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Suppose $\mathcal A_1$ and $\mathcal A_2$ are unital C$^*$-algebras with faithful states $\varphi_1,\psi_1$ of $\mathcal A_1$ and $\varphi_2, \psi_2$ of $\mathcal A_2$.

Are the reduced free products $(\mathcal A_1, \varphi_1) * (\mathcal A_2, \varphi_2)$ and $(\mathcal A_1,\psi_1) * (\mathcal A_2, \psi_2)$ isomorphic as C$^*$-algebras?

Because of the faithfulness of the states both of these algebras are C$^*$-norm completions of the algebraic free product $\mathcal A_1 \circledast_\mathbb C \mathcal A_2$. The question then is whether this free product can be thought of as the unique spatial free product akin to the spatial tensor product being unique.

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    $\begingroup$ That seems very unlikely -- I think that already free product of two copies of $\mathbb{C}^2$ depends on the state. If you start with a symmetric measure on two points then you get the group algebra of $\mathbb{Z}_2 \ast \mathbb{Z}_2$, but if you take a non-symmetric measure then you end up with a Hecke-von Neumann algebra, which is a deformation of it. It is not yet a counterexample, but they are both interpolated free group factors, so distinguishing them (or showing that they are isomorphic) is not a simple matter. $\endgroup$
    – Mateusz Wasilewski
    Commented May 16, 2017 at 16:08
  • $\begingroup$ Have you looked at Avitzour, "Free products on $C^*$-Algebras. There is something on page 429. $\endgroup$
    – Matthew Daws
    Commented May 16, 2017 at 19:40
  • $\begingroup$ @MatthewDaws Excellent! I had not noticed this before. For those following along at home, in comment 2.4 of Avitzour he provides two non-$*$-isomorphic reduced free products of the same algebras. This is obtained by showing that they don't have the same traces. $\endgroup$
    – Chris Ramsey
    Commented May 16, 2017 at 19:48

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