Revisions.

src/harmonic/rep/unit.tex

@@ -21,7 +21,7 @@
     Let $G$ be a locally compact group, $(H_1, \pi_1)$ and $(H_2, \pi_2)$ be unitary representations of $G$, and $T \in L(H_1; H_2)$, then $T$ is an \textbf{intertwining operator} for $\pi_1$ and $\pi_2$ if the following diagram commutes 
     \[
     \xymatrix{
-    H_1 \ar@{->}[d]_{\pi_1(x)} \ar@{->}[r]^{T} & H_2 \ar@{->}[d]^{\pi_1(x)} \\
+    H_1 \ar@{->}[d]_{\pi_1(x)} \ar@{->}[r]^{T} & H_2 \ar@{->}[d]^{\pi_2(x)} \\
     H_1 \ar@{->}[r]_{T} & H_2
     }
     \]
@@ -103,9 +103,9 @@
 \begin{proof}[Proof, {{\cite[Theorem 3.5]{FollandHarmonic}}}. ]
     (1): $(H, \pi)$ is irreducible if and only if $\pi(G)'$ contains no non-trivial projections. Since $\pi(G)'$ is a von Neumann algebra, \autoref{theorem:vn-projection-norm-dense} implies that the span of projections in $\pi(G)'$ is dense in $\pi(G)'$. Therefore $(H, \pi)$ is irreducible if and only if $\pi(G)' = \complex I$. 
 
-    (2): For any $T \in \mathcal{C}(\pi, \tau)$, $T^* \in \mathcal{C}(\tau, \pi)$, so $T^*T \in \mathcal{C}(\pi, \pi) = \pi(G)'$, and $TT^* \in \mathcal{C}(\tau, \tau) = \tau(G)'$. Given that $(H, \pi)$ and $(K, \tau)$ are irreducible, there exists $\lambda, \mu \in \complex$ such that $T^*T = \lambda I_H$ and $TT^* = \mu I_K$. As such, either $T = 0$ or $\lambda^{-1/2}T$\footnote{Pick any branch.} is unitary, so $\mathcal{C}(\pi, \tau)$ is non-zero if and only if $(H, \pi)$ and $(K, \tau)$ are unitarily equivalent. 
+    (2): For any $T \in \mathcal{C}(\pi, \tau)$, $T^* \in \mathcal{C}(\tau, \pi)$, so $T^*T \in \mathcal{C}(\pi, \pi) = \pi(G)'$, and $TT^* \in \mathcal{C}(\tau, \tau) = \tau(G)'$. Given that $(H, \pi)$ and $(K, \tau)$ are irreducible, there exist $\lambda, \mu \in \complex$ such that $T^*T = \lambda I_H$ and $TT^* = \mu I_K$. As such, either $T = 0$ or $\lambda^{-1/2}T$\footnote{Pick any branch.} is unitary, so $\mathcal{C}(\pi, \tau)$ is non-zero if and only if $(H, \pi)$ and $(K, \tau)$ are unitarily equivalent. 
 
-    If $\mathcal{C}(\pi, \tau) \ne \bracsn{0}$, then $\mathcal{C}(\pi, \tau)$ consists of scalar multiples of unitary operators. In which case, for any unitary operators $S, T \in \mathcal{C}(\pi, \tau)$, $S^{-1}T = S^*T \in \mathcal{C}(\pi)$. Thus there exists $\lambda \in \complex$ such that $S^{-1}T = \lambda I_H$, so $T = \lambda S$, and $\mathcal{C}(\pi, \tau)$ is one-dimensional.
+    If $\mathcal{C}(\pi, \tau) \ne \bracsn{0}$, then $\mathcal{C}(\pi, \tau)$ consists of scalar multiples of unitary operators. In which case, for any unitary operators $S, T \in \mathcal{C}(\pi, \tau)$, $S^{-1}T = S^*T \in \mathcal{C}(\pi, \pi)$. Thus there exists $\lambda \in \complex$ such that $S^{-1}T = \lambda I_H$, so $T = \lambda S$, and $\mathcal{C}(\pi, \tau)$ is one-dimensional.
 \end{proof}
 
 \begin{corollary}