## 63.15 Preliminaries and sorites

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

$1\to G\to \tilde\Gamma \to \mathbf{Z}\to 1$

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

$\text{Tr}_{\Lambda }\left(1; \left(P \otimes _ A M\right)_ G\right) \in \Lambda ^\natural .$

The element $1\in \mathbf{N}$ will correspond to the Frobenius.

Lemma 63.15.1. Let $e\in G$ denote the neutral element. The map

$\begin{matrix} \Lambda [G] & \longrightarrow & \Lambda ^{\natural } \\ \sum \lambda _ g\cdot g & \longmapsto & \lambda _ e \end{matrix}$

factors through $\Lambda [G]^\natural$. We denote $\varepsilon : \Lambda [G]^\natural \to \Lambda ^\natural$ the induced map.

Proof. We have to show the map annihilates commutators. One has

$\left(\sum \lambda _ g g\right)\left(\sum \mu _ g g\right)-\left(\sum \mu _ g g\right)\left(\sum \lambda _ g g\right) = \sum _ g\left(\sum _{g_1g_2=g} \lambda _{g_1}\mu _{g_2}-\mu _{g_1}\lambda _{g_2}\right)g$

The coefficient of $e$ is

$\sum _ g\left(\lambda _ g\mu _{g^{-1}}-\mu _ g\lambda _{g^{-1}}\right) = \sum _ g\left(\lambda _ g\mu _{g^{-1}}-\mu _{g^{-1}}\lambda _ g\right)$

which is a sum of commutators, hence it zero in $\Lambda ^\natural$. $\square$

Definition 63.15.2. Let $f : P\to P$ be an endomorphism of a finite projective $\Lambda [G]$-module $P$. We define

$\text{Tr}_{\Lambda }^ G(f; P) := \varepsilon \left(\text{Tr}_{\Lambda [G]}(f; P)\right)$

to be the $G$-trace of $f$ on $P$.

Lemma 63.15.3. 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 63.15.4. The map $A\to \Lambda$ defines an $A$-module structure on $\Lambda ^\natural$.

Proof. This is clear. $\square$

Lemma 63.15.5. 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

$\left(\sum \lambda _ g g\right)\left(p \otimes m\right) = \sum g p \otimes \lambda _ g g m.$

Proof. For any $\Lambda [G]$-module $N$ one has

$\mathop{\mathrm{Hom}}\nolimits _{\Lambda [G]}\left(P \otimes _ A M, N\right)= \mathop{\mathrm{Hom}}\nolimits _{A[G]}\left(P, \mathop{\mathrm{Hom}}\nolimits _{\Lambda }(M, N)\right)$

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 63.15.6. With assumptions as in Lemma 63.15.5, 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

$\begin{matrix} \varphi : & A[G]\otimes _ A M & \cong & \left(A[G]\otimes _ A M\right)' \\ & g \otimes m & \longmapsto & g \otimes g^{-1}m \end{matrix}$

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,

$\varphi \circ (u \otimes v) \circ \varphi ^{-1} = \sum _ ga_ gR_ g \otimes g^{-1}v.$

Working out explicitly both sides of the equation, we have to show

$\text{Tr}_\Lambda ^ G\left(\sum _ g a_ gR_ g \otimes g^{-1}v\right) = a_ e\cdot \text{Tr}_\Lambda (v; M).$

This is done by showing that

$\text{Tr}_\Lambda ^ G\left(a_ gR_ g \otimes g^{-1}v\right) = \left\{ \begin{matrix} 0 & \text{ if } g\neq e \\ a_ e\text{Tr}_\Lambda \left(v; M\right) & \text{ if }g = e \end{matrix} \right.$

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 63.15.7. 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 63.15.3. $\square$

Lemma 63.15.8. 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 63.15.6. $\square$

Lemma 63.15.9. 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

$\text{Tr}_\Lambda (1; P_ G) = \sum \nolimits '_{\gamma \mapsto 1} \text{Tr}_\Lambda ^{Z_\gamma }(\gamma , P)$

where $\sum _{\gamma \mapsto 1}'$ means taking the sum over the $G$-conjugacy classes in $\Gamma$.

Sketch of proof. We first prove this after multiplying by $\# G$.

$\# G\cdot \text{Tr}_\Lambda (1; P_ G) = \text{Tr}_\Lambda (\sum \nolimits _{\gamma \mapsto 1} \gamma , P_ G) = \text{Tr}_\Lambda (\sum \nolimits _{\gamma \mapsto 1} \gamma , P)$

where the second equality follows by considering the commutative triangle

$\xymatrix{ P^ G \ar[rd]_ a & & P_ G \ar[ll]^ c \\ & P \ar[ur]_ b }$

where $a$ is the canonical inclusion, $b$ the canonical surjection and $c = \sum _{\gamma \mapsto 1} \gamma$. Then we have

$(\sum \nolimits _{\gamma \mapsto 1} \gamma ) |_ P = a \circ c \circ b \quad \text{and}\quad (\sum \nolimits _{\gamma \mapsto 1} \gamma ) |_{P_ G} = b \circ a \circ c$

hence they have the same trace. We then have

$\# G\cdot \text{Tr}_\Lambda (1; P_ G) = {\sum _{\gamma \mapsto 1}}' \frac{\# G}{\# Z_\gamma }\text{Tr}_\Lambda (\gamma , P) = \# G{\sum _{\gamma \mapsto 1}}' \text{Tr}_\Lambda ^{Z_\gamma }(\gamma , P).$

To finish the proof, reduce to case $\Lambda$ torsion-free by some universality argument. See [SGA4.5] for details. $\square$

Remark 63.15.10. Let us try to illustrate the content of the formula of Lemma 63.15.8. 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 63.15.8 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 (63.14.3.1) can in some cases be viewed as a highly nontrivial trace formula for a certain type of group, see Section 63.30.

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