most recent 30 from http://mathoverflow.net 2013-05-24T07:56:25Z http://mathoverflow.net/feeds/user/12517 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58096/computations-in-group-cohomology mister_jones 2011-03-10T17:48:57Z 2011-03-11T01:07:49Z

<p>Hello,</p> <p>Given a finitely presentable group $G$, I'm interested in the cup-product from $H^1$ to $H^2$ with real coefficients. I want to know if this is explicitly computable (with a computer) with a presentation of the group. More precisely, I want a program that takes the generators and relations as entries and returns the dimension of the $H^1$ and a finite generating set of linear relations between the cup-products of every couple of elements in a basis of $H^1$. (I am not really interested in all the $H^2$) Does this seem possible ?</p> <p>I precise that I am not really familiar with group cohomology and I ask this question because it is certainly known if such a problem cannot be resolved with an efficient algorithm.</p> <p>The problem comes in the study of Kähler groups where this cup-product plays an important role.</p> <p>Thank you.</p>

http://mathoverflow.net/questions/57811/isotropic-subspaces-in-cohomology mister_jones 2011-03-08T11:21:08Z 2011-03-08T19:37:06Z

<p>Hello,</p> <p>Here is a problem I encountered in the study of Kähler manifolds but there is a natural generalisation of this for topological spaces. If $X$ is a topological space, denote by $g_\mathbb{R}$ the real genus of $X$, that is the maximal dimension of an isotropic subspace in $H^1(X,\mathbb{R})$ (isotropic means that the cup-product restricted to this space is $0$). We can in the same way define $g_\mathbb{C}$ (here we take the complex dimension).</p> <p>Now the question is : $g_\mathbb{C} = g_\mathbb{R}$ ?</p> <p>This seems totally obvious : if one has a real isotropic space, then its complexification is a complex isotropic subspace but conversely I don't see how to construct a real isotropic subspace from a complex one.</p> <p>This is true if $H^2$ has dimension $0$ or $1$ : $0$ is clear and for $1$ one can see the cup-product as a standard symplectic form (maybe degenerate), but in general ?</p> <p>Thank you for your answers and sorry if I just missed something obvious.</p>

http://mathoverflow.net/questions/53385/groups-with-large-negative-deficiency mister_jones 2011-01-26T18:54:42Z 2011-01-27T10:39:05Z

<p>Hello, I've just learnt the notion of deficiency of a group but I don't know how work with it. I want to construct a group with large negative deficiency ; naively I think that $(Z_2)^n$ will work because we need $n$ generators and $n$ relations for the square of the elements being $1$, plus $n(n-1)/2$ relations of commutation ; but maybe we can do better ... Another problem is to control the deficiency of a direct product. Still naively we take generators and relations in both groups and add relations of commutation but it certainly does not work. So my two questions :</p> <ul> <li>Does there exist finite groups with arbitrarily large negative deficiency ?</li> <li>If I fix a finitely presentable group G and consider the product of G with groups of arbitrarily large negative deficiency, can I obtain products with arbitraly large negative deficiency ?</li> </ul> <p>Thank you, mister_jones</p>

http://mathoverflow.net/questions/53345/cohomological-dimension-of-a-group-fibration-and-local-coefficients mister_jones 2011-01-26T11:45:19Z 2011-01-26T19:55:52Z

<p>Hello, I want to show that the cohomological dimension (say over Z or R) of some group $K$ is 1. $K$ occurs in an exact sequence $1 \to K \to \pi_1(X) \to \pi_1(C) \to 1$, where $\pi_1(X)$ has cohomological dimension 3 (in the same coefficients) and $C$ is a curve of genus greater than 2.</p> <p>So I want a kind of additivity but this is not true in general. If I look at the associated fibration $BK \to B\pi_1(X) \to C$ and use Leray-Serre spectral sequence, I have some information on the cohomology of $BK$ and in fact can solve the problem if I assume that the action of the fundamental group of $B$ on the cohomology of the fiber is trivial. But I'm not familiar with cohomology with local coefficients and don't manage to show the general case.</p> <p>Someone can help me ? (or solve this problem more directly ?) (or this is false in general ?)</p> <p>mister_jones</p>

http://mathoverflow.net/questions/58096/computations-in-group-cohomology/58098#58098 mister_jones 2011-03-11T12:03:41Z 2011-03-11T12:03:41Z

Thank you for this precise answer. I will try to do this and post again if there is a problem.

http://mathoverflow.net/questions/57811/isotropic-subspaces-in-cohomology/57878#57878 mister_jones 2011-03-09T19:20:14Z 2011-03-09T19:20:14Z

This is a nice example and seems to show that in high dimensions such examples are not at all pathological. Do you think than we can obtain in this way a result like : &quot;a generic finitely presentable group (with sufficiently generators and relations) does not verify $g_\mathcal{R} = g_\mathcal{C}$&quot; ?

http://mathoverflow.net/questions/57811/isotropic-subspaces-in-cohomology mister_jones 2011-03-08T14:25:42Z 2011-03-08T14:25:42Z

You are right, in full generality this is a question of linear algebra but maybe one can say something when the question is restricted to some class of manifolds. For instance this seems to be true for K&#228;hler manifolds and I want to see if this has something to do with Hodge decomposition.

http://mathoverflow.net/questions/53345/cohomological-dimension-of-a-group-fibration-and-local-coefficients mister_jones 2011-01-26T21:02:41Z 2011-01-26T21:02:41Z

I'm sorry M. Kent, I made a mistake in writing : What I want is to prove that there is no K&#228;hler group of cohomological dimension three.

http://mathoverflow.net/questions/53345/cohomological-dimension-of-a-group-fibration-and-local-coefficients mister_jones 2011-01-26T18:02:52Z 2011-01-26T18:02:52Z

I don't think so because in a way I try to prove something stronger. In fact we can adapt the proof in this article to show that if the cohomology of G satisfies 3-dimensional Poincar&#233; duality, then we have a contradiction. What I want to prove is that there is no K&#228;hler group of cohomological dimension one, without assumptions of Poincar&#233; duality.

http://mathoverflow.net/questions/53345/cohomological-dimension-of-a-group-fibration-and-local-coefficients mister_jones 2011-01-26T16:35:59Z 2011-01-26T16:35:59Z

Yes and in fact this is my original problem, where G is a K&#228;hler group and the epimorphism is induced by the Albanese map (G has 1-dimensional Albanese image).