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Assume that $A$ is a unital $C^*$ algebra. Is there a $C^*$ embedding of $A$ in some $B(H)$ whose image is a hereditary $C^*$ subalgebra of $B(H)$?

If not, is the answer affirmative when $A$ is commutative?

If the answer of the later question is affirmative, we consider the following definition:

We say that a compact Hausdorff topological space has the property $P$ if for every Hilbert space $H$ and every two hereditary $C^*$ embeddings $\alpha, \beta: C(X) \to B(H)$, there is an automorphism $\phi$ of $B(H)$ with $\phi \circ \alpha=\beta$. Of course, this property is a topological invariant.

What are some examples of topological spaces which satisfy this property $P$? Does every finite set $X$ satisfy $P$? What are some other examples? Is this property preserved by disjoint union, product or wedge sum? Is this property identical to a well known classical property of topological spaces?

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If $A\subseteq B(H)$ is hereditary and has unit $p$, then we must have $A=pB(H)p\cong B(pH)$. So the answer is, such an embedding exists if and only if $A\cong B(H)$ for some $H$. In particular, the only commutative example is $A=\mathbb C$.

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  • $\begingroup$ Thank you very much for your answer. Just some more questions. For a given $C^{*}$ algebra A, is there a Hilbert space H such that $A$ is isomorphic to a subalgebra of $B(H)$ and all injective representaions of A on $H$ are $\phi$ related, for some automorphism $\phi$ of $B(H)$? If not, what about if $A$ is commutative? $\endgroup$
    – Ali Taghavi
    Feb 28, 2018 at 7:37
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    $\begingroup$ No. Take $A=C_0(\mathbb N)$. Every infinite-dimensional Hilbert space admits one faithful representation of $A$ having $\operatorname{rank}(\delta_1)=1$, and another having $\operatorname{rank}(\delta_1)=2$. But $\dim (\delta_1 B(H)\delta_1)$ is invariant under automorphisms. $\endgroup$
    – t.c.
    Feb 28, 2018 at 9:49

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