In what sense (if any) does the cohomology of profinite groups commute with projective limits?

2010-04-08T18:53:43Z

Background:

Let $G$ be a profinite group. If $M$ is a discrete $G$-module, then $M=\varinjlim_U M^U$, where the direct limit is taken with respect to inclusions over all open normal subgroups of $G$, and one naturally has $H^n(G,M)\simeq\varinjlim H^n(G/U,M^U)$, where the cohomology groups on the right can be regarded as the usual abstract cohomology groups of the finite groups $G/U$ (this is sometimes, as in Serre's Local Fields, taken as the definition of $H^n(G,M)$).

More generally if one has a projective system of profinite groups $(G_i,\varphi_{ij})$ and a direct system of abelian groups $(M_i,\psi_{ij})$ such that $M_i$ is a discrete $G_i$-module and the pair $(\varphi_{ij},\psi_{ij})$ is compatible in the sense of group cohomology for all $i,j$, then $\varinjlim M_i$ is canonically a discrete $\varprojlim G_i$-module, the groups $H^n(G_i,M_i)$ form a direct system, and one has $H^n(\varprojlim G_i,\varinjlim M_i)\simeq\varinjlim H^n(G_i,M_i)$. The statement and straightforward proof of this more general result can be found, for instance, in Shatz' book on profinite groups.

Question:

In general, I'm wondering if there are, under appropriate hypotheses, any similar formulae for projective limits of discrete $G$-modules. Now, given a projective system of discrete $G$-modules $(M_i,\psi_{ij})$, it isn't even obvious to me that the limit will again be a discrete $G$-module, and at any rate, while each $M_i$ is discrete, the limit (in its natural topology) will be discrete if and only if it is finite. So, for the sake of specificity, I'll give a particular situation in which I'm interested. If $R$ is a complete, Noetherian local ring with maximal ideal $\mathfrak{m}$ and finite residue field and $M$ is a finite, free $R$-module as well as a discrete $G$-module such that the $G$-action is $R$-linear, then the canonical isomorphism of $R$-modules $M\simeq\varprojlim M/\mathfrak{m}^iM$ is also a $G$-module isomorphism (each $M/\mathfrak{m}^iM$ is a discrete $G$-module with action induced from that of $M$). Moreover, in this case, one can see that the limit is a discrete $G$-module (because it is isomorphic to one as an abstract $G$-module!). There is a natural homomorphism $C^n(G,M)\rightarrow\varprojlim C^n(G,M/\mathfrak{m}^iM)$ where the projective limit is taken with respect to the maps induced by the projections $M/\mathfrak{m}^jM\rightarrow M/\mathfrak{m}^iM$, and this induces similar map on cohomology. I initially thought the map at the level of cochains was trivially surjective, just because of the universal property of projective limits. However, given a ``coherent sequence" of cochains $f_i:G\rightarrow M/\mathfrak{m}^iM$, the property gives me a map $f:G\rightarrow M$ that is continuous when $M$ is regarded in its natural profinite topology, which is, as I noted above, most likely coarser than the discrete topology, so this might not be a cochain. So, what I'd really like to know is whether or not the map on cohomology is an isomorphism.

Why I Care: The reason I'd like to know that the map described above is an isomorphism is to apply it to the particular case of $G=\hat{\mathbb{Z}}$. It is well known (and can be found, for instance, in Serre's Local Fields) that $H^2(\hat{\mathbb{Z}},A)=0$ for $A$ a torsion abelian group. In particular the higher cohomology of a finite $\hat{\mathbb{Z}}$-module vanishes, and I'd like to be able to conclude that the same is true for my $M$ above, being a projective limit of finite abelian groups.

Thanks!

Keenan

Answer by Leonid Positselski for In what sense (if any) does the cohomology of profinite groups commute with projective limits?

2010-10-07T13:23:05Z

For example, the profinite group cohomology $H^2(\hat{\mathbb Z}, \mathbb Z_p)$, where $\mathbb Z_p$ is considered as a trivial discrete $\hat{\mathbb Z}$-module, is isomorphic to $H^1(\hat{\mathbb Z},\mathbb Q_p/\mathbb Z_p)$ (since $H^i(\hat{\mathbb Z},\mathbb Q_p)=0$ for $i>0$). Which is isomorphic to $\mathbb Q_p/\mathbb Z_p$, hence nonzero.

Answer by Ahmed Matar for In what sense (if any) does the cohomology of profinite groups commute with projective limits?

2010-11-11T21:48:11Z

Hi Keenan,

You're right that the projective limit of discrete $G$-modules is not necessarily discrete. To take the cohomology of such "topological $G$-modules" you can use continuous cochain cohomology and this continuous cochain cohomology commutes with inverse limits under certain conditions. See section 7 of chapter II of Cohomology of Number Fields by Neukirch, Schmidt & Wingberg.

In what sense (if any) does the cohomology of profinite groups commute with projective limits? - MathOverflow

In what sense (if any) does the cohomology of profinite groups commute with projective limits?

Answer by Leonid Positselski for In what sense (if any) does the cohomology of profinite groups commute with projective limits?

Answer by Ahmed Matar for In what sense (if any) does the cohomology of profinite groups commute with projective limits?