Lemma 64.15.1. 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 $.
$\square$
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