Bourbaki

Lemma 4

Tag 03EP, permanent. Cite it as [Bourbaki, Tag 03EP].

Théories spectrales, chapter V, § 4. Représentations des groupes compacts

Let π\pi be an irreducible representation of G. The mapping f↦π(f)f\mapsto \pi (f) defines by passing to the subspaces an isometric isomorphism of ϱG(π‾)\boldsymbol{\varrho }_G(\overline{\pi}) into End(Eπ_{\pi}).

Put επ(g,h)u=π(g)∘u∘π(h−1)\varepsilon_{\pi}(g, h)u=\pi (g)\circ u\circ \pi (h^{-1}) for every (g,h)∈G×G(g, h)\in G\times G and for every u∈u\in End(Eπ_{\pi}). The mapping επ\varepsilon_{\pi} is a continuous representation of G×GG\times G in End(Eπ_{\pi}). It is unitary. In fact, since π\pi itself is unitary, one obtains

∥επ(g,h)u∥22=\|\varepsilon_{\pi}(g, h)u\|^2_2= dim(π\pi ) Tr((π(g)uπ(h−1))∗π(g)uπ(h−1))((\pi (g)u\pi (h^{-1}))^*\pi (g)u\pi (h^{-1}))

= dim(π\pi ) Tr(π(h)u∗uπ(h)−1\pi (h)u^*u\pi (h)^{-1}) == dim(π\pi ) Tr(u∗uu^*u) =∥u∥22=\|u\|^2_2 for every (g,h)∈G×G(g, h)\in G\times G and every u∈u\in End(Eπ_{\pi}).

The mapping Ψ defined by f↦π(f)f\mapsto \pi (f) is a (G×(G\times G)-morphism of ϱG(π‾)\boldsymbol{\varrho }_G(\overline{\pi}) into End(Eπ_{\pi}), since

π(ϱG(g,h)f)=∫Gf(g−1xh)π(x)dμ(x)=π(g)π(f)π(h−1) \pi (\boldsymbol{\varrho }_G(g, h)f) =\int_Gf(g^{-1}xh)\pi (x)d\mu(x) =\pi (g)\pi (f)\pi (h^{-1})

for every (g,h)∈G×G(g, h)\in G\times G and every f∈ϱG(π‾)f\in \boldsymbol{\varrho }_G(\overline{\pi}). Since ϱG(π‾)\boldsymbol{\varrho }_G(\overline{\pi}) is an irreducible representation (th. 1, a) of V, p. 462), there exists λ∈C∗\lambda \in \mathbf{C}^* such that the mapping λΨ\lambda \Psi is zero or isometric (cor. 5, a) of V, p. 388).

Let f=f= dim(π\pi )χ‾π∈ϱG(π‾)\overline{\chi}_{\pi}\in \boldsymbol{\varrho }_G(\overline{\pi}). Then π(f)=1Eπ\pi (f) = 1_{E_{\pi}} (cor. 2 of V, p. 465), whence ∥π(f)∥2=\|\pi (f)\|_2= dim(π\pi ) =∥f∥=\|f\| (cor. 3 of V, p. 459). Consequently, the mapping Ψ is isometric. Since ϱG(π‾)\boldsymbol{\varrho }_G(\overline{\pi}) and End(Eπ_{\pi}) are of the same dimension, Ψ is an isometric isomorphism.

We denote by F2(G^)F^2(\widehat{G}) the Hilbert sum of the Hilbert spaces End(Eπ_{\pi}). The norm of an element x∈F2(G^)x\in F^2(\widehat{G}) is again denoted by ∥x∥2\|x\|_2.

In LIE, IX, p. 79, this space is denoted L2(G^)L^2(\widehat{G}), a notation which we prefer to avoid here so as not to create confusion with the space ℓ2(G^)\ell^2(\widehat{G}).

Let (uπ)(u_{\pi}) be an element of F2(G^)F^2(\widehat{G}). Since

π∈G^∑π∈G^∥uπ∥22=∑π∈G^^{\pi\in\widehat{G}}\sum_{\pi\in\widehat{G}}\|u_{\pi}\|^2_2=\sum_{\pi\in\widehat{G}} dim(π\pi ) Tr(uπ∗uπu^*_{\pi}u_{\pi})<+∞<+\infty

and ∥uπ∥2⩽\|u_{\pi}\|^2\leqslant Tr(uπ∗uπu^*_{\pi}u_{\pi}) ((cf. EVT, V, p. 52, formula (33)), the norm in L(Eπ)\mathscr{L}(E_{\pi}) of the endomorphism uπu_{\pi} tends to 0 at infinity. One can therefore identify F2(G^)F^2(\widehat{G}) with a subspace of F0(G^)F_0(\widehat{G}).

Let π∈G^\pi \in \widehat{G} and u∈u\in End(Eπ_{\pi}). We denote by Fπ(u)\mathscr{F}_{\pi}(u) the function on G defined by Fπ(u)(g)=⟨π(g)∣u⟩=\mathscr{F}_{\pi}(u)(g) =\langle \pi (g)|u\rangle = dim(π\pi)Tr(π(g)∗u\pi (g)^*u) for every g∈Gg\in G. This is a continuous function on G. If G is a compact real Lie group, then the function Fπ(u)\mathscr{F}_{\pi}(u) is analytic on G (LIE, III, § 8, no1^o1, th. 1).

Since G is compact, one can identify L2(G)L^2(G) with a subspace of L1(G)L^1(G).

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Machine translation of the English, by gpt-5.4, against glossary version 34. Not checked by a person.