Notation: We fix the notation for this section. We denote by $A$ a commutative ring, $\Lambda $ a (possibly noncommutative) ring with a ring map $A\to \Lambda $ which image lies in the center of $\Lambda $. We let $G$ be a finite group, $\Gamma $ a monoid extension of $G$ by $\mathbf{N}$, meaning that there is an exact sequence
and $\Gamma $ consists of those elements of $\tilde\Gamma $ which image is nonnegative. Finally, we let $P$ be an $A[\Gamma ]$-module which is finite and projective as an $A[G]$-module, and $M$ a $\Lambda [\Gamma ]$-module which is finite and projective as a $\Lambda $-module.
Our goal is to compute the trace of $1 \in \mathbf{N}$ acting over $\Lambda $ on the coinvariants of $G$ on $P \otimes _ A M$, that is, the number
The element $1\in \mathbf{N}$ will correspond to the Frobenius.
Lemma 64.15.1. Let $e\in G$ denote the neutral element. The map factors through $\Lambda [G]^\natural $. We denote $\varepsilon : \Lambda [G]^\natural \to \Lambda ^\natural $ the induced map.
\[ \begin{matrix} \Lambda [G]
& \longrightarrow
& \Lambda ^{\natural }
\\ \sum \lambda _ g\cdot g
& \longmapsto
& \lambda _ e
\end{matrix} \]
Proof. We have to show the map annihilates commutators. One has
The coefficient of $e$ is
which is a sum of commutators, hence it zero in $\Lambda ^\natural $. $\square$
Definition 64.15.2. Let $f : P\to P$ be an endomorphism of a finite projective $\Lambda [G]$-module $P$. We define to be the $G$-trace of $f$ on $P$.
\[ \text{Tr}_{\Lambda }^ G(f; P) := \varepsilon \left(\text{Tr}_{\Lambda [G]}(f; P)\right) \]
Lemma 64.15.3. Let $f : P\to P$ be an endomorphism of the finite projective $\Lambda [G]$-module $P$. Then
\[ \text{Tr}_{\Lambda }(f; P) = \# G \cdot \text{Tr}_\Lambda ^ G(f; P). \]
Proof. By additivity, reduce to the case $P = \Lambda [G]$. In that case, $f$ is given by right multiplication by some element $\sum \lambda _ g\cdot g$ of $\Lambda [G]$. In the basis $(g)_{g \in G}$, the matrix of $f$ has coefficient $\lambda _{g_2^{-1}g_1}$ in the $(g_1, g_2)$ position. In particular, all diagonal coefficients are $\lambda _ e$, and there are $\# G$ such coefficients. $\square$
Lemma 64.15.4. The map $A\to \Lambda $ defines an $A$-module structure on $\Lambda ^\natural $.
Proof. This is clear. $\square$
Lemma 64.15.5. Let $P$ be a finite projective $A[G]$-module and $M$ a $\Lambda [G]$-module, finite projective as a $\Lambda $-module. Then $P \otimes _ A M$ is a finite projective $\Lambda [G]$-module, for the structure induced by the diagonal action of $G$.
Note that $P \otimes _ A M$ is naturally a $\Lambda $-module since $M$ is. Explicitly, together with the diagonal action this reads
Proof. For any $\Lambda [G]$-module $N$ one has
where the $G$-action on $\mathop{\mathrm{Hom}}\nolimits _{\Lambda }(M, N)$ is given by $(g\cdot \varphi )(m) = g \varphi (g^{-1} m) $. Now it suffices to observe that the right-hand side is a composition of exact functors, because of the projectivity of $P$ and $M$. $\square$
Lemma 64.15.6. With assumptions as in Lemma 64.15.5, let $u\in \text{End}_{A[G]}(P)$ and $v\in \text{End}_{\Lambda [G]}(M)$. Then
\[ \text{Tr}_\Lambda ^ G \left(u \otimes v; P \otimes _ A M\right) = \text{Tr}_ A^ G(u; P)\cdot \text{Tr}_\Lambda (v;M). \]
Sketch of proof. Reduce to the case $P=A[G]$. In that case, $u$ is right multiplication by some element $a = \sum a_ gg$ of $A[G]$, which we write $u = R_ a$. There is an isomorphism of $\Lambda [G]$-modules
where $\left(A[G]\otimes _ A M\right)'$ has the module structure given by the left $G$-action, together with the $\Lambda $-linearity on $M$. This transport of structure changes $u \otimes v$ into $\sum _ ga_ gR_ g \otimes g^{-1}v$. In other words,
Working out explicitly both sides of the equation, we have to show
This is done by showing that
by reducing to $M=\Lambda $. $\square$
Notation: Consider the monoid extension $1 \to G\to \Gamma \to \mathbf{N} \to 1$ and let $\gamma \in \Gamma $. Then we write $Z_\gamma = \{ g\in G | g\gamma = \gamma g\} $.
Lemma 64.15.7. Let $P$ be a $\Lambda [\Gamma ]$-module, finite and projective as a $\Lambda [G]$-module, and $\gamma \in \Gamma $. Then
\[ \text{Tr}_{\Lambda }(\gamma , P) = \# Z_\gamma \cdot \text{Tr}_\Lambda ^{Z_\gamma }\left(\gamma , P\right). \]
Proof. This follows readily from Lemma 64.15.3. $\square$
Lemma 64.15.8. Let $P$ be an $A[\Gamma ]$-module, finite projective as $A[G]$-module. Let $M$ be a $\Lambda [\Gamma ]$-module, finite projective as a $\Lambda $-module. Then
\[ \text{Tr}_{\Lambda }^{Z_\gamma }(\gamma , P \otimes _ A M) = \text{Tr}_ A^{Z_\gamma }(\gamma , P)\cdot \text{Tr}_\Lambda (\gamma , M). \]
Proof. This follows directly from Lemma 64.15.6. $\square$
Lemma 64.15.9. Let $P$ be a $\Lambda [\Gamma ]$-module, finite projective as $\Lambda [G]$-module. Then the coinvariants $P_ G = \Lambda \otimes _{\Lambda [G]} P$ form a finite projective $\Lambda $-module, endowed with an action of $\Gamma /G = \mathbf{N}$. Moreover, we have where $\sum _{\gamma \mapsto 1}'$ means taking the sum over the $G$-conjugacy classes in $\Gamma $.
\[ \text{Tr}_\Lambda (1; P_ G) = \sum \nolimits '_{\gamma \mapsto 1} \text{Tr}_\Lambda ^{Z_\gamma }(\gamma , P) \]
Sketch of proof. We first prove this after multiplying by $\# G$.
where the second equality follows by considering the commutative triangle
where $a$ is the canonical inclusion, $b$ the canonical surjection and $c = \sum _{\gamma \mapsto 1} \gamma $. Then we have
hence they have the same trace. We then have
To finish the proof, reduce to case $\Lambda $ torsion-free by some universality argument. See [SGA4.5] for details. $\square$
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 and Therefore, Lemma 64.15.8 says 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 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.
\[ H_*\left(\left(P_\bullet \right)_ G\right) = \text{Tor}_*^{\Lambda [G]}\left(\Lambda , \Lambda \right) = H_*(G, \Lambda ) \]
\[ \text{Tr}_\Lambda ^{Z_\gamma }\left(\gamma , P_\bullet \right) =\frac{1}{\# Z_\gamma }\text{Tr}_\Lambda (\gamma , P_\bullet )=\frac{1}{\# Z_\gamma }\text{Tr}(\gamma , \Lambda ) = \frac{1}{\# Z_\gamma }. \]
\[ \text{Tr}_\Lambda (1 , P_ G) = \text{Tr}\left(1 |_{H_*(G, \Lambda )}\right) = {\sum _{\gamma \mapsto 1}}'\frac{1}{\# Z_\gamma }. \]
\[ 1 = \sum \nolimits _{ \frac{\sigma \text{-conjugacy}}{\text{classes}\langle \gamma \rangle } } \frac{1}{\# Z_\gamma } \mod \ell . \]
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