If a C$^{\star}$- algebra is separable, is there a representation in a ,also separable, Hilbert space ? Probably it's not hard to adapt the proof of the Gelfand-Naimark Theorem, but can someone give me some references ?
Thank you.
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1$\begingroup$ Do you want the representation to be faithful? I'm guessing the zero representation won't satisfy you. $\endgroup$– Nate EldredgeJun 18, 2013 at 14:54
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1$\begingroup$ If $A$ is separable, then $A$ has a faithful state $\pi$. The GNS representation given by $\pi$ acts faithfully on a separable Hilbert space. $\endgroup$– MichaelJun 18, 2013 at 15:23
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$\begingroup$ Thank you all for the comments. Meanwhile I found some online notes on operator algebras that satisfy what i needed. $\endgroup$– thetruthJun 18, 2013 at 15:25
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$\begingroup$ @thetruth Then I suggest you answer your own question. $\endgroup$– JulienJun 18, 2013 at 20:22
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As julien suggested, i'll answer : Let $\mathbb{A}$ be a separable C$^*$-algebra.
Let $\lbrace D_{i}\rbrace_{i\in\mathbb{N}}\subset A$ be a dense and countable subset and following the GNS-Construction, we have a cyclic representation $\pi^{d}:\mathbb{A}\rightarrow L(H_{d})$, for each $d\in D$. Let $\xi_{i}\in H_{d_{i}}$ be the associated cyclic vectors. First we note that $\lbrace\pi^{d_{i}}(d_{j})\xi_{i} : j\in\mathbb{N}\rbrace$ is countable and dense (since $D$ is dense and $\pi$ is continuous - so $\pi(\overline{D})=\overline{\pi(D)}$ - and $\xi$ is cyclic). Therefore, each $H_{d_{i}}$ is separable. Now, we just need to consider $\pi:\mathbb{A}\rightarrow L(H)$, with $H=\bigoplus_{d\in D}H_{d}$ and such that $\pi(x)=\bigoplus_{d\in D}\pi(x)$. It's clear that $H$ is separable and since $D$ is dense and $\pi$ continuous and following the fact that, by consequence of GNS-Construction, we have that $\forall d\in D : ||\pi(d)||=||d||$, we can conclude that $\pi$ is an isometry.
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$\begingroup$ I think you need to be bit more precise. For every $d\in D$, there exists a pure state $f$ such that $f(d^*d)=\|d\|^2$ (a bit tricky). Now GNS applied to this state yields $(\pi^d,\xi_d)$ (irreducible) such that $\|\pi^d(d)\xi_d\|=\|d\|$. Then $\pi=\bigoplus_{d\in D}\pi^d$ is a $*$-representation of $A$ which is isometric on $D$ dense, whence isometric on $A$. As you observed, every $\pi^d$ acts on a separable space, so does $\pi$. $\endgroup$– JulienJun 18, 2013 at 22:01
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$\begingroup$ Since injective *-homomorphisms are isometric, it suffices to use a countable dense set of states, which is possible since the (one-point compactification of the) set of states is compact metrizable. The faithful state Michael mentioned is obtained by taking a suitably weighted sum of those. // The GNS representation for each state is separable because it's a completion of a quotient of $\mathbb{A}$ with respect to a continuous seminorm. $\endgroup$– MartinJun 18, 2013 at 22:20