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The center of $K(H)$ is 0 and $K(H)$ has no nonzero finite dimensional representation. Can we conclude that if the center of a $C^*$-algebra $A$ is zero, then $A$ has no nonzero finite dimensional representation?

If the above conclusion is not true, does there exist a counterexample?

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    $\begingroup$ (Snarky answer: No, we cannot infer the general statement from the fact that it holds for $K(H)$, and yes, if the conclusion is not true then there must be a counterexample. But I give a real answer below.) $\endgroup$
    – Nik Weaver
    Mar 7, 2019 at 14:58
  • $\begingroup$ I guess $K(H)$ is the (non-unital) algebra of compact self-operators of an infinite-dimensional Hilbert space $H$. $\endgroup$
    – YCor
    Mar 7, 2019 at 15:11

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Let $P$ be a rank $1$ projection in $K(H)$ and let $\mathcal{A}$ be the set of continuous functions $f: [0,1]\to K(H)$ with $f(0) = \alpha P$ for some $\alpha \in \mathbb{C}$. Operations are pointwise. The map $f \mapsto \alpha$ is then a complex homomorphism, i.e., a one dimensional representation, but the center of $\mathcal{A}$ is trivial. (Any nonzero $f \in \mathcal{A}$ must satisfy $f(t) \neq 0$ for some $t > 0$; find $B \in K(H)$ which doesn't commute with $f(t)$ and let $g(s) = sB$. This belongs to $\mathcal{A}$ and doesn't commute with $f$.)

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  • $\begingroup$ How to define the adjoint of $f$? $\endgroup$
    – math112358
    Mar 8, 2019 at 19:27
  • $\begingroup$ Operations are pointwise. $\endgroup$
    – Nik Weaver
    Mar 8, 2019 at 19:46

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