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For a unital C*-Algebra we know that the state space is weak--compact and convex. The first one is due to the Banach-Alaoglu theorem and weak--closure and the second due to the fact that in unital C*-Algebras we have $\varphi(1)=\|\varphi\|$ for positive functionals $\varphi$.

The latter result is - in my opinion - especially surprising, since the state space $\mathcal{S}(\mathcal{A})$ is a subset of the dual sphere and one would not expect sphere-like things to be convex...

Now in the non-unital case things change:

(1) In $C_0(\mathbb{R})$ we can take the evaluations at $n$ for $n\in \mathbb{N}$ which converge weak-* to $0$ which is not a state.

Question 1: Do all non-unital C*-algebras have a non-closed state space?

(2) For convexity we cannot argue as in the unital case. However I think that we can embed a non-unital C*-algebra in its algebra with adjoint unit and use Hahn-Banach theorem for states. Then we should get $\mathcal{S}(\mathcal{A})=\mathcal{S}(\mathcal{\tilde A})$ (where the latter is the state space of the algebra with adjoint unit). This would mean that the state space is always convex... As I pointed out above, I consider this as a very surprising fact, therefore I cannot believe that it should hold in general and therefore it would be nice to get a confirmation that I didn't made a mistake. So

Question 2: Is the state space convex even for non-unital C*-algebras?

If the last gets a positive answer then I would like to get some intuition on the geometry of the state space. When I think about a convex portion of a sphere then I think about something like $\mathbb{R}^2$ with $1$-norm and then the part of it in some quadrant of the plance.

Question 3: Is this the right intuition for the geometry of the state space?

This should be enough for the moment.

Kind regards, Sebastian

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    $\begingroup$ The state space is always convex, which is easily seen by using an approximate unit. $\endgroup$
    – user42761
    Oct 29, 2016 at 14:13

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  1. Let $A$ be non-unital. Take any strictly positive $a\in A$ (all C$^*$-algebras have them). Then $C^*(a)\subset A$ is an abelian non-unital C$^*$-algebra (it cannot be unital because $a$ is strictly positive). So $C^*(a)\simeq C_0(X)$ with $X$ locally compact, non-compact. Now you can play the same game you played in $\mathbb R$ to show that there are sequences of states that converge $\sigma$-weakly to zero.

  2. If $\varphi,\psi$ are states, we can extend them by Hahn-Banach to the unitization of $A$. But the extensions are still positive (easy exercise) and have $\|\tilde\varphi\|=\|\tilde\psi\|=1$; thus $\tilde\varphi(1)=\tilde\psi(1)=1$. Then a convex combination is a state in the unitization of $A$, and this forces it to be a state in $A$.

  3. I cannot give you a lot of intuition. But notice that we are talking about positive guys here. So, if you make the analogy with the unit circle, then we are saying that the set $\{1\}$ is convex.

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  • $\begingroup$ Ad 1: Do we assume that $A$ is separable ? Otherwise, it might not be possible to find a strictly positive element. If such an element would always exist, any $\mathrm{C}^*$-algebra would have a countable approximate unit. $\endgroup$
    – user42761
    Dec 28, 2016 at 17:13
  • $\begingroup$ You are totally right. I'll review that when I have time. $\endgroup$ Dec 28, 2016 at 18:19
  • $\begingroup$ @MartinArgerami Does this mean that a separable nonunital $C^*$-algebra doesn't necessarily have a faithful state? $\endgroup$ Apr 17, 2017 at 0:17
  • $\begingroup$ Not a all. A separable C$^*$-algebra always has a faithful state: let $\{a_n\}$ be dense in $\{a\in A:\ \|a\|=1,\ a\geq0\}$. For each $n$, let $f_n$ be a state with $f_n(a_n)=1$. Then $f=\sum_n2^{-n}\,f_n$ is faithful. $\endgroup$ Apr 17, 2017 at 1:20

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