1
$\begingroup$

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?

$\endgroup$
5
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
3
$\begingroup$

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]

$\endgroup$
6
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.