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I am looking for an elementary example of a problem, for which one does not need many things to understand the question, but which can be solved with group homology or cohomology.

My background is, that I am looking for an introductional problem to motivate a talk about group homology and cohomology in a beginner's course.

Thank a lot!

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The best place for info is Ken Brown's book Cohomology of Groups, and it will contain many applications, including the classification of low-dimensional cohomology groups in terms of group extensions. –  Chris Gerig May 4 '13 at 18:18

5 Answers 5

Is there any finite simply-connected CW complex on which $\mathbb{Z}/2$ acts freely? No, since then the classifying space $B(\mathbb{Z}/2)$ would have finite homological dimension, which would imply that the group $\mathbb{Z}/2$ has finite homological dimension. But one can compute $H_p(\mathbb{Z}/2,\mathbb{Z})=\mathbb{Z}/2$ for odd $p$.

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The CW complex should be assumed to be connected and simply connected. Otherwise there are easy counterexamples: a two point space or the action on $S^1$ via the antipodal map. –  yup May 4 '13 at 12:23
    
Right, I've edited it. –  Martin Brandenburg May 4 '13 at 16:37
    
But $\mathbb Z/2$ acts freely on any sphere $S^n$ via the antipodal map. Am I missing something? –  Jim May 24 '13 at 9:36

A more or less tautological example that shows that group cohomology is "needed" is the fact that taking invariants under finite groups in short exact sequences does not preserve exactness.

It is easy to find examples of surjections $f:M\to N$ of $G$-modules such that the induced map $M^G\to N^G$ on the invariant subspaces is not surjective.

For example, if your students are familiar with basic algebraic topology (or de Rham cohomology) you can study in detail the following situation.

If $G$ is a finite group which acts properly discontinuously on a manifold $M$, then the quotient $M/G$ is also a manifold and its de Rham cohomology $H^\bullet(M/G)$ is just the invariant subspace $H^\bullet(M)^G$ for the natural action of $G$ on $H^\bullet(M/G)$. This can be proved completely by hand.

But if the group is not finite, then this is no longer true. A minimal example is $G=\mathbb Z$ acting on $M=\mathbb R$ by translations. Then $M/G$ is a circle, which has non-trivial $1$-cohomology, so $H^1(M/G)\not\cong H^1(M)^G$.

What's wrong with this example is that $\mathbb Z$ has cohomology inpositive degrees, so we don't get isomorphisms but short exact sequences involving group cohomology. In this particular case one can be complete explicit about this.

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You can find the details of the calculation of the cohomology of $M/\mathbb Z$ for a properly discontinuous action of $\mathbb Z$ on a manifold $M$ in these notes (which are in Spanish); this is done without mentioning group cohomology. –  Mariano Suárez-Alvarez May 4 '13 at 17:12
    
I didn't expect to see such a hands-on and geometric answer to this question. Thank you! –  Adam Saltz May 4 '13 at 17:48

Another application is called "Galois descent". Roughly, for a Galois extension $K/k$, if two structures are isomorphic over $K$, we may ask whether they are isomorphic over $k$. Galois descent provides an answer in terms of Galois cohomology.

As an example, let $k$ be a field and let $M$ be a matrix with entries in $k$ and let $K/k$ be a Galois extension. Then, $\operatorname{GL}_n(K)$ conjugacy class of matrices in $\operatorname{M}_n(k)$ containing $M$ is a union of $\operatorname{GL}_n(k)$ conjugacy class of matrices. The set containing those conjugacy classes are in one to one correspondence between $H^1(\operatorname{Gal}(K/k), Z_{\operatorname{GL}_n(K)}(M))$.

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If a group $G$ acts on an abelian group $N$, we can form the semidirect product $N\rtimes G$ and there is a canonical surjection $p:N\rtimes G\to G$.

This surjection is split, and in fact split surjections are of this form with abelian kernel.

Now, a split surjection admits many different splittings. As soon as you try to classify them, you end up with $H^1(G,N)$.

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The construction of cross product algebras is a very natural problem.

It is very easy to arrive at the $2$-cocycle condition for assocativity and to the condition that such cocycles be cohomologous for the algebras to be isomoorphic.

With sufficient hand waving, this example can be concluded by mentioning the Brauer group of fields and the amazing fact that they are "just" group cohomology groups.

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