Section 59.58 (0DVG): Tate's continuous cohomology

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 stacks.project@gmail.com.

Comments (4)

Comment #7013 by Joshua Ruiter on February 02, 2022 at 14:12

I believe the assertion regarding isomorphisms when $G$ 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 Johan on February 02, 2022 at 15:35

Are you sure? Because that reference has a different definition for $H^i_{cont}(G, -)$ , see Definition 4.2.2 in your reference. Moreover, the definition of $H^i(G, -)$ that is used there is certainly different from what is our $H^i(G, -)$ ; theirs doesn't use the topology on $G$ and ours does.

In fact, I think that the fact that our $H^i_{cont}$ is equal to their $H^i_{cont}$ 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!

Comment #8409 by Johan on May 18, 2023 at 06:35

A comment on the cohomology $H^1(G, M)$ used on this page. By Definition 59.57.2 it is given as the derived functors of taking invariants on the abelian category $\text{Mod}_G$ of discrete $G$ -modules defined in Definition 59.57.1. If $M$ is an object of this category, then $M$ has the discrete topology and the action of $G$ on $M$ is continuous. This means that every element of $M$ has an open stabilizer. If $G$ is $S^1$ this means that every element is fixed by all of $S^1$ and hence $H^0(S^1, M) = M$ . Thus for $S^1$ we are taking the derived functors of an exact functor and it follows that $H^1(S^1, M) = 0$ for all $M$ in $\text{Mod}_{S^1}$ .

Comment #9519 by Giancarlo Castellano on July 22, 2024 at 19:29

When $G$ is profinite and $M$ is discrete with a continuous $G$ -action, the $H^i_{cont}(G, M)$ as defined on this page form a universal cohomological $\delta$ -functor, and therefore, the canonical map $H^i(G, M) \to H^i_{cont}(G, M)$ is an isomorphism. A reference for this — though probably not the original one, and certainly not a very recent one — is Stephen S. Shatz' book Profinite groups, Arithmetic, and Geometry, Section II.§2; specifically, one needs to combine Proposition 5 and Theorem 8. (Note that Shatz writes $H^i$ for what this page denotes by $H^i_{cont}$ .)

Let me also weigh in on comments 7013–7014 above. Gille & Szamuely define $H^i_{\mathrm{cont}}(G, M) := \varinjlim H^i(G/U, M^U)$ where $U$ ranges over the open normal subgroups of $G$ . (I use the notation $H^i$ on the right-hand side because the way these groups are defined by Gille & Szamuely agrees with 59.57.2.(2).) On the other hand, in Shatz' book, it is proved (see Corollary 1 of Theorem 7 in Section II.§2) that $H^i_{cont}(G, M) = \varinjlim H^i_{cont}(G/U, M^U)$ where $U$ ranges over the open normal subgroups of $G$ . But each $G/U$ is discrete — even finite — so we can identify $H^i_{cont}(G/U, M^U) = H^i(G/U, M^U)$ for all $U$ ; therefore, Gille–Szamuely's definition is equivalent to the one on this page, after all.

What I think Remark 4.2.4 from Gille & Szamuely's book actually tells us is that, if $G_{\mathrm{disc}}$ denotes $G$ equipped with the discrete topology, as opposed to its given profinite topology, then $H^i(G_{\mathrm{disc}}, M) \ne H^i_{cont}(G, M)$ in general.

The Stacks project

59.58 Tate's continuous cohomology

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