The Stacks project

Lemma 59.57.3. Let $G$ be a topological group. Let $R$ be a ring. For every $i \geq 0$ the diagram

\[ \xymatrix{ \text{Mod}_{R, G} \ar[rr]_{H^ i(G, -)} \ar[d] & & \text{Mod}_ R \ar[d] \\ \text{Mod}_ G \ar[rr]^{H^ i(G, -)} & & \textit{Ab} } \]

whose vertical arrows are the forgetful functors is commutative.

Proof. Let us denote the forgetful functor $F : \text{Mod}_{R, G} \to \text{Mod}_ G$. Then $F$ has a left adjoint $H : \text{Mod}_ G \to \text{Mod}_{R, G}$ given by $H(M) = M \otimes _\mathbf {Z} R$. Observe that every object of $\text{Mod}_ G$ is a quotient of a direct sum of modules of the form $\mathbf{Z}[G/U]$ where $U \subset G$ is an open subgroup. Here $\mathbf{Z}[G/U]$ denotes the $G$-modules of finite $\mathbf{Z}$-linear combinations of right $U$ congruence classes in $G$ endowed with left $G$-action. Thus every bounded above complex in $\text{Mod}_ G$ is quasi-isomorphic to a bounded above complex in $\text{Mod}_ G$ whose underlying terms are flat $\mathbf{Z}$-modules (Derived Categories, Lemma 13.15.4). Thus it is clear that $LH$ exists on $D^-(\text{Mod}_ G)$ and is computed by evaluating $H$ on any complex whose terms are flat $\mathbf{Z}$-modules; this follows from Derived Categories, Lemma 13.15.7 and Proposition 13.16.8. We conclude from Derived Categories, Lemma 13.30.2 that

\[ \text{Ext}^ i(\mathbf{Z}, F(M)) = \text{Ext}^ i(R, M) \]

for $M$ in $\text{Mod}_{R, G}$. Observe that $H^0(G, -) = \mathop{\mathrm{Hom}}\nolimits (\mathbf{Z}, -)$ on $\text{Mod}_ G$ where $\mathbf{Z}$ denotes the $G$-module with trivial action. Hence $H^ i(G, -) = \text{Ext}^ i(\mathbf{Z}, -)$ on $\text{Mod}_ G$. Similarly we have $H^ i(G, -) = \text{Ext}^ i(R, -)$ on $\text{Mod}_{R, G}$. Combining everything we see that the lemma is true. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 59.57: Group cohomology

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DVD. Beware of the difference between the letter 'O' and the digit '0'.