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

Question

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 1

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.

Answer 2

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.