User misha - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T02:49:30Z http://mathoverflow.net/feeds/user/21684 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130725/locally-symmetric-space-and-global-symmetric-space/130733#130733 Answer by Misha for locally symmetric space and global symmetric space Misha 2013-05-15T15:38:25Z 2013-05-17T23:52:24Z <p>No, take $Z$ to be a non-orientable compact connected hyperbolic surface. </p> <p>Edit: Here is an easy orientable example. Let $S$ be a non-orientable hyperbolic surface. Then take $Z$ to be the orientable 2-fold cover of $S\times S$. The point is is that fundamental group of $Z$ (regarded as a deck-transformation group) still contains non-holomorphic automorphisms of the bidisk. I am quite sure that there are even stably parallelizable examples of similar to this, but I do not see a sufficient motivation for constructing them. </p> <p>I looked briefly at one of the papers you have links to: They are just sloppy in their formulations. Whenever they say "locally symmetric space" you should just read "quotient $\Gamma\backslash X$, where $\Gamma&lt; G^0$''. The correct statement is that for a lattice (actually, any subgroup) $\Gamma$ in the Lie group $G=Isom(X)$, the intersection $\Gamma \cap G^0$ is a finite-index subgroup of $\Gamma$, where $G^0$ is the identity component of $G$. </p> http://mathoverflow.net/questions/130985/cayley-graphs-of-finitely-generated-infinite-groups-quasi-isometrically-embeddabl/130994#130994 Answer by Misha for Cayley graphs of finitely generated infinite groups quasi-isometrically embeddable in R^3 Misha 2013-05-17T18:47:08Z 2013-05-17T18:47:08Z <p>Yes, the theorem is:</p> <p>Suppose that $G$ is a finitely-generated group whose Cayley graph quasi-isometrically (qi) embeds in $R^3$. Then $G$ is commensurable to a free abelian group of rank $\le 3$. (The converse is, of course, also true.) </p> <p>In other words, $G$ contains a free abelian subgroup $A$ of finite index, so that $A$ has rank $\le 3$. </p> <p>Here is a proof. First, you note that $R^3$ has polynomial growth, equivalent to $x^3$. Thus, $G$ also has growth at most $x^3$ since growth of a qi embedded subspace can only be lower than the one for the ambient space. By Gromov's polynomial growth theorem, it follows that $G$ is virtually nilpotent. For nilpotent groups there is a precise formula for growth in terms of their derived series, due to Bass and Guivarch. This formula implies that the group has to be virtually abelian of rank $\le 3$. </p> <p>You can find definitions and proofs of most of the results in <a href="https://www.math.ucdavis.edu/~kapovich/EPR/kapovich_drutu.pdf" rel="nofollow">this book</a>. </p> <p>There is a bit more general result, due to Scott Pauls that if a nilpotent group $G$ qi embeds in, say, a Hilbert space, then $G$ is virtually abelian. However, in the 3d case you are interested in, you do not need this. </p> http://mathoverflow.net/questions/130196/jordan-curve-theorem-can-every-point-on-the-curve-be-reached-from-each-region/130544#130544 Answer by Misha for Jordan curve theorem: Can every point on the curve be reached from each region? Misha 2013-05-14T05:00:35Z 2013-05-14T05:00:35Z <p>The answers by Kent and Eremenko are, of course, absolutely correct. Here is the more general statement: Let $M$ be a closed connected topological $n-1$-dimensional manifold embedded in ${\mathbb R}^n$ (note that $M$ can be wild for $n\ge 3$). Let $U$ be a component of ${\mathbb R}^n -M$. Then for every point $x\in M$ there exists a continuous path $p: [0,1)\to U$, so that</p> <p>$$\lim_{t\to 1} p(t)=x$$</p> <p>In particular, <code>$({\mathbb R}^n -M) \cup \{x\}$</code> is path-connected. The proof is based on the following observation:</p> <p>Observation. There exists a function $\phi(r), r\in [0, \infty)$, satisfying $\phi(r)\ge r$ and<br> $$\lim_{r\to 0+} \phi(r)=0$$ so that: For every $r>0$ the map $$\tilde{H}_0(U \cap B(x, r))\to \tilde{H}_0(U \cap B(x, \phi(r)))$$ (induced by inclusion) is zero. In other words, any two points in $U \cap B(x, r)$ can be connected by a path in $U \cap B(x, \phi(r))$. </p> <p>The above observation follows from the Poincare/Alexander duality in ${\mathbb R}^n$ (in conjunction with the fact that $M$ is a manifold).</p> <p>Given the above observation, take a sequence $x_k\in U\cap B(x, \frac{1}{k})$ and construct the path $p$ by concatenating paths $p_i$ in $$B(x, \phi(\frac{1}{i}))\cap U$$ connecting points $x_i, x_{i+1}$, $i\in {\mathbb N}$. </p> http://mathoverflow.net/questions/130431/mapping-class-group-of-once-punctured-torus Mapping class group of once-punctured torus Misha 2013-05-12T19:02:51Z 2013-05-13T16:16:14Z <p>Let $T$ be the 2-dimensional torus and let $S$ be $T$ minus one point. Then Birman exact sequence of mapping class groups becomes an isomorphism $$\beta: Map(S)\to Map(T)=GL(2, {\mathbb Z}).$$ It is then essentially immediate that $\beta$ preserves Thurston's classification of elements of the mapping class group into three types: $\beta(f)$ is Anosov if and only if $f$ is pseudo-Anosov, etc. </p> <p><em>Question.</em> Did anybody bother to record this elementary observation in the literature? </p> <p><em>I just need a reference, since anybody who knows anything about the mapping class group knows how to prove it (in several ways).</em> (Please, do not write proofs, I know at least 4.) </p> <p>I was nearly sure that Farb and Margalit have it, but they do not. Same for Casson and Bleiler, same for Ivanov. Of course, maybe this is one of the cases when it is easier to write a proof then to find a reference. </p> http://mathoverflow.net/questions/130306/the-role-of-the-automatic-groups-in-the-history-of-geometric-group-theory/130309#130309 Answer by Misha for The role of the Automatic Groups in the history of Geometric Group Theory Misha 2013-05-10T22:37:07Z 2013-05-11T00:04:39Z <p>My (admittedly subjective) take on it that development of mathematics is driven by proofs of hard theorems and (in most cases linked to it) development of new technique. Just think of, say, Mostow Rigidity Theorem, Gromov's Polynomial Growth theorem or Rips work on group actions on trees, Thurston/Perelman Geometrization theorems, etc. Most "easy" results in AGT were proven by early 1990s; hard problems are still there, it is just that none of them were solved (not for the lack of trying) and no new technique was introduced (also not for the lack of trying). So, people moved onto something else. </p> <p>Edit: My feeling for the first question (historic role) is the same as Zhou Enlai's about French Revolution. (The entire field is way too unsettled.) However, if you were to press me for a definition answer I'd say "not particularly significant so far". The key reason: Lack of truly deep theorems/powerful techniques. All this, of course, might suddenly change if one finds a way to bring, say, number theory (or ergodic theory, or model theory, or nonlinear PDEs...) into the picture, addressing, say, the problem of automaticity of uniform lattices of higher rank. </p> http://mathoverflow.net/questions/130038/homomorphisms-of-lie-groups-preserving-regularity Homomorphisms of Lie groups preserving regularity Misha 2013-05-07T23:32:11Z 2013-05-08T18:17:19Z <p>Let $G_1, G_2$ be connected semisimple Lie groups, let us assume for simplicity that both groups are complex (even though, I am interested in the real Lie groups as well). Let $f: G_1\to G_2$ be a monomorphism which sends regular semisimple elements to regular semisimple elements. Does it follow that $f$ also sends regular unipotent elements to regular ones? </p> <p>I suspect that the answer is well-known, but I could not find it. (Actually, I do not even know if there is a standard name for homomorphisms preserving regularity.) This question is motivated by study of discrete subgroups of higher rank Lie groups, but explaining the precise motivation will take us a bit too far. </p> <p>Edit: I should have thought a bit more before asking, since there is an obvious counter-example: The reducible (faithful) representation $SL(2)\to SL(3)$ preserves regularity of semisimple elements but does not preserve regularity of unipotent elements. However, the reducible representation $SL(n)\to SL(n+1), n\ge 3$, fails to preserve regularity of semisimple elements, so maybe there is a hope to classify all counter-examples. </p> http://mathoverflow.net/questions/129571/construct-seifert-fibration-on-mapping-torus-of-surface-with-monodromy-a-periodic/129576#129576 Answer by Misha for construct Seifert fibration on mapping torus of surface with monodromy a periodic mapping class Misha 2013-05-03T19:40:12Z 2013-05-03T19:40:12Z <p>Just use the suspension flow of the periodic diffeomorphism $f: S\to S$ in the periodic mapping class. Then all flow lines will be periodic (i.e., circles) and you are done; the base will be the quotient $S/f$.</p> <p>For the second question, the answer is yes; again, just suspend the invariant multicurve. </p> http://mathoverflow.net/questions/129321/does-a-topological-manifold-have-an-exhaustion-by-compact-submanifolds-with-bou/129331#129331 Answer by Misha for Does a *topological* manifold have an exhaustion by compact submanifolds with boundary? Misha 2013-05-01T17:16:46Z 2013-05-01T17:16:46Z <p>Since topological manifolds of dimension $\le 3$ are smoothable, the question is about manifolds of dimension $\ge 4$. Kirby and Siebenmann proved for $n\ge 6$ that every topological $n$-manifold admits a handle decomposition; this was extended to $n=5$ by Freedman and Quinn (I think, it is Quinn's paper "Ends of maps, III"). This applies to noncompact manifolds as well. Using this handle decomposition you can easily construct the required exhaustion (just use finitely many handles). This settles the problem in all dimensions but 4. </p> <p>Handle decomposition is known to fail in dimension 4, but there is an alternative argument: Take $N^5=M^4\times R$, construct an exhaustion of $N$ as above by compact submanifolds $S_i$. Now, Quinn proved in 1988 a topological transversality theorem in all dimensions ("Topological transversality holds in all dimensions"), which allows you to perturb each $S_i$ to $S_i'$ whose boundary is transversal to $M\times 0$. Then $S_i'\cap M$ will be the required exhaustion. </p> http://mathoverflow.net/questions/129144/purely-parabolic-kleinian-groups/129176#129176 Answer by Misha for Purely parabolic Kleinian groups Misha 2013-04-30T03:57:36Z 2013-04-30T03:57:36Z <p>Suppose that $G$ is a (not necessarily discrete) nonelementary group of isometries of the hyperbolic $n$-space. Then pairs of fixed points of loxodromic elements of $G$ are dense in $L(G)\times L(G)$, where $L(G)$ is the limit set of $G$. See Lemma 3.24 in "Hyperbolic Manifolds and Discrete Groups". I am quite sure that this is also proven in Maskit's book and in Beardon's book. The same argument works for arbitrary convergence groups acting on compacts, e.g. on ideal boundaries of proper geodesic Gromov-hyperbolic spaces. </p> <p>Just as a curiosity: There are 2-generated infinite purely elliptic (nondiscrete, of course) isometry groups of the hyperbolic $4$-space, but not of the hyperbolic 3-space. The existence is a cute application of the Tits alternative for $SU(2)$. </p> http://mathoverflow.net/questions/128803/tracial-triples/128833#128833 Answer by Misha for tracial triples Misha 2013-04-26T15:49:22Z 2013-04-26T15:49:22Z <p>Let $x, y, z$ be traces of $A, B, AB$ respectively. Define $$k(x,y,z)= x^2+y^2 + z^2 -xyz -2.$$ Then a triple of real traces $(x, y, z)$ is realizable in $SL(2,R)$, unless it is realizable in $SU(2)$, the latter happens if and only if $x, y, z\in [-2,2]$ and $k(x,y,z)\le 2$. See Goldman's paper "Topological components of spaces of representations", Inventiones, 1988. </p> http://mathoverflow.net/questions/128665/do-quasi-convex-hyperbolic-subgroups-remain-quasi-convex-after-adding-redundant-g/128669#128669 Answer by Misha for Do quasi convex hyperbolic subgroups remain quasi convex after adding redundant generators? Misha 2013-04-25T00:40:31Z 2013-04-25T00:40:31Z <p>First, one should decide what quasiconvexity (qc) means in the context of subgroups $H\subset G$, where $G$ is merely a semihyperbolic group, e.g., a RAAG. (I am assuming that generating sets are fixed.) Here are eight competing definitions:</p> <ol> <li><p>$H$ is qc in $G$ if there exists a constant $C$ so that every geodesic in $G$ connecting elements of $H$ is within distance $\le C$ from $H$. </p></li> <li><p>$H$ is qc in $G$ if there exists a constant $C$ so that for all $x, y\in H$ there exists a geodesic in $G$ connecting $x, y$, which is within distance $\le C$ from $H$. </p></li> <li><p>See e.g. <a href="http://berstein.wordpress.com/2011/02/24/well-behaved-subgroups-of-non-positively-curved-groups/" rel="nofollow">here</a> for a discussion of the following definition: $H$ is quasiconvex with respect to a fixed combing of $G$ if there exists a constant $C$ so that every combing path in $G$ connecting elements of $H$ is within distance $\le C$ from $H$. </p></li> <li><p>$H$ is qc in $G$ if $H$ is quasi-isometrically embedded in $G$.</p></li> <li><p>$H$ is qc in $G$ if there exists an $L$-Lipschitz retraction $G\to H$ (which is not assumed to be a homomorphism). </p></li> <li><p>Assuming that some "reasonable" notions of boundary for $G$ and $H$ are fixed: $H$ is qc in $G$ if there exists an equivariant injective continuous map $\partial H\to \partial G$. </p></li> <li><p>Assume that $G$ is a CAT(0) group, so it acts geometrically on a CAT(0) space $X$. Then $H$ is qc in $G$ if its image under the orbit map is qc in $X$ (in the sense of Definition 1). </p></li> <li><p>Same setup as in 7, only assume that the convex hull of an orbit of $H$ is within bounded distance from the orbit itself. </p></li> </ol> <p>All these notions are equivalent if $G$ is hyperbolic (and $X$ above is $CAT(-1)$) but not otherwise. (There are other related notions but I will limit myself to these eight.) </p> <p>As the question does not specify the notion of quasiconvexity, I will use Definition 1. Here is an example of instability of quasiconvexity for nonelementary hyperbolic subgroup $H\subset G$, where $G$ is a RAAG. </p> <p>Let $G=F_2\times Z$ and $H=F_2$ embedded in $G$ as the first factor. First, let's give $G$ the obvious generating set $a, b, c$, where $c$ is the central generator and $a, b\in H$. Then, $H$ is qc in $G$. Now, let's add the generator $d=ac$. Then $H$ is no longer qc in $G$. </p> http://mathoverflow.net/questions/128574/visual-boundaries-of-universal-covers-of-finite-volume-nonpositively-curved-manif/128647#128647 Answer by Misha for Visual boundaries of universal covers of finite-volume nonpositively curved manifolds Misha 2013-04-24T18:35:14Z 2013-04-24T18:35:14Z <p>Here is the proof in the case of locally symmetric spaces; you will have to combine it with "rank rigidity theorem" and Igor's comments to get a complete proof. </p> <p>Let $G$ be a semisimple Lie group, $K\subset G$ a maximal compact subgroup, $B\subset G$ a Borel subgroup; $X=G/K$ is the symmetric space of $G$. Let $r$ denote the real rank of $G$, i.e., the rank of $X$. Let $\Gamma\subset G$ be a lattice. Consider the action of $\Gamma$ on $G/B$, the complete flag manifold of $G$, also known as Furstenberg boundary of $X$. (Geometrically, one thinks of elements of $G/B$ as Weyl chambers at infinity in the Tits boundary of $X$.) The action of $\Gamma$ on $G/B$ is known to be minimal, i.e., every orbit is dense. You should be able to find a proof in Mostow's book on strong rigidity. The other ingredient you need is a theorem of Prasad and Raghunathan (Annals of Math., 1972) that $\Gamma$ contains a semisimple subgroup $\Lambda$ isomorphic to ${\mathbb Z}^r$: Such subgroup stabilizes (unique) maximal flat $F$ in $X$ and acts on $F$ cocompactly. It is elementary that end-points of axes of elements of $\Lambda$ are dense in the sphere $S^{r-1}$, the geometric boundary of $F$ (rational directions are dense in the unit sphere). </p> <p>Now, we can put all this together: Given a point $\xi$ in the geometric boundary of $X$, let $\sigma$ be a Weyl chamber at infinity containing $\xi$. Let $\tau$ be a chamber at infinity stabilized by the abelian subgroup $\Lambda\subset \Gamma$ as above. Let $\gamma_n\in \Gamma$ be a sequence such that $\lim_n \gamma_n(\tau)=\sigma$. Since ends of axes of elements $\lambda$ of $\Lambda$ are dense in $\tau$, their images under the sequence $\gamma_n$ will accumulate to $\xi$ as well. These images are ideal points of axes of conjugates $\gamma_n \lambda \gamma_n^{-1}$. qed </p> http://mathoverflow.net/questions/128332/fixed-point-set-of-an-isometric-group-action-on-an-hyperbolic-manifold/128469#128469 Answer by Misha for Fixed point set of an isometric group action on an hyperbolic manifold Misha 2013-04-23T12:27:35Z 2013-04-23T13:11:40Z <p>Here is the setup with which you are dealing: You have a smooth closed manifold $M$ and a finite cyclic branched covering $p: M\to M'$ ramified over a codimension 2 totally geodesic submanifold $V'\subset M'$, where $M'$ is hyperbolic. Let $V=p^{-1}(V')$. Furthermore, you know (from the construction) that $p: V\to V'$ is a diffeomorphism. In particular, $V$ is diffeomorphic to a closed hyperbolic manifold $V'$. Let $\beta$ denote the generator of the deck group of $p$. Now, you are trying to prove that the topological manifold $M$ does not admit a hyperbolic metric $g$. Suppose it does (smooth structure might change at this point). Then, by Mostow Rigidity theorem, $\beta: M\to M$ is homotopic to an isometry $\alpha: (M,g)\to (M,g)$. In particular, the fixed-point set $F$ of $\alpha$ is a closed hyperbolic submanifold. Note that $V$ is typically not connected, so you have to argue for each component $V_j$ of $V$. First, you note that each inclusion map $V_j\to M$ is $\pi_1$-injective (since this is the case for $V'$). Next, let's lift everything to the universal cover of $M$, which is supposed to be the hyperbolic $n$-space $H^n$, $M=H^n/\Gamma$. Then the lifts $\tilde\beta, \tilde\alpha$ can be chosen so that they agree on the sphere at infinity. The subgroup $\pi_1(V_j)\subset \Gamma$ consists of elements which commute with $\tilde\beta$. The same applies to $\pi_1(F_j)$ and to $\tilde\alpha$. Therefore, we have a natural isomorphism $\pi_1(V_j)\cong \pi_1(F_j)$. By construction, $V$ is diffeomorphic to $V'$, so $V_j$ is diffeomorphic to a hyperbolic manifold. Now, it follows from Mostow rigidity (if $dim(V)\ge 3$) or classification of surfaces (if $dim(V)=2$) that $V_j$ is diffeomorphic to $F_j$, since they are both diffeomorphic to hyperbolic manifolds with isomorphic fundamental groups. </p> <p>With a bit more care, the same argument goes through if you merely assume that $M$ is homotopy-equivalent to a hyperbolic manifold. </p> <p>Now, one can ask if something stronger holds: Let $M$ be a closed hyperbolic manifold and $\beta: M\to M$ be a finite order diffeomorphism. Let $\alpha: M\to M$ be an isometry homotopic to $\beta$. Is it true that fixed-point sets of $\alpha, \beta$ are diffeomorphic? Note that both orbifold $M/(\alpha), M/(\beta)$ are aspherical and have isomorphic fundamental groups. In dimension 3 it suffices to conclude that they are diffeomorphic. In dimensions $\ge 5$, it probably follows from some equivariant form of Farrel and Jones but I am not sure (I have not looked at their papers for a long time). Situations in smooth category and dimension 4 seem totally unclear. </p> <p>Lastly: The paper you are reading has two constructions of negatively curved manifolds not homotopy-equivalent to locally symmetric ones. The second construction is less known since people usually get tired from reading the first one. Some details for the second construction could be found <a href="http://www.math.ucdavis.edu/~kapovich/EPR/gt.pdf" rel="nofollow">here</a>. Namely, I explain why manifolds in the 2nd construction are not homotopy-equivalent to hyperbolic ones. </p> http://mathoverflow.net/questions/127367/is-the-group-of-integer-points-of-rm-son-1-maximal/127663#127663 Answer by Misha for Is the group of integer points of ${\rm SO}(n,1)$ maximal? Misha 2013-04-15T23:22:03Z 2013-04-15T23:22:03Z <p>Here is a proof for even $n$, I am not quite sure about odd $n$ since $G=SO(n,1)$ is not an adjoint group and weak approximation might fail in this case (I do not know enough about this, you would need to ask experts, like Andrei Rapinchuk). First, note that the commensurator of $G({\mathbb Z})$ in $G({\mathbb R})$ is $G({\mathbb Q})$ (as it has to preserve the set of rational lines in the light cone: They are the fixed points of unipotent elements of $G({\mathbb Z})$). </p> <p>Claim. Let $G=SO(n,1)$, $n$ is even. Then $G({\mathbb Z})$ is a maximal subgroup of $G({\mathbb Q})$ and, hence, in $G({\mathbb R})$. </p> <p>Proof. It is known that <code>$$G({\mathbb Z}_p)= G({\mathbb Q}_p)\cap SL_{n+1}({\mathbb Z}_p)$$</code></p> <p>is a maximal compact subgroup of $G({\mathbb Q}_p)$ for all $p$. By weak approximation (since $G$ is an adjoint group), $G({\mathbb Q})$ is dense in $G({\mathbb Q}_p)$, which implies that $G({\mathbb Z})$ is dense in $G({\mathbb Z}_p)$. If $G({\mathbb Z})$ is not a maximal subgroup of $G({\mathbb Q})$ and $G({\mathbb Z})&lt;\Gamma&lt; G({\mathbb Q})$ is a larger discrete subgroup, then $\Gamma$ is not contained in $G({\mathbb Z}_p)$ for some prime $p$. The closure of $\Gamma$ in $G({\mathbb Q}_p)$ is compact (since $\Gamma$ is a finite extension of $G({\mathbb Z})$) and strictly contains $G({\mathbb Z}_p)$, which contradicts maximality of $G({\mathbb Z}_p)$. qed </p> http://mathoverflow.net/questions/127530/surfaces-filled-densely-by-a-geodesic/127552#127552 Answer by Misha for Surfaces filled densely by a geodesic Misha 2013-04-14T18:48:30Z 2013-04-14T18:48:30Z <p>Donnay and Pugh proved that every embedded surface $S\subset R^3$ can be $C^0$-perturbed so that the new metric has ergodic geodesic flow, see <a href="http://repository.brynmawr.edu/cgi/viewcontent.cgi?article=1009&amp;context=math_pubs" rel="nofollow">here</a>. In particular, the new metric will have dense geodesics (moreover, "generic" geodesics will be dense in the unit tangent bundle). </p> http://mathoverflow.net/questions/127371/quasi-conformal-area-preserving-homomorphisms-of-the-disc/127414#127414 Answer by Misha for quasi conformal, area preserving homomorphisms of the disc Misha 2013-04-12T20:55:57Z 2013-04-13T22:11:00Z <p>Claim. A QS homeomorphism $f$ of the circle extends to a QC area preserving map of the disk if and only if $f$ is BL (bi-Lipschitz). </p> <p>Proof. </p> <ol> <li><p>Suppose that $f: S^1\to S^1$ is a QS map which admits an area-preserving quasiconformal extension $F: D^2\to D^2$. Then it immediately follows from the definition of quasiconformality that $F$ has to be (globally) bi-Lipschitz. In particular, $f$ was bi-Lipschitz to begin with. </p></li> <li><p>Suppose now that $f$ is a BL homeomorphism of the unit circle. I claim that it extends to a BL area preserving homeomorphism $F$ of the disk. By looking at the BL flow from the identity to $f$, we see that the problem reduces to the case of Lipschitz vector fields $v$ tangent to the circle which we have to extend to Lipschitz divergence-free vector fields $V$ on the disk. By removing a point on the circle, we reduce the question to the case when $v=u'$, where $u$ is a $C^{1,1}$ function on the circle (maybe minus a point). Now, extend $u$ to a harmonic function $U$ on the disk. Then the gradient $V$ of $U$ has zero divergence and is Lipschitz. Furthermore, $V$ will be Lipschitz at the boundary (except maybe for one point), to it will extend to the remaining point as well. Qed</p></li> </ol> <p>I was assuming here that $f$ is orientation-preserving. However, by composing with a symmetry of the disk, the general case reduces to this one, provided that you allow QC maps to reverse orientation. </p> http://mathoverflow.net/questions/126807/when-are-brieskorn-manifolds-homeomorphic/126891#126891 Answer by Misha for When are Brieskorn Manifolds Homeomorphic? Misha 2013-04-08T18:31:59Z 2013-04-09T00:48:26Z <p>This is by no means a complete answer but, rather, a DIY suggestion:</p> <p>Let $B$ be a $2k+1$-dimensional Brieskorn manifold. Then $B$ is $k-1$-connected. C.T.C. Wall wrote in <a href="http://www.sciencedirect.com/science/article/pii/0040938367900201#" rel="nofollow">"Classification problems in differential topology—VI. Classification of (s−1)-connected (2s+1)-manifolds"</a> a complete list of invariants (up to diffeomorphism) for such manifolds (there are few exceptions, but one should be able to deal with them on case-by-case basis), assuming, of course, that $k\ge 2$, so the fundamental group is trivial. The invariants are (mostly) of homological nature, so you should compare them with computations done by Hirzebruch's and Milnor, to see if you get enough information from there to determine a complete list of diffeomorphism invariants for Brieskorn manifolds in terms of the parameters $a_i$. </p> <p>It is quite possible that such analysis was made by Alan Durfee in his 1971 thesis "Diffeomorphism classification of isolated hypersurface singularities". You may want to ask Durfee (he is at Mount Holyoke College) for a copy. I have no idea why his thesis was never published, but people refer to it quite a bit. </p> <p>Now, if $k=1$, then the situation is quite different and $B$ is a Seifert manifold. Topology of such manifolds is completely determined by "Seifert invariants" which were first computed by Milnor <a href="http://www.maths.ed.ac.uk/~aar/papers/milnbries.pdf" rel="nofollow">here</a> if there are no singular fibers (which reduces to computing genus of the base ond Euler number of the fibration) and, in general by Neumann and Raymond <a href="http://www.maths.ed.ac.uk/~aar/papers/neumraym.pdf" rel="nofollow">here</a>, in terms of the parameters $a_i$, following an earlier paper by Neumann which I do not have access to. So, answering your question in this case, is still a DIY project, working through the formulae in the paper by Neumann and Raymond. </p> <p>Addendum: <a href="http://www.math.columbia.edu/~neumann/preprints/neumann011.pdf" rel="nofollow">Here</a> is the link to a scan of Neumann's thesis. Since it is a scan, it is harder to read, but, unlike the paper of Neumann and Raymond, it deals specifically with Brieskorn manifolds, not with complete intersections of such. </p> <p>Here is the description (taken fron Neumann's thesis) of a complete set of topological invariants of $\Sigma=\Sigma(a_0,a_1,a_2)$ from the vector $(a_0,a_1,a_2)$ in the <em>generic case</em> (for nongeneric cases see Corollary 9.2 in Neumann's thesis). This is not at all pretty (to say the least), but it is what it is. Define numbers $$d=gcd(a_0,a_1,a_2),$$ $$a_i'= \frac{1}{a_i} lcm(a_0,a_1,a_2),$$ <code>$$t_i= gcd(a_j', a_k'), \{i,j,k\}=\{0,1,2\}$$ $$s_i= \frac{1}{d} gcd(a_j, a_k), \{i,j,k\}=\{0,1,2\}$$</code> $$g= \frac{1}{2}(d^2 s_0 s_1 s_2- d(s_0+s_1+s_2)) +1.$$</p> <p>Genericity assumption: $t_0, t_1, t_2$ are all $\ne 1$. Now, find integers $\beta_i'$ so that $$0\le \beta_i'&lt; t_i, \quad \beta_i'a_i' = 1 (mod\ t_j)$$ and set $$b= \frac{d}{t_0t_1t_2}(1- \sum_{i=0}^2 \beta_i'a_i')$$</p> <p>Then the tuple <code>$$(g; b; \{ds_0(t_0, \beta_0'), ds_1(t_1, \beta_1'), ds_2(t_2, \beta_2')\})$$</code> is a complete topological invariant of $\Sigma$. Here $$ds_i(t_i, \beta_i')= (ds_i t_i, ds_i \beta_i').$$ Topological meaning of some of the quantities in this tuple:</p> <ol> <li><p>$g$ is the genus of the base-orbifold $O$ of the Seifert fibration on $\Sigma$. </p></li> <li><p>Under our genericity assumptions, the base-orbifold $O$ will have $3$ singular points of the orders $$ds_i t_i, i=0, 1, 2.$$ The numbers $$ds_i \beta_i'$$ define the second set of invariants for the Seifert fibration at the singular fibers. </p></li> <li><p>The number $b$ is responsible for the Euler number of the Seifert fibration (I did not bother to write a precise formula for the transition between these invariants, maybe it literally is the Euler number). </p></li> </ol> <p>Given how complex this description is, it is very likely that the complete sets of topological/smooth invariants in higher dimensions is much messier. </p> <p>Now, consider the special case where the numbers $a_0, a_1, a_2$ are pairwise coprime, and greater than $1$. Then $$g=0, t_i=a_i, d=1, s_i=1, ds_it_i=a_i.$$ In particular, the numbers $a_i$ are orders of cone-points of the base-orbifold. In particular, for coprime numbers $a_i$, the vector $(a_0, a_1, a_2)$ is the complete topological invariant of $\Sigma$. This was the original comment made by myself and Bruno: We both missed the coprimality condition. </p> <p>In this setting, your "genus" equals $2$ (I still do not know why do you call it "genus"; I would write instead: $$\frac{1}{2}\left(\frac{d}{\tau} -l\right) + 1,$$ then, at least in the coprime case it matches the genus of the base-orbifold.) Now, it is clear that this number is insufficient to determine the topology of $\Sigma$. </p> http://mathoverflow.net/questions/126654/every-continuous-function-is-homotopic-to-a-locally-lipschitz-one/126785#126785 Answer by Misha for Every continuous function is homotopic to a locally Lipschitz one Misha 2013-04-07T16:14:30Z 2013-04-07T16:14:30Z <p>Such approximation is possible under some mild assumptions about domain and range.</p> <p>For the domain you want to have structure of a finite dimensional structure of a metric simplicial complex of locally bounded geometry. For example, a Riemannian manifold or Alexandrov space would do. For the target you should impose some conditions implying local linear contractibility, for instance, a space which is locally CAT(k), where $k&lt;\infty$ would suffice. The proof is based on barycentric maps of smplices, which you can find in the paper of Bruce Kleiner, "The local structure of length spaces of curvature bounded above", Math. Z. 1999. </p> <p>The construction of Lipschitz approximation is the same as cellular approximation in algebraic topology. First, approximate your map on the set of vertices. Then extend to simplices by induction on skeleta, using barycentric simplices as in Kleiner's paper. </p> <p>Some of this might even work if domain is infinite dimensional, but you would need to control the Lipschitz constant for the barycentric maps. </p> http://mathoverflow.net/questions/126489/is-there-non-simple-connected-projective-varietyover-c-with-trivial-etale-funda/126774#126774 Answer by Misha for Is there non-simple-connected projective variety(over C) with trivial etale fundamental group? Misha 2013-04-07T14:07:08Z 2013-04-07T15:03:49Z <p>There are several classes of spaces for which this question can be asked, here are the answers:</p> <ol> <li><p>Compact complex-projective manifolds (also frequently called <em>manifolds admitting flat complex-projective structures</em>): These are n-manifolds admitting an atlas where transition maps are elements of $PGL(n+1, C)$. For such manifolds, either the fundamental group is trivial or it admits a nontrivial projective-linear representation (holonomy of the structure). It follows from Malcev's theorem, that the group always contains a finite-index subgroup (different from the original group). Thus, the answer to OP's question in this setting in NO. </p></li> <li><p>Complex-projective varieties. It was observed by Serre (and many others) that every finitely-presented group $\pi$ appears as fundamental group of such a variety. One does not need Simpson's paper for this (he was answering a much harder question). Here is the construction. Take finite simplicial complex with the given fundamental group. Embed it as a subcomplex of a standard affine k-simplex. Replace each face of this subcomplex with its complex-projective span. Take the union of these subspaces of $CP^k$. That's your variety. Then, as Ricky noted, the answer to OP's question is YES. </p></li> <li><p>Smooth complex-projective varieties. Then it becomes a well-known open problem. It is also open for compact Kahler manifolds. </p></li> <li><p>Compact complex manifolds. It was proven by Taubes in 1992 that every finitely-presented group is the fundamental group of such a 3-dimensional manifold. Thus, the answer again is, YES. </p></li> </ol> http://mathoverflow.net/questions/126217/name-for-a-class-of-parabolic-subgroups Name for a class of parabolic subgroups Misha 2013-04-02T00:18:05Z 2013-04-02T02:00:39Z <p>This is a reference request for a (the) name of the following class of parabolic subgroups of a complex simple Lie group $G$: </p> <p>Recall that parabolic subgroups of $G$, containing fixed Borel subgroup, are determined by faces $c$ of the positive Weyl chamber. Interiors of <em>some</em> of these faces contain coroots of the group $G$. Is there a name for the class of parabolic subgroups $P$ and the corresponding partial flag manifolds $G/P$ coming from such faces? </p> <p>(I do not know the name even in the special case when such $P$ is maximal parabolic. Such maximal parabolics exist except for the type $A$ and $D_3$ groups: Most of the time they correspond to the highest (co)root.) My tentative name for $P$'s and $G/P$'s is "root type parabolic subgroups and partial flag manifolds". In a paper I am cowriting, such subgroups and manifolds behave better than other parabolics. </p> http://mathoverflow.net/questions/126174/intersection-of-conjugates-of-subgroups-in-free-groups/126209#126209 Answer by Misha for Intersection of conjugates of subgroups in free groups Misha 2013-04-01T20:58:59Z 2013-04-01T20:58:59Z <p>Here is the proof of Fact 1 which uses only facts that were known already to Klein and Poincare (or at least to Dehn and Nielsen):</p> <ol> <li><p>Every finitely generated free group can be realized as a discrete group of isometries $F$ of the hyperbolic plane $H^2$, so that the quotient $H^2/F$ has finite area but noncompact; thus, $F$ contains parabolic elements. </p></li> <li><p>If $A\subset F$ is a finitely generated infinite index subgroup then the limit set of $A$ is a proper subset of the unit circle. Indeed, being finitely generated Fuchsian group, $A$ is geometrically finite, i.e., the quotient of the convex hull of its limit set by $A$ has finite area. (This is the only mildly nontrivial ingredient in the entire proof.) In our setting, this would mean that covering map $H^2/A\to H^2/F$ is between two surfaces of finite area and so has to be finite. This contradicts the assumption that $|F:A|=\infty$. Thus, the limit set $L(A)$ of $A$ has empty interior in $S^1$. </p></li> <li><p>Fixed points of parabolic elements of $F$ are dense in $S^1$. </p></li> <li><p>Now, suppose that $A, B$ are finitely generated subgroups of infinite index. By combining 2 and 3 we can find a parabolic element $g\in F$ whose fixed point $p$ is not in the union of the limit sets $L(A) \cup L(B)$ of the groups $A$ and $B$. (Similarly, one can use hyperbolic elements of $F$ since their fixed pairs are dense in $S^1\times S^1$.) Since $g^n, n\to infinity$ converges to $p$ uniformly on compacts away from $p$, there exists $n$ so that $$g^p(L(A))\cap L(B)=\emptyset.$$ It follows that for $f=g^n$ the groups $fAf^{-1}, B$ have trivial intersection (since fixed points of infinite order elements of a Fuchsian group belong to its limit set). </p></li> </ol> http://mathoverflow.net/questions/125406/relation-between-the-character-variety-of-a-knot-k-subset-m-and-that-of-m/125472#125472 Answer by Misha for Relation between the Character variety of a knot $K\subset M$ and that of $M$ Misha 2013-03-24T17:54:49Z 2013-03-24T17:54:49Z <p>I do not think there is a particularly nice answer to this. Below is a suggestion on how you could proceed.</p> <p>Consider representations $\rho$ of a finitely generated group $\pi$ to $Sl(2, C)$. First of all, the condition that $Tr(\rho(g))=2$ is necessary but not sufficient for $\rho(g)=1$. You can see this by looking at unipotent matrices in $Sl(2, C)$. If $\rho$ maps $g$ to the identity, then for every element $h$ in the normal closure $H$ of $g$ in the ambient group $\pi$, you have $Tr(\rho(h))=2$. One can show that this condition is necessary and sufficient for $\rho(g)=1$. By the Nullstellensatz, it suffices to check this condition only for finitely many elements $h$ in the normal closure $H$. </p> <p>Now, it should be possible to find an explicit set of such elements of $H$. Doing so is a worthwhile task, I do not think anybody computed this set, it will depend on the word which represents $g$ in terms of generators of $\pi$. If you manage to do this, you can then apply the answer to the special case of generalized knot groups as in your question where $g=m$. Even in the case of knots in the sphere the answer is very likely to be very ugly, and will depend on the knot diagram. However, this is the best I can suggest. </p> http://mathoverflow.net/questions/125217/when-are-isometry-groups-of-hyperbolic-3-manifolds-finite/125278#125278 Answer by Misha for When are isometry groups of hyperbolic 3-manifolds finite? Misha 2013-03-22T14:27:57Z 2013-03-22T14:50:42Z <p>Here is the detailed answer. First, you have to assume that your hyperbolic manifold is complete and has finitely generated fundamental group, otherwise you will get no answer except for the tautological one. Thus, you are dealing with a finitely generated torsion free Kleinian group G in SO(3,1). If such a group is elementary then the normalizer of the group is not discrete, consider for instance the case of an abelian discrete group of isometries. </p> <p>Thus, let us exclude elementary groups G. Then the normalizer is discrete (consider its action on the limit set of G). Now, if G is also geometrically finite, then the normalizer is a finite extension of G. To see this, consider convex core of your hyperbolic manifold and observe that its thick part is compact. In the geometrically infinite case, there is one example you should be aware of: Take finite volume hyperbolic 3-manifold which fibers over the circle and take M to be its infinite cyclic cover associated with the fiber. Then the isometry group of M contains infinite cyclic group. Covering Theorem of Thurston and Canary in addition to the positive solution of tameness conjecture by Agol, Calegari and Gabai shows that such M's are the only examples where the isometry group is infinite. Moreover, the isometry group in this case is a finite extension of an infinite cyclic group. (The latter is immediate.)</p> <p>A side note: If you consider higher dimensional hyperbolic manifolds, the situation in geometrically finite case is essentially the same except you have to assume that the limit set is not contained in a round sphere of codimension 2. In geometrically infinite case, there are more examples one has to exclude and there is no even conjectural classification of exceptions. </p> http://mathoverflow.net/questions/124965/are-all-free-groups-linear-i-e-admit-a-faithful-representation-to-gln-k-for/124972#124972 Answer by Misha for Are all free groups linear, i.e., admit a faithful representation to GL(n,K) for some field K ? Misha 2013-03-19T15:29:03Z 2013-03-20T17:21:03Z <p>Free group of rank $c$ embeds in $Sl(2, F(t))$ where $F$ is a field of cardinality $c$. </p> <p>Edit: Here is the detailed argument which, as Yves noted in his comment, proves a stronger result. </p> <p>Theorem. Let $L$ be a field which is not an algebraic extension of a finite field and let $c$ be the cardinality of $L$. Then the free group of rank $c$ embeds in $SL(2, L)$. </p> <p>Proof. Let $P$ be the prime field of $L$; then $L$ has the form $$P\subset E \subset L$$ where $E$ is a purely transcendenetal extension of $P$ and $L$ is an algebraic extension of $E$. Under our assumptions, $E$ and $L$ have the same cardinality, thus, it suffices to consider the case when $L=E$. Then $L$ is isomorphic to the functional field $L=F(t)$, where $F$ is a subfield of $L$. I will consider the case when $F$ is infinite since otherwise $L$ is countable and everything is clear (as the question reduces to the case of free groups of finite rank). Thus, $F$ has the same cardinality $c$ as $L$. </p> <p>Let $T$ be the Bruhat-Tits building associated with $G=SL(2, L)$: This building is a simplicial tree with the path-metric $d$, where every edge has unit length. The group $G$ acts on $T$ by simplicial automorphisms with the kernel $\pm 1$. Detailed description and properties of $T$ and the action of $G$ are in Serre's book "Trees." </p> <p>Let $v\in T$ be the vertex stabilized by $K=SL(2, O)$, where $O=F[t]$ is the ring of polynomial functions in $t$. Then the link $L_v$ of $v$ in $T$ is naturally identified with the projective line over $F$ (so that $K$ acts on $L_v$ by linear-fractional transformations). In particular, the group $K$ acts transitively on pairs of distinct points in $L_v$. Let $g\in G\setminus K$ be a diagonal matrix with the axis $\gamma\subset T$. Then $\gamma$ contains $v$ and $g$ acts on $\gamma$ as a translation by some even integer distance $\ge 2$. In view of transitivity of the action of $K$ on pairs noted above, there exists a subset $K_o\subset K$ of cardinality $c$ so that the elements $g_k=kgk^{-1}$, $k\in K_o$ have axes $k(\gamma)$ with the property that the 2-point sets $$k(\gamma) \cap L_v, k\in K_o,$$ are pairwise disjoint. (Call this property D.) Now, I claim that the elements $g_k, k\in K_o$, are free generators of a free subgroup of $G$. The proof is rather standard. For each $k\in K_o$ let $D_k\subset T$ denote the Dirichlet fundamental domain for the cyclic group $\langle g_k \rangle$: <code>$$D_k=\{ x\in T: d(x, g_k^m(v))&gt; d(x, v), \forall m\in {\mathbb Z} \setminus 0\}.$$</code> Since each $g_v$ translates $v$ at least by $2$, and in view of Property D above, the domains $D_k$ have pairwise disjoint complements. Thus, Tits' ping-pong argument (from his proof of the Tits alternative) applies in this setting and the subgroup of $G$ generated by the elements $g_k$ is indeed free with free generators $g_k$. qed. </p> <p>Note that one has to exclude fields $L$ with are algebraic extensions of finite fields, since in this case the group $GL(n, L)$ is torsion (for every finite $n$) and, hence, cannot contain a free subgroup. </p> http://mathoverflow.net/questions/123198/can-the-sl-2-character-variety-of-a-three-manifold-be-nonreduced/123259#123259 Answer by Misha for Can the SL_2 character variety of a three-manifold be nonreduced? Misha 2013-02-28T19:22:32Z 2013-03-18T20:07:15Z <p>There is actually an old (ca 1986) example of nonreduced $SL(2, {\mathbb C})$-representation scheme of a 3-manifold group. Take an oriented Seifert manifold $M$ which fibers over the $S^2(3,3,3)$ orbifold (sphere with three cone points of order 3). The fundamental group of the base-orbifold is von Dyck group with presentation $$\Gamma=\langle a, b, c | a^3, b^3, c^3, abc\rangle.$$ It is an old observation of Lubotzky and Magid (in their book "Representation varieties of finitely-generated groups") that the $SL(2, {\mathbb C})$-representation scheme of $\Gamma$ is nonreduced at a representation $\rho_0$ whose image is isomorphic to ${\mathbb Z}_3$ ($\rho_0$ sends all generators to an order $3$ element). Namely, $H^1(\Gamma, Ad \rho_0)$ is $1$-dimensional, while the representation $\rho_0$ s locally rigid. There is a nice geometric explanation of this phenomenon: Take a spherical equilateral triangle in $S^3=SU(2)$ contained in a great circle. Then this triangle is locally rigid but admits a nontrivial 1st order deformation in $S^3$. Now, $\rho_0$ lifts to an $SL(2, {\mathbb C})$-representation $\tilde\rho_0$ of the central extension $\pi=\pi_1(M)$ of the group $\Gamma$ (killing the center of $\pi$). This is your example. It is locally rigid but has 1-st order nontrivial infinitesimal deformations. A drawback of this example is that the image has large centralizer. </p> <p>Below is a more difficult "universality" result:</p> <p>Theorem 1. Let $X$ be an affine scheme over ${\mathbb Q}$ and $x\in X$ be a rational point. Then there exists an open subscheme $X'\subset X$ containing $x$, a natural number $N$, a closed 3-dimensional manifold $M$ with fundamental group $\pi$, a unitary representation $\rho: \pi\to SU(2)\subset SL(2, {\mathbb C})$ (whose image is dense in $SU(2)$) and an open subscheme $$R'\subset Hom(\pi, SL(2, {\mathbb C}))$$ containing $\rho$, so that $R'$ admits a regular etale covering over $X'\times SL(2, {\mathbb C})^N$ (with abelian group of deck transformations) and the covering sends $\rho$ to $x$. In particular, the centralizer of $\rho(\pi)$ is ${\pm 1}$. </p> <p>For instance, the analytic germ of the character scheme $$Hom(\pi, SL(2, {\mathbb C}))//SL(2, {\mathbb C})$$ at $[\rho]$ could be isomorphic to the germ at $0$ of the scheme <code>$$\{x^{100}=0\} \times {\mathbb C}^{k}$$</code> (for some $k$). In particular, the character scheme of 3-manifold groups could be nonreduced at points of Zariski density. </p> <p>Proof of Theorem 1 could be now found <a href="http://arxiv.org/abs/1303.2347" rel="nofollow">here</a>. The proof is a combination of my old work with Millson (see <a href="http://www.math.ucdavis.edu/~kapovich/EPR/ihes.pdf" rel="nofollow">here</a>) with the recent theorem of Panov and Petrunin, which deserves to be better known:</p> <p>Theorem 2. For every finitely presented group $\Gamma$ there exists a closed 3-dimensional orbifold $O$ so that the fundamental group of the underlying space of $O$ is isomorphic to $\Gamma$. </p> <p>It is a difficult open problem if Theorem 1 holds for 3-manifolds which are homology spheres. </p> http://mathoverflow.net/questions/124753/are-there-results-from-gauge-theory-known-or-conjectured-to-distinguish-smooth-fr/124801#124801 Answer by Misha for Are there results from gauge theory known or conjectured to distinguish smooth from PL manifolds? Misha 2013-03-17T17:37:02Z 2013-03-18T05:21:18Z <p>First, it follows from the work of Kirby and Siebenmann that in dimensions $\le 6$ PL and DIFF categories are equivalent. In particular, if you are working in dimension 4 (where gauge-theoretic invariants are mostly used) then the answer to your your question is negative. Starting in dimension 7, there are smooth manifolds which are PL-equivalent but not diffeomorphic. Milnor's exotic 7-spheres are the first examples of such manifolds. The smooth structures on Milnor's spheres are distinguished via index and 1st Pontryagin class. Whether you consider such invariants gauge-theoretic or not, depends on how broadly you interpret gauge theory. For instance, characteristic classes of smooth manifolds can be defined via differential forms, i.e., Chern-Weil theory, or as indices of some elliptic operators. Does this qualify as gauge theory? (You can lift forms from, say, tangent bundle to the principal bundle- frame bundle, if you so desire.) The point of usage of gauge theory in dimension 4 is that the "traditional" topological invariants turned out to be insufficient, so one considers spaces of connections satisfying some differential equation (like self-duality) and uses such spaces to derive some smooth invariants of 4-manifolds. As far as I know, nobody used this viewpoint in higher (i.e., at least 7) dimensions, since there was no need for it. </p> http://mathoverflow.net/questions/124725/nilpotent-subgroups-of-uniform-finite-index/124734#124734 Answer by Misha for Nilpotent subgroups of uniform finite index Misha 2013-03-16T23:11:36Z 2013-03-17T17:59:19Z <p>Start with this paper: A. I. Mal′cev, On a class of homogeneous spaces, Izvestiya Akad. Nauk. SSSR. Ser. Mat. 13 (1949), 9–32 (look also <a href="http://www.math.utoronto.ca/~vtk/NILPnew1.pdf" rel="nofollow">here</a>). This reduces your problem to the case when $\Gamma$ is a lattice in $N\rtimes K$. Next, observe that projection of $\Gamma$ to $K$ has to be a finite group (this should be in M.Raghunathan, "Discrete subgroups of Lie groups"), the result is due to Auslender, you can read his original paper <a href="http://www.jstor.org/discover/10.2307/1969945?uid=3739560&amp;uid=2&amp;uid=4&amp;uid=3739256&amp;sid=21101863815891" rel="nofollow">here</a>. Now, use <a href="http://en.wikipedia.org/wiki/Jordan%E2%80%93Schur_theorem" rel="nofollow">Jordan-Schur Theorem</a>: For every compact group $K$ there exists a number $j=j(K)$ so that every finite subgroup of $K$ contains an abelian subgroup of index $\le j$. This is also in Raghunathan's book, see also wikipedia article <a href="http://en.wikipedia.org/wiki/Jordan%E2%80%93Schur_theorem" rel="nofollow">here</a> and Tao's blog <a href="http://terrytao.wordpress.com/2011/10/05/the-jordan-schur-theorem/" rel="nofollow">here</a>. </p> http://mathoverflow.net/questions/124396/finite-index-subgroup-of-a-fuchsian-group/124684#124684 Answer by Misha for finite index subgroup of a fuchsian group Misha 2013-03-16T11:04:59Z 2013-03-16T14:28:25Z <p>On Yves' request, here is a proof of residual finiteness of non-finitely generated Fuchsian groups. (I do not know a reference for this: In general, people do not like working with infinitely generated groups, so, it is possible that nobody bothered writing a proof, even though many people could. The result could be in Beardon's or Maskit's book though, I did not check.) A side remark: Residual finiteness of Fuchsian groups was a partial motivation for <a href="http://mathoverflow.net/questions/89439/non-residually-finite-matrix-groups" rel="nofollow">this question</a>. </p> <p>A Fuchsian group is a discrete group $\Phi$ of isometries of hyperbolic plane $H^2$. Then $\Phi$ contains an index 2 subgroup $\Gamma$ which preserves orientation on the hyperbolic plane and, hence, $\Gamma &lt; PSL(2, R)$. Since residual finiteness is a commensurablity invariant, it suffices to consider the case of discrete subgroups of $PSL(2,R)$. Let $O=H^2/\Gamma$ be the quotient orbifold. Its singular set is a discrete subset <code>$\Sigma=\{x_i: i\in I\}$</code> of the underlying space $X$ of $O$. I will assume that $\Gamma$ is not finitely generated, thus, $O$ is noncompact. </p> <p>Let $T\subset X$ be an infinite locally finite, 1-ended properly embedded tree (with geodesic edges) which contains $\Sigma$. Let $U$ be a small neighborhood (with smooth boundary) of $T$ in $X$ which admits a retraction to $T$. Every boundary component of $U$ is contractible. Let $O_U$ be the suborbifold of $O$ supported by $U$. Thus, by Seifert-van Kampen theorem for orbifolds applied to this situation, the fundamental group $\Gamma$ of $O$ splits as a free product $$\pi_1(O_U) * \prod^*_i \pi_1(S_i)$$ where $S_i$'s are components of $X\setminus U$, and $\prod^*$ means free product. Now, since each surface $S_i$ is noncompact, by a theorem of Whitehead, it is homotopy-equivalent to a graph (see the discussion <a href="http://mathoverflow.net/questions/18454/fundamental-groups-of-noncompact-surfaces" rel="nofollow">here</a>). Hence, $\Gamma \cong \pi_1(O_U) * F$, where $F$ is a free group. Lastly (again by Seifert-van Kampen theorem applied to $O_U$), the group $\pi_1(O_U)$ is a free product of finite cyclic groups $Z_{n_i}$, where $n_i$ is the order of the singular point $x_i\in \Sigma$. Thus, $\Gamma$ is the free product of countably many cyclic groups (some of which might be finite and some infinite). </p> <p>The converse is also true: If $\Gamma$ is a countable product of cyclic groups $C_i$, then $\Gamma$ is isomorphic to a discrete subgroup of $PSL(2, R)$. Namley, make each $C_i$ act isometrically on $H^2$ with the fundamental domain $D_i$, so that the sets $cl(H^2 -D_i)$ are pairwise disjoint. Now, take the subgroup $\Phi$ of $PSL(2, R)$ generated by the subgroups $C_i$. By Klein combination theorem (also known as the "ping-pong argument") $\Phi$ is discrete (its fundamental domain is the intersection of $D_i$'s) and isomorphic to $\Gamma$. </p> <p>Now, suppose that $\Gamma$ is a group which is a countable free product of residually finite groups $\Gamma_i, i\in {\mathbb N}$. Let $S\subset \Gamma$ be a finite subset not containing the identity. Then $$S\subset \prod^*_{j\in J} \Gamma_j$$ where $J\subset I$ is a finite subset. Hence, $S$ projects bijectively to a subset $P$ in the quotient $$\Lambda=\Gamma/\langle \langle \bigcup_{i\notin J} \Gamma_i\rangle \rangle.$$ Now, $\Lambda$ is residually finite (as a finite free product of residually finite groups). Hence, $\Lambda$ contains a finite-index subgroup $\Lambda'$ disjoint from $P$. Taking preimage of $\Lambda'$ under the epimomorphism $\Gamma \to \Lambda$, we obtain a finite index subgroup in $\Gamma$ disjoint from $S$. This argument, of course, applies to free products of cyclic groups, thereby proving residual finiteness of non-finitely generated Fuchsian groups. </p> http://mathoverflow.net/questions/124667/topological-space-with-certain-properties/124671#124671 Answer by Misha for Topological space with certain properties Misha 2013-03-16T03:55:16Z 2013-03-16T04:19:02Z <p>See Theorem 1.4 <a href="http://www.math.nus.edu.sg/~matberic/jonhAJB03306.pdf" rel="nofollow">here</a>: For every $n>1$ there exists an acyclic group $G_n$ with finite $n$-dimensional $X_n=K(G_n,1)$ so that cohomological dimension $cd(G_n)$ of $G_n$ equals $n$. Now, take the complex $X_n$ as an example for your question. If $X_n$ were homotopy-equivalent to a complex of dimension $n-1$, then $cd(G_n)$ would be less than $n$, which is a contradiction. </p> http://mathoverflow.net/questions/124013/siegel-set-in-son-1-modulo-integer-points/124019#124019 Answer by Misha for Siegel set in SO(n,1) modulo integer points? Misha 2013-03-08T21:04:55Z 2013-03-08T21:04:55Z <p>The number of cusps could be more than 1, see <a href="http://www.ma.utexas.edu/users/areid/allflat_new.pdf" rel="nofollow">here</a>, remark on page 294. </p> http://mathoverflow.net/questions/131175/closed-geodesic-loops-around-points-in-compact-manifolds/131181#131181 Comment by Misha Misha 2013-05-20T02:08:48Z 2013-05-20T02:08:48Z Anton: You need to use minimax argument here since you are looking for an unstable critical point of energy. http://mathoverflow.net/questions/131115/is-there-any-result-concerning-on-the-metric-dimension-of-inverse-limit/131118#131118 Comment by Misha Misha 2013-05-19T13:20:07Z 2013-05-19T13:20:07Z Good, now it works. http://mathoverflow.net/questions/130959/reference-request-affine-transforms-circle-inversion/131005#131005 Comment by Misha Misha 2013-05-19T12:40:47Z 2013-05-19T12:40:47Z Thank you, Robert, this looks quite plausible. At least, this seems to show that for every $k$, every orbit is dense in $(S^n)^k$ minus diagonals. http://mathoverflow.net/questions/131115/is-there-any-result-concerning-on-the-metric-dimension-of-inverse-limit/131118#131118 Comment by Misha Misha 2013-05-19T04:29:03Z 2013-05-19T04:29:03Z This construction does not seem to be an inverse limit of metric spaces in Petrunin's sense (<a href="http://mathoverflow.net/questions/15948/inverse-limit-in-metric-geometry" rel="nofollow" title="inverse limit in metric geometry">mathoverflow.net/questions/15948/&hellip;</a>). http://mathoverflow.net/questions/131115/is-there-any-result-concerning-on-the-metric-dimension-of-inverse-limit Comment by Misha Misha 2013-05-19T02:37:09Z 2013-05-19T02:37:09Z What is your notion of inverse limit of metric spaces? Do you mean this one: <a href="http://mathoverflow.net/questions/15948/inverse-limit-in-metric-geometry" rel="nofollow" title="inverse limit in metric geometry">mathoverflow.net/questions/15948/&hellip;</a> ? http://mathoverflow.net/questions/131076/complex-morse-function-on-a-four-manifold Comment by Misha Misha 2013-05-18T19:42:04Z 2013-05-18T19:42:04Z kaavek: Then $f$ is a fibration on open and dense subset with constant genus of the fiber on this subset (I am assuming your function is also proper, otherwise, genus is not even defined). Is this what you wanted to know? http://mathoverflow.net/questions/131049/topological-classification-of-a-real-valued-functions-on-manifold Comment by Misha Misha 2013-05-18T19:37:34Z 2013-05-18T19:37:34Z OK, good luck with that. http://mathoverflow.net/questions/131049/topological-classification-of-a-real-valued-functions-on-manifold Comment by Misha Misha 2013-05-18T15:32:52Z 2013-05-18T15:32:52Z What is &quot;topological conjugacy of functions on a manifold&quot;? Do you just precompose them with a homeomorphism of the domain? Diffeomorphism of the domain? Symplectomorphism (if the manifold is symplectic)? Postcompose? Both? Or, maybe, your manifold is the real line? The most classical example, of course, is the implicit function theorem, followed by the Morse theorem on local classification of functions with nondegenerate Hessian at the critical point. This is generalized to the area of mathematics devoted to classification of singularities of hypersurfaces. http://mathoverflow.net/questions/131039/fx-y-min-max Comment by Misha Misha 2013-05-18T12:26:51Z 2013-05-18T12:26:51Z chana: What you should do is to read the FAQ for this site, which is not suitable for such questions. http://mathoverflow.net/questions/130988/why-are-schur-multipliers-of-finite-simple-groups-so-small Comment by Misha Misha 2013-05-18T05:10:17Z 2013-05-18T05:10:17Z Stefan: That would be way too close to asking for a conceptual reason for classification of finite simple groups. http://mathoverflow.net/questions/130932/is-there-a-contractible-bounded-homogeneous-space/131001#131001 Comment by Misha Misha 2013-05-18T05:05:28Z 2013-05-18T05:05:28Z Todd: This is a variation on the solution of the Hilbert's 5th problem or, more precisely, its generalization to transitive group actions on various classes of spaces (Gleason-Montgomery-Zippin-Yamabe et al), so not completely unexpected. http://mathoverflow.net/questions/130959/reference-request-affine-transforms-circle-inversion/131005#131005 Comment by Misha Misha 2013-05-17T22:56:32Z 2013-05-17T22:56:32Z Robert: It seems likely that the &quot;conformal+projective&quot; group is transitive on $k$-point sets for all $k$. Do you know if this is indeed the case? Another side remark is that &quot;conformal+projective&quot; transformations frequently appear in proofs of rigidity results, like Mostow rigidity. http://mathoverflow.net/questions/131004/relation-between-volume-entropy-and-hausdorff-dim-of-limit-set Comment by Misha Misha 2013-05-17T22:33:59Z 2013-05-17T22:33:59Z See also <a href="http://math.univ-lyon1.fr/~remy/smf_sec_18_09.pdf" rel="nofollow">math.univ-lyon1.fr/~remy/smf_sec_18_09.pdf</a> and references therein. http://mathoverflow.net/questions/131004/relation-between-volume-entropy-and-hausdorff-dim-of-limit-set Comment by Misha Misha 2013-05-17T22:33:30Z 2013-05-17T22:33:30Z In this context, limit set is the ideal boundary of the universal cover. Classically, one uses this in the case of Kleinian groups. Then the right formula is that volume entropy equals Hausdorff dimension of the conical limit set of the group (which could be less than dimension of the full limit set). See Nicholls' book &quot;Ergodic theory of discrete groups&quot;. Normalization in variable curvature was worked out in papers by Besson, Courtois and Gallot (maybe also Hammenstadt) and you only have an inequality. Equality if I remember correctly is only in curvature -1 case. http://mathoverflow.net/questions/130959/reference-request-affine-transforms-circle-inversion Comment by Misha Misha 2013-05-17T22:23:33Z 2013-05-17T22:23:33Z Bill: You cannot, since you have to take compositions as well.