Proposition 7

Let $f$ be a mapping from G to K. The following assertions are equivalent:

(i) There are an element $\lambda$ of $\widehat{G}$ and an element $a$ of K such that $f=a\chi_{\lambda}$.

(ii) For every pair $(g, g')$ of elements of G, we have

Let $\lambda$ be an element of $\widehat{G}$; for every endomorphism $u$ of $V_{\lambda}$, set

The endomorphism $u^{\natural}$ of $V_{\lambda}$ is K[G]-linear. By Schur’s lemma (VIII, p. 47, Theorem 1)$,u^{\natural}$ is a homothety. Since $u$ and $u^{\natural}$ have the same trace, we therefore have

$u^{\natural}=d^{-1}_{\lambda}$ Tr($u$) $1_{V_{\lambda}}$.

Let $u$ and $v$ be endomorphisms of $V_{\lambda}$; it follows that

(36) Tr($u$) Tr($v$) $=d_{\lambda}$ Tr($u^{\natural}v$) $=d_{\lambda}|G|^{-1}\sum_{h\in G}$ Tr($\pi_{\lambda}(h)u \pi_{\lambda}(h^{-1})v$).

Take $u=\pi_{\lambda}(g)$ and $v=\pi_{\lambda}(g')$ in formula (36); relation (34) for $f=\chi_{\lambda}$ follows.

Conversely, suppose that assertion (ii) holds. If we have $f(1) = 0$, then it follows that $f(g)f(g') = 0$ for every pair $(g, g')$ of elements of G, and therefore $f= 0$. We can therefore assume $f(1)\not= 0$. Taking $g'= 1$ in (34), we obtain the relation

for every $g\in G$, which implies that $f$ is a central function. By Proposition 5, there exists a family $(a_{\lambda})$ of elements of K such that $f=\sum_{\lambda\in\widehat{G}}a_{\lambda}\chi_{\lambda}$. Let us replace $f$ with this expression in formula (34); taking into account that each of the characters $\chi_{\lambda}$ also satisfies this relation, we find

for $g, g'\in G$. This relation can also be written as

for $g, g'\in G$. Now, the functions $\chi_{\lambda}$, for $\lambda \in \widehat{G}$, are linearly independent (Proposition 5 of VIII, p. 411); it follows that

for $\lambda , \mu\in \widehat{G}$. In particular, $a_{\lambda}a_\mu= 0$ whenever $\lambda \not=\mu$. Consequently, there exists at most one element $\lambda$ of $\widehat{G}$ such that $a_{\lambda}\not= 0$, and we have $f=a_{\lambda}\chi_{\lambda}$, and therefore (i).