Remark 64.15.10. Let us try to illustrate the content of the formula of Lemma 64.15.8. Suppose that $\Lambda$, viewed as a trivial $\Gamma$-module, admits a finite resolution $0\to P_ r\to \ldots \to P_1 \to P_0\to \Lambda \to 0$ by some $\Lambda [\Gamma ]$-modules $P_ i$ which are finite and projective as $\Lambda [G]$-modules. In that case

$H_*\left(\left(P_\bullet \right)_ G\right) = \text{Tor}_*^{\Lambda [G]}\left(\Lambda , \Lambda \right) = H_*(G, \Lambda )$

and

$\text{Tr}_\Lambda ^{Z_\gamma }\left(\gamma , P_\bullet \right) =\frac{1}{\# Z_\gamma }\text{Tr}_\Lambda (\gamma , P_\bullet )=\frac{1}{\# Z_\gamma }\text{Tr}(\gamma , \Lambda ) = \frac{1}{\# Z_\gamma }.$

Therefore, Lemma 64.15.8 says

$\text{Tr}_\Lambda (1 , P_ G) = \text{Tr}\left(1 |_{H_*(G, \Lambda )}\right) = {\sum _{\gamma \mapsto 1}}'\frac{1}{\# Z_\gamma }.$

This can be interpreted as a point count on the stack $BG$. If $\Lambda = \mathbf{F}_\ell$ with $\ell$ prime to $\# G$, then $H_*(G, \Lambda )$ is $\mathbf{F}_\ell$ in degree 0 (and $0$ in other degrees) and the formula reads

$1 = \sum \nolimits _{ \frac{\sigma \text{-conjugacy}}{\text{classes}\langle \gamma \rangle } } \frac{1}{\# Z_\gamma } \mod \ell .$

This is in some sense a “trivial” trace formula for $G$. Later we will see that (64.14.3.1) can in some cases be viewed as a highly nontrivial trace formula for a certain type of group, see Section 64.30.

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).