This is a question that is a bit outside my usual mathematical comfort zone, but I feel like an expert might know the answer.

Recall that a dimension group is an ordered abelian group $G$ with positive cone $G^+$ that arises as a limit of ordered groups of the form $\mathbb Z^d$ with the componentwise order. It is called simple if every element in $G^+\setminus\{0\}$ is an order unit.

If one considers a distinguished order unit $u\in G^+$, then one defines the state space $\mathcal{S}(G,G^+,u)$ as the set of all states (duh), meaning all order-preserving homomorphisms $s: G\to\mathbb R$ with $s(u)=1$. If $G$ is countable, then the topology of pointwise convergence will give the state space the structure of a metrizable Choquet simplex.

It is a well-known fact that every metrizable Choquet simplex $T$ can be realized as the state space of a simple dimension group $(G,G^+,u)$. However, it is also well-known that this is highly non-unique, and in fact there are many possible choices one can make in general. I am interested to know whether one can ask the following in addition:

Let $T$ be a metrizable Choquet simplex and let $f: T\to (0,1)$ be an affine continuous function. Does there exist a simple dimension group $(G,G^+,u)$ with distinguished order unit together with another element $a\in G^+$ such that $T$ is affinely homeomorphic to $\mathcal{S}(G,G^+,u)$ in such a way that $f$ corresponds to the assignment $[s\mapsto s(a)]$?

If this happens to be false in general, are there any natural sufficient conditions on the function $f$?

A note for C*-algebraists: What this question really asks is whether there exists a unital simple AF algebra $A$ having $T$ as its tracial state space and a projection $p\in A$ such that the function $f$ corresponds to the point evaluation $[\tau\mapsto\tau(p)]$.


1 Answer 1


Yes. Form Aff $(T)$; since $T$ is metrizable, Aff $(T)$ contains a countable dense set. Adjoin $f$ and the constant function $1$ to the countable dense set (creating a larger countable dense set), and let $G$ be the subgroup of Aff $(T)$ generated by this enlarged set. This is of course countable, and dense.

Therefore, equipped with the strict ordering ($G^+$ consisting of $0$ together with all the elements in $G$ that are bounded below away from zero on $T$), $G$ is a simple partially ordered group, and it is well known that dense subgroups of Aff of a Choquet simplex with respect to the strict ordering satisfy interpolation. Hence $G$ is a simple dimension group [using the result that a countable unperforated partially ordered group with interpolation is a dimension group, i.e., a limit of simpicial groups], and with respect to the order unit $u = 1$, $S(G,u)$ is naturally identified with $T$ (that is, the normalized traces of $G$ are the points of $T$).

It is easy to check that with $u = 1$ (the constant function), the map $(G,u) \to ({\rm Aff\ }S(G,u),1)$ given by $g \mapsto \hat g$ where $\hat g(s) = s(g)$ is equivalent to the inclusion $(G,1) \subset {\rm Aff } (T),1)$, and in such a way that $\hat f$ is sent to $f$.

More generally, we can do this for any countable family of $f$s, rather than just a single one.

For all of the unjustified statements here, see Goodearl's book on partially ordered abelian groups. The use of countable subgroups as here is a standard construction.

  • $\begingroup$ Thanks, David! I am happy to hear that this type of construction is known. I will sure go and check out Goodearl's book. $\endgroup$ Nov 30, 2017 at 7:33

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