User henrikrüping - MathOverflow

Can Lipschitz maps increase the Lebesgue dimension ?

2010-08-24T09:57:36Z

Given a map $f:X\rightarrow Y$ of compact metric spaces, such that there is a $C\in \mathbb{R}$ with $d(f(x),f(x'))\le C\cdot d(x,x')$.

Does this already imply, that the Lebesgue dimension of $f(X)$ is at most the Lebesgue dimension of $Y$?

My motivation were space filling curves. All constructed examples were not rectifiable, so I was wondering, whether the additional assumption of being Lipschitz rules out the existence of space filling curves.

Answer by HenrikRüping for For any entourage $U,V$ there's an entourage $W$ such that $U\circ W\subseteq V\circ U$

2013-04-03T13:32:15Z

The second statement follows from the first one by passing to inverses (i.e. reflecting along the diagonal).

Now consider the uniform structure on $\mathbb{R}$ consisting of all subsets of $\mathbb{R}^2$ that contain an open neighborhood of the diagonal.

Just to avoid confusion with left and right, I want to use the notation $U\circ V = \{(x,z)| \exists y: (x,y)\in U \wedge (y,z)\in V\}.$

Let $U:=\{(x,y)| |y-x|<1 \}\cup \mathbb{R}\times {0}$. $V:=\{(x,y)| |y-x|<1 \}$. Then $V\circ U=\{(x,y)| |y-x|<2 \}\cup \mathbb{R}\times {0}$.

Now let $W$ be any entourage and let $Z$ be some open neighborhood of $0$ such that $\{0\}\times Z\subset W$. Then $U\circ W$ will contain $\mathbb{R}\times Z$. Thus $U\circ W$ will not be contained in $V\circ U$. Since $W$ was arbitrary this should be a counterexample.

Does the poset of free factors of a free group form a lattice?

2013-03-25T14:06:18Z

Let $F_n$ denote a free group of rank $n$. The set of its free factors is partially ordered by inclusion. Recall that a psoet is called a lattice if any two elements have a smallest upper bound and a greatest lower bound.

Is this true for this poset?

Are these groups isomorphic (Cancellation in torsionfree, virtually Abelian groups)

2013-01-30T13:17:43Z

I wondered whether it is possible to find two finitely generated, virtually Abelian, torsionfree groups $G,H$ that are not isomorphic but that become isomorphic after crossing with $\mathbb{Z}$. I have the following candidates:

Consider $K:=(\mathbb{Z}[t]/(t^5+1))\rtimes_{\cdot t} \mathbb{Z}$. Let $\varphi$ be the automorphism of $K$ given by $\cdot t$ on $\mathbb{Z}[t]/(t^5+1)$ and $(0,s)\mapsto (1,s)$ , where $s$ denotes a generator of the other copy of $\mathbb{Z}$. My candidates are $G:=K\rtimes\mathbb{Z}$ and $H:=K\rtimes_{\varphi^3}\mathbb{Z}$.

Are these two groups isomorphic ?

Each one contains a finite index subgroup isomorphic to the other one. Crossing with $\mathbb{Z}$ gives $K\rtimes \mathbb{Z^2}$ where a basis of $\mathbb{Z}^2$ acts by $\varphi,\mbox{id}$ respectively $\varphi^3,\mbox{id}$. The isomorphism is given by a base change of $\mathbb{Z}^2$.

$G$ contains a finite index subgroup isomorphic to $H$ and vice versa. If it turns out that they are actually isomorphic, one might still hope to get an example by replacing $5$ with a bigger number (that is coprime to 3 to make the base change work).

Related: Hirshon, some cancellation theorems with applications to nilpotent groups (The example given there is torsionfree, nilpotent, but maybe not virtually Abelian).

Does this group act geometrically on a Median space?

2012-12-10T12:57:50Z

Let $G$ be the semidirect product of $\mathbb{Z}^2$ with $\mathbb{Z}/6$ where $\mathbb{Z}/6$ acts by the order 6 element of $SL_2(\mathbb{Z})$. We can think of this group as the group of order preserving isometries of the tesselation of $\mathbb{R^2}$ with regular triangles.

Does this group acts properly, isometrically and cocompactly on a median space??

Let for two points in a metric space $[x,y]=\{z|d(x,z)+d(z,y)=d(x,y)\}$. If $X$ is a geodesic metric space than this is just the set of all points lying on some geodesic from $x$ to $y$. $X$ is called a median space if for every triple of points $x,y,z$ we have that $[x,y]\cap[x,z]\cap[y,z]$ consists of exactly one point - the median of $x,y,z$. Examples for median spaces are trees and $\mathbb{R}^n$ with the $l^1$- metric.

The motivation is that the one skeleton of a CAT(0) cube complex is a median graph. If a group acts geometrically on this CAT(0)-cube complex it also acts that way on that graph. For example this group acts properly and isometrically on $\mathbb{R}^3$. This gives a proper and isometric action on a median space, but this action is not cocompact. So I was wondering whether there is a better action. The problem seems to be that the automorphism of $\mathbb{Z}^2$ does not extend to a cube-complex automorphism of $\mathbb{R}^2$, but I could not make this precise.

Answer by HenrikRüping for Artin groups whose graphs differ by one edge and coverings

2012-09-27T14:36:44Z

Let $A_{\Gamma'}$ be the group asoociated to any graph $\Gamma'$. Concerning your first question I can show that there is a short exact sequence $$1\rightarrow F \rightarrow A_{\Gamma \setminus e}\rightarrow A_\Gamma\rightarrow 1.$$

Let me work in the case of graph products (which includes both the right angled Artin and the right andgled Coxeter cases).

We obtain a presentation for $A_\Gamma$ by adding a relation to a presentation for $A_{\Gamma\setminus e}$.

So there is a canonical group homomorphism from $A_{\Gamma\setminus e}$ to $A_\Gamma$. So maybe you are wondering what the Kernel looks like.

Instead of answering this question let me instead consider the homomorphism $A_{\Gamma\setminus e}\rightarrow A_{\Gamma\setminus v}$ where $v$ is one of the endpoints of $e$ (Of course we also have to remove all edge to $v$). By the argument of Section 4 in Holt,Rees its Kernel is a free product of copies of the vertex group. We have a factorization $A_{\Gamma\setminus e}\rightarrow A_\Gamma \rightarrow A_{\Gamma\setminus v}$. So especially the Kernel of the map we were originally interested in is contained in that Kernel.

In the Artin case we get immediately that this kernel is free (as it is a subgroup of a free group) and in the Coxeter case it is also a free group. To show this it is enough to show that it is torsionfree (using the action of a free product on a tree). But this is also clear since any torsion element in a free product is conjugate to an element in one of the factor groups. Such elements cannot lie in the Kernel of $A_{\Gamma\setminus e}$ to $A_\Gamma$ (but to show this I would have to make the isomorphism from the kernel to such a free product explicit. It is in the paper mentioned above but writing it down would take quite some notation).

Finding hyperbolic metrics by approximation

2012-04-16T09:33:12Z

Given a presentation $ < X ; R >$ of a group $G$. Suppose we know for some reason that $G$ is the fundamental group of a three-dimensional finite volume manifold.

Then there is a injective group homomorphism $G\rightarrow Isom(\mathbb{H}^3)$. Mostows rigidity theorem then tells us that it is unique up to composition with inner automorphisms of $\mathbb{H}^3$ from the left (and probably automorphisms of $G$ from the right).

As usual to speficy a homomorphism from a group with a specific presentation to another group we have to pick a image for each generator, such that all relators get mapped to the neutral element. My hope is that there might be an algorithm that starts with some first choice of the images of the generators. Note that $Isom(H^3)$ can be viewed as a subgroup of $GL_4(\mathbb{R})$. Then I would like to iteratively minimize the sum of some norm of the images of the relators (hoping that it will converge to a homomorphism). Of course there are also noninjective homomorphisms (like the trivial one) so I cannot hope that the sequence always converges to an injective homomorphism. But maybe it does with high probability for a reasonably random first choice of generators.

Has someone already done this ? If so I am interested in the convergence properties. They might also depend on the given presentation. So one could also hope that any fundamental group of a hyperbolic 3-manifold has a nice presentation.

Order of ring automorphisms of localizations of polynomial rings over finite fields

2012-07-31T14:21:16Z

Suppose that $F$ is a finite field and $S\subset F[t]$ is a (finite) set of primes. Is is true that any ring automorphism of $R:=F[t][S^{-1}]$ has finite order?

A ring automorphism of $R$ is uniquely determined by an automorphism of $F$ ($F$ is the set of ring elements which are algebraic over the prime ring, so $F$ has to be mapped to itself) and the image of $t$.

Say I want to pick $id_F$ and some image $x\in R$. Under which conditions does this choice really give an automorphism (and not only a homomorphism)?

Dual of a semilinear morphism

2012-07-23T15:19:02Z

Let $R$ be a commutative ring and let $M$ and $N$ be $R$-modules. Let $\sigma:R\rightarrow R$ be a ring automorphism.

Let $f: M\rightarrow N$ be a $\sigma$-semilinear map, i.e. a map of abelian groups satisfying $f(rm)=\sigma(r)f(m)$. Let $Hom_{R,\sigma}(M,N)$ denote the set of all $\sigma$-semilinear maps from $M$ to $N$.

The usual definition of the dual of $f$ does not give the right result, as it is a map $N^*=Hom_{R,id}(N,R)\rightarrow Hom_{R,\sigma}(M,R)\qquad n^* \mapsto n^* \circ f$. To force the target to be $Hom_{R,id}(M,R)$ I would like to postcompose with $\sigma^{-1}\in Hom_{R,\sigma^{-1}}(R,R)$. But of course the $\sigma$ is not uniquely determined by the map $f$. For example the zero map is $\sigma$-semilinear for any $\sigma$.

So my question is: Is this well defined, i.e. Given any $f\in Hom_{R,\sigma}(M,N)\cap Hom_{R,\sigma'}(M,N)$ and any $n^* \in N^*$. Do we have

$\sigma^{-1} \circ n^* \circ f= \sigma '^{-1}\circ n^* \circ f$?

Answer by HenrikRüping for Which topological spaces admit a nonstandard metric?

2012-07-07T22:46:29Z

I once considered the following situation: $[0;1]^X$ is metrizable (with values in $\mathbb{R}$), if $X$ is countable. So I wondered whether $[0;1]^\mathbb{R}$ is nonstandard-metrizable for some totally ordered field $F$.

However it turns out that it does not admit a totally ordered local basis, so it can't be nonstandard-metrizable. I could give more details, if somebody is interested in them.

Answer by HenrikRüping for Projective modules over non-rational group rings

2012-06-13T13:39:43Z

I would guess that the map on $K_0$ is an isomorphism, butI could only show the surjectivity right now:

The inclusion of rings $RG\rightarrow QG$ induces a map on $K_0$. Given a projective $RG$-module - say it is a submodule of $RG^n$ - to its $Q$-span. It is a projective submodule of $QG^n$.

So let us first show that this map is surjective, ie. every projective $QG$ module arises this way. Given any such $P'$ the obvious candidate for a preimage would be $RG^n\cap P'$.

First note that it is as a $R$-module a direct summand of $RG^n$. $R$ is a PID and hence one just has to verify that the quotient is $R$-torsionfree. But $RG^n/(RG^n\cap P')$ embeds into the $Q$-vectorspace $QG^n/P'$ and hence it is $R$-torsionfree.

So we have a section of $R$-modules $s:RG^n/(RG^n\cap P')\rightarrow RG^n$. It need not be a $RG$-map. So let us make it equivariant by setting $s'(x):=\frac{1}{|G|}\sum_{g\in G}gs(g^{-1}x)$. Note that it is still a section (project down again; it is a $RG$ map).

So we have found a $RG$-complement of $RG^n\cap P'$; hence $RG^n\cap P'$ is a projective $RG$-module. So the map $K_0(RG)\rightarrow K_0(QG)$ is surjective.

Is there a natural topology on the set of open sets ?

2012-04-02T07:45:01Z

Given a topological space $(X,\mathcal{O})$ can one assign a natural topology to $\mathcal{O}$ such that

1) The intersection of a compact set of open sets is again open,

2) The maps $\cap,\cup:\mathcal{O}^2\rightarrow \mathcal{O}$ are continuous,

3) For any continuous map $f: (X,\mathcal{O})\rightarrow (X',\mathcal{O}')$ is the induced map $f^{-1}: \mathcal{O}'\rightarrow \mathcal{O}$ also continuous?

Since these properties would allow taking the discrete topology and I am looking for a more interesting one let me also add:

4) For $X=\mathbb{R}$ the subset $\{]a,b[|a\le b\}$ of intervals has the topology coming from $\{(a,b)\in \mathbb{R}^2| a\le b\}/\Delta \mathbb{R}$.

Which spaces are inverse limits of discrete spaces ?

2010-02-16T10:34:29Z

There is the following theorem:

"A space $X$ is the inverse limit of a system of discrete finite spaces, if and only if $X$ is totally disconnected, compact and Hausdorff."

A finite discrete space is totally disconnected, compact and Hausdorff and all those properties pass to inverse limits. I guess the other direction might be proved by taking the system of all decompositions of $X$ into disjoint clopen sets. The inverse limit should give $X$ back.

So what happens, if I dismiss the finiteness condition. As mentioned above every inverse limit of discrete spaces is totally disconnected, Hausdorff. So the question is:

"Which totally disconnected Hausdorff spaces are inverse limits of discrete spaces?"

For example I think it is impossible to write $\mathbb{Q}$ as an inverse limit of discrete spaces, but I don't have a proof.

infinite dimensional CAT(0) groups

2012-01-05T13:23:59Z

Usually a CAT(0) group is defined to be a group acting properly isometrically and cocompactly on a CAT(0) space, but I would like to consider only those groups that act properly, isometrically and cocompactly on a finite-dimensional CAT(0) space.

So is there a group I have to leave out?

Not every CAT(0) space with a proper isometric cocompact group action is finite-dimensional. For example the trivial group acts on the compact CAT(0)-space $[0;1]^\mathbb{N}$.

Torsion in some specific module over the Laurent polynomials

2012-03-19T14:37:38Z

Consider the ring $\mathbb{Z}[t,t^{-1}]$ of Laurent-polynomials over $\mathbb{Z}$. The abelian group $M:=\prod_\mathbb{Z}\mathbb{Z}$ becomes a module over this ring via $t\cdot (x_*):=x_{*+1}$.

Is there a irreducible polynomial $p\in \mathbb{Z}[t,t^{-1}]$ with leading coefficient not equal to $\pm 1$ such that $M$ has $p$-torsion ?

Are these abelian groups free?

2012-03-08T15:32:34Z

Suppose we have a countable, torsion-free abelian group $A$ with the property that for each element $a\neq 0$ the set $D_a=\{x\in A|\exists n\in \mathbb{Z}:nx=a\}$ is finite. Is $A$ already a free abelian group?

If one drops the condition "countable" the infinite direct product of countably many copies of $\mathbb{Z}$ is a counterexample.

Orbits of exterior products

2012-02-27T12:20:27Z

In linear algebra one learns a lot of normal forms (Which I want to think of as a classification of the orbits of a group action on some set). For example if $V$ is a $k$-vectorspace $GL(V)$ acts on $V$ with exactly two orbits - the orbit of $0$ and the other orbit.

Now what happens if we let $GL(V)$ act diagonally on $V^{\wedge n}$ or $V^{\otimes n}$? Can one give a normal form for this group action or at least find the number of orbits ?

The motivation for this question comes from surgery theory. The number of isomorphism types of fake $n$-tori (for $n\ge 5$) is given by the number of orbits of $GL_n(\mathbb{F}_2)$ acting on $(\mathbb{F}_2^n)^{\wedge n-3}$.

Analogues of the dihedral group

2012-01-03T13:17:24Z

A virtually-$\mathbb{Z}$ group $G$ admits either a epimorphism onto $\mathbb{Z}$ or a epimorphism onto $D_\infty$.

So what happens if one replaces $\mathbb{Z}$ by another group $F$ (like the free group or $\mathbb{Z}^n$). My first guess would be that any virtually-$F$ group $G$ maps surjectively onto one of the groups

$F\rtimes H$, where $H\le $Aut$(F)$ is any finite subgroup.

This is clear in the case where $G$ is semidirect product of $F$ and a finite group $K$; one can simply take $H$ to be the image of $K\rightarrow $ Aut $ (F) $.

So my questions are:

1) Is it still true that there is an epimorphism even in the non-split case?

2) Is it also true that two groups $F\rtimes H_1$ and $F\rtimes H_2$ (with $H_i\subset $ Aut $(F)$ finite) cannot surject onto each other unless $H_1$ and $H_2$ are conjugated (in which case the groups are isomorphic) ?

I doubt that this is true for all groups $F$ but maybe one can find sufficient conditions that guarantee this.

Hyperbolizing geodesic spaces

2012-02-01T13:38:21Z

Consider the Poincare half plane model for the n-dimensional hyperbolic space $\mathbb{H}^n$. $\mathbb{H}^n$ can be constructed out of $\mathbb{R}^{n-1}$ by crossing it with $(0;\infty)$ and equpping the product with the following metric:

Let $\gamma=(\gamma_1,\gamma_2)$ be a path $[0;t]\rightarrow \mathbb{R}^{n-1}\times (0;\infty)$ such that $\gamma_1$ is parametrized by arclength. Then define its length to be

$\int_0^t\frac{1}{\gamma_2(t')}\sqrt{1+\dot\gamma_2(t')}$.

Then define the distance of two points as the infimum over the length of all paths connecting them. (Hopefully it is really a metric).

So one could perform this construction on any geodesic metric space. Has this construction already been studied before?

Does this construction turn CAT(0) spaces into CAT(-1) spaces ?

Is a inverse limit of compact spaces again compact ?

2010-03-18T13:35:11Z

Then one can construct a model for the inverse limit by taking all the compatible sequences. This is a subspace of a product of compact spaces. This product is compact by Tychonoff. If all the spaces are Hausdorff, then this is even a closed subspace.

However, if the spaces are not Hausdorff, it needn't be a closed subspace. If you take a two point space with the trivial topology as $X_n$ and constant structure maps, you will get as the inverse limit the space of all constant sequences, which is not a closed subspace of the infinite product, as the infinite product also has the trivial topology.

But the space is again compact. So I am wondering, whether there is a generalization of the proof of Tychonoff's theorem, that applies directly to inverse limits.

Answer by HenrikRüping for Finding lattice with short basis-vectors containing given lattice

2012-01-16T23:52:25Z

I think you are asking for a short basis of the direct summand of $\mathbb{Z}^n$ spanned by $v_1,\ldots,v_n$ (The submodule spanned by them need not be a direct summand; so you consider the direct summand W spanned by them. It can be defined as the preimage of the torsion subgroup of $\mathbb{Z}^n/\langle v_1,\ldots,v_n\rangle$).\

So the first question arising might be: How large can the smallest (nonzero) vector of $W$ at most be?

A bound can be obtained from Minkowski's theorem. This gives that there has to be a vector $w\in W$ of length $\le 2 (vol(W))^{\frac{1}{rk(W)}}$. The volume of W can be defined as the square root of the determinant of the matrix $(\langle w_i,w_j\rangle)_{i,j}$, where $w_1\ldots,w_{rk(W)}$ is any $\mathbb{Z}$-basis of $W$. Unfortunately the proof of Minkowski's theorem is not constructive and does not yield an algorithm to find a shortest vector. A quick google search let me to the wikipedia page Lattice problem saying that finding a shortest vector in a lattice is NP-hard and giving some names of approximation algorithms.

As far as I know the bound given in Minkowski's theorem is not optimal and can be improved, though this is a hard task.

I see that this still does not answer your question, but it was too long for a comment, so I put it as an answer.

Group rings of infinite products of groups

2010-11-22T13:56:00Z

Given a infinite family of groups $(G_i)$ for $i\in I$. Is there a ring theoretic construction, that produces $R[\prod_{i\in I} G_i]$ using only the rings $(R[G_i])_{i\in I}$ ?

For the case of a finite family, we have $R[G\times H]\cong R[G][H]$ and for commutative $R$ we have $R[G\times H]\cong R[G]\otimes R[H]$. Neither of those constructions generalizes to the infinite case, e.g.

The map $R[\prod_i G_i]\rightarrow \mbox{invlim}_{I'\subset I, |I'|<\infty}R[\prod G_i]$ is not surjective (This product runs over $i\in I'$). The same holds for the map into the infinite tensor product (assuming that $R$ is commutative).

So I am hoping, that there is a better contruction in a more elaborate category (like $R$-Algebras with an augmentation), that produces $R[\prod_{i\in I} G_i]$ out of the group rings $(R[G_i])_{i\in I}$ .

Are subgroups of hyperbolic groups quasiisometrically embedded ?

2011-12-07T13:25:49Z

Given a finitely generated subgroup of a finitely generated hyperbolic group. Is it true that the inclusion of each subgroup is a quasiisometric embedding ?

The first example for a group that does not have this property is a Baumslag-Solitar group $BS(1,m)= \langle a,b| bab^{-1}=a^m\rangle$. We have $a^{m^k}=b^kab^{-k}$ for each $k$. This shows for example that the inclusion of the subgroup generated by $a$ is not a quasiisometric embedding.

Then one can consider the class of groups with the property that the inclusion of any subgroup is a quasiisometric embedding. Has this class been studied?

Is every solvable subgroup of $GL(n,\mathbb{Z})$ polycyclic?

2011-10-25T12:12:59Z

Is every solvable subgroup of $GL(n,\mathbb{Z})$ polycyclic?

The first solvable group that is not polycyclic is $\mathbb{Z}[1/2]\rtimes \mathbb{Z}$ (where the automorphism is given by multiplication with 2) and I do not see a way of embedding it into $GL_n(\mathbb{Z})$ for some $n$.

When is a extension of $\mathbb{Z}$ by a free group a CAT(0) group?

2011-02-01T11:11:17Z

The question has an easy answer, if one replaces free by free abelian: Then the resulting group is always solvable and a solvable subgroup of a CAT(0) group is virtually abelian. If the resulting was CAT(0), then the chosen automorphism $\varphi$ in $\mathbb{Z}^n\rtimes_\varphi \mathbb{Z}$ would have finite order - otherwise the group would not be virtually abelian.

Now one can ask the same question for the free group instead or the free abelian group. I would like to know for which automorphisms $\varphi$ of the free group $F_n$ the group $F_n\rtimes_\varphi \mathbb{Z}$ is CAT(0).

I only know, that $F_n \times \mathbb{Z}$ is CAT(0). I think that if the chosen automorphism has finite order, then the result should be CAT(0) (although I don't have a proof). And I do not know automorphism, that gives a non-CAT(0) group.

Structure theorem of f.g. modules over a (non) PID

2010-11-16T13:37:07Z

I am looking for an example of a commutative ring with $1$, in which every ideal is generated by a single element, for which the conclusion of the structure theorem for finitely generated modules is wrong. (The ring is allowed to have zero divisors, so it is not a PID).

Are there any examples? What happens if one drops the other conditions instead (commutativity, $1\in R$)? Does then the structure theorem still fail ?

Is this the CAT(0) metric on an affine building?

2011-08-29T11:59:10Z

Let $R$ be a discrete valuation ring qith quotient field $Q$ and let $t\in R$ be a generator of the unique maximal ideal in $R$. Let $V$ be a finite-dimensional $Q$-vector space. Then one can consider the set of all homothety classes of $R$-lattices in $Q$ (i.e. finitely generated $R$-submodules of the same rank).

The affine building is the simplicial complex whose vertices are homothety classes of lattices (meaning $L\sim \lambda L$ for $\lambda\in Q\setminus {0}$) and a set $\{[L_0],\ldots ,[L_m]\}$ of homothety classes spans a simplex if there are represenatatives $L_0,\ldots,L_n$ with $L_0\subset L_1\subset \ldots \subset L_m\subset t^{-1}L_0$.\

I read several times that there is a Euclidean structure on this building that turns it into a CAT(0)-space but I never read explicitly what the metric on the simplex is.

In general for any maximal simplex (aka chamber) one can find a basis $e_1,\ldots,e_n$ of $V$ such that $e_1,\ldots,e_i, te_{i+1},\ldots te_n$ is an $R$-basis for $L_i$ (for $i=0,\ldots, n-1$).

So the question is: what is the metric on this simplex. My guess is that the lattice $L_i$ corresponds to the point $(0,\ldots,0,1,\ldots,1)\in \mathbb{R}^n$ (with $i$ zeros) and that its homothety class corresponds to the orthogonal projection of this point onto the plane $\{x\in \mathbb{R}^n| \sum_i x_i=0\}$.

Then we can take the convex hull of those $n$- points in that plane to get an Euclidean simplex which gives the building the structure of an Euclidean simplicial complex.

However I could not verify that it is a CAT(0) space. One would have to show that the links are CAT(1) (with the induced spherical metric) and I do not have a clue how this works. However I had the impression that this is rather well known and I just don't see it.

On which space does $GL_n(F_p[X])$ act nicely?

2011-06-27T12:10:15Z

The group $GL_n(\mathbb{Z})$ acts properly and isometrically on the space of homothety classes of scalar products on $\mathbb{R}^n$. This is a Riemannian manifold of nonpositive sectional curvature.

Is there a similar space for the case of $GL_n(F_p[x])$. Maybe one can construct a building or something like this. I guess if the space I am looking for exists it should be rather well known. Otherwise I apologize for the vagueness of this question.

Question concerning h-cobordisms

2011-06-07T15:48:05Z

Suppose we have a cobordism $W$ of manifolds $M_0$ and $M_1$ and suppose the inclusion of $M_0$ into $W$ is a homotopy equivalence. Is the same true for the inclusion of $M_1$ (ie. is $W$ already an h-cobordism)?

Using Poincare Lefschetz duality one can show that this map induces isomorphisms on homology. Hence it suffices to show that the inclusion $M_1\rightarrow W$ induces an isomorphism on $\pi_1$.

Borromean braids

2010-11-11T10:04:52Z

Consider the Kernel $K_n$ of the natural group homomorphism from the $n$-th braid group to the symmetric group. Then one can delete the $m$-th braid. This is a well defined homomorphism $d_m:K_n\rightarrow K_{n-1}$. So is there for every $n\in \mathbb{N}$ a braid $1\neq b\in K_n$ with $d_m(b)=0$ for all $m$.

This is clearly true for $n=2$, as $K_1$ is trivial and it is also true for $n=2$ (The "standard" braid does the job). What about higher $n$. Is there a nice construction, that works for every $n$ ?

Comment by HenrikRüping

2013-04-03T16:46:13Z

@CC Thank you comments. I agree with your first comment. However your second comment suggests that your definition of $U\circ V$ is my definition of $V \circ U$. (To avoid this I included my definition of $U\circ V$). I still think this is a counterexample.

Comment by HenrikRüping

2013-04-02T13:38:25Z

It looks like you need an additional assumption on the action. If the action was trivial the LHS is independent of $G$, where the RHS is not.

Comment by HenrikRüping

2013-03-26T12:27:59Z

Thus it is $F(a,b)$ and hence $L=F(a,b,c)$.

Comment by HenrikRüping

2013-03-26T12:27:29Z

One difference to the free Abelian case is the following: In the free abelian case the rank is additive, i.e. the sum of the ranks of the greatest lower bound and the least upper bound of two elements equals the sum of the ranks of those elements. This does not hold in this case. The elements $c$ and $[a,b]c$ generate free factors of the free group generated by $a,b,c$. Their intersection is trivial. Their least upper bound $L$ contains $[a,b]$. Thus $L\cap F(a,b)$ is a free factor that contains $[a,b]$. But this commutator is not contained in rank $1$ free factor.

Comment by HenrikRüping

2013-03-26T12:17:44Z

Let me emphasize that my assumption that the group in consideration is free (almost) not needed at all. The argument that the intersection of two free factors is again a free factor does not need it; thus we always have greatest lower bounds of finite sets of free factors. The least upper bound really just uses that this poset does not have infinite descending chains; which is true for any finitely generated group: The statement about intersection can be used to pruduce from such a chain a free product decomposition of the whole group with infinitely many, nontrivial factors.

Comment by HenrikRüping

2013-02-27T21:18:01Z

$X=S^1\vee S^1$ should do the job.

Comment by HenrikRüping

2013-01-31T14:50:49Z

Let me just mention, that the groups above seem to be $\mathbb{Z}/10\oplus \mathbb{Z}/10$-manifolds, so the theory from chapter 3 there does not apply directly. Still it would be nice to find an elementary argument that shows that these two groups are not isomorphic.

Comment by HenrikRüping

2013-01-31T14:49:07Z

@Igor: Thank you. The precise place where this is mentioned is the remark after Theorem 3.10 on p. 29. In that chapter $\mathbb{Z}/p$-flat manifolds are studied. Flat manifolds are isometrically covered by $\mathbb{R}^n$. A $\mathbb{Z}/p$ manifold is a manifold where the quotient of its fundamental group by the subgroup of translations is $\mathbb{Z}/p$. Maybe its possible to cook up easier groups by unraveling the theory. I will have a look at it.

Comment by HenrikRüping

2012-12-11T13:33:29Z

The usual definition of a $G$-CW-complex also requires that a group element fixing a cell setwise also fixes it pointwise. For example the unit interval with the $\mathbb{Z}/2$-action given by reflection at $1/2$ and the usual G-CW structure is not a G-CW-complex. I never came across the notion of a $G$-complex.

Comment by HenrikRüping

2012-11-27T09:46:11Z

The sphere is compact and $T^*M$ is compact only for zero-dimensional $M$.

Comment by HenrikRüping

2012-11-01T14:18:33Z

Let m consider only the case when N is geodesic. Let us compute the distance of two points $(t,x)$ and $(t',x')$. If you $N$ was the real line then your warped product is hyperbolic space. Now use the Poincare half plane model to compute the distance from $(t,0)$ to $(t,d(x,x'))$. And this is the distance between $(t,x)$ and $(t',x')$. A geodesic in the warped product will be mapped to a geodesic in $N$ (maybe parametrized in a weird way), so we just have to pick a geodesic from $x$ to $x'$ first and consider the preimage of it. It will be some part of $\mathbb{H}^2$ ...

Comment by HenrikRüping

2012-09-28T16:12:49Z

Dear Talon, I had a look at your paper. It mihgt still have a mistake (I did not check all the tables) but it seems to make sense to be. I strongly think that someone qualified should review this. I have not checked all the tables. I am author of only one article on the arXiv, so I think I cannot endorse you. Hopefully you will find someone on this problem not prejudging it wrong just by the topic.

Comment by HenrikRüping

2012-09-27T19:51:28Z

@Misha: thank you.

Comment by HenrikRüping

2012-09-03T20:52:15Z

Maybe it is a bit tautological but if a norm comes from a scalar product you can reconstruct the scalar product via $s(v,w)=(||v+w||^2-||v||^2-||w||^2)/2$. So one could require that this expression is bilinear. Now I wonder whether 3. is sufficient. I don't think so but couldn't find a counterexample.

Comment by HenrikRüping

2012-08-16T15:32:16Z

You got a polite comment on your first question saying that this might be the wrong site for this. Now you posted the same question three times in a row. Seriuosly ?