresponding to $L$.  More precisely, $\nu$ is affine and $\nu_*\mathcal{O}_Y$ is the  $\mathbb{Z}/d\mathbb{Z}$-graded $\mathcal{O}_X$-algebra $$\nu_*\mathcal{O}_Y = \mathcal{O}_X \oplus L^\vee \oplus \dots \oplus (L^\vee)^{\otimes (d-1)}.$$
Of course we have to say what is the multiplication rule on this algebra.  But there is a unique multiplication rule that is $\mathbb{Z}/d\mathbb{Z}$-graded and that is compatible with the multiplication rule $s:(L^\vee)^{\otimes d} \to \mathcal{O}_X$.  Now, for every $q$, $\nu_*\Omega^p_Y$ has a natural $\mathbb{Z}/d\mathbb{Z}$-grading, and the graded pieces turn out to be expressible in terms of tensor products of $\Omega^r_X$ with powers $(L^\vee)^{\otimes s}$.  Now assume that $X$ is a $\mathbb{C}$-scheme.  Then the Hodge theorem gives surjectivity of the various projections $H^r(Y^{\text{an}};\mathbb{C}) \to H^q(Y,\Omega^p_Y)$.  Both groups have a $\mathbb{\mu}_d$-action that is equivalent to a $\mathbb{Z}/d\mathbb{Z}$-grading.  In particular, vanishing of certain graded pieces of $H^r(Y^{\text{an}};\mathbb{C})$ will imply vanishing of certain cohomology groups $H^q(X,\Omega^p_X\otimes (L^\vee)^{\otimes s})$.  In this way, one can prove the Kodaira-Akizuki-Nakano vanishing theorem.  
So in this case, the "covering trick" is to use a geometric result on the cyclic cover (the Hodge decomposition) combined with computation of the graded pieces to deduce a geometric result on the original scheme.  In a similar way, one can prove extensions of the Kodaira vanishing theorem, e.g., the Kawamata-Viehweg vanishing theorem.  This is discussed in detail in the book of Kollár and Mori.<|endoftext|>
TITLE: Aubin's book - construction of Green's function on compact manifold
QUESTION [8 upvotes]: In Aubin's book (nonlinear problems in Riemannian Geometry), starting from p. 106, it is shown that a Green's function of a compact manifold without boundary satisfies
$$G(P,Q) \leq k \rho(P,Q)^{2-n},$$
where $\rho(P,Q)$ denotes the geodesic distance between $P$ and $Q$. 
a) I do not understand their proof. In fact, I think they proved that near the diagonal, such estimate holds. However, from what he stated, this looks global (which, for my purposes, is great). Anyone knows that proof and can help me get it?
b) When one considers a manifold with boundary, it seems that the same estimate holds (p. 112). However, their constant $k$ now depends on the distance between P and the boundary of the manifold. Is there any way to avoid that ? Is there any way to get the explicit dependence ?

REPLY [9 votes]: Your have two nice texts of Frederic Robert (a descendant of Aubin) about the construction and estimate of Green function
one in french: http://www.iecl.univ-lorraine.fr/~Frederic.Robert/ConstrucGreen.pdf
an other in english where you will find an answer to (b) when assuming Neumann boundary condition: http://www.iecl.univ-lorraine.fr/~Frederic.Robert/NotesGreenNeumannRobert.pdf
Else you can also have a look to appendix A of the book: Blow-up theory for elliptic PDEs in Riemannian geometry, Olivier Druet, Emmanuel Hebey and Frederic Robert,
Mathematical Notes, Princeton University Press, Volume 45.<|endoftext|>
TITLE: How did Hankel determinants get the name Hankel-Hadamard?
QUESTION [5 upvotes]: My question concerns the name for determinants of Hankel-matrices $H = (s_{i+j})_{i,j = 0}^n$.
In the classical textbook of Shohat and Tamarkin (1943) "The Problem of Moments", these determinants are defined without a name (on page viii).
In several math articles, I found the name Hankel determinant. In the field of physics (especially methods of moments), where I work, however, the common name seems to be "Hankel-Hadamard determinant"
Does anyone can give me a hint on how Hadamard is connected to this? Any help would be highly appreciated!

REPLY [2 votes]: Some background is given in the appendix of this 1988 paper by Handy and Bessis, who apparently introduced this terminology: Hankel-Hadamard matrix, Hankel-Hadamard determinant, Hankel-Hadamard positivity, Hankel-Hadamard inequality. It refers to a class of matrices of the Hankel form (constant diagonals) with a determinant that satisfies a generalized Hadamard inequality. The inequality is not quite the original Hadamard inequality (it's a lower bound rather than an upper bound on the determinant), but the name stuck.<|endoftext|>
TITLE: Why is there no stack of $\ell$-adic sheaves on a curve?
QUESTION [9 upvotes]: One of the main players in the categorical geometric langlands correspondence is the moduli stack of rank n integrable connections on a complex curve.  The reason for considering such objects is that they are the de Rham analogue of lisse $\ell$-adic sheaves, which have an obvious Galois interpretation.  I have read in many places that there is no moduli stack of lisse $\ell$-adic sheaves on a curve over a finite field, making the naive $\ell$-adic formulation of the categorical geometric Langlands correspondence problematic.  I have two questions

Why do $\ell$-adic sheaves not form a stack?
Is it possible that they form some kind of higher stack?

REPLY [13 votes]: Let's work over the complex numbers for simplicity. Let $X$ be a Riemann surface, fix a base point $x \in X$, and let $\Gamma$ be the fundamental group $\pi_{1}(X,x)$.
The data of an $n$-dimensional local system on $X$, together with a trivialization at the point $x$, is equivalent to the data of an $n$-dimensional representation of $\Gamma$. This is true in either the $\ell$-adic setting (in which case you'd study representations into $GL_n( \mathbf{Z}_{\ell} )$) or in the setting of "usual" local systems (where you'd study representations into $GL_n( \mathbf{C})$). 
In particular, you can topologize the set of representations by picking a set of generators for $\Gamma$ and thereby embedding into some power of $GL_n(k)$. In the case $k = \mathbf{C}$, you can do even better: $GL_n( \mathbf{C} )$ is again an algebraic variety over $\mathbf{C}$, so that the collection of representations of $\Gamma$ into $GL_n( \mathbf{C} )$ inherits the structure of an algebraic variety. (In fact, it has multiple inequivalent "algebraic structures"; the one that I've described is the "Betti version", which is different from the one used in geometric Langlands).
Working with $\ell$-adic coefficients, you can still talk about whether or not two representations are "close" to one another, but the space of representations no longer bears any resemblance to an algebraic variety defined over $\mathbf{C}$ (for example, it's totally disconnected). This has nothing to do with stacks versus higher stacks: it's a disconnect between the "coefficient field" for your local systems and the field that your variety is defined over.<|endoftext|>
TITLE: Proper-class sized "ring" with no maximal ideals
QUESTION [5 upvotes]: Suppose I have a collection of "elements" together with operations that satisfy the axioms for a commutative ring with identity --- except that these elements form not a set, but a proper class.
Must such a thing contain a maximal ideal (where an "ideal" is allowed to be proper-class-size)?  Obviously, the usual Zorn's Lemma argument is not available.

REPLY [6 votes]: The axiom of global choice is sufficient. More generally, any proper class ring whose elements can be enumerated along the ordinals will have a maximal ideal. The argument is the usual one...
Suppose $x_\alpha$, $\alpha \in \mathrm{Ord}$, enumerates the elements of the proper class ring $R$. Without loss of generality $x_0$ is the zero of the ring $R$. Define $h(0) = 0$ and for each $\alpha \gt 0$, let $h(\alpha)$ be the first ordinal $\eta$ (if any) such that $x_\eta$ is not in $I_\alpha = \sum_{\beta\lt\alpha} Rx_{h(\beta)}$ and $Rx_\eta + I_\alpha \neq R$. These $R$-ideals are all uniformly definable from $R$, $\langle x_\alpha\rangle_{\alpha\in\mathrm{Ord}}$ and $\langle h(\beta)\rangle_{\beta\lt\alpha}$, so there is no trouble carrying out this construction in NBG. The elements indexed by the function $h$ will enumerate a (not necessarily proper) class of generators for a maximal $R$-ideal.
Since the issue is choice, a counterexample in the absence of global choice could be done by extending the ideas of Hodges in Six Impossible Rings [J. Algebra 31 (1974), 218–244; MR0347814] to proper classes. This seems plausible but I strongly suspect that nobody has ever done it in public...<|endoftext|>
TITLE: What makes the Cartier operator "tick"?
QUESTION [21 upvotes]: Let $C$ be a smooth curve over a finite field of characteristic $p$. Let $t$ be a local parameter at a point. If $f$ is a regular function on a neighbourhood of the point, one can write uniquely 
$$f = \sum_{i=0}^{p-1} f_i^p t^i $$
for functions $f_i$. Using this decomposition one can define an operator on differential forms on $C$ via 
$$ f \mathrm d t \mapsto f_{p-1} \mathrm d t.$$
Amazingly (?) this does not depend on the choice of coordinate, producing a map of sheaves $\Omega_C \to \Omega_C$. This is the Cartier operator.
Question: What is so special about $\Omega_C$ here? Is there a similar construction on quadratic differentials? On spin curves, i.e. on a square root of $\Omega_C$?
I am told that the Cartier operator is analogous to the residue of a logarithmic differential form. So I guess a similar question (that I don't know how to answer either) is what makes the residue tick.

REPLY [18 votes]: I came to this question hoping to learn more, but I can record what I've learned. I find the Cartier operator much more natural in a larger context. This is a combination of material from Brion and Kumar's book, Section I.3, and things I saw asserted in papers and worked out for myself. Of course, there is a risk that any of this is wrong.$\def\cF{\mathcal{F}}$ $\def\cC{\mathcal{C}}$
Notation: Let $k$ be a perfect field of characteristic $p$ and $A$ a $k$-algebra. Let $\Omega^r$ be the $r$-th wedge power of the Kahler differentials of $A/k$. Let $Z^r = \mathrm{Ker}(d: \Omega^r \to \Omega^{r+1})$ (the de Rham co-cycles) and $B^r = \mathrm{Im}(d: \Omega^{r-1} \to \Omega^r)$ (the de Rham co-boundaries). Let $H^r = Z^r/B^r$. Let $A^p$ denote the ring of $p$-th powers in $A$. Note that $d$ is a map of $A^p$ modules and hence $Z^r$, $B^r$ and $H^r$ are all $A^p$ modules.
The inverse of the Cartier operator is more basic than the Cartier operator. 
Theorem There is a unique map $\cF: \Omega^r \to H^r$ such that

$\cF(f) = f^p$ for $f \in \Omega^0 = A$.
$\cF(\alpha+\beta) = \cF(\alpha)+\cF(\beta)$.
$\cF(d u) = u^{p-1} du$ for $u \in A$.
$\cF(\alpha \wedge \beta) = \cF(\alpha) \wedge \cF(\beta)$.

The only hard part to check is that $(u+v)^{p-1} (du+dv)$ is equivalent to $u^{p-1} du+ v^{p-1} dv$ modulo $B^1$. To this end, note that 
$$(u+v)^{p-1} (du+dv) - u^{p-1} du - v^{p-1} dv = d \sum_{k=1}^{p-1} \frac{1}{p} \binom{p}{k} d(u^k v^{p-k}). \quad (\ast)$$
Here $\frac{1}{p} \binom{p}{k}$ must be interpreted as an integer which we reduce modulo $p$ to make an element of $k$. (Morally, equation $(\ast)$ comes from the relation $d(u+v)^p = \sum_{k=0}^{p} \binom{p}{k} d(u^k v^{p-k})$. We can't deduce $(\ast)$ directly from this because we can't divide by $p$, but $(\ast)$ is still true in characteristic $p$.)
If $A$ has a sufficiently nice deformation $\tilde{A}$ over the ring of Witt vectors $W(k)$, and $F: \tilde{A} \to \tilde{A}$ is a lift of Frobenius, then $\cF$ for $\omega \in \Omega^r$ is given by reduction modulo $p$ of $\frac{1}{p^r} (F^{\ast})^r \omega$. I don't know exactly what "sufficiently nice" means, but $\tilde{A}$ smooth over $W(k)$ is certainly enough. This is my favorite way to think of $\cF$.
$\cF$ respects localization, and thus makes sense on schemes. It also respects etale extension and completion, for those who prefer other topologies.
We now have Theorem I.3.4 in Brion and Kumar (presumably not original, but they don't cite it): If $A$ is regular, than $\cF$ is an isomorphism $\Omega^{\bullet} \to H^{\bullet}$. Proof sketch: The claim is that a map of finitely generated $A^p$ modules is an isomorphism; this can be checked on completions. So we are reduced to proving the claim in a power series algebra. This is a combinatorial exercise, very similar to the proof of the Poincare lemma for formal power series in characteristic zero. 
Note that this is interesting even for $r=0$: It says that $df=0$ for $f \in A$ if and only if $f$ is a $p$-th power.
For $A$ regular, the Cartier operator $\cC: H^{\bullet} \to \Omega^{\bullet}$ is the inverse to $\cF$.
In particular, if $A$ regular of dimension $1$, then we can take the composition $\Omega^1 \to H^1 \to \Omega^1$ to get the definition you gave.
$\cC$ on a curve is analogous to residue in several senses: 

It is the composition of $\Omega^1 \to H^1$ and an isomorphism between $H^1$ and an explicit free $H^0$-module of rank $1$. In the case of residue, working in the Laurent series ring $k((x))$, we have $H^0 = k$ and we have a canonical isomorphism $H^1 \cong k$; the residue map is $\Omega^1 \to H^1 \to k$. In the Cartier case, $H^0 = A^p$ and $H^1$ is a free $A^p$-module of rank $1$. It doesn't have a canonical generator, but what you can do canonically is use $\cC$ to turn $A^p$ into $A$ and $H^1$ into $\Omega^1$.
We have $\cC(df) = 0$ and $\cC(du/u) = du/u$, for $u$ a unit of $A$.
Concretely, $\mathrm{res}(\cC(\omega)) = \mathrm{res}(\omega)^{1/p}$.<|endoftext|>
TITLE: Irreducibility of the trinomial over Q
QUESTION [9 upvotes]: I'm trying to find an algebraic proof of irreducibility of the polynomial $x^n-x-1$ over rational numbers (or integers, which the same). I've read the Selmer's paper "On the irreducibility of certain trinomials", and I got his idea, but... In his proof he uses analysis methods. I wonder, if the pure algebraic proof exists? I mean, without making graphics and curves, but studying polynomial as the element of Q[x], not as a function.

REPLY [3 votes]: For an algebraic approach to the irreducibility, see the sketch I wrote as an answer to https://math.stackexchange.com/questions/393646/irreducibility-of-xn-x-1-over-mathbb-q.<|endoftext|>
TITLE: Taylor Series and Fourier Series
QUESTION [7 upvotes]: Taylor series expansion of function, f, is a vector in the vector space with basis: {(x-a)^0, (x-a)^1, (x-a)^3, ..., (x-a)^n, ...}. This vector space has a countably infinite dimension. When f is expressed as linear combination of the basis vector the scalar multiple for the n-th basis vector is Diff_n{f}(a)/n! 
Fourier series expansion of function, f, is a vector in the vector space with basis: {sin(1x), cos(1x), sin(2x), cos(2x), ..., sin(nx), cos(nx), ...}. This vector space has a countably infinite dimension. When f is expressed as linear combination of the basis vector the scalar multiple for the n-th basis vectors are Int{f.sin(nx)} and Int{f.cos(nx)}.
Questions:

The vector space for the Fourier series has an inner product, Int{f.g}, and it's this inner product that provides the above expressions like Int{f.sin(nx)} and Int{f.cos(nx)}. Is there a similar inner product based derivation of the scalar multiples for the vector space of spanned by the polynomial basis in Taylor series? 
What is the relationship, if any, between the vector space produced by Taylor Series and that of Fourier Series? E.g. is one a subspace of the other? 
When Fourier series is taught, why isn't Taylor Series re-explained in the vector space framework used for Fourier series? And would this approach not lead the discussion of the implication of the choice of basis (and perhaps the choice of inner product) for function spaces?
Just as Fourier series get generalized to Fourier Transform (the summation of the series becomes an integral), is there something equivalent to Taylor series?
Are there any recommended resources (books, courses, etc.) available which can help clarify my thinking regarding these issues?

REPLY [2 votes]: In your question you only discuss the formal analogy, disregarding the questions of convergence,
and what exactly a "function" is. This is a reasonable setting indeed, if one restricts oneself to
finite sums. Then there is an exact correspondence, if we extend the set of Taylor series to Laurent
series. Finite Laurent series $\sum a_n z^n$ are in one-to one correspondence with
finite trigonometric sums $\sum a_n\exp(int)$. (These are essentially the same as the series in
sines and cosines). The subspace of polynomials consists of those sums for which $n\geq 0$.
One can give a condition on the series of sines and cosines to correspond to a Laurent series
with $n\geq 0$.
If one wants to pass to infinite sums, one has to use some topology.
The simplest case is when the topology is defined by the standard scalar product. Completion of the
space of polynomials in this topology gives the space $H_2$ which can be thought of as a space
of trigonometric (Fourier) series, and as a space of Power series. And for the coefficients of
the expansion you have two formulas: one involving integration, another differentiation.
These formulas give the same result in view of Cauchy formula. 
On your other questions. Fourier series are generalized to Fourier transform.
Taylor series are similarly generalized to Laplace transform. 
(With an intermediate step of Dirichlet series).
For recommended reading... there are a lot of books, the appropriate one depends on your prerequisites.
Hoffman's book Banach Spaces of Analytic Functions begins with a very clear explanation
of relations
between Taylor and Fourier series.<|endoftext|>
TITLE: Is the ideal of compact operators strongly Borel?
QUESTION [6 upvotes]: Let $H$ be a separable infinite dimensional Hilbert space. Denote by $\mathcal{B}(H)$ the space of bounded operators on $H$, and $\mathcal{K}(H)$ the ideal of compact operators. When endowed with the strong (or weak) operator topology, is $\mathcal{K}(H)$ Borel in $\mathcal{B}(H)$?
Remarks: 

It is well known that, unlike in the norm topology, $\mathcal{K}(H)$ is not closed; the identity operator is a strong limit of finite rank projections.
$\mathcal{K}(H)$ is analytic: $\mathcal{K}(H)$ is Polish in the norm topology, and the inclusion map $\mathcal{K}(H)\to\mathcal{B}(H)$ is norm-weak continuous.

REPLY [7 votes]: Yes.
This can be deduced from the argument given for Corollary 3.2 in G. A. Edgar, Measurability in a Banach space, Indiana Univ. Math. J. 26 (1977), 663-677, MR542944.
The proof (attributed to Talagrand) is easy, so I reproduce it: 
Choose a countable norm-dense subset $\lbrace d_k : k \in \mathbb{N}\rbrace \subseteq \mathcal{K}(H)$ and let $B$ be the unit ball in $\mathcal{B}(H)$. Then $B$ is compact in the weak operator topology and
$$
\begin{align*}
\mathcal{K}(H) & = \bigcap_{n \in \mathbb{N}} \left(\mathcal{K}(H) + \tfrac{1}{n}B\right) \supseteq \bigcap_{n\in\mathbb{N}}\bigcup_{k \in \mathbb{N}}\left(d_k + \tfrac{1}{n}B\right) \supseteq \mathcal{K}(H)
\end{align*}
$$
shows that $\mathcal{K}(H)$ is a $K_{\sigma\delta}$ in $\mathcal{B}(H)$ with the weak operator topology.
Therefore $\mathcal{K}(H)$ is Borel with respect to all the usual topologies considered on $\mathcal{B}(H)$.<|endoftext|>
TITLE: Embedding a hypercube into the Erdos-Renyi random graph
QUESTION [8 upvotes]: Let $C_n=\{0,1\}^n$ be the hypercube and denote by $\operatorname{G}(N,p)$ the Erdos-Renyi random graph (edges appear independently with probability $p$). Assume that $N=2^n$. Could one pin down $p=p(n)$ such that we can embed the hypercube into $\operatorname{G}(N,p)$ and vice versa? Is there anything known about similar problems?

REPLY [6 votes]: And $1/4$ is indeed the correct threshold probability: Oliver Riordan proved this in 2000 ('Spanning subgraphs of random graphs') by a second moment argument. His paper is much more general, and is still the best place to look for bounds on the appearance threshold for most graphs.
Incidentally, $G_{N,p}$ does embed in $C_n$ if $p$ is small enough, but at the point where the random graph becomes acyclic (above this point it rapidly acquires many cycles, about half of which are odd and therefore don't embed) and at this point $G_{n,p}$ is not so interesting.<|endoftext|>
TITLE: Strict applications of deformation theory in which to dip one's toe
QUESTION [25 upvotes]: I hesitate to ask a question like this, but I really have tried finding answers to this question on my own and seemed to come up short. I readily admit this is due to my ignorance of algebraic geometry and not knowing where to look... Then I figured, that's what this site is for!
Here's the short of it:

What are some examples of strict applications of deformation theory? That is, what are examples of problems that can be stated without mentioning deformation theory or moduli spaces and one of whose solutions uses deformation theory? Please state the problem precisely in your answer, and provide a reference if at all possible :)

Here's the long of it:
I really want to swim in the Kool-Aid fountain of deformation theory and taste of its sweet, sweet purple love, but I'm having trouble. When I wanted to learn about K-theory, I learned about it through the solution to the Hopf invariant one problem, the solution to the vector fields on spheres problem, and through the Adams conjecture. When I wanted to learn some equivariant stuff, it was nice to have the solution to the Kervaire invariant one problem as a guiding force. I have trouble learning things in a bubble; I need at least a slight push.
Now, I know that deformation theory is useful for building moduli spaces, but the trouble is that, aside from the ones that appear in homotopy theory, I haven't fully submerged in this sea of goodness either. The exception would be any example of a strict application that used deformation theory to construct some moduli space and then used this space to prove some tasty fact.
To give you all an idea, here are the only examples I have found (from asking around) that fit my criteria:

Shaferavich-Parshin. Let $B$ be a smooth, proper curve over a field and fix an integer $g \ge 2$. Then there are only finitely many non-isotrivial (i.e. general points in base have non-isomorphic fibers) families of curves $X \rightarrow B$ which are smooth and proper and have fibers of genus $g$. 
Given $g\ge 0$, then every curve of genus $g$ has a non-constant map to $\mathbb{P}^1$ of degree at most $d$ whenever $2d - 2 \ge g$. 
There are finitely many curves of a given genus over a finite field.
The solution to the Taniyama-Shimura conjecture uses deformations of Galois reps.

1, 2, and 3 are stolen from Osserman's really great note: https://www.math.ucdavis.edu/~osserman/classes/256A/notes/deform.pdf 
I really like the theme of 'show there are finitely many gadgets by parameterizing these gadgets by a moduli space with some sort of finite type assumption, then showing no point admits nontrivial deformations.' Any examples of this sort would be doubly appreciated. (I guess Kovács and Lieblich have an annals paper where they do something along these lines for the higher-dimensional version of the Shaferavich conjecture, but since they end up counting deformation types of things instead of things, it doesn't quite fit the criteria in my question... but it's still neat!)
Galois representations are definitely a huge thing, and I'd be grateful for any application of their deformation theory that's more elementary than, say... the Taniyama-Shimura conjecture.
So yeah, that's it. Proselytize, laud, wax poetic- make Pat Benatar proud.

REPLY [7 votes]: Here are few well-known examples which are not of algebro-geometric nature, where a problem was solved via a reduction to a deformation problem/moduli space problem:

Donaldson's work on intersection forms of smooth simply-connected 4-manifolds (definite forms must be diagonalizable); the moduli space in this case is the space of instantons (self-dual connections). 
Thurston's work on hyperbolization of Haken 3-manifolds. The moduli space in question was the character variety, i.e., moduli space of $SL(2,C)$-representations of the fundamental group. The problem of hyperbolization was reduced by Thurston to a certain fixed-point problem (actually, two slightly different problems depending on existence of fibration over the circle) for a weakly contracting map and solved this way. 
Margulis' arithmeticity theorem: Every rreducible lattice in a higher rank semisimple Lie group $G$ is arithmetic. The very first step of the proof (actually, due to Selberg) is to look at the character variety, which is defined over $Z$ and observe that isolated points are fixed by a finite index subgroup of the absolute Galois group. This implies that the lattice is conjugate to a subgroup of $G(F)$, where $F$ is a number field. The moduli space in this case is again a character variety. 
Any of hundreds (if not thousands) of papers on application of gauge theory to low-dimensional ($\le 4$) topology, or even higher-dimensional topology as in Ciprian Manolescu's recent disproof of the triangulation conjecture.<|endoftext|>
TITLE: Certain partial integrations in quantum mechanics
QUESTION [6 upvotes]: In classical quantum mechanics (and specifically in the introductury texts on this topic) while calculating expectation values of certain operators in the Schrödinger approach we often have to do partial integrations such as:
$$\int_{\mathbb{R}} \overline{\psi(x)} \psi''(x) \mathrm{d}x = - \int_{\mathbb{R}} \overline{\psi'(x)} \psi'(x) \mathrm{d}x$$
or more general
$$\int_{\mathbb{R}} \overline{A\psi(x)} (B\psi)'(x) \mathrm{d}x = - \int_{\mathbb{R}} \overline{(A\psi)'(x)} B\psi(x) \mathrm{d}x$$
for certain operators $A$ and $B$.
The dropping of the boundary term which usually features partial integration is most often wrongly justified by the false belief that 
$\psi$ is square integrable implies that it is decaying to zero at infinity.
(even in cases that $\psi$ ís square integrable and does decay to zero, $\psi' \psi$ does not have to satisfy any of both conditions).  
Question: is there some "natural" condition or observation on the properties of $\psi$ (or the set of functions where it comes from) by which we can generically justify the dropping of the boundary terms in the above partial integrations?
With "natural" I mean: can we give this property a physical interpretation within the q.m. framework?
(generalisation to other domains than just $\mathbb{R}$ is of course also welcome)

REPLY [2 votes]: A key-word/phrase relevant here is "essential self-adjoint-ness": if we start with the Laplacian on test functions on $\mathbb R^n$, certainly the integration by parts formula is correct, because the boundary terms are literally zero. The "essential" self-adjointness is the (non-trivial) provable assertion that the (graph-) closure of this operator (densely defined on $L^2$) is its unique self-adjoint extension. Further, necessarily this graph-closure is identical with the Friedrichs extension. 
This does not preclude a variety of self-adjoint extensions of restrictions or of perturbations of it.
A useful auxiliary point is that, yes, while square integrability does not imply pointwise decay, various Sobolev inequalities prove growth restrictions... whose details depend on the dimension and how many derivatives we have in $L^2$. This is not exactly the same as the question of the validity of the integration by parts "extension", perhaps addressing more details than are strictly necessary to understand the self-adjointness of the Laplacian. On the other hand, indeed, that self-adjointness does not address details of pointwise behavior, such as may be needed for applications.<|endoftext|>
TITLE: Is the first-order theory (with =) of real numbers with addition and multiplication complete and decidable?
QUESTION [13 upvotes]: Due to Andreas Blass's answer to my question "Is the feasibility of a system of nonlinear, non-convex equations (inequalities) decidable?", i have now investigated real closed fields (RCF), because i had never studied them before. My background is logic-based cognitive robotics, not model theory or foundations of mathematics.
So, the theory of RCF is complete and decidable, but i'm still not sure whether my problem language is a RCF:
Literature about RCFs rarely give examples. I would like to know the exact definition of the theory of real numbers which is a RCF.
On the Wikipedia page Tarski's axiomatization of the reals, it is mentioned that Tarski showed that the theory of real-closed fields completely axiomatizes the first-order theory of the structure $\langle \mathbb{R}, +, *, < \rangle$.
However, i need to know whether a certain fragment of the first-order logic is decidable, or equivalently, whether a certain class of systems of equations (including the products of at most two variables at a time) is feasible.
Is $\langle \mathbb{R}, +, -, *, < \rangle$ (where $\mathbb{R}(x)$ is the predicate stating that $x$ is a real number, or $\mathbb{R}$ is the set of real numbers) a model in the RCF?
Is the first order language employed in the theory assumed to include equality (=)?
So, do sentences like the following fall within the RCF (assume $x$ and $y$ with subscripts are real number variables)?
$(\exists x_1,x_2,x_3,y_1,y_2,y_3)\quad (x_1*y_2 + x_2*y_3 = 0.223) \; \land$
$\quad \lnot( x_1*y_1 + x_3*y_2 = 0.928) \; \land $
$\quad ((x_1 + x_2 + x_3 = 1) \lor (x_1 + x_2 + x_3 = 0))$.
Moreover, is truth of sentences of this kind decidable and complete in the structure $\langle \mathbb{R}, +, -, *, < \rangle$?
Andreas hinted at to the affirmative, but due to my unfamiliarity with the subject, i need to confirm it. I would appreciate a reference which plaintly states that the first-order theory (with =) of real numbers with addition and multiplication is complete and decidable.

REPLY [23 votes]: Yes, all those assertions are expressible in the language of real closed fields, and all such assertions are decidable by Tarski's theorem. You can refer to addition, subtraction, multiplication, division, etc., plus the order; you can refer to any specific rational number (such as .223 etc.) because you have $1$ in the language and can therefore form any rational number; you can form any polynomial and indeed any rational function, with rational coeffficients, in any specific finite number of variables; and you can quantify over the real numbers and apply any kind of logical connective.
What Tarski proved is that in fact every such assertion is equivalent to a (much longer) quantifier-free assertion, and this is ultimately why they are decidable. 
Theorem. (Tarski) The first-order theory of real-closed fields, which is the same as the theory of the structure $\langle\mathbb{R},+,\cdot,0,1,\lt\rangle$ is complete and decidable, and admits quantifier-elimination. 
What you cannot do while remaining under the decidability result is quantify over the integers or the rational numbers. So you cannot necessarily decide questions like, "does this specific polynomial $p(\vec x)$ have an integer solution?". In particular, if you add a predicate for the integers to the structure, forming $\langle\mathbb{R},+,\cdot,\mathbb{Z},0,1,\lt\rangle$, where $\mathbb{Z}$ is a predicate that is true of exactly the real numbers that are integers, then the theory is no longer decidable. For this reason, the theory of the structure $\langle\mathbb{R},+,\cdot,\sin(x),0,1,\lt\rangle$ is not decidable, because in it we can define the integers. So while you can refer to specific polynomial or rational functions over the integers, you may not quantify over the class of such polynomials.
Similarly, you also cannot quantify over the dimension (which amounts to an integer quantifier), and ask something like: does every upper triangular $n\times n$ matrix (for every $n\geq 1$) have a certain property? That is, in order to stay within Tarski's language, you can in effect quantify over $\mathbb{R}^3$ or $\mathbb{R}^{86}$ or $\mathbb{R}^n$ for a specific $n$ by using $n$ individual real quantifiers, but you cannot treat $n$ itself as a variable here. 
My opinion---probably overblown---is that Tarski's theorem is one of the great pinnacles of mathematical achievement. One of the consequences of his theorem, for example, is that Cartesian geometry is decidable. Tarski's theorem provides an algorithm that will decide by rote any question you can put to it about lines, circles, ellipses, cones, spheres, ellisoids, planes, cubes and so on, in any specific dimension. All such questions are expressible in the language of real-closed fields. (OK, I admit that the algorithm takes double exponential time, and is not feasible. But still, I find the existence of such an algorithm for a topic that has been studied by mathematicians for thousands of years to be amazing.)<|endoftext|>
TITLE: Zariski's main theorem in the complex analytic category
QUESTION [7 upvotes]: Hello,
I am looking for a reference to something like that: if $f\colon X\to Y$ is a finite (i.e., proper with finite fibers) morphism of reduced and irreducible normal (or at least smooth) complex spaces such that $f$ is 1-1 over $U\subset Y$, where $U$ is open and dense, then $f$ is an isomorphism.
Could somebody help me?
Thanks in advance,
Serge

REPLY [6 votes]: One reference is Proposition 14.7 in Remmert's paper Local Theory of complex analytic spaces, Several complex variable VII, Encyclopaedia of Math. Sci. vol 74. For the reader's convenience I will restate the result here.
Recall that a  finite, surjective, holomorphic map $\eta \colon X \to Y$ between complex spaces is called a one sheeted (analytic) covering if there exists a  (critical) thin set $A \subset Y$ such that $\eta^{-1} A$ is thin in $X$ and $\eta \colon X \setminus \eta^{-1}(A) \to Y \setminus A$ is biholomorphic. Then Remmert's statement essentially says that normalizations "dominate" all one-sheeted coverings:

Proposition. Let $\eta \colon X \to Y$, $\xi \colon Z \to Y$ be one-sheeted coverings. If $X$ is normal, there exists a unique holomorphic map $g \colon X \to Z$ such that $\eta= \xi \circ g$. If moreover $Z$ is normal, the map $g$ is biholomorphic.

In your setting, take $\eta=f$ and $g=1_Y$. Then if both $X$ and $Y$ are normal Remmert's Proposition tells us that $f$ is a biholomorphism.<|endoftext|>
TITLE: Are principal bundles isotrivial?
QUESTION [7 upvotes]: Let $U$ be a $k$-scheme, where $k$ is a field. Let $G$ be a smooth affine $k$-group. Recall that a principal $G$-bundle over $U$ is a smooth surjective $U$-scheme $E$ with an action of $G$ on $E$ such that the action commutes with the projection to $U$ and the obvious morphism $G\times E\to E\times_UE$ is an isomorphism.
It is easy to see that every principal $G$-bundle is locally trivial in etale topology. However Serre in his Espaces fibres algebriques required a stronger condition called local isotriviality: for all $u\in U$ there is a a finite etale morphism $T'\to T$, where $T$ is a Zariski neighborhood of $u$ such that the pull-back of $E$ to $T'$ is trivial. I wonder if these definitions are known to coincide (I think I can prove it in some generality but the proof is not trivial).
Sketch of a proof. Let us embed $G$ into $GL(n)$. Consider the associated space $E'=E\times^GGL(n)$. It is a scheme because $GL(n)$ is affine and affine schemes can be glued in any reasonable topology. Moreover, it is a principal $GL(n)$-bundle, so passing to a Zariski cover we can assume it is trivial. The original bundle can be obtained from E' via reduction of the structure group, that is, it is a pull-back of the principal $G$-bundle $GL(n)\to GL(n)/G$. It remains to use the isotriviality of the latter bundle.

REPLY [4 votes]: This answer is coming late, but since I have also been struggling to find a reference, I hope this can be helpful to other people.
The answer is yes. Precisely, you can found a proof in 
Raynaud, Michel
Faisceaux amples sur les schémas en groupes et les espaces homogènes. (French)
Lecture Notes in Mathematics, Vol. 119 Springer-Verlag, Berlin-New York 1970 ii+218 pp. 
http://link.springer.com/book/10.1007%2FBFb0059504
Lemma XIV 1.4
Let $k$ be a field $G/k$ a smooth affine algebraic group $X/k$ a scheme $P$ a fpqc $G_X$-torsor. Then $P$ is representable and $P$ is locally isotrivial.
Remarks :
0) in fact "semi-locally isotrivial" in the original statement but this implies locally isotrivial,
1) the principle of the proof is the one you give,
2) this seems due to A.Grothendieck,
3) this is false is $G$ is not affine. There is a classical example also in Raynaud's book (XIII 3.1) where $X$ is a nodal curve and $G$ an abelian variety, see also 
Brion, Michel
Some basic results on actions of nonaffine algebraic groups. (English summary) Symmetry and spaces, 1–20,
Progr. Math., 278, Birkhäuser Boston, Inc., Boston, MA, 2010. 
and remark 3.1 in
Brion, Michel(F-GREN-F)
On automorphism groups of fiber bundles. (English summary)
Publ. Mat. Urug. 12 (2011), 39–66. 
5) your definition of a torsor is a bit strange.<|endoftext|>
TITLE: Cellular model structures on continuous functors
QUESTION [8 upvotes]: The category of enriched functors from finite based CW complexes to based topological spaces 
has a projective model structure. The fibrations
are the objectwise Serre fibrations and the weak equivalences are the objectwise 
weak homotopy equivalences. 
The cofibrations are defined by the 
left lifting property. In particular every cofibration is an objectwise Hurewicz cofibration
of based spaces, but probably not an objectwise Serre cofibration. 

Is this model structure cellular in the sense of Definition 12.1.1 of Hirschhorn's Model Categories and Their Localisations? It is known to be cofibrantly generated.

REPLY [5 votes]: I am glad that by putting our heads together in Munster this week we finally have a proof and can answer this question (a full 2 years after it was asked!). The answer is yes, and I want to sketch here the email I sent you in case it might be of use to others who are trying to prove things are cellular. Let Top mean compactly generated spaces. Let I (and J) denote the generating (trivial) cofibrations of Top$_\ast$. Let W(I) (and W(J)) denote the generating (trivial) cofibrations of your category of functors, so these sets are generated by functors of the form CW$_\ast(X,-) \wedge i^n_+$ where $i^n_+$ runs through the maps in I (resp. J). We have three conditions to check.
(1) We must show the domains of maps in W(J) are small (for some $\kappa$) relative to cofibrations in Fun(CW$_\ast$,Top$_\ast$). This is easy, because Top is cellular, so domains of maps in J are small relative to cofibrations and the representable functors CW$_\ast(X,-)$ preserve filtered colimits (since finite CW complexes are compact, i.e. small relative to $\kappa = \aleph_0$). So smallness of domains of J relative to I-cell implies smallness of domains of W(J) relative to WI-cell. An easier way to see this is to note that maps in W(J) are objectwise inclusions, colimits are computed objectwise in Fun(CW$_\ast$, Top$_\ast$), and all spaces are small relative to inclusions.
(2) We must show cofibrations are effective monomorphisms. We know the generators are objectwise closed inclusions, i.e. for all $Y$, $CW_\ast(X,Y)\wedge i_\ast^n$ is a closed inclusion, since it's a compact space smashed with a closed inclusion. But in Top, closed inclusions are effective monomorphisms (indeed, these two classes coincide!) so we're done.
(3) We must show the domains and codomains of $W(I)$ are compact in the sense of Hirschhorn relative to $W(I)$. This is somewhat painful. It means that there is some regular cardinal $\kappa$ such that for every relative $W(I)$-cell complex $\eta:F\to T$ (a map in our category, hence a natural transformation) and for every presentation of $\eta$ by a chosen collection of cells then every natural transformation $CW(X,-)\wedge K \to T$, where $K$ is a sphere or disk, factors through a subcomplex of size at most $\kappa$. A presentation of $\eta$ is a realization of $\eta$ as the colimit of a $\lambda$-sequence of maps which are pushouts of coproducts of cells (i.e. maps in $W(I)$). A subcomplex of the given presentation of $\eta$ is a $\lambda$-sequence formed by pushouts of coproducts of a subset of cells. The size is the cardinality of the set of cells. For us $\kappa$ will be taken to be larger than $\gamma_+$ where $\gamma$ is the cardinal making Top cellular. We'll follow the roadmap given by Hovey's proof of Proposition A.8 in his paper on Spectra and Symmetric Spectra in General Model Categories. In particular, we'll use his reduction that it is sufficient to consider cells which are transfinite compositions of pushouts of maps in $W(I)$ rather than coproducts of maps in $W(I)$.
The idea is to reduce to considering maps in Top, use the cellularity of Top to identify which cells we need in our sub-presentation of $T$, then prove that $\eta$ actually factors through that sub-presentation. First, evaluate $\eta$ on a given $Y$ and get $\eta_Y:CW(X,Y)\wedge K \to T(Y)$. Be careful here: it is NOT true that CW(X,Y) is cofibrant (see e.g. this MO question). However, with the compact open topology these CW(X,Y), are compact (this follows from Ascoli's theorem, for example), hence so are the spaces $CW(X,Y)\wedge K$, since K is also compact.
Write $T(Y)$ as the filtered colimit of the cells that make up $T$, evaluated at $Y$, e.g. as a transfinite composition of pushouts of maps $C_{\beta}(Y)\to D_{\beta}(Y)$. Note that Hovey demonstrates it's enough to consider such pushouts rather than pushouts of coproducts of maps in $W(I)$. The filtered colimit building $T(Y)$ consists of closed inclusions because cofibrations in Fun(CW$_\ast$,Top$_\ast$) are objectwise closed inclusions. Since we're working in compactly generated spaces, closed inclusions are automatically closed $T_1$-inclusions, so Hovey's 2.4.2 tells us that maps into such a colimit $T(Y)$ from a compact space factors through a finite subcomplex. This implies only finitely many cells have been used.
Use Hovey's method of writing the $\lambda$-sequence as a retract of another $\lambda$-sequence with the same cells (technically, a set of cells in bijection with your cells), but now formed out of maps in $I$ rather than maps in $W(I)$ evaluated at $Y$. This relies on Hirschhorn 10.5.25 (beware: Hovey was referencing a different draft of Hirschhorn's book than the published one, so his numbering does not match Hirschhorn's), but at the end allows you to apply cellularity in Top to find a subcomplex with lesser cardinality through which $\eta_Y$ factors, though verifying this carefully requires a transfinite induction following the same steps as in Proposition A.8. Let $S_Y$ be the set of $W(I)$ cells which appear (evaluated at $Y$) in the subcomplex for $\eta_Y$.
Next, form a subcomplex $R$ of $T$ which contains all the cells $S_Y$. We know this is a subcomplex, because for each $Y$ there were fewer than $\kappa$ many cells and only had the cardinality of $CW$ many Y’s, which is much less than $\kappa$. Next, use that $R(Y)$ contains all the cells in $S_Y$ so contains the subcomplex through which $CW(X,Y)\wedge K \to T(Y)$ factored through. This provides a map of functors $CW(X,-) \wedge K \to R$, though we don't automatically know it's a natural transformation. We can make it natural by "going further along if necessary", which I now make rigorous. Let $f:Y \to Z$ in CW, i.e. $f$ is a cellular map. We have to check that the naturality square commutes in Top (i.e. that $\eta_Y \circ CW(X,f) \wedge K = R(f) \circ \eta_X$). Because we started with a natural transformation, we know that the requisite naturality diagram commutes eventually in the colimit defining $T$ (i.e. replacing $R$ by $T$ above). Because f has a bounded number of cells (finite even), we can make this occur in $R$ by adding in finitely many more cells to $R$ from our presentation of $T$ (using here that we know the square will commute eventually). Even if you have to add these finitely many more cells for each map f, you still won’t reach the regular cardinal $\kappa$, so $R$ will still be a subcomplex and by construction will admit a natural transformation from $CW(X,-)\wedge K$ factoring $\eta$. This proves the domains are compact as required.
Incidentally, I think this proof method can be used to prove something more general, namely a statement of the form: "if $F:\mathcal{M} \leftrightarrows \mathcal{N}:G$ is an adjunction through which the model structure on $\mathcal{M}$ is transferred to $\mathcal{N}$, and if the model structure on $\mathcal{M}$ is cellular, and if $F$ and $G$ behave sufficiently well with respect to filtered colimits, then $\mathcal{N}$ is cellular.'' I'll try to write this up formally and fold it into a future paper.<|endoftext|>
TITLE: Topology on the Unitary Dual
QUESTION [10 upvotes]: Suppose I have a locally compact topological group G. The unitary dual of G is the set of equivalence classes of irreducible unitary representations of G. Now, it seems to me that the sensible way of putting a topology on this space is as follows:

Fix a hilbert space Hn of cardinality n.
Consider the set R(G,Hn), the set of unitary representations of G on Hn. We can give it the topology of uniform convergence on compact sets. Specifically, reps pn approach p if for any compact K in G and v in Hn, pn(g)v -> p(g)v uniformly on K.
Now take the subspace I(G,Hn) of irreducible representations, with the subspace topology. Then quotient by unitary equivalence, and give the resulting space the quotient topology. 
Finally, take a disjoint union over all (countable) n.

I am not sure, however, if this is commonly done. The popular topology on the unitary dual seems to be the Fell topology. Is what I described equivalent? If not, what advantages does the Fell topology have? Also, there is the perspective that the unitary dual is more importantly a measure space than a topological space- is a topological structure significant or important?
Thanks.

REPLY [3 votes]: As you pointed out in a comment, with the topology that you describe the space of $n$-dimensional representations is closed. This means that the unitary dual with that topology would be just the disjoint union of pieces, each corresponding to a dimension $n$. On the contrary, the nice aspect of the Fell topology is that it mixes all representations with all dimensions, finite and infinite.
It makes sense in the Fell topology to say that a sequence of representations of various dimensions converge to the trivial (one-dimensional) representation. In fact, it is a serious question to ask if the trivial representation is isolated in this topology or not: for discrete groups, that's the definition of Property T.
It also makes sense to say that some finite-dimensional representations converge to some infinite-dimensional ones. This happens for instance here in an elaborate context, but there are also various simple examples. Fell's topology is just a weak topology that considers all the representations as a whole.<|endoftext|>
TITLE: Consistency strength of projective determinacy (PD)
QUESTION [12 upvotes]: Let PD stand for projective determinacy, and consider the two claims:
(1) For each n=1,2,..., Con(ZFC+PD) implies Con(ZFC + there are n Woodin cardinals)
(2) Con(ZFC+PD) implies Con(ZFC + there are infinitely many Woodin cardinals).
Clearly (2) implies (1).
As far as I know, (1) is true, and I have seen casual claims that (2) is true, but I am not sure such claims can be substantiated.
Can someone confirm these please?  References to the literature would be of great help.

REPLY [19 votes]: (2) is false. (1) is true. In fact, $\mathsf{ZFC}+\mathsf{PD}$ (with $\mathsf{PD}$ stated as an axiom schema) implies that for every $n$ there is an inner model of $\mathsf{ZFC}$ with $n$ Woodin cardinals. Also, $\mathsf{ZFC}+\mathsf{PD}$ is equiconsistent with the theory that result from adding to $\mathsf{ZFC}$ the axiom schema, the $n$-th axiom of which states that there are $n$ Woodin cardinals. 
This theory, in turn, is strictly weaker than $\mathsf{ZFC}+$"there are infinitely many Woodin cardinals", where of course the latter can be stated as $\forall n\,($there are $n$ Woodin cardinals$)$. This theory is equiconsistent with $\mathsf{ZFC}+L(\mathbb R)\models\mathsf{AD}$ or, if you prefer, with $\mathsf{ZF}+\mathsf{AD}$. 
You can think of the difference this way: $\mathsf{PD}$ is only determinacy for sets of reals in $L_{\omega+1}(\mathbb R)$, which is far from determinacy for all sets of reals in $L(\mathbb R)$. 
The paper 

Peter Koellner, and W. Hugh Woodin. Large cardinals from determinacy. In Handbook of set theory. Vols. 1, 2, 3, Matthew Foreman, and Akihiro Kanamori, eds, pp. 1951–2119, Springer, Dordrecht, 2010. MR2768702,

states the difference nicely, in Theorems 8.2 and 8.3:

8.2 Theorem. The following are equivalent:
   1. $\mathsf{PD}$ (Schematic). 
   2. For every $n \lt \omega$, there is a fine-structural, countably iterable inner model $M$ such that $M\models$ There are $n$ Woodin cardinals.
8.3 Theorem. The following are equivalent: 
   1. $\mathsf{AD}^{L(\mathbb R)}$.
   2. In $L(\mathbb R)$, for every set $S$ of ordinals, there is an inner model $M$ and an $\alpha<\omega_1$ such that $S\in M$ and $M \models \alpha$ is a Woodin cardinal. 

This section of the paper is a survey, but proofs of this can be found in several places. I suggest

Ralf Schindler, and John Steel. The core model induction. Preprint, April 12, 2013,

though it has many prerequisites and is perhaps not the easiest source for these results, but it illustrates precisely the role of inner model theory at the level of Woodin cardinals in establishing these equivalences (or any consistency strength lower bounds that lie at the level of determinacy), and how both $\mathsf{PD}$ and $\mathsf{AD}^{L(\mathbb R)}$ are but natural stopping points of a much longer hierarchy.<|endoftext|>
TITLE: Topological Generalization of Whitney's Extension Theorem
QUESTION [9 upvotes]: From Wikipedia:

In mathematics, in particular in mathematical analysis, the Whitney extension theorem is a partial converse to Taylor's theorem. Roughly speaking, the theorem asserts that if $A$ is a closed subset of a Euclidean space, then it is possible to extend a given function off $A$ in such a way as to have prescribed derivatives at the points of $A$. It is a result of Hassler Whitney. A related result is due to McShane, hence it is sometimes called the McShane–Whitney extension theorem.

I need a stronger theorem which allows for both the set and the function to vary continuously, and for the resulting extended function to vary continuously.
Consider Euclidean space $\mathbb R^d$ (for any $d \ge 1$), and let $\mathcal K$ denote the metric space of compact subsets of $\mathbb R^d$, equipped with the Hausdorff metric. Select $\alpha \in \mathbb N$, and let $X = C^\alpha(\mathbb R^d, \mathbb R)$ be the Fréchet space of $C^\alpha$-smooth functions. 
For any compact set $K \in \mathcal K$, let $X_K = C^\alpha(K, \mathbb R)$ be the Banach space of $C^\alpha$-smooth functions defined on $K$, and let $\rho_K : X \to X_K$ be the restriction map. 
Define $Y_K$ to be the Fréchet space of functions which are real analytic on $\mathbb R^d - K$, and $C^\alpha$-smooth on $K$. Let $\iota_K : Y_K \to X$ be the inclusion map, which forgets that a function is analytic and just preserves its $C^\alpha$-smooth character.
The classical Whitney theorem implies that for each $K \in \mathcal K$, there is a function $w_K : X_K \to Y_K$ for which the following diagram commutes: $$\begin{array}[ccc] ~Y_K &  \rightarrow^{\iota_K}  & X \\  \uparrow {w_K} & & \downarrow {\rho_K} \\ X_K & \rightarrow_{1_{X_K}} & X_K \end{array}$$
where $1_{X_K}$ denotes the identity function on $X_K$. Equivalently, $1_{X_K} = \rho_K \circ \iota_K \circ w_K$. This means that we may extend a function from $K$ analytically, embed it into $X$, then restrict back to $K$, and we're left with the original function.

I would like to ensure not only that the Whitney map is continuous in the extending function, but also that the compact set may be allowed to vary. 
Define the coproduct $\mathcal X = \coprod_{K \in \mathcal K} \{K\} \times X_K$, and refine the topology so that the projection map $\mathcal X \to \mathcal K$ is continuous. This way, $\mathcal X$ is kind of like a fibration over $\mathcal K$, except the fibers aren't necessarily homotopically equivalent. A point in $\mathcal X$ encodes both a compact set $K$ and a $C^\alpha$-smooth function defined on that set. (I think this is an admissible construction topologically. Tell me if this is not well-defined.)
Similarly, let $\mathcal Y = \coprod_{K \in \mathcal K} \{K\} \times Y_K$ encode the analytically extended functions, with continuous projection map $\mathcal Y \to \mathcal K$.
Consider the product space $\mathcal K \times X$, along with the inclusion map $\iota : \mathcal Y \to \mathcal K \times X$ and the restriction map $\rho : \mathcal K \times X \to \mathcal X$.
Now for the Whitney-type question. Does there exist a continuous map $w : \mathcal X \to \mathcal Y$ so that the following diagram commutes? $$\begin{array}[ccc] ~\mathcal Y &  \rightarrow^{\iota}  & \mathcal K \times X \\  \uparrow {w} & & \downarrow {\rho} \\ \mathcal X & \rightarrow_{1_{\mathcal X}} & \mathcal X \end{array}$$

REPLY [4 votes]: This is not an answer but a suggestion of how to formulate the question.
I will simplify things as follows.  Let $B$ be a closed ball in $\mathbb{R}^d$ (or $B$ could be be any other compact metric space).  Let $\mathcal{K}$ be the space of closed subsets of $B$ with the Hausdorff metric.  Let $\mathcal{X}$ be the set of pairs $(K,f)$, where $K\in\mathcal{K}$ and $f:K\to [-1,1]$ is continuous.  Let $\mathcal{Y}$ be the set of pairs $(K,f)$ where $K\in\mathcal{K}$ and $f:B\to [-1,1]$ is continuous.  There is a map $\rho:\mathcal{Y}\to\mathcal{X}$ given by $\rho(K,f)=(K,f|_K)$, and the Tietze extension theorem says that this is surjective.
For any point $(K,f)\in\mathcal{X}$ we can put 
$$ G(f)=\{(x,f(x)):x\in K\}\subset K\times [-1,1], $$
and define $d((K_0,f_0),(K_1,f_1))$ to be the Hausdorff distance between $G(f_0)$ and $G(f_1)$.  This defines a metric on $\mathcal{X}$.  We can introduce a similar metric on $\mathcal{Y}$, and I think it is equivalent to the obvious product metric.  The map $\rho$ is continuous, and one can ask whether there is a continuous section.  I think that that is true.  Indeed, one just needs to show that the ingredients in the standard proof of the Tietze extension theorem depend continuously on the input data with respect to suitable metrics.  
One of the main ingredients is Urysohn's lemma, which says that if $P$ and $Q$ are disjoint closed subsets of $B$ then there is a continuous map $f:B\to [0,1]$ with $f=0$ on $P$ and $f=1$ on $Q$.  As we have a metric on $B$ we can just take $f(x)=d(x,P)/(d(x,P)+d(x,Q))$, and this depends continuously on the pair $(P,Q)$.
The other main ingredient is the map $\phi_a:\text{Map}(B,[-1,1])\to\mathcal{K}$ given by $\phi_a(f)=f^{-1}([a,1])$.  Given $f$ and $\epsilon>0$ I think one can find $\delta>0$ such that the sets $d(f^{-1}([a-\delta,1]),f^{-1}([a+\delta,1]))<\epsilon$, and then $d(\phi_a(f),\phi_a(g))<\epsilon$ whenever $d(f,g)<\delta$.  Thus, $\phi_a$ is also continuous.
For the Whitney theorem you probably want to define $J_\alpha(f)(x)\in\mathbb{R}^N$ (for suitable $N$) to be the vector of all derivatives of $f$ at $x$ of orders less than or equal to $\alpha$, then let $G_\alpha(f)\subset K\times\mathbb{R}^N$ be the graph of $J_\alpha(f)$.  You can then use the Hausdorff distance between these sets as a metric on $\mathcal{X}$.  If you are willing to work with $C^\alpha$ extensions rather than real analytic ones then there is an obvious parallel metric on $\mathcal{Y}$.  This could probably be adapted to cover the analytic case as well.<|endoftext|>
TITLE: The modular arithmetic contradiction trick for Diophantine equations
QUESTION [19 upvotes]: It is a slick, and seemingly ad-hoc, technique often used to prove that a Diophantine equation has no solutions.
The equation $f(x_1,\ldots, x_k)=0$, with variables $x_i\in\mathbb{Z}$ and some elementary function $f$, is taken modulo some ingeniously chosen $n\in\mathbb{Z}^+$. Then by evaluating $f$ on the $k^n$ possible $k$-tuples $(x_1,\ldots,x_k)$ of residues modulo $n$, it is shown that $0$ is achieved in no case. Thus no solution exists in $\mathbb{Z}$ either.
A traditional classroom example is $x^2+y^2-3z^2=0$. Assume $\gcd(x,y)=1$ without loss of generality, and use modulo $3$.
The trouble is in coming up with the brilliant $n$. An example of a good heuristic principle is choosing modulo $20$ if there are powers of $4$ as, out of to $20$ possible residues, the only quartic residues are $0,1,5,16$. Are there similar mods for other powers?
In general, has there been any work done on an objective way of finding $n$? Perhaps good heuristics, if nothing else? Restrict $f$ to some non-trivial case, say certain polynomials of given degree, if needed. Paper references would be nice.

REPLY [16 votes]: Alex says in his answer that for a polynomial that is an integral or rational quadratic form, congruence conditions explain completely when it has no rational zeros (other than $(0,\dots,0)$). Congruence conditions need not be sufficient to explain a lack of integral  solutions in some other situations. 

Consider $2x^2 + 7y^2 = 1$. It has no integral solutions, but has the rational solutions $(1/3,1/3)$ and $(3/5,1/5)$.  Because the denominators in the two solutions are relatively prime, they can be used to produce solutions to $2x^2 + 7y^2 \equiv 1 \bmod m$ for any $m \geq 2$. (See Diophantine equation with no integer solutions, but with solutions modulo every integer for more examples in this direction.)
Consider a Mordell equation $y^2=x^3+k$ with $k$ a nonzero integer. Depending on $k$ there may or may not be an integral solution (e.g., there is if $k = 1$ and there is not if $k = 6$), but you can't rule out the possibility of an integral solution using modular arithmetic alone since $y^2 \equiv x^3 + k \bmod m$ has a solution for every $m \geq 2$.
It suffices by the Chinese remainder theorem to show $y^2 \equiv x^3 + k \bmod p^r$ is solvable for every prime power $p^r$, or what amounts to the same thing, there is a $p$-adic integer solution for every prime $p$. There is a 2-adic integer solution $(x,k+1)$ for some $x$ and a 3-adic integer solution $(1−k,y)$ for some $y$. In the 5-adic integers there is a solution $(0,y)$ if $k \equiv 1, 4 \bmod 5$, $(1,y)$ if $k \equiv 0, 3 \bmod 5$, and $(-1,y)$ if $k \equiv 2 \bmod 5$. In the 7-adic integers there is a solution $(0,y)$ if $k≡1,2,4 \bmod 7$, $(1,y)$ if $k≡0,3 \bmod 7$, $(−1,y)$ if $k≡5 \bmod 7$, and $(x,0)$ if $k ≡ 6 \bmod 7$. For $p \geq 11$, let $N_p$ be the number of mod $p$ solutions to $y^2 \equiv x^3 + k \bmod p$, so $N_p=p+S_p$ where $S_p=0$ if $p|k$ and $|S_p| \leq 2\sqrt{p}$ by the Hasse bound if $(p,k)=1$. Then $N_p \geq 4$, so there is a solution to $y_0^2 \equiv x_0^3 + k \bmod p$ where $y_0 \not\equiv 0 \bmod p$, and this solution lifts $p$-adically to $(x_0,y)$ for some $p$-adic integer $y$.<|endoftext|>
TITLE: algebraic de rham cohomology of a curve
QUESTION [8 upvotes]: Let $X$ be a smooth projective curve over a field $k$ of characteristic zero. The algebraic de Rham cohomology of $X$ is, by definition, the hypercohomology of the complex of Kähler differentials for the Zariski topology:
$H^k_{DR}(X)=\mathbb{H}(X, \mathcal{O}_X \to \Omega^1_X \to \Omega^2_X \to \cdots)$ 
In "Hodge cycles on abelian varieties", p. 24, Deligne claims
"For a complete smooth curve $X$ and an open affine subset, the map
$H^1_{DR}(X) \to \Gamma(U, \Omega^1_X) / d\Gamma(U, \mathcal{O}_X)$
is injective with image the set of classes represented by forms whose residues are all zero (such forms are said to be of the second kind)."
I have several questions regarding this quote: 
1) How does one prove the statement?
2) How to use this to determine $H^1_{DR}(X)$?
3) I'm a bit confused with the terminology "second kind". I thought this was reserved for $H^1(X, \mathcal{O}_X)$ whereas "first kind" are differentials in $H^0(X, \Omega^1_X)$. What does it mean?
4) related to 3) how does one see the Hodge decomposition in this setting?
Any help would be appreciated. Thanks a lot!

REPLY [8 votes]: For question 1: consider de localization long exact sequence
\begin{equation}
...\to H^1_{DR,Z}(X)\to H^1_{DR}(X)\to H^1_{DR}(U)\to ... 
\end{equation}
where $H^1_{DR,Z}(X)$ is the de Rham cohomology with support on $Z=X\setminus U$. Since $Z$ has dimension $0$ we can prove that $H^1_{DR,Z}(X)$ (which is Poincaré dual to $H^1_{DR}(Z)$) is trivial. Then you just have to note that the de Rham cohomology of $U$ is given by the cohomology of the complex $\Gamma(U,\Omega^\bullet_{U/k})$ since $U$ is affine. 
You will find all the details in the paper by Hartshorne: \emph{On the de Rham cohomology of algebraic varieties}, Pub. Math. IHES 1975.
For question 2: let me just note that from the above sequence you get the following exact sequence
\begin{equation}
0\to H^1_{DR}(X)\to H^1_{DR}(U)\to H^0_{DR}(Z)\to H^2_{DR}(X)\to 0 \ .
\end{equation}
For question 3: I don't know the answer. What I know is that the de Rham cohomology of $U$ can be computed using the de Rham complex of differentials forms on $X$ with log poles along the complement $Z$. This means that a class 
\begin{equation}
[\omega]\in H^1_{DR}(U)=\Gamma(U,\Omega^1_{U/k})/d\Gamma(U,\Omega^0_{U/k})
\end{equation}
can be represented by a 1-form $\omega\in \Gamma(U,\Omega^1_{U/k})$ that we can extend to a meromorphic 1-form on $X$ having only simple poles at any point of $Z$. In particular you can take $\omega$ to be holomorphic on $X$. From what I understand the injection of point 1 gives you all the classes $[\omega]$ such that the sum of the residues of $\omega$ at the points of $Z$ is $0$.<|endoftext|>
TITLE: A Sequence of Real numbers 
QUESTION [5 upvotes]: Consider the sequence $\lbrace \frac{\phi(i)}{i}\rbrace_{i=1}^\infty$ where $\phi$ is the Euler's function. The Sequence is clearly dense in $[0,1]$. What can be said about the limsup of its average sequence ?? I mean the sequence $\frac 1n\sum_{i=1}^n \frac{\phi(i)}{i}$. Its value or an upper bound would be helpful.

REPLY [7 votes]: It is well known that $\sum_{k\le x} \phi(k)=\frac{3}{\pi^2}x^2+O(x\log(x))$. In a similar way we obtain
$$
\sum_{k\le x} \frac{\phi(k)}{k}=\frac{6}{\pi^2}x+O(\log(x)),
$$
by using the Moebius function, i.e., $\sum_{k\le x} \frac{\phi(k)}{k}=\sum_{k\le x}\sum_{d\mid k}\frac{\mu(d)}{d}=\sum_{d\le x}\frac{\mu(d)}{d}\lfloor \frac{x}{d}\rfloor$. Then
$$
\sum_{k\le x} \frac{\phi(k)}{k}=x\sum_{d\le x}\frac{\mu(d)}{d^2}-\sum_{d\le x}\frac{\mu(d)}{d}((x/d))=\frac{6}{\pi^2}x+O(\log(x)).
$$
Edit: The term $O(1)$ (which I hade before from the paper of Carella) is not correct, please see the valuable comment of Peter. 
The value $6/\pi^2$ is  reasonably good for the question posed, if $n$ is large.<|endoftext|>
TITLE: Solvability of finite groups of order coprime to 15 -- proof without using CFSG?
QUESTION [8 upvotes]: Is the solvability of finite groups of order coprime to 15
essentially easier to prove than the entire Classification of Finite Simple Groups?

REPLY [4 votes]: As Derek Holt has pointed out, the answer to the question is yes. --
Thompson proved that the only finite simple groups of order coprime to 3 are the
Suzuki groups, and Glauberman later extended this to a classification of
simple groups that do not have ${\rm S}_3$ as a subgroup.
Both of these results are pre-classification, though they might not have been published.<|endoftext|>
TITLE: series representation in injective tensor products
QUESTION [6 upvotes]: All books on tensor products of Banach spaces contain the well-known theorem of Grothendieck that every element of the completed projective tensor product 
$X \tilde{\otimes}_ \pi Y$ has a representation as a series $\sum\limits_{n=1}^\infty x_n \otimes y_n$ which converges with respect to the $\pi$-norm (in an appropriate sense, this is even uniform for compact sets).
Knowing this, it is most natural to ask whether the same is true for the injective tensor product $X \tilde{\otimes}_\varepsilon Y$. 
The only thing I have found in this direction is that if $X$ and $Y$ have Schauder bases $(e_n)_{n\in\mathbb N}$ and $(f_n)_{n\in\mathbb N}$ then one can order $e_n \otimes f_m$ in a suitable way to obtain a Schauder basis of the injective (as well as the projective) tensor product. This of course answers the question and I believe that it would be enough that one of the spaces has a Schauder basis.
Moreover, it seems that the question rather easily reduces to the following problem about finite rank operators between the Banach spaces $Y^*$ and $X$: Can every finite rank operator $T$ be written as a sum $\sum\limits_{k=1}^n T_k$ of one-dimensional operators such that for all $m\le n$
$$ \| \sum\limits_{k=1}^m T_k\| \le c \| T\| $$
(where the constant $c$ is independent of $T$)?

REPLY [5 votes]: Given $T: Y\to X$ of finite rank,  let $(x_i,x_i^*)$ be an Auerbach basis (meaning they are biorthogonal and both the vectors and the biorthogonal functionals all have norm one) for the range of $T$.  For $1\le i \le n$ let $T_i y  = n^{-1} x_i^*(Ty) x_i$. For $j = kn +i$,  $1 \le k \le n$ and $1\le i \le n$, let $T_j = T_i$.<|endoftext|>
TITLE: Checking whether modules are isomorphic, via a computer algebra software
QUESTION [5 upvotes]: Hi
Let $R = K[X_1,\ldots, X_n]$ where $K$ is a computable field.
Suppose we are given two modules with presentations
$$ R^n \rightarrow R^m \rightarrow M \rightarrow 0 $$
and
$$ R^l \rightarrow R^p \rightarrow N \rightarrow 0 $$
Then is it possible to verify whether $M$ is isomorphic to $N$ (using a computer algebra software)?
Longtime ago (in 2003) this was not possible. I do not know whether it is possible now.

REPLY [3 votes]: You can almost do. There is a theorem (due essentially to Bongartz, but in this form perhaps can be found in Yongwei Yao's thesis) that if $\rm{length}(M\otimes L)) =  \rm{length}(N\otimes L))$ for all finite length module $L$, then $M\cong N$ (this is a local result, but under reasonable assumptions, for example if both $M,N$ are presented by matrices in $m=(x_1,\dots,x_n)$, it still should be OK). 
So, one can generate a bunch of random modules $L$ supported at the origin and compare the lengths. If they are equal for say, 1000 of them, then the universe has to  really conspire against you for the modules not to be isomorphic! $\text{}$<|endoftext|>
TITLE: Dimension of eigenspaces of Laplacian on a compact Riemannian manifold
QUESTION [9 upvotes]: Let $M$ be a compact smooth manifold, let $g$ a riemannian metric and let $\Delta_{g}$ the Laplacian operator on functions induced  by $g$.  Is there a (topological?) bound on the dimension of  $n$-th eigenspace of $\Delta_{g}$?
Does the answer change if $M$ is a compact complex manifold and $g$ is a kahler metric?

REPLY [13 votes]: In dimension $n\geq 3$, there cannot be any sort of bound on the multiplicities which does not depend on some geometric input. This is because it is a theorem of Colin de Verdière (mathscinet and article (in French)) that:

If $M^n$ is a closed (smooth) manifold of dimension $n\geq3$, then any sequence $0=\lambda_1<\lambda_2\leq \lambda_3\leq\dots\leq \lambda_m$ is the first $m$ values of the spectrum of $\Delta_g$ for some metric $g$ (taken with multiplicity).

In fact, I'll remark that a theorem of Lohkamp has generalized this to the following result (mathscinet and article)

Given $M^n$ closed with $n\geq 3$ and $0=\lambda_1<\lambda_2\leq \lambda_3\leq\dots\leq \lambda_m$, as well as constants $V > 0$ and $K < 0$ there is a metric $g$ whose first $m$ elements of the spectrum agrees with the given sequence (with multiplicity) and also has
  $Vol(M,g) = V$ and $Ric_g \leq Kg$. 

On the other hand, in dimension $n=2$, there is such a bound, depending only on the topological type of the surface, due to Nadirashvili (mathscinet and article).<|endoftext|>
TITLE: A question in Fourier analysis
QUESTION [14 upvotes]: I recently came across this problem:
Let $T=\mathbb R /2 \pi \mathbb Z$ the circle, with its proabability Haar measure $\mu$. Any integrable function $f : T \rightarrow \mathbb C$
has Fourier coefficients $c_n(f)= \int_T f(t) e^{-int} \frac{dt}{2\pi}$, and we define
$$||f||_{A(T)}  = \sum_{n \in \mathbb Z} |c_n(f)| \leq + \infty$$
its algebra norm. (This definition is standard)
Now let $E$ be a measurable subset of $T$. I want to compute, or estimate,
$$ \alpha(E) = \inf \{  ||f||_{A(T)}\  | \ f : T \rightarrow \mathbb R, \  f_{|E} \leq 0, \int_T f = 1  \}$$
An easy lower bound for $\alpha(E)$ is $\alpha(E) \geq \frac{1}{1-\mu(E)}$, but Id like to know as many thing as possible about $\alpha(E)$. For example, here is a basic question:

When $E$ is an interval such that $\mu(E) \rightarrow 1$, is that lower bound sharp?

Also, has this problem, or something related, been studied ? Does $\alpha(E)$ have a name ?

REPLY [14 votes]: Yemon Choi (henceforth "YC") answered the question to within a constant factor
with a function $f_\delta$ showing that $\alpha(E_\delta) \leq 2/\delta$
where $E_\delta$ is an interval of measure $1-\delta$ (YC's notation).
That's optimal asymptotically $-$ and even exactly if $2/\delta \in {\bf Z}$
$-$ for functions $f$ with nonnegative Fourier coefficients $c_n(\phantom.f)$.
But without this additional hypothesis on the $c_n(\phantom.f)$ the constant
can be improved somewhat: we give another piecewise linear function
(still vanishing on $E_\delta$) that reduces the constant from $2$
to below $1.9$, and that's surely not optimal either.  But it
certainly can't go all the way down to Joël's lower bound of $1/\delta$:
the constant must be at least $3/2$, and I think it must be at least $\pi/2$
if $f=0$ on $E_\delta$ as YC suggested.
If $\| \phantom. f \|_{A(T)} \lt \infty$ then $\sum_n c_n(\phantom.f) e^{int}$
converges absolutely to a continuous function with Fourier coefficients $c_n$,
so we may assume $f$ is continuous.  If moreover each $c_n(\phantom.f) \geq 0$ then
$\| \phantom. f \|_{A(T)} = \sum_{n\in{\bf Z}} c_n = f(0)$.  Now suppose
$f(t) \leq 0$ for $t \in E_\delta$, and let $N$ be the integer
$\lfloor 2/\delta \rfloor$.  We have $f(m/N) \leq 0$ for all integers
$m \not\equiv 0 \bmod N$.  Therefore
$$
f(0) \geq \sum_{m \phantom. \bmod \phantom. N} f(m/N)
 = N \sum_{N\phantom.|\phantom.n} c_n(\phantom.f) \geq N c_0(\phantom.f) = N
$$
because $c_0 = \int_T f \phantom. d\mu = 1$; this proves our claim.
To give an example with smaller $\| \phantom. f \|_{A(T)}$,
consider the following generalization:
YC's function $f_\delta$ is the self-convolution
of uniform meausre on an interval of length $\delta/2$;
for $0 \lt b \lt 1$ define $f_{\delta,b}$ to be the convolution of
uniform measures on intervals of length $b\delta$ and $(1-b)\delta$.
Then
$$
c_n(\phantom.f_{\delta,b}) =
  \frac{\sin(\pi b \delta n)}{\pi b \delta n}
  \frac{\sin(\pi (1-b) \delta n)}{\pi (1-b) \delta n}.
$$
Hence $\delta \sum_n \left| c_n(\phantom.f_{\delta,b}) \right|$ is
a Riemann sum for
$$
\int_{-\infty}^\infty
  \left|
   \frac{\sin(\pi b s)}{\pi b s}
   \frac{\sin(\pi (1-b) s)}{\pi (1-b) s}
  \right| \phantom. ds.
$$
Numerical calculation shows this integral decreasing from $2$ at $b=1/2$
(YC's function) down to about 1.9 at $b \sim 0.38$, then up slightly and
back down to just below $1.9$ at $b \sim 0.29$ before increasing and
crossing $2$ at $b\sim 0.14$ (it must blow up at $b \rightarrow 0$).
We do a bit better by averaging $f_{\delta,0.29}$ with $f_{\delta,0.38}$,
getting just below $1.875 = \frac{15}{8}$.
This kind of analysis suggests that the asymptotic value of
$\delta \cdot \alpha(E_\delta)$ should be the infimum,
call it $\alpha_0$, of 
$\int_{-\infty}^\infty \bigl|\phantom.\hat f(s)\bigr| ds$
over functions $f: {\bf R} \rightarrow {\bf R}$ such that
both $f$ and its Fourier transform
$$
\hat f(s) = \int_{-\infty}^\infty e^{-2\pi i s t} f(t) dt
$$
are absolutely integrable and satisfy $\hat f(0) = 1$ and
$f(t) \leq 0$ for all $t$ with $|t| \geq 1/2$.  I don't quite
have a proof of this, but all examples and bounds so far
translate immediately to $\alpha_0$.  For example, if we
also assume that $\hat f(s) \geq 0$ for all $s$ then
$f(0) \geq 2$ follows from Poisson summation:
$$
f(0) \geq \sum_{m \in {\bf Z}} f(m/2)
 = 2 \sum_{s \in {\bf Z}} f(2n) \geq 2\hat f(0) = 2.
$$
This is the 1-dimensional case of the inequality in Henry Cohn's thesis
on upper bounds of sphere-packing densities.
The dictionary between bounds on
$\delta \alpha(E_\delta)$ as $\delta \rightarrow 0$
and bounds on $\alpha_0$ also works for the following
improved lower bound, which I'll give only in the $\alpha_0$ setting.
Here Joël's "easy lower bound" is $\alpha_0 \geq 1$, proved by noting that
$$
1 = \int_{-\infty}^\infty f(t) dt \leq \int_{-1/2}^{1/2} f(t) dt
$$
implies $f(t) \geq 1$ for some $t \in [-1/2, +1/2]$, and thus
$\bigl\| \phantom. \hat f \bigr\|_1 \geq 1$.  But clearly
equality cannot hold here.  To exploit this, we write
$$
\int_{-1/2}^{1/2} f(t) dt
 = \langle \phantom. f, \chi \rangle
 = \langle \phantom. \hat f, \hat\chi \rangle
 \leq \bigl\| \phantom. \hat f \bigr\|_1 \bigl\| \hat\chi \bigr\|_\infty
$$
where $\chi$ is the characteristic function of $[-1/2,1/2]$.
Now $\bigl\| \hat\chi \bigr\|_\infty = 1$,
but only because $\hat\chi(0) = 1$.
So we can improve the estimate by writing 
$\langle \phantom. f, \chi \rangle \leq \langle \phantom. f, \psi \rangle$
for any function (or even any measure) $\psi$ such that 
$\hat\chi - \hat\psi$ is nonnegative and supported outside $(-1/2, 1/2)$;
if $\bigl\| \hat\psi \bigr\|_\infty \lt 1$ then
$1 \bigl/ \bigl\| \hat\psi \bigr\|_\infty \bigr.$
is an improved lower bound on $\alpha_0$.
A choice that gives $\alpha_0 \geq 3/2$ is
$$
\hat\psi(s) = \hat\chi(s) - \frac13 \cos(\pi s)
 = \frac{\sin \pi s}{\pi s} - \frac13 \cos(\pi s)
 = \frac23 - \frac1{180} (\pi s)^4 + - \cdots,
$$
so
$
   \langle \phantom. f, \psi \rangle 
 = \langle \phantom. f, \chi \rangle
 - \frac16\bigl( \phantom.f(1/2) + f(-1/2)\bigr)
 \geq \langle \phantom. f, \chi \rangle \geq 1
$; then $\bigl\| \hat\psi \bigr\|_\infty \lt 2/3$,
so $\alpha_0 \geq 3/2$ as claimed.<|endoftext|>
TITLE: Increasing functions on the set of all non-empty proper subsets of a finite set
QUESTION [8 upvotes]: For a finite set $X$ let $X^* $  be the set of all non-empty proper subsets of $X$. Let $f : X^* \longrightarrow X^* $ be an increasing function such that for some $A \in X^* $, $|f(A)| \not = |A|$. It is true that $f$ must have a fixed point ? 
(By increasing I mean when $A\subseteq B$ then $f(A)\subseteq f(B)$ )

REPLY [6 votes]: Take $A$ maximal among sets such that $|f(A)|>|A|$. For each one-element set $\{x\}$, $f(\{x\})\subseteq f(A)$, because otherwise $|f(A \cup \{x\})| > |A \cup \{x\}|$. Draw a directed graph with one arrow leaving each element $x\in X$ pointing to an element $y\in X$ with $y \in f(\{x\})$. Let $T$ be the set of points in a cycle of that graph, then $T$ is proper because no point in $f(A)^c$ is in the image of any arrow, and $T\subseteq f(T)$ . A set $S$ maximal among sets such that $S\subseteq f(S)$ is a fixed point.<|endoftext|>
TITLE: Schubert calculus, as lowbrow as possible
QUESTION [35 upvotes]: Starting in a week I'm going to be an instructor at a summer program for exceptionally mathematically talented high school students, and I'm going to be teaching a class on Schubert calculus. The students will definitely know linear algebra, and many (but not all) of them will know what a ring is. They will definitely all have enough exposure to rigorous mathematics to know how to recognize a rigorous proof when they see it. They will also have seen quite a bit about the theory of symmetric functions, including Schur functions and the Pieri rule.
I want to present the basics of Schubert calculus in as believable a way as possible. I'll have maybe four or five days, one hour per day, to get to, say, the Pieri rule for Schubert cycles, or to any other way to connect Schubert calculus to symmetric functions.
I definitely plan to talk about the Grassmannian and the Plücker embedding and to convince them that what we're looking for is "just" the number of points on the Grassmannian (and therefore in projective space) that satisfy some polynomial equations. I'd like to talk about the product on the cohomology ring of the Grassmannian, but probably without calling it that, and with as little hand-waving as possible.
So my question is:

What is the best way to convince this audience that multiplying Schubert cycles is a sensible thing to do, that it counts solutions to Schubert problems, and that it formally looks like multiplying Schur functions?

I know it probably won't be possible to be rigorous about everything, but I'd like the presentation to be as clean and convincing as possible because otherwise I think it will be very unsatisfying.

EDIT: Some of the responses have inspired me to be more specific. I'll definitely be able to show that if you intersect two Schubert varieties of complementary dimension with respect to opposite flags, you get one point if they're dual and no points otherwise. Brushing under the rug the question of the multiplicity of that point, this means I could show the Pieri rule by counting the number of points in the intersection of three Schubert varieties w.r.t. flags in general position, one of which is a "special" Schubert corresponding to a partition with one row, provided I could convince them that:

There is some notion of "generic" such that, given finitely many subvarieties of the Grassmannian, I can take generic $GL_n$ translates of each of them and intersect them, and there's some meaningful equivalence relation such that the thing I get is "equivalent" provided the translates were "generic" enough.
Schuberts span the cohomology ring. Happy to just state this without proof if I can find a way to.

Clearly I have to black-box something; I'm not going to develop the entire modern theory of intersection cohomology for high-schoolers in two hours. The question is how to make it not seem like I'm pushing 100% of the motivation under the rug.

REPLY [5 votes]: Although this is certainly a too late answer on this question, maybe this paper http://arxiv.org/abs/math/0608784 (Schubert Calculus according to Schubert, by Felice Ronga) is worth to be mentioned. It gives a nice elementary introduction to Schubert calculus.<|endoftext|>
TITLE: Covering a hexagon
QUESTION [19 upvotes]: For $\epsilon > 0$ sufficiently small, can a regular hexagon with sides of length $1 + \epsilon$ be covered by seven equilateral triangles with sides of length $1$?
Motivation: Conway and Soifer showed that an equilateral triangle with sides $n + \epsilon$ can be covered with $n^2 + 2$ triangles. They conjectured that this is best possible, i.e. that it can not be covered by $n^2 + 1$ such triangles. This is fairly clear for $n=1$ and $n=2$ but the problem seems to be open even for $n=3$. The hexagon I've asked about is a substructure of the $n=3$ case that might be more tractable, but might still capture some of the difficulties of the $n=3$ and larger cases.

REPLY [10 votes]: The case of hexagon is exactly the problem that is Alexander and I posed in:
Karabash, D., and Soifer, A., On Covering of Trigons, Geombinatorics XV(1) (2005), 13–17.
Hexagon is type of 6-trigon (n-trigon is n connected triangles from triangulation); n-trigons for n<6 are trivial counting of vertecies and hence 6-trigon is the simplest hard case.
I gave this question to several students as one of possible problems, but I consider it a hard problem even though I see a clear non-elegant solution to the hexagon problem; one just has to work hard: 7 covering triangles can be described by 21 variables (x_i,y_i,r_i) for i=1,...,7 where x_i,y_i are coordinates of the center and r_i is rotation.  Then line intersections with sides define a clear regions that one has to analyze; one can write computer program that checks all the regions and that hexagon is never covered via checking conditions.  This is of course not something I would expect a high-school student to do, that is why I gave this problem with 2 stars saying it is most likely not a good project unless you get some very new idea because I have given this quesiton to guys with IMO gold, putnam fellows, and top computer science guys without any progress.
If you get any progress I will be very interested--it is also a "50 dollar problem" :).<|endoftext|>
TITLE: Looking for a special rank 2 vector bundle
QUESTION [5 upvotes]: Let $E\to C$ be a rank $2$, degree $2g-2$, holomorphic vector bundle over a curve of genus $g$.
By Riemann-Roch theorem, 
$$H^0(E)-H^1(E)= \deg(E)+2.(1-g)=0. $$
Question: For which $g$, there is such $E$ with $H^0(E)=0$ (and thus $H^1(E)=0$ as well)?

REPLY [5 votes]: Every curve admits a degree g-1 line bundle with $h^0=h^1=0$ -- in fact a generic degree g-1 line bundle has this property, since the space of degree g-1 divisors is g-1 dimensional, but the space of line bundles is g dimensional.  Taking the direct sum of two such line bundles will give you a bundle of the type you are seeking.<|endoftext|>
TITLE: Equivalent form of the Univalence Axiom
QUESTION [11 upvotes]: I'm reading the new HoTT book and I'm wondering about a potential equivalent form of the Univalence Axiom: $(A \simeq B) \simeq (A = B)$.
For simplicity, I'm tacitly working in a fixed universe. It is known that the univalence axiom implies function extensionality $$\mathsf{funext}:(f \sim g) \to (f = g)$$ where $f,g$ are from the dependent product type $\prod_{x:A} B(x)$ and $$(f \sim g) :\equiv {\textstyle\prod_{x:A} (f(x) = g(x))}$$ is the type of homotopies from $f$ to $g$. 
Assuming this, it seems that 
$$\mathsf{isequiv}(f) :\equiv \Big(\sum_{g:B\to A} (f \circ g \sim \mathsf{id}_B)\Big) \times \Big(\sum_{h:B\to A} (h \circ f \sim \mathsf{id}_A)\Big)$$
boils down to saying that $f$ is an isomorphism:
$$\mathsf{isisom}(f) :\equiv \sum_{g:B\to A} (f \circ g = \mathsf{id}_B)\times(g \circ f = \mathsf{id}_A).$$
Therefore, still assuming function extensionality, there is an equivalence between equivalence $$A \simeq B :\equiv \sum_{f:A \to B} \mathsf{isequiv}(f)$$ and isomorphism $$A \cong B :\equiv \sum_{f:A \to B} \mathsf{isisom}(f).$$
Assuming that the above is correct, the Univalence Axiom should be equivalent to the conjunction of functional extensionality with the weaker statement: $(A \cong B) \simeq (A = B)$. I like this conceptual splitting of the Univalence Axiom. Is the above correct or is there a flaw in my informal reasoning? Is it known that $(A \cong B) \simeq (A = B)$ is weaker than the Univalence Axiom?

As Mike pointed out in his answer, the correct definition of $A \cong B$ should be $$A \cong B :\equiv \sum_{f:A \to B} \mathsf{biinv}(f)$$ where $$\mathsf{biinv}(f) :\equiv \Big(\sum_{g:B\to A} (f \circ g = \mathsf{id}_B)\Big)\times\Big( \sum_{h:B\to A} (h \circ f = \mathsf{id}_A)\Big)$$
With this revised definition, it does appear to be the case that the Univalence Axiom splits into function extensionality and the formally weaker $(A \cong B) \simeq (A = B)$. The second question above should be corrected with this new formulation in mind: is $(A \cong B) \simeq (A = B)$ actually weaker than the Univalence Axiom?

REPLY [4 votes]: Still not an answer to the second question, but I wanted to add something else that's missing: in fact the bare statement $(A=B)\simeq (A\simeq B)$ is not known to be a correct form of the univalence axiom.  The correct statement is that the canonical map $(A=B) \to (A\simeq B)$ is an equivalence.  The statement $(A=B)\simeq (A\simeq B)$ can be regarded as an abusive abbreviation of this, in the same way that mathematicians often write "$A\cong B$" to mean that some canonical map $A\to B$ is an isomorphism, but to be precise we have to be careful.
There is now a much weaker-looking known equivalent form of the univalence axiom, see here.  However, it does still use pointwise homotopies in the equivalences that are to be made into equalities.  I suspect that the corrected form "the canonical map $(A=B) \to (A\cong B)$" of your second question is indeed strictly weaker than univalence in the absence of funext, but I don't know.<|endoftext|>
TITLE: On Applications of Murnaghan Nakayama Rule
QUESTION [10 upvotes]: This question is crossposted at math.stackexchange here and may be beyond the usual scope of the site. The question is located below. In short, I am looking for an accessible explanation of the Murnaghan Nakayama rule in relation to the following problem. Pardon the long setup.
Let $Y$ be a standard Young tableau of shape $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_n)$. At the risk of being pedantic, standard here means $\lambda_1\geq\lambda_2\geq \ldots \geq \lambda_n$ and the entries are strictly increasing rightward along rows and downard along columns (with each number from $1$ to $\left|\lambda\right|$ occuring only once). Let $f_\lambda$ denote the number of standard Young tableaux of shape $\lambda$. It is well known that $f_\lambda$ can be calculated via the Hook formula:
$$f_\lambda=\frac{|\lambda|!}{\prod_{i,j}h(i,j)}.$$ 
Now suppose I remove some boxes on the outer edge of the shape (i'll call this a pattern from now on), giving a new shape $\delta_i$ (I'll explain the $i$ in a moment). Here's an example: let $\lambda$ be the triangular shape $(n-1,n-2,\ldots,1)$. Now I want to remove three adjacent boxes in an "L" shape:

Here I've indexed the location of the left box as $i$ which gives the $x$ coordinate of the box ($i=3$ in this example). To repeat, I'm calling the new shape $\delta_i$, and one easily finds $f_{\delta_i}/f_\lambda$ as the ratio of the particular hooks that change. I'm interested in the following. In the above example, I removed three adjacent boxes at location $i$. I want to evaluate sums of the form:
$$\sum_{i=1}^{n-2}\frac{f_{\delta_i}}{f_\lambda}$$.
In full generality, I am interested in removing any pattern of boxes indexed by $i$ (the sum's range changes appropriately depending on the pattern). Going along with my example, one can show that 
$$f_{\delta_i}/f_\lambda=\frac{1}{3N(N-1)(N-2)}a_ia_{n-i-1},$$
where $N:=\binom{n}{2}$ and $a_i:=\frac{(2i+1)!!}{(2i-2)!!}$. One can then evaluate the sum using relatively straightforward generating functions for $a_i$, giving a tidy answer of $\frac{1}{2(N-2)}$. 

Question: I have heard that calculating $\sum\limits_{i=1}^{n-2}\frac{f_{\delta_i}}{f_\lambda}$ can be done via the Murnaghan Nakayama rule. In particular, I've been told my $L$ example gives $-\frac{\chi^{\lambda}(\pi)}{\chi^\lambda(\mbox{id})}$ where $\pi$ is a 3-cycle. Can someone provide an accessible reference for this rule in the context above? I am not an expert in representation theory and the usual references I've seen (Enumerative Combinatorics, Vol II by Stanley) are somewhat beyond me at the moment (and in much greater generality). In short, I am looking for an accessible explanation of the Murnaghan Nakayama rule for these types of questions.

REPLY [11 votes]: For a very nice bijective proof of the Murnaghan-Nakayama rule I recommend Section 3.3 of Nick Loehr's article Abacus proofs of Schur function identities, SIAM J. Disc. Math. 24 (2010) 1356-1370. Chapter 11 of Loehr's recent textbook Bijective combinatorics gives an expanded version of his article with minimal prerequisites.
To explain the connection: the Murnaghan-Nakayama rule states that if $\lambda$ is a partition of $m$, $\pi$ is a $k$-cycle and $\sigma$ is a permutation of the remaining $m-k$ points then
$$\chi^\lambda(\pi \sigma) = \sum (-1)^{\ell(\lambda/\mu)} \chi^\mu(\sigma) $$
where the sum is over all partitions $\mu$ obtained from $\lambda$ by removing a $k$-hook from the border of $\lambda$, and $\ell(\lambda/\mu)$ is leg-length of the hook (i.e. the number of rows of the Young diagram occupied by the hook, minus one). In particular, if $\lambda = (n-1,n-2,\ldots,1)$, $\sigma = 1$ and $\pi$ is a $3$-cycle, then we get
$$\chi^{(n-1,\ldots,1)}(\pi) = -\sum \chi^{\delta_i}(1) $$
since every $3$-hooks that can be removed from $(n-1, \ldots, 1)$ is an $L$-shape of leg-length $1$. Now divide by $\chi^{(n-1,\ldots,1)}(id)$ and use the hook formula to get the sum in the question.
For the intended application we only need the Murnaghan-Nakayama rule for cycles. However my feeling is that any proof of the rule for cycles will do almost all the work needed for the general version. For instance, stated using symmetric polynomials, the Murnaghan-Nakayama rule for cycles is $s_\mu p_k = \sum_\lambda (-1)^{\ell(\lambda/\mu)} s_\lambda$, where the sum is over all partitions $\lambda$ obtained from $\mu$ by adding a $k$-hook. Given this, the full result follows in a few lines by taking $\mu = \varnothing$ and repeatedly multiplying by power-sum symmetric polynomials.
There is however an alternative route to the final answer that might be of interest, using 
explicit formulae for the characters of symmetric groups on cycles. These can be deduced from the Frobenius formula for character values: see 4.10 and Exercise 4.15 of Fulton & Harris, Representation theory: a first course for the method. The relevant formula for $3$-cycles is (5.2) in R. E. Ingham, Some characters of the symmetric group, Proc. Amer. Math. Soc. 1 (1950) 358-369. It states that
$$\frac{\chi^{\lambda}(\pi)}{\chi^\lambda(id)} m(m-1)(m-2) = 3K_3 (\lambda) - \frac{3m(m-1)}{2} $$
where $\lambda$ is a partition of $m$ and $\pi$ is a $3$-cycle. Here $K_3(\lambda)$ can be defined as follows: put zeros in the boxes in the leading diagonal of the Young diagram of $\lambda$. In each row, put $1^2$, $2^2$, $\ldots$ in the boxes right of the $0$. In each column, put $1^2$, $2^2$, $\ldots$ in the boxes below the $0$. Then $K_3(\lambda)$ is the sum of all the numbers in the boxes of the Young diagram. Calculation shows that
$$K_3(n-1,\ldots,1) = \frac{(n+1)n(n-1)(n-2)}{12} = \frac{N(N-1)}{3} $$
where $N = \binom{n}{2}$, as in the question. Therefore 
$$\frac{\chi^{(n-1,\ldots,1)}(\pi)}{\chi^{(n-1,\ldots,1)}(id)} N(N-1)(N-2) = -\frac{N(N-1)}{2} $$
and so
$$\sum_{i=1}^{n-2} \frac{f_{\delta_i}}{f_{(n-1,\ldots,1)}} = -\frac{\chi^{(n-1,\ldots,1)}(\pi)}{f_{(n-1,\ldots,1)}} = \frac{1}{2(N-2)}. $$
This differs from the answer claimed in the question but seems to be correct at least for $n=3,4,5$. (For example, if $n=3$ then we have $\delta_1 = \varnothing$ and the sum in the question is $1/f_{(2,1)} = 1/2$.)<|endoftext|>
TITLE: A question about Weil algebra
QUESTION [5 upvotes]: Let $G$ ba a compact Lie group with Lie algebra $\mathfrak{g}$, $\mathfrak{g}^{*}$ be the dual of $\mathfrak{g}$. We known the Weil algebra is
$$W(\mathfrak{g})=\wedge(\mathfrak{g}^{*})\otimes S(\mathfrak{g}^{*}).$$
Choose a basis $e_{1},\cdots,e_{n}$ for $\mathfrak{g}$ and let 
$e_{1}^{*},\cdots,e_{n}^{*}$ be the dual basis of $\mathfrak{g}^{*}$. 
Let $\theta^{1},\cdots,\theta^{n}$ be the dual basis of $\mathfrak{g}^{*}$ generating the exterior algebra $\wedge(\mathfrak{g}^{*})$ and let $\phi^{1},\cdots,\phi^{n}$ be the dual basis of $\mathfrak{g}^{*}$ generating the symmetric algebra $S(\mathfrak{g}^{*})$.
On $(S(\mathfrak{g}^{*})\otimes\Omega^{*}(M))^{G}$, there is the Cartan model. 
The exterior differential in Cartan model of equivariant cohomology is defined by 
$$D=1\otimes d-\sum_{j=1}^{n}\phi^{j}\otimes\iota_{j}$$ 
here, $\iota_{j}$ is the contraction operator.
If we choose $\widetilde{e}_{1},\cdots,\widetilde{e}_{n}$ as the another basis of 
$\mathfrak{g}$, then the dual basis of $\mathfrak{g}^{*}$ generating the symmetric algebra 
$S(\mathfrak{g}^{*})$ is $\widetilde{\phi}^{1},\cdots,\widetilde{\phi}^{n}$.
How to express the operator $\sum_{j=1}^{n}\phi^{j}\otimes\iota_{j}$
in the basis $\widetilde{e}_{1}^{*},\cdots,\widetilde{e}_{n}^{*}$
and Why is it?

REPLY [3 votes]: If $\tilde e_i = \sum_j A_{ij} e_j$ then for the dual basis we have 
$\tilde e_k^\ast = \sum_\ell ((A^t)^{-1})_{kl} e_l^\ast$, so the the base change matrix cancels out of the formula.<|endoftext|>
TITLE: Direct proof of the separation theorem of Hahn-Banach
QUESTION [10 upvotes]: The "extension" (or "analytic") form of the theorem of Hahn-Banach has a natural and yet elegant proof. In just any textbook I have ever seen, it is proved first; the "separation" (or "geometric") version of Hahn-Banach's theorem is proved as a kind of corollary of the former.
Question: Are the two theorems actually equivalent? If so, is any direct proof of the analytic version known that is instead based on the geometric one?

REPLY [14 votes]: Yes, the two theorems are equivalent in the sense that one can easily be deduced from the other and both have direct proofs from scratch.
A standard textbook starting with a direct proof of the geometric version is Schaefer's Topological Vector Spaces, Chapter II, Section 3.
The statement Schaefer proves is:

Let $F$ be a subspace of a topological vector space $E$ and let $U$ be a nonempty open convex subset, disjoint from $F$. Then there is a closed hyperplane $H$ containing $F$ and disjoint from $U$.

The usual reductions via translation and taking the difference of the convex sets then yield the separation theorems of an open convex set from a point and of an open convex set from a compact convex set.

The proof starts by a simple geometric observation: Let $U$ be an open and convex subset of a Hausdorff topological vector space of dimension $\geq 2$. If $U$ does not contain $0$, then there is a one-dimensional subspace disjoint from $U$. This is easily reduced to the two-dimensional case, where it is rather clear.
To establish the above statement, a straightforward application of Zorn's lemma shows that there is a maximal (hence closed) subspace $M$ containing $F$ and disjoint from $U$. Since $U$ is non-empty, $E/M$ has dimension at least $1$. If the dimension of $E/M$ is $1$, then $M$ is a hyperplane and we're done. Suppose towards a contradiction that the dimension is at least $2$. The image of $U$ in $E/M$ does not contain $0$ and is open since the canonical projection $\pi \colon E \to E/M$ is open. Since $M$ is closed, $E/M$ is Hausdorff. Therefore there is a one-dimensional subspace $L$ of $E/M$ not meeting the image of $U$. The pre-image of $L$ contains $M$, is strictly larger and does not meet $U$, contradicting the maximality of $M$.

In order to get the analytic form, identify a linear functional on a subspace with its graph in $E \times \mathbb{R}$. Endow $E \times \mathbb{R}$ with the product topology induced by the sublinear functional $p$ and the usual topology on $\mathbb{R}$. The set $U = \lbrace (x,t) \mid p(x) \lt t\rbrace$ is an open convex cone in $E \times \mathbb{R}$, not containing $(0,0)$. A linear functional $f$ is dominated by $p$ iff $\operatorname{graph}(f)$ is disjoint from $U$. A maximal closed hyperplane $M$ containing $\operatorname{graph}(f)$ and disjoint from $U$ is then seen to be the graph of a linear functional $F$ which is obviously an extension of $f$ and dominated by $p$.

REPLY [2 votes]: The separating version is ubiquitous in economics, and in my experience, most textbooks in mathematical economics prove it directly (though sometimes only in the finite dimensional case).<|endoftext|>
TITLE: What's the normal cone to a union?
QUESTION [5 upvotes]: I have a reducible variety, $X = \bigcup_{i \in I} X_i \subset Y$, and I want to understand the normal cone $C_X Y$. 
I write $X_J$ for $\cap_{j \in J} X_j$. 


Can $C_X Y$ be expressed in terms of the $C_{X_J} X_K$? 


(I take the convention $X_\emptyset = Y$ so that the $C_{X_i} Y$ are among the above.)
In fact in the example that has prompted this question, all the $X_J$ are smooth and connected, (though I do not know that they intersect transversely), and I am actually only interested in numerical information; i.e., 

Can the Segre class $S(X,Y)$ be expressed in terms of the Segre classes $S(X_J, X_K)$ for $J \subset K$ and perhaps also the intersection multiplicities of $X_M$ and $X_N$ along their intersection?

REPLY [2 votes]: No.  Take $X$ to be the union of two smooth, genus $0$ curves, say $X_1$ and $X_2$, that intersect in two nodes, say $p$ and $q$.  Then $\text{Pic}(X)$ is an extension of $\mathbb{Z}^2$ by $\mathbb{G}_m$.  So let $Y$ be a geometric $\mathbb{A}^1$-bundle over $X$ whose restriction to each of $X_1$ and $X_2$ is trivial, yet that is nontrivial on $X$.  Identify $X$ with the zero section of this $\mathbb{A}^1$-bundle.  The normal cone is a normal bundle, essentially just the data of $Y$ once again.  Yet you cannot recover $Y$ from its restriction to the two irreducible components.
Edit. Vivek clarified that he would also like to keep track of the restriction maps between the cones on different strata.  That would certainly make a difference in the example above.  However, if one allows non-transversal intersections, as Vivek explicitly allows, then there are still counterexamples.  For instance, consider a union of three concurrent lines in $\mathbb{P}^2$.  The irreducible components are each $\mathbb{P}^1$, and the intersections between components are all one (reduced) point.  Now there is simply no way to encode all invertible sheaves by the restriction on strata and restriction homomorphisms between them.  (Of course if one allows nilpotent thickenings of the strata, one could encode these invertible sheaves.)<|endoftext|>
TITLE: Is there a known example of a curve X of genus > 1 over Q such that we know the number of points of X over the n-th cyclotomic field, for every n?
QUESTION [18 upvotes]: By Falting's theorem, these numbers are of course finite. Is there an example where we can explicitly compute them for every $n$?
Thank you!

REPLY [29 votes]: Let $n > 1$ be odd.
The curve:
$$X:  \ x^n + 2  y^n + 4 z^n = 0$$
does not have any points over $K/\mathbf{Q}_2$ unless the ramification index $e(K/\mathbf{Q}_2)$ is divisible by $n$. Proof: At least two of the terms $x^n$, $2 y^n$, $4 z^n$ must have the same $2$-adic valuation. On the other hand, the ramification index $e$ of any abelian extension $K/\mathbf{Q}_2$ is a power of two. So $X$ has no points over any Galois extension of $\mathbf{Q}$ whose decomposition group at any prime above $2$ is abelian. In particular, it has no point over any cyclotomic extension. It has genus $>1$ if $n>3$.
You can ask whether any smooth projective curve $X/\mathbf{Q}$ has at least one rational point over any solvable extension. Since local Galois groups are solvable, there are no longer any local obstructions. This is an open problem, and a positive answer would have various nice consequences including (generalizations of) Serre's conjecture, etc. There's no particular reason why it should be true, however.<|endoftext|>
TITLE: Comparing different layered structures for fibered 3-manifolds: example request.
QUESTION [7 upvotes]: Let's consider a fibering hyperbolic 3-manifold obtained as a mapping torus over some hyperbolic surface with pseudo-Anosov monodromy, and let's suppose that the surface is punctured at the singular points of the two invariant foliations.
Ian Agol has introduced a canonical layered ideal triangulation of such manifolds through a periodic splitting sequence of train tracks. Such triangulations are characterized by a combinatorial property called veeringness.
Another layered structure can be obtained by considering the flow of incomplete euclidean metrics associated to the Teichmuller line determined by the monodromy. Given one of these metrics, you consider the Dirichlet decomposition with respect to the set of singular points . This changes finitely many times along the geodesic segment determined by the monodromy, and allows us to produce a layered cell decomposition (it is not clear from the definition that it should always be a triangulation), see http://ldtopology.wordpress.com/2009/08/20/canonical-triangulations-of-surface-bundles/.
If we restrict to the case where the fiber is a once punctured torus, the two constructions are the same, and produce the usual Floyd-Hatcher monodromy ideal triangulation.
I am looking for an example of a manifold such that Agol's construction and the Dirichlet domain construction produce different results.
Thank you very much!

REPLY [5 votes]: The two triangulations (or cellulations) do not coincide in general.
Context:
In fact, given a centrally symmetric convex shape  $S$ in the plane, one could at any time $t$ on the Teichmüller line look for maximal copies of $S$ (up to scaling and translation, but not rotation) disjoint from the singularities. Such maximal copies will have some number of singularities in their boundary. The convex hull of those singularities gives a polygon (let us say a triangle since this is the generic case), and the union of these triangles (over all maximal copies of $S$ at time $t$) is a triangulation of the surface at time $t$. Letting $t$ vary, we will see (generically) a sequence of diagonal exchanges, forming a layered triangulation. The two examples considered so far are for $S$ a square (Agol) and a circle (Dirichlet). For other shapes $S$ we would get even more different triangulations. Also, if all singularities have even number of prongs we can drop the condition that $S$ be centrally symmetric.
Now the counter-example. Consider a flat surface $F$ obtained by gluing together opposite sides of a regular Euclidean $8$-gon. Call the vertices $N, NE, E, SE, S, SW, W, NW$ (like the cardinal directions); they are the singularities in the quotient. Stretch slightly in the $NE, SW$ direction, so that the disk circumscribed to $N, NE, E$ does not meet any other points: the triangle $N, NE, E$ will therefore appear in the Dirichlet decomposition. However, it is not in the Agol decomposition (in fact not even the edge from $N$ to $E$ is there) since any rectangle containing $N$ and $E$ contains $NE$.
It only remains to make sure we can find a pseudo-Anosov whose Teichmüller line passes through (or near) this metric. Since $F$ is a Veech surface, this is not difficult: there is a whole (non-cocompact) lattice of $SL(2, R)$ inducing mapping classes of $F$, which can therefore be chosen pseudo-Anosov with eigendirections close to vertical and horizontal (or to any other given pair of directions).<|endoftext|>
TITLE: Can $\zeta(s)$ for $\Re(s)>1$ be split into two factors that each can be analytically continued?
QUESTION [5 upvotes]: Assuming the RH and $s \in \mathbb{C}, \rho_n =\frac12 \pm i\gamma_n$, the following (altered) Hadamard product: 
$$\displaystyle \displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{\frac12+ (-1)^n i \gamma_n} \right) \left(1- \frac{s}{\frac12+ (-1)^{n+1} i \gamma_n} \right) = \frac{\xi_{rie}(s)}{\xi_{rie}(0)}$$
runs through the alternating non-trivial zeros $\rho_n$ with $\xi_{rie}(s)= \frac12 s(s-1) \pi^{-\frac{s}{2}} \Gamma\left(\frac{s}{2}\right) \zeta(s)$.
Contrary to the factors of the original Hadamard product, these alternating factors do converge.
This question suggest that many similarities exist between infinite (Hadamard/Weierstrass) products using $\gamma_n=n$ and $\gamma_n=\Im(\rho_n)$, and this question shows that a closed form for alternating factors using $\gamma_n=n$ does exist. I therefore like to conjecture that also a closed form exists for the alternating formula above.
Let's call the closed form for each factor $A_-$ and $A_+$ and it is easy to see that:
$$\displaystyle A_-A_+=\frac{\xi_{rie}(s)}{\xi_{rie}(0)}=s(s-1) \pi^{-\frac{s}{2}} \Gamma\left(\frac{s}{2}\right) \zeta(s)$$
is the entire function to be split into two factors.
Splitting this function is easy to do for the Gamma-part: $G_-G_+=s(s-1) \pi^{-\frac{s}{2}} \Gamma\left(\frac{s}{2}\right)$, that for instance (there are more ways) could be factored into:
$$\displaystyle G_-=s\Gamma\left(\frac{s}{4}\right) \pi^{-\frac{s}{4}}2^{\frac{s}{4}-\frac32} \text{       and      } 
G_+=(s-1)\Gamma \left(\frac{s}{4}+\frac12\right) \pi^{-(\frac{s}{4}+\frac12)}2^{\frac{s}{4}+\frac12}$$
But what to do with $\zeta(s)$? 
The poles of $G_-$ (-4,-8,...) and $G_+$ (-2,-6,...) might provide some hints, since they need to be annihilated by the zeros of the 'to be found' $\zeta(s)$-factors. It is also clear that a $\zeta(s)$-factor must now induce alternating non-trivial zeros only i.e.: $\frac12+14.134...i,\frac12-21.022...i,\frac12+25.010...i, \dots$ (and its complement). Dividing the infinite product factors $A_-$ en $A_+$ (using n=699) by $G_-$ and $G_+$ respectively, one gets the following graphs of what the (absolute) $\zeta(s)$-factors might look like:
 
Question (apologies for the long intro):
The graphs of the two potential factors for $\zeta(s)$ above, could both be seen as analytically continued functions across $\mathbb{C}/1$, that have been derived "bottom up" from their alternating zeros. This would imply that in the domain $\Re(s)>1$ the two factors must multiply into:
$$\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s} = \prod _{p \in \mathbb{P}}(1-p^{-s})^{-1}$$
Hence my question: are there any ways to split, the known analytically "discontinued" expressions for $\zeta(s)$ in the domain $\Re(s)>1$, into two factors that each can be analytically continued again?
P.S.:
(1) I have f.i. tried splitting the Euler product into its $(p \mod 4 = 1)$ and $(p \mod 4 = 3)$ factors, however did not see any way to analytically continue these. 
(2) Also hoped to find some 'natural' connection with the alternating Zeta: $\eta(s)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s}$ that is valid over the domain $\Re(s)>0$, however so far have been unsuccessful in factoring it further.

REPLY [3 votes]: Haven't found a full answer yet, however did discover a nice way (very different from the approach in the question) to split $\zeta(s)$ into two factors that are valid in $\mathbb{C}{/1}$ and that each induce the alternating conjugates of the $\rho$s.
With $\displaystyle \chi(s)=\pi^{-\frac{s}{2}}\,\Gamma\left(\frac{s}{2}\right)$ and $K(s)=\Psi\left(\frac{s}{2}\right)-\ln(\pi)$, with $\Psi\left(s\right)$ the Digamma function, then:
$$\zeta(s):= -\dfrac{2}{\chi(s)}\cdot\dfrac{\chi(1-s)\cdot\zeta'(1-s)+\chi(s)\cdot\zeta'(s)}{K(1-s)+K(s)}\qquad(1)$$
This allows the $\zeta(s)$ function to be split into (assuming principal branches for the squared roots):
$$\zeta_1(s):= i\cdot \sqrt{\dfrac{2}{\chi(s)}}\cdot\dfrac{\sqrt{\chi(1-s)\cdot\zeta'(1-s)}+i\cdot\sqrt{\chi(s)\cdot\zeta'(s)}}{\sqrt{K(1-s)}+i\cdot\sqrt{K(s)}}$$
and
$$\zeta_2(s):= i\cdot\sqrt{\dfrac{2}{\chi(s)}}\cdot\dfrac{\sqrt{\chi(1-s)\cdot\zeta'(1-s)}-i\cdot\sqrt{\chi(s)\cdot\zeta'(s)}}{\sqrt{K(1-s)}-i\cdot\sqrt{K(s)}}$$
so that $\zeta(s)=\zeta_1(s)\cdot\zeta_2(s)$.
The graph below shows a plot of both functions and how the zeros alternate. The additional pair of roots induced by the numerator at $\frac12 \pm 6.28...$ is exactly cancelled by the denominator.

Note that multiplying both sides of (1) by $\frac12\, s(s-1)\, \chi(s)$ gives the Riemann $\xi(s)$ function as:
$$\xi(s)=\xi(1-s)=s\,(1-s)\cdot\dfrac{\chi(1-s)\cdot\zeta'(1-s)+\chi(s)\cdot\zeta'(s)}{K(1-s)+K(s)}$$
Of course this can also be split into factors with e.g. $\sqrt{s\,(s-1)}$ and I had hoped to then link these to each alternating factor of the Hadamard product in the question (but no connection found).
The main issue I encountered is that although the $\zeta_1(s)$ and $\zeta_2(s)$ do nicely behave in the complex plane, in the real domain there are discontinuities induced by the real roots of $\zeta'(s)$ in the numerator and by roots of $K(s)$ in the denominator, that don't nicely annihilate each other (since squared roots are taken).<|endoftext|>
TITLE: How can one interpret homology and Stokes' Theorem via derived categories?
QUESTION [16 upvotes]: I am very far removed from being an expert on derived categories. Every few months, however, I read a different introductory text with the hope that eventually I will have some basic grasp on this concept.
This has the added benefit that it puts various aspects of the technicalities of homological algebra, algebraic geometry and algebraic topology into a uniform context. For example, Verdier duality generalizes both Poincare duality and Serre duality, and it is easy to see how the proof of the DeRham Theorem for real manifolds is analogous to that of the DeRham Theorem for affine varieties.
In all of my readings so far, however, only cohomology (as opposed to homology) was considered. I therefore wonder how homology is treated via derived categories. In order to ask a more precise question:
Question
Is there a simple proof of Stokes' Theorem via derived categories? What is the set up of this proof, and in particular how is homology and its relationship with cohomology treated? Are there references about this that you can recommend?

REPLY [9 votes]: There is an interpretation of homology using derived categories, and you can find a brief treatment around remark 3.3.10 in Kashiwara-Schapira "Sheaves on Manifolds".  If $M$ is an $n$-manifold and $a: M \to \ast$ is the tautological map to a point, then the homology of $M$ with coefficients in an abelian group $F$ is the homology of the complex $Ra_! a^! F$.  If we write $or_M = H^{-n}(a^!F)$ for the orientation sheaf, then the counit of the adjunction induces an "integration" map $Ra_! a^! F \to F$ that induces a map $H^n_c(M,or_M) \to F$.  When $F$ is a real vector space and $M$ is smooth, you can relate this to differential forms, because the orientation sheaf is quasi-isomorphic to a suitable version of the de Rham complex.  Softness of the de Rham sheaves implies you get an isomorphism $H^n_c(M, or_M) \to \frac{\Gamma_c(M, \Omega^n \otimes or_M)}{d\Gamma_c(M,\Omega^{n-1} \otimes or_M)}$.  Stokes's theorem on $M$ implies a comparison between the "integration" map and "honest integration" of forms (or densities), in particular, the fact that $d\Gamma_c(M,\Omega^{n-1} \otimes or_M)$ is annihilated by honest integration.
I don't think Stokes's theorem is a statement whose proof is made easier with derived categories.  Instead, I would rather say that it implies the fact that honest integration actually yields a morphism in a derived category.
Here's the setup: If we are given a smooth manifold $M$ and a $k$-form $\omega$, we can obtain a real number from any smooth map $f$ from the standard $k$-simplex $\Delta$ to $M$, by integrating $f^\ast\omega$ along $\Delta$.  This produces a map of graded real vector spaces $\int_M: \Omega^\ast(M) \to \operatorname{Hom}(\operatorname{Sing}_\ast(M),\mathbf{R})$.
Here's Stokes's theorem: $\int_M$ is in fact a map of cochain complexes.
If you want to prove the theorem efficiently, you can use naturality of pullback to reduce to a simpler statement about forms on $\Delta$ itself.  There will always be a step where you have to use properties of integrals.  It is common to invoke Fubini's theorem (e.g., the short proof in Guillemin-Pollack), but you can do a lower-tech proof using the fact that polynomial forms uniformly approximate smooth forms on the simplex.<|endoftext|>
TITLE: What evidence is there that $\mathbb{Q}^{ab}$ is ample?
QUESTION [12 upvotes]: A field $K$ is called ample if every smooth curve over $K$ that has a $K$-point has infinitely many $K$-points. Examples include fraction fields of henselian rings. (This is far from trivial, but was proven in the paper "henselian implies large". "Large" is a synonym for "ample".) For example the field of $p$-adics, for any prime $p$, is ample.
I have known for a long time that $\mathbb{Q}^{ab}$ is conjectured to be ample, but I don't know how much evidence there is for this conjecture. Can you direct me to some references supporting this conjecture, or suggest heuristic arguments that would explain why this conjecture is reasonable?

REPLY [7 votes]: It seem like most of the evidence is rather indirect. The ampleness of $\mathbb{Q}^{ab}$ would settle other open conjectures/questions:

By a result of Pop, it would imply the following conjecture of Shafarevich:

Conjecture: The Galois group of $\mathbb{Q}^{ab}$ is a free profinite group.

By a result of Fehm and Petersen, the ampleness conjecture would imply an affirmative answer to the following question of Frey and Jarden:

Question: Is the rank of $A(\mathbb{Q}^{ab})$ infinite for every non-trivial abelian variety $A/\mathbb{Q}$?
There are partial results and evidence for both of these conjectures/question that are independent of the ampleness conjecture. For example the answer to the above question is known to be "yes" for elliptic curves, and for some other families of abelian varieties.
You can read more about this in this survey of Fehm and Bary-Soroker on ample fields and in the references therein.<|endoftext|>
TITLE: On Deligne's determinant of motives
QUESTION [7 upvotes]: This is a question about Deligne's conjecture on special values of L-functions. I have to confess that I've never understood the definition of the determinant which is supposed to give the right special value up to some power of $2\pi i$ and some rational number. Can anybody work out the following example? 
Let $X$ be a (smooth projective) curve of genus $g$ over $\mathbb{Q}$ and $M=h^1(X)$ the Chow motive corresponding to $H^1(X)$. The factor at infinity is simply
$L_\infty(M, s)=[2 (2\pi)^{-s}\Gamma(s)]^g$ 
and the dual of $M$ being $h^1(X)(1)$, one has
$L_\infty(M^\vee, s)=L_\infty(M, s+1)=[2(2\pi)^{-s}\Gamma(s+1)]^g$
By definition, an integer $n$ is critical if neither $L_\infty(M, s)$ nor $L_\infty/M^\vee, 1-s)$ have poles at $s=n$. So the only critical integer is $s=1$. Deligne's conjecture says that $L(M, 1)$ is a rational multiple of $c^\pm(M)$.So...
What is $c^\pm(M)$ in this case? 
Can one give a explicit description in terms of a basis of $H^0(X, \Omega^1_X)$ or $H^1(X, \mathcal{O}_X)$ and the homology $H_1(X(\mathbb{C}), \mathbb{Q})$? 
What is known about the conjecture in this case?

REPLY [5 votes]: You can compute $c^{\pm}(M)$ in this case by pairing $H_1(X(\mathbf{C}),\mathbf{Q})^{\pm}$ with $\Omega^1(X)$, which are all $\mathbf{Q}$-vector spaces of dimension $g$. Here $(H_1)^{\pm}$ means the $(\pm 1)$-eigenspace with respect to the action of complex conjugation on $X(\mathbf{C})$.
Moreover, the motive $M=h^1(X)$ coincides with the motive of the Jacobian $J$ of $X$. So Deligne's conjecture on $L(h^1(X),1)$ can be cast in terms of $J$, and in fact it amounts to the BSD conjecture for $J$ (more precisely, the BSD conjecture up to a non-zero rational factor). Note that the BSD conjecture is wide open for general abelian varieties over $\mathbf{Q}$.<|endoftext|>
TITLE: Groups in which all characters are rational.
QUESTION [27 upvotes]: The Symmetric groups $S_n$ has interesting property that all complex irreducible characters are rational  (i.e. $\chi(g)\in \mathbb{Q}$ for all $\mathbb{C}$-irreducible characters $\chi$,$\forall g\in S_n$). 
Question: What are other families of (finite) groups where all complex irreducible characters are rational? Are such (finite) groups characterised?

REPLY [16 votes]: All groups that be constructed from symmetric groups via cross products and wreath products have this property.  See Section 3 of my paper "Mass formulas for local Galois representations to wreath products and cross products" http://arxiv.org/pdf/0804.4679v1.pdf
So, for example, $((S_7 \wr S_4) × S_3) \wr S_8$ has a rational character table.  In fact, taking cross products and wreathing with $S_n$ preserves the property you are asking about (see above reference).
This includes several of the examples given: $(\mathbb Z/2\mathbb Z)^n,$  hyperoctahedral groups, and Sylow 2-subgroups of $S_n$.  I am not sure if the index 2 subgroups of hyperoctahedral groups can be constructed from symmetric groups via cross products and wreath products.<|endoftext|>
TITLE: Goodstein's theorem without transfinite induction
QUESTION [18 upvotes]: Goodstein's theorem is an example of a theorem that is not provable from first order arithmetic. All proofs of the theorem seem to deploy transfinite induction and I've wondered if one could prove the theorem without transfinite induction. Some time ago I came across this old usenet post where Torkel Franzen writes


Goodstein's
    theorem does not necessarily require transfinite induction for its
    proof, but it's not provable in elementary arithmetic. It can be
    proved by ordinary induction on a statement involving quantification
    over sets...


What is the proof/statement Franzen is referring to?

REPLY [4 votes]: I am quite convinced that Goodstein's theorem can be proven with the constructive omega rule. Or "meta-logic". For every individual 'n' Goodstein's theorem can be proven in elementary arithmetic (the problem is the quantification). In a meta-proof you can prove that for every 'n' such valid proof can be generated. I believe that this "meta-proof" only requires elementary arithmetic (if the meta-proof would require second order logic, then we wouldn't have achieved anything).
Of course, I can only claim this by detailing this out, but given my understanding of the Goodstein theorem, I am quite convinced that it should be possible.<|endoftext|>
TITLE: Is a C* completion of a nuclear Fréchet algebra a nuclear C* algebra?
QUESTION [7 upvotes]: I am sure that this is well known in the right places, but: Is the C* completion of a star nuclear Fréchet algebra a nuclear C* algebra? (Suppose that the C* norm is continuous with respect to the Fréchet topology.)
[Basic, but commutative, example to explain topology - smooth functions on a compact manifold completed to continuous functions.]
To explain more: I expect that if the result was true, all that would be needed would be that the product and star were continuous. If we take continuous seminorms $\|.\|_1 \le \|.\|_2 \le \dots$ then likely we should have $\|a\,b\|_n\le C\, \|a\|_m\,\|b\|_m$ where $m,C$ just depend on $n$. By C* completion, this would be via some star representation of the algebra on a Hilbert space $H$, so that the map from the algebra to $B(H)$ was continuous.

REPLY [3 votes]: Counterexample.
I think the $C^\star$ algebra of the Free group on two generators $F_2$ is not a nuclear $C^\star$ algebra, same for the reduced group $C^\star$ algebra of $F_2$.
The finite support functions $c_f (F_2)$ is a dense subagebra of $C^\star(F_2)$, but is not Frechet.
Let $\ell$ be the word length of $F_2$.  Let $\omega=e^\ell$, the exponentiated word length function on $F_2$.  Then $\omega(gh) \leq \omega(g) \omega(h)$ for $g, h \in F_2$, and $\omega(e) = 1$.   
Form rapidly vanishing complex-valued functions on $F_2$ using this "weight" $\omega$.  I'll denote them by ${\cal S}^e(F_2)$, ${\cal S}$ for "Schwartz functions".  There is a natural Fr'echet space topology on ${\cal S}^e(F_2)$, given by norms $\| \varphi \|_n = sup_{g \in F_2} w(g)^n |\varphi(g)|$.  ${\cal S}^e(F_2)$ is naturally a Fr'echet algebra for convolution multiplication, because $\omega$ has the submultiplicative property.    Also ${\cal S}^e(F_2)$ is nuclear, since the exponentiated word lenght function satisfies a summability condition $\sum_{g \in F_2} {1\over{e^{pl(g)}}} <  \infty$ for some $p>0$.  And ${\cal S}^e(F_2)$ is dense in the $C^\star$-algebra since it contains $c_f(F_2)$.<|endoftext|>
TITLE: Is the counit of geometric realization a Serre fibration?
QUESTION [28 upvotes]: Recall that a Serre fibration between topological spaces is a map which has the homotopy lifting property (HLP) for all CW complexes (equivalently for all disks $D^k$). The Serre fibrations are the fibrations in a model category structure on topological spaces in which the weak equivalences are the weak homotopy equivalences. The cofibrant objects include the CW-complexes. 
The geometric realization and singular simplicial set functors from an adjunction
$$|-|: sSet \leftrightarrows Top: Sing_*$$
which is in fact a Quillen equivalence between the above model structure and the standard model structure on simplicial sets. 
The geometric realization of every simplicial set is a CW-complex and the counit map $$\epsilon_X: |Sing(X)| \to X$$ is a weak homotopy equivalence. Hence this gives a nice functorial CW-approximation for every space. Moreover $|Sing(X)|$, being a CW-complex, is cofibrant. 
I would like this to be a cofibrant replacement of $X$, however for that I should further require that the counit map $\epsilon_X$ is also a Serre fibration. 
This is not so obvious to me because to check the lifting property I have to map into a geometric realization (which is a bit subtle). However using cellular approximation I have been able to show that this map has a weaker fibration property; it is a sort of "Serre fibration version" of a Dold fibration (where we only have a weak homotopy lifting property). This is good enough for many applications, but it still leads me to ask my question:

Is the counit map $\epsilon_X$ a Serre fibration? Are there conditions on X (such as paracompactness) which will make this hold true? If it is not a Serre fibration generally, what is the easiest counterexample?  

Another related question is whether in Top there is a functorial cofibrant replacement by CW-complexes. The usual approach would probably be to use the small object argument, which would result in a cellular space (a space like a CW-complex but with the cells possibly attached out of order). I suspect that a little clever handicrafting can make this yield an actual CW-complex. Is that intuition correct?

REPLY [21 votes]: $\newcommand{\real}[1]{\left\lvert #1 \right\rvert}$$\newcommand{\Sing}[1]{\operatorname{Sing}(#1)}$$\newcommand{\counit}{\epsilon}$$\newcommand{\To}{\longrightarrow}$$\newcommand{\proj}{\mathrm{proj}}$$\newcommand{\NN}{\mathbb{N}}$$\newcommand{\RR}{\mathbb{R}}$Yes, the map $\real{\Sing{X}} \to X$ is a Serre fibration.
[Disclaimer: This answer is very long. A lot of what I will write is contained in Oscar's answer and in the comments. I present it here for completeness and convenience.]
The present answer adapts the constructions given by Oscar Randall-Williams in his answer. The missing point is to prove that the maps generalizing the ones appearing in Oscar's answer are always inclusions of retracts. We will actually prove that they are trivial cofibrations, which will fundamentally require the fact that finite cell complexes are Euclidean neighbourhood retracts. Please upvote Oscar's answer.
Setup
Let $X$ be a topological space, and let $\counit_X:\real{\Sing{X}}\to X$ be the counit map of the adjunction between the singular complex functor and the geometric realization functor. We want to show that $\counit_X$ is a Serre fibration. Let then $h:\real{\Delta^n} \to \real{\Sing{X}}$ and $H:\real{\Delta^n} \times I = \real{\Delta^n \times \Delta^1}\to X$ be continuous maps. We need only prove that it is possible to provide a diagonal lift for the diagram
$$ \begin{array}{ccc}
\real{\Delta^n} & \overset{h}{\To} & \real{\Sing{X}} \\
\Big\downarrow\rlap{\scriptstyle \iota_0} & & \Big\downarrow\rlap{\scriptstyle \counit_X} \\
\real{\Delta^n}\times I & \underset{H}{\To} & X
\end{array} $$
Constructions (adapted from  Oscar Randall-Williams' answer)
The space $C$.
Since simplices are compact, the image of $h:\real{\Delta^n} \to \real{\Sing{X}}$ is contained in the realization $\real{K}$ of some finite sub-simplicial set $K$ of $\Sing{X}$. Then $h$ restricts to a map $h:\real{\Delta^n}\to\real{K}$, and we can take the mapping cylinder of $h$:
$$ C = M_h = \real{K} \coprod_{\real{\Delta^n}} \bigl( \real{\Delta^n}\times I \bigr) $$
This mapping cylinder generalizes the space $C$ described by Oscar Randall-Williams in his answer.
The space $D$.
The preceding space $C$ includes naturally into the mapping cylinder
$$ D = M_{\proj_{\real{K}}} = \real{K} \coprod_{\real{K}\times\real{\Delta^n}} \bigl( \real{K}\times\real{\Delta^n}\times I \bigr) $$
of the projection map $\proj_{\real{K}} : \real{K}\times\real{\Delta^n} \to \real{K} $. Importantly, observe that since geometric realization preserves colimits and finite products, the space $D$ is also the geometric realization
$$ D = \real{M_{\proj_K}} $$
of the simplicial mapping cylinder $M_{\proj_K} = K \coprod_{K\times\Delta^n} (K\times\Delta^n\times\Delta^1)$ of the projection $\proj_K: K\times\Delta^n \to K$.
The space $D$ plays here the role of the join appearing in Oscar Randall-Williams' answer: note that the join $\real{K}\ast\real{\Delta^n}$ is naturally a quotient of $D$.
The maps.
The inclusion map $j:C \to D$ is given by:

$j$ restricted to $\real{K}$ is the canonical inclusion $\real{K}\hookrightarrow D$ of the end of the mapping cylinder;
$j([x,t]) = [h(x),x,t]$ for $(x,t)\in \real{\Delta^n} \times I$ (where we see $D$ as a quotient of $\real{K}\times\real{\Delta^n}\times I$).

There is a further map $G:C\to X$ determined by:

$G$ restricted to $\real{K}$ coincides with $\counit_X$;
$G$ restricted to $\real{\Delta^n}\times I$ coincides with $H$.

Main argument: $j$ is a trivial Hurewicz cofibration
Now that we have adapted Oscar's construction, the main part of the argument consists of showing that $j: C\to D$ is a trivial cofibration.
First, the map $j$ is easily seen to be injective. Since $C$ is compact (because both $\real{K}$ and $\real{\Delta^n}\times I$ are compact) and $D = \real{M_{\proj_K}}$ is Hausdorff, it follows that $j: C\to D$ is a closed map, and in particular a homeomorphism onto its image.
Second, the map $j$ is a homotopy equivalence. Simply note that the composition of $\real{K} \hookrightarrow C \overset{j}{\to} D$ is the canonical inclusion $\real{K} \hookrightarrow M_{\proj_{\real{K}}} = D$ into the mapping cylinder, and thus a homotopy equivalence. Similarly, $\real{K}\hookrightarrow C$ is the inclusion into the mapping cylinder, and thus a homotopy equivalence. By the two-out-of-three property for homotopy equivalences, $j$ is itself a homotopy equivalence.
It remains to show that the inclusion of $j(C)$ into $D$ is a Hurewicz cofibration. Observe that both $C$ and $D$ are finite cell complexes, and in particular are Euclidean neighbourhood retracts (ENRs). This can be proved in the same way as corollary A.10 in the appendix to Allen Hatcher's book "Algebraic topology", which states that finite CW-complexes are ENRs. The desired result is now encoded in the following lemma.
Lemma: Assume $Y$ is a closed subspace of the topological space $Z$. Assume also that $Y$ and $Z$ are both ENRs. Then the inclusion of $Y$ into $Z$ is a Hurewicz cofibration.
This result actually holds in general for ANRs, and is stated as proposition A.6.7 in the appendix of the book "Cellular structures in topology" by Fritsch and Piccinini. Nevertheless, for completeness, I will provide a proof of the lemma at the end of this answer which uses only the results in the appendix of Hatcher's book when $Z$ is compact.
I will now conclude the proof that $\counit_X$ is a Serre fibration.
Conclusion
First, observe that $j:C\to D$ is the inclusion of a strong deformation retract because it is a homotopy equivalence and a Hurewicz cofibration. In particular, $j$ admits a right inverse $r:D\to C$. The composite $\overline{G} = G\circ r:D\to X$ is then an extension of $G:C\to X$ along $j$, i.e. $\overline{G}\circ j = G$.
Now we use the description of $D$ as the realization of the simplicial set $M_{\proj_K}$ to give a diagonal lift for the square diagram at the beginning of this answer. Note that to give a map $\overline{G}:D=\real{M_{proj_K}} \to X$ is by adjunction the same as giving a map $F:M_{proj_K}\to\Sing{X}$. I claim that the composite
$$ \widetilde{H} : \real{\Delta^n}\times I \hookrightarrow C \overset{j}{\To} D \overset{\real{F}}{\To} \real{\Sing{X}} $$
is such a diagonal lift:

$\counit_X \circ\real{F}= \overline{G}$ by construction of $F$ via adjunction. Consequently, $\counit_X \circ \real{F}\circ j = G$.
In particular, $\counit_X \circ \real{(F|_K)} = \counit_X \circ (\real{F})|_{\real{K}} = \counit_X \circ \real{F} \circ j|_{\real{K}} = G|_{\real{K}} = \counit_X$. This implies by adjunction that $F|_K$ is the inclusion of $K$ into $\Sing{X}$. Therefore, the restriction $(\real{F})|_{\real{K}} = \real{(F|_K)}$ is the inclusion of $\real{K}$ into $\real{\Sing{X}}$.
It follows that $\widetilde{H}\circ \iota_0 = \real{F}\circ j|_{\real{K}} \circ h = (\real{F})|_{\real{K}} \circ h = h$, i.e. the map $\widetilde{H}$ makes the upper triangle commute.
Furthermore, it follows from item 1 that $\counit_X \circ\widetilde{H} = \counit_X \circ\real{F}\circ j|_{(\real{\Delta^n}\times I)} = G|_{(\real{\Delta^n}\times I)} = H$. So $\widetilde{H}$ makes the lower triangle commute.


Proof of the lemma
We will use (a simplification of) the characterization of Hurewicz cofibrations given in theorem 2 of Arne Strøm's article "Note on cofibrations" (published in Mathematica Scandinavica 19 (1966), pages 11-14).

The inclusion of a closed subspace $Y$ of a metrizable topological space $Z$ is a cofibration if there exists a neighbourhood $U$ of $Y$ in $Z$ which deforms in $Z$ to $Y$ rel $Y$. More explicitly, there must exist a homotopy $F:U\times I\to Z$ such that $F(x,0)=x$ and $F(x,1)\in Y$ for $x\in U$, and $F(y,t)=y$ for $(y,t)\in Y\times I$.

This characterization of closed cofibrations is very similar to (and follows easily from) the usual characterization in terms of NDR-pairs, except that it does not demand the homotopy to be defined on the whole space.
Assuming that $Y$ and $Z$ are ENRs, we will now prove the existence of the neighbourhood $U$ and the homotopy $F$ as above.
A space $A$ is a ENR exactly when:

$A$ is homeomorphic to a closed subspace of $\RR^N$ for some $N\in\NN$;
for any $N\in\NN$, if $B$ is a closed subspace of $\RR^N$ which is homeomorphic to $A$, then some neighbourhood of $B$ in $\RR^N$ retracts onto $B$.

[For reference, the aforementioned appendix of Allen Hatcher's book "Algebraic topology" explains these two points when $A$ is compact.]
So we may assume without loss of generality that $Z$ is a closed subspace of $\RR^N$, and that some neighbourhood $V$ of $Z$ in $\RR^N$ retracts to $Z$ via a retraction $r_Z:V\to Z$. Since $Y\subset Z\subset\RR^N$ is closed in $\RR^N$ and $Y$ is a ENR, there also exists a neighbourhood $W$ of $Y$ in $\RR^N$ which admits a retraction $r_Y:W\to Y$. Using the convexity of $\RR^N$, we can produce a straight line homotopy $SLH:W\times I\to\RR^N$ between the identity of $W$ and $r_Y$. We may assume without loss of generality (by shrinking $W$ if necessary) that the image of the homotopy $SLH$ is contained in the open $V$. Define then the desired neighbourhood by $U=W\cap Z$ and the homotopy by $F = r_Z \circ SLH$.<|endoftext|>
TITLE: Lyndon-Schützenberger for torsion-free hyperbolic groups
QUESTION [7 upvotes]: Given a torsion-free hyperbolic group $G$, does there exist a number $n(G)$ such that for any $x,y,z\in G$, $x^n y^n z^n =1$ implies that $x$, $y$, and $z$ commute pairwise?
Some musings/questions...
When $G$ is free, the result is true for $n=2$. 
Clearly, this does not generalize to hyperbolic groups (e.g. non-orientable surface groups of genus 3). 
Somewhat related: given any 3 non-commuting elements $x,y,z\in G$ there is a number $n(x,y,z)$ such that $\langle x^n, y^n , z^n \rangle $ is free.  Is this true??..definitely in case of two elements (Gromov).

REPLY [8 votes]: The second statement is true if $x,y,z$ are not torsion as proved by Arzhantseva and, independently, by Kapovich and Weidmann (see Arzhantseva, Goulnara N.
A dichotomy for finitely generated subgroups of word hyperbolic groups.  Topological and asymptotic aspects of group theory, 1–10, 
Contemp. Math., 394, Amer. Math. Soc., Providence, RI, 2006 and Kapovich, Ilya; Weidmann, Richard
Nielsen methods and groups acting on hyperbolic spaces. 
Geom. Dedicata 98 (2003), 95–121. This also implies the first statement (about the equation $x^ny^nz^n=1$) in the torsion-free case.<|endoftext|>
TITLE: Is there a version of supersymmetry for homogeneous spaces?
QUESTION [11 upvotes]: The notion of "supersymmetry" that I am aware of proceeds as follows.  One fixes a spacetime $\mathbb R^n$ and signature; I will write $\mathrm{SO}(n)$ for the corresponding group of orthogonal transformations, as if I had chosen Euclidean signature, but of course this story also works with $\mathrm{SO}(n-1,1)$, ....  In any case, $\mathrm{SO}(n)$ has a double cover $\mathrm{Spin}(n)$, which is usually the universal cover.  The usual symmetry of usual spacetime is the group of rigid transformations $\mathrm{ISO}(n) = \mathrm{SO}(n) \ltimes \mathbb R^n$, and it is convenient to work instead with the double cover $\mathrm{ISpin}(n) = \mathrm{Spin}(n) \ltimes \mathbb R^n$.  I realize that these are not quite the standard names, but no matter; for example, you may replace the "I" in $\mathrm{SISO}$ with a "U", and the "O" with a "Y", if such changes make you feel better.
Now, suppose that you choose for yourself some real spin representation $\mathbb S$ of $\mathrm{Spin}(n)$ and a symmetric $\mathrm{Spin}(n)$-equivariant pairing $\Gamma: \mathbb S \otimes \mathbb S \to \mathbb R^n$.  Exactly what flavor of choices this requires depends on the value of $n$ mod 8, and on the signature, and perhaps other things.  As usual, "spin representation" means it extends to the appropriate clifford algebra.
Let $\pi \mathbb S$ denote the supervector space which is the "parity reversal" of $\mathbb S$.  Then $\Gamma$ makes sense as an antisymmetric pairing on $\pi\mathbb S$.  Indeed, it defines a nonabelian super Lie algebra structure on the supervector space $\pi\mathbb S \oplus \mathbb R^n$, whose only nontrivial bracket is $\Gamma : (\pi\mathbb S)^{\wedge 2} \to \mathbb R^n$.  Lacking a good notation, I will call this Lie algebra $\mathfrak t$.  It is a central extension of $\pi\mathbb S$ by $\mathbb R^n$, and fits into a nontrivial short exact sequence $\mathbb R^n \to \mathfrak t \to \pi\mathbb S$.  Exponentiating of finite-dimensional super Lie algebras is trivial if the non-super part can be exponentiated, and so we end up with a nonabelian super translation group $T$, also given as a central extension, now of super groups, of $\pi \mathbb S$ by $\mathbb R^n$.  By definition, everything is $\mathrm{Spin}(n)$-equivariant.  The super isometry group is the extension $\mathrm{SISO}(n,\mathbb S) = \mathrm{Spin}(n) \ltimes T$.

My question is: What happens when flat space $\mathbb R^n$ is replaced by some other geometry?  Simple cases include the round sphere, and de Sitter and anti de Sitter spaces.  More complicated examples include other symmetric or homogeneous spaces for other Lie groups.  Any "$G_2$ supersymmetry" or "$E_8$ supersymmetry" out there?

The motivation for my question comes from an observation that I learned from Dyson's excellent paper Birds and Frogs: $\mathrm{ISO}(n)$ is simply a limit as some "curvature" parameter goes to $\infty$ of $\mathrm{SO}(n+1)$, which is in the technical sense a more simple group.  Note that there will no longer be a well-defined "super translation group", as the round sphere is not (usually) a group, but perhaps there is still be a simple deformation of $\mathrm{SISO}(n)$, acting on some super homogeneous space?

REPLY [3 votes]: Homogeneous superspaces generalizing the "Super-Poincaré\Lorentz =
Superspace" exist. These superspaces are given as coset spaces of super-Lie groups $G/H$. Please see for example, the following article  by: 
Schunck and Wainwright containing an explicit construction of the
supersphere $S^{2|2}$ as the supercoset space $UOSP(1|2)/U(1)$. There are also treatments of the super-Ads case, please see for example: N. Alonso-Alberca, E.Lozano-Tellechea, T. Ortin  . 
In super-coset spaces, the standard construction of the (super) frame bundle and (super) $H-$ connection  applies. However, In the contrast to the nonsupersymmetric case, where the algebra of $G$ is realized as an algebra of isometries of the Killing-Cartan metric; this metric does not have a natural generalization to supermanifolds, but we can still talk about isometries leaving the super-frame fields and the super-$H$ connection invariant up to a gauge transformation. The action of these super-isometries on functions (superfields) or more generally sections of super-vector bundles can be realized by means of differential operators.
There is another type of generalization of "flat supersymmetry", to an
arbitrary spin manifold $(M, g)$, (which can be applied to homogeneous
spaces having spin structures) by means of the construction of a super-cotangent bundle $T^{*}M \oplus \pi TM$, please,see for example the following work by: J.P. Michel. The super-cotangent bundle is a symplectic supermanifold and the generators of the supersymmetry algebra can obtained as a quantization of the corresponding symbols realized in the Poisson algebra of the super-cotangent bundle. The Dirac operator plays the role of the supersymmetry generator (it Poisson squares to the Laplacian). These supermanifold describe the phase spaces of a spinning particle moving on $(M, g)$.<|endoftext|>
TITLE: H*(braid group, irrep of symmetric group) = ?
QUESTION [10 upvotes]: As in the title, say $\lambda$ is some irrep of the symmetric group $S_n$, and $Br_n$ the braid group on $n$ strands, 

What is $H^*(Br_n, \lambda)$?

REPLY [18 votes]: I claim that the dimension of $H^*(Br_n;\lambda)$ is twice the dimension of the subspace of $\lambda$ fixed by the subgroup $S_2\leq S_n$.
Indeed, there is a spectral sequence
$$ H^p(S_n;H^q(PBr_n;\lambda)) \Longrightarrow H^{p+q}(Br_n;\lambda). $$
As a $PBr_n$-module, $\lambda$ is just a sum of finitely many copies of $\mathbb{Q}$ with trivial action.  This means that $H^q(PBr_n;\lambda)=H^q(PBr_n)\otimes\lambda$, which is a $\mathbb{Q}[S_n]$-modules and thus injective over $\mathbb{Z}[S_n]$.  This means that the spectral sequence collapses to an isomorphism
$$ H^*(Br_n;\lambda) = (H^*(PBr_n)\otimes\lambda)^{S_n}, $$
so you just want to know the isotypical pieces of $H^*(PBr_n)$.  The classifying space of $PBr_n$ is the configuration space of distinct $n$-tuples in $\mathbb{C}$, and there is an elegant argument based on this and the Lefschetz fixed point theorem to determine the character of the $S_n$-action on $H^*(PBr_n)$.  The answer is that if we take two copies of the trivial representation of $S_2$ and induce up to $S_n$, we get $H^*(PBr_n)$.  The claim now follows by Frobenius reciprocity.
This tells you the total dimension of $H^*(Br_n;\lambda)$, but not the degrees of the generators.  For that, you would need to look in more detail at the structure of $H^*(PBr_n)$.  There are generators $\alpha_{ij}\in H^1$ for $1\leq i,j\leq n$ with $i\neq j$, subject only to the relations $\alpha_{ij}=\alpha_{ji}$ and $\alpha_{ij}^2=0$ and
$$ \alpha_{ij}\alpha_{jk} + \alpha_{jk}\alpha_{ki} + \alpha_{ki}\alpha_{ij} = 0. $$<|endoftext|>
TITLE: How to estimate of $\prod_{k=a}^N \frac{1}{e^{k\kappa}-1}$ for large $N$?
QUESTION [5 upvotes]: I have to estimate the expression $\prod_{k=a}^N \frac{1}{e^{k\kappa}-1}$ for $\kappa$ very small $\kappa \sim 10^{-19}$ and $N$ very large $N\sim 10^{26}$ and $a$ arbitrary $a=1, \ldots, N$. I do not really need an exact expression, just the leading order expression in $N$.

REPLY [4 votes]: Taking logarithm gives
$$
-\sum_{k=a}^N \log(e^{k \varkappa}-1)=
-\frac1\varkappa \sum_{k=a}^N \log\left(e^{k \varkappa}-1\right)\varkappa.
$$
The last sum is a Riemann sum of the integral
$$
\int_{\varkappa a}^{\varkappa N}\log(e^x-1)\,dx=
\text{Li}_2(e^{\varkappa a})-
\text{Li}_2(e^{\varkappa N})+
i \pi  \varkappa( a-N),
$$
where $\text{Li}_2(x)$ is the polylogarithm.
So the product is approximately equal to
$$
e^{\frac1\varkappa (\text{Li}_2(e^{\varkappa N})-
\text{Li}_2(e^{\varkappa a}))+i \pi  (N-a)}.
$$

REPLY [3 votes]: I think I would go for a Euler-Maclaurin expansion of the second term of
$$
-\sum_{k=a}^N \log(e^{k\kappa}-1) = -\kappa\frac{(N+a)(N-a+1)}{2}-\sum_{k=a}^N\log(1-e^{-k\kappa}).
$$<|endoftext|>
TITLE: Inequalities for Hadamard products of complex symmetric matrices
QUESTION [8 upvotes]: Consider a complex symmetric matrix $$ C= C_R + i C_I $$ with $C_R,C_I \in \text{Mat}_{n\times n}(\mathbb R)$ symmetric, and assume that the eigenvalues of $C_R$ are all strictly positive. Then, $C$ is invertible, and the real part of its inverse is $$ (C^{-1})_R = \Big(C_R + C_I (C_R)^{-1} C_I\Big)^{-1}. $$ Let now $S\in \text{Mat}_{n\times n}(\mathbb R)$ be symmetric matrix with nonnegative eigenvalues and diagonal entries all equal to $1$. Let $$\tilde C = C \circ S $$ be the Hadamard (i.e. entrywise) product of $C$ and $S$. It is easy to show that the eigenvalues of its real part $\tilde C_R = C_R\circ S $ are convex combinations of the eigenvalues of $C_R $, and so $\tilde C $ is invertible with real part $$ (\tilde C^{-1})_R = \Big(\tilde C_R + \tilde C_I (\tilde C_R)^{-1} \tilde C_I\Big)^{-1}. $$ Here, $\tilde C_I=C_I\circ S $. Based on numerical simulations, and exact computations for special cases (such as $C_R= 1$ or $S $ a block matrix), I expect that the diagonal values $$ \Big(C_R + C_I (C_R)^{-1} C_I\Big)_{ii} \geq\Big(\tilde C_R + \tilde C_I (\tilde C_R)^{-1} \tilde C_I\Big)_{ii}.  $$ (Note that the $()^{-1}$ is gone). Since $(C_R)_{ii} = (\tilde C_R)_{ii} $ (as $S$ has $1$'s on the diagonal), this is really a statement about the diagonal values of $C_IC_R^{-1}C_I $. Numerically, it is even suggested that $$\Big( C_I^aC_R^{-b}C_I^a\Big)_{ii}\geq \Big( \tilde C_I^a\tilde C_R^{-b}\tilde C_I^a\Big)_{ii}  $$ for any $a,b\in \mathbb N $. 
I know this is a rather specific question, but any hints or suggestions from people with background in such problems would be greatly appreciated! This Lemma would very much simplify part of my research, which is in mathematical physics.

REPLY [4 votes]: The first inequality is true while the second one is false. The proof below suggests a version of the second inequality that does hold.
To ease notation, let $A = C_R$, and $B=C_I$. We wish to show that
\begin{equation*}
  (BA^{-1}B)_{ii} \ge [(S\circ B)(S\circ A)^{-1}(S\circ B)]_{ii}.
\end{equation*}
Since $A \succ 0$ (symmetric, positive definite), and $B$ is symmetric, using Schur complements it follows that
\begin{equation*}
  \begin{bmatrix}
    BA^{-1}B & B\\
    B & A 
  \end{bmatrix} \succeq 0.
\end{equation*}
Since $S$ is a correlation matrix, it is semidefinite; also $
\begin{bmatrix}
  S & S\\
  S & S
\end{bmatrix} = 11^T \otimes S \succeq 0$, whereby using the Schur product theorem we see
\begin{equation*}
  \begin{bmatrix}
  S & S\\
  S & S
\end{bmatrix} \circ \begin{bmatrix}
    BA^{-1}B & B\\
    B & A 
  \end{bmatrix} = \begin{bmatrix}
    S \circ BA^{-1}B & S \circ B\\
    S \circ B & S\circ A 
  \end{bmatrix} \succeq 0.
\end{equation*}
Now using Schur complements again, this implies that (in Loewner order)
\begin{equation*}
    S \circ BA^{-1}B \succeq (S\circ B)(S\circ A)^{-1}(S\circ B).
\end{equation*}
Since $S$ is a correlation matrix, it has 1s on its diagonals, so that $[S \circ BA^{-1}B]_{ii} = (BA^{-1}B)_{ii}$, which yields
\begin{equation*}
  (BA^{-1}B)_{ii} \ge [(S\circ B)(S\circ A)^{-1}(S\circ B)]_{ii},
\end{equation*}
as desired. 
To prove the other inequality, we proceed similarly, by starting from (using $a, b \in \mathbb{N}$),
\begin{equation*}
  \begin{bmatrix}
    B^aA^{-b}B^a & B^a\\
    B^a & A^b
  \end{bmatrix} \succeq 0.
\end{equation*}
This ultimately leads to the inequality
\begin{equation*}
  (B^aA^{-b}B^a)_{ii} \ge [(S\circ B^a)(S\circ A^b)^{-1}(S\circ B^a)]_{ii}.
\end{equation*}

NOTE: The original version written by the OP, i.e.,
\begin{equation*}
 (B^aA^{-b}B^a)_{ii} \ge [(S\circ B)^a(S\circ A)^{-b}(S\circ B)^a]_{ii}
\end{equation*}
is false (Matlab quickly yields a counterexample; I can include it here in case anyone needs the gory details).<|endoftext|>
TITLE: Solving for special rational triangles
QUESTION [6 upvotes]: I ran into a need for isosceles triangles that (1) have the two equal
integer side lengths $a$ (but the base $x \in \mathbb{R}$),
and (2) the apex angle $\gamma$ is a rational multiple of $\pi$. 
         


I found a few such triangles for particular $x$, 
but saw no general technique to solve this problem:

Given the base $x$, find all isosceles triangles over that base such that 
  side length $a$ is an integer
  and $\gamma$ is a rational multiple of $\pi$. For which $x$ are there solutions?

My underlying question is for non-isosceles triangles, with the side lengths $a$ and $b$
incident to $\gamma$ both
integers, but it seems already difficult for isosceles triangles.
The question arose in the design of gears :-)

REPLY [3 votes]: Here is an expanded take on my comment:
Imagine that we have a triangle of the sort you seek and the base angles are $\beta=\frac{p}{q}2\pi$ with the fraction in lowest terms (and $q \ge 6$). Then $x=(2\cos\beta)a$
We know that $2a \cos \beta$ is an integer for $q=6.$  Otherwise it is an algebraic integer of degree $\frac{\phi(q)}{2}$ (short proof below).
So a not totally satisfactory answer to

Given a value $x$, determine if it can be a base and, if so, find all isosceles triangles over that base such that
side length $a$ is an integer

is: If $x$ is rational then it must be an integer and it arises only from an equilateral triangle. Otherwise,

Verify that $x$ is an algebraic integer and determine the degree $d.$ (This means that you somehow know an exact value of $x.$)
Find the (finite list) of integers with Euler totient function $\phi(q)=2d.$ (See the comment on this below.)
Compute $2 \cos(\frac{2p \pi}{q})$ for all cases.
Check if $x$ is an integral multiple of any of them (and if so, of which). Voila!

That does not immediately shed any light (that I can see) on questions such as "Is there an $x$ which arises in over $7$ ways." But it does suggest picking a value $2d$ so that there are numerous solutions to $\phi(q)=2d$ and then computing the various $\cos(\frac{2p \pi}{q})$ and looking for rational ratios. Maybe some nice patterns will become clear. (What solutions do you have, if one may ask.)
The same procedure offers, in some sense, an answer to the modified question

For which $x$, algebraic of degree $d$,  are there solutions?

If one wanted a huge $d$ that might not be very practical.
Further comments:
Maple has the function invphi to compute the  solutions of  $\phi(q)=2d$.  It wasn't always perfect,  but that particular bug has been fixed. See also this question  .
The fact about the degree of $2 \cos \beta$ must be a standard fact involving Chebyshev polynomials, but this seems to work:
Let $\zeta=e^{\frac {2\pi i}{q}}.$ Then $2\cos\beta=\zeta^p+\zeta^{-p}$ is a sum of two algebraic integers and hence itself an algebraic integer. The dimension of $\mathbb{Z}[\zeta^p]=\mathbb{Z}[\zeta]$ relative to $\mathbb{Z}$ is $\phi(q)$ but $\mathbb{Z}[\zeta^p]$ has dimension $2$ over  $\mathbb{Z}[2\cos\beta] \subset \mathbb{R}$ as $t=\zeta^p$  satisfies $t^2-2\cos\beta t+1=0.$
I am not sure that you can have an $x$ in two ways in the isosceles case and feel that it might be fairly straightforward . But I have no proof. Do you have any examples? There seem to be none of degrees $4$ or $6$.<|endoftext|>
TITLE: Line bundles on symmetric product of a curve
QUESTION [7 upvotes]: Let $C$ be a complex smooth projective curve of genus $g$, $C_d$ its d-fold symmetric product and $C^d$ its d-fold fiber product. Let $L$ a line bundle on $C$, and $L\boxtimes\cdots\boxtimes L$ (d copies) the coresponding line bundle on $C^d$. Since $L\boxtimes\cdots\boxtimes L$ is invatiant under the action of the symmetric group $S_d$,  it descends to a line bundle $\widetilde{L}$ on $C_d$. 
My question is: 

Can $\widetilde{L}$ be expressed by $\theta$ and $x$, here $\theta$ is the class of the pull back of the theta divisor and $x$ is the class of $C_{d-1}$ in $C_d$? 
Is $\widetilde{K_C}$ the canonical bundle of $C_d$?

REPLY [5 votes]: Here are the answers to your questions.

The numerical class of the line bundle $\widetilde{L}$ is ${\rm deg}(L) \cdot x.$  To see why, note that the pullback of this class to $C^{d}$ must be ${\rm deg}(L)(X_{1}+ \cdots +X_{d}),$ where the divisor class $X_{i}$ on $C^{d}$ represents pullback of a point from the $i-$th projection. 
The canonical bundle on $C_{d}$ is not $\widetilde{K}_{C},$ but rather $\widetilde{K_{C}}(-{\Delta}/2),$ where $\Delta$ is the diagonal divisor on $C_{d}.$  EDIT:  Here is a justification.  If $\pi : C^{d} \rightarrow C_{d}$ is the quotient map, then $\Delta$ is the branch divisor of $\pi.$  By relative duality we have $$({\pi}_{\ast}\mathcal{O}_{C^d})^{\vee} \cong {\pi}_{\ast}(K_{C^d} \otimes {\pi}^{\ast}{K}_{C_d}^{\vee}) \cong {\pi}_{\ast}\pi^{\ast}(\widetilde{K}_{C} \otimes K_{C_d}^{\vee}) \cong \widetilde{K}_{C} \otimes {K}_{C_d}^{\vee} \otimes {\pi}_{\ast}\mathcal{O}_{C^d}$$
We then must have that $\mathcal{O}_{C_d}({\Delta}/2) \cong \det(\pi_{\ast}\mathcal{O}_{C^d})^{\vee} \cong \widetilde{K}_{C} \otimes K_{C_d}^{\vee}.$<|endoftext|>
TITLE: Existence of a possible counterexample in automaton semigroups
QUESTION [5 upvotes]: In an attempt to resolve a question posed by Cain in his paper on Automaton Semigroups (open problem 6.12), I would like to know if there exists a finite semigroup $S$ satisfying the following properties:

$S$ is self-dual (anti-isomorphic to itself)
$S\neq S^2$
$S^2$ is a band
$S$ has a faithful left-regular representation (i.e.all rows in the Cayley Table of $S$ are distinct)

I'm having no luck constructing one (or proving that one doesn't exist) so any help/suggestions would be greatly appreciated!

REPLY [3 votes]: I believe I have an example but you should check the details of whether it works.  Maybe GAP can be used.  
Let $X=\{1,2,3,4,5\}$ and $X'=\{1',2',3',4',5'\}$.  Let $a,b\colon X\to X$ be given by $$a=\begin{pmatrix}  1 & 2 & 3& 4 &5\\ 2& 3& 3 &4& 5 \end{pmatrix}\qquad b=\begin{pmatrix}  1 & 2 & 3& 4 &5\\ 4& 5& 4 &4& 5 \end{pmatrix}.$$  Let $T=\langle a,b\rangle$ where we view $a,b$ as functions acting on the right of $X$.   Then one checks $a\neq a^2=a^3$, $b^2=b$, $ba=b$ and I suppose these are defining relations.Anyway $T=\{a,a^2,b,ab,a^2b\}$.   In particular $T^2$ is a band and $T^2\neq T$.  Crucial is that $ab\neq a^2b$.
Let $T'=\langle a',b'\rangle$ be the dual semigroup obtained by reversing the multiplication of $T$.  So $a'b'=b'$ and still $a'\neq (a')^2=(a')^3$, $(b')^2=b'$. Also $b'a'\neq b'(a')^2$. We view $T'$ as functions acting on the left of $X'$ in the obvious way (replace $i$ by $i'$ for each $i$ in $X$ to get $a',b'$ from $a$ and $b$). Let $\overline {T}$ be the subsemigroup of $T'\times T$ generated by $(a',a)$ and $(b',b)$.  It has $11$ elements and satisfies $\overline{T}^2$ is a band, $\overline{T}^2\neq \overline{T}$ and $\overline{T}$ is self-dual via the obvious involution.  $\overline{T}$ almost acts faithfully on the left of itself except that elements of the form $(b'a',t)$ and $(b'(a')^2,t)$ act the same for any $t\in \{b,ab,a^2b\}$.
To remedy that let $R=X'\times X$ with the rectangular band multiplication $(i,j)(k,l)=(i,l)$.  Then $S=\overline T\cup R$ is a semigroup using the products in $\overline{T}$ and $R$ already defined and by putting $(u,v)(i,j)=(u(i),j)$ and $(i,j)(u,v)=(i,jv)$ for $u\in T', v\in T$, $i\in X'$ and $j\in X$. So $R$ is the minimal ideal of  $S$.  Note $S^2$ is a band, $S^2\neq S$ and still $S$ is self-dual (using the obvious involution on $X'\times X$ and the involution on $\overline{T}$).
Because $T'$ acts faithfully on the left of $X'$ we can now distinguish $(b'a',t)$ and $(b'(a')^2,t)$ (with $t$ as above) by the action on the left of $R$.  Clearly if $i\neq k$ then $(i,j)$ and $(k,l)$ do not act the same on the left of $R$.  On the other hand if $\{j,k\}\neq \{2,3\}$, then $$(i,j)(a',a)=(i,ja)\neq (i,ka)=(i,k)(a',a).$$  On the other hand $$(i,2)(b',b)=(i,5)\neq (i,4)=(i,3)(b',b).$$ Thus the action of $S$ on the left of itself is faithful.  Note that $S$ has $36$ elements. 
I hope this is correct and helps.<|endoftext|>
TITLE: Simultaneous Powers Far From 1
QUESTION [6 upvotes]: I'm looking for a reference or proof of the following. Let $K/\mathbb{Q}$ be a finite Galois extension of degree $n$. Let $a_1,\ldots,a_n$ be Galois conjugate elements in the ring of integers of $K$ with $a_1, \ldots, a_k$ ($1 \le k < n$) lying on the unit circle (hence are not roots of unity). Show there are infinitely many positive integers $m$ such that $a_1^m, \ldots, a_k^m$ are all simultaneously far from $1$ (where ``far" needs to be that the angle is at least $\pm\pi/6$).
It seems that Kroenecker's generalization of Dirichlet's simultaneous approximation should work as a hammer for this nail (see: Simultaneous diophantine approximation) but I've been unable to finish it.
Thank you,
Ben

REPLY [9 votes]: There is no proof because the desired result is false!
Indeed, for any $\theta \gt 0$ there exists an algebraic integer
of degree $n$ with $k \lt n$ conjugates $a_1,\ldots,a_k$ on the unit circle
such that for each $m$ at least one of $a_1^m,\ldots,a_k^m$ 
is within $\theta$ of $1$.
Suppose $\theta \geq \pi/r$ for some integer $r$.
Note that an algebraic integer with a conjugate $x$ on the unit circle
must be a unit, because $\bar x$ is an algebraic conjugate of $x$
(and thus also an algebraic integer) such that $x \bar x = 1$.
Let $b \in {\bf R}$, then, be an algebraic unit of degree $2r+4$,
with $2r+2$ conjugates $b_1,\ldots,b_{2r+2}$ on the unit circle,
and $b$ the unique conjugate such that $|b| \gt 1$;
and let $K$ be the splitting field of ${\bf Q}$.
Assume that the subgroup of ${\rm Gal}(K/{\bf Q})$
that fixes $b$ acts transitively on $b_1,\ldots,b_{2r+2}$.
Then take $a = b/b_1$.
[We shall see later how to construct such $b$; that's where I needed
the clarification on whether ${\bf Q}(a)$ is allowed to be a
non-Galois extension of ${\bf Q}$, though it may be possible to have
${\bf Q}(a)/{\bf Q}$ Galois.]
The conjugates of $a$ are the quotients $\beta/\beta'$
where $\beta,\beta'$ are conjugates of $b$ with $\beta' \neq \beta^{\pm 1}$.
I claim that for each $m$ at least one of these conjugates is
within $\pi/r$ of $1$, and thus a fortiori within $\theta$ of $1$.
Indeed the $2r+2$ numbers $b_j^m$ come in $r+1$ conjugate pairs,
and none equals $\pm 1$.  Therefore $r+1$ of the $b_j^m$ are on
the open arc $\lbrace e^{i\psi}: 0 \lt \psi \lt \pi \rbrace$
of length $\pi$.  We conclude that two of them are within $\pi/r$
of each other, and their ratio is a conjugate of $a^m$
whose angular distance from $1$ is less than $\pi/r \leq \theta$,
as claimed.
[Remark: $a$ has $n = (r+1)(2r+4)$ conjugates.  Indeed there are
$2r+4$ choices of $\beta$, and for each one $2r+2$ choices of $\beta'$;
but $\beta / \beta' = ({\beta'}^{-1}) / (\beta^{-1})$, so each conjugate
arises at least twice.  But $a = \beta/\beta'$ only for
$(\beta,\beta') = (b,b_1)$ and $(b_1^{-1},b^{-1})$, because
$|b| = |a| = |\beta|/|\beta'|$, and $|\beta|=1$ for all
$\beta \neq b^{\pm 1}$.  It follows that the number $k$ of
conjugates of norm $1$ is $\frac12(2r+2)2r = 2(r^2+r)$.]
It remains to find our unit $b$.  Let $F \subset {\bf R}$ be any totally real
number field of degree $r+2$ whose normal closure has Galois group $S_{r+2}$.
Choose positive $c \in F$ all of whose other embeddings are negative,
and assume that $F' := F(c^{1/2})$ has ${\rm Gal}(F'/{\bf Q})$ the full
hyperoctahedral group $\lbrace \pm 1 \rbrace^{r+2} \rtimes S_{r+2}$
(which is the usual case).
Let $\sigma$ be the Galois involution of $F'/F$,
which permutes the two real embeddings of $F'$
and acts as Galois conjugation on the $r+1$ complex embeddings.
By Dirichlet, the group of units of $F'$ has rank $r$,
and its $\sigma$-invariant subgroup has rank $r-1$.
Hence there is a rank-$1$ subgroup of units inverted by $\sigma$.
Let $b \in F'$, then, be a unit $b$ of infinite order
such that $b^{\sigma} = b^{-1}$.  Then all conjugates of $b$
other than $b^{\pm 1}$ lie on the unit circle, and are permuted
transitively by ${\rm Gal}(K/{\bf Q}(b))$ because
${\rm Gal}(K/{\bf Q}))$ is as large as possible.
Thus $a = b/b_1$ works as claimed, QED.
To get explicit examples for small $r$, we can let $b$ be a
Salem number
(which usually has hyperoctahedral Galois group, though that's
not guaranteed).  For example, for $r=3$ we can use for $b$
Lehmer's number, the larger real root of
$y^{10} + y^9 - y^7 - y^6 - y^5 - y^4 - y^3 + y + 1$.
This makes $a$ a unit of degree $n=40$ with $k=24$
conjugates on the unit circle such that for each $m$
at least one conjugate pair of conjugates $a_k$ satisfies
$|a_k^m - 1| \lt 1$.  The supremum over $m$ of $\prod_{j=1}^{24} |a_j^m - 1|$
still exceeds $1$, though probably not by as much as it would for a
typical unit with $24$ conjugates on the unit circle:
the value is apparently $2^{24}/5^5 = 5368.70912$,
nearly attained when the eight $b_k^m$ are approximately at
$1$, $1$, $-1$, $-1$, and the four roots of $5z^4 + 6z^2 + 5$.
Numerically, for $m \leq 10^7$ the largest product observed is
$5359.938\ldots$ for $m=953110$.<|endoftext|>
TITLE: Conjecture on homotopy groups of moduli space of self-dual connections
QUESTION [5 upvotes]: In the paper Stability in Yang-Mills Theories (1983), Taubes puts a (topological) bound on the Hessian of the YM-action on $S^4$. He consequently conjectured:
"The inclusion $\mathcal{M}_n\hookrightarrow\mathcal{B}_n$ induces an isomorphism of the pointed homotopy groups $\pi_k(\cdot)$ for $k\le 2|n|$."
Here $\mathcal{M}_n$ is the moduli space of self-dual connections on a principal $SU(2)$-bundle over $S^4$ of degree $n\in\mathbb{Z}$, and $\mathcal{B}_n$ is the space of pairs $(P,A)$ where $P$ is a principal $SU(2)$-bundle over $S^4$ satisfying $n=-c_2(P\times_{SU(2)}\mathbb{C}^2)$ (instanton number) and $A$ is a smooth connection on $P$.
This was 30 years ago, at a time when Yang-Mills was being squeezed for useful information. I want to guess this has been proven by now, but I cannot immediately locate a proof.
Is this still an open problem?
If not, where is the proof? If so, what is currently known about the problem (with references)?

REPLY [6 votes]: On afterthought I now asked Taubes himself, who pointed me to the paper The Topology  of Instanton Moduli Spaces (Boyer-Hurtubise-Mann-Milgram). This conjecture is a form of the Atiyah-Jones Conjecture.  The paper is essentially what is known about this problem, and in particular the above conjecture is still open.  Atiyah-Jones showed that $B_n\simeq \Omega^3_nSU(2)$, and this paper proves the partial result:
The (forgetful) inclusion map $\mathcal{M}_n\to \Omega^3_nSU(2)$ is a homotopy equivalence through dimension $\lfloor\frac{n}{2}\rfloor-2$.
I also asked Milgram, who confirmed that there are no further (significant) results.
I also asked Hurtubise, who remembers improving the stability range from $\frac{n}{2}$ to $n$ by computing a spectral sequence differential, but cannot remember the actual reference.<|endoftext|>
TITLE: What (classes of) Banach spaces are known to have Schauder basis?
QUESTION [14 upvotes]: Motivation:
I am trying to see for what class of Banach spaces the following result is true:

There exists an increasing sequence of finite dimensional subspace {$V_n$} of a Banach space X (with some property) and corresponding projections $P_n: X \to V_n$ such that
  a) $\cup V_n$ is dense in $X$
  b) $\sup_n ||P_n|| < \infty$

I know if the space has Schauder basis, then the result is automatic. Hence I would appreciate if you can let me know:
i) a reference for the above result (if it exists)
ii) positive results for large class of Banach spaces which has a Schauder basis
I have tried Googling and Wikipedia, but couldn't find systematic information about existence of Schauder basis. The only counterexample I found was given by Per H. Enflo.
Thank you! I apologize if there is any inappropriate etiquette in my post as I am relatively new to the forum.

REPLY [11 votes]: The property you define is usually called the $\pi$ property.
For a good expository article, read Casazza's contribution in the Handbook of the Geometry of Banach Spaces, vol. 1.
Most of the classical separable Banach spaces are known to have a Schauder basis. The book of Albiac and Kalton is good place to start. Singer's two volumes "Bases in Banach Spaces" probably contains more than you would ever want to know.
In [JRZ], which you mentioned in a comment, you will find some results on when the $\pi$ property implies the existence of a finite dimensional decomposition and some results on the existence of bases, such as a separable complemented subspace of an $L_p$ space must have a Schauder basis.<|endoftext|>
TITLE: Generalized central limit theorem
QUESTION [10 upvotes]: I am looking for a generalized central limit theorem for non-square integrable stationary sequences. More precisely I suspect that when $(X_j)_{j\geqslant 1}$ is a stationary sequence such that $X_i$ belongs to the domain of attraction of the $\alpha$-stable law then
$$n^{1/\alpha}\sum_{i=1}^n X_i$$
converges as soon as the dependence of $X_i$ is not too strong (maybe some mixing condition). 
Are there results of this sort known? Or maybe I am wrong and my conjecture is false.

REPLY [7 votes]: One of the most recent results in this area is 
Katarzyna Bartkiewicz, Adam Jakubowski, Thomas Mikosch, Olivier Wintenberger, Stable limits for sums of dependent infinite variance random variables, Probability Theory and Related Fields 150, 3 (2011) 337-372
which is available here.
It gives sufficient conditions on $(X_j)$ in order to ensure the convergence of partial sums divided by $a_n$, where the conditions involve $a_n$. It seems these ones involve $\alpha$-mixing coefficients. Work on $\phi$-mixing coefficient has been performed in 
Thomas Mikosch, Daniel Straumann, Stable limits of martingale transforms with application to the estimation of GARCH parameters, Annals of Statistics 2006, Vol. 34, No. 1, 493-522,
which can be downloaded here.<|endoftext|>
TITLE: stable reduction and blow down
QUESTION [6 upvotes]: Let $R$ be a regular local ring, not necessarily of dimension one. Let $X \to Spec R$ be a nodal curve, i.e. all the geometric fibers are reduced and has at most nodal singularities. Assume $X$ admits a stable model $Y$ over $Spec R$. Is it necessairily true that there is morphism $X \to Y$ over $Spec R$, which is "contraction of  unstable $\mathbb{P}^1$'s" over each geometric point of $Spec R$?

REPLY [4 votes]: As user61789 points out, there are two natural questions related to this post.  First, under the given hypotheses on the (proper) morphism $\pi_X:X\to \text{Spec}(R)$, and assuming that the generic fiber is stable (in the sense of Deligne, Mumford, Mayer, Grothendieck, Knudsen, ...), does there exist a stable model $\pi_Y:Y\to \text{Spec}(R)$ of the generic fiber.  Second, assuming that this stable model does exist, then does there exist an $R$-morphism $\nu:X\to Y$ that is an isomorphism on the generic fiber?
I claim that both of these questions have positive answers.  I am certain there are references, and it would be best to give one.  However, the basic idea (for at least one proof) is that it suffices to produce the invertible sheaf $\mathcal{L}$ that would be $\nu^* \omega_{Y/R}^{\otimes d}$ for $d$ sufficiently large and divisible, assuming that $Y$ and $\nu$ actually exist.  For then $Y$ would be the (relative) Proj (over $R$) of the corresponding graded ring of global sections of powers $\mathcal{L}^{\otimes n}$ for $n\geq 0$.  Of course it is necessary to prove that this ring is actually a finitely generated $R$-module, that $\mathcal{L}$ is globally generated (relative to $R$), and that the morphism $\nu$ contracts the unstable rational curves.  However, all of these things can be checked after base change, e.g., to geometric fibers.  
So why should such an invertible sheaf $\mathcal{L}$ exist?  Of course there are constructions (and I am sure they are written up in the references).  But if you believe that a stable model $Y\to \text{Spec}(R)$ and a morphism $\nu$ exists fppf locally, then you should believe that $\mathcal{L}$ exists already over $\text{Spec}(R)$.  That is because the obstruction to descending an invertible sheaf is an element in the Brauer group (interpreted broadly, soas to include the second group cohomology of any automorphism groups of the stable models of fibers), and this is a torsion group.  Thus, up to passing from the original invertible sheaf to some sufficiently divisible, positive tensor power, the sheaf will descend.  We gave ourselves that freedom through the integer $d$.  
Also, since we are working étale locally, we may assume that the family is pulled back from a "versal" family.  Then it suffices to produce the invertible sheaf for this versal family.  This addresses the issue user61789 raises: what if $X$ is not regular?  For a versal family, $X$ is regular, the generic fiber is smooth, fibers with $\geq r$ nodes occur in codimension $\geq r$ over $\text{Spec}(R)$, etc.  Also, now the invertible sheaf $\mathcal{L}$ can be constructed on an open subset whose complement is codimension $\geq 2$: via Hartog's theorem / Riemann extension / the S2 property of $X$, this invertible sheaf will extend uniquely to all of $X$.  Of course in codimension $1$ in $\text{Spec}(R)$, when there is at most one node on the "special fibers", it is easy to construct $\mathcal{L}$; namely, define $\mathcal{L}$ to be the relative dualizing sheaf $\omega_{X/R}$ twisted down by the (reduced) Cartier divisor in $X$ that is the union of all rational tails of fibers.  
This does not mean that we are done!  Why should the invertible sheaf $\mathcal{L}$ defined in this way have all the necessary properties?  However, if you believe that such an invertible sheaf can be constructed from a stable model $Y$ and morphism $\nu$ after fppf base change, then you must accept that the invertible sheaf defined above (the unique extension of the invertible sheaf defined in codimension $1$) does have these properties.
So why should a stable model $Y$ and a morphism $\nu$ exist fppf locally?  I am not going to reprove the existence of a proper Deligne-Mumford stack of stable curves -- that is delicate to prove (particularly in mixed characteristic, although the Artin-Winters proof does simplify the original argument).  However, if you believe in the existence of such a proper Deligne-Mumford stack, then, after allowing a base change, you should believe that such a projective model $Y$ and morphism $\nu$ can be obtained after a blowing up of $\text{Spec}(R)$.  Using Zariski's Main Theorem, to prove that the blowing up is not necessary, it suffices to prove that for any positive-dimensional fiber of the blowing up, the corresponding family of stable curves is actually Zariski locally trivial.  Since all of these stable curves are obtained by blowing down rational components of the corresponding at-worst-nodal fiber of $\pi_X$, this is obvious.  So, if you believe in the existence of a proper Deligne-Mumford stack, you should believe in the existence of $\mathcal{L}$.
Of course maybe you are trying to go through the proof of the existence of the proper Deligne-Mumford stack of stable curves!  In that case, the logic above is pretty much backwards: you construct the stable model from $\mathcal{L}$ and not the other way around.  However, my impression is that the question is really about why it is unnecessary to make a base change to construct $Y$ and $\nu$, assuming that a stable model exists after the base change, rather than about how one proves that always a stable model exists after a base change.
Edit.  I tracked down a reference, although there is still a little work to go from the reference to the claims above.
MR1487226 (99f:14028) Reviewed 
de Jong, A. J.(1-PRIN); Oort, F.(NL-UTRE-MI) 
On extending families of curves. 
J. Algebraic Geom. 6 (1997), no. 3, 545–562. 
14H10 (14K10)
This is available on de Jong's webpage.  They prove a stronger theorem.  Via the argument about passing to an fppf cover and then realizing the original family as the base change of a versal family, then one may assume that the generic fiber is smooth, the "boundary divisor" in the base is a simple normal crossings divisor, and that the generic points of this divisor parameterize curves with a single node.  In this case, the main theorem of de Jong and Oort is that the family of stable curves over the interior extends over the entire family if it extends over every generic point of the boundary.  The family is already stable except at the generic points where the fiber has a rational tail.  If one defines $\mathcal{L}$, as above, to be $\omega_{X/R}(D)$, where $D$ is the Cartier divisor of rational tails, then the relative Proj of the graded algebra of $\mathcal{L}$ gives a stable model over these generic points.<|endoftext|>
TITLE: Question about product topology
QUESTION [16 upvotes]: Suppose $S\subset\mathbb{R}$ is dense without interior point, and for every open interval $I,J\subset\mathbb{R}$, $I\cap S$ is homeomorphic to $J\cap S$.
Is $S\times S$ homeomorphic to $S$?
By Luzin scheme,  if $S$ is the set of rationals or irationals , I can see this statement is true.

REPLY [5 votes]: In the following paper, van Engelen constructs a strongly homogeneous, zero dimensional, Borel, dense subspace $Y$ of $2^{\mathbb{N}}$ such that while $Y$ does not admit a topological group structure, $Y^2$ does. I haven't read the proof but maybe you can use it to construct a counterexample in the context of $\mathbb{R}$?<|endoftext|>
TITLE: Largest subset of $GL_n(p)$ in which pairwise subtraction is also in $GL_n(p)$
QUESTION [12 upvotes]: Suppose $X\subset \mathrm{GL}_n(p)$ is a set of invertible matrices such that for every $A,B\in X$ then also $A-B\in \mathrm{GL}_n(p)\cup \{0\}$. (If anyone knows a name for such sets I would be grateful).
I was trying to figure out what is the size of the largest such $X$. Since $\mathbb{F}_{p^n}$ can be embedded in $\mathrm{GL}_n(p)\cup \{0\}$ then obviously the size is at least $p^n-1$.
My interest in this question was motivated from a computer science question (ideal secret sharing schemes), and from there it appears that $p^n-1$ must also be the upper bound.
I was wondering if anyone knows a more direct way of showing that this holds (something more algebraic or direct combinatorial arguments).

REPLY [13 votes]: The upper bound can be obtained as follows: Fix a nonzero vector $v$ from $\mathbb F_p^n$. Then the vectors $Bv$ for $B\in X$ are pairwise distinct and nonzero, for if $Bv=B'v$, then $B-B'$ is not invertible. Thus the map $X\to\mathbb F_p^n\setminus\{0\}$, $B\mapsto Bv$ is injective, hence $\lvert X\rvert\leq p^n-1$.
Your sets with $\lvert X\rvert=p^n-1$ do have a name, they are called spreadsets and they are closely related to finite translation planes. See here.<|endoftext|>
TITLE: Is “problem solving” a subject to be taught?
QUESTION [27 upvotes]: I am witnessing a new curriculum change in my country (Iran). It includes the change of all the mathematics textbooks at all grades. The peoples involved has sent me the textbook for seven graders (13 years old students) to have my advice about the book. The book starts with Polya's famous four steps of problem solving. Then it continues with "teaching" a bunch of problem solving strategies. That is the first chapter of the book. Personally I am against separating "problem solving" from "solving problems" (at least, for such young students). However, obviously as a MO question I am not up to discussion. Instead, I am looking for the facts.
I am aware that years ago it was a common approach in the USA to have a separate chapter like the one I described in the mathematics textbooks. Are such textbooks still in use? If yes, at what grade (or at what age), and in which country?
PS. Probably you are not living in a country with a centralized system like mine (in which all students at a certain grade use the same book). If this is the case, it would be great if you just mention the book you are aware of, or you have experienced.
PPS. I initially posted the question on https://math.stackexchange.com/questions/429492/is-problem-solving-a-subject-to-be-taught. Surprisingly after about one day I had just one comment (that was indeed useful, though not enough)
PPPS. Please do not rush to vote this question as off topic. Consider that next year about tens of thousands of students would read a textbook with or without such a chapter. And in this case, for coming to a decision, a piece of fact might be worth more than tens of reasons. That is why I didn't phrase the question directly as a research question.

REPLY [5 votes]: I realize Mathematics Education posts are often of questionable admissibility on MO. I will try to do the question here some justice by answering from within the field of Math Education, but I cannot speak to how widespread my own views on the matter are. If my response seems somewhat long, then I might suggest one consider its ratio to how broad the question title is; unless, of course, a length:breadth comparison only confuses further.
First, here is one concrete answer: James Stewart's tome on the Calculus sequence (2012) contains a section on Polya's problem solving strategies. This book is widely used in the United States at the tertiary (undergraduate) level. I cannot say Stewart has made a concerted effort to incorporate a discussion of Polya's four steps or the use of heuristics into latter parts of his text, so at least the sections on problem-solving and on Calculus are separate.
An issue of ambiguity now arises, for the question title ("Is problem solving a subject to be taught?") is somewhat different in spirit from the actual question (related to the organization of textbooks). If you restrict yourself to what appears in mathematical textbooks, then you may be doing the title-question a disservice. Probably some teachers make it a pedagogical goal to use the problem solving section as a reference that can be returned to repeatedly while teaching Calculus; probably others gloss over or completely skip the section. If you wish to explore more deeply issues related to teacher adherence to curricular materials, then the term fidelity is what should let you comb the literature.
Second, you asked a separate question that was framed as "an attempt to get an indirect answer for [this question]," and I had left an answer there with four Math Ed references some time ago. I hope that my general point about the difficulty of directly teaching heuristics was not lost, even as the question was ultimately closed, and that the relation to this post is apparent.
Third, I see elsewhere a mention of the Common Core State Standards for Mathematics (CCSSM pdf). Consider the Standards as a document, the forthcoming textbooks that will be designed to satisfy them in some quantifiable way, how teachers actually adjust (or don't) to these new texts, the corresponding professional development to implement them (nationally, I cannot say this will occur) and Standards-aligned examinations to evaluate students based on CCSSM (this will occur and already sample tests have been administered). The interplay between these components - and many others - is nontrivial, and I would be hesitant to conclude anything about how problem solving actually finds its way into the classroom, even after the next batch of textbooks is published.
From a historical perspective, the very issue you raise has been discussed and led to various curricular shifts every decade or so since at least 1980. An early document of relevance in the United States is the National Council of Teachers of Mathematics (NCTM) published piece An Agenda for Action, where the first recommendation, verbatim, is that "Problem solving be the focus of school mathematics in the 1980s." Subsequent documents of relevance include two more NCTM pieces: Curriculum and evaluation standards for school mathematics (1989) and Principles and standards for school mathematics (2000) before CCSSM was released in 2010. Plenty is written on each of these, and I'm sure a search through google scholar would be more useful than my attempt at a broad summary.
If I am to venture a guess as to the relevance of all this to your question: Assuming for a moment (perhaps unwisely) that we do not start over with a new set of standards in the near future, I expect the focus not to be on problem solving as a separate subject to be taught, but instead as a "Standard for Mathematical Practice" (CCSSM, p. 6) to be integrated with the more broadly-defined goal of sense-making in mathematics. (Not sense-making: A classic example discusses asking elementary school students, given that the farmer has 20 sheep and 10 cows, how old is the farmer? A shocking proportion of students will try to answer this question with a number: usually 30. This is no anomaly; even more extreme examples exist in which a numerical answer is given when no question at all was posed.)
I suspect that CCSSM will have a stronger effect than its predecessors, for two reasons: 1, its near nationwide acceptance by governors, which coincides not surprisingly with a shift towards a more centralized approach to education, and presages tests (hence accountability) of some sort or another; and 2, the realization that the Standards might have a long-term presence has led to some who might otherwise oppose such a document to try and make the best of the situation. I recently attended a colloquium given by Alan Schoenfeld at Teachers College, where he talked about related issues, and his work to help teachers with CCSSM despite the shortcomings it might have. (This talk has since been written up as a short article: Schoenfeld, A. Mathematical Modeling, Sense Making, and the Common Core State Standards. The Journal of Mathematics Education at Teachers College, 4(2).) Henry Pollak, who was involved in the School Mathematics Study Group (SMSG) of the 1950s and 60s behind New Math, remarked on the wonder of seeing others helping to promote and improve mathematics curricula with which they might not agree. (It is perhaps a lack of this sort of support that derailed New Math, and led to its condemnation and the following Back to the Basics movement, though its detractors would no doubt be surprised to realize the atavistic re-emergence of some wonderful materials developed around that time in "new" textbooks.)
Let me end somewhat abruptly here, for the question of how to teach problem solving or incorporate it into a curriculum is rather general, and perhaps all that was desired was a textbook reference or two.<|endoftext|>
TITLE: Existence of a generalized matrix inverse over an arbitrary field?
QUESTION [6 upvotes]: Let $A\in M_n(K)$ be a square matrix over a field $K$. The notion of inverse matrix was generalized by Moore and Penrose for real and complex matrices
(also called pseudo-inverse $A^{\dagger}$ of $A$, satisfying 
$AA^{\dagger}A=A, A^{\dagger}AA^{\dagger}=A^{\dagger}$ and $AA^{\dagger}$  Hermitian). This was again generalized to arbitrary fields with involutory automorphism of $K$ by Pearl, but the existence depends on additional rank conditions.
Question: What is known on the existence of a generalized inverse $A'$, which just satisfies $AA'A=A$, over an arbitrary field ? 
Do we also need an additional assumption, perhaps on the characteristic of $K$, or again some rank condition ?

REPLY [6 votes]: The condition $AA'A=A$ means that the semigroup of matrices is regular. This is really true: see Clifford - Preston, The Algebraic Theory of Semigroups, sec.2.2, ex. 6(g).<|endoftext|>
TITLE: What is the homotopy fiber of a fold map?
QUESTION [8 upvotes]: If $X$ and $Y$ are based spaces, let $p_X: X\vee Y\to X$ be the fold map, or projection, onto $X$.

What is the homotopy fiber $F$ of $p_X$?

I think I have an argument that $F$ is the half-smash product $$Y\rtimes\Omega X := (Y\times\Omega X)/(\ast\times \Omega X).$$
But I'm not entirely sure of this and would like a reference, if one exists.

REPLY [6 votes]: The synthetic summary of pretty much every approach described (this is just to give an easier forumalation of the actual calculation) is sometimes called Mather's Cube theorem, sometimes Ganea's (fiber) theorem, sometimes "flattening"; I like to call it distributivity, because it decategorifies to the equation $ x ( y + z) = (xy) + (xz)$.
Call the fiber you want $F$. The wedge sum $X\vee Y$ is a pushout
\begin{array}{c}
\bullet & \to & X \\
\downarrow & \ulcorner & \downarrow\\
Y & \to & X\vee Y
\end{array}
and the map $X\vee Y\to X$ (usually called "pinch") restricts as $1_X$ on $X$ and as $0: Y\to X$ on $Y$.  By the Pasting Lemmas, the fibers of the composites $Y \to X$ and $ X \to X $ are the pullbacks of $F$ along $Y \to X\vee Y$ and along $X \to X\vee Y$, respectively, and of course the fiber of $ \bullet\to X$ is $\Omega X$, either way you factor it; but these fibers we know because they are easy to describe; by the universality of this last fiber, there is a coherent cube:
\begin{array}{c}
\Omega X & & & \to & & & \bullet \\
& \searrow & & & & \swarrow \\
& & \bullet & \to & X \\
\downarrow & &\downarrow & \ulcorner & \downarrow & & \downarrow \\
& & Y & \to & X\vee Y\\
& \nearrow & & & & \nwarrow \\
Y \Omega X & & & \to & & & F
\end{array}
What distributivity (flattening/Mather/Ganea) says is that, because the bottom square (small, inner square) is a pushout, and the vertical (distorted) squares are all pull-backs, and the whole cube is coherent, so the upper square (large) is, as the bottom square, a pushout.  That is, $F\sim Y \Omega X / \Omega X $ is the half-smash you were suspecting.
It is important to make sure you understand the maps among the known fibers, when doing these calculations; not all maps $A B \to B$ are the obvious projection!  In the present case, there just aren't too many maps $\Omega X \to \bullet$, and you can choose the structure of $\Omega X$ via the fiber-of $Y\Omega X \to Y$, as in the leftmost square.
As "some guy on the street", I once asked here how broadly does the distributive law hold, and Jacob Lurie's answer was that, according to Rezk, among homotopical categories those with the distributive law are probably homotopy toposes.  I don't know how much that helps you, but searching for "Mather cube", or "Ganea fiber" or "flattening" will provide lots of relevant references.<|endoftext|>
TITLE: Quasi-isomorphisms in exact categories
QUESTION [9 upvotes]: I am trying to understand quasi-isomorphisms in an exact category as defined via the mapping cylinder. I would like to know whether these form a category of weak equivalences in the sense of Waldhausen.
Some background.
Recall that in an exact category in the sense of Quillen we have a good notion of a long exact sequence that is simply a chain complex whose differentials factor through short exact sequences whose kernels and cokernels are also objects of the exact category. Call such sequences acyclic.
By the Gabriel-Quillen embedding theorem we can consider every exact category $\mathcal{E}$ as a subcategory of some abelian category $\mathcal{A}$, with short exact sequences inherited from $\mathcal{A}$. In general it is not the case that the acyclic sequences of $\mathcal{E}$ are exactly the long exact sequences of $\mathcal{A}$ whose objects are in $\mathcal{E}$ (consider the exact category of free $R$-modules sitting inside the abelian category of all $R$-modules, for example).
In an abelian category a morphism of chain complexes is a quasi-isomorphism iff its mapping cone is exact. Since we don't have homology in exact categories it is sensible to define quasi-isomorphisms via the mapping cone. But now there are two possible classes of quasi isomorphism I can choose, either $q_{1}$, the morphisms between chain complexes of exact categories whose mapping cones are acyclic in $\mathcal{E}$, or $q_{2}$, the morphisms whose mapping cone is a long exact sequence when viewed in the ambient abelian category.
Clearly $q_{1} \subseteq q_{2}$ but I see no reason for them to be equal. So, after all this set-up, my question is: is there a standard choice here? Do either form a category of weak equivalences in the sense of Waldhausen? I'd like to work with $q_{1}$ as it avoids extrinsic information about the exact category but I can pass to $q_{2}$ if necessary. 
Any insights would be appreciated.
A bonus question, in case both form weak equivalences, do the corresponding $K$-theories differ? How?

REPLY [3 votes]: If $\mathcal E$ is karoubian, i.e. idempotent complete, then $q_1=q_2$ on bounded complexes. This was noticed in 1.11.8 of Thomason-Trobaugh's paper, see also 1.11.5. On unbounded complexes, $q_1\subsetneq q_2$ even in the karoubian case, e.g. in the category $\mathcal E$ of free $\mathbb Z/4$-modules, the complex
$$\cdots\rightarrow \mathbb Z/4\stackrel{2}\rightarrow \mathbb Z/4\stackrel{2}\rightarrow \mathbb Z/4\rightarrow \mathbb \cdots$$
is $\sim 0$ in $q_2$ but not in $q_1$. If $\mathcal E$ is the category of projective objects in an abelian category where any object has projective dimension $\leq d$, then you can check as an exercise that $q_1=q_2$. 
If you're interested in $K$-theory then you're probably interested just in bounded complexes, but also  in non-karoubian categories. You can combine the fact that $q_1=q_2$ on karoubian categories with the cofinality theorem to deduce that both weak equivalences yield the same $K$-theory in all cases.<|endoftext|>
TITLE: Growth of Poincaré duality groups
QUESTION [11 upvotes]: Can one prove that Poincaré duality groups cannot have intermediate growth?

REPLY [13 votes]: The question is well beyond of what is currently known about Poincare duality groups and groups of intermediate growth. The only known case is of PD(2) groups, since they are virtually surface groups. The answer is unclear already for PD(3) groups (conjecturally, they are 3-manifold groups, so they should not have intermediate growth). Note that before Perelman, it was unknown if 3-manifold groups could have intermediate growth. On the other side, there are no known examples of finitely presented (or even $FP_2$) groups of intermediate growth. For all what we know, they are never $FP_2$, while $PD(n)$ groups are $FP_n$ by definition.  
Edit: It is an old theorem (due to Avez, 1970) that amenable fundamental group of a closed nonpositively curved  manifold has to be virtually abelian. The same holds under the mere CAT(0) assumption (no smoothness is needed); this is proven by Burger and Schroeder, Math. Annalen, 1987. Groups of intermediate growth are, of course, amenable.<|endoftext|>
TITLE: An easy-to-state elusive combinatorial problem
QUESTION [6 upvotes]: Let $a_1,a_2 \in \mathbb{Q}:a_1≥1,a_2≥1$. What should be the minimum value of $x\in\mathbb{R}$: $n∈[1,x]$ to ensure that $4k−3≤na_1≤4k−1$ such that $k∈N$ and $4l−3≤na_2≤4l−1$ such that $l∈N$ for all $a_1, a_2$?
Numerical computations suggest the answer to this is $3$ but I'm out of ideas how to prove this formally.

REPLY [4 votes]: The bound is $3$. For readability, I'll change $(a_1, a_2)$ to $(x,y)$. Without loss of generality, $x \geq y$. We break into cases.
Case 1 (the main case): $y \geq 7/3$. In this case, there exists an integer $\ell$ such that $y \leq 4 \ell-3 < 4 \ell-1 \leq 3y$. As $r$ ranges from $(4 \ell-3)/y$ to $(4 \ell-1)/y$, the value of $rx$ increases by $2 (x/y) \geq 2$. Therefore, for some $r$ in this range, $rx$ must lie in an interval of the form $(4m-3,4m-1)$. For this $r$, we have $r \in [1,3]$ and $(rx, ry)$ of the desired form.
So, from now on, assume $y \leq 7/3$. Since this means $1 \leq y \leq 3$, if $4 \ell-3 \leq x \leq 4 \ell - 1$, we are done. So we may also assume that $4m-1 \leq x \leq 4m+1$ for some integer $m$.
Case 2: $y \leq 3x/(4m+1)$. In this case, take $r = (4m+1)/x$ to achieve $rx=4m+1$ and $ry \leq 3$. Note that $1 \leq r \leq 4m+1/4m-1 \leq 5/3 < 3$.
Note that, if $m \geq 2$, then $3x/(4m+1) \geq 3 (4m-1)/(4m+1) \geq 7/3$. So Cases $1$ and $2$ together cover all possible values for $y$ if $m \geq 2$. We are thus left to deal with $m=1$.
More precisely, we are left to deal with the triangle $T$ bounded by $x \geq 3$, $y \leq 7/3$ and $y \geq (3/5) x$. The vertices of $T$ are at $(3,9/5)$, $(3,7/3)$ and $(35/9, 7/3)$. It is easy to check that, for any point in this triangle, we can rescale it by a factor of $\leq 3$ to land in the square $[9,11] \times [5,7]$. 
The equality is tight on $\{ 3 \} \times (9/5, 7/3)$ and on the mirror image $(9/5, 7/3) \times \{ 3 \}$, as suggested by Doug Zare.<|endoftext|>
TITLE: Any reference to an algorithm for finding the largest empty circle on a sphere (with great-circle distance)?
QUESTION [7 upvotes]: Given a set $S$ of 2D points in the plane, there are known algorithms for finding the largest empty circle ($LEC$) of the set of points.
The $LEC$ problem is stated in this way: find a $LEC$ whose center is in the closed Convex Hull of S, empty in that it contains no points in its interior, and largest in that there is no other such circle with strictly larger radius.
O'Rourke describes a simple $O(n^2)$ algorithm in his book "Computational Geometry in C" (the algorithm is attributed to G.T. Toussaint, 1983. Computing Largest Empty Circles with Location Constraints. International Journal of Parallel Programming, v12.5, pp 347-358.), the algorithm goes like this:

Compute the Voronoi Diagram $VD$ of the set of points
Compute the Convex Hull $CH$ of the set of points
The center of the $LEC$ is at one of the $VD$ vertexes inside the $CH$ or it is at an intersection between one of the $VD$ edges and $CH$.
For each candidate center compute the radius of the circle centered on it and update the maximum radius.

Now, on the plane, with $P=(P_x,P_y)$, the radius is computed with the distance $dist(P,C)=\sqrt{(P_x-C_x)^2+(P_y-C_y)^2}$ for a proper $P$.
I am interested in an algorithm for a similar problem: I have a set $S'$ of 3D points lying on a sphere of radius $R$ and I would like to find the equivalent of the $LEC$. I can assume that all points of $S'$ lie on a hemisphere.
In the sphere problem, the distances are measured by means of the great-circle distance.
With $S$ the 2D circle is the locus of points equidistant from a fixed point on the plane; with $S'$ the locus of points on the sphere equidistant (distance measured by the great-circle distance) from a fixed point on the sphere is the spherical curve usually called small circle. So instead of the largest empty circle maybe I can call what I am looking for the largest empty small-circle.
Is there any published algorithm for my problem?

REPLY [8 votes]: I believe the same algorithm works.
Note the remarkable fact, first understood by Kevin Brown,
that the Voronoi diagram of points $P$ on a sphere is combinatorially identical 
to the (3D) convex hull of $P$: each face of the hull corresponds to a Voronoi vertex
(and to an empty circle). This leads to
an $O(n \log n)$ algorithm, as explained in this paper:

Hyeon-Suk Na, Chung-Nim Lee, Otfried Cheong,
  "Voronoi diagrams on the sphere." 
  Computational Geometry: Theory and Applications, 23 (2002) 183–194.
  (ACM link)

Here is a recent implementation article:

Zheng, Xiaoyu, et al. "A Plane Sweep Algorithm for the Voronoi Tessellation of the Sphere." Electronic-Liquid Crystal Communications [online] (2011).
  (PDF download.)

I can't resist re-sharing this image of a Vornoi diagram on a sphere from an earlier MO question:
          

          

Mathematica image by Maxim Rytin.<|endoftext|>
TITLE: Examples of families of stable genus 2 curves
QUESTION [9 upvotes]: Given a smooth genus 2 curve $C$, it is canonically a two-fold cover of a $\mathbb{P}^1$, branched at six points. Allowing stable curves allows degenerations in two directions: the six points are allowed to collide (in certain ways), and $\mathbb{P}^1$ can "break" into a degenerate conic in $\mathbb{P}^2$ (here a union of two distinct lines). For instance, a smooth genus one curve with two of its points glued together is an example of the former situation, while two genus one curves joined to each other along a point is an example of the latter.
It is possible to construct explicit families (ideally, over high-dimensional bases and without appealing to stable reduction results) of stable genus 2 curves  that include degenerations of both kinds? (I would even be interested in degeneration of smooth curves to the second locus of unions of elliptic curves.) One difficulty here is that, if $C$ is a union of elliptic curves, then the map from $C$ to the quotient of $C$ by its "hyperelliptic" involution (multiplication by $-1$ on each elliptic curve) is not flat at the singular point.

REPLY [4 votes]: An explicit example of such a family with $1$-dimensional basis can be found in my paper  On surfaces of general type with $p_g=q=1, K^2=3$. The relevant statement is Proposition 6.1, that  I will restate here for the reader's convenience.

Proposition. There exists exactly one irreducible family of minimal surfaces $S$ of
  general type  with $p_g=q=1$, $K^2=3$ and a rational pencil $|G|$
  of curves of genus $2$, and this family is parametrized by the
  coarse moduli space of elliptic curves. Moreover:

the pencil $|G|$ is base point free, hence it defines a fibration $f \colon S \to \mathbb{P}^1$ whose general fibre is a smooth genus $2$ curve;
$|G|$ is the only genus $2$ pencil on $S$;
$|G|$ contains $13$ singular elements; six of these are genus $1$ curves
  with an ordinary double point, the other seven consist of two
  smooth elliptic curves intersecting transversally at a single
  point.


As far as I know this family was first constructed by Xiao Gang in his book Surfaces fibrées en courbes de genre deux (Lecture Notes in Mathematics 1137, Springer 1985), see in particular Chapter 3. The costruction follows from the classification of non-isotrivial genus $2$ fibrations $f \colon S \to C$ ($S$ smooth surface, $C$ smooth curve) whose associated Jacobian fibration has a fixed part $E \times C$. 
In Xiao's book one can also find many other examples, see in particular the table at page $52$. The family of surfaces in the Proposition corresponds to the case $d=4$ in that table. These surfaces are explicitly mentioned in the Corollaire $4$, page $51$, where they are called fibrations $f(E, 4)$.<|endoftext|>
TITLE: Does the forgetful G-Groups to Groups (nonabelian) have a left adjoint?
QUESTION [7 upvotes]: Motivation
For a fixed group $G$, consider the category of $G$-groups, whose objects consist of (not necessarily abelian) groups $A$ together with an action $G\to \mathrm{Aut}(A)$. Now we can construct a nonabelian $H^1$, a functor from $G$-groups to pointed sets by mimicking the abelian construction. This construction is very common and there are Čech cohomology variants for instance. (Other variants of nonabelian cohomology $H^n(G,*)$ for all $n$ have also been constructed via crossed modules; see Labesse et al. in the Asterisque "Cohomologie, stabilisation et changement de base".)
One approach to try and construct higher $H^n(G,-)$ functors for the category of $G$-groups, would be to find a similar "nonabelian bar resolution". In the abelian case, the forgetful functor $U:G\text{-Mod}\to\text{Ab}$ has a left-adjoint $F:\text{Ab}\to G\text{-Mod}$ given by $A\mapsto \mathbb{Z}G\otimes_{\mathbb{Z}} A$. In this case we can then use the corresponding cotriple to construct the bar resolution in the usual fashion.
The Question:
A naive thing to do would be to look for a left adjoint to the forgetful functor $U:G\text{-Group}\to\text{Group}$, sending a $G$-group $A$ to the group $A$, forgetting the $G$-action. So my question is, does this forgetful functor have a left-adjoint? If so, what is it? My guess would be no, but I'm not sure. I've tried various things with free products but I can neither construct one nor prove it doesn't exist.

REPLY [6 votes]: Yes. I believe there's a general theorem to the effect that if $T_1 \to T_2$ is any morphism of Lawvere theories, then the corresponding functor $\text{Prod}(T_2, \text{Set}) \to \text{Prod}(T_1, \text{Set})$ between the categories of models has a left adjoint (I think this follows from the general adjoint functor theorem).
$\text{Grp}$ is the category of models of a Lawvere theory given by taking all operations you can write down from the group axioms; more explicitly, it's the opposite of the category of free groups $F_n$. $G\text{-Grp}$ is the category of models of a more complicated Lawvere theory where, in addition to the operations you can write down from the group axioms, there is one additional unary operation for each element of $G$, and these operations satisfy all of the relations in $G$ and satisfy the relations making them group homomorphisms. 
The general construction of the left adjoint is that one freely applies all possible operations in the Lawvere theory of $G\text{-Grp}$ to a group $A \in \text{Grp}$, then quotients by all relations coming from the fact that $A$ is already a group. Somewhat more explicitly, take the free monoid on symbols $(g_i a_i)$ where $g_i \in G, a_i \in A$, and quotient by all relations of the form $(g a)(g b) = (g (ab))$ and $(g \text{ id}_A) = (\text{id}_G \text{ id}_A)$. Multiplication is concatenation, inverses are given symbolwise by $(g a)^{-1} = (g a^{-1})$, the identity is $\text{id}_G \text{ id}_A$, and the $G$-action is given symbolwise by $g(ha) = (gh(a))$. 
Edit: Equivalently, take the free product $\bigsqcup_{g \in G} gA$ over copies of $A$ indexed by $G$ (note that $\mathbb{Z}[G] \otimes A$ can be described analogously when $A$ is abelian) with $G$ acting on the indexing set.<|endoftext|>
TITLE: Determinants in Graph Theory
QUESTION [26 upvotes]: In graph theory, we work with adjacency matrices which define the connections between the vertices. These matrices have various linear-algebraic properties. For example, their trace can be calculated (it is zero in the case of a loopless graph, i.e., an irreflexive symmetric binary relation). And we can also calculate their determinants. 
How would you interpret the determinant in the context of a graph?
For example, I teach network theory and the calculation of 'eigenvector centrality' requires the use of determinants. But the general question always comes up: what does the determinant mean in the context of the network (or graph)? Does it tell me of a property of the network that is useful?
In essence, I am trying to find a user-friendly interpretation of determinants in the context of networks or graphs.
I would be grateful for any assistance.

REPLY [12 votes]: If your graph is directed and each edge has weight $1$ then the determinant counts the number of not-necessarily-connected-cycles (that is subgraphs being disjoint unions of connected cycles) passing through every vertex of the graph. The cycle is counted as $-1$ if the number of its components has different parity than the number of vertices of the graph, otherwise it is counted as $1$.<|endoftext|>
TITLE: Caccioppoli-Leray Inequality for De Giorgi's theorem proof
QUESTION [6 upvotes]: I am studying De Giorgi's proof of Holder continuity of solutions of elliptic equations with bounded measurable coefficients.
This is the translation of the original paper
De Giorgi paper
At page 163, De Giorgi refers to a "Lemma by Caccioppoli and Leray" but I can't find it anywhere, the referenced book is very hard to come across.
If anyone has it ("Equazioni alle derivate parziali di tipo ellittico" by C.Miranda) and can look at what this lemma at page 153 is it would be great.
The inequality I am struggling with, is, in any case:
$$\int_{A(k)\cap B(y,\varrho_2)} (u(x)-k)^2 dx\geq (\varrho_2-\varrho_1)^2 \frac{\tau_1}{\tau_2}\sqrt{\int_{A(k)\cap \partial B(y,\varrho _1)}(u(x)-k)^2d\mu_{n-1}\cdot \int_{A(k)\cap \partial B(y,\varrho _1)}|\nabla u(x)|^2d\mu_{n-1}}$$
where $A(k)$ is the subset of the domain where the solution of the elliptic equation (with constants $\tau_1, \tau_2$) $u(x)$ is greater than $k$, $B(x,r)$ is the n-dimensional ball centred at $x$ of radius $r$ and $\partial$ indicates the boundary.
Thank you very much for any hint, reference or idea!

REPLY [11 votes]: I made a trip to the library and scanned the relevant pages from Miranda's 1955 book:
page 152-153 and page 154-155
the references are:
[3] J. Leray, J.Math. pures et appl. 17, 89-104 (1938)
[8] R. Cacciopoli, Rend.Acc.Lincei 22, 305-310 and 376-379 (1935).
[11] R. Cacciopoli, Giorn.Mat.Battaglini, 80, 186-212 (1950-51)<|endoftext|>
TITLE: wrapping M5-branes on a Riemann surface
QUESTION [7 upvotes]: AdS/CFT gives us a way to use geometry to study field theory! I am trying to wrap M5-branes on a Riemann surface $\Sigma_{g}$. In my problem, for a Riemann surface in 11d, the normal bundle is max $SO(5)$. Here is my question: How do we put $SO(2)$ in $SO(5)$?  

Urs Schreiber suggests the following mathematically precise interpretation of the question, which probably addresses the concerns of those who commented or who had voted to put the OP on hold: 

There is a famous construction of (N=2)-supersymmetric 4-dimensional Yang-Mills field theories and their Seiberg-Witten theory from the N=(2,0)-superconformal 6-dimensional field theory on the worldvolume of M5-branes: by Kaluza-Klein-compactifying the latter on a Riemann surface. This construction was revived more recently in 2009 by the influential article

Davide Gaiotto, N=2 dualities (http://arxiv.org/abs/0904.2715) 

On page 22 of this article, around the displayed formula (2.27), the author mentions that the Kaluza-Klein compactification of the 6d theory on a Riemann surface involves a “well known twisting procedure” of the holonomy of the Riemann surface by choosing an SO(2)-subgroup of the SO(5) group that is the “R-symmetry” group of the 6-dimensional supersymmetric field theory (the group under which its supercharges transform).
Question: What is this “well known twisting procedure” exactly, and how does it work? Of course I know how to find $SO(2)$-subgroups of $SO(5)$, but what does such a choice amount to in the context of the construction of an N=2, D=4 SYM from the 6d-field theory on the 5-brane?  Where is this twisting procedure discussed in the literature?

REPLY [6 votes]: Now that the question is open again (now in my paraphrasing), maybe I'll repost my reply from the nForum with some brief comments thrown in:
That the 6-dimensional (2,0)-superconformal QFT on the worldvolume of the M5-brane yields N=2 D=4 super Yang-Mills theory under Kaluza-Klein compactification on a 2d Riemann surface was known since about the mid 90s. Edward Witten had famously advertized this in the Proceedings to Graeme Segal's 60th birthday conference that by this construction the remaining invariance under Moebius transformations of that compactification manifold geometrically explains the "Montonen-Olive"/"electric-magnetic" S-duality invariance of (super) Yang-Mills theory. 
Later he realized further compactification of this down to 2-dimensions as a geometric realization of geometric Langlands duality. In the course of this the N=2 D=4 super Yang-Mills theory is "topologically twisted" in a way analogous to the well-known twisting of N=4 SYM that goes back to the work that won him the Fields medal. The twisting of the N=2 theory then also showed up in the more recent work by Gaiotto et al. that lead to the AGT correspondence. 
While the details for the topological twisting of the N=2 supersymmetric field theory are a tad more involved than those of the N=4 theory, the basic idea is the same: one picks an embedding of the spacetime rotation symmetry into the R-symmetry group (the one under which the supercharges transform) and then asks for a linear combination $Q$ of the supercharges that is held fixed by the resulting external+internal symmetry transformations. The cohomology of this $Q$ is then seen to pick inside the quantum observables of the origional super gauge field theory those of a topological field theory. That is the topologically twisted theory. 
Pointers to more details on this topological twisting that the above questing is after are collected here:
http://ncatlab.org/nlab/show/topologically+twisted+D=4+super+Yang-Mills+theory
Pointers specifically concerning the twistied Kaluza-Klein compactification of the M5-brane on a Riemann surface to a topologically twisted N=2 super Yang-Mills theory are here:
http://ncatlab.org/nlab/show/N=D2+D=4+super+Yang-Mills+theory#ReferencesConstructionFrom5Branes<|endoftext|>
TITLE: Density of prime pairs whose gap is less than the average gap
QUESTION [8 upvotes]: By the prime number theorem we know that the "average gap" between the first $n$ primes is $\ln p_n$. I would like to know the density of consecutive prime pairs whose gap is less than the average gap between the first $n$ primes? In other words, does the following limit exist and what is its value?
$$
\lim_{n \to \infty}\frac{1}{n}\#\Big\{m \le n : \frac{p_{m+1} - p_m}{\ln p_m} < 1\Big\}
$$

REPLY [20 votes]: It is conjectured  (see e.g. Goldston-Pintz-Yildirim http://arxiv.org/abs/1103.5886) that 
$$
 \lim_{n \to \infty}\frac{1}{n}\#\Big\{m \le n : \alpha<\frac{p_{m+1} - p_m}{\ln p_m} < \beta \Big\} = \int_\alpha^\beta e^{-t} dt,
$$
and thus in particular your value should be $1-e^{-1}$. (GPY proved that this is positive for any $0=\alpha<\beta$ when $\lim_{n \to \infty}$ is replaced by $\liminf_{n \to \infty}$). Gallagher proved this result under the Hardy-Littlewood $k$-prime conjecture, but since GPY also proved conditional results under the Elliot-Halberstam conjecture which are still weaker than this conjecture, it seems unlikely that the method used to prove Zhang's recent breakthrough result is sufficient to prove such a result (for reasons why Zhang's method can not give anything better than what can be proved under Elliot Halberstam, see this question Does Zhang's theorem generalize to $3$ or more primes in an interval of fixed length? )<|endoftext|>
TITLE: Kazhdan-Lusztig Polynomials and Intersection Cohomology
QUESTION [8 upvotes]: I hope this question has not been asked before.
I would like to know which Ideas led (Deligne), Kazhdan and Lusztig believe, that  Kazhdan–Lusztig polynomials can be expressed  via intersection cohomology groups of Schubert varieties?
Is there nowadays more insight why intersection cohomology groups (or perverse sheaves) appear in representation theory? (I should mention that I know about Koszul-Duality and localisation, so if there is another viewpoint I would appreciate to hear about it)

REPLY [8 votes]: First I'd comment that there are quite a few questions on MO related to this one, but apparently not quite identical.   (It's hard to search the site efficiently.)
In any case I won't attempt a detailed answer but will rather suggest a reference.
While Kazhdan and Lusztig themselves must have interesting views on this question, Steve Kleiman (MIT) has published a detailed history of intersection cohomology along with Kazhdan-Lusztig theory.   What I think is the last revised version of this article, The Development of Intersection Homology Theory, published as Pure Appl. Math. Q. 3 (2007), no. 1, 225--282, can be found here.   Obviously it's difficult to sort out the history and motivation to everyone's satisfaction, but Kleiman got enough feedback along the way to be reliable in his version.<|endoftext|>
TITLE: Zeros of polynomials with real positive coefficients
QUESTION [29 upvotes]: The following problem arose in collaborative work with Subhro Ghosh:
Question: To any polynomial $P_n(z)=\sum_{i=0}^n a_i z^i =a_n \prod_{i=1}^n(z-z_i)$, attach the empirical measure of zeros
$L_n=n^{-1}\sum \delta_{z_i}$ (a probability measure on the complex plane). 
Let ${\cal M}$ denote the collection of limit points (in the weak topology)
of those empirical measures obtained when all $a_i$
are constrained to be non-negative reals (but otherwise, arbitrary $a_i$), with limit points being probability
measures. Can ${\cal M}$ be characterized?
Remarks: of course, probability measures in $\cal M$ are symmetric about the real axis, so enough to look at restriction of measures to the (closed) upper half plane. Probability measures supported on the negative half plane belong to $\cal M$. On the negative side,
by a theorem of Obrechkoff, any probability measure in $\cal M$ charges the symmetric cone of total width $\alpha$ around the real axis with mass at most $\alpha/\pi$. Finally, for random polynomials with $a_i$ iid with bounded moments of some order, the limit point is the uniform measure on the circle, and from this (or using products of polynomials $\sum_{i=0}^k  z^i$) one easily shows that any radially symmetric probability measure belongs to $\cal M$. Results of Barnard et als on factoring polynomials with positive coefficients allow one to construct further examples of measures in $\cal M$.
There are more examples, but we look for a characterization of $\cal M$.

REPLY [20 votes]: I don't have a complete proof yet, but I have a plausible conjecture. Let $\mu$ be a probability measure in the plane, define the potential
$$u(z)=\int\log|1-z/t|d\mu(t).$$
Then I conjecture that $\mu\in M$ iff $u$ satisfies $u(z)\leq u(|z|)$ for all $z$.
It is evident that this condition is necessary.
It seems that it is strictly stronger than the Obrechkoff condition.
I don't think that this condition can be restated as a simple
property of $\mu$ itself.
To prove the sufficiency, I am first going to restrict to a dense subclass of $\mu$
with convenient properties (it is clear that it is enough to prove sufficiency for a dense
subclass). The convenient properties I have in mind is that $\mu$ does not charge
some small angular sector $|\arg z|<\epsilon$ and that it behaves nicely near $0$ and $\infty$, say has some small atom at $-\epsilon$ and nothing else in the disc
$|z|<100\epsilon$, and similarly at infinity. In addition, I want to require that
$u(|z|)>u(z)$ for all $z$ except on the positive ray.
Then I am going to discretize the measure to obtain a polynomial, whose $(1/n)\log|P_n|$
approximates $u$ nicely near the positive ray, and apply the saddle point method
to the integral
$$\int_{|z|=r}\frac{P_n(z)}{z^k}\frac{dz}{z},$$
with $n\to\infty$,
using the nice behavior near the positive ray, and obtain an asymptotic for the coefficients which will show that they are positive.
The difficulty is that the asymptotics must be uniform in $k$, but I hope to achieve this
by the arrangement near $0$ and $\infty$ described above.
In fact, there is an (unpublished and unproved) conjecture of Alan Sokal
that if a polynomial
satisfies $|P(z)|<P(|z|)$ then some sufficiently high power has positive coefficients.
This of course would imply sufficiency of my condition.
ADDED on July 19. The above outline is correct; we are writing a proof which will soon be posted on arxiv.
ADDED on August 23. Here is the precise statement. A probability measure $\mu$
is a limit measure if and only if it is symmetric with respect to complex conjugation, and $u(z)\leq u(|z|)$ where
$$u(z)=\int_{|\zeta|\leq 1}\log|z-\zeta|d\mu(\zeta)+\int_{|\zeta|>1}\log|1-z/\zeta|d\mu(\zeta).$$
(The potential I wrote earlier may be divergent for some probability measures,
so it has to be modified a little bit). A proof of this
is now available: https://arxiv.org/abs/1409.4640.
UPDATE on September 10, 2014. What I called "Sokal's Conjecture" above (Theorem 1 in the preprint cited above) turned out to be known before. It was proved by V. de Angelis, MR1976089.
This was found as a result of David Handelman's answer to another MO question:
Stability of real polynomials with positive coefficients.<|endoftext|>
TITLE: Vanishing theorems in positive characteristic
QUESTION [12 upvotes]: In the paper
Deligne, Pierre; Illusie, Luc (1987), "Relèvements modulo $p^{2}$ et décomposition du complexe de De Rham", Inventiones Mathematicae 89 (2): 247–270, doi:10.1007/BF01389078
I found the following

Theorem. Let $X$ be a smooth and projective scheme over a field $k$ with $\textrm{char}(k) = p > 0.$  Assume that $X$ admits a lift to $W_{2}(k)$. If $L$ is an ample line bundle on $X$ then
  $$H^{i}(X,L^{-1}) = 0 \: \: \textrm{for} \: \: i < \min(p, \, \dim X).$$

I just wanted to ask if anyone knows a theorem like this under weaker hyphoteses. More precisely I have the following

Question. Does such result hold for $X$ normal and $L$ big and nef?

REPLY [10 votes]: In the book Lectures on Vanishing Theorems by Esnault and Viehweg  (Birkhäuser 1992) this is stated as an open problem. See in particular Problem 11.7 page 132.
However, they are able to prove the result when $\dim X=2$, since in this case one can perform the embedded resolution of singularities for curves on surfaces. See in particular Proposition 11.5 p. 129 and the subsequent remarks.<|endoftext|>
TITLE: What is the difference between internal presheaves and presheaves on a total space?
QUESTION [12 upvotes]: Suppose that $\mathbb{C}$ is a category with finite limits and that $\mathcal{D}$ is a category internal to $\mathbb{C}$. We can also represent $\mathcal{D}$ as a fibration $\mathbb{D}\to\mathbb{C}$.
My main question is this: What is the difference between an internal presheaf on $\mathcal{D}$ and an ordinary (external) presheaf on $\mathbb{D}$?
Other related questions (or the same question phrased differently):

If $F$ is an internal presheaf on $\mathcal{D}$, we should be able to represent it as an internal fibration $\mathcal{F}\to\mathcal{D}$; how does this differ from an ordinary fibration over $\mathbb{D}$.
Given an internal presheaf, (I think) we should be able to externalize the corresponding internal fibration $\mathcal{F}\to\mathcal{D}$ to an external fibration $\mathbb{F}\to\mathbb{D}$. Is that correct? If so, what further properties does $\mathbb{F}$ satisfy, in addition to being an (external) fibration?
We can raise the same questions for internal sheaves and stacks (say for the regular topology). Do any additional subtleties arise in that situation?

REPLY [9 votes]: Let me first recall some basic facts.
Fix a category $\mathbb{C}$ with finite limits. Denote by $\mathbf{Cat}(\mathbb{C})$ the 2-category of $\mathbb{C}$ internal categories. It is well-known that $\mathbf{Cat}(\mathbb{C})$ is a full 2-subcategory of the 2-category $\mathbf{Cat}^{\mathbb{C}^{op}}$ of $\mathbf{Cat}$-valued presheaves on $\mathbb{C}$ via the usual externalization functor:
$$\mathit{fam}(-) \colon \mathbf{Cat}(\mathbb{C}) \rightarrow \mathbf{Cat}^{\mathbb{C}^{op}}$$
By so-called fibred Yoneda lemma (a variant of Yoneda lemma for weak 2-categories) $\mathbf{Cat}^{\mathbb{C}^{op}}$ is weakly 2-equivalent to the category $\mathit{PsFn}(\mathbb{C}^{op}, \mathbf{Cat})$ of pseudofunctors $\mathbb{C}^{op} \rightarrow \mathbf{Cat}$, also called $\mathbb{C}$-indexed categories or variations on $\mathbb{C}$.
Moreover, under assumption of AC in our meta-theory, the Grothendieck construction and its inverse induce equivalence between $\mathbb{C}$-indexed categories and fibrations over $\mathbb{C}$. Fibrations that actually correspond to $\mathbf{Cat}$-valued presheaves are called split; similarly a split cartesian functors between split fibration is a cartesian functor that corresponds to a strict natural transformation.
This means that under sufficiently strong meta-foundations one may equivalently speak of:

$\mathbb{C}$-internal categories
$\mathbf{Cat}$-valued presheaves on $\mathbb{C}$ satisfying suitable conditions
$\mathbb{C}$-indexed categories satisfying suitable conditions
fibrations over $\mathbb{C}$ satisfying suitable conditions

These conditions are called "smallness" (which may by characterised as "having generic object and generic morphisms").
Now, if we agree that the above views are equivalent, we may check how internally defined concepts transport along the equivalences. I shall start with fibrations internal to $\mathbf{Cat}(\mathbb{C})$.
Because the concept of an internal fibration is universally characterisable by (weighted-)limits (actually comma objects) and adjunctions, and since the externalization functor $\mathit{fam}(-) \colon \mathbf{Cat}(\mathbb{C}) \rightarrow \mathbf{Cat}^{\mathbb{C}^{op}}$ preserves limits and adjunctions (every 2-functors preserves adjunctions, since adjunctions are equationally defined), a $\mathbf{Cat}(\mathbb{C})$-internal fibration is mapped to an internal fibration between $\mathbf{Cat}$-valued (small) presheaves on $\mathbb{C}$. Therefore, it suffices to characterise fibrations in $\mathbf{Cat}^{\mathbb{C}^{op}}$, or by the above equivalences (to directly address your question), it suffices to characterise fibrations internal to fibrations over $\mathbb{C}$. 
So let us assume that there are two fibrations $p, q$ together with a cartesian functor $G$ (that is, a functor that maps cartesian arrows from one fibration to cartesian arrows in the other fibration) over $\mathbb{C}$:
$$\require{AMScd}
\begin{CD}
\mathbb{A} @>{G}>> \mathbb{B}\\
@VpVV @VVqV \\
\mathbb{C} @= \mathbb{C}
\end{CD}$$
Then $G$ is a fibration from $p$ to $q$ internal to fibrations over $\mathbb{C}$ iff $G$ is a fibration in the usual sense between categories $\mathbb{A}$ and $\mathbb{B}$. In terms of "fibrewise" structure it means that for each object $C \in \mathbb{C}$ the fibre $p_C$ of $p$ over $C$ is fibred (in our case: by the restriction of $G$) over the fibre $q_C$ of $q$ over $C$ and the Beck-Chevalley condition (the stability condition) holds.
Now, as you said, an internal presheaf $F \colon \mathcal{D}^{op} \rightarrow \mathbb{C}$ over $\mathcal{D}$ corresponds to a discrete internal fibration $\pi_F \colon \int F \rightarrow \mathcal{D}$. Its externalised version:
$$\require{AMScd}
\begin{CD}
\mathit{Fam}(\int F) @>{\mathit{fam}(\pi_F)}>> \mathit{Fam}(\mathcal{D})\\
@V{\mathit{fam}(\int F)}VV @VV{\mathit{fam}(\mathcal{D})}V \\
\mathbb{C} @= \mathbb{C}
\end{CD}$$
consists of a fibred collection of discrete fibrations (these fibrations are discrete almost by definition of an internal discrete fibration; in fact, the converse is also true). In other words, for each $C \in \mathbb{C}$ the restriction $\mathit{fam}(\pi_F)_C$ of $\mathit{fam}(\pi_F)$ to $C$ is a discrete fibration from fibre $\mathit{fam}(\int F)_C$ to fibre $\mathit{fam}(\mathcal{D})_C$, and so corresponds to a $\mathbf{Set}$-valued preseaf over fibre $\mathit{fam}(\mathcal{D})_C$.
Of course $\mathit{fam}(\pi_F) \colon \mathit{Fam}(\int F) \rightarrow \mathit{Fam}(\mathcal{D})$ thought of as an ordinary fibration between categories is not (generally) discrete, therefore cannot be thought of as a single $\mathbf{Set}$-valued presheaf (by Yoneda the "best $\mathbf{Set}$-valued presheaf approximation" to $\mathit{fam}(\pi_F)$ is the presheaf associated to the discrete fibration $\mathit{fam}(\pi_F)_1$).<|endoftext|>
TITLE: Do all the roots of the polynomial lie in the unit disk?
QUESTION [14 upvotes]: How to prove (or to disprove) that all the roots of the polynomial of degree $n$ $$\sum_{k=0}^{k=n}(2k+1)x^k$$ belong to the disk $\{z:|z|<1\}?$ Numerical calculations confirm that, but I don't see any approach to a proof of so simply formulated statement. It would be useful in connection with an irreducibility problem.

REPLY [6 votes]: The idea is taken from this other question Polynomial with the primes as coefficients irreducible?
Show instead that $f(1/x)$ has all roots lying outside of the unit disk. For that, multiply by $(x-1)$ and equate to $0$ obtaining:
$$x^{k+1}+\sum_1^k 2x^j=2k+1$$
Take absolute values and apply the triangular inequality and one obtains:
$$|x^{k+1}|+\sum_1^k 2|x^j|\geq \left|x^{k+1}+\sum_1^k 2x^j\right|=2k+1$$
This is clearly non possible if 
$|x|<1$. Moreover, if $|x|=1$ there is an equality which means that all the terms are aligned. In particular $2x^2/2x$ is real, so the only possibility is $x=1,-1$. But neither is a root of $f(1/x)$ so you are done.<|endoftext|>
TITLE: Why did Voiculescu develop free probability?
QUESTION [24 upvotes]: I was recently asked why Voiculescu developed free probability theory. I am not very expert in this and the only answer I was able to provide is the classical one: he was challenging the isomorphism problem of whether $II_1$-factors associated to two different free groups are isomorphic or not. First: is this true or just legend? Were there any other motivations? In particular I would be interested in more down-to-earth motivations, something that could theoretically be explained to someone with basic knowledge in probability theory and operator theory (without necessarily knowing what a $II_1$-factor is).
Thanks in advance!
Valerio

REPLY [12 votes]: Here is a more detailed account on the history by Dan Voiculescu himself. This is from his article "Background and Outlook" in the Lectures Notes
"Free Probability and Operator Algebras", see
http://www.ems-ph.org/books/book.php?proj_nr=208

Just before starting in this new direction, I had worked with Mihai Pimsner,
  computing the K-theory of the reduced $C^*$-algebras of free groups. From the
  K-theory work I had acquired a taste for operator algebras associated with free
  groups and I became interested in a famous problem about the von Neumann
  algebras $L(\mathbb{F}_n)$ generated by the left regular representations of free groups,
  which appears in Kadison's Baton-Rouge problem list. The problem, which
  may have already been known to Murray and von Neumann, is:are $L(\mathbb{F}_m)$ and $L(\mathbb{F}_n)$ non-isomorphic if $m \not= n$?
This is still an open problem. Fortunately, after trying in vain to solve it,
  I realized it was time to be more humble and to ask: is there anything I can
  do, which may be useful in connection with this problem? Since I had come
  across computations of norms and spectra of certain convolution operators on
  free groups (i.e., elements of $L(\mathbb{F}_n)$), I thought of finding ways to streamline
  some of these computations and perhaps be able to compute more complicated
  examples. This, of course, meant computing expectations of powers of such
  operators with respect to the von Neumann trace-state $\tau(T) = \langle T e_e,e_e\rangle$, $e_g$
  being the canonical basis of the $l^2$-space.
The key remark I made was that if $T_1$, $T_2$ are convolution operators on $\mathbb{F}_m$
  and $\mathbb{F}_n$ then the operator on $\mathbb{F}_{m+n} = \mathbb{F}_m \ast \mathbb{F}_n$ which is $T_1 + T_2$, has moments $\tau((T_1 + T_2)^p)$ which depend only on the moments $\tau(T_j^k)$, $j = 1, 2$ , but not
  on the actual $T_1$ and $T_2$. This was like the addition of independent random
  variables, only classical independence had to be replaced by a notion of free
  independence, which led to a free central limit theorem, a free analogue of
  the Gaussian functor, free convolution, an abstract existence theorem for one
  variable free cumulants, etc.<|endoftext|>
TITLE: Every weakly compact cardinal is Mahlo
QUESTION [7 upvotes]: This is a reference question. Does anyone know any book or paper that has the proof that every weakly compact cardinal is Mahlo, using only combinatorics?
I know the definition of weak compactness has a lot of equivalent forms. In this particular case I am referring to a proof that is based only on the combinatorial definition: $$\kappa \text{ is weakly compact iff }\kappa\to (\kappa)^2_2,$$
i.e. for every partition of the 2-element subsets of $\kappa$ into two sets, there is a homogeneous set of size $\kappa$.

REPLY [8 votes]: This may help:
Hajnal, Kanamori, and Shelah analyzed regressive partition relations for infinite cardinals in their paper
Regressive partition relations for infinite cardinals. Trans. Amer. Math. Soc. 299 (1987), no. 1, 145–154.
They characterize Mahlo cardinals (even n+1-Mahlo cardinals for each $n<\omega$) in terms of these regressive partition relations:
For example, Theorem 3.4 tells us that $\kappa$ is Mahlo if and only if
For any closed unbounded $C\subseteq\kappa$ and regressive coloring $f$ of $[C]^4$, there is a min-homogeneous set of size $\omega$.
There are some comments in the paper about how weakly compact cardinals satisfy these regressive partition relations, but it's not clear to me (yet) if there's a direct proof or if the proofs need to go through the tree property.  I feel like Asaf --- I need to look at it after I get some other work done!  
Edit:  One can prove directly that if $\kappa\rightarrow (\kappa)^5_2$ then $\kappa$ satisfies the regressive partition relation mentioned above that is equivalent to being Mahlo. 
This is still unsatisfying, but what I'm wondering is if this can be put together with the result from the paper to get something like:
If $\kappa$ is not Mahlo, then there is a 2-coloring of the 5-tuples from $\kappa$ with no homogeneous set of size $\kappa$.
Another edit based on another strategy:
If $\kappa>\omega$ is not $\omega$-Mahlo, then $\kappa\nrightarrow(\kappa)^2_2$ by virtue of results on negative square-brackets partition relations: If $\kappa$ has a non-reflecting stationary subset this follows from Todorcevic's result that $\kappa\nrightarrow[\kappa]^2_\kappa$ for such $\kappa$. Otherwise, $\kappa$ must be weakly inaccessible with a stationary subset that does not reflect in an inaccessible cardinal, and by results in Shelah's Cardinal Arithmetic, we have $\kappa\nrightarrow[\kappa]^2_\theta$ for every $\theta<\kappa$.
In either case, we certainly have $\kappa\nrightarrow[\kappa]^2_2$, which is equivalent to $\kappa\nrightarrow(\kappa)^2_2$, so $\kappa$ cannot be weakly compact.
So the explicit "bad coloring" of pairs is there to see. It's just really really complicated.
Final Edit
Suppose $\kappa$ is weakly inaccessible and not Mahlo.  Let $\langle C_\delta:\delta<\kappa\rangle$ be a $C$-sequence in the sense of Todorcevic, so $C_\delta$ is club in $\delta$ of order-type ${\rm cf}(\delta)$. Let $c(\alpha,\beta)$ be the length of the minimal walk from $\beta$ down to $\alpha$ (so $c:[\kappa]^2\rightarrow\omega$).  The fact that $\kappa$ isn't Mahlo implies that the $C$-sequence is non-trivial in the sense of Todorcevic, and his work shows that for any $H\subseteq\kappa$ of size $\kappa$ the range of $c$ restricted to the pairs from $H$ is infinite.   In particular, $\kappa\nrightarrow(\kappa)^2_{\omega}$ and so $\kappa$ is not weakly compact.
Still not as simple as we'd like, but I don't see how to do better!  This is why the tree property is a good thing...<|endoftext|>
TITLE: Introducing division strategically in operads to accommodate formulas like the Baker-Campbell-Hausdorff formula
QUESTION [13 upvotes]: I am not as familiar with operad terminology as I'd like to be, so I might be missing some well-known term in the area. If so, I'd appreciate any pointers to the correct terms.
Consider the following two objects:

The free associative operad with one binary operation over $\mathbb{Q}$. What I mean is that we are working with $\mathbb{Q}$-vector spaces, and consider the operad where the generating set is an operation of arity 2 and the relation is the associativity relation.
The free associative operad with one binary operation over $\mathbb{Z}$. What I mean is that we are working with $\mathbb{Z}$-modules (aka abelian groups), and consider the operad where the generating set is an operation of arity 2 and the relation is the associativity relation.

(2) can be viewed as a $\mathbb{Z}$-suboperad of (1). I am interested in defining an intermediate $\mathbb{Z}$-suboperad of (1). The goal is to allow for a limited amount of division based on the degree.
The intermediate suboperad I want to define is the suboperad of (1) generated by the following infinite set of operations: for each prime number $p$, consider:
$$(x_1,x_2,\dots,x_p) \mapsto \frac{1}{p}(x_1x_2 \dots x_p)$$
The representations of the operad (1) correspond to associative $\mathbb{Q}$ algebras and the representations of the operad (2) correspond to associative $\mathbb{Z}$-algebras (also known as associative rings). The representations of the intermediate operad defined above are associative rings with additional "multiply them all and divide by $p$" operations associated to strings of length $p$ for every prime $p$.
The intermediate object I have defined above is closely related to the nature of denominators that appear in the Baker-Campbell-Hausdorff formula. If we perform a similar construction for the Lie operad instead of the associativity operad (UPDATE: There is a slight issue, in that the assumption of the alternating condition rather than skew symmetry is a non-operadic identity in so far as it involves a repeated variable, and this could be an issue if we are working over a base where 2 is non-invertible), then the representations of that operad will be the "Lie algebras" in a suitable generalization of the Lie correspondence or Lazard correspondence (there are some messy details I am not getting into here).
[As a quick illustration, note that the Baker-Campbell-Hausdorff formula has a denominator of 12 for its degree three terms. To make sense of a denominator of 12, note that we can separately make sense of denominators of 3 and 4 for length three Lie products: the denominator of 3 makes sense directly setting $p = 3$, and the denominator of 4 makes sense by consider $\frac{1}{2}[x,\frac{1}{2}[y,z]]$. We can combine these using the Euclidean algorithm to get a denominator of $1/12$ (in this case, $1/12 = 1/3 - 1/4$). There are general bounds on the denominators that appear in the Baker-Campbell-Hausdorff formula that guarantee that the denominators can always be achieved.]
Additionally, the general theme behind this sort of operad seems well-suited to situations where we want to introduce divisibility by certain primes but only in high degrees dependent on the prime. This might make it suitable to the study of things like the representation theory of symmetric groups, though I might be barking up the wrong tree here.
I'd like to know if operads of this sort have been studied before, and if so, I'd like any pointers to the appropriate terms and references. If not, I would appreciate any views on whether this is useful as an intermediate framework between the extremes of $\mathbb{Z}$-operads and $\mathbb{Q}$-operads.

REPLY [7 votes]: Nice question! I've enjoyed thinking of this operad you define. I don't know of any reference where this sort of operad has been considered. Let me just record here some properties I have found out. I will work in the context of nonsymmetric operads, unlike Neil.
For $k$ a commutative ring, let $\mathtt{Ass}_{k}$ be the associative operad in the category of $k$-modules. This operad is given by $\mathtt{Ass}_{k}(n)=k$, $n>0$, and $\mathtt{Ass}_{k}(0)=0$. Composition laws are given by multiplication in $k$. This operad has a presentation with a single generator $\mu_2$ in degree $2$ and two relations: $\mu_2\circ_1\mu_2=\mu_2\circ_2\mu_2$. Notice that this operad is not free in any sense. I will write $\mu_n$ for the $(n-1)^{\text{th}}$ power of $\mu_2$. Due to the defining relation, we need not specify in which way this power is taken.
We pull $\mathtt{Ass}_{\mathbb Q}$ back to the category of operads of abelian groups. As you point out, there is a morphism (inclusion) of operads $\mathtt{Ass}_{\mathbb Z}\subset \mathtt{Ass}_{\mathbb Q}$ given aritywise by the usual inclusion $\mathbb Z\subset\mathbb Q$. Let me denote your operad by $\mathtt{O}$, $\mathtt{Ass}_{\mathbb Z}\subset\mathtt{O}\subset \mathtt{Ass}_{\mathbb Q}$, generated by $\frac{1}{n}\mu_n$, $n\geq 2$.
The first interesting question is, what abelian group is $\mathbb Z\subset\mathtt{O}(n)\subset\mathbb Q$, $n>0$? Obviously, $\mathtt{O}(0)=0$. Notice also that $\mathtt{O}(1)=\mathtt{Ass}_{\mathbb Z}(1)=\mathbb Z$. Moreover, $\mathbb Z\subset\mathtt{O}(2)=\left(\frac{1}{2}\right)\subset\mathbb Q$ is the abelian subgroup generated by $\frac{1}{2}$. The description of $\mathtt O$ in terms of generators shows that $\mathtt O(3)$ contains $\frac{1}{3}\mu_3$, $\left(\frac{1}{2}\mu_2\right)^2=\frac{1}{4}\mu_3$ and their integer multiples, so it is the subgroup generated by $\left(\frac{1}{3},\frac{1}{4}\right)=\left(\frac{1}{12}\right)$. Recall that for $a,b\in\mathbb Z$, $\left(\frac{1}{a},\frac{1}{b}\right)=\left(\frac{1}{\text{least common multiple of }a\text{ and }b}\right)$. Reasoning in this way, one can easily check that $\mathtt{O}(n)=\left(\frac{1}{a_n}\right)$, where $a_n$ is the least common multiple of the numbers $p_1\cdots p_r$, $r\geq 1$, such that $p_1+\cdots+p_r-r+1=n$ and $p_i\geq 2$. I haven't found any easy way of computing these numbers. I've calculated them for small values of $n$, $n=1,2,3,4,5,6,7$, and the outcome is
$$a_n=1,2,12, 24, 720, 1440, 60480.$$
At a first glance, there seems to be some relation between these numbers and the denominators in the Baker-Campbell-Hausdorff formula, but I cannot precise what. Looking this sequence up in The On-Line Encyclopedia of Integer Sequences® produces only one outcome: http://oeis.org/search?q=2%2C12%2C24%2C720&language=english&go=Search I haven't checked much, but I don't see any immediate link between the description of the sequence therein and $\{a_n\}_{n\geq 1}$. 
Anyway, $\mathtt O$ can be presented as an operad of abelian groups in the following way: a sequence of generators $\{m_n\}_{n\geq 2}$, which correspond to $m_n=\frac{1}{a_n}\mu_n$, and relations $m_p\circ_im_q=\frac{a_{p+q-1}}{a_p\cdot a_q}m_{p+q-1}$, $p,q\geq 2$, $1\leq i\leq p$. Notice that he coefficient $\frac{a_{p+q-1}}{a_p\cdot a_q}$ is always an integer. I don't think this operad can be presented with finitely many generators. An $\mathtt O$-algebra is therefore an abelian group $B$ together with $n$-ary operations, $(b_1,\dots,b_n)\mapsto m_n(b_1,\dots,b_n)\in B$ such that
$$m_p(b_1,\dots,b_{i-1},m_q(b_i,\dots,b_{q+i-1}),b_{q+i},\dots,b_{p+q-1})=\frac{a_{p+q-1}}{a_p\cdot a_q}m_{p+q-1}(b_1,\dots,b_{p+q-1}).$$
I don't know if any of this can be of help for you. If you happen to find any sense to the $a_n$'s, please let me know. I'm curious about it!<|endoftext|>
TITLE: Restriction of highest-weight representations to Heisenberg subalgebras
QUESTION [6 upvotes]: Let $\mathfrak{g}$ be a finite-dimensional complex simple Lie algebra and $\tilde{\mathfrak{g}}=\mathfrak{g}((t))\oplus \mathbf{C}K\oplus \mathbf{C}d$ its Kac-Moody extension ($K$ is the level and $d$ is the energy operator). $\tilde{\mathfrak{h}}=\mathfrak{h}\oplus \mathbf{C}K\oplus \mathbf{C}d$ is the Cartan subalgebra.
Pick an element $\sigma$ of the Weyl group $W$ of order $k$. Then the Heisenberg subalgebra $\hat{\mathfrak{a}}\subset \tilde{\mathfrak{g}}$ is the subalgebra of fixed points of $\mathfrak{h}((t^{1/k}))\oplus \mathbf{C}K$ under the action of $\sigma$, where $\sigma$ acts on $t$ by $t^{1/k}\mapsto \exp(2\pi i/k) t^{1/k}$. For example, for $\sigma=e$ one gets the homogeneous Heisenberg subalgebra $\widehat{\mathfrak{a}}\cong\mathfrak{h}((t))\oplus \mathbf{C} K$, while for $\sigma$ the Coxeter element one gets the principal Heisenberg subalgebra. Let $\mathfrak{a}_+\subset\hat{\mathfrak{a}}$ be the subalgebra containing nonnegative powers of $t$.
Consider $L(\lambda)$, an irreducible integrable highest-weight representation of $\tilde{\mathfrak{g}}$ of highest weight $\lambda\in \tilde{\mathfrak{h}}^*$. What is known about the restriction of $L(\lambda)$ to the Heisenberg $\hat{\mathfrak{a}}$? For example, is it true that the space of invariants $L(\lambda)^{\mathfrak{a}_+}$ is finite-dimensional?
If $\mathfrak{g}$ is simply-laced, $\hat{\mathfrak{a}}$ the homogeneous Heisenberg and $L(\lambda)$ is the level 1 basic representation, then a theorem of Frenkel-Kac and Segal identifies the restriction with a direct sum of Fock modules over the root lattice; in particular, $L(\lambda)^{\mathfrak{a}_+}$ is one-dimensional. It is also one-dimensional for the principal Heisenberg (using the principal realization of Kac-Kazhdan-Lepowsky-Wilson).

REPLY [2 votes]: The answer to your second question is no. Take for instance $\Lambda=k \Lambda_0$, where the level $k$ is two or more. I'll consider only the homogeneous Heisenberg subalgebra.  In this case, the space of invariants you're interested is very important in 2-dimensional conformal field theory and is known as the $parafermionic$ subspace $K({ \frak g},k)$ (actually a vertex algebra). It is known that it contains the conformal vector
$$\omega_{paraf}=\omega_{Sugawara}-\omega_{Heisenberg},$$
so it carries a representation of the Virasoro algebra of central charge $c_{paraf}=\frac{dim({\frak g})k}{k+h^\vee}-rank({\frak g})$. As long as $\omega_{paraf} \neq 0$ the parafermionic space is infinite-dimensional. I believe this happens precisely if the level is at least two for simply-laced algebras. For non-simply laced, it can be infinite-dimensional even if the level is one (e.g. $G_2$). As $\omega_{paraf} \neq 0$ can be difficult to check directly, you can simply look at $c_{paraf}$. For example, let $\frak{g}$$=A_n$. Then
$c_{paraf}=\frac{k(n^2+2n)}{n+1+k}-n$. This is nonzero and positive for all $n$ and $k \geq 2$. But there are no finite dimensional Virasoro modules with nonzero central charge so you're done.
For a general dominant weight $\Lambda$, it is more interesting to consider
a slightly bigger space $L(\Lambda)^{t \mathbb{C}[t] \otimes \frak{h} }$ (the coset subspace), a $K({\frak g},k)$-module. Then you study decomposition of $L(\Lambda)$ 
as $K({\frak g},k) \otimes M(1)$-module (where $M(1)$ is the vacuum space for the Heisenberg, also a VOA). Thus $L(\Lambda)=\oplus_{\mu} M_{\mu} \otimes M(1,\mu)$, where  $\mu \in \frak{h}^*$. If $\mu=0$ appears in the decomposition you get $L(\lambda)^{\mathbb{C}[t] \otimes \frak{h}}=M_0$ and infinite-dimensional if $c_{paraf} \neq 0$. 
Same thing happens if $\sigma$ is the Coxeter element. Already for $k=2$ and ${\frak g}={\frak sl}_2$ the invariant space is infinite-dimensional.
I should also mention that $K({\frak g},k)$ is quite difficult to describe in terms of generators.<|endoftext|>
TITLE: An NP-hard $n$ fold integral
QUESTION [17 upvotes]: We are given rational numbers $[c_1, c_2, \ldots, c_n]$ and $v$ from the interval $[0,1]$. 
Consider the $n$-fold integral 
$$
J = \int_{\theta_1 \in I_1, \theta_2 \in I_2 \ldots, \theta_n \in I_n} d\theta_n\ldots d\theta_2 d\theta_1
$$
whose intervals are defined by 
$$
I_j = \begin{cases}
[0,1] & j = 1\\
[\max(c_j,\theta_{j-1}),1] & 2\leq j\leq n  
\end{cases}
$$
We want to check if $J$ is equal to $v$. 

Is this problem $\mathsf{NP}$-hard?

Informally, each $\max$ in the lower limits of the intervals leads to a two-way split in evaluating the integral, and thus to $2^{n-1}$ integrals that sum to $J$. 
Note that $J$ is also the volume of an $n$ dimensional polytope define by the following inequalities:
$$c_j \leq x_j \leq 1 \ (\text{for } 1 \leq j \leq n),$$
$$0 \leq x_1 \leq x_2 \leq \ldots \leq x_n \leq 1.$$
Is there a result in computational complexity or computational geometry about volume of such shapes?
This question was originally posted CS.SE.

REPLY [11 votes]: The integral essentially asks for the probability that, for $n$ independent "events" uniformly distributed in $[0,1]$, at least one happens after $c_n$, at least two happen after $c_{n-1}$, etc (thinking of the unit interval as time). Let $P_n(c_1,\dots,c_n) = n!\cdot J$ denote this probability (the integral $J$ also requires the $n$ events to occur in a specified order). 
If we condition on the number $k$ of events in the interval $[c_n,1]$, then the remaining events will be uniformly distributed in $[0,c_n]$, and we can write $P_n$ as $$P_n(c_1,\dots,c_n) = Pr(k=1)\cdot P_{n-1}\left(\frac{c_1}{c_n},\dots,\frac{c_{n-1}}{c_n}\right) + Pr(k=2)\cdot P_{n-2}\left(\frac{c_1}{c_n},\dots,\frac{c_{n-2}}{c_n}\right)+\dots.$$
To complete a recursive computation, we only need to compute the $O(n^2)$ integrals of the form $$P_i\left(\frac{c_1}{c_j},\dots,\frac{c_i}{c_j}\right)$$ for $1\leq i < j\leq n$. I'm not an expert on the complexity of rational arithmetic, but it seems to me that this should be doable in polynomial time.
EDIT: 
Just for completeness, yes, the integral can be evaluated in $O(n^2)$ arithmetical operations, but so can $2^{2^n}$ (in fact in a linear number of operations), and that doesn't mean we can output (let alone compute) the digits in polynomial time. Even if the problem is to check an identity (so that the potential answer has to be part of the input), I don't see a general reason why that should be possible in polynomial time just because the number of arithmetical operations is polynomial (but people must have thought of that before, feel free to comment).
In this case however, the integral $P_n(c_1,\dots,c_n)$ is a degree $n$ polynomial in $c_1,\dots,c_n$, with coefficients that grow "only" exponentially. It follows that all the denominators (and numerators) of the partial results will have size (number of digits) polynomial in the size of the input.<|endoftext|>
TITLE: Three questions on $\operatorname{hocolim}$
QUESTION [12 upvotes]: I posted this on math.stack.exchange but didn't get a helpful response, so please let me try it here.
Let $D$ be a small category and $F:D\to sSets$ a functor.
There is a bisimplicial set indicated by 
$$
...\begin{array}{c}\to \\ \to\\\to\end{array}\coprod_{d_1\to d_0}F(d_1)\begin{array}{c}\to\\\to\end{array}\coprod_{d_0}F(d_0)
$$
which i like to call $sres(F)_{\bullet\bullet}$. By considering either the horizontal or the vertical index first, I get two functors $H:\Delta^{op}\to sSets$ and $V:\Delta^{op}\to sSets$ (the ''vertical'' direction is the one of the simplicial sets $F(d)$, i.e. each object in the diagram above should be imagined as a vertical column).
The homotopy colimit $\operatorname{hocolim}F$ of $F$ can be defined in different ways (up to weak equivalence) and one way is $(\operatorname{hocolim}F)_n=sres(F)_{n n}$, the diagonal of $sres(F)_{\bullet\bullet}$.

Is there a weak equivalence between between $\operatorname{hocolim}F$ and $\operatorname{hocolim}H$? (I don't think there is a weak equivalence between $\operatorname{hocolim}F$ and $\operatorname{hocolim}V$, or is it?)

Let me call a simplicial set $A$ $k$-dimensional, if it is equal to its $k$-skeleton $\mathbf{sk}_k$ (If I am not mistaken, $k$ should be the smallest integer such that all elements of $A_{k+1}$ are degenerated simplices).

Suppose that the nerve $N(D)$ of $D$ is weakly equivalent to a simplicial set $A$ of dimension $k$. Is it true that $\operatorname{hocolim}F$ is weak equivalent to the homotopy colimit of the diagram
  $$
\coprod_{d_{k+1}\to ...\to d_0}F(d_{k+1})\begin{array}{c}\to \\ \vdots\\\to\end{array}...\begin{array}{c}\to \\ \to\\\to\end{array}\coprod_{d_1\to d_0}F(d_1)\begin{array}{c}\to\\\to\end{array}\coprod_{d_0}F(d_0),
$$
  (the truncated $H$) or in other words, can I ''stop'' at the $k+1$th stage of the diagram in the horizontal direction to calculate the homotopy colimit? If yes, why?

In particular, if $D$ is linear (for example $\mathbb{N}$ or $\cdot\to\cdot\to\cdot$), this would mean, that $\operatorname{hocolim}F$ is the homotopy colimit of the simple diagram
$$
\coprod_{d_1\to d_0}F(d_1)\begin{array}{c}\to\\\to\end{array}\coprod_{d_0}F(d_0).
$$

My third question is a little vague and like the second one but not ''for the source'' of $F$ but ''for the target''. Please don't hesitate to post an answer only to the first two questions, if this third one is not well formulated. I wondered, why one takes only two-limits of stacks and not $k$-limits. The nerve of a category is a 2-coskeletal simplicial set and this is where the reason comes from, I guess.

Suppose the functor $F$ factorize through the nerve $F:D\to CAT\xrightarrow{N} sSets$. Is it true that $\operatorname{hocolim}F$ is weakly equivalent to the homotopy colimit of the diagram
  $$
\coprod_{d_2\to d_1\to d_0}F(d_{2})\begin{array}{c}\to \\ \to\\\to\end{array}\coprod_{d_1\to d_0}F(d_1)\begin{array}{c}\to\\\to\end{array}\coprod_{d_0}F(d_0)
$$
  and what is the reason? If not, what did I misunderstand here?

REPLY [14 votes]: First question$\newcommand{\op}[1]{{#1}^{\mathrm{op}}}$$\newcommand{\sSet}{\mathrm{sSet}}$$\newcommand{\Grpd}{\mathrm{Grpd}}$$\newcommand{\Cat}{\mathrm{Cat}}$$\newcommand{\NN}{\mathbb{N}}$$\newcommand{\sres}{\mathrm{sres}}$$\newcommand{\hocolim}{\operatorname{hocolim}}$$\newcommand{\diag}{\operatorname{diag}}$$\newcommand{\To}{\longrightarrow}$$\newcommand{\real}[1]{\lvert #1 \rvert}$
The homotopy colimit of a functor $X:\op{\Delta}\to\sSet$ is equivalent to its geometric realization $\real{X}$. This holds because every functor $\op{\Delta}\to\sSet$ is cofibrant in the Reedy model structure. In particular, geometric realization of simplicial simplicial sets preserves objectwise weak equivalences. You may consult corollary 15.8.8 and theorem 18.7.4 in Hirschhorn's book "Model categories and their localizations" for these results. The equivalence between the homotopy colimit and the geometric realization is given explicitly by the Bousfield–Kan map (in section 18.7 of Hirschhorn's book).
On the other hand, it is a very interesting well-known fact that the geometric realization of a simplicial simplicial set $X:\op{\Delta}\to\sSet$ is canonically isomorphic to the diagonal $\diag X$ of the associated bisimplicial set. You will find this as theorem 15.11.6 of Hirschhorn's book, or as exercise IV.1.4 in the book "Simplicial homotopy theory*" by Goerss and Jardine. [Further, if you are very categorically inclined, you may see this quite formally as a consequence of geometric realization being a coend whose weight is the Yoneda embedding.]
From the preceding paragraphs, we conclude that there is a natural weak equivalence $\hocolim X\overset{\sim}{\To}\diag X$ for every simplicial simplicial set $X$. This is stated explicitly as corollary 18.7.7 of Hirschhorn's book.
What I said so far is in part briefly described in section XII.3.4 of the book "Homotopy limits, completions and localizations" by Bousfield and Kan.

Answer to first question: We conclude from the previous discussion that the homotopy colimits of both $H$ and $V$ are naturally equivalent to their diagonals. Each of these diagonals is by definition also the diagonal of $\sres(F)$, i.e. your proposed definition for the homotopy colimit of $F$.

[As a consistency check, the definition of homotopy colimit of a diagram $F:D\to\sSet$ given in both Bousfield–Kan and Hirschhorn is isomorphic to your definition as the diagonal of $\sres(F)$. You can see a brief explanation of this fact in section 4.13 of Dan Dugger's notes "A primer on homotopy colimits". These are very readable notes, and I highly recommend them. You can also read about the Bousfield–Kan map in section 17.2 of those notes.]
Second question

The answer to your second question is no.

Take the very simple category $D$ described schematically as $0\overset{a}{\to} 1\overset{b}{\to} 2$. Then the category $D$ has a terminal object so its nerve is contractible, and we may take $k=0$ in your second question. Consider the terminal functor $F=\ast : D\to \sSet$ whose value is always the terminal simplicial set with a single vertex, and no non-degenerate higher dimensional simplices. The homotopy colimit of this diagram is equivalent to the nerve of $D$, and thus contractible. On the other hand, the $1$-truncated simplicial object you write in your question is levelwise discrete, so we can think of it as a simplicial set with:

three vertices $0$, $1$, and $2$
three non-degenerate edges: $a$ from $0$ to $1$, $b$ from $1$ to $2$, and $ab$ from $0$ to $2$.

Non-coincidentally, this is just the $1$-skeleton of the nerve of $D$. It is quite easy to see that its realization is a circle. Thus the homotopy colimit of your $1$-truncated simplicial object is not contractible. The problem is of course that we "erased" the non-degenerate $2$-simplex from the nerve of $D$, thus changing its homotopy type.

On a positive note, I believe a formula like the one you state does hold. Under the stronger assumption that the nerve of $D$ is $k$-skeletal, then the homotopy colimit of $F:D\to\sSet$ is equivalent to the homotopy colimit of a $k$-truncated simplicial object like the one you write (except you write a $(k+1)$-truncated simplicial object instead). [By a $k$-truncated simplicial object, I mean a functor indexed by the full subcategory of $\op{\Delta}$ spanned by the $n$-simplices with $n\leq k$.]

Sketch of proof: If the nerve of $D$ is $k$-skeletal, $\sres(F)$ is also (horizontally) $k$-skeletal. Consequently, the diagonal of $\sres(F)$ coincides with the geometric realization of the horizontal $k$-truncation of $\sres(F)$ (in your question you actually write down the $(k+1)$-truncation). This is equivalent to the homotopy colimit of the $k$-truncation for the same reasons as before: every $k$-truncated simplicial object in simplicial sets is Reedy cofibrant (which actually follows from the analogous result for simplicial simplicial sets).
Intermezzo
I will make a couple of simple comments regarding your examples in between the second and third questions.
The homotopy colimit of any functor $F$ indexed by a category $D$ with a terminal object $d$ is weakly equivalent to $F(d)$. So the homotopy colimit of any functor $F$ indexed by $0\to 1\to 2$ is weakly equivalent to $F(2)$.
Moreover, any filtered colimit in simplicial sets is actually a homotopy colimit, essentially because filtered colimits commute with homotopy groups in $\sSet$. In other words, homotopy colimits in $\sSet$ along filtered categories (such as $\NN$) are no harder than ordinary colimits.
Third question
For this question, the relevant part about stacks is not that they have values in categories. It is that they have values in groupoids. Moreover, one needs to take the homotopy colimit in groupoids and not in simplicial sets. I will elaborate below.
First, here is a counter-example to your third question as it stands. Consider any category $C$ whose nerve is equivalent to $S^3$, for example. [Any simplicial set is weakly equivalent to the nerve of some category, for example its category of simplices. Here, however, is an explicit example for $S^3$: simply take the join of categories $C = 2\ast 2 \ast 2 \ast 2$, where $2$ denotes a discrete category with two objects.] Then the homotopy colimit along $C$ of the nerve of the constant terminal functor $\ast:C\to\Cat$ is simply
$$ \hocolim N(\ast) = \hocolim \ast \simeq BC \simeq S^3 $$
However, the homotopy colimit of the corresponding $2$-truncated simplicial set you write will be equivalent to its geometric realization, which is $2$-dimensional. In particular, it cannot have non-trivial homology in dimension $3$, and cannot be equivalent to $S^3$.
What is true is that the homotopy colimit in groupoids (with respect to the natural model structure) of a functor $F:D\to\Grpd$ (from $D$ to the category of groupoids) is equivalent to the homotopy colimit in groupoids of the $2$-truncated simplicial object you write down.
If you want to further compose $F:D\to\Grpd$ with the nerve functor, then you will have to apply the fundamental groupoid functor to the homotopy colimit of $N\circ F$ in simplicial sets to recover the homotopy colimit of $F$ in groupoids. This holds because the fundamental groupoid functor is part of a Quillen adjunction between simplicial sets and groupoids. The same holds for the homotopy colimit of the $2$-truncated simplicial object you write down.<|endoftext|>
TITLE: Range of $2^n$ mod $n$
QUESTION [9 upvotes]: Is there any integer $n\ge 2$ such that $2^n\equiv 3 \bmod n$?  I understand that $n$ must be an odd non-prime.  I checked up to a million with no success (but $2^n\equiv 5 \bmod n$ and $2^n\equiv 7 \bmod n$ have solutions).

REPLY [14 votes]: The smallest such $n$ is $n=4700063497$. A few others are known. J. Crump found $n=8365386194032363$ in 2000. Max Alekseyev found $n=3468371109448915$. Joe Crump found $n=10991007971508067$. Some information on these is at http://www.cs.ucla.edu/~klinger/newno.html<|endoftext|>
TITLE: What is $Aut(Ell)$?
QUESTION [19 upvotes]: Consider the stack $Ell$ (of groupoids) of elliptic curves. I'm interested in the autoequivalence 2-group of $Ell$, the objects of which consists of transformations $Ell \Rightarrow Ell: Ring \to Gpd$ valued in equivalences of groupoids. The arrows are isomorphisms of such transformations.
In a chat opinions ranged from optimistic that it would be large to hunches it would be small (in fact $\mathbb{Z}/2 \rightrightarrows \ast$)
Even if this 2-group is very small, I would also be interested in knowing if it's possible to calculate much of the endomorphism monoidal groupoid of $Ell$, given that we know a very explicit presentation of it (using a Hopf algebroid built from finitely generated polynomial rings). Here we would take all transformations, not just those valued in equivalences.
EDIT, Will Jagy: to get fuller context, you can scroll arbitrarily, and conveniently, back in the transcript http://chat.stackexchange.com/transcript/9417/2013/6/28/5-16 where David's announcement of this question occurs pretty late in this segment. I will check later today, the terminal hour marker '16' may change, that is how the system works.

REPLY [3 votes]: The 2-group is $B\mathbb Z/2$. In other words, the automorphism $1$-group of $M_{1,1}$ is trivial, and the identity functor $M_{1,1} \to M_{1,1}$ has exactly one non-identity invertible natural transformation to itself: the one which sends a family of elliptic curves $\xi \colon E \to S$ to $\xi \circ i \colon E \to S$, where $i$ is inversion in the group structure of $E$.
The claim about the $1$-group was explained in the comments: the automorphism $1$-group of $\overline M_{1,1}$ is $\mathbb G_m$, since $\overline M_{1,1} \cong \mathbb P(4,6)$. None of these automorphisms fix the point at infinity. So we need only to determine the natural equivalences from the identity to itself. 
Such a natural equivalence would assign to any $\newcommand{\id}{\mathrm{id}}\xi \colon E \to S$ an automorphism $a_\xi \colon E \to E$ over $S$, and it should satisfy the conditions of a natural transformation. For any $E \to S$ the automorphism group of $E$ over $S$ has two distinguished elements, $\id$ and $-\id$, and these are stable under pullback. In particular given $S' \to S$ and $\xi' \colon E' \to S'$ the pullback of $\xi$, if $a_\xi$ is trivial or inversion in the group, then the same holds for $a_{\xi'}$. The converse holds by descent if $S' \to S$ is étale. 
Now let $X \to M_{1,1}$ be an étale cover by a scheme, let $\eta \colon C \to X$ be the pullback of the universal family. There is an open dense $U \subset X$ such that the only automorphism of $C$ over $U$ is inversion in the group. Then the same holds globally on $X$, so $a_\eta = \pm \id$, because the isomorphism scheme of $C$ over $X$ is separated. Let $\xi \colon E \to S$ be arbitrary. There is an étale cover $S' \to S$ such that $\xi'$ is pulled back from $\eta$. Then $a_{\xi'}$ is $\pm \id$. Then the same is true for $a_\xi$.<|endoftext|>
TITLE: Terminology for sequences/functions that approach each other
QUESTION [6 upvotes]: What do I call two sequences $a, b$ such that $\lim_{n\to\infty} |a_n - b_n| = 0$?  Or what do I call two functions $f, g$ such that $\lim_{x\to c} |f(x) - g(x)| = 0$?  (For my purposes, these are essentially the same thing, and I will happily extend a term for one to the other.)
This seems like such a straightforward condition that it must have a standard term, but I can't find it (in either context).  I looked through the Wikipedia article on all of the variations of big-$O$ and the like, but these are all too weak.  If $\lim_n a_n$ (hence $\lim_n b_n$) existed, then I could call $a$ and $b$ ‘coterminal’, but that limit might not exist.  In an incomplete space, I have seen $a$ and $b$ called ‘co-Cauchy’ under the weaker assumption that one (hence both) is Cauchy, but I don't want to assume that either.  I could call $\exp f$ and $\exp g$ ‘asymptotic’ (as $x \to c$), but I want to refer to $f$ and $g$ directly.
Surely somebody knows a term for this?

REPLY [4 votes]: You could say $ a $ and $ b $ are mutually asymptotic. Example: "Since $ n -1/n$ and $n+1/n $ are mutually asymptotic. .." And usually even drop the word "mutually".<|endoftext|>
TITLE: Matching on sphere to create cycle with chords
QUESTION [6 upvotes]: Imagine a number of chords of a sphere $S$ which nearly, but not quite, pass through
the center of $S$, in such a way that no pair of chords intersect:

     


I would like to connect pairs of chord endpoints by noncrossing paths on the surface of $S$ so
that the paths + the chords form a cycle. I can prove (for example, by induction) that
this is always possible if I am permitted
to use arbitrary paths on the surface. But what I would really like 
to achieve is connecting
the chord endpoints by noncrossing arcs of great circles. 
Henceforth consider the chords to pass exactly through the center of $S$, but pretend they do not intersect there. An example is shown left below, where three axes-chords are connected into a cycle. The example right below of coplanar chords shows that what I want to achieve is not always possible.

   



Q. Under what conditions can a set of chords through the center of $S$ be connected into
  a single cycle by noncrossing arcs of great circles? In particular, can this be achieved if no three chord endpoints lie on a great circle, 
  i.e., are in general position in this sense?

There is a considerable literature on noncrossing geometric matchings in the plane, but I don't
see that it applies to my question. Any pointers, ideas, or counterexamples welcomed!
(This arose in an investigation related (nonobviously) to an earlier question, "Untangling entwined rigid chains in 3-space".)

REPLY [4 votes]: Among the perfect matchings (with great-circle arcs) that give a single cycle when combined with the chords, consider the one that has minimum total length. Suppose that two of the arcs are crossing, say AB and CD cross at the point X. Then consider replacing these two arcs either by AC+BD or by AD+BC. One of these possibilities (say AC+BD) must again give a single cycle when combined with the chords. And except in "degenerate" cases, the total length of AC+BD will be smaller than that of (AX+XC)+(BX+XD) = (AX+XB)+(CX+XD). The contradiction shows that if the points are in "general position", the desired non-crossing matching with arcs of great circles is possible.<|endoftext|>
TITLE: Notion strictly between stationary and club
QUESTION [7 upvotes]: Let $\kappa$ be an infinite cardinal.
We know that the notion of being closed and unbounded (club) in $\kappa$ is strictly stronger than being stationary in $\kappa$.
My question: Do you know any notion that is strictly between being a club and being a stationary set?

REPLY [7 votes]: Really what you're seeing in most of the above examples is that whenever you have a proper ideal $I$ extending the non-stationary ideal on $\kappa$, then ``not being in $I$'' gives you a notion intermediate between stationary and club. So realy we might ask instead for examples of natural proper ideals $I$ on $\kappa$ that extend the non-stationary ideal.
One family of such ideals I haven't seen mentioned yet arises from the theory of club-guessing.  Concentrating on one particular case, suppose $\kappa$ is a regular cardinal greater than $\omega_1$, and let $S$ be the stationary subset of $\kappa$ consisting of ordinals of countable cofinality. 
There is sequence $\langle C_\delta:\delta\in S\rangle$ such that each $C_\delta$ is club in $\delta$ of order-type $\omega$, and for every closed unbounded $E\subseteq\kappa$ there are stationarily many $\delta\in S$ with $C_\delta\setminus E$ finite.  (So $C_\delta$ is almost contained in $E$ for stationarily many $\delta\in S$.)  
There is a natural way to get a normal filter out of the above set-up: For $E\subseteq\kappa$ club, let $S_E=\{\delta\in S: C_\delta\subseteq^* E\}$. These sets generate a normal filter $F$ concentrating on the stationary set $S$, and the condition ``positive with respect to $F$'' gives you a nice intermediate notion.
Now look at the collection of $A\subseteq\kappa$ with the property that for $F$-almost all $\delta\in S$, $A\cap C_\delta$ is finite.  (These are sets that ``run away'' from the sequence $\langle C_\delta:\delta\in S\rangle$.)
This collection generates a proper ideal $I$ on $\kappa$, and any $I$-positive set is also stationary:  if $A\notin I$ and $E\subseteq\kappa$ is club, we can find a $\delta$ for which $C_\delta\subseteq^* E$ and $A\cap C_\delta$ is infinite, so $A\cap E\neq\emptyset$.<|endoftext|>
TITLE: Is there a monad on Set whose algebras are Tychonoff spaces?
QUESTION [13 upvotes]: Compact Hausdorff spaces are algebras of the ultrafilter monad on Set.
Is the category of Tychonoff spaces also monadic over Set?

REPLY [4 votes]: Even easier.  Let $I$ be the unit interval and $E$ be the equivalence relation that identifies all the rationals.  $I/E$ is certainly not Hausdorff.  Is it a regular category?  I'm not sure.<|endoftext|>
TITLE: Possible mistake in De Giorgi's paper on Holder's regularity
QUESTION [6 upvotes]: $\mu_{n-1}$ is the $n-1$ dimensional measure and $\operatorname{meas}$ is the $n$-dimensional one.
$I(\varrho)$ is the ball of radius $\varrho$ around a fixed point $y$ in the domain $\Omega\subset \mathbb{R}^n$ and $A(t,\varrho)$is the subset of $I(\varrho)$ where a certain function $u$ is greater than $t$.
Surely $I(\varrho) \geq 2\tau(t;\varrho)$ infact, $$\operatorname{meas}{A(t;\varrho)}+\operatorname{meas}(I(\varrho)\setminus A(t;\varrho))= 
\operatorname{meas}I(\varrho)$$
It seems to me that it is claimed $x > y $ implies $ A-x > A-y$
Fortunately, it turns out the inequality still holds but you need a bit of work (EDIT)
Original article: De Giorgi's original article
pages 5-6
EDIT:
Let us assume that $\operatorname{meas}(I(\varrho))=1$ then $\tau(t;\varrho)=x\leq \frac{1}{2}$ what we need to prove is that for $\alpha < 1$ $$(1-x)^{\alpha} + x^{\alpha} -1 \geq 2x^{\alpha}-(2x)^{\alpha}$$
but considering $$f(x)=\left( \frac{1-x}{x}\right) ^{\alpha} -\frac{1}{x^{\alpha}}$$  
we can prove that it is decreasing in $(0,\frac{1}{2})$ and $f(\frac{1}{2})=1-2^{\alpha}$
From this we obtain the inequality and hence justify the line.
Due to the several mistakes the question has changed a lot since I posted it, just to be clear I am asking the following now:
1) Is my "fixing" of the proof alright or am I oversimplifying things?
2) Do you think it was an actual mistake that fortunately did not lead to a problem or it is something obvious enough to be omitted in an article?
P.S. sorry if the question is not specific enough, I think it was when I first posted it!

REPLY [7 votes]: Your explanation looks correct, but there are other ways to see it.  For example, it's a special case of Karamata's majorization inequality for the concave function $f(z) := z^\alpha$.  When $0 \le x \le 1/2$, the sequence $1,x$ majorizes $1-x,2x$, so $f(1) + f(x) \le f(1-x) + f(2x)$, which is the inequality you want.  (And Karamata's inequality is overkill, since the version for sequences of length 2 is easier to prove.  All you need to check is that $f(z)+f(y-z)$ is an increasing function of $z$ for $z \le y/2$, which follows immediately from concavity.  Then apply this with $y=1+x$.)
I don't see any reason to think De Giorgi made a mistake here, and I would assume he had an argument based on concavity.  Maybe there's a one-line version he thought was obvious.
However, I wouldn't say it's obvious the way he has written it, and he could have made it easier to read with a little more explanation.  Even a hint like "and it follows from the concavity of $z \mapsto z^{1-1/r}$ that..." could help.<|endoftext|>
TITLE: Etale cohomology of the completion of a Henselian local ring
QUESTION [10 upvotes]: Let $\pi: R\to S$ be a local morphism of Henselian local rings. Let $f: R \to \hat{R}$ and $g: S \to \hat{S}$ be their completions. Let $\mathcal F$ be a constructible $l$-adic sheaf on $\operatorname {Spec} S$ for $l$ invertible in $R$. Is the natural base change map $f^*R^i\pi_* \mathcal F \to R^i \hat{\pi}_* g^* \mathcal F$ an isomorphism?
Motivation: For most purposes, Henselian local rings are as good as complete ones, and completions are as good as Henselizations. But are they cohomologically the same? $H^i$ is the same because global sections are just the stalk at the closed point in both cases, but the behavior of pushforwards and higher pushforwards are not so clear.
EDIT: Thanks to user61789, I realize my assumptions are much too general! Thus I will ask about a much more specific case. Let $R$ be the etale local ring of $\mathbb A^n_k$ at the origin, let $S$ be the etale local ring of $\mathbb A^m_k$ at the origin for $m\geq n$, and let $\pi$ be induced by the natural projection $\mathbb A^m_k \to \mathbb A^n_k$. Then is the map an isomorphism?

REPLY [14 votes]: You probably meant to assume $R$ and $S$ are noetherian. The answer is "no" to the initial hypergeneral part of the question.  EDIT: In the 2nd half (below the long line), I now give a proof of an affirmative answer to the added part involving maps of affine spaces.
Counterexamples to the initial hypergeneral part can be made using 2-dimensional regular excellent local rings built from henselization and completion of local rings at $k$-points on smooth schemes over any field $k$.
Let $R$ be a noetherian henselian local ring, and 
$S = \widehat{R}$, so $\widehat{S}=S$.  Let $\pi:{\rm{Spec}}(\widehat{R}) \rightarrow {\rm{Spec}}(R)$ be the natural map.  Then in this case an affirmative answer to your question says exactly that 
$\pi_{\ast}$ is exact and $\pi^{\ast} \circ \pi_{\ast} \rightarrow {\rm{id}}$ is an isomorphism of functors (all on the category of constructible abelian etale sheaves on ${\rm{Spec}}(\widehat{R})$). In particular, every such sheaf on ${\rm{Spec}}(\widehat{R})$ would be the $\pi$-pullback of one on ${\rm{Spec}}(R)$.  Clearly this cannot be true in general, so it is just a game to find a suitable counterexample to that.
Let $Z' \subset {\rm{Spec}}(\widehat{R})$ is a Zariski-closed set and $j':U' \hookrightarrow {\rm{Spec}}(\widehat{R})$ the open subscheme complementary to $Z'$. Consider the extension-by-zero $F' := j'_{!}(\mathbf{Z}/(n))$ for an integer $n > 1$. Suppose $F' = \pi^{\ast}(F)$ for an abelian etale sheaf $F$ on ${\rm{Spec}}(R)$.  
Let $Z$ be the set of points $x \in {\rm{Spec}}(R)$ such that $F_x = 0$. This satisfies $Z' = \pi^{-1}(Z)$, so $Z$ is closed since $\pi$ is topologically a quotient map (as it is fpqc). Hence, if there is a closed set $Z'$ that is not the preimage of a closed set in ${\rm{Spec}}(R)$ then we have a counterexample. 
Suppose $R$ is excellent, so the fpqc map $\pi$ is a "regular morphism" (flat and its fiber algebras that are regular and remain so after finite extension of the base field). Thus, if $Z$ is a reduced closed set in ${\rm{Spec}}(R)$ then the scheme-theoretic preimage $\pi^{-1}(Z)$ is reduced.  Consider any radical ideal $J'$ in $\widehat{R}$ and let $Z' := {\rm{Spec}}(\widehat{R}/J')$. If $Z' = \pi^{-1}(Z)$ topologically for a closed set $Z$ in ${\rm{Spec}}(R)$ then by using the reduced scheme structure on $Z$ we would have $\pi^{-1}(Z) = Z'$ as schemes.  In other words, for the radical ideal $J$ in $R$ corresponding to such a $Z$ we would necessarily have $J' = J\widehat{R}$.  
Note that $J \otimes_R \widehat{R} \rightarrow J\widehat{R}$ is an isomorphism, so if $J'$ is invertible then necessarily $J$ is invertible.  In particular, if $J' = r'\widehat{R}$ for $r'$ that is not a zero-divisor then necessarily $J = rR$ for some $r \in R$ that is not a zero-divisor.  In such a situation, $r'$ would have to be a unit multiple of $r$ in $\widehat{R}$.  Thus, if we can find $r' \in \widehat{R}$ that isn't a zero-divisor such that no unit multiple of $r'$ lies in the subring $R$ then we have our counterexample (using $Z' = {\rm{Spec}}(\widehat{R}/(r'))$ and the associated $F'$ as above).
In the special case that $R$ is regular, so $\widehat{R}$ is also regular, we're just looking for a nonzero $r' \in \widehat{R}$ having no unit multiple in $R$.  Obviously in the dvr case this cannot be arranged, so we go on to dimension 2. 
Take $R = k[x,y]_{(x,y)}^{\rm{h}}$ for a field $k$, so $\widehat{R} = k[\![x,y]\!]$. The ring $R$ is excellent, since passage to henselization preserves excellence (18.7.6, EGA IV$_4$). Since $k(\!(x)\!)$ is not algebraic over $k(x)$ (lazy way is to use $e^x$ and assume characteristic is 0), we can find $h \in x k[\![x]\!]$ that is not algebraic over $k(x)$.  For such $h$, the irreducible element $r' = y - h(x)$ in $\widehat{R}$ will do the job.
Indeed, suppose that $r'$ admits a unit multiple $r \in R$. We will show that $h$ must be algebraic over $k(x)$. Clearly $R/(r) \rightarrow \widehat{R}/(r') = k[\![x]\!]$ is a completion map, so $R/(r)$ is a dvr.  Since henselization is compatible with quotients, it follows by the link between transcendence degree and dimension for finitely generated domains over $k$ that $R/(r)$ is a direct limit of a directed system of local-etale extensions of local rings at $k$-points on regular curves over $k$, and these curves must all be quasi-finite over the affine $x$-line over $k$ (since $x \in k[\![x]\!]$ is transcendental over $k$).  Thus, everything in $R/(r)$ is algebraic over $k(x)$, including the class of $y$.  But mapping that into $k[\![x]\!]$ with $y \mapsto h(x)$ in there, we conclude that $h$ is algebraic over $k(x)$.

Now let's give an affirmative proof under some additional "smoothness" hypotheses as follows.  The key point is to use Artin-Popescu approximation and to work at the level of derived categories. Also, one has to be extremely careful because certain schemes will intervene that (as far as I know) might fail to be noetherian or excellent, even though our original setup will involve only excellent schemes and smooth maps. (Maybe for some reasons which escape me, the non-noetherian concerns in the argument below cannot really happen?) 
Setup: Let $\mathcal{X} \rightarrow \mathcal{Y}$ be a smooth map of noetherian schemes with $\mathcal{Y}$ excellent. Choose $y \in Y$ and $x \in X_y$ a $k(y)$-rational point.   Let $X = {\rm{Spec}}(O_{\mathcal{X},x}^{\rm{h}})$, $Y = {\rm{Spec}}(O_{\mathcal{Y},y}^{\rm{h}})$.  Let $X'$ and $Y'$ denote Spec of the corresponding completed local rings, and let $\pi:X \rightarrow Y$, and $\pi':X' \rightarrow Y'$ be the natural maps. 
Claim: Let $F$ be a torsion etale abelian sheaf on $X$ whose torsion-orders are invertible on $Y$. Let $G \rightsquigarrow G'$ be shorthand for pullback on derived categories (of abelian etale sheaves) from $Y$ to $Y'$ or from $X$ to $X'$. Then the natural base change morphism $$R\pi_{\ast}(F)' \rightarrow R\pi'_{\ast}(F')$$ is an isomorphism for any torsion $F$ on $X$ whose torsion-orders are invertible on $Y$.
Proof: By Artin-Popescu approximation, since $\mathcal{Y}$ is excellent we know that the map $Y' \rightarrow Y$ is a limit of smooth maps.  Hence, if we let $T = X \times_Y Y'$ and $p:T \rightarrow Y'$ be the projection then by the smooth base change theorem and standard limit stuff with etale sheaf theory, the natural map $R\pi_{\ast}(F)' \rightarrow Rp_{\ast}(G)$ is an isomorphism, where $G$ is the pullback of $F$ along the first projection. 
I don't know if $T$ is noetherian, by the way, so I am tacitly using the good behavior of the limit formalism in etale sheaf theory without noetherian hypotheses (but even the proof of the smooth base change theorem in SGA4.5 leaves the noetherian framework due to such kind of fiber products intervening, so I suppose you don't mind).
The $y$-fiber of $p$ is identified with $X_y$, so naturally $x \in T$, and $O_{T,x}$ is a "partial henselization" of the local ring at a rational point on the special fiber of a finite type (even smooth) $Y'$-scheme.  So even though I/we do not know if $T$ is noetherian, we do know that $O_{T,x}$ is noetherian, by the same reasoning which proves that passage to henselization preserves the noetherian property (namely, EGA 0$_{\rm{III}}$, 10.3.1.3).  
Note that the henselization of $O_{T,x}$ coincides with that of a local ring $R$ on a finite type (even smooth) $Y'$-scheme, and $Y'$ is excellent (as for any complete local noetherian ring: IV$_2$, 7.8.3(iii)), so $R$ is excellent. Consequently $R^{\rm{h}}$ is excellent (IV$_4$, 18.7.6), so $O_{T,x}$ has excellent henselization. Now it is not true that excellence descends from the henselization (counterexample in IV$_4$, 18.7.7), but this is for geometric reasons related to failure of being universally catenary, and we don't actually care about that.  
Indeed, what matters for the following is geometric regularity of formal fibers of $O_{T,x}$, which is to say that the flat map of noetherian schemes ${\rm{Spec}}(O_{T,x}^{\wedge}) \rightarrow {\rm{Spec}}(O_{T,x})$ is a regular morphism (since Artin-Popescu approximation is about regular morphisms being a limit of smooth morphisms, which in practice is easiest to remember under the banner of "excellence" but is not strictly a necessary condition). So all we care is to know that geometric regularity of formal fibers descends from the henselization, and that is fine; see IV$_4$, 18.7.4.
The upshot is that $O_{T,x}$ is noetherian and its completion morphism is regular. 
Now comes the crux of the matter: I claim that the map $h:X' \rightarrow T$ (or rather, its factorization through ${\rm{Spec}}(O_{T,x})$) is computing the completion of the noetherian local ring $O_{T,x}$. (This is the step at which our argument breaks down in the setting of the counterexamples above! Indeed, in that setting the map in the role of $h$ would be akin to a graph morphism, super-far from an isomorphism.)
In view of the local structure theorem for smooth morphisms (applied to a smooth $Y'$-scheme whose "partial henselization" at a suitable point computes $T$), our
hypothesis that $k(x) = k(y)$, and the fact that completion of a local noetherian ring is insensitive to "partial henselization", justifying our assertion about $h$ amounts to the following down-to-earth observation: if $A$ is a henselian (e.g., complete) local noetherian ring, $B = A\{x_1,\dots,x_N\}$ is the henselization at the origin of the special fiber of an affine space over $A$, and $B'$ is the local ring at the origin on the special fiber of the ring $\widehat{A} \otimes_A B$ (a ring I/we do not know to be noetherian, but the local ring $B'$ certainly is, as explained above), then the natural map $B' \rightarrow \widehat{A}[\![x_1,\dots,x_N]\!]$ is the completion of $B'$.  This verification is a simple exercise using the compatibility of henselization and quotients.
From our description of $h$ via completion of a local noetherian ring having geometrically regular formal fibers, $h$ is a regular morphism.  Thus, by Popescu's theorem, $h$ is a limit of smooth maps. Hence, by the acyclicity theorem for smooth maps and the usual limit games (which do not require knowing $T$ to be noetherian), it follows that the natural map $G\rightarrow Rh_{\ast}(h^{\ast}G)$ is an isomorphism.  But $h^{\ast}(G)=F'$, so we have a natural isomorphism
$$R\pi_{\ast}(F)' \simeq Rp_{\ast}(G)=Rp_{\ast}Rh_{\ast}(F')=R(p \circ h)_{\ast}(F').$$
But $p\circ h=\pi'$, and as such it is a standard exercise to check that the composite isomorphism we have made is the base change morphism of initial interest. 
QED<|endoftext|>
TITLE: Switching from pure mathematics (e.g. geometry) to more applied areas (e.g imaging) after Ph.D., as postdoc and chance of getting such a postdoc?
QUESTION [16 upvotes]: Before I start my question, I should probably mention that this question might not be the right question to ask here, but I tried academiabeta, and stackoverflow, but without getting any to-the-point responses. I also noticed that there were some questions of similar interest asked here, but mine is little different and more subject/situation-specific. Also I am afraid my question might be a little too long, pardon my slightly unnecessary detail!
Here is my situation: I am a pure mathematician finishing Ph.D., working in Riemann surfaces, Teichmueller theory and differential geometry (with published papers in decent journals), having backgrounds in these topics and some PDE and topology and have some introductory knowledge about programming. I have been wanting to switch to little more applied areas (described below) because: 1) I believe getting posts and grants are slightly easier in there, 2) my postdoc hunting in pure mathematics have resulted in vein (despite getting shortlisted twice in Europe). I wanted to switch later anyway but reason 2) is forcing me to think of it now.
By applied areas, I meant areas like medical imaging, computer vision or even theoretical biology (there are interesting works of Prof. Robert Penner on moduli spaces and protein folding, for example), where they use Riemann surface and differential geometry a lot, and I want to work in more theoretical problems, but would also like to learn more computational techniques (although secondary). 
Without further ado, here is my question:
Since I have not published any single papers in those applied areas, or haven't had any formal training in them, what are my chances of getting a postdoc in these areas, being a pure mathematician all my life so far? If any of you have done so before, or are working in these applied areas, I would very much appreciate if you inform me! By informal communication with some people (who are Ph.D. students) working in these areas, I was informed that my backgrounds are good, but one or two professors I have contacted already mentioned that they want somebody with actual research experience in say imaging, which I do not have. So, I am not sure where I stand compared to other applicants. Your responses and honest opinion (even though not in my favor) will be highly appreciated! Thanks a lot in advance!

REPLY [7 votes]: There is a newish emerging area, `applied algebraic topology'. It includes Topological Data Analysis (see work by Gunnar Carlsson and others).  There is also new work again in Applied Algebraic Topology looking at Configuration spaces and probablity theory. Look for research assistant jobs at Post Doc level on specific topics in applied geometry and topology.
There is also an ESF programme (http://acat.lix.polytechnique.fr/) that gives some idea of what may be available. Some of this would provide a half way house between where you are and where you want to be!  Good Luck.  (Soon there is s summer school in Slovenia on these areas. Look at the program schedule and you may get some ideas on who to contact to see what they think, whetehr there are any positions coming up, short visits are a good way to start if they have money for such and seem interested. I think there is quite a good chance and all of this area is growing fast so it could be very interesting to get into it at the moment. (The topic of big data is even hitting the newspapers, and that forms part of this.)<|endoftext|>
TITLE: Strongly continuous semigroups that cannot be contractions
QUESTION [8 upvotes]: Let $X$ be a Banach space, and $(P_t)_{t \ge 0}$ a strongly continuous semigroup of bounded operators on $X$.  Using the uniform boundedness principle, it's simple to prove that there are constants $M, \omega$ such that $$\|P_t\| \le M e^{\omega t} \quad (*)$$ for all $t$.  Moreover, if $M=1$ you can get a contraction semigroup by studying $e^{-\omega t} P_t$, and I'm having trouble thinking of an example where that's not the case.

What is a simple example of a semigroup $P_t$ for which (*) cannot hold with $M=1$?

If possible, I'd like to see an example where $X$ is a separable Hilbert space.

REPLY [11 votes]: There are many examples constructed with weighted shifts. The following is a Hilbert space example.
Let us consider the Hilbert space $L^2\big((0,1),\mu\big)$, where $\mu$ denotes the measure defined by
$$\mu(A):=2\lambda(A\cap(0,\tfrac12))+\lambda(A\cap(\tfrac12,1)).$$
for all Lebesgue measurable sets A. Here $\lambda$
is the Lebesgue-measure. Furthermore, let $T(t)$ be the nilpotent left shift semigroup. Obviously, $T$ satises the semigroup property and, since the norm
$\|\cdot\|_{\mu}$ is equivalent to the norm $\|\cdot\|_{\lambda}$, $T$ is strongly continuous. 
Clearly, $\|T(t)\|\leq 2$.
In addition we see that $T(t) = 0$ for all $t>1$.
Finally, consider the function 
$$f_t = \frac{1}{\sqrt{t}}\chi_{(\frac12,\frac12+t)}$$
for $t\in(0,\tfrac12)$.
Clearly, $\|f_t\|_{\mu} = 1$ and
$$\|T(t)f_t\|_{\mu} = 2.$$
ADDED:
Theoretically, you can always introduce an equivalent norm in your space which makes your semigroup a contraction semigroup, see Lemma II.3.10 in 
Engel, Klaus-Jochen; Nagel, Rainer, One-parameter semigroups for linear evolution equations, Graduate Texts in Mathematics. 194. Berlin: Springer. xxi, 586 p. (2000). ZBL0952.47036.
The construction is, however, in most cases highly non-constructive and can only used for theoretical purposes.
It is more important that there are examples in Hilbert spaces where it is impossible to find an equivalent Hilbert space norm (i.e., the semigroup is not similar to a contraction semigroup), see
Packel, E. W., A semigroup analogue of Foguel’s counterexample, Proc. Am. Math. Soc. 21, 240-244 (1969). ZBL0175.13802.<|endoftext|>
TITLE: Can we prove that there are countably many isomorphism classes of compact Lie groups without the classification of simple Lie algebras?
QUESTION [5 upvotes]: This is an old math.SE question of mine that was never answered:

It is a nontrivial fact that there are only countably many isomorphism classes of compact Lie groups. One can prove this by a series of reductions: first to the connected case, then to the simply connected case, then by classifying simple Lie algebras. Of course, this proof actually gives a much stronger classification result.
If I only want to prove that there are countably many isomorphism classes of compact Lie groups, can I work without appealing to the classification of simple Lie algebras? I have some ideas involving Tannaka's theorem but I haven't worked out a proof yet.

The idea I had was to classify the possible symmetric monoidal [more adjectives if necessary] categories of representations of compact Lie groups; I think these categories are all "finitely presented" in a suitable sense, and from here it should be possible to show that there are only countably many presentations. Such presentations are given for the classical groups in Baez's Higher-Dimensional Algebra II: 2-Hilbert Spaces.

REPLY [2 votes]: In comments it is mentioned that a step in the proof can be " to prove is that the Lie algebras of compact Lie groups are infinitesimally rigid in the sense that they have no first-order deformations". This was first done, without the use of cohomological arguments, in Segal "A class of operator algebras which are determined by groups", Duke Math. Journ. 18, 1951 pages 256-257.
Once one takes for granted rigidity the fact that there are only countably many isomorphism classes of simple Lie algebras seems to me to follow from algebraic arguments. The variety of Lie algebra laws on a vector space of fixed dimension is an algebraic variety, with finitely many components, and each semisimple Lie algebra is a Zariski open dense subset of a component. Thus you have finitely many of them, without relying on classification.
The passage from local to global, then, should be as described by others in comments.<|endoftext|>
TITLE: Direct product decomposition for infinite abelian groups with constrained torsion
QUESTION [9 upvotes]: Let $g$ be a positive integer, and let $G$ be a commutative group with the following constraint on its torsion subgroup: there is an injection $G[\operatorname{tors}] \hookrightarrow (\mathbb{Q}/\mathbb{Z})^{2g}$.  Must there be subgroups $G_1,\ldots,G_g$ of $G$ such that   
(i) $G = G_1 \times \ldots \times G_g$ (internal direct product), and
(ii) For all $1 \leq i \leq g$, there is an injection $G_i[\operatorname{tors}] \hookrightarrow (\mathbb{Q}/\mathbb{Z})^2$?  
Motivation: If this is true, then it reduces the "Inverse Mordell-Weil Problem for Abelian Varieties" to the "Inverse Mordell-Weil Problem for Elliptic Curves".
Thus although the given question certain has an affirmative answer in many special cases -- e.g. it is a triviality if $G$ is finitely generated -- I am not really interested in that.  But it would be "lucky for me" if the answer turns out to be affirmative in the general case, so it's worth asking.  
Added: This previous question contains some information on when the torsion subgroup of a commutative group is a direct summand.

REPLY [3 votes]: The answer is negative.
Define $S=\bigoplus_p C_p$, where $C_p$ is cyclic of order $p$ and $p$ ranges over prime numbers. Fix an integer $n\ge 1$. Define $P=\prod_p C_p$, and consider a maximal $\mathbf{Z}$-free family in $P^n/S^n$, generating a free abelian group (of continuum rank) $L/S^n$. Fix a prime $q$ and let $U/S^n$ be (*) a subgroup of $P^n/S^n$ isomorphic to $\mathbf{Z}_q$ (the group of $q$-adic numbers), with $L\subset U$.
Let's check that for every direct decomposition $U=V\oplus W$, either $V$ or $W$ is finite. Since $T(U)\simeq S^n$, this implies that $U$ answers negatively your question (as soon as $n\ge 3$).
First observe that $\mathbf{Z}_q$ is indecomposable (i.e. has no nontrivial decomposition as direct product, easy exercise (**)). Since $U/T(U)\simeq\mathbf{Z}_q$, it follows that either $V$ or $W$, say $W$, is torsion. So $V/T(V)\simeq\mathbf{Z}_q$.
Write $T(V)=\bigoplus A_p$ and $T(W)=\bigoplus B_p$, so that $A_p\oplus B_p=C_p^n$ for all $p$.
We have $P^n=(\prod A_p)\times (\prod B_p)$. So $P^n/T(V)=(\prod A_p/\bigoplus A_p)\times \prod B_p$. Since there is no nonzero homomorphism $\mathbf{Z}_q\to C_p$ for $p\neq q$ and since the subgroup $V/T(V)$ of $P^n/T(V)$ is isomorphic to $\mathbf{Z}_q$, the projection of $V/T(V)$ into $\prod B_p$ is contained in $B_q$. Thus in the above decomposition, $V\subset (\prod A_p)\times B_q$ and $W\subset (\bigoplus A_p)\times (\bigoplus B_p)$. So $U\subset (\prod A_p)\times (\bigoplus B_p)$. Since $P^n/U$ is torsion (because it is a quotient of $P^n/M$ which is torsion), it follows that $\prod B_p/\bigoplus B_p$ is torsion. This means that $B_p=0$ for large $p$. It follows that $W$ is finite.
Here are the two easy verifications:
(*): pick in $\mathbf{Z}_q$ a maximal $\mathbf{Z}$-free family, generating an abelian free subgroup $Z$; since $P^n/S^n$ is a rational vector space, any isomorphism $Z\to L/S^n$ (it exists since both groups are free abelian of continuum rank) can be extended to an injective homomorphism $\mathbf{Z}_q\to P^n/S^n$.
(**): $\mathbf{Z}_q$ is $p$-divisible for all $p\neq q$ and $\mathbf{Z}_q/q\mathbf{Z}_q$ has order $q$, so for every direct decomposition of $\mathbf{Z}_q$, at least one factor is divisible. But $\mathbf{Z}_q$ does not contain any copy of $\mathbf{Q}$.<|endoftext|>
TITLE: For which metric spaces is Gromov-Hausdorff distance actually achieved?
QUESTION [10 upvotes]: Question

For which pairs $M,N$ of compact metric spaces does there exist a metric space $K$ along with isometric embeddings $i:M \to K$ and $j:N \to K$ so that the Hausdorff distance between $i(M)$ and $j(N)$ in $K$ equals the Gromov-Hausdorff distance between $M$ and $N$ exactly? 

We always have $d_\text{GH}(M,N) \leq d_\text{H}(i(M),j(N))$ by definition, so it suffices to decide if there exist any constraints intrinsic to $M$ and $N$ which force the reverse inequality to also hold for some judicious choice of $i,j,K$.
Update
Bill Johnson's nice answer appears to settle the question for compact $M$ and $N$ affirmatively: the Gromov-Hausdorff distance can always be realized by an ultraproduct construction. In light of that answer -- and also in light of the fact that I did not intend to assume compactness in the original question but stupidly wrote it down anyway -- here is a modified question:

Can the GH-distance be similarly achieved if we only assume that $M$ and $N$ are bounded, rather than compact?

In general, I am interested in the weakest hypotheses on $M$ and $N$ which are known to guarantee embeddings into a common target space $K$ so that the Gromov-Hausdorff distance between $M$ and $N$ equals the Hausdorff distance of their isometric images in $K$. It is not clear to me that a modification of Bill's argument works when we drop compactness.
Background
Given a compact metric space $M$ and two subspaces $A, B \subset M$, their Hausdorff distance -- denoted $d_\text{H}(A,B)$ -- is defined to be the smallest $\epsilon > 0$ so that $B$ is contained in $A$ thickened by $\epsilon$ and vice-versa. 
The Gromov-Hausdorff distance $d_\text{GH}(M,N)$ between two compact metric spaces $M$ and $N$ is defined to be the infimum over triples $(i,j,K)$ of $d_\text{H}(i(M),j(N))$ where $K$ is a metric space and $i,j$ are isometric embeddings of $M, N$ into $K$. The question asks when this infimum can be explicitly realized.

REPLY [2 votes]: Bounded is not sufficient to ensure that the Gromov-Hausdorff distance is actually achieved. Here is a counterexample:
Let $M$ be a metric space whose underlying set is $\mathbb{N}\times\{0,1\}$ with the following metric: $d((a,i),(b,j))=1$ if $a\neq b$ and $d((a,0),(a,1))=2^{-a}$. Let $N$ be $M$ with a single new point $x$ such that $d(x,y)=1$ for all $y\neq x$. $d_{GH}(M,N)=0$ but this cannot be realized in any common embedding because $M$ and $N$ are not isomorphic.
For bounded metric spaces we can use the formalism of continuous logic to state a few situations in which we can ensure that the distance is actually achieved (although these conditions are probably not very useful).
The first one is for any bounded metric spaces $M$ and $N$ there are elementary extensions $M^\prime \succeq M$ and $N^\prime \succeq N$ and a common embedding $i:M^\prime \rightarrow K $ and $j:N^\prime \rightarrow K$ such that $d_H(i(M^\prime),j(M^\prime))=d_{GH}(M,N)$. You can achieve this with exactly the same argument as in Bill's answer, specifically an ultrapower construction, the only difference is that for non-compact spaces the ultrapower will grow.
The second one is that if $M$ and $N$ are "resplendent" in the model theoretic sense then there will always exist $i:M\rightarrow K$ and $j:N\rightarrow K$ such that $d_H(i(M),j(N))=d_{GH}(i(M),j(N))$. Nobody has bothered to write up the details of how resplendence works in continuous logic but I'm fairly confident that the most common conditions ensuring it will work the same as they do in discrete logic, in particular if you know that $M$ and $N$ have the same density character and are either saturated or special then the Gromov-Hausdorff distance between them will be achieved.
In general though I imagine that for an arbitrary pair of bounded metric spaces this problem is very hard.<|endoftext|>
TITLE: Intuition for failure of Implicit Function theorem on Frechet Manifolds
QUESTION [28 upvotes]: When dealing with moduli spaces of, say connections or metrics, I am using the notions of Frechet spaces/manifolds/groups. I have become familiar with Banach manifolds (I think), but Frechet manifolds less so. In my studies, we want to apply the Inverse and Implicit Function Theorems, which work on Banach spaces, but apparently fail for Frechet spaces (and as a result we pass to Sobolev completions).
Why does it fail (intuitively), or why should I expect it to fail? What is the hinge that allows it to work for Banach but not Frechet? Do I have to do a lot more to get an analogous result in the land of Frechet? This should also help me see the differences between a Banach and Frechet manifold, because right now it seems subtle.

REPLY [6 votes]: In a Banach setting the proof of the Implicit Function Theorem (IFT) is based on Newton iteration. Each iteration step requires the solution of a linearized equation. When applying this, for example, to a partial differential equation $F(u)=f$, problems arise, even if the base manifold is compact, if the linearizations $DF(u)$ (assumed invertible) are not elliptic. The reason is "loss of derivatives" in the iterative corrections $DF(u)^{-1}F(u)$. Application by $F$ decreases (Sobolev or Hölder) differentiability by the order of $F$, and this loss is regained by $DF(u)^{-1}$ only if $DF(u)$ is elliptic. If $DF(u)$ is not elliptic, then the techniques of Nash and Moser are used to obtain an IFT. Here the iteration scheme is modified by mollification: In each iteration step $u$ is replaced by $S_\theta u$ for a suitable mollification parameter $\theta$, $S_\theta u\to u$ in each Hölder space as $\theta\to\infty$. It is essential that norm estimates such as
$$\|S_\theta u\|_\alpha\lesssim\theta^{\alpha-\beta}\|u\|_\beta,\quad\alpha\geq\beta,$$
allow control of higher order derivatives of the mollifications of $u$ by lower order derivatives of $u$. (It would be of no help to assume $u\in C^\infty$ from the beginning because this gives no control of higher order derivatives.) The method of Nash was simplified and applied by Hörmander; see chapter 3 of the book of Alinhac and Gerard (GSM 82) for an exposition and references. Hamilton abstracted the estimates as tameness properties and gave a generalization of the IFT to Frechet spaces. 
So, my intuition about IFT in a Frechet setting comes from the special, but significant, situation which I just described: The non-ellipticity of the derivative/linearization of the differential operator and the resulting loss in derivatives during iteration are the cause of trouble. A remedy if is approximation by suitable mollications.<|endoftext|>
TITLE: Historical use of figures in geometry
QUESTION [20 upvotes]: I was surprised to learn from John Stillwell's comment in answer to the
question,
"Can the unsolvability of quintics be seen in the geometry of the icosahedron?",
that

There is not a single picture in the whole ... 

of Felix Klein's 1884 book,
Lectures on the Icosahedron and the Solution of Equations of the Fifth Degree.
And now that I have it in my hands, I can verify the lack of figures (but its
remarkable clarity nonetheless).
By the time of David Hilbert's and Stephan Cohn-Vossen's 1932
Geometry and the Imagination (citing an earlier MO question), 
the use of figures had reached a high art.
This high art is continued today in
Tristan Needham's Visual Complex Analysis, with its stunning figures, e.g.,
(from this MSE answer): 
      
To return to ancient times, it is clear that figures were
valued to illustrate Euclid by 100 AD, and likely earlier:

Finally, my question:

Has the use of figures to illustrate geometry waxed and waned over history
  in a fashion that could almost be graphed with respect to time, or am I plucking out
  idiosyncratic examples that do not point to any recognizable trends?

Perhaps this needs to be mapped country by country, different in France
(during the Bourbaki period) than in Germany, etc.? Or perhaps any such attempt to
capture gross trends is hopelessly historically naive?

REPLY [4 votes]: The Sangaku  problems of Japan's Edo period are worth mentioning here. Also see Sangaku: Reflections on the Phenomenon for some cultural history about it.<|endoftext|>
TITLE: Math research in the app store
QUESTION [10 upvotes]: What apps can be found in the Apple app store, Google Play, Blackberry World etc. showcasing specific mathematics research?
(Edit - Since the first version of the question got closed, examples should showcase specific work of specific mathematicians. Tools such as Wolfram Alpha, bibliography managers, arXiv app, etc. are no longer allowed.)

REPLY [10 votes]: Akio Kawauchi, Ayaka Shimizu, and Kengo Kishimoto have created an app called Region Select, based on a knot theory paper by Shimizu. She found an algorithm to unknot a knot diagram by "region crossing change" moves, and then they realized that it was tricky and interesting enough that it was worth turning it into a game. The game got big-time press coverage in Japan. I wrote a blog post about it here. The game became popular enough that it was developed into an honest to goodness android app.
It's playable, a lot of fun, and there's a genuine theorem and research paper hiding behind it- can you figure out a strategy to win the game from any starting configuration?<|endoftext|>
TITLE: How big is the proper class of all sets?
QUESTION [11 upvotes]: Let $\operatorname{ZFC}^{-}$ be the theory of $\operatorname{ZFC}$ minus the axiom of foundation and define the proper classes $G$ and $V$ as follows: 
$G:=$ The proper class of all sets.
$V:=$ The proper class of Von neumann cumulative heirachy.
And let the statements $G=V$ and $|G|=|V|$ be:
$G=V~:~~\forall x~\exists y~(\operatorname{ord}(y)~ \wedge "x\in V_{y}")$ 
$|G|=|V|~:~~\exists F\colon G \longrightarrow V~~$ a one to one function.
Using the axiom of foundation it is clear that we have $G=V$ and $|G|=|V|$ means that $G$ is as "small" as Von Neumann's cumulative hierarchy.
Now the question is: "How big is the proper class of all sets in the absence of the axiom of foundation? " 
In the other words which one of the following statements are true?
(1) $\operatorname{Con}(\operatorname{ZFC}) \longrightarrow \operatorname{Con}(\operatorname{ZFC}^{-} + G\neq V)$
(2) $\operatorname{Con}(\operatorname{ZFC}) \longrightarrow \operatorname{Con}(\operatorname{ZFC}^{-} + |G|\neq|V|)$

REPLY [10 votes]: It seems that most of the usual anti-foundational set theories, when combined with the axiom of choice, imply that $|V|=|G|$. For example, in the case of Aczel's anti-foundation axiom AFA, we have
Theorem. ZFC-foundation + AFA proves $|V|=|G|$. 
Proof. Since $V$ is contained in $G$, it suffices to find an injection of $G$ into $V$. Consider any set $x$, not necessarily well-founded. Let $y$ be the transitive closure of $\{x\}$, so $y$ has $x$ and all the hereditarily elements of $x$, and let $\langle y,{\in},x\rangle$ be the corresponding accessible pointed graph, using the $\in$ relation. By AC, this graph has isomorphic copies $\langle h,\to,a\rangle$ in $V$, since we may well-order the nodes and thus find an isomorphic copy built on ordinals. Let $F(x)$ be the collection of all such graphs isomorphic to $\langle y,{\in},x\rangle$ chosen of $\in$-minimal rank in $V$. This is a set in $V$, and furthermore, $F$ is injective, since we can recover $y$ and hence $x$ from any graph isomorphic to $\langle y,{\in},x\rangle$. So we have injections both ways $V\to G$ and $G\to V$, and so they are bijective. QED
The same argument works with many of the other anti-foundational set theories, provided that equality of sets is determined by properties of the underlying $\in$-graph on the hereditary closure of the set. One prominent exception to this is that the Boffa AFA does not have this property.<|endoftext|>
TITLE: Is this a description of the $\aleph_1$-localizing subcategory generated by a compact generator?
QUESTION [5 upvotes]: This should be obvious but I'm not seeing it:
The $\mathfrak T$ be a triangulated category with coproducts and with a compact generator $A$ (that is, the functor $\mathfrak T(A,\_)$ preserves coproducts and the localizing subcategory $\langle A\rangle$ of $\mathfrak T$ generated by $A$ is all of $\mathfrak T$.)
For instance, $\mathfrak T$ could be the derived category of a ring or a ring spectrum.
Let $\langle A\rangle_{\aleph_1}\subseteq\mathfrak T$ be the $\aleph_1$-localizing subcategory of $\mathfrak T$ (that is, the smallest triangulated subcategory containing $A$ and being closed under countable coproducts).
Certainly, if $B$ belongs to $\langle A\rangle_{\aleph_1}$, then $\mathfrak T_*(A,B)$ is a countably generated module over $\mathfrak T_*(A,A)$.
Question: Does the converse hold, that is, do we have
$$
\langle A\rangle_{\aleph_1}=\{B\in\mathrm{Ob}(\mathfrak T)\mid\text{$\mathfrak T_*(A,B)$ is countably generated over $\mathfrak T_*(A,A)$}\}\;?
$$

REPLY [7 votes]: I don't think what you say is true. Let $k$ be a field, $V$ a $k$-vector space of uncountable dimension and $R=k\oplus V$ the $k$-algebra where $V$ is a square-zero ideal. Consider $\mathfrak T=D(R)$ the derived category of $R$, and $A=R$. Take a non-trivial vector $0\neq v\in V$. The complex 
$$B=\cdots\rightarrow 0\rightarrow R\stackrel{v}\longrightarrow R\rightarrow 0\rightarrow \cdots$$
is in $\langle A\rangle_{\aleph_1}$, in fact it is in $\langle A\rangle_{\aleph_0}$. In this case $\mathfrak T_*(A,A)=R$ concentrated in degree $0$ and $\mathfrak T_*(A,B)=R/(v)\oplus V[1]$, which is not countably generated since $\dim_kV$ is uncountable.
What you claim is true under some transfinite coherence hypothesis, e.g. it is proven in the literature under the hypothesis that $\mathfrak T_*(A,A)$ is countable.<|endoftext|>
TITLE: Mathematics of privacy?
QUESTION [17 upvotes]: I wonder to which extent the current public debate on privacy issues (not only by state sniffing, but e.g. by microtargetting ads too an issue) offers interesting questions in mathematics? 
Can we expect proven and durable software solutions? 
Question stimulated by AdLeaks

REPLY [2 votes]: Proving and verifying statements about private data without releasing that data is a nice candidate question/problem with some known solutions. Let me hereby name interactive proof/argument systems, zero knowledge, commitments, oblivious transfer as well as implementations like U-Prove, DAA, Idemix.
"How to prove a theorem so no one else can claim it", Manuel Blum, International Congress of Mathematicians, 1986.<|endoftext|>
TITLE: The Irreducible Corepresentations of the eight-dimensional Kac-Paljutkin Quantum Group
QUESTION [6 upvotes]: I asked this question on Math.Stack but have not had any answers.
Question
What are the irreducible corepresentations of the eight-dimensional Kac-Paljutkin Quantum Group, $A$?
The trivial corepresentation is given by $\Delta_{|\mathbb{C}1_A}$ where $1_A$ is just the unit. 
By my reckoning there should be seven more one dimensional invariant subspaces and hence irreducible corepresentations.
The eight-dimensional Kac-Paljutkin Quantum Group
Here we give the defining relations and the main structure of an eight-dimensional quantum group introduced by Kac and Paljutkin. This is actually the smallest finite quantum groups that is not a group algebra. In other words, it is the non-commutative C*-Hopf algebra of smallest dimension.
Consider the multi-matrix algebra $$A=\mathbb{C}\oplus\mathbb{C}\oplus\mathbb{C}\oplus\mathbb{C}\oplus M_2(\mathbb{C}),$$
with the usual multiplication and involution. We shall use the basis $e_1=(1,0,0,0,0)$ (with $e_2,\,e_3,\,e_4$ defined in the same way) and
$$ a_{11}=0+ 0+ 0+ 0+\left(\begin{array}{cc}1&0\\0&0\end{array}\right),$$
with the other $a_{ij}$ defined in the same way. The algebra $A$ is an eight-dimensional C*-algebra. Its unit is of course  $1_A=e_1+e_2+e_3+e_4+a_{11}+a_{22}$. The following defines a comultiplication on $A$,
$ \scriptsize{  \Delta(e_1)=e_1\otimes e_1+e_2\otimes e_2+e_3\otimes e_3+e_4\otimes e_4+\frac{1}{2}a_{11}\otimes a_{11}+\frac{1}{2}a_{12}\otimes a_{12}+\frac{1}{2}a_{21}\otimes a_{21}+\frac{1}{2}a_{22}\otimes a_{22}}$
$ \scriptsize{ \Delta(e_2)=e_1\otimes e_2+e_2\otimes e_1+e_3\otimes e_4+e_4\otimes e_3+ \frac{1}{2}a_{11}\otimes a_{22}+\frac{1}{2}a_{22}\otimes a_{11}+\frac{i}{2}a_{21}\otimes a_{12}-\frac{i}{2}a_{12}\otimes a_{21}}$
$ \scriptsize{  \Delta(e_3)=e_1\otimes e_3+e_3\otimes e_1+e_2\otimes e_4+e_4\otimes e_2+ \frac{1}{2}a_{11}\otimes a_{22}+\frac{1}{2}a_{22}\otimes a_{11}-\frac{i}{2}a_{21}\otimes a_{12}+\frac{i}{2}a_{12}\otimes a_{21}}$
$ \scriptsize{  \Delta(e_4)=e_1\otimes e_4+e_4\otimes e_1+e_2\otimes e_3+e_3\otimes e_2+ \frac{1}{2}a_{11}\otimes a_{11}+\frac{1}{2}a_{22}\otimes a_{22}-\frac{1}{2}a_{12}\otimes a_{12}-\frac{1}{2}a_{21}\otimes a_{21}}$
$ \scriptsize{  \Delta(a_{11})=e_1\otimes a_{11}+a_{11}\otimes e_1+e_2\otimes a_{22}+a_{22}\otimes e_2+e_3\otimes a_{22}+a_{22}\otimes e_3+e_4\otimes a_{11}+a_{11}\otimes e_4}$
$ \scriptsize{  \Delta(a_{12})=e_1\otimes a_{12}+a_{12}\otimes e_1+ie_2\otimes a_{21}-ia_{12}\otimes e_2-ie_3\otimes a_{21}+ia_{21}\otimes e_3-e_4\otimes a_{12}-a_{12}\otimes e_4}$
$ \scriptsize{  \Delta(a_{21})=e_1\otimes a_{21}+a_{21}\otimes e_1-ie_2\otimes a_{12}+ia_{12}\otimes e_2+ ie_3\otimes a_{12}-ia_{12}\otimes e_3-e_4\otimes a_{21}-a_{21}\otimes e_4}$
$ \scriptsize{  \Delta(a_{22})=e_1\otimes a_{22}+a_{22}\otimes e_1+e_2\otimes a_{11}+a_{11}\otimes e_2+e_3\otimes a_{11}+a_{11}\otimes e_3+e_{4}\otimes a_{22}+a_{22}\otimes e_4}$     
The counit is given by (looking at this we can see the relationship between the counit and the comultiplication with respect to the unit and multiplication in a commutative, group algebra, case. To encode $ ge=g=eg$ any time there is a term of the form $ x\otimes e_1$, there must be a term of the form $ e_1\otimes x$ --- to capture the left and right symmetry $ R_\varepsilon=(I_A\otimes\varepsilon)\circ\Delta=(\varepsilon\otimes I_A)\circ \Delta=L_\varepsilon$. This also shows that, in this case, $\Delta(x)$ must contain $ x\otimes e_1$ and $ e_1\otimes x$ to encode $L_\varepsilon=I_A=R\varepsilon$):
$$ \varepsilon\left(x_1+x_2+x_3+x_4+\left(\begin{array}{cc}c_{11} &c_{12}\\ c_{21}&c_{22}\end{array}\right)\right)=x_1.$$
The antipode is the transpose map, i.e.
$$ S(e_i)=e_i\text{, and }S(a_{jk})=a_{ji}.$$
Background
A corepresentation $\chi$ of a finite quantum group $(A,\Delta)$ on a complex vector space $V$ is a linear map $\chi:V\rightarrow V\otimes A$ that satisfies
$$(\chi\otimes I_A)\circ \chi=(I_V\otimes \Delta)\circ\chi\text{ and }$$
$$(I_V\otimes \varepsilon)\circ \chi=I_V$$
where $\varepsilon$ is the counit of $(A,\Delta)$.
The dimension of the vector space is called the dimension of $\chi$ and is denoted by $d_\chi$. 
If $W$ is a vector space which has the property that $\chi(W)\subset W\otimes A$ then $W$ is said to be invariant and $\chi_{|W}$ is called a subrepresentation.
It can be shown that $\chi$ is equivalent to a direct sum of irreducible unitary corepresentations.
Two corepresentations $\chi_1$ and $\chi_2$ are equivalent as corepresentations if they admit and invertible intertwiner: an invertible linear map $T:V_1\rightarrow V_2$ such that
$$\chi_2\circ T=(T\otimes I_A)\circ\chi_1.$$
The space of intertwiners from $\chi_1$ to $\chi_2$ is denoted by $\text{Hom}(\chi_1,\chi_2)$.
Further Background
I had been under the impression that the comultiplication plays the role of the regular representation in the representation theory of finite groups so that all of the irreducible corepresentations could be found by finding subspaces $W$ of $A$ invariant under the comultiplication in the sense that $\Delta(W)\subset W\otimes A$... however I am either having a stupid misunderstanding or else this isn't as easy as I thought.
I am trying to apply the philosophy of a use of the representation theory of finite groups to the corepresentation theory of finite quantum groups and perhaps I have confused myself a little in the process!

REPLY [7 votes]: The Kac-Paljutkin Quantum Group $A$ is self-dual, i.e. the dual space $A^*$ (which is of course again finite quantum group = finite-dimesional $C^*$-Hopf algebra, with the multiplication being the dual of the coproduct of $A$ and the coproduct the dual of the multiplication of $A$) is isomorphic to $A$.
This must have been known already to Kac and Paljutkin. You can find the isomorphism, e.g., in Franz and Gohm, Random Walks on Finite Quantum Groups, in 
"Quantum Independent Increment Processes II", Lecture Notes in Mathematics Volume 1866, 2006, pp 1-32 .<|endoftext|>
TITLE: How does one Segal-subdivide a 2-category?
QUESTION [5 upvotes]: Let $\mathcal{C}$ be a small category. Then, its Segal subdivision $\text{sd }\mathcal{C}$ is a new category whose objects are morphisms of $\mathcal{C}$, and a morphism from $f:x \to y$ to $g: w \to z$ is a pair of morphisms $a: x \to w$ and $b: y \to z$ in $\mathcal{C}$ so that $f = b\circ g \circ a$. It is easy to see (by a contractible-fiber argument) that there is a homotopy equivalence of classifying spaces $B\text{Sd }\mathcal{C} \simeq B\mathcal{C}$. My first encounter with this construction was in the preprint on Morse theory and classifying spaces (link here).
Here's the question:

What is the analogue of Segal subdivision for small (strict) $2$-categories?

In particular, from a given small $2$-category $\mathcal{D}$, I would like to construct a new $2$-category $\text{sd}_2\mathcal{D}$ whose one-skeleton coincides with the ordinary Segal subdivision of the one-skeleton of $\mathcal{D}$. Some mysterious and powerful $2$-morphisms should exist, and their addition should magically make the classifying spaces of $\text{sd}_2\mathcal{D}$ and $\mathcal{D}$ homotopy-equivalent. 
What, if anything, is the precise collection of $2$-morphisms which achieve the homotopy equivalence, and where can I find this written down?

REPLY [2 votes]: Another, somewhat similar, suggestion, in a cubical setting, which produces a double category:

In the picture I wrote the one morphisms as subdivisions, but if you want to see it as a double category, then these should be the vertical morphisms, and the top and bottom faces are the horizontal morphisms.<|endoftext|>
TITLE: Recovering a Weighted Graph from Shortest Path Distances
QUESTION [7 upvotes]: I am interested in the following problem (A) and its related formulation (B). 
(A) Suppose that $G = (V,E,w)$ is an unknown weighted graph on the vertex set $V$ and that one has access to $d_G(v,v'), \forall v,v'\in V$, where $d_G(\cdot,\cdot)$ is the shortest path metric with respect to $G$. Can one recover the structure (i.e., the edge set $E$ and the weights $w$) of $G$ from just this distance information?
(B) Given a finite metric $(V,d)$, find the sparsest weighted graph $G=(V,E,w)$ that is consistent with $d$ in the sense that $d_G$ is the same as $d$. 
I am curious to know if these problems or similar ones have been studied and have interesting answers.

REPLY [2 votes]: Andreas Blass has produced a simple counterexample to (A) in his answer. I suspect that if your graph is flat in the sense of Andrew Stacey's answer here, then you can in fact recover it from the pairwise distances alone by embedding it a sufficiently high-dimensional Euclidean space.
You can find a nice approximation scheme for question (B) in the paper "Representing graph metrics with fewest edges" by Feder et. al., available here. The basic idea is to add new "Steiner" vertices while removing superfluous edges. The introduction, which is remarkably well-written, contains an overview of how the problem is NP hard, and who did what in which paper, etc. Note that the scheme suggested by Andreas in the comments to his answer does not guarantee that the remaining graph is sparsest possible independent of the order in which the edges are processed.
This is a fairly old problem, and the literature goes back at least to Hakimi and Yau's 1964 paper "The distance matrix of a graph and its realizability". I wasn't able to find a free copy, but you can admire the paywall here.<|endoftext|>
TITLE: Finite groups with no elements of order $p^2q$
QUESTION [5 upvotes]: Let $G$ be a finite group.  What can be said if $G$ has the following Property P: $G$ has no element of order $p^2q$ for any two distinct primes $p,q$?

In particular, which finite simple groups satisfy Property P?

For instance, the alternating group $\text{Alt}_n$ has Property P if and only if $n\le 8$. The question for $\text{PSL}_2$ of finite fields seems not obvious.
As a side remark, a result of Malle, Moreto and Navarro states the following:

Suppose that $p$ and $q$ are distinct primes. If $G$ does
  not have any elements of order $pq$, then one of the following holds:

The Sylow $p$-subgroups or the Sylow $q$-subgroups of $G$ are abelian.
$G/O_{\{p,q\}′} (G)$ = $M$ and $\{p, q\} = \{5, 13\}$ or $\{7, 13\}$.


Here $M$ is the Monster sporadic group, and $O_{\{p,q\}′} (G)$ is the largest normal subgroup of $G$ that is divisible by neither $p$ nor $q$.

REPLY [4 votes]: It seems hard to give a complete answer to this, but at the very least one should be able to specify what the composition factors of $G$ might be. To do this it is enough to classify which simple groups $G$ have property $P$, in which case:

if $G$ is alternating, then @Yves asserted that $n\leq 8$;
if $G$ is sporadic, then we can consult the ATLAS and do each group one by one. For instance, of the Mathieu groups, $M_{11}, M_{12}, M_{22}$ and $M_{23}$ have property P, but $M_{24}$ does not;
suppose next that $G$ is a quasisimple group of type $A_4$ over a field of characteristic $p$, i.e. is isomorphic to a quasisimple cover of $PSL_5(p^a)$, and suppose that $p^a>3$. Since $p^a>3$, there is a non-trivial element of a split torus of order $q$, that has a centralizer equal to a quasisimple cover of $PSL_4(p^a)$. Since the Sylow $p$-subgroup of $PSL_4(p^a)$ has exponent $>p$ and we obtain an element of order $p^2q$. We can consult the ATLAS to see that $PSL_5(2)$ contains an element of order 12. Thus any quasisimple group of type $A_4$ contains an element of order $p^2q$ for some primes $p$ and $q$, i.e. it does not satisfy property P.
Now any simple group of Lie type which contains a subgroup of type $A_4$ will also fail to satisfy P. This includes, in particular, $E_6, {^2E_6}, E_7, E_8$ plus all of the classical groups of dimension at least 10. 
We've dealt with all groups of Lie type or large rank ($\geq 5$), and we should be able to deal with `of medium rank' groups (say rank 3 or 4) with ad hoc methods. For instance the ATLAS tells us that the Tits group ${^2F_4}(2)'$ contains an element of order $12$ and so we can immediately rule out $G={^2F_4}(q)$ and $G=F_4(q)$ and that deals with all of the exceptional groups of (twisted) rank at least $3$. 
We are left with the situation when $G$ is a low rank group of Lie type (say rank 1 or 2). In this case (as the comments indicate) the question is subtle and somewhat number-theoretic. For instance for $PSL_2(q)$ and for ${^2B_2}(q)$ one needs to determine when the order of a maximal torus of $G$ is divisible by $p^2q$ for some primes $p$ and $q$.<|endoftext|>
TITLE: Why isn't $BG$ a group, for $G$ not abelian?
QUESTION [8 upvotes]: If $G$ is a discrete or topological group, $G$ is a closed subgroup of $EG$, and normal iff $G$ is abelian, according to Segal, Cohomology of topological groups, Symposia Mathematica IV (1970) (a reference I found from two answers by Chris Schommer-Pries: Classifying Space of a Group Extension and Good functorial model for BG). Hence for $G$ not abelian, $G$ is not normal in $EG$, hence $BG=EG/G$ is not a group.
On the other hand, we also learn from Segal that the reason $EG$ is a group is that the universal bundle functor $E$ is monoidal with respect to the Cartesian monoidal structure. Any monoidal functor takes group objects in one category to group objects in another simply by functoriality. But the classifying space functor $B$ is also monoidal (see Good functorial model for BG or Peter May's answer at Group structure on Eilenberg-MacLane spaces). Hence $BG$ should be a group?

REPLY [2 votes]: [As a non-homotopy theorist, I am not certain about this answer.]
If your topological group $G$ has a $2$-fold loop space structure, but not an infinite loop space structure, then $BG$ is naturally a group, but there is some $n \geq 2$ for which $B^nG$ is not.  In this case, $G$ is not strictly abelian, despite $BG$ being a group.  In fact, even in the infinite loop space case, you can have groups that are not strictly abelian, like the infinite unitary group $U$.
As other answers have pointed out, this behavior is not possible when $G$ is discrete.  Indeed, 2-fold loop structure automatically strictifies to the abelian property in the discrete case.<|endoftext|>
TITLE: Does the family of graphs closed under taking unions and adding dominating vertices have a name?
QUESTION [6 upvotes]: Let $\mathcal{F}$ be the smallest family of (simple, finite) graphs such that:

The empty graph is in $\mathcal{F}$.
If $G \in \mathcal{F}$, then $G_{\bullet}$, the graph obtained from $G$ by adding another vertex $v$ connected by an edge to each vertex in $G$, is also in $\mathcal{F}$.
If $G,H \in \mathcal{F}$, then $G \cup H \in \mathcal{F}$.

There are natural bijections between $\{G \in \mathcal{F}: \textrm{$G$ has $n$ vertices}\}$ and the set of rooted (but unlabeled) forests on $n$ vertices and between $\{G \in \mathcal{F}: \textrm{$G$ has $n$ vertices and is connected}\}$ and the set of rooted (but unlabeled) trees on $n$ vertices.
Does $\mathcal{F}$ have a name? I want to call it something like the set of "generalized threshold graphs" because the rules that define it are close to those of threshold graphs.

REPLY [9 votes]: It appears that you are looking at the family of quasi-threshold graphs. These are the graphs which can be built, starting from $K_1$, by repeatedly taking the disjoint union of two quasi-threshold graphs, or the join of a quasi-threshold graph with a single new vertex.
By the way, if you relax your conditions a bit more, you get the extremely well-studied family of cographs. These are the graphs which can be built, starting from $K_1$, by repeatedly taking the disjoint union or join of two smaller cographs.<|endoftext|>
TITLE: An Euler-proof that cannot be repaired?
QUESTION [11 upvotes]: Ed Sandifer writes on Euler's wonderful work on prime numbers: The sum of the series of reciprocals of prime numbers $\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...$ is infinitely large, and is infinitely less than the harmonic series $1 +\frac{1}{2}+\frac{1}{3}+...$. Moreover, the first sum is almost the logarithm of the second sum. Euler’s proof of this again on adding and subtracting series that do not converge, and so does not meet modern standards of rigor. Unfortunately, Euler’s proof can’t really be “repaired”.
http://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2029%20infinitely%20many%20primes.pdf
Is it really true that this proof can't be repaired with modern tools? If it can be repaired, however, what has to be done?

REPLY [16 votes]: This result of Euler is the last theorem in http://eulerarchive.maa.org/docs/originals/E072.pdf, and if you prefer English to Latin look at the last theorem in http://eulerarchive.maa.org/docs/translations/E072en.pdf.  I don't know why Sandifer says the proof can't be repaired. Euler's goal is to show $\sum_{p} 1/p$ diverges, and that in some sense $\sum_{p \leq n} 1/p$ diverges like $\log(\log n)$. (Euler didn't speak directly about asymptotic approximations, and he wrote the last conclusion as $\sum_p 1/p = l(l(\infty))$, using $l$ for the natural logarithm.)
Euler's proof: for each positive integer $k$, set $A_k = \sum_p 1/p^k$.  Then 
$$
e^{\sum_{k \geq 1} A_k/k} = \sum_{n \geq 1} \frac{1}{n}
$$
and $\sum_{k \geq 2} A_k/k < \infty$, so $A_1$, which is $\sum_p 1/p$, diverges. 
Since $A_1 = \infty$ while $\sum_k A_k/k < \infty$, we have 
$$
e^{A_1} = \sum_{n \geq 1} \frac{1}{n}, 
$$
so 
$$
\sum_p \frac{1}{p} = A_1 = \log\left(\sum_{n \geq 1} \frac{1}{n}\right).
$$
QED
The last part is kind of a bogus calculation, but the rest can be fixed using the zeta-function.  

In place of $A_k$ use $\sum_{p} 1/p^{ks}$, so the replacement of the first displayed equation above is 
$$
e^{\sum_{p,k} 1/kp^{ks}} = \sum_{n \geq 1} \frac{1}{n^s}
$$
for ${\rm Re}(s) > 1$.  In other words, this is saying $\sum_{p,k} 1/kp^{ks}$ is a logarithm of $\zeta(s)$ for ${\rm Re}(s) > 1$. Series are absolutely convergent, so all manipulations of terms in the series are justified. Euler does all his work at $s = 1$, where expressions may be divergent.
Euler's claim that $\sum_{k \geq 2} A_k/k < \infty$ is correct, and the modern replacement for this is the more general observation that $\sum_{k \geq 2} \sum_p 1/kp^{ks}$ is convergent for ${\rm Re}(s) > 1/2$. Euler's statement is at $s=1$.
Let $s$ be real and greater than 1, so the displayed equation can be rewritten as 
$$
\sum_{k \geq 1} \sum_{p} \frac{1}{kp^{ks}} = \log\left(\sum_{n \geq 1} \frac{1}{n^s}\right).
$$
As $s \rightarrow 1^+$, $\sum_n 1/n^s \rightarrow \infty$, so the right side above tends to $\infty$ as $s \rightarrow 1^+$. On the left side above, split off the terms where $k=1$ from the terms where $k \geq 2$.  Since $\sum_{k \geq 2} \sum_p 1/kp^{ks}$ is convergent for ${\rm Re}(s) > 1/2$, this series has a limit as $s \rightarrow 1^+$ (Dirichlet series are continuous on open half-planes where they converge). Therefore the sum of the terms with $k = 1$, $\sum_p 1/p^s$, must tend to $\infty$ as $s \rightarrow 1^+$.  For $s > 1$ we have $\sum_p 1/p > \sum_p 1/p^s$, and the right side tends to $\infty$ as $s \rightarrow 1^+$, so $\sum_p 1/p = \infty$.<|endoftext|>
TITLE: A question on an elliptic fibration of the Enriques surface
QUESTION [8 upvotes]: Let $S$ be an Enriques surface over complex numbers. It is known that $S$ admits an elliptic fibration over $\mathbb{P}^1$ with $12$ nodal singular fibers and $2$ double fibers. How can I see this well-known fact? Is it possible to explicitly construct such a fibration via the Enriques lattice $NS(S)\cong U\oplus E_8$?

REPLY [2 votes]: A complete and detailed treatment of elliptic pencils on Enriques surfaces (included the result you are quoting) can be found in the book by Barth-Hulek-Peters-Van de Ven Compact Complex Surfaces (Ergebnisse der Mathematik und ihrer Grenzgebiete 4, Springer).
See in particular Chapter VIII, Section 17 "Elliptic pencils".<|endoftext|>
TITLE: When does a pair of homotopic Lipschitz functions fail to admit a Lipschitz homotopy?
QUESTION [14 upvotes]: Let $(M,d)$ and $(N,\rho)$ be metric spaces. A function $f: M \to N$ is Lipschitz if there exists some constant $\kappa \geq 0$ so that $\rho(f(x),f(y))$ is smaller than $\kappa d(x,y)$ for all points $x,y \in M$.
If $f$ and $g$ are two Lipschitz maps from $M$ to $N$, a Lipschitz homotopy between them is a Lipschitz map $H: M \times [0,1] \to N$ for which $H(x,0) = f(x)$ and $H(x,1) = g(x)$ for each $x \in M$. Here we assume that the metric $\Delta$ on $M \times [0,1]$ is given by
$$\Delta\left((x,t),(y,s)\right) = \max\left\{d(x,y),|s-t|\right\}.$$
Here's the question which has been driving me nuts:

Is there an example of metric spaces $M, N$ and Lipschitz functions $f,g:M \to N$ which are homotopic, but not Lipschitz-homotopic?

If there is any justice in the world, the answer is an easy "yes". And in this case, I ask

Does the answer continue to be "yes" if $M$ and $N$ are assumed compact?

REPLY [2 votes]: I agree with Eremenko that 
there is no "justice" in the world of arbitrary metric spaces:-)
However, some justice is provided by the following beautiful result of 
Lang and Schlichenmaier, Theorem 1.5 in [1]. From this result one can conclude some results about Lipschitz homotopy of maps.

Theorem. Suppose that $X$ and $Y$ are metric spaces such that 
  the Nagata dimension of $X$ satisfies $\dim_N X\leq n$ and  $Y$ is Lipschitz $(n-1)$-connected. Then there
  is a constant $C\geq 1$ such that for any closed set $Z\subset X$ and
  any $L$-Lipschitz map $f:Z\to Y$ there is a $CL$-Lipschitz extension
  $F:X\to Y$.

The definitions of the Nagata dimension and Lipschitz $k$-connectivity are provided below:
Definition. A metric space $Y$ is Lipschitz $n$-connected for some integer
$n\geq 0$ if there is a constant $\gamma\geq 1$ such that for each 
$k\in \{0,1,\ldots,n\}$, every $L$-Lipschitz map 
$f:\mathbb{S}^k\to Y$ admits a $\gamma L$-Lipschitz extension 
$F:\mathbb{B}^{k+1}\to Y$.
Definition. The Nagata dimension $\dim_N X$ of a metric space $X$ is the least integer
$n$ with the property that there is $C>0$ such that for any $s>0$,
there is a covering $X=\bigcup_{i\in I} A_i$ such that

$\operatorname{diam} A_i\leq Cs$ for all $i\in I$;
Every ball $\mathbb{B}(x,s)$ intersects at most $n+1$ sets $A_i$.

If no such integer $n$ exists, then $\dim_N X=+\infty$.
The Nagata dimension can be regarded, in some sense, as a quantitative version of the topological dimension.
[1] U. Lang, T. Schlichenmaier,
Nagata dimension, quasisymmetric embeddings, and Lipschitz extensions. 
Int. Math. Res. Not. 2005, no. 58, 3625-3655.<|endoftext|>
TITLE: Proving that some principal series representations of SL(2,F) are irreducible
QUESTION [8 upvotes]: I am sorry in advance if this question is not "research level".
Let $F$ be a p-adic field.
I saw, in Bumps book, a proof which I liked, showing which principal series representations of $GL(2,F)$ are irreducible. I want to modify it to use for $SL(2,F)$, but run into trouble.
The proof in Bump, shows that the twisted Jacquet ( = Whittaker) functor of our principal series rep. $V$ is at most 1-dimensional. Then assuming that $V$ has some non-trivial sub-object, we get that this sub-object (or the quotient) has trivial twisted Jacquet functor. 
Then, we realize that a representation which has trivial twisted Jacquet functor, "coincides" with its Jacquet functor. This is because a representation of $N$ is the same as a sheaf on $N^*$ (the Pontryagin dual of $N$), and the twisted Jacquet functor is the fiber of this sheaf at a non-zero point (all fibers are "the same" since $A$ acts transitively - this is the crucial point).
My question is: If we deal with $SL(2,F)$, then there are "two" twisted Jacquet functors (two orbits of $A$ on $N^* - 0$). So one can not do the proof exactly as above. How does one modify the proof? Is there a reference?
Thank you,
Sasha

REPLY [3 votes]: Paul Sally explains this in these notes:
http://link.springer.com/article/10.1023%2FA%3A1007583108067#page-1
I am afraid he excludes residue characteristic two there:(
But this seems to be more general: http://www.jstor.org/stable/2373536?seq=1<|endoftext|>
TITLE: Heegaard genus of covers of cusped hyperbolic 3-manifold
QUESTION [10 upvotes]: Is there a cusped (non-compact) orientable hyperbolic 3-manifold $M$ which has Heegaard genus $g$, and has a finite-sheeted cover with Heegaard genus $<g$?
Moreover, if the cover is index $n$, then it should have fewer than $kn$ cusps,
where $k$ is the number of cusps of $M$ (so it does not induce the trivial
$n$-fold cover of each cusp component).
Remarks: I believe this is an open question. To clarify, by Heegaard genus $g$ of a cusped hyperbolic manifold $M$, I mean that $M$ is the interior of a compact manifold $N$ with torus boundary components, so that $N$ has Heegaard genus $g$, i.e. there is a surface of genus $g$ in $N$ splitting it into two compression bodies. For closed manifolds, there are examples of Alan Reid. One may modify these examples to get cusped examples which are the trivial cover of the cusps. For more discussion in the closed case, see this paper of Peter Shalen.
I would also be happy with a 1-cusped manifold of Heegaard genus 3, and which has a $k$-fold cover with a maximal embedded horocusp of volume $\leq \pi$ and $<k$ cusps (e.g. an irregular 3-fold cover with 2 cusps). (although Nathan's answer answers the original version of this part of the question, I realized I hadn't given the correct formulation).

REPLY [6 votes]: Here's an answer to your alternate question:  The manifold m135, which is the punctured torus bundle $-LLRR$, has Heegaard genus 3 and max cusp volume < 2.83. There are plenty of examples of this kind, even if you want the lower bound on Heegaard genus to be certified by the rank of the homology, e.g. m136, m306, m307, s298, s299, s595 which have max cusp volumes in the range [2.8,3.12].<|endoftext|>
TITLE: Do cocontinuous SET-valued functors separate points?
QUESTION [6 upvotes]: Let $C$ be a category.  For the purposes of this question, I would like to avoid cases where the answer might be "no" simply because $C$ is "too large", and so I will ask that $C$ has a set of generators, i.e. a set of objects $\{X_\alpha\}_{\alpha \in A}$ such that for any two morphisms $f,g : Y \to Z$ that are not equal, there exists an $\alpha\in A$ such that $\hom(X_\alpha,f) \neq \hom(X_\alpha,g)$ as maps $\hom(X_\alpha,Y) \to \hom(X_\alpha,Z)$.  
(I am most interested in the specific cases where $C$ is either small or presentable.  More generally, $C$ could be a "site" or even a "colimit sketch", and this question would still be apt.)
I would like to advocate for the following terminology: a cosheaf on $C$ is a covariant functor $F : C \to \mathrm{Set}$ that takes colimits to colimits.  (Similarly, a sheaf is contravariant functor $C \to \mathrm{Set}$ that takes colimits to limits.)

Given two nonequal morphisms $f,g: Y \to Z$, does there necessarily exist a cosheaf $F$ such that $F(f) \neq F(g)$ as maps $F(Y) \to F(Z)$?

Note that the usual version of the Yoneda lemma doesn't suffice: the corepresentable functor $\hom(X,-)$ preserves limits, not colimits, and the representable cofunctor $\hom(-,X)$ is a sheaf, not a cosheaf.  Using representable sheaves, one immediately has the answer "yes" if $\mathrm{Set}$ were replaced by $\mathrm{Set}^{\mathrm{op}}$.  Perhaps there is a faithful cocontinuous functor $\mathrm{Set}^{\mathrm{op}} \to \mathrm{Set}$ that I am not aware of?
Some applications of cosheaves that I care about are listed at http://ncatlab.org/nlab/show/cosheaf; the proposition there also provides one reason I care about this question.

REPLY [7 votes]: No. Let $C$ be any category with a zero object. If $F : C \to \text{Set}$ is a cosheaf, then in particular it sends the zero object to the empty set. But since the zero object is also the terminal object, every object in $c$ is equipped with a morphism $c \to 0$, from which it follows that $F(c)$ is equipped with a morphism to the empty set. Hence $F(c)$ is itself empty, so $F$ is the trivial cosheaf. Now take $C$ to be, say, the category of finite-dimensional vector spaces. 
Edit: In fact the free example suffices. Let $C$ be the free category on a pair of parallel morphisms $f, g : c \to d$. Then the coproduct $d \sqcup d$ exists and both inclusions $d \to d \sqcup d$ are isomorphisms. It follows that if $F$ is a cosheaf then $F(d)$ is the empty set, from which it follows that $F(c)$ must also be the empty set and $F(f) = F(g)$ must be the unique morphism $\emptyset \to \emptyset$, so $F$ is again the trivial cosheaf. 
Edit #2: Here's another example. Let $C = \text{CRing}$. If $F : C \to \text{Set}$ is a cosheaf, then in particular it sends tensor products to disjoint unions, sends $\mathbb{Z}$ to $\emptyset$, and preserves epimorphisms. Hence it sends $\mathbb{F}_p$ for all primes $p$ to $\emptyset$. Hence it sends $\mathbb{F}_p \otimes \mathbb{F}_q \cong 0$ (where $p \neq q$) to $\emptyset$. And now it follows as above that $F$ is the trivial cosheaf.<|endoftext|>
TITLE: Is there something wrong with Hörmander's theorem on stationary phase method
QUESTION [11 upvotes]: It is well-know that the Bessel function has the asymptotic expansion $J_n(\omega) \sim \left( \frac 2 {\pi \omega} \right)^{1/2} \left( \cos \left(\omega -\frac 1 2 n \pi - \frac 1 4 \pi\right) - \sin \left(\omega -\frac 1 2 n \pi - \frac 1 4 \pi\right) \frac {4n^2 - 1} {8 \omega} ...\right)$.
We also have the integral representation $$J_n(\omega) = \int_0^1 \cos (n \pi x - \omega \sin \pi x) dx = \Re  \int_0^1 e^{n \pi i x} e^{ - \omega \sin \pi x } dx.$$
Now we look at Hörmander's Theorem 7.7.5(The Analysis of Linear Partial Differential Operators I). The phase function is $f(x) = - \sin \pi x $, and has stationary point $x_0 = 1/2$. The weight function is $u(x) = e^{n \pi i x} $.  Hörmander's theorem says \begin{equation}
\begin{split}
\int_0^1 u(x) e^{i \omega f(x)} d x & \sim e^{i \omega f(1/2)} (\det (\omega f''(1/2)/ 2 \pi i))^{-1/2} (u (1/2) + L_1 u \omega^{-1} )\\
& = \left( \frac 2 {\pi x} \right)^{1/2} e^{- i \omega + \pi i /4}  (e^{n \pi i /2} + L_1 u \omega^{-1} ).
\end{split}
\end{equation}
The first term is right. Let us look at the second term.
We have $f(1/2) = -1$, $f''(1/2) = \pi^2$, so $g_{1/2} (x) = - \sin \pi x + 1 - (\pi^2/2) (x-1/2)^2$. We note that $g_{1/2} (x)$ vanishes not only of third order but of fourth order at $1/2$, that is, $g(1/2) = g'(1/2) = g''(1/2) = g'''(1/2) = 0$, and $g''''(1/2) = - \pi^4$.  $L_1 u $ is defined as
$$i^{-1} \sum_{\nu - \mu = 1} \sum_{2\nu \geq 3 \mu} 2^{-\nu} \langle \pi^{-2} D, D \rangle^{\nu} (g_{1/2}^{\mu} u) (1/2) /\mu! \nu! .$$
We see that 
$$L_1 u = i^{-1} \left( 2^{-1} \pi^{-2} (n \pi i)^2 e^{n \pi i /2} + 2^{-3} \pi^{-4} (- \pi^4 )  e^{n \pi i /2} + 0\right) = i e^{n \pi i /2} \left( \frac 1 2 n ^2 + \frac 1 8 \right).$$
Therefore the second term is $ \sin \left(\omega -\frac 1 2 n \pi - \frac 1 4 \pi\right) \frac {4n^2 + 1} {8 \omega}$ according to Hömander instead of $ - \sin \left(\omega -\frac 1 2 n \pi - \frac 1 4 \pi\right) \frac {4n^2 - 1} {8 \omega}$. My guess is that there is a factor $(-1)^{\nu}$ missing in Hörmander's theorem, then also in  Hörmander's Lemma 7.7.3., but this lemma looks right.
Could anyone tell me what goes wrong here? I would really appreciate it!

REPLY [9 votes]: Looking closely again, I am pretty sure my comment gives the answer. Note that Hörmander uses the notation $D = - i \partial$. So in particular
$$ Du = -i (n\pi i) u $$
when $u = e^{n\pi i x}$. This gives an extra minus sign in the first term ($\nu = 1$ and $\mu = 0$) of the definition for $L_1 u$, while leaving the second term ($\nu = 2$ and $\mu = 1$) unchanged, providing the extra $(-1)^\nu$ you are looking for.<|endoftext|>
TITLE: Correspondence between operads and $\infty$-operads with one object
QUESTION [13 upvotes]: Given a simplicial operad one can form its category of operators. This is a simplicial category with a functor to the category of finite pointed sets which is a bijection on objects and whose hom-spaces have a particular product decomposition. Assuming the spaces in the operad are fibrant we can apply  the coherent nerve construction to obtain an $\infty$-operad with 'one-object' (by this I mean the $\infty$-category associated to the operad has a single equivalence class).  This is Proposition 2.1.1.27 of Lurie's Higher Algebra. 
Q1: Does every $\infty$-operad with 'one-object' come from this construction (up to equivalence)?
EDIT: To clarify, I mean an operad in the classical sense of May and a weak equivalence of such operads (a singly colored simplicial operad, for the multicolored analogue of this question Urs' response says the answer would be yes). So one side I'm allowing arbitrary weak equivalences of $\infty$-operads and on the other I'm only considering weak equivalences between simplicial singly-colored operads.
Let $\mathcal{C}_\mathcal{O}$ be the category of operators associated to an operad $\mathcal{O}$. We know the unit of the adjunction $\mathfrak{C}N\mathcal{C}_\mathcal{O}\rightarrow \mathcal{C}_\mathcal{O}$ is a weak equivalence of simplicial categories. 
Q2: Is $\mathfrak{C}N\mathcal{C}_\mathcal{O}$ the category of operators associated to another simplicial operad $\mathcal{O}^\prime$?

REPLY [14 votes]: The answer to Q1 is indeed yes. The construction you describe (category of operators followed by coherent nerve) gives a functor from the category of fibrant colored simplicial operads to the category of $\infty$-operads (in the sense of Lurie). This functor preserves weak equivalences and induces an equivalence on the level of homotopy categories. The category of simplicial operads (with one color, or object) is a full subcategory of the category of colored such guys, so (as long as we only care about things up to weak equivalence) we only have to identify the essential image of this subcategory. "Having one object" is a property that you can see on the level of underlying categories. The construction we're talking about is compatible with "taking underlying categories"; i.e. taking the underlying simplicial category of a simplicial operad and then taking the coherent nerve produces the same thing as first running the construction you describe and then taking the fiber over $\langle 1 \rangle$. The statement therefore reduces to "every $\infty$-category with one object is (up to equivalence) the coherent nerve of a fibrant simplicial category with one object", which is true.
The answer to Q2 is no, at least if you're really asking it "up to isomorphism". For example, consider the space of morphisms lying over the inert morphism $\langle 3 \rangle \rightarrow \langle 1 \rangle$ which sends 2 and 3 to the basepoint and 1 to 1. In the category of operators of a simplicial operad, this space is isomorphic to the space of morphisms lying over the identity $\langle 1 \rangle \rightarrow \langle 1 \rangle$. This need not be the case in a category of the form $\mathfrak{C}N\mathcal{C}_{\mathcal{O}}$. Already the trivial operad gives a counterexample. In this case, the first space I mentioned has non-degenerate 1-simplices, corresponding to factorizations $\langle 3 \rangle \rightarrow \langle 2 \rangle \rightarrow \langle 1 \rangle$ of the given morphism into two inerts, whereas the space of morphisms lying over $\langle 1 \rangle \rightarrow \langle 1 \rangle$ is just a 0-simplex.<|endoftext|>
TITLE: Pick's Theorem for rational points of bounded height
QUESTION [6 upvotes]: I wonder if the various lattice-point theorems, such as 
Pick's Theorem or 
Minkowski's Lattice Theorem,
have been generalized to the collection of points
with rational coordinates no more than height $h$?
A rational $a/b$ in lowest terms has height $\max( |a|,|b| )$.
For example, here are the positive rationals of height $\le 5$:
$$\left\{\frac{1}{5},\frac{1}{4},
\frac{1}{3},\frac{2}{5},\frac{1}{2},\frac{3}{5},\frac
   {2}{3},\frac{3}{4},\frac{4}
   {5},1,\frac{5}{4},\frac{4}{
   3},\frac{3}{2},\frac{5}{3},
   2,\frac{5}{2},3,4,5 \right\}$$
(Incidentally, the length of this list is given 
by integer sequence A018805.)
Let us call rationals of height $\le h$, $h$-rationals, 
and points with both coordinates $h$-rationals,
$h$-rational points
(my own terminology).
Generalizations from lattice points to $h$-rational points
seems natural, and likely
has been explored...? If so, I would appreciate a reference!
That's my primary question.
Here is a specific, Pick's Theorem -like question:

Can the number $i$ of $h$-rational points inside a polygon $P$
  be expressed in a form that is not tantamount to
  enumerating each interior point?
  Assume we are
  given the vertex coordinates of $P$, and perhaps the number
  of boundary points on each edge.

If $P$ is an axis-aligned rectangle, then the number of interior points $i$
can be computed from the number of points $b_x$ lying on the
bottom side and the number lying on the left side $b_y$:
$i = (b_x - 2)(b_y - 2)$. So, for the rectangle with lowerleft
corner $(1,1)$ below, $b_x=8$ and $b_y=7$ leads to $i=30$.
But just knowing the total number of boundary points $b=26$,
a la Pick's Theorem, is inadequate to determine $i$.
It seems feasible that there is a nonlinear transformation that
maps the $h$-rational points to the integer lattice, allowing
Pick's Theorem to be applied there.
Answering this question when $P$ is a triangle should lead to a result for arbitrary
$P$ via triangulation.

REPLY [2 votes]: One can at least give an asymptotic behaviour of this number, when the height bound grows to infinity.
As shown by Emmanuel Peyre in his paper, Hauteurs et mesures de Tamagawa sur les variétés de Fano (Duke Math. J, 79), there are cases of algebraic varieties over a number field where one can prove an equidistribution statement for the points of bounded height, and the projective space is such a case.
Here is a consequence of that theorem, for the case of the projective line $\mathbf P^1_{\mathbf Q}$. Since you normalized the height of a point $p/q\in\mathbf P^1(\mathbf Q)$ to be $\max(|p|,|q|)$, when written in lower terms, let us define a measure on $\mathbf P^1(\mathbf R)=\mathbf R\cup\{\infty\}$  by $\mu=\mathrm dt/\max(1,|t|)^2$. (For an explanation of the exponent $2$ in the definition of the measure $\mu$, observe that the differential form $\mathrm dt$ has a pole of order $2$ at infinity.)
For any subset $U$ of $\mathbf P^1(\mathbf R)$, let $N(U;B)$ be the number of points of $\mathbf P^1(\mathbf Q)$ of height $\leq B$ which belong to $U$.
Peyre's Theorem asserts that for any open subset $U$ of $\mathbf P^1(\mathbf R)$ whose boundary has  $\mu$-measure $0$, the following limit formula holds:
   $$ \frac{N(U;B)}{N(\mathbf P^1(\mathbf R);B)} \to \frac{\mu(U)}{\mu(\mathbf P^1(\mathbf R))} $$
when $B\to\infty$.
The actual statement holds at an adelic level, and for more general normalization of the height, for any number field, as well as for more general varieties.<|endoftext|>
TITLE: Example of 4-manifold with $\pi_1=\mathbb Q$
QUESTION [28 upvotes]: This might be well known for algebraic topologist. So I am looking for an explicit example of a 4 dimensional manifold with fundamental group isomorphic to the rationals $\mathbb Q$.

REPLY [58 votes]: Since any compact manifold has the homotopy type of a finite CW-complex (see this MathOverflow question: Are non-PL manifolds CW-complexes?)  and $\mathbb{Q}$ is not finitely presented, the manifold $X$ you are looking for is necessarily non-compact.
An explicit construction of a non-compact three-manifold $M$ with $\pi_1(M)=\mathbb{Q}$   can be found in the paper
B. Evans and L. Moser: Solvable Fundamental Groups of Compact 3-Manifolds, Transactions of the American Mathematical Society 168 (1972), see in particular page 209.
Now it suffices to take $X=M \times \mathbb{R}$.<|endoftext|>
TITLE: Random path in a graph
QUESTION [6 upvotes]: Consider a finite graph $G$. I would like to define a random path between two vertices $s$ and $t$ of the graph $G$ by looking at a measure $\mu$ on all spanning trees. Then the probability of a given path $p$ (without cycles) between $x$ and $y$ to be chosen is $\mathbb{P}(p) = \mu(T \mid p \subset T)$.
For example, if $\mu$ is the uniform measure on all spanning trees, it seems this should coincide with the random path starting at $x$ and stopping at $y$ with potential cycles deleted. [I actually have no idea if this is the case and, if so, why, and would be grateful to anyone who has a nice argument or reference]. (EDIT: see explanation in J.W.'s answer below.)
It was pointed out to me that there is another interesting measure on the space of spanning trees, namely the "minimal" measure. My question is (apologies in advance if it is well-known)
$\mathbf{Question}$: what's a random path associated to the minimal measure on spanning trees?
I'll make an attempt at describing the "minimal measure" [EDIT: corrected, thanks to J.W.'s comment below]. Take an injective function $w$ from the set of edges $E$ to $[0,1]$. Let $T_w$ be the tree of minimal cost, if $w$ is thought of as the cost of an edge. Many constructions are possible. For example, an edge $e$ belongs to $T_w$ if any path between its endpoints ($e$ excluded) contains an edge $e'$ with $w(e')>w(e)$. 
Next consider $w$ given by independent uniform $[0,1]$ random variables (one on each edge). The minimal measure on the spanning trees is that of $T_w$ (where $w$ is random), i.e. $\mu(T) = \mu(w \mid T = T_w)$.

REPLY [3 votes]: You might be interested in Loop Erased Random Walks and David Wilson's algorithm, see for instance http://research.microsoft.com/en-us/um/people/schramm/memorial/usf-talk.ps<|endoftext|>
TITLE: On Joyal's completeness theorem for first order logic
QUESTION [24 upvotes]: In 1978, in a series of unpublished conferences in Montréal, A. Joyal announced a remarkable theorem that unified several completeness theorems for fragments of first order logic, as well as first order logic itself. Given a coherent category $T$, we have the evaluation functor $e_T: T \to \mathcal{S}et^{\mathcal{M}od_c(T)}$, where $\mathcal{M}od_c(T)$ is the category of coherent models of $T$, and the evaluation functor sends an object $A$ to the set-valued functor from $\mathcal{M}od_c(T)$ that simply evaluates a model $M$ in $A$. The theorem is question is the following:
Theorem (Joyal): If $T$ is a Heyting category, then $e_T$ is a conservative Heyting functor.
The power and elegance of Joyal's formulation resides in the fact that it subsumes at least three different completeness theorems: that of coherent logic, that of classical first order logic, and, finally, that of intuitionistic first order logic. Indeed, the first follows from the conservativity of $e_T$, while the second is just the particular case when $T$ is a Boolean category (in which case coherent models are all Boolean). It is the third completeness theorem the most unexpected and amazing, and about which my question refers. The key fact here is that a Kripke model for an intuitionistic first order theory $T$ is the same thing as a Heyting functor $\mathcal{C}_T \to \mathcal{S}et^P$, where $P$ is the underlying poset of the model and $\mathcal{C}_T$ is the (Heyting) syntactic category of the theory. Now, the category $\mathcal{M}od_c(T)$ is not necessarily a poset (though it can be taken to be small), but Joyal's theorem provides us anyway with a generalized Kripke model which only differs in that aspect. In fact, the notion of forcing at a node $M$ coincides precisely with the satisfaction in the model $M$.
Now my question. In a paper by Makkai, the author mentions that "from the formulation of Joyal's theorem, it is quite easy to prove the existence of a conservative Heyting embedding of the form $e_T: T \to \mathcal{S}et^P$ with a poset $P$, with possible further conditions on $P$ (for example, regarding its size, or that it be a forest -a tree with many roots)". How exactly can this be achieved? The size is not an issue, since due to Löwenheim-Skolem theorem one can always cut down the category of models to restrict the cardinality. From the proof of Joyal's theorem, which can be found in Makkai-Reyes "First order categorical logic", I can also see that it is enough to consider only elementary embeddings as the morphisms in the category of models. Yet, this still doesn't necessarily make it into a poset. I also have an idea of which models can be taken as the roots of the forest, and how to derive a tree from a general poset with a bottom node, but cannot see how to bring the category of models into a posetal one. So, which is the argument Makkai is referring to?

EDIT
Andreas has given the right answer to the question. The non trivial thing in his construction of the poset $P$ is to prove that the functor $E^*: \mathcal{S}et^M \to \mathcal{S}et^P$ preserves $\forall$. Here is the calculation:
For a natural transformation $f: F \to G$ in $\mathcal{S}et^M$ and a subfunctor $A$ of $F$, we need to show that $E^*(\forall_fA)$ is the same subfunctor of $E^*(G)$ as $\forall_{E^*(f)}E^*(A)$. By definition, for any object $p$ in $P$ and $y \in E^*(G)(p)=G(E(p))$, we have $y \in \forall_{E^*(f)}E^*(A)(p)$ if and only if for all arrows $l: p \to q$ in $P$ one has:
$$E^*(f)_q^{-1}(G(E(l))(y)) \subseteq E^*(A)(q)$$
$$\iff (\forall x E^*(f)_q(x)=G(E(l))(y) \implies x \in E^*(A)(q))$$
$$\iff (\forall x f_{E(q)}(x)=G(E(l))(y) \implies x \in A(E(q))) (1)$$
On the other hand, also by definition, for $y \in G(E(p))$ one has $y \in \forall_fA(E(p))$ if and only if for all arrows $t: E(p) \to r$ in $M$ one has:
$$f_r^{-1}(G(t)(y)) \subseteq A(r)$$
$$\iff (\forall x f_r(x)=G(t)(y) \implies x \in A(r)) (2)$$
But because the functor $E$ is surjective (both on objects and arrows), we can find $q, l \in P$ such that $r=E(q)$ and $t=E(l)$, from which we deduce that $(1)$ and $(2)$ above are equivalent. Hence, $E^* \forall=\forall E^*$, as we wanted.

REPLY [11 votes]: The following is based on vague memories from long ago, so it comes with no guarantees.  If it turns out to be totally wrong, then someone will surely say so and I'll delete it.
I think the poset $P$ that you want, to replace the category $M=\mathcal Mod_c(T)$, consists of finite composable sequences of morphisms of $M$, i.e., chains $A_0\to A_1\to\cdots\to A_n$ in $M$. One such sequence is below another in $P$ if the former is an initial segment of the latter.  There's a functor from $E:P\to M$ sending each chain to the last object in it and sending any morphism $f$ of $P$ to the composite of the morphisms of $M$ that are in the codomain minus the domain of $f$.  Then, given Joyal's version of $e_T$, compose it with $\mathcal Set^E$, and (if my memory is right) you get Makkai's $e_T$.
The actual definition of this construction can be found in Mac Lane and Moerdijk's Sheaves in Geometry and Logic, Th. 9.1. This is used to prove the existence of the Diaconescu cover.<|endoftext|>
TITLE: Omitting primes from a Hecke algebra
QUESTION [8 upvotes]: Let $N \ge 1, k \ge 2$ be integers, and $M_k(\Gamma_1(N))$ the space of weight k modular forms of level $\Gamma_1(N)$. Let $\mathbb{T}$ be the $\mathbb{Z}$-subalgebra of $\operatorname{End} M_k(\Gamma_1(N))$ generated by the diamond operators $\langle d \rangle$ and the Hecke operators $T_\ell$ for $\ell \nmid N$. (This is sometimes called the "anemic Hecke algebra" because we've omitted the Hecke operators at primes dividing $N$).
If we make $\mathbb{T}$ still more anemic, by omitting the $T_\ell$'s for some finite subset $\Sigma$ of the primes $\ell \nmid N$, is the resulting subalgebra $\mathbb{T}^{\Sigma}$ the whole of $\mathbb{T}$?
(This is certainly true after tensoring with $\mathbb{Q}$ by the strong multiplicity one theorem, and I managed to convince myself it should be true integrally by a complicated argument using Galois representations and Chebotarev density; but is there a slick proof that doesn't use such heavy machinery?)

REPLY [7 votes]: It seems to me that the proof you describe is fine except maybe at the prime 2, the problem being that the argument you are sketching (presumably) relies on expressing the missing operators via traces of Frobenius morphisms modulo arbitrary powers of maximal ideals of the Hecke algebra, but this works only for Hecke operators $T_{\ell}$ with $\ell$ prime to the residual characteristic of the maximal ideal you are considering, so that one needs a supplementary argument at the residual characteristic of the maximal ideal. This is readily provided by $q$-expansion and the duality between Hecke operators and cusp forms except if the residual characteristic is 2.  
Consequently, I'm quite sure that your $\mathbb T^\Sigma$ is equal to $\mathbb T$ if $2|N$ (in which case there are no operator at 2 to begin with) or if $2\notin \Sigma$ but otherwise the index of $\mathbb T^{\Sigma}$ in $\mathbb T$ might be a power of 2 (and actually is sometimes, see here for an example).
The argument with $q$-expansion can be found in 
Multiplicities of $p$-finite mod $p$ Galois representations in $J_{0}(Np)$ (Ken Ribet)
Bol. Soc. Brasil. Mat. (N.S.) 21 (1991), no. 2, 177–188.<|endoftext|>
TITLE: Homotopy limit-colimit diagrams in stable model categories
QUESTION [5 upvotes]: It is shown in Remark 7.1.12 of (a newer version of) Mark Hovey's book Model Categories that, in a stable model category, homotopy pullback squares coincide with homotopy pushout squares. The argument goes as follows: given a square $S$ of the form
$$
\begin{matrix}
W & \to & X\\
\downarrow_h & & \downarrow_g\\
Z & \to & Y 
\end{matrix}
$$
one shows that $S$ is a pullback square if and only if the induced map $\mathrm{HoFibre}(h)\to\mathrm{HoFibre}(g)$ is an isomorphism in the homotopy category. Similarly, $S$ is a pushout square if and only if the induced map $\mathrm{HoCofibre}(h)\to\mathrm{HoCofibre}(g)$ is an isomorphism in the homotopy category. One concludes by observing that $\mathrm{HoCofibre}(h)$ is the suspension of $\mathrm{HoFibre}(h)$ and similarly for $g$.
Question: Is it possible to generalize this statement to say something like as follows?
"In a diagram of the form
$$
\begin{matrix}
W & & \to &&  X\\
 &\searrow & \\
\downarrow &&Y&&\downarrow\\
&&&\searrow\\
Z & &\to && V,  
\end{matrix}
$$
$V$ is a homotopy colimit of
$$
\begin{matrix}
W & & \to &&  X\\
 &\searrow & \\
\downarrow &&Y&&\\
&&&\\\
Z & & &&   
\end{matrix}
$$
if and only if 
$W$ is a homotopy limit of
$$
\begin{matrix}
 & &  &&  X\\
 & & \\
 &&Y&&\downarrow\\
&&&\searrow\\
Z & &\to && V."  
\end{matrix}
$$

REPLY [9 votes]: This is not true. For example, take $W = X = Y = 0$, and the map $Z \rightarrow V$ to be an isomorphism. Then you've got a homotopy colimit diagram, but it is only a homotopy limit diagram if $Z$ is weakly contractible.<|endoftext|>
TITLE: Reconstructing a group from deformed classifying space
QUESTION [7 upvotes]: Suppose that we have a finite group $G$.  Its classifying space $BG$ is not naturally pointed, so if we would like to consider it as a spectrum there are two possible approaches.  The first is to take any point in the classifying space and let it be the basepoint when constructing the suspension spectrum; the second is to add a disjoint basepoint.
My question is: is it possible to reconstruct $G$ from these spectra?  If so, how?  What information do we know about $G$ if we are given one of these spectra?

REPLY [8 votes]: Let $X_+$ be a space $X$ together with a disjoint base-point. Since $\Sigma^\infty(BG_+)=\Sigma^\infty(BG)\vee S$, where $S$ is the sphere spectrum, there's not much difference between the two spectra you propose. In Example 5.2 of Martino–Priddy's 'Stable homotopy classification of $BG^{\hat{}}_p$' you can find non-isomorphic finite groups whose classifying spaces have the same suspension spectrum, so the answer to your first question in its current form is no. In that paper you can also find very interesting results concerning your last question.<|endoftext|>
TITLE: Difference Sets
QUESTION [7 upvotes]: Suppose
$$
P \subseteq \{1,2,\dots,N\},\quad |P| = K
$$
We calculate the differences as: $$d=p_i-p_j\mod N,\quad i\ne j$$
Now let $a_d$ denote the number of occurrence of $d$ (for $d = 1, 2, \dots , N −
1$), then we have a set $$A=\{a_1,a_2,...,a_{N-1} \}$$
As you know given $P$ and $N$, it is easy to build $A$. Although given $A$ and $K$ there could be many $P$'s leading to $A$ or even no possible $P$ leading to $A$.
I want to know, given $A$ and $K$, is there any algorithm returning $P$ ?

REPLY [13 votes]: A keyword in this area is homometric, and a key paper this one:

Joseph Rosenblatt and Paul D. Seymour.
  "The Structure of Homometric Sets."
  SIAM. J. on Algebraic and Discrete Methods, 3, 343-350, 1982.
  (PDF download)

The precise question you pose---an algorithm---is called the beltway reconstruction
problem in this paper:

Skiena, Steven S., Warren D. Smith, and Paul Lemke. "Reconstructing sets from interpoint distances." Proceedings 6th Symposium on Computational Geometry. ACM, 1990. (ACM link). Later (much later!) published in Discrete and Computational Geometry. Springer. Berlin, Heidelberg, 2003. 597-631.

They provide a worst-case exponential algorithm which nevertheless runs fast under probabilistic assumptions,
and proved many variants NP-hard, but not your particular variant. 
Another, later paper on this topic is below, but I don't think it settled the
1-dimensional reconstruction problem:

R.J. Gardner, P. Gritzmann, D. Prangenberg, On the computational complexity of reconstructing lattice sets from their X-rays. Discrete Math. 202 (1999), no. 1-3, 45-71; DOI: 10.1016/S0012-365X(98)00347-1.

It may be that the computational complexity of the beltway reconstruction problem remains open...?
See also the earlier MO question, "Largest pair of homometric Golomb rulers?"<|endoftext|>
TITLE: A question on elementary submodels
QUESTION [7 upvotes]: Suppose $\chi$ is a regular cardinal, and $N$ is an elementary submodel of the structure $\langle H(\chi),\in\rangle$, where $H(\chi)$ is those sets hereditarily of cardinality less than $\chi$.
Can we characterize when the transitive collapse of $N$ is $H(\lambda)$ for some $\lambda<\chi$?
The motivation is idle curiosity: I just ran across an argument where such $N$ were used in a critical manner, and so I'm wondering if there's a relatively simple way to detect when a given $N$ has this property.

REPLY [8 votes]: This hypothesis is equivalent to an instance of $\kappa$, the least point not in $N$, having some degree of extendibility. After all, if $j$ is the inverse of the collapse, then we have an elementary embedding $j:H(\lambda)\to H(\chi)$ with critical point $\kappa$. 
A cardinal $\kappa$ is $\eta$-extendible if there is an elementary embedding $j:V_{\kappa+\eta}\to V_\theta$ for some ordinal $\theta$. Although extendibility is usually defined this way in terms of the von Neumann hierarchy, there is a parallel characterization in terms of the hereditary-size hierarchy, and the two hierarchies are interleaved. For example, $V_{\kappa+\eta}$ amounts essentially to $H(\beth_{\kappa+\eta})$, since the von Neumann hierarchy takes the power set each step. There is a good case to be made that the original definition of extendibility should perhaps have referred to the $H(\theta)$ hierarchy instead of the $V_\theta$ hierarchy, since $H(\theta)$ generally satisfies a better theory and has better closure properties. A similar issue affects the strong cardinals, which are also defined by reference to the $V_\theta$-hierarchy. 
But meanwhile, if one is in a situation where the least cardinal $\kappa$ not in $N$ is known not to have extendibility properties (which are quite strong in the large cardinal hierarchy), then you know $N$ does not collapse to an $H(\lambda)$. Conversely, when $\kappa$ does have nontrivial extendibility, then you can expect to find such $N$, where $\kappa$ is the least ordinal missing from $N$, and the existence of these elementary submodels for every $\chi$ is equivalent to $\kappa$ being an extendible cardinal.<|endoftext|>
TITLE: Cohomology of the genus 2 mapping class group
QUESTION [8 upvotes]: Is the cohomology of the genus 2 mapping class group (that is, the cohomology of the moduli stack $M_2$ of genus 2 curves) known? I'd be interested in references. The rational cohomology is known to be trivial, but I am interested in torsion phenomena, especially at the prime $2$.

REPLY [9 votes]: The possible automorphism groups of a curve of genus two are known very classically, this is usually attributed to Bolza. In his list we find only three prime factors in the orders of the groups: 2,3 and 5. So this is the only possible torsion.
I haven't read Benson and Cohen, "Mapping class groups of low genus and their cohomology", but the Mathematical Review suggests that it is exactly what you are looking for. They compute the ring structure on the mod 3 and mod 5 cohomology and the Poincaré series of the mod 2 cohomology.<|endoftext|>
TITLE: Reference to a "simple" consistency proof
QUESTION [6 upvotes]: Bartoszynski has written a "simplified" account of Shelah's proof of consistency of "Covering of null ideal has countable cofinality" in his article in handbook. In a subsequent paper, Shelah also constructed a model of "The null ideal restricted to some non null set is $\omega_1$-saturated". The proof uses iterations of "partial random reals" and makes several references to his countable cofinality paper and I could not follow it starting page 17. Has someone written a simplified account of this proof that maybe does not refer to Shelah's countable cofinality paper?

REPLY [3 votes]: Shelah and I have a preprint that has a generalization of this. The main result is that the null ideal restricted to a non null set of reals could be isomorphic to a large class of sigma ideals - For example it could be isomorphic to the non stationary ideal on $\omega_1$.<|endoftext|>
TITLE: Better bounds for exact-intersection Erdős–Ko–Rado system?
QUESTION [5 upvotes]: What is the largest possible number of subsets of a $4n$-element set $X$, such that each subset contains precisely $2n$ elements, and such that each of the pairwise intersections of the subsets has precisely $n$ elements?

The best bounds I have for the maximal size $N$ of such a set system are $\lfloor \log_2 n \rfloor \le N \le 8n-1$ (corrected: was $2n-1$).  The lower bound is by a tree-like construction based on the characteristic vectors of the subsets relative to $X$, while the upper bound comes from the Plotkin bound applied to the Hamming distances between these vectors.  Are better bounds known?
The usual Erdős–Ko–Rado setup involves the size of the intersections being of at least some size, whereas here they are precisely $n$.  The EKR upper bounds are exponential in $n$ as they allow many possible set systems where the intersections have different sizes, while a linear upper bound appears to be available for this case.  So this seems like it should be a well known example, and I would appreciate a pointer if this is the case.
This is a reformulation of a question from Math.SE, Number of binary strings of length $n$ with Hamming distance $n/2$, based on my answer to that question.  I am interested in such set systems for general values; the $4n/2n/n$ case avoids some of the distracting details.

Edit: upon investigation of the history of the Frankl-Wilson theorem, it seems that a better upper bound is due to Ryser: $N \le 4n$.  This applies as long as the common size of intersections is at least 2 and less than the size of the subsets.  Thanks go to Joshua Erde, Gerry Myerson and Ben Millwood for suggesting Frankl-Wilson for the upper bound.

H. J. Ryser, An Extension of a Theorem of de Bruijn and Erdös on Combinatorial Designs, Journal of Algebra 10 246–261, 1968.  doi:10.1016/0021-8693(68)90099-9

Thanks also to Benoît Kloeckner for pointing out a rescaling error.
Benoît also pointed out that a better lower bound of $4n-1$ exists if $n$ is a power of 2, by Sylvester's construction of Hadamard matrices, and that the existence of Hadamard matrices of every multiple of 4 (which is the Hadamard conjecture) would imply a general lower bound of $4n-1$.
So, assuming the Hadamard conjecture, $4n-1 \le N \le 4n$.  If the conjecture is false, the maximum number of rows of a Hadamard-like matrix with $4n$ columns provides a lower bound for $N$.
It is interesting that there remains a gap between lower and upper bounds, even assuming the Hadamard conjecture.
A set system with $N=4n$ would correspond to a Hadamard-like matrix with $4n$ columns and $4n+1$ rows.

REPLY [3 votes]: Here is a proof that $|X| \leq 4n-1$. Given your system $X$, let $\chi_1 , \ldots, \chi_N \in \mathbb{R}^{4n}$ be the characteristic vectors of the elements of $X$. Then 
\begin{equation*}
\chi_i \cdot \chi_k = 
\begin{cases} 2n & i = k
\\
n & i \neq k .
\end{cases}
\end{equation*}
Let $j \in \mathbb{R}^{4n}$ denote the all ones vector and consider the vectors $$(2 \chi_1 - j) , \ldots , (2 \chi_N - j), j.$$ These are vectors with $\pm 1$ entries and an easy calculation reveals they are pairwise orthogonal. Thus these are $N+1$ vectors in $\mathbb{R}^{4n}$ that are linearly independent so $N+1 \leq 4n$. 
The tensor product of two Hadamard matrices is again a Hadamard matrix. As mentioned before, one can construct a Hadamard matrix for all powers of 2 by taking the tensor product of $$\left(\begin{array}{rr}
1 & 1 \\
1 & -1
\end{array}\right)$$
Many constuctions of Hadamard matrices are known and they are known to exist if $4n < 668$ and it is an important conjecture in combinatorics that they exist for all values of $4n$. One easy construction I will mention here is take $p \equiv 3 \mod 4$ and define the $p \times p$ matrix, $Q$, as follows $$Q_{ij} =  \left(\frac{i-j}{p}\right).$$ Then $$\left(\begin{array}{rr}
1 & j^T \\
j & Q - I
\end{array}\right),$$ is a Hadamard matrix.<|endoftext|>
TITLE: The nerve of categories preserves weak equivalence?
QUESTION [5 upvotes]: Does the nerve functor $\operatorname{Cat}\to \operatorname{sSet}$ from small categories to simplicial sets preserve weak equivalences? 
If $f\colon C\to D$ and $g\colon D\to C$ are functors of small categories such that $\alpha\colon f\circ g\sim 1_D$ and $\beta\colon g\circ f \sim 1_C$, then we have a pair of maps between the corresponding simplicial sets. What can we say about $\alpha$ and $\beta$ under the nerve functor? (Homotopic in some sense?)

REPLY [6 votes]: Such questions are typically framed in terms of Classifying Spaces, but the answer is yes. It follows from, for instance, from Proposition 2.1 in Graeme Segal's article 
Classifying spaces and spectral sequences. Publications Mathématiques de l'IHÉS, 34 (1968), p. 105-112, available here.
The proposition in question says: assume $C, D$ are topological categories and $F,G$ continuous functors from $C$ to $D$. Now, any natural transformation $\eta:F \implies G$, induces a homotopy between the continuous functions $BF, BG$ on classifying spaces $BC \to BD$.
In order to use this proposition, treat $C$ and $D$ as topological categories with the discrete topology on each hom-set, so all functors are continuous.<|endoftext|>
TITLE: What is a Kelley ring?
QUESTION [22 upvotes]: I've heard that in some book by someone named Kelley, perhaps an early edition of John L. Kelley's General Topology, the author gave a definition of a ring which turned out to be weaker than the usual definition.  I seem to recall the problem involved taking some shortcut in stating the distributive law.  I've heard that as a mean joke, people called this new mathematical structure a Kelley ring.  
Is this true, and if so, can someone tell me what a Kelley ring is?

REPLY [9 votes]: As you can see from the other answers, a so-called "Kelley ring" is a ring (without identity) in which the usual distributive laws are replaced by the identity $(u+v)(x+y)=ux+uy+vx+vy$. Toru Saito calls them $c$-rings in the note listed below. Here is a short bibliography of this topic:

John L. Kelley, General Topology, D. Van Nostrand Company, Inc., Princeton, New Jersey, 1955. The following quotation is from p. 18 of the July, 1957, reprinting:


A ring is a triple $(R,+,\cdot)$ such that $(R,+)$ is an abelian group and $\cdot$ is a function on $R\times R$ to $R$ such that: the operation is associative, and the distributive law $(u+v)\cdot(x+y)=u\cdot x+u\cdot y+v\cdot x+v\cdot y$ holds for all members $x$, $y$, $u$, and $v$ of $R$.


D. W. Jonah, Problem 4784, Amer. Math. Monthly 65 (1958), 289:


In John L. Kelley, General Topology, p. 18, a definition of a ring is given in 
  which the left and right distributive laws are replaced by the composite distributive 
  law: $(u+v)(x+y)=ux+uy+vx+vy$.
  (a) Show by an example that such a system is not necessarily a ring.
  (b) Show that if such a system contains an element $a$ such that $a0=0$ (in 
  particular, if the system has a multiplicative identity), then the system is a 
  ring.


R. A. Beaumont, Postulates for a ring (Solution of Problem 4784), Amer. Math. Monthly 66 (1959), 318.


Summary by me: For part (a) Beaumont takes an additive group of order $3$ and defines the product of every pair of elements to be the same nonzero element $u$. For part (b) he uses the "composite distributive law" to show that $a0=0$ implies $b0=0b=0$ for all $b$.


Toru Saito, Note on the distributive laws, Amer. Math. Monthly 66 (1959), 280-283.


Summary by me: The author defines $w$-rings and $c$-rings. A $c$-ring is a "Kelley ring". A $w$-ring is a system $(S,+,\cdot)$ which is an abelian group with respect to addition, a semigroup with respect to multiplication, and contains fixed elements $e_1,e_2$ such that
  $$x(y+z)=xy+xz-e_1,\quad(y+z)x=yx+zx-e_2,\quad\text{for all }x,y,z\in S.$$
  He shows that, in a $w$-ring $S$, we have $e_1=00=e_2$; this element is called the defining element of the w-ring $S$. The order of the defining element with respect to the additive group of $S$ is called the the order of $S$. After proving some results on the existence and structure of $w$-rings, he relates $c$-rings (Kelley rings) to $w$-rings with the following:
THEOREM 3. A $c$-ring is a $w$-ring of order $3$ or $1$ according as $00\ne0$ or $00=0$ and $00$ is the defining element of the $w$-ring. Conversely, a $w$-ring of order $3$ or $1$ is a $c$-ring.


bof, an answer to the question Counterexamples in Algebra? on Math Overflow.<|endoftext|>
TITLE: Canonical differential on Tate curve
QUESTION [7 upvotes]: I am starting studying the theory of (algebraic) modular forms, and I have some trouble in understanding completely the construction of the Tate curve. My problem is the following: as far as I know the Tate curve $T(q)$ is an elliptic curve over $\mathbb Z((q))$ with an affine equation given by $y^2+xy=x^3+b(q)x+c(q)$ for some $b(q),c(q)\in \mathbb Z[[q]]$ and an invariant differential $\omega_{can}=dx/(2y+x)$. But often I read the equality $\omega_{can}=dx/(2y+x)=dt/t$ and I can't really understand what does it mean formally. For example, say that I consider the curve $T(q)$ base changed to $\mathbb F_p((q))$ (which is the case I am interested in). If I look at the affine part of the curve described by the equation above, an invariant differential should be a basis of the $R$-module $H^0(T(q),\Omega^1_{T(q)/\mathbb F_p((q))})$ ($R$ the coordinate ring of $T(q)$) which I know being free with basis $dx/(2y+x)$. So how should I interpret this $dt/t$ in such a setting? I know that if $E=\mathbb C/\Lambda$ is an elliptic curve over $\mathbb C$ for some lattice $\Lambda$, then using the exponential map we can view it as a quotient of $\mathbb C^*$ by a discrete subgroup $q^{\mathbb Z}$ for some $q$ with $|q|<1$ and in such a case we get an isomorphic complex manifold $\mathbb C^*/q^{\mathbb Z}$ which has a canonical differential $dt/t$. But what does this "$dt/t$" mean when we deal with a scheme over $\mathbb Z[[q]]$?
I'm sorry if it is a silly question, thank you in advance to everybody willing to help me!

REPLY [5 votes]: You can view it in (at least) two ways. 
You can view $\mathbb{F}_p((q))^*/q^{\mathbb{Z}}$ as an analytic manifold over the valued field $\mathbb{F}_p((q))$, analogously to $\mathbb{C}^*/q^{\mathbb{Z}}$ or $\mathbb{Q}_p^*/q^{\mathbb{Z}}$, the latter two for element $q$ in the respective field with $|q|<1$. You can imagine a theory of manifolds over a valued field and differentials. This is better done with the theory of rigid analytic spaces and was in fact the motivation for them.
But another more direct way of viewing this is to look at the ring of power series in two variables $\mathbb{F}_p[[q,t]]$, which has a differentiation $\partial/\partial t$. Then you have two power series $x(q,t),y(q,t)$ that satisfy the equation of the elliptic curve and $t\partial x/\partial t = 2y+x$ is an identity that you can verify.<|endoftext|>
TITLE: A conjecture by Euler about $8n+3$
QUESTION [12 upvotes]: Euler's conjecture: For any positive integer $n$, $8n+3$ can be represented as a sum 
$$8n+3=(2k-1)^2+2p,$$
where $k$ is a positive integer, and $p$ is a prime.
I want to know whether there has been progress  on the problem. Could you recommend some references? Thanks a lot.

REPLY [16 votes]: Barry Mazur discussed this problem in his presentation of January 2012, Why is it plausible? Mazur says there that it is still unsettled.<|endoftext|>
TITLE: Strictly commutative elements of $E_\infty$-spaces
QUESTION [14 upvotes]: Let $X$ be an $E_\infty$-space (not necessarily grouplike). Let $x \in \pi_0 X$ be an element; say that $x$ is strictly commutative if there is a map of $E_\infty$-spaces $\mathbb{Z}_{\geq 0} \to X$ that takes $1 \mapsto x$. (The terminology is abusive, as for an element to be strictly commutative is extra data than a condition.) 
There is also a natural space of strictly commutative elements in $X$, given by the (derived) mapping space (in the homotopy theory of $E_\infty$-spaces) $\hom(\mathbb{Z}_{\geq 0}, X)$. I do not know of a simple presentation of $\mathbb{Z}_{\geq 0}$ as an $E_\infty$-space (the free $E_\infty$-space on one object is $\bigsqcup_{n \geq 0} B \Sigma_n$), so I am not sure how to write this space down in terms of $X$. If $X$ is grouplike, so that it can be identified with a  connective spectrum, then this is the mapping space in spectra $\hom( H \mathbb{Z}, X)$. 
What are examples of strictly commutative elements? For instance, I am interested in the following example:  given an $E_\infty$-ring $R$, what is the space of strictly commutative elements in  the infinite loop space $\Omega^\infty R$ with multiplicative structure? (Equivalently, what is the space of maps $S^0[\mathbb{Z}_{\geq 0}] \to R$ in $E_\infty$-rings?) One reason is that the $E_\infty$-ring $S^0[\mathbb{Z}_{\geq 0}]$ is easier to compute with than the free $E_\infty$-ring on a generator in degree zero, but seems to be less nice formally, and I'd like to know conditions under which an element in $\pi_0 R$ can be hit by a map from the monoid algebra.

REPLY [18 votes]: In the "easier" grouplike case, as you say, this is related to spaces of units, and Jacob and Neil have mentioned things about $gl_1$.  This thing exhibits strange behaviour, and was an object of close study (along with some serious calculation) a number of years ago.
Here's an example of something that may seem counterintuitive about the interaction between $gl_1$ and "genuinely commutative" phenomena.
Let $R$ be the graded ring $\mathbb{Z}/2[x]/x^3$, where $|x|=1$.  Taking zero differential, we can view this as a commutative DGA, hence giving rise to an $E_\infty$ ring spectrum.  Take $gl_1(R)$: it's a connective spectrum with homotopy groups $0, \mathbb{Z}/2, \mathbb{Z}/2$, and then zeros.
There are two such spectra: one is equivalent to a product of Eilenberg-Mac Lane spectra, and in the other the class in degree two is a multiple of $\eta$.  It turns out that it's the latter, and so there is no possibility of describing it in chain-complex terms.
Why?  Here's a sketch of the argument.

Suppose that $gl_1(R)$ is a product of Eilenberg-Mac Lane spaces.
Then $\pi_1$ splits off by a map $\Sigma H\mathbb{Z}/2 \to gl_1(R)$.
This is equivalent to an infinite loop map $K(\mathbb{Z}/2,1) \to GL_1(R)$ which is an equivalence in degree 1.
This is equivalent to a map of $E_\infty$ ring spectra $\Sigma^{\infty}_+ K(\mathbb{Z}/2,1) \to R$ (which hits the class in degree 1).
Since $R$ is an $H\mathbb{Z}/2$-algebra, this is equivalent to a map of $H\mathbb{Z}/2$-algebras $H\mathbb{Z}/2 \wedge \Sigma^\infty_+ K(\mathbb{Z}/2,1) \to R$ which is an isomorphism in degree 1.
On homotopy groups, this is a ring map $H_*(K(\mathbb{Z}/2,1); \mathbb{Z}/2) \to R$, where the former has the Pontrjagin product, which is an isomorphism in degree one.
The class in degree one in $H_*(K(\mathbb{Z}/2,1); \mathbb{Z}/2)$ squares to zero.<|endoftext|>
TITLE: Is first etale cohomology of a variety always (dual to) a Tate Module?
QUESTION [9 upvotes]: The two examples I have in mind are curves and abelian varieties. To be precise, if $C$ is a smooth projective algebraic curve over a number field $K$, then, for a prime $l$ 
$H^1_{et}(C_{\bar{K}},\mathbb{Q}_l)\cong T_l(J(C))\otimes\mathbb{Q}_l,$
where $J(C)$ is the Jacobian of $C$. The isomorphism is as $G_K$-modules, where $G_K$ is the absolute Galois group of $K$. A similar statement is true for abelian varieties.
What about smooth projective varieties of dimension $d$? Is there some generalization of the Jacobian with Tate module dual to the first etale cohomology group? When applied to an abelian variety, does this construction return the original abelian variety?

REPLY [6 votes]: You have an exact sequence of sheaves for the étale topology:
$0\rightarrow \mu_{l^{n}} \rightarrow \mathbb{G}_m \rightarrow \mathbb{G}_m \rightarrow 0$.
If you take an injective resolution of this complex, you can see immediately that the p^power commute the diagramme in the first arrow of this complex. 
By applying global section and the fact that we are in characteristic 0:
we have a short exact sequence of etale group cohomology:
$0\rightarrow H^1_{et}(X_{\overline{K}},\mu_{l^{n}})\rightarrow H^{1}_{et}(X_{\overline{K}}, \mathbb{G}_m)  \rightarrow H^{1}_{et}(X_{\overline{K}},\mathbb{G}_m) \rightarrow 0$. Wa have also a duality between $H^1_{et}(X_{\overline{K}},\mu_{l^{n}})$ and $H^1_{et}(X_{\overline{K}},\mathbb{Z}/l^{n})$, and it is easy to see using complex Cech that $H^{1}_{et}(X_{\overline{K}}, \mathbb{G}_m)$ is the picard group and $H^1_{et}(X_{\overline{K}},\mu_{l^{n}})$ classifies the invertible sheaf with trivialisation of its $l^{n}$ tensor power. Thus, by taking the projective limits over $n$ you find that $H^1_{et}(X_{\overline{K}},\mathbb{Z}_p)$ is the dual of the tate module of the Picard scheme of $X$.<|endoftext|>
TITLE: Van Vleck-Morette Determinant
QUESTION [5 upvotes]: There seems to be something curious about the so-called Van-Vleck-Morette determinant, as I cannot find any source that properly defines it in terms of expressions previously defined in that source and that actually proves properties.
To my understanding, a possible definition is
$$ \triangle(p, q) = (\det(g_{ij}(x(p)))\det(g_{ij}(y(q)))^{-1/2} \det \left(- \frac{\partial^2\sigma}{\partial x^i \partial y^j}\right),$$
where $\sigma = d(p, q)^2/2$ and the expression means the following: Choose charts $x$ and $y$ such that $p$ and $q$ are in the respective domains of these charts, then we denote by $g_{ij}(x)$ is the matrix of the metric in the chart $x$ at the point $p$ and $g_{ij}(y)$ the corresponding matrix with respect to $y$ at $q$. One can check easily be the usual transformation laws under coordinate changes that this expression is well-defined.
Define on the other hand
$$ \mu(p, q) = \det_g (d \exp_p|_X),$$
where $X$ is such that $\exp_p(X) = q$. Now if $x$ is a geodesic chart about $p$, then $\mu(p, q) = \det(g_{ij}(x))^{1/2}$.
Now my question is: These two functions should be related, if one trusts the literature, if I am not mistaken, $\triangle = \mu^{-1}$. However, I cannot find a proof anywhere and I couldn't figure out a proof myself. This should be a straightforward calculation of Riemannian geometry; however, the definition of $\triangle$ might be wrong. Does anybody know how to show this?

REPLY [6 votes]: The Van-Vleck Morette determinant often arises in semiclassical analysis, for example in computations of asymptotic expansions of heat kernels or Feynman path integrals. In these applications it is not easy to trace its origin as a transformation Jacobian.
The space of initial data of the geodesic initial data problem is the tangent bundle $TM$. (Equivalently the cotangent bundle $T^{*}M$ due to the existence of a metric. The space of boundary data of the geodesic boundary value problem is $M \times M$. Trading the dependence on the initial momenta by the dependence on the final positions defines a  map $T^{*}M  \rightarrow M \times M$ whose Jacobian as will be shown below is the Van-Vleck Morette deterinant. From the derivation it will be explicit that it is the inverse of the Jacobian of the exponential map.
In the following, the initial $t=0$ and final $ t=1$ positions along the geodesic will be denoted by $x(0)$ and $x(1)$, and the corresponding momenta by $p(0)$ and $p(1)$.
The Liouville volume element of $T{*}M$ is given by:
$$d\mu_L = \wedge dx^{i}(0)\wedge_idp_i(0)$$
The Hamilton-Jacobi function is equal to the action integral along the geodesic between the initial and final points;
$$ \sigma(x(0), x(1)) =  \int_{x(0), t=0}^{x(1), t=1} dt \sum_{ij} g_{ij}(x(t)) \dot{x}^{i}(t)\dot{x}^{i}(t)$$
$\sigma$ satisfies the Hamilton-Jacobi equation:
$$ g^{ij}(x(0)) \frac{\partial\sigma}{\partial x^{i}(0)} \frac{\partial\sigma}{\partial x^{j}(0)} = E$$
Where $E = \sum_{ij} g^{ij}(x(0)) p_i(0){p_j(0)}$, is the ( invariant) energy.
The initial momentum along the geodesic can be derived from the Hamilton-Jacobi function by:
$$p_i(0) = \frac{\partial\sigma}{\partial x^{i}(0)}$$
Thus the transformed Liouville measure on $M\times M$ is given by:
$$ \wedge_i dx^{i}(0)\wedge_i dp_i(0)  = \det\big ( \frac{\partial p(0)}{\partial x(1)} \big) \wedge_i  dx^{i}(0) \wedge_i dx^{i}(1) $$
$$ =  \frac{\det( \frac{\partial^2 \sigma}{\partial x(0)^i\partial x(1)^j}) }{ \sqrt{\det(g(x(0)) \det(g(x(1))}}  \sqrt{\det(g(x(0))} \wedge_i  dx^{i}(0)  \sqrt{\det(g(x(1))}\wedge_i  dx^{i}(1) $$ 
$$ = \Delta(x(0), x(1)) \det(g(x(0))) \sqrt{\det(g(x(0))} \wedge_i  dx^{i}(0)  \sqrt{\det(g(x(1))}\wedge_i  dx^{i}(1).$$
Now by definition: 
$$x(1) = \mathrm{Exp}_{x(0)}(v(0))$$
Where the velocity:  $v_0 \in T_{x(0)}$ is given by:
$$v^i(0)= \sum_j g^{ij}(x(0))p_j(0)$$
Thus:
$$\det_g( d \mathrm{Exp}_{x(0)}) =   \sqrt{\det(g(x(0))}^{-1} \sqrt{\det(g(x(1))} \det(\frac{\partial x(1)}{\partial v(0)}) = \sqrt{\det(g(x(0))}^{-1}   \sqrt{\det(g(x(1))} \det(g(x(0)) \det(\frac{\partial x(1)}{\partial p(0)}) = \Delta(x(0), x(1))^{-1}$$
Using the intermediate step of the evaluation of the Van-Vleck Morette determinant.<|endoftext|>
TITLE: Brownian motion on Metric spaces
QUESTION [7 upvotes]: Is there a generalization of Brownian motion to general metric spaces (which should probably be length spaces)?
This should be a process satisfying
$$d(B_t, B_s) \sim \mathcal{N}(0, t-s)$$
and such that $d(B_t, B_s)$ and $d(B_s, B_r)$ are independent; however, this will be not enough in general (this not even defines Brownian motion on $\mathbb{R}^n$).
\Edit: 1) It was pointed out by Did in the comments that $d(B_t, B_s)$ is not normal distributed even in $\mathbb{R}^n$. So this requirement should be different.
2) I wanted to know, if it is possible to define Brownian motion on general metric spaces and if there is some widely accepted definition. Benoît Kloeckner suggested below that one defines the process $B^\varepsilon_{\tau_j}$ for some subdivision $\tau$ of $[0, \infty)$, by jumping uniformly in a ball of radius $\varepsilon$. Then take the limit $\varepsilon \longrightarrow 0$ by simultaneously refining the subdivision in an appropriate way.
To me, this seems reasonable and probably is the answer that I was looking for. However, there is no answer, so I cannot accept.
However, he claimed that one should be more precise about the type of metric space. To me, this sounded like there may be other definitions that give different results. It would be interesting to learn about this.

REPLY [6 votes]: The correct point of view on Brownian motion that generalizes on metric spaces is to see the Brownian motion as the Hunt process in $\mathbb{R}^n$ generated by the Dirichlet form $\int \| \nabla f \|^2 (x) dx$. In a similar way, the Brownian motion on a Riemannian manifold $\mathbb{M}$ is  the Hunt process in $\mathbb{R}^n$ generated by the Dirichlet form $\int_{\mathbb{M}} \| \nabla f \|^2 (x) dx$. More generally, once we have a "natural and nice" Dirichlet form on a metric space, one can construct an associated continuous Markov process. You will find some details and relevant references in the paper by Sturm about diffusion processes on metric measure spaces
https://projecteuclid.org/euclid.aop/1022855410
If you use this construction on fractals for instance, you will get the Brownian motion on the fractal, which is a very fun object to study. You can for instance have a look into the short survey by Kumagai
http://www.proba.jussieu.fr/bulletin/Bmfrac.pdf
and the references therein.<|endoftext|>
TITLE: Laurent expansion of inverse of vandermonde determinant
QUESTION [6 upvotes]: I wish to find the coefficients of the Laurent expansion of the inverse of the Vandermonde determinant, that is, the Laurent expansion at 0 of
$$\prod_{1\leq i<j \leq n}(x_j-x_i)^{-1}.$$
We can iteratively do a series expansion in $x_1,x_2,$ and so on,
(the full expansion depends on the order of expansion).
An answer I would be satisfied with is if 
I can say that the coefficient $c_a(k)$ of
$x_1^{a_1+k}x_1^{a_2+k}\cdots x_{n-1}^{a_{n-1}+k} x_n^{a_n-k}$ 
in the Laurent expansion of
$$\prod_{1\leq i<j \leq n}(x_j-x_i)^{-d}.$$
is a polynomial in $k,$ for every fixed positive integer $d$.
A paper stating this, or a short argument of this fact would be terrific.

REPLY [11 votes]: In your question, the exponent of $x_2$ is $a_2+k$. I am assuming that is a typo, and you intended  $c_a(k)$ to be the coefficient of 
$$x_1^{a_1+k} x_2^{a_2} x_3^{a_3} \cdots x_{n-2}^{a_{n-2}} x_{n-1}^{a_{n-1}} x_n^{a_n-k}.$$
If so, I can answer all of your questions.
It will be convenient to switch the sign in your Vandermonde determinant, so I'll set $\Delta = \prod_{1 \leq i < j \leq n} (x_j-x_i)$. Let $\rho$ be the vector $(0,1,2,\ldots, n-1)$. So
$$\Delta = x^{\rho} \prod_{1 \leq i < j \leq n} ( 1- x_i/x_j)$$
and
$$\Delta^{-1} = x^{-\rho} \prod_{1 \leq i < j \leq n} \frac{1}{1-x_i/x_j} 
=x^{-\rho} \prod_{1 \leq i < j \leq n} \sum_{c_{ij}=0}^{\infty} \left( \frac{x_i}{x_j} \right)^{c_{ij}}.$$
So the coefficient of $x^a$ in $\Delta^{-1}$ is the number of ways to write $a+\rho$ in the form $\sum_{1 \leq i < j \leq n} c_{ij} (e_i-e_j)$, where $e_k$ is the $k$-th basis vector and the $c_{ij}$ are nonnegative integers. This is called the Kostant partition function of $a+\rho$; I'll denote it by $K(a+\rho)$.
If you do the same argument for $\Delta^{-d}$, you are counting the number of ways to write
$$a+\rho = \sum_{1 \leq i < j \leq n} \sum_{1 \leq k \leq d} c_{ij}^k (e_i-e_j)$$
where $c_{ij}^{k}$ are nonnegative integers. Let's call this $K^d(a+\rho)$.
As I will explain below, the values of $K^d(\beta)$ are piecewise polynomial, with domains of polynomiality being convex polyhedral cones. The line $a+\rho + k (e_1-e_n)$ will, for $k$ sufficiently large, eventually lie in just one cone, so it will be polynomial for $k$ sufficiently large. I'll discuss the switch between "eventually polynomial" and "polynomial" below.
The piecewise polynomiality of $K^d(\beta)$ is a special case of the following theorem of Blakley; I like the presentation of it in this paper of Sturmfels
Let $v_1$, $v_2$, ..., $v_N$ be a finite list of vectors in $\mathbb{Z}^M$, all lying in an open half space. Let $\Lambda$ be the lattice generated by the $v_i$. For $\lambda \in \Lambda$, let $c(\lambda)$ be the number of ways to write $\lambda$ as $\sum c_r \lambda_r$, with $c_r \in \mathbb{Z}_{\geq 0}$. Then can partition $\mathbb{R} \Lambda$ into finitely many convex polyhedral cones $K_j$, so that $c$ is a quasi-polynomial on $\Lambda \cap K_j$.
Here a function $\phi$ is called quasi-polynomial means that we can find a finite index sublattice $M$ of $\Lambda$ so that $\phi$ is polynomial on cosets of $\Lambda/M$. However, you don't have to worry about the word "quasi" because of a refinement of this result, also stated in Sturmfels' paper: Suppose that, whenever $v_{i_1}$, $v_{i_2}$, ..., $v_{i_D}$ is a $\mathbb{Q}$-basis for $\mathbb{Q} \Lambda$, then $v_{i_1}$, $v_{i_2}$, ..., $v_{i_D}$ is a $\mathbb{Z}$-basis for $\Lambda$. Then we get honest polynomials instead of quasi-polynomials. In this case, we say that the $v$'s are unimodular.  The $v$'s in our setting are unimodular, so we get honest polynomials. See the top of page 305 in Sturmfels' paper for more on this.
For discussion of practical computation of these polynomials for the Kostant partition function, see de Loera and Sturmfels and de Loera's website.

For notational simplicity, set $\beta = a+\rho$, so $\sum \beta_i=0$. You asked for $K^d(\beta+k(e_1-e_n))$ to be polynomial, not just eventually polynomial, in $k$. In a dumb sense, this is false. If $\beta = (-10, 0,0,\cdots, 0, 10)$, then $K(\beta + k(e_i-e_j))$ is $0$ for $k < 10$, and then becomes $\geq 0$. However, I'm guessing you might have meant the following:

Suppose that $\beta$ is in the positive integer span of $e_i-e_j$ for $i<j$ (so that $x^{\beta-\rho}$ is an exponent in the Vandermonde inverse). Then $K^d(\beta+k (e_1-e_n))$ is polynomial in $k$ for $k \geq 0$.

I expect this is false for $n=4$ and $d=1$, with $\beta$ of the form $a (e_1-e_2) + b (e_2-e_3) + c (e_3-e_4)$, and $b \gg a > c$. Take a look at Figure 1 in the de Loera-Sturmfels paper, where they draw the chamber complex for $n=4$. I am describing a point in the region they label $3$. For $k \gg 0$, the point $\beta + k (e_1-e_4)$ will be in the region they label $4$. The polynomials on regions $3$ and $4$ are different. Now, different polynomials can become equal when restricted to a line, and I haven't had time to check that this doesn't happen in your case, but I know of no reason it should happen. I'll try to do some examples this evening.
UPDATE (algebra errors in the earlier counter-example now corrected) I now have a specific counterexample. Let $n=4$ and let $\beta = (0,m,-m,0)$. I get that (for $k$, $m \geq 0$) 
$$K(k,m,-m,-k) = (k+1)(k+2)(2k+3)/6 - \min(0, (k-m+1)(k-m)(k-m-1)/6).$$ 
So, for $m \geq 2$, this is not polynomial in $k \geq 0$.<|endoftext|>
TITLE: Homotopy-theoretic derived Morita equivalences
QUESTION [20 upvotes]: Recall that two $k$-algebras $A, B$ are Morita equivalent iff their categories of left modules are equivalent. However, this relation turns out to be rather fine and one introduces a coarser equivalence relation of derived Morita equivalence by using (bounded) derived categories of modules, along with their triangulated structure.
(Note that I am no expert in this matters and I was mostly exposed to this viewpoint through algebraic geometry, where one instead works with bounded derived categories of coherent sheaves on a variety.)
However, as I understand it, the derived category of an algebra arises as a homotopy category of the stable model category of chain complexes. The latter may be seen as presenting a homotopy theory (ie. an $(\infty, 1)$-category), for example through the process of simplicial localization (and probably also some more direct, dg-theoretic methods?).
One could then say that two algebras are "higher derived Morita equivalent" if their $(\infty, 1)$-categories of (bounded?) complexes are equivalent as higher categories. The question is as follows: What can we say about this new equivalence relation? How far is this relation from derived Morita equivalence? How far is it from ordinary Morita equivalence? 
I have no intuition about this and I can imagine answers that completely equate "higher derived Morita equivalence" with either of these two, although it would be probably most interesting if it was somewhere between them. 
Note that one can imagine that somehow the derived category of an algebra remembers all the "higher homotopy", as it happens to be the case for some other homotopy categories of stable model categories. For example, in "The stable homotopy category is rigid" by S. Schwede it is proven that any stable model category $\mathcal{C}$ that satifies $ho(\mathcal{C}) \simeq \mathcal{SHC}$ (as triangulated categories, where the latter is the stable homotopy category) is in fact Quillen equivalent to model category of spectra, so they present the same homotopy theory. 
I ask the question since I am currently studying higher categories and this led me to wonder what is their possible strength as invariants of other mathematical objects.

REPLY [17 votes]: Fernando's answer is excellent, but I can't resist mentioning what is perhaps the simplest counterexample to a generalization to your question.  As Fernando says, there are counterexamples if you generalize from rings concentrated in degree 0 to dg-algebras.  These examples are somewhat complicated, but if you generalize further to $A_\infty$ ring spectra, there is a very easy example.
Fix a prime $p$ and an integer $n>0$ and consider the Morava K-theory spectrum $K(n)$, which can be given an $A_\infty$ structure.  The homotopy groups $\pi_*K(n)=\mathbb{F}_p[v_n^{\pm1}]$ are a graded field, and it follows that every $K(n)$-module is free (a wedge of suspensions of $K(n)$).  In particular, the homotopy category of $K(n)$-modules is semisimple, and is actually equivalent (as a triangulated category) to the category of graded $\mathbb{F}_p[v_n^{\pm1}]$-vector spaces.  This category is also equivalent to the homotopy category of $H\mathbb{F}_p[v_n^{\pm1}]$-modules (or equivalently, dg-modules over the graded ring $\mathbb{F}_p[v_n^{\pm1}]$).  However, the corresponding $(\infty,1)$-categories are not equivalent.  Indeed, the space of endomorphisms of a simple $K(n)$-module is $\Omega^\infty K(n)$, while the space of endomorphisms of a simple $H\mathbb{F}_p[v_n^{\pm1}]$-module is $\Omega^\infty H\mathbb{F}_p[v_n^{\pm1}]\simeq \prod_i K(\mathbb{F}_p,2i(p^n-1))$.  These spaces have isomorphic homotopy groups, but are otherwise quite far from being homotopy equivalent.<|endoftext|>
TITLE: Why is this operator compact?
QUESTION [9 upvotes]: Let $D$ be the Dirac-Operator on $\mathbb{R}^n$ or more generally the Dirac spinor bundle $\mathcal{S}\to M$ of a (semi-)Riemannian spin manifold $M$. Then we consider $D$ as an unbouded Operator on $\mathcal{H}=L^2(\mathbb{R}^n)$ with domain $C^\infty_c(\mathbb{R}^n,\mathbb{C}^N)$. Then it is said that the operator $f\langle D\rangle^{-n}$ is compact, where $f\in C^\infty_c(\mathbb{R}^n,\mathbb{C})$ is considered as a multiplication operator on $\mathcal{H}$ and $\langle D\rangle:=\sqrt{D^\dagger D+ DD^\dagger}$.
Since I am not really a crack in functional analysis, it is not even obvious for my how exactly $\langle D\rangle$ works. I suspect that the Operator $D^\dagger D+DD^\dagger$ is (essentially) self-adjoint and then the spectral theorem is used for defining $\langle D\rangle$ and its powers $\langle D\rangle^{-n}$.
But what is even more mysterious to me is the claim that $f\langle D\rangle^{-n}$ is actually compact (note that $f$ has compact support, however). Why is this true?

REPLY [2 votes]: As discussed in the comments, the statement probably needs to be modified in order for $\langle D \rangle^{-n}$ to be defined.  I'm guessing that the correct statement should fit into the following framework:

Proposition: Let $D$ be an essentially self-adjoint first order elliptic operator on a possibly non-compact manifold $M$, let $f \in C_c^\infty(M)$, and let $g \in C_0(\mathbb{R})$.  Then $m(f) g(D)$ is compact where $m(f)$ is the multiplication operator by $f$.
I'm not going to be able to drudge up all of the gory details, but in the end the proof can be pieced together using standard elliptic analysis.  First consider the resolvent function $g(t) = (i + t)^{-1}$.  Let $K$ denote the support of $f$ and let $v$ be a vector in the domain of the closure $\overline{D}$ of $D$, so that $v' = m(f)v$ is in the Sobolev space $L_1^2(K)$.  Garding's inequality estimates the Sobolev $1$-norm of $v'$ in terms of the $L^2$ norm of $v'$ and of $\overline{D}v'$; from this it follows that $m(f)(i + \overline{D})^{-1}$ maps $L^2(M)$ continuously into $L_1^2(K)$.  But the Rellich lemma asserts that the inclusion of $L_1^2(K)$ into $L^2(M)$ is compact, so the result is proved for the specific $g$ above.  Now, the set of all $g$ for which the result is true is closed under linear combinations, pointwise multiplication, complex conjugation, and uniform limits, so by the Stone-Weierstrass theorem the result is true for any $g \in C_0(\mathbb{R})$.  Notice, however, that the proposition does not apply to $g(t) = t^{-n}$, hence my concerns in the comments.

Now, the Dirac operator on a complete Riemannian manifold is essentially self adjoint and therefore fits into the proposition above.  This ultimately follows from the fact that its symbol is the Clifford multiplication endomorphism which is not only invertible (away from the $0$ section) but bounded in norm on the unit cosphere bundle.  In other words, it has "finite propagation speed". I'm not quite sure how this works out in the semi-Riemannian case, but my guess is that it does as long as you have some counterpart of the completeness assumption: note that the Dirac operator on $(0,1)$ is not essentially self-adjoint.<|endoftext|>
TITLE: local systems with finite monodromy
QUESTION [5 upvotes]: This is a question on a sentence in the paper "Faisceaux pervers", p. 163. 
The say that if $j: U \hookrightarrow X$ is a Zariski open subset and $L$ is a local system on $U$ with finite monodromy, then $L$ is a direct factor of $\pi_\ast \mathbb{C}$ for $\pi: \tilde{X} \to X$ a finite covering.
I don't understand how to construct $\tilde{X}$ out of the data of $L$ on $U$. 
Can anybody explain the procedure to me?

REPLY [5 votes]: First, suppose $L$ is simple.  $L$ is trivialied by some finite etale map $\tilde U\to U$ (this is what it means to have finite monodromy); then applying Grauert-Remmert (or SGA I, as mentioned in Georges Elencwajg's answer to this question), we may extend the map to a (possibly ramified) map $\pi: \tilde X\to X$. That is, $\pi^*L|_{\tilde U}=\underline{\mathbb{C}}^n$.  Then $\operatorname{Hom}_U(L, \pi_*(\underline{\mathbb{C}}^n)|_{U})\simeq\operatorname{Hom}_{\tilde U}(\pi^*L|_{\tilde U}, \underline{\mathbb{C}}^n)$ is non-zero, so $L$ is a factor of $\pi_*\underline{\mathbb{C}}^n$ (using simplicity). Again, as $L$ is simple, it is in fact a factor of $\pi_*\underline{\mathbb{C}}$.  
Now if $L$ is not simple, decompose it into simple factors $L=\oplus L_i$, and let $\pi: {\tilde X_i}\to X$ be covers trivializing the $L_i$.  Setting $\tilde X=\bigsqcup {\tilde X_i}$ does the trick.
BTW, just a silly but perhaps cute remark; one may apply this same argument to the map $\pi: *\to BG$, for $G$ a finite group, to obtain that every irreducible representation of $G$ is a subrepresentation of the regular representation (this is basically the same as the usual argument using the adjointness of induction and restriction).<|endoftext|>
TITLE: Binomial distribution conjecture
QUESTION [10 upvotes]: Conjecture: Let $m$ and $n$ be fixed positive integers and let $f(k)$ be the probability that a Binomial($k(m+n)$, $p$) random variable is less than $kn$. Then for sufficiently small $p$, $f(k)$ is an increasing function of $k$.
Unless I've made a mistake, the conjecture holds if you replace the binomial with its normal approximation and $p < n/(m+n)$. This suggests the conjecture should be true for large $m$ and $n$. But is it true for all $m$ and $n$, i.e. for small values? How large can $p$ be?

REPLY [5 votes]: Unless I've made a mistake: the conjecture is true for $p<kn/(kn+km+1)$ and possibly other values as well.
Requiring $k(m+n)$ to be an integer, $f(k) = \displaystyle\sum_0^{\lfloor kn\rfloor}(km+kn)Cr p^r (1-p)^km+kn-r$ and a similar expression with the same limits gives a lower bound for $f(k +1/(m+n))$. Considering the ratio of equivalent terms in this sum: if this is greater than $1$, $f(k + 1/[m+n])>f(k)$ which happens if $p<kn/(km+kn+1)$.
Hope this helps!<|endoftext|>
TITLE: Does a smooth homeomorphism of closed manifolds preserve cobordism fundamental class?
QUESTION [10 upvotes]: Let $f:M\to N$ be a smooth map of closed oriented smooth manifolds which is also a homeomorphism.  Let $[M]\in H_\bullet(M;\mathbb Z)$ denote the fundamental class (and similarly for $N$).  It is clear that $f_\ast[M]=N$.  Is the same true if $[M],[N]$ instead denote the fundamental classes in oriented bordism?

REPLY [6 votes]: Let $X$ be an oriented $d$-manifold, and consider the homomorphism
$$\rho_i : \Omega_d(X) \to H^{4i}(X;\mathbb{Z})$$
which sends $f : M \to X$ to $f_!(p_i(TM))$, the pushforward along $f$ of the $i$th Pontrjagin class of $M$. This is cobordism-invariant by the usual argument.
Now $\rho_i(id_M) = p_i(TM)$, and if $f : N \to M$ is a homeomorphism then $\rho_i(f) = (f^{-1})^*(p_i(TN))$.
So if the cobordism fundamental class were homeomorphism invariant, the integral Pontrjagin classes would be. This is false cf. chapter 4.4 of the Novikov conjecrure book by Kreck and Lueck.<|endoftext|>
TITLE: Examples of non Quillen-equivalent model categories having equivalent homotopy categories
QUESTION [12 upvotes]: I am looking for examples (references) of pairs of non Quillen-equivalent model categories having the same homotopy categories. 
The motivation is of course that I have two model categories and all the attempts to prove that they are Quillen equivalent have failed so far, but the homotopy categories are equivalent as category. I cannot find in this case any zig-zag of adjunctions between them. Reading such examples could help me to understand better the problem.

REPLY [16 votes]: The ones which usually come up to my mind as soon as I think of this:

The categories of modules over $\mathbb Z/p^2$ and $\mathbb F_p[\epsilon]/(\epsilon^2)$, $p$ a prime integer. These rings are quasi-Frobenius, so their module categories have a model structure where cofibrations are monomorphisms, fibrations are epimorphisms and weak equivalences are homomorphisms which become isomorphisms in the stable module category, obtained from the module category by killing projective-injective objects. The homotopy category is this stable module category, which in both cases is the category of $\mathbb F_p$-vector spaces. This example is equivalent to Rasmus'.
The category of DG-modules over $\mathbb F_p[v_n^{\pm1}]$, $|v_n|=2p^n-2$, $d(v_n)=0$, and the category of modules over Morava's $K(n)$.
The model category of spectra localized at $K_{(p)}$-equivalences, where $K$ is complex $K$-theory and $p$ is an odd prime, and Franke's algebraic model category $C^{(T,N)}(\mathcal A)$ defined in http://www.math.uiuc.edu/K-theory/0139/. This doesn't happen for $p=2$ by results of Constanze Roitzheim.

In all these cases the model categories are stable and the homotopy categories are not only equivalent as categories but as triangulated categories.

REPLY [14 votes]: Daniel Dugger and Brooke Shipley give an example in their paper
A curious example of triangulated-equivalent
model categories which are not Quillen equivalent
available here.<|endoftext|>
TITLE: Product of commuting nonnegative operators
QUESTION [8 upvotes]: Let $V$ be a real vector space with an inner product and $A,B : V \to V$ linear maps which are self-adjoint nonnegative-definite, i.e. $\langle Ax,y \rangle = \langle x,Ay \rangle$ and $\langle Ax,x \rangle \geq 0$, likewise for $B$. Assume that $A$ and $B$ commute (but not that $A,B$ are bounded). Can we conclude that then also $AB$ is nonnegative-definite?
This is true when $V$ is complete, because then $A$ is bounded (Hellinger-Toeplitz theorem), hence has a square root, and the claim follows easily from $A \, B = \sqrt{A} \, B \, \sqrt{A}$. It is also true when $A$ (or $B$) is essentially selfadjoint, i.e. when its closure in the completion of $V$ is self-adjoint (Theorem 3.1. in Sebestyen, Stochel, On products of unbounded operators, Acta Math. Hungar. 100(2003), 105-129).
What about the general case?

REPLY [4 votes]: The answer is no. Let me sketch the proof. 
Statement 1: Let $\sum^2\subseteq\mathbb R[x,y]$ denote the set of finite sums of squares of polynomials. Then the set $C=\sum^2+x\cdot\sum^2+y\cdot\sum^2$ is a convex cone in $\mathbb R[x,y].$ (Clear)
Statement 2: $C$ is closed in the finest locally-convex topology. (The proof is non-trivial. See e.g. Schmüdgens book "Unbounded Operator Algebras ..." for the proof of similar statements).
Statement 3: $xy\notin C.$ (Easy proof by contradiction).
Statement 4: There exists a linear functional $\varphi:\mathbb R[x,y]\to\mathbb R$ such that $\varphi(xy)<0$ and $\varphi(C)\geq 0.$ (Hahn-Banach separation).
Statement 5: Let $\varphi:\mathbb R[x,y]\to\mathbb R$ be a linear functional such that $\varphi(\sum^2)\geq 0$. Then there exists a GNS-representation of $\varphi.$ That is, there exists a real inner-product space $V,$ a homomorphism $\pi:\mathbb R[x,y]\to L(V)$ such that every $\pi(f),\ f\in\mathbb R[x,y]$ is symmetric, and a vector $\xi\in V$ such that $\varphi(\cdot)=\langle\pi(\cdot)\xi,\xi\rangle.$ Thereby $\xi$ is cyclic, that is $\{\pi(f)\xi,\ f\in\mathbb R[x,y]\}=V.$ (See Schmüdgens book "Unbounded Operator Algebras" Chapter 8.6)
Proof. Let $\varphi$ be as in Statement 4 and $(V,\pi,\xi)$ be as in Statement 5. Since $\xi$ is cyclic, for every $v\in V$ there exists $f_v\in \mathbb R[x,y]$ such that $v=\pi(f_v)\xi.$ Hence
$$\langle\pi(x)v,v\rangle=\langle\pi(x)\pi(f_v)\xi,\pi(f_v)\xi\rangle=\langle\pi(xf_v^2)\xi,\xi\rangle=\varphi(xf_v^2)\geq 0,\ \langle\pi(y)v,v\rangle\geq 0,$$
i.e. $A=\pi(x)$ and $B=\pi(y)$ are nonnegative. On the other hand $\langle\pi(xy)\xi,\xi\rangle=\varphi(xy)<0.$ That is, $AB=\pi(x)\pi(y)=\pi(xy)$ is not nonnegative.
I believe there exists a constructive solution, see my question.<|endoftext|>
TITLE: Volumes of convex vs non-convex polyhedra with prescribed facets areas
QUESTION [8 upvotes]: It is known that given a set of Areas $A_f$ and normals $\vec{n}_f$ if $\sum_f A_f \vec{n}_f=0$ exist a unique convex polyhedron with given face areas and normals. (Minkowski theorem - See Alexandrov book on Convex Polyhedra).
Obviously here I'm identifying all the isometric polyhedra.
In principle with the same set of areas and normals one can build "others" polyhedra if we relax the convexity requirement. 
What I want to prove is that in the collection of all the possible polyhedra one can build from a given set of Areas $A_f$ and normals $\vec{n}_f$ the convex one is the one with bigger volume.
Thank you for your help.
Pietro

REPLY [3 votes]: The convex polytope has the largest volume. This was proved in
K. Boroczky, I. Bárány, E. Makai Jr. & J. Pach: Maximal volume enclosed by plates and proof of the chessboard conjecture, Discrete Math. 69 (1986) 101–120.
For measures and convex bodies, a proof is provided by
Zhang, Gaoyong: The affine Sobolev inequality. J. Differential Geom. 53 (1999), no. 1, 183–202.<|endoftext|>
TITLE: Does a free action always induce a diffeomorphism?
QUESTION [8 upvotes]: Suppose that $G$ is a Lie group with a transitive action on a smooth manifold $M$. The regular theory of Lie groups tells us that $G$ and $M$ are diffeomorphic if the isotropy group is trivial.
The proof that I know goes, as usual, by proving that the canonical map $G/H \rightarrow M$ is bijective and its differential is injective. This implies that the canonical map is a diffeomorphism, but heavily relies on the fact that both manifolds are finite dimensional.
Suppose now that both $G$ and $M$ are infinite dimensional manifolds. Is it still true that $G$ and $M$ are diffeomorphic if the isotropy group is trivial? Or asked differently, is there a direct proof that the differential of $G/H \rightarrow M$ is surjective?
I just realized that the proof I know also depends on the SMOOTH exponential map, which can be defined in the infinite dimensional case, but according to "INFINITE DIMENSIONAL LIE GROUPS
WITH APPLICATIONS TO
MATHEMATICAL PHYSICS" by Rudoph Schmid is not always differential (the counterexample is the group of $s$-Sobolev diffeomorphisms).

REPLY [10 votes]: The answer is no, in general, even for both $G$ and $M$ connected.
A simple example is the following:
Take $G=Diff(S^1)$, the diffeomorphism group with the the $C^\infty$-topology, which is a regular Frechet Lie group. Let $M=Diff(S^1)$ (smooth diffeomorphisms), but now with the Sobolev $H^1$-topology, which is a topological group modelled on pre-Hilbert spaces.
Then $G$ acts smoothly on $M$ but $G$ and $M$ are not diffeomorphic. 
Of course you can say, that $M$ carries the wrong topology, and that I just forced this outcome. But this is the possibility in infinite dimensions.<|endoftext|>
TITLE: Is an algebraic space over a DVR, whose special fibre and generic fibre are schemes, actually a scheme?
QUESTION [7 upvotes]: Is an algebraic space over a DVR, whose special fibre (and all its infinitesimal neighborhood) and generic fibre are schemes, actually a scheme?

REPLY [10 votes]: Olivier's example is perfect, but let me just point out that counterexamples are easier to construct if you allow nonseparated spaces.
For example, start with two DVRs $R\hookrightarrow R'$ (with spectra $X'\to X$) such that $R'$ is finite free of rank $2$ over $R$, and the fraction field extension $K\hookrightarrow K'$ is separable. 
The constant group scheme $(\mathbb{Z}/2\mathbb{Z})_X$ acts naturally on $X'$. Let $G$ be the open subgroup scheme obtained by removing the  nontrivial point over the closed point of $X$. Put $Y:=X'/G$. The morphism $Y\to X$ is an isomorphism on the generic points, but has the same closed fiber as $X'$.
If $R\hookrightarrow R'$ is split unramified, then $Y$ is the familiar ``$X$ with the closed point doubled'' wich is a scheme. Otherwise, the closed fiber of $Y\to X$ is a one-point scheme, which has no affine (or even separated) Zariski neighborhood in $Y$. In particular, $Y$ is not a scheme.
You can check that $Y$ is locally separated (in Artin's sense: the diagonal map is an immersion) iff $R\hookrightarrow R'$ is unramified.<|endoftext|>
TITLE: Products of primitive roots of the unity
QUESTION [24 upvotes]: Let $m>2$ be an  integer and $k=\varphi(m)$ be the number of $m$-th primitive roots of the unity. Let $\Phi = \{ \xi_1, \ldots, \xi_{k/2}\} $ be a set of $k/2$ pairwise distinct primitive $m$-th roots of the unity such that $\Phi \cap \overline{\Phi} = \emptyset$, where
$\overline\Phi = \{ \overline \xi_1, \ldots, \overline\xi_{k/2}\}$ is the set of complex conjugate primitive roots.

Question 1. Does there exist an integer $N$, independent of $m$ and $\Phi$, such that  $1$  belongs to $\Phi^N$, the set of products of $N$
  non-necessarily distinct elements in $\Phi$ ?


The real question behind this concerns   automorphisms of complex Abelian varieties
and symmetric differentials. Let $A$ be a complex Abelian variety, and $\varphi : A \to A$ 
be an automorphism of finite order. 

Question 2. Does there exist an integer $N$, independent of $A$ and
  $\varphi$, such that there exists a non-zero holomorphic section
  $\omega \in Sym^N \Omega^1_A$ satisfying $\varphi^* \omega = \omega$ ?


Update (07/05/2013) : If $m$ is prime then we can take $N=3$ is Question 1 above. Indeed, the product of any two elements in $\Phi$ is still a primitive root of the unity. If we denote the set of all these products by  $\Phi^2$ then it  must intersect $\overline \Phi$. Otherwise $\Phi^2 \subset\Phi$ and, inductively, we deduce that $\Phi^m\subset\Phi$. Thus $1 \in \Phi$ contradicting the choice of $\Phi$.
In general $N=3$ does not work. If we take $m=6$ then  $\varphi(6)=2$ and for  $\Phi=\{\xi_6\}$ we need to take $N=6$. If we take $m=12$ then $\varphi(12)=4$ and for $\Phi=\{\xi_{12}, \xi_{12}^7 \}$ we also  need to take $N=6$. 
Concerning Question 2 for the particular case $A=Jac(C)$ is the jacobian of a curve and $\varphi$ is induced by an automorphism of $C$ then  I know that $N = 2^2 \times 3 \times 5$ suffices. Indeed, there always exists  $N\le 6$ that works for a given pair $(A=Jac(C),\varphi)$ with $\varphi$ induced by an automorphism of $C$.The argument in this cases consists in looking at the quotient orbifold and orbifold symmetric differentials on it.

REPLY [12 votes]: Theorem If $m$ is relatively prime to $30$, then $N=3$ works. 
I'll use additive notation. The cyclic group of order $n$ is denoted $C(n)$, the generators of this cyclic group are denoted $U(n)$.
Lemma Let $p \geq 7$ be prime and let $X$, $Y$ and $Z$ be subsets of $C(p)$ of size at least $(p-1)/2$. Then $0 \in X+Y+Z$.
Proof By the Cauchy-Davenport theorem (thanks quid!), we have $|X+Y+Z| \geq \min(3 (p-1)/2 -2,p)$, which equals $p$ once $p \geq 7$. $\square$
We now prove the Theorem by induction on $m$; the case $m=1$ is vacuously true. Let $p$ be a prime dividing $m$ and let $\pi$ be the projection map $C(m) \to C(m/p)$. Let $A \subset U(m)$ so that $U(m)$ is the disjoint union of $A$ and $-A$. We want to show that $0 \in A+A+A$.
For $b \in U(m/p)$, we have $|\pi^{-1}(b) \cap A|+|\pi^{-1}(-b) \cap A| = p-1$ or $p$; the first case occurs when $GCD(p,m/p)=1$ and the second case occurs otherwise. Choose a subset $B$ of $U(m/p)$ so that $U(m/p)$ is the disjoint union of $B$ and $-B$ and, for each $b \in B$, we have $|\pi^{-1}(b) \cap A| \geq (p-1)/2$.
Inductively, we can find $b_1$, $b_2$ and $b_3$ in $B$ (not necessarily distinct) so that $b_1+b_2+b_3=0$. Choose lifts $c_i$ of the $b_i$ to $C(m)$ so that $c_1+c_2+c_3=0$. Let $X_i$ be $(\pi^{-1}(b_i) \cap A) - c_i$ for $i=1$, $2$, $3$. Use the Lemma to find $x_1$, $x_2$, $x_3$ with $x_i \in X_i$ and $x_1+x_2+x_3=0$. Then $c_i+x_i \in \pi^{-1}(b_i) \cap A \subseteq A$ and $\sum (c_i+x_i) = 0$. $\square$

As noted in comments below, $12$ always works. Indeed, if we want to have a single $N$ for all $\Phi$, then $12$ is the best possible because $m=12$, $\Phi = \{1,7 \}$ forces $6|N$ and $m=12$, $\Phi = \{ 1,5 \}$ forces $4 | N$. However, it seems to me that a better statement (which I will now prove) is that, for any $\Phi$, either $N=4$ or $N=6$ works. 
Once our inductive claim is that either $4$ or $6$ works, the prime $5$ behaves like any other prime, so we may reduce to numbers of the form $2^a 3^b$. Also, if $9 | m$ and the claim holds for $m/3$, then it holds for $m$, as we get to replace the bound $(p-1)/2$ by $p/2$. So we are reduced to $m$ of the form $2^a$ or $2^a \times 3$. 
gcousin has already done $2^a$. For $m=2^a \times 3$, consider the map $\pi: C(2^a \times 3) \to C(2^a)$. If all of the fibers of this map have size $0$ or $2$, then we may reduce to the case of $2^a$ as above, since then we have subsets of $C(3)$ of size $2$ in the lemma. 
Suppose, then, that there is some $x \in U(2^a)$ such $\pi^{-1}(x) \cap A$ and $\pi^{-1}(-x) \cap A$ nonzero; let $a$ and $a'$ occupy these sets. Then $a+a'$ is $3$-torsion, so $N=6$ works.<|endoftext|>
TITLE: On complemented von Neumann algebras
QUESTION [20 upvotes]: Edit: according to Narutaka Ozawa, question 3) is still open in the type $\mathrm{II}_1$ case. In other terms, it is not known whether every topologically complemented type $\mathrm{II}_1$ factor in $B(H)$ is injective.
Let $M$ be a von Neumann algebra sitting in $B(H)$. I will say that $M$ is B-complemented (resp. CB-complemented, 1-complemented) if there exists a bounded (resp. completely bounded, norm $1$) linear idempotent from $B(H)$ onto $M$. Of course, 1-complemented is just a synonym for injective. Pisier (Corollaire 5) and Christensen-Sinclair proved that CB-complemented implies 1-complemented. The converse follows from Tomiyama. So CB-complemented=1-complemented=injective.
1) Haagerup-Pisier proved that the free group factors $L(F_n)$ ($2\leq n\leq \infty$) are not B-complemented in $B(H)$. Are there other known examples among finite factors? Other than those in which the factors $L(F_n)$ are B-complemented.  And other than non injective McDuff factors.
2) If I am not mistaken, their Corollary 4.6 also implies that every non injective McDuff factor and every non injective properly infinite von Neumann algebra is not B-complemented. Was this known before by other means?
3) Are there examples of B-complemented not 1-complemented (injective) von Neumann algebras?
4) If no to 3), what is known about B-complemented implies 1-complemented in general?
5) More generally, if $M$ is semi-finite or finite and sits in a von Neumann algebra $N$, Pisier proved that $M$ CB-complemented in $N$ implies 1-complemented in $N$. He says it should not be too hard, but has it been shown that the assumption semi-finite or finite can be removed?
6) When $M$ is B or CB-complemented, what can be said about the complement other than it is a closed subspace of $B(H)$?
I apologize if all this is well-known, but I couldn't find the answers in the literature I am aware of. Sorry also for all these questions in one, but I thought it would be inappropriate to post several questions on such related topics. Thank you.

REPLY [13 votes]: (i) Obviously, a vN subalgebra $N$ of a B-complemented vN algebra $M$ is also B-complemented if there exists a conditional expectation from $M$ onto $N$. (ii) Haagerup and Pisier's proof actually says if $N$ is a B-complemented vN algebra which contains a (possibly non-unital) copy of $M_2(N)$ with a conditional expectation, then $N$ is CB-complemented and hence is injective. (iii) Existence of a conditional expectation is automatic if the vN algebras are finite. With these observations, one can see that "almost all" known non-injective vN algebras are not B-complemented, but it is still open whether that holds for all vN algebras. For Question (5), Christensen and Sinclair (Proc LMS 1995) proved the theorem in that generality. There is another proof by Haagerup (unpublished).<|endoftext|>
TITLE: Open Problems in Algebraic Topology and Homotopy Theory
QUESTION [26 upvotes]: Some time ago (I see it was initially written before 1999?) Mark Hovey assembled a list of open problems in algebraic topology. The list can be found here. Some of the problems I know about have been worked on quite a bit since the time of writing. The list is very good as is, but there must also be a few good additions since 1999. Can someone point me to a more recent list of open problems in algebraic topology? My googling isn't turning up much else. Thank you!

REPLY [7 votes]: The website discussed in the comments went live a while back, and has a bunch of problems:
http://topology-octopus.herokuapp.com/problemsinhomotopytheory/show/HomePage<|endoftext|>
TITLE: Complexity of the union of randomly rotated unit cubes
QUESTION [10 upvotes]: It is a remarkable fact that the union of congrent cubes
has only at most near-quadratic combinatorial complexity,
$O^*(n^2)$ for $n$ cubes, known to be almost tight.
This contrasts with the union of congruent rectangular boxes, which
can have cubic complexity, $\Omega(n^3)$.
The "fatness" of cubes in comparison to boxes accounts for the lower complexity.
(In response to Igor's reasonable request: By combinatorial complexity I mean the 
the total number of vertices, edges, and faces of the nonconvex polyhedron
that is the union of the cubes. Of course, $\{V, E, F\}$ are interrelated by 
Euler's formula. An edge of a cube is in general fractured into many edges in
the polyhedron that constitutes the union. Where a cube edge penetrates another cube
face, it constitutes a vertex of the union. Most faces of the union are nonconvex.)
The upperbound was first established in this paper:

János Pach, Ido Safruti, and Micha Sharir.   "The union of congruent cubes in three dimensions." Proceedings 17th Symposium Computational Geometry. ACM, 2001.
  (ACM link)

I am considering the special case of congrent cubes all centered on the origin.
Still I believe the quadratic complexity can be realized,
as illustrated left below.
But I wonder if the complexity of a random union, right below, has lower
complexity, perhaps $O(n \log n)$?
   
By a "random union" I mean that each cube is rotated about
the origin by a
random orthogonal matrix,
chosen uniformly.
If anyone can see a simple argument to establish bounds on expected complexity,
or can
point me to related work in this direction, I'd appreciate
it—Thanks!

REPLY [5 votes]: I wonder where $\log n$ came from. I'm getting just $n$. Can you spot any errors in the argument below?
I'll assume that the distance from the origin to each cube vertex is $1$.
Claim 1: Let $x$ be any point with $|x|<1$. Assume that $r\approx \frac{1-|x|}{10}$. Then the probability that the surface of the randomly rotated cube intersects the ball $B(x,r)$ and the probability that its edge intersects the ball are at most $Cr^2\approx C(1-|x|)^2$, and the probability that it covers the ball is at least $cr^2\approx c(1-|x|)^2$
Proof: just look at the part of the sphere of radius about $1-|x|$ that is inside the cube.
Claim 2: It is enough to count the edge ends.
Proof: $V+F+E=2E+2$ and nobody cares about $2$.
Claim 3: The ball $B(0,1)$ can be covered by balls $B(x,r)$ such that 
a) All $r$ are of the kind $r=2^{-k}\in (0,1)$
b) For every ball, $r\approx \frac{1-|x|}{10}$ 
b) There are about $r^{-2}$ balls of any fixed radius $r$ in the covering.
Proof: Look up "Whitney decomposition" on Google.
Claim 4: Every edge end inside $B(0,1)$ is either an intersection of 3 faces or of an edge and a face.
Proof: What else can it be?
Claim 5: The expected number of edge ends in $B(x,r)$ is at most $Cn^2 r^4(1-c r^2)^{n-2}+Cn^3 r^6(1-c r^2)^{n-3}$.
Proof: Either one cube must have an edge crossing the ball, the surface of another one should intersect the ball, and the rest should not cover the ball, or 3 cube surfaces should intersect the ball and the rest should not cover it.
Claim 6: The total expected number of edge ends is at most $8n+Cn\sum_r\left[n r^2(1-c r^2)^{n-2}+n^2 r^4(1-c r^2)^{n-3}\right]$.
Proof: The additivity of expectation ($8n$ is the number of vertices on the unit sphere).
Claim 8: For every $k\ge 1, n\ge 2k+2$, we have
$$
\sum_r n^kr^{2k}(1-cr^2)^{n-k-1}\le C_k
$$
Proof: Geometric growth for $r^2<n^{-1}$, double exponential decay beyond that. Thus, only a constant number of terms with $r^2\approx n^{-1}$ matter whence the estimate.
Claim 9: The average complexity is of order $n$.
Proof: We certainly have $8n$ vertices. On the other hand, we have at most $Cn$ edge ends for $n\ge 6$. So, we are fine from both below and above.<|endoftext|>
TITLE: Does this lattice have a name (and literature)?
QUESTION [7 upvotes]: The "lattice" in the title appears to be a lattice. At least it's a poset, which I define now.
Fix a partition $\lambda$ of $n$ and consider the set of all standard Young tableaux (each of $1,\dots,n$ occurs once in the tableau, and rows and columns are increasing) with shape $\lambda$. We define the poset by saying that $T'$ covers $T$ if $T'$ can be obtained by swapping the positions of entries $i$ and $i+1$ in $T$ for some $i$, such that $i$ moved to a strictly lower row.
Thus the $T'$:s covering a fixed $T$ are those given by swapping $i,i+1$ not in the same row or column such that $i$ is in a higher row than $i+1$.
I drew this poset for a few $\lambda$ and observed that it was a lattice.

REPLY [5 votes]: Your poset is either the same or closely related to the one in http://www.fh-bingen.de/fileadmin/user_upload/Lehrende/Winkel_Rudolf/06.pdf, which is a lattice. 
      

      (Figure added by J.O'Rourke.)<|endoftext|>
TITLE: Explicit formula for associator of commutative power series
QUESTION [7 upvotes]: Perhaps this question is too elementary, but if it's written down anywhere, I'd love to know about it. Suppose I have a power series $f\in R[[x,y]]$ for some commutative, unital ring. I've recently been trying to write down an explicit formula for the power series in three variables, $f(f(x,y),z)-f(x,f(y,z))$, written down in terms of the coefficients of $f$, but am wondering if this classical-seeming computation has already been done somewhere else. I'm perfectly happy assuming $f(x,y)=f(y,x)$.  Additionally, I'm really only interested in power series such that $f(x,0)=x$ and $f(0,y)=y$.  Note that the so-called associator is also a the Gerstenhaber circle product of $f$ with itself, if that helps at all.
Thanks

REPLY [3 votes]: Okay, so I'm pretty sure I have a sort-of answer for this for $f=x+y+\sum_{i,j>0}a_{ij}x^iy^j$, though it's not a closed form at all. 
With a bit of fiddling one can see that $$ f\circ f = \sum_{i,j>0}a_{ij}\left(x+y+\sum_{l,k>o}a_{lk}x^ly^k\right)^iz^j-\sum_{i,j>0}a_{ij}x^i\left(y+z+\sum_{l,k>o}a_{lk}y^lz^k\right)^j,$$ and also that all of the monomials of less than 3 variables cancel out. 
Now, if one attempts to work this out degree by degree, one quickly sees that the number of coefficients gets enormous quite quickly. As far as I can tell, this is because they are controlled by certain kinds of partitions of integers. Let me make a definition:
Suppose we have a collection of $n$ indistinguishable red stones and $m$ indistinguishable blue stones. Then I'm interested in partitions of this collection which have a certain form.  Specifically, the only monochrome blocks have precisely one stone, e.g. $r\vert r\vert b$ and $r\vert rb$ are valid partitions of $\{r,r,b\}$ but $rr\vert b$ is not.  I'll call these "restricted two-colored partitions" for the purposes of this answer.  I think there are things called colored partitions in combinatorics, and I don't know if they're connected to such ideas. It should be clear that a partition is unchanged by switching two stones of identical color, but changed into a new partition by switching two stones of different color. Let the set of all restricted partitions of of $n$ red stones and $m$ blue stones be denoted by $Q(n,m)$ (since $P(n,m)$ means something else!). 
Now, in the above expression for $f\circ f$, let's attempt to determine the coefficient of a given monomial $x^\alpha y^\beta z^\gamma$. On the left hand side of the expression, we can fix $j=\gamma$ and on the right hand side we can fix $i=\alpha$.  Let's start with $j=\gamma$. So we're interested in determining all the ways we can get $x^\alpha y^\beta$ out of the product $\left(x+y+\sum_{l,k>o}a_{lk}x^ly^k\right)^i$, and what the associated coefficients will be. I claim that these are determined by the set $Q(\alpha,\beta)$. That is, given a partition $P\in Q(\alpha,\beta)$, there is a map which takes a spot in the partition say, $\vert r^pb^q\vert$ to the monomial $a_{pq}x^py^q$, and multiplies all such spots together, then multiplies that entire monomial by the coefficient $a_{b_P,\gamma}$, where $b_P$ is the number of spots in the partition $P$.  In other words, we've defined a map (which I claim is a bijection) $\phi:Q(\alpha,\beta)\to\{\mathrm{monomials~in~}x,y\mathrm{~of~degree~(\alpha,\beta)}\},$ where perhaps the target should be better explained, but I think it's clear. Thus if we sum all the coefficients that exist in the image of $\phi$, we'll get the total coefficient of $x^\alpha y^\beta z^\gamma$ coming from fixing the power of $z$, which I'll denote $A^{\alpha,\beta}_\gamma$.  We must do an analogous process with fixed power of $x$, and get another coefficient $A^{\beta,\gamma}_{\alpha}$.  So the coefficient of $x^\alpha y^\beta z^\gamma$ is $A^{\alpha,\beta}_\gamma-A^{\beta,\gamma}_{\alpha}$, where I've packed quite a bit of meaning into those little symbols. 
Anyway, the reason we're having to use so-called "restricted colored partitions" is because our polynomials are restricted, in the sense that there are no monomials of the form $a_{n0}x^n$ in them, except when $n=1$.  However, I suspect that this is really unnecessary, and that if we ignored this restriction, we could actually just get everything in terms of all possible colored partitions of $n$ blue stones and $m$ red stones, or whatever.  The cool thing about that is that if you set all your coefficients to be $1$, I suspect you get a sort of generating function for these types of partitions, which happen to subsume non-colored partitions.  So I imagine this sort of power series actually has some kind of combinatorial meaning, and I'd be surprised if it wasn't described somewhere in that discipline's literature, which I know nothing about. 
Anyway, I'll leave this answer up, though it's pretty unsatisfying. Basically what it tells me is that this power series is really hard to deal with.  I tried computing all the coefficients of the monomial $x^3y^3$ and it gets really big really fast.  Luckily a huge amount of stuff cancels out, so when you just have a computer calculate these terms, they get (relatively...) simpler.<|endoftext|>
TITLE: Is quantum game theory reducible to classical game theory?
QUESTION [14 upvotes]: Quantum game theory is an extension of classical game theory to the quantum domain. It differs from classical game theory in three primary ways:

Superposed initial states,
Quantum entanglement of initial states,
Superposition of strategies to be used on the initial states.

This theory is based on the physics of information much like quantum computing.

I wondered if QGT is reducible to classical GT, i.e., whether any quantum game can be transformed to some classical game.

Related issues: To prove the opposite, would we need a space-like separation between players' acts? Would one need Bell's theorem? Should players' acts be outside each other's light-cones? Do we have to appeal to physics (e.g., QM itself and/or GR)? Would we need counterfactual definiteness? Would we need to dismiss superdeterminism? Are, some or all, such issues already covered (by hidden assumptions) in classical game theory or even economics?

Can anyone perhaps point to relevant literature that specifically deals with this (the title) question?

This question was asked on MSE, but didn't get any answers. It couldn't be migrated after 60 days, but it will be closed and a link to this question will be supplied.

REPLY [9 votes]: There are several versions of quantum game theory.  First, there are Eisert-Wilkens type models in which players submit strategies that are quantum superpositions of classical strategies.  In these models, a classical game {\bf G} is replaced by a quantum game {\bf G}, which is itself a classical game with a much larger strategy space than {\bf G}.  So here the answerr to your question is:  Yes, the "quantum game" is itself in fact a classical game.
Next, there are models in which players use quantum technology to randomize among (classical) strategies, where these technologies might include observations of entangled particles.  As long as players have no private information, this is equivalent to allowing players to choose their strategies contingent on classical correlated random variables.  (E.g. perhaps you and I each choose our strategies contingent on whether we see the sun shining; whether or not the sun shines near my house is correlated with whether or not it shines near yours.)  So here all we've done by adding quantum strategies is imbed our classical game {\bf G} into a larger (classical) game.  Every equilibrium in the expanded game is a correlated equilibrium in the original game (but not, in general, vice versa).  Here again, the answer to your question is that a "quantum game" is nothing but a particular classical game.  
However, genuinely new phenomena occur when players have access to private information, e.g. when they have only partial information about the payoff matrix (and one player's information might differ from the other's).  Here each player has a set of basic strategies (say "Confess (C)" and "Not Confess" (N)), and a piece of information $x$ chosen from some information set $P$.  We can then model a (classical) mixed strategy as either
a)  A map from $P$ to probability distributions over $\lbrace N,C\rbrace $ or
b)  A probability distribution over maps from $P$ to $\lbrace N, C\rbrace $
It turns out not to matter which model you choose; the two models (usually called "the game of behavioral strategies" and "the game of mixed strategies") are equivalent in an appropriate sense and make all the same predictions about payoffs.  Whichever model you choose, you've got a new (classical) game to study.
But in the quantum environment, the equivalence between mixed and behavioral strategies breaks down, and genuinely new phenomena emerge.  In this case, the quantum version of the behavioral-strategy game has equilibria that cannot be recognized as correlated equilibria in any naturally defined classical game.  So in this case, the answer to your question becomes no.
In particular:  If we allow players to choose a map from the information set $P$ to the set of quantum strategies over $\lbrace N,C\rbrace$ --- where a ``quantum strategy'' is a strategy contingent on the realization of a quantum observable --- and if the observables available to the two players are entangled --- then one gets equilibria that are not in any sense equivalent to correlated equilibria in the original classical game.  
This is all spelled out in careful detail in Dahl and Landsburg, to which Carlo has already referred.  But since the key diagrams are missing from that arxiv'd version (due to insane policies on the part of the arxiv), you might prefer to read it here.<|endoftext|>
TITLE: A convex curve inside the unit circle
QUESTION [6 upvotes]: Has this theorem a specific name; and I need some references for general form. 
Suppose ${\gamma}$ is a convex polygon contained in the interior of the unit circle. then the length of $\gamma$ is not more than the length of the unit circle.

REPLY [3 votes]: In general (say in $\mathbb R^n$) the perimeter of a set $E$ containing a convex set $C$ is not smaller than the perimeter of $C$.
This is a consequence of the following facts about the projection $\pi$ on the convex set $C$:

$\pi(\partial E) \supset \partial C$
$\pi$ is a non-expanding map (i.e. $1$-lipschitz)
the surface area $\mathcal H^{n-1}$ of a non-expanding image of a set is not larger than the surface area of the set.<|endoftext|>
TITLE: Topology of boundaries of hyperbolic groups
QUESTION [13 upvotes]: For many examples of word-hyperbolic groups which I have seen in the context of low-dimensional topology, the ideal boundary is either homeomorphic to a n-sphere for some n or a Cantor set. So, I was wondering if this is generally true or there are some examples of hyperbolic groups whose boundaries are neither a sphere of some dimension nor a Cantor set. If there exist such examples, is there any known topological classification of boundaries of hyperbolic groups?

REPLY [14 votes]: There are plenty of other possibilities. Here are a few examples:

The boundary of the fundamental group of an acylindrical hyperbolic 3-manifold with totally geodesic boundary is homeomorphic to a Sierpinski carpet. (This appears in the Kapovich and Kleiner paper mentioned below, but must be standard---the point is the topological fact that any planar continuum with no local cut points is homeomorphic to a Sierpinski carpet.  Indeed, they prove a converse result, modulo the Cannon Conjecture.)
The boundary of a random group is homeomorphic to a Menger sponge (Dahmani--Guirardel--Przytycki).
Bowditch proved that cyclic splittings of $\Gamma$ correspond to local cut points on the boundary.  In particular, the boundary of a graph of free groups with cyclic edge groups can be decomposed along cut pairs into Cantor sets.

The classification of 1-dimensional Polish spaces implies that this is a complete list of boundaries of 2-dimensional hyperbolic groups, in the sense that if the boundary is connected with no local cut point (ie the group has no splitting over a virtually cyclic subgroup) then it must be a Sierpinski carpet or Menger sponge (Kapovich--Kleiner).
I imagine that a classification in higher dimensions is completely out of reach, though I'm no expert.  (Misha Kapovich is active on MO and can provide an authoritative answer.)
I'll add full references tomorrow.<|endoftext|>
TITLE: Are subfactor planar algebras hard to classify at index 6?
QUESTION [7 upvotes]: Given a finite index inclusion, $N\subset M$, of $II_1$ factors we can construct two towers of finite dimensional algebras known as the $\textit{standard invariant}$.  For low index, this has allowed for a complete calculation of subfactors. However, at index 6 this breaks downs because there is an uncountable family of non-isomorphic factors with the same standard invariant. 
Recent research in this direction has centered on Jones' planar algebra formulism. I have seen many talks by Vaughan where he says that things are hopeless at index 6 and cites the result above. 
However, if one doesn't care above classifying subfactors and only wants to classify standard invariants (planar algebras) the above problem does not come up. So my questions is
How hard it is to classify (subfactor) planar algebras at index 6? And why?

REPLY [4 votes]: If a subfactor $N \subset M$ at index 6 is composed, i.e., it admits a non-trivial intermediate subfactor $N \subset P \subset M$, then the two components are necessarily of indices $2$ and $3$. There exists one subfactor at index $2$ given by the graph $A_{3}$, and two at index $3$ given by $A_{5}$ and $D_{4}$.
Then, the standard invariant of a composed subfactor at index $6$ is necessarily given by a quotient of $A_{3} \star A_{5}$ or $A_{3} \star D_{4}$ (see here and here), the second is given by $\mathbb{Z}_{2} \star \mathbb{Z}_{3}$ (see Noah's answer). In this case, there exists many quotients with property (T), related to a continuous family of non-isomorphic subfactors classified by cocycles (Bisch -Nicoara-Popa).
Conclusion: the composed subfactors are completely classified by these quotients and cocycles.   
So now, it remains to classify all the maximal subfactors at index 6, here are some examples:
$R^{S_{6}} \subset R^{S_{5}} $ , $R^{A_{6}} \subset R^{A_{5}} $, ... (with $S_{n}$ and $A_{n}$ the symmetric and alternating groups).   

Problem:
List all the group-subgroup maximal subfactors at index $6$. 
Questions:
1. Is there a non group-subgroup maximal subfactor at index $6$ (other than $A_{\infty}$) ?
2. Are there only finitely many maximal subfactors of a fixed finite index ?<|endoftext|>
TITLE: What is a branched Riemann surface with cuts?
QUESTION [10 upvotes]: Edit: Let me restate the main claim being made in these two papers,

Consider the "branched" Riemann surface which has "n" sheets stuck along the intervals, $[z_i, z_{i+1}]$ for $i=1,..,2N$ then it is of genus $(n-1)(N-1)$ and is represented by the equation, $y^n = \prod_{i=1...N} \frac{z-z_{2i-1} }{z-z_{2i}}$ 
If the same intervals are labelled as, $[u_i,v_i]$ and $i=1,..,N$ then this is represented by the curve, $y^n = \prod_{i=1}^{N-1}(z-u_k)(z-v_k)^{n-1}(z-u_N)$ and has genus $(n-1)(N-1)$
(..and $v_N$ has somehow been sent to infinity using conformal maps..) 

I would like to understand why the above claims are true and their derivation and why these are the same things. (...to begin with I can't see how a "genus" can even be defined for such an object - it isn't something compact and orientable!..) 
I can't see how I can use any of the standard Riemann surface theory (like in the books by Griffiths or Jurgen Jost) on this weird structure! 
==================================================================================
The two paper references where this concept is mentioned which I would like to understand, 

http://arxiv.org/abs/1303.7221 (the branched Riemann surface that has been constructed at the top of page 6 and its algebraic curve and genus) 
http://arxiv.org/abs/1209.2428 (the construction around equation 3.16 on page 16) 

I guess both of them are constructing the same "branched Riemann surface" and writing down an algebraic curve for it and calculating its genus. I can't understand this construction and how the Riemann-Hurwitz formula is being used here. 
If I go back to my usual references of the books by Jurgen Jost or Phillip Griffiths then I don't see anything there called "n sheeted branched Riemann surface with N cuts" and how one calculates it genus or its algebraic curve. (or am I missing something and hence not able to recognize the concept?)

REPLY [6 votes]: Since no one has answered the genus computation, here goes:
The Riemann-Hurwitz formula states that for a map of Riemann surfaces $f : C_1 \to C_2$, that we have
$$
\chi(C_1) = n\chi(C_2) - \deg R
$$
where $n$ is the degree of the map, and $R$ the ramification divisor.
In the case you have, since we are covering $\mathbb{C}$, we are secretly covering $\mathbb{P}^1$, so the Euler characteristic of the base is 2. The degree is clearly $n$ (the number of sheets), and it is hopefully clear that the map is only ramified at the $2N$ points listed, with ramification index $(n - 1)$. Thus the Riemann-Hurwitz formula now reads
$$
2 - 2g(C) = 2n - 2N(n - 1)
$$
and solving for the genus gives you the desired answer.
As for why the genus is well-defined---well, here is the slight-of-hand that I used to say that it's secretly a map to $\mathbb{P}^1$. If you take the $n$ given sheets and glue them together you get a non-compact Riemann surface. To make it compact, you just have to add in a few points. The easiest way to do it is to replace the copies of $\mathbb{C}$ that you have with copies of $\mathbb{P}^1$. You can still perform the cutting-and-gluing construction to build a surface (this time with a map to $\mathbb{P}^1$). In this case, you will find that you have a compact surface. It is orientable, as at no point are we using anything but holomorphic transition functions.

REPLY [5 votes]: @user6818: To reiterate earlier replies, the Riemann surface constructed from slit planes is smooth and compact. Your authors' equation should be viewed as defining a locus in $S^2 \times S^2$, viewed as the componentwise compactification of $\mathbb{C}^2$. In this setting:

As Alexandre Eremenko and Paul Garrett noted, the "Riemann surface of $\sqrt{z}$" is the sphere $S^2$, equipped with the squaring mapping $f(w) = w^2$, namely, the (compact, smooth) locus of $z = w^2$ in $S^2 \times S^2$. The branch points (images of critical points of $f$) are $z = 0$ and $\infty$. Since $f$ is two-to-one, the classical construction is to take two copies of $\mathbb{C}$ (with coordinate $z$), make a slit between the branch points (e.g., by removing the negative real axis), and to "cross-glue" the edges of the slits. Here, the result is another copy of $\mathbb{C}$ (with coordinate $w$). Adding the branch point $z = \infty$ compactifies the $w$ plane to $S^2$.
Igor Khavine's example of the "(complex) circle" $w^2 + z^2 = 1$ (again, a compact, smooth locus in $S^2 \times S^2$) may be viewed as "the Riemann surface of $\sqrt{1 - z^2}$". The picture is identical to the squaring map (rotate the "$z$ sphere" to bring $0$ and $\infty$ to $-1$ and $-1$, say); the branch points are $\pm 1$, and the Riemann surface can be constructed by slitting two copies of $\mathbb{C}$ along the closed interval $[-1, 1]$ and "cross-gluing" the edges. In the diagram below, the slits are the upper and lower halves of the circle in the vertical plane above the real axis; the two "sheets" ($z$ planes) of the surface are the two "tilted planes". The locus is smooth even at the branch points $z = \pm 1$, i.e., $w = 0$. (The vertical axis is the real part of $w$. At risk of posting gratuitous eye candy, there's an animation loop at http://mathcs.holycross.edu/~ahwang/epix/misc/complex/quadric.gif that shows the result of rotating the $w$ plane while keeping $z$ fixed.)

Andy

REPLY [2 votes]: For convenience I've added some figures from those two papers:




Especially since you are talking about Entanglement Entropy, I thought you were asking what are Riemann surfaces in a more philosophical sense.<|endoftext|>
TITLE: Does there exist a non-square number which is the quadratic residue of every prime?
QUESTION [23 upvotes]: I want to know whether there exist a non-square number $n$ which is the quadratic residue of every prime.
I know it is very elementary, and I think those kind of number are not exist, but I don't know
how to prove.

REPLY [27 votes]: This is actually an elementary consequence of quadratic reciprocity,
generalizing the familiar proof à la Euclid that that are infinitely many
primes of the form $4k+3$ 
(i.e. primes of which $-1$ is not a quadratic residue).
We want to show that there exists $p$ such that $(n/p) = -1$.
[As stated Paul's question asks only for $(n/p) \neq +1$, but this is
trivial (if $n = -1$, take $p=3$; else let $p$ be a factor of $n$), 
so we'll exclude the finitely many prime factors $p$ of $n$.]
By QR there exists a nontrivial homomorphism 
$\chi: ({\bf Z} / 4n{\bf Z})^* \rightarrow \lbrace 1, -1 \rbrace$
such that $(n/p) = \chi(p)$ for all primes $p \nmid 2n$.
Let $a$ be any positive integer coprime to $4n$ such that $\chi(a) = -1$.
Then we have a prime factorization $a = \prod_j p_j$,
and $\prod_j \chi(p_j) = \chi(a) = -1$.
Therefore $\chi(p_j) = -1$ for some $j$, QED.
As in Euclid we can iterate this argument to construct
infinitely many distinct $p$ for which $(n/p) = -1$.<|endoftext|>
TITLE: Deformation theory of octonion algebras?
QUESTION [9 upvotes]: In Grothendieck's Brauer group papers, he uses deformation theory to bootstrap the theory of central simple algebras over a field to the theory of Azumaya algebras over rings (and schemes). I am surprised to be unable to find any parallel discussion of octonion algebras in the literature, and wonder if someone may know a reference.  The lack of associativity is what seems to make it hard to adapt Grothendieck's calculations.
Is it known (better: proved somewhere citable?) that any two octonion algebras over an artin local ring with algebraically closed residue field (of any characteristic) are necessarily isomorphic?
The non-associativity is very disorienting to me for dealing with the cohomological obstruction. I'd prefer not to reinvent a useful non-associative wheel. 

Some remarks to avoid confusion:
Remark 1: An octonion algebra over a scheme $S$ is a rank-8 non-degenerate quadratic space $(V,q)$ equipped with a (not necessarily associative) $O_S$-algebra structure admitting 2-sided identity $e \in V(S)$ such that $q(xy) = q(x)q(y)$ (which forces $q(e)=1$).  It can be deduced from this that for the perfect bilinear $B_q(x,y) := q(x+y) - q(x)-q(y)$ and the linear form $t = B_q(\cdot,e)$, the operation $x \mapsto x^{\ast} := t(x)-x$ is an anti-involution of the algebra, $x x^{\ast} = x^{\ast} x = q(x)$, $B_q(x,y) = t(x y^{\ast})$, and every algebra automorphism automatically preserves $q$ (hence also $t$ and the conjugation).
Remark 2: There is a recent paper developing a low-degree cohomology/obstruction theory for left alternative algebras (which has a hypothesis of alg. closed ground field of char. 0 that is never used in the main proofs). In that theory, vanishing of ${\rm{H}}^2$ is equivalent to triviality of all deformations as an alternative algebra, but it isn't clear that this framework should be helpful adequate for dealing with octonion algebras. It seems unlikely, since in the same paper it is shown that the "left alternative" ${\rm{H}}^2$ for a $2 \times 2$ matrix algebra is nonzero, so that matrix algebra has nontrivial deformations as an alternative algebra (which are irrelevant if one cares about deforming it as an associative algebra).
Remark 3: By computing derivations over a field, we know that the automorphism scheme of an octonion algebra over a field is smooth, even connected semisimple of type ${\rm{G}}_2$. By a trick with Dedekind domains I know that the automorphism scheme of an octonion algebra over a Dedekind domain is actually flat, so (by the theory over fields) it is a semisimple group scheme of type ${\rm{G}}_2$.  I do not know what happens for a more general base ring.
The question above is equivalent to asking for smoothness of the Isom-scheme between a pair of octonion algebras over a general base (and it is equivalent to flatness as well, without which it isn't clear how to show the Isom-scheme is a torsor for the automorphism scheme of either of the given octonion algebras). It is also equivalent to asking that every octonion algebra becomes "split" etale-locally on the base, which is a classical fact for fields but I have no idea if it is true over the dual numbers  over a field.
For example, in lieu of knowing a positive answer to my question, it isn't clear (to me) that the study of octonion algebras is completely controlled by the study of ${\rm{G}}_2$-bundles.

REPLY [7 votes]: I can answer my own question, upon finding the 1959 paper "The arithmetics of octaves and the group ${\rm{G}}_2$" by van der Blij and Springer in the library this morning (pp. 406-418 in volume 21 of Indag. Math.).  They work over a field, but their style of calculation on pages 407-412 (ignoring everything about "radical", and ignoring the top half of page 410, up to and including 2.3(i)) can be adapted to work over any local ring. The upshot is that one can prove the following theorem:
Theorem: Let $A$ be an octonion algebra over a local ring $R$, with quadratic norm form $q:A \rightarrow R$. If the non-degenerate quadratic space $(A,q)$ is split then $(A,q)$ is a "split" octonion algebra (i.e., isomorphic to a certain explicit construction built from the vector cross product on $R^3$ and the determinant on $3 \times 3$ matrices).
Since we can split any non-degenerate quadratic space of even rank over a ring by passing to an etale cover (OK for odd rank if 2 is a unit, but we need rank 8), it follows from the above Theorem (and "spreading out" from local rings) that any octonion algebra over a ring becomes split over an etale cover.
Thus, any two octonion algebras over a scheme become isomorphic etale-locally over the base, and the automorphism scheme of any such algebra is flat (hence even semisimple of type ${\rm{G}}_2$) due to comparison with the split octonion algebra that begins life over the Dedekind base Spec($\mathbf{Z}$). Hence, isomorphism classes of octonion algebras over a scheme $S$ are classified by isomorphism classes of ${\rm{G}}_2$-torsors over $S$.
So that affirmatively answers everything I was wondering about, and happily without any need to develop an octonionic cohomological deformation theory. And a posteriori, we see it is the same as the deformation theory of ${\rm{G}}_2$-torsors.  
It is impractical to get into the gritty details of the argument over local rings here, but it is relatively straightforward if one reads the first two sections of the paper of van der Blij and Springer. 

For anyone who wants to look at the above paper and carry out the same exercise I did of adapting their arguments to work over a local ring, let me make some additional remarks, using notation as in that paper.

To fill in various details that van der Blij and Springer omit (over a field), one has to bring in the two additional Moufang identities which they don't discuss (but which can be proved exactly as in the theory over a field; see the proof of Prop. 1.4.1 in the book of Springer and Veldkamp) in order to see that the associator is alternating. This underlies their computation of $x_0(x_1x_2)$ and $(x_1x_2)y_1$ on page 411, and at the bottom of page 411 their formula for the multiplication law in the split octonions should have the minus signs replaced with plus signs (but leave the sign in their formula for the norm form $Q$). [EDIT: actually, all of the necessary identities are explained in the early sections of the Springer--Veldkamp book on Octonions and Jordan algebras.]
The role of being over a local ring is that at certain steps one can use unit constructions over the residue field to check certain constructions in the local ring are units, and more importantly one can carry out their direct sum decompositions of the octonion algebra on the top of page 411 (denoted as $F \oplus G$ in their paper) without having bases, so one gets the existence of the required bases of the modules $F, G, F_0, G_0$ they build due to the fact that a direct summand of a finite free module over a local ring is itself finite free. 
In section 1.8 of the book of Springer and Veldkamp there is a very short proof over fields that an octonion algebra with isotropic norm form is split. That proof is too field-specific, and does not seem to adapt to working over local rings.  It is the proof in the paper of van der Blij and Springer that is sufficiently robust that it adapts to working over local rings.<|endoftext|>
TITLE: Which sheaves satisfy cohomological purity?
QUESTION [10 upvotes]: The absolute cohomological purity theorem in étale cohomology is as follows.

Let $X$ be a regular scheme over $\mathbb{Z}[1/n]$, and $i \colon Z \to X$ the inclusion of a regular
  closed subscheme everywhere of codimension $c$.  Let $\Lambda$ be the constant
  sheaf $\mathbb{Z}/n\mathbb{Z}$ on $X$.  Then $R^{2c} i^! \Lambda \cong \Lambda(-c)$, and $R^{p} i^! \Lambda$ is trivial for $p \neq 2c$.

Here $i^!$ is the functor taking a sheaf $\mathcal{F}$ on $X$ to $\mathrm{ker}(\mathcal{F} \to j_* j^* \mathcal{F})$, considered as a sheaf on $Z$, where $j$ is the inclusion of $X \setminus Z$.  There are several other ways to state the result: chapter 16 of Milne's online Lectures on Etale Cohomology explains nicely how to go between them.  As far as I understand, the theorem has been proved by Gabber in at least two ways.
My question is this:

What other sheaves $\mathcal{F}$ may replace $\Lambda$ in the purity theorem?

The problem seems to be in defining the morphism $R^{2c} i^! \mathcal{F} \to i^* \mathcal{F}(-c)$ in the first place: once that's done, proving that it's an isomorphism is purely local and ought to follow from the case of $\Lambda$, at least for $\mathcal{F}$ something nice like a flat sheaf of $\Lambda$-modules.  Maybe one can define the morphism just by tensoring the original one with $\mathcal{F}$, but I'm not sufficiently happy with the formal properties of étale cohomology to understand how $R^p i^!$ works with tensor products.

REPLY [6 votes]: Having had a few up-votes but no answers, let me answer my own question in case anybody finds it useful in future. This is basically a formal calculation, similar to the proof of the projection formula, and I'm sure it's well known to experts in étale cohomology.  But I couldn't find the details in the literature.
Everything happens in the category of sheaves of $\Lambda$-modules on $X$.  Note that cohomology in this category is the same as cohomology in the category of sheaves of Abelian groups on $X$.$\DeclareMathOperator{\Hom}{Hom}\newcommand{\F}{\mathcal{F}}$
Step 1. Whenever $\F$ is flat, there is a "projection formula" $i^!(\mathcal{G} \otimes \F) \cong i^! \mathcal{G} \otimes i^* \F$ for any sheaf $\mathcal{G}$ on $X$.  To see this, take the short exact sequence $0 \to i_* i^! \mathcal{G} \to \mathcal{G} \to j_* j^* \mathcal{G}$ and tensor with $\F$.  The usual projection formula gives $(j_* j^* \mathcal{G}) \otimes \F \cong j_*(j^* \mathcal{G} \otimes j^* \F) \cong j_*j^*(\mathcal{G} \otimes \F)$ and $(i_* i^! \mathcal{G}) \otimes \F \cong i_*(i^! \mathcal{G} \otimes \F)$.  We obtain

$0 \to i_* (i^! \mathcal{G} \otimes i^* \F) \to \mathcal{G} \otimes \F \to j_* j^* (\mathcal{G} \otimes \F)$.

Comparing this with the definition of $i^!(\mathcal{G} \otimes \F)$ gives the claimed isomorphism.
Step 2. If $\F$ is locally free of finite rank, then there is a natural isomorphism $- \otimes \F\to \Hom(\F^\vee,-)$, where $\F^\vee$ denotes $\mathrm{Hom}(\F,\Lambda)$; this can be checked on stalks and so follows from duality for $\mathbb{Z}/n\mathbb{Z}$-modules.  It follows that

$\Hom(A, B\otimes\F) \cong \Hom(A,\Hom(\F^\vee,B)) \cong \Hom(A\otimes\F^\vee, B).$  

So the functor $- \otimes \F$ admits an exact left adjoint $- \otimes \F^\vee$; therefore it is exact and takes injectives to injectives.
Step 3. Using these two facts, we can deduce that $R^p i^! \F \cong (R^p i^! \Lambda) \otimes \F$ for all $p$.  To do so, consider an injective resolution of $\Lambda$.  Tensoring with $\F$ gives an injective resolution of $\F$ by (2).  Now apply $i^!$ and take homology; by the projection formula proved in (1), we get the claimed isomorphism.
Step 4. Now we can deduce the purity result for $\F$ by taking the isomorphism from the purity theorem and tensoring with $\F$.
So the purity theorem hold as least for $\Lambda$-modules $\F$ which are locally free of finite rank.<|endoftext|>
TITLE: Riemannian and symplectic structures
QUESTION [5 upvotes]: Let $(\mathcal M,g)$ be a smooth Riemannian manifold and $\Delta$ be the standard (positive) Laplace operator given in coordinates by the usual
$$
\Delta=-\vert g\vert^{-1/2}\partial_j(\vert g\vert^{1/2} g^{jk}\partial_k(\cdot)).
$$
The cotangent bundle of $\mathcal M$ has a standard symplectic structure, given in coordinates by 
$
\omega =d(\xi\cdot dx).
$
The principal symbol of $\Delta$ is
$
p(x,\xi)=g^{jk}(x)\xi_k\xi_j.
$
Let $\phi$ be a smooth function defined on the base $\mathcal M$. I can calculate
the Poisson bracket
$$
\{p,\phi\}=2g(X,\nabla \phi),\quad X=g^{-1}\xi,
$$
but I am in trouble with the calculation of $\{p,\{p,\phi\}\}$. I think that
$$
\frac14 \{p,\{p,\phi\}\}=D_XD_X \phi-\frac12 D_{\nabla g(X,X)}\phi=(\nabla^2\phi)(X,X),
$$
but I am not able to prove the second equality. Maybe a simpler question would be about the proof (if correct !)
of
$$
\nabla \bigl(g(X,X)\bigr)=2D_X X.
$$
Thanks in advance for your help.

REPLY [6 votes]: I am going to shoot for coordinate-free.  First, some notation: let $\pi:T^*\mathcal{M}\to\mathcal{M}$ be the bundle projection, set $E:=\pi^{-1}T\mathcal{M}$, the pullback of the tangent bundle of $\mathcal{M}$, so that $E^*=\pi^{-1}T^*\mathcal{M}$ is the bundle along the fibres.  Equip both with the pullback $D$ of the Levi-Civita connection of $\mathcal{M}$.
Now $E^*$ has a tautological section $\xi$ given by $\xi(\alpha)=\alpha$ and then the Liouville form is $\langle\xi,d\pi\rangle$.  Thus the symplectic form is
$$
\omega=d\langle\xi,d\pi\rangle=\langle D\xi\wedge d\pi\rangle+\langle \xi,d^Dd\pi\rangle=\langle D\xi\wedge d\pi\rangle
$$
since $d^Dd\pi=0$, $D$ being the pullback of a torsion-free connection.
The connection gives us a splitting into vertical and horizontal bundles $T(T^*\mathcal{M})=\mathcal{V}\oplus\mathcal{H}$ where $\mathcal{V}=\ker d\pi$ and $\mathcal{H}=\ker D\xi:T(T^*\mathcal{M})\to E^*$. Clearly $d\pi:\mathcal{H}\cong E$.  Note that both $\mathcal{V}$ and $\mathcal{H}$ are Lagrangian subbundles.  Let $X$ be the section of $\mathcal{H}$ corresponding to $g^{-1}\xi\in\Gamma E$.  Thus $X$ is the vector field on $T^*\mathcal{M}$ satisfying
$$
D_X\xi=0,\qquad d\pi(X)=g^{-1}\xi.
$$
Now let's get to work: the function $p=g(\xi,\xi)$ and, since we want to compute Poisson brackets with $p$, we should compute the Hamiltonian vector field of $p$: $dp=2g(D\xi,\xi)$ so that $$dp(Y)=2g(D_Y\xi,\xi)=2\langle D_Y\xi,g^{-1}\xi\rangle=2\langle D_Y\xi,d\pi(X)\rangle=2\omega(Y,X).$$
Thus $p$ has Hamiltonian vector field $2X$.
Now let $\phi$ be a function on $\mathcal{M}$ then
$$
\{p,\phi\circ\pi\}=d(\phi\circ\pi)(2X)=2d\phi(g^{-1}\xi)=2g(\xi,d\phi),
$$
in agreement, modulo the violence I have done to their notation, with OP.
Now
$$
\tfrac14\{p,\{p,d(\phi\circ\pi)\}\}=d_Xg(\xi,d\phi)=g(D_X\xi,d\phi)+g(\xi,D_X(d\phi))=g(\xi,D_X(d\phi)),
$$
since $X$ is horizontal.  This pretty much bakes the cake: we are viewing $d\phi$ is a section of $E^*$ (so we should really have written $(d\phi)\circ\pi$) and $D$ is the pullback of the Levi-Civita $\nabla$ so the defining property of pullback connections tells us that $D_X((d\phi)\circ\pi)=(\nabla_{d\pi(X)}d\phi)\circ\pi$.  Putting it all together now yields
$$
\tfrac14\{p,\{p,d(\phi\circ\pi)\}\}=\pi^*(\nabla^2\phi)(X,X)=\nabla^2\phi(g^{-1}\xi,g^{-1}\xi).
$$<|endoftext|>
TITLE: Size of a certain sumset in $\mathbb{Z}/p^2\mathbb{Z}$
QUESTION [13 upvotes]: We are interested in estimating the size of a certain sumset in $\mathbb{Z}/p^2\mathbb{Z}$. Let $p$ be an odd prime, $g$ a primitive root modulo $p^2$, and $A=\langle g^p\rangle$ the unit subgroup of order $p-1$. Experimentation suggests strongly that if $p \geq 61$, then the size of the sumset $3A = A + A + A$ approximates the order of the ring. This observation is due to Felipe Voloch (see the original question). In particular, we conjecture that
$$|3A|\geq p^2-1,$$
for such $p$. It is known that, as sets, $4A = \mathbb{Z}/p^2\mathbb{Z}$. Additionally, numerical evidence suggests that $3A \supseteq (\mathbb{Z}/p^2\mathbb{Z})\setminus\{0\}$, and that $p\equiv 1\pmod{3}$ implies strict containment (i.e., $3A = \mathbb{Z}/p^2\mathbb{Z}$). Perhaps these observations could provide a foothold toward the result. Is there work that addresses this phenomenon, or any existing progress toward an explanation of the estimate above?
(Edit: The remark about "strict containment" refers to the $\supseteq$ relation. That is, $3A \supsetneq (\mathbb{Z}/p^2\mathbb{Z})\setminus\{0\}$ whenever $p\equiv 1\pmod{3}$, which implies that $3A=\mathbb{Z}/p^2\mathbb{Z}$.)

REPLY [2 votes]: Using some results of Shkredov et al (following Heath-Brown and Konyagin), it's possible to show that $|A+A+A|\gg |A|^{11/6}/(\log|A|)^{1/4}$. I posted more details in the old thread. Shkredov's results are rather complicated, so it would be really nice to see a slick proof.<|endoftext|>
TITLE: Are overlaps among {algebraic geometry, arithmetic geometry, algebraic number theory} growing?
QUESTION [17 upvotes]: From a naive outsider's viewpoint, just watching the MO postings
in those three fields scroll by, and hearing of breakthroughs in the news,
it appears there might be increasing overlap among the fields,
crudely like this:
     
Thus work surrounding
the ABC conjecture,
schemes and stacks,
the Grothendieck Riemann Roch theorem,
and the Langlands conjectures,
all arguably fall within field overlaps.
Of course I recognize this depends on how broadly one construes these field categorizations!
If this picture is at all accurate,
it runs counter to the oft-expressed lament over the increasing
specialization of mathematics research
(e.g., Morris Kline:
"Now [1977] most mathematicians work only in small corners of mathematics...")
I am interested to know if modern advances are increasingly
drawing from these field intersections, i.e.: Is my picture
a gross misrepresentation of the reality of research in these fields today?
But to make this question more specific, let me ask if
there are counterexamples to this overlap hypothesis:
Modern technical achievements which clearly fall in one but
do not draw upon the other two of these fields?
Finally, let me close with this quote from
Michael Atiyah (2005 Notices of AMS):

"I really do
  not believe in the division of mathematics into
  specialities."

REPLY [27 votes]: I am not sure I really agree with the following quote (which is the opening paragraph of Modular forms and Galois cohomology by H.Hida) because I suspect that a mathematician valuing creativity and versatility in approaches (surely true about Hida) will tend to consider his own field as uniquely amenable to this way of thinking, and conversely for a mathematician valuing the overall coherence of a given theory (Grothendieck comes to mind). Nevertheless, I think it is beautifully written and thought-provoking.
Traditionaly, mathematical research has been classified by the method mathematicians exploit to study their research areas. For example, algebraists study mathematical questions related to abstract algebraic systems in a purely algebraic way (only allowing axioms defining their algebraic systems), differential geometers study manifolds via infinitesimal analysis, and algebraic geometers study geometry of algebraic varieties (and its siblings) via commutative algebras and category theory. There are no central techniques which distinguish number theory from other subjects, or rather, number theorists exploit any techniques available to hand to solve problems specific of number theory. In this sense, number theory is a discipline in mathematics which cannot be classified by methodology from the above traditional viewpoint but it is just a web of rather specific problems (or conjectures) tightly and subtly knit to each other. We just study numbers, those simple ones, like integers, rational numbers, algebraic numbers, real and complex numbers and $p$-adic numbers, and that is it.
Certainly, my impression is that the growing overlap you describe largely reflects the fact that a result will be classified as arithmetic geometry if it has an impact on some problems which are already classified as arithmetic geometry, and this however remote the former statement seems to be from arithmetic geometry. To take a specific example, Peter Scholze proved that the adic spectra of a  perfectoid algebra and its tilt (whatever that means) are homeomorphic, a result which surely belongs to analytic topology by any reasonable standard. Because this is a crucial step in the proof of part of the weight-monodromy conjecture, a defining problem in arithmetic geometry, and in the construction of Galois representations attached to automorphic forms, another defining problem in arithmetic geometry, this will surely be classified as standard arithmetic geometry in the future. 
It turns out that the defining problems of arithmetic geometry seem to require a subtle mix of advanced techniques coming from many different subfields (sometimes to the surprise of the very mathematicians who formulated them in the fist place) so that people motivated by these problems will 1) come from rather diverse backgrounds and 2) necessarily strive to master all the required areas, therefore increasing the overlap. In particular, I think that the very nature of the field makes it very hard to find a recent significant advance in arithmetic geometry which would not draw extensively upon several different other fields. 
As to whether modern advances are increasingly drawn from these field intersections, I think that this may reflect the sociological structure of the mathematical society. Partly for historical reasons, partly because of sociological inertia (talented famous people have talented students) and probably partly because some of these problems at least did prove incredibly fruitful (but as I work in this field, I don't want to mistake my feelings for an objective assessment), the defining problems of arithmetic geometry have reliably attracted some of the best mathematical minds of each generations. A handful of them have even gained general cultural status. So certainly a disproportionately high number of Abel prizes, Fields medals and NYT articles about maths are about problems lying in the intersections you drew.   
That said, what is true for arithmetic geometry is less so for algebraic geometry, as far as I can tell, and I know plenty of outstanding mathematicians in that field who do very exciting work but who do not seem to be really moving closer to arithmetic geometry in any meaningful sense. Claire Voisin comes to mind. So I suspect more knowledgable people than me would find counterexamples in that field.<|endoftext|>
TITLE: When does localization preserve homotopy type of classifying spaces?
QUESTION [13 upvotes]: Let $\mathcal{C}$ be a small category and $\Sigma$ a collection of morphisms in $\mathcal{C}$. Denote by $F_\Sigma:\mathcal{C} \to \mathcal{C}[\Sigma^{-1}]$ the usual quotient functor from $\mathcal{C}$ to its localization about $\Sigma$.

Are there conditions on $\Sigma$ which guarantee that $F_\Sigma$ induces a homotopy equivalence $B\mathcal{C} \sim B\mathcal{C}[\Sigma^{-1}]$ of classifying spaces?

For example: if $\mathcal{C}$ consists of two objects with a single arrow from one to the other, then localization about that single arrow preserves homotopy type of classifying spaces: everything is contractible before and after localization. On the other hand, see this paper for a counter-example to the conjecture that the group completion of a monoid has the same classifying space as that monoid. Clearly, we can't just shamelessly start inverting arrows all over the place without destroying homotopy type.
One always has Quillen's Theorem A: if the under categories $F_\Sigma \downarrow c$ are all contractible, then $BF_\Sigma$ is a homotopy-equivalence of classifying spaces. So, one possible answer would highlight those conditions on $\Sigma$ which magically give contractible over/under categories. Is there a known result that does the trick?

REPLY [9 votes]: You can find some sufficient conditions in terms of simplicial localization of Dwyer and Kan. In Prop. 3.7 of Simplicial Localizations of Categories they prove that it holds when $\Sigma$ is free and $\mathcal{C} = \mathcal{D} * \Sigma$ where $*$ denotes the coproduct of categories with a fixed set of objects (aka the free product). More generally it follows from 4.3 of the same paper that a sufficient condition is that $L^H(\mathcal{C}, \Sigma) \to \mathcal{C}[\Sigma^{-1}]$ is a DK-equivalence i.e. that $L^H(\mathcal{C}, \Sigma)$ has homotopy discrete mapping spaces. (As pointed out in Prop 7.2 of Calculating Simplicial Localizations this is implied by a calculus of fractions which generalizes the classical criterion of a monoid satisfying the Ore conditions.)<|endoftext|>
TITLE: Depth of ideals in a commutative ring
QUESTION [9 upvotes]: Let $I, J \subset S = k[x_1,\dots,x_n]$ be two monomial ideals and $k$ a field. If every element of $S$ which is $S/J$-regular is also $S/I$-regular is it true that depth$_S S/I \geq$ depth$_S S/J$ ?

REPLY [6 votes]: $R = k[a,b,c,d]$, $I= (a, b)\cap (c,d)$ and $J = (ac, bd) = (a, b) \cap (a, d)\cap (c, b)\cap (c, d)$. Then $\mathrm{Ass}R/I \subseteq \mathrm{Ass}R/J$. Hence if $x$ is a regular element of $R/J$ then $x$ is a regular element of $R/I$ but $\mathrm{depth}R/I = 1$ and $\mathrm{depth}R/J = 2$.<|endoftext|>
TITLE: Why is the path fibration a strong Hurewicz fibration?
QUESTION [6 upvotes]: In May and Sigurdsson "Parametrized homotopy theory" there is a general treatment of Hurewicz style model structures in Chapter 4, see definitions 4.2.1 and 4.2.2. I am trying to adapt these to a more general setting. There is an "observation" after Lemma 4.2.4 stating that $X \to \mathrm{Cyl}(X)$ and $\mathrm{Cocyl}(X) \to X$ are strong Hurewicz cofibration and fibration, respectively. The text makes it sound as if this holds for a purely formal reason, i.e., no geometry of the interval is used. Is this really the case? I do not see this, and it is crucial for what I am trying to do to know the exact geometric requirement on the interval for this to hold. I think I am missing some fairly obvious use of the lifting properties, and my excuse is that I am not a homotopy theorist.

REPLY [7 votes]: Andrej, thank you for your reading of May and Sigurdsson, which I wish we
had made more accessible.   I don't have time to check anything, but my 
recollection is that the verification of the observation is easy, whether 
or not it is formal, and works equally well in homological analogues.  I 
think we inserted it only for reassurance, since it is of course 
to be expected.  It is obviously true when spaces are restricted to be compactly 
generated since then there is no ``strong'' distinction. (MS could not just
restrict like that because of point-set problems with parametrized function spaces).
The paper of Schw\"anzl and Vogt noted in Karol's answer is the original source 
for the distinction between strong and ordinary Hurewicz cofibrations and fibrations 
and may well be helpful.   A crucial non-formal thing, noted just after the 
observation you refer to and emphasized more strongly in Schw\"anzl and 
Vogt is the need for a good cylinder object:  $\partial I \longrightarrow I$ has the LLP 
wrt $h$-acyclic $h$-fibrations.  (Of course, that fails for simplicial sets.)
In case you are thinking about model structures, I'll say a bit about that
(probably repeated from another comment or answer).  There is a mistake in the 
general proof of the factorization axioms in the paper of Cole that Karol refers 
to and in MS (4.4.2) and MP ("More Concise'').  A fix is given in: T. Barthel and E. Riehl. 
On the construction of functorial factorizations for model categories. Algebraic 
& Geometric Topology 13 (2013), 1089--1124.  (That is the paper referred to by Omar, 
but their fix does not seem directly relevant to the precise question you ask).   They and I 
are just finishing a related paper "Six model structures for DG-modules over DGAs'', 
in which their fix is again utilized and is described in full categorical generality. 
If anyone is interested, I will be happy to say why ``six'' and why that is also to be 
expected in topological contexts of interest (but I sometimes get scolded for digressing). 
Ok, Karol, but briefly:  Six projective type model structures should appear whenever has a category $\mathcal M$ of structured objects enriched in a  category $\mathcal V$ with two model structures (like the $h$- and $q$-model structures on spaces, $R$-modules, and, conjecturally, certain categories of spectra).  The category $\mathcal M$ then has three natural subcategories of weak equivalences, the structure preserving homotopy equivalences ($h$-equivalences), the homotopy equivalences of underlying objects in $\mathcal V$ ($r$-equivalences, where $r$ stands for relative), and the weak equivalences of underlying objects in $\mathcal V$ ($q$-equivalences).  These can be expected to yield $q$-, $r$-, and $h$-model structures with accompanying mixed 
$(r,h)$-, $(q,h)$-, and $(q,r)$-model structures.  And they are interesting!<|endoftext|>
TITLE: How to prove Lambert's W function is not elementary?
QUESTION [24 upvotes]: Liouville's theorem gives such a proof for antiderivatives of functions like $e^x/x$ or $e^{x^2}$, and differential Galois theory extends that to Bessel functions, say. But what tools exist for implicit functions like Lambert's W?

REPLY [13 votes]: In [Ritt 1948], the method of J. Liouville is given for the Kepler equation.
[Ritt 1948] Ritt, J. F.: Integration in finite terms. Liouville's theory of elementary methods. 1948, page 56
A further method is the method of Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22. It is a byproduct of Liouville's theory of integration in finite terms. It is written in the language of Differential algebra, but it can be represented also without that.
This method is applicable only for functions satisfying a differential equation that is simple enough.
A reference for Kepler's equation is Zarzuela Armengou, S.: About some questions of differential algebra concerning to elementary functions. Pub. Mat. UAB 26 (1982) (1) 5-15.
A reference for Lambert W function is Bronstein/Corless/Davenport/Jeffrey 2008 from the answer of Igor Khavkine above.
The branches of Lambert W are the local inverses of the Elementary function $f$ with $f(z)=ze^z$, $z \in \mathbb{C}$.
The incomprehensibly unfortunately hardly noticed theorem of Joseph Fels Ritt in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 answers which kinds of Elementary functions can have an inverse which is an Elementary function.
And Ritt's theorem shows that no antiderivatives, no differentiation and no differential fields are needed for defining the Elementary functions.
Ritt's theorem is proved also in Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math 101 (1979) (4) 743-759.
By extension of Risch's structure theorem for the elementary functions, Ritt's theorem could possibly be extended to other and to larger classes of functions, as I proposed in my question here: How to extend Ritt's theorem on elementary invertible bijective elementary functions.<|endoftext|>
TITLE: When was the continuum hypothesis born?
QUESTION [12 upvotes]: The question Solutions to the Continuum Hypothesis states that the continuum hypothesis was posed by Cantor in 1890. In http://en.wikipedia.org/wiki/Continuum_hypothesis the year 1878 is quoted without source. What is the correct date?

REPLY [22 votes]: Finally, a good use for the newly purchased copy of "Zermelo's Axiom of Choice".
Moore writes that Cantor formulated the following problem in 1878:

Every infinite subset of $\Bbb R$ is either denumerable or has the power of the continuum.

The given reference is:

Cantor G. "Ein Beitrag zur Mannigfaltigkeitslehre." Journal für die reine und angewandte Mathematik 84 (1878), pp. 242-258.

Although searching for the reference, it seems that it may have published in 1877, so it's unclear which one is the correct date. The paper was written in 1877, though. Moore points that the hypothesis was given on page 257, but I don't read German very well (or at all, for that matter), so I can't tell.
Moore also adds that in 1895 Cantor pointed out that in $\aleph$ notation, which introduced in the paper below, that this is equal to $2^{\aleph_0}=\aleph_1$. Cantor already made this observation in 1882 in a letter to Dedekind, where he used the terminology "numbers of the second number-class" to talk about $\aleph_1$.

Cantor G. "Beiträge zur Begründung der transfiniten Mengenlehre." Mathematische Annalen 46 pp. 481-512.

However it is good to note that this equivalence requires the axiom of choice. The question you cite refers to this statement, and not the first statement.
(Both the statements appear on page 41 of the book "Zermelo's Axiom of Choice: Its Origins, Developments & Influence" by Gregory H. Moore)

REPLY [10 votes]: The main theorem of Cantor's "Beitrag" (contribution) is a bijection between $\mathbb R$ and $\mathbb R^n$, as well as a bijection between $\mathbb R$ and $\mathbb R^{\aleph_0}$.   He uses (in modern language) Hilbert's Hotel to find a bijection between the real numbers and the irrational numbers (in the unit interval), and then uses continued fractions to find a bijection to Baire space $\mathbb N^{\mathbb N}$.  The bijection between Baire space and its square is then easy. 
At the end of the paper he introduces what we now call the continuum hypothesis: 
Cantor's original text, dated July 11, 1877: (from the link quoted by Asaf)
[...] Verstehen wir unter einer linearen Mannigfaltigkeit jeden denkbaren Inbegriff unendlich vieler, voneinander verschiedener reeller Zahlen, fragt es sich, in wie viel und in welche Klassen die linearen
Mannigfaltigkeiten zerfallen, wenn Mannigfaltigkeiten von gleicher Mächtigkeit in eine und
dieselbe Klasse, Mannigfaltigkeiten von verschiedener Mächtigkeit in
verschiedene Klassen gebracht werden. [...]
Durch ein Inductionsverfahren, auf dessen Darstellung wir hier nicht näher eingehen, wird der Satz nahe gebracht, dass die Anzahl der nach diesem Eintheilungsprinzip sich ergebenden Klassen eine endliche, und zwar, dass sie gleich  zwei ist. [...]  
Eine genaue Untersuchung dieser  Frage verschieben wir auf eine spätere Gelegenheit.
My translation
...If we define a linear manifold to be any embodiment of infinitely many different real numbers, the question appears, in how many (and which) classes the linear manifolds can be divided, if manifolds of the same power are placed in the same class, manifolds of different power in different classes. [...]
Using an inductive method, the presentation of which we will not address here, the theorem suggests itself that the number of classes resulting from this division is finite, and in fact two. [namely, the countable sets and those which can be bijected onto the unit interval]
We defer a closer investigation of this question to a later occasion.

REPLY [3 votes]: Since your question includes the reference-request tag, I would point you towards the following two sources as well (also written by Moore):
Moore, G. H. (1988). The origins of forcing, Logic Colloquium ’86 (Frank R. Drake and John K. Truss,  editors). Studies in Logic and the Foundations of Mathematics, North-Holland, Amsterdam,    143-173.
Moore, G. H. (1989). Towards a history of Cantor’s continuum problem. The history of modern   mathematics, 1, 79-121.
The latter is probably most hopeful for the question you have asked here, though I personally enjoyed the former, which I read through carefully when thinking about some of the issues outlined in this MO post. (I have listed two other sources on Cohen's related work and forcing there.)
Another nice source on CH and its "solution" is:
Yandell, B. (2002). The honors class: Hilbert's problems and their solvers.
I use scare quotes because the Continuum Hypothesis was ultimately shown to be independent from ZF(C), which means, in some sense, you need Z and F (and C) before you can talk about the Hypothesis in its resolved-form.
The Z part probably dates back to around 1908. See:
Zermelo, E. (1908). Untersuchungen über die Grundlagen der Mengenlehre. I. Mathematische  Annalen, 65(2), 261-281.
The F part probably dates back to around 1922.
For other dates and references, I think the Axiom of Choice book Asaf mentions is a nice source.<|endoftext|>
TITLE: When does the categorical definition of a module work?
QUESTION [8 upvotes]: $\DeclareMathOperator{\ab}{Ab}\DeclareMathOperator{\qcoh}{QCoh}$
This entry in the nlab shows that for $A$ a (commutative unital) ring, the category $\mathsf{Mod}_A$ of $A$-modules is equivalent to the category $\ab(\mathsf{CRing}/A)$ of abelian group objects in the slice category of rings over $A$. A module $M$ is associated with the square-zero extension $A\oplus M$, with the familiar 
$$
  (a,m) (b,n) = (a b, a n + b m)
$$
Kahler differentials, and the cotangent complex are easily defined in this setting, and this question asks about geometric intuition. 
Of course, the functor $M\mapsto \mathcal{O}_X \oplus M$ works for quasi-coherent modules on an arbitrary scheme $X$, so we get an embedding $\qcoh(X)\hookrightarrow \ab(\mathsf{Sch}^{op}/X)$.
My question is this: is this an equivalence of categories? If so, does the definition of Kahler differentials at the nlab recover $\Omega^1$? More generally, does this work for stacks?

REPLY [10 votes]: It is not true, but something similar is true. If $X$ is a scheme, the functor you describe is actually $\mathsf{Qcoh}(X) \to \mathsf{Ab}(\mathsf{QAlg}(X)/\mathcal{O}_X)$, where $\mathsf{QAlg}(X)$ denotes the category of quasi-coherent algebras on $X$, and $\mathsf{QAlg}(X)/\mathcal{O}_X$ is the slice category consisting of homomorphisms $A \to \mathcal{O}_X$ (which is anti-equivalent to the category of affine morphisms $Y \to X$ together with a section $X \to Y$; it is not just $\mathsf{Sch}^{\mathrm{op}}/X$!). It is equivalent to the category of non-unital quasi-coherent $\mathcal{O}_X$-algebras (i.e. semigroup objects in $\mathsf{Qcoh}(X)$), by taking the kernel of $A \to \mathcal{O}_X$. Those non-unital qc algebras $B$ for which the addition map $B \times B \to B, (a,b) \mapsto a+b$ is a homomorphism are precisely those with trivial multiplication, i.e. which are just qc modules. And this is the only map which could make $B$ an abelian group object.
Thus, the same proof as in the affine case works. Besides, we don't really use schemes or quasi-coherence here: If $X$ is an arbitrary ringed space, then $\mathsf{Mod}(X) \cong \mathsf{Ab}(\mathsf{Alg}(X)/\mathcal{O}_X)$.
The connection with Kahler differentials is fine: If $M$ is some module on a ringed space $X$, then homomorphisms $\mathcal{O}_X \to \mathcal{O}_X \oplus M$ in $\mathsf{Alg}(X)/\mathcal{O}_X$ correspond 1:1 to derivations $\mathcal{O}_X \to M$.
Finally a remark about the nlab, which I couldn't resist to include here. In my opinion, the statements at the nlab about "The correct definition of the notion of module ..." and "... the correct definition of derivations and Kähler modules" should not be taken too seriously. This is a quite subjective point of view, which may explain some aspects for modules quite elegantly, but not all of them. Besides, there are lots of notions of modules, let alone modules for monads.<|endoftext|>
TITLE: A question about the Axiom of Choice
QUESTION [8 upvotes]: Let AC denote the Axiom of Choice. Let PP denote the so-called "Partition Principle" which states that "If S is a non-empty set and T is a non-empty set of pairwise disjoint subsets
of S, then S can be mapped onto T". It is well known that "AC implies PP" is provable in ZF,
but the question of whether "PP implies AC" is provable in ZF has long been an open problem.
What is the present status of this problem? Has any progress been made on it? Or-on the other
hand-have any models of ZF been constructed in which pp is true and some consequence of AC
-such as the existence of non-measurable sets of real numbers-is false? If "pp implies AC"
could be proved in ZF, this would seem to provide a powerful philosophical argument for
accepting AC. In my opinion, any set theory in which pp can be disproved yields a really
counter-intuitive picture of the "Set-theoretical Universe".

REPLY [12 votes]: To my best knowledge, the Banaschewski-Moore paper from some twenty years ago is pretty much the last recorded progress on the topic.
The two main papers on the subject are the Banaschewski-Moore paper and a paper by Higasikawa, both are from more or less twenty years ago.


Bernhard Banaschewski, Gregory H. Moore, The dual Cantor-Bernstein theorem and the partition principle, Notre Dame J. Formal Logic 31 (3), (1990), 375–381.

Masasi Higasikawa, Partition principles and infinite sums of cardinal numbers. Notre Dame J. Formal Logic 36 (1995), no. 3, 425–434.



You can find a nice diagram of implications in Gregory Moore's book "Zermelo's Axiom of Choice" (which, oddly enough, is the second time I refer to on this site today).
I have some master plan on how to prove its independence from the axiom of choice, but it's a wild and vague dream at the moment which doesn't worth much mentioning except for the fact that I believe, at the moment, that PP does not imply the axiom of choice. For whatever that is worth.
One interesting fact on $\sf PP$ is that it implies the existence of non-measurable sets of real numbers all by itself. $\sf PP$ implies $\sf DC$, as well $\aleph_1\leq2^{\aleph_0}$ and therefore implies the existence of a non-measurable set. But in fact even much weaker versions of $\sf PP$ imply the existence of non-measurable, for example $\sf WPP$ which asserts $A\leq^\ast B\rightarrow B\nless A$, or in other words: if $B$ can be mapped onto $A$ then it cannot have a strictly smaller cardinality.
The reason that $\sf WPP$ implies the existence of a non-measurable set is that $\Bbb R$ can always be mapped onto $[\Bbb R]^\omega$, the set of countably infinite sets of real numbers, and of course can be mapped into that set injectively.
Since in $\sf ZF$ we have $\Bbb R\leq[\Bbb R]^\omega\leq^\ast\Bbb R$, in $\sf ZF+WPP$ we have that $\Bbb R$ is equipotent with $[\Bbb R]^\omega$. Sierpinski proved from this assumption that there exists a non-measurable set.

Sierpinski, W. "L’axiome de M. Zermelo et son rôle dans la théorie des ensembles et l’analyse." Bulletin de l’Académie des Sciences de Cracovie, Classe des Sciences Math., Sér. A (1918), 97-152.<|endoftext|>
TITLE: Infinitely many primes, and Mobius randomness in sparse sets
QUESTION [20 upvotes]: Problem 1: Find a (not extremely artificial) set A of integers so that for every $n$, $|A\cap [n]| \le n^{0.499}$, ($[n]=\{1,2,...,n\}$,) where you can prove that $A$ contains infinitely many primes.
Problem 2: Find a (not extremely artificial) set $A$ of integers so that for every $n$, $ |A\cap [n]| \le n^{0.499}$, where you can prove that
$ \sum \{\mu(k): k \le n, k \in A\} = o(|A \cap [n]).$
Variation
For problem 1 it makes sense to ask not just about infinitely many primes but about a result regarding the density of primes which is of the full strength of the prime number theorem. (In fact, I thought about Problem 2 as a weak form of problem 1 but this is not the case the way I formulated it.) 
Problem 3: Find a (not extremely artificial) set $A$ of integers so that for every $n$, $|A\cap [n]| \le n^{0.499}$, ($[n]=\{1,2,...,n\}$,) where you can prove that the density of primes in $A$ in the interval $[n]$ is $1/\log n+o(1)$. 
Perhaps the best way to formulate and think about Problem 3 is in terms of orthogonality with the Van Mangold function.
Motivations:
This question is motivated by various recent results on Mobius randomness and infinitude of primes in various exotic sets, and also on this question: Why so difficult to prove infinitely many restricted primes?.
(The set of $p_{n^5}$ is extremely artificial and I suppose that a set that can be ordered in a sequence (not necessarily increasing) such that $a_n$ can be provably computed in $poly (\log |a_n|)$ steps is not extremely artificial.) 
Examples:
1) A very natural example is an interval of the form $[n,n+t]$ and here indeed the best known absolute results is when $t=n^{0.535..}$. RH allows $t=\sqrt n \log n$ and it looks that here the $n^{1/2}$ is a viable barrier. (Here I base the info on the paper: A Survey of Results on Primes in Short Intervals by Cem Yalçın YILDIRIM.)
2) There is a result by  Friedlander and Iwaniec that there are infinitely many primes of the form $a^2+b^4$. Here the density is above the square-root barrier, but I don't know if there are any insights regarding improvement to, say, $a^3+b^7$.
3)  There is a result by Elkies about the infinitude of supersingular primes. Those primes are conjectured to be of density $n^{1/2}$ among the first $n$ numbers but provably it is only known that they are of density less or equal $n^{3/4}$ I don't know if there are Elkies-like results that can lead (provably or conjecturally) to sparser sets of primes.
4) There is a "PNT for majority functions" result of Bourgain that implies that there are infinitely many m-digits primes with more than a fraction $c$ of their binary digits are ones, for some $c=1/2+m^{-\rho}$, for some $\rho<1$ Having this for $c=0.9$ will cross the square root barrier. (Update based on Christian's answer: )Eric Naslund used the results about primes in intervals to prove it for $c=0.737..$.

REPLY [11 votes]: Let me first comment on your examples:
1) The record is acually $t=n^{0.525}$, due to Baker, Harman and Pintz.
2) A slightly thinner sequence of this type is $n=a^3+2b^3$, investigated by 
Heath-Brown, and also cubic polynomials in two variables, 
by Heath-Brown and Moroz,
eprints.maths.ox.ac.uk/00000163/01/morozcub2.pdf
Other polynomials in two or more variables, like those that you suggest,
are certainly not known by today's methods.
4) Here is a sketch to obtain primes with a density of 
0.7375 one bits (compare also this paper by Eric Naslund
http://front.math.ucdavis.edu/1211.2455
and his answer on mathoverflow, to a question by Gil Kalai
Primes with more ones than zeroes in their Binary expansion
their-binary-expansion/97345#97345
By Baker-Harman-Pintz one can fix the first 0.475n bits as one.
With $x=2^n$ 
There are $\gg x/\log x$ many primes in the interval [2^n-(2^n)^0.525, 2^n].
For the remaining 0.525n bits about 50% must be 0 and 50% ones.
If the proportion of one bits was smaller, then estimating the tail 
(sum of binomial coefficients) would show the number of primes is too small.
So altogether the asymptotic density of one bits can be as large as
 $(0.475+1/2\,  0.525)n=0.7375n$.
Here is a paper, where a similar method was used to find quadratic non-squares
or even primitive roots, with only few one-bits (less bits than the least non-square or least primitive root might need).
http://arxiv.org/abs/1207.0842
Now an example for a sequence with quite small
 counting function (Problem 1).
Let us first look at large and sparse, but finite sets.
Let $F_n=2^{2^n}+1$ be the $n$-th Fermat number.
Let $D(F_n)$ be the set of its divisors.
Observe that the number of divisors is (by Wigert's bound)
$d(F_n)=O(exp (c \frac{\log F_n}{\log \log F_n})=O(\exp(c' 2^n/n)$.
Let $S(k)=\cup_{n=1}^k D(F_n)\subset [1,2^{2^k}+1]$.
Observe that $|S(k)|= O(k \exp(c' 2^k /k)< (F_k)^{0.499}$.
The set of divisors of the Fermat numbers contains 
also the prime factors (for each $F_n$ at least one new prime number,
as the Fermat numbers are coprime).
Well, to extend this to an infinite set we need to take care of the fact
that with $F_{n+1}$ one possibly also adds smaller elements, so that 
the counting becomes a bit more tricky.
But these added elements are not very small, 
as one can show that the prime factors of $F_n$ are 
at least of size $2^n$.
So maybe one can take the union over a thin subset of
the Fermat numbers, such as $\cup_{n=1}^k D(F_{2^n})$ or similar,
Or one only takes the second smallest element $s_n$ of $D(F_n)$, 
(which must be a prime!).
$S= \cup_{n=1}^{\infty} s_n$.
Note again, that the $s_n$ are distinct and are (at least)
 exponentially increasing, $s_n>2^n$, by properties of the Fermat numbers.
In case anybody thinks this is an artificial sequence, equivalent to writing
down some prime numbers:
well, the prime divisors of sequences such as
$s_n=n^2+1$ or $s_n=\lfloor 2^n/n\rfloor$ 
might not work, as these might be  too dense.<|endoftext|>
TITLE: A question on maps from $\mathbb{Z}/p\mathbb{Z}$ to itself
QUESTION [37 upvotes]: Let $p\geq 3$ be a prime number, and let $u:\mathbb{Z}/p\mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}$ be a map such that, for all $l\in \mathbb{Z}/p\mathbb{Z}$,$l\neq 0$, the map $k\mapsto u(k+l)-u(k)$ is a permutation. Is $u$ a polynomial of degree $2$? 
Note that the property clearly holds when $u$ is a polynomial of degree $2$. Explicit computations seem to show that the converse holds -- that is, the answer is positive -- for $p$ at most $13$.
This is (in a non-obvious way) a special case of this other question, but presumably the statement here is much easier. The question came up quite naturally when thinking about some aspects of complex Hadamard matrices.

REPLY [24 votes]: The question was an open problem on planar functions for many years, which was settled independently in three papers around 1990. See the papers by Gluck, Ronyai and Szonyi, and Hiramine. Elkies' answer is similar to Gluck's proof. The proof by Hiramine avoids Segre's theorem, it is based on quite complicated computations instead.<|endoftext|>
TITLE: Geometry of numbers for three by three matrices?
QUESTION [15 upvotes]: While trying to use Minkowski's theorem to calculate the (left) class number of a noncommutative ring, I ran into the following problem:

What is the volume of the largest symmetric convex subset $S$ of $\mbox{Mat}_3(\mathbb{R})$ such that every matrix in this subset has determinant at most $1$? In particular, is there such a set with volume greater than 885?

The best subset that I have been able to find so far is
$S_1 = \{M\in\mbox{Mat}_3(\mathbb{R})\mid \forall O\in O(3,\mathbb{R}), \mbox{ Tr}(MO) \le 3\}$.
If we apply Gram-Schmidt to the rows of $M$ and use the resulting orthonormal basis for the columns of an orthogonal matrix $O$, then by the AM-GM inequality we can conclude that $\det M \le 1$ from $\mbox{Tr}(MO)\le 3$. Furthermore, if $S$ is any symmetric convex set of matrices of determinant at most $1$ which contains an orthogonal matrix $O$, then every $M\in S$ must satisfy $\mbox{Tr}(MO^{-1}) \le 3$ by considering the tangent hyperplane to the collection of matrices with determinant $1$. Thus $S_1$ is the largest such set which contains all orthogonal matrices.
Unfortunately, it's incredibly difficult to calculate the volume of $S_1$. All I've been able to do so far is to note that it contains the set $S_r$ of matrices such that the sum of the norms of the rows is at most $3$, and that the volume of $S_r$ is $\frac{972\pi^3}{35}\approx 861.089$. If we let $S_c$ be the similar set obtained by considering the columns, then Monte-Carlo integration indicates that the volume of $S_r\cup S_c$ is around $1050$, but I haven't been able to prove this.

What is the volume of $S_1$? If this is too hard, what is the volume of $S_r\cup S_c$?

Motivation
The ring I am dealing with is $\mathbb{Z}\langle a,x\rangle/(a^3 = a-1, x^3 = 3x-1, xa = a(x^2-2)+2-x)$. It is an order in the central simple algebra over $\mathbb{Q}$ with invariants $1/3$ at $3$, $2/3$ at $2$, and $0$ at every other prime. The discriminant of this ring is $-6^6$ (as one can verify by working out the trace form by hand).
Since the associated central simple algebra has nontrivial invariants at $2$ and $3$, and since $\mbox{Nm}(x+1) = -3$ and $(a+xa-1)^3 = 2$, all left ideals of norm at most $4$ are principal. If we tensor this ring up to $\mathbb{R}$ it becomes $\mbox{Mat}_3(\mathbb{R})$, so we can think of it as a lattice in $\mbox{Mat}_3(\mathbb{R})$ of covolume $\sqrt{6^6}$. By Minkowski's bound, every left ideal class contains a representative of norm at most $\left(\frac{2^9\sqrt{6^6}}{\mbox{Vol}(S)}\right)^{1/3}$, so if we can find a set $S$ with volume at least $885$ then we can conclude that this ring has left class number $1$.

REPLY [11 votes]: The volume of $S_1$ is $\frac{16767 \pi^4}{560} \approx 2916.53$.
The idea is to realize that the set $S_1$ can be equivalently described as the nuclear norm (sum of the singular values $\sigma_i$) of $M$ being less than or equal to 3, i.e., $S_1 = \{M : \sum_i \sigma_i(M) \leq 3 \}$. 
This is an orthogonally invariant set, so from the singular value decomposition $M = U \Sigma V^T$ we can do the integral pretty explicitly, using the fact that the Jacobian is given by $(\sigma_1^2-\sigma_2^2) (\sigma_1^2-\sigma_3^2)(\sigma_2^2-\sigma_3^2)$, and the integral is over the polytope $\{(\sigma_1,\sigma_2,\sigma_3) : \sigma_1 \geq \sigma_2 \geq \sigma_3 \geq 0, \, \sigma_1 + \sigma_2 + \sigma_3 \leq 3\}$ (this integral is $\frac{16767}{17920}$). The other factor corresponds to the orthogonal matrices $U,V$, and is $\frac{1}{2^n} Vol(O^n)^2= \frac{1}{2^n} (\prod_{i=1}^n Vol(S^{i-1}))^2 = \frac{1}{2^n} (\prod_{i=1}^n \frac{2 \pi^{i/2}}{\Gamma(i/2)})^2$, which for $n=3$ gives $32 \pi^4$.<|endoftext|>
TITLE: vanishing of vector field in infinite dimensions
QUESTION [6 upvotes]: A simple fact: Given a vector field on a compact manifold with boundary, if the vector field points inward along the boundary, then it must vanish somewhere in the interior. (EDIT: As pointed out in the accepted answer and in a comment, the Euler characteristic must be nonzero for this to be true.)
My question: Is there an analog of this fact in infinite dimensions? Perhaps for Banach manifolds?

REPLY [5 votes]: The simple fact in question is false in any dimension greater than one.
Consider the strip $ \mathbb{R} \times [-\pi/2,\pi/2] \subset \mathbb{R}^2$.  At a point $(x, y)$ take the vector $(-sin(y), cos(y))$.  This does not depend on $x$ so descends to a vector field on the annulus $\mathbb{R}/\mathbb{Z} \times [-\pi/2, \pi/2]$.  It is obviously nowhere vanishing and points inwards at the boundaries.<|endoftext|>
TITLE: Relation between Lee and Yang' s "circle theorem", zeta functions and Weil conjectures?
QUESTION [6 upvotes]: Ruelle mentions ( http://www.ihes.fr/~ruelle/PUBLICATIONS/%5B94%5D.pdf ) Lee and Yang' s "circle theorem", which comes from statistical mechanics and shall have not yet explored connections with zeta functions and Weil conjectures. Does one know now more about that? (Thanks to Alexandre Eremenko for the hint to that interesting article in an other MO thread)

REPLY [5 votes]: I do not know about any connection with Weil's conjectures, and this should be considered
an extended comment rather than an answer.
Actually Ruelle was wrong when he said that this result "remains isolated" in mathematics.
A new proof which fits very well into "manstream" mathematics was given in 1981 in
MR0623156 (83c:82008)
Lieb, Elliott H.; Sokal, Alan D.
A general Lee-Yang theorem for one-component and multicomponent ferromagnets. 
Comm. Math. Phys. 80 (1981), no. 2, 153–179.
It uses a result of Takagi (early 20-s century) on the distribution of zeros of polynomials, which does not seem to be related to physics. Subsequent papers of A. Sokal
further explore this idea.<|endoftext|>
TITLE: How hard is reconstructing a permutation from its differences sequence?
QUESTION [34 upvotes]: My interest in combinatorially motivated computational problems led me to search for simple problems that turn out to be computationally hard. In this pursuit, I came up with a problem which I hope is NP-complete. I searched the literature without finding an equivalent or close problem.
Originally motivated by this post, Can you identify the sum of two permutations in polynomial time? , and my interest in computational properties of permutations.
A differences sequence $a_1, a_2, \ldots a_n$  of a permutation $\pi$ of numbers $1, 2, \ldots n+1$ is formed by finding the difference between every two adjacent numbers in the permutation $\pi$. In other words, $a_i= |\pi(i+1)-\pi(i)|$ for $1 \le i \le n$
For example, sequence $1, 1, 3$  is the differences sequence of permutation $2 3 4 1$.  While, sequences $2, 2, 3$ and $ 3, 1, 2$ are not the differences sequence of any permutation of numbers $1, 2, 3, 4$. 

Is there an efficient algorithm to determine whether a given sequence is the differences sequence for some permutation $\pi$, or is it NP-hard?

We get computationally equivalent problem if we formulate the problem using circular permutations.
Cross posted on TCS SE without an answer. Efficient algorithm for existence of permutation with differences sequence?
EDIT Marzio's nice NP-completeness proof has been published in the Electronic Journal of Combinatorics.

REPLY [11 votes]: I tried to give a formal proof of the NP-completeness of the problem.
For the reduction details see my answer on cstheory.stackexchange.com<|endoftext|>
TITLE: Representation of surface group
QUESTION [9 upvotes]: Is there a faithful representation of a surface group of genus $>2$ into $GL(n,\mathbb{C})$ for some $n$ for which, for each conjugacy class of each embedded loop in the fundamental group, the image in $GL(n,\mathbb{C})$ has a power that is a unipotent element?

REPLY [4 votes]: Misha has said it is an open problem, so I hesitate to say this is an "answer". But, the hypothesis that the rep is faithful means that (after replacing the surface group by a finite index subgroup if necessary) the Zariski closure is connected and not solvable. Hence it has a simple factor, call it $G$. The fact that "all" elements are quasi-unipotent means that the traces (of the adjoint representation of $G$)are all bounded, for all embeddings of the trace field into $\mathbb C$. This is impossible, because that means that the image is finite (this argument is used for example, in Tits' well known paper on the existence of free groups in linear non-solvable by finite groups), and connectedness means that the image is trivial. 
Now your hypotheses do not mean all elements have bounded trace, but only that all EMBEDDED loops do; if there is a Zariski dense subset of these embedded loops, we are still OK, I think.<|endoftext|>
TITLE: Visualization of the real projective plane
QUESTION [5 upvotes]: Consider a closed (compact and without boundary) and non-orientable 2-manifold $M$. By Whitney embedding theorem, one can embed $M$ in $\mathbb{R}^4$. $M$ cannot be embeded in $\mathbb{R}^3$ and just can be immersed in $\mathbb{R}^3$.
We can not imagine a four-dimensional object. Can we use different figures of immersed $M$ in $\mathbb{R}^3$ and obtain imagination of $M$ in $\mathbb{R}^4$ (even a bit)? Can knowing of different figures of immersed $M$ help to imagine embedded $M$ in higher dimension? For example, the real projective plane is a closed and non-orientable 2-manifold, so it can be embedded in $\mathbb{R}^4$ and we cannot imagine this. But different figures of immersed real projective plane in $\mathbb{R}^3$ are within our reach. (following figures, from left to right respectively: cross-cap, Roman's surface, Boy's surface)

REPLY [6 votes]: You could slice the 4D object with a 3D 3-flat that varies in the 4th dimension.
Here are snapshots of such a slicing of the hypercube down a diagonal:
     
     (Image above from Fleischfilm.)

Here is a video of a (much!) more complex object, a "4D quaternion Julia set," being sliced by a moving 3-flat: video link. And here is a still from the animation:

(Image above from Creative Applications Network.)<|endoftext|>
TITLE: Indescribability of cardinals and categoricity of $V_\kappa$
QUESTION [8 upvotes]: If $\kappa$ is an inaccessible cardinal then $V_\kappa$ is a model of $\sf ZFC_2$ ($\sf ZFC$ with a second-order replacement axiom).
If there are many inaccessible cardinals then there are many models of $\sf ZFC_2$, but one can add all sort of $\varphi$ which describe $V_\kappa$ completely. For example if $\varphi$ is "there is no inaccessible cardinal" then $\sf ZFC_2+\varphi$ is only satisfied by $V_\kappa$ if $\kappa$ is the least inaccessible cardinal. We can continue and state that there is exactly one, or two, or $\alpha$ inaccessible cardinals for every "small enough" $\alpha$. We can continue by adding more and more information (e.g. there is no inaccessible which is the limit of inaccessibles; there is only one inaccessible which has a stationary set of inaccessibles; and so on).
Let us say that $\varphi$ in $n$-order logic is a categorizer for $\kappa$ if $V_\kappa$ is the unique model of $\sf ZFC_2+\varphi$. We say that $\kappa$ is $\Pi^n_m$-categorical (or $\Sigma^n_m$-categorical) if there is a $\Pi^n_m$ ($\Sigma^n_m$) sentence $\varphi$ which is a categorizer for $\kappa$.
On the other hand, we say that $\kappa$ is $\Pi^n_m$-indescribable if for every $R\subseteq V_\kappa$ and a $\Pi^n_m$ sentence $\psi$ such that $\langle V_\kappa,\in,R\rangle\models\psi$ there is some $\alpha<\kappa$ such that $\langle V_\alpha,\in,R\cap V_\alpha\rangle\models\psi$. (Similarly, of course, we define $\Sigma^n_m$-indescribable cardinals.)
Note that inaccessible cardinals are $\Pi^1_0$-indescribable, but the least inaccessible is $\Pi^0_n$-categorical for some $n$ (because the statement "there are no inaccessible cardinals is a first-order statement).

Question: Is there a [deep?] connection between the two notions?

REPLY [9 votes]: Categoricity should perhaps be conceived of not as a large
cardinal notion, but rather as an anti-large cardinal notion,
since most large cardinal concepts express some degree of
reflection, which is the opposite of categoricity.
For example, if $\kappa$ is $\Pi^n_m$-categorical, then clearly it
is not $\Pi^n_m$-indescribable. This is one connection between
your concepts.
But let me argue that there cannot be an equivalence here between
levels of categoricity and failure of indescribability. First, let's observe the following.
Lemma. Every $\Pi^n_m$-categorical cardinal is
$\Delta_2$-definable in set theory.
Proof. If $\kappa$ is $\Pi^n_m$-categorical, using formula
$\varphi$, then $\kappa$ is the unique cardinal such that
$V_\kappa$ satisfies $\varphi$, and this is a first-order
expressible fact $\varphi^\ast$ about $V_{\kappa+n}$ (since the
second order quantifiers of $\varphi$ are interpreted as intended
as first-order quantifiers inside $V_{\kappa+n}$. So $\kappa$ is
the unique ordinal such that $V_{\kappa+n}\models\varphi^\ast$.
This is $\Pi_2$ expressible as follows: a given ordinal $\xi$ is
the $\kappa$ we are talking about if and only if $\forall Z\ (Z=V_{\xi+n}\to Z\models\varphi^\ast)$. This is a $\Pi_2$
assertion, since saying that $Z=V_{\xi+n}$ has complexity $\Pi_1$---one must say that $Z$ computes the power sets of its members
correctly---and the latter part has all quantifiers bounded by $Z$.
Similarly, $\xi=\kappa$ if and only if $\exists Z\ Z=V_{\xi+n}$
and $Z\models\varphi^\ast$, which is a $\Sigma_2$ formulation. QED
Now, recall that a cardinal $\delta$ is $\Sigma_2$-correct, if
$V_\delta\prec_{\Sigma_2} V$. The reflection theorem shows that
there is a closed unbounded class of $\Sigma_2$-correct cardinals.
A $\Sigma_2$-reflecting cardinal is a regular $\Sigma_2$-correct
cardinal, and these have the consistency strength strictly weaker
than a Mahlo cardinal. Meanwhile, every strong cardinal, every
strongly unfoldable cardinal (a transfinite continuation of the
totally indescribable cardinals), every supercompact cardinal is
$\Sigma_2$-reflecting.
Theorem. Every $\Pi^n_m$-categorical cardinal is smaller than
every $\Sigma_2$-correct ordinal.
Proof. If $\kappa$ is $\Pi^n_m$-categorical, then it is
$\Sigma_2$-definable by some formula $\psi$. In particular, the
assertion "$\exists\kappa\ \psi(\kappa)$" is true in $V$. Since
this is a $\Sigma_2$-assertion, it follows that if $\delta$ is
$\Sigma_2$-correct, then $V_\delta$ will agree that there is such
a $\kappa$. Since $\kappa$ is unique with $\psi(\kappa)$, it
follows that $\kappa\lt\delta$. QED
So from a large cardinal perspective, the categorical cardinals must lie low in the hierarchy of
ordinals.
Notice that since there are only countably many formulas,
and in set theory we have a uniformly expressible account of
whether a cardinal is $\Pi^n_m$-categorical by a given formula, it follows that there
will be only countably many $\Pi^n_m$-categorical cardinals. True
reflection does not even begin until you are above them all. But
meanwhile, the indescribability and non-indescribability phenomena
are unbounded in the ordinals.
Lastly, let me mention that the strongly unfoldable cardinals are best understood as a transfinite extension of the indescribability notions of indescribable cardinals. That is, they in effect replace $\Pi^n_m$ with $\Pi^\alpha_m$ for transfinite $\alpha$. But at this level (and even at the finite levels in my opinion), it is better to characterize the property in terms of embeddings than higher-order logic.<|endoftext|>
TITLE: Prequantization and Hilbert space
QUESTION [6 upvotes]: In the definition of pre-quantization on a symplectic manifold $(M,\omega)$, we represent a function $f\in C^{\infty}(M)$(with Lie algebra structure) to $\hat{f}$ in the Hilbert space $L^2(M,L,\mu)$ associated to $M$. But we always assume that the Hilbert space should not be too big. But why do we assume this?. Also we consider only irreducible representations for definition of prequantization ? 
PS: see page 1 , line 9 of this reference

REPLY [4 votes]: I'll add a "general abstract" reason to the "phenomenological reasons".
Phenomenologically, as others pointed out here, the processes of cutting down the space of sections of the prequantum line bundle on phase space to only those which respect a polarization corresponds precisely to wave functions depending only on canonical coordinates and not on canonical momenta.
This is known since the 1930s to yield the correct description of observable physics, and that "proves" it in as far as we regard geometric quantization as a theory of physics.
But of course one may ask why this step is natural just from the point of view of mathematics? Is there maybe a deeper mathematical reason for this?
And, yes, that turns out to be the case. In the modern understanding, geometric quantization in the case that there exists a Kähler polarization is really the push-forward of generalized cohomology of the prequantum line bundle in complex topological K-theory, hence the (de-categorified) index of the spin^c-Dirac operator twisted by the prequantum bundle.
For pointers to details of how this works see on the nLab the page geometric quantization by push-forward.
And from this perspective, the fact that geometric quantization "cuts down" the space of sections in the hallmark of index theory: from a space of sections of some bundle we pass to just the graded kernel of some graded differential operator on these. 
By another way: some references relating geometric quantization to deformation quantization are listed here.<|endoftext|>
TITLE: Where was it first stated that there are no 4-transitive finite groups other than symmetric, alternating and Mathieu groups?
QUESTION [8 upvotes]: It seems to be well-known that the six-transitive finite groups are the symmetric and alternating groups, and the only other four-transitive finite groups are the Mathieu groups (the statement can be found in Cameron's 1999 "Permutation Groups", p 110), but I can't seem to figure out where the result was first stated (Cameron is not much help, though he does give a vague sketch of how one might go about proving this result).

REPLY [5 votes]: Peter Cameron has an article "Finite permutation groups and finite simple groups. He states that Wielandt showed that any 6-transitve group is alternating or symmetric, given the Schreier's conjecture is true. Schreier's conjecture is that the outer automorphism group of a finite simple group is solvable, and the truth of this is a consequence of the classification.
I found Cameron's article by googling on "Wielandt Schreier". Wielandt's result was published as H. Wielandt, "Uber den Transitivitatsgrad von Permutationsgruppen", Math. Z. 74 (1960), 297-298.<|endoftext|>
TITLE: $\Pi$, $\Sigma$, and identity types without $\eta$ in comprehension categories
QUESTION [7 upvotes]: In comprehension categories, dependent sums are defined as a choice of left adjoints for all reindexing functors along display maps, satisfying a Beck-Chevalley condition.  Dependent products are right adjoints of the same functors, and identity types are left adjoints of reindexing functors along diagonal maps.
These definitions are too strong, however, when one wants to model type formers without the $\eta$ rule.
In categories with attributes (i.e. full split comprehension categories), one can replicate the usual syntactic definitions to get suitably weak notions of $\Pi$, $\Sigma$ and identity types (as Hofmann does), but I don't know how to generalise this approach.
Has someone given a definition of type formers without $\eta$ for general comprehension categories? I am mostly interested in identity types, but it would be nice to find a pattern that works for all type formers.
I suspect that somehow "weakening" the adjunctions should do the trick, but I'm not sure how to make this precise.

REPLY [4 votes]: I'm not sure exactly what you're asking, but the general pattern for "positive" types is to assert that every "structured" display map over the type has a specified structured section.  This matches the induction principle of a positive type former in type theory.  One then needs to require a sort of "pullback-stability" condition as well in order to model substitution in type theory.  A forthcoming paper of Lumsdaine and Warren will describe a general process of "strictification" of "weakly stable" structure in a comprehension category into "strictly coherent" structure in a fully strict structure.  (Edit: this paper is now available, as Lumsdaine–Warren 2015 The local universes construction…)
For identity types, the categorical structure is essentially a well-behaved weak factorization system in which the display maps are (or generate) the right class: see for instance this paper.
Edit: One can be more precise about the notion of "structured section", for instance to say that a "strongly homotopy initial object" in a display-map category is an object $X$ such that every display map with codomain $X$ has a section.  Since all ordinary categorical left universal properties can be reformulated as initial objects in some category, this gives a way to translate them into an $\eta$-free version as long as the category in question inherits a natural notion of display map (which is generally induced directly from the underlying category of types/contexts).  Similarly, a strongly homotopy initial object is "weakly preserved" by a functor if its image under that functor is another strongly homotopy initial object (not necessarily the "chosen" such object in the codomain of the functor, if there is one).  Weak stability then means that pullback functors weakly preserve the relevant SHIOs.  I think this is as close to a "general pattern" as you can get.<|endoftext|>
TITLE: Is an affine fibration over an affine space necessarily trivial?
QUESTION [37 upvotes]: Let $X$ be an algebraic variety over an alg. closed field with zero char. and let $f:X\to \mathbb{A}^n$ be a smooth surjective morphism, such that all fibers (at closed points) are isomorphic to $\mathbb{A}^m$. Does it follow that $X\cong \mathbb{A}^{n+m}$? 
If not, is it true with some additional assumptions? I know that every vector bundle on $\mathbb{A}^n$ is trivial (this is Serre's problem, right?) and that it is even enough to ask it to be locally trivial in etale toplogy. Is every "affine bundle" on $\mathbb{A}^n$ is trivial? I guess it is. The main question is about a general "fibration".

REPLY [27 votes]: I feel like I already answered this question, but it might have been a variant with fibers isomorphic to tori.  Let the base $B$ be $\mathbb{A}^2$ with coordinates $s$ and $t$.  Begin with $B\times \mathbb{P}^3$, where homogeneous coordinates on $\mathbb{P}^3$ are $[x,y,z,w]$.  Let $S$ be the Cartier divisor in $B\times \mathbb{P}^3$ with defining equation $yz-(sx+tw)^2=0$.  Let $L$ be the Cartier divisor in $S$ with defining equation $y+z-2(sx+tw)=0$.  Let $U$ be the complement of $L$ in $S$.  Then $U$ is affine, the morphism $U\to B$ is smooth, and the fiber over every point other than $(0,0)$ is isomorphic to $\mathbb{A}^2$.  Of course the fiber over $(0,0)$ is isomorphic to a disjoint union of two copies of $\mathbb{A}^2$.  Thus, define $V\subset U$ to be the open subscheme obtained by removing one of these two copies of $\mathbb{A}^2$, i.e., remove the closed subscheme with defining equations $s=t=z=0$.  Then $V$ is quasi-affine, and the affine hull is $U$; this follows by Hartog's theorem / the Riemann extension theorem / S2 extension.  Therefore $V$ is not isomorphic to an affine space.  However, the projection $V\to B$ has all of the requisite properties.    
Edit.  The older answer I mention above was similar, but a little bit different.  That answer was in response to the following similar question, When is a holomorphic submersion with isomorphic fibers locally trivial?.<|endoftext|>
TITLE: Fold-and-cut problem in three dimensions
QUESTION [11 upvotes]: The fold-and-cut theory states that "Any shape with straight sides can be cut from a single (idealized) sheet of paper by folding it flat and making a single straight complete cut. Such shapes include polygons"
My question

Question 1 What is the time complexity for finding such a crease pattern?
Question 2 How do we express; $\chi(P)$ ie the minimum number of  folds required to make a crease pattern from which the polygon $P$ can be cut out?
  Question 3 Does the fold-and-cut-theorem have a three dimensional analog?

Would also request for a reference to the proof.

REPLY [5 votes]: In response to @redhound.
Now updated with step-by-step instructions online:

         


         

The turtle nearly fully folded.
The black lines all align.


         


         

The hole remaining after 1-cut.<|endoftext|>
TITLE: Relationship between basic sets and attractors
QUESTION [9 upvotes]: Definition: Let be $f:M\to M$ a diffeomorphism of a compact manifold. We say that $A\subset M$ is an attractor  when there exists a neighborhood $U\supset A$ such  that $f( \overline{U})\subset int (U)$  and 
$$
A=\bigcap_{n\geq 0}f^n(U)
$$
$U$ is called an  basin of attraction of $f$.
Theorem: Let $M$ be a compact  manifold and let $f:M→M$ a diffeomorphism. If 
$\overline{Per f}$ has hyperbolic structure, then 
can be partitioned into a finite number of compact, invariant and topologically transitive sets, called basic sets:
    $$\overline{Per(f)}=⋃_{i=1}^{m}Λ_i$$    
Definition: Le be $\Lambda_i$ a basic set of $f$, then we define $$W^s(\Lambda_i)=\{x: d(f^n(x), \Lambda_i)\to 0,~n\to\infty   \}$$ 
Question: Supose that the chain recurrent set of $f$, $\mathcal{R}(f)$ has hyperbolic structure,  I would like to see that if $int (W^s(\Lambda_i))\neq \varnothing$ then the basic set $\Lambda_i$ is an attractor.

REPLY [2 votes]: Ok, here is the answer assuming that the basic set $\Lambda$ has local product structure.
Step 1. We will prove that $\forall x\in\Lambda$ $W^u(x)\subset \Lambda$. Then any point $y$ sufficiently close to $\Lambda$ belongs to a local stable manifold of some point in $\Lambda$, hence $\Lambda$ is an attractor.
Step 2. So we need to prove that local unstable of every point in $\Lambda$ is in it. Basic set is transitive and closed; local unstable manifolds vary continuously. Therefore it sufficient to show that local unstable manifold of a point $z\in\Lambda$ is in $\Lambda$, where $z$ is a point whose tranjectory is dense.
Step 3. It is easy to show that 
$$
W^s(\Lambda)=\bigcup_{x\in\Lambda}W^s(x)
$$
See for example Proposition 9.1 of Shub's book. (Here local product structure is essential.) 
Step 4. Therefore if $p$ is in the interior of the above set. Then $\exists q\in \Lambda$ such that $p\in W^s(q)$. Take a transvesral to $W^s(q)$ that passes through $p$ and apply the $\lambda$-lemma to it to conclude that $W^u(q)$ is in $W^s(\Lambda)$. Hence $W^u(q)\subset W^s(\Lambda)$. Note that the same reasoning applies to points sufficiently close to $q$. In particular to some iterate of $z$. Hence we are done.<|endoftext|>
TITLE: Frobenius density theorem
QUESTION [14 upvotes]: As mentioned by @MichaelZieve in his comment re Quadratic residue, Chebotarev's density theorem was preceded by an allegedly much easier theorem of Frobenius (Mike Zieve is certainly not the only one to mention that the Frobenius theorem is much easier than Chebotarev) -- the difference between the two theorems is that Frobenius talks of conjugacy in $S_n,$ while Chebotarev of conjugacy in the Galois group itself. Does anyone know what the "easier" argument is? Is there a good reference?
EDIT If you look at the theorem in Janusz's book it proves the theorem for the Dirichlet density. Does proving it for the natural density require the full Chebotarev machine?

REPLY [7 votes]: Regarding the edit: I think natural density should still be easier for Frobenius. Let $G$ be $\mathrm{Gal}(F/\mathbb{Q})$ and let $c$ be a class function  $G \to \mathbb{C}$.
Cebatarov with natural density is the statement
$$ \sum_{p \leq N} c(\mathrm{Frob}(p)) \sim \left( \sum_{g \in G} c(g) \right) \cdot \pi(N) $$
Frobenius is the special case of a class function where $c(g)=c(h)$ whenever $g$ and $h$ generate the same cyclic subgroup.
In both cases, the set of such class functions is a vector space, so it is enough to prove this for a spanning set of class functions. In the Cebatarov case, the spanning set is inductions of one dimensional characters of subgroups $H$ of $G$. In the Frobenius case, it is enough to induct only trivial characters.
Let $H$ be a subgroup of $G$, let $L$ be the fixed field and let $\chi: H \to \mathbb{C}^{\ast}$ be a character. Let $K \subset H$ be the kernel of $\chi$ and let $M$ be the fixed field of $K$, so $M/L$ is an abelian extension. 
Then I believe (this is the part I am not sure of) that Cebatarov for $\mathrm{Ind}_H^G \chi$ should reduce to showing that $L(M/L, \chi)(s)$ has no zeroes or poles on $\mathrm{Re}(s)=1$, except for a simple pole at $s=1$ for $\chi$ trivial. (For Dirichlet density, we only need to show this at $s=1$.) Frobenius amounts proving this only in the case that $\chi$ is trivial. In this case, we are trying to show that $\zeta_L(s)$ has no zeroes or poles on $\mathrm{Re}(s)=1$, except a simple pole at $s=1$.
So the key question is, are the easiest proofs that $\zeta_L(s)$ has no zeroes or poles on the critical line easier than the corresponding proofs for $L(M/L, \chi)(s)$? I think the answer is yes. The proofs that I am aware involve two main parts
(1) Show that the function in question has no poles (other than the stated one at $s=1$) on $\mathrm{Re}(s)>0$.
(2) Write down some clever product of various $L$-functions which is manifestly non-vanishing, and conclude that none of the factors can vanish. This argument requires (1) to have been already done, to rule out the possibility that a zero and pole cancel.
Now, I don't know whether (2) gets significantly easier when $\chi$ is trivial. But (1) definitely does! The fact that $\zeta_H(s)$ extends to $\{ \mathrm{Re}(s)>0,\ s \neq 1 \}$ follows from the same sort of computations that lead to the class number formula for number fields. For $L$ functions, the way I know how to do it requires first proving that the character $\chi$ on $\mathcal{O}_L$ is periodic, so that we can group together sums over a complete period and guarantee that they converge. This is equivalent to Artin reciprocity, which as far as I know is as hard as all of class field theory.
Chebatarov's original proof (see the last section of Lenstra and Stevenhagen) avoids this. To my limited understanding of this, Cebatarov first proves the result when $M$ is a cyclotomic extension of $L$, where Artin reciprocity is easy, and then some how cleverly reduces to this case. But, even so, it seems to me that it must be easier when we work with just trivial characters.<|endoftext|>
TITLE: Three-dimensional compact Kähler manifolds
QUESTION [20 upvotes]: Consider the problem of trying to identify which $n$-dimensional compact complex manifolds can be endowed with a Kähler metric.
$\underline{n = 1}:$ Any hermitian metric on a Riemann surface is a Kähler metric as the Kähler form is a two-form, and all two-forms are closed. (This is also true of non-compact Riemann surfaces).
$\underline{n = 2}:$ A necessary and sufficient condition for a complex surface to admit a Kähler metric is that its odd Betti numbers are even. 
The condition that odd Betti numbers are even is necessary for a complex manifold to admit a Kähler metric, but not sufficient for $n \geq 3$. To see it is not sufficient, consider the example of Hironaka which gives a deformation of three-dimensional Kähler manifolds to a non-Kähler manifold. The Betti numbers are invariant under diffeomorphisms, so the central fibre of the deformation has odd Betti numbers even (as it is diffeomorphic to a Kähler manifold) but it is not itself Kähler.
In general, for $n \geq 3$, it is not easy to determine when a compact complex manifold can be equipped with a Kähler metric. In the simplest of these cases, $n = 3$, are we any closer to solving this problem?

What conditions (necessary or sufficient) are there for a three-dimensional compact complex manifold to admit a Kähler metric? How far are we from a single necessary and sufficient condition (like we have for $n = 1$ and $2$)?

Of course I'm interested in results which apply for all $n$, but I'm guessing that there are some results specifically for $n = 3$.

REPLY [19 votes]: The main obstruction to existence of Kahler metric (in addition to Lefschetz
SL(2)-action and Riemann-Hodge relations in cohomology)
is homotopy formality: the cohomology ring of a Kahler manifold is related to its de Rham algebra by a chain of homomorphisms of differential graded algebras inducing isomorphisms on cohomology. This is proven by Deligne-Griffiths-Morgan-Sullivan in 1970-ies.
This is a very strong topological condition; for instance, no nilmanifold (except torus) is homotopy formal. There are symplectic nilmanifolds satisfying hard Lefschetz and the rest of Riemann-Hodge conditions for cohomology. 
Another obstruction is existence of a positive, exact current. As shown by Peternell, all non-Kahler Moishezon manifolds admit a positive, exact (n-1,n-1)-current, hence they are not Kahler. However, Moishezon manifolds are homotopy formal ([DGMS]), and often satisfy the Riemann-Hodge. This argument
is also used to prove that twistor spaces of compact Riemannian 4-manifolds are not Kahler, except CP^3 and flag space (Hitchin).
The sufficient condition in this direction is obtained by Harvey-Lawson: they proved that a manifold is Kahler if and only if it does not admit an exact (2n-2)-current with positive, non-zero (n-1, n-1)-part. 
Finally, Izu Vaisman has shown that any compact locally conformally Kahler manifold (a manifold with Kahler metric taking values in a non-trivial 1-dimensional local system) is non-Kahler. 
Also, a complex surface is Kahler if and only if its $b_1$ is even. This was known from Kodaira classification of surfaces, and the direct proof was obtained in late 1990-ies by Buchdahl and Lamari using the Harvey-Lawson criterion.<|endoftext|>
TITLE: Is SL_n/S(GL_k x GL_n-k) symmetric?
QUESTION [5 upvotes]: Background: a symmetric variety is a homogeneous space $G/H$ associated to an involution $\theta$ of a semisimple algebraic group $G$ and $\{g | \theta(g) = g\} = G^\theta \subset H \subset N_G(G^\theta) = \{h| hG^\theta h^{-1} = G^\theta \}$.
A spherical variety is a homogeneous space $G/H$ which contains an open orbit under the action of a Borel subgroup $B$. It is known that symmetric varieties give examples of spherical varieties.
Let $G = SL_n(\mathbb{C})$ and $H = S(GL_k \times GL_{n-k}) = \{(g,g') | \det g \det g' = 1 \}$. It is known that $G/H$ is spherical.  $G/H$ is a non compact affine variety. In fact if I write $Gr_{k,n}(\mathbb{C}) = G/P$ then there is a Levi decomposition $P = H U$ so that $G/H$ is a $U$-fiber bundle over $Gr_{k,n}(\mathbb{C})$.
QUESTION
Is $G/H$ symmetric? If so, what is the associated involution?
The answer to the first question seems to be yes (at least for small values of $k$) as this particular homogeneous spaces makes numerous appearances in http://arxiv.org/pdf/1012.4171.pdf (see for example figures 3,4,5 starting on page 15)
In fact the cited paper even gives a name to the involution (AIII) referencing a list of the possible involutions of a simple algebraic group. My confusion is that this involution seems to be a composition of usual transposition (swap across the diagonal) with a swap along the anti-diagonal and this does not seem to exhibit $G/H$ as a symmetric space.
Further, I thought the rank of a symmetric variety was supposed to agree with its rank as a spherical variety. In my particular example, the rank of $G/H$ seems to be 1 as a spherical variety while the paper seems to say that the rank of $G/H$ as a symmetric variety is in general greater than 1. I'm likely missing a basic point and if it can be pointed out to me I would greatly appreciate it.
UPDATE: José Figueroa-O'Farrill answered the question over the real numbers and pointed out that the rank is $\min\{k,n-k\}$ and the involution can be taken to be conjugating by the matrix $\left(\begin{array}{cc} -I_k & \\ & I_{n-k} \end{array}\right)$.  The negative of this also works as Lev Soukhanov points out below.
The same involution seems to work over $\mathbb{C}$ but I'm not certain the statement about ranks still holds.

REPLY [2 votes]: It seems that you just take the involution which is the conjugation with the diagonal matrix with k-dimensional subspace with eigenvalue 1 and n-k dimensional subspace with eigenvalue -1.
Dunno about ranks at all.<|endoftext|>
TITLE: Is there any large cardinal beyond Kunen inconsistency?
QUESTION [24 upvotes]: First fix the following notations:‎
‎
‎$‎AF:=‎$ ‎The ‎axiom ‎of ‎foundation‎
‎
‎$‎ZFC^{-}‎:=‎ZFC‎\setminus ‎\left\lbrace AF ‎\right\rbrace ‎‎‎‎‎$‎‎‎‎‎
‎
‎$‎G:=‎$ ‎The ‎proper ‎class ‎of ‎all ‎sets‎
‎
‎$‎V:=‎$ ‎The ‎proper ‎class ‎of ‎Von ‎neumann's ‎cumulative ‎hierarchy‎
‎
‎$‎L:=‎$ ‎The ‎proper ‎class ‎of ‎Godel's cumulative ‎hierarchy‎
‎
‎$‎G=V:~‎‎\forall x‎~‎\exists ‎y~(ord(y) ‎‎\wedge ‎``x\in ‎V_{y}")‎$ ‎‎
‎
‎$‎G=L:~‎‎\forall x‎~‎\exists ‎y~(ord(y) ‎‎\wedge ‎``x\in L_{y}")‎$‎‎ ‎‎
‎‎
Almost ‎all ‎of ‎‎$‎ZFC‎$ ‎axioms ‎have a‎ ‎same "‎nature" in some sense. ‎They ‎are ‎"generating" ‎new ‎sets ‎which form the world of mathematical objects ‎$‎G‎$‎. ‎In ‎other ‎words‎ ‎they are ‎complicating ‎‎our ‎mathematical chess by increasing its playable and legitimated nuts. ‎But ‎‎$‎AF‎$‎ ‎is ‎an ‎exception ‎in ‎‎$‎ZFC‎$. ‎It ‎is "simplifying" ‎our mathematical ‎world ‎by ‎removing ‎some ‎sets which innocently are accused to be ‎"ill ‎founded". Even ‎$‎AF‎$‎ is regulating ‎$‎G‎$ ‎by ‎‎$‎V‎$ ‎and ‎says ‎‎$‎G=V‎$. So ‎‎it‎‎ ‎is "miniaturizing" the "real" size of ‎$‎G‎$ ‎by the ‎‎"very ‎small" ‎cumulative ‎hierarchy ‎‎$‎V‎$ ‎as ‎same ‎as ‎the ‎assumption ‎of ‎constructibility ‎axiom ‎‎$‎G=L‎$. ‎In ‎fact‎ ‎"minimizing" the size of mathematical universe ‎is ontological ‎"nature" ‎of ‎all ‎constructibilty ‎kind ‎of ‎axioms ‎like ‎‎$‎G=W‎$ ‎which ‎‎$‎W‎$ ‎is ‎an ‎arbitrary ‎cumulative ‎hierarchy. ‎But ‎in ‎the ‎opposite direction ‎the ‎large ‎cardinal ‎axioms ‎says a‎ ‎different ‎thing ‎about ‎‎$‎G‎$‎. We know that any large cardinal axiom stronger than "‎$‎‎0^{\sharp}$ ‎exists" ‎implies ‎‎$‎G\neq L‎$‎. This illustrates the "nature" of large cardinal axioms. They implicitly say the universe of mathematical objects is too big and is "not" reachable by cumulative hierarchies. So it is obvious that any constructibility kind axiom such as ‎$‎AF‎$, ‎imposes a‎ ‎limitation ‎on ‎the ‎height ‎of ‎large ‎cardinal ‎tree‎. ‎One ‎of ‎the‎se serious limitations is Kunen inconsistency theorem in ‎$‎‎ZFC^{-}+AF$‎.‎
‎
Theorem (Kunen inconsistency) There is no non trivial elementary embedding ‎$‎‎j:\langle V,\in\rangle\longrightarrow ‎\langle V,\in\rangle $ ‎(or equivalently ‎$‎‎j:\langle G,\in\rangle\longrightarrow ‎\langle G,\in\rangle$‎)‎
‎‎
The ‎proof ‎has ‎two ‎main ‎steps as follows:‎
Step (1): ‎By ‎induction ‎on ‎Von ‎neumann's ‎"rank" ‎in ‎‎$‎V‎$‎ one can prove any non trivial elementary embedding  ‎$‎‎j:\langle V,\in\rangle\longrightarrow ‎\langle V,\in\rangle‎$ has a critical point ‎$‎‎‎\kappa‎$‎ on ‎$‎Ord‎$‎.
‎
‎Step ‎(2): ‎By ‎iterating ‎‎$‎j‎$ ‎on this critical point one can find an ordinal ‎$‎‎\alpha‎$ ‎such ‎that  ‎‎$‎‎j[‎\alpha‎]=‎\lbrace‎ j(‎\beta‎)~|~‎\beta ‎\in ‎‎\alpha ‎\rbrace‎‎‎ \notin V (=G)‎$ ‎which ‎is a‎ ‎contradiction.‎
‎
Now ‎in ‎the ‎absence ‎of ‎‎$‎AF‎$ ‎we ‎must ‎notice ‎that ‎the Kunen inconsistency ‎theorem ‎splits ‎into ‎two ‎distinct ‎statements and the orginal proof fails in both of them‎‎‎‎. ‎
‎
Statement (1):(Strong version of Kunen inconsistency) There is no non trivial elementary embedding ‎‎‎$‎‎j:\langle G,\in\rangle‎\longrightarrow ‎\langle G,\in\rangle$‎.‎ 
Statement (2):(Weak version of Kunen inconsistency) There is no non trivial elementary embedding ‎‎‎$‎‎j:\langle V,\in\rangle\longrightarrow \langle V,\in\rangle$‎.
‎‎
In statement ‎(1)‎, step (1) collapses because without ‎$‎AF‎$‎ we have not a "rank notion" on ‎$‎G‎$ ‎and ‎the ‎induction ‎makes ‎no ‎sense. So we can not find any critical point on ‎$‎Ord‎$ ‎for ‎‎$‎j‎$ ‎by "‎this ‎method". ‎
‎
In statement ‎(2)‎, step (2) fails because without ‎$‎AF‎$‎ we don't know‎ $‎‎G=V$ and so $j[‎\alpha‎]\notin V‎$ ‎is ‎not a‎ ‎contradiction.‎
‎
‎But ‎it is clear that ‎in ‎‎$‎ZFC^{-}‎$ ‎the ‎original ‎proof ‎of ‎Kunen ‎inconsistency theorem ‎works ‎for ‎both of the ‎following ‎propositions:‎
‎
Proposition (1): There is no elementary embedding ‎‎‎$‎‎j:\langle G,\in\rangle\longrightarrow ‎\langle G,\in\rangle $‎ with a critical point on ‎$‎Ord‎$‎.
‎Proposition (2): Every non trivial elementary embedding ‎‎‎$‎‎j:\langle V,\in\rangle\longrightarrow \langle V,\in\rangle$‎ has a critical point on ‎$‎Ord‎$‎.
 ‎‎
Now ‎the ‎main ‎questions ‎are‎: ‎‎
‎‎Question ‎(1): ‎Is ‎the ‎statement ‎"‎There is a non trivial elementary embedding $‎‎j:\langle V,\in\rangle\longrightarrow ‎\langle V,\in\rangle$‎" an acceptable large cardinal axiom in ‎the absence of ‎$‎AF‎$($‎G=V‎$‎)‎‎? 
What about other statements by replacing ‎$‎V‎$ ‎with a‎n ‎arbitrary ‎cumulative ‎hierarchy ‎‎$‎W‎$‎?(In this case don't limit the definition of a cumulative hierarchy by condition ‎$‎‎W_{‎\alpha +1‎}\subseteq P(W_{‎\alpha‎})$‎)‎
‎
Note that such statements are very similar to ‎the ‎statment "‎‎$‎‎0^{\sharp}$ exists" that ‎is ‎equivalent ‎to ‎‎existence of a non trivial elementary embedding $‎‎j:\langle L,\in\rangle\longrightarrow ‎\langle L,\in\rangle$ ‎and ‎could ‎be an ‎"acceptable" ‎large ‎cardinal ‎axiom ‎in ‎the ‎"absence" ‎of ‎‎$‎‎G=L$‎‎. So if the answer of the question (1) be positive, ‎we can go "beyond" weak version of Kunen inconsistency by removing ‎$‎AF‎$ ‎from ‎$‎ZFC‎$‎  and so we can find a family of "Reinhardt shape" cardinals correspond to any cumulative hierarchy ‎$‎W‎$ by a similar argument to proposition (2) dependent on "good behavior" of "rank notion" in ‎$‎W‎$‎‎.
‎
Question ‎(2): ‎Is ‎‎$‎AF‎$ ‎necessary to prove "strong" version of ‎Kunen ‎inconsistency ‎theorem? ‎In ‎the ‎other ‎words ‎is ‎the‎ ‎statement ‎"$Con(ZFC)‎\longrightarrow Con(ZFC^{-}+ ‎\exists‎$‎ a‎ ‎non ‎trivial ‎elementary ‎embedding ‎‎$‎‎j:\langle G,\in\rangle‎\longrightarrow \langle G,\in\rangle)‎‎$"‎‎ ‎true?‎ 
It seems to go beyond Kunen inconsistency it is not necessary to remove‎ $‎AC‎$ ‎which possibly "harms" our powerful tools and changes the "natural" behavior of objects.It simply suffices that one omit ‎$‎AF‎$‎'s limit on largeness of "Cantor's heaven" and his "set theoretic intuition". َAnyway whole of the set theory is not studying $L$, $V$ or any other cumulative hierarchy and there are many object "out" of these realms. For example ‎without limitation of ‎‎$‎G=L‎$ ‎we ‎can see more large cardinals that are "invisible" in small "scope" of $L$.‎In the same way without limitation of $AF$ we ‎can probably discover ‎more stars in the mathematical universe out of scope of $V$. Furthermore we can produce more interesting models and universes and so we can play an extremely ‎exciting ‎mathematical ‎chess beyond inconsistency, beyond imagination!

REPLY [15 votes]: The answer to question 2 is yes, and one can even have nontrivial automorphisms. For example, the theory $\mathit{ZFC}^-+{}$“there are two urelements (i.e., sets $x$ satisfying $x=\{x\}$) and the whole universe is obtained from them by iterated power set” is consistent relative to ZFC, and one can define uniquely in this theory an automorphism swapping the two urelements.
For a theory rich in elementary embeddings and automorphisms, Boffa’s set theory (introduced in [1]) is relatively consistent wrt ZFC. The theory proves that any class endowed with a set-like binary relation satisfying the axiom of extensionality (but not necessarily well founded) is isomorphic to $\langle T,\in\rangle$ for some transitive class $T$. (For example, such a transitive collapse of the diagonal on the universe gives you a proper class of urelements, and you can construct even weirder objects. More to the point, any ultrapower of the universe gives you an elementary embedding into a transitive class.) Also, every isomorphism $f\colon\langle t,\in\rangle\to\langle s,\in\rangle$ of transitive sets $t,s$ can be extended to an automorphism of the universe.
Boffa’s theory consists of $\mathit{ZFC}^-$ + global choice + the following axiom:

If $t$ is a transitive set, and $\langle x,e\rangle$ a structure satisfying extensionality which is an end-extension of $\langle t,\in\rangle$, then there exists a transitive set $s\supseteq t$ and an isomorphism $f\colon\langle x,e\rangle\to\langle s,\in\rangle$ identical on $t$.

[1] Maurice Boffa, Forcing et négation de l’axiome de Fondement, Académie Royale de Belgique, Classe des Sciences, Mémoires, Collection 8o, 2e Série, tome XL, fasc. 7, 1972, 52pp.<|endoftext|>
TITLE: A reformulation of the Riemann Hypothesis
QUESTION [8 upvotes]: I am studying Sieve theory from Iwaniec's notes. I have come across a theorem which estimates $\varphi(x,N)=\#\{1\leq n \leq x:(n,N)=1\}$, where $N$ is product of distinct primes. 
Let's define $R(x,N)=\sum_{d|N}\mu(d)\{\frac{x}{d}\}=x\sum_{d|N}\frac{\mu(d)}{d}-\sum_{d|N}\mu(d)[\frac{x}{d}]=x\frac{\varphi(N)}{N}-\varphi(x,N)$.
Also say, $K(N)=\displaystyle\sum_{n=1}^{\infty}(\frac{R(n.N)}{n})^2$.
Now let's introduce a weighted $l^2$ space $\mathcal{H}$ with norm defined as $||x||^2:=\sum_{n=1}^{\infty}\frac{x(n)^2}{n^2}$ (provided convergent). Consider a Hilbert space $\mathcal{M}$ generated by $\left\{\gamma_n|\gamma_n(k)=\left\{\frac{k}{n}\right\},k=1,2,...;n>1\right\}$ inside $\mathcal{H}$.
It is known from Bagchi's result that the RH is true if and only if $\gamma=(1,1,...)\in\mathcal{M}$. It can be deduced that, above is also equivalent to the statement that if $x_m=\sum_{n=2}^{m}\mu(n)\gamma_n$
 then $x_m\to_{strongly} \gamma$ as $m\to\infty$.
If we define, $\bar x_m=\sum_{n|m}\mu(n)\gamma_n$ we see $K(N)=||x_N||^2$.
I think that it will be easy to show, $\bar x_m\to_{strongly} \gamma$ if and only if $x_m\to_{strongly} \gamma$.
My questions are:

Can this methodology create some other reformulation of RH? Like, "RH is true if and only if $K(N)\to1$"? 
Does there exist any sieve method which gives a good bound of $K(N)$? Honestly, I know a very little of sieve methods. Whatever sieve bounds I have seen those force upper bounds of $K(N)$ to go to infinity. 
Has any work been done in this way? If not, will it be a good reformulation? If yes, how one can approch further? As, $K(N)=\displaystyle\sum_{n=1}^{\infty}[\frac{\varphi(n,N)}{n}-\frac{\varphi(N)}{N}]^2$ looks like a 'good' arithmetical as well as a probabilistic function.

REPLY [5 votes]: Extended comment. Robin's criterion, equivalent to RH, is fairly widely known. First, however, his adviser, J.L. Nicolas, came up with THIS as pdf. The description of this on wikipedia is poor.
First, in a procedure invented by Ramanujan for his "superior highly composite numbers," it is easy to show that the smallest value of $\frac{\phi(n)}{n^\delta}$ for $0 < \delta < 1$ occurs when $n = n_\delta$ is a primorial, the product of consecutive primes beginning with 2.   
Let's see, Rosser and Schoenfeld gave some effective bounds, the way I am writing this comes out
$$ \frac{\phi(n)}{n} > \frac{1}{e^\gamma \log \log n + \frac{3}{\log \log n}}  $$
So the reasonable question comes, we know we get surprisingly small values of $\frac{\phi(n)}{n}$ when $n$ is a primorial. In that case, is it possible to replace the $3$ by a $0,$ giving
 $$ \frac{\phi(n)}{n} > \frac{1}{e^\gamma \log \log n } ? $$
And here is the answer: If RH is true, we cannot drop the 3. On the other hand, if RH is false, the inequality (no 3) is true for infinitely many primorials and false for infinitely many primorials. So, we have a statement equivalent to RH. 
Next, it is much easier to compute this comparison than Robin's. All you do is let $P$ be a primorial, and calculate 
$$ \frac{e^\gamma \log \log P \phi(P)}{P}  $$
which can be update fairly nicely as each $P$ is multiplied by the next prime. For all known primorials, this quantity strictly increases with $P.$ Since RH says it is below 1, we see the ratio increasing to 1, very pretty. i wrote out my own C++ program. There is, however, an amusing catch. Michael Planat and colleagues showed that Cramer's conjecture on prime gaps would be violated if the sequence increased forever. 
Enough for now, let me see if I can find the C++ program and post some early output. 
    p   phi(P) / P        log(P)      exp(gamma) * loglog(P) *  phi(P) / P
   2  0.5                0.6931471805599453  -0.3263930268425172
   3  0.3333333333333333  1.791759469228055  0.3462393386356046
   5  0.2666666666666667  3.401197381662155  0.5814026130255172
   7  0.2285714285714286  5.347107530717468  0.6825296531368984
  11  0.2077922077922078  7.745002803515839  0.7575980132430825
  13  0.1918081918081918  10.30995216097738  0.7970469070005248
  17  0.1805253569959452  13.14316550503359  0.8282264738589192
  19  0.1710240224172113  16.08760448420003  0.8462108841813194
  23  0.1635881953555934  19.22309870012918  0.8613001326390455
  29  0.1579472231019522  22.59039453011566  0.8770078674522567
  31  0.1528521513889861  26.0243817346008  0.8872418245081134
  37  0.1487210121622567  29.63529964724503  0.8976791721551992
  41  0.1450936704022016  33.34887171394934  0.9062933133274125
  43  0.1417193989974993  37.1100718296429  0.9121906541657894
  47  0.1387040926358503  40.96021943135295  0.9171685846758989
  53  0.1360870342842305  44.93051134490508  0.9222875682673444
  59  0.133780474381108  49.0080487888108  0.9273537165832041
  61  0.1315873518502701  53.11892265298411  0.9310291166644691
  67  0.1296233615241467  57.32361527237508  0.9347206020080276
  71  0.1277976803759193  61.58629514941639  0.9378817320183697
  73  0.1260470272200848  65.87675459056479  0.94015320110454
  79  0.1244514952299571  70.24620244303181  0.9424874828967286
  83  0.1229520796247769  74.6650430508284  0.9444916553953424
  89  0.1215705955840491  79.15367942056054  0.9465200179408284
  97  0.1203172904749352  83.72839039906393  0.9488025715708337
 101  0.1191260301732031  88.34351091590518  0.9507925028376726
 103  0.1179694667734633  92.97823990413482  0.9523051177952127
 107  0.1168669483924029  97.65106873859673  0.9536116556434188
 109  0.1157947745539405  102.3424166208259  0.9545403985983615
 113  0.114770042035764  107.0698044395382  0.9553238060578663
 127  0.1138663409173722  111.9139915259968  0.956775584934687
 131  0.1129971322080793  116.789188849198  0.9580534856815344
 137  0.1121723356226188  121.7091697750261  0.9593043754635934
 139  0.1113653404023122  126.6436437081568  0.9602858875196447
 149  0.1106179220103504  131.6475900141022  0.9614757581574626
 151  0.1098853529904143  136.6648698509172  0.9624286793532265
 157  0.109185446283469  141.7211156562655  0.9633634421768695
 163  0.1085155969197667  146.8148658570722  0.9642779835121893
 167  0.1078658029262352  151.932859669489  0.9650870024015282
 173  0.1072423011752165  157.0861512639868  0.9658796096962042
 179  0.1066431821742376  162.2735370698275  0.9666545741957177
 181  0.1060539933224463  167.4720341010934  0.9672701979733297
 191  0.1054987368129047  172.72430752914  0.9680084010810988
 193  0.1049521112335632  177.9869977180449  0.9686032079042936
 197  0.1044193593998903  183.2702014467829  0.9691265093804561
 199  0.1038946390008959  188.5635062715074  0.9695253328152903
 211  0.1034022473468632  193.9153644049834  0.9700846945685031
 223  0.1029385601390297  199.3225361764435  0.9707768755899375
 227  0.1024850863058181  204.7474861939249  0.9714019207735292
 229  0.1020375531778451  210.1812081974792  0.9719201381994437
 233  0.1015996237650647  215.6322466510449  0.9723820734575654
 239  0.101174520736759  221.1087102029764  0.9728329395456006
 241  0.1007547094473948  226.5935071364671  0.9731934250226916
 251  0.1003532962623454  232.1189600755988  0.9736223342629629
 257  0.09996281651035187  237.6680361604941  0.9740401207640036
 263  0.09958272975555965  243.2401901926718  0.9744468690918663
 269  0.09921253373416351  248.8349015722737  0.9748426972126225
 271  0.09884643582370535  254.4370203931534  0.9751650771767573
 277  0.09848958948499162  260.0610378993407  0.9754797735926634
 281  0.09813909272525856  265.6993925686745  0.9757574820666572
 283  0.09779231147887955  271.3448394663177  0.9759715944322618
 293  0.09745854932366153  277.0250120753348  0.9762367602929667
 307  0.09714109476560401  282.751859822922  0.9765970572556302
 311  0.09682874397857634  288.4916527351012  0.9769226892226303
 313  0.09651938696906012  294.2378559256413  0.9771919557217343
 317  0.09621490940764353  299.9967576995186  0.9774309460927161
 331  0.09592422992302829  305.7988760748956  0.9777507384622824
 337  0.09563958829120922  311.618959005248  0.9780609321048802
 347  0.09536396988114811  317.4682837851949  0.9784009755317026
 349  0.09509072068378092  323.3233557073974  0.9786926522404308
 353  0.09482134187164556  329.1898237643306  0.9789569560335052
 359  0.09455721557116745  335.0731461528189  0.9792133789754222
 367  0.09429956648241768  340.9785080008735  0.9794794850413051
 373  0.09404675263125838  346.9000864205173  0.9797375117033696
 379  0.09379860816521284  352.8376226255997  0.9799876894087147
 383  0.09355370318305824  358.7856576147804  0.980214493257318
 389  0.09331320523143084  364.7492369583989  0.9804344186018505
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Wed Jul 10 15:50:20 PDT 2013<|endoftext|>
TITLE: An "advanced beginner's" book on algebraic topology?
QUESTION [45 upvotes]: It has so happened that I have come this far knowing nothing on the subject of algebraic topology (as in homology theories of topological spaces and their applications). I've decided to finally read up on that during the summer.
Seemingly, however, the authors of most books for beginners are hesitant to make use of nontrivial homological algebra and category theory, which, if I'm not mistaken, could be used to speed up and at the same time clarify the presentation. I, on the other hand, would dare say to be somewhat familiar with these disciplines. (I'm, to different degrees, acquainted with derived functors, spectral sequences, derived categories as well as sheaf cohomology and Lie algebra/group cohomology.)
Thus, what I'm looking for is an introduction to algebraic topology the author of which readily employs the above concepts when appropriate.

REPLY [5 votes]: You have many good suggestions already. Another book which you might enjoy is "Cohomology Operations and Applications in Homotopy Theory" by Mosher and Tangora.
This is not really a beginner's book per-se, as it assumes a basic knowledge of ordinary cohomology from the start. However it has a lot to recommend it, including brevity, affordability and concreteness (the focus is on applications of cohomology theory to calculations of the homotopy groups of spheres). It also seems to meet your criteria in that it gets quickly to the deeper applications of homological algebra and spectral sequences in homotopy theory.<|endoftext|>
TITLE: Unicity up to homotopy of simplicial enrichments
QUESTION [8 upvotes]: On the one hand, in their paper Simplicial structures on model categories and functors, Rezk, Schwede and Shipley proved that a simplicial model category structure on a given model category is unique up to simplicial Quillen equivalence.
On the other hand, we know that every model category can be simplicially enriched (for instance taking a cosimplicial resolution of the source or a simplicial one of the target),
but in a way which does not give in every case a honest simplicial model category, but simplicial mapping spaces with good  homotopy invariance properties and a composition defined up to homotopy (see for instance the category of chain complexes over a field).
Now, let us consider a model category $M$. Suppose that $M$ is equipped with two simplicial mapping space functors $Map(-,-)$ and $\underline{Map}(-,-)$, both homotopy invariant under weak equivalences of a cofibrant source or a fibrant target, and with composition defined at least up to homotopy (so they both induce mapping spaces with a well defined composition on the homotopy category $Ho(M)$ of $M$).
Suppose moreover that we have
$\pi_0Map(X,Y)\cong[X,Y]\cong\pi_0\underline{Map}(X,Y)$
where $X$ is a cofibrant object of $M$, $Y$ a fibrant object and $[-,-]$ denotes the set of homotopy classes.
Do we then have $\pi_nMap(X,Y)\cong\pi_n\underline{Map}(X,Y)$ for every $n$ ?
A more general idea underlying my question is that I wonder if a result similar to the result of Rezk, Schwede and Shipley, in a weaker version, could hold under weaker assumptions (especially, when axiom SM7 of simplicial model categories is not fulfilled anymore).

REPLY [6 votes]: If, given any fixed cofibrant object $A$, there is a funtor $map(A,-)$ from $M$ to simplicial sets which preserves weak equivalences between fibrant objects and commutes with homotopy limits up to canonical weak equivalences (e.g. $map(A,-)$ is a right Quillen functor), and such that $\pi_0(map(A,X))=[A,X]$ (functorially) for any fibrant object $X$, then, for any fibrant object $X$, we have a canonical isomorphism $map(A,X)\simeq Map(A,X)$ in the homotopy category of simplicial sets, where $Map(A,X)$ denotes the usual simplicial enrichment. This is because both $map(A,X)$ and $Map(A,X)$ must represent the same presheaf. For (a more precise statement and) a proof, see for instance Remark 6.14 in this paper (but this is in fact a very formal and easy fact which can be found in many ways and disguises in the literature).<|endoftext|>
TITLE: Nonexistence of an approximately distance-preserving map between discrete cubes
QUESTION [7 upvotes]: Let $n>m$. Can there exist a map $\phi:\{0,1\}^n\to\{0,1\}^m$ that approximately preserves Hamming distance? I'm defining Hamming distance slightly nonstandardly by dividing by the dimension, so that the maximum distance between two sequences is 1, whatever the dimension. By "approximately preserves" I mean that for every $x,y\in\{0,1\}^n$, $d(\phi(x),\phi(y))$ should be within $\delta$ of $d(x,y)$, where $\delta$ is some small constant like $1/100$. 
It seems to me that the answer ought to be no, for the following reason. The fact that distances are approximately preserved implies a kind of continuity of $\phi$ (because in particular small distances go to small distances). But then the Borsuk-Ulam theorem would suggest that there will probably be two antipodal points in $\{0,1\}^n$ that map very close together in $\{0,1\}^m$, contradicting the approximate distance-preserving property.
My question is, is that thought the basis for a well-known argument? Or does the assertion follow easily from a known result?
Edit. As asked the question is not a great one, since, as Felipe points out below, if $m=n-1$ then one can just ignore the last coordinate. There are also arguments based on the pigeonhole principle if $n$ is a fair amount bigger than $m$: then there must be large sets of points in $\{0,1\}^n$ that have the same image, or, more generally, images in the same not too large Hamming ball. Amongst such sets there must be (if the sets are reasonably large) almost antipodal points (see Eoin's comment).
I'm actually interested in the case $m=n/2$. It's not clear to me (but I need to do some proper calculations) that that can be proved by pigeonhole type arguments. However, what is clear is that a topological proof of the kind I speculated about is unlikely to exist.

REPLY [5 votes]: I guess the answer would be no if $n$ is only slightly bigger than $m$, as Felipe mentioned in his post (just by projecting onto subcubes). However, you're probably more interested in what happens when $n$ is reasonably separated from $m$. I think here you can prove what you would like using concentration of measure. 
We must have that the preimage of some $x\in \{0,1\}^m$ under $f$ has size at least $2^{n-m}$. Now find some $\alpha \in [0,1/2]$ such that $2^{n-m} \approx \sum _{i= 0}^{\alpha n} \binom {n}{i}$. By an old result of Kleitman, any subset of $\{0,1\}^n$ of this size must contain two points at Hamming distance about $2\alpha n$ (this can be proved pretty easily using compressions). It's easy to see that for say $n/m \to \infty $ we have $ \alpha \to 1/2 $, and so two points which were almost antipodal get mapped to the same point.
In case you're interested in a better dependence between $n$ and $m$, I think that taking `preimage of a dense Hamming ball' of an appropriate radius in $\{0,1\}^m$ in place of $x$ in the above argument should create a worse distortion in this case.<|endoftext|>
TITLE: English translation of Steinitz 1910?
QUESTION [5 upvotes]: Does there exist an English translation of Steinitz' 1910 work "Algebraische Theorie der Körper"?
http://www.digizeitschriften.de/dms/img/?PPN=GDZPPN002167042

REPLY [4 votes]: You may already know this, but the best reference about Steinitz $1910$ work I could find is the following summary by Peter Roquette: 
http://www.rzuser.uni-heidelberg.de/~ci3/STEINITZ.pdf. Some English references are given there, and apparently a lot of Steinitz $1910$ work can be found in the book "Modern Algenbra" by van der Waerden, but I don't know about a full translation.
Edit: I think, as Jim has pointed out, that most likely there is no translation. Certainly Peter Roquette would have mentioned a translation in his discussion, and I think he has searched for it.<|endoftext|>
TITLE: Square root algorithm
QUESTION [5 upvotes]: I would like an efficient algorithm for square root of a positive integer.  Is there a reference that compares various square root algorithms?

REPLY [6 votes]: If you want the square root of an odd perfect square then one approach is to find the square root 2-adically. 
Suppose you want the square root of $n$. You can use the usual Newton-Raphson algorithm for the equation $f(x)=x-n/x=0$, working modulo $2^N$, for some large enough $N$. That's the iteration $x \rightarrow -x(-3n+x^2)/(2n)$. However, that requires division by $n$. You can achieve this by first using Newton-Raphson to solve for $g(x) = 1/x-n = 0$ to find $n^{-1}$ modulo $2^N$. That's $x\rightarrow x(2-nx)$. So two Newton-Raphson solves are required.
Requires only addition, subtraction and multiplication modulo $2^N$. Note also that it may converge to $-\sqrt n$.<|endoftext|>
TITLE: Decomposing tensor products of modules for the orthogonal/symplectic groups in characteristic zero
QUESTION [6 upvotes]: I would like to know if there is a perfect analogue of the classical Littlewood-Richardson rule for decomposing tensor products of simple modules for the orthogonal/symplectic groups in characteristic zero, i.e. a "tableaux counting algorithm" which determines the solution.
I know that there are tableaux counting rules due to King and De Concini, there is the Littelmann path algorithm, and the theory of crystal bases and probably many more descriptions besides.  There are also results due to Koike and others which express the answer in terms of a sum or products of Littlewood-Richardson coefficients (this isn't really what I'm looking for).
So my question is: is there some summary of how all these different methods coincide?  Do any of these methods give "perfect solutions"?  I know the Littelmann path approach gives a complete answer, but can it be expressed in terms of something like the classical tableaux counting procedure?  Do the King/De Concini rules provide (positive) solutions in all cases?  And is there some bible in which all these different approaches are compiled and compared?

Post Jim's comment:
The main question is: under what conditions (probably on $n$) do (any) of the tableaux-based approaches coincide with the Littelmann path approach to answer the problem?

And what is the best reference for these tableaux?  I was told to look for 
R. C. King: "Weight multiplicities for the classical groups"
but can't get a hold of it.  Does anyone have a pdf?  If not, does anyone have another good reference?  There seems to be a LOT of literature to wade through

REPLY [5 votes]: I counted 8 questions asked in this post. I will answer one: Before Littelmann invented the LS-path mode, he worked out a generalized LR-rule using the language of generalized tableaux for most simple complex Lie algebras (icnluding all classical ones), see here. Since he was extending the work of Lakshmibai and Seshadri, I am pretty sure, the formalism Littelmann was using is equivalent to the LS-path model, just he did not have a geometric picture at the time.<|endoftext|>
TITLE: Do Measurable Cardinals Exist? (assuming ZFC)
QUESTION [10 upvotes]: In Appendix B of his Uniform Central Limit Theorems (1999), Dudley writes:

It is consistent with the usual axioms of set theory (including the axiom f choice) that there are no measurable cardinals, in other words all cardinals are of measure 0; see, for example, Drake (1974, pp. 67-68, 177-178). It is apparently unknown whether existence of measurable cardinals is consistent (Drake, 1974, pp. 185-186). So, for practical purposes, a probability measure defined on the Borel sets of a metric space is always concentrated in some separable subspace.
The continuum hypothesis implies that the cardinality $c$ of the continuum (that is, of $[0,1]$) is of measure 0 (RAP, Appendix C).

Has there been any progress on the existence of measurable cardinals (assuming ZFC) in the 14 years since this book was published? i.e., do they exist? What makes the problem difficult?
The Wikipedia article talks about measurable cardinals but doesn't point out the open problem (this should be fixed). You can prove that measurable cardinals do exist if you assume ZF+AD.

From a few paragraphs prior in Appendix B, here are the definitions:

A cardinal number $\zeta$ is said to be measurable if for a set $S$ of cardinality $\zeta$, there exists a probability measure $P$ defined on all subsets of $S$ which has no point atoms; in other words, $P(\{x\}) = 0$ for all $x \in S$. If there is no such $P$, $\zeta$ is said to be of measure 0.

In the absence of any news, can somebody formulate the problem of existence of measurable cardinals in the structural set theory of ETCS? I would accept that as answer.

REPLY [17 votes]: This is not really a problem.
If $\kappa$ is a measurable cardinal then $V_\kappa$, or the set of sets which are hereditarily have size smaller than $\kappa$, is a model of $\sf ZFC$. This means that $\sf ZFC$ cannot even prove the consistency of $\sf ZFC+\exists\kappa\text{ measurable}$, because of the incompleteness theorem. Well, at least if $\sf ZFC$ is consistent to begin with.
Furthermore, if $\kappa$ is the least measurable cardinal, and if there is a measurable cardinal then there is a least measurable cardinal, then in $V_\kappa$ there are no measurable cardinal, therefore $V_\kappa\models\sf ZFC+\lnot\exists\kappa\text{ measurable}$.
Regarding the assumption on determinacy, two points are relevant here: $\sf ZF+AD$ proves that the axiom of choice fails. And moreover the consistency strength of $\sf ZF+AD$ is much higher than that of $\sf ZFC$, or even $\sf ZFC+\exists\kappa\text{ measurable}$. This means that if we assume that $\sf ZF+AD$ is consistent then we can prove a lot more than we can from $\sf ZFC$ or $\sf ZFC+\exists\kappa\text{ measurable}$.
So we know that $\sf ZFC$ cannot prove that a measurable cardinal exist, or even the consistency of one, and we know that if $\sf ZFC+\exists\kappa\text{ measurable}$ is consistent, then so is $\sf ZFC+\lnot\exists\kappa\text{ measurable}$. 

I am not too familiar with structural set theory or $\sf ETCS$, but Misha Gavrilovich has a characterization of measurable cardinals using his model categorical construction $\rm QtNaamen$. You can find the papers in Misha's homepage, in particular "Exercises de style: A homotopy theory for set theory. Part II" with Assaf Hasson.<|endoftext|>
TITLE: Algorithm to quickly compute the individual inverses of a linear sequence of matrices
QUESTION [6 upvotes]: Fix $n \times n$ real symmetric positive definite matrices $A$ and $B$. Fix vectors $x$ and $y$ in $\mathbf{R}^n$.  I want to compute the following bilinear products quickly: $\{x^T (A+mB)^{-1} y\}_{m=0}^M$.
A naive but practical method would involve inverting each matrix individually, thus requiring $O(Mn^3)$.
Are there practical improvements for $M > n > \log(M)$ ? (e.g. Hopefully $O(n^3 + n^2M\log(n))$ or something).

REPLY [3 votes]: The following ideas are somewhat similar to that in Suvrit's answer.
Idea 1. Instead of first computing the square root of $A$, just solve the generalized eigenvalue problem $Ax = \lambda B x$ directly. This is especially tractable since according to the OP both $A$ and $B$ are symmetric and positive definite. You get a matrix $U$ such that $U^T B U = I$, $U^T A U = \mathrm{diag} (\lambda_i)$, and
\begin{align}
  U^T (A + mB) U &= \mathrm{diag} (\lambda_i + m) , \\
  U^{-1} (A + mB)^{-1} (U^{-1})^T &= \mathrm{diag} (\frac{1}{\lambda_i + m}) .
\end{align}
The formula that you want is then
$$
  x^T (A + mB)^{-1} y = x^T U \mathrm{diag} (\frac{1}{\lambda_i + m}) U^{T} y .
$$
The cost of solving the generalized eigenvalue problem is still $O(n^3)$ and the cost of each of the above multiplications once $U$ is known is $O(n^2)$. Thus, the overall cost is $O(n^3 + Mn^2)$.
Idea 2. Now, an eigenvalue problem requires iteration. If you'd prefer an algorithm with a definite number of steps, you could instead simultaneously tridiagonalize $A$ and $B$, that is, find a matrix $U$ such that $U^T A U = T_A$ and $U^T B U = T_B$, where both $T_A$ and $T_B$ are tridiagonal. Tridiagonalization is in any case a common preconditioning step in eigenvalue problem algorithms. Similar to above, the formula that you want is then
$$
  x^T (A + mB)^{-1} y = x^T U (T_A + m T_B)^{-1} U^{T} y .
$$
Since $T_A + m T_B$ is tridiagonal, you can compute its inverse in linear time, though slower by a constant factor than a diagonal matrix.
The cost of a tridiagonalization is once again $O(n^3)$ and the cost of each of the above inversions + multiplications is still $O(n^2)$ (dense matrix-vector multiplication dominates tridiagonal inversion). Thus, the overall cost is again $O(n^3 + Mn^2)$, though without requiring iteration.<|endoftext|>
TITLE: Etale cohomology of and forms of algebraic groups
QUESTION [8 upvotes]: Let $k$ be a field, and $K$ its separable closure.  Consider two different $k$-schemes, $X$ and $Y$, which become isomorphic upon extension of scalars to $K$: $X_K \cong Y_K$.  Then the etale cohomologies (and indeed, etale homotopy types) of $X_K$ and $Y_K$ will be equivalent, but may differ at $k$.  I'd like to know some sample computations in which $H^*_{et}(X_k)$ and $H^*_{et}(Y_k)$ differ, or agree.  I'm happy to take any sort of coefficients here, but being an algebraic topologist, I'm lazy, so I would prefer that they're constant sheaves.
To give some focus to the question, let's consider the case $k=\mathbb{R}$, and $K=\mathbb{C}$.  Take $X = GL_2(\mathbb{R})$, and let $Y$ be the nonzero quaternions, $\mathbb{H}^{\times}$.  These become isomorphic (to $GL_2(\mathbb{C})$) upon extension of scalars to $\mathbb{C}$.  Do they have the same etale cohomology over $\mathbb{R}$?
Let's take coefficients such that $2$ is invertible; first note that by a comparison to the analytic topology, $H^*_{et}(GL_2(\mathbb{C}))$ is an exterior algebra on two generators in dimensions 1 and 3, $\Lambda[x_1, x_3]$. There are actions of $\mathbb{Z} / 2 = Gal(\mathbb{C} / \mathbb{R})$ on $H^*_{et}(X_{\mathbb{C}})$ and $H^*_{et}(Y_{\mathbb{C}})$, and the invariants are $H^*_{et}(GL_2(\mathbb{R}))$ and $H^*_{et}(\mathbb{H}^{\times})$, respectively.  I presume that the action of $\mathbb{Z} / 2$ on $\Lambda[x_1, x_3]$ must differ for the two examples, but I'm not really sure how to compute these actions.
Lastly, are there any criteria on a scheme under which the etale cohomology over $\mathbb{R}$ bears any resemblance to the singular cohomology of the real points?  I have read that this is related to the Sullivan conjecture in homotopy theory, but don't know much more than that slogan.

REPLY [13 votes]: For your first question: the two forms of $GL_2$ differ by modifying the Galois action by an automorphism of $GL_2$. The outer automorphism group of $GL_2$ is cyclic of order $2$: an automorphism is inner if and only if it is trivial on the center of $GL_2$. Since the form of $GL_2$ you're considering has the same center as the standard form, it follows that it's an inner twist of $GL_2$. Now $GL_2$ is connected, so any inner automorphism acts trivially on its cohomology. It follows that the two Galois actions are the same (both act by a sign on the class in degree $1$, and preserve the class in degree $3$).
As for your second question: the Sullivan conjecture says that if $X$ is a finite $\mathbf{Z}/2 \mathbf{Z}$-complex, then the $2$-adic completion of the fixed set of $X$ is homotopy equivalent to the homotopy fixed set of the $2$-adic completion of $X$ (here you need to be careful about the meaning of "$2$-adic completion" if the fundamental group is nontrivial: you should really take a $2$-profinite completion, rather than a Bousfield localization). If $Y$ is an algebraic variety defined over $\mathbf{R}$, then $Y(\mathbf{C})$ is a
finite $\mathbf{Z}/2\mathbf{Z}$-complex having fixed set $Y(\mathbf{R})$.
It follows that the $2$-adic completion of $Y( \mathbf{R})$ can be recovered as the homotopy fixed points of complex conjugation on the $2$-adic completion of
$Y(\mathbf{C})$. And the latter can be recovered in a "purely algebraic" way using etale homotopy theory. So in this sense, etale homotopy theory "knows"
the $2$-adic completion of $Y(\mathbf{R})$ (and, in particular, invariants like the $\mathbf{F}_2$-cohomology of $Y(\mathbf{R})$).<|endoftext|>
TITLE: Least prime $p$ such that an irreducible polynomial of degree $n$ has no root modulo $p$?
QUESTION [12 upvotes]: This question is inspired by an old question of Greg Kuperberg, about how small is the first prime $p$ which makes a given monic polynomial $P$ with integral coefficient have a (simple) root modulo $p$. Now "small" makes sense w.r.t. something, some notion of size of polynomials. In Greg's question the size is measured by the degree $n$ of $P$ and the $L^1$-norm (or $L^\infty$-norm of its coeffcients, but I will rather use as measure of size the degree $d$ and the product $M$ of the primes $p$ dividing the discriminant of $P$ (in short $M$ is the radical of the discriminant of $P$): this is essentially equivalent to Greg's notions but more in line with the literature on the subject.
This being said the result of Weinberger alluded to by Greg is, under GRH, the following:

Theorem : For any $\epsilon>0$, and for any monic polynomial $P$ with integral coefficients, of degree $n$ and radical of discriminant $M$, there is always a prime $p = O(n^{2+\epsilon} (\log M)^{2+\epsilon})$ not dividing $M$ such that $P$ mod $p$ has a root (necessarily simple).

The theorem is proven in this paper of Weinberger (though not stated this way -- near the end of the argument the author stops calculating and just says that the bound is polynomial. I hope my calculation of his bound is correct), and the proof is surprisingly simple. In fact it is so simple that it puzzles me, avoiding all the complications of Lagarias-Odlyzko's effective Chebotarev's theorem (to which the result is closely related, as Greg says) by working with the field $K=\mathbb Q[X]/(P(x))$ even if it is nor Galois (assuming $P$ is irreducible, which one can of course do) rather than its normal closure $L$, which is Galois but can have degree as large as $n!$, which is bad for the polynomial estimate. 
To understand better the scope of the method, I ask the following, kind of opposite, question:

Is it true (under GRH or any standard conjecture) that for any monic irreducible polynomial $P$  with integral coefficients, of degree $n$ and radical of discriminant $M$, there is always a prime $p$ not dividing $M$ less than a fixed polynomial in $n$ and $\log M$, such that $P$ mod $p$ has no simple roots ?

(or at least less than a function of $n$ and $\log M$, polynomial in $n$ - I don't really care about the dependence in $M$)?

REPLY [11 votes]: Let $K$ be the number field obtained by adjoining some root of $f(x)$ to 
${\Bbb Q}$, and let $d_K$ be its discriminant.  Let $k$ be the degree of $K$, and let 
$\zeta_K(s) = \sum_{n=1}^{\infty} a(n) n^{-s}$ denote the Dedekind zeta function, and write 
$-\frac{\zeta_K^{\prime}}{\zeta_K}(s) = \sum_{n=1}^{\infty} \Lambda_K(n)n^{-s}$ so that 
$\Lambda_K(n)$ is supported on prime powers, and we have $\Lambda_K(p) = a(p) \log p$, and 
$\Lambda_K(p^{\ell}) \ll k \log p$ for all $\ell\ge 1$.  The problem observes that bounding the least prime $p$ 
for which $a(p)$ is non-zero can be solved on GRH without passing to the normal closure of $K$ (a result 
of Weinberger) and asks for other properties of $a(p)$ that can be obtained similarly (that is with a 
polynomial bound in $k$ rather than a $k!$ type bound).   The method below shows that 
any property that can be identified using a polynomial in $a(p)$ of small degree can be studied in 
such a way.  This is illustrated by solving the specific example in the question of bounding the 
least prime $p$ for which $a(p)$ equals zero.  
From GRH for $\zeta_K$ it follows that 
$$ 
\sum_{n\le x}  \Lambda_K(n) (2-2n/x) = x+ O(x^{\frac 12} \log d_K ). 
$$ 
This is a mild smoothing, so that I don't have to worry about $\log \log $ powers in the remainder 
term, but one could also deal just with $n\le x$ without the weight $(2-2n/x)$.  As observed in the problem, 
the method of Weinberger in finding non zero value of $a(p)$ proceeds in this manner, omitting the prime 
powers which make a smaller contribution (no more than $\ll k\sqrt{x}$): thus 
$$
\sum_{p\le x} a(p)\log p (2-2p/x) = x+ O(x^{\frac 12} (k+ \log d_K)).  
$$
We now wish to evaluate $\sum_{p\le x} a(p)^2 \log p (2-2p/x)$.  This is a 
Rankin-Selberg type problem, which could be understood by studying the Rankin-Selberg 
function for $\zeta_K(s)$.  Since we are assuming the Artin conjecture, we may write 
$$ 
\zeta_K(s) = \prod_{j=1}^{r} L(s,\rho_j)
$$ 
where the $\rho_j$ are irreducible representations of the Galois group of the normal 
closure of $K$ (occurring possibly with multiplicity) and $\sum_{j} \text{dim}(\rho_j) = k$.
Then the Rankin-Selberg $L$-function is 
$$ 
\zeta_{K\times K}(s) = \prod_{1\le j , \ell \le r}  L(s,\rho_j \otimes \overline{\rho_\ell}).
$$ 
Note that the tensor product $\rho_j\otimes \overline{\rho_\ell}$ may be decomposed in terms 
of irreducible representations of the Galois group, and so by Artin's conjecture the $L(s,\rho_j \otimes{\overline \rho_\ell})$ 
have a meromorphic continuation (holomorphic except for a pole at $1$) and functional equation, and there is a simple pole at $1$ 
precisely when $\rho_j$ and $\rho_\ell$ are the same.   Thus if the $\rho_j$ consist of $t$ distinct representations 
occurring with multiplicities $m_1$, $\ldots$, $m_t$ (so that $r=m_1+ \ldots+m_t$) then $\zeta_{K\times K}(s)$ has a 
pole of order $\sum m_i^2 =R \ge r$ at $s=1$.  Further, we may bound the conductor of $\zeta_{K\times K}$ by 
$d_K^k$.  Thus applying GRH we find that 
$$ 
\sum_{n\le x} \Lambda_{K\times K}(n) (2-2n/x)  = Rx + O( kx^{\frac 12} \log d_K). 
$$ 
Here $\Lambda_{K\times K}(n)$ denotes the Dirichlet series coefficients of $-\frac{\zeta_{K\times K}^{\prime}(s)}{\zeta_{K\times K}(s)}$, 
and note that $\Lambda_{K\times K}(p) = a(p)^2 \log p$, and $0\le \Lambda_{K\times K}(p^j) \le k^2 \log p$.
Thus from the above estimate we find that 
$$ 
\sum_{p\le x} a(p)^2 \log p (2-2p/x) =  Rx + O(kx^{\frac 12} (k+\log d_K)).
$$ 
Now suppose that $a(p) \ge 1$ for all $p\le x$.  From our first estimate we see that 
$$ 
\sum_{p\le x} (a(p)-1)\log p (2-2p/x) = O(x^{\frac 12 } (k+\log d_K)).
$$ 
Since $a(p)\le k$ for all $p$ (and $a(p)\ge 1$ by assumption) it follows that 
$$ 
\sum_{p\le x} a(p)(a(p)-1) \log p (2-2p/x) = O(kx^{\frac 12} (k+\log d_K)).
$$ 
But we also know that 
$$ 
\sum_{p\le x} a(p)(a(p)-1) \log p (2-2p/x) = (R-1) x + O(kx^{\frac 12} (k+\log d_K)).
$$ 
It follows that 
$$ 
x\ll \Big(\frac{k(k+\log d_K)}{R-1}\Big)^2. 
$$<|endoftext|>
TITLE: Does a self map from the wedge sum of two spheres have either a fixed point or a point of period 2?
QUESTION [12 upvotes]: Let $X$ be the wedge sum of two $2$-dimensional spheres and $f$ a continuous function from $X$ into $X$. Does $f$ have either a fixed point or a periodic point of order 2?
Thanks

REPLY [12 votes]: Will's comment above shows that the action of $f$ on $H^2$ is of order 3
(the eigenvalues are the nontrivial third roots of unity) and something like
$\begin{pmatrix}0&-1\\1&-1\end{pmatrix}$. So I thought 
one should try a construction using third roots of unity realizing this.
Write the wedge as $S^1 \times [0,2] / \sim $ where one identifies
$(s,0)\sim (s',0),  (s,1)\sim (s',1), (s,2)\sim (s',2)$.
Let $\zeta$ be a third root of unity, and
define 
$f: S^1 \times [0,2] / \sim\  \to \ S^1 \times [0,2] / \sim$
by 
$f(s,t)=(\zeta s, 2-2t)$ for $t\le 1$ and $f(s,t)=(\zeta s, t-1)$ for $t\ge 1$
Note that the three points where $t=0,t=1,t=2$ are permuted cyclically.
This implies $f$ and $f^2$ have no fixed points (for all other points the $\zeta$ respectively $\zeta^2$ factor works).
Note also that multiplication by $\zeta$ is not really necessary, almost any rotation of the circle would be fine. Multiplication by $\zeta$ allows $f^3$ to be close to the identity. If one uses a different rotation, the three points with $t=0,t=1,t=2$ are the only fixed points.<|endoftext|>
TITLE: étale covers and torsion line bundles
QUESTION [10 upvotes]: Let $n \geq 2$ be an integer, $X$ a smooth variety over a field $k$ containing $\mu_n$ and $G$ a cyclic group of order $n$ acting on it. Assume that the action is free. Then the morphism $\pi: X \to Y=X/G$ is etale. I'm trying to understand why $\pi_\ast \mathcal{O}_X$ decomposes as 
$$
\pi_\ast \mathcal{O}_X=\mathcal{O}_Y \oplus L \oplus L^2 \oplus \cdots \oplus L^{n-1} 
$$ 
where $n$ is the order of $G$ and $L$ is a line bundle on $Y$ such that $L^n \simeq \mathcal{O}_X$. 
Can anybody help me please?

REPLY [8 votes]: By construction the sheaf $\pi_* \mathcal{O}_X$ is a $G$-equivariant vector bundle of rank $n$ on $Y$ and, since $G$ is a cyclic group, the representation of $\pi_* \mathcal{O}_X$ as a $G$-module splits into direct summands which are all line bundles. 
Now take any isomorphism $G \cong \mathbb{Z}/ n \mathbb{Z}$, and call $L$ the eigensheaf of $\pi_* \mathcal{O}_X$ corresponding to the generator $\bar{1} \in G$. Clearly $L$ is a $n$-torsion line bundle on $X$ and moreover for any $k \in \mathbb Z$ the eigensheaf corresponding to $\bar{k} \in G$ is precisely $L^k$. 
So you obtain the desired splitting.<|endoftext|>
TITLE: Is the dimension given by Klee trick ever sharp?
QUESTION [13 upvotes]: The Klee Trick allows one to find an $\mathbb{R}^m$ where two embeddings of same compact metric space have homeomorphic complements. More precisely, given two embeddings of a compact metric space $K$ into $\mathbb{R}^n$, $f_{n,1}$, $f_{n,2}$, we can construct two embeddings of $K$ into $\mathbb{R}^{2n}$ such that the images of $K$ are equivalent under a homeomorphism of $\mathbb{R}^{2n}$. (The trick itself produces an isotopy of the the two embeddings in $\mathbb{R}^{2n}$.) 
However, I was wondering if there is an example of a compact metric space $K$ and a pair of embeddings $f_{n,1},f_{n,2}:K \rightarrow \mathbb{R}^n$ such that the embeddings are not equivalent under a homeomorphism of $\mathbb{R}^{n+m}=\mathbb{R}^n \times\mathbb{R}^m$ for all $m < n$. 
To clarify with a non-example, any two (tamely) embedded knots in $\mathbb{R}^3$ will be isotopic in $\mathbb{R}^4=\mathbb{R}^3 \times\mathbb{R}$, and so for embeddings of $S^1$ into $\mathbb{R}^3$, the isotopy obtained from the Klee trick is not optimal.

REPLY [6 votes]: Are the embeddings required to be isometric embeddings?  If not, then what about including three points into $\mathbb{R}$ in two ways, so that the middle point of the three changes?  More explicitly, let $K=\{a,b,c\}$ and define $f(a)=0$, $f(b)=1$, $f(c)=2$, whereas $g(a)=1$, $g(b)=0$, $g(c)=2$.  There isn't any self-homeomorphism of $\mathbb{R}$ whose composition with $f$ is equal to $g$.<|endoftext|>
TITLE: Iterated function system on the plane
QUESTION [10 upvotes]: Let $r_1, r_2, r_3$ be three nonnegative real numbers with $r_1^2+r_2^2+r_3^2 <1$. Can you find three similitudes $f_1,f_2,f_3$ on $\mathbb{R}^2$ with similarity ratios $r_1,r_2,r_3$ resp. and a nonempty open set $O \subset \mathbb{R}^2$ such that $f_i(O) \subset O$ and $f_i(O) \cap f_j(O)=\emptyset$ for $i \neq j$ ?
(Natural generalizations thinkable but I stuck at this stage.)

REPLY [3 votes]: A good solution may involve (the interior of) sets with boundary of fractal dimension. However, using just right triangles one can get around $\frac{4}{5}$ of the possibilities (in a sense made precise below) and very slightly more using rectangles as well. 
I'll focus on the squared ratios $\rho_i=r_i^2$ with $\rho_1+\rho_2+\rho_3 \le 1.$ So I will allow $r_i=0$ and allow the sum to actual equal $1$. I find it easier to consider the triples in all $6$ possible orders. So the  set of possible $\rho$ triples can be thought of as a body $\mathbf{B}$ comprising part, or perhaps all, of the pyramid $\mathbf{P}$  in $\mathbb{R}^3$ with corners at the origin and the points $(1,0,0)$ , $(0,1,0)$ and $(0,0,1).$ The volume of $\mathbf{P}$  is $\frac{1}{6}.$ My claim is that the volume of $\mathbf{B}$ is over $\frac{4}{5}$ of this.  If a point $P=(\rho_1,\rho_2,\rho_3)$ is in $\mathbf{B}$ so are all the points of the box determined by the origin and $P$. So $\mathbf{B}$ could be specified were one able to described the surface made of the points in each radial direction furthest from the origin. 
Certainly the the points with $\rho_1+\rho_2+\rho_3=1$ are of interest. As was kindly pointed out to me, we can achieve the triples $(t,1-t,0)$ using (the interior of) a right triangle with legs of lengths $\sqrt{t},\sqrt{1-t}$ and the usual division by the perpendicular to the hypotenuse. Iterating this for one of the sub-triangles (picture at the end) gives us triples of the form $(t,t^2,1-t-t^2)$ where $0 \le t \le \frac{\sqrt{5}-1}{2}.$ Here is a sketch of these points making up the boundary of the equilateral triangle and $6$ internal curves. 

I can't really picture the body made up of all the boxes determined by these points and the origin, but the volume , roughly estimated by counting the included points of the form $(\frac{a}{100},\frac{b}{100},\frac{c}{100})$ , is somewhere between $0.83$ and $0.86$ of the total volume of $\mathbf{P}$ depending on if I use the floor or round in estimating. All the points with $\rho_1+\rho_2+\rho_3 \le 0.75$ appear to be achieved however then things start to fall off.
Certainly the point $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ is not accounted for. That can be achieved using a rectangle with sides $3 \times \sqrt{3}$ partitioned into three  rectangles of with sides $1 \times \sqrt{3}.$ That increases the volume of $\mathbf{B}$ but not by very much. 
Below is an sketch of possibly missing points with $\rho_1+\rho_2+\rho_3=0.9.$ 

The small triangle missing in the center comes from the rectangle just mentioned. So a good goal (as might have been obvious) would be triples like $(0.48.0.48.0.04)$, $(0.94,0.03,0.03)$ and $(0.32,0.32,0.36)$ with two equal components close to but not equal to $\frac{1}{2}$ or $0$ or $\frac{1}{3}$  and sum equal to, or nearly equal to , $1$.
Another interesting rectangle partition is a $4 \times 2\sqrt{2}$ rectangle partitioned in half and then one half again partitioned in half. However this gives the same result $(\frac{1}{2}, \frac{1}{4},\frac{1}{4})$ as an isosceles right triangle. 
I thought of a clever (I thought) ways to pack a figure with similar copies which fill it mostly but not completely. However the $\rho$-triple is $(t^2,t^4,t^4)$ for $t=\frac{\sqrt{5}-1}{2}$ so this is not as good as the triangle construction $(1-t^2-t^4 ,t^2,t^4).$ The proportion covered increases and the pieces rotate if we hold one leg fixed and shrink the other, but this seems likely to be turning into the triangle situation. I don't think varying the angle would help.<|endoftext|>
TITLE: Identifying the little disk operad with parenthesized braids
QUESTION [8 upvotes]: Let $D_2$ be the topological operad of little disks. This operad can be modelled "combinatorially" in terms of an operad of groupoids called $\newcommand{\PaB}{\mathbf{PaB}}\PaB$, the operad of parenthesized braids. An object of the groupoid $\PaB(n)$ is a complete parenthesization of a permutation of the symbols $\{1,...,n\}$, e.g.
$$ ((24)(13)5) \in \mathrm{ob} \,\PaB(5),$$
and morphisms are $n$-strand braids, such that the start- and endpoints of each strand are labeled by the same element of $\{1,...,n\}$. Composition in this operad is given by "cabling". The connection to the little disk operad is that one can quite easily write down a morphism of operads 
$$ \PaB \to \Pi_1(D_2),$$
where $\Pi_1(-)$ denotes the fundamental groupoid, such that $\PaB(n) \to \Pi_1(D_2)(n)$ is an equivalence of categories for all $n$.
If we denote by $N$ the composition of the nerve functor with the geometric realization, then $N\PaB$ is a topological operad.
Fact: $N\PaB$ is equivalent to $D_2$, i.e. there is a zig-zag of operad morphisms between $N\PaB$ and $D_2$, each of which is arity-wise a weak equivalence.
In the literature this fact is in several places proven by appealing to Fiedorowicz's recognition principle, see e.g. Dmitry Tamarkin, Formality of chain operad of little discs, Section 2. I don't understand why and I guess I am missing something: it seems to me that there is a much more direct argument. Indeed the map $\PaB \to \Pi_1(D_2)$ gives a homotopy equivalence 
$$N\PaB \to N\Pi_1(D_2),$$
moreover, the unit of the adjunction between $\Pi_1$ and $N$ gives a morphism 
$$D_2 \to N\Pi_1(D_2)$$
which is also an equivalence since $X \to N\Pi_1(X)$ is a homotopy equivalence for any $K(\pi,1)$-space, such as $D_2(n)$. What am I missing?

REPLY [4 votes]: It seems to me that the operad morphism from PaB to $\Pi_1(D_2)$ works in positive arity  but does not respect composition with the $0$-ary operation on the object level.<|endoftext|>
TITLE: How does it End?
QUESTION [21 upvotes]: A recent project has forced my colleague and me to take a rather abstract approach to dynamical systems, and the following definition arose naturally in that context.
Let $\mathcal{C}$ be a category. Its endomorphism category $\text{End}(\mathcal{C})$ is defined as follows. The objects are the endomorphisms $f:x \to x$ of $\mathcal{C}$ and a morphism from $f:x \to x$ to $g: y \to y$ is a morphism $h:x \to y$ in $\mathcal{C}$ so that the commuting relation $h \circ f \equiv g \circ h$ holds in $\mathcal{C}$.
That's all we really need for the project, but now a perverse question springs to mind. Note that $\text{End}$ itself defines a dynamical system. After all, we have $\text{End}(\text{End}(\mathcal{C}))$ and so on by iterating the process. I'm interested in the asymptotic behavior. So, here are some questions:

Which categories are fixed points of $\text{End}$, i.e., for which $\mathcal{C}$ is $\text{End}(\mathcal{C})$ equivalent to $\mathcal{C}$?

Slightly more interesting is the question of when $\text{End}(\mathcal{C})$ is homotopy equivalent to $\mathcal{C}$. And while I suspect that the answer to the following question is no, I was unable to prove it:

Is it possible to have non-trivial periodic orbits, for instance, non-equivalent categories $\mathcal{C}$ and $\mathcal{D}$ so that $\text{End}(\mathcal{C})$ is equivalent to $\mathcal{D}$ and vice-versa?

REPLY [8 votes]: Here are some easy examples within groupoids.$\DeclareMathOperator\End{End}$
For a group $G$, $\End(BG)$ is equivalent to $\coprod_{x\in I} BC_x$ (which is skeletal) where $I$ is a set containing one representative of each conjugacy class of $G$ and $C_x$ denotes the centralizer of $x$ in $G$ (in fact, $\End(BG)$ is isomorphic to $\coprod_{x \in I} BC_x \times E_{[G:C_x]}$ where $E_n$ is the groupoid on $n$ objects and a unique isomorphisms between every pair of them). From this we see, for example, that:

The only finite groupoids that are fixed points for $\End$, or even belong to an $\End$-cycle, are the finite discrete categories: when you apply $\End$ to a groupoid $C$ you always get a copy of $C$ in the result and, unless, $C$ is discrete, you get a groupoid with a strictly larger skeleton.
Among groupoids there are no cycles that are not simply fixed points (again, because $BC$ always contains a copy of $C$, so the multiplicity of any given group in $\End^n(C)$ is a monotone function of $n$).
Given any $\mathcal{G}$ is any class of groups such that whenever $x\in G \in \mathcal{G}$ then also $C_x \in \mathcal{G}$, we can form the groupoid $C := \coprod_{G \in \mathcal{G}} \coprod_{\mathbb{N}} BG$ and this will be an $\End$-fixed point. In particular, for any Abelian group $G$, $\coprod_{\mathbb{N}}BG$ is a fixed point. (And notice too, that the equivalence $C \simeq \End(C)$ is not canonical in these examples.)
As a simple, explicit example, $\End(BS_3) = BS_3 \sqcup B\mathbb{Z}/2 \sqcup B\mathbb{Z}/3$, and $\End(BG) = |G|\cdot BG$ for any Abelian $G$ (as was already mentioned by Benjamin Steinberg --and where $n \cdot C$ is the coproduct of $n$ copies of the category $C$), so we get that $\End^n(BS_3) = BS_3 \sqcup (2^n-1)\cdot B\mathbb{Z}/2 \sqcup \frac{1}{2}(3^n-1) \cdot B\mathbb{Z}/3$. (And similarly, for any fixed group, working out the formula for the $n$-th iterate $\End^n(BG)$ is just a a matter of patience.)

ADDED: It might be worth mentioning that when you stick to just groupoids, the question is very homotopy theoretical: equivalence is the same as homotopy equivalence and $\End$ is just the free loop space functor. In other words, the question for groupoids becomes "study the dynamical system given by the free loop space functor on the homotopy category of 1-types". A reasonable analogue of this question for $\infty$-groupoids would be to study the free loop space functor on the homotopy category (of all spaces) as a dynamical system. An analogue of the above description of $\End(BG)$ works for topological groups too: the free loop space $\mathop{Map}(S^1,BG)$ is the homotopy quotient $G/G$ where $G$ acts on itself by conjugation — this is analogous because the description given above of (the isomorphism type of) $\End(BG)$ for discrete $G$ is just the action groupoid of $G$ acting on itself by conjugation.<|endoftext|>
TITLE: What tools cannot work for orbifolds?
QUESTION [24 upvotes]: Consider all of your basic constructions/tools/theorems for manifolds: fundamental group, Euler characteristic, triangulations, orientation, smoothness, bundle structure, cobordisms, etc.. Viewing orbifolds as the natural generalization of manifolds by quotients of group actions (or just locally $\mathbb{R}^n/G_i$), it seems that we can translate all of our notions for manifolds onto orbifolds with slight adjustments in the definitions. I am curious as to the extent of that statement's validity (which is important, because orbifolds are arising everywhere for me):
What basic constructions that exist on manifolds, cannot (or currently do not) have analogs for orbifolds?
Why? ($\leftarrow$ in case there is more to say than "orbifolds have singularities")
As pointed out nicely below, there is the question of whether such constructions will be useful to (all) orbifolds. So this thread also seeks known constructions which are either intractable or trivial.

REPLY [13 votes]: The integral cohomology ring of orbifolds does not satisfy Poincaré duality. Indeed, orbifolds do not even have finite cohomological dimension.<|endoftext|>
TITLE: Force-directed graph drawing in 1D?
QUESTION [6 upvotes]: I have a real-world graph that I wish to draw in one dimension.  Here's the graph:

I'd like to draw it using some kind of force-directed graph drawing method.  I'm supposing this is both possible and practical, but it may be that there's some obstacle I'm unaware of.  So, I'll begin with:

Question:  Is 1D force-directed graph drawing possible and practical?

And if so:

Question:  Does there exist software for force-directed graph drawing in 1D?

Gephi (among other packages) can provide a force-directed graph drawing in 2D, but I don't believe it can draw in 1D.
(Motivation: I'd like to see how similar the 1D force-directed drawing is to the drawing above, which places the vertices equispaced along the line, and in the order in which they are generated.)

REPLY [3 votes]: As an alternative to force-directed layout,
you might investigate the MinLA problem: the minimum linear arrangement problem for
nodes on a line. It plays a role in VLSI design. 
It is NP-hard but there are good approximation algorithms.
Here is one paper that will lead to many others:

Yehuda Koren and David Harel.
  "A Multi-Scale Algorithm for the Linear Arrangement Problem."
  Graph-Theoretic Concepts in Computer Science. Springer Berlin Heidelberg, 2002.
  (ACM link)

From its Introduction:<|endoftext|>
TITLE: Is the conjugacy problem in $\mathbb{Q}^n \rtimes \mathbb{Z}^m$ solvable
QUESTION [6 upvotes]: Given two elements in $\mathbb{Q}^n \rtimes_\phi \mathbb{Z}^m$, is there an algorithm that decides if they are conjugate?  Just to be explicit, $\phi$ is a homomorphism from $\mathbb{Z}^m \to Aut(\mathbb{Q}^n)$, through which elements of $\mathbb{Z}^m$ act on $\mathbb{Q}^n$.  If the answer is unknown, can anyone point me in the direction of texts/papers that deal with decision problems in infinitely generated groups?

REPLY [4 votes]: This is not an answer, but you will get an answer by doing this. Look at the paper by Noskov, http://link.springer.com/content/pdf/10.1007%2FBF01138933.pdf. As in that paper, reduce the problem to a commutative algebra problem, then use "Constructive" Commutative Algebra.<|endoftext|>
TITLE: To what extent can fields be classified?
QUESTION [21 upvotes]: The study of algebraic geometry usually begins with the choice of a base field $k$. In practice, this is usually one of the prime fields $\mathbb{Q}$ or $\mathbb{F}_p$, or topological completions and algebraic extensions of these. One might call such fields $0$-dimensional. Then one could say that a field $K$ is $d$-dimensional if it has transcendence degree $d$ over a $0$-dimensional field.
But is there a way to make this less ad hoc? Is there a reason I've never seen any such definition of $0$-dimensional fields? Am I missing something?
To what extent has the classification of abstract fields been considered?

REPLY [26 votes]: I think you are lumping too many disparate kinds of fields together under the heading "zero-dimensional".  As Jason says in his answer, there are some precise definitions of dimensions of fields (e.g. cohomological dimension but also other definitions of a field-arithmetic nature).
Another important comment is that in modern algebraic / arithmetic geometry there is no restriction on the kind of field that can be taken as a "ground field", i.e., over which to define algebraic varieties.  Although you have indeed listed some common ground fields, some people think about other special cases as well as the general case.  (For instance I have recently been thinking about elliptic curves defined over transfinitely iterated function fields.)  
The basic classification of fields, as I understand/view it, is as follows: 
Step 1: Every field has a characteristic, either $0$ or a prime number.  There are no homomorphisms between fields of different characteristics, so fields of a given characteristic are somehow different worlds (one can think of the characteristic as a connected component in the category of fields, essentially).
Step 2: For a fixed characteristic $p \geq 0$ there is a unique minimal field, called the prime subfield, say $k_0$.  This is precisely the initial object in the category of fields of characteristic $p$: it's $\mathbb{Q}$ in characteristic $0$ and $\mathbb{F}_p$ in characteristic $p$.  This means that the absolute theory of fields is reduced to the theory of field extensions, and one can define "absolute invariants" on a field $K$ by giving invariants of $K/k_0$.  In particular:
Step 3: The absolute transcendence degree of a field $K$ is the cardinality of a transcendence basis for $K/k_0$.  This means that there is a subextension $k_0 \subset F \subset K$ such that $F/k_0$ is purely transcendental: a rational function field, perhaps in infinitely many variables, and $K/F$ is algebraic.  
Step 4: The algebraic extension $K/F$ is in many ways the most interesting part. For instance, as Qiaochu Yuan mentions, algebraic geometry in dimension $n$ is the study of finite degree field extensions of $\mathbb{C}(t_1,\ldots,t_n)$.
When $n = 1$ we get precisely the compact Riemann surfaces, or (replacing $\mathbb{C}$ with any field $k$ and looking at finitely generated field extensions of transcendence degree $1$) of complete, regular, integral algebraic curves over $k$.  So a classification here is the (rich) classical story of the genus, the moduli spaces $\mathcal{M}_g$, and so forth.  When $k$ is not algebraically closed, Galois cohomology enters the picture.
When $n \geq 2$ we are studying birational algebraic geometry only, but that is still very rich.  When $n = 2$ there is a "classification of algebraic surfaces" which ends up tossing most of them into a very large box called "general type".  To the best of my knowledge there is no explicit description of the connected components of the (infinite type) moduli space of all complex algebraic surfaces like there is in dimension one.  
In fact, I believe that probably starting even in dimension $2$ it is in some sense hopeless to try to give an algorithmic classification, although I cannot recall having seen a precise impossibility theorem along these lines analogous e.g. to the algorithmic impossibility of classifying compact $4$-manifolds because this problem -- via the fundamental group -- contains the word-problem for finitely presented groups which Novikov proved is algorithmically unsolvable.  Bjorn Poonen discussed similar (open) problems in the last of a series of three lectures on undecidability that he gave at UGA a few years ago; notes are available on his webpage.  
Step 5: A vaguely dual role to purely transcendental extensions of the prime subfield is played by the algebraically closed fields.  Here there is a precise classification: the characteristic and the absolute transcendence degree determine an algebraically closed field up to isomorphism.  If the field is uncountable, then the absolute transcendence degree is just its cardinality, so that classifies the field: e.g. the only algebraically closed field of characteristic $0$ and continuum cardinality is $\mathbb{C}$.  This has been used for some sneaky purposes: e.g. the algebraic closure of $\mathbb{Q}_p$ must then also be isomorphic to $\mathbb{C}$.  (However for countable fields cardinality is not enough: e.g. $\overline{\mathbb{Q}}$ is not isomorphic to $\overline{\mathbb{Q}(t)}$.)  This uncountable categoricity means that the first order theory of algebraically closed fields of given characteristic $p \geq 0$ is complete, which has various pleasant consequences.  Algebraically closed fields of infinite transcendence degree have some further nice properties which makes them suitable for use as "ground fields" in algebraic geometry, as was exploited by Weil in his pre-(scheme-theoretic) foundations.  From a model-theoretic perspective, these large algebraically closed fields enjoy good saturation properties.  
Fields which are "close" to being algebraically closed tend to be better understood than fields which are farther away from being algebraically closed.  One can measure this (at least in characteristic $0$) by the size / complexity of the absolute Galois group $\operatorname{Aut}(\overline{K}/K)$.  Thus for instance the absolute Galois group of $\mathbb{Q}_p$ is known completely -- i.e., it is a topologically finitely generated compact totally disconnected Hausdorff group, and explicit generators and relations are known, which in some sense means we understand every algebraic extension of $\mathbb{Q}_p$.  This needs to be taken with a grain of salt: a certain filtration on the absolute Galois group provides much coarser information -- e.g. you can break the group up into three pieces, two of which are commutative and one of which is pro-$p$, so the group is pro-solvable.  That's useful: the pro-$p$-part is the hard part and knowing generators and relations for it does not in practice seem to take away the mystery.  For instance, the Local Langlands Correspondence is a deep theorem about representations of the absolute Galois group of a $p$-adic field, and it was certainly not proved by looking at its explicit structure as a topologically finitely presented group!
Let me note finally that every compact totally disconnected topological group occurs up to isomorphism as the automorphism group of an algebraic Galois extension of fields (the Leptin-Waterhouse Theorem; I discovered this independently as a graduate student, and at the time I knew neither that others had published the result before nor even that such a result would be publishable)...so algebraic field extensions can be awfully complicated.<|endoftext|>
TITLE: Prospects for reverse mathematics in Homotopy Type Theory
QUESTION [22 upvotes]: Reverse mathematics, as I mean here, is the study of which theorems/axioms can be used to prove other theorems/axioms over a weak base theory.  Examples include

Subsystems of Second Order Arithmetic (SOSOA).  Theorems expressible in second order arithmetic are compared over $\mathsf{RCA}_0$ (which is roughly the theory of computable sets of natural numbers). 
Constructive Reverse Mathematics.  Bridges and others use Bishop-style constructive mathematics as a base theory to compare various nonn-constructive principles.

Here are my questions.

Would it be possible to do reverse mathematics over Homotopy Type Theory (HoTT)?
Would HoTT make for a good base theory (roughly corresponding to computable objects in mathematics)? My understanding is that HoTT is a constructive theory, this makes me believe the answer is yes.
Would the univalence axiom need to be removed from the base theory?  My understanding is that the univalence axiom is incompatible with the law of excluded middle.  This would be problematic.
Would the results be similar to those in Constructive Reverse Mathematics?  (Would they be exactly the same?)
Would computability theoretic ideas be of use as they are in SOSOA reverse mathematics?  Computability theorists are drawn to SOSOA because of its computability theoretic nature.
Would the computer implementations of HoTT, in say Coq, be helpful in proving reverse-mathematics-like results.  It would be nice to be able to do reverse mathematics with the "safety net" of a proof checker.
Has reverse mathematics in HoTT been done already in any sense (formally or informally)?

Note:
There may be some relationship with this question: Forcing in homotopy type theory

REPLY [8 votes]: At present, homotopy type theory is purported to be the internal language (i.e. syntax) of $(\infty,1)$-topoi. That univalence is an axiom is to say that there is a class of higher topoi that satisfy univalence and a class of higher topoi that do not. For instance, the topos of sets, together with trivial higher morphisms, do not satisfy univalence.
Suppose we wish to pursue the reverse mathematics of any theorem that has been proven in homotopy type theory. Take any theorem, say, like in my comment, that the fundamental group of the circle is the integers. In this paper by Shulman and Licata, they mention that univalence is required to prove this theorem. As I see it, the circle in the topos of sets, where univalence does not hold, is actually just the singleton set. Every path and higher path is trivial. Hence the fundamental group of the circle, in the model of the topos of sets, is trivial. 
A reverse mathematician interested to study the strength of this topogical result would want to see how much of univalence is required. Is there a higher topos in which univalence fails yet the fundamental group of the circle is still the integers? Can we characterize those higher topoi in which the fundamental group of the circle is the integers? Can we characterize these topoi syntactically?<|endoftext|>
TITLE: Relation between math and piano music
QUESTION [17 upvotes]: What, if any, is the relation between Cantor's function and Ligeti studio: Devil's Staircase?

REPLY [22 votes]: There is a Masters thesis by Lauren Halsey entitled,
"An examination of rhythmic practices and influences in the keyboard works of György Ligeti"
(UNCG link), which addresses your question:

"...the idea for this etude emerged: “an endless 
  climbing, a wild apocalyptic vortex, a staircase it was almost impossible to ascend.”28
  This etude shares the name and characteristics of the mathematical concept of a “devil’s 
  staircase.” This phenomenon, based on Cantor Sets, involves the relationship of 
  disproportional segments combining to create a self-similar group.29 This concept is also 
  used in the “mode locking” features of clocks and pendulums.30 Ligeti expresses this 
  concept with the inclusion of groups of two and three eighth notes that, when combined, 
  create a self-similar rhythmic set. This grouping structure creates pulse streams and 
  defines the formal boundaries of this piece. The structures seem to spiral infinitely up the 
  piano, suddenly falling down to the lowest octaves.31


     

And here is a graph from Wikipedia's Cantor function article of the Devil's Stairase:<|endoftext|>
TITLE: Rankings of mathematical conferences and journals
QUESTION [25 upvotes]: Just to keep it simple: What rankings of mathematical conferences and journals are available in the internet?
(I'm only interested in rankings, not about any discussion about rankings.)

REPLY [3 votes]: Here are the rankings of (not just math) journals employed by some countries:

Brazil (in Portuguese)
Finland
Norway<|endoftext|>
TITLE: Origin and first uses of $\ell_p$ norms?
QUESTION [26 upvotes]: When exactly were $\ell_p$ norms first defined and used?
(Here is what I know, or think I know: Lebesgue and/or Riesz had something to do with them, but in some sense they go back to Minkowski, since Minkowski's inequality is (in essence) the statement that an $\ell_p$ norm is a norm.)
Here is what is really my main question: how were $\ell_p$ norms ($p\geq 1$ arbitrary) first used? What was their motivation? It is clear that $\ell_1$, $\ell_2$ and $\ell_\infty$ norms are very natural, and their use long predates the formal definition of "form". The $\ell_4$ norm also pops up on its own sometimes. In contrast, $\ell_p$ norms for other $p$ seem to arise most often in the course of a proof, as a tool, when one needs some notion of "size" that falls between an $\ell_1$ and an $\ell_2$ norm (for example). Did the first uses of $\ell_p$ norms fit this framework? Can you think of some interesting (and preferably early) instances that do not obey this pattern?

REPLY [13 votes]: Toeplitz in his review of Riesz [1913] laments the lack of explicit motivation in generalizing from $\ell_2$ to $\ell_p$:

The considerations of the 3rd Chapter require not the convergence of the sum of squares of the unknowns, but the more general convergence of $\sum|x_a|^p$, where $1<p<\infty$ (even the limiting cases $p = 1$, $p = \infty$ are discussed), a generalization on which the author seems to place considerable value, but whose deeper analytic significance he doesn't further motivate.

But isn't the motivation simply that this increases the chances for a system $\sum a_{ik}x_k=c_i$ to have a solution? Namely (if $p>2$ say) we are allowed to look for solutions in the wider space $\ell_p$ — but the price to pay is that we must know that each "row" $a_{i\,\cdot}$ is in $\ell_q$ where $\frac1p+\frac1q=1$. That seems to be the point of the theorem that Riesz [1913, p. 47] attributes to Landau [1907]: 

If $\sum a_kx_k$ converges for all $x\in\ell_p$, then $a\in\ell_q$ (and $|\sum a_kx_k|\leqslant\|a\|_q\|x\|_p$).$(*)$

This, in retrospect, is essentially the proof that $\ell_p$ has dual $\ell_q$, and I would say it qualifies as a use of $\ell_p$ norms predating Riesz. But as to who first used this to solve a concrete problem...? I don't know.
Another pre-Riesz $\ell_p$ result is the Hausdorff-Young inequalities for Fourier series $f(e^{i\theta})=\sum c_ne^{in\theta}$, proved by Young in [1912a] (resp. [1912b]) for $q\in 2\mathbf Z$ and later by Hausdorff in general:

If $\frac1p+\frac1q=1$ and $1<p\leqslant 2$, then $\|c\|_q\leqslant\|f\|_p$ (resp. $\|f\|_q\leqslant\|c\|_p$). 

More secondary sources — in addition to the Dieudonné and Pietsch texts cited by András: 

Riesz [1913, p. 45] attributes the Minkowski inequality to Minkowski [1907, p. 95], but there is no such thing there. According to this paper the correct reference is Minkowski [1896, p. 116].
Encyklopädie der math. Wiss., Vol. II.3.2 (1927) reports on both Riesz's $\ell_p$ theory with forerunners Hölder and Landau (p. 1445, no mention of Minkowski), and on Hausdorff-Young (pp. 1211-1212). 
A more recent survey: The Hausdorff-Young theorems of Fourier analysis and their impact (1994).
According to this paper, Hölder's inequality $(*)$ goes back to Grolous (1875) and Besso (1879).

Yet again, none of these references really address the motivation for generalizing from $\ell_2$ to $\ell_p$. Maybe it was a case of "because we can..."?<|endoftext|>
TITLE: counting points on unit sphere mod p
QUESTION [12 upvotes]: Let $f(n)$ be the number of points on the unit sphere $x^2 + y^2 + z^2 = 1\; \pmod n$ with $x,y,z \in \mathbb{Z}/n\mathbb{Z}$  
This is sequence A087784 in the Online Encyclopedia of Integer sequences:

1, 4, 6, 24, 30, 24, 42, 96, 54, 120...

There is a (due to Bjorn Poonen) indicating some regularity to the solutions to this congurence
$$f(n) = n^2* \left\{\begin{array}{cl}3/2&\text{if}\quad\quad 4|n \\
1 &\text{otherwise}
\end{array}\right\}*\prod_{\substack{p|n \\ 1 \mod 4}} \left( 1 + \frac{1}{p}\right)* \prod_{\substack{p|n \\ 3 \mod 4}} \left( 1 - \frac{1}{p}\right)$$
What are some proofs to this identity ?  

Sequence A060968 is the number of points on the unit circle $x^2 + y^2  \equiv 1\; \pmod n$ 

1, 2, 4, 8, 4, 8, 8, 16, 12, 8, 12,...

with a similar multiplicative formula:
 $$g(n) = n* \left\{\begin{array}{cl}2&\text{if}\quad\quad 4|n \\
1 &\text{otherwise}
\end{array}\right\}*\prod_{\substack{p|n \\ 1 \mod 4}} \left( 1 + \frac{1}{p}\right)* \prod_{\substack{p|n \\ 3 \mod 4}} \left( 1 - \frac{1}{p}\right)$$
Perhaps there is a tower of such identities.

The multiplicative structure of these formulas could have an algorithmic interpretation.  The formula for the Euler phi function
\[ \phi(n) =  n \prod_{p|n} \left( 1 - \frac{1}{p} \right) \]
This suggests a sieving algorithm to generate the list of numbers relatively prime to n

write down the numbers $\{ 1, 2, \dots, n \}$
for reach prime $p|n$ cross out multiples

I'd be especially interested if this type of algorithm existed for $f(n), g(n)$.

REPLY [9 votes]: Since this old question has resurfaced, let me sketch two ways to prove the stated formulæ using algebraic geometry:
The first way is fairly elementary.  Let us stick for definiteness with the number $f(n) := \#\{(x,y,z) \in (\mathbb{Z}/n\mathbb{Z})^3 : x^2+y^2+z^2=1\}$ of points on the ($2$-)sphere in $3$-dimensional affine space $\mathbb{A}^3_{\mathbb{Z}/n\mathbb{Z}}$ over $\mathbb{Z}/n\mathbb{Z}$.  As others have pointed out, by the Chinese Remainder Theorem, $f$ is multiplicative.  Furthermore, as $\{(x,y,z)\in\mathbb{A}^3_{\mathbb{Z}} : x^2+y^2+z^2=1\}$ is smooth outside of the prime $2$ (because the equation and its partial derivatives generate the unit ideal in $\mathbb{Z}[\frac{1}{2}][x,y,z]$), an appropriate form of Hensel's lemma or the definition of (formal) smoothness ensures that $f(p^{k+1}) = p^2\cdot f(p^k)$ whenever $p$ is an odd prime (the exponent $2$ in $p^2$ is the dimension of the tangent space).
So, leaving aside the case of the prime $2$ (I'm merely sketching the argument here), we are left with proving that the number $f(p)$ of points over $\mathbb{F}_p$ is $p(p+1)$ or $p(p-1)$ according as $p$ is congruent to $1$ or $3$ mod $4$.  Now stereographic projection $\varphi\colon (x,y,z) \mapsto (\frac{x}{1-z}, \frac{y}{1-z})$ defines a birational map between the sphere and the plane, with inverse $(u,v) \mapsto (\frac{2u}{1+u^2+v^2}, \frac{2v}{1+u^2+v^2}, \frac{-1+u^2+v^2}{1+u^2+v^2})$, so it is "almost" a bijection, and we just have to analyse the points where it fails: $\varphi$ is defined everywhere except when $z=1$ which, when $p\equiv 3\pmod{4}$, is just one point $(0,0,1)$, whereas when $p\equiv 1\pmod{4}$, consists of the $(x,\pm\sqrt{-1}\,x,1)$, so, $2p-1$ of them; as for the image of $\varphi$, it consists of those $(u,v)$ such that $1+u^2+v^2 \neq 0$, which is the complement of a conic having $p-1$ or $p+1$ points according as $p$ is congruent to $1$ or $3$ mod $4$, so the image of $\varphi$ has $p^2-p+1$ or $p^2-p-1$ points, and finally the sphere has $p^2+p$ or $p^2-p$ in each case.
The second, more sophisticated way, works in any dimension.  Once again leaving aside the troublesome prime $2$, the idea is to write the ($(n-1)$-)sphere in dimension $n$ as $\mathit{SO}(n)/\mathit{SO}(n-1)$ where $\mathit{SO}(n)$ refers to the orthogonal group of the quadratic form $x_1^2 + \cdots + x_n^2$.  Now the latter is a "standard" quadratic form over the reals, but over $\mathbb{F}_p$, when $n=2m$ is even, this form is said to be of "plus type" (= Witt index $m$) for $p\equiv 1\pmod{4}$ or $m$ even, and of "minus type" (= Witt index $m-1$) for $p\equiv 3\pmod{4}$ and $m$ odd; for $n$ odd, there is only one nondegenerate quadratic form over $\mathbb{F}_q$.  Now the order of the groups $\mathit{SO}(2m,{+})$ (for a quadratic form of "plus" type), $\mathit{SO}(2m,{-})$ ("minus" type) and $\mathit{SO}(2m+1)$ over $\mathbb{F}_q$ are known (up to trivial factors, they are what group theorists write $D_m$, $^2D_m$ and $B_m$), and are one half of:

$\#\mathit{Spin}(2m,{+})(\mathbb{F}_q) = q^{m(m-1)} (q^m-1) \prod_{i=1}^{m-1}(q^{2i}-1)$
$\#\mathit{Spin}(2m,{-})(\mathbb{F}_q) = q^{m(m-1)} (q^m+1) \prod_{i=1}^{m-1}(q^{2i}-1)$
$\#\mathit{Spin}(2m+1)(\mathbb{F}_q) = q^{m^2} \prod_{i=1}^{m}(q^{2i}-1)$

(See any number of books on finite simple groups or algebraic groups for a proof and explanation of these formulæ.)  So by taking the ratio of these quantities we get the number of points on the sphere: for example, for the ($2$-)sphere in dimension $3$, we have $\#\mathit{Spin}(3)(\mathbb{F}_p) = p^3 - p$ and $\#\mathit{Spin}(2,{+})(\mathbb{F}_p) = p - 1$ and $\#\mathit{Spin}(2,{-})(\mathbb{F}_p) = p + 1$, giving $f(p) = p(p+1)$ for $p\equiv 1\pmod{4}$ and $f(p) = p(p-1)$ for $p\equiv 3\pmod{4}$.  But of course, the algebraic group method makes it possible to compute many more things.<|endoftext|>
TITLE: What should the morphisms in the Category of Directed Sets be?
QUESTION [8 upvotes]: Directed sets are defined to be sets equipped with a preorder that admit (finitary) upper bounds e.g. pairs $(D, \preceq)$ such that $\forall p,q \in D$ there exists $r \in D$ such that $p \preceq r$ and $q \preceq r$. Equivalently, they may be defined as thin categories in which every finite diagram admits a cocone.
In either case, there is some intuition as to what a morphism of directed sets ought to be: in the first case, perhaps monotone functions; in the second, a functor such that there exists a cocone over any finite diagram that maps to a cocone over the image.
However, these aren't the only two descriptions of directed sets nor are the two suggested definitions equivalent.  Finally, I haven't been able to find a source which describes a (the?) category of directed sets.  Is there a consensus on the 'right' definition of a morphism of directed sets?  Further, are there any good resources on the properties of the category of directed sets?

REPLY [5 votes]: Here is a partial answer.
Given a directed set $D$, we say that a subset $A\subseteq D$ is cofinal if for each $d\in D$ there is some $a\in A$ with $d\leq a$. We say that $A\subseteq D$ is bounded if $A\subseteq\downarrow d=\{x\in D|x\leq d\}$ for some $d\in D$ and we say that $A$ is unbounded if it is not bounded.  Let $D,E$ be two directed sets. Then we say that a function $f:D\rightarrow E$ (not necessarily order preserving) is a cofinal map (also called a convergent map) if the image of every cofinal subset of $D$ is a cofinal subset of $E$. A function $g:E\rightarrow D$ (not necessarily order preserving) is said to be unbounded (also called a Tukey map) if the image of every unbounded subset of $E$ is an unbounded subset of $D$. If $D,E$ are posets, then there is an unbounded map $f:D\rightarrow E$ if and only if there is a cofinal map $g:E\rightarrow D$, and in either case we say that $D$ is Tukey reducible to $E$ and we write $D\leq_{T}E$. The Tukey ordering $\leq_{T}$ in a sense measures how big your directed set is. Clearly the class of directed sets can be made a category where the morphisms are either the cofinal maps or the unbounded maps, and these categories are enough for us to define the Tukey ordering. I conjecture that one can take a quotient category of one of these categories to get a more natural category of directed sets(i.e. where a directed set is isomorphic to each of its cofinal subsets), but I don't see how exactly to go about this. Perhaps someone else will give a better and more complete answer to this question.<|endoftext|>
TITLE: Does Anyone Know Anything about the Determinant and/or Inverse of this Matrix?
QUESTION [20 upvotes]: The matrix I am inquiring about here is the $n \times n$ matrix where the entry $A_{ij}$ is $\frac{1}{(i+j-1)^2}$. The $2 \times 2$ matrix looks like 
$$
\begin{pmatrix}
1 & 1/4 \\
1/4 & 1/9
\end{pmatrix}.
$$
The $3 \times 3$ case looks like 
$$
\begin{pmatrix}
1 & 1/4 & 1/9 \\
1/4 & 1/9 & 1/16 \\
1/9 & 1/16 & 1/25 
\end{pmatrix}.
$$
The $n \times n$ matrix looks like
$$
 \begin{pmatrix}
1 & 1/4 & \cdots & 1/(n^2) \\
1/4 & 1/9 & \cdots & 1/{(n+1)^2} \\
\vdots & \vdots & \cdots & \vdots \\
1/(n^2) & 1/{(n+1)^2} & \cdots & 1/{(2n-1)^2}
\end{pmatrix}.
$$
If anyone has any information or knows of any papers talking about this matrix please let me know. Thanks!

REPLY [10 votes]: Just a couple observations, building upon and explaining why Igor Rivin found that the denominators are often perfect squares or close to being such.
Let $A$ be the matrix with components $A_{ij} = 1/(x_i + y_j)^2$.  Expressing its determinant as a sum over permutations and putting everything on the same denominator gives
$$
\det A = \frac{P(\{x_i\}, \{y_i\})}{\prod_{i,j} (x_i + y_j)^2} \,,
$$
for some polynomial $P$ with integer coefficients.  Also, if two $x_i$ or two $y_i$ are equal, then the corresponding rows/columns are equal, and the determinant vanishes.  Hence
$$
\det A = \frac{Q(\{x_i\}, \{y_i\}) \prod_{i\neq j} (x_i - x_j) (y_i - y_j)}{\prod_{i,j} (x_i + y_j)^2} \,.
$$
Actually, as pointed out in other answers, the Cauchy formula provides $Q$ as the permanent of some matrix, but all I need here is to know $Q$ has integer coefficients.
In your particular case, the products give
$$
\frac{\prod_{i\neq j} (i - j) ((i - 1) - (j - 1))}{\prod_{i,j} (i + j - 1)^2}
=
\frac{\prod_{i < j} (j - i)^4}{\prod_{i,j} (i + j - 1)^2}
=
\prod_{j=1}^N \left[(j - 1)!^4 \frac{(j - 1)!^2}{(j + N - 1)!^2} \right]
=
\frac{\prod_{j=0}^{N-1} j!^8}{\prod_{j=0}^{2N-1} j!^2}
$$
Note that $j!^8$ divides $(2j)!(2j+1)!$, so the numerator divides the denominator.  All in all,
$$
\det A = \frac{Q}{\prod_{j=0}^{2N-1} j!^2 \bigg/ \prod_{j=0}^{N-1} j!^8}
$$
for some integer $Q$.  The denominator appearing in this formula is typically rather close to the ones given in Igor Rivin's answer, and are trivially squares: $1$, $12^2$, $2160^2$, $6048000^2 = 672000^2 \cdot 9^2$, etc.  The discrepancy is of course accounted for by cancellations between $Q$ and the product of factorials.  Again using Igor Rivin's data, I find (indices denote the size of the matrix)
$$
\begin{aligned}
  Q_1 &= 1 \\
  Q_2 &= 7 \\
  Q_3 &= 647 = 2^3 \cdot 3^4 - 1 = 8 \cdot (16 \cdot (6 - 1) + 1) - 1 \\
  Q_4 &= 878769 = 16 \cdot (12 \cdot (32 \cdot (144 - 1) + 1) - 1) + 1 \\
  Q_5 &= 18203480001 = 40000 \cdot ( 48 \cdot ( 120 \cdot (80 - 1) + 1) - 1) + 1 \\
  Q_6 &= 5850859031888599 \,.
\end{aligned}
$$
Here I've included some expressions which may or may not generalize.  In any case, it is certain that $Q$ will not be expressed as a (simple) product, as it involves some very large prime factors.<|endoftext|>
TITLE: Which values can attain the minimum solid angle in a simplex
QUESTION [9 upvotes]: Given a simplex $S$ with a vertex $v$ by the solid angle at this vertex I mean the value $\hbox{vol}(B \cap S)/\hbox{vol}(B)$ where $B$ is a small enough ball centered at $v$ (for example, in the plane, 1/4 corresponds to $90^°$). Let $\rho_{min}(S)$ denotes the minimum of the solid angles of $S$. 

My question is what values can $\rho_{min}(S)$ attain when it ranges over simplices in dimension $d$?

Comments: Since it is easy to let $\rho_{min}(S)$ to be close to $0$, the question above is equivalent with asking what is the least upper bound on $\rho_{min}(S)$.
In the plane, the question is trivial (the least upper bound is 1/6), thus more interesting cases occur when $d \geq 3$. In higher dimensions, it is relatively easy to show that $\rho_{min}(S) \leq \frac1{2(d+1)}$. But I suspect that this value should be much smaller.
A natural candidate for the least upper bound is the size of the solid angle in the regular simplex. Is this value the least upper bound?
Actually, I would be also happy with a weaker upper bound (than the one from the regular simplex) such as $\rho_{min}(S) \leq \frac{1}{2^d}$ (this is the solid angle in the $d$-cube).

REPLY [6 votes]: This answer is based on a comment by J. Kynčl. It gives a bound roughly $\rho_{min}(S) \leq \left(\frac{1}{1.074}\right)^d$ which is already exponentially decreasing, but perhaps still far from the optimum. Thus, a better bound would still be of interest.
Let $AB$ one of the longest edges of $S$. Without loss of generality, at least half of the remaining vertices of $S$ are not farther from $B$ than from $A$. Let $V_1, \dots, V_k$ be such vertices ($k \geq (d-1)/2$) and $U_1, \dots, U_{\ell}$ be the remaining vertices (closer to $A$ than to $B$). As J. Kynčl points out, the angles $V_iAB$ are at most $60^°$. We also need that angles $U_iAB$ are at most $90^°$ since $BU_i$ is at most as long as $AB$.
Let $h$ be the hyperplane perpendicular to $AB$ passing through $A$ and $h^+$ be the halfspace containing $B$ with the boundary hyperplane $h$. Let $C$ be the cone with apex $A$ determined by $S$ (that is, our task is to determine which fraction of $C$ belongs to a ball $B(A,\varepsilon)$ with center $A$ and small radius $\varepsilon$. From the discussion above it follows that $C$ is fully contained in $h^+$. Furthemore, let $\kappa$ be the affine $(k+1)$-space determined by $A$, $B$, $V_1$, $\dots$, $V_k$. We also need another $(k+1)$-dimensional cone $C_{60}$ which is formed by all points $X$ in $\kappa$ such that the angle $XAB$ is at most $60^°$. From the discussion above it follows that $V_i \in C_{60}$, and consequently $C \cap \kappa \subseteq C_{60}$. It is not too difficult to compute that $\hbox{vol}_{k+1}(B(A,\varepsilon) \cap C_{60})/\hbox{vol}_{k+1}(B(A,\varepsilon)) \leq \left(\frac{\sqrt 3}2\right)^{k+1}$, where $\hbox{vol}_{k+1}$ is the $(k+1)$-dimensional volume in $\kappa$. (See the left picture below.)

Now, let us consider a small enough ball $B(A, \varepsilon)$ and let us estimate $\hbox{vol}(B(A, \varepsilon) \cap C)/ \hbox{vol}(B(A, \varepsilon))$.
For this let us consider a $(k+1)$-space $\kappa'$ parallel with $\kappa$. The task is to show that 
$$(*) \hskip{2cm} \frac{\hbox{vol}_{k+1}(B(A, \varepsilon) \cap C \cap \kappa')}{\hbox{vol}_{k+1}(B(A, \varepsilon) \cap \kappa')} \leq \left(\frac{\sqrt 3}2\right)^{k+1}.$$
As soon as we show $(*)$ we get the same bound on $\hbox{vol}(B(A, \varepsilon) \cap C)/ \hbox{vol}(B(A, \varepsilon))$ by the Fubini theorem.
In order to show $(*)$, let us first reailize that $C \cap \kappa'$ is either empty or it equals to  $(C \cap \kappa) + Y$, where $Y$ is the intersection point of $\kappa'$ and the $\ell$-dimensional cone determined by $A$ and $U_1, \dots, U_\ell$ (here, for simplicity, we assume that $A$ is the origin). Thus, in particular, $Y \in h^+$ and $C \cap \kappa' \subseteq Y + C_{60}$. (See the right picture above.)
The final step is thus to show that $\hbox{vol}_{k+1}(B(A, \varepsilon) \cap (Y + C_{60})) \leq \hbox{vol}_{k+1}(B(A, \varepsilon) \cap (Z + C_{60}))$ where $Z$ is the center of $B(A, \varepsilon) \cap \kappa'$. This inequality is best shown by the picture below. (First shift $Y$ to $Y'$ on $h$. Then bound $\hbox{vol}_{k+1}(B(A, \varepsilon) \cap (Y' + C_{60}))$ by decomposing $B(A, \varepsilon) \cap (Y' + C_{60})$ into two parts as on the middle and right picture.)

Finally, $\hbox{vol}_{k+1}(B(A, \varepsilon) \cap (Z + C_{60}))/\hbox{vol}_{k+1}(B(A, \varepsilon)) \leq \left(\frac{\sqrt 3}2\right)^{k+1}$ as in the case of $C_{60}$. This gives the final bound
$$\rho_{min}(S) \leq \left(\frac{\sqrt 3}2\right)^{(d+1)/2} \leq \left(\frac{1}{1.074}\right)^d.$$<|endoftext|>
TITLE: Interesting behaviour of Brion's formula under a degenerate change of variables
QUESTION [6 upvotes]: This is, probably, a question for those knowledgeable on the subject of Brion's theorem and its applications.
Lately, I've been dealing with situations of the following sort. Suppose we are given a polytope $P\subset\mathbb{R}^n$ as well as a linear map $\varphi:\mathbb{R}^n\rightarrow\mathbb{R}^m$ with $m<n$. Bases are fixed in both spaces and $\varphi$ maps integral points to integral points (in other words, has an integral matrix with respect to these bases). Also $P$ is rational and thus subject to Brion's theorem.
Now, $\varphi$ can be viewed as a change of variables $F$: $$F:x_i\rightarrow\prod\limits_{j=1}^m y_j^{\varphi_{ji}}, 1\le i\le n.$$ Here $x_i$ are the exponents of basis vectors in $\mathbb{R}^n$ (the variables in Brion's formula), $y_j$ -- the exponents of basis vectors in $\mathbb{R}^m$. Thus, here $\exp(\bar v)$ is simply substituted by $\exp(\varphi(\bar v))$.
In my case $F$ is applicable to the identity provided by Brion's theorem (the denominators do not vanish identically) and I noticed the following to hold. The summands (vertex cones' integer point transforms) corresponding to vertices of $P$ not mapped to vertices of $\varphi(P)$ vanish under $F$!
I see that the naive generalization is far from being correct... But isn't there, just maybe, a known theorem stating something of the sort?
The above is what I'm chiefly after. However, another interesting trait of my polytopes is that the summands which don't vanish, turn into neat products superficially resembling integer point transfroms of simplicial cones, namely $F$ transforms them into $$\exp(\varphi(v))\prod\limits_{i=1}^n \frac 1{1-\exp(\varphi(e_i))},$$ where $\{e_i\}$ is a subset of edges of the (non-simplicial) vertex cone at vertex $v$.
Any known reasons for such behaviour to take place? Unfortunately, I can't grasp any geometrical sense of the set $\{e_i\}$.
Update. Yes, here is a small nontrivial example. Pyramid with vertices $(0,0,0)$,$(2,0,0)$,$(0,2,0)$,$(2,2,0)$ and $(1,1,1)$, map $(a,b,c)\rightarrow(a,b)$ (projection onto the base). Then the integer point transform of the cone at the apex is $$f=\frac{(1+xy)(1-z^{-2})}{(1-xyz^{-1})(1-x^{-1}y^{-1}z^{-1})(1-xy^{-1}z^{-1})(1-x^{-1}yz^{-1})}$$ and $F$ is given by $x\rightarrow t,y\rightarrow w,z\rightarrow 1$. Thus, $f$ is seen to vanish under F.
Update 2. Actually, just like in the above example, each of the vertex cones $C$ the integer point transfrom of which vanishes, has two edges $e_1$ and $e_2$ such that $\varphi(e_1)=-\varphi(e_2)$. This forces $\varphi(C)$ to have an apex of positive dimension and sometimes gives me the feeling that the rest should be obvious via some formal argument...
Update 3. Maybe, I didn't make clear enough what $F$ is... The definition I give is a bit formal as compared to natural. 
The natural way might be to first introduce $F$, substituting each variable with a Laurent monomial in some 
other (smaller) set of variables. Then the matrix $φ$ can be defined via 
$F$.
And, just in case, in general, $\mathrm{IPT}(φ(C))\not=F(\mathrm{IPT}(C))$! (Integer Point Transform.)

REPLY [3 votes]: It seems that when your map $\phi$ maps a vertex of $P$ to a non-vertex of $\phi(P)$, you're looking at a cone in the image of $\phi$ that now contains a whole line.  Thus, by general Brion-type tangent cone analysis, any cone generating function containing a whole line must vanish.<|endoftext|>
TITLE: Is there a graph that is Ramsey for $P_{2n}$ but is $C_{2n+1}-$free
QUESTION [6 upvotes]: Write $F\to G$ to mean that for every two coloring of the edges of $F$, there exists a monochromatic copy of $G$. Nešetřil  and Rödl proved that for every graph $G$, there exists a graph $F$ such that $F\to G$ and $\omega (F)=\omega (G)$ (where $\omega$ is the size of the largest clique in the graph).
My question is an extension of that. Is there a graph $F$ so that $F \to P_{2n}$ but so $F$ is $C_{2n+1}-$free? This is trivial for $n=1$. For $n=2$, the graph $F:= C_6\cup \{(x,v_i)|i=1,3,5\}$ works (where $v_1,\dots,v_6$ are the vertices of the $C_6$ and $x\not \in C_6$).
The motivation is that if this is true, I can construct another graph with a special property (using this lemma) that will help me solve a much more interesting problem (depending on how interesting this itself is!) 
So my question is, does this generalize to all $n$? It seems that it should.

REPLY [6 votes]: A simple proof of much more: for each $n$ and $g$, there is a graph $G$ with girth at least $g$ that is Ramsey for $P_n$. As shown by Erdos, there are graphs of arbitrarily large girth and chromatic number. Let $G$ be a graph with girth at least $g$ and chromatic number at least $4n+1$, so it contains a subgraph of minimum degree at least $4n$. Any two-coloring of the edges of $G$ thus contains a monochromatic subgraph of average degree at least $2n$ and hence a subgraph of minimum degree at least $n$. One can then greedily build a monochromatic path of length $n$, showing that $G$ is Ramsey for $P_n$.<|endoftext|>
TITLE: The fundamental group of a closed surface without classification of surfaces?
QUESTION [16 upvotes]: The fundamental group of a closed oriented surface of genus $g$ has the well-known presentation 
$$
\langle x_1,\ldots, x_g,y_1,\ldots ,y_g\vert \prod_{i=1}^{g} [x_i,y_i]\rangle.
$$
The proof I know is in two steps: 1. draw your favorite presentation of the surface onto a sheet of paper and compute 
the fundamental group using Seifert-van Kampen. 2. Appeal to the classification of surfaces to prove that any surface 
is diffeomorphic to what you drew.
Here is my question: 
''Can one avoid using the classification of surfaces? More specifically, can one prove that a discrete subgroup of $PSL_2 (R) $ that acts freely and cocompactly on the upper half plane must be of the above form - using a group-theoretic argument and without refering to the classification of surfaces or something that comes close to it?''
By ''something that comes close to it'', I mean an argument using Morse theory or another device that decomposes a 
surface into simpler parts.
Background: while contemplating again about the well-known paper by Earle and Eells ('A fibre bundle description of 
Teichmueller theory'), I realized that their main arguments can be upgraded slightly to give at once

closed surfaces are determined up to diffeomorphism by their fundamental groups
each isomorphism of fundamental groups is realized by a diffeomorphism which is unique up to isotopy (Dehn-Nielsen-Baer-Epstein)
the group of diffeomorphisms homotopic to the identity is contractible (the original Earle-Eells result)

and I would like to know whether this also gives the classification cheaply. According to the above two step argument, 
I would be happy with an argument that proves that two surfaces with the same genus (defined by the relation $\chi = 2 -2g$) 
must have isomorphic fundamental groups.

REPLY [8 votes]: I will answer the question of whether this also gives the classification cheaply.
No.
It gives the classification at the expense of proving that every surface group has a free cocompact action on the upper half plane (or on the euclidean plane, or on the 2-sphere). This in turn depends upon proving: 

Every surface has a triangulation (Rado's Theorem, and now you've done 90% of the work of the classification theorem)...
Every triangulated surface has a compatible smooth structure...
Every smooth surface has a compatible conformal structure...

and, finally

Every conformal structure has a compatible hyperbolic metric, euclidean metric, or spherical metric (the uniformization theorem).<|endoftext|>
TITLE: positive elements in tensor product
QUESTION [7 upvotes]: Let x be  a positive  element in the spatial tensor product of two non unital C* algebras
A  and B. Is there a single element $a \otimes b \geq x$?
How can we noncommutativize the following proof, in the commutative case:
Let $F^2$ be  a positive function on $X\times Y$. Define $f(x)=\sup_{y\in Y} F(x,y)$  and $g(y)=\sup_{x\in X} F(x,y)$,
then $F^2 \leq fg$.

REPLY [12 votes]: Yes. For a self-adjoint element $y$, denote $s(y)=\sup{\rm Sp}(y)$. Then for $\gamma \geq s(y)$, one has
$$\inf \lbrace s(y - \gamma(e\otimes f)) : 0\le e\le 1,\ 0\le f\le 1\rbrace \le 0.$$
Indeed, if $e_n$ and $f_n$ are approximate units, then so is $g_n:=e_n\otimes f_n$ and 
$y - \gamma g_n \le y-g_n^{1/2} y g_n^{1/2} \to 0$.
Now let $y_0 := x \le 1$ and find $0 \le e_n \le 1$ and $0 \le f_n \le 1$ recursively so that the elements
$y_{n+1} := y_n - 4^{-n}(e_n \otimes f_n)$ satisfy $s(y_{n}) \le 4^{-n}$ for all $n$.
Let $e = \sum_{n=0}^\infty 2^{-n}e_n$ and likewise for $f$. Then, one has
$$x = y_0 \le \sum_n 4^{-n}e_n\otimes f_n \le e\otimes f.$$
If $A$ and $B$ have strictly positive elements, one can arrange $(e\otimes f) - x$ is strictly positive. 
I think with more efforts one can find $e$ and $f$ such that $\| e \| \| f \| = \| x \|$.<|endoftext|>
TITLE: Are simplicial sets the intended model of HoTT?
QUESTION [11 upvotes]: While thinking about Jason Rute's question, I wondered if there was an intended model for HoTT. The main candidate for the intended model are simplicial sets, where Vladimir Voevodsky first observed the univalence phenomenon. However, it is not clear that HoTT is intended to describe this model as opposed to groupoid models or a broader class of models.
A related question is whether there is a notion of standard model for HoTT. That is, a notion comparable in role to ω-models for second-order arithmetic and transitive/well-founded models for set-theory.

REPLY [13 votes]: My impression is as follows.
While I cannot speak for Voevodsky, he certainly gives the impression that simplicial sets is his favorite, if not the intended model. For example, he would suggest axioms on the basis of them being valid in simplicial sets (such as excluded middle for mere propositions).
If one takes HoTT as the internal language of something, then the something will be something like $(\infty, 1)$-toposes, but this has not been worked out yet in precise detail. This point of view will probably appeal to categorical logicians and others close to category theory. One certainly would not worry about the intended model under this view.
Classical first-order logic and model theory often use the concept of "standard model" or "intended model", despite the fact that many results explain that belief in such a model requires a certain amount of blind faith (for example, faith in ones own mathematical instinct). I find it intriguing that Martin-Löf also seems to have an intended model of his theory. He talks about it explicitly. So even though many participants of the HoTT project are of the categorical kind, others are of a more orthodox type-theoretic kind. My impression is that for them a more profound foundational shift is hapenning with HoTT. One consequence of this is the urgency to resolve the computational content of the Univalence axiom, without which the axiom is the forbidden fruit.
During the Univalent year at the Institute for Advanced Study there were several discussions where the two views arose in opposition to each other. This was a very fruitful situation, as it made everyone think harder. When we wrote the book, we decided at a very early stage not to speak about models or other "meta" topics, such as the meta properties of the underlying type system. After all, if HoTT aspires to be a foundation of mathematics, then it cannot place itself on top of any notion outside HoTT. It must build directly on top of reader's premathematical knolwedge. At the same time, it has to respect existing informal mathematical tradition as much as possible, or else it is just a logicians' hobby.
Thus, while the question of the intended or standard model is important, I think it is perhaps not the right question to ask about a proposed new foundation, because it immediately makes it harder to take the new system at face value, directly and not through the eyes of a logician. I understand of course the logician's urge to view the landscape from the meta-level, and it is good that there are such people (me being one of them). My point is that to understand the intent and the value of a new foundation one actually has to descend and truly live with it for a while, to see what sort of mathematical intuitions it begets. I think anthropologists could teach us a lesson.<|endoftext|>
TITLE: What Approximation Property does the space of Schatten-p class operators have?
QUESTION [5 upvotes]: Background
This is a follow-up question to:
What (classes of) Banach spaces are known to have Schauder basis?
In the previous question, I asked about what spaces are known to have Schauder basis. It seems that not a lot of positive results are available in that area. So I am restricting my question to one of the spaces that is in particular relevant to my research:

Does $C_p$, the Banach space of all Schatten-p operators on a separable infinite-dimensional Hilbert space, have property $\pi$? Or does it have a (Schauder) basis?

Thank you!
Reference
Good reference for different approximation properties: Handbook of the Geometry of Banach Spaces, vol. 1 -- contribution by Pete Casazza

REPLY [8 votes]: It has a Schauder basis, namely $e_i\otimes e_j$, where $(e_i)$ is an orthonormal basis of the Hilbert space. This holds for $1\le p<\infty$. For the Schatten ideal of compact operators, I do not know the answer.<|endoftext|>
TITLE: Small quadrilaterals containing a given convex region
QUESTION [9 upvotes]: It is easy to prove that
(*) Every convex planar set of area 1 is contained in a quadrilateral of area 2.
It is also easy to see that statement (*) remains true if the constant 2 is replaced with a somewhat smaller one. Contest: Find such a constant, the smaller the better.
Update:
Reaching $\sqrt 2$ and even a strictly smaller value was proved by Chakerian (1973) and Kuperberg (1983) and the research challenge offered is to improve it even further, and perhaps even to verify the conjecture that the minimum is attained by a regular pentagon. But any nice arguments for bounds below 2 are welcome.

REPLY [5 votes]: G. D. Chakerian, Minimum area of circumscribed polygons, Elem. Math. 28 (1973), 108–111, MR0322682 (48 #1044) proved that if $K$ is a convex body of area 1 in the plane then $K$ is contained in a quadrilateral of area at most $\sqrt2$. 
W. Kuperberg, On minimum area quadrilaterals and triangles circumscribed about convex plane regions, Elem. Math. 38 (1983), no. 3, 57–61, MR0703939 (85a:52009), proved that the infimum of the area ratio is strictly less than $\sqrt2$, and suggested the infimum might be attained when $K$ is a regular pentagon.<|endoftext|>
TITLE: sufficient conditions on Non-Haken manifolds
QUESTION [5 upvotes]: Is there an algorithm to detect the  Non-Haken Manifold?
Or, is there a sufficient condition for a manifold to be 
a non-Haken manifold? (off course,  I hope that condition is not the ones in its 
definition.)
Note: a Non-Haken manifold means that either it is reducible or it contains no 2-sided properly embedded
incompressible surface.

REPLY [10 votes]: Yes, there is an algorithm due to Jaco-Oertel to detect if an irreducible manifold is Haken, and therefore to detect a non-Haken manifold (if it is given to be irreducible). The phrasing is a bit confusing (are you assuming the given manifold is irreducible?), but given Rubinstein's solution to sphere recognition, there is also an algorithm to detect if a manifold is irreducible. See Jaco and Tollefson's exposition for improvements to these algorithms, and the papers of Ben Burton (arXiv, MathSciNet) for implementations.
Addendum: Sufficient conditions for being Haken are easier than for being non-Haken. For example, if a manifold has $b_1>0$, or has positive-dimensional $SL_2\mathbb{C}$ character variety, then it is Haken. In fact, verifying a manifold is Haken is likely NP. However, showing something is non-Haken is more difficult. There are also techniques to analyze when a group has no fixed-point free action on a tree. Some version of this was implemented by Fenley (Laminar free hyperbolic 3-manifolds, Comment. Math. Helv. 82 (2007), no. 2, 247–321) to find laminar-free 3-manifolds, and in principle could be used to prove that a 3-manifold is non-Haken. However, it seems that this requires an exponential search.
Work of Dani Wise implies that a compact 3-manifold is Haken or reducible if and only if it surjects an infinite virtually-free group.<|endoftext|>
TITLE: Has negative cardinality been considered?
QUESTION [12 upvotes]: I'm not a set theorist. Perhaps the answer to my question is so well known that it is not appropriate to answer here or my question is absurd. Nevertheless I'll try: I have heard the rumor that, similar to the extension of positive integers to negative integers, recently(?) also attempts have been made to investigate negative cardinalities.
My question: What would be the intuitive picture behind negative cardinality and can also in that domain countable and uncountable cardinals exist?

REPLY [3 votes]: See here.
More specifically, see here.<|endoftext|>
TITLE: Strength of Bishop style constructive mathematics vs $\mathsf{RCA}_0$
QUESTION [7 upvotes]: This question came out of this other MO question of mine.  My question is

Is there a formal comparison between $\mathsf{RCA}_0$ and $\mathsf{BISH}$ (Bishop style constructive mathematics as used in constructive reverse mathematics)?

More specifically,

Is $\mathsf{BISH}$ strictly weaker than $\mathsf{RCA}_0$ (or $\mathsf{RCA_0}$ plus full induction) when formalized and restricted to second order sentences of arithmetic?
(Update: I wasn't entirely clear, I meant to ask if everything provable in $\mathsf{BISH}$ is provable in $\mathsf{RCA_0}$ plus full induction.  Obviously, nonconstuctive principles like LEM are provable in $\mathsf{RCA_0}$ but not in $\mathsf{BISH}$.)

I am sure people have thought of this, but I couldn't find a resource.
Also, I imagine there could be a lot of caveats.  I don't know much about Bishop-style constuctivism, but I gather the community doesn't like formal theories or models, which are generally needed for such comparisons.  However, I know others are interested in such things, and I believe there are formalizations of $\mathsf{BISH}$ that at least get close to the intuitive idea.
Also, this question can be answered without formal theories:

Is there a theorem known to be constructively provable (in the informal style of Bishop), that is not provable in $\mathsf{RCA}_0$ (or $\mathsf{RCA}_0$ plus full induction)?

REPLY [3 votes]: Maybe it is interesting that for $CT$ (Church thesis) $BISH+CT$ is consistent, but $RCA_0\vdash \neg CT$. By $CT$ I mean axiom scheme of $CT$ for every formula.
$CT-scheme$: For every first order formula $\psi(x,y)$, $\forall x \exists y \psi(x,y)\rightarrow \exists e\forall x\exists y(\psi(x,y)\land \phi(e,x,y)$ where $\phi(e,x,y)$ is kleene predicate.
Also let $U(e,x,y)$ be a first order formula such get godel number of a formula like $\eta(x,y)$ in $e=\ulcorner \eta(x,y) \urcorner$, gets $x,y$ and simulates $\eta(x,y)$, then $CT$ can be formalized in one axiom like this:
$$CT:=\forall u\exists e(\forall x\exists y U(u,x,y)\rightarrow \forall x\exists y(U(u,x,y)\land \phi(e,x,y)))$$
Define $$h(n)=\left\{\begin{matrix} 1,\exists x\phi(n,n,x)\\  0, \forall x \neg \phi(n,n,x) \end{matrix}\right.$$
This function can represent by formula $H(n,y):=(\exists x\phi(n,n,x)\land y=1)\lor(\forall x\neg\phi(n,n,x)\land y=0)$, trivially $RCA_0\vdash \forall n \exists y H(n,y)$, therefore $RCA_0\vdash \neg CT$, but $BISH+CT\nvdash \bot$, similar arguments work if we take axiom scheme of $CT$ instead of $CT$<|endoftext|>
TITLE: Review papers in mathematics
QUESTION [9 upvotes]: Are there review papers, literature reviews in mathematics that describe the recent developments in various fields for a newcomer? Or is the prerequisite knowledge always provided in research monographs and textbooks? That's one can read research papers and make contributions to a field after reading some textbooks or monographs? In physics, all textbooks in my area of interest seem outdated and I should read review articles (see string wiki) that summarize more recent development. I wonder if there are review articles that describe the most important recent achievements in geometry, Analysis, algebra etc.

REPLY [6 votes]: For the last ten years or so, David Eisenbud has organized a "Current Events Bulletin" session at the January joint math meetings, in which, typically, four speakers survey recent work in various fields, with a booklet of the survey papers available at the time of the meeting (and permanently at the website).  Here is a portion of Eisenbud's description:

The Current Events Bulletin Session at the Joint Mathematics Meetings,
  begun in 2003, is an event where the speakers do not report on their
  own work, but survey some of the most interesting current developments
  in mathematics, pure and applied.  The wonderful tradition of the
  Bourbaki Seminar is an inspiration, but we aim for more accessible
  treatments and a wider range of subjects.<|endoftext|>
TITLE: Enumeration of graphs with a given and bounded degree sequence
QUESTION [5 upvotes]: What is the best known asymptotic formula for the number of graphs with a given degree sequence $(d_1, ... ,d_n)$, when the degrees are bounded by a constant and the number of vertices $n$ goes to infinity?
There are several papers of Mckay et al that provide bounds when the degrees are all $O(n^{\frac{1}{2}})$, but I am interested in bounded degrees, and I was hoping for more exact bounds for this special case. There are works on this as well, but they all seem to be pretty old, so I was wondering what the current state of the art is.

REPLY [4 votes]: In my comment I was misreading the question, sorry.  The situation for the real question is as follows. For very low degrees (say, at most 3) it isn't hard to get the exact number as a single or double summation.  For the more general case of bounded maximum degree, the best asymptotic formula is my result with Wormald (Combinatorica, 11 (1991) 369-382), which in this case has an error term $O(1/n)$ where $n$ is the number of vertices of non-zero degree. It would be possible (though somewhat tedious) to find an explicit bound on the error term, but as far as I know nobody has done it.  It would also be possible (in this special case of bounded degree) to obtain the precise term of order $1/n$ and have an error term of size $O(1/n^2)$.  But, again, I think nobody has done it.<|endoftext|>
TITLE: Is every module the colimit of its finitely generated submodules? (for algebraic spaces or stacks)
QUESTION [8 upvotes]: For (quasi-compact and quasi-separated) schemes there is a categorical way to characterise quasi-coherent sheaves of finite type using purely the abelian category $\operatorname{QCoh}(X)$. In an abelian category one can define a categorically finitely generated object as being an object M such that for any directed family of subobjects $M_\alpha \subset M$, such that the sum $\sum_\alpha M_\alpha$ is equal to the original $M$, then there exists an index $\alpha_0$ such that $M_{\alpha_0} = M$.
Now, in the category of $R$-modules this is easy to see, and the argument essentially relies on the fact that any module is the colimit of its finitely generated submodules.
For quasi-compact and quasi-separated schemes the only reference I could find is Daniel Murfet's excellent notes. This is Corollary 64 from here.
I would like to know whether this result still holds for algebraic spaces or more generally for algebraic stacks. (I imagine that knowing this is essentially equivalent to proving Lemma 61 in Murfet's notes for a scheme but using the étale toplogy)
EDIT: Perhaps one cook up a variant of Proposition 15.4 of Laumon-Moret-Bailly. Unfortunately the argument there seems to use at the very end that a submodule of a finite type module is of finite type, which does not hold in the non noetherian case.
EDIT2: By the way, for qcqs algebraic spaces this was already known to Raynaud-Gruson [RG Proposition 5.7.8].

REPLY [8 votes]: Note that on any qcqs algebraic space or Artin stack, a quasi-coherent sheaf is finite type if its pullback to any fppf scheme cover is finite type in the usual sense on schemes.  We can similarly define the notion of "finite type" for a quasi-coherent sheaf of modules over any quasi-coherent sheaf of algebras on such an algebraic space or stack in the same way, and this is the notion that will be used below; it satisfies the categorical condition you want, so that is good enough.
In this paper, absolute noetherian approximation is proved for qcqs algebraic spaces, building on the version for schemes proved by Thomason and Trobaugh: according to Theorem 1.2.2, every qcqs algebraic space $X$ is an inverse limit (with affine transition maps) of finitely presented algebraic spaces over $\mathbf{Z}$. In particular, $X$ is affine over an algebraic space $X_0$ of finite presentation over $\mathbf{Z}$.  
Hence, we have an affine morphism $f:X \rightarrow X_0$ to a noetherian algebraic space, so quasi-coherent $O_X$-modules are the "same" as quasi-coherent sheaves of modules over $X_0$ over the quasi-coherent $O_{X_0}$-algebra $A := f_{\ast}(O_X)$. Consequently, it suffices to show that for any noetherian algebraic space $X_0$ and quasi-coherent $O_{X_0}$-module $A$, every quasi-coherent $A$-module $F$ is the direct limit of its directed system quasi-coherent $A$-submodules of finite type.  The underlying $O_{X_0}$-module $F'$ of $F$ is the direct limit of its directed system of coherent $O_{X_0}$-submodules $F'_i$.  Hence, the image $F_i$ of the natural $A$-linear map $A \otimes_{O_{X_0}} F'_i \rightarrow F$ is a quasi-coherent $A$-submodule of $F$ of finite type (over $A$), and it contains $F'_i$, so clearly the inclusion $\varinjlim F_i \rightarrow F$ is an isomorphism.
That settles the case of qcqs algebraic spaces, and to prove the same for qcqs Artin stacks you just need absolute approximation for qcqs Artin stacks (which would give that any such stack is affine over a noetherian one, so you can bootstrap from the known noetherian case exactly as above). In the case that the Artin stack is qcqs with quasi-finite diagonal (e.g., any Deligne--Mumford stack), this is proved in this paper (which includes as a special case qcqs algebraic spaces, so it is an alternative to the initial reference at the top).<|endoftext|>
TITLE: Grothendieck trace on local cohomology?
QUESTION [10 upvotes]: Let R be an augmented regular local ring over a field $k$ with maximal ideal m. There is the Grothendieck residue symbol:
$$Res: H^n_m(\Omega^n) \to k$$
If $k=\mathbb{C}$ and $R$ is affine space, this map has a well-known interpretation by way of an integral formula. 
If X is a projective scheme over $\mathbb{C}$, there is the Serre-duality trace 
$$ tr: H^n(\omega) \to \mathbb{C}$$ which at least when $X$ is smooth can be represented similarly in terms of integrals. My question is whether there is a way to combine these two ideas namely suppose $X$ is a smooth quasi-projective variety over $\mathbb{C}$ and $Z$ a proper subscheme, can we define a trace:
$$ tr/Res: R^n\Gamma_{Z}(\Omega_X^n) \to \mathbb{C} $$
Is there a description for this map in terms of integrals?
Most of the literature on residues, Grothendieck duality and the like is way too heavy for me. Where is this written up in a friendly way? 
Edit: Proposition 2.3.2 or Remark 2.3.8 of http://www.math.purdue.edu/~lipman/papers/formal-duality.pdf seem close to what I want (or maybe are exactly what I would like). Still, I would be happy if there is a more simple minded reference with some analytic treatment.

REPLY [3 votes]: Perhaps you should look at Lipman's book:
"Dualizing sheaves, differentials and residues on algebraic varieties",
Asterisque 117
It gives a treatment of the topic you asked for. In brief, there is a commutative triangle:
$\require{AMScd}$
\begin{CD}
\bigoplus_{P \in X} H^n_P(\Omega^n) @>can>> H^n(X,\Omega^n)\\
@V\oplus_{P \in X}res_PVV \swarrow \tiny{\int_X} \\
k 
\end{CD}
Expressing the fact that the integral is computed through residues. The treatment is very down to earth. In particular, he avoids derived categories.<|endoftext|>
TITLE: What if the Riemann Hypothesis were false?
QUESTION [66 upvotes]: There are lots of known and interesting consequences of the Riemann Hypothesis being true. Are there any known and interesting consequences of the Riemann Hypothesis being false?

REPLY [46 votes]: An explicit zero $\rho$ for $\zeta(s)$, off the critical line, would give an explicit lower bound on the class number $h(-d)$ for $\mathbb Q(\sqrt{-d})$, for a range of $-d$ in terms of $\text{Im}(\rho)$.  This is the 'Deuring-Heilbronn phenomenon,' with results due to these two and others beginning in the 1930's.  For an elementary account, see
http://arxiv.org/abs/1201.0713<|endoftext|>
TITLE: Is Hopf property a quasi-isometry invariant?
QUESTION [17 upvotes]: Recall that a group $G$ is called Hopfian if every surjective endomorphism $G\to G$ is injective. Malcev observed that all finitely-generated (f.g.) residually finite groups are Hopfian. It is well-known that residual finiteness is not a coarse invariant, i.e. a residually finite f.g. group can be quasi-isometric to a non-residually finite one. For instance, Burger and Mozes proved that 
$F_2\times F_2$ is quasi-isometric to a simple group. Earlier examples, due to Deligne, were of non-residually finite central extensions of residually finite groups, with kernel of order 2. Deligne's examples imply that residual finiteness is not even a virtual isomorphism invariant. 
Question 1: Is Hopfian property of preserved by quasi-isometries of f.g. groups?  
Natural candidates would be examples of non-Hopfian CAT(0) groups constructed by Dani Wise in his thesis. However, I do not know if such groups are quasi-isometric to, say, residually finite groups. A subquestion of Question 1 is:
Question 2. Suppose that $G$ is a group acting geometrically on a product of simplicial trees of finite valence. Can $G$ be non-Hopfian? 
Note that such $G$ is necessarily quasi-isometric to a product of free groups and such products are residually finite.

REPLY [23 votes]: The answer to Question 1 is "no". The group $\langle x, y \mid  x^{12}y = yx^{18}\rangle$
is Hopfian but contains a non-Hopfian subgroup of finite index (see  Baumslag, Gilbert; Solitar, Donald Some two-generator one-relator non-Hopfian groups. Bull. Amer. Math. Soc. 68 1962 199–201.).
 Edit.  Since the paper of Baumslag and Solitar contains almost no proofs, and some results of it turned out to be wrong, it is probably better to refer to Proposition A of  Collins, Donald J.; Levin, Frank
Automorphisms and Hopficity of certain Baumslag-Solitar groups. 
Arch. Math. (Basel) 40 (1983), no. 5, 385–400 in combination with  S.Meskin, On residually finite one-relator groups. Trans. Amer. Math. Soc. 164, 105--114 
(1972). The simplest Hopfian group with non-Hopfian subgroup of finite index would then be 
$\langle a,b \mid ab^2=b^4a\rangle$. Collins and Levin proved that this group is Hopfian and  Meskin proved that it contains a non-Hopfian subgroup of finite index.<|endoftext|>
TITLE: Small eigenvalues and spectral clustering
QUESTION [7 upvotes]: Let $L$ be the discrete Laplacian associated to an undirected graph.  It is well-known that the spectral gap of $L$, i.e. the smallest nonzero eigenvalue, is a measure of how well connected the graph is; the smaller the spectral gap, the fewer cuts are needed to disconnect the graph.
I have observed experimentally that the number of "small" eigenvalues corresponds to the number of "loosely connected clusters" in the graph.  For example, take four graphs A, B, C, D, each with four vertices, such that every vertex is connected to every other vertex within each of the four graphs.  Then form a single connected graph G by connecting a vertex in A to a vertex in B, a vertex in B to a vertex in C, and a vertex in C to a vertex in D.  The first few eigenvalues of the Laplacian of G are:
0, .1126, .3542, .5688, 4, 4,...
I suspect that if this observation were formulated correctly then it would not be hard to turn into a theorem, but if anyone happens to know a reference I would appreciate it.  My main question is: can one formulate a quantitative version of my observation?  In other words if I give you the spectrum of the Laplacian of a graph, can you tell me how many "loosely connected clusters" there are in the graph?  I don't want to commit to any precise meaning of the phrase "loosely connected clusters"; I'm hoping for answers that include an appropriate definition.

REPLY [3 votes]: It seems that this is a rather difficult problem, some aspects of which are still open.  Upon further research I think the best answer currently available is in this paper:
http://arxiv.org/abs/1111.1055<|endoftext|>
TITLE: cover and hide with squares
QUESTION [5 upvotes]: I am studying two numbers, related to squares, that can characterize a polygon P:

MinCoverNumber = the minimum number of axis-aligned squares required to exactly cover P (the covering squares may overlap, but may not cover points outside P). For example: a square has MinCoverNumber=1. A 4-by-5 rectangle has MinCoverNumber=2, as it can be covered by 2 overlapping 4-by-4 squares. An L-shape has MinCoverNumber=3, if it is fat enough. A triangle has MinCoverNumber=$\infty$, because it cannot be exactly covered by a finite number of axis-aligned squares.
MaxHideNumber = the maximum number of dots that can be placed inside P, such that no two dots can be covered by a single square. For example, a square has MaxHideNumber=1, a 4-by-5 rectangle has MaxHideNumber=2, etc.

Obivously, MinCoverNumber is an upper bound for MaxHideNumber, for example: if MinCoverNumber=3 (such as a fat L-shape), then for every 4 dots, at least 2 of them are covered by one square, therefore MaxHideNumber$\leq$3. However, I don't know if this is also a lower bound. 
If MinCoverNumber=2, then obviously MaxHideNumber=2, but if MinCoverNumber=3 (i.e. a polygon that cannot be covered by 2 squares), I haven't managed to prove that MaxHideNumber=3 (i.e. it is possible to hide 3 dots such that no 2 are coverable by a single square). I also haven't managed to find a counter-example.
So, my question is:
Is it possible to find an axis-aligned polygon P, such that MaxHideNumber(P) < MinCoverNumber(P)?
An alternative presentation of the question for MinCoverNumber=3:
Is it possible to find an axis-aligned polygon P, such that P cannot be covered by 2 squares, but in every set of 3 dots in P, there is a subset of 2 dots that are covered by a single square contained in P?

REPLY [7 votes]: Problem 1 is solved in a paper I coauthored:

Covering orthogonal polygons with squares
  L. J. Aupperle and H. E. Conn and J. M. Keil and Joseph O'Rourke
  Proc. 26th Allerton Conf. Commun. Control Comput., pp. 97-106, 1988.
  (link to PDF scan link).


That paper establishes a time complexity of $O(n^{5/2})$, but it was subsequently
improved to $O(n^{3/2})$. Here the polygon $P$ is assumed to have vertices on
the integer lattice, and $n$ is the number of unit squares ("pixels") inside $P$.
Problem 2 is called the maximum hidden vertex set problem, although usually visibility
is just by straight lines-of-sight, rather than being enclosed in a square, which
is a type of restricted visibility. I am not recalling work resolving your exact question,
but my guess is (a) it exists and (b) you could find it by searching on that phrase, e.g., 

Antonio L. Bajuelos, Santiago Canales, Gregorio Hernández, A. Mafalda Martins.
  "Estimating the Maximum Hidden Vertex Set in Polygons."
  2008. (ACM link)

See also the related MO question, "Upper bounds on art gallery problems using independent witnesses," and that alternative terminology for the same concept.
(After clarifications):
I think (but am not certain) that your question is addressed and answered positively
in this paper:

Michael O. Albertson and Claire J. O’Keefe.
  "Covering Regions with Squares."
  SIAM. J. on Algebraic and Discrete Methods, 2(3), 240–243.
  (SIAM journal link.)

Here's the abstract:

A unit square in $R^2 $ whose corners are integer lattice points is called a block. A board consists of a finite set of blocks. Given a board B, its graph $G(B)$ has vertices corresponding with the blocks of B, and two vertices of $G(B)$ are joined by an edge provided the corresponding blocks are contained in a square subset of B. If B is simply connected, then $G(B)$ is perfect.<|endoftext|>
TITLE: Metrically homogeneous subsets of the plane
QUESTION [12 upvotes]: A metric space $M$ is metrically homogeneous if for every pair of points $x, y \in M$ there is an isometry $f$ of $M$ onto $M$ such that $f(x)=y$. What is known about metrically homogeneous spaces? Any references? In particular, is there an explicit description of all metrically homogeneous subspaces (under the Euclidean metric) of the plane $\mathbb{R}^2$?
Remark. The same question in one dimension has an easy answer: A subset $M$ of $\mathbb{R}^1$ is metrically homogeneous if and only if $M$ is the union of two cosets $G+a\ $ and $G+b\ $ of an additive subgroup $G$ of $\mathbb{R}^1$.

REPLY [9 votes]: Here is an description of homogeneous subsets of ${\mathbb R}^2$ using group theory, similar to the one you have in dimension 1. Let $M\subset {\mathbb R}^2$ be homogeneous and let $G< I=Isom({\mathbb R}^2)$ be its group of isometries. Then $G$ fits into short exact sequence
$$
1\to T\to G\to S \to 1
$$
where $T$ is a group acting by translations on the plane and $S$ is a subgroup of $O(2)$. Unlike in 1-dimensional case, this sequence need not split, but you can still find a subset $C\subset G$ of coset representatives of $S$ so that $G=TC$ (I am using action on the left). Then $M=G\cdot m$ for some $m\in M$ and, hence, 
$$
M= T\cdot (Cm),
$$ 
i.e., $M$ is a disjoint union of $T$-orbits, where $T$ is a subgroup of ${\mathbb R}^2$. This is an analogue of the description that you liked in the 1-dimensional case, except now $S$ is typically infinite. Thus, you have a (typically) infinite disjoint union, indexed by the set $S/G_m$, where $G_m$ is the projection of the stabilizer of $m$ in $G$ to the group $S$. I think, this is the best one can do without introducing further restrictions on the set $M$. Even in the case when $S$ is a finite dihedral group, there will be infinitely many possibilities for choosing a subgroup $T\subset {\mathbb R}^2$ normalized by $S$.  
In order to get something more geometrically appealing, let's assume that $M$ is a closed subset of ${\mathbb R}^2$. This immediately implies that the group $G$ is a closed subgroup of the Lie group $I=Isom({\mathbb R}^2)$. Basic Lie theory tells you that $G$ is a Lie subgroup of $I$. Now, the problem essentially reduces to classification of Lie subgroups of $I$. Here it is:

$dim(T)=2$, then $T$ acts transitively on ${\mathbb R}^2$ and, hence, $M={\mathbb R}^2$.
$dim(T)=1$. Then: 

a. Either $T$ is either isomorphic to ${\mathbb R}$, acting via translations along a line $L\subset {\mathbb R}^2$, or 
b. $T\cong {\mathbb R}\times  {\mathbb Z}$, where the first factor acts by translations along a line $L$ and the second acts by translations in the orthogonal direction. 
In either case, clearly, $M$ is a product ${\mathbb R}\times M_1$, where $M_1$ is a discrete homogeneous subset of the real line (there are four possibilities for $M_1$ which you already know: single point, two points, orbit of the discrete group of translations or union of two such orbits). 
3. $dim(T)=0$. Thus, $G$ is a discrete group of Euclidean isometries. 
There are again 3 subcases here, depending on the rank of the free abelian group $T$ ($0, 1$, or $2$). If $T=0$ then $G< O(2)$ and, hence, $G$ is either 1-dimensional (in which case $M$ is a round circle) or finite cyclic or dihedral group; in the finite case everything reduces to vertex sets of regular or semiregular planar convex polygons. If $rank =1$ then $S$ is a subgroup of $Z_2\times Z_2$ and $M$ is either contained in one line or in two parallel lines.  
The most interesting case is when $T$ has rank 2 and, hence, $G$ is a Euclidean crystallographic group (Russians call them "Fedorov groups"). If you so desire, you can now go through the well-known list of such groups and identify $G$-orbits in ${\mathbb R}^2$. Some will give you esthetically pleasing vertex sets of regular and semiregular "floor-tiling" patterns on ${\mathbb R}^2$. If I were Joseph O'Rourke, I would add some nice pictures here, but I am not.<|endoftext|>
TITLE: Is the Alexander-Pontryagin duality applicable to stratified spaces
QUESTION [5 upvotes]: If $D$ is the discriminant of the space of all planar curves of a fixed degree, and $D'$ is the subspace whose only singularities are nodes or cusps, then is it possible to apply Alexander-Pontryagin duality to the pair $(D,D\setminus D')$ to conclude that $\mathrm{H}_N(D,D\setminus D';\mathbb{Z}/2)=\mathrm{H}^0(D';\mathbb{Z}/2)$? Here $N$ is the dimension of $D$. 
I have seen this being used, but I am not sure why one is permitted to use the duality here. After all, $D$ is not a manifold, but does it form a nice enough space for which the Alexander-Pontryagin duality is applicable? Can the duality be applied to stratified spaces? 
Thank you!

REPLY [6 votes]: Set $X=D,Y=D\setminus D',n=\dim D$. I presume you are interested in real algebraic curves (in the complex case everything simplifies). The problem is that $X\setminus Y$ is not smooth. But there is an ad hoc way to handle this. Let us triangulate $X$ so that $Y$ is a subcomplex. Assume that $H_n(X,\mathbb{Z}/2)=\mathbb{Z}/2$ and that $X$ is compact. (Actually, in the real case there are two possible definitions of smoothness: given a plane real curve, one may require that the corresponding complex curve is smooth or that it has no real singularities so depending on the definition of $D$ these assumptions may or may not be true.) The "fundamental class" of $X$ mod 2 is represented in the simplicial chain group by the sum of all the simplices: if we take a combination of all simplices but one, it will come from a subcomplex homotopy equivalent to something of dimension $< n$. So each $n-1$-simplex is in the boundary of an even number of $n$-simplices.
So the relative homology group $H_n(X,Y)$ is freely generated by the sums $\sum_{Int(\sigma)\subset C}\sigma$ where $C$ is a component of $X\setminus Y$ and $Int$ denotes the interior. There is one such element for each component, which proves the isomorphism in the original posting.
Remarks:

This does not generalize to higher (co)homology or to other coefficients. The general Alexander duality theorem is the following statement: suppose $X\supset Y$ is a pair of nice spaces (say, the one point compactification of $X$ can be made into a CW-complex so that the closure of $Y$ is a subcomplex) such that $X\setminus Y$ is a manifold. Let's assume it orientable. Then $$H^{BM}_*(X,Y)\cong H^{BM}_*(X\setminus Y)\cong H^{n-*}(X\setminus Y)$$
where $d$ is the real dimension of $X$, $H^{BM}$ denotes the Borel-Moore homology (it coincides with the usual homology for compact spaces) and the second isomorphism is the Poincar\'e-Lefschetz duality. So $H^0(X\setminus Y)\cong H_n^{BM}(X,Y)$. Assuming that in addition to the above $H^{BM}_{n-1-i}(X)=H_{n-i}^{BM}(X)=0$ we get $H^i(X\setminus Y)\cong H_{n-1-i}^{BM}(Y)$ from the long exact sequence for the Borel-Moore homology. In the case $X=\mathbb{R}^n$ we get the usual Alexander duality; for $i=0$ these assumptions are not quite satisfied: $H^{BM}_n(\mathbb{R}^n)=\mathbb{Z}$, so we get an exact sequence $$0\to \mathbb{Z}\to H_n^{BM}(X,Y)\to H^{BM}_{n-1}(Y)\to 0$$ instead.
There is a duality theory for singular spaces called the Verdier duality. Informally, it says the following. Let $X$ be a nice space (e.g. a finite CW complex) and $A$ a commutative ring. Let $D^b(X)$ be the derived category of the category of complexes of sheaves of $A$-modules on $X$ whose cohomology sheaves are constructible with respect to some stratification. Then there is an anti-autoequivalence $F\mapsto F^\vee$ of $D^b(X)$ such that $H^*(X,F)\cong H^{-*}(X,F^\vee)$ where $H^i(X,F)$ denotes $\mathop{\mathrm{Hom}}\nolimits_{D^b(X)}(\underline{A},F[i])$, $\underline{A}$ being the constant sheaf with stalk $A$. If $X$ is an orientable manifold (in the usual sense), $\underline{A}^\vee$ is again constant (up to a shift), which gives one the usual Poincar\'e duality, but in general $\underline{A}^\vee$ needn't even be a sheaf (it may have "components" in different degrees).<|endoftext|>
TITLE: What happened to online articles published in K-theory (Springer journal)?
QUESTION [54 upvotes]: As most people probably know, the journal "K-theory" used to be published by Springer, but was discontinued after the editorial board resigned around 2007. The editors (or many of them) started the new "Journal of K-theory" in collaboration with Cambridge University Press. 
Maybe I'm being very stupid or missing something obvious, but I am quite sure that articles from "K-theory" used to be available online via Springer, and as far as I can see they are no longer there. Did I miss something here, or did they just disappear? I can't believe Springer would do that, but then I can't find anything by searching, and the article links from MathSciNet are dead: 

Sorry, the page you requested is unavailable. The link you requested
  might be broken, or no longer exist. SpringerLink is providing
  researchers with access to millions of scientific documents from
  Journals, Books, Protocols and Reference works.

In case it helps, here is the journal ISSN: 0920-3036 (print version) 1573-0514 (electronic version).
Grateful for any clarification.

REPLY [5 votes]: I don't have a grasp of the ins and outs of things here, but as of 2021 it seems most of the links in the above answers are broken. Currently it appears that archives of the journal "K-Theory" are available here through Portico. "Journal of K-Theory" doesn't appear to be available through Portico: here's what you get if you just search Portico for journals called "K-Theory".<|endoftext|>
TITLE: Can the Lawvere fixed point theorem be used to prove the Brouwer fixed point theorem?
QUESTION [77 upvotes]: The Lawvere fixed point theorem asserts that if $X, Y$ are objects in a category with finite products such that the exponential $Y^X$ exists, and if $f : X \to Y^X$ is a morphism which is surjective on points in the sense that the induced map $\text{Hom}(1, X) \to \text{Hom}(1, Y^X)$ is surjective, then $Y$ has the fixed point property: for every morphism $g : Y \to Y$ there exists a point $y : 1 \to Y$ such that $g \circ y = y$. 
The Brouwer fixed point theorem asserts that the closed $n$-disks, all of which I will denote by $D$ for ease of notation, have the fixed point property as objects of $\text{Top}$. 
Seeing these two theorems together, it is tempting to try to prove the latter from the former by finding a topological space $X$ such that the exponential $D^X$ exists, together with a surjective continuous map $X \to D^X$. Does there in fact exist such an $X$? 
Edit, 4/13/17: I'm still interested in this question, and so are some people associated with MIRI (at least when $n = 1$); for some details about why see here.

REPLY [2 votes]: EDIT: Pre-print available here. https://arxiv.org/abs/2005.01563 Note that I suspect that the use of the hypothesis of contractibility and the use of a homotopy is not really necessary, probably the hypothesis on the metric is enough.
Consider the "generalised Cantor space", the carrier set being $2^{\omega_{1}}$ and the basic open sets being sets consisting of all bit-strings of length $\omega_{1}$ with a fixed countable well-ordered bit-string as initial fragment. Denote this space by $A'$. Consider space with same carrier set with product topology obtained by viewing it as a product of $\aleph_1$ discrete spaces of cardinality two. Denote this space by $A''$. 
I define a space $X$ to be a ``nice space" if:
First, the space $X$ is compact and contractible and is the image of $A''$ under a continuous surjection. Secondly, the space $X$ is the disjoint union of an open dense subset $U$ and the complement $V$, and both $U$ and $V$ admit a ``homogenous" metric, where what we mean by this is as follows. Firstly, with regard to $U$, there exists some $\epsilon>0$, with the property that, for all $\delta$ such that $0<\delta<\epsilon$, every pair of distinct open balls of radius $\delta$ centred at a point in $U$ such that the closure of the balls does not intersect $V$, has the property that the balls in the pair are isometric. Then, with regard to $V$, we require that sufficiently small open balls in $X$, centred at points of $V$, of the same radius, are isometric.
Every closed ball in a finite-dimensional Euclidean space is indeed a nice space. The space $A''$ is a $k$-space, and so is every nice space, so exponential topologies exist here. I claim that for each nice space $X$ a continuous surjection $A' \rightarrow X^{A''}$ exists with the following property. Take an open subset of $X$ and take the pre-image under the evaluation map in $A'' \times X^{A''}$, then take projection onto the second factor and then pre-image of that in $A'$ under the continuous surjection $A' \rightarrow X^{A''}$. The result subset of $A'$ is open in $A'$, and its image under the continuous map $A' \rightarrow A''$ which fixes every point is open in $A''$. This is enough to apply the method of proof of Lawvere to obtain that every continuous endomorphism of $X$ has a fixed point. Thus Brouwer can be recovered as a corollary of a suitable generalisation of Lawvere.
The question of the existence of a space $A$ with a continuous surjection $A \rightarrow X^{A}$ for every nice space $X$, or for every space $X$ in some class which includes every closed ball in a finite-dimensional Euclidean space, still remains an open problem.
Brief outline of method of proof for showing that the continuous surjection exists. 
It is possible to construct an $\omega$-sequence $S:=\{C_{n}:n \in \omega\}$ of coverings of $X$ by finitely many open balls, each covering $C_{n} \in S$ being such that it can be partitioned into two collections of open balls with each collection having the property that all of the balls in it are pairwise isometric, and also such that the mesh of the covering $C_{n}$ tends towards zero as $n$ goes to infinity. Let $T$ be the collection of all centres of open balls occurring in some covering $C_{n}$. 
An element of $X^{A''}$ can be coded for, non-uniquely, by a mapping from a countable collection of countable bit-strings $B$, closed under taking initial fragments, and such that every element of $2^{\omega_1}$ has some element of $B$, maximal under the "initial fragment" relation, as an initial fragment, into $X$, with bit-strings of successor length being mapped to elements of $T$.  
One can further require that the trace of the mapping into $X$ along each branch of $B$ is ``generalised Cauchy"; it satisfies an obvious generalisation of the Cauchy criterion for each fragment of the branch of limit length, relative to the metric on $X$ which we have been holding fixed throughout, and with the speed of convergence having a uniform lower bound across all fragments of branches of $B$ of a given limit length.
Consider set of all mappings from such a set $B$ into $X$ satisfying the ``generalised Cauchy criterion" in question; this is in one-to-one correspondence with an appropriate set of countable well-ordered bit-strings $D$ under an appropriate coding scheme. Every element of $X^{A''}$ is coded for by at least one such mapping which in turn can be coded for by a countable well-ordered bit-string; further argument is required to show that every countable well-ordered bit-string does indeed code for an element of $X^{A''}$. To show that part of it, need to use a transfinite induction argument showing that from every mapping from an appropriate set $B$ into $X$ satisfying the appropriate constraints we can recover a continuous map $2^{\alpha} \rightarrow X$ for each countable ordinal $\alpha$ ($2^{\alpha}$ having the product topology), and that when $\alpha$ is sufficiently large, this map $2^{\alpha} \rightarrow X$ determines the map $2^{\beta} \rightarrow X$ (also continuous) for all $\beta$ such that $\alpha \leq \beta \leq \omega_1$. To get this transfinite induction to work we need to use both the "generalised Cauchy criterion" which is assumed to hold for our map $B \rightarrow X$ and also the given hypotheses on the space $X$, including the existence of an appropriate kind of metric on $X$.
Then, having done that, you need to show that the coding scheme for such maps $B \rightarrow X$, each such map being represented by a countable well-ordered bit-string from $D$, can indeed be constructed in such a way that it induces a surjection $A' \rightarrow X^{A''}$, continuous relative to generalised Cantor space topology on $A'$, and exponential topology on $X^{A''}$ arising from product topology on $A''$, which does indeed have all the properties I claimed.
That in very brief outline is the proof. Perhaps this post will still be thought inappropriate since I have not given all details. Or on the other hand maybe it can be accepted as a partial answer to the question.<|endoftext|>
TITLE: Distribution of moduli of quadratic residues
QUESTION [9 upvotes]: Let $D$ be a fixed positive squarefree integer. For a positive integer $x$, define
$S(D,x) = \{ q < x : D \text{is a quadratic residue} \pmod q \}$.
Here $q$ can be any integer, not necessarily a prime. Are elements of $S(D,x)$ evenly distributed? In other words,
let $0 < a < b < 1$ be constants and consider an interval $I = [ax,bx]$. Ideally I would like to see some result
that says that the number of elements of $S(D,x)$ in $I$ is proportional to the length of $I$ on the average as $x$ goes
to infinity (is this true?).
I am aware of classical 1917-1918 results of Vinogradov and Polya (and some later developments) about
distribution of quadratic residues, which in particular imply that quadratic residues modulo a fixed prime $p$ are
evenly distributed in the interval $[0,p]$ in the same sense as I described above. What I need, however, is a
result on the distribution of moduli with respect to which a fixed integer is a quadratic residue, and I cannot
find anything like this in the literature.
In other words, I am wondering how the divisors of $x^2-D$ are distributed on the average as $x$ goes to infinity.
It is a well-known fact that for an arbitrary integer $y$, there are unproportionally many (on the average) small
and large divisors of $y$ as $y$ goes to infinity, i.e., divisors are not uniformly distributed. But what if we take
$y$ in the special form $x^2-D$?
Any thoughts on the subject are much appreciated!

REPLY [8 votes]: For $D=-1$, Landau proved that
$$\# S(-1, x) \sim K \frac{x}{\sqrt{\log x}}$$
where $K = \frac{1}{\sqrt{2}} \prod_{p \equiv 3 \bmod 4} \frac{1}{\sqrt{1-p^{-2}}}$.
This shows that, for fixed $0<a<b$, 
$$\# S(-1,x) \cap [ax,bx] \sim K \left( \frac{bx}{\sqrt{\log(bx)}} -  \frac{ax}{\sqrt{\log(ax)}} \right)$$
and
$$ K \left( \frac{bx}{\sqrt{\log(bx)}} -  \frac{ax}{\sqrt{\log(ax)}} \right) = K \left( \frac{bx}{\sqrt{\log x + \log b}} -  \frac{ax}{\sqrt{\log x + \log a}} \right)$$
$$= K \left( \frac{bx}{\sqrt{\log x}} +O\left( \frac{x}{(\log x)^{3/2}} \right) - \frac{ax}{\sqrt{\log x}} - O\left( \frac{x}{(\log x)^{3/2}} \right) \right) \sim K \frac{(b-a)x}{\sqrt{\log x}}$$
So they are equidistributed in that $\frac{\#(S(-1,x) \cap [ax,bx])}{\#S(-1,x)}$ approaches $b-a$ but, if you ask for the denominator to be $x$, then the limit just goes to zero.
I believe the same should be true for any nonsquare $D$. Let 
$$Z_D(s) = \prod_{\left( \frac{D}{p} \right) = 1} \frac{1}{1-p^{-s}} = \sum_{D \ \mbox{is a square modulo}\ n} \frac{1}{n^s}.$$
The proof of Landau's theorem in Leveque's "Topics in number theory" comes down to analyzing the behavior of $Z_D(s)$ on $Re(s)=1$. The key facts are that $Z_{-1}(s) = C (s-1)^{1/2} + \cdots$ and $Z_{-1}(s)$ is otherwise bounded on $Re(s)=1$. These results hold for any nonsquare $D$. Let $\zeta_D(s)$ be the zeta function of $\mathbb{Q}(\sqrt{D})$. Then 
$$\frac{Z_D(s)^2}{\zeta_D(s)} =\left( \mbox{factors for primes dividing}\ 2D \right) \times \prod_{\left( \frac{D}{p} \right) = -1} \left( 1- \frac{1}{p^{2s}} \right) .$$
The right hand side is clearly convergent to the right of $Re(s)=1/2$, and $\zeta_D(s)$ is well known to have a simple pole at $s=1$ and no other poles on the line $Re(s)=1$.<|endoftext|>
TITLE: Solving a SDE with quadratic drift
QUESTION [5 upvotes]: I am wondering whether the following SDE can be solved explicitly? 
$$
d X_t = X_t^2 d t + X_t d B_t
$$
where $B_t$ is a standard Brownian motion. If not, can we say some thing about the moments of the solution, i.e., $E(|X_t|^n)$?
Thank you very much for any hints!
Anand

REPLY [7 votes]: Actually your SDE may be solved explicitly. Look at the more general SDE,
\begin{align}
dX_{t} = (a X_{t}^{n} + b X_{t}) dt + c X_{t} dW_{t}
\end{align}
where $n > 1$ and $a,b,c \in \mathbb{R}$. It has a solution given by
\begin{align}
dX_{t} = \Theta_{t} \Bigl( X_{0}^{1-n} + a(1-n)\int_{0}^{t} \Theta^{n-1}_{s} ds \Bigr)^{\frac{1}{1-n}}
\end{align}
with
\begin{align}
\Theta_{t} = e^{ (b-\frac{1}{2}c^2) t + c W_{t}}
\end{align}
See p. 125 in "Numerical Solution of Stochastic Differential Equations", 1995, Peter E. Kloeden and Eckhard Platen.<|endoftext|>
TITLE: Surface in 3D that realizes all pairs of principal curvatures
QUESTION [11 upvotes]: This is a question that Willie Wong raised in comments after he answered my question,
Surface analog of clothoid: curvatures covering $\mathbb{R}^3$. Willie's question is
more interesting (and challenging) than mine, essentially requiring the 
curvatures to map onto $\mathbb{R}^2$ rather than only onto $\mathbb{R}$:

Is there a surface $S \subset \mathbb{R}^3$ that realizes all pairs of principal curvatures in the sense that, for every $(\kappa_1,\kappa_2) \in \mathbb{R}^2$, 
  there is a point
  $p \in S$ such that the principal curvatures at $p$ are $\kappa_1$ and $\kappa_2$?

He also suggested that it might be natural to restrict to $\kappa_1 \ge \kappa_2$.

REPLY [7 votes]: Here is one view of Manfred's Angel's Curl surface, using $g(u) = e^{u^2}$:
          

Hopefully I've computed this correctly.
I've let the parameters $u$ and $v$ range over $\pm 4$.<|endoftext|>
TITLE: Borel kernel over an analytic set implies existence of a Borel map
QUESTION [6 upvotes]: Let $X$ and $Y$ be standard Borel spaces, and let $A\subseteq X\times Y$ be an analytic set with a full projection on $X$: that is $\pi_X(A) = X$. Suppose that there exists a Borel-measurable kernel $\mu:X\to\mathcal P(Y)$ such that $\mu(x,A_x) = 1$ for all $x\in X$ where 
$$
  A_x = \{y\in Y:(x,y)\in A\}
$$ 
is an $x$-section of $A$. Does it follow that there exist a Borel map $f:X\to Y$ such that whose graph is a subset of $A$, i.e. $\mathrm{Gr}[f]\subseteq A$?
Some comments:

There always exists a universally measurable $g:X\to Y$ whose graph is a subset of $A$.
Clearly, the converse fact always hold true. Given the existence of such $f$ one can come up with a Borel kernel $\mu(C|x) = \delta_{f(x)}(C)$.
In case $A$ is Borel, the existence of Borel $f$ follows from the existence of Borel $\mu$ as proven e.g. in Controlled Markov processes or can be also found here.

REPLY [4 votes]: The following is a construction of an analytic set $A\subseteq\omega^\omega\times\omega^\omega$ whose vertical sections exclude at most a single point, and which does not admit a Borel uniformization. This is as stated in jonathanverner's answer, although that did not include a construction - only mentioning exercise 36.25 in Kechris: Classical Descriptive Set Theory. The construction below is after viewing the hint in Kechris's book. I'll use the notation $\mathcal{N}$ for Baire space $\omega^\omega$.

Construction. Let $S$ be a universal subset of $\mathcal{N}\times\mathcal{N}^3$ (i.e., $S$ is closed and every closed subset of $\mathcal{N}^3$ is a vertical section $S_x$ of $S$ for some $x\in\mathcal{N}$). Then, let $A\subseteq\mathcal{N}^2$ consist of the points $(x,y)$ such that whenever, for any such $x$, there is a unique $(u,v)\in\mathcal{N}^2$ with $(x,x,u,v)\in S$ we have $y\not=u$. The set $A$ satisfies the required properties.

It is clear the sections $A_x$ exclude at most a single point. More precisely, it excludes the point $y$ if and only if there is a unique $(u,v)\in\mathcal{N}^2$ with $(x,x,u,v)\in S$ and $y=u$.
$A$ does not have a Borel section: Suppose that $\Gamma\subseteq\mathcal{N}^2$ is Borel. I'll now use the fact that every Borel subset of a Polish space is a continuous bijective image of a closed subset of $\mathcal{N}$ (Kechris, Theorem 13.7). By taking the graph of such a continuous bijection, there exists a closed set $C\subseteq\mathcal{N}^2\times\mathcal{N}$ whose projection onto $\mathcal{N}^2$ is one-to-one and has image $\Gamma$. Write $C=S_x$ for some $x\in\mathcal{N}$. Then, $(x,y)\in\Gamma$ if and only if $(x,x,y,v)\in S$ for some $v$, which is then unique. If $\Gamma_x$ consists of the single point $y$ then $(y,v)$ is unique such that $(x,x,y,v)\in S$ and, by construction, $(x,y)\not\in A$. So, $\Gamma$ is not a section of $A$.
$A$ is analytic: The set $T=\lbrace(x,y,z)\colon(x,x,y,z)\in S\rbrace$ is a closed subset of $\mathcal{N}^3$ and $(x,y)\in A$ iff whenever there is unique $(u,v)\in\mathcal{N}^2$ with $(x,u,v)\in T$ we have $y\not=u$. We can write $A=B\cup(\mathcal{N}^2\setminus C)$ where
$$
\begin{align}
&B = \left\lbrace(x,y)\in\mathcal{N}^2\colon(x,y^\prime,z)\in T, {\rm some\ }y^\prime\not=y\right\rbrace\\
&C=\left\lbrace(x,y)\in\mathcal{N}^2\colon\exists ! z\in\mathcal{N}{\rm\ with\ }(x,y,z)\in T\right\rbrace.
\end{align}
$$
As $B$ is the projection onto $\mathcal{N}^2$ of the (Borel) set of $(x,y,y^\prime,z)\in\mathcal{N}^2\times\mathcal{N}^2$ with $(x,y^\prime,z)\in T$ and $y\not=y^\prime$, it is analytic. Finally, $\mathcal{N}^2\setminus C$ is analytic by the following (surprising) result from Kechris (originally by Lusin).
Theorem (Kechris, Thm 18.11). Let $X,Y$ be standard Borel spaces and $B\subseteq X\times Y$ be Borel. Then,
$$
\left\lbrace x\in X\colon\exists ! y\in Y{\rm\ s.t.\ }(x,y)\in B\right\rbrace
$$
is coanalytic.<|endoftext|>
TITLE: How metric is Riemannian geometry
QUESTION [14 upvotes]: Let $(M, g)$ be a finite-dimensional Riemannian manifold. It is well-known, that the Riemannian metric induce a metric on the manifold by 
$$d(x, y) = \text{inf} \int_a^b \| \dot\gamma(t) \| \, dt\,,$$ where the infinum is taken over all $C^1$-curves connecting $x$ and $y$.
I'm interested in the basic properties of the usual constructions which solely depend on the metric and not on $g$. So what can be said about the smoothness of geodesics, exponential map,... if one forgets $g$ and only consider $(M, d)$ as a smooth metric manifold.
Clearly one can define geodesics only with respect to $d$ and assuming that the space is locally uniquely geodesic one can define the exponential map $\text{exp}$ as a map from (a subset of) local geodesics $\mathcal{G}$ to a neighborhood of $M$. This is basic topology of metric spaces, but I could not find any exposition discussing the smoothness of this constructions if $M$ is a manifold. So for example: Can $\text{exp}$ be considered as a map from the tangent bundle to the manifold (thus is the map $\gamma \rightarrow \dot\gamma(0)$ a bijection of $\mathcal{G}$ with some open subset of $TM$) and is it smooth? Does there exists a (adapted) definition of Jacobi-Fields?
(I'm not so interested in results which first recover the Riemannian metric $g$ [I think there is an old paper of Palais discussing this issue] and then run along the basic route to define geodesics, exponential map and Jacobi-Fields. The reason is, that I have a generalization of Riemannian geometry in mind where the standard procedure is definitely not possible.)

REPLY [17 votes]: If you forget about the Riemannian metric, you should also forget about the tangent bundle and manifold structure. Then you end up with metric spaces of inner type or Aleksandrov spaces, where you can find a continuous curves with arc-length the distance.
Look at "M. Gromov: Metric structures for Riemannian and Non-Riemannian Spaces, Birkhaeuser 1999". 
Added in edit: Interesting remarks by Thomas Richard and Igor Belegradek. 
Answering a remark by the OP: You can define a "geodesic structure" by requiring an exponential mapping with some properties. If you differentiate this you end up with the geodesic spray, a certain vector field on $TM$. If you differentiate this again and flip coordinates, you get a vector field on $TTM$ whose integral curves project to Jacobi fields, velocity fields of geodesics, etc. 
See 22.6-22.9 of here, and also this paper. But this is not the route of low regularity.<|endoftext|>
TITLE: Which complex manifolds embed into tori?
QUESTION [12 upvotes]: If $X$ is a compact Kahler manifold then it's well-known that $X$ can be embedded into a projective space if and only if it admits an ample line bundle. Suppose now that we look for other things to embed $X$ into, like complex tori, and ask for necessary and sufficient conditions for this to happen. 
If $X$ is a submanifold of a torus $A$, then the short exact sequence
$$
0 \longrightarrow N_{X/A}^* \longrightarrow \Omega_{A|X}^1 \longrightarrow
\Omega_X^1 \longrightarrow 0
$$
over $X$ shows that the cotangent bundle $\Omega^1_X$ is globally generated, because $A$ is a torus and so $\Omega^1_A$ is trivial. It's not hard to see that the converse has some truth to it, that is if $\Omega_X^1$ is globally generated then the Albanese morphism $\alpha : X \to \operatorname{Alb} X$ is a local immersion (by looking at its differential), which seems to me like it should imply that the image of $\alpha$ in $\operatorname{Alb} X$ is smooth (locally the image looks like $X$) and that $\alpha : X \to \alpha(X)$ is a finite morphism (by compactness of everything).
Is there a simple condition on $X$ that ensures that the Albanese morphism is actually injective and thus an embedding when $\Omega_X^1$ is globally generated?

REPLY [3 votes]: See:
@article {MR0350057,
    AUTHOR = {Matsushima, Yozo},
     TITLE = {Holomorphic immersions of a compact {K}\"ahler manifold into
              complex tori},
   JOURNAL = {J. Differential Geometry},
  FJOURNAL = {Journal of Differential Geometry},
    VOLUME = {9},
      YEAR = {1974},
     PAGES = {309--328},
      ISSN = {0022-040X},
   MRCLASS = {32C10 (32J99)},
  MRNUMBER = {0350057 (50 #2550)},
MRREVIEWER = {B. Smyth},
}<|endoftext|>
TITLE: Hecke $L$-series exercise in Silverman's Advanced Topics in Arithmetic of EC
QUESTION [5 upvotes]: This has been posted on SE, but I haven't gotten a reply, so I thought I'll try my luck here.
I would like to refer you to 2.30 & 2.32 in Silverman's book Advanced Topics in the Arithmetic of Elliptic Curves.
2.30(b)[(c) in errata]: Suppose $\mathfrak{P}$ remains inert in $L'$, say $\mathfrak{P}R_{L'}=\mathfrak{P}'$. Prove that 
$$q_{\mathfrak{P}}^2=q_{\mathfrak{P}'},\quad a_{\mathfrak{P}}=0,\quad\psi_{E/L'}(\mathfrak{P}')=-q_{\mathfrak{P}}.$$
2.32(a) Prove that the local $L$-series of $E$ at $\mathfrak{P}$ is given by
$$ L_{\mathfrak{P}}(T,E/L) = \left\{ 
  \begin{array}{l l}
    \cdots & \quad \mathfrak{P}\text{ splits}\\
    1-\psi_{E/L'}(\mathfrak{P}')T & \quad \text{$\mathfrak{P}$ inert}\\
    \cdots & \quad \mathfrak{P}\text{ ramifies}\\
  \end{array} \right.$$
In the beginning of the chapter, we have that if $E$ has good reduction (in this case, we have good reduction because of 2.30(c))$$L_{\mathfrak{P}}(T,E/L) = 1-a_{\mathfrak{P}}T+q_{\mathfrak{P}}T^2$$
So is the question missing a power of 2 in $T$? Should it be $1+\psi_{E/L'}(\mathfrak{P}')T^2\text{when is $\mathfrak{P}$ inert}$? Thanks for the help! (If it helps, I can print the question; say so in the comments and I'll do it.)
Edit: In Deuring's paper, he has $L_0(s,Kk_1,\mathbf{P})=1+N\mathbf{p}^{1-2s}$. So the presence of $2s$ seems to me to tally with $T^2$.

REPLY [3 votes]: There is nothing wrong with the question. The subtlety is that $T$ is not constant with respect to the fields $L'$ and $L$. My mistake was that I took $T$ to be the same in both fields when in reality it is not.
\begin{align*}
L(s,E/L)&=\prod_{\text{$\mathfrak{P}$ in $L$}}\left(1-a_{\mathfrak{P}}q_{\mathfrak{P}}^{-2}+q_{\mathfrak{P}}^{1-2s}\right)^{-1}\\
&=\prod_{\text{$\mathfrak{P}$ in $L$}}\left(1+q_{\mathfrak{P}}^{1-2s}\right)^{-1}\\
&=\prod_{\text{$\mathfrak{P}$ in $L$}}\left(1+q_{\mathfrak{P}}q_{\mathfrak{P}}^{-2s}\right)^{-1}\\
&=\prod_{\text{$\mathfrak{P}$ in $L$}}\left(1+q_{\mathfrak{P}}q_{\mathfrak{P}'}^{-s}\right)^{-1}\\
&=\prod_{\text{$\mathfrak{P}'$ in $L'$}}\left(1-\psi_{E/L'}(\mathfrak{P}')q_{\mathfrak{P}'}^{-s}\right)^{-1}\\
&=L(s,\psi_{E/L'})
\end{align*}<|endoftext|>
TITLE: Polynomial identities for mod p matrices
QUESTION [5 upvotes]: Can there be a polynomial over the field $F_p$ of $p$ elements ($p$ prime) in non-commuting variables $X_1,..., X_r$ such that:
1) $f(A_1,...,A_r)=0$ for every $n \times n$ matrices $A_1,...,A_r$ over $F_p$; and
2) $f$ has a non-trivial monomial of total degree less than $n$ ?

REPLY [2 votes]: I tried to just comment, but it seems that I do not have enough "reputation" for that. So I am commenting here.
The answer would be "no" if your field were infinite. Since in that case, each homogeneous component of a polynomial identity would also be an identity (see [1], prop. 4.2.3). And that would imply that you would have a multilinear identity of the same degree of such monomial (use the multilinearization process (see [1])). 
Finally, it is easy to show that there does not exist polynomial identities of degree less than 2n for M_n(F) (staircase argument see [1] Exercise 7.1.2 - this is true for any field).
Ok, but it was not asked about infinite fields.
For matrices over finite fields, there are descriptions for bases of identities for matrices of order up to 4. (see [2,3,4])
I checked the papers for 2 and 3 (unfortunately the case $n=4$ is in russian), and some of the generators of the ideal of identities of such algebras are polynomials which contains monomials of degree exactly n (2 or 3, in this case), but not less than n. So, it does not answer your question.
At first sight I thought the answer to your question would be "no" even if you were asking for the degree of your monomial to have degree less than 2n. For 2n, the papers [2] and [3] show I was wrong.  
References:
[1] V. Drensky, Free Algebras and PI-algebras: Graduate Course in Algebra, Springer, Singapore, 1999.
[2] Yu. N. Mal'tsev, E. N. Kuz'min, A basis for the identities of the algebra of second-order matrices over a finite field, Algebra Log. 17 (1978) 18–21
[3] G. Genov, Basis for identities of a third order matrix algebra over a finite field, Algebra Log. 20 (1981) 241–257.
[4] G. Genov, P. Siderov, A basis for identities of the algebra of fourth-order matrices over a finite field. I, II, Serdica 8 (1982) 313–323, 351–366 (in Russian).<|endoftext|>
TITLE: Repeated random two-steps in $\mathbb{R}^3$: unbounded?
QUESTION [19 upvotes]: I created a random isometry $T$ of $\mathbb{R}^3$ by generating
a random orthogonal matrix $M$,
uniformly distributed among all such,
and a random displacement $v$, whose coordinates
are drawn from a zero-mean normal distribution.
For a point $p \in \mathbb{R}^3$, $T(p) = M . p + v$.
Let $T_1$ and $T_2$ be two such random isometries.
Now, starting with $p_0=(0,0,0)$, I formed the sequence of points
$p_1 = T_1(p_0)$,
$p_2 = T_2(p_1)$,
$p_3 = T_1(p_2)$,
$p_4 = T_2(p_3)$,
and so on, altering the two isometries one after the other, so
$$p_{2i} = T_2 \circ T_1 \circ T_2 \circ T_1 \circ T_2 \circ \ldots \circ T_1 ( p_0 ) \;.$$
Because the isometries are random, and include a random translation,
I expected the $p_i$ to wander off to infinity, and indeed this is common:

         

Above, the red segments connect $p_i$ to $p_{i+1}$.
Essentially these look like two spirals of points forming an unbounded V.
However, (very) roughly 40% of the time, instead
spirograph-like bounded rings are formed 
(or at least: apparently bounded, apparently rings):

     

So far inspecting the transformations numerically has not led me to
an understanding of this phenomenon.
If anyone sees intuitively why this process might lead to essentially
two qualitatively different shapes, I'd appreciate an explanation.
A specific question is:

Is there a positive probability that $|p_i|$ remains bounded
  as $i \to \infty$?

I thought the answer should be No but my simulations indicate Yes.

Answered. Both Benoît Kloeckner and Ofer Zeitouni answered the question,
using rather different language but ultimately equivalently. One way to summarize
the computational consequence their analyses is this: 
If $\det(M_1 \cdot M_2)=1$, the sequences is unbounded;
if instead this determinant is $-1$, the sequence is bounded. As Benoît says, the
analysis holds for any number of transformations ($k$-steps rather than two-steps),
in which case the $\pm 1$ determinant of the product of the $k$ matrices signifies bounded/unbounded. Here is an example of a bounded 4-step sequence.

          

Now I can generate an infinite variety of these elegant figures!
Thanks for everyone's help!

REPLY [11 votes]: I assume your random $M_i$s are from $O(3)$, not $SO(3)$. In fact, I suspect that the answer depends on whether $M_1M_2$ has eigenvalue $1$ or eigenvalue $-1$.
If I did not make a mistake, when you expand $p_{2i}$, you will get essentially 
$$\sum_{k=0}^{i-1} U^k x_1+ V\left(\sum_{k=1}^{i-1} U^i\right)^T x_2$$
Here $U=M_2M_1$ and $V$ is also an orthogonal matrix.
Now, $U$ is an orthogonal matrix, and has a conjugate pair of eigenvalues as well as an eigenvalue that is either $1$ or $-1$. Call these $r_1, r_1^*$ and $r_2$.
If $r_2=1$, obviously you run off to infinity.
If $r_2=-1$, you get an oscillating sum, which remains bounded.
I have not written down a formal proof but it seems to me that the terms due to $\sum r_1^k$ (and the conjugate of that) are essentially harmless - even if they run off to infinity, they do it VERY slowly (if the angles corresponding to 
the $r_1$ are rational, you would get a bounded sum; in reality you are rational with probability $0$, but on a computer....).
Note added: Of course, $\sum_{k=0}^i r_1^k$ remains bounded as long as $|1-r_1|$ is bounded below, which occurs with probability $1$. I should have noted that instead of the nonsense I wrote above concerning not writing a formal proof....

REPLY [11 votes]: This phenomenon is to be expected: the point is to recall the classification of isometries and that indirect isometries almost always have fixed points, while direct ones almost always have no fixed points.
The fact that there are two isometries is not relevant to the boundedness of the dynamic: let us look at $U=T_2\circ T_1$. This is a random isometry, whose law may be involved but is certainly absolutely continuous.
Writting $U\cdot p = A\cdot p + w$, it has a fixed point if and only if $w$ is in the image of $A-I$. There are two cases. 
First, $A$ is a direct isometry, aka a rotation. Then $1$ is an eigenvalue and $A-I$ has rank $2$. Then with probability $1$, $w$ is not in the image of $A-I$, $U$ has no fixed point and its orbits are all unbounded.
Second, $A$ is an indirect isometry (an anti-rotation with full (conditional) probability). Then $A-I$ is invertible and $U$ has a fixed point, so that all its orbits are bounded.
Probably more interesting things happen if you choose randomly at each step whether to apply $T_1$ or $T_2$; such random dynamical systems have already been studied in different contexts.<|endoftext|>
TITLE: A simple and good reference on surgery theory
QUESTION [7 upvotes]: Can anyone help me to find a simple and good reference (a book, lecture notes or a website) for learning the surgery theory and its applications? I seek a reference together with many examples and figures for more intuition.
Thanks.

REPLY [4 votes]: Scorpan's "the wild world of 4-manifold" is, as its title indicate, restricted to dimension 4 but it has far more picture than one would think possible, including about some surgery constructions. The book is nicely written, and combine an overview of the subject with non-mandatory, small font technical details.<|endoftext|>
TITLE: Generalizations of the Tietze extension theorem (and Lusin's theorem)
QUESTION [15 upvotes]: I am reasking a year-old math.stackexchange.com question asked by someone else.
(For my needs every space $X$ and $Y$ will be Polish---that is a completely separably metrizable space.)
The Tietze extension theorem says that if $X$ is a Polish space (even a normal space) and $Y=\mathbb{R}^n$, then a continuous function $f:C \rightarrow Y$ on a closed set $C \subseteq X$ can be extended to a continuous function $g:X \rightarrow Y$.
It seems important to the theorem in general that $Y = \mathbb{R}^n$, however there are some examples of pairs $X,Y$ where the theorem also holds.  For example, it is true if $X, Y \in \{2^\mathbb{N},\mathbb{N}^\mathbb{N}\}$.  (Although, other pairs like $X=\mathbb{R}, Y = 2^\mathbb{N}$ do not hold.)

Is there a characterization of the pairs $X,Y$ (or $X,Y,C$) for which the Tietze extension theorem holds?
If not, what extensions are known, especially those that include my $2^\mathbb{N}$ example above?


One motivation for asking this question is Lusin's theorem:
If $(X,\mathcal{B},P)$ is a Borel probability measure on a Polish space $X$ (even a Radon measure on a finite measure space) and $f:X \rightarrow Y$ is a measurable map (again $Y$ is Polish, or even second countable), then for all $\varepsilon > 0$, there is a closed set $C$ of measure $1-\varepsilon$ such that $f$ is continuous on $C$.
If $Y$ is $\mathbb{R}^n$, we can apply the Tietze extension theorem to find some continuous $g:X \rightarrow Y$ such that $g = f$ on $C$.  Wikipedia currently (12 July 2013) has a false statement that for any locally compact $X$ we can find such a continuous $g:X \rightarrow Y$.  For a counterexample, take $X=[0,1]$ and $Y=2^\mathbb{N}$ and $f$ to be the bit representation of the reals.
I am interested in which cases this stronger version of Lusin's thoerem (with the continuous $g:X \rightarrow Y$) holds.

REPLY [6 votes]: A compact space $Y$ is called an absolute extensor in dimension $0$ abbreviated $AE(0)$ if the pair $X,Y$ satisfies Tietze extension theorem for every compact zero-dimensional $X$. Any compact (retract of a) topological group is $AE(0)$ (Schepin/Uspenski). In particular $2^\mathbb{N}$ is $AE(0)$. In fact, Schepin proved that the only zero-dimensional $AE(0)$ space with character $\kappa$ at every point is $2^\kappa$.<|endoftext|>
TITLE: Number theory underlying Euler's theory of music
QUESTION [12 upvotes]: I've recently been studying Euler's theories on music, and I came across Euler's concept of gradus suavitatis or 'degree of pleasure' of a rational number representing the ratio of two tones. (I found this on http://www.mathematik.com/Piano/)
The formula is $G(p/q)=1+\Sigma e_i (p_i-1)$ where $p,q$ are relatively prime, the $p_i$ are the prime factors of $pq$ and $e_i$ is the multiplicity of $p_i$.
This formula seemed familiar. Is this formula used in number theory, and, if so, what is its mathematical significance?

REPLY [7 votes]: This is not directly an answer to your question (graph theoretic relation, rather than number theoretic), but it's too long for a comment.
Euler's formula is a special case of a disharmonicity function
$$D(x)=\sum_i |e_i| g(p_i)$$
with $g(p_i)>0$ a function of the prime factors $p_i$ of a rational number $x=p_1^{e_1} p_2^{e_2}\cdots $ (each with positive or negative multiplicity $e_i$). Euler's disharmonicity has $g(p_i)=p_i-1$ (and adds unity to the sum over $i$). An alternative disharmonicity, due to Barlow, takes $g(p_i)=(2/p_i)(p_i-1)^2$.
Disharmonicity functions can be used to define the notion of a "harmonic distance" of two rational numbers, and to formulate the problem of the harmonization of a musical scale as a problem in graph theory.
See Musical scale rationalization – a graph-theoretic approach, by Albert Gräf.<|endoftext|>
TITLE: Where should I search for resolutions?
QUESTION [9 upvotes]: In my research, to test some conjectures or just to illustrate some facts, I often need to compute some explicit examples of derived functors (in the sence of Quillen's model categories). Mainly I work in the category of differential non-negatively graded algebras over some field $k$ of characteristic zero (denote this category $DGA^+_k$). I will regard that differentials have degree $-1$.
It seems like the main problem when computing these functors is to find an appropriate cofibrant resolution. To be more precise, for an algebra $A\in DGA^+_k$ a map $\phi\colon R\twoheadrightarrow A$ is a cofibrant resolution if $\phi$ is surjective in each degree quasi-isomorphism  and $R$ is semi-free. Here $R\in DGA^+_k$ is called semi-free if it's underlying graded algebra $R_\#$ is free.
It's not a big deal to construct some resolution. For example, the cobar-bar adjunction gives a resolution $\Omega B(A)\twoheadrightarrow A$. But this resolution is "huge". So the problem is how to construct resolutions that are "small".
I know a couple of examples. For exapmle, the (commutative) polynomial algebra $A=k[x,y]$ has a resolution in $DGA^+_k$ of the form $R=k\langle x,y,t\rangle$, where $\deg x=\deg y=0$, $\deg t=1$ and $d(t)=xy-yx=[x,y]$.
Algebra $A=k[x,y,z]$ has resolution of the form $R=\langle x,y,z;\xi,\lambda,\theta;t\rangle$ with $\deg x,y,z=0$, $\deg \xi,\lambda,\theta=1$ and $\deg t =2$. The differential is defined by $d(\xi)=[y,z],d(\lambda)=[x,y],d(\theta)=[z,x]$ and $d(t)=[x,\xi]+[y,\theta]+[z,\lambda]$.
So my questions are the following. 
1) Do you know any other nice examples of algebras with simple resolutions? What are resolutions of symmetric algebras, matrix algebras?
2) How can you come up with such nice resolutions? I understand that probably there is no general recipe, but maybe there are some techniques, or hints, how to do that.
I hope this question is appropriate to ask here.
Thanks a lot for your help!

REPLY [5 votes]: There is a way to approach it going back to one of the most efficient general computational methods for associative algebras. Namely, if you know a "Groebner basis" of your algebra (a confluent presentation via generators and relations, if that sounds less technical?) then you can build a resolution out of it, sort of. This approach is explained in (sorry for this self-advertising) in the full generality of operads here, and a preprint predecessor of that paper which discusses first as a toy example the case of associative algebras (maybe preferable for you) is here.<|endoftext|>
TITLE: Einstein field equations in perspectives from PDE and functional analysis
QUESTION [15 upvotes]: The Einstein field equations have been subject of research in theoretical physics, and differential geometry, apparently with methods from classical analysis and geometry. In particular, solutions in closed form have been of interest.
It seems that the classical programme of the PDE community, i.e., (i) existence (ii) uniqueness (iii) regularity, heavily employing concepts from functional analysis, has not found prominent application in general relativity.
Why is this the case? Is it simple due to the Sobolev theory on manifolds still being rather fresh, or do serious technical obstacles exist? Or have I overlooked something?

REPLY [18 votes]: The statement

It seems that the classical programme of the PDE community, i.e., (i) existence (ii) uniqueness (iii) regularity, heavily employing concepts from functional analysis, has not found prominent application in general relativity.

is just plain wrong. You are overlooking quite a lot of stuff. For a modern presentation of the mathematics I would second Ben Whale in recommending Hans Ringstrom's book The Cauchy Problem in General Relativity. Here let me give a bit of historical remarks.
The local wellposedness (after imposing gauge conditions, for reasons already described in Rafe Mazzeo and Peter Michor's answers) of Einstein's equations is a result more than 60 years old! The original proof was published by Yvonne Choquet-Bruhat in 1952 (before she took her current last name), and yes, more modern presentations usually use heavily the notion of Sobolev spaces on manifolds.
Insofar as the local Cauchy problem is concerned, the Einstein system is actually, in some aspects, easier than some of the elliptic and parabolic problems on manifolds. This is principally due to the fact that in appropriate gauge conditions the equations of motion are manifestly hyperbolic (in fact a quasilinear wave equation). Such equations automatically enjoy finite speed of propagation, and thus we can more easily localise the analysis onto individual coordinate patches. That is to say, for the local problem we don't need much of the machinery of global analysis where the geometry and topology of the underlying manifold in the large may come into play.
Perhaps the only conceptually tricky bit of the Cauchy problem for the Einstein system is that, from the very get-go, there is no preferred notion of time on the solution manifold. So whereas in classical non-geometric PDEs the notion of a local existence theorem is stated in the form "there exists $t> 0$ such that a solution exists on $(0,t) \times D$ where $D$ is some domain", the corresponding theorem in general relativity would more naturally look like "there exists a manifold $M$ into which the initial data embeds as a codimension 1 Riemannian manifold with the initial conditions satisfied." This lack of a preferred notion of time also makes it more difficult to interpret what is meant by "global in time solution" to the Cauchy problem.
In 1969 Choquet-Bruhat and Robert Geroch showed that, essentially due to the hyperbolic nature of the equations, one can define the maximal manifold $M$ into which all solutions embed. With this notion one can then formulate the question of "global Cauchy problem" as a question of studying the geometric properties of this maximal solution.
It is well known that this maximal solution can be, in general, quite bad. There are plenty of explicit solutions to Einstein's equation which illustrate this, none-the-least the classical families of black hole space-times. In particular, the classical Schwarzschild solutions are geodesically incomplete in its maximum extension, while the classical Minkowski space is geodesically complete. A natural question, for example, is to ask about the genericity of "geodesic completeness" as a property for maximal solutions of the Cauchy problem, in terms of the initial data. That geodesic incompleteness is generic (in the sense that there exists open sets of initial data that are close to Schwarzschild initial data that leads always to incomplete solutions) turns out to be something that can be established geometrically with minimum amounts of PDE theory. One can read the result off of the incompleteness theorem of Penrose, which establishes a sufficient condition for geodesic incompleteness (the existence of trapped surfaces) and that the existence of trapped surface is an "open condition" and hence is preserved for small perturbations by classical Cauchy stability of the initial value problem at finite times. (A more detailed account is given in Mihalis Dafermos' exposé "The formation of black holes in general relativity (after D. Christodoulou)" for the Seminaire Bourbaki; see  Astérisque (2012) 64, Exp. No. 1051.)
The genericity of the case of completeness turns out to be much harder, in terms of the PDE portion. Whereas for the incompleteness case the geometric criterion of Penrose allows us to use "finite time Cauchy stability", which is automatically true in view of the local wellposedness theorems, for the completeness case we have no such criterion and in fact needs to prove global Cauchy stability. In terms of the classical theory of nonlinear PDEs, this corresponds to, roughly speaking, proving global in time existence, with suitable decays, for small data to an initial value problem. Even for simple nonlinear wave equations this is not understood until at least the 70s and 80s. The developments in those two decades led to the understanding that in the physically relevant case of 3 space and 1 time dimensions, the appropriate small data global existence for nonlinear wave equations is not true in general. For Einstein's field equations (which, as described above, can be cast as a quasilinear wave equation, at least locally), this means one needs to search for geometric structure which affords one extra cancellations. As alluded to by Peter Michor, this was done by Demetrius Christodoulou and Sergiu Klainerman in their 500-page opus. (An alternative proof was more recently obtained by Igor Rodnianski and Hans Lindblad; the comparative shorter length testifies to how much the technology for nonlinear wave equations have evolved in the past two decades.)
Along this line of investigation, a further development was produced by Christodoulou in the past few years. As discussed above by Penrose's theorem one automatically has that initial data close to that of Schwarzschild space-time will lead to geodesically incomplete space-times. One may ask whether one can construct data initially far from that of Schwarzschild and still get singularity formation. And the answer turns out to be yes.
The result of Christodoulou and Klainerman on the global stability of Minkowski space actually yielded much more information then just geodesic completeness. It gives precise asymptotic convergence of such a "small data" solution to Minkowski space, and gave a complete description of the "conformal infinity" for such solutions. The aforementioned "quick" proof of stability of geodesic incompleteness by way of Penrose's theorem does not yield nearly as much information. And an ongoing active program of research (involving too many people to be reasonably listed here) is to demonstrate that similar asymptotic stability result with convergence and so forth can be had also for the classical black hole space-times. This is the motivation for the recent bloom in the study of the linear wave equation on black hole space-times.
There are also many other aspects of the global existence problem in general relativity. For example, one can try to consider situations where the initial manifold has compact topology or situations where the initial manifold has symmetries. The former is carefully studied in cosmological settings (I refer again to Ringstrom's book); though oftentimes with high degrees of symmetry so that the system reduces to that of a nonlinear system of ODEs. A spectacular example of the latter is the analysis by Choquet-Bruhat and Vincent Moncrief of solutions with an $U(1)$ symmetry. Under this particular symmetry assumption, and with the so-called CMC gauge condition, the evolution equations of the Einstein vacuum problem reduces to a coupled system of (a) an ODE on the cotangent bundle of Teichmuller space (describing the geometry of the "constant time" slices) and (b) a wave(map) equation.
Coming back to the local well-posedness problem in 3+1 dimensions: in the regime of "classical" solutions, existence, uniqueness, and regularity follows from a very similar analysis to that of Hughes, Kato, Marsden. To close the argument one needs to work in the Sobolev space $H^{5/2+}$ (or classically $H^3$ if you don't want to deal with fractional numbers of derivatives). One may ask about the optimal regularity for the local well-posedness statements. The scaling regularity should be $H^{3/2}$. However, the equation is quasilinear, and it is generally the case that compared to semilinear problems we cannot expect to go all the way down to scaling critical (see for example these two papers of Hans Lindblad 1 2). This problem motivated a body of literature studying the below-classical-regularity local existence problem for quasilinear wave equations. Between the groups of Klainerman-Rodnianski and that of Hart Smith and Daniel Tataru (esp. this paper and this other one), for general quasilinear wave equations the sharp exponent $H^{2+}$ was obtained. For the Einstein vacuum equations specifically, though, one may do better: a recently posted series of pre-prints by Klainerman, Rodnianski, and Jeremie Szeftel obtains a local existence theorem at the regularity level $H^2$ (these are arXiv items: http://arxiv.org/abs/1204.1772 http://arxiv.org/abs/1204.1767 http://arxiv.org/abs/1204.1768 http://arxiv.org/abs/1204.1769 http://arxiv.org/abs/1204.1770 http://arxiv.org/abs/1204.1771 http://arxiv.org/abs/1301.0112).
Some further reading:

Rodnianski's ICM address: http://www.icm2006.org/proceedings/Vol_III/contents/ICM_Vol_3_22.pdf
Alan Rendall's survey article: http://arxiv.org/pdf/gr-qc/0505133.pdf<|endoftext|>
TITLE: Kodaira Spencer map and versal deformation
QUESTION [6 upvotes]: First I want to clarify what I mean by the Kodaira-Spencer map.
Let's have a family of deformations $\pi:\mathcal{X}\rightarrow B$ of a complex manifold $X=\mathcal{X}_0:=f^{-1}(0)$ (by that I mean that $f$ is an holomorphic proper submersion between the complex manifolds $\mathcal{X}$ and $B$).
Then the Kodaira-Spencer map $\rho:T_{B,b}\rightarrow H^1(\mathcal{X}_b,T_{\mathcal{X}_b})$ is the map induced in cohomology by the exact sequence of sheaves
$$ 0\rightarrow T_{\mathcal{X}_b}\rightarrow T_{\mathcal{X}|\mathcal{X}_b}\rightarrow f^*T_B|\mathcal{X}_b\rightarrow 0 $$
By "versal deformation" $Def(X)$ I mean that any other deformation $\mathcal{X}\rightarrow B$ of $X$ can be obtained as the pullback deformation for a certain map $g:B\rightarrow Def(X)$ and the differential of $g$ is unique.
My question is: why is the Kodaira-Spencer map for the versal deformation bijective? 
This is not clear to me. Also, some authors define the versal deformation as the deformation for which the Kodaira-Spencer map is bijective.

REPLY [2 votes]: This is basically a tautology. A tangent vector $v$ to $Def(X)$ at $0$ is a map $D:=Spec(k[t]/(t^2))\to Def(X)$ mapping the closed point to 0, i.e. a family $X_v\to D$, i.e. a first order deformation of $X_0$. These are classified by the vector space $H^1(X, T_X)$, so we get a bijection $T_0 Def(X) \to H^1(X, T_X)$ which turns out to be the Kodaira-Spencer map.<|endoftext|>
TITLE: When are MCM ideals principal?
QUESTION [12 upvotes]: Suppose $(R,\mathfrak m, k)$ is a $d$-dimensional Cohen-Macaulay local ring with canonical module $\omega_R$ and $d>1$.  Suppose $I\subset R$ is an ideal which is MCM (=maximal Cohen-Macaulay, i.e., its depth as a module over $R$ is $d$).  My main question is: 

Under which assumptions on $R$ can we conclude that every MCM ideal $I$ is principal? 

Of course, if $d=1$ there are too many MCMs, so we must take $d>1$. Clearly, this is true for regular local rings (as any MCM is then free). However, if $R$ is a domain (or more generally, generically Gorenstein), then $\omega_R$ is isomorphic to a MCM ideal of $R$, and this ideal will be principal precisely when $R$ is Gorenstein. So we must at least impose that $R$ is Gorenstein. It is not too hard to see that any MCM ideal is principal if $R$ is a unique factorization domain (for $I$ must be unmixed of height one by the depth lemma, whence principal). Using Graham's observation below, we see that in a two-dimensional local Gorenstein ring, every MCM ideal is principal if and only if the ring is a unique factorization domain. Is this true in higher dimensions?
Sometimes we can prove that a certain ideal is MCM: for instance, if $R$ is two-dimensional and $\bar R$ is its integral closure, then the conductor ideal $I=\text{Hom}(\bar R,R)$ is MCM by the depth lemma. 
More generally, one could ask when are there only finitely many different isomorphism types of MCM ideals. Do we have Brauer-Thrall-like behavior?

REPLY [6 votes]: On question 1, for what rings all MCM ideals are principal, we can say quite a bit more if one knows that $R$ is parafactorial (that is, the Picard group of the punctured spectrum $Spec^o(R):=Spec(R)-\{m\}$ is trivial). For instance:

If R is a local complete intersection which is locally a UFD in codimension $3$, then any MCM ideal $I$ is free. 

Proof: Suppose it is not, then there is a prime $P$ such that $I_P$ is not $R_P$- free. Then the height of $P$ is at least $4$, and $I_P$ is locally free on $Spec^o(R_P)$, so is an element in the Picard group of $Spec^o(R_P)$. But $R_P$ is a complete intersection of dimension at least $4$, so by SGA, it is parafactorial, and $I_P$ is free. 
The last question, whether there are finitely many MCM module of rank one, is well-known among certain people, and has resurfaced from time to time. The most ambitious conjecture is perhaps

If $R$ is a normal local domain with an algebraically closed residue field, and the class group $Cl(R)$ finitely generated, then it contains finitely many MCM elements. 

For references and some positive results in this direction, see Section 4 of my paper with Kurano and the subsection 6 (on page 3) in this paper by Kollár.<|endoftext|>
TITLE: Is $π$ definable in $(\Bbb R,0,1,+,×,<,\exp)$?
QUESTION [35 upvotes]: (This question is originally from Math.SE, where it didn't receive any answers.)
Is there a first-order formula $\phi(x) $ with exactly one free variable $ x $ in the language of ordered fields together with the unary function symbol $ \exp $ such that in the standard interpretation of this language in $\Bbb R $ (where $ \exp $ is interpreted as the exponential function $ x \mapsto e^x $), $\phi (x) $ holds iff $ x=\pi $?
(Since a negative answer to this problem would imply that $e$ and $\pi$ are algebraically independent, I cannot expect anyone to give a complete proof that it isn't possible. However, in the case that one suspects strongly that the answer to this question is negative, I would already be pleased if someone could give intuitive arguments why one shouldn't believe in the definability of $\pi$.
However because there are such intricate connections between exponential and trigonometric functions, I don't think that $\pi$ should be undefinable.)

REPLY [38 votes]: It seems to me that Schanuel's conjecture (which is a kind of article of faith in transcendental number theory, but of course very far from proven itself) ought to imply that $\pi$ is not definable in this structure. The linked Wikipedia article contains the assertion 

Euler's identity states that $e^{\pi i} + 1 = 0$. If Schanuel's conjecture is true then this is, in some precise sense involving exponential rings, the only relation between $e$, $π$, and $i$ over the complex numbers. [2]

where reference [2] is unfortunately not something I have access to: 

Terzo, Giuseppina (2008). "Some consequences of Schanuel's conjecture in exponential rings". Communications in Algebra 36 (3): 1171–1189. doi:10.1080/00927870701410694. 

(link), courtesy of Dominik in a comment below. 
See in particular theorem 2.5.1 on page 37, which gives a more precise answer to the OP (under the assumption of Schanuel's conjecture). The bottom line is that a claim that $\pi$ should be definable in this structure would amount to an assertion that SC is false (which would be a pretty big deal, if you know a little about the stature and importance of SC). 
Related of course is Wilkie's theorem, which says that any first-order unary formula $\phi(x_1)$ in the language of ordered rings with an exponential function is equivalent to 
$$\exists x_{2}\ldots\exists x_n \;\; f_1(x_1,\ldots,x_n,e^{x_1},\ldots,e^{x_n})=\cdots= f_r(x_1,\ldots,x_n,e^{x_1},\ldots,e^{x_n})=0$$ 
for some exponential polynomials $f_i$ with integer coefficients. So a putative definition of $\pi$ can't be that intricate in principle, at least not in terms of logical complexity.<|endoftext|>
TITLE: Conjectural identities for Young symmetrizers and Young-Jucys-Murphy elements
QUESTION [20 upvotes]: The following questions I have found in my own notes from about 3 years ago. Unfortunately, I lost much of the context; I believe I made these conjectures reading Okounkov-Vershik, arXiv:0503040v3, but I don't remember any details, including whether I was lacking a proof or I had a proof and was looking for a better one. What is sure that I am not able to prove them now.
Preliminaries and notation:
In the following, "Young tableau" will always mean "Young tableau of straight shape".
Let $n$ be a nonnegative integer. An $n$-permutational tableau will mean a Young tableau with $n$ cells, each of them being filled with an integer from the set $\left\lbrace 1,2,...,n\right\rbrace$, in such a way that every integer from the set $\left\lbrace 1,2,...,n\right\rbrace$ appears exactly once. Such a tableau doesn't, a priori, have to be standard.
When $T$ is an $n$-permutational tableau, we can define two subgroups $R_T$ and $C_T$ of the symmetric group $S_n$ as follows:

We let $R_T$ be the group of all $\sigma \in S_n$ such that for every $i \in \left\lbrace 1,2,...,n\right\rbrace$, the integers $i$ and $\sigma\left(i\right)$ lie in the same row of $T$.
We let $C_T$ be the group of all $\sigma \in S_n$ such that for every $i \in \left\lbrace 1,2,...,n\right\rbrace$, the integers $i$ and $\sigma\left(i\right)$ lie in the same column of $T$.

Now, we define three elements $a_T$, $b_T$ and $c_T$ of $\mathbb{Q}\left[S_n\right]$ by
$a_T = \dfrac{1}{\left|R_T\right|} \sum\limits_{r\in R_T} r$, $b_T = \dfrac{1}{\left|C_T\right|} \sum\limits_{c\in C_T} \left(-1\right)^c c$, and $c_T = a_T b_T$.
(Today I learned that not everybody is using the convention that $c_T = a_T b_T$, and some (particularly in England) prefer to set $c_T = b_T a_T$ instead. Apparently, there is even a disagreement about how to multiply permutations. I will stick to defining $c_T$ as $a_T b_T$, and defining products of permutations by $\left(\sigma \pi\right)\left(j\right) = \sigma\left(\pi\left(j\right)\right)$, since these are the notations I've been using for years. The questions at hand are confusing enough without nonstandard notations adding to the mess.)
It is known that for every $n$-permutational tableau $T$, the elements $a_T$ and $b_T$ of $\mathbb{Q}\left[S_n\right]$ are idempotents, while $c_T$ is a quasi-idempotent (that is, $c_T^2 = \lambda c_T$ for some rational $\lambda$). The element $c_T$ (or, occasionally, a scalar multiple of $c_T$ which actually is idempotent) is called the Young symmetrizer corresponding to the tableau $T$, and is sometimes denoted by $e_T$. Its main significance is that the left $\mathbb{Q}\left[S_n\right]$-module $\mathbb{Q}\left[S_n\right] c_T$ is (isomorphic to) the irreducible representation of $\mathbb{Q}\left[S_n\right]$ corresponding to the shape of $T$.
Now, let us talk about standard tableaux. If $T$ is a standard $n$-permutational tableau, then for every $i \in \left\lbrace 0,1,...,n\right\rbrace$, we can define the $i$-restriction $T\mid_{\leq i}$ to be tableau obtained by only keeping the cells of $T$ which are filled with integers $\leq i$, while removing all the other cells (along with the integers in them). This $T\mid_{\leq i}$ is a standard $i$-permutational tableau. Clearly, $T\mid_{\leq n} = T$.
One last notation. If $\lambda$ is a partition of $n$, the initial $\lambda$-tableau $T_{\lambda}$ will mean the tableau obtained by writing the integers $1$, $2$, ..., $n$ into the cells of the Young diagram of $\lambda$ in the usual order in which books are written in the Western world (i. e., filling the first row from left to right, then the second row from left to right, and so on). Formally, this can be defined as the tableau whose $i$-th row is $\left(k_{i-1}+1, k_{i-1}+2, ..., k_i\right)$ for every $i$, where $\lambda=\left(\lambda_1,\lambda_2,\lambda_3,...\right)$ and $k_i = \lambda_1+\lambda_2+...+\lambda_i$. Of course, this initial $\lambda$-tableau $T_{\lambda}$ is a standard $n$-permutational tableau.
If you have heard about Young symmetrizers, but you know them as being indexed by partitions rather than by tableaux, you are most likely used to only considering the $c_{T_{\lambda}}$'s. (There is not much lost by this restriction except for clarity, since all other $c_T$'s have the form $\omega c_{T_{\lambda}} \omega^{-1}$ for some $\omega\in S_n$ and some partition $\lambda$ of $n$. Indeed, any $n$-permutational tableau $S$ and any $\pi\in S_n$ satisfy $c_{\pi S} = \pi c_S \pi^{-1}$, where $\pi S$ means the tableau obtained by applying $\pi$ to all entries of $S$.)
Facts:
In the following, whenever $m\leq n$, we view $\mathbb{Q} \left[S_m\right]$ as a subring of $\mathbb{Q} \left[S_n\right]$ in the standard way.
Scroll down to "Conjectures" if you don't care for the little that has been shown.
Lemma 1. Let $n$ be a nonnegative integer. Let $S$ and $T$ be two $n$-permutational tableaux such that the shape of $S$ is greater than the shape of $T$ in lexicographic order (this makes sense because shapes of Young tableaux are partitions). Then, $a_S \mathbb{Q} \left[S_n\right] b_T = 0$ and $b_T \mathbb{Q} \left[S_n\right] a_S = 0$.
Proof sketch. Lemma 4.41 in Etingof et al, arXiv:0901.0827v5 shows that if two partitions $\lambda$ and $\mu$ of $n$ satisfy $\lambda > \mu$ in lexicographic order, then $a_{T_{\lambda}} \mathbb{Q} \left[S_n\right] b_{T_{\mu}} = 0$. From this it is easy to deduce that $a_S \mathbb{Q} \left[S_n\right] b_T = 0$ (recalling that every $c_T$ has the form $\omega c_{T_{\lambda}}$ for some $\omega\in S_n$, where $\lambda$ is the shape of $T$), or alternatively one can notice that $a_S \mathbb{Q} \left[S_n\right] b_T = 0$ follows by the same argument as Lemma 4.41.
Applying the antipode of the Hopf algebra $\mathbb{Q} \left[S_n\right]$ to the equality $a_S \mathbb{Q} \left[S_n\right] b_T = 0$, we obtain $b_T \mathbb{Q} \left[S_n\right] a_S = 0$, whence Lemma 1 follows.
Lemma 2. Let $n$ be a nonnegative integer. Let $S$ and $T$ be two $n$-permutational tableaux having different shapes. Then, $c_S \mathbb{Q} \left[S_n\right] c_T = 0$.
Proof sketch. Either the shape of $S$ is greater than the shape of $T$ in lexicographic order, or the shape of $S$ is smaller than the shape of $T$ in lexicographic order. In the first case, Lemma 2 follows from $a_S \mathbb{Q} \left[S_n\right] b_T = 0$, which is a direct application of Lemma 1. In the second case, Lemma 2 follows from $b_S \mathbb{Q}\left[S_n\right] a_T = 0$, which in turn follows from Lemma 1 with $S$ and $T$ switched. In either case, Lemma 2 is thus proven.
Proposition 3. Let $n$ be a nonnegative integer. Let $\lambda$ be a partition of $n$, and let $S$ be a standard $n$-permutational tableau distinct from $T_{\lambda}$. Then, $c_S c_{T_{\lambda}} = 0$.
Proof sketch.

In the case when the shape of $S$ is distinct from $\lambda$, the product $c_S c_{T_{\lambda}}$ is $0$ because Lemma 2 yields $c_S \mathbb{Q} \left[S_n\right] c_{T_{\lambda}} = 0$.
In the case when the shape of $S$ is $\lambda$, one can show the stronger claim that $b_S a_{T_{\lambda}} = 0$. This follows from proving that there exist two distinct integers which are in the same row of $T_{\lambda}$ and in the same column of $S$ (akin to the proof of Lemma 4.41 in Etingof et al, arXiv:0901.0827v5). Alternatively, this case follows from Lemma 3.1.20 in James/Kerber, "The Representation Theory of the Symmetric Group", 1981.

Proposition 4. Let $n$ be a nonnegative integer. Let $\lambda$ be a partition of $n$, and let $S$ be a standard $n$-permutational tableau distinct from $T_{\lambda}$. Then, $c_{S\mid_{\leq n}} c_{S\mid_{\leq n-1}} ... c_{S\mid_{\leq 1}} \cdot c_{T_{\lambda}} = 0$.
Proof sketch. Again, the case when the shape of $S$ is distinct from $\lambda$ can be easily dealt with: in this case, Lemma 2 yields $c_S \mathbb{Q}\left[S_n\right] c_{T_{\lambda}} = 0$, but since $S = S\mid_{\leq n}$, this becomes $c_{S\mid_{\leq n}} \mathbb{Q}\left[S_n\right] c_{T_{\lambda}} = 0$. So we are done in this case.
What remains is the case when the shape of $S$ is $\lambda$. Let $w_S$ denote the word obtained by reading the entries of the tableau $S$ row by row, from top to bottom, where the English convention is used in describing Young tableaux (so the top row is the longest). Since $S \neq T_{\lambda}$, we have $w_S \neq 1 2 ... n$. Thus, there exists a $k\in\left\lbrace 1,2,...,n\right\rbrace$ such that the $k$-th letter of $w_S$ is $\neq k$. Pick the smallest such $k$. Of course, the first $k-1$ letters of $w_S$ are $1 2 ... \left(k-1\right)$ then. This yields easily that the shape of $T_{\lambda}\mid_{\leq k}$ is greater than the shape of $S\mid_{\leq k}$ in the lexicographic ordering. Hence, Lemma 1 yields $a_{T_{\lambda}\mid_{\leq k}} \mathbb{Q} \left[S_k\right] b_{S\mid_{\leq k}} = 0$ and $b_{S\mid_{\leq k}} \mathbb{Q} \left[S_k\right] a_{T_{\lambda}\mid_{\leq k}} = 0$.
Now, we need to prove that $c_{S\mid_{\leq n}} c_{S\mid_{\leq n-1}} ... c_{S\mid_{\leq 1}} \cdot c_{T_{\lambda}} = 0$. Of course, in order to do this, it is enough to show that $c_{S\mid_{\leq k}} c_{S\mid_{\leq k-1}} ... c_{S\mid_{\leq 1}} \cdot c_{T_{\lambda}} = 0$. Thus, it is enough to show that $c_{S\mid_{\leq k}} \mathbb{Q} \left[S_k\right] c_{T_{\lambda}} = 0$ (since all the factors in $c_{S\mid_{\leq k-1}} c_{S\mid_{\leq k-2}} ... c_{S\mid_{\leq 1}}$ lie in $\mathbb{Q} \left[S_k\right]$). Since $c_{S\mid_{\leq k}} = a_{S\mid_{\leq k}} b_{S\mid_{\leq k}}$ and $c_{T_{\lambda}} = a_{T_{\lambda}} b_{T_{\lambda}}$, this boils down to proving that $b_{S\mid_{\leq k}} \mathbb{Q} \left[S_k\right] a_{T_{\lambda}} = 0$.
But if $H$ is a subgroup of a group $G$, then the sum of all elements of $H$ divides the sum of all elements of $G$ (both from the left and from the right) in the group algebra $\mathbb{Q}\left[G\right]$. Hence, $a_{T_{\lambda}\mid_{\leq k}}$ divides $a_{T_{\lambda}}$ in the group algebra $\mathbb{Q}\left[S_n\right]$ (since $R_{T_{\lambda}\mid_{\leq k}}$ is a subgroup of $R_{T_{\lambda}}$). Thus, $b_{S\mid_{\leq k}} \mathbb{Q}\left[S_k\right] a_{T_{\lambda}} = 0$ follows immediately from $b_{S\mid_{\leq k}} \mathbb{Q}\left[S_k\right] a_{T_{\lambda}\mid_{\leq k}} = 0$. Proposition 4 is proven.
Conjectures:
Conjecture 5. Let $n$ be a nonnegative integer. Let $\lambda$ be a partition of $n$. Let $T = T_{\lambda}$. Then, $c_{T\mid_{\leq 1}} c_{T\mid_{\leq 2}} ... c_{T\mid_{\leq n}}$ is an integer multiple of $c_T$.
A stronger conjecture: $c_{T\mid_{\leq n-1}} c_T = \kappa_{T\mid_{\leq n-1}} c_T$, where $\kappa_S$ denotes the product of the hook lengths of the shape of a tableau $S$.
Unless I have done a mistake, in proving the latter (stronger) conjecture one can WLOG assume that $T$ is a two-column tableau, with both columns having the same length. I am not fully sure about it, and more vexingly, I am not able to prove it in this seemingly simple case!
EDIT: Conjecture 5 follows from Theorem 1.2 in Claudiu Raicu, Products of Young symmetrizers and ideals in the generic tensor algebra, 1301.7511v2 (after some straightforward transformations like transposing tableaux). I would still enjoy a simpler proof (Raicu's one is about 10 pages long).
Conjecture 6. Let $n$ be a nonnegative integer. Let $\lambda$ be a partition of $n$. Let $T = T_{\lambda}$. Then, $c_{T\mid_{\leq n}} c_{T\mid_{\leq n-1}} ... c_{T\mid_{\leq 1}}$ commutes with all the Young-Jucys-Murphy elements $y_1$, $y_2$, ..., $y_n$ in $\mathbb{Q}\left[S_n\right]$. Here, the Young-Jucys-Murphy elements $y_1$, $y_2$, ..., $y_n$ are defined by
$y_i = \left(1,i\right) + \left(2,i\right) + ... + \left(i-1,i\right)$ (a sum of $i-1$ transpositions)
for every $i \in \left\lbrace 1,2,...,n\right\rbrace$. As a consequence, $c_{T\mid_{\leq n}} c_{T\mid_{\leq n-1}} ... c_{T\mid_{\leq 1}}$ lies in the commutative subalgebra of $\mathbb{Q}\left[S_n\right]$ generated by $y_1$, $y_2$, ..., $y_n$. (This is a "consequence" because that commutative subalgebra is maximally commutative. But how to write it explicitly as a polynomial in the $y_j$ ?)
EDIT: Conjecture 6 can be strengthened to the statement that $c_{T\mid_{\leq n}} c_{T\mid_{\leq n-1}} ... c_{T\mid_{\leq 1}}$ is a scalar multiple of the seminormal idempotent $e\left(T\right)$ (where I am using the notations of Adriano Garsia, Young seminormal representation, Murphy elements and content evaluations). Thanks to Andrew Mathas for suggesting this statement! It can be proven by induction over $n$ using Conjecture 5 (actually, using the statement that $c_{T\mid_{\leq k}} c_T$ is a scalar multiple of $c_T$ for every $k\leq n$; but this statement isn't hard to derive from Conjecture 5).
So, technically, all conjectures above are settled; but any ideas for short proofs are welcomed.
[Obsolete remark: It seems possible that §3.2 of James/Kerber (which is about Young's seminormal form) has something to tell about this, but my knowledge of that part of $S_n$ theory is negligible.]
Remark: It is not generally true that $c_S \mathbb{Q}\left[S_n\right] c_{T_{\lambda}} = 0$ in the context of Proposition 4.

REPLY [10 votes]: I think that these conjectures are essentially equivalent to results in the literature, but the translation is not straightforward. I know corresponding results for the Hecke algebras of the symmetric groups. When you work in this generality you have to change some of the statements a little in order to get them to work. Primarily this is because your column stabliser $C_T$ will not usually be a standard parabolic subgroup so there is no corresponding natural subalgebra of the Hecke algebra to work with. To get around this you take the corresponding standard parabolic, for which the Hecke algebra does have a subalgebra, and then in the Hecke algebra you conjugate by $T_d$ where $d$ is a minimal length coset representative which makes the two parabolics conjugate. Ultimately, this will prove the results that you want, but it gives an additional layer of complications, which add to the notational confusion.
I think that the easiest way to approach these questions is using seminormal forms. As you probably know, Okounkov and Vershik [1] have given a really nice approach to this subject.
The main fact we need is that over $\Bbb{Q}$ the Specht module $S^\lambda$ has a seminormal basis $\{v_T\}$ indexed by the standard $\lambda$-tableaux with the property that
$$ y_k v_T = c_k(T) v_T.\qquad\qquad(1)$$
Here $c_k(T)$ is the content of $k$  the tableau $T$. (This conflicts slightly with your notation, but it shouldn't cause any problems.) That is, $c_k(T)$ is equal to the column index of $k$ in $T$ minus the row index of $k$ in $T$. (If I have misread your conventions above then you will need to replace $c_k(T)$ with $-c_k(T)$.) 
Each seminormal basis element $v_T$ is a simultaneous eigenvector for $y_1,\dots,y_n$. So $v_T$ is uniquely determined up to multiplication by a non-zero scalar. The exact choice of seminormal basis $\{v_T\}$ doesn't matter in what follows.
Now, for each standard tableau $S$ define
$$F_S = \prod_{k=1}^n\prod_{c\ne c_k(S)}\frac{y_k-c}{c_k(S)-c},$$
where in the second product $c\in\{c_m(V)\}$ where $1\le m\le n$ and $V$ ranges over all of the standard tableaux of size $n$ (so any shape). Acting on our seminormal basis using (1) shows that
$$F_S v_T = \delta_{ST}v_T.$$
To see this you only need to observe that if $S$ and $T$ are two standard tableaux, not necessarily of the same shape, then $S=T$ if and only if $c_m(S)=C_m(T)$ for 
$1\le m\le n$. In turn, this is easily proved by induction on $n$. Letting $\lambda$ range over all partitions of $n$, and using (1) again, you can now prove that $\{F_S\mid S\text{ standard}\}$ is a complete set of pairwise orthogonal primitive idempotents in $\Bbb{Q}S_n$. 
To connect with what you are saying let me set $a_\lambda=a_T$ and $b_\lambda=b_T$, where $T=T_\lambda$ is your initial tableau. Also let $T^\lambda$ be the final tableau, that is, the unique standard $\lambda$-tableau which has the numbers $1,2,\dots,n$ entered in order from top to bottom and then left to right down the columns of $\lambda$. (To add to the notational confusion, in my papers, I call these tableaux $t^\lambda$ and $t_\lambda$ respectively.)
The connection with your questions is that $a_\lambda b_\lambda$ is a scalar multiple of $F_{T^\lambda}$ and the multiple is exactly the product of the hook lengths. As far as I am aware, the first place that this appears in the literature is in Murphy's paper [3]. After translation, (I believe!) this follows from the displayed formula at the bottom of page 512. In Prop. 4.4 of my paper [2] you will find a statement more in keeping with the notation above, except that this paper considers the cyclotomic Hecke algebra case (so more notational complications).
Once you have this fact, then your conjecture 6 follows and conjecture 5 should follow from the definition of $F_s$. In the mathematical-physics literature, I think that your conjecture 5 is often referred to as a fusion system -- actually, this is not quite the same thing, but it is very similar. 
Unfortunately, translating all of the notation back to your setting is slightly painful. Perhaps there are some (old?) papers in the representation theory of the symmetric groups that deal with seminormal forms which just give the result you are interested in, but I am not aware of any. I suspect that it might be easier to just re-prove these things in your setting using the seminormal idempotents $\{F_T\}$.
References

A. Okounkov and A. Vershik, A new approach to representation theory of symmetric groups, Selecta Math. (N.S.), 2 (1996), 581–605.
Mathas, Andrew Matrix units and generic degrees for the Ariki-Koike algebras. J. Algebra 281 (2004), no. 2, 695–730. arXiv:0108164
G.E. Murphy, On the representation theory of the symmetric groups and associated Hecke algebras, J. Algebra 152 (1992) 492–513.<|endoftext|>
TITLE: The Icosahedron Equation
QUESTION [5 upvotes]: $$1728 V^5 + F^3 = E^2 \;.$$
Can anyone point me to a concise, modern derivation and explanation of
the significance of the icosahedron equation, more modern and
concise than Klein's description in his book?

Lectures on the Ikosahedron and the solution of equations of the fifth degree.
  Felix Klein, 1888.

The equation first appears in Klein on p.62 (Dover edition) as
$$T^2 = -H^3 + 1728 f^5 \;,$$
where $f$, $H$, and $T$ are "forms."
$H$ represents the "Hessian form," and $T$ the "functional determinant."
I cannot find the equation in Jerry Shurman's 1997 book, 
Geometry of the Quintic (PDF download),
although I admit I have only scanned his book.
Incidentally, here is a (crude—Sorry!) plot of $x^3 + y^5 + z^2 = 0$,
equivalent to the above by scaling variables:
     
Perhaps this equation has been studied in its own right?

REPLY [5 votes]: The equation is at the bottom of p.61 of Geometry of the Quintic.  It is the syzygy on the three degenerate icosahedral forms.<|endoftext|>
TITLE: Analogy between Jacobian of curve and Ideal class group
QUESTION [5 upvotes]: It is excerpt from "Algebraic Geometry Codes Basic Notions"(https://www.google.ru/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CCoQFjAA&url=http%3A%2F%2Fwww.math.umass.edu%2F~hajir%2Fm499c%2Ftvn-book.pdf&ei=CtLiUc2aAums4ATDloGoBA&usg=AFQjCNH8m6i46UGeRvF8J0nV_cMriSYSww&sig2=-doWN37rrQ2BMFnyUj3c1g&bvm=bv.48705608,d.bGE&cad=rjt) pg 190: 
"The fractional ideal a of the field $\mathbb{K}$ is the same divisor $D$ written multiplicatively. Then the space $L(D)$ corresponds to a
$a^{-1}$. The ideal class group
$Cl_{K}$ is almost the group of $\mathbb{F}_r$-points of the Jacobian of the curve; itscardinality
corresponds in fact to the product of the class number of the number field by its regulator."
So I want to understand accurate statement. Is it true that for every curve $C$ over finite field there is number field $\mathbb{K}$ that Jacobian $C$ is isomorphic $Cl_{\mathbb{K}}$?
If is it true, how to build it number field for some curve?

REPLY [4 votes]: I think this excerpt from Tsfasman-Vladut-Nogin should not be taken as a literal statement--rather it is an explication of (part of) the number field/function field dictionary.  Most of this dictionary comes from the fact that these fields are the function fields of Dedekind schemes.
In particular, suppose $X$ is a Dedekind scheme, with function field $K$ (this may be either a function field or a number field).  If $X$ is a projective curve over a field, rational points of its Jacobian correspond to degree $0$ line bundles $\mathcal{L}$ on $X$; a choice of meromorphic section of $\mathcal{L}$ embeds $\mathcal{L}$ into the constant sheaf $\underline{K}$ on $X$.  On any affine open $\operatorname{Spec} \mathcal{O}$, this lets one view $\mathcal{L}$ as a fractional ideal.  Conversely, fractional ideals for a Dedekind subring $R$ of $K$ correspond to line bundles (locally free sheaves of rank one) on $\operatorname{Spec} R$; that is, line bundles defined on open subsets of $X$.
That is, you should not think of this dictionary as assigning to each function field a number field; rather, it reinterprets the algebraic notion of a fractional ideal as a geometric notion--a line bundle--in the function field case.  This analogy goes quite far; for example, if $X$ is a curve over a finite field $\mathbb{F}_q$, with zeta function $\zeta_X(t)$, one has
$$\operatorname{res}_{t=1} \zeta_X(t)=\frac{\#|\operatorname{Jac} X(\mathbb{F}_q)|}{1-q}$$
which is a function field analogue of the analytic class number formula.  Here the numerator counts line bundles and the denominator counts units; this analogy (and in particular the lack of a regulator term) explains why Tsafaman-Vladut-Nogin refer to the rational points of the Jacobian as corresponding to the size of the class group times the regulator.<|endoftext|>
TITLE: Reference request: expository text on the structure of reductive groups over non-archimedean local fields
QUESTION [10 upvotes]: I am interested in an expository text in English, which summarizes the main results and aspects of the structure theory of reductive groups over local fields, in a hopefully not very technical manner (full proofs are not necessary, though sketches of the main arguments would be nice). Specifically covering the following topics:

The apropriete BN-pair (with the Iwahori subgroup)
The affine root system
The affine Weyl group
the Bruhat decomposition of the affine flag variety.
Classification and structure of standard parabolic (or parahoric?) subgroups.

The emphasis is really being on readability and not thoroughness, so well written notes of some course or seminar on the subject would be great. I could not find any textbook on the subject, and the standard reference everywhere seems to be the original french papers of Bruhat and Tits from the 70's, but if there is such a textbook It would be optimal.
Another point is that I really don't know much about buildings, and even though It seems a fundamental part of the theory, It would be much easier for me to approach this at first from a direction not relaying heavily on the theory of buildings (if it is at all possible). 
Finally, I am mostly interested in the case of the field $\mathbb{C}((t))$ (and not, say, $\mathbb{Q}_p$), which is not a local field in the strict sense (not locally compact), but is a complete non-archimedean DVR (which some people still call a local field), so I am looking for a source that applies for this case.

REPLY [2 votes]: Paul Garrett's book and lecture notes provide a reasonable approach to the subject, for which there are few textbook options.   There are of course other lecture notes, usually slanted in some way which you might or might not be interested in.  As Vishal indicates, the slender 1971 paperback volume by I.G. Macdonald (also referenced in Paul's book), based on his 1970 lectures in Madras, provides a nice shorter exposition  with emphasis on spherical functions.  Macdonald bases his account on an axiomatic version of the Bruhat-Tits theory (emphasizing properties of the BN-pair), in order to avoid assuming too much about algebraic groups or algebraic geometry.     
There's also something to be said for going back to the historical origins in the 1965 IHES paper by Iwahori and Matsumoto here.   They give the first systematic development of affine (and extended affine) Weyl groups, in a separate first section, followed by a detailed treatment of the BN-pair in a split semisimple $p$-adic group (obtained via Chevalley's reduction mod $p$ process from a semisimple Lie algebra over $\mathbb{C}$).  Though some of their notation is obsolete, their approach to the structure theory is concrete and doesn't yet involve directly the theory of Bruhat-Tits buildings which came into play soon afterward.  But they do emphasize the (Iwahori)-Hecke algebra, which originated in Iwahori's earlier work on finite Chevalley groups.       
To get started with just the rank 1 case, there is a short exposition (with details) in my old lecture notes Arithmetic Groups (Lecture Notes in Math. 789, Springer, 1980), $\S15$.  A description of the building is also given, which is just a tree in this case.   Anyway, it's definitely best to start with the split groups, before the notation thickens.<|endoftext|>
TITLE: What is the shape of the $n$-gon which gives the maximum of a function?
QUESTION [9 upvotes]: What is the shape of the $n$-gon $P_1P_2\cdots P_n$ which gives the maximum of $A_n$? The quantity $A_n$ is defined by
$$ A_n = \frac{{\sum_{i\lt{j}\le{n}}{\lvert P_i P_j\rvert}^2}-{\sum_{i=1}^{n}{\lvert P_i P_{i+1}\rvert}^2}}{{\sum_{i=1}^{n}{\lvert P_i P_{i+1}\rvert}^2}} $$
Here, $\lvert P_i P_j \rvert$ is the Euclidean length of the line segment from $P_i$ to $P_j$. Note that $P_{n+1}=P_1$ and that the $n$-gon can be either convex or concave.
I'm interested in $A_n$ because I knew the fact that the maximum of $A_4$ is $1$ only if $P_1P_2P_3P_4$ is a parallelogram. I've already proved the case $n=4$, but I don't have any good idea for $n\ge5$.
Could you tell me how to solve this problem?

REPLY [13 votes]: The optimal shape is the regular $n$-gon and all its affine images.
I am going to optimize the ratio
$$
 \frac{\sum_{i<j} |P_iP_j|^2}{\sum_i|P_iP_{i+1}|^2}
$$
which differs from $A_n$ by 1. For the regular $n$-gon this ratio equals
$$
 \frac{n}{2(1-\cos\frac{2\pi}n)} .
$$
Let us prove that this is the maximum.
The maximum ratio equals the maximum $\lambda\in\mathbb R$ such that the quadratic form
$$
 P\mapsto Q_\lambda(P) := \sum |P_iP_j|^2-\lambda\sum|P_iP_{i+1}|^2
$$
is nonnegative definite. Here $P$ denotes a polygon $P_1\dots P_n$ regarded as a point in $\mathbb R^{2n}$, so $Q_\lambda$ is a quadratic form in $\mathbb R^{2n}$. If $\lambda$ is the maximum and $P$ is an optimal polygon, then $Q_\lambda(P)=0$.
By the Pythagorean theorem, the coordinate expression for $Q_\lambda(P)$ splits into 2 independent summands: one for $x$-coordinates of the vertices and a similar one for $y$-coordinates. Since the form is nonnegative on all configurations of points, it is also nonnegative on the projections to $x$- and $y$-lines, and both summands must be zero separately.
Thus we have reduced the problem to the 1-dimensional case. It remains to solve the similar problem for the from $Q_\lambda$ in $\mathbb R^n$:
$$
 Q_\lambda(x) = \sum (x_i-x_j)^2-\lambda\sum(x_i-x_{i+1})^2
$$
where $x=(x_1,\dots,x_n)\in\mathbb R^n$. By the way, this reduction also shows that, whatever the optimal shape is, it comes with all its affine images.
Now assume that $\lambda$ is the maximum ratio and $x=(x_i)$ is an optimal configuration on the line. Differentiating $Q_\lambda(x)$ with respect to $x_i$ yields
$$
 2n(x_i-a) -2\lambda(2x_i-x_{i-1}-x_{i+1})
$$
where $a$ is the arithmetic mean of $x_1,\dots,x_n$. The derivative must be zero, and by translation we may assume that $a=0$. This gives us a linear recurrence equation
$$
 x_{i+1}=-x_{i-1}+(\tfrac{n}{\lambda}-2)x_i .
$$
The solution must be $n$-periodic, so it is a linear combination of two geometric progressions whose ratios are two conjugate $n$th roots of unity. In other words, we have
$$
 x_k = A \cos k\varphi + B\sin k\varphi
$$
for some real constants $A$ and $B$, where $\varphi$ is a multiple of $\frac{2\pi}n$. (Notation $i$ is changed to $k$ to prevent confusion with complex numbers.) Then
$$
 \lambda = \frac{n}{2(1-\cos\varphi)} .
$$
Checking the condition $Q_\lambda(x)=0$ leaves the only possibility $\varphi=\frac{2\pi}n$ (for larger multiples, the quantity is strictly positive).
Thus the maximum ratio equals $\frac{n}{2(1-\cos\frac{2\pi}n)}$ and the  optimal polygons are those whose $x$- and $y$-coordinates are sequences of the above form. These are precisely affine images of a regular $n$-gon. The solution is the same in all dimensions, not only in the plane.<|endoftext|>
TITLE: the convolution of integrable functions is continuous?
QUESTION [10 upvotes]: The question is simple but I still can't prove it or contradict it. Here it goes:

Suppose $f$ and $g$ are defined on the circle
  (or, equivalently, $2\pi$ periodic functions) and Lebesgue integrable, 
  is their convolution $(f*g)(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x-y) g(y) dy $ continuous?

In the case when two functions are bounded, it is proved in Elias Stein's 
Fourier Analysis (page 47) that their convolution is truly continuous.
However, for unbounded functions, I have tried tools in real analysis, say, Lusin's theorem, transition continuity of $L_1$ functions, etc., but can't figure it out.

REPLY [15 votes]: A less explicit answer: Salem and Zygmund proved that convolution $L^1(\mathbb T) \times L^1(\mathbb T) \to L^1(\mathbb T)$ is onto.
This was shown to hold for all locally compact groups by Paul Cohen in 1959. This result was the starting point of an entire industry establishing "factorization theorems".
A nice survey on this topic is Jan Kisynski, On Cohen's proof of the Factorization Theorem, Annales Polonici Mathematici 75, 2 (2000), 177-192.<|endoftext|>
TITLE: Are there uniformly discrete paradoxical subsets in $\mathbb{R}^3$?
QUESTION [6 upvotes]: I think there aren't any discrete paradoxical subsets in $\mathbb{R}^2$ (any isometry that mapped a discrete subset into itself would have to either be a glide-reflection, a translation or a rotation by $2\pi/n$, and hence the subgroup of the isometry group on $\mathbb{R}^2$ generated by isometries that map a discrete set into itself would have only translations and glide-reflections as elements of infinite order and thus no free subsemigroup of rank 2, hence the subgroup would be supramenable).  
Say that a subset in a metric space is uniformly discrete if there is an $r>0$ such that every pair of points has distance greater than $r$.

Are there any uniformly discrete paradoxical subsets in $\mathbb{R}^3$?

By definition, given a group $G$ acting on a set $X$, a subset $E\subseteq X$ is $G$-paradoxical if $E\neq\emptyset$ and there are $n,m$ and pairwise disjoint subsets $U_1,\dots,U_{n+m}$ of $E$ such that $E=\bigcup_{i=1}^ng_iU_i=\bigcup_{i=n+1}^{n+m}g_iU_i$. That is, one can divide $E$ into pieces that can be reassembled into $E$ twice over. In the above question, $G$ is meant to be the isometry group of the Euclidean space.

REPLY [10 votes]: I think it is not possible for a uniformly discrete set $E$ in any ${\bf R}^d$ to be paradoxical, because one can create an invariant (or almost invariant) mean on such a set.  Indeed, for any $\varepsilon>0$ and $C>0$, one can use the pigeonhole principle to find a large radius $R$ such that $|E \cap (B(0,R+C) \backslash B(0,R-C))| \leq \varepsilon |E \cap B(0,R)|$, where $|A|$ denotes the cardinality of $A$, because the uniformly discrete nature of $E$ forces $|E \cap B(0,R)|$ to grow at most polynomially.  This makes the probability measure $\mu(A) := |A \cap E \cap B(0,R)| / |E \cap B(0,R)|$ approximately invariant (up to error $\varepsilon$) with respect to isometries $T$ that involve a translation by at most $C$, and which map $A$ to a subset of $E$, in the sense that $|\mu(TA)-\mu(A)| \leq \varepsilon$ for such sets.  This should be inconsistent with any putative paradoxical decomposition if one chooses $C,\varepsilon$ appropriately with respect to this decomposition.  (One could also create a genuinely invariant mean by sending $R \to \infty$ and taking an ultralimit or by using a suitable compactness theorem, although this does not seem necessary for this particular application.)
(There should also be some way to use the Tits alternative or something related to this alternative (e.g. Jordan's theorem, Gromov's theorem on groups of polynomial growth, or the Solovay-Kitaev argument) to show the stronger assertion that the semigroup of isometries that map a uniformly discrete set to a subset should be "virtually nilpotent" (or maybe even virtually abelian) in some sense, although I was not able to make any of these fancier tools work here (largely because one only has a semigroup of isometries to play with rather than a group), whereas the more elementary counting argument above seems to already suffice for the purposes of answering the stated question.)<|endoftext|>
TITLE: Adjoining an arrow to a CCC
QUESTION [5 upvotes]: I just started reading Lambek and Scott's book "Introduction to higher-order categorical logic".
Right now I am reading Part I, section 5 (Polynomial categories). They explain two ways of adjoining an inderterminate arrow $x : A_0 \to A$ to a category $\mathcal{A}$ (called $\mathcal{A}[x]$):

Take the underlying graph of $\mathcal{A}$ adjoin $x : A_0 \to A$ to it, and then form the cartesian closed category freely generated by the new graph.
"Equivalently"; form a deductive system (some kind of graph) with objects as the objects of $\mathcal{A}$ and arrows freely generated from the arrows of $\mathcal{A}$ and the new arrow $x : A_0 \to A$, using the application, "initial arrow", pairing, projections, eval and currying, and then they impose the appropiate equations of a CCC and those of $\mathcal{A}$.

I don't see these constructions are infact equivalent. For example, suppose I have two objects $A$ and $B$ in $\mathcal{A}$. Using the construction 1, you end up with two forms of $A^B$ in $\mathcal{A}[x]$: the original one (already avaiable in $\mathcal{A}$) and the one generated freely. Using construction 2, I think you end up with only one form of $A^B$: the original one.
Why do they mean by equivalently? That those categories are equivalent? What's the standard way of constructing this category theory?

REPLY [5 votes]: They mean this: given a cartesian closed $\mathcal{A}$ and objects $A, B$ of $\mathcal{A}$, the inclusion $i: \mathcal{A} \to \mathcal{A}[x]$ is universal with respect to strict cartesian closed functors $F: \mathcal{A} \to \mathcal{C}$ to cartesian closed categories $\mathcal{C}$ that come equipped with a specified arrow $\phi: F(A) \to F(B)$. (I don't have their book to hand, but I think they always deal with chosen products, chosen exponentials, and strict preservation of structure, as they find that more convenient for their purposes. However, such strictness can be relaxed so as to be 2-categorically more appropriate.) In other words, $\mathcal{A}[x]$ comes equipped with a specified "indeterminate" arrow $x: i(A) \to i(B)$, and $F$ can be extended uniquely to a strict cartesian closed functor $\hat{F}: \mathcal{A}[x] \to \mathcal{C}$ that takes $x$ to $\phi$. 
Since they are working strictly, considering ccc's as objects of a 1-category, equivalence here actually means isomorphism (a strict ccc functor that is invertible).  
In any case, the inclusion $i: \mathcal{A} \to \mathcal{A}[x]$ is required to be a strict cartesian closed functor. So it preserves exponentials (strictly, in their setting): $i(A^B) = i(A)^{i(B)}$. In other words, the "two" exponentials coincide. 
Their "functional completeness theorem" gives a very elegant construction of the polynomial ccc $\mathcal{A}[x]$ as a Kleisli category construction. You can find some details on this in the nLab, although the strictness assumptions are not used there.<|endoftext|>
TITLE: Isomorphism between Spin(3,2) and Sp(4, R)
QUESTION [12 upvotes]: I've been using the fact that Spin(3,2) is isomorphic to Sp(4, R) for a while, but I've never seen a proof. Can anyone point me in the direction of a good reference?

REPLY [22 votes]: Here is a coordinate-free version.  Let $(V, \psi)$ be a 4-dimensional symplectic space over a field, and let $\omega \in (\wedge^2 V)^{\ast} = \wedge^2(V^{\ast})$ be the nonzero 2-form arising from $\psi$. Then on the 6-dimensional vector space $W = \wedge^2(V)$ the kernel of $\omega$ is a hyperplane $H$ of dimension 5, and on $W$ there is a natural non-degenerate quadratic form $q$ valued in the line $\wedge^4(V)$ via $q(w) = (1/2)(w \wedge w)$ for $w \in W$ (e.g., if $w = e_1 \wedge e_2 + e_3 \wedge e_4$ for a basis $\{e_i\}$ of $W$ then $q(w) = e_1 \wedge e_2 \wedge e_3 \wedge e_4$); the definition of $q$ uses base change from the $\mathbf{Z}_{(2)}$-flat case for $(V,\psi)$ if $2$ isn't a unit.
By computing in linear coordinates of $V$ that "standardize" $\psi$ we see that $q$ is a split quadratic form on $W$, and the action of ${\rm{SL}}(V)$ on $W$ clearly preserves $q$ while the action of its subgroup ${\rm{Sp}}(V,\psi)$ preserves $H$.  Hence, the action on $H$ defines a map 
$${\rm{Sp}}(V,\psi) \rightarrow {\rm{O}}(q|_H),$$
so this lands inside ${\rm{SO}}(q|_H)$ and as such defines a homomorphism $${\rm{Sp}}_4 = {\rm{Sp}}(V,\psi) \rightarrow {\rm{SO}}(q|_H) = {\rm{SO}}_5.$$
This map kills the center $\mu_2$ inside ${\rm{Sp}}_4$, and thereby identifies ${\rm{Sp}}_4$ as the degree-2 "simply connected" central cover of ${\rm{SO}}_5$ (in the sense of algebraic groups).  Hence, this uniquely lifts to an isomorphism of ${\rm{Sp}}_4$ onto ${\rm{Spin}}_5$.  
A nice feature of this conceptual construction is that it kills two birds with one stone: if we don't restrict to $H$ and instead work with the entire 6-dimensional $W$ then a similar construction defines the isomorphism of ${\rm{SL}}_4 = {\rm{SL}}(V)$ onto ${\rm{Spin}}_6$.<|endoftext|>
TITLE: Average of Fourier coefficients of a cusp form of half integral weight
QUESTION [7 upvotes]: Suppose $f$ is a cusp form of half integral weight $k$ w.r.t. the group $\Gamma_0(4)$ ($k$ is not very low, can assume $k \ge 11/2$), and $a_n$ is its Fourier coefficient. The Linnik bound says that if $f$ is square-free then we have the estimate
$$
|a_n| \ll_\epsilon n^{k/2-2/7+\epsilon}
$$
My question is, is there a nontrivial bound for $\left| \sum_{n=N}^{N+H} a_n \right|$? Here nontrivial means better than simply adding up the Linnik bound individually. Also will averaging take care of the square-free issue in the Linnik bound?
An equivalent problem is to estimate
$$
\sum_{n=N}^{N+H} \sum_{4|c} \frac{K(m,n,c)}{c}
$$
Note: $m$ is assumed to be fixed and is not averaged. To give some order of magnitude, $H$ is supposed to be around $N^{1/5}$, so it is a rather short sum.

REPLY [2 votes]: The paper of Kohnen-Zagier, "Values of L-series of Modular Forms at the Center of the Critical Strip" (http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN002097249&IDDOC=172193), Invent. math. 64 (1981) 175-198 discusses the (weighted) average of the square absolute value of Fourier coefficients of half-integral weight cusp forms.  See for example Corollary 6.<|endoftext|>
TITLE: Why does CH imply that there is a unique ultrapower of $\mathbb{N}$?
QUESTION [19 upvotes]: I've read these words: "How many ultra products $∏_Uℕ$ exist up to isomorphism, where $U$ is a non-principal ultrafilter over $ℕ$? If continuum hypothesis(CH) holds, then obviously just one ..."
i don't know why "obviously" just one? CH just tell the cardinality, not the structral of ultrafilter. even $U_1$ and $U_2$ have same cardinality c, how to make sure their structural can't be different?
where can I find the proof detail or the insight which make this "obvious"?

REPLY [8 votes]: Under the continuum hypothesis, one only obtains a result about the uniqueness of the ultrapower, but the continuum hypothesis does not imply that there is up-to-isomorphism one ultrafilter on a countable set. 
The notion of Rudin-Keisler equivalence formalizes the notion of whether two ultrafilters are structurally equivalent. In fact, two ultrafilters are Rudin-Keisler equivalent if and only if they are isomorphic in the category of ultrafilters. It is well known that for every infinite cardinal $\kappa$, there are $2^{2^{\kappa}}$ Rudin-Keisler types of ultrafilters on sets of size $\kappa$.
Furthermore, the uniqueness of the ultrapower only follows if you have countably many function, relation, and constant symbols. However, we do not obtain a uniqueness result if you consider uncountably many function symbols, relation symbols, and constant symbols.
Let $\mathfrak{C}(\mathbb{N})$ be the structure over the natural numbers where function is a fundamental operation and every relation is a fundamental relation. Then there are $2^{\aleph_{0}}$ function and relation symbols. In this case, there are $2^{2^{\aleph_{0}}}$ non-isomorphic ultrapowers $\mathfrak{C}(\mathbb{N})^{\mathcal{U}}$. This is because there are $2^{2^{\aleph_{0}}}$ distinct Rudin-Keisler types of ultrafilters on a countable set, and if $\mathcal{U},\mathcal{V}$ are ultrafilters on a countable set, then
$\mathfrak{C}(\mathbb{N})^{\mathcal{U}}$ and $\mathfrak{C}(\mathbb{N})^{\mathcal{V}}$ are isomorphic if and only if the ultrafilters $\mathcal{U}$ and $\mathcal{V}$ are Rudin-Keisler equivalent.<|endoftext|>
TITLE: Serre-Tate 1964 Woods Hole notes
QUESTION [8 upvotes]: I am not sure if this is the right venue to ask this. Apologies in advance.
I would like to clarify the following. When people give as reference: 
J.-P. SERRE and J. TATE.-Mimeographed notes from the 1964 A.M.S. Summer Institute in Algebraic Geometry at Woods Hole;
which notes do they exactly refer to? I downloaded a 1964 Woods Hole notes from Milne's site: http://www.jmilne.org/math/Documents/ but didn't find notes which were co-authored by Serre and Tate. There is one written solely by Tate and another solely by Serre. Another part contains notes by the two of them together with Lubin.

REPLY [7 votes]: They are probably referring to the sections "Serre discussed..." and "Tate discussed..." in the Seminar report by Lubin, Serre, Tate, which outline what has become known as Serre-Tate theory. But without knowing the context, I can only guess. It is quite likely the author didn't have access to the notes of conference, because only a very small number were produced, and they have become available on the internet only fairly recently.