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?

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    $\begingroup$ Uniform answer (including finite groups with discrete topology as profinite groups) is to work in category of discrete $G$-modules, so take injective resolutions there. In general, for $G$-module $M$ its "discretization" is maximal discrete $G$-submodule (i.e., elts with open stabilizer). This has univ. mapping property, so discretization of injective $G$-mod is inj. in category of discrete $G$-mods. So latter category has enough injectives. For finite discrete $G$ it's the old thing. For general $G$ recovers "cont. cochain" construction, via work; dropping continuity does matter. $\endgroup$ – BCnrd May 9 '10 at 8:12
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    $\begingroup$ For a group (no topology), I define the cohomology using injective resolutions in the category of all G-modules, and I prove that it can be computed using cochains (no conditions on G). For a profinite group, I define the cohomology using injective resolutions in the category of all discrete G-modules, and state that it can be computed using continuous cochains. A finite group can be regarded as a profinite group with the discrete topology, in which case the two definitions coincide (obviously). $\endgroup$ – JS Milne May 9 '10 at 19:53
  • $\begingroup$ Yes, that make sense. My main problem is that I didn't realize that a module could be injective in the category of discrete G-modules but not in the category of G-modules, which is what lead me to wonder whether cochains only worked for finite groups. This last wonder was in part justified by the fact that I hadn't looked in detail at the proof that cochains worked, which is only in the appendix. Thanks for clearing up the confusion! $\endgroup$ – David Corwin May 10 '10 at 3:31

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.

  • $\begingroup$ At first, I was confused by the following: If we compute the cohomology of profinite groups using injective modules in the category of discrete modules in order to get the theory using continuous cochains, then how is this theory any different from the normal cohomology theory? (But the two clearly are different, for example in the case of $H^1$). $\endgroup$ – David Corwin May 9 '10 at 15:16
  • $\begingroup$ But I think the answer to my concerns is the following: An injective object in the category of discrete $G$-modules is not necessarily injective in the category of $G$-modules. Hence, when we compute cohomology of profinite groups in the category of discrete modules, the injective resolution isn't injective in the category of $G$-modules, so the cohomology we get isn't the same as normal group cohomology. At the very least, we can conclude that the category of discrete $G$-modules doesn't have enough injectives-which-are-injective-in-the-category-of-$G$-modules-as-well. Is all this true? $\endgroup$ – David Corwin May 9 '10 at 15:20
  • $\begingroup$ I disagree with the last part of the answer ("if G is an arbitrary topological group, there is not much left"): there is a very developed theory of continuous cohomology for topological groups and, in particular, Lie groups and p-adic Lie groups. See the books of Guichardet and Borel-Wallach. The key technical notion in Guichardet is a "strong relatively injective resolution". Of course, it would be facile to expect that the cohomology of discrete and t.d. groups had an exact analogue for arbitrary topological groups, but much is known for the Lie case. $\endgroup$ – Victor Protsak May 9 '10 at 19:13
  • $\begingroup$ yes, I should have written "there is not much left that I know of" ;) @Davidac: you are right, injective is relative to the category you're working with. $\endgroup$ – Homology May 9 '10 at 20:11

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.

  • $\begingroup$ Actually, I see you're right - you don't need that $G$ is finite to show that tensor powers of $\mathbb{Z}[G]$ are free. $\endgroup$ – David Corwin May 9 '10 at 15:07

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