3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.