Nondegenerate representations of the full group C star algebra are unitary representations

The integrated form of $\pi$ extends uniquely to a nondegenerate star-representation ${\pi }_{C^{\ast }}$ of $C^{\ast }(G)$. Since $\mathrm{\parallel }f{\mathrm{\parallel }}_{C^{\ast }}={\sup }_{{\pi }^{\mathrm{\prime }}}\mathrm{\parallel }{\pi }^{\mathrm{\prime }}(f)\mathrm{\parallel }\ge \mathrm{\parallel }\pi (f)\mathrm{\parallel }$, the map $f\mapsto \pi (f)$ is contractive for the full seminorm, so it kills $N$ and descends to a contractive linear map on the dense subalgebra $L^1(G)\mathrm{/}N\subseteq C^{\ast }(G)$; by [F4] it has a unique bounded linear extension to $C^{\ast }(G)$, which is multiplicative and star-preserving because these identities hold on the dense subalgebra and both sides are continuous, and it is nondegenerate because the closed span of $\{\pi (f)\xi \}$ is $H$ by [F1].

The restriction of $\rho$ to $L^1(G)$ (composed with $q$) is a nondegenerate star-representation of $L^1(G)$: it is complex-linear, multiplicative and star-preserving because $q$ and $\rho$ are, and bounded because $\rho$ is contractive by [F4]; it is nondegenerate because $q(L^1(G))$ is dense in $C^{\ast }(G)$ and $\rho$ is bounded: the $\rho$-images of $\{\rho (x)\xi :x\in C^{\ast }(G)\}$ are approximated by $\rho (q(f))\xi$ with $q(f)\to x$, so the closed span of $\{\rho (q(f))\xi \}$ equals the closed span of $\rho (C^{\ast }(G))K$, which is $K$.

The two constructions are mutually inverse. Starting from a unitary representation $\pi$, restricting the extension ${\pi }_{C^{\ast }}$ of step 1.1 to $L^1(G)$ returns the integrated form $\pi (\cdot )$, so [F3] returns $\pi$ itself. Starting from a nondegenerate star-representation $\rho$ of $C^{\ast }(G)$, step 1.2 gives a nondegenerate star-representation $\sigma :=\rho \circ q$ of $L^1(G)$, whose reconstructed unitary representation $U$ satisfies ${\pi }_U(f)=\sigma (f)=\rho (q(f))$ for all $f\in L^1(G)$ by [F3]; hence the extension of ${\pi }_U$ to $C^{\ast }(G)$, which is unique by the argument of step 1.1, coincides with $\rho$ on the dense subalgebra and therefore everywhere by continuity.

A closed subspace $M\subseteq K$ is invariant under the unitary representation $U$ corresponding to $\rho$ if and only if it is invariant under $\rho (C^{\ast }(G))$; since step 2.1 identifies the two sides of the correspondence, this gives the irreducibility statement. Indeed, if $\rho (x)M\subseteq M$ for all $x\in C^{\ast }(G)$, then in particular $\sigma (f)M\subseteq M$ for all $f\in L^1(G)$, and $U(g)\xi ={\lim }_{\lambda }\sigma (L_g{e}_U)\xi \in M$ for $\xi \in M$ by [F1] and closedness of $M$; conversely if $U(g)M\subseteq M$ for all $g$, then for $f\in C_c(G)$ the Bochner integral representing $\sigma (f)\xi$ is a norm limit of finite linear combinations of the vectors $U(g)\xi \in M$, hence lies in the closed subspace $M$; for general $f\in L^1(G)$ the contractivity of the integrated form and density of $C_c(G)$ give $\sigma (f)M\subseteq M$, and then $\rho (C^{\ast }(G))M\subseteq M$ by continuity. Hence $M$ is a nontrivial closed invariant subspace for $U$ exactly when it is one for $\rho$, so irreducibility corresponds.

Steps 1.1, 1.2 and 2.1 establish the claimed bijection respecting unitary equivalence (an intertwiner of unitary representations intertwines the integrated forms, and conversely by [F3]), and step 3.1 upgrades it to preserve irreducibility. The Axiom of Choice is inherited from the full-norm completion and the $L^1$ correspondence (The Axiom of Choice).