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According to the article on nLab the Alfsen Shultz theorem states that the space of states of a given $C^*$-algebra depends on somehow weaker structure namely on the so called Jordan algebra structure. This article gives reference for this theorem: however I've checked the cited paper and I don;t see how the main theorem (namely theorem 9.5 in this paper) implies the above statement. For convienience I quote this theorem below: Let $A$ be a JB algebra (see the Introduction of the paper). Then there is unique Jordan ideal $J$ such that $A/J$ has faithful, isometric Jordan representation as an JC algebra and every factor representation of $A$ not anihilating $J$ is onto the exceptional algebra $M_3^8$ .

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  • $\begingroup$ The fact that you can get back the Jordan multiplication follows because you can recover the squaring map because the state space of a unital C$^*$-algebra is strongly spectral, a notion defined here, although perhaps the place to look is the two books by Alfsen and Shultz called State Spaces of Operator Algebras and Geometry of State Spaces of Operator Algebras. $\endgroup$ – Robert Furber Aug 16 '20 at 0:37
  • $\begingroup$ This would give the one direction: that the state space is enough to recover the Jordan product, right? How about the converse implication? $\endgroup$ – truebaran Aug 16 '20 at 21:06
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    $\begingroup$ Wait -- isn't the fact that the state space depends only on the Jordan structure pretty much trivial? A state is a positive linear functional, and "positive" and "linear" depend only on the Jordan structure. It's the other direction -- recovering the Jordan algebra from the state space -- which is hard. $\endgroup$ – Tim Campion Aug 16 '20 at 21:07
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    $\begingroup$ @truebaran An element $x$ of a Jordan algebra is positive iff it is a square, i.e. there is a $y$ such that $x = y \circ y$. (This agrees with the fact that positive elements of C$^*$-algebras are the squares of self-adjoint elements.) The positive cone and unit allow you to define when a linear functional is a state, and get the state space. $\endgroup$ – Robert Furber Aug 16 '20 at 21:10
  • $\begingroup$ Thank you, you are both right $\endgroup$ – truebaran Aug 16 '20 at 21:12

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