A state space of a Boolean algebra is a Choquet simplex but not all Choquet simplices can be viewed as state spaces of Boolean algebras. Is it known which Choquet simplices are precisely state spaces of Boolean algebras?

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    $\begingroup$ Since it's not obviously documented, would you say what the state space is? I guess it's the set of functions $f:A\to [0,1]$ such that $f(1)=1$ and $f(xy)=f(x)+f(y)$ for all $x,y$ such that $xy=0$, with the compact topology induced by inclusion into $[0,1]^A$? $\endgroup$ – YCor Oct 4 '18 at 21:33
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    $\begingroup$ @YCor You probably mean $f(x\lor y)$ in place of $f(xy)$? $\endgroup$ – მამუკა ჯიბლაძე Oct 5 '18 at 17:00
  • $\begingroup$ Yes (I wanted to write $x+y$, which amounts to the same as $x\vee y$ when $xy=0$). $\endgroup$ – YCor Oct 5 '18 at 17:11
  • $\begingroup$ You are right, this is what is meant by the state space. $\endgroup$ – Miroslav Korbelar Oct 5 '18 at 19:49
  • $\begingroup$ @GerryMyerson sorry, I thought a block of 9 was better than 3 blocks of 3... (PS we should eventually erase these comments to unspam this post's comments) $\endgroup$ – YCor Oct 17 '18 at 12:15

The state space of a unital Boolean algebra is characterized (as a Choquet simplex) by the extremal boundary (the set of extreme points) being both compact and totally disconnected. Once stated, this result is pretty obvious, as is the proof. [Just observe that continuous functions on the extremal boundary extend (uniquely) to affine functions on the whole C simplex; take the indicator functions of clopen sets in the ext boundary, extend these; the set of them recovers the boolean algebra; the reverse is based on the extremal traces of a lattice being compact, etc]

  • $\begingroup$ In the definition here encyclopediaofmath.org/index.php/Choquet_simplex of Choquet simplex, compactness is part of the definition. $\endgroup$ – YCor Oct 5 '18 at 8:32
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    $\begingroup$ @Ycor Compactness of the simplex: yes; compactness of the extremal boundary: no; the latter defines Bauer simplex. But I can see how the wording might have been ambiguous, so I will rewrite it. $\endgroup$ – David Handelman Oct 5 '18 at 14:24
  • $\begingroup$ OK, I misread, thanks for the clarification. $\endgroup$ – YCor Oct 5 '18 at 16:31
  • $\begingroup$ Many thanks for the answer. If I may ask, what precisely means that Boolean algebra is unital? $\endgroup$ – Miroslav Korbelar Oct 10 '18 at 16:59
  • $\begingroup$ %Miroslav It depends on the definition of boolean algebra; a non-unital one would be the lattice of finite subsets of an infinite sets (where only relative complements are defined); but the usual definition precludes this, and so would require the boolean algebra to be unital. $\endgroup$ – David Handelman Oct 10 '18 at 22:54

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