Normalized positive type and pointed cyclic unitary representations

Let $(\pi ,H,\xi )$ be a strongly continuous unitary triple with $\xi$ cyclic and $\mathrm{\parallel }\xi \mathrm{\parallel }=1$, and set $\phi (g)=⟨\pi (g)\xi ,\xi ⟩$. By [F2] and [F5], $\phi$ is continuous and of positive type. Also [F5] gives $\phi (e)=\mathrm{\parallel }\xi {\mathrm{\parallel }}^2=1$, so $\phi \in P_1(G)$ by [F1].

Identity maps, inverses, and compositions of pointed unitary intertwiners show that the stated relation is an equivalence relation. If $U:H\to H^{\mathrm{\prime }}$ is a unitary intertwiner with $U\xi ={\xi }^{\mathrm{\prime }}$, then the squared-norm identities $\mathrm{\parallel }u+v{\mathrm{\parallel }}^2-\mathrm{\parallel }u-v{\mathrm{\parallel }}^2=4\mathrm{Re}⟨u,v⟩,\mathrm{\parallel }u+iv{\mathrm{\parallel }}^2-\mathrm{\parallel }u-iv{\mathrm{\parallel }}^2=4\mathrm{Im}⟨u,v⟩$ and [F3, F8] show that $U$ preserves the inner product. Thus, for every $g\in G$, $⟨{\pi }^{\mathrm{\prime }}(g){\xi }^{\mathrm{\prime }},{\xi }^{\mathrm{\prime }}⟩=⟨U\pi (g)\xi ,U\xi ⟩=⟨\pi (g)\xi ,\xi ⟩\mathrm{.}$ So the map in the statement is well-defined.

Let $\phi \in P_1(G)$. Then $\phi$ is continuous and of positive type, and $\phi (e)=1$ by [F1]. By AC and [F6], its GNS triple $({\pi }_{\phi },H_{\phi },{\xi }_{\phi })$ is strongly continuous and cyclic, has coefficient $\phi$, and satisfies $\mathrm{\parallel }{\xi }_{\phi }{\mathrm{\parallel }}^2=\phi (e)=1$. Nonnegativity of the Hilbert norm [F9] gives $\mathrm{\parallel }{\xi }_{\phi }\mathrm{\parallel }=1$, so this triple is in the stated domain [F4]. Hence every member of $P_1(G)$ is attained.

If two domain triples have the same image $\phi$, then they have the same diagonal coefficient at every $g\in G$. They are cyclic, so [F7] supplies a unique unitary intertwiner taking the first distinguished vector to the second. Thus the triples are equivalent and the coefficient map is injective on equivalence classes.

Step 1.2 proves the forward assignment is well-defined, step 1.1 places its values in $P_1(G)$, step 1.3 constructs a GNS class for every member of $P_1(G)$, and step 1.4 proves that class is unique. Therefore the coefficient map and the GNS assignment are inverse bijections. The zero function has value $0$ at $e$, so it is not in $P_1(G)$; its zero GNS vector is not a unit vector.

AC is used through the GNS construction [F6] to realize each $\phi \in P_1(G)$ and through pointed uniqueness [F7] to identify any two cyclic triples with the same coefficient. The coefficient calculation and the invariance under a given unitary intertwiner use no choice.