User mauricio - MathOverflow

Diffeomorphism coming from the s-cobordism theorem

2012-12-20T22:56:41Z

Let $f:M_0\rightarrow M_1$ be a diffeomorphism between two compact $n$-dimensional manifolds. Let $W^{n+1}$ be an $h$-cobordism between $M_0$ and $M_1$. Assume that the cobordism has no torsion and its dimension is high enough so that, by the $s$-cobordism theorem, there is a diffeomorphism $F:M_0\times I\rightarrow W$. Can $F$ be taken so that $F|_{M_0\times 0}=id_{M_0}$ and $F|_{M_0\times1}=f$?

Thanks in advance!

representatives of the group of homotopy 7-spheres

2012-05-16T00:42:10Z

In Milnor's paper "On manifolds homeomorphic to the 7-sphere" it is proven that there are manifolds homeomorphic but not diffeomorphic to the standard 7-sphere. His construction involves sphere bundles over $S^4$. Also, Kervaire and Milnor proved that there are exactly 28 h-cobordism (therefore diffeomorphism) classes of homotopy spheres in dimension 7.

Does every class contain an exotic sphere arising as the total space of an $S^3$-bundle over $S^4$?, if not, how can one determine the number of classes with such representatives?, how do they look like?

Geometric meaning of L-genus

2012-03-01T18:13:43Z

Is there any reasonable geometric meaning of the L-genus for smooth manifolds? or perhaps easier, for complex algebraic surfaces?

The question came up after a friend and I realized that we don't understand why one would expect to have such a formula for the signature in terms of Pontrjagin classes (i.e. the signature theorem). Any insights about this will be appreciated.

Answer by Mauricio for Actions of finite groups on exotic smooth manifolds of dimension >4

2012-05-09T19:28:35Z

I think the answer is no. But I don't know the reason. I take the fact that exotic spheres in dimension 7 do not admit orientation reversing diffeomorphisms (but the standard sphere does) as a clue to say that one shouldn't expect an affirmative answer to your question.

EDIT: I gave a wrong answer and I'm not sure if I can delete the post. I decided to change my answer by a comment and make it CW. My apologies to those who read this expecting to see a right answer...

Answer by Mauricio for Finite groups admitting free isometric actions on round spheres

2012-05-09T17:44:51Z

After all these comments, a possible answer to your question goes as follows.
If $n$ is even, then the only group that can act freely and isometrically on $S^n$ is $\mathbb{Z}/2$. One way to see this is as in Max's answer above, i.e. by looking at the behavior of eingenvalues of orthogonal matrices. Here is another reason: an action of $G$ on $S^n$ means that there is a group homomorphism $G\rightarrow\mbox{Homeo}(S^n)$. But any homeomorphsim has degree $\pm 1$. So we get a homomorphism $G\rightarrow\mathbb{Z}/2$. But the action has no fixed points, then the degree of the image of every non trivial element in $G$ is $(-1)^{n+1}=-1$, therefore the map has a trivial kernel and therefore it is an isomorphism.
As macbeth already pointed out, when $n$ is odd, the problem is not so easy. In that case, we consider $S^n$ as the universal cover of a complete Riemannian manifold $M$ with constant sectional curvature. A simple argument using lifting properties of covering spaces shows that there is an isometry between $M$ and $S^n/G$ (whenever the action is free and properly discontinuous). So the problem of finding the subgrups with that particular action on the sphere is the same as the classification of complete Riemannian manifolds with constant sectional curvature (=1). That is done in Wolf's book Spaces of Constant Curvature.

Answer by Mauricio for whitehead group of product of groups

2012-03-21T00:41:46Z

This is a partial answer. Let $G$ be a group. Then by identifying $\mathbb{Z}G[t,t^{-1}]$ with $\mathbb{Z}[G\times\mathbb{Z}]$ and using Bass-Heller-Swan decomposition you will get $K_1(\mathbb{Z}[G\times\mathbb{Z}])=K_1(\mathbb{Z}G)\oplus\tilde{K}_0(\mathbb{Z}G)\oplus NK_1(\mathbb{Z}G)^2$. Therefore

$Wh(G\times\mathbb{Z})=Wh(G)\oplus\tilde{K}_0(\mathbb{Z}G)\oplus NK_1(\mathbb{Z}G)^2$.

As for the example you are looking for, take $G=\mathbb{Z}/4\mathbb{Z}$ and $H=\mathbb{Z}^{3}$. The Whitehead group of these two groups vanishes but the Whitehead group of its product does not. The reason is that $G\times H$ is the fundamental group of $L\times T\ ^3$ where $L$ is a 3-dimensional lens space and $T\ ^3$ is a 3-dimensional torus. Farrell and Hsiang (F. T. Farrell and W. C. Hsiang, Bull. Amer. Math. Soc. Volume 73, Number 5 (1967), 741-744) proved that there is a non trivial h-cobordism with that space as one of its boundary components.

Is there a Galois correspondence for ring extensions?

2011-05-02T21:19:29Z

Given an ring extension of a (commutative with unit) ring, Is it possible to give a "good" notion of "degree of the extension"?. By "good", I am thinking in a degree which allow us, for instance, to define finite ring extensions and generalize in some way the Galois' correspondence between field extensions and subgroups of Galois' group.

I suppose one can call a ring extension $A\subset B\ $ finite if $B$ is finitely generated as an $A$-module, and the degree would be the minimal number of generators, but is that notion enough to state a correspondence theorem?

Thanks in advance!

When is a finitely generated group finitely presented?

2011-02-10T02:10:38Z

I think the question is very general and hard to answer. However I've seen a paper by Baumslag ("Wreath products and finitely presented groups", 1961) showing, as a particular case, that the lamplighter group is not finitely presented. To prove this, he gives conditions to say if a wreath product of groups is finitely presented. The question is: which ways (techniques, invariants, etc) are available to determine whether a finitely generated group is also finitely presented? For instance, is there another way to show that fact about the lamplighter group?

Thanks in advance for references and comments.

Answer by Mauricio for Open Questions in Riemannian Geometry

2011-02-10T19:40:22Z

You can try one of these: http://www.aimath.org/WWN/nnsectcurvature/nnsectcurvature.pdf. All of them concern with nonnegatively curved Riemannian manifolds and Alexandrov geometry. In the same context I know a couple of surveys: http://arxiv.org/abs/0707.3091 and http://arxiv.org/abs/math/0701389. It has been conjectured (you can check in those papers) that any nonnegatively curved manifold is rationally elliptic. This is an important open problem in Riemannian geometry.

Smoothness of distance function in Riemannian Manifolds

2010-04-14T03:46:43Z

Let $(\mathcal{M},g)$ be a $C^{\infty}$-Riemannian manifold. A basic fact is that $g$ endows the manifold $\mathcal{M}$ with a metric space structure, that is, we can define a distance function $d:\mathcal{M}\times\mathcal{M}\longrightarrow\mathbb{R}$ (the distance between two points will be the infimum of the lengths of the curves which join the points) which is compatible with the topology of $\mathcal{M}$. Of course $d$ is continuous function, but what can we say about the differentiability of $d$?, is it smooth?. If not, Is there some criterion to know when it is?

Thanks in advance.

Answer by Mauricio for Commutator of Lie derivative and codifferential?

2010-04-10T07:48:34Z

You can find a formula for the commutator of the codifferential and the Lie derivative in http://arxiv.org/pdf/gr-qc/0306102. The important point to get the commutator is to notice that the codifferential is a derivation on the Schouten-Leibniz algebra. You can see the details in the paper above. I hope this is what you are looking for.

Comment by Mauricio

2012-12-21T16:41:22Z

@Tom Goodwillie: If one adds the condition that $f$ is homotopic to the identity, is the answer pretty trivially no again?...

Comment by Mauricio

2012-12-01T21:09:38Z

@algori: I'll delete my question to redo it...thanks for pointing out how badly it was written...

Comment by Mauricio

2012-12-01T20:33:30Z

@algori: What a shame!...I'm editing again...

Comment by Mauricio

2012-12-01T20:12:37Z

If $F:R^N→R^N$ is such an extension, I want $F|_{R^N−S^{n+k}}:R^N−S^{n+k}→F(R^N−S^{n+k})$ to be homotopic to the identity map on $R^N−S^{n+k}$.

Comment by Mauricio

2012-12-01T19:03:24Z

@Sergei Ivanov, @algori: Thanks for pointing this out...I missed a condition on the extension when I typed. I'm editing...

Comment by Mauricio

2012-05-31T01:10:27Z

What is $N_0$?, what is a topological generator?, where did you read it?

Comment by Mauricio

2012-05-09T23:52:27Z

Thanks for that link. I honestly don't know a different source to read about that stuff, sorry.

Comment by Mauricio

2012-05-09T23:28:49Z

In Milnor's original paper or these notes nd.edu/~lnicolae/MS.pdf by Liviu Nicolaescu.

Comment by Mauricio

2012-05-09T17:38:22Z

Yes, it acts on $S^1$, but how can you extend that action to a free action on $S^n$. Probably I'm missing something...

Comment by Mauricio

2012-05-09T16:37:14Z

In your last line you should have said "any \textit{finite} cyclic group admits..."

Comment by Mauricio

2012-05-09T16:03:50Z

Your answer is still wrong. The antipodal map is an isometry.

Comment by Mauricio

2012-05-09T15:53:59Z

It is not true that only $\mathbb{Z}/2$ and the trivial group act freely on spheres. Remember that you get lens spaces as quotients of odd dimensional spheres by a finite group of orthogonal transformations that have no fixed points

Comment by Mauricio

2012-05-09T14:29:14Z

@aglearner: The formulation was precise, I think. But the classification of such finite groups when n is even is quite simple. Anyhow you might be interested in these [notes](math.utah.edu/~keenan/actions.pdf).

Comment by Mauricio

2012-05-09T14:02:59Z

Well, if the dimension of the sphere is even, then $\mathbb{Z}/2$ is the only nontrivial group acting freely.

Comment by Mauricio

2012-05-03T23:46:20Z

I found your lecture notes very useful. Thank you for bringing them to my attention. By the way, I think that on page 91 there is a typo at the very bottom. You didn't use $\epsilon$ to define the regions $D^{\pm}$.