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Let $\mathcal{A}$ be a $C^*$ algebra. Let $(\pi, \mathcal{H})$ be a faithful, irreducible, unitary, Hilbert space representation of $\mathcal{A}$; i.e., $\pi:\mathcal{A}\rightarrow\mathcal{B}(\mathcal{H})$ is an injective *-homomorphism from $\mathcal{A}$ into a subset of the bounded operators $\mathcal{B}(\mathcal{H})$ on a Hilbert space $\mathcal{H}$, such that the resulting representation is irreducible. (A representation is irreducible if whenever $\pi(\mathcal{A})$ acts invariantly on a subspace $\mathcal{H}^\prime \subseteq \mathcal{H}$, then $\mathcal{H}^\prime = \mathbf{0}$ or $\mathcal{H}^\prime = \mathcal{H}$.)

Let $U \in \mathcal{B}(\mathcal{H})$ be any unitary ($U^* = U^{-1}$) bijection on $\mathcal{H}$, not necessarily in $\pi(\mathcal{A})$.

Does it follow that $U\pi(\mathcal{A})U^{*} = \pi(\mathcal{A})$?

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This may be a stupid comment, but if $U \pi (\mathcal{A}) U^* = \pi (\mathcal{A} )$ isn't U necessarily a scalar multiple of the identity, by some version of Schur's lemma? –  Mark Feb 9 '11 at 22:26
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I presume that $U\pi(A) U^*=\pi(A)$ is supposed to hold as sets and not pointwise for all elements in $A$. –  Andreas Thom Feb 9 '11 at 22:41

1 Answer 1

up vote 4 down vote accepted

The answer is no.

Consider the Toeplitz algebra $\mathcal T$ with its canonical representation on $\ell^2 \mathbb N$, which is generated as a $C^\star$-algebra by the shift $S(e_n)=e_{n+1}$. It is well-known that the Toeplitz algebra contains all compact operators; hence the representation is irreducible.

It is well-known that any other isometry $V$ of Fredholm index $-1$ is conjugate to $S$. Now, it is easy to see that if $V$ is in $\mathcal T$, then $[S,V]$ is compact. Indeed, the Toeplitz algebra modulo the compact operators is just $C(S^1)$, which is commutative. It is easy to construct such $V$ with $[S,V]$ not compact.

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