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

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

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. $\square$

Comment #3622 by Owen B on

I find the discussion involving the 'commutative triangle' somewhat obscure... the trace map $\sum_{\gamma\mapsto 1}\gamma$ is a map $P\rightarrow P^G$ which factors through $P_G$.

Writing $P\xrightarrow{i} P'\xrightarrow{j}P$ as a direct summand of a free $\Lambda[G]$ module $P'$, $i_G$ and $j_G$ write $P_G$ as a direct summand of the free $\Lambda$-module $\Lambda\otimes_{\Lambda[G]}P'$, and so we verify it is a projective $\Lambda$-module.

The surjection of $\Lambda$-modules $P\xrightarrow{r}\Lambda\otimes_{\Lambda[G]}P$ allows us to write $P=P_G\oplus\ker r$, and it is clear from the discussion above that $\sum_{\gamma\mapsto1}\gamma$ acts by $0$ on $\ker r$. So indeed $\text{Tr}_\Lambda(\sum_{\gamma\mapsto1}\gamma,P)=\text{Tr}_\Lambda(\sum_{\gamma\mapsto1}\gamma,P_G)$.

Comment #3725 by on

OK, the arrow $c$ was pointing in the wrong direction (fixed here). Does that help?

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