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Let $A$ (resp. $B$) be a unital $C^\ast$-algebra, $\mathcal{Q}(A)$ (resp. $\mathcal{Q}(B)$) the compact convex subset of $A^\ast$ equipped with the $\sigma(A^\ast, A)$ (resp. $\sigma(B^\ast, B)$) topology. Suppose $\mathcal{Q}(A)$ and $\mathcal{Q}(B)$ are isomorphic in the sense that there exists a bijective affine homeomorphism $\varphi$ from $\mathcal{Q}(A)$ onto $\mathcal{Q}(B)$, does it follow that $A$ and $B$ are isomorphic as $C^\ast$ algebras?

In the abelian case (i.e. both $A$ and $B$ are commutative), the answer is affirmative, as we can simply look at the compact subspace of pure states and conclude by the Gelfand duality. But in the non-abelian case, the argument above fails. In fact, I strongly suspect that there exists non-isomorphic unital $C^\ast$-algebras $A$ and $B$ with $\mathcal{Q}(A)$ and $\mathcal{Q}(B)$ being isomorphic. Can anyone give a concrete example of this?

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For any C$^*$-algebra $A$, we can define its opposite algebra $A^{\mathrm{op}}$, which is the algebra where $ab$ is defined to be $ba$, as calculated in $A$. Let's restrict to unital algebras for simplicity. Then the identity mapping $A \rightarrow A^{\mathrm{op}}$ is linear, positive and unital, and is its own inverse. Therefore it induces a continuous, affine, origin-preserving isomorphism of quasi-state spaces. To get a counterexample, all you need is a C$^*$-algebra that is not isomorphic to its opposite. As far as I know the first example was a von Neumann algebra constructed by Connes. Since then, there have been several other examples, such as this separable example.

The missing piece of structure is what is defined and called an orientation in this article. The second author of that article is sometimes here on this forum. If you want to apply that article to the non-unital case, just remember that the quasi-state space of a C$^*$-algebra $A$ is isomorphic to the state space of the unitization $\tilde{A}$ in the obvious manner.

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