diff --git a/src/op/c-star/gns.tex b/src/op/c-star/gns.tex index a73d3d7..0014c1c 100644 --- a/src/op/c-star/gns.tex +++ b/src/op/c-star/gns.tex @@ -106,7 +106,41 @@ For each $x, y \in A$, $\dpn{x, y}{\phi} = \dpn{\pi_\phi(x)(1_A + N_\phi), \pi_\phi(y)(1_A + N_\phi)}{H_\phi}$ by well-definedness of the inner product on $H_\phi$. \end{proof} -\begin{theorem}[Gelfand-Naimark-Segal] +\begin{proposition} +\label{proposition:gns-irreducible} + Let $A$ be a unital $C^*$ algebra, $\phi \in S(A)$ be a state, and $(H, \pi, \xi)$ be its GNS triple, then $\phi$ is a pure state if and only if $(H, \pi)$ is irreducible. +\end{proposition} +\begin{proof} + ($\Rightarrow$): Suppose that $(H, \pi)$ is reducible, then there exists a non-trivial invariant subspace $M \subset H$. Let $P \in B(H)$ be the orthogonal projection onto $M$, $\xi_1 = P\xi/\norm{P\xi}_H$, and $\xi_2 = (\xi - P\xi)/\norm{\xi - P\xi}_H$. For each $x \in A$, let + \[ + \dpn{x, \phi_1}{A} = \dpn{\pi(x)\xi_1, \xi_1}{H} \quad \dpn{x, \phi_2}{A} = \dpn{\pi(x)\xi_2, \xi_2}{H} + \] + + then $\phi_1$ and $\phi_2$ are vector states on $A$. Since $M$ and $M^\perp$ are invariant for $\pi$, + \begin{align*} + \dpn{x, \phi}{A} &= \dpn{\pi(x)\xi, \xi}{A} = \dpn{\pi(x)P\xi, P\xi}{A} + \dpn{\pi(x)(I - P)\xi, (I - P)\xi}{A} \\ + &= \norm{P\xi}_H^2\dpn{\pi(x)\xi_1, \xi_1}{H} + \norm{(1 - P)\xi}_H^2 \dpn{\pi(x)\xi_2, \xi_2}{H} \\ + &= \norm{P\xi}_H^2 \dpn{x, \phi_1}{A} + \norm{(1 - P)\xi}_H^2 \dpn{x, \phi_2}{A} + \end{align*} + + so $\phi$ is a strict convex combination of two states, and is not pure. + + ($\Leftarrow$): Suppose that $(H, \pi)$ is irreducible. Let $t \in (0, 1)$ and $\phi_1, \phi_2 \in S(A)$ with $\phi = t\phi_1 + (1 - t)\phi_2$. Let $T: \pi(A)H \to H$ be defined by + \[ + \dpn{T\pi(x)\xi, \pi(y)\xi}{H} = \dpn{x, y}{\phi_1} = \dpn{y^*x, \phi_1}{A} + \] + + then since $\phi_1 \le t^{-1}\phi$, $T$ extends continuously into a positive operator in $B(H)$. Moreover, for any $x, y, z \in A$, + \begin{align*} + \dpn{\pi(z)T\pi(x)\xi, \pi(y)\xi}{H} &= \dpn{T\pi(x)\xi, \pi(z^*y)\xi}{H} \\ + &= \dpn{y^*zx, \phi_1}{A} = \dpn{T\pi(z)\pi(x)\xi, \xi}{H} + \end{align*} + + so $T \in \pi(A)'$. By \hyperref[Schur's Lemma]{lemma:cstar-schur}, there exists $\lambda \in \complex$ such that $T = \lambda I$. However, since $\dpn{T\xi, \xi}{H} = \dpn{1_A, \phi_1}{A} = 1$, $\lambda = 1$, $T = I$, and $\phi_1 = \phi$. By symmetry, $\phi_2 = \phi$ as well. +\end{proof} + + +\begin{theorem}[Gelfand-Naimark] \label{theorem:gns} Let $A$ be a unital $C^*$-algebra, then: \begin{enumerate} diff --git a/src/op/c-star/homomorphism.tex b/src/op/c-star/homomorphism.tex index 1fa99c7..ceea065 100644 --- a/src/op/c-star/homomorphism.tex +++ b/src/op/c-star/homomorphism.tex @@ -50,7 +50,6 @@ so $\bracsn{\pi(x)\eta|x \in A, \eta \in H} \subset \bracsn{\xi}^\perp$, and cannot be dense in $H$. \end{proof} - \begin{proposition} \label{proposition:star-homomorphism-contractive} Let $A, B$ be unital $C^*$-algebras and $\phi: A \to B$ be a unital *-homomorphism, then for each $x \in A$, @@ -108,4 +107,75 @@ +\begin{definition}[Invariant Subspace] +\label{definition:cstar-invariant-subspace} + 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: + \begin{enumerate} + \item For each $x \in A$, $\pi(x)(M) \subset M$. + \item $P \in \pi(A)'$. + \end{enumerate} + + If the above holds, then $M$ is a \textbf{invariant subspace} for $\pi$. +\end{definition} +\begin{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 $P$. +\end{proof} + +\begin{definition}[Intertwining Operator] +\label{definition:cstar-intertwining} + 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 + \[ + \xymatrix{ + H_1 \ar@{->}[d]_{\pi_1(x)} \ar@{->}[r]^{T} & H_2 \ar@{->}[d]^{\pi_1(x)} \\ + H_1 \ar@{->}[r]_{T} & H_2 + } + \] + + 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. +\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 + \[ + \pi: A \to B([l^2(I); H_i]) \quad \pi(x)(\xi)_i = \pi_i(x)(\xi_i) + \] + + is the \textbf{direct sum} of $\bracsn{(H_i, \pi_i)}_{i \in I}$, denoted $\bigoplus_{i \in I}\pi_i$. +\end{definition} + +\begin{proposition} +\label{proposition:cstar-rep-invariant-gymnastics} + Let $A$ be a $C^*$-algebra, $(H, \pi)$ be a unitary representation of $G$, and $M \subset H$ be an invariant subspace, then: + \begin{enumerate} + \item $M^\perp$ is also an invariant subspace. + \item $(H, \pi)$ is the direct sum of $(M, \pi_M)$ and $(M^\perp, \pi_{M^\perp})$. + \end{enumerate} + + Moreover, + \begin{enumerate}[start=3] + \item $(H, \pi)$ is a direct sum of cyclic representations. + \end{enumerate} +\end{proposition} +% Proof omitted. + +\begin{lemma}[Schur] +\label{lemma:cstar-schur} + Let $A$ be a $C^*$-algebra and $(H, \pi), (K, \tau)$ be unitary representations of $G$, then + \begin{enumerate} + \item $(H, \pi)$ is irreducible if and only if $\pi(A)' = \complex I$. + \item 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, $\mathcal{C}(\pi, \tau) = \bracsn{0}$. + \end{enumerate} +\end{lemma} +\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. + + 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. +\end{proof} + +