Question about computing group cohomology using cochains - MathOverflow

Question about computing group cohomology using cochains

2010-05-09T06:00:44Z

In Milne's notes on Class Field Theory (http://www.jmilne.org/math/CourseNotes/CFT.pdf), he initially defines group cohomology in terms of injective resolutions, then he talks about computing cohomology using cochains. I don't see him mention anywhere that the group has to be finite in order for cochains to work, but this seems to be the case?

Later, he discusses profinite groups, in which he says that cohomology of profinite groups can be computed using continuous cochains. What isn't clear is the following: is the cohomology using continuous cochains a modified cohomology theory, different from the one using injective resolutions? In this case, then, do we get the cohomology theory using injective resolutions if we use all cochains, not just continuous ones? Or do the continuous cochains give the same cohomology as injective resolutions, and cochains which are not necessarily continuous only give cohomology in the case of finite groups? I.e., is there only one such cohomology theory? At the very least, using general cochains versus continuous cochains in the case of infinite profinite groups is different, for in one case $H^1$ is $\mathrm{Hom}(G,M)$ when M is trivial, and in the other case $H^1$ is $\mathrm{Hom}_{\mathrm{cts}}(G,M)$.

Assuming that set-theoretic cochains only work for finite groups, why is it the case? It seems that the proof that cochains compute cohomology (i.e. by looking at a projective resolution of $\mathbb{Z}$) fails because the modules used in the case of finite groups, i.e. tensor powers of $\mathbb{Z}[G]$, aren't necessarily projective when $G$ is infinite (in the case when $G$ is finite, they are free). Is this correct?

Answer by Robin Chapman for Question about computing group cohomology using cochains

2010-05-09T06:38:26Z

You are basically correct. The point is that for applications to class field theory one does use a modified theory, the "cohomology of profinite groups", which is not the same as plain cohomology of groups.

Cochains do work for cohomology of arbitrary groups. Tensor powers of $\mathbb{Z}[G]$ are free $\mathbb{Z}[G]$-modules on the basis $1\otimes g_2\otimes\cdots\otimes g_n$. But in class field theory we don't use the plain theory for infinite Galois group. The theory we do use is either defined by taking direct limits of the theory for finite groups, or by using continuous cochains.

Answer by Homology for Question about computing group cohomology using cochains

2010-05-09T09:42:50Z

You should take a look at the beginning of chapter 2 of Serre's Galois Cohomology. He explains there that if G is a profinite group, then the category of discrete abelian groups with a continuous action of G has enough injectives (but not enough projectives in general), and that cohomology can be "computed" as a direct limit of cohomology of a finite group. To sum up:

-if G is discrete (i.e. no topology), then the category has enough injectives and projectives (it is the category of left $\mathbb{Z}[G]$-modules), and using a projective resolution for $\mathbb{Z}$ gives you the equivalence between the derived functor definition and the cochains definition (using the fact that Ext can be computed two ways). You can find this in Serre's Local Fields.

-if G is profinite, and we consider the category of discrete modules with a continuous action of G, then there are enough injectives (this can be seen quite easily from the discrete case), but not enough projectives. Luckily though, the two definitions agree (thanks to the "direct limit computability"). You can find this in Serre's Galois Cohomology.

-if G is an arbitrary topological group, there is not much left. There aren't enough injectives nor projectives in general, and if you define cohomology with cochains, you don't get an homological functor (only the beginning of the long exact sequence exists). However, see the end of J.-M. Fontaine and Yi Ouyang's book (it's a pdf, I found it on Fontaine's web page) about p-adic representations, they mention that if you have a continuous set-theoretic section in your short exact sequence, you get a long exact sequence. I haven't read the reference they provide, though.