User igor belegradek - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T15:00:00Z http://mathoverflow.net/feeds/user/1573 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/131144/volume-growth-of-covers-and-growth-of-deck-transformation-groups Volume growth of covers and growth of deck-transformation groups Igor Belegradek 2013-05-19T13:36:00Z 2013-05-19T14:33:01Z <p>It is well-known that if $\widetilde M\to M$ is a Galois cover of a compact Riemannian manifold $M$ with deck-transformation group $G$, then the growth of $G$ equals the volume growth of $\widetilde M$ (in the pullback metric).</p> <p><b>Question.</b> Is the same true when $M$ is a finite volume complete Riemannian manifold and $G$ is finitely generated?</p> <p>The usual proof (a la Svarc-Milnor) argues that $G$ and $\widetilde M$ are quasiisometric (since $M$ is compact), and then uses that growth is a quasiisometry invariant. One could hope that the reasoning extends to the finite volume case when quasi-isometry is replaced by measure equivalence (ME), but growth type is not an invariant of ME. On the other hand, I do not have counterexamples for the above question.</p> http://mathoverflow.net/questions/129144/purely-parabolic-kleinian-groups Purely parabolic Kleinian groups Igor Belegradek 2013-04-29T20:10:31Z 2013-04-30T03:57:36Z <p>What can be said about a discrete finitely generated subgroup $G$ of $PSL(2,\mathbb C)$ whose nontrivial elements are parabolic? If $G$ is geometrically finite, one can show that $G$ must be elementary so the real question is: Can $G$ be geometrically infinite? </p> http://mathoverflow.net/questions/128130/what-does-a-mathematician-expect-from-mathematics-education/128136#128136 Answer by Igor Belegradek for What does a mathematician expect from mathematics education? Igor Belegradek 2013-04-19T21:29:39Z 2013-04-19T21:29:39Z <p>I think the expectation is that math educators at a math department will be leaders on innovative ways to teach and assess students' progress, will train and mentor TAs, take part in and initiate pilot projects, and help with technology training. (Math and non-math) educators at my institution are excellent, and do all the above. Like everything else, teaching gets better with practice and faculty and TAs could sure benefit from professional advice. </p> http://mathoverflow.net/questions/123771/hyperbolic-groups-with-infinitely-generated-commutator-subgroups Hyperbolic groups with infinitely generated commutator subgroups Igor Belegradek 2013-03-06T15:17:01Z 2013-03-08T11:06:28Z <p>I am trying to get a sense of how often the commutator subgroup $[G,G]$ of a (Gromov) hyperbolic group $G$ is infinitely generated.</p> <p><b>Remarks: </b></p> <ol> <li><p>$[G,G]$ is infinitely generated if is $G$ is free noncyclic, and of course $[G,G]$ is finitely generated when $G$ has finite abelianization. </p></li> <li><p>There are examples when a hyperbolic group has an finitely generated, <b>normal</b> (infinite) subgroup (of infinite index), obtained via various versions of the <a href="http://berstein.wordpress.com/2011/02/27/the-rips-construction-i/" rel="nofollow"> Rips constructions</a>, or Morse theory considerations of Bestvina-Brady and Brady, see <a href="http://www.math.ou.edu/~nbrady/papers/sub.ps" rel="nofollow"> here </a>. </p></li> </ol> <p>More generally, how often is the kernel of a surjection $G\to\mathbb Z$ infinitely generated (or infinitely presented)? Here is a specific:</p> <p><b>Question.</b> Suppose the abelianization of $G$ has rank $>1$, so that there are infinitely many surjections $G\to \mathbb Z$. Is it true that there are infinitely many surjections $G\to\mathbb Z$ whose kernel is not finitely generated?</p> http://mathoverflow.net/questions/122466/non-tame-3-manifolds-covered-by-the-euclidean-space Non-tame 3-manifolds covered by the Euclidean space Igor Belegradek 2013-02-20T22:19:06Z 2013-02-22T23:44:22Z <p>An open 3-manifold is <i>tame</i> if it is homeomorphic to the interior of a compact manifold. Is there a (known) example of an open 3-manifold that is not tame, has finitely generated fundamental group and universal cover homeomorphic to $\mathbb R^3$?</p> http://mathoverflow.net/questions/121851/linear-groups-that-are-nonlinear-over-the-integers Linear groups that are nonlinear over the integers Igor Belegradek 2013-02-14T23:26:15Z 2013-02-15T08:02:07Z <p>What are sources of finitely generated $\mathbb C$-linear groups that are not $\mathbb Z$-linear?</p> <p>Recall that a group is <i>$R$-linear</i> if it is isomorphic to a subgroup of $GL(n,R)$ for some $n$, where $R$ is a ring.</p> <p>I know only one source: any solvable $\mathbb Z$-linear group is polycyclic. For example, the Baumslag-Solitar group $B(1,2)$ is solvable, $\mathbb C$-linear, and it contains dyadic rationals, and therefore, is not polycyclic (abelian subgroups of polycyclic groups are finitely generated). </p> <p>My personal motivation for the question is an attempt to digest recent applications of virtual Haken conjecture implying that many $3$-manifold groups are $\mathbb Z$-linear.</p> http://mathoverflow.net/questions/77410/image-of-the-hilbert-space-under-a-continuous-bijection Image of the Hilbert space under a continuous bijection Igor Belegradek 2011-10-06T23:54:28Z 2013-02-13T16:49:38Z <p>Consider a continuous bijection $X\to Y$ such that $X$ is homeomorphic to a separable infinite dimensional Hilbert space. I wonder what can be said about topological properties of $Y$. </p> <p>To exclude silly pathologies (e.g. $Y$ is $X$ with trivial topology), let's suppose that $Y$ is separable and metrizable.</p> <p><b>Motivation.</b> Suppose you study a space that appears naturally, and you manage to bijectively parametrize with a continuous parameter that takes values is a nice parameter space (such as $l^2$). If you discover that the parametrization isn't a homeomorphism, you may wonder if you have proved anything at all.</p> <p><b> Remarks </b> (to the original question):</p> <ol> <li><p>$Y$ is path-connected; in fact, no countable (or finite) subset can separate $Y$. (Here is why: the preimage of a countable subset of $Y$ is a countable subset of $X$. Then it is known that the complement in $l^2$ of a countable subset is homeomorphic to $l^2$, and hence the image of the complement in $Y$ is path-connected.)</p></li> <li><p>$Y$ can be homeomorphic to $X$ even when the bijection is not a homeomorphism. In fact, there is a <a href="http://www.springerlink.com.www.library.gatech.edu:2048/content/n587u1h31v224642/" rel="nofollow">result </a> of Savkin that any infinite dimensional Banach manifold admits a continuous self-bijection that is not a homeomorphism. See also a recent <a href="http://www.jstor.org/pss/10.4169/000298910X521698" rel="nofollow"> paper </a> by Creswell in Monthly.</p></li> <li><p>The same question is interesting when $X$ is $\mathbb R^n$. (I briefly thought that the answer in this case follows from the invariance of domain, and Bill kindly corrected me in comments). </p></li> </ol> http://mathoverflow.net/questions/121571/connected-sum-of-topological-manifolds/121642#121642 Answer by Igor Belegradek for Connected sum of topological manifolds Igor Belegradek 2013-02-12T20:05:08Z 2013-02-12T22:24:27Z <p>I just taught this in my undergraduate topology class (in the topological category), and in fact I usually give a version of this as a homework exercise (with hints). </p> <p>The idea is to define connected sum along the <i>coordinate balls</i>, i.e. the balls that are mapped via coorinate chats to standard balls in $\mathbb R^n$. For such balls the argument is easy, and no annulus theorem is needed. For notational convinience let's write a coordinate ball as $B_x$ where $x$ is the point in $M$ that cooresponds to the center of the corresponding ball in $\mathbb R^n$. Fix $y\in M$ and consider the subset $Y$ of $M$ consisting of points $x$ such that there is a homeomorphism of $M$ taking $B_x$ to $B_y$. It is easy to show that $Y$ is open and closed (the point is that given two metric balls in $\mathbb R^n$ there is a homeomorphism that maps one ball into the other one, and is the identity outside a compact set; such a homeomorphism can be constructed with bare hands, and this is where the work is). Thus if $M$ is connected, then $Y=M$. The same argument shows that any two coordinate balls are ambiently isotopic. </p> http://mathoverflow.net/questions/121038/hyperbolic-3-manifolds-with-no-geometrically-finite-structure Hyperbolic 3-manifolds with no geometrically finite structure Igor Belegradek 2013-02-07T03:40:02Z 2013-02-07T17:57:15Z <p>Does there exist a compact hyperbolic 3-manifold $M$ that is not diffeomorphic to a geometrically finite hyperbolic manifold? If yes, can such $M$ have incompressible boundary?</p> <p>I think the answer should be yes to both questions but I cannot find this in the literature.</p> <p><b>Remarks:</b> as usual, a <i>compact hyperbolic manifold</i> is a compact manifold whose interior carries a complete hyperbolic structure. The structure is <i> geometrically finite</i> if it is obtained as the quotient of the hyperbolic 3-space by a geometrically finite group. Thurston's hyperbolization theorem implies:</p> <ol> <li><p>A compact 3-manifold with non-empty boundary is hyperbolizable if and only if it is irreducible and atoroidal. </p></li> <li><p>Any compact, atoroidal, pared 3-manifold is diffeomorphic to a geometrically finite one.</p></li> <li><p>Any compact hyperbolic 3-manifold is homotopy equivalent to a geometrically finite one.</p></li> </ol> http://mathoverflow.net/questions/118693/is-the-volume-functional-contiunuous-for-compact-manifolds-with-lower-bounds-on-v/118694#118694 Answer by Igor Belegradek for Is the volume functional contiunuous for compact manifolds with lower bounds on volume? Igor Belegradek 2013-01-12T02:49:56Z 2013-01-12T13:23:26Z <p>I assume you talk about continuity in Gromov-Hausdorff topology. </p> <p>In general, volume is not continuous, as easy examples show (see e.g. Colding's paper "Large manifolds with positive Ricci curvature", examples 1-2 which have a lower volume bound).</p> <p>On the other hand, for manifolds with a lower Ricci curvature bound, Colding shows the for each $r$ the volume of $r$-balls is continuous in Gromov-Hausdorff topology; see his paper "Ricci curvature and volume convergence".</p> http://mathoverflow.net/questions/74064/frechet-manifolds-vs-ilh-manifolds Fréchet manifolds vs ILH manifolds Igor Belegradek 2011-08-30T13:59:53Z 2013-01-10T03:38:50Z <p>What is the precise relation between ILH manifolds and <a href="http://en.wikipedia.org/wiki/Fr%C3%A9chet_manifold" rel="nofollow"> Fréchet manifolds</a>? Specifically:</p> <ol> <li><p>Does any ILH manifold has a canonical structure of a Fréchet manifold?</p></li> <li><p>If so, is it true that any ILH submanifold is a Fréchet sumanifold?</p></li> </ol> <p><b> Background.</b> The notion of an ILH (Inverse Limit Hilbert) manifold was developed by Omori in his studies of diffeomorphism groups. Very roughly it is a manifold modelled on a topological vector space that is the inverse limit of a countable family of Hilbert spaces. This object is easier to deal with than the usual Fréchet manifolds due to the avaiability of the inverse function theorem. </p> <p>For all I know the inverse limit of Hilbert spaces need not be Fréchet. More precisely, it seems that the inverse limit inherits a countable family of seminorms from the Hilbert spaces, but I see no reason for the inverse limit to be complete (as a uniform space), and I am not sure if the inverse limit is always Hausdorff.</p> <p>The ILH manifold I am trying to understand is the diffeomorphism group of a compact manifold, so in particular, it is what Omori called a <i> strong ILH manifold,</i> which perhaps makes a difference in answering 1-2. The diffeomorphism group is also a Fréchet manifold (indeed, it is the so called <i>Fréchet-Lie group</i>), but I am not sure how ILH and Fréchet manifold structures interact.</p> http://mathoverflow.net/questions/116312/4-dimensional-h-cobordisms/116335#116335 Answer by Igor Belegradek for 4-dimensional h-cobordisms Igor Belegradek 2012-12-14T01:03:21Z 2012-12-14T16:07:28Z <p>I think both questions are open. The somewhat sad state of affairs is that there are nontrivial TOP 4d s-cobordisms that are either nonsmoothable or not known to be smoothable, and there are smooth 4d s-cobordisms that may well be products. No h-cobordisms with nontrivial torsion seems to be known.</p> <p>It seems the state of the art is described in the introduction to a paper by Weimin Chen <a href="http://arxiv.org/abs/math/0403396" rel="nofollow"> "Smooth s-cobordisms of elliptic 3-manifolds" </a>, JDG (2006), where references can be found.</p> <p>Convention: all cobordisms below are of dimension 4 (i.e. have 3-manifold boundaries).</p> <ol> <li><p>There are only finitely many orientable TOP s-cobordisms with the boundary the same elliptic 3-manifold and in some cases there is a complete classification (Cappell-Shaneson, Kwasik-Schultz).</p></li> <li><p>There are infinitely many non-orientable TOP s-cobordisms (Matsumoto-Siebenmann, Kwasik).</p></li> <li><p>Kwasik gave (modulo now known elliptization conjecture) a list of finite groups such that any 4-dimensional topological h-cobordism with the fundamental group on the list must have trivial Whitehead torsion, see "On four-dimensional h-cobordism". Of course, the Whitehead group itself of those finite groups is often nontrivial.</p></li> <li><p>Cappell-Shaneson constructed examples of smooth s-cobordsims with ellipltic 3-manifold boundaries, but it is unknown whether the cobordisms aren't products, and partial results of Akbulut indicate they are probably smooth products.</p></li> <li><p>Chen proved that a symplectic s-cobordism with elliptic boundaries is a product, and conjectured that a smooth s-cobordism is a product if and only if its universal cover is a product.</p></li> </ol> http://mathoverflow.net/questions/114528/are-homeomorphic-open-subsets-of-mathbbrn-also-diffeomorphic/114580#114580 Answer by Igor Belegradek for Are homeomorphic open subsets of $\mathbb{R}^n$ also diffeomorphic? Igor Belegradek 2012-11-26T20:19:42Z 2012-11-26T20:26:26Z <p>Adding to Andy's answer: there are lots of contractible open subset of $\mathbb R^n$ that are not homeomorphic to $\mathbb R^n$. For example, any compact contractible manifold of dimension $n>4$ embeds into $\mathbb R^n$: the double of any compact contractible manifold is simply-connected and hence a homotopy sphere, which after removing a point becomes $\mathbb R^n$. For constructions of compact contractible manifolds, see Kervaire's paper <a href="http://dx.doi.org/10.2307%2F1995269" rel="nofollow">Smooth homology spheres and their fundamental groups.</a></p> <p>You mention a confusion about a "sloppy definition of a manifold" and ask which manifolds have one chart. By a chart you seem to mean any open subset of a manifold together with a homeomorphism onto an open subset of $\mathbb R^n$, which I think is a valid definition. Any atlas of charts whose transition functions are smooth defines a smooth structure on your manifold, which by definition is the set of all atlases compatible the given one. If there is only one chart, then the (only) transition function is the identity, which is smooth. However, this merely implies that your manifold with one chart it diffeomorphic to an open subset of $\mathbb R^n$. </p> http://mathoverflow.net/questions/109687/is-it-true-that-the-orbit-space-of-a-free-finite-group-action-on-a-cw-complex-is/109763#109763 Answer by Igor Belegradek for Is it true that the orbit space of a free finite group action on a CW-complex is also a CW-complex? Igor Belegradek 2012-10-15T21:09:03Z 2012-10-15T21:09:03Z <p><b>Lemma</b> If $X$ is a countable locally finite CW-complex and $G$ acts freely and properly discontinuously on $X$, then $X/G$ is homotopy equivalent to a CW-complex.</p> <p><b>Proof</b> Any metrizable ANR is homotopy equivalent to a CW-complex (I am not sure who proved it first but see Theorem 3.6.1 <a href="http://image.diku.dk/aasa/oldpage/aasa.pdf" rel="nofollow">here</a>. Since $X$ is countable and locally finite, it is a metrizable separable ANR. As Misha remarks in comments averaging the metric over the group action implies that $X/G$ is metrizable. Also a countable dense subset of $X$ projects to a countable dense subset of $X/G$. Finally, if a metrizable separable space is locally ANR, it is an ANR (see Borsuk's "Theorey of Retracts", Corollary 10.4, Chapter IV). It follows that $X/G$ is a metrizable ANR as desired.</p> <p><b> Remark</b> In seeing whether $X/G$ is homeomorphic to a CW-complex, even the case when $X$ is a PL manifold is unclear. The difficulty is that it seems unknown which topological manifolds are homeomorphic to CW-complexes (Kirby-Siebenmann prove this for compact manifolds of dimension $\ge 6$ (or maybe $\ge 5$?, but certainly not $4$). So there might exist manifolds not homeomorphic to CW-complexes but whose finite covers are PL. </p> http://mathoverflow.net/questions/107825/intuition-behind-thom-class/107829#107829 Answer by Igor Belegradek for Intuition behind Thom class Igor Belegradek 2012-09-22T11:45:01Z 2012-09-22T11:45:01Z <p>Thom class gives an orientation covector in every fiber $F\cong\mathbb R^n$ (of an oriented vector bundle) which can thought of a generator in $H^n(F-0)$ . Using local trivializations such covectors are defined locally. One needs to prove that these covectors glue to a cohomology class on the total space (with the zero section deleted), and this is where Mayer-Vietoris becomes relevant. How else would you glue? Read the exposition in Milnor-Stasheff or Bott-Tu.</p> http://mathoverflow.net/questions/107716/are-negatively-pinched-manifold-locally-conformally-flat/107729#107729 Answer by Igor Belegradek for Are negatively pinched manifold locally conformally flat? Igor Belegradek 2012-09-21T00:52:29Z 2012-09-21T01:01:47Z <p>Regarding vanishing rational Pontryagin classes (which do vanish for conformally flat manifolds):</p> <ol> <li><p>Recent result of Ontaneda gives examples in each dimension $\ge 4$ of closed manifolds with nonzero rational Pontryagin classes that are pinched arbitrary close to $-1$, see Corollary 4 of his paper <a href="http://front.math.ucdavis.edu/1110.6374" rel="nofollow">"Pinched smooth hyperbolization"</a>. </p></li> <li><p>On the other hand, if you restrict topology of your negatively pinched $n$-manifolds in a suitable way, then one can prove vanishing of Pontryagin classes for pinching close enough to $-1$. For example, for closed manifolds of uniformly bounded simplicial volume, if the pinching is close enough to $-1$, the manifold is diffeomorphic to a hyperbolic one (this is due to Gromov), and hence has zero Pontryagin classes. Long ago I proved similar results in the noncompact case, e.g. if you fix the fundamental group, and the dimension, and assume the metric is complete and the fundamental group is hyperbolic, then the Pontryagin classes vanish for pinching close to $-1$, see <a href="http://front.math.ucdavis.edu/0001.5132" rel="nofollow">here</a>. (I should mention that my proof depends on an accessibility result of Delzant-Potyagailo in which a gap was recently discovered by Louder-Touikan who announced a fix for hyperbolic groups, see <a href="http://www.d503.net/apps/Louder-Research-Statement.pdf" rel="nofollow"> here</a>. Without the fix I can only handle the case of hyperbolic groups that do not split over {1} ot $\mathbb Z$.).</p></li> </ol> http://mathoverflow.net/questions/103985/groups-with-finitely-generated-center Groups with finitely generated center Igor Belegradek 2012-08-05T00:09:54Z 2012-09-19T04:40:49Z <p>Does every group with a finite classifying space have finitely generated center? </p> <p><b>Remarks:</b></p> <ol> <li><p>If $G$ is a finitely generated group with infinitely generated center $Z(G)$, then the quotient $G/Z(G)$ is not finitely presented (as follows from a result of B.H Newmann).</p></li> <li><p><i>Finite classifying space</i> means that the group is the fundamental group of a finite aspherical cell complex.</p></li> <li><p>I suspect the above question is a well-known open problem, but cannot find it stated in the literature, so a reference would be appreciated.</p></li> <li><p>Alperin-Shalen (Inventiones, 1982) showed that the answer is yes for every subgroup of $GL_n(K)$ where $n>0$ and $K$ is a field of characteristic zero.</p></li> <li><p>The answer is also yes for elementary amenable groups. (I know a proof, but have no reference). </p></li> </ol> http://mathoverflow.net/questions/107458/manifold-whose-universal-covering-is-a-sphere-but-which-is-not-a-space-form/107461#107461 Answer by Igor Belegradek for Manifold whose universal covering is a sphere but which is not a space form? Igor Belegradek 2012-09-18T12:32:03Z 2012-09-18T12:57:49Z <p>There are lots of fake lens spaces, and fake spherical space forms (search on these keywords). In particular, a construction of fake lens spaces is in chapter 12 of Milnor's <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.bams/1183527946" rel="nofollow"> "Whitehead torsion"</a>. Here are details: start with any lens space $L$ with fundamental group of order different from 2,3,4, 6, so that its Whitehead group is infinite. Then there are infinitely many distict manifolds that are h-cobordant to $L$. On the other hand, Corollary 12.13 implies that any h-cobordism between lens spaces is a product. (Of course, I assume dimension $\ge 5$ here).</p> http://mathoverflow.net/questions/106169/closed-manifold-has-no-nontrivial-totally-convex-subset/106172#106172 Answer by Igor Belegradek for Closed manifold has no nontrivial totally convex subset? Igor Belegradek 2012-09-02T11:23:35Z 2012-09-02T11:30:11Z <p>The inclusion from a closed totally convex subset to the ambient manifold is a homotopy equivalence. Details can be found in <a href="http://www.intlpress.com/JDG/archive/1981/16-2-333.pdf" rel="nofollow">"Totally convex sets in complete Riemannian manifolds"</a> by Bangert, JDG, 1981. </p> <p>For the second assertion, Cheeger-Gromoll prove in their paper on the soul theorem that any closed totally convex subset is a manifold with boundary, so if the boundary is non-empty, the manifold has zero top-dimensional homology, hence it cannot be homotopy equivalent to a closed manifold of that top dimension.</p> http://mathoverflow.net/questions/105858/when-is-a-manifold-a-tangent-bundle/105863#105863 Answer by Igor Belegradek for When is a manifold a tangent bundle? Igor Belegradek 2012-08-29T17:40:24Z 2012-08-29T17:46:03Z <p>Suppose $M$ is an open $2n$-manifold that is homotopy equivalent to a closed smooth $n$-manifold $N$, and suppose $n>2$. Then Haefliger's embedding theorem ensures that the homotopy equivalence $N\to M$ is homotopic to a smooth embedding. Moreover, by Siebenmann's open collar recognition theorem $M$ is diffeomorphic to the normal bundle to this embedding if and only if $M$ is the interior of a compact manifold with boundary such that the inclusion of the boundary induces an isomorphism on the fundamental group. Now it remains to check whether the normal bundle and tangent bundle to the embedding are isomorphic, which of course rarely happens.</p> <p>A good example is when $N$ is an orientable $3$-manifold and $M=N\times \mathbb R^3$, which is precisely $TN$ because orientable $3$-manifolds are parallelizable. By above arguments, any two homotopy equivalent orientable $3$-manifolds have diffeomorphic tangent bundles. Specific examples can be found among lens spaces, such as $L(7,1)$ and $L(7,2)$.</p> <p>The case $n=2$ seems more delicate.</p> http://mathoverflow.net/questions/104974/nontrivial-examples-of-non-trivial-principal-circle-bundles/105034#105034 Answer by Igor Belegradek for Nontrivial examples of non-trivial principal circle bundles Igor Belegradek 2012-08-19T11:45:53Z 2012-08-19T12:29:35Z <p>Orientable circle bundle with torsion Euler class have been studied systematically. There are exactly the flat $SO(2)$-bundles, see <a href="http://projecteuclid.org/euclid.tjm/1255958322" rel="nofollow">"A Remark on Torsion Euler Classes of Circle Bundles"</a> by Miyoshi or <a href="http://www.emis.de/journals/HOA/IJMMS/Volume2005_21/384804.pdf" rel="nofollow"> "Flat circle bundles, pullbacks, and the circle made discrete"</a> by Oprea-Tanré.</p> <p>It is a standard fact that any flat $G$-bundle over a (connected) finite cell complex $X$ can be written as $(\tilde X\times G)/\pi_1(X)$ where $\tilde X$ is the universal cover and $\pi_1(X)$ acts by deck transformations on the first factor, and via some homomorphism $\pi_1(X)\to G$ on the second factor. Thus all examples look like the one given by Anton.</p> <p>As a caution I wish to point out that many people also studied flat circle bundles with $G=Diff(S^1)$. The answer there is different, namely one gets the so called Milnor-Wood inequality as a condition on the Euler class.</p> http://mathoverflow.net/questions/104451/irreducible-homology-3-spheres-that-bound-smooth-contractible-manifolds Irreducible homology 3-spheres that bound smooth contractible manifolds Igor Belegradek 2012-08-11T00:24:51Z 2012-08-14T07:55:35Z <p>Some examples of irreducible homology 3-spheres that bound <b> smooth</b> contractible 4-manifolds are listed in the comment to problem 4.2 in <a href="http://math.berkeley.edu/~kirby/problems.ps.gz" rel="nofollow"> Kirby's problem list</a>, and all of them happen to occur among the <a href="http://www.maths.ed.ac.uk/~aar/papers/milnbries.pdf" rel="nofollow"> Brieskorn spheres</a> $\Sigma(p,q,r)$ modelled on $\widetilde{SL}_2(\mathbb R)$, i.e. such that $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}&lt;1$ (except for the standard $S^3$, of course). </p> <p><b> Question.</b> Are there any known homology 3-spheres that bound smooth contractible 4-manifolds and are modelled on other geometries, e.g. NIL or the hyperbolic space?</p> <p>EDIT: Glazner in the paper "Uncountably many contractible 4-manifolds" constructed some other examples but I cannot recognize the geometry. (Glazner's six page paper is easily googlable by title, and it gives an explicit representation for the fundamental group, denoted $G_n$ on page 40). </p> http://mathoverflow.net/questions/104467/a-homotopy-equivalence-between-total-spaces-in-a-hurewicz-fibration-which-is-no/104469#104469 Answer by Igor Belegradek for A homotopy equivalence between total spaces in a (Hurewicz) fibration which is not a fiber homotopy equivalence Igor Belegradek 2012-08-11T11:37:01Z 2012-08-11T11:37:01Z <p>There are lots of such examples, but here is the simplest one I know. Let $M_{p,q}$ be the total space of the principal circle bundle over $S^2\times S^2$ with Euler class $(p, q)$. </p> <p>If $p,q$ are relatively prime, then it is known that $M_{p,q}$ is diffeomorphic to $S^2\times S^3$. Namely, Smale proved in his paper <a href="http://www.jstor.org/stable/1970417" rel="nofollow">"On the structure of 5-manifolds" </a> that the diffeomorphism type of closed, simply-connected, spin 5-manifolds is determined by the second cohomology which is $\mathbb Z$ for $M_{p,q}$ and also for $S^2\times S^3$. (If you have trouble showing $M_{p,q}$ satisfies the above conditions, see Wang-Ziller's paper <a href="http://www.intlpress.com/JDG/archive/1990/31-1-215.pdf" rel="nofollow"> "Einstein metrics on principal torus bundles."</a></p> <p>On the other hand, the fiber homotopy equivalence in this case preserves the Euler class (up to sign).</p> http://mathoverflow.net/questions/103301/subgroups-of-gl-n-mathbb-z-with-finite-coinvariants Subgroups of $GL_n(\mathbb Z)$ with finite coinvariants Igor Belegradek 2012-07-27T12:17:43Z 2012-07-30T18:30:15Z <p>Is there a finite index <b>torsion-free</b> subgroup $G$ of $GL_n(\mathbb Z)$, where $n\ge 3$, such that the coinvariants group $\mathbb Z^n_G$ is finite? </p> <p>Here $G$ acts on $\mathbb Z^n$ in the standard way, and $\mathbb Z^n_G$ by definition is the quotient of $\mathbb Z^n$ by the subgroup generated by the set <code>$\{gz-z: g\in G, z\in\mathbb Z^n\}$</code>. </p> <p>If $G=GL_n(\mathbb Z)$, then $\mathbb Z^n_G$ is finite because G contains $-I_n$, but then $GL_n(\mathbb Z)$ isn't torsion-free.</p> http://mathoverflow.net/questions/98812/coefficients-in-hirzebruch-polynomial-and-divisibility-of-bernoulli-numbers-refe Coefficients in Hirzebruch polynomial and divisibility of Bernoulli numbers: reference request Igor Belegradek 2012-06-04T21:03:01Z 2012-06-04T21:27:30Z <p>I seek a reference for the fact that "coefficients of the Hirzebruch $L$-polynomial have odd denominators". The coefficients are $$\frac{2^{2k}(2^{2k-1}-1)B_k}{(2k)!}$$ where $B_k$ is the Bernoulli number, but I cannot locate the appropriate divisibility property of $B_k$. Of course, $2^{2k-1}-1$ is odd, so it can be ignored. </p> http://mathoverflow.net/questions/98576/subgroups-of-amenable-periodic-groups Subgroups of amenable periodic groups Igor Belegradek 2012-06-01T15:00:42Z 2012-06-02T12:04:10Z <p>Does every countable, infinite, amenable, periodic group $G$ contain an infinite locally finite subgroup? </p> <p>Remarks:</p> <ol> <li><p>I would be happy with an infinitely generated counterexample as long as it is countable.</p></li> <li><p>A counterexample cannot be elementary amenable, as elementary amenable periodic groups are locally finite.</p></li> <li><p>First Grigorchuk's group is not a counterexample: it is a 2-group, and hence it contains an infinite abelian group (by a result of Held, "On abelian subgroups of infinite 2-groups").</p></li> <li><p>Amenability of $G$ is essential, due to existence of Tarski monsters whose subgroups are finite cyclic (constructed by Olshanskii).</p></li> <li><p>Any infinite locally finite group contains an infinite abelian subgroup (see a paper of Hall-Kulatilaka <a href="http://jlms.oxfordjournals.org/content/s1-39/1/235.extract" rel="nofollow">here</a>, so it is equivalent to ask whether $G$ contains an infinite abelian subgroup.</p></li> <li><p>This question was asked <a href="http://www.math.osu.edu/~bergelson.1/Amenab.pdf" rel="nofollow">here</a> in 2008, so perhaps it is an open problem.</p></li> </ol> http://mathoverflow.net/questions/51876/reference-request-arzela-ascoli-theorem-for-smooth-holder-norms Reference request: Arzela-Ascoli theorem for smooth Hölder norms Igor Belegradek 2011-01-12T21:45:40Z 2012-05-29T14:58:27Z <p>Could anyone suggest a textbook account of the Arzela-Ascoli theorem for $C^{k,\alpha}$ norms?</p> http://mathoverflow.net/questions/97598/smooth-thickenings-of-non-smoothable-manifolds Smooth thickenings of non-smoothable manifolds Igor Belegradek 2012-05-21T19:43:39Z 2012-05-23T16:54:26Z <p>It is known that any closed topological manifold is homotopy equivalent to an open smooth manifold.</p> <p><b> Question 1.</b><i> What can be said about the <b>smallest</b> dimension of a smooth manifold that is homotopy equivalent to a given closed topological manifold?</i></p> <p>The following somewhat heavy-handed argument yields a smooth manifold of roughly twice the dimension, namely, use </p> <ul> <li> <a href="http://www.maths.ed.ac.uk/~aar/papers/west.pdf" rel="nofollow"> West's solution </a> of Borsuk's conjecture that any compact ANR of dimension $n$, and in particular any closed $n$-manifold, is homotopy equivalent to a finite polyhedron of dimension $\max\{n, 3\}$; <li> <a href="http://www.pef.uni-lj.si/repovs/clanki/1993/BullAustralMathSoc.pdf" rel="nofollow"> Stallings-Dranishnikov-Repovs's </a> embedding up to homotopy type theorem that any finite $n$-dimensional polyhedron is homotopy equivalent to a finite $n$-dimensional subpolyhedron of $\mathbb R^{2n}$, so its regular neighborhood is the desired open smooth manifold. </ul> <p>Here is a specific question that shows the state of my ignorance on this matter:</p> <p><b> Question 2. </b> <i> Is there a closed $n$-manifold which is <b>not</b> homotopy equivalent to a smooth $(n+1)$-manifold? </i></p> <p>The naive idea to look at the product of a non-smoothable manifold of dimension $\ge 5$ with $\mathbb R$ fails, because such a product is also non-smoothable (by the topological product structure theorem of Kirby-Siebenmann).</p> <p><b>Edit:</b> Misha kindly corrects me that a $5$-manifold is smoothable if and only if its Kirby-Siebenmann invariant vanishes; in particular, this apples to products of a $\mathbb R$ and a closed $4$-manifold $M$. Thus $M\times\mathbb R$ is smoothable iff $M$ has zero KS invariant. Smooth $4$-manifolds have zero KS invariant, but amazingly so do some non-smoothable ones.</p> http://mathoverflow.net/questions/7921/failure-of-smoothing-theory-for-topological-4-manifolds/7966#7966 Answer by Igor Belegradek for Failure of smoothing theory for topological 4-manifolds Igor Belegradek 2009-12-06T05:00:13Z 2012-05-23T16:14:59Z <p>Take any two closed simply-connected homeomorphic smooth closed 4-manifolds that are not diffeomorphic. Then their products with $\mathbb R$ are diffeomorphic because the smooth structure on a such a product is unique. (Indeed, since PL/O is 6-connected, it is enough to show that the associated PL structure is unique, but the set of PL-structures on a PL-manifold $M$ of dimension $\ge 5$ is bijective to the set of homotopy classes of maps from $M$ to $TOP/PL$, and the latter space is $K(\mathbb Z_2, 3)$, so the set of PL structures on $M$ is bijective to $H^3(M,\mathbb Z_2)$, which vanishes by Poncare duality if $M$ is homotopy equivalent to a simply-connected $4$-manifold; in fact the argument shows that all we need is $H_1(M;\mathbb Z_2)=0$). </p> <p>It follows that the original closed simply-connected $4$-manifolds are tangentially homotopy equivalent, i.e. there is a homotopy equivalence that pulls stable tangent bundles to each other. </p> http://mathoverflow.net/questions/97477/characteristic-classes-for-block-bundles Characteristic classes for block bundles Igor Belegradek 2012-05-20T13:46:51Z 2012-05-21T02:10:01Z <p>Block bundles are PL analogs of vector bundles, see e.g. Rourke-Sanderson's <a href="http://projecteuclid.org/euclid.bams/1183528512" rel="nofollow"> article</a> in Bulletin of AMS, 1966. There is a classifying space $B\widetilde{PL}_q$ for rank $q\$ block bundles, which is a PL analog of $BO_q$, and there are similar classifying spaces $B\widetilde{SPL}_q$, $BSO_q$ for oriented bundles.</p> <p><b>Question.</b> <i> Is the rational homotopy type of $B\widetilde{SPL}_q$ recorded in the literature?</i></p> <p>Long time ago Colin Rourke explained to me how to compute the rational homotopy type of $B\widetilde{SPL}_q$, but I cannot find the correspondence. If memory serves, the result was as follows. </p> <ol> <li><p>If $q\ge 3$ is even, then $H^*(B\widetilde{SPL}_q;\mathbb Q)$ is a polynomial algebra on the Pontryagin classes and the Euler class. The Euler class occurs in degree $q$, while the Pontryagin classes occur in all degrees divisible by $4$ (which is different from $BSO_q$ where there is no Pontryagin classes in degrees $\ge 2q$).</p></li> <li><p>If $q\ge 3$ is odd, then $H^*(B\widetilde{SPL}_q;\mathbb Q)$ a polynomial algebra on the Pontryagin classes, which occur in all degrees divisible by $4$, and a new class in degree $2q-2$ which arises from works of Haefliger and Hirsch. </p></li> <li><p>If $q\le 2$, then $BSO_q\to B\widetilde{SPL}_q$ is a rational homotopy equivalence.</p></li> </ol> <p>As usual, the Pontryagin classes are stable, i.e. they survive in $B\widetilde{SPL}\approx BPL$, while the Euler class and the Haefliger-Hirsch classes die in $B\widetilde{SPL}_{q+1}$. I recall that the proof of 1-2 was not too hard, and I can probably reconstruct it, but it would be much nicer if it were written somehwere. </p> http://mathoverflow.net/questions/131175/closed-geodesic-loops-around-points-in-compact-manifolds Comment by Igor Belegradek Igor Belegradek 2013-05-21T03:56:53Z 2013-05-21T03:56:53Z Anton: the argument in my first comment surely works. The argument in the second comment does not give complete detail, but I think the point is that the gradient flow of the energy functional gives a deformation retraction of the manifold of broken geodesics loops onto a small neighborhood of a constant loop at $p$, and that small neighborhood is contractible. http://mathoverflow.net/questions/131175/closed-geodesic-loops-around-points-in-compact-manifolds Comment by Igor Belegradek Igor Belegradek 2013-05-20T14:44:13Z 2013-05-20T14:44:13Z In fact, one needs very little from the Morse theory because there is no critical points (e.g. the index theorem is not needed); basically, replace the loop space by the finite dimensional manifold of piecewise geodesic loops, and note that the energy functional has no critical points, so the finite dimensional manifold is contractible. The rest is elementary algebraic topology. What could be easier? http://mathoverflow.net/questions/131175/closed-geodesic-loops-around-points-in-compact-manifolds Comment by Igor Belegradek Igor Belegradek 2013-05-20T02:59:43Z 2013-05-20T02:59:43Z Suppose $p$ is totally convex, so there is no geodesic loop based at $p$. By the main theorem of the Morse theory the loop space $\Omega_p M$ has a homotopy type of a CW complex with cells correponding to geodesic loops at $p$, so $\Omega_p M$ is contractible, i.e. $\pi_i(M)=0$ for $i&gt;0$. Since $M$ is connected, it is contractible. No closed manifold is contractible by Poincare diality. I do not think there is a proof without Morse theory in disguise. http://mathoverflow.net/questions/131144/volume-growth-of-covers-and-growth-of-deck-transformation-groups/131146#131146 Comment by Igor Belegradek Igor Belegradek 2013-05-19T14:50:43Z 2013-05-19T14:50:43Z This is very helpful. I see now that I have had some silly confusions about growth. http://mathoverflow.net/questions/131144/volume-growth-of-covers-and-growth-of-deck-transformation-groups Comment by Igor Belegradek Igor Belegradek 2013-05-19T14:34:01Z 2013-05-19T14:34:01Z By &quot;equal&quot; I mean that the growth functions have the same growth type, i.e. they dominate each other, where $g$ dominates $f$ if and only if $f(t)\le Ag(At+B)+B$ for all $t$ and some constants $A, B$. http://mathoverflow.net/questions/130367/the-center-of-a-derived-subgroup-in-an-amenable-group Comment by Igor Belegradek Igor Belegradek 2013-05-11T21:17:14Z 2013-05-11T21:17:14Z Juschenko-Monod recently constructed a f.g. simple infinite amenable group $G$. Clearly it has trivial center and its derived subgroups all equal $G$. http://mathoverflow.net/questions/130081/converse-to-milnors-theorem-on-manifolds-with-nonnegative-ricci-curvature Comment by Igor Belegradek Igor Belegradek 2013-05-09T00:32:00Z 2013-05-09T00:32:00Z The introduction of <a href="http://arxiv.org/abs/math/0109167" rel="nofollow">arxiv.org/abs/math/0109167</a> lists a number of examples (maybe all known ones?). I have not been following the subject for the last several years but I do not recall any new examples since that paper was written. http://mathoverflow.net/questions/129321/does-a-topological-manifold-have-an-exhaustion-by-compact-submanifolds-with-bou/129331#129331 Comment by Igor Belegradek Igor Belegradek 2013-05-01T23:01:34Z 2013-05-01T23:01:34Z Misha, it wasn't me. http://mathoverflow.net/questions/129144/purely-parabolic-kleinian-groups/129176#129176 Comment by Igor Belegradek Igor Belegradek 2013-04-30T14:53:06Z 2013-04-30T14:53:06Z Yes, It does extend to the Karlsson-Noskov setting. http://mathoverflow.net/questions/129144/purely-parabolic-kleinian-groups/129176#129176 Comment by Igor Belegradek Igor Belegradek 2013-04-30T12:47:26Z 2013-04-30T12:47:26Z I see. This should also extend to the setting of Karlsson-Noskov paper on isometry groups of spaces with contractive bordifications, such as e.g. visibility spaces. http://mathoverflow.net/questions/129144/purely-parabolic-kleinian-groups Comment by Igor Belegradek Igor Belegradek 2013-04-30T03:48:16Z 2013-04-30T03:48:16Z The original question was actually a toy case of the same question for purely parabolic isometry groups of Hadamard manifolds. Neither argument extends to this setting, as far as I can see. http://mathoverflow.net/questions/129144/purely-parabolic-kleinian-groups Comment by Igor Belegradek Igor Belegradek 2013-04-29T22:11:08Z 2013-04-29T22:11:08Z Thanks, Misha! I do not understand what you mean by the Zariski closure argument (what's is special about the Zarisky closure of a purely parabolic group?), but the other one I get: the group generated by high powers of parabolic elements with disjoint fixed points at infinity is geometrically finite (it visibly has a fundamental polyhedron with 4 faces), and hence it contains a hyperbolic element. http://mathoverflow.net/questions/128574/visual-boundaries-of-universal-covers-of-finite-volume-nonpositively-curved-manif Comment by Igor Belegradek Igor Belegradek 2013-04-24T22:59:16Z 2013-04-24T22:59:16Z Sven, I forgot to mention that the lower curvature bound is never needed. http://mathoverflow.net/questions/128574/visual-boundaries-of-universal-covers-of-finite-volume-nonpositively-curved-manif Comment by Igor Belegradek Igor Belegradek 2013-04-24T14:15:37Z 2013-04-24T14:15:37Z (cont) The hyperbolic element is some word in the above hyperbolic elements. Doesn't this work? http://mathoverflow.net/questions/128574/visual-boundaries-of-universal-covers-of-finite-volume-nonpositively-curved-manif Comment by Igor Belegradek Igor Belegradek 2013-04-24T14:14:46Z 2013-04-24T14:14:46Z I do not know enough of the locally symmetric case to see why the answer is yes (as Misha says), but if so, then it seems you can combine the above to conclude that the answer is always yes, i.e., use rank rigidity to decompose the universal cover as a product of locally symmetric or rank one factors. Fix a point at infinity $z$, pick a tangent vector $v$ in its direction, project it to factors, and approximate the projections by axis of hyperbolic elements, which commute and stabilize the product of axes, which is a flat. Arguing in that flat find a hyperbolic element with axis ending at $z$.