Schur lemma for complex unitary representations

AC implies Countable Choice by restricting a choice function to any countable subfamily, so [A3] supplies adjoints. If $S$ is a self-intertwiner of $\pi$, apply its relation at $g^{-1}$ and take adjoints; [A1] and [A3] give $\pi (g)S^{\ast }=S^{\ast }\pi (g)$. If $T:H\to K$ intertwines $\pi$ with $\rho$, the same operation gives $\pi (g)T^{\ast }=T^{\ast }\rho (g)$, so $T^{\ast }$ intertwines in the reverse direction.

Let $A=A^{\ast }$ be a bounded self-intertwiner of $\pi$. By [A9], $A$ is normal. If $\sigma (A)=\{\lambda \}$, the coordinate function on this singleton equals the constant $\lambda$; [A8] then gives $A=z(A)=\lambda I_H$. Hence a nonscalar $A$ has two distinct spectral points $\lambda ,\mu$. Choose disjoint nonempty relatively open neighborhoods $U,V$ of them in $\sigma (A)$. By [A7], $E_A(U)$ and $E_A(V)$ are nonzero; [A6] makes them orthogonal projections with $E_A(U)E_A(V)=E_A(\mathrm{\varnothing })=0$, so $P:=E_A(U)$ is nonzero and not $I_H$: if $P=I_H$, then $E_A(V)=P{E}_A(V)=0$, contrary to [A7].

Since $A$ belongs to the commutant and is self-adjoint, each $\pi (g)$ commutes with both $A$ and $A^{\ast }$. By [A4] and [A5], it commutes with $P={\mathbf{1}}_U(A)$. The range of the orthogonal projection $P$ is nonzero and proper, and is closed because $\mathrm{ran}P=\ker (I-P)$ for an idempotent bounded operator; commutation gives $\pi (g)\mathrm{ran}P\subseteq \mathrm{ran}P$, and applying the same inclusion to $g^{-1}$ gives equality. This contradicts [A2]. Therefore every bounded self-adjoint self-intertwiner of $\pi$ is scalar.

If $S$ is any bounded self-intertwiner, then by step 1.1 its adjoint also intertwines. The operators $A=(S+S^{\ast })\mathrm{/}2$ and $B=(S-S^{\ast })\mathrm{/}(2i)$ are self-adjoint self-intertwiners, so step 2.1 makes both scalar. Since $S=A+iB$, $S$ is scalar.

Let $T:H\to K$ be a nonzero bounded intertwiner. By step 1.1, $T^{\ast }$ intertwines in reverse, and $T^{\ast }T$ is a bounded self-intertwiner of $\pi$. Step 3.1 gives $T^{\ast }T=c{I}_H$ for some $c\in \mathbb{C}$.

The adjoint identities make $T^{\ast }T$ self-adjoint, hence $c$ is real. For any $x$, $\mathrm{\parallel }Tx{\mathrm{\parallel }}^2=⟨Tx,Tx⟩=⟨x,T^{\ast }Tx⟩=⟨x,cx⟩=c\mathrm{\parallel }x{\mathrm{\parallel }}^2$ by [A12]; since $T\mathrm{\ne }0$, this forces $c>0$. Let $s=\sqrt{c}>0$ by [A10] and set $U=s^{-1}T$. Then $\mathrm{\parallel }Ux{\mathrm{\parallel }}^2=c^{-1}\mathrm{\parallel }Tx{\mathrm{\parallel }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}^2$ for every $x$, so the nonnegative norms are equal, $U$ is an isometry, and it still intertwines.

The range of $U$ is closed: if $U{x}_n$ converges, the isometry identity makes $(x_n)$ Cauchy, and completeness [A13] gives a limit whose image is the given range limit. Its range is nonzero because $H\mathrm{\ne }\{0\}$ and $U$ is an isometry. The intertwining identity and surjectivity of each $\pi (g)$ give $\rho (g)\mathrm{ran}U=U\pi (g)H=\mathrm{ran}U$, so the range is invariant. By irreducibility of $\rho$ it is all of $K$; thus $U$ is a unitary intertwiner and $\pi ,\rho$ are unitarily equivalent.

If the irreducible representations are inequivalent, a nonzero bounded intertwiner would produce the unitary equivalence in step 6.1, a contradiction. The self-intertwiner assertion is step 3.1.