User andy putman - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T14:27:47Z http://mathoverflow.net/feeds/user/317 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/133146/homeomorphisms-and-disjoint-unions/133596#133596 Answer by Andy Putman for Homeomorphisms and disjoint unions Andy Putman 2013-06-13T06:09:06Z 2013-06-13T06:09:06Z <p>The result you want is <em>false</em>. Counterexamples are given in</p> <p>Yamamoto, Shuji and Yamashita, Atsushi, A counterexample related to topological sums. Proc. Amer. Math. Soc. 134 (2006), no. 12, 3715–3719.</p> <p>These counterexamples are compact subsets of $\mathbb{R}^4$.</p> http://mathoverflow.net/questions/130431/mapping-class-group-of-once-punctured-torus/130432#130432 Answer by Andy Putman for Mapping class group of once-punctured torus Andy Putman 2013-05-12T19:21:19Z 2013-05-12T19:21:19Z <p>This is (essentially) contained in Corollary 1.3 of</p> <p>Birman, Joan S. Mapping class groups and their relationship to braid groups, Comm. Pure Appl. Math. 22 1969 213–238.</p> <p>A special case of this corollary (the case $m=0$ and $g=1$ and $n=1$) is that the Birman kernel map $\pi_1(T) \rightarrow \text{Mod}(S)$ has kernel the entire group $\pi_1(T)$.</p> <p>It would not surprise me if the result were also contained in</p> <p>Magnus, W., Uber Automorphismen uon Fundamentalgruppen berandeter Flachen, Math. Ann. 109, 1934, pp. 617-646.</p> <p>But I haven't had a chance to look (the kids are screaming and I have to go supervise them).</p> http://mathoverflow.net/questions/130157/generators-of-sections-of-free-groups/130188#130188 Answer by Andy Putman for Generators of sections of free groups Andy Putman 2013-05-09T16:54:46Z 2013-05-09T16:54:46Z <p>This is really a cohomological question and has a simple cohomological answer. Recall that if a group $G$ acts on an abelian group $M$, then $M_G$ denotes the <em>coinvariants</em> of the action, that is, the quotient of $M$ by the subgroup generated by <code>$\{\text{$m-g(m)|m \in $M,$g \in G$}\}$</code>. The group $F$ acts on $H$ by conjugation, and thus there is an induced action of $F$ on $H_1(H)$.</p> <p><strong>Key Observation</strong> : $H/[H,F] \cong (H_1(H;\mathbb{Z}))_F$ and $H/[H,F]H^p \cong (H_1(H;\mathbb{Z}/p))_F$.</p> <p>Indeed, we have $H/[H,H] \cong H_1(H;\mathbb{Z})$ and $H/[H,H]H^p \cong H_1(H;\mathbb{Z}/p)$ by definition, and quotienting by $[H,F]$ just kills off the $F$-action.</p> <p>The other needed ingredient is the 5-term exact sequence in group homology. Given a short exact sequence</p> <p>$$1 \longrightarrow A \longrightarrow B \longrightarrow C \longrightarrow 1$$</p> <p>of groups and a ring of coefficients $R$, this 5-term exact sequence takes the form</p> <p>$$H_2(B;R) \longrightarrow H_2(C;R) \longrightarrow (H_1(A;R))_B \longrightarrow H_1(B;R) \longrightarrow H_1(C;R) \longrightarrow 0.$$</p> <p>Letting $Q = F/H$, we will apply this to the short exact sequence</p> <p>$$1 \longrightarrow H \longrightarrow F \longrightarrow Q \longrightarrow 1.$$</p> <p>The key simplification that occurs is that $H_2(F;R) = 0$ since $F$ is free. We thus get exact sequences</p> <p>$$0 \longrightarrow H_2(Q;\mathbb{Z}) \longrightarrow H/[F,H] \longrightarrow H_1(F;\mathbb{Z}) \longrightarrow H_1(Q;\mathbb{Z}) \longrightarrow 0$$</p> <p>and</p> <p>$$0 \longrightarrow H_2(Q;\mathbb{Z}/p) \longrightarrow H/[F,H]H^p \longrightarrow H_1(F;\mathbb{Z}/p) \longrightarrow H_1(Q;\mathbb{Z}/p) \longrightarrow 0.$$</p> <p>If you understand $Q$ enough to calculate its first and second homologies, these short exact sequences allow you to determine $H/[F,H]$ and $H/[F,H]H^p$.</p> http://mathoverflow.net/questions/129758/when-does-an-even-dimensional-manifold-fiber-over-an-odd-dimensional-manifold/129774#129774 Answer by Andy Putman for When does an even-dimensional manifold fiber over an odd-dimensional manifold? Andy Putman 2013-05-05T20:36:57Z 2013-05-05T20:36:57Z <p>Being the total space of a fiber bundle is not invariant under homotopy equivalences, so I doubt there is a criterion of the type you state (eg in terms of homological invariants).</p> <p>However, one can say something a little weaker. If a manifold $M^k$ fibers over a manifold $N^{\ell}$, then the fibers form a codimension $\ell$ foliation of $M^k$. A deep theorem of Thurston says that if $\chi(M^k)=0$, then $M^k$ supports a codimension $1$ foliation (the converse is an easy exercise). By the way, to appreciate how deep this is, it implies in particular that an odd-dimensional closed oriented manifold always supports a codimension $1$ foliation. This is nontrivial to prove even in dimension $3$ (where it was first proved by Lickorish).</p> http://mathoverflow.net/questions/126425/homotopy-equvalence-from-contractibility-of-fiber/126449#126449 Answer by Andy Putman for Homotopy equvalence from contractibility of fiber. Andy Putman 2013-04-03T21:10:04Z 2013-04-03T21:10:04Z <p>In his paper</p> <p>MR0087106 (19,302f) Smale, Stephen A Vietoris mapping theorem for homotopy. Proc. Amer. Math. Soc. 8 (1957), 604–610. </p> <p>Smale proved the following theorem:</p> <p><strong>Theorem</strong> : Let $X$ and $Y$ be connected, locally compact separable metric spaces. Assume also that $X$ is locally contractible. Consider a proper surjective continuous map $f : X \rightarrow Y$. Assume that for all $y \in Y$, the space $f^{-1}(y)$ is contractible and locally contractible. Then $f$ is a weak homotopy equivalence.</p> <p>To see how this fits into your situation, remember that (for instance) finite CW complexes are locally compact and locally contractible. So you need to impose conditions on the fibers to ensure that they are also locally contractible. </p> http://mathoverflow.net/questions/111097/textbook-for-undergraduate-course-in-geometry Textbook for undergraduate course in geometry Andy Putman 2012-11-01T03:07:28Z 2013-03-24T02:58:39Z <p>I've been assigned to teach our undergraduate course in geometry next semester. This course originally was intended for future high-school teachers and focused on axiomatic, Euclid-style geometry (planar, spherical, and hyperbolic). Rice University has changed a lot since this course began being taught (many, many years ago); we now have very few students who want to be high school teachers, and in general the level of our students is such that most of our math majors perceive the course to be beneath them.</p> <p>My assignment is to redesign the course. I have almost complete freedom except that I cannot require any prerequisites beyond multivariable calculus and ODE's.</p> <p><strong>Question</strong> : What textbook should I use?</p> <p>Here are my thoughts about what I am looking for.</p> <ol> <li><p>As I said, I cannot require any prerequisites beyond multivariable calculus and ODE's. However, our undergraduate students are very strong (based on test scores and high school grades, they are pretty similar to the students at eg Cornell or Northwestern). So I want a book that has plenty of meat in it.</p></li> <li><p>It should contain a mixture of proofs and computation, but plenty of proofs.</p></li> <li><p>There are no topics that I am required to cover, though of course it has to be geometric (in particular, this course is not a prerequisite for anything else). </p></li> <li><p>I find axiomatic treatments of geometry boring.</p></li> <li><p>I don't want to develop any machinery unless it has an immediate payoff. However, I am not at all adverse to developing some tools from scratch as long as they lead to something cool.</p></li> <li><p>I want there to be lots of good problems.</p></li> </ol> <p>Does anyone have any suggestions?</p> http://mathoverflow.net/questions/44021/which-manifolds-are-homeomorphic-to-simplicial-complexes/44025#44025 Answer by Andy Putman for Which manifolds are homeomorphic to simplicial complexes? Andy Putman 2010-10-28T22:03:55Z 2013-03-12T01:43:27Z <p>I don't know about dimension 4, but for high dimensions this is a well-known open problem. I don't think much progress has been made on it for a while. I recommend Ranicki's lecture notes from Siebenmann's retirement conference for a good summary about what is known about this and related problems: <a href="http://www.maths.ed.ac.uk/~aar/slides/orsay.pdf" rel="nofollow">http://www.maths.ed.ac.uk/~aar/slides/orsay.pdf</a></p> <p>EDIT : Hot off the press is a <a href="http://arxiv.org/abs/1303.2354" rel="nofollow">paper</a> of Manolescu claiming to disprove the conjecture of Galewski-Stern and construct manifolds in all dimensions $\geq 5$ which are not homeomorphic to simplicial complexes.</p> http://mathoverflow.net/questions/17578/triangulating-surfaces Triangulating surfaces Andy Putman 2010-03-09T05:36:15Z 2013-03-01T00:48:52Z <p>I've had a few undergraduate students ask me for references for the classical fact (due to Rado) that closed topological surfaces can be triangulated. I know two sources for this, namely Ahlfors's book on Riemann surfaces and Moise's book "Geometric topology in dimensions 2 and 3". Both of these strike me as being a bit much for a bright undergraduate. Question : in the 30+ years since Moise's book, has anyone written a more accessible account?</p> http://mathoverflow.net/questions/122499/topologically-embedding-curves-in-jacobian/122502#122502 Answer by Andy Putman for Topologically embedding curves in Jacobian Andy Putman 2013-02-21T05:03:40Z 2013-02-21T05:10:15Z <p>I'll do things a little more abstractly than Will.</p> <p>Of course, things are trivial in genus $1$. To simplify things a bit, I'll restrict myself to a surface $\Sigma_g$ of genus $g \geq 3$ (the case $g=2$ can be handled with a bit more care). Both the surface $\Sigma_g$ and the torus $(S^1)^{2g}$ are Eilenberg-MacLane spaces. Recall that (modulo basepoint issues) homotopy classes of maps between Eilenberg-MacLane spaces are in bijection with homomorphisms between their fundamental groups. Fixing an identification of $\pi_1((S^1)^{2g})$ with $H_1(\Sigma_g;\mathbb{Z})$, there thus exists a unique homotopy class of continuous maps $\phi : \Sigma_g \rightarrow (S^1)^{2g}$ inducing the surjection $\pi_1(\Sigma_g) \rightarrow H_1(\Sigma_g;\mathbb{Z})$; since the target is abelian, there is no need to worry about basepoints. Since the dimension of $(S^1)^{2g}$ is greater than $5=2 \cdot 2 + 1$, the usual Whitney argument shows that we can homotope $\phi$ a little bit to assure that $\phi$ is an embedding.</p> http://mathoverflow.net/questions/122175/diffm-and-connectedness/122203#122203 Answer by Andy Putman for Diff(M) and connectedness Andy Putman 2013-02-18T17:52:11Z 2013-02-18T20:16:01Z <p>Let $M$ be a compact oriented manifold. The following hold if and only if $M$ is connected.</p> <p>1) $\text{Diff}_0(M)$ is simple. </p> <p>This was proven by Thurston if $M$ is connected; see</p> <p>MR1445290 (98h:22024) Banyaga, Augustin(1-PAS) The structure of classical diffeomorphism groups. (English summary) Mathematics and its Applications, 400. Kluwer Academic Publishers Group, Dordrecht, 1997. xii+197 pp. ISBN: 0-7923-4475-8 </p> <p>If $M$ is not connected, then $\text{Diff}_0(M)$ contains normal subgroups consisting of elements that fix some connected components and don't fix others.</p> <p>2) $\text{Diff}_0(M)$ does not decompose as a direct product.</p> <p>If $M$ is the disjoint union of submanifolds $M_1$ and $M_2$, then it is clear that $\text{Diff}_0(M) = \text{Diff}_0(M_1) \times \text{Diff}_0(M_2)$.</p> <p>If $M$ is connected, then one can show that $\text{Diff}_0(M)$ does not decompose as a direct product by exhibiting elements $f \in \text{Diff}_0(M)$ whose centralizers consist only of $\langle 1, f, f^2, \ldots \rangle$. There are many such constructions; for instance, see</p> <p>MR0985855 (90i:58151a) Palis, J.(BR-IMPA); Yoccoz, J.-C.(F-POLY) Rigidity of centralizers of diffeomorphisms. Ann. Sci. École Norm. Sup. (4) 22 (1989), no. 1, 81–98. </p> <p>A famous conjecture of Smale says that such elements should in fact be generic. This was recently proven by Bonatti-Crovisier-Wilkinson for $C^1$ diffeomorphisms; see</p> <p>MR2511588 (2010g:37035) Bonatti, Christian(F-DJON-IM); Crovisier, Sylvain(F-PARIS13-AG); Wilkinson, Amie(1-NW) The C1 generic diffeomorphism has trivial centralizer. (English summary) Publ. Math. Inst. Hautes Études Sci. No. 109 (2009), 185–244. </p> http://mathoverflow.net/questions/121060/does-this-subgroup-of-even-braids-have-a-name/121073#121073 Answer by Andy Putman for Does this subgroup of "even braids" have a name? Andy Putman 2013-02-07T14:22:19Z 2013-02-08T21:34:07Z <p>I don't know if these groups have been studied before, but I can say something about their cohomology rings, at least over $\mathbb{Q}$. Namely, we have $H^k(E_n;\mathbb{Q}) = \mathbb{Q}$ if $k=0,1$ and $H^k(E_n;\mathbb{Q}) = 0$ for $k \geq 2$. Of course, this is the same as the cohomology of the ordinary braid group as computed by Arnold in</p> <p>V. I. Arnold, On some topological invariants of algebraic functions, Trudy Moscov. Mat. Obshch. 21 (1970), 27-46 (Russian), English transl. in Trans. Moscow Math. Soc. 21 (1970), 30-52.</p> <p>Recall that if $H$ is a finite-index normal subgroup of $G$, then $G$ acts on $H^k(H;\mathbb{Q})$ and using the transfer map we have that $H^k(G;\mathbb{Q})$ is equal to the invariants of this action.</p> <p>For braid groups, the action of $B_n$ on $H^k(PB_n;\mathbb{Q})$ factors through an action of the symmetric group $S_n$, so $H^k(PB_n;\mathbb{Q})$ is a <em>representation</em> of $S_n$ and $H^k(B_n;\mathbb{Q})$ is the trivial subrepresentation <code>$\{\text{$v \in H^k(PB_n;\mathbb{Q})|\sigma v = v$for all$\sigma \in S_n$}\}$</code>.</p> <p>Let's now consider $E_n$. In this case, the above argument shows that $H^k(E_n;\mathbb{Q})$ is the subrepresentation <code>$\{\text{$v \in H^k(PB_n;\mathbb{Q})|\sigma v = v$for all$\sigma \in A_n$}\}$</code>. </p> <p>Now, representations of finite groups over $\mathbb{Q}$ decompose into direct sums of irreducible representations. The only two irreducible representations of $S_n$ that restrict to the identity on $A_n$ are the trivial representation and the alternating representation. As we said above, the trivial representation corresponds to $H^k(B_n;\mathbb{Q})$, so we conclude that $$H^k(E_n;\mathbb{Q}) = W \oplus H^k(B_n;\mathbb{Q}),$$ where $W \subset H^k(PB_n;\mathbb{Q})$ is the direct sum of all alternating subrepresentations.</p> <p>The above calculation is thus equivalent to the assertion that the alternating representation does not occur in $H^k(PB_n;\mathbb{Q})$. This follows from the calculation of $H^k(PB_n;\mathbb{Q})$ as a representation of $S_n$ which was done by in the paper "Coxeter group actions on the complement of hyperplanes and special involutions" by Felder-Velesov; see <a href="http://arxiv.org/abs/math/0311190" rel="nofollow">here</a>.</p> <hr> <p>The above ref to Felder-Velesov was suggested by Vladimir Dotsenko; I originally included the argument below, which only works for $n \gg k$.</p> <p>It's quite hard to decompose $H^k(PB_n;\mathbb{Q})$ into irreducibles; however the paper "Representation Theory and Homological Stability" (see <a href="http://arxiv.org/abs/1008.1368" rel="nofollow">here</a>) by Church and Farb introduces a recipe that they call "representation stability" which describes how the decomposition of $H^k(PB_{n+1};\mathbb{Q})$ into irreducibles can be constructed from the decomposition of $H^k(PB_n;\mathbb{Q})$ into irreducibles, at least for $n$ large. Their results are hard to summarize briefly, but they do imply that the alternating representation does not occur (it is not "stable" in their sense), again at least for $n$ large.</p> http://mathoverflow.net/questions/119953/definition-of-area/119960#119960 Answer by Andy Putman for Definition of area Andy Putman 2013-01-26T20:10:40Z 2013-01-26T20:10:40Z <p>For students, probably the most elementary thing to do would be to restrict yourself to regions in the plane that can be triangulated with finitely many triangles (with straight sides). Accepting the area of a triangle as known, you then define the area of a region by adding up the areas of the triangles in a finite triangulation. The one thing you have to check is that this is well-defined. For this, I would first prove that any two finite triangulations of a region in the plane have a common subdivision (if you're reasonably clever about it, this can be done very quickly; certainly in 3-5 pages), and then prove that your notion of area is invariant under subdivisions.</p> <p>The nice thing about this is that all the main properties you want (eg that areas behave correctly under linear maps and translations) come for free from the analogous properties of triangles, which are easy.</p> http://mathoverflow.net/questions/119726/discovering-and-selecting-conferences/119766#119766 Answer by Andy Putman for Discovering and selecting conferences Andy Putman 2013-01-24T15:21:24Z 2013-01-24T15:54:16Z <p>Jon McCammond maintains a list of conferences in geometric group theory <a href="http://www.math.ucsb.edu/~jon.mccammond/geogrouptheory/conferences.html" rel="nofollow">here</a>.</p> <p>Jesse Johnson maintains a list of conferences in low-dimensional topology <a href="http://ldtopology.wordpress.com/conferences/" rel="nofollow">here</a>.</p> <p>I hear about many conferences from the geometry listserv <a href="http://listserv.utk.edu/archives/geometry.html" rel="nofollow">here</a>.</p> <p>There are also interesting conferences (slanted towards algebraic topology, but with a broad focus) on the list <a href="http://nilesjohnson.net/algtop-conf/" rel="nofollow">here</a>.</p> <p>As far as your question of how I decide which conferences to attend, at this point in my career (on the tenure track, but not yet tenured; I started this policy when I was a postdoc) I mostly only go to conferences where I am speaking. Sometimes I'll attend a conference where I'm not speaking if the conference has an unusually strong lineup of speakers, but I already travel far too much.</p> <p>When I was a graduate student, I basically went to conferences suggested by my advisor.</p> http://mathoverflow.net/questions/117523/manifolds-covered-by-an-n-dimensional-torus/117539#117539 Answer by Andy Putman for Manifolds covered by an n-dimensional torus Andy Putman 2012-12-29T17:17:10Z 2012-12-30T04:43:43Z <p>People have already mentioned the Bieberbach theorems, which imply that your manifold is homotopy equivalent to a Euclidean manifold. In fact, it is homeomorphic to the Euclidean manifold, at least if the dimension is at least $5$. This is predicted by the Borel conjecture, which asserts that homotopy equivalent aspherical manifolds are homeomorphic in high dimensions. The necessary case of the Borel conjecture is proven in the following paper of Farrell and Jones (building on lots of earlier work of theirs): </p> <p>F. T. Farrell and L. E. Jones. Topological rigidity for compact non-positively curved manifolds. In Differential geometry: Riemannian geometry (Los Angeles, CA, 1990), pages 229– 274. Amer. Math. Soc., Providence, RI, 1993.</p> <hr> <p>EDIT : The question inspired me to do a little more reading about the history of the Borel conjecture, and I learned that for virtually abelian fundamental groups like the ones in this question, the Borel conjecture was proven in the earlier paper</p> <p>MR0704219 (84k:57017) Farrell, F. T.(1-MI); Hsiang, W. C.(1-PRIN) Topological characterization of flat and almost flat Riemannian manifolds Mn (n≠3,4). Amer. J. Math. 105 (1983), no. 3, 641–672. </p> http://mathoverflow.net/questions/117457/manifolds-with-two-coordinate-charts/117475#117475 Answer by Andy Putman for Manifolds with two coordinate charts Andy Putman 2012-12-29T04:55:50Z 2012-12-29T16:58:54Z <p>I'll only discuss the first question (EDIT : Actually, I address the second question at the end). As Agol pointed out in the comments, for $n \geq 5$ this is an easy consequence of Newman's 1966 proof of the Poincare conjecture in the topological category.</p> <p>I don't know if it was explicitly stated earlier than this. However, it can easily be derived from the main result of the paper</p> <p>MR0126835 (23 #A4129) Brown, Morton The monotone union of open n-cells is an open n-cell. Proc. Amer. Math. Soc. 12 1961 812–814. </p> <p>In fact, this works in all dimensions (including $3$ and $4$). </p> <p>Brown's theorem is as follows. Assume that $M$ is a topological $n$-manifold and that for all compact $K \subset M$, there exists some open set $U \subset M$ with $K \subset U$ and $U \cong \mathbb{R}^n$. Then $M \cong \mathbb{R}^n$. Brown's proof is clever, but completely elementary.</p> <p>To get the desired result from this, assume that $X = U_1 \cup U_2$ with $U_i \cong \mathbb{R}^n$ and that $X$ is compact. Let $\phi : \mathbb{R}^n \rightarrow U_1$ be a homeomorphism. It is enough to prove that <code>$X \setminus \{\phi(0)\} \cong \mathbb{R}^n$</code>. We will do this with Brown's theorem. Consider a compact set <code>$K \subset X \setminus \{\phi(0)\}$</code>. To verify Brown's criteria, it is enough to construct a homeomorphism <code>$\psi : X \setminus \{\phi(0)\} \rightarrow X \setminus \{\phi(0)\}$</code> such that $\psi(K) \subset U_2$. </p> <p>For $r>0$, let $B(r) \subset \mathbb{R}^n$ be the ball of radius $r$. The set $U_1 \setminus U_2$ is compact, so there exists some $R>0$ such that $U_1 \setminus \phi(B(R)) \subset U_2$. Also, there exists some $\epsilon > 0$ such that $K \cap \phi(B(\epsilon)) = \emptyset$. It is easy to construct a homeomorphism $f : \mathbb{R}^n \rightarrow \mathbb{R}^n$ such that $f(B(\epsilon)) = B(2R)$ and $f(0)=0$ and $f|_{\mathbb{R}^n \setminus B(3R)} = \text{id}$. We can therefore define a homeomorphism <code>$\psi : X \setminus \{\phi(0)\} \rightarrow X \setminus \{\phi(0)\}$</code> by $\psi(p) = \phi \circ f \circ \phi^{-1}(p)$ for <code>$p \in U_1 \setminus \{\phi(0)\}$</code> and $\psi(p) = p$ for $p \notin U_1$. Clearly $\psi(K) \subset U_2$.</p> <hr> <p>EDIT : Lee suggested that this might be able to address his second question too. I thought a bit about it, and I believe that it can. The key is the following "relative" version of Brown's theorem, which can be proven exactly like Brown's theorem.</p> <p>Theorem : Let $(M,N)$ be a pair consisting of a topological $n$-manifold $M$ and a closed submanifold $N \subset M$. Assume that for all compact $K \subset M$, there exists some open set $U \subset M$ such that $K \subset U$ and such that the pair $(U,U \cap N)$ is homeomorphic to the pair $(\mathbb{R}^n,\mathbb{R}^{n-1})$ (the second embedded in the standard way). Then $(M,N) \cong (\mathbb{R}^n,\mathbb{R}^{n-1})$.</p> <p>To apply this, assume that $X$ is a compact manifold with boundary and that $X = U_1 \cup U_2$ with $(U_i,\partial U_i) \cong (\mathbb{R}^n_{\geq 0},\mathbb{R}^{n-1})$. Double $X$ to get a closed manifold $Y$, and let $Y' \subset Y$ be the image of the boundary of $X$. The open sets $U_i$ double to give an open cover $Y = V_1 \cup V_2$. Letting $V_i' = V_i \cap Y'$, we have $(V_i,V_i') \cong (\mathbb{R}^n,\mathbb{R}^{n-1})$. Let $(M,M')$ be the result of deleting the image of $0$ in $(V_1,V_1')$. It is enough to prove that $(M,M') \cong (\mathbb{R}^n,\mathbb{R}^{n-1})$, and this can be proven just like above.</p> <hr> <p>Of course, Agol answered the second question first -- it follows from the topological Schonfleiss theorem applied to the double, which was proven by Brown in</p> <p>MR0117695 (22 #8470b) Reviewed Brown, Morton A proof of the generalized Schoenflies theorem. Bull. Amer. Math. Soc. 66 1960 74–76. 54.00 (57.00)</p> <p>Mazur had earlier proven a weaker result. This requires the sphere to be bicollared, but this holds. Indeed, from the assumptions the sphere is locally bicollared, and Brown proved in</p> <p>MR0133812 (24 #A3637) Brown, Morton Locally flat imbeddings of topological manifolds. Ann. of Math. (2) 75 1962 331–341. </p> <p>that this implies that the sphere is bicollared. See </p> <p>MR0267588 (42 #2490) Connelly, Robert A new proof of Brown's collaring theorem. Proc. Amer. Math. Soc. 27 1971 180–182. </p> <p>for a super-easy proof of Brown's collaring theorem.</p> http://mathoverflow.net/questions/116896/liouvilles-theorem-with-your-bare-hands/116902#116902 Answer by Andy Putman for Liouville's theorem with your bare hands Andy Putman 2012-12-20T20:00:27Z 2012-12-20T20:00:27Z <p>I think the most illuminating proof of Liouville's theorem uses Riemann surfaces. Let $f : \mathbb{C} \rightarrow \mathbb{C}$ be a bounded holomorphic function, and set $g(z) = f(1/z)$. Then $g : \mathbb{C} \setminus 0 \rightarrow \mathbb{C}$ is a bounded holomorphic function, so Riemann's removable singularities theorem says that $g$ can be extended over $0$. Translated into the language of Riemann surfaces, this says that $f$ extends to a holomorphic function $F : \mathbb{P}^1 \rightarrow \mathbb{C}$. Since $\mathbb{P}^1$ is compact, $F$ must have a global maximum. However, the maximum modulus principle says that a nonconstant holomorphic function cannot have a local maximum, so $F$ must be constant.</p> http://mathoverflow.net/questions/116556/status-of-the-isomorphism-problem-for-automatic-groups/116557#116557 Answer by Andy Putman for Status of the Isomorphism problem for automatic groups? Andy Putman 2012-12-17T01:16:24Z 2012-12-17T01:16:24Z <p>It's still an open problem. The isomorphism problem for hyperbolic groups (a much smaller class) was only solved recently by Dahmani and Guirardel (see <a href="http://arxiv.org/abs/1002.2590" rel="nofollow">here</a>), following work of Sela.</p> <p>In the same vein, the conjugacy problem is also still open for automatic groups. It has been solved for biautomatic groups, but it is also still open whether all automatic groups are biautomatic.</p> http://mathoverflow.net/questions/116304/if-x-is-a-simplicial-complex-is-their-a-characterization-of-the-links-of-the/116329#116329 Answer by Andy Putman for If $X$ is a simplicial complex, is their a characterization of the links of the vertices of $X$ that is equivalent to the statement "$|X|$ is a manifold Andy Putman 2012-12-13T22:54:31Z 2012-12-13T22:54:31Z <p>I don't think there will be a simple necessary and sufficient condition.</p> <p>For exotic triangulations, the links $L$ of vertices will not be manifolds. One thing you need (not sufficient) is for $L \times \mathbb{R}$ to be a manifold. It's kind of shocking that this is possible for non-manifolds; the first example of it is Bing's <a href="http://en.wikipedia.org/wiki/Dogbone_space" rel="nofollow">dogbone space</a>. A weaker version of the phenomena is the <a href="http://en.wikipedia.org/wiki/Whitehead_manifold" rel="nofollow">Whitehead manifold</a> $W$, which is a contractible $3$-manifold which is not homeomorphism to $\mathbb{R}^3$ but such that $W \times \mathbb{R}$ is homeomorphism to $\mathbb{R}^4$.</p> <p>To get an idea of the kinds of things that are involved here, I recommend reading the introduction to Edwards's paper "Suspensions of homology spheres", available <a href="http://arxiv.org/abs/math/0610573" rel="nofollow">here</a>. That paper also contains a very detailed account of the fact that the double suspension of Mazur's homology 3-sphere is a 5-sphere (with lots of pictures). Another nice account of this is in Steve Ferry's notes on geometric topology, available <a href="http://www.math.rutgers.edu/~sferry/ps/geotop.pdf" rel="nofollow">here</a>.</p> http://mathoverflow.net/questions/115632/representation-theory-of-discrete-subgroups-of-lie-groups/115635#115635 Answer by Andy Putman for Representation theory of Discrete Subgroups of Lie groups Andy Putman 2012-12-06T19:05:03Z 2012-12-06T19:05:03Z <p>The result you are looking for is the Margulis superrigidity theorem. See Chapter 13 of the <a href="http://people.uleth.ca/~dave.morris/books/IntroArithGroups.html" rel="nofollow">book</a> "Introduction to Arithmetic Groups" by Dave Witte Morris for more details.</p> http://mathoverflow.net/questions/114528/are-homeomorphic-open-subsets-of-mathbbrn-also-diffeomorphic/114533#114533 Answer by Andy Putman for Are homeomorphic open subsets of $\mathbb{R}^n$ also diffeomorphic? Andy Putman 2012-11-26T14:52:59Z 2012-11-26T14:52:59Z <p>In fact, there exist uncountably many <em>small</em> exotic smooth $\mathbb{R}^4$'s, i.e. smooth manifolds $X$ which are homeomorphic to $\mathbb{R}^4$ but not not diffeomorphic to it and which can be smoothly embedded as open subsets of $\mathbb{R}^4$. There are discussions of this in many places; I recommend first reading the appropriate part of Scorpan's book "The Wild World of 4-Manifolds" for a brief survey (it includes a nice bibliography of more detailed sources). </p> http://mathoverflow.net/questions/30549/deligne-mumford-space-defined-in-complex-geometry-category/30554#30554 Answer by Andy Putman for Deligne-Mumford space defined in complex geometry category Andy Putman 2010-07-04T20:43:42Z 2012-11-24T21:52:22Z <p>I'm not sure where to point you for full details of this, but quite a few details are in some old research announcements of Bers. See his papers</p> <p>MR0361051 (50 #13497) Bers, Lipman Spaces of degenerating Riemann surfaces. Discontinuous groups and Riemann surfaces (Proc. Conf., Univ. Maryland, College Park, Md., 1973), pp. 43--55. Ann. of Math. Studies, No. 79, Princeton Univ. Press, Princeton, N.J., 1974. </p> <p>and</p> <p>MR0361165 (50 #13611) Bers, Lipman On spaces of Riemann surfaces with nodes. Bull. Amer. Math. Soc. 80 (1974), 1219--1222. </p> <p>and</p> <p>MR0374496 (51 #10696) Bers, Lipman Deformations and moduli of Riemann surfaces with nodes and signatures. Collection of articles dedicated to Werner Fenchel on his 70th birthday. Math. Scand. 36 (1975), 12--16. </p> <hr> <p>EDIT : Sorry to resurrect this ancient thread, but I heard a lovely talk from Sarah Koch a few weeks ago in which she described a recent paper that she wrote with John Hubbard in which they give a complex-analytic construction of the DM compactification of the moduli space of curves and prove that (as a complex analytic space) it is isomorphic to the analyticification of the usual one. In particular, this gives all the missing details in Bers's papers above (along with much more). See their paper "An analytic construction of the Deligne-Mumford compactification of the moduli space of curves" available from Sarah's webpage <a href="http://www.math.harvard.edu/~kochs/papers.html" rel="nofollow">here</a>.</p> http://mathoverflow.net/questions/114205/random-walks-on-coxeter-groups/114311#114311 Answer by Andy Putman for Random walks on Coxeter groups Andy Putman 2012-11-24T04:12:21Z 2012-11-24T04:12:21Z <p>Not an answer, but too long for a comment. There is actually a general recipe for computing the growth function of any Coxeter group (which implies, in particular, that it is always a rational function). I haven't done the calculation, but your purported calculation of the growth function for the Coxeter group you wrote down can be derived from this (assuming that it is true).</p> <p>This is very old stuff. It appears as an exercise in Bourbaki's book "Groups et algebres de Lie"; see exercises 15-26 of Section 4.1. The details of these exercises appear in the paper</p> <p>MR1170370 (93g:20081) Paris, Luis(CH-GENV-SM) Growth series of Coxeter groups. Group theory from a geometrical viewpoint (Trieste, 1990), 302–310, World Sci. Publ., River Edge, NJ, 1991.</p> <p>The book this paper appears in is extremely hard to track down; as far as I can tell, it is not available anywhere at any price. It contains a number of important papers in geometric group theory, so I scanned it a long time ago. I just posted a scan of the above paper <a href="http://www.math.rice.edu/~andyp/ParisCoxeter.djvu" rel="nofollow">here</a>.</p> http://mathoverflow.net/questions/114210/where-did-nachman-aronszajn-prove-the-existence-of-aronszajn-trees/114212#114212 Answer by Andy Putman for Where did Nachman Aronszajn prove the existence of Aronszajn trees? Andy Putman 2012-11-23T05:47:36Z 2012-11-23T07:09:35Z <p>I believe that these first appeared in the paper</p> <p>Georges Kurepa, Ensembles linéaires et une classe de tableaux ramifiés (tableaux ramifiés de M.Aronszajn), Publ. Math. Univ. Belgrade, 6 (1936) 129-160.</p> <p>The author attributes them to a letter that Aronszajn wrote to him (see page 132). This paper is available <a href="http://elib.mi.sanu.ac.rs/files/journals/publ/6/14.pdf" rel="nofollow">here</a>.</p> <hr> <p><strong>Edit by A. Caicedo</strong>: The following is from Stevo Todorcevic's essay introducing the papers on "Theory of partially ordered sets", part A of <strong>Selected papers of Ðuro Kurepa</strong>, Edited and with commentaries by Aleksandar Ivić, Zlatko Mamuzić, Žarko Mijajlović and Stevo Todorčević, Matematički institut SANU, Belgrade, 1996. <a href="http://www.ams.org/mathscinet-getitem?mr=1429393" rel="nofollow">MR1429393 (97m:01106)</a>.</p> <p>The papers it refers to are </p> <ul> <li>A[35] <em>Ensembles ordonnés et ramifiés</em>, Publ. Math. Univ. Belgrade <strong>4</strong> (1935), 1-138.</li> <li>A[37] <em>Ensembles lineaires et une classe de tableaux ramifies (Tableaux ramifies de M. Aronszajn)</em>, Publ. Math. Univ. Belgrade <strong>6/7</strong> (1937/38), 129-160.</li> <li><code>[1]</code> J. E. Baumgartner, <em>Decomposition and embedding of trees</em>, Notices Amer. Math. Soc. <strong>17</strong> (1970), 967.</li> <li>[6] K. J. Devlin, <em>Note on a theorem of J. Baumgartner</em>, Fund. Math. <strong>76</strong> (1972), 255-260. <a href="http://www.ams.org/mathscinet-getitem?mr=540759" rel="nofollow">MR0540759 (58 #27476)</a>.</li> <li>[12] W. P. Hanf, <em>Incompactness in languages with infinitely long expressions</em>, Fund. Math. <strong>53</strong> (1964), 309-324. <a href="http://www.ams.org/mathscinet-getitem?mr=160732" rel="nofollow">MR0160732 (28 #3943)</a>.</li> </ul> <blockquote> <p>The most important publication in this group is Kurepa's thesis A[35] written under the direction of M. Fréchet. It was the first systematic study of trees and ramified partially ordered sets and of their close relationship to linear orderings. It was the source of many crucial notions and problems in this area such as, for example, the notions of <em>Aronszajn</em> and <em>Souslin tree</em>. It is the source of the problem whether inaccessible cardinals have the <em>tree property</em> i.e., whether they satisfy the analogue of König's infinity lemma, which was later proved by Hanf, Tarski and others ([12]) to be equivalent to the large cardinal property of <em>weak compactness</em>. Trees are classified in §8 A11 as "large", "étroit" and "ambigu" according to their heights and widths. The very thin and tall trees ("étroit") always have cofinal branches i.e., chains intersecting every level (Theorem $5^{bis}$). This is a fine result representing a recurring theme in applications of trees, especially in the partition calculus. This result was also a source of the problem whether the same fact is true about the class of slightly wider trees ("ambigu") i.e., the trees of height equal to some cardinal $\theta$ and whose levels are now only assumed to be of size $\lt \theta$ (rather than $\lt\lambda$ for some cardinal $\lambda&lt;\theta$ as it was the case with the trees in Theorem $5^{bis}$). This is the problem known today as the problem whether <em>$\theta$ has the tree property</em>. For $\theta= \omega_1$ the problem was solved in June 1934 by N. Aronszajn and appeared (with an acknowledgement) as Theorem 6 of A[35]. According to the footnote on the same page, Aronszajn constructed his tree as a subtree of the tree of all $1-1$ sequences from $\mathbb Q^{\lt\omega_1}$, while Kurepa's version of the proof presented in A[35] (and also in A[37]) was to build such a tree inside the tree $\sigma\mathbb Q$ (denoted by $\sigma_0$) of all (nonempty bounded) well-ordered subsets of rationals. This is the construction most frequently used in subsequent expositions of this result. In the sequel A[37;§27] he modified his construction in order to produce an Aronszajn tree with the surprising property that it is a union of countably many antichains, thus introducing yet another remarkable notion, the notion of a <em>special Aronszajn tree</em>. (It is known today that some care is needed to get such a tree, as it is possible to have nonspecial Aronszajn subtrees of both $\sigma\mathbb Q$ and the set of all $1-1$ sequences from $\mathbb Q^{\lt\omega_1}$; see <code>[1]</code> and [6].) </p> </blockquote> <p>The essay goes on to describe additional work (by Kurepa and others) on the tree property. </p> http://mathoverflow.net/questions/114052/primitive-subwords-in-a-free-group-of-rank-2/114099#114099 Answer by Andy Putman for Primitive subwords in a free group of rank 2 Andy Putman 2012-11-21T20:43:09Z 2012-11-21T20:43:09Z <p>As Misha pointed out, your conjecture is false.</p> <p>However, I thought I'd point out that there is a fairly nice way of parameterizing primitive elements of $F_2$. Let $\pi : F_2 \rightarrow \mathbb{Z}^2$ be the abelianization map. It is then standard that if $x \in F_2$ is primitive, then $\pi(x)$ is primitive (i.e. nonzero and not divisible by any integer other than $\pm 1$). Moreover, if $v \in \mathbb{Z}^2$ is primitive, then there is an element $x \in F_2$ such that $\pi(x) = v$, and $x$ is unique up to conjugation.</p> <p>It is easy to enumerate the primitive elements of $\mathbb{Z}^2$, so this leads to the question of determining for a primitive $v \in \mathbb{Z}^2$ the (unique up to conjugacy) primitive element $x \in F_2$ with $\pi(x) = v$. There is an elegant geometric solution to this in the following paper.</p> <p>MR0608526 (82i:20042) Osborne, R. P.; Zieschang, H. Primitives in the free group on two generators. Invent. Math. 63 (1981), no. 1, 17–24.</p> http://mathoverflow.net/questions/111943/does-combinatorial-formula-for-the-pontrjagin-classes-exist/111947#111947 Answer by Andy Putman for Does combinatorial formula for the Pontrjagin classes exist? Andy Putman 2012-11-09T22:35:32Z 2012-11-09T22:35:32Z <p>I believe there is a formula like the one you seek together with a good survey of the previous literature in the paper "Local formulae for combinatorial Pontrjagin classes" by Gaifullin, available <a href="http://arxiv.org/abs/math/0407035" rel="nofollow">here</a>.</p> http://mathoverflow.net/questions/110772/convergence-action-on-the-boundary-of-hyperbolic-groups/110775#110775 Answer by Andy Putman for convergence action on the boundary of hyperbolic groups Andy Putman 2012-10-26T18:24:15Z 2012-10-26T18:24:15Z <p>Bowditch proved much more. Namely, if a group $\Gamma$ acts properly discontinuously on a $\delta$-hyperbolic space $X$, then $\Gamma$ acts as a convergence group on $\partial X$. See Lemma 1.11 of his paper </p> <p>B.H. Bowditch, Convergence groups and configuration spaces, in Geometric Group Theory Down Under, proceedings of a Special Year in Geometric Group Theory, Canberra, Australia'' (ed. J.Cossey, C.F.Miller III, W.D.Neumann, M.Shapiro), de Gruyter (1999), 23-54.</p> <p>which is available <a href="http://www.warwick.ac.uk/~masgak/papers/bhb-convergence.dvi.gz" rel="nofollow">here</a>. To get the result you want, consider the action of $\Gamma$ on its Cayley graph.</p> http://mathoverflow.net/questions/110457/bass-stable-range-condition-for-principal-ideal-domains/110472#110472 Answer by Andy Putman for Bass' stable range condition for principal ideal domains Andy Putman 2012-10-23T20:47:22Z 2012-10-24T16:28:21Z <p>EDIT : Here's a proof that works for $R$ a a PID, which implies that the condition of generating the unit ideal is the same as having gcd equal to $1$.</p> <p>For some $n \geq 2$ consider a tuple $(a_1,\ldots,a_{n+1})$ of elements of $R$ whose gcd is $1$. We want to find $r_1,\ldots,r_n \in R$ such that $\text{gcd}(a_1+r_1 a_{n+1},\ldots,a_n + r_n a_{n+1}) = 1$. </p> <p>There are three cases. If $a_{n+1}=0$, then there is nothing to do. If $a_i=0$ for some $1 \leq i \leq n$, then we can take $r_i=1$ and $r_j=0$ for $j \neq i$.</p> <p>The most interesting case is when none of the $a_i$ equal $0$. In this case, we will only need $r_1$ (the rest of the $r_i$ can be taken to be $0$). Set <code>$b = \text{gcd}(a_2,\ldots,a_n)$</code>, and let $p_1,\ldots,p_k$ be the distinct primes dividing $b$. For each $i$, we know that $p_i$ cannot divide both $a_1$ and $a_{n+1}$. This implies that there exists some <code>$c_i \in \{0,1\}$</code> such that $$a_1 + c_i a_{n+1} \neq 0 \quad (\text{mod } p_i).$$ By the Chinese remainder theorem, there exists some $r_1 \in R$ such that $$r_1 = c_i \quad (\text{mod } p_i)$$ for $1 \leq i \leq k$, which implies that $$a_1 + r_1 a_{n+1} \neq 0 \quad (\text{mod } p_i)$$ for all $1 \leq i \leq k$. We conclude that the gcd of $a_1+r_1 a_{n+1}$ and $b$ equals $1$, and thus that the gcd of $a_1+r_1 a_{n+1},a_2,\ldots,a_n$ is $1$.</p> <hr> <p>Here is what was my original answer:</p> <p>This does not exactly answer your question, but it is much easier to prove that the complexes that van der Kallen needs are highly connected for $\mathbb{Z}$ than for general rings. This was originally done by Maazen in his unpublished thesis. I happen to have a scan of this which I posted <a href="http://www.math.rice.edu/~andyp/MaazenThesis.djvu" rel="nofollow">here</a>. There is also a different proof of this connectivity in Step 2 of the proof of Theorem B in my paper "The complex of partial bases for $F_n$ and finite generation of the Torelli subgroup of $\text{Aut}(F_n)$" with Matt Day, available on my <a href="http://www.math.rice.edu/~andyp/papers/" rel="nofollow">webpage</a>.</p> http://mathoverflow.net/questions/109941/publication-and-career-as-a-fresh-ph-d/109948#109948 Answer by Andy Putman for Publication and Career as a fresh Ph.D Andy Putman 2012-10-17T20:28:49Z 2012-10-17T20:28:49Z <p>(I wanted to write more than the comments allowed, so I marked this as community wiki so that I don't feel guilty about getting reputation for it).</p> <p>Here is the advice I give to postdocs when they are trying to figure out what projects to work on and what publication rate is appropriate for their goals.</p> <ol> <li><p>People tend to judge you based on some combination of your best work and your total paper counts. The better the mathematician, the more they will judge you by your best work as opposed to number of papers. So it is better to write a few strong papers than lots of mediocre papers.</p></li> <li><p>However, it's much better to write mediocre papers than to not write strong papers! In other words, you need to be aware that people are going to judge you based on what you produce during your postdoc. If you are doing a standard 3 year postdoc, then you will be applying for jobs during your third year, so you have exactly two years in which to prove yourself. You need to have things to show for those two years!</p></li> <li><p>The upshot of the above is that you should have a mixture of long-term and short-term projects, but during your postdoc you should concentrate the majority (though not all) your attention on projects that you can realistically complete before you go on the job market. And you should also work on multiple projects at once rather than concentrating all your attention on one goal (which might or might not pan out).</p></li> <li><p>As far as figuring out how productive you need to be, I advise you to do the following. Make a list of people who are in your field (different fields have different rates of publication) and who have gotten tenure-track jobs at places you aspire to be at in the last 5-10 years. Next, go to their webpages and look at their CV's. You want to have a publication record (both in terms of number of papers and in terms of quality of journals) after your postdoc which is similar to them when they were on the tenure-track job market. If you are way off, then you might want to reconsider your career goals.</p></li> </ol> http://mathoverflow.net/questions/107851/extending-jordan-loops/107855#107855 Answer by Andy Putman for Extending Jordan loops Andy Putman 2012-09-22T17:30:59Z 2012-09-22T17:30:59Z <p>What you're asking is equivalent to asking whether any homeomorphism $g : S^1 \rightarrow S^1$ can be extended to a homeomorphism of the disc. This is easy -- write the disc in polar coordinates $(t,\theta)$ with $\theta \in S^1$, and define an extension $G(t,\theta) = (t,g(\theta))$.</p> <p>The question about whether this can be done smoothly if $g$ is smooth is more subtle. Observe that the above also works for $S^k$ with $k > 1$. The smooth version fails in higher dimensions and is responsible for the existence of exotic spheres. However, for $k=1,2$ there is no problem. For $k=1$, this is a theorem of Smale; see</p> <p>Smale, Stephen Diffeomorphisms of the 2-sphere. Proc. Amer. Math. Soc. 10 1959 621–626.</p> <p>For $k=2$, it is a much deeper theorem of Hatcher; see</p> <p>Hatcher, Allen E. A proof of the Smale conjecture, Diff(S3)≃O(4). Ann. of Math. (2) 117 (1983), no. 3, 553–607. </p> http://mathoverflow.net/questions/107651/free-subgroup-of-diff0-1/107656#107656 Answer by Andy Putman for Free subgroup of Diff([0,1])? Andy Putman 2012-09-20T06:26:11Z 2012-09-20T06:26:11Z <p>Much more is true. The compactly supported diffeomorphism group of any (positive-dimensional, nonempty) manifold contains free subgroups of uncountable rank. In fact, there are such subgroups that are generated by sets which are arcwise connected! See the paper</p> <p>MR0974661 (90b:58031) Grabowski, Janusz(PL-WASW) Free subgroups of diffeomorphism groups. Fund. Math. 131 (1988), no. 2, 103–121. </p> <p>which is available online <a href="http://matwbn.icm.edu.pl/ksiazki/fm/fm131/fm131110.pdf" rel="nofollow">here</a>.</p> http://mathoverflow.net/questions/4011/whats-a-non-abelian-totally-ordered-group/4015#4015 Comment by Andy Putman Andy Putman 2013-06-18T15:46:35Z 2013-06-18T15:46:35Z @Salvo Tringali : To me, a surface group is the fundamental group of a closed oriented surface (I also often require that the group be nonabelian, though I'm not doing that here). This is a common usage in geometric and combinatorial group theory, though I suppose I should have explained it here. As far as Thurston goes, I have spoken to people with whom he discussed it, but I'm pretty sure that there is no written record of it and that he never gave talks about it. http://mathoverflow.net/questions/133773/linear-algebra-linear-transformation Comment by Andy Putman Andy Putman 2013-06-14T18:34:03Z 2013-06-14T18:34:03Z MO is not for homework or for undergraduate level topics. See the FAQ. http://mathoverflow.net/questions/71512/nonfree-differential-calculus Comment by Andy Putman Andy Putman 2013-06-14T16:17:03Z 2013-06-14T16:17:03Z I rolled back the edit. The &quot;differential calculus&quot; tag does not seem appropriate here. I also (along with Misha) voted to close as &quot;no longer relevant&quot;. http://mathoverflow.net/questions/133597/what-would-remain-of-current-mathematics-without-axiom-of-power-set Comment by Andy Putman Andy Putman 2013-06-13T06:18:32Z 2013-06-13T06:18:32Z @Nik Weaver : Ah, I see what you are saying. So he is just really saying that he doesn't believe in uncountable sets. http://mathoverflow.net/questions/133597/what-would-remain-of-current-mathematics-without-axiom-of-power-set Comment by Andy Putman Andy Putman 2013-06-13T06:11:51Z 2013-06-13T06:11:51Z You are very confused. Just because there are only countably many formulas does not mean that such formulas can only define countable sets. http://mathoverflow.net/questions/133245/action-of-an-isomorphism-in-cohomology-as-the-intersection-with-the-class-of-the/133335#133335 Comment by Andy Putman Andy Putman 2013-06-11T21:30:26Z 2013-06-11T21:30:26Z @David Speyer : I believe that Bredon's book on algebraic topology does it too. Neither Hatcher nor May discuss smooth manifolds at all, so they really can't discuss this interpretation of cup products. http://mathoverflow.net/questions/132366/is-there-a-math-concept-map Comment by Andy Putman Andy Putman 2013-05-30T20:56:18Z 2013-05-30T20:56:18Z This is not appropriate for MO. Read the FAQ. http://mathoverflow.net/questions/130625/research-on-the-structure-of-a-non-goldbach-number Comment by Andy Putman Andy Putman 2013-05-14T20:25:45Z 2013-05-14T20:25:45Z Questions about whether anything is known about a very specific topic are fine, but the topic in question has to be much more specific and better thought out than your question. http://mathoverflow.net/questions/130625/research-on-the-structure-of-a-non-goldbach-number Comment by Andy Putman Andy Putman 2013-05-14T20:16:41Z 2013-05-14T20:16:41Z This is far too vague for MO. http://mathoverflow.net/questions/130431/mapping-class-group-of-once-punctured-torus/130432#130432 Comment by Andy Putman Andy Putman 2013-05-12T21:57:41Z 2013-05-12T21:57:41Z @Misha : Ah, so that's the statement you are after! I had assumed that you wanted a reference for the fact that the Birman exact sequence becomes an isomorphism for a once-punctured torus. I don't think I've ever seen the statement you want in print, though of course everyone (suitably interpreted) knows it. My inclination would just be to do as you suggest and say that it is well-known and easy to prove. http://mathoverflow.net/questions/130306/the-role-of-the-automatic-groups-in-the-history-of-geometric-group-theory/130309#130309 Comment by Andy Putman Andy Putman 2013-05-11T00:14:10Z 2013-05-11T00:14:10Z @Misha : I largely agree with you on the significance of automatic group theory within geometric group theory. But they do have a few triumphs. For instance, I know of no proof that mapping class groups have quadratic isoperimetric inequalities that does not use Mosher's theorem they they are automatic (or at least its proof). http://mathoverflow.net/questions/130306/the-role-of-the-automatic-groups-in-the-history-of-geometric-group-theory Comment by Andy Putman Andy Putman 2013-05-10T22:33:20Z 2013-05-10T22:33:20Z I think that lots of people know about automatic groups; certainly I spent lots of time with ECHLPT in grad school. But the remaining open questions are really hard, so not too many people are actively working on them. That happens to a lot of subjects. Eventually someone will come along with a big idea and the subject will start moving again... http://mathoverflow.net/questions/130298/show-f-is-in-l1-d-space-and-f-d-over-xlimnfn-d-over-x Comment by Andy Putman Andy Putman 2013-05-10T20:23:43Z 2013-05-10T20:23:43Z MO is not for homework. See the FAQ. http://mathoverflow.net/questions/130296/show-that-is-continuous-xyxy-etc Comment by Andy Putman Andy Putman 2013-05-10T20:22:45Z 2013-05-10T20:22:45Z MO is not for homework. See the FAQ. http://mathoverflow.net/questions/130157/generators-of-sections-of-free-groups/130188#130188 Comment by Andy Putman Andy Putman 2013-05-10T16:43:24Z 2013-05-10T16:43:24Z It's nontrivial to compute it, but there is a huge literature on group cohomology, so there are many tools available. To help your search, you should be aware that $H_2(G;\mathbb{Z})$ is also known as the Schur multiplier of $G$. For a particular finite group of reasonable size, by the way, you should be able to compute $H_2$ using GAP. The relevant packages are cohomolo (see <a href="http://www.gap-system.org/Packages/cohomolo.html" rel="nofollow">gap-system.org/Packages/cohomolo.html</a>) and hap (see <a href="http://www.gap-system.org/Packages/hap.html" rel="nofollow">gap-system.org/Packages/hap.html</a>).