User igor belegradek - MathOverflow

Volume growth of covers and growth of deck-transformation groups

2013-05-19T13:36:00Z

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).

Question. Is the same true when $M$ is a finite volume complete Riemannian manifold and $G$ is finitely generated?

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.

Purely parabolic Kleinian groups

2013-04-29T20:10:31Z

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?

Answer by Igor Belegradek for What does a mathematician expect from mathematics education?

2013-04-19T21:29:39Z

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.

Hyperbolic groups with infinitely generated commutator subgroups

2013-03-06T15:17:01Z

I am trying to get a sense of how often the commutator subgroup $[G,G]$ of a (Gromov) hyperbolic group $G$ is infinitely generated.

Remarks:

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

More generally, how often is the kernel of a surjection $G\to\mathbb Z$ infinitely generated (or infinitely presented)? Here is a specific:

Question. Suppose the abelianization of $G$ has rank $>1$, so that there are infinitely many surjections $G\to \mathbb Z$. Is it true that there are infinitely many surjections $G\to\mathbb Z$ whose kernel is not finitely generated?

Non-tame 3-manifolds covered by the Euclidean space

2013-02-20T22:19:06Z

An open 3-manifold is tame if it is homeomorphic to the interior of a compact manifold. Is there a (known) example of an open 3-manifold that is not tame, has finitely generated fundamental group and universal cover homeomorphic to $\mathbb R^3$?

Linear groups that are nonlinear over the integers

2013-02-14T23:26:15Z

What are sources of finitely generated $\mathbb C$-linear groups that are not $\mathbb Z$-linear?

Recall that a group is $R$-linear if it is isomorphic to a subgroup of $GL(n,R)$ for some $n$, where $R$ is a ring.

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).

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.

Image of the Hilbert space under a continuous bijection

2011-10-06T23:54:28Z

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 $Y$.

To exclude silly pathologies (e.g. $Y$ is $X$ with trivial topology), let's suppose that $Y$ is separable and metrizable.

Motivation. 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.

Remarks (to the original question):

$Y$ 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.)
$Y$ can be homeomorphic to $X$ even when the bijection is not a homeomorphism. In fact, there is a result of Savkin that any infinite dimensional Banach manifold admits a continuous self-bijection that is not a homeomorphism. See also a recent paper by Creswell in Monthly.
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).

Answer by Igor Belegradek for Connected sum of topological manifolds

2013-02-12T20:05:08Z

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).

The idea is to define connected sum along the coordinate balls, 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.

Hyperbolic 3-manifolds with no geometrically finite structure

2013-02-07T03:40:02Z

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?

I think the answer should be yes to both questions but I cannot find this in the literature.

Remarks: as usual, a compact hyperbolic manifold is a compact manifold whose interior carries a complete hyperbolic structure. The structure is geometrically finite if it is obtained as the quotient of the hyperbolic 3-space by a geometrically finite group. Thurston's hyperbolization theorem implies:

A compact 3-manifold with non-empty boundary is hyperbolizable if and only if it is irreducible and atoroidal.
Any compact, atoroidal, pared 3-manifold is diffeomorphic to a geometrically finite one.
Any compact hyperbolic 3-manifold is homotopy equivalent to a geometrically finite one.

Answer by Igor Belegradek for Is the volume functional contiunuous for compact manifolds with lower bounds on volume?

2013-01-12T02:49:56Z

I assume you talk about continuity in Gromov-Hausdorff topology.

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).

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".

Fréchet manifolds vs ILH manifolds

2011-08-30T13:59:53Z

What is the precise relation between ILH manifolds and Fréchet manifolds? Specifically:

Does any ILH manifold has a canonical structure of a Fréchet manifold?
If so, is it true that any ILH submanifold is a Fréchet sumanifold?

Background. 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.

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.

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 strong ILH manifold, which perhaps makes a difference in answering 1-2. The diffeomorphism group is also a Fréchet manifold (indeed, it is the so called Fréchet-Lie group), but I am not sure how ILH and Fréchet manifold structures interact.

Answer by Igor Belegradek for 4-dimensional h-cobordisms

2012-12-14T01:03:21Z

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.

It seems the state of the art is described in the introduction to a paper by Weimin Chen "Smooth s-cobordisms of elliptic 3-manifolds" , JDG (2006), where references can be found.

Convention: all cobordisms below are of dimension 4 (i.e. have 3-manifold boundaries).

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).
There are infinitely many non-orientable TOP s-cobordisms (Matsumoto-Siebenmann, Kwasik).
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.
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.
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.

Answer by Igor Belegradek for Are homeomorphic open subsets of $\mathbb{R}^n$ also diffeomorphic?

2012-11-26T20:19:42Z

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 Smooth homology spheres and their fundamental groups.

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 $\mathbb R^n$.

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?

2012-10-15T21:09:03Z

Lemma 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.

Proof Any metrizable ANR is homotopy equivalent to a CW-complex (I am not sure who proved it first but see Theorem 3.6.1 here. 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.

Remark 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.

Answer by Igor Belegradek for Intuition behind Thom class

2012-09-22T11:45:01Z

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.

Answer by Igor Belegradek for Are negatively pinched manifold locally conformally flat?

2012-09-21T00:52:29Z

Regarding vanishing rational Pontryagin classes (which do vanish for conformally flat manifolds):

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 "Pinched smooth hyperbolization".
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 here. (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 here. Without the fix I can only handle the case of hyperbolic groups that do not split over {1} ot $\mathbb Z$.).

Groups with finitely generated center

2012-08-05T00:09:54Z

Does every group with a finite classifying space have finitely generated center?

Remarks:

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).
Finite classifying space means that the group is the fundamental group of a finite aspherical cell complex.
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.
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.
The answer is also yes for elementary amenable groups. (I know a proof, but have no reference).

Answer by Igor Belegradek for Manifold whose universal covering is a sphere but which is not a space form?

2012-09-18T12:32:03Z

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 "Whitehead torsion". 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).

Answer by Igor Belegradek for Closed manifold has no nontrivial totally convex subset?

2012-09-02T11:23:35Z

The inclusion from a closed totally convex subset to the ambient manifold is a homotopy equivalence. Details can be found in "Totally convex sets in complete Riemannian manifolds" by Bangert, JDG, 1981.

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.

Answer by Igor Belegradek for When is a manifold a tangent bundle?

2012-08-29T17:40:24Z

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.

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 $L(7,2)$.

The case $n=2$ seems more delicate.

Answer by Igor Belegradek for Nontrivial examples of non-trivial principal circle bundles

2012-08-19T11:45:53Z

Orientable circle bundle with torsion Euler class have been studied systematically. There are exactly the flat $SO(2)$-bundles, see "A Remark on Torsion Euler Classes of Circle Bundles" by Miyoshi or "Flat circle bundles, pullbacks, and the circle made discrete" by Oprea-Tanré.

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.

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.

Irreducible homology 3-spheres that bound smooth contractible manifolds

2012-08-11T00:24:51Z

Some examples of irreducible homology 3-spheres that bound smooth contractible 4-manifolds are listed in the comment to problem 4.2 in Kirby's problem list, and all of them happen to occur among the Brieskorn spheres $\Sigma(p,q,r)$ modelled on $\widetilde{SL}_2(\mathbb R)$, i.e. such that $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}<1$ (except for the standard $S^3$, of course).

Question. 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?

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).

Answer by Igor Belegradek for A homotopy equivalence between total spaces in a (Hurewicz) fibration which is not a fiber homotopy equivalence

2012-08-11T11:37:01Z

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 $(p, q)$.

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 "On the structure of 5-manifolds" 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 "Einstein metrics on principal torus bundles."

On the other hand, the fiber homotopy equivalence in this case preserves the Euler class (up to sign).

Subgroups of $GL_n(\mathbb Z)$ with finite coinvariants

2012-07-27T12:17:43Z

Is there a finite index torsion-free subgroup $G$ of $GL_n(\mathbb Z)$, where $n\ge 3$, such that the coinvariants group $\mathbb Z^n_G$ is finite?

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 $\{gz-z: g\in G, z\in\mathbb Z^n\}$ .

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.

Coefficients in Hirzebruch polynomial and divisibility of Bernoulli numbers: reference request

2012-06-04T21:03:01Z

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.

Subgroups of amenable periodic groups

2012-06-01T15:00:42Z

Does every countable, infinite, amenable, periodic group $G$ contain an infinite locally finite subgroup?

Remarks:

I would be happy with an infinitely generated counterexample as long as it is countable.
A counterexample cannot be elementary amenable, as elementary amenable periodic groups are locally finite.
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").
Amenability of $G$ is essential, due to existence of Tarski monsters whose subgroups are finite cyclic (constructed by Olshanskii).
Any infinite locally finite group contains an infinite abelian subgroup (see a paper of Hall-Kulatilaka here, so it is equivalent to ask whether $G$ contains an infinite abelian subgroup.
This question was asked here in 2008, so perhaps it is an open problem.

Reference request: Arzela-Ascoli theorem for smooth Hölder norms

2011-01-12T21:45:40Z

Could anyone suggest a textbook account of the Arzela-Ascoli theorem for $C^{k,\alpha}$ norms?

Smooth thickenings of non-smoothable manifolds

2012-05-21T19:43:39Z

It is known that any closed topological manifold is homotopy equivalent to an open smooth manifold.

Question 1. What can be said about the smallest dimension of a smooth manifold that is homotopy equivalent to a given closed topological manifold?

The following somewhat heavy-handed argument yields a smooth manifold of roughly twice the dimension, namely, use

West's solution 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 $\max\{n, 3\}$;
Stallings-Dranishnikov-Repovs's 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.

Here is a specific question that shows the state of my ignorance on this matter:

Question 2. Is there a closed $n$-manifold which is not homotopy equivalent to a smooth $(n+1)$-manifold?

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).

Edit: 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.

Answer by Igor Belegradek for Failure of smoothing theory for topological 4-manifolds

2009-12-06T05:00:13Z

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 $H_1(M;\mathbb Z_2)=0$).

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.

Characteristic classes for block bundles

2012-05-20T13:46:51Z

Block bundles are PL analogs of vector bundles, see e.g. Rourke-Sanderson's article 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.

Question. Is the rational homotopy type of $B\widetilde{SPL}_q$ recorded in the literature?

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.

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 $\ge 2q$).
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.
If $q\le 2$, then $BSO_q\to B\widetilde{SPL}_q$ is a rational homotopy equivalence.

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.

Comment by Igor Belegradek

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.

Comment by Igor Belegradek

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?

Comment by Igor Belegradek

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 $i>0$. 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.

Comment by Igor Belegradek

2013-05-19T14:50:43Z

This is very helpful. I see now that I have had some silly confusions about growth.

Comment by Igor Belegradek

2013-05-19T14:34:01Z

By "equal" 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 $A, B$.

Comment by Igor Belegradek

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 $G$.

Comment by Igor Belegradek

2013-05-09T00:32:00Z

The introduction of arxiv.org/abs/math/0109167 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.

Comment by Igor Belegradek

2013-05-01T23:01:34Z

Misha, it wasn't me.

Comment by Igor Belegradek

2013-04-30T14:53:06Z

Yes, It does extend to the Karlsson-Noskov setting.

Comment by Igor Belegradek

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.

Comment by Igor Belegradek

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.

Comment by Igor Belegradek

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.

Comment by Igor Belegradek

2013-04-24T22:59:16Z

Sven, I forgot to mention that the lower curvature bound is never needed.

Comment by Igor Belegradek

2013-04-24T14:15:37Z

(cont) The hyperbolic element is some word in the above hyperbolic elements. Doesn't this work?

Comment by Igor Belegradek

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 $z$.