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This commit is contained in:
Bokuan Li
2026-09-26 16:29:13 -04:00
parent a67e55f88d
commit 1cea2d3530
2 changed files with 14 additions and 14 deletions
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src/op/c-star/homomorphism.tex
+11 -11
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@@ -44,7 +44,7 @@
$\neg (2) \Rightarrow \neg (1)$: Let $\xi \in H \setminus \bracsn{0}$ with $\pi(x)\xi = 0$ for all $x \in A$, then for any $x \in A$ and $\eta \in H$,
\[
\dpn{\pi(x)\eta, \xi}{H} = \dpn{\eta, \pi(x)\xi}{H} = 0
\dpn{\pi(x)\eta, \xi}{H} = \dpn{\eta, \pi(x^*)\xi}{H} = 0
\]
so $\bracsn{\pi(x)\eta|x \in A, \eta \in H} \subset \bracsn{\xi}^\perp$, and cannot be dense in $H$.
@@ -64,7 +64,7 @@
(2): By (1) and \autoref{corollary:c-star-unique-norm},
\begin{align*}
\norm{\phi(x)}_B^2 &= \sup\bracsn{|\lambda|\ | \lambda \in \sigma_B(\phi(x^*x))} \\
&\ge \sup\bracsn{|\lambda|\ | \lambda \in \sigma_A(x^*x)} = \norm{x}_A^2
&\le \sup\bracsn{|\lambda|\ | \lambda \in \sigma_A(x^*x)} = \norm{x}_A^2
\end{align*}
\end{proof}
@@ -77,14 +77,14 @@
Let $y \in \ol{\Phi(A)} \cap B_{sa}$, then there exists $x \in A_{sa}$ such that $\norm{y - \Phi(x)}_B \le \norm{y}_B/2$. Let
\[
f: \complex \to \complex \quad z \mapsto \begin{cases}
z & |z| \le 2\norm{y}_F \\
2\norm{y}_F \cdot \sgn z = 2\norm{y}_F \cdot \frac{z}{|z|} & |z| \ge 2\norm{y}_F
z & |z| \le 2\norm{y}_B \\
2\norm{y}_B \cdot \sgn z = 2\norm{y}_B \cdot \frac{z}{|z|} & |z| \ge 2\norm{y}_B
\end{cases}
\]
then $f \in C(\complex; \complex)$. Since $\norm{\Phi(x)}_B \le \norm{y}_B + \norm{y - \Phi(x)}_B \le 2\norm{y}_B$, $\sigma_B(\Phi(x)) \subset \ol{B_\complex(0, 2\norm{y}_B)}$, and $f|_{\sigma_B(\Phi(x))}$ is the identity. Thus by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $\Phi(x) = f(\Phi(x)) = \Phi(f(x))$. By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-continuous}, $\sigma_A(f(x)) = f(\sigma_A(x))$. By \autoref{theorem:c-star-normal-spectral-radius}, $\norm{f(x)}_A = [f(x)]_{sp} \le \norm{f}_u = 2\norm{y}_F$.
The above setup implies that for every $y \in \ol{\Phi(A)} \cap B_{sa}$, there exists $z \in A_{sa}$ such that $\norm{y - \Phi(z)}_{B} \le \norm{y}_B/2$, and $\norm{z}_A \le 2\norm{y}_B$. By the \hyperref[method of successive approximations]{theorem:successive-approximation}, $\phi(A_{sa}) = \ol{\Phi(A)} \cap B_{sa}$. Therefore $\Phi(A) = \ol{\Phi(A)}$.
The above setup implies that for every $y \in \ol{\Phi(A)} \cap B_{sa}$, there exists $z \in A_{sa}$ such that $\norm{y - \Phi(z)}_{B} \le \norm{y}_B/2$, and $\norm{z}_A \le 2\norm{y}_B$. By the \hyperref[method of successive approximations]{theorem:successive-approximation}, $\Phi(A_{sa}) = \ol{\Phi(A)} \cap B_{sa}$. Therefore $\Phi(A) = \ol{\Phi(A)}$.
\end{proof}
\begin{definition}[Representation of $C^*$-Algebra]
@@ -94,7 +94,7 @@
\begin{definition}[Unitary Equivalence]
\label{definition:representation-unitary-equivalent}
Let $A$ be a $C^*$-algebra and $(H_1, \pi_1), (H_2, \pi_2)$ be representations of $A$, then $(H_1, \pi_1)$ and $(H_2, \pi_2)$ are \textbf{unitarily equivalent} if there exists an isometry $U \in L(H_1; H_2)$ such that the following diagram commutes
Let $A$ be a $C^*$-algebra and $(H_1, \pi_1), (H_2, \pi_2)$ be representations of $A$, then $(H_1, \pi_1)$ and $(H_2, \pi_2)$ are \textbf{unitarily equivalent} if there exists a unitary map $U \in L(H_1; H_2)$ such that the following diagram commutes
\[
\xymatrix{
H_1 \ar@{->}[r]^{U} \ar@{->}[d]_{\pi_1(x)} & H_2 \ar@{->}[d]^{\pi_2(x)} \\
@@ -126,19 +126,19 @@
Let $A$ be a $C^*$-algebra, $(H_1, \pi_1)$ and $(H_2, \pi_2)$ be representations of $A$, 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
\[
\xymatrix{
H_1 \ar@{->}[d]_{\pi_1(x)} \ar@{->}[r]^{T} & H_2 \ar@{->}[d]^{\pi_1(x)} \\
H_1 \ar@{->}[d]_{\pi_1(x)} \ar@{->}[r]^{T} & H_2 \ar@{->}[d]^{\pi_2(x)} \\
H_1 \ar@{->}[r]_{T} & H_2
}
\]
for all $x \in A$. The set $\mathcal{C}(\pi_1, \pi_2)$ is the \textbf{space of intertwining operators} for $\pi_1$ and $\pi_2$.
In particular, $\mathcal{C}(\pi_1, \pi_1) = \pi(G)'$, and hence is a von Neumann algebra.
In particular, $\mathcal{C}(\pi_1, \pi_1) = \pi(A)'$, and hence is a von Neumann algebra.
\end{definition}
\begin{definition}[Direct Sum]
\label{definition:cstar-rep-direct-sum}
Let $H$ be a $C^*$-algebra and $\bracsn{(H_i, \pi_i)}_{i \in I}$ be representations of $A$, then the representation
Let $A$ be a $C^*$-algebra and $\bracsn{(H_i, \pi_i)}_{i \in I}$ be representations of $A$, then the representation
\[
\pi: A \to B([l^2(I); H_i]) \quad \pi(x)(\xi)_i = \pi_i(x)(\xi_i)
\]
@@ -172,9 +172,9 @@
\begin{proof}[Proof, {{\cite[Theorem 3.5]{FollandHarmonic}}}. ]
(1): $(H, \pi)$ is irreducible if and only if $\pi(A)'$ contains no non-trivial projections. Since $\pi(A)'$ is a von Neumann algebra, \autoref{theorem:vn-projection-norm-dense} implies that the span of projections in $\pi(A)'$ is dense in $\pi(A)'$. Therefore $(H, \pi)$ is irreducible if and only if $\pi(A)' = \complex I$.
(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 $TT^* \in \mathcal{C}(\tau, \tau) = \tau(A)'$. 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 $T^*T \in \mathcal{C}(\pi, \pi) = \pi(A)'$, and $TT^* \in \mathcal{C}(\tau, \tau) = \tau(A)'$. 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}