Let $A$ be a unital C*-algebra, and let $\mathcal R$ be a separating family of irreducible representations of $A$. Each vector state of a representation in $\mathcal R$ is a pure state, and the span of the set $X$ of such states is weak* dense in $A^*$. Is the closed convex hull of $X$ the entire state space of $A$? Does the closure of $X$ contain every pure state of $A$? This is probably textbook material, so please suggest a reference.

  • $\begingroup$ If $A$ is separable, the answer to the second question should be yes by an application of Glimm's lemma. Are you interested in the non-separable case? $\endgroup$ – Caleb Eckhardt Aug 25 '17 at 22:15
  • $\begingroup$ I find the question to be interesting and natural in the general case too, but I would accept an answer for separable $A$. $\endgroup$ – Andre Kornell Aug 26 '17 at 4:24

Good question. I don't know a reference, but I think the answer is yes. The direct sum $\pi$ of the irreps in $\mathcal{R}$ is faithful, so $\|\pi(x)\| = \|x\|$ for all $x \in A$. But the norm of an element of a direct sum is the sup of its norms in the summands, so $\|x\| = \sup \|\pi_\alpha(x)\|$ where $\pi_\alpha$ ranges over $\mathcal{R}$. If $x$ is self-adjoint then for any $\alpha$ there are unit vectors $v$ in the Hilbert space of $\pi_\alpha$ such that $|\langle \pi_\alpha(x)v,v\rangle|$ gets arbitrarily close to $\|\pi_\alpha(x)\|$. This means that there are pure states $f$ in your set $X$ for which $|f(x)|$ gets arbitrarily close to $\|x\|$. Now if the (weak*) closed convex hull of $X$ were not the entire state space then there would be a pure state $g$ outside it, and a self-adjoint element $x$ of $A$ and $a \in \mathbb{R}$ such that $|g(x)| > a \geq |f(x)|$ for all $f \in X$. By what I said above, this would force $|g(x)| > \|x\|$, a contradiction.

  • $\begingroup$ I had actually originally written the first part of the question as a statement, but I realized I couldn't remember the argument. It's really the second part that I'm after. $\endgroup$ – Andre Kornell Aug 24 '17 at 12:50
  • $\begingroup$ I didn't see that! Let me think some more. $\endgroup$ – Nik Weaver Aug 24 '17 at 13:10
  • $\begingroup$ (For future readers: this is the right argument for the first part of the question. Thanks Nik!) $\endgroup$ – Andre Kornell Aug 24 '17 at 14:24

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