Proposition 32.3.5.label Let $G$ be a locally compact group, $\phi \in S(G)$ be a state, and $(H_{\phi}, \pi_{\phi}, \xi_{\phi})$ be the GNS triple associated with $\phi$, then $(H_{\phi}, \pi_{\phi})$ is irreducible if and only if $\phi$ is an extreme point of $S(G)$.
Proof. ($\Rightarrow$): Suppose that $(H_{\phi}, \pi_{\phi})$ is irreducible. Let $\psi_{1}, \psi_{2} \in S(G)$ and $t \in (0, 1)$ such that $\phi = t\psi_{1} + (1 - t)\psi_{2}$. Define
where $f, g \in L^{1}(G; \complex)$. Since both $\psi_{1}$ and $\psi_{2}$ are positive-definite functions, $\phi - t\psi_{1}$ is also positive-definite. In particular, $\dpn{f, f}{\psi_1}\le t^{-1}\dpn{f, f}{\phi}$ for all $f \in L^{1}(G; \complex)$, so $T$ extends continuously into a positive operator on $H_{\phi}$.
For any $x \in G$ and $f, g \in L^{1}(G; \complex)$,
so $T \in \pi_{\phi}(G)'$. By Schur’s Lemma, there exists $\lambda \in \complex \setminus \bracsn{0}$ such that $T = \lambda I$.
Using Proposition 31.4.10, let $\angles{g_\alpha}_{\alpha \in A}\subset L^{1}(G; \complex)$ be an approximate identity for $L^{1}(G; \complex)$, then for any $f \in L^{1}(G; \complex)$,
Since $\phi, \psi_{1} \in S(G)$, $\lambda = 1$, and $\phi = \psi_{1}$. As the argument is symmetric, $\phi = \psi_{1} = \psi_{2}$.
($\Leftarrow$): Suppose that $(H_{\phi}, \pi_{\phi})$ is reducible. By Schur’s Lemma, $\pi_{\phi}(G)'$ is non-trivial. Since $\pi_{\phi}(G)'$ is a von Neumann algebra, it admits a non-trivial projection $P \in B(H_{\phi})$ by Theorem 39.6.2.
Let $\xi_{1} = P\xi_{\phi}/\normn{P\xi_\phi}_{H_\phi}$ and $\xi_{2} = (1 - P)\xi_{\phi}/\normn{\xi_\phi - P\xi_\phi}_{H_\phi}$. For each $f \in L^{1}(G; \complex)$, let
then $\psi_{1}, \psi_{2} \in S(G)$. Since $P \in \pi_{\phi}(G)'$, for any $f \in L^{1}(G; \complex)$,
so $\phi$ is a strict convex combination of two states. If $\psi_{1} = \phi$, then for any $f, g \in L^{1}(G; \complex)$,
so $I - \normn{P\xi_\phi}_{H_\phi}^{-2}P = 0$, which contradicts the fact that $P$ is a non-trivial projection.$\square$
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