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/Part 7: Operator Algebras/Chapter 38: $C^{*}$-Algebras

38.5 *-Homomorphisms

Definition 38.5.1 (*-Homomorphism).label Let $A, B$ be involutive algebras over $\complex$ and $\phi: A \to B$, then $\phi$ is a *-homomorphism if:

  1. (SH1)

    For each $x, y \in A$ and $\lambda \in \complex$, $\phi(\lambda x + y) = \lambda \phi(x) + \phi(y)$.

  2. (SH2)

    For each $x, y \in A$, $\phi(xy) = \phi(x)\phi(y)$.

  3. (SH3)

    For each $x \in A$, $\phi(x^{*}) = \phi(x)^{*}$.

If $A$ and $B$ are unital, then $\phi$ is unital if:

  1. (U)

    $\phi(1_{A}) = 1_{B}$.

Definition 38.5.2 (*-Representation).label Let $A$ be an involutive Banach algebra, then a *-representation of $A$ is a pair $(H, \pi)$, where $H$ is a complex Hilbert space, and $\pi: A \to B(H)$ is a *-homomorphism.

Definition 38.5.3 (Non-Degenerate).label Let $A$ be an involutive Banach algebra and $(H, \pi)$ be a *-representation of $A$, then the following are equivalent:

  1. (1)

    $\text{span}\bracsn{\pi(x)\xi|x \in A, \xi \in H}$ is dense in $H$.

  2. (2)

    There exists no non-zero $\xi \in H$ such that $\pi(x)\xi = 0$ for all $x \in A$.

If the above holds, then $(H, \pi)$ is non-degenerate.

Proof. $\neg (1) \Rightarrow \neg (2)$: Let $\xi \in \bracsn{\pi(x)\eta|x \in A, \eta \in H}^{\perp}$, then for any $x \in A$ and $\eta \in H$,

\[\dpn{\pi(x)\xi, \eta}{H}= \dpn{\xi, \pi(x^*)\eta}{H}= 0\]

so $\pi(x)\xi = 0$ for all $x \in A$.

$\neg (2) \Rightarrow \neg (1)$: Let $\xi \in H \setminus \bracsn{0}$ with $\pi(x)\xi = 0$ for all $x \in A$, then for any $x \in A$ and $\eta \in H$,

\[\dpn{\pi(x)\eta, \xi}{H}= \dpn{\eta, \pi(x^*)\xi}{H}= 0\]

so $\bracsn{\pi(x)\eta|x \in A, \eta \in H}\subset \bracsn{\xi}^{\perp}$, and cannot be dense in $H$.$\square$

Proposition 38.5.4.label Let $A, B$ be unital $C^{*}$-algebras and $\phi: A \to B$ be a unital *-homomorphism, then for each $x \in A$,

  1. (1)

    $\sigma_{B}(\phi(x)) \subset \sigma_{A}(x)$.

  2. (2)

    $\norm{\phi(x)}_{B} \le \norm{x}_{A}$.

Proof. (1): Since $\phi$ is unital, $\phi(G(A)) \subset G(B)$, so $\sigma_{B}(\phi(x)) \subset \sigma_{A}(x)$.

(2): By (1) and Corollary 38.4.5,

\begin{align*}\norm{\phi(x)}_{B}^{2}&= \sup\bracsn{|\lambda|\ | \lambda \in \sigma_B(\phi(x^*x))}\\&\le \sup\bracsn{|\lambda|\ | \lambda \in \sigma_A(x^*x)}= \norm{x}_{A}^{2}\end{align*}

$\square$

Theorem 38.5.5.label Let $A, B$ be unital $C^{*}$-algebras and $\Phi: A \to B$ be a unital *-homomorphism, then $\Phi(A)$ is closed.

Proof, [Theorem 11.1, Zhu93]. 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

\[f: \complex \to \complex \quad z \mapsto \begin{cases}z&|z| \le 2\norm{y}_{B} \\ 2\norm{y}_{B} \cdot \sgn z = 2\norm{y}_{B} \cdot \frac{z}{|z|}&|z| \ge 2\norm{y}_{B}\end{cases}\]

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 continuous functional calculus, $\Phi(x) = f(\Phi(x)) = \Phi(f(x))$. By the Spectral Mapping Theorem, $\sigma_{A}(f(x)) = f(\sigma_{A}(x))$. By Theorem 38.4.3, $\norm{f(x)}_{A} = [f(x)]_{sp}\le \norm{f}_{u} = 2\norm{y}_{B}$.

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 method of successive approximations, $\Phi(A_{sa}) = \ol{\Phi(A)}\cap B_{sa}$. Therefore $\Phi(A) = \ol{\Phi(A)}$.$\square$

Definition 38.5.6 (Representation of $C^{*}$-Algebra).label Let $A$ be a $C^{*}$-algebra, then a representation of $A$ is a pair $(H, \pi)$, where $H$ is a Hilbert space, and $\pi: A \to B(H)$ is a *-homomorphism.

Definition 38.5.7 (Unitary Equivalence).label 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 unitarily equivalent if there exists a unitary map $U \in L(H_{1}; H_{2})$ such that the following diagram commutes

\[\xymatrix{ H_1 \ar@{->}[r]^{U} \ar@{->}[d]_{\pi_1(x)} & H_2 \ar@{->}[d]^{\pi_2(x)} \\ H_1 & H_2 \ar@{->}[l]^{U^*} }\]

for all $x \in A$.

Definition 38.5.8 (Invariant Subspace).label Let $A$ be a $C^{*}$-algebra, $(H, \pi)$ be a representation of $A$, $M \subset H$ be a closed subspace, and $P \in B(H)$ be the orthogonal projection onto $M$, then the following are equivalent:

  1. (1)

    For each $x \in A$, $\pi(x)(M) \subset M$.

  2. (2)

    $P \in \pi(A)'$.

If the above holds, then $M$ is an invariant subspace for $\pi$.

Proof. Since $\pi(A)$ is a self-adjoint subspace of $B(H)$, $M$ is invariant for $\pi(A)$ if and only if it is a reducing subspace for $\pi(A)$, if and only if $\pi(A)$ commutes with $P$.$\square$

Definition 38.5.9 (Intertwining Operator).label 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 intertwining operator for $\pi_{1}$ and $\pi_{2}$ if the following diagram commutes

\[\xymatrix{ H_1 \ar@{->}[d]_{\pi_1(x)} \ar@{->}[r]^{T} & H_2 \ar@{->}[d]^{\pi_2(x)} \\ H_1 \ar@{->}[r]_{T} & H_2 }\]

for all $x \in A$. The set $\mathcal{C}(\pi_{1}, \pi_{2})$ is the space of intertwining operators for $\pi_{1}$ and $\pi_{2}$.

In particular, $\mathcal{C}(\pi_{1}, \pi_{1}) = \pi(A)'$, and hence is a von Neumann algebra.

Definition 38.5.10 (Direct Sum).label Let $A$ be a $C^{*}$-algebra and $\bracsn{(H_i, \pi_i)}_{i \in I}$ be representations of $A$, then the representation

\[\pi: A \to B([l^{2}(I); H_{i}]) \quad \pi(x)(\xi)_{i} = \pi_{i}(x)(\xi_{i})\]

is the direct sum of $\bracsn{(H_i, \pi_i)}_{i \in I}$, denoted $\bigoplus_{i \in I}\pi_{i}$.

Proposition 38.5.11.label Let $A$ be a $C^{*}$-algebra, $(H, \pi)$ be a unitary representation of $G$, and $M \subset H$ be an invariant subspace, then:

  1. (1)

    $M^{\perp}$ is also an invariant subspace.

  2. (2)

    $(H, \pi)$ is the direct sum of $(M, \pi_{M})$ and $(M^{\perp}, \pi_{M^\perp})$.

Moreover,

  1. (4)

    $(H, \pi)$ is a direct sum of cyclic representations.

Lemma 38.5.12 (Schur).label Let $A$ be a $C^{*}$-algebra and $(H, \pi), (K, \tau)$ be unitary representations of $G$, then

  1. (1)

    $(H, \pi)$ is irreducible if and only if $\pi(A)' = \complex I$.

  2. (2)

    If $(H, \pi)$ and $(K, \tau)$ are irreducible, then $(H, \pi)$ and $(K, \tau)$ are unitarily equivalent if and only if $\mathcal{C}(\pi, \tau)$ is one-dimensional. Otherwise, $\mathcal{C}(\pi, \tau) = \bracsn{0}$.

Proof, [Theorem 3.5, Fol16]. (1): $(H, \pi)$ is irreducible if and only if $\pi(A)'$ contains no non-trivial projections. Since $\pi(A)'$ is a von Neumann algebra, Theorem 39.6.2 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 $T^{*}T \in \mathcal{C}(\pi, \pi) = \pi(A)'$, 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 $\lambda^{-1/2}T$[1] 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 $S, T \in \mathcal{C}(\pi, \tau)$, $S^{-1}T = S^{*}T \in \mathcal{C}(\pi, \pi)$. 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.$\square$

  1. Pick any branch.keyboard_return

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