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
\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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