Extreme normalized positive type is equivalent to irreducible GNS

For ${\psi }_1,{\psi }_2\in P_1(G)$ and $r\in [0,1]$, every test matrix of $r{\psi }_1+(1-r){\psi }_2$ is a convex combination of positive semidefinite matrices and is therefore positive semidefinite; its value at $e$ is $1$. Thus $P_1(G)$ is convex by [F1]. The normalized correspondence [F3] identifies the canonical GNS triple as the pointed cyclic class associated with $\phi$, while [F4] gives its coefficient and $\mathrm{\parallel }{\xi }_{\phi }{\mathrm{\parallel }}^2=\phi (e)=1$, so $H_{\phi }\mathrm{\ne }\{0\}$.

Suppose ${\pi }_{\phi }$ is irreducible and write $\phi =s{\phi }_1+(1-s){\phi }_2$ with $0<s<1$ and ${\phi }_1,{\phi }_2\in P_1(G)$.

If ${\phi }_1={\phi }_2$, the convex identity gives ${\phi }_1={\phi }_2=\phi$. In the remaining case assume ${\phi }_1\mathrm{\ne }{\phi }_2$. The matrix test in [F1] shows that $s{\phi }_1$ and $(1-s){\phi }_2=\phi -s{\phi }_1$ are of positive type, so $0\le s{\phi }_1\le \phi$. By [F6] there is a unique positive contraction $T$ in the commutant with coefficient $s{\phi }_1$.

Suppose instead that ${\pi }_{\phi }$ is reducible. By [F4] its Hilbert space is nonzero, so [F5] supplies a proper nonzero closed invariant linear subspace $M\subset H_{\phi }$. AC gives Countable Choice by [F14]; apply [F9] and [F10] to obtain the unique orthogonal projection $P=P_M$ onto $M$.

If $y\in M^{\perp }$, $m\in M$, and $g\in G$, unitarity and invariance give $⟨{\pi }_{\phi }(g)y,m⟩=⟨y,{\pi }_{\phi }(g{)}^{-1}m⟩=0$, since ${\pi }_{\phi }(g{)}^{-1}m\in M$. Applying this for $g^{-1}$ shows ${\pi }_{\phi }(g)M^{\perp }=M^{\perp }$.

In the distinct-summand case of step 1.3, $T$ is a bounded self-intertwiner of the irreducible representation, so [F8] gives $T=\lambda I$. Evaluating its coefficient at $e$ and using $\mathrm{\parallel }{\xi }_{\phi }\mathrm{\parallel }=1$ gives $\lambda =s$; for each $g$, $s{\phi }_1(g)=⟨{\pi }_{\phi }(g)T{\xi }_{\phi },{\xi }_{\phi }⟩=s\phi (g)$, so ${\phi }_1=\phi$ and then ${\phi }_2=\phi$, contradicting that case. Together with the equal-summand case in step 1.3, every strict convex decomposition is trivial, and [F2] makes $\phi$ extreme.

For $x=m+n$ with $m\in M$ and $n\in M^{\perp }$, both summands remain in their respective subspaces under ${\pi }_{\phi }(g)$. Uniqueness in [F9] therefore gives $P{\pi }_{\phi }(g)x={\pi }_{\phi }(g)m={\pi }_{\phi }(g)Px$ for every $x,g$, so $P\in {\pi }_{\phi }(G{)}^{\mathrm{\prime }}$. From [F10], $P=P^{\ast }=P^2$, and $I-P$ is also self-adjoint and idempotent. Orthogonality of $Px$ and $(I-P)x$ gives $⟨Px,x⟩=\mathrm{\parallel }Px{\mathrm{\parallel }}^2\ge 0$ and $⟨(I-P)x,x⟩=\mathrm{\parallel }(I-P)x{\mathrm{\parallel }}^2\ge 0$; thus [F13] makes $P$ and $I-P$ positive.

The projection is nonscalar: if $P=\lambda I$, idempotence yields ${\lambda }^2=\lambda$, so $\lambda =0$ or $1$; its range would then be $\{0\}$ or $H_{\phi }$, contrary to $\mathrm{ran}P=M$ being proper and nonzero. By [F7], this nonscalar positive contraction yields $s\in (0,1)$ and distinct ${\phi }_1,{\phi }_2\in P_1(G)$ with $\phi =s{\phi }_1+(1-s){\phi }_2$. The definition [F2] then shows $\phi$ is not extreme.

Steps 1.2, 1.3 and 2.1 prove irreducibility implies extremality, and steps 1.4, 1.5, 2.2 and 3.1 prove that reducibility implies non-extremality. These give both implications of the stated equivalence.

AC is declared because [F3], [F4], [F6], [F7] and [F8] assume it, and because [F14] supplies Countable Choice for the orthogonal decomposition and projection in [F9] and [F10] and for the positive-operator definition [F13]. After the subspace in [F5] is fixed, the decomposition and projection are unique; the invariant-complement and commutation arguments use no further choice.