Proposition 6

Let $\pi$ be an irreducible unitary representation of G in a Hilbert space E. Let $\chi$ be the central character of $\pi$. Then $\pi$ is square-integrable modulo the center if and only if there exist non-zero elements $x_0$ and $y_0$ of E such that the function on $G/C$ deduced from $|f_{x_0,y_0}|$ by passing to the quotient belongs to $\mathscr{L}^2(G/C)$.

Suppose that there exist non-zero elements $x_0$ and $y_0$ of E such that $|f_{x_0,y_0}| \in \mathscr{L}^2(G/C)$. It is enough to prove that $\pi$ is then square-integrable modulo the center.

Let F be the set of elements $x\in E$ such that $|f_{x,y_0}|$ belongs to $\mathscr{L}^2(G/C)$. It is a vector subspace of E; it contains $x_0$, hence is non-zero. Relation (10) above implies that F is stable under $\pi$; since the representation $\pi$ is irreducible, the space F is therefore dense in E.

Let $x\in F$. Since $f_{x,y_0}$ belongs to $\mathscr{F}_{\chi}(G)$ and is $\mu$-measurable, and since $|f_{x,y_0}| \in \mathscr{L}^2(G/C)$, one has $f_{x,y_0}\in \mathscr{L}_{\chi}^2(G)$ (Prop. 5 of V, p. 413 applied to Z = C). Let $u$ denote the partial operator from E into $L^2_{\chi}(G)$ whose domain is F and which associates to $x\in F$ the class of $f_{x,y_0}$ in $L^2_{\chi}(G)$.

Let us prove that the partial operator $u$ is closed. Let $(x_n, u(x_n))_{n\in\mathbf{N}}$ be a sequence of elements of the graph of $u$ which converges in $E\times L^2_{\chi}(G)$. Let $x$ be the limit of $(x_n)$ and let $f\in \mathscr{L}_{\chi}^2(G)$ be a function whose class is the limit of the sequence $(u(x_n))$.

The function $u(x_n)$ is the class of the matrix coefficient $f_{x_n,y_0}$. For every $g\in G$, one has

when $n\rightarrow +\infty$. Hence $f_{x,y_0}\in \mathscr{L}_{\chi}^2(G)$ and $f=f_{x_0,y}$ almost everywhere modulo C (cor. of Prop. 3 of V, p. 409); this means that $u(x)$ is the class of $f$ in $L^2_{\chi}(G)$. Thus we have proved that $u$ is closed.

The domain F of $u$ is stable under $\pi$, and relation (10) proves that $u$ satisfies $u\circ \pi (g) =\boldsymbol{\gamma }_{G,\chi}(g)\circ u$ for all $g\in G$. Consequently one also has the equality $u^*\circ \boldsymbol{\gamma }_{G,\chi}(g) =\pi (g)\circ u^*$ for all $g\in G$ (lemma 6 of V, p. 381), whence $(u^*\circ u)\circ \pi (g) =\pi (g)\circ (u^*\circ u)$. Now the partial operator $u^*\circ u$ is self-adjoint (prop. 12 of IV, p. 241), and in particular closed (prop. 7 of IV, p. 236), hence cor. 1 of V, p. 387 implies that the domain of $u^*\circ u$ is equal to E. A fortiori, one has F = E, that is to say that the function $|f_{x,y_0}|$ belongs to $\mathscr{L}^2(G/C)$ for all $x\in E$.

Let $(x, y)\in E\times E$. One has $f_{x,y_0}\in \mathscr{L}_{\chi}^2(G)$. One proves mutatis mutandis, using the representation $\boldsymbol{\delta }_{G,\chi}$ instead of $\boldsymbol{\gamma }_{G,\chi}$, that the set of $y\in E$ such that the function $|f_{x,y}|$ belongs to $\mathscr{L}^2(G/C)$ is equal to E. The proposition follows.

In the remainder of this No., we fix a closed subgroup Z of C such that $C/Z$ is compact.