Separable reducing subspace of a representation

Question

I'm looking for a hint or a reference to understand what's going on in the following problem. Suppose that $A$ is a unital $C^*$-algebra, and let $\pi : A \rightarrow B(H)$ be a representation.

Problem. If $A$ is separable and $\pi$ is injective, then, for each $a \in A$, there exists a separable reducing subspace $M_a \subset H$ of $\pi$ such that $||P_{M_a}\pi(a)_{\downharpoonright M_a}|| = ||a||$, where $P_{M_a}$ denotes the orthogonal projection in $H$ onto the subspace $M_a$, and $\pi(a)_{\downharpoonright M_a}$ the restriction of $\pi(a)$ to $M_a$.

Any help will be appreciated.

Where is the problem from? Is $M_a$ supposed to be associated with $a$ in some way? I don't think the statement is as intended. — Jonas Meyer, Mar 12 '13 at 05:01
ragrigg: Partly, yes, thanks. Where is the problem from? [I deleted comments based on misunderstanding.] — Jonas Meyer, Mar 12 '13 at 14:47

score 1 · Accepted Answer · answered Mar 12 '13 at 20:12

1

Let $(x_n)_n$ be a sequence of unit vectors in $H$ such that $\|\pi(a)x_n\|\to\|a\|$. Then $M_a$ can be taken to be the closed linear subspace of $H$ generated by $\pi(A)\{x_n\}_n$.

answered Mar 12 '13 at 20:12

Jonas Meyer

53,602

Separable reducing subspace of a representation

1 Answers1