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59.58 Tate's continuous cohomology

Tate's continuous cohomology ([Tate]) is defined by the complex of continuous inhomogeneous cochains. We can define this when $M$ is an arbitrary topological abelian group endowed with a continuous $G$-action. Namely, we consider the complex

\[ C^\bullet _{cont}(G, M) : M \to \text{Maps}_{cont}(G, M) \to \text{Maps}_{cont}(G \times G, M) \to \ldots \]

where the boundary map is defined for $n \geq 1$ by the rule

\begin{align*} \text{d}(f)(g_1, \ldots , g_{n + 1}) & = g_1(f(g_2, \ldots , g_{n + 1})) \\ & + \sum \nolimits _{j = 1, \ldots , n} (-1)^ jf(g_1, \ldots , g_ jg_{j + 1}, \ldots , g_{n + 1}) \\ & + (-1)^{n + 1}f(g_1, \ldots , g_ n) \end{align*}

and for $n = 0$ sends $m \in M$ to the map $g \mapsto g(m) - m$. We define

\[ H^ i_{cont}(G, M) = H^ i(C^\bullet _{cont}(G, M)) \]

Since the terms of the complex involve continuous maps from $G$ and self products of $G$ into the topological module $M$, it is not clear that this turns a short exact sequence of topological modules into a long exact cohomology sequence. Another difficulty is that the category of topological abelian groups isn't an abelian category!

However, a short exact sequence of discrete $G$-modules does give rise to a short exact sequence of complexes of continuous cochains and hence a long exact cohomology sequence of continuous cohomology groups $H^ i_{cont}(G, -)$. Therefore, on the category $\text{Mod}_ G$ of Definition 59.57.1 the functors $H^ i_{cont}(G, M)$ form a cohomological $\delta $-functor as defined in Homology, Section 12.12. Since the cohomology $H^ i(G, M)$ of Definition 59.57.2 is a universal $\delta $-functor (Derived Categories, Lemma 13.16.6) we obtain canonical maps

\[ H^ i(G, M) \longrightarrow H^ i_{cont}(G, M) \]

for $M \in \text{Mod}_ G$. It is known that these maps are isomorphisms when $G$ is an abstract group (i.e., $G$ has the discrete topology) or when $G$ is a profinite group (insert future reference here). If you know an example showing this map is not an isomorphism for a topological group $G$ and $M \in \mathop{\mathrm{Ob}}\nolimits (\text{Mod}_ G)$ please email

Comments (2)

Comment #7013 by Joshua Ruiter on

I believe the assertion regarding isomorphisms when is profinite is false. A counterexample is given in Gille & Szamuely's Central Simple Algebras and Galois Cohomology remark 4.2.4.

Comment #7014 by on

Are you sure? Because that reference has a different definition for , see Definition 4.2.2 in your reference. Moreover, the definition of that is used there is certainly different from what is our ; theirs doesn't use the topology on and ours does.

In fact, I think that the fact that our is equal to their is the result that you claim is wrong and that the text above claims is true.

Sorry, but this is just incredibly confusing; so can you please very carefully match the definitions used in both locations and then comment again. Thanks a bunch!

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