added math overflow

fetch_mathoverflow.py

@@ -10,6 +10,8 @@ from itertools import groupby
 import dataclasses
 from tqdm import tqdm
 from bs4 import BeautifulSoup
+import sys
+from pathlib import Path
 
 """
 Author: E.W.Ayers
@@ -166,20 +168,28 @@ def text_of_post(post):
             
     return text, post.Score, post.Id, answered
 
-def pull_and_format(url, save_dir)
+def get_and_format(url, save_dir):
+    Path(save_dir).mkdir(exist_ok=True, parents=True)
     archive_path = os.path.join(save_dir, "archive.7z")
     os.system(f"wget -O {archive_path} {url}")
 
-    sys.exit()
+    global DATA_DIR
+    DATA_DIR = os.path.join(save_dir, "xml")
+    os.system(f"7z e {archive_path} -o{DATA_DIR}")
 
+    print("parsing xml...")
     qs = questions()
 
-    qs_texts = [text_of_post(qs[key]) for key in tqdm(list(qs.keys())[1:100])]
+    qs_texts = [text_of_post(qs[key]) for key in tqdm(qs.keys())]
 
     for post, score, eyed, answered in tqdm(qs_texts): 
         if score >= 5 and answered: 
-            with open(os.path.join("stack_exchange", str(eyed) + ".txt"), "w") as f: 
+            with open(os.path.join(save_dir, str(eyed) + ".txt"), "w") as f: 
                 f.write(post)
+
+    os.system(f"rm -r {DATA_DIR}")
+    os.remove(archive_path)
+
 if __name__ == '__main__':
-    pull_and_format("https://archive.org/download/stackexchange/mathoverflow.net.7z", 
+    get_and_format("https://archive.org/download/stackexchange/mathoverflow.net.7z", 
             "stack_exchange/math_overflow")

stack_exchange/math_overflow/100015.txt

@@ -0,0 +1,45 @@
+TITLE: Smooth bases of matroids
+QUESTION [15 upvotes]: Motivated by algebraic geometry, I've come up with a purely
+combinatorial definition within the theory of matroids. 
+The question is: is this concept known?
+If you like matroids but not algebraic geometry, skip to the definition below.
+Let $n\choose k$ denote the collection of all $k$-element subsets of
+$[1,n]$ (rather than the number thereof). We can and will identify
+this collection both with the set of $T$-fixed points on the
+$k$-Grassmannian $Gr(k,n)$, where $T$ is the $n$-torus that acts
+(unfaithfully), and also with the set of Plücker coordinates.
+Let $C \subseteq {n\choose k}$ be a subcollection. Then, following 
+Neil White, we can define a subscheme $\Pi_C$ of $Gr(k,n)$ by killing
+all Plücker coordinates $p_S, S \notin C$. This subscheme is $T$-invariant,
+and its $T$-fixed points are exactly $C$.
+Easy fact: if $\Pi_C$ is irreducible, then $C$ is a matroid. The
+non-Pappus matroid shows the converse is false. (This is my own motivation
+for matroids -- they serve as combinatorial stand-ins for subvarieties
+of Grassmannians.)
+I'm interested in the smooth points of $\Pi_M$, where $M$ is a matroid.
+Perhaps the most efficient way to describe $M$ is by listing
+its connected flats $F$, and for each, giving the rank. (Saying that
+$rank(F) \leq r$ means that for each $S$ that intersects $F$ too much,
+$p_F = 0$. I'm pretty sure that the connected flats gives the 
+shortest list of $F$s to give all the $S$.)
+
+If $M \subseteq {n\choose k}$ is a matroid, call $\lambda \in M$
+  a smooth base if for any connected flat $F$, $rank(F) = |\lambda \cap F|$.
+
+Note that $\geq$ is required for $\lambda$ to be a base at all.
+It's pretty easy to prove that $\lambda$ is a smooth point of $\Pi_M$
+iff $\lambda$ is a smooth base of $M$ in the sense above.
+
+Is this concept known to matroid theorists? Is this characterization
+  of smooth points known to anybody?
+
+Example: let $M$ be the Schubert matroid for a $\lambda \in {n\choose k}$,
+i.e. for each $i \notin \lambda, i+1 \in \lambda$, we have a connected flat
+$[1,i]$ with rank $|[1,i] \cap \lambda|$. Then $\lambda$ is a smooth
+base of $M$. And indeed, $\lambda$ is the point in the Bruhat cell whose
+closure is the Schubert variety $\Pi_M$.
+
+REPLY [7 votes]: I hope this question gets a good answer. In the mean time I'll mention a concept I've seen which seems somewhat related to your condition of smoothness.
+When you have a matroid $\mathcal M$ with a base $B$ with the property that all cyclic flats $F$ are spanned by $F\cap B$, this is called a fundamental transversal matroid. The base $B$ is called a fundamental base. 
+Schubert matroids are a special case of fundamental transversal matroids, and the base $\lambda$ is a fundamental base in the sense above. At some point I was convinced that your definition implies at least that the matroid is transversal, but now I'm not so sure anymore.
+Some further characterizations in terms of some rank inequalities, or affine representations on simplices are proved in "Characterizations of transversal and fundamental transversal matroids" by J.E. Bonin, J.P.S. Kung and A. de Mier. See also these slides for pictures. I should also point out that I haven't seen the geometric connection which you explain being mentioned in the literature on fundamental transversal matroids.

stack_exchange/math_overflow/10002.txt

@@ -0,0 +1,5 @@
+TITLE: Is every left fibration of simplicial sets with nonempty fibers a trivial kan fibration?
+QUESTION [7 upvotes]: In Lemma 2.1.3.4 of Higher Topos Theory, the statement of the lemma requires that the fibers are not only nonempty but contractible.  However, in the proof, I don't see where contractibility is directly used, only the fact that the fibers are nonempty.  There is one other place where contractibility is mentioned, "Since the boundary of this simplex maps entirely into the contractible kan complex $S_t$, it is possible to extend $f'$ to $X(n+1)$."  However, I don't see how contractibility directly factors in, since that would only attest to the uniqueness of the extension.  The existence of the extension comes from the fact that the inclusion $\partial \Delta^n \times \Delta^1 \subseteq X(n+1)$ is left anodyne and $S_t$ is a nonempty Kan complex and the fact that the map f' factors through the inclusion of $S_t$.  
+Please correct me if I'm wrong.  Also, there is a relevant post on meta where I first asked if this question is appropriate, and I was greenlighted by Anton.
+
+REPLY [5 votes]: Whenever I get confused about quasicategories I think back to ordinary categories.  A left fibration between (nerves of) ordinary categories is the same as an opfibration with groupoid fibers, and a trivial fibration between ordinary categories is the same as an equivalence that is surjective on objects.  Clearly an opfibration in groupoids with nonempty fibers need not be an equivalence unless the fibers are contractible.

stack_exchange/math_overflow/100033.txt

@@ -0,0 +1,11 @@
+TITLE: Interesting mathematical documentaries
+QUESTION [174 upvotes]: I am looking for mathematical documentaries, both technical and non-technical. They should be "interesting" in that they present either actual mathematics, mathematicians or history of mathematics. I am in charge of nourishing our departmental math library (PUCV) and I would like to add this kind of material in order to attract undergraduates toward mathematics. For this reason, I am not looking for videos of conferences or seminar talks, but rather for introductory or "wide public" material.
+Here are some good examples. 
+
+"Dimensions", by Leys, Ghys & Alvarez, http://www.dimensions-math.org/ which explains actual maths and is excellent.
+"Julia Robinson and Hilbert's tenth problem", https://www.vismath.eu/en/films/julia-robinson, about the life of some great mathematicians.
+BBC documentary on "Fermat's last theorem" (by the way, any information about how to purchase it would be welcome, it does not seem to be possible to do it from the BBC site http://www.bbc.co.uk/iplayer/episode/b0074rxx/horizon-19951996-fermats-last-theorem. Maybe http://vimeo.com/18216532 ?). 
+
+Are there more examples? Thanks, Ricardo.
+
+REPLY [4 votes]: Recently a documentary on Maryam Mirzakhani was released called Secrets of the Surface: The Mathematical Vision of Maryam Mirzakhani. It both described her personal life/journey through mathematics, as well as an introduction to part of her research aimed at a general audience.

stack_exchange/math_overflow/100049.txt

@@ -0,0 +1,23 @@
+TITLE: Can every $\mathcal{L}_{\omega_1,\omega}$ formula be expressed as a type? What about canonical forms?
+QUESTION [6 upvotes]: If $\mathcal{L}$ is a countable, first-order language, it is easy to see that every $n$-type $p$ (over $\emptyset$) can be expressed as an $\mathcal{L}_{\omega_1,\omega}$-formula, namely $\bigwedge_{\psi\in p} \psi$. Does the converse hold? Namely, given an $\mathcal{L}_{\omega_1,\omega}$ formula $\psi(v_1,\ldots,v_n)$ is there an $n$-type $p$ such that $\mathfrak{A}\models\psi(\bar{a})$ if and only if $\bar{a}$ realizes $p$ in $\mathfrak{A}$?
+On a related note, are there any nice canonical forms for such infinitary formulas?
+
+REPLY [8 votes]: As noted in the other answers, not every $L_{\omega_1,\omega}$ formula is expressible as a type. 
+Nevertheless, there is some sense in which $L_{\omega_1,\omega}$ is equivalent to omitting types:
+Theorem: For every $L_{\omega_1,\omega}$ sentence $\phi$, there exists a countable first-order theory $T$ and a countable set of types $\{p_n:n<\omega\}$ such that any model $A$ satisfies $\phi$ if and only if $A$ can be expanded into a model of $T$ which omits every $p_n$.
+One can construct $T$ and the $p_n$ as follows: introduce a new relation symbol $R_\psi(x_1,\dots,x_n)$ for every subformula $\psi(x_1,\dots,x_n)$ of $\phi$. If $\psi$ is atomic or constructed by the usual first-order operations from other subformulas, include in $T$ a corresponding axiom: for example,
+$$R_{\exists x_{n+1}\,\chi(x_1,\dots,x_{n+1})}(x_1,\dots,x_n)\leftrightarrow\exists x_{n+1}\,R_{\chi(x_1,\dots,x_{n+1})}(x_1,\dots,x_{n+1}).$$
+The only problem is to deal with countable conjunctions and disjunctions. If for instance $\psi(\vec x)=\bigwedge_{n<\omega}\psi_n(\vec x)$, we include in $T$ the axioms $R_\psi(\vec x)\to R_{\psi_n}(\vec x)$ for all $n$, and we include the type
+$$p_\psi(\vec x)=\{R_{\psi_n}(\vec x):n<\omega\}\cup\{\neg R_\psi(\vec x)\}$$
+as one of the $p_n$'s. Notice that $A$ omits $p$ iff it validates the implication
+$$\bigwedge_{n<\omega}R_{\psi_n}(\vec x)\to R_\psi(\vec x).$$
+The rest of the proof is easy.
+If we work only with infinite models, a single type $p$ is sufficient instead of countably many. This can be seen as follows. Introduce a new predicate $N(x)$ and constants $\{c_n:n<\omega\}$, and consider the type
+$$p(x)=\{N(x)\}\cup\{x\ne c_n:n<\omega\}.$$
+Then for each $\psi(\vec x)=\bigwedge_{n<\omega}\psi_n(\vec x)$ introduce a new predicate $S_\psi(u,\vec x)$ together with the axioms
+$$\begin{gather*}
+R_{\psi_n}(\vec x)\to S_\psi(c_n,\vec x),\\
+\forall u\,(N(u)\to S_\psi(u,\vec x))\to R_\psi(\vec x),
+\end{gather*}$$
+which will serve together with $p$ as a replacement for $p_\psi$.
+This result has various interesting consequences: for example, Hanf numbers of $L_{\omega_1,\omega}$ and of FO with omitting types are the same. (The Hanf number of a logic $L$ is the smallest cardinal $\kappa$ such that for every $L$-sentence $\phi$, if $\phi$ has a model of size at least $\kappa$, then it has models of arbitrarily large cardinality. The axiom of replacement implies that every logic whose formulas do not form a proper class has a Hanf number.) As a matter of fact, both Hanf numbers equal $\beth_{\omega_1}$, but this beautiful result of Morley has a considerably more difficult proof.

stack_exchange/math_overflow/100057.txt

@@ -0,0 +1,10 @@
+TITLE: Open immersions of open manifolds
+QUESTION [31 upvotes]: For concreteness, I will work in the category of smooth manifolds, but my question makes sense in topological and PL category as well. Recall that a manifold $M$ is called open if every connected component of $M$ is non-compact. 
+Question. Is it true that for every $n$ there exists a compact $n$-dimensional manifold $N^n$ so that every open $n$-dimensional manifold $M^n$ admits an immersion in $N^n$? (In this context an immersion is just a local diffeomorphism.) 
+I think that the answer is positive and that the manifolds $N$ are connected sums of products of projective spaces of various dimensions. 
+Edit: Igor noted that real-projective spaces are not enough, one should include complex-projective spaces as factors in the products. (The reason I think that products real and complex projective spaces are the right thing to use is that real and complex projective spaces generate rings of unoriented and oriented cobordisms.) 
+Some background: As we know very well, not every manifold admits an open embedding in a compact manifold. For instance, the infinite connected sum of 2-dimensional tori does not. However, it is easy to prove that every open surface admits an immersion in ${\mathbb R}P^2$. Whitehead proved that every open oriented 3-dimensional manifold admits an immersion in ${\mathbb R}^3$  (and, hence, to any 3-dimensional manifold). I also convinced myself (although I do not have a complete proof) that every open non-orinentable  3-manifold admits an immersion in ${\mathbb R}P^2\times S^1$.  More generally, every open paralelizable $n$-manifold admits an immersion in ${\mathbb R}^n$. This is a special case of the Hirsch-Smale immersion theory, which reduces existence of immersion from an open manifold to a homotopy-theoretic question about maps of tangent bundles.
+
+REPLY [8 votes]: This is the special case announced in the comment. We assume $M$ to be an oriented 4-manifold with the homotopy type of a 2-complex. We prove that $M$ immerses in $\mathbb{C}P^2$.
+In the case of compact 4-manifolds with boundary I have a more direct and elementary proof in my paper which does not depend on Phillips' theorem.
+Anyway, a bundle map $TM \to T\mathbb{C}P^2$ (needed to apply the theorem of Phillips) can be constructed in this way. Firstly, endow $M$ with an almost-complex structure (which exists by our assumptions). There is a nowhere vanishing vector field on $M$, hence $TM$ splits as a Whitney sum of complex rank 1 bundles $TM = \xi \oplus \varepsilon^1$ with $\varepsilon^1$ trivial and $\xi$ a pullback of the complex universal bundle. By cellular approximation, $\xi$ is a pullback of the canonical bundle $\gamma^1_1$ on $\mathbb{C}P^1$, so $TM$ is the pullback of $\gamma^1_1 \oplus \varepsilon^1$. Now, rank 4 real oriented vector bundles over $\mathbb{C}P^1$ are classified by $\Bbb Z_2$ (via the second Stiefel-Whitney class), hence $T\mathbb{CP}^2_{|CP^1} = \gamma^1_1 \oplus \varepsilon^1$ (as real vector bundles). So $TM$ is a pullback of $T\mathbb{C}P^2$ by a suitable map $M \to \mathbb{C}P^2$ (with values in $\mathbb{C}P^1$).

stack_exchange/math_overflow/100072.txt

@@ -0,0 +1,26 @@
+TITLE: Constant Mean Curvature hypersurfaces "condensing" onto a minimal submanifold
+QUESTION [7 upvotes]: Let $M$ be Riemannian manifold and $S\subset M$ a minimal submanifold, with $\dim S<\dim M-1$. According to a few references (e.g., Mahmoudi, Mazzeo & Pacard), it should not be hard to see that:
+
+''The closer a constant mean curvature (CMC) hypersurface of $M$ is to $S$ (in the Hausdorff metric), the larger its mean curvature must be.''
+
+I was wondering if this claim is indeed not hard to see, given that I am still unable to find a simple/elementary proof. Any suggestions?
+Moreover, I was wondering if anything is known about the speed in which the value of the mean curvature $H(t)$ of a family $N_t$ of CMC hypersurfaces of $M$ "condensing" (i.e., collapsing) onto $S$ diverges to $+\infty$. For instance, is it always the case that both $H$ and $H'$ diverge to $+\infty$ as those hypersurfaces collapse onto the minimal limit submanifold? Perhaps, in some special situation (e.g., if the sequence of CMC hypersurfaces collapsing is a solution to the Mean Curvature Flow (MCF)), this is implied by some property of the MCF?
+
+REPLY [8 votes]: Suppose that N is the CMC hypersurface which is $\epsilon$ close to S.
+Take the tube of radius $\epsilon$ around S. One can check that this has
+(typically variable) mean curvature on the order of $1/\epsilon$. If one
+chooses this tube correctly, it is tangent to N and this gives a lower bound 
+for the mean curvature of N at the point of contact. Since the mean
+curvature of N is constant, this means that it is everywhere bigger than
+$1/\epsilon$.
+The situation is now a bit better understood than when that paper was written.
+First, Rosenberg proved a theorem that if N is a surface in a 3-manifold and has mean curvature H and H is very large, then one of the two regions N bounds in the ambient
+manifold has inradius less than C/h.
+Second, it is only true that the set on which the $N_t$ collapse must be minimal
+if the norm of the second fundamental form of N is comparable to the mean curvature.
+If this fails -- and a typical scenario is a sequence of spheres joined by very small
+necks, then it seems to be the case that the collapsing set is not necessarily minimal.
+Finally, regarding your question -- by the same comparison result (using geodesic tubes) it is certainly the case that H is increasing as $N_t$ collapses, so this gives some sort of
+average increase of H'(t).  There maybe could be some exotic examples where the mean curvature wobbles and H' is not necessarily tending to $+\infty$.  
+There are lots of other good questions here -- Harold Rosenberg had several questions which remain unanswered. For example, if  T is a  `geodesic triangle' in a 3-manifold,  then is there a CMC tube which condenses to T? The problem is whether it is possible for the CMC
+surfaces to turn the corners at the vertices of T.

stack_exchange/math_overflow/100082.txt

@@ -0,0 +1,12 @@
+TITLE: Covering a (hyper)cube with lines
+QUESTION [7 upvotes]: Let $K_n$ be the sets of vectors $x \in  \mathbb{Z}^d $ with each coordinates $x_i$ between $1$ and $n$. For any subset $A$ of $K_n$, let $S(A)$ be the set of points $x \in K_n$ which are on some line containing at least two points of $A$ (in other words, $S(A)$ is the union of the lines passing through - at least - two points of $A$).
+Such a set $A$ is said to generate $K_n$ if $S(A) = K_n$. Now let $r_d(n)$ be the smallest size of a generating subset of $K_n$.
+Question : What are the best known bounds on $r_d(n)$ ? (The first non trivial case is $d=2$)
+This problem may be "well-known" ; I'm almost sure this question has already been studied, but I didn't find any reference, and Google gives nothing.
+The trivial bound is $r_d(n) \gg_d n^{\frac{d}{2}- \frac{1}{2}} $ : taking a generating subset of size $r_d(n)$, there are at most $O(r_d(n)^2)$ lines to consider, each one intersecting $K_n$ in at most $n$ points, so that $|K_n| \ll r_d(n)^2 \times n$.
+A refinement of this argument (a typical line contains much less than $n$ points of $K_n$) gives a lower bound $r_d(n) \gg_d n^{\frac{d}{2}- \frac{1}{4} - \frac{1}{4(2d-1)} } $.
+
+REPLY [9 votes]: I think the probabilistic method gives an A of size $O_d(n^{d/2}\sqrt\log n)$. (Formula updated according to js's comment.)
+Put every point to A independently with probability p. What is the probability that a point x will be in S(A)? For any x, we can find $\Omega(n^d)$ pairs of points that are all different such that x lies on the line of any pair. (This is true because e.g. for d=2 if x is in the bottom-left part of $K_n$, then we can take the $n/4\times n/4$ grid upper-right from it, contract the $n/8 \times n/8$ bottom-left of this grid, and double each point from x to get its pair.)
+The probability that both points of a fixed pair are in A is $p^2$, the probability that no such pair exists is $(1-p^2)^{n^d}$. So if $n^d(1-p^2)^{n^d}<1$, then we are done using the union bound. Unless I am mistaken this is true if $p>\Omega_d(n^{-d/2}\sqrt\log n)$.
+Now of course we cannot be sure about how big A is. But if we replace the above $<1$ with a $<1/2$, then we can even add the condition that A should be at most $pn^d$, for which the probability is $\ge 1/2$. So we get $O_d(n^{d/2}\sqrt\log n)$ points. Maybe this can be further improved with some more advanced probabilistic methods.

stack_exchange/math_overflow/100085.txt

@@ -0,0 +1,27 @@
+TITLE: RS to RSK correspondence
+QUESTION [5 upvotes]: The RS correspondence is a correspondence which associates to each permutation a pair of standard Young tableaux of the same shape.
+The RSK correspondence associates to each integer matrix (with non-negative entries) a pair of semistandard Young tableaux of the same shape.
+Given an integer matrix, replace it by a permutation matrix whose rows and columns, when partitioned according to the row and column sums of the original matrix, have block sums equal to the entries of the original matrix. There is a unique such permutation matrix with the property that there are no descents within any of the blocks (each block is a partial permutation).
+For example, if
+$A=\begin{pmatrix} 2 & 1\\ 1 & 0\end{pmatrix}$
+then the corresponding permutation matrix is 
+$\tilde A  =\begin{pmatrix} 1 & 0  & 0 & 0\\
+                             0 & 1  & 0 & 0\\
+                             0 & 0  & 0 & 1\\
+                             0 & 0 & 1 & 0\end{pmatrix}$
+Here the row and column partitions are both $(3,1)$.
+It seems to be well-known (for example, it is implicit in Fulton's matrix ball construction) that to obtain the SSYT's for $A$, one may substitute for each entry in the SYT's for $\tilde A$ the integers corresponding to the blocks the rows and columns corresponding to these entries belong.
+In the above example, the SYT's associated to $\tilde A$ are 
+$P = Q = \begin{array}{cc} 1 & 2 & 3 \\ 4 & &\end{array}$
+into which we would saubstitute $1$ for $1,2,3$ and $2$ for $4$ to get the SSYT's for $A$:
+$P = Q = \begin{array}{cc} 1 & 1 & 1 \\ 2 & & \end{array}$.
+Is there a nice reference for this result?
+
+REPLY [14 votes]: [Responding particularly to Bruce...]  You may want to take a look at my thesis, which was the first place that the Knuth versions of RSK were "Fominized".  There are lots of examples, which others have told me they've found helpful in understanding this material.  (I'm sure Fomin already understood that this could be done, but it doesn't appear in his papers before 1991.)  I put a scan on the web at:
+http://www.math.uconn.edu/~troby/research.html
+Scroll down to: 
+Applications and Extensions of Fomin's Generalization of the Robinson-Schensted Correspondence to Differential Posets, Ph.D. Thesis, Massachusetts Institute of Technology, 1991. 
+The key idea is just that RS commutes with "standardization" of words or SSYT, where one adds subscripts from Left to Right in the word and corresponding tableaux. See EC2, Lemma 7.11.6.  
+Thanks to Tricia Hersh for mentioning this thread at the SIAM DM conference.  This is my first posting to MathOverflow, so I'm not allowed to comment.  
+Hope this helps!
+Tom

stack_exchange/math_overflow/100087.txt

@@ -0,0 +1,13 @@
+TITLE: Are there infinite groups which have only a finite number of irreducible representations?
+QUESTION [9 upvotes]: If I know that a certain group has only finitely many irreducible representations (let's say over the complex numbers), is that group necessarily finite?
+In the following cases, you can see that there will be an infinite number of irreducible representations(irreps).
+
+If the group is a direct product of infinitely many (non-trivial)groups. 
+If the group is finitely generated, infinite, and abelian.
+If there exists a normal subgroup N such that G/N satisfies conditions 1 or 2.
+
+I'm stuck. Any ideas? Given an arbitrary infinite group, can you construct infinitely many irreps?
+
+REPLY [3 votes]: This is not an answer, but it's a bit too long for a comment.
+If I'm not mistaken, any group $G$ which embeds densely into a compact Hausdorff group $H$ has the property that it is finite if and only if it has finitely many irreducible representations. This includes the residually finite groups, which embed densely into their profinite completions. To see this, use the fact that by Peter-Weyl $L^2(H)$ decomposes as a direct sum $\bigoplus_i n_i V_i$ where $V_i$ are the finite-dimensional unitary irreducible representations of $H$ and $n_i = \dim V_i$. It follows that there are infinitely many $V_i$ if and only if $H$ is infinite if and only if $G$ is infinite; moreover, by density the $V_i$ are irreducible representations of $G$, any two which are inequivalent as $H$-representations are inequivalent as $G$-representations. 
+So to find a counterexample we should look for groups that are not residually finite. Infinite simple groups certainly have this property, but I don't know any of them well enough to analyze their irreducible representations.

stack_exchange/math_overflow/10010.txt

@@ -0,0 +1,21 @@
+TITLE: is localization of category of categories equivalent to |Cat|
+QUESTION [9 upvotes]: It might be a stupid question. 
+Suppose There is a category of categories,denoted by CAT,where objects are categories, morpshims are functors between categories
+Take multiplicative system S={category equivalences}. Then we take localization at S. Then we get localized category S^(-1)CAT 
+Another Category, denoted by |CAT|,where objects are categories, morphism are isomorphism classes of functors between categories. 
+I want to prove this two categories are equivalent. I want to prove this problem in two ways:
+
+Naive prove
+Because either |CAT| or S^(-1)CAT has same objects as CAT. So one can prove the morphism class of these two category has same equivalent relationship. For the S^(-1)Cat, suppose F and G are two functors in the same equivalent class, we have sF=sG,for s belongs to S. Then I can prove F and G belongs to the same equivalent class in |CAT|. But on the other hand, if F and G are isomorphic functor. I can not prove sF=sG,for some s belongs to S. What I can only prove is sF is isomorphic to sG.
+Using adjoint functors
+Because I want to prove the projection functor CAT--->|CAT| is a localization functor at S. So I want to construct an adjoint functor |CAT|--->CAT which is fully faithful. But what is this functor. It seems I can not make it well defined.
+
+This is an exercise in Toen's lecture: Lectures on DG-categories
+I attach the lecture notes on DG-categories here Toen notes
+page 5,exercise 2
+
+REPLY [5 votes]: Bertrand Toën is using a different definition of localization than what you describe in your question. Let $\mathcal{C}$ be a category and let $\Sigma$ be a class of morphisms in $\mathcal{C}$. A category $\mathcal{D}$ with a functor $L\colon \mathcal{C} \rightarrow \mathcal{D}$ is called a localization of $\mathcal{C}$ at $\Sigma$ if $L(f)$ is an isomorphism for every $f \in \Sigma$, and $L$ is universal with this property, that is, whenever $F \colon \mathcal{C} \rightarrow \mathcal{E}$ is a functor such that $F(f)$ is an isomorphism for all $f\in \Sigma$, then there exists a unique functor $G \colon \mathcal{C} \rightarrow \mathcal{E}$ with $F=G \circ L$. If a localization of $\mathcal{C}$ at $\Sigma$ exists, then it is unique up to isomorphism and ususally denoted by $\Sigma^{-1}\mathcal{C}$ or $\mathcal{C}[\Sigma^{-1}]$.
+Note that we do not make any assumptions about the class $\Sigma$. As with all universal properties, such an object need not exist in the most general case: usually the resulting category might be "too big", that is, there would be a proper class of morphisms between certain objects in the localization. If the class $\Sigma$ admits a right calculus of fractions in the sense of (e.g.) Borceux, Handbook of Categorical Algebra, Volume I, Chapter 5, then the localization of $\mathcal{C}$ at $\Sigma$ does exist. However, there are lots of examples of categories with classes $\Sigma$ which do have a localization but don't admit a right calculus of fractions. In many of these cases, the functor $L \colon \mathcal{C} \rightarrow \mathcal{D}$ will then not have a right adjoint.
+One of the main sources of such categories are Quillen model categories $\mathcal{M}$, with $\Sigma$ the class of weak equivalences. The localization at $\Sigma$ is then called the homotopy category of $\mathcal{M}$. In fact, the notion of Quillen categories was invented mainly to deal with the difficulties of formally inverting a class of arrows in a category. There is a model structure on the category of small categories (sometimes called the folk model structure) whose weak equivalences are the equivalences of categories, so the desired localization does exist. Moreover, there is a notion of left and right homotopy in a model category. For nice objects (objects which are both fibrant and cofibrant), these notions agree and define an equivalence relation on maps. In the model category of categories, all objects are fibrant and cofibrant. Two functors are homotopic if and only if they are naturally isomorphic.
+The usual construction of the homotopy category of a model category (see e.g. Mark Hovey's book "Model Categories") shows that the localization of $\mathbf{Cat}$ at the equivalences is given by the category whose objects are again categories and whose morphisms are homotopy classes of morphisms. Thus we do indeed find that the localization of $\mathbf{Cat}$ at the class of equivalences is the category $\vert \mathbf{Cat} \vert$ of categories and isomorphism classes of functors, as Toën claimed.
+Charles Rezk has a nice writeup of the folk model structure here.

stack_exchange/math_overflow/100102.txt

@@ -0,0 +1,4 @@
+TITLE: Bitopological spaces and algebraic topology
+QUESTION [5 upvotes]: Is it possible to introduce the concept of bitopological spaces such as $(X,T_1,T_2)$ (introduced by J.C.Kelly see Proc. London Math. Soc. (3) 13 (1963) 71–89 MR0143169, J.C. Kelly) in the homotopy theory or homology theory?
+
+REPLY [4 votes]: In some sense, the (relatively new) field of directed algebraic topology represents an attempt to do this. This article by Marco Grandis includes a discussion (on page 8) of why bitopological spaces are too general to admit a good (directed) homotopy theory.

stack_exchange/math_overflow/100110.txt

@@ -0,0 +1,12 @@
+TITLE: What do we mean by contractible for simplicial objects in a category?
+QUESTION [5 upvotes]: EDIT: removed cruft from this question.
+Recall that extra degeneracies for an augmented simplicial set $X$ are maps $s_0\colon X_n \to X_{n+1}$ for $n=-1,0,1,2,\ldots$ which satisfy the usual simplicial identities with respect to the existing $d_i$, $s_i$. This definition clearly works for simplicial objects in pretty much an arbitrary category.
+A simplicial set, augmented by $X_{-1} = \ast$, with extra degeneracies is contractible. In fact, given enough structure on an ambient category $C$, one can sensibly talk about homotopy of simplicial objects in $C$ (for example, one can say when $sC$ is a category with cofibrant objects, and has a notion of homotopy of maps). [EDIT: By this I mean there is some sort of model structure around relative to which we can talk about homotopy]
+So my question is this: is it reasonable to think of simplicial objects in $C$ with extra degeneracies as being contractible for any category $C$ with terminal object? Certainly, ignoring size issues, we can think of such things as being contractible after we embed them in the category of simplicial sets in $Pre(C)$, if not some smaller (co)completion category.
+Secondarily, can I get away with saying a simplicial object with extra degeneracies "is a contractible simplicial object?"
+
+REPLY [3 votes]: Actually the issue is not that simple. Here is a talk by Mike Barr on showing there are three different notions of 'contractible' for simplicial objects in a category that do not coincide even for simplicial sets. They even give explicit examples.
+
+Michael Barr (joint with John F. Kennison, R. Raphael), Contractible simplicial objects, talk 9 October 2018 in the Logic and Categories seminar, McGill University, abstract, slides.
+
+They find "Strong extra degeneracies" implies "Extra degeneracies" implies "Homotopic to a constant simplicial object", and these are strict implications. See the slides for precise definitions. Beware that the 'extra degeneracies' given in my question are not necessarily the same as Barr et al's definition.

stack_exchange/math_overflow/100111.txt

@@ -0,0 +1,30 @@
+TITLE: Elementary Equivalence =?  Homotopy Equivalence
+QUESTION [12 upvotes]: One of the most interesting novelties in recent foundational studies is Voevodsky's Homotopical Type Theory project (see here). 
+Finally homotopy theory ideas have entered in a royal fashion the foundational arena! 
+I wonder if other areas of logic and foundational studies can be tackled from an homotopical standpoint. For instance, in model theory, one encounters the central notion of elementary equivalence:  
+
+two structures M and N of the same
+  signature $\sigma$ are called
+  elementarily equivalent if they
+  satisfy the same first-order
+  σ-sentences.σ-sentences.
+
+The question:
+
+take elementary embedding as a notion
+  of weak equivalence, what kind of
+  structure has the associated  homotopy
+  category?  Perhaps dreaming a little,
+  can one even manage to identify a
+  Quillen model structure on the
+  category of $\sigma$ -structures? 
+
+NOTE: Andreas Blass has (rightly) asked why I mention elementary embedding in my question, whereas the title talks about homotopy equivalences.
+ Point well taken: I should reformulate the question in a broader form, as: can you choose some maps in the category as weak equivalences, so that we can have a homotopy category, possibly with a good amount of homotopy limits and colimits to do some real computations?
+PS: The dream behind this question is that perhaps one could think of an elementary substructure as a homotopical retract of the ambient structure , and more generally one could come up with a notion of "continuous deformation" of structures, just like in the topological category
+
+REPLY [3 votes]: The dream behind this question is that perhaps one could think of an elementary substructure as a homotopical retract of the ambient structure
+I have tried and failed to do something similar for models of a sufficiently nice theory, say a first order categorical theory such as ACF; here is what my thoughts were. My desired weak equivalences were: add a finite tuple to a model M and get a model Ma prime(primary) and minimal over $M\cup a$ (acyclic cofibration); represent a model $N$ as the union $\cup M_i$ of an increasing chain of elementary submodels of strictly smaller cardinality (acyclic fibration). 
+The first one is an elementary equivalence between a model and its substructure, and thus fits your suggested definition. However, the second one requires one to extend your category of models and consider the category of families of models; from the categorical point of view, you formally add new limits ignoring those limits you already have in your category. 
+It is then easy to define a model "pre"structure that satisfies some of the axioms of a model category; Lowenheim-Skolem theorem then means that every morphism can be decomposed as a cofibration and an acyclic fibration, and existence of prime and minimal models (that holds for sufficiently nice theories) means that every morphism decomposes as an anyclic cofibration and a fibration (axiom M2 of Quillen). But that's all: I was not able to construct a model category for an interesting theory, say non-locally modular or even ACF itself. 
+We were only able to construct a model category for a trivial theory T with the empty language; then this all becomes set theory and is described in here and here

stack_exchange/math_overflow/100129.txt

@@ -0,0 +1,47 @@
+TITLE: What is the first interesting matric Toda bracket in the stable homotopy of the sphere?
+QUESTION [12 upvotes]: Feel free to gloss ‘interesting’ as you see fit.  One way:
+
+1. What is the lowest degree matric Toda bracket in $\pi_\ast(S)$ that doesn't contain zero?
+
+By ‘degree’ I mean total homotopical degree, i.e. the $\ast$ in $\pi_\ast$.  By ‘matric’ I mean to exclude ordinary Toda brackets, that is matric Toda brackets all of whose entries are one-by-one matrices, and also to exclude brackets that are trivially determined by ordinary Toda brackets.
+I'm also interested in the title question with ‘first’ replaced by ‘simplest’.  For instance:
+
+2. What is the lowest order matric Toda bracket in $\pi_\ast(S)$, all of whose matrix entries are sums of products of Hopf elements, that doesn't contain zero?
+
+By ‘order’ I mean the number of entries in the bracket, i.e. whether it is 3-fold or 4-fold or 5-fold or...
+Now I'd like to know the same, but with the proviso that the bracket be detectable in the classical Adams spectral sequence:
+
+3. What is the first or simplest, interesting matric Toda bracket in $\pi_\ast(S)$ that is detectable in the $\mathrm{E}_2$ term of the classical Adams spectral sequence?
+
+Some remarks:
+
+I believe the matric Massey product $\langle h_2^2, h_0, \left(\begin{array}{cc} h_1 h_3 & h_2^2 \end{array} \right), \left(\begin{array}{c} h_1 \\ h_2 \end{array}\right)\rangle $ in the $\mathrm{E}_2$ term of the Adams spectral sequence is the class $e_0$, but that is not a permanent cycle.
+Kochman lists the matric Toda bracket $\langle \sigma, \left(\begin{array}{cc} A[31] & \nu\end{array}\right),\left(\begin{array}{cc} \eta & 0 \\ 0 & \eta \end{array}\right),\left(\begin{array}{c} \nu \\ \eta A[30] \end{array}\right)\rangle$ in degree 44 as containing an element of order 8, where $A[30]$ and $A[31]$ refer to certain elements of order 2 in those degrees.
+
+REPLY [16 votes]: I know this post is quite old, but in case you are still interested, or anyone else is, I thought about sharing my recent thoughts about the topic. After all, this is the second result on "matric toda bracket" on google, and it frustrated me several times that this is still unanswered.
+The bracket André suggested sadly uses relations which are only valid in $tmf$: in $\pi(S)$, we have $\nu^3 + \eta \epsilon = \eta^2 \sigma \neq 0$.
+We want to construct a nontrivial matric Toda bracket, so we have to use a relation which cannot be written as a single product. The first one of those is given by $4\nu + \eta^3 = 0$.
+Multiplying this by two relates this in some sense to the easier relation $8\nu = 0$, so we could wonder about this enabling us to exhibit the Toda bracket $\langle \nu, 8, \nu\rangle$ as twice the matric Toda bracket 
+$$\left\langle \left(\begin{array}{cc} \nu & \eta\end{array}\right), \left(\begin{array}{cc} 4 & \eta^2\\ \eta^2 & 0\end{array}\right), \left(\begin{array}{c} \nu \\ \eta\end{array}\right)\right\rangle$$
+This yields a degree $7$ class of indeterminacy $\nu \cdot \pi_4 + \pi_4 \cdot \nu + \eta \cdot\pi_6 + \pi_6 \cdot \eta = 0$.
+We want to multiply the middle term by $2$. In order to make sense of that, consider the following setting: In a good model for spectra, we can think of bimodules over a ring spectrum $E$, i.e. there are maps $E \wedge X\rightarrow X$ and $X\wedge E \rightarrow X$ plus respective commutative diagrams. Then we can talk about Toda brackets $\langle a, x, b\rangle$, where $a,b\in \pi_* E$ and $x\in \pi_* X$. Now given a bimodule map $f:X \rightarrow Y$, a defining system for the Toda bracket $\langle a, x, b\rangle$ can be pushed forward using $f$ to obtain a defining system for the Toda bracket $\langle a, f_* x, b\rangle$, so we obtain
+$$
+f_* \langle a, x, b \rangle \subseteq \langle a, f_* x, b\rangle
+$$
+This statement immediately generalizes to matric Toda brackets.
+Notice that we can use this here: $S\xrightarrow{2} S$ has a representative which commutes with the right and the left action of $S$ on itself, since it can be factored naturally through the pinching map $S\rightarrow S\vee S$. 
+This can be applied as above to see
+$$
+2\cdot\left\langle \left(\begin{array}{cc} \nu & \eta\end{array}\right), \left(\begin{array}{cc} 4 & \eta^2\\ \eta^2 & 0\end{array}\right), \left(\begin{array}{c} \nu \\ \eta\end{array}\right)\right\rangle \subseteq \left\langle \left(\begin{array}{cc} \nu & \eta\end{array}\right), \left(\begin{array}{cc} 8 & 0\\ 0 & 0\end{array}\right), \left(\begin{array}{c} \nu \\ \eta\end{array}\right)\right\rangle
+$$
+This actually is an equality, since the indeterminacy is zero on both sides.
+In fact, the right hand side equals the usual Toda bracket $\langle \nu, 8, \nu\rangle$.
+This element is usually known as $8\cdot \sigma$, which is nonzero. We obtain
+$$
+\left\langle \left(\begin{array}{cc} \nu & \eta\end{array}\right), \left(\begin{array}{cc} 4 & \eta^2\\ \eta^2 & 0\end{array}\right), \left(\begin{array}{c} \nu \\ \eta\end{array}\right)\right\rangle = 4\sigma
+$$
+What this example tells us is that matric Toda brackets really should be expected all over the place: Whenever a simple relation (consisting of a single product) lifts to somewhere as a more complicated relation (with more than one summand), we should be able to lift Toda brackets built from the simple relation to matric Toda brackets by adding additional rows which account for the new summands, in our case the $\eta^3$. Note also that the exact way in which we did that did not matter that much, we could also have chosen something like
+$$\left\langle \left(\begin{array}{cc} \nu & \eta^2\end{array}\right), \left(\begin{array}{cc} 4 & \eta\\ \eta & 0\end{array}\right), \left(\begin{array}{c} \nu \\ \eta^2\end{array}\right)\right\rangle
+$$
+or similar. The only way this generally might go wrong is that you are unable to do this in a way which actually reduces back to the bracket you wanted (we used $2\cdot \eta^2$ to end up with a diagonal matrix).
+If you look closely into the reference André posted, there a similar thing happened, only that the lifting did not happen along an actual map, but from the $E^{\infty}$-page of a spectral sequence to the actual thing, because of nontrivial extensions.

stack_exchange/math_overflow/100155.txt

@@ -0,0 +1,10 @@
+TITLE: Second homotopy group of Cayley complex
+QUESTION [5 upvotes]: Is there a good reference for information about the second homotopy group of the Cayley complex or Presentation complex of a finitely presented group, especially a hyperbolic group?  I'm looking for an argument that the second homotopy group of the Cayley complex of a hyperbolic group $G$ is finitely generated as a $G$-module in particular, but I'd welcome other interesting starting points around the second homotopy group of the Cayley complex too.
+
+REPLY [10 votes]: If $\langle X,R \rangle$ is a finite presentation of a group $G$, then there exists an exact sequence of $\mathbb ZG$-modules
+$$0 \to \pi_2(Z) \to \mathbb{Z} G^{\oplus R} \to \mathbb Z G^{\oplus X} \to \mathbb Z G \to \mathbb{Z} \to 0,$$
+where $Z$ is the presentation $2$-complex of the presentation above. If one knows in addition that $G$ is of type $FP_3$, then $\pi_2(Z)$ must be finitely generated as a $\mathbb Z G$-module. It is well-known that hyperbolic groups are $FP_{\infty}$, using the Rips complex.
+Any example of a finitely presented group which is not of type $FP_3$ gives a counterexample, i.e. $\pi_2$ is not finitely generated. Brady constructed a subgroup of a hyperbolic group with this property in
+Brady, N. Branched Coverings of Cubical Complexes and Subgroups of Hyperbolic Groups J. London Math. Soc. (1999) 60(2): 461-480.
+Much earlier, Stallings gave an example where the third homology is not finitely generated as a module over the group ring of a finitely presented group.
+Stallings, J. A finitely presented group whose 3-dimensional integral homology is not finitely generated. Amer. J. Math. 85 (1963), 541–543.

stack_exchange/math_overflow/100157.txt

@@ -0,0 +1,17 @@
+TITLE: isomorphic spectral sequences => quasi-isomorphic filtered chain complexes?
+QUESTION [7 upvotes]: Let $(C,\partial)$ and $(C',\partial')$ be chain complexes of $R$-modules where $R$ is a (commutative) ring. Let $F$ and $F'$ be finite filtrations of $C$ and $C'$ respectively, i.e., $$\varnothing = F_0C \subset F_1C \subset \ldots \subset F_nC = C$$ and similarly for $F'$. There exist spectral sequences associated to $F$ and $F'$, let's call them $E_{\ast,\ast}^\ast$ and ${E'}_{\ast,\ast}^\ast$. 
+It is quite easy to see that an isomorphism from $F$ to $F'$ induces an isomorphism from $E$ to $E'$. What is completely unclear to me (and sorry if this is not a research level question but I couldn't find an answer) is the converse, namely:
+
+Does an isomorphism from $E$ to $E'$ imply a (quasi-)isomorphism between $F$ and $F'$?
+
+More precisely, if two filtrations give rise to isomorphic spectral sequences, what is the strongest statement that can be made about them? Does this statement depend on the finiteness of the filtrations, on the nature of the ring $R$, on convergence of the spectral sequences, etc?
+
+REPLY [14 votes]: Say your two filtered chain complexes are concentrated in degree zero.  Then the spectral sequences degerate, and your questions become: If you have two filtered abelian groups and an isomorphism between the associated graded modules, can you deduce that the abelian groups are isomorphic?  The answer is no; you can take
+$$
+0 \subset 2 \mathbb{Z} \subset \mathbb{Z}
+$$
+and
+$$
+0 \subset \mathbb{Z} \subset \mathbb{Z} \oplus \mathbb{Z}/2.
+$$
+As Ralph says, you usually need something on the chain level, unless you're in very degenerate cases (where the spectral sequence determines the isomorphism type of the object in the derived category).

stack_exchange/math_overflow/100158.txt

@@ -0,0 +1,9 @@
+TITLE: Negative objects in categories
+QUESTION [8 upvotes]: I seem to remember a proof that if a category $C$ has coproducts and a $0$ object, then necessarily if we had objects of $C$, say $a$ and $-a$, such that $a \oplus -a \simeq 0$, then $a\simeq 0\simeq -a$.  
+But right now, I can't place this, nor am I 100% sure that that is the correct property.  I am able to show that, using the various universal properties at play, that the morphisms in such a category are necessarily quite boring, but not that it completely collapses (i.e. if I assume that all objects have a corresponding 'negative' object).
+
+REPLY [10 votes]: There are three proofs:
+1) If $C$ has infinite coproducts, you may use the Eilenberg swindle:
+$$a \cong a \oplus (-a \oplus a) \oplus (-a \oplus a) \oplus \dotsc \cong (a \oplus -a) \oplus (a \oplus -a) \oplus  \dotsc \cong 0.$$
+2) Here is my personal summary of Philippe's proof: (a) It is enough to prove the dual statement $a \times a^{-1} \cong 1 \Rightarrow a \cong 1$ in categories with products. (b) This may be reduced, by the Yoneda embedding which preserves products, to the case $\mathsf{Set}$. (c) In the case of $\mathsf{Set}$ it is clear. 
+3) If $\mathcal{L}$ is an invertible object of a symmetric monoidal category, then $\mathcal{L} \otimes -$ is cocontinuous since $\mathcal{L}^{-1} \otimes - $ is right adjoint to it. In particular $\mathcal{L} \otimes -$ preserves initial objects. Now apply this to $(C,\oplus,0)$. Thus if $a$ is invertible w.r.t. $\oplus$, we have $a \oplus 0 \cong 0$. On the other hand, the left hand side is also $a$.

stack_exchange/math_overflow/100165.txt

@@ -0,0 +1,7 @@
+TITLE: Number of double cosets
+QUESTION [7 upvotes]: If $G$ is a finite group and $H \leq G$ is a subgroup, then $|G/H| = \frac{|G|}{|H|}$.
+Is there an easy way to compute $|K \backslash G / H|$, for $K \leq G$ also a subgroup?
+
+REPLY [3 votes]: Take $K=H$ and consider the diagonal action of $G$ on $\Omega\times\Omega$, where $\Omega$ is the set of  the right cosets $H\backslash G$. Let the number of orbits of this action be $d$. Then $d$ is the number of double cosets $H\backslash G/H$:
+If we denote the orbits by $\Omega_1,\Omega_2,\ldots, \Omega_d$, let $g_1,\ldots,g_d$ be representatives of the respective orbits. Then the map sending $(Hg,Hg^{\prime})$ to $Hg^{\prime}g^{-1}H$ is easily shown to be a bijection between the set of orbits and double cosets.
+More details can be found under the topic of Schur bases.

stack_exchange/math_overflow/100173.txt

@@ -0,0 +1,10 @@
+TITLE: Bounding the minimum entry of an inverse matrix
+QUESTION [5 upvotes]: Suppose $A$ is an $n\times n$ stochastic matrix, that is, entrywise nonnegative and row sums are all $1$. If $A$ is invertible, is it true that the minimum diagonal entry of $A^{-1}$ is no larger than $1$? 
+Small matrices support this claim, but for larger ones, I don't know how to (dis)prove it. 
+Edited I forgot to add the condition that the diagonal entries of $A$ are all zero.
+
+REPLY [8 votes]: Here's a counterexample:
+$$ A = \frac{1}{5}
+     \left[ \matrix{ 0 & 1 & 3 & 1 \cr 2 & 0 & 1 & 2 \cr 3 & 1 & 0 & 1 \cr 0 & 3 & 2 & 0}\right], \quad
+ A^{-1}=    \frac{1}{3} \left[ \matrix{ 7 & -9 & 11 & -6 \cr -8 & 6 & -4 & 9 \cr 12 & -9 & 6 & -6 \cr -13 & 21 & -14 &9} \right].
+$$

stack_exchange/math_overflow/100183.txt

@@ -0,0 +1,6 @@
+TITLE: Homotopy invariance of vector bundles by parallel transport: reference needed for my students.
+QUESTION [8 upvotes]: Let $M$ be a smooth manifold and $V \to [0,1] \times M$ be a smooth vector bundle. The homotopy invariance states that the restrictions $V_0$ and $V_1$ to the bottom and top of the cylinder are isomorphic.
+One can prove that using parallel transport: pick a connection on $V$. For each $x \in M$, take the curve $c_x:t  \mapsto (t,x)$. Parallel transport defined by the connection along the curves $c_x$ gives an isomorphism $V_0 \cong V_1$.
+This is of course an easy exercise, and my question is to find a self-contained reference that is accessible for undergraduates who know the notions of manifolds, vector bundles and integral curves of vector fields on manifolds (I wish to assign this as a seminar talk).
+
+REPLY [2 votes]: You can check pages 33-35 of my course notes. Also you can check section 3.4 of Husemoller's book  Fiber Bundles, Graduate Texts in Math, vol. 20, Springer Verlag, 1994.  In my notes  I prove the result for arbitrary compact spaces, while Husemoller does this for paracompact spaces. In both cases no smoothness is assumed.

stack_exchange/math_overflow/100196.txt

@@ -0,0 +1,11 @@
+TITLE: Indeterminacy of long games
+QUESTION [7 upvotes]: Hello, all,
+Several months ago I sat in on a seminar on AD+, which was incredibly wonderful even though I could barely follow it at all. AD+ is a technical variant of AD, the axiom of determinacy, which asserts that $$ \text{For all countable } X, A\subseteq X^\omega, \text{ the game with payoff set A is determined.}$$
+It is easy to show that $ZFC+AD$ is contradictory; however, assuming some large cardinals, $ZF+AD$ is consistent, and in fact has the natural model $L(\mathbb{R})$.
+One of the results mentioned was the fact that a natural strengthening of AD, in which $\omega$ is replaced by an arbitrary ordinal, is inconsistent with ZF (even though assuming large cardinals, many classes of long games are determined; this I gleaned from the amazon.com preview of Itay Neeman's "Determinacy of Long Games," which I suspect has the answers to my questions, but I don't have access to it). Having lost my notes from the seminar, I have two questions:
+1) What is the proof of this fact?
+2) When/by whom was it proved?
+Thanks to all in advance!
+
+REPLY [11 votes]: Your statement of AD is incorrect.  $X$ should be at most $\omega$ (and at least 2), not an arbitrary set.  Specifically, for $X=\aleph_1$, determinacy of games in $X^\omega$ is inconsistent with ZF.  
+Concerning your actual question, determinacy of games of length $\omega_1$, even with $X=2$, is inconsistent.  I believe that this result and the one in the first paragraph of my answer are in Mycielski's first big paper on AD, "On the axiom of determinateness" in Fundamenta Mathematicae 1963 or 1964.  For the proof, fix a coding of the well-orderings of $\omega$ as $\omega$-sequences of 0's and 1's, and consider the following game of length $\omega_1$.  Player I must, at one of his moves, play 1 (if he always plays 0, he loses).  As soon as he plays a 1, say at move $\alpha$, player II must, in his next $\omega$ moves, play a code for a well-ordering of $\omega$ of length $\alpha$.  Clearly, player I cannot have a winning strategy, since every $\alpha$ has a code.  Determinacy of this game would mean that player II has a winning strategy, but that would give a function assigning to each countable ordinal $\alpha$ one of its codes.  The existence of such a function (or of any set of reals of size $\aleph_1$) contradicts AD (which is, of course, a consequence of length-$\omega_1$ determinacy).

stack_exchange/math_overflow/100201.txt

@@ -0,0 +1,48 @@
+TITLE: Integral representation of a determinant
+QUESTION [9 upvotes]: In a paper by Mathai, he uses the following integral representation of a determinant,
+(or, really, what I give is a simple special case of what he gives), without any explication.
+All matrices are real $p\times p$ symmetric positive definite.
+\begin{equation}
+    | I-U |^{-a} = \frac{1}{\Gamma_p(a)} \int_{T>0} |T|^{a-(p+1)/2} \exp(-\text{Tr}(I-U)T) \;dT
+\end{equation}
+where $U$ satisfies $0\lt U \lt I$ (in the cone order in the cone of positive definte matrices),
+the integral is over the cone of positive definite matrices and $\Gamma_p(a)$ is the
+generalized gamma function in dimension $p$, and $\Re(a) > (p-1)/2$.
+Any references for this?
+
+REPLY [12 votes]: OLDER EDIT. (Elementary derivation) I realized that my original answer was actually overkill for the question. The said integral in question follows from the definition of the multivariate Gamma function
+\begin{equation*}
+\Gamma_p(a) := \int_{A > 0} \exp(-\mbox{tr}(A))\det(A)^{a-(p+1)/2}(dA),
+\end{equation*}
+where $\Re(a)>(p-1)/2$. 
+From this it follows (by a change of variables) that for a positive definite matrix $S$, 
+\begin{equation*}
+\int_{A > 0} \exp(-\mbox{tr}(S^{-1}A))\det(A)^{a-(p+1)/2}(dA) = \Gamma_p(a)\det(S)^a,
+\end{equation*}
+so that with $S=(I-U)^{-1}$ we obtain the integral in question. 
+Of course, to complete the picture it may be helpful to express $\Gamma_p(a)$ in more elementary terms. Chapter 2 of Muirhead's book provides these details. I cite the result that provides this expression.
+
+Theorem (Muirhead (1982), Thm 2.1.2) Let $\Re(a) > (p-1)/2$. Then,
+  $$
+\Gamma_p(a) = \pi^{p(p-1)/4}\prod_{j=1}^p \Gamma(a - (j-1)/2)
+$$
+
+(Hint: To prove the above, write the Cholesky decomposition $A=T'T$, with these change of variables, the original Gamma integral factorizes into a product of Gaussian and Gamma integrals.)
+
+The part that I recalled below provides yet another representation that expresses the multiplicative determinantal lhs in terms of an infinite sum.
+OLDER STUFF
+This is actually somewhat classical knowledge. Here are two related pointers.
+A partition $\tau=(t_1,\ldots,t_m)$ is a vector of nonnegative integers listed in increasing order, and $|\tau|$ denotes $t_1+\cdots+t_m$. The generalized Pochhammer symbol $(a)_\tau$ is defined as
+\begin{equation*}
+\newcommand{\risingf}[2]{{{#1}}^{\overline{{#2}}}}
+  (a)_\tau := \frac{\Gamma_d(a+\tau)}{\Gamma_d(a)} = \prod_{l=1}^m \risingf{\bigl(a - \tfrac{1}{2}(l-1)\bigr)}{t_l}
+\end{equation*}
+Let $C_\tau(X)$ be the Zonal Polynomial with signature partition $\tau$. Then, the following representation exists
+For a matrix $U$ satisfying $\|U\| < 1$, we have the following "binomial-theorem"
+\begin{equation}
+  \frac{1}{|I-U|^a} = \sum_{k\ge 0}\sum_{|\tau| = k} \frac{(a)_\tau C_\tau(U)}{k!}.
+\end{equation}
+Using representations for these Zonal polynomials, one can obtain the integral representation mentioned in the original post.
+More directly, you can look at Chapter 7 of R. Muirhead, "Aspects of Multivariate Statistical Theory", where you'll see that actually, $|I-U|^{-a}={}_1F_0(a;U)$, a matrix argument hypergeometric function. I've to run now, if I get a chance I'll clean up my answer and fill in the details.
+
+REPLY [2 votes]: A very similar (but different) expression for log of the determinant is given by Du and Ji in their paper "an integral representation of the determinant of a matrix and its applications". I am guessing a slight adaptation of their thing can get your formula.

stack_exchange/math_overflow/100217.txt

@@ -0,0 +1,13 @@
+TITLE: Generalized Grassmannians that parameterize the submodules of a module
+QUESTION [7 upvotes]: I'm looking for something like a Grassmannian, but which parameterizes the submodules of a module rather than the subspaces of a vector space. Most specifically, I'm looking for something which parameterizes the submodules of specifically $\mathbb{Z}^n$. So another way to say it is that I'm looking for a space parameterizing for the subgroups of a free abelian group. (A moduli space?)
+I've seen some references to the concept of a "Grassmannian of submodules" here and there (like the papers on the first page of https://www.google.com/search?q=%22grassmannian+of+submodules%22) but can't figure out if this handles modules like $\mathbb{Z}^n$ or not.
+Does anyone know if such an object exists and if so, how to construct it? Where I can get more information on this?
+EDIT: to give a bit more info, the only specific application I really care about is parameterizing the free subgroups of a free abelian group, or the "lattices" in the $\mathbb{Z}$-module $\mathbb{Z}^n$, etc. A solution which works only for that, but which doesn't handle more exotic modules would be fine for my purposes. (And if it doesn't work out for all free abelian groups in general, then having a solution for at least finitely generated free abelian groups would even be a great start.)
+I framed the question in terms of the "submodules of a module" in general just because I saw some references to there being a "Grassmannian of submodules" before, so I thought such a construction might be widely known.
+
+REPLY [6 votes]: Building on Donu Arapura's answer and Qiaochu Yuan's comment:
+If the quotient is torsion then $k$ changes as a functioin of the characteristic, so no $Grass_{k,n}$ will do. If it is non-torsion then classifying the $\mathbb Z$-module is equivalent to classifying the $\mathbb Q$-subsoace it generates, so a point on the Grassmanian will be a fine enough invariant. It's also clear that every rational subspace contains a rank-k $\mathbb Z$-submodule whose quotient is torsion-free, so for the torsion-free case that is sufficient.
+The next-simplest case is probably when the quotient is finite, since every other case can be viewed as a combination of this and the other one (by splitting the quotient into rank and torsion.) As Quiaochu points out, this is choosing a surjection $\mathbb Z^n\to A$, which is equivalent to choosing a set of $n$ generators of $A$. But since only the kernel of the map matters, and composing with an automorphism of $A$ doesn't change the kernel, you need to quotient out by the automorphism group of $A$. (The automorphism group of $A$ is not the image of $GL_n(\mathbb Z)$, for instance $GL_n(\mathbb Z) \to GL_n(\mathbb F_p)$ is not a surjection.)
+For instance, there is only one sublattice of $\mathbb Z^n$ whose quotient is $(\mathbb Z/p)^n$: that being $p\mathbb Z^n$.
+We can break this finite quotient into a product of $p$-torsion parts and consider those separately. If the $p$-torsion part is $(\mathbb Z/p)^k$, then $Grass_{k,n}(\mathbb F_P)$ will classify the choices of generators. If the torsion involves prime powers then I am not sure what to do.
+But a deeper problem is that I don't think one can make a scheme over $\mathbb Z$ whose $\mathbb Z$-points have a natural bijection to the $\mathbb F_p$ points of an arbitrary scheme over $\mathbb F_p$ like $Grass_{k,n}(\mathbb F_p)$, other then silly stuff like the disjoint union of one copy of $\textrm{Spec} \mathbb Z$ for each point on the $\mathbb F_p$-scheme.

stack_exchange/math_overflow/100224.txt

@@ -0,0 +1,22 @@
+TITLE: Which immersed plane curves bound an immersed disk?
+QUESTION [17 upvotes]: I am looking for a nice answer to the following question.
+
+Which immersed plane curves bound an immersed disk?
+
+Comments.
+
+I am not sure what is a nice answer, but for sure I could make a stupid algorithm.
+I am aware that there are plane curves which bound few "different" immersed disks. For example this Bennequin’s curve is bounding five different immersed disks. It suggests that there is no "nice" answer to the question.
+
+REPLY [17 votes]: The answer was given by Samuel Blank in his Brandeis 1967 phd dissertation, on which Poenaru gave a Bourbaki seminar.
+Then Peter Shor and C. J. Van Wyk gave a polynomial time algorithm to decide if there is an extension.
+EDIT: Blank's method already gave a polynomial algorithm, but with an exponent too large to make it practical, which is needed for applications (for instance to integrated circuit design). 
+The answer is in term of existence of a chain of "reductions" of a certain kind for the cyclically reduced word in the free group $F_n$ on $n$ generators determined by the immersion, where $n$ is the number of bounded components of the complement of the curve (assumed to have only transverse double points).
+
+REPLY [3 votes]: Dear Anton,
+Some time ago I was looking at this question (which is important for Chekanov's invariants of legendrian links) and the literature is rather scattered (however, look at the work of Charles Titus in MathSciNet). 
+I found the following paper, but I haven't really taken a good look at it yet.
+"When Does a Curve Bound a Distorted Disk?
+Jack E. Graver and Gerald T. Cargo
+Consider a closed curve in the plane that does not intersect itself; by the Jordan–Schoenflies theorem, it bounds a distorted disk. Now consider a closed curve that intersects itself, perhaps several times. Is it the boundary of a distorted disk that overlaps itself? If it is, is that distorted disk essentially unique? In this paper, we develop techniques for answering both of these questions for any given closed curve in the plane.
+Read More: http://epubs.siam.org/doi/abs/10.1137/090767716"

stack_exchange/math_overflow/100226.txt

@@ -0,0 +1,7 @@
+TITLE: Is there a uniform Dugundji theorem
+QUESTION [5 upvotes]: A theorem of Dugundji states that if $X$ is a separable metric space and $A \subseteq X$ is closed then any continuous function $f$ from $A$ to some normed linear space $L$ may be extended to a continuous map $\bar{f} \colon X \to L$. I got this formulation from van Mill's book "Infinite-Dimensional Topology". Does there exist a version of Dugundji's Theorem where we assume that $f$ is uniformly continuous an conclude the existence of a uniformly continuous extension? Since my main interest is when $L$ is a Banach space (or even a $C^*$-algebra), I would be very happy to add that assumption to get a positive answer.
+
+REPLY [3 votes]: Note that the case of real-valued functions  is easy. A function $f:A\to\mathbb{R}$ has a uniformly continuous extension to any metric space $X\supset A$  iff it has a sub-linear modulus of continuity (that is, that verifies $\omega(t)\le a|t| + b$).
+Rmk. Note that these general extensions problems may be approached as selection problems for multivalued functions. Precisely, given  $f:A\to L$ and a concave modulus of continuity $\omega$, consider the multi $F:X\to 2^L$, taking $x\in X$ to the set $F(x)\subset L$ of all admissible values for an $\omega$-continuous extension of $f$ to $A\cup \{x\}$, that is $$F(x)=\cap _ {a\in A} \bar B\big(a,\omega(d(a,x))\big)   \, .$$
+Then, any extension $\tilde f$ of $f$ to $X$ has to be a selection of $F$, $f(x)\in F(x)$; note that $F(a)=\{a\}$ for $a\in A$. If $\omega$ is not too small and $L$ is compact enough, you may hope to prove that $F(x)$ is a not-empty bounded closed set, for all $x$. You may then correspondingly look for a modulus of continuity for $F$ seen as a map valued into non-empty, bounded, closed subsets of $L$ (even convex, if $L$ is a normed space) with respct to the Hausdorff distance. Finally, you may construct $\tilde f$ as a composition $c\circ F$, where $c$ is a map that picks a point $c(S)$ within every such subset $S$. This program can be achieded in a very satisfactory way in the case of the Kierszbraun theorem, where the Lipschitz constant is preserved: Hilbert spaces $L$ are a nice setting, because in this case there is a 1-Lipschitz function $c$, taking a bounded convex set $S$ to the center of the smallest closed ball containing $S$. Slightly more genreal, I guess, uniformly convex Banach spaces $L$ could work, but I guess the modulus of continuity of the extension in general will be larger. 
+If $L$ is a metric space, then of course there are in general topological obstructions even for continuous extensions.

stack_exchange/math_overflow/10023.txt

@@ -0,0 +1,9 @@
+TITLE: What do you call this ring?
+QUESTION [10 upvotes]: I want a ring $R$ of "numbers" such that:
+For any sequence of congruences $x\equiv a_1 \pmod{n_1}, x\equiv a_2 \pmod{n_2},\dots$ with $a_i\in \mathbb{Z}$ and $n_i\in \mathbb{N}$ such than any finite set of these congruences has a solution $x\in\mathbb{Z}$, there is a $r\in R$ such that $r\equiv a_1 \pmod{n_1}, r\equiv a_2 \pmod{n_2},\dots$ 
+and
+For any $r\in R$ and $n\in\mathbb{N}$ there is a $a, 0\leq a< n $ such that $r\equiv a \pmod{n}$.
+I think that $R$ has to be the product set of the p-adic integers over all primes p, but what do you call this ring?
+(Perhaps there should be a "terminology" tag? Edit: It already exists but it is called "names")
+
+REPLY [3 votes]: Another (fancy) name is "the free profinite group of rank 1"

stack_exchange/math_overflow/100233.txt

@@ -0,0 +1,20 @@
+TITLE: Question on Hilbert Manifolds
+QUESTION [9 upvotes]: I have a very basic question on Hilbert manifolds.
+Consider the Hilbert space 
+$$
+\mathcal{H}:= L^2(S^1)
+$$
+with $S^1$ the unit circle. 
+On $\mathcal{H}$ let us introduce the equivalence relation 
+$$
+f\sim g : \Leftrightarrow f(\cdot ) = g(\cdot + \alpha)\quad
+\mbox{for some  }\alpha \in S^1.
+$$
+Now define the factor space 
+$$
+\overline{\mathcal{H}}:= \mathcal{H}/\sim.
+$$
+What is the structure of $\overline{\mathcal{H}}$? Is it a Hilbert manifold? If so, how to construct the smooth structure? 
+I am particularly interested in computing a (Riemannian) distance between two elements of $\overline{\mathcal{H}}$.
+
+REPLY [3 votes]: It probably is NOT a smooth manifold. I think finding a chart around the point corresponding to constants, namely, the fixed points of the action of the group of rotation, is problematic.More precisely, at a fixed point, there is not a well-defined tangent space.

stack_exchange/math_overflow/100239.txt

@@ -0,0 +1,9 @@
+TITLE: Is NBG set theory stronger than ZFC?
+QUESTION [7 upvotes]: It is well known, that every statement involving only set-variables is provable in NBG set theory if and only if it is provable in ZFC. What confuses me however is that NBG has a global axiom of choice. Global choice implies that every set is ordinal definable (V=OD).
+So the statement V=OD seems to be a counterexample: It only involves set-variables, it follows from NBG, but it is known to be independent of ZFC. Where am i wrong?
+
+REPLY [18 votes]: It is not true that global choice implies $V=OD$. 
+Global choice is the assertion that there is a class well-ordering of the universe. This is equivalent to the assertion that there is a global choice function, which selects from every non-empty set an element. It is not part of the axiom, however, that this class is definable. Rather, the class well-ordering of $V$ is simply one of the classes that is available in the second order part of the model, and there is no reason to suppose that all such classes are necessarily first-order definable, even from set parameters. 
+The axiom $V=HOD$, on the other hand, is equivalent to the assertion that there is a definable global choice function class, or equivalently, a definable well-ordering of the universe. 
+So the critical difference is in the question of whether the global well-ordering is definable or not. It might be interesting to note that there can be a global well-ordering that is definable from set parameters, and so still counts as a class in ZFC, without having $V=HOD$, simply because those parameter may be essential, and it may not be possible to define the order without them. This is what happens, for example, after adding a generic Cohen real over $L$, since in $L[c]$ we do not have $V=HOD$, and in fact $\text{HOD}^{L[c]}=L$ by the homogeneity of the forcing, but nevertheless one can use $c$ as a parameter to define in $L[c]$ a well-ordering of the universe, essentially the usual $L$-order relativized to $c$.
+The proof that NGBC is conservative over ZFC is not difficult. Given any model $M$ of ZFC, one first forces global choice by class forcing:  consider the class partial order consisting of well-ordered sequences of longer and longer set length. This forcing is $\kappa$-closed for every $\kappa$, and hence adds no new sets, but the generic filter $G$ provides a class well-ordering of $M$. Now, to form the GBC model $M[G]$, one considers as classes all the classes that are definable from set parameters and this new global well-order class. One can verify that the resulting model satisfies all the GBC axioms (and this amounts essentially to the verification that this kind of class forcing works as expected). Since the sets of $M[G]$ are the same as $M[G]$, it follows that any statement that is true in every GBC model is also true in every model of ZFC, and so the extension is conservative.

stack_exchange/math_overflow/100245.txt

@@ -0,0 +1,10 @@
+TITLE: No injective groups with more than one element?
+QUESTION [16 upvotes]: There are several claims in the literature that there are no injective groups (with more than one element), but I have not found a proof.  For example, Mac Lane claims in his Duality from groups paper in the 1950 Bull. A.M.S. that Baer showed him an elegant proof, but gives no hint of what it might be.  Does anyone know of an actual published proof?
+
+REPLY [9 votes]: The argument in the slides by Maria Nogin (née Voloshina) linked to by a-fortiori in a comment was published as:
+Maria Nogin, A short proof of Eilenberg and Moore’s theorem, Central European Journal Of Mathematics Volume 5, Number 1 (2007), 201–204. Also available on her homepage.
+
+Added: The above paper was also mentioned in Jonas Meyer's answer to the same question on math.SE. As Steve D. points out in a comment there, the result  appears as Exercise 7 on page 9 of D.L. Johnson's Topics in the Theory of Group Presentations, Cambridge University Press 1980:
+
+
+The argument by Eilenberg and Moore appears on pages 21/22 of Foundations of Relative Homological Algebra, Memoirs of the AMS, Volume 55 (1965). Here's a scan of the relevant passage for the convenience of the readers:

stack_exchange/math_overflow/100263.txt

@@ -0,0 +1,14 @@
+TITLE: Why is Heegaard Floer Homology defined in terms of Sym$^g\Sigma_g$ instead of Pic$^g\Sigma_g$?
+QUESTION [11 upvotes]: Recall the definition of Heegaard Floer homology: $\Sigma_g$ is a closed surface, and $\{\alpha_1,\ldots,\alpha_g\}$ and $\{\beta_1,\ldots,\beta_g\}$ are sets of attaching circles.  Then Heegaard Floer homology is (more or less) the Lagrangian intersection Floer homology of $\mathbb T_\alpha=\prod_{i=1}^g\alpha_i$ and $\mathbb T_\beta=\prod_{i=1}^g\beta_i$ in $\operatorname{Sym}^g\Sigma_g$.
+Now if we think of $\Sigma_g$ as a complex curve, then there is a birational map $\phi:\operatorname{Sym}^g\Sigma_g\to\operatorname{Pic}^g\Sigma_g$.
+
+What happens if instead we consider the Lagrangian intersection Floer homology of $\phi(\mathbb T_\alpha)$ and $\phi(\mathbb T_\beta)$ inside $\operatorname{Pic}^g\Sigma_g$?  Are the resulting groups trivially the same, trivially different, or at least interesting?  (if they're not the same, then I guess there may be no good reason why they would even be invariants of the underlying three-manifold).
+
+There is at least one concrete reason (and one philosophical reason) why one might try this definition instead of the original:
+
+There are no holomorphic spheres in $\operatorname{Pic}^g\Sigma_g$ (because it is an abelian variety; in fact the map $\phi$ is exactly contracting all the embedded $\mathbb P^n$'s in the symmetric product).  This means we don't have to worry about some types of bubbling.
+$\operatorname{Pic}^g\Sigma_g$ is a complex torus; in particular its topology is very concrete and easy to understand.  Also it is perhaps algebrogeometrically more natural than the symmetric product.
+I could imagine that maybe there is some general statement whereby blowing down all the $\mathbb P^n$'s always does something understandable (perhaps nothing) to the Lagrangian intersection Floer homology.
+
+REPLY [7 votes]: There is a tacit assumption behind this question, which I don't think is justified: that the  Abel-Jacobi images of the Heegaard tori $\mathbb{T}_{\alpha}$ and $\mathbb{T}_{\beta}$ are Lagrangian with respect to some reasonable symplectic form on the Jacobian torus. 
+One can make the Heegaard tori Lagrangian by using a Kaehler form on the symmetric product that is product-like outside some neighbourhood of the diagonal. And one can probably find symplectic forms for which Abel-Jacobi is a symplectomorphism outside a neighbourhood of the theta-divisor (this is certainly true in the genus 2 case). Doing both of these things at once would suffice to make the images Lagrangian, but this might be tricky to achieve  - and it's perhaps not very natural?

stack_exchange/math_overflow/100272.txt

@@ -0,0 +1,20 @@
+TITLE: Second homotopy group of the mod 2 Moore spectrum
+QUESTION [13 upvotes]: Let $S/2$ be the mod 2 Moore spectrum (i.e. the cofiber of $2: S \to S$). Then multiplication by 2 acts nontrivially on this spectrum: the homotopy groups of $S/2$ are  all $\mathbb{Z}/4$-modules by a formal argument, but not $\mathbb{Z}/2$-modules. For instance, $\pi_2(S/2) = \mathbb{Z}/4$.
+The long exact sequence in homotopy groups enables one to determine the homotopy groups of $S/2$ (in the range that the stable homotopy groups of spheres are known) up to extension, but the extension problems are generally nontrivial: $\pi_2(S/2)$ is a case in point (as the aforementioned exact sequence $0 \to \mathbb{Z}/2 \to \pi_2(S/2) \to \mathbb{Z}/2 \to 0$ shows). 
+Is there a general technique for resolving these kinds of extension problems? I can get it in this case using some computation in the cobar complex to get the $t-s = 1$ line of the Adams spectral sequence for $S/2$ and see that the Bockstein acts nontrivially. I'm curious about a more efficient method of doing this. (Another example I had in mind was $bo \wedge \mathbb{CP}^2$: this is $bu$ by the "theorem of Reg Wood" and this is visible in the ASS, but directly computing $KO$-groups of complex projective spaces yields nontrivial extension problems, I think.)
+
+REPLY [5 votes]: Often these extension problem are solved using Toda brackets (as Peter May already mentioned). I will first give the general statement, not only for spectra but also for module spectra. The statement may sound a bit complicated, but is quite useful.
+
+Let $R$ be a strictly associative ring spectrum. Let $x\in \pi_n R$ be an element in the coefficients and denote by $Cx$ the cone of $\Sigma^n R \xrightarrow{x} R$. Then we have a long exact sequence
+  \begin{eqnarray*}\cdots\to\pi_* \Sigma^n R\to \pi_* R \to \pi_* Cx \to \pi_{*-1} \Sigma^n R\to \pi_{*-1}R\to\cdots\end{eqnarray*}
+  which splits into short exact sequences of the form
+  \begin{eqnarray*}0\to \pi_* R/x\pi_*R \xrightarrow{\alpha} \pi_* Cx \xrightarrow{\beta} \{\pi_{*-n-1}R\}_x\to 0\end{eqnarray*}
+  where $\{\pi_{*-n}R\}_x$ denotes all elements which are annihilated by $x$.
+Let $y\in \pi_m R$ and $z\in \pi_k R$ be elements in the coefficients of $R$ with $xy=0$ and $yz=0$. Let $\widetilde{y} \in \pi_*Cx$ be an element with $\beta(\widetilde{y}) = y$. Let $w\in \pi_*R$ be an element such that the projection of $w$ is mapped to $\widetilde{y}z$ under $\beta$. Then $w\in \langle x,y,z\rangle$.
+
+Back to your example: We take $R = \mathbb{S}$ and $x = 2\in\pi_0 \mathbb{S}$. We have now a short exact sequence
+$$ 0 \to \pi_2\mathbb{S} \to \pi_2 \mathbb{S}/2 \to \pi_1\mathbb{S} \to 0 $$
+where the outer groups are isomorphic to $\mathbb{Z}/2$ and are generated by $\eta^2$ and $\eta$, respectively. Let now $y = \eta$ and $z =2$. Lift $\eta$ to an element $\widetilde{\eta} \in \pi_2 \mathbb{S}/2$. Then the statement above tells us that $2\cdot \widetilde{\eta}$ is in the image of $\langle 2, \eta, 2 \rangle = \eta^2$. In particular, $2\cdot \widetilde{\eta}$ is non-zero. This solves the extension problem.
+This technique can be applied to many cases, including $KO\wedge C\eta$. I did some calculations in the category of $TMF$-modules with this.  
+The problem remains how to compute Toda brackets. In general, this might be difficult, but often methods to compute $\pi_*R$ give also methods to compute the Toda brackets in $\pi_* R$. For example, Massey products in the $E^2$-term of the Adams spectral sequence converge to Toda brackets (if there are no "crossing differentials"). The Massey product $\langle 2,\eta, 2\rangle$ can, for example, be computed via cobar representatives. 
+Regarding references: The statement about Toda brackets and cofiber sequences follows more or less directly from the definition of Toda brackets in the framework of triangulated categories. See for example, Section 4.6 of my Thesis -- although I was a little bit lazy there with signs. Regarding the convergence of Massey products to Toda brackets one finds a statement in Kochman's Bordism, Stable Homotopy, and Adams Spectral Sequences. Note that he uses another definition of Toda bracket, which agrees with the one in triangulated categories in the case of triple brackets (and also else if one ignores indetermenancy) as proven in some paper by Kochman.

stack_exchange/math_overflow/100276.txt

@@ -0,0 +1,11 @@
+TITLE: Can one prove complex multiplication without assuming CFT?
+QUESTION [20 upvotes]: The Kronecker-Weber Theorem, stating that any abelian extension of $\mathbb Q$ is contained in a cyclotomic extension, is a fairly easy consequence of Artin reciprocity in class field theory (one just identifies the ray class groups and shows that each corresponds to a cyclotomic extension). However, one can produce a more direct and elementary proof of this fact that avoids appealing to the full generality of class field theory (see, for example, the exercises in the fourth chapter of Number Fields by Daniel Marcus). In other words, one can prove class field theory for $\mathbb Q$ using much simpler methods than for the general case.
+The theory of complex multiplication is similar to the theory of cyclotomic fields (and hence the Kronecker-Weber Theorem) in that it shows that any abelian extension of a quadratic imaginary field is contained in an extension generated by the torsion points of an elliptic curve with complex multiplication by our field. To prove this, one normally assumes class field theory and then shows that the field generated by the $m$-torsion (or, more specifically, the Weber function of the $m$-torsion) is the ray class field of conductor $m$.
+My question is: Can one prove that any abelian extension of an imaginary quadratic field $K$ is contained in a field generated by the torsion of an elliptic curve with complex multiplication by $K$ without resorting to the general theory of class field theory? I.e. where one directly proves class field theory for $K$ by referring to the elliptic curve. Is there a proof in the style of the exercises in Marcus's book?
+Note: Obviously there is no formal formulation of what I'm asking. One way or another, you can prove complex multiplication. But the question is whether you can give a proof of complex multiplication in a certain style.
+
+REPLY [10 votes]: (modified)
+Historically, Complex Multiplication precedes Class Field Theory and many of the main theorems of CM for elliptic curves were proved directly. See Algebren (3 volumes) by Weber or Cox's book for an exposition.
+Please also read Birch's article on the beginnings of Heegner points where he points this out explicitly (page three, paragraph beginning "Complex multiplication ...).
+But not all. so the answer is no (unlike what I had mistakenly presumed at first and the comments below alerted me).  
+The actual history is quite complicated; see Schappacher.

stack_exchange/math_overflow/100281.txt

@@ -0,0 +1,41 @@
+TITLE: Does the curvature determine the metric?
+QUESTION [41 upvotes]: Hello,
+I ask myself, whether the curvature determines the metric.
+Concretely: Given a compact Riemannian manifold $M$, are there two metrics $g_1$ and $g_2$, which are not everywhere flat, such that they are
+not isometric to one another, but that there is a diffeomorphism which preserves the curvature?
+If the answer is yes:
+Can we chose $M$ to be a compact 2-manifold?
+Can we classify manifolds, where the curvature determines the metric?
+What happens, if we also allow semi-riemannian manifolds for the above questions?
+Thank you for help.
+
+REPLY [46 votes]: It should not be surprising that, for a $2$-dimensional manifold, the Gauss curvature $K:M\to\mathbb{R}$ does not determine a unique metric $g$.  After all, the former is locally one function of $2$ variables while the latter is locally three functions of $2$ variables.  It would be remarkable if $K$ determined $g$, even up to isometry, and, of course, except for constant $K$, it does not, as the arguments of Weinstein and Kulkarni show.
+Meanwhile, if one regards the curvature $\mathsf{R}(g)$ of of the Levi-Civita connection of the metric $g$  on a surface $M$ as a section of the rank $3$ bundle $\bigl(T\otimes T^\ast)_0\otimes \Lambda^2(T^\ast)={\frak{sl}}(T)\otimes \Lambda^2(T^\ast)$, then the equation $\mathsf{R}(g) = {\mathsf{R}}_0$ for a given nonzero section ${\mathsf{R}}_0$ of ${\frak{sl}}(T)\otimes \Lambda^2(T^\ast)$ is a determined second-order equation for $g$.  (Here, I am writing "$T$" for "$TM$", etc., to save space.) This then becomes a more reasonable question, one that has an easy answer:  Namely, if $M$ is compact and connected and $g_1$ and $g_2$ are metrics on $M$ such that $\mathsf{R}(g_1) = \mathsf{R}(g_2)$ and such that these curvature tensors are nowhere vanishing, then $g_2 = c g_1$ for some constant $c>0$, and conversely.  The reason for this is simple:  If ${\mathsf{R}}_0$ is a nonvanishing section of ${\frak{sl}}(T)\otimes \Lambda^2(T^\ast)$, then requiring that $\mathsf{R}(g) = {\mathsf{R}}_0$ determines the conformal class of $g$ purely algebraically (because using $g$ to 'lower the first index' of ${\mathsf{R}}_0$ has to result in a tensor that is skewsymmetric in the first two indices).  Thus, if $\mathsf{R}(g_1) = \mathsf{R}(g_2) = {\mathsf{R}}_0$, where the latter is nonvanishing, then $g_2 = e^u\ g_1$ for some function $u$ on $M$.  Plugging this back into the equation $\mathsf{R}(g_1) = \mathsf{R}(g_2)$ and computing shows that $u$ must be harmonic with respect to the conformal structure determined by $g_1$ (which is also the conformal structure determined by $g_2$).  If $M$ is compact and connected, then $u$ has to be constant.
+As Misha points out, in higher dimensions, the situation is rather different and is well-described in the works of Kulkarni and Yau that he cites.  However, just qualitatively, it's worth pointing out that specifying the sectional curvature of a metric is generally a very overdetermined problem in higher dimensions, which is why the rigidity results of Kulkarni and Yau should not be surprising.
+To see the nature of this, recall that the Riemann curvature tensor $\mathsf{Rm}(g)$ of a metric $g$ on $M$ is got from $\mathsf{R}(g)$ by 'lowering an index'. By the first Bianchi identity, $\mathsf{Rm}(g)$ is a section of the subbundle ${\mathsf{K}}(T^\ast)\subset {\mathsf{S}}^2\bigl(\Lambda^2(T^\ast)\bigr)$ that is the kernel of the natural map
+induced by wedge product
+$$
+\mathsf{S}^2\bigl(\Lambda^2(T^\ast)\bigr)\to \Lambda^4(T^\ast)
+$$
+In fact, this map has a natural $\mathrm{GL}(T)$-equivariant right inverse, so that one has a canonical bundle splitting
+$$
+\mathsf{S}^2\bigl(\Lambda^2(T^\ast)\bigr) = {\mathsf{K}}(T^\ast)\oplus \Lambda^4(T^\ast)
+$$
+and these two subbundles are $\mathrm{GL}(T)$-irreducible.  (I'm using $\mathrm{GL}(T)$ to denote the 'gauge' group of bundle automorphisms of $T = TM$.  I don't say that it's great notation, but it will be OK for this discussion.)  An important interpretation of this splitting is the following one:  Obviously, one can regard a section of $\mathsf{S}^2\bigl(\Lambda^2(T^\ast)\bigr)$ as a quadratic form on the bundle $\Lambda^2(T)$.  Then the elements of $\Lambda^4(T^\ast)\subset \mathsf{S}^2\bigl(\Lambda^2(T^\ast)\bigr)$ are exactly the quadratic forms that vanish on all of the decomposable elements in $\Lambda^2(T)$, i.e., the elements of the form $x\wedge y$ for $x,y\in T_xM$ for some $x$.  In particular, a section $Q$ of ${\mathsf{K}}(T^\ast)$ is completely determined by its values on the 'cone bundle' $\mathsf{C}\subset \Lambda^2(T)$ that consists of the decomposable elements.
+Now, any metric $g$ on $M$ determines a unique section $\Lambda^2(g)$ of ${\mathsf{K}}(T^\ast)$ with the property that 
+$$
+\Lambda^2(g)\bigl(x\wedge y\bigr) = |x\wedge y|^2_g = g(x,x)g(y,y)-g(x,y)^2,
+$$
+and the sectional curvature of $g$ is simply the function $\sigma_g:\mathsf{Gr}(2,T)\to\mathbb{R}$ defined as the ratio
+$$
+\sigma_g\bigl([x\wedge y]\bigr) 
+= \frac{\mathsf{Rm}(g)\bigl(x\wedge y\bigr)}{\Lambda^2(g)\bigl(x\wedge y\bigr)}
+\quad\text{for $x,y\in T_xM$ with $x\wedge y\not=0$},
+$$
+where we regard $\mathsf{Gr}(2,T)$ as the projectivization of the cone bundle $\mathsf{C}$.
+The fact that the sectional curvature function is the ratio of two quadratic forms that are sections of $\mathsf{K}(T^\ast)$ shows that it is a very constrained function on $\mathsf{Gr}(2,T)$.  In fact, unless the sectional curvature is constant on $\mathsf{Gr}(2,T_x)$, the numerator and denominator of this ratio at $x\in M$ are uniquely determined up to a common multiple.  In particular, the set of such functions on $\mathsf{Gr}(2,T)$ can be regarded as the set of sections of a bundle over $M$ whose fibers are isomorphic to a space of dimension $d_n = \tfrac12 n(n{+}1) + \tfrac1{12}n^2(n^2-1) -1$ that is singular along the $1$-dimensional curve of constant functions.  
+In particular, specifying a candidate sectional curvature function $\sigma:\mathsf{Gr}(2,T)\to\mathbb{R}$ that is not constant on any fiber $\mathsf{Gr}(2,T_x)$ is equivalent to specifying $d_n$ functions of $n$ variables for the $\tfrac12 n(n{+}1)$ coefficients of $g$, which is highly overdetermined when $n>2$.  This is why one has the level of rigidity for the sectional curvature that is indicated in Kulkarni's and Yau's results.  In fact, as this analysis shows, if $\sigma_{g_1} = \sigma_{g_2} = \sigma$, where $\sigma$ is not constant on any fiber $\mathsf{Gr}(2,T_x)$, then one must have $g_2 = e^u g_1$ for some function $u$ on $M$ and that function $u$ will have to satisfy $\mathsf{Rm}(e^u g_1) = e^{2u} \mathsf{Rm}(g_1)$ which is, itself, a very highly overdetermined second order equation for the single function $u$.   (It's kind of remarkable that there are any nonflat metrics $g$ that admit nonzero solutions $u$ to $\mathsf{Rm}(e^u g) = e^{2u} \mathsf{Rm}(g)$ at all, even without the compactness assumption.  That is what makes Yau's counterexample in dimension $3$ so interesting.  Indeed, though, it turns out that, up to constant scaling and diffeomorphism, there is exactly a $1$-parameter family of such exceptional metrics in dimension $3$, so they are extremely rare indeed. )
+The upshot of all these observations is that specifying a nonconstant sectional curvature function $\sigma_g$ in dimensions above $2$ very nearly determines $g$ in all cases, so that, except for a very small set of degenerate cases, the sectional curvature does, indeed, determine the metric.  However, it's such an overdetermined problem that this result is not all that surprising.
+In particular, specifying $\sigma_g$ is specifying a greater quantity of information about $g$ than specifying, say, $\mathsf{Rm}(g)$.  When $n>3$, even specifying $\mathsf{Rm}(g)$ is overdetermined, and, generally speaking, you'd expect $\mathsf{Rm}(g)$ to determine $g$ as well when $n>3$.  
+When $n=3$, specifying $\mathsf{Rm}(g) = \mathsf{R}$ is locally $6$ second order equations for $6$ unknowns, so it's a determined system.   Back in the 80s, by using the Cartan-Kähler theorem, I showed that, when $\mathsf{R}$ is appropriately nondegenerate and real-analytic, the equation $\mathsf{Rm}(g) = \mathsf{R}$ is locally solvable for $g$, and the general solution depends on $5$ functions of $2$ variables.  Later, Dennis DeTurck and Deane Yang proved this result in the smooth category as well (see Local existence of smooth metrics with prescribed curvature, in Nonlinear problems in geometry (Mobile, Ala., 1985), 37--43, Contemp. Math., 51, Amer. Math. Soc., Providence, RI, 1986).
+In higher dimensions, the natural determined curvature problem is to specify the Ricci tensor $\mathsf{Rc}(g)$ as a section of $S^2(T^\ast)$ (or some trace-modified version of it), and, there, Dennis DeTurck has the best results.

stack_exchange/math_overflow/100288.txt

@@ -0,0 +1,9 @@
+TITLE: Bayesian statistics for pure mathematicians
+QUESTION [25 upvotes]: Could someone please recommend reading on Bayesian statistics presented from a pure mathematical point of view?  That is, works that start assuming a good knowledge of measure theoretic probability.  The usual reference works for statisticians gloss over the fine details, but for me this just leads to more confusion!
+I already know of the following question
+Statistics for mathematicians
+and will follow up some of the suggestions there, but I am specifically looking for a treatment of the Bayesian approach, so if there are any recommendations that apply just to this I would appreciate hearing about them.
+
+REPLY [3 votes]: There is a book of 2018 on the topic, that also presents some recent results of the author:
+
+S. Watanabe, Mathematical Theory of Bayesian Statistics, Chapman and Hall/CRC, Apr 2018

stack_exchange/math_overflow/100289.txt

@@ -0,0 +1,21 @@
+TITLE: About the axiom of choice, the fundamental theorem of algebra, and real numbers
+QUESTION [9 upvotes]: About fundamental theorem of algebra, there is a large collection different demonstrations.
+I ask: is there some proof that avoids AC (choice axiom)?
+In a general topos (with natural number object) there are the two constructions of real numbers (generalizations of the classical Dedekind and Cauchy constructions) that are different. 
+In ZF theory, are the Dedekind and Cauchy constructions different? (In the "Cauchy" reals,  operates on a real number $r$ through a choice of a Cauchy sequence converging to $r$.)
+
+REPLY [23 votes]: The fundamental theorem of algebra is, unless I miscounted quantifiers, a $\Pi^1_2$ sentence of second-order arithmetic and therefore absolute between the full universe and Gödel's constructible universe by the Shoenfield absoluteness theorem.  So, since it's provable in ZFC, it is necessarily provable in ZF as well.
+I believe, though, that this metamathematical argument can be avoided, because I don't see any serious use of the axiom of choice in usual proofs by methods of analysis and (plane) topology.  
+The Dedekind and Cauchy constructions of the reals are equivalent in ZF; no choice is needed.  Although AC is often involved when one wants to prove general results about sequences of reals, in the present situation one only needs to consider sequences of rationals.  So one can  fix an enumeration of the rationals and thereafter, whenever a rational needs to be chosen, taking the first suitable one in the enumeration.
+
+REPLY [12 votes]: Regarding Cauchy and Dedekind reals. The fact that every Dedekind real has a Cauchy representation is provable in very weak systems of intuitionistic analysis. The converse fact that every Cauchy real has a Dedekind representation is equivalent to the Weak König Lemma (WKL) — that every infinite tree of $\lbrace0,1\rbrace$-sequences has an infinite path. (See note.) This very weak form of König's lemma is provable in ZF, so there is no difference between Cauchy reals and Dedekind reals in ZF.
+Breaking this down further, the ability to compare a Cauchy real to a given rational number is equivalent to the Lesser Limited Principle of Omniscience (LLPO), which can be formulated as follows
+$$\lnot(\exists n A(n) \land \exists n B(n)) \rightarrow (\forall n \lnot A(n) \lor \forall n \lnot B(n)),$$ where $A(n)$ and $B(n)$ are simple decidable statements (e.g., primitive recursive predicates). Note that LLPO is a consequence of the Law of Excluded Middle, and therefore true in every classical system such as ZF, but not necessarily in intuitionistic systems such as topoi. To extract a Dedekind cut from a Cauchy real, it is necessary to simultaneously compare a real with every rational number, so we need to repeat LLPO countably many times. WKL is precisely equivalent to countably many repetitions of LLPO. In set theory and topos theory, such repetitions come for free through comprehension. So the only thing that can prevent every Cauchy real to have a Dedekind cut in a topos is the failure of LLPO.
+
+This is not quite true as I originally stated it. It is the existence of a "uniform method" for converting Cauchy reals into Dedekind reals which is equivalent to WKL. In classical systems, one can get around the uniformity requirement through clever use of the Law of Excluded Middle. In weak classical systems such as RCA0 one can prove that every Cauchy real has a Dedekind cut by providing separate methods for rational and irrational numbers, thus sweeping a lot of the computation cost in deciding whether a real is rational or not. The equivalence with WKL is recovered when one asks to simultaneously find Dedekind cuts for an infinite sequence of Cauchy reals, as demonstrated by Hirst [Representations of reals in reverse mathematics, Bull. Pol. Acad. Sci. Math. 55 (2007), 303–316; MR2369116].
+
+REPLY [8 votes]: In classical logic, using the ZF axioms, one can show that there is an isomorphism between the Dedekind cuts in the rationals, and the equivalence classes of Cauchy sequences.   One can also (without AC) choose a representative from each equivalence class of Cauchy sequences. (EDIT: In high school one might chose decimal fractions, and represent $\pi$ by the sequence $(3, 3.1, 3.14,\ldots)$; a more sophisticated approach might choose continued fractions and represent 
+$\pi$ by the sequence $(3, 22/7, 333/106,\ldots)$. 
+The fundamental theorem of algebra (every non-constant polynomial with real or complex coefficients has a zero in the complex numbers) can be proved without AC. In fact, there is a quite explicit algorithm that computes, for each irreducible polynomial (with leading coefficient 1, to be on the safe side), 
+the roots of this polynomial continuously from the coefficients.  
+In constructive mathematics or intuitionistic logic, the story is different, as Francois Dorais explains in his answer.

stack_exchange/math_overflow/10029.txt

@@ -0,0 +1,12 @@
+TITLE: A local-to-global principle for being a rational surface
+QUESTION [14 upvotes]: Let $k$ be a number field and $F$ a $1$-variable function field over $k$ (a finitely generated extension of $k$, of transcendence degree $1$, in which $k$ is algebraically closed).  If $F$ becomes the rational function field over every completion $k_v$ of $k$, then $F$ is the rational function field over $k$.  This is a restatement of the local-to-global principle for the existence of rational points on $k$-conics.
+Now suppose that $F$ is a $2$-variable function field over $k$, and that $F$ becomes the rational functional field over every completion $k_v$ of $k$.  Does it follow that $F$ is the rational function field over $k$ ?  (In other words, if a $k$-surface is birational of ${\mathbb P}_2$ over every $k_v$, is it $k$-birational to ${\mathbb P}_2$ ?)
+I don't think the answer is known, but perhaps someone can put together what is known about the group of $\bar k$-automorphisms of  the rational function field ${\bar k}(x,y)$ (the Cremona group) to decide one way or the other.
+
+REPLY [5 votes]: It seems to me that there are irrational surfaces over $\mathbb Q$ that are $\mathbb Q_v$-rational for all $v$. (I couldn't find them in the literature, but didn't look very hard. Almost certainly they are to be found there, in papers by either Iskovskikh or Colliot-Thelene.) 
+Take the affine surface $S$ given by $y^2+byz+cz^2=f(x)$, where $f$ is an irreducible cubic and $b^2-4c$ equals the discriminant $D(f)$ of $f$, up to a square in $\mathbb Q^*$, and $D(f)$ is not a square. According to Beauville-Colliot--Thelene--Sansuc--Swinnerton-Dyer $S$ is not $\mathbb Q$-rational, but is stably rational. (Irrationality is Iskovskikh I think, in fact.) Via projection to the $x$-line a projective model $V$ of $S$ is a conic bundle over $\mathbb P^1$ with $4$ singular fibers (one is at infinity). There is an embedding of $V$ into a weighted projective space $\mathbb P(2,2,1,1)$; the defining equation is $Y^2+bYZ+cZ^2=F(X,T)T$, where $F$ is the homogeneous version of $f$. By construction the Galois action on the $8$ lines that comprise the singular fibers is via the symmetric group $S_3$: the two lines in the fiber at infinity are conjugated, and the other six are permuted transitively. 
+Claim: Assume that $D(f)$ is square-free and prime to $6$. Then $S$ is $\mathbb Q_v$-rational for all $v$. 
+Proof: Suppose that the decomposition group $G_v$ at $v$ is cyclic. Whatever its order ($1,2$ or $3$) there are at least $2$ disjoint lines among the $8$ that are $G_v$-conjugate, so they can be blown down to give a conic bundle over $\mathbb P^1$ with at most $2$ singular fibres and a $\mathbb Q_v$-point; it is well known that such a surface is $\mathbb Q_v$-rational.
+Now suppose that $G_v= S_3$. Then $v$ is non-archimedean and $V$ has bad reduction there. In fact, exactly two of the singular fibers are equal modulo $v$; it follows that $G_v=S_3$ is impossible, and we are done.
+E.g., $f=x^3+x+1$, of discriminant $-31$, $c=8$, $b=1$.
+(This doesn't use stable rationality, but rather the fact that these surfaces, although irrational, are very close to being rational, in the sense that the action of $Gal_{\mathbb Q}$ on the lines is as small as possible subject to the surface being irrational, and the action of the decomposition groups is even smaller.)

stack_exchange/math_overflow/100316.txt

@@ -0,0 +1,17 @@
+TITLE: Raoul Bott's quote on Morse Theory cited by Bestvina and Kahle: where is it from?
+QUESTION [17 upvotes]: I wanted to properly cite the following awesome quote:
+
+Every mathematician has a secret weapon. Mine is Morse theory. - Raoul Bott
+
+Now this has been attributed to Bott in precisely two places that I can find: the older of the two is Bestvina's work on PL Morse theory here and the more recent occurrence is in the 14th slide of Kahle's talk available as a rather large pdf file here.
+Question:
+
+What is the original reference for this quote?
+
+I've combed through the likely suspect, Morse theory indomitable available here but the quote does not appear to be in that paper
+
+REPLY [7 votes]: Here is Mladen's response to my email asking this question:
+
+I heard him say it in a colloquium talk in 2001 (I think).
+
+Case closed, unless Bob MacPherson has a different answer.

stack_exchange/math_overflow/100322.txt

@@ -0,0 +1,8 @@
+TITLE: elements of the pure braid group seen as Galois automorphisms
+QUESTION [7 upvotes]: Let $Y_{n}$ be what I'm calling the ordered configuration space, the topological space of all ordered subsets of $\mathbb{C}$ of cardinality $n$. This can be viewed as $$\mathrm{Spec}(\mathbb{C}[t_{1}, ... , t_{n}, \{(t_{i} - t_{j})^{-1}\}_{1 \leq i < j \leq n}]).$$
+The (topological) fundamental group of this is the pure braid group $P_{n}$.  So by a general version of the Riemann Existence Theorem, the elements of $P_{n}$ must be Galois automorphisms of the maximal Galois extension of the function field $\mathbb{C}(t_{1}, ... , t_{n})$ which is unramified except at the primes $(t_{i} - t_{j})$.  How may I describe algebraically the Galois automorphisms corresponding to, say, the generators of $P_{n}$?
+I have a guess (something along the lines of each generator $A_{i,j}$ of $P_{n}$ sending $(t_{i} - t_{j})^{1/n} \mapsto \zeta_{n}(t_{i} - t_{j})^{1/n}$, except that seems to make the Galois group abelian) but am not sure how to prove anything systematically, and I can't find this explicitly described in any source.  Thank you for any help you can give.
+
+REPLY [3 votes]: Your confusion comes from the fact that your description of the Galois action is, I think, correct, but you are not giving the entire Galois action because you are not having it act on the correct field. The covering generated by $(t_i-t_j)^{1/n}$ is an abelian covering (and by Kummer theory is the maximal abelian covering), so of course the Galois action on it is abelian. But there are other unramified covers whose function fields are not generated by elements of the form $(t_i-t_j)^{1/n}$).
+To describe the full Galois action, you need to consider all of these covers. However, as far as I know, there is no well-behaved way to do so. For instance, each such scheme fibers over $\mathbb P^1$ minus three points: take the map $(t_1,..,t_n) \to (2t_1-t_2-t_3:-t_1+2t_2-t_3)$. It is well-defined since if both of those are $0$ then $t_1-t_2$ is zero as well, and the inequalities $t_1\neq t_2$, $t_1\neq t_3$, $t_2\neq t_3$ give three points that are not in the image. So take any unramified cover of $\mathbb P^1$ minus three points and take the fibered product with this map. You get an unramified covering of this scheme.
+Writing down equations for all the umramified covers of $\mathbb P^1-\{0,1,\infty\}$ is no easy task, and those aren't even all the covers of this scheme! Thus finding an explicit representation for the Galois action is difficult.

stack_exchange/math_overflow/100323.txt

@@ -0,0 +1,13 @@
+TITLE: What's the name for the analogue of divided power algebras for x^i/i?
+QUESTION [16 upvotes]: I recently came across divided power algebras here: http://amathew.wordpress.com/2012/05/27/lazards-theorem-ii/ It interests me because the free divided power algebra on one variable $x$, where $x^{(i)}$ models $x^i/i!$, seems like a good way of handling exponential and exponential-type series. (Divided power algebras that aren't generated by a single variable are more complicated to axiomatize, so I will stick to the one-variable case).
+Quick summary: A system of divided powers for an element $x$ in an associative unital ring $R$ associates to each nonnegative integer $n$ a ring element $x^{(n)}$ with $x^{(0)} = 1$ and $x^{(1)} = x$ and satisfying the following condition for all $i,j \ge 0$:
+$$x^{(i)}x^{(j)} = \binom{i + j}{i} x^{(i + j)}$$
+As we can see, if $R$ is an algebra over the rationals, the obvious (and unique!) choice is to set $x^{(i)} = x^i/i!$ for all $i$.
+Suppose I am interested in finding some $x^{[i]}$ that models $x^i/i$ instead of $x^i/i!$, i.e., I am interested in the kind of terms that appear in the expansions of logarithmic and inverse trig series. My best guess for the right analogue to consider is this: $x^{[i]}$ is defined for all positive integers $i$ with $x^{[1]} = x$ and satisfy the following condition. For each $i,j > 0$, write $ij/(i + j)$ as a reduced fraction $r/s$. Then, we must have:
+$$rx^{[i]}x^{[j]} = sx^{[i + j]}$$
+I'd like to know whether this structure has been studied in the past, and/or what names have been given to the structure in question. Also, I'd be interested if anybody has opinions on whether the above is a reasonable way of trying to model $x^i/i$.
+
+REPLY [12 votes]: One possible way to generalize divided powers is to model sequences $x^n/a_n$ where $a_{n+m}/(a_na_m)$ is integral (an integer, or at least integral in some sense). You can model these in the obvious way, like divided powers. Choosing $a_n = n!$ gives you divided powers, and choosing $a_n=p^n$ gives you a sequence which, from the point of view of $p$-adic analysis, is much more "regular" in its growth. You can come up with other examples.
+As you may know, one uses divided powers to construct Fontaine's ring of periods $B_{cris}$ and using $a_n=p^n$ instead of $a_n=n!$ gives a ring $B_{max}$ that is similar in nature to $B_{cris}$ but much better behaved (see III.2 of Colmez' 1998 paper "Théorie d'Iwasawa des représentations de de Rham d'un corps local").
+More generally, tweaking divided powers to get various convergence conditions is something that happens often in $p$-adic Hodge theory. See for instance 5.2.3 of Fontaine's "Le corps des périodes $p$-adiques" for one of many examples of custom made convergence conditions.
+In the $p$-adic setting, I would say that your log series belongs to the "space of holomorphic functions on the $p$-adic open unit disk, having order of growth $\leq 1$". Using precise analytic conditions rather than "divided powers" seems the right way to study the analytic series that interest you.

stack_exchange/math_overflow/10033.txt

@@ -0,0 +1,14 @@
+TITLE: Everywhere locally isomorphic abelian varieties
+QUESTION [19 upvotes]: Is there a standard example of two abelian varieties $A$, $B$ over some number field $k$ which are $k_v$-isomorphic for every place $v$ of $k$ but not $k$-isomorphic ?
+
+REPLY [26 votes]: (If you upvote this answer, please consider upvoting the answers by Felipe Voloch and David Speyer too, since this answer builds on their ideas.)
+The smallest examples are in dimension $2$.  Let $E$ be any elliptic curve over $\mathbf{Q}$ without complex multiplication, e.g., $X_0(11)$.  We will construct two twists of $E^2$ that are isomorphic over $\mathbf{Q}_p$ for all $p \le \infty$ but not isomorphic over $\mathbf{Q}$.
+Let $K:=\mathbf{Q}(\sqrt{-1},\sqrt{17})$.  Let $G:=\operatorname{Gal}(K/\mathbf{Q}) = (\mathbf{Z}/2\mathbf{Z})^2$.  Let $\alpha \colon G \to \operatorname{GL}_2(\mathbf{Z}) = \operatorname{Aut}(E^2)$ be a homomorphism sending the two generators to the reflections in the coordinate axes of $\mathbf{Z}^2$, and let $A$ be the $K/\mathbf{Q}$-twist of $E^2$ given by $\alpha$.  Define $\beta$ and $B$ similarly, but with the lines $y=x$ and $y=-x$ in place of the coordinate axes.  The representations $\alpha$ and $\beta$ of $G$ on $\mathbf{Z}^2$ are not conjugate: only the former is such that the lattice vectors fixed by nontrivial elements of $G$ generate all of $\mathbf{Z}^2$.  Thus $A$ and $B$ are not isomorphic over $\mathbf{Q}$.
+On the other hand, every decomposition group $D_p$ in $G$ is smaller than $G$ since $-1$ is a square in $\mathbf{Q}_{17}$ and $17$ is a square in $\mathbf{Q}_2$.  Also, the restrictions of $\alpha$ and $\beta$ to any proper subgroup of $G$ are conjugate: any single line spanned by a primitive vector in $\mathbf{Z}^2$ can be mapped to any other by an element of $\operatorname{GL}_2(\mathbf{Z})$.  Thus $A$ and $B$ become isomorphic after base extension to $\mathbf{Q}_p$ for any $p \le \infty$. $\square$
+Remark: The abelian surfaces $A$ and $B$ constructed above are isogenous even over $\mathbf{Q}$, because the $\mathbf{Z}^2$ with one Galois action can be embedded into the $\mathbf{Z}^2$ with the other Galois action: rotate $45^\circ$ and dilate.
+Remark: The nonexistence of examples in dimension $1$ follows from these two well-known facts:
+1) Twists of an elliptic curve over a field $k$ of characteristic $0$ 
+are classified by
+$H^1(k,\mu_n)=k^\times/k^{\times n}$ where $n$ is 2, 4, or 6.  
+2) If $n<8$, the map $k^\times/k^{\times n} \to \prod_v k_v^\times/k_v^{\times n}$ is injective.  
+[Edit: This answer was edited to simplify the construction and to add those remarks at the end.]

stack_exchange/math_overflow/100331.txt

@@ -0,0 +1,22 @@
+TITLE: On infinite-dimensional unitary representations of Kazhdan groups
+QUESTION [7 upvotes]: Let $G$ be a finitely generated discrete group that satisfies Kazhdan's property T. Rapinchuk has proved that every unitary representation $\rho$ of $G$ on some finite-dimensional Hilbert space is locally rigid: this means that if another representation is sufficiently closed to $\rho$ (on a fixed finite set of generators) then it must be conjugated to $\rho$. The proof is nice and goes as follows:
+
+Consider the adjoint action $\rho^{\rm ad}$ on the Lie algebra $gl_n$. As proved by André Weil, if $H^1(G, \rho^{\rm ad})=0$ then the representation $\rho$ is locally rigid.
+Since $G$ is Kazhdan, every action on a Hilbert space has a fixed point. Using cocycles, this is equivalent to say that $H^1(G, \eta)=0$ for any unitary representation $\eta$ on any Hilbert space.
+If $H^1(G, \rho^{\rm ad})\neq 0$ we get a contradiction, because it is easy to find an invariant positive-definite scalar product on $gl_n$ which turns the adjoint action $\rho'$ into a unitary action.
+
+Points 1 and 3 heavily rely on the fact that the representation $\rho$ is finite-dimensional, where point 2 does not. I am interested in infinite-dimensional Hilbert spaces:
+
+Can a Kazhdan group $G$ have a non locally rigid unitary representation into some infinite-dimensional Hilbert space?
+
+The question might depend on which notion of local rigidity one uses in infinite-dimension, i.e. on which topology (or distance) one puts on the set of all bounded (actually, unitary) operators. I don't know if there is a standard accepted notion. I suspect that there should be plenty of non-rigid representations but I don't know any example. 
+A related question on non-rigidity in infinite dimension is here.
+
+REPLY [7 votes]: The situation in infinite dimensions is different for Kazhdan groups.
+If $\Gamma$ contains a non-abelian free group, then the left-regular representation $\lambda \colon \Gamma \to U(\ell^2 \Gamma)$ admits a deformation $\lambda_t$ (for $t \in [0,1]$ say), such that $\lambda_t$ is a unitary representation, $$\sup_{g \in \Gamma} \|\lambda_t(g) - \lambda_s(g)\| \leq |s-t|,$$ $\lambda_0=\lambda$ and $\lambda_t$ not equivalent to $\lambda$ for $t \neq 0$. This is a very strong condition on a deformation; typically one does not require a uniform deformation and usually also using the strong operator topology.
+An explicit construction of such a deformation for the free group itself goes back to Pytlik-Swarc in
+T. Pytlik and R. Swarc, An analytic family of uniformly bounded representations of tree groups, Acta Math. 157(3-4):289-309, 1986.
+The idea is then to induce this deformation to $\Gamma$ and check that the continuity is preserved (this is easy because of the strength of the assumption) and that the resulting representations remain inequivalent.
+A similar behaviour is conjectured to hold for all non-amenable groups. These ideas have appeared in
+Marc Burger, Narutaka Ozawa, and Andreas Thom, On Ulam stability, 
+to appear in Israel J. Math., http://arxiv.org/abs/1010.0565

stack_exchange/math_overflow/100349.txt

@@ -0,0 +1,19 @@
+TITLE: Constant scalar curvature metrics in a conformal class
+QUESTION [11 upvotes]: Let $(M,g)$ be a compact Riemannian manifold, then by the resolved Yamabe-problem, there exists a metric $\tilde{g}$ of constant scalar curvature in the conformal class $[g]$ of $g$. By normalizing volume, we have $s_{\tilde{g}}=Y(g)$ where $Y(g)=Y([g])$ is the Yamabe functional.
+This might not be the only constant scalar curvature metric in $[g]$. So my question is: is there anything known about the set of constants $c$ where $c\equiv scal_{\tilde{g}}$ for a metric $\tilde{g}\in [g]$ of unit volume?
+
+REPLY [12 votes]: Yes, as the OP mentions, in many cases the solution to the Yamabe problem may not be unique (we obviously ignore the effect of rescaling a metric), in the sense that there may be constant scalar metrics different from the minimizer of the total scalar curvature. Recall that the solution to the Yamabe problem was proved by finding a minimizer of the Hilbert-Einstein functional (total scalar curvature) on unit volume metrics in the conformal class $[g]$, however any critical point of this functional is a constant scalar curvature metric.
+Perhaps a quick review of some uniqueness/non-uniqueness results could be of interest.
+For example, in the conformal class of an Einstein metric (except the round sphere), the solution is unique. Moreover,
+
+Anderson proved that on generic conformal classes, the solution is unique.
+
+On the other hand, there are many non-uniqueness results:
+
+Ambrosetti and Malchiodi (JFA, 1999) and Berti and Malchiodi (JFA, 2001) proved non-uniqueness results for conformal classes of deformations of the round metric on spheres;
+Pollack (CAG, 1993) proved existence of arbitrarily small $C^0$ perturbations of any given metric (on any closed manifold!), with arbitrarily large number of solutions in its conformal class;
+Schoen (1991) proved that taking the product $S^1\times S^n$ with the product of round metrics, and making the radius $r$ of $S^1$ go to $+\infty$, the number of solutions also goes to $+\infty$ as $r\to+\infty$;
+Recently, de Lima, Piccione and Zedda obtained a sort of generalization, proving existence of infinitely many bifurcation points for many $1$-parameter family of product metrics, obtained by scaling one of the factors. In particular, this yields existence of a countable set of metrics in this family in which conformal class there exist at least three distinct solutions to the Yamabe problem;
+More recently, in a joint paper with Piccione, we obtained a similar result (infinitely many bifurcations) for families of homogeneous metrics on spheres. Recall that the homogeneous metrics on spheres are given by scaling the round metric in the direction of the Hopf fibration, e.g., take $S^3\to S^{4n+3}\to HP^n$ and shrink the round metric on $S^{4n+3}$ in the vertical directions by a factor of $t^2$, obtaining a metric $g_t$. Then, as $t\to 0$ (i.e., as the fibers collapse), our result implies that there are infinitely many values of $t$ for which $[g_t]$ has at least 3 constant scalar curvature metrics. In a subsequent paper, we extend this result to any homogeneous fibration $K/H\to G/H\to G/K$ where $K/H$ has positive scalar curvature and either $H$ is normal in $K$ or $K$ is normal in $G$.
+
+In most cases above, where explicit non-uniqueness of solutions to the Yamabe problem is proven, it is easy to give a qualitative description of the values of the (constant) scalar curvatures of the various solutions. For general results, however, the best available are the type of compactness results of Brendle, Khuri, Marques, Schoen etc mentioned in macbeth's answer.

stack_exchange/math_overflow/10036.txt

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+TITLE: Which is the correct version of a quantum group at a root of unity?
+QUESTION [29 upvotes]: By this I mean the specialisation of the quantum group Uq(g) with q a root of unity, and the 'correct' meaning of 'correct' (enclosed in quotations since there isn't necessarily a correct answer) is likely to mean that its category of representations is the 'correct' one.
+My suspicion is that I want to take an integral form of Uq(g) defined over ℤ[q±1/2] and base change to the appropriate ring of integers in a cyclotomic field. Having heard of the 'small quantum group' and Lusztigs algebra U dot (notation in his quantum groups book), I suspect the existence of multiple approaches, which diverge at least when an integral form is desired, and hence turn to mathoverflow for clarification and enlightenment.
+Afficionados of this type of question can consider it as a continuation in a sequence initiated by this question on the universal enveloping algebra in positive characteristic.
+
+REPLY [57 votes]: There are (at least) five interesting versions of the quantum group at a root of unity.
+The Kac-De Concini form:
+This is what you get if you just take the obvious integral form and specialize q to a root of unity (you may want to clear the denominators first, but that only affects a few small roots of unity).  This is best thought of as a quantized version of jets of functions on the Poisson dual group.  It's most important characteristic is that it has a very large central Hopf subalgebre (generated by the lth powers of the standard generators).  In particular, its representation theory is sits over Spec of the large center, which is necessarily a group and turns out to be the Poisson dual group.  It also has a small quotient Hopf algebra when you kill the large center.
+The main sources for the structure of the finite dimensional representations are papers by subsets of Kac-DeConcini-Procesi (the structure of the representations depends on the symplectic leaf in G*, in particular there are "generic" ones coming from the big cell) as well as some more recent work by Kremnitzer (proving some stronger results about the dimensions of the non-generic representations) and by DeConcini-Procesi-Reshetikhin-Rosso (giving the tensor product rules for generic reps).  The main application that I know of this integral form is to invariants of knots together with a hyperbolic structure on the compliment and to invariants of hyperbolic 3-manifolds due to Kashaev, Baseilhac-Bennedetti, and Kashaev-Reshetikhin.  The hope is that these invariants will shed some light on the volume conjecture.
+The Lusztig form:
+Here you start with the integral form that has divided powers.  Structurally this has a small subalgebra generated by the usual generators (E_i, F_i, K_i)  since E^l = 0.  The quotient by this subalgebra gives the usual universal enveloping algebra via something called the quantum Frobenius map.  The main representation that people look at are the "tilting modules."  Tilting modules have a technical description, but the important point is that the indecomposable tilting modules are exactly the summands of the tensor products of the fundamental representations.  Indecomposable tilting modules are indexed by weights in the Weyl chamber.  The "linkage principle" tells you that inside the decomposition series of a given indecomposable tilting module you only need to look at the Weyl modules with highest weights given by smaller elements in a certain affine Weyl group orbit.
+It is the Lusztig integral form (not specialized) that is important for categorification.  The Lusztig form at a root of unity is important for relationships between quantum groups and representations of algebraic groups and for relationships to affine lie algebras.  The main sources are Lusztig and HH Andersen (and his colaborators).  I'm also fond of a paper of Sawin's that does a very nice job cleaning up the literature.
+The Lusztig integral form is also the natural one from a quantum topology point of view.  For example, if you start with the Temperley-Lieb algebra (or equivalently, tangles modulo the Kauffman bracket relations) and specialize q to a root of unity what you end up with is the planar algebra for the tilting modules for the Lusztig form at that root of unity.
+The small quantum group:
+This is a finite dimensional Hopf algebra, it appears as a quotient of the K-DC form (quotienting by the large central subalgebra) and as a subalgebra of the Lusztig form (generated by the standard generators).  I gather that the representation theory is not very well understood.  But there has been some work recently by Roman Bezrukavnikov and others.  I also wrote a blog post on what the representation theory looks like here for one of the smallest examples.
+The semisimplified category:
+Unlike the other examples, this is not the category of representations of a Hopf algebra!  (Although like all fusion categories it is the category of representations of a weak Hopf algebra.)  You start with either the category of tilting modules for the Lusztig form or the category of finite dimensional representations of the small quantum group and then you "semisimplify" by killing all "negligible morphisms."  A morphism is negligible if it gives you 0 no matter how you "close it off." Alternately the negligible morphisms are the kernel of a certain inner product on the Hom spaces.  The resulting category is semisimple, its representation theory is a "truncated" version of the usual representation theory.  In particular the only surviving representations are those in the "Weyl alcove" which is like the Weyl chamber except its been cut off by a line perpendicular to l times a certain fundamental weight (see Sawin's paper for the correct line which depends subtly on the kind of root of unity).
+This example is the main source of modular categories and of interesting fusion categories.  Its main application is the 3-manifold invariants of Reshetikhin-Turaev (where this quotient first appears, I think) and Turaev-Viro.  For those invariants its very important that your braided tensor category only have finitely many different simple objects.
+The half-divided powers integral form:
+This appears in the work of Habiro on universal versions of the Reshetikhin-Turaev invariants and on integrality results concerning these invariants.  This integral form looks like the Lusztig form on the upper Borel and like the K-DC form on the lower Borel.  The key advantage is that in the construction of the R-matrix via the Drinfeld double you should be looking at something like U_q(B+) \otimes U_q(B+)* and it turns out that the dual of the Borel without divided powers is the Borel with divided powers and vice-versa.  There's been very little work done on this case beyond the work of Habiro.

stack_exchange/math_overflow/100364.txt

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+TITLE: A good reference to grok hypergeometric functions?
+QUESTION [15 upvotes]: When I was introduced during my degree to special functions, I made friends with a number of nice functions - Laguerre, Legendre, Hermite, Bessel, and whatnot - but I made only a passing acquaintance with the hypergeometric and confluent hypergeometric functions, mostly limited to looking them up in Arfken and recoiling in horror at the dryness of the material and the lack of physical content in the calculations.
+I know, of course, that this lack of physical content is also accompanied by an astounding generality. After a while I did get the core of the idea, which I believe is "explore all special functions whose series coefficients are rational functions of $n$", and I do find it appealing, but I've not had the energy nor the motivation to follow that up and see what's interesting about the thing.
+However, it appears that the long-delayed moment is here and some pretty hairy integrals (think $\int_0^\infty x^k e^{-\alpha x^2}J_m(\beta x)dx$) have pushed some ugly "${}_1 F_1$" symbols onto my page. So my question is, then: what's a good introduction to hypergeometric and confluent hypergeometric functions? I'd like one where I can get an intuitive understanding of what to expect from them in different circumstances, what nice properties they have, and generally why it really is worth it to deal with them instead of their more specific cases like Laguerre, Legendre, Hermite, Bessel, etc.
+
+REPLY [2 votes]: I would also recommend "Special Functions and their Application" by Lebedev and Silverman (http://www.amazon.com/Special-Functions-Their-Applications-Mathematics/dp/0486606244/ref=sr_1_1/179-2267788-6339652?s=books&ie=UTF8&qid=1422720681&sr=1-1) and the series "Higher Transcendental Functions" (part of the Bateman Manuscript Project, edited by Erdelyi) which has the benefit of being free online via Caltech:
+http://authors.library.caltech.edu/43491/

stack_exchange/math_overflow/10039.txt

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+TITLE: Characterizing faces of 3-dimensional polyhedra. (Related to Victor Eberhard's Theorem [1890]:)
+QUESTION [11 upvotes]: Originally asked by Ali Dino Jumani;  EXTENSIVELY EDITED by David Speyer. The previous version was a bit confused, but Steven Sivek and Graham, in the comments, figured out what was going on.
+
+G. C. Shephard, in his paper "Twenty Problems on Convex Polyhedra: Part I", associates to a three dimensional polyhedron the sequence $(p_3, p_4, p_5, p_6, \ldots)$, with $p_k$ being the number of facets that are $k$-gons. He poses the problem of characterizing all sequences of integers which arise in this way.
+Are there any developments and references on this problem?
+the original question as appeared in Shephard's paper "Given any finite sequence (f3, f4, ...., fm) of non-negative integers, find a necessary and sufficient condition for it to be assocoated with some convex polyhedron". Following Grunbaum's Convex Polytope 2e notation, section 13.3 under Eberhard's Therorem heading, f3 is triangle renamed as p3 and f4 as p4 is square and fm as pk is n-gon. A convex polyhedron containing these faces satisfies the Eberhard's criterion.
+Revisions and corrections are highly appreciated, Thanks.
+
+REPLY [20 votes]: Eberhard Theorem
+Consider a simple 3-polytope P, (so every vertex has 3 neighbors). If $p_k$ is the number of faces of P which are k-gonal, Euler's theorem implies that
+$$(*) ~~\sum_{k \ge 3} (6-k)p_k=12.$$
+Note that 6-gonal faces do not contribute to the LHS.
+One way to think about it is that polygonal faces with 7 and more sides contribute "negative curvature", small faces  namely triangles quadrangles and pentagons, contribute positive "curvature" and hexagons are "flat".
+Eberhard's theorem asserts that if you have a sequence of numbers $p_k, k \ne 6$ such that $\sum_{k \ge 3} (6-k)p_k=12$ then you can find a simple 3-polytope with $p_k$ k-gonal faces. (But you have no control on $p_6$).
+Extensions of Eberhard theorem
+There are various results extending Eberhard's theorem in various directions. Chapter 13 in Grunbaum's book "Convex Polytopes" and especially the supplemantary material at the end of the chapter in the new 2nd edition is a good source. Another general source is my chapter from the "Handbook of Discrete and Computational geometry" on garphs and skeleta of polytops.
+A relatively recent paper on the subject is by Stanislav Jendrol "On the face vectors of trivalent convex polyhedra". Another paper by Jendrol which deals with general 3-polytopes from the same year is "On face vectors and vertex vectors of convex polyhedra" Discrete Math 118 (1993)119-144. There are analogs of Eberhard theorem for 4-regular planar graphs, for toroidal graphs and in other directions.
+A far as I know, there is no good answer known for the question posed by Shephard of characterizing all sequences $(p_3,p_4,\dots)$, and no such characterization is known even for the simple case.
+High dimensions
+In higher dimensions and even in four dimensions there are various different ways to extend these problems, these problems become very difficult and very little is known.
+2-dimensional faces
+You can ask again about the numbers of k-gonal 2-dimensional faces. While the formula above implies that in dimension 3 and more $p_3+p_4+p_5>0$ it is known that in dimensions 5 and more $p_3+p_4>0$.
+Types of facets
+Perhaps an even more natural extension is to consider the type of facets a given d-simensional polytope have. You can ask for a simple 4 polytope (a 4-polytope whose graph is 4-regular) what are the number of facets $p_Q$ isomorphic to a given 3-polytope Q. This gives you a vector indexed by combinatorial types of simple 3-polytopes, but I am not aware of any Eberhard type theorem and I do not know even which 3-polytopes should be considered as the analogs of the hexagons in the above formula. Dually stated and extented to triangulations of 3-spheres the question is to associate to a triangulated 3-simensional sphere (or just simplicial 3-polytope) the list of links of vertices (with multiplicities) it has. A related MO question is this one.
+Type of facets according to their number of their facets
+Rather than classifying the facets according to their full combinatorial type we can classify them according to their own number of facets. Here, for dimension greater than 4 there is no analog for (*). It is possible that under wide circumstences the numbers of facets with k facets can be prescribed, but I am not aware of results in this direction.
+Type of facets according to their f-vectors
+Precscribing the entire vector of face numbers of the individual facets, may well be the most interesting extension from 3 to higher dimensions.
+There are some reasons to jump from dimension 3 directly to dimension 5. The nature of the problem is different in even and odd dimensions. Let me try to elaborate on that.
+Suppose you have a simple 5-polytope P and denote by $p_{a,b}$ the number of facets F so that $f_3(F)=a$ and $f_2(F)=b$. (For a polytope or some other cell complex X, $f_i(X)$ denotes the number of $i$-dimensional faces of $X$.) Dually, we can consider a triangulation $K$ of the 4-simensional sphere $S^4$ and let $p_{a,b}$ be the number of vertices whose link has $a$ vertices and $b$ edges.
+Now the Dehn-Sommervill relations imply that
+$$(**) \sum p_{a,b} (b-6a+30) = 60 .$$
+So now we can consider 4 dimensional facets as "small" if b-6a+30 is negative, as "large" if $b-6a+30$ is positive, and as neutral if $b=6a-30$. An analog of Eberhard theorem will say that you can prescribe the types of small and large facets and realize the polytope (or the triangulated sphere) by allowing to add many "neutral" facets.
+Facets and non facets
+A related question that was studied in several papers is: for which d-polytopes P there is a (d+1) polytopes Q such that all faces of Q are isomorphic to P. Such polytopes P are called "facets". An intriguing open problem is if the icosahedron is a facet.
+Cubical complexes
+Simple polytopes are dual to simplicial polytopes, and there is some special interest in these problems for cubical complexes (or their duals). Indeed there is an analog of Eberhard's theorem for 3-polytopes with all vertex degrees 4. There should be an analog for (**) and this may be related to interesting questions about curvature of cubical complexes.
+Stacked polytopes
+A stack polytope is a polytope obtained by gluing simplices along facets. (They are related to Appolonian circle packing.) Stacked polytopes are simplicial. I do not know if there is an analog of Eberhard's  theorem when we consider only simple 3-polytope which are dual to stacked 3-polytopes. This is also interesting for other dimensions. All facets of stacked 5 polytopes are stacked and their number of edges is four times the number of vertices minus 10. Therefore, for a dual-to-stacked 5 polytope formula (**) reduces to $$\sum p_{a,b}(10-a)=30.$$ (Here, $b=4a-10$ so we can omit the subscript $b$.) So in this case, the distinction is between facets with $a<10$, those with $a>10$ and those with $a=10$. Again we can look for some analogs for Eberhard's theorem, which in this case, might be easier.

stack_exchange/math_overflow/100392.txt

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+TITLE: Simply connected simplicial complexes
+QUESTION [13 upvotes]: Let $Z$ be a simply connected, two dimensional simplicial complex.
+Let $X\subset Z$ be a finite subcomplex with nontrivial $\pi_{1}$.
+Must there exist a finite, simply connected subcomplex $Y\subset Z$ such that $Y\supset X$?
+(Motivation: the fact that Whitehead conjecture remains unproven indicates that there are probably some very weird things that can happen in two dimensional complexes.  This question attempts to locate some of these pathologies.)
+
+REPLY [11 votes]: The answer is negative unless you assume that $Z$ is aspherical (and even in this case I am not sure). Take Whitehead manifold $W$, which is a contractible open 3-manifold, not homeomorphic to ${\mathbb R}^3$ and let $Z$ be the 2-dimensional skeleton of a triangulation of $W$. Then $Z$ is a counter-example.
+Proof.  
+
+I claim that $W$ does not admit an exhaustion by simply-connected compact submanifolds with boundary. Indeed, if such an exhaustion $W_i$ exists, then each $W_i$ is bounded by some 2-spheres, all but one of which bounds a 3-ball in $W$. Filling in these spheres by 3-balls in $W$, we obtain an exhaustion $W_i'$ of $W$ by closed 3-balls. This would imply that $W$ is homeomorphic to the 3-space. Contradiction. 
+Your question is equivalent to asking if $Z$ (provided that it is locally finite) is exhaustable by finite simply-connected subcomplexes. Note that my $Z$ is locally finite. Suppose that such an exhaustion $Z_i$ does exist, we can then enlarge each $Z_i$ by adding to it 3-simplices in $W$ whose boundary is contained in $Z_i$. This does not change $\pi_1$, of course. The resulting subcomplexes $V_i$ will exhaust $W$ and will be simply-connected. Taking $W_i$ to be a regular neighborhood of $V_i$ we obtain an exhaustion of $W$ by simply-connected compact submanifolds with boundary. This contradicts Part 1.

stack_exchange/math_overflow/10041.txt

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+TITLE: A parametrization of Heronian triangles
+QUESTION [8 upvotes]: Let $a,b,c$ be integers which are the sides of a triangle with integral area, a so called Heronian triangle. This website attributes to Gauss the result that there must then exist integers $m,n,p,q$ such that
+$a = mn(p^2+q^2)$
+$b = (mp)^2+(nq)^2$
+$c = (m+n)(mp^2-nq^2)$
+(where I left out a $4pq$ factor designed to make the radius of the circumscribed circle integral as well). It's not hard to see that the triangle defined by these formulas is indeed Heronian, however I could neither prove nor find a reference for the fact that this parametrization is exhaustive. 
+Can someone do one of these two things?
+Thanks!
+(Note: I'm communicating this question on behalf of my dad, who is really the person who looked into that but is not easily capable of asking it himself over here. I may be slow to respond on his behalf if questions come up).
+
+REPLY [5 votes]: Let your triangle $\triangle{ABC}$ have side lengths $a,b,c \in \mathbb{Q}$ and rational area. Assume WLOG that $c$ is the longest side and drop the altitude from $C$ with length $h\in Q$. The triangle is divided into two right triangles one with hypotenuse $a$ and legs $d,h$, and one with hypotenuse $b$ and legs $e,h$. We have $d+e=c\in \mathbb{Q}$. Also notice that $$d-e=\frac{d^2-e^2}{d+e}=\frac{a^2-b^2}{c}\in \mathbb{Q}$$ so we conclude that $d,e$ are rational. From the pythagorean triples we have relations $$a-d=r,\quad a+d=\frac{h^2}{r},\quad b-e=s,\quad b+e=\frac{h^2}{s}$$ and therefore $$a=\frac{1}{2r}(h^2+r^2),\quad b=\frac{1}{2s}(h^2+s^2),\quad c=\frac{r+s}{2rs}(rs-h^2)$$
+This is exactly your parametrization up to scaling.

stack_exchange/math_overflow/100451.txt

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+TITLE: Lagrangian Kleinian bottles
+QUESTION [7 upvotes]: I remember some talks some time ago about proofs of nonexistence of Lagrangian Kleinian bottles in ${\mathbb C}^2$ for the standard symplectic structure, mentioning that this were the only compact surface for which the problem were open.
+Question 1: Where to find references for the other surfaces?
+Question 2: What is the current status for Kleinian bottles? Do there exist written proofs now?
+
+REPLY [8 votes]: While working on my thesis, I (by accident) constructed a Lagrangian Klein bottle in $(S^2\times S^2, \omega\oplus\omega)$.  The construction works in $S^2\times D^2$ as long as the area of $D^2$ is strictly greater than half the area of $S^2$.  The method is to pick a Hamiltonian on $S^2$ that rotates $S^2$ halfway around, and realize it as the monodromy around $\partial D^2$, of a symplectic fibration over $D^2$ with fiber $S^2$. There is a circle that goes through the two fixed points of the rotation, which then traces out the Klein bottle.  See McDuff's ``Geometric Variants of the Hofer Norm'' for the precise version of how to realize such a monodromy and the constraints: the area of the base is an easy way to think of it, but it's really a ratio of volume to fiber area.
+As I understand it the paper by Nemirovskii called ``homology class of a Lagrangian Klein bottle'', is flawed, because it predates the accepted proofs of non-existence as described by others.  I would love to know if its main claim is true, that a Lagrangian Klein bottle in a symplectic 4-manifold must realize a nontrivial second homology class with mod 2 coefficients. This is true for the example above. (Might need to ask this as a question.)

stack_exchange/math_overflow/100461.txt

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+TITLE: The ring of continuous real-valued functions on a Stone space
+QUESTION [7 upvotes]: Let $X$ be a topological space and consider the ring $C(X,\mathbb{R})$ of continuous real-valued functions on $X$ (where ring-addition and multiplication are defined in the obvious point-wise way).
+It is well-known that this ring uniquely determines the topological space, if the latter is compact and Hausdorff (Gelfand-Duality).
+My question is the following: What properties does the ring $C(X,\mathbb{R})$ have, if $X$ is a Stone space (that is, if $X \in Comp_2$ is also totally disconnected)? Ideally, I would like to have a statement along the following line:
+$X$ is a Stone space (if and) only if the ring $C(X,\mathbb{R})$ has the property [nice property given to me by the lovely folks from MO].
+
+REPLY [3 votes]: The answer is given by the following theorem (Davidson's $C^\*$-Algebras by Example, III, 2.5).
+
+Let $A$ be a commutative C*-Algebra. Then the following statements are equivalent:
+
+$A$ is an AF-algebra, i.a. a colimit of a sequence of finite-dimensional C*-algebras
+$A$ is separable and the projections in $A$ generate a dense subspace.
+The spectrum of $A$ is totally disconnected.
+
+
+Thus, the category of locally compact totally disconnected Hausdorff spaces is equivalent to the category of commutative AF-algebras; the inverse functors are $C_0(-,\mathbb{C})$ and $\mathrm{Spm}$. Under this equivalence, compact corresponds to unital. The finite-dimensional commutative algebras are just powers of $\mathbb{C}$, corresponding to finite sets, but colimits produce interesting examples:
+For example, if $X$ is the Cantor set, then it is easy to see that $C_0(X,\mathbb{C})$ is the colimit of the sequence $\mathbb{C} \to \mathbb{C}^2 \to \mathbb{C}^{2^2} \to \dotsc$, where $\mathbb{C}^{2^n} \to \mathbb{C}^{2^{n+1}}, a \mapsto (a,a)$. In general, AF-algebras are classified via so-called Bratteli-diagrams (see loc. cit).
+By the way, there is a nice connection to Stone duality, which says that $C_0(-,\mathbb{F}_2)$ exhibits an equivalence of categories between the category of locally compact totally disconnected Hausdorff spaces and the category of boolean rings. In the diagram you just exchange $\mathbb{C}$ with $\mathbb{F}_2$. I would love to see a purely algebraic functor $\mathbb{F}_2 \otimes_{\mathbb{C}} (-)$, which doesn't use the spectrum as an intermediate step ...