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 $M=\Lambda$. $\square$

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