Proposition 4

Let $f\in L^2(G)$. Then $f$ is central if and only if $u(f)$ is a homothety for all $u\in \widehat{G}$. In that case

By Prop. 1 (no. 1), to say that $f$ is central means that $u(\gamma (s)f) =$ $u(\delta (s^{-1})f)$ for all $s\in G$ and all $u\in \widehat{G}$; but this can also be written as $u(s)u(f) =u(f)u(s)$ for all $s\in G$ and all $u\in \widehat{G}$ (formulas (12) and (13)), hence the first assertion of Prop. 4 (Schur’s lemma). If $u(f)$ is a homothety, then $u(f) =\lambda_u\varepsilon_u$ with

$\lambda_u=\frac{1}{d(u)}$ Tr($u(f)$) $=\frac{1}{d(u)}\int_Gf(g$)Tr($u(g)$)$dg=\frac{1}{d(u)}\int_Gf(g)\chi_u(g)dg$.

Consequently, for $f\in$ ZL$^2(G)$ we have

so

with

$uu$

Conversely, if $\varphi$ is a square-integrable complex function on $\widehat{G}$, the element $(\frac{\varphi(u)}{d(u)}\varepsilon_u)_{u\in\widehat{G}}$ of $F( \widehat{G})$ belongs to $L^2( \widehat{G})$, and we have (formula (9)) $(\mathscr{F}_u(\frac{\varphi(u)}{d(u)}\varepsilon_u))(g) =d(u$)Tr $(\frac{\varphi(u)}{d(u)}\varepsilon_uu(g)^{-1})=\varphi (u)\overline{\chi}_u(g)$,

so

$u\in \widehat{G}$

Note, in particular, that formulas (20) and (21) give, for $u, v$ in $\widehat{G},^8$

$u(\overline{\chi}_v) = 0$ if $u\not=v$, (23)

$u(\overline{\chi}_u) =\frac{\varepsilon_u}{d(u)}\in$ End(E$_u$), (24)

$\overline{\mathscr{F}}(\chi_u) =\frac{\varepsilon_u}{d(u)}\in$ End(E$_u$)$\subset F( \widehat{G})$. (25)