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Assume $S$ and $T$ are unital, separable operator systems, and assume further that we have an affine homeomorphism between the state space of $S^*$ and the state space of $T^*$. Does that mean that $S^*$ and $T^*$ are isometrically isomorphic?

I know that each element $\varphi\in S^*$ can be written uniquely as $\varphi_1+i\varphi_2$ where $\varphi_k$ are self adjoints. Now each $\varphi_k$ can be decomposed uniquely as $(\varphi_k)_+-(\varphi_k)_-$ such that $\|\varphi_k\|=\|(\varphi_k)_+\|+\|(\varphi_k)_-\|$ for $(\varphi_k)_{\pm}\geq 0$, and eventually we can present each $\varphi$ as a linear combination of elements in $S^*$ (same goes for $T^*$), and assuming the norm condition, even uniquely. However, I don't see a way to extend this nicely to an isometric isomorphism. Is this even true? any help would be appreciated.

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    $\begingroup$ Two comments. First, your condition does not imply that $S$ and $T$ are completely isometrically isomorphic. That fails for any C*-algebra that is not isomorphic to its opposite algebra. (The two will be isometric and have isometrically isomorphic duals, but not completely isometric.) $\endgroup$ Commented Nov 19 at 17:43
  • $\begingroup$ Second, the affine homeomorphism between the two state spaces should extend to an algebraic isomorphism between $S^*$ and $T^*$, and since the real part of the unit ball is $\{\phi - \psi\ : \phi$ and $\psi$ are states$\}$, this isomorphism is an isometry between the real parts of $S^*$ and $T^*$. But does that determine the norms on all of $S^*$ and $T^*$? I don't know. $\endgroup$ Commented Nov 19 at 17:45
  • $\begingroup$ @NikWeaver Isn't it true that if we let $\varphi=\varphi_1+i\varphi_2$ be the decomposition to real and imaginary part, then $\|\varphi_1\|=\|\varphi\|$ (assuming it is non-zero)? $\endgroup$ Commented Nov 19 at 19:12
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    $\begingroup$ There is a well-known example of an order isomorphic but not isometric operator systems: $\{ a, bz, c\bar{z} : a,b,c \in \mathbb{C}\}\subset C(\mathbb{T})$ and $\left[\begin{smallmatrix} a & 2b \\ 2c & a \end{smallmatrix}\right]\subset M_2(\mathbb{C})$. In both cases, the element is positive iff $c=\bar{b}$ and $a\geq 2|b|$. I cannot remember (an easy argument) why they are not isometric. $\endgroup$ Commented Nov 19 at 23:41
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    $\begingroup$ $\ell_\infty^2$ embeds isometrically into the second (spanned by $b$ and $c$), but not into the first. $\endgroup$ Commented Nov 21 at 1:53

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