# When do infinitesimals split in dimension groups?

Let $G$ be a dimension group (i.e. a directed, unperforated abelian group satisfying the Riesz interpolation property) with order unit $u\in G^{+}$. There is a canonical positive group homomorphism $\theta\colon G\to \operatorname{Aff}(S(G,u))$, where $S(G,u)$ is the state space of $G$ and $\operatorname{Aff}(S(G,u))$ are the real-valued affine, continuous functions on $S(G,u)$ defined by $\theta(g)(\sigma) = \sigma(g)$ for all $\sigma\in S(G,u)$.

There is the obvious short exact sequence (of abelian groups): $$0 \to \ker\theta \to G\to \theta(G)\to 0.$$ When $G$ is finitely generated, $\theta(G)$ is a finitely-generated, torsion-free abelian group, and thus free. Thus the short exact sequence splits. Also when $G = K_0(C(X))$ for a compact space $X$, we see that $\theta(G) \cong C(X,\mathbb{Z})$, which is also free. And so the short exact sequence splits in this case as well.

My questions are:

1. Does this short exact sequence always split? Does it split when $G$ is simple?
2. If not, is there a necessary and sufficient condition under which it does split?
3. If not, is there a concrete example of a group $G$ where it does not split?

Edit: Corrected (I hope) a mistake noted in the answer.

Since every countable torsion-free abelian group can arise as the underlying group of a dimension group, even with unique trace (state), the answer to the first question is no, even when $G$ is also simple. A stationary example is given by almost any $2 \times 2$ strictly positive integer matrix both of whose eigenvalues are integers (that is, non-splitting is generic, even in this very special case).
Giving reasonable/computable necessary and sufficient conditions for splitting is next to impossible; for example, let $H$ be a countable torsion-free abelian group and $T:H \to {\bf Z}[1/n]$ (where $n > 1$ is an integer) an (onto) group homomorphism. Then splitting of this for an essentially random choice of $H$ is highly unlikely, and when it occurs, is not so easy to determine. However, the pair $(H,T)$ describes a simple dim group (where $H^+ \setminus 0 = T^{-1}(>0)$) with unique trace, and this completely describes all simple dimension groups with unique trace whose value group is ${\bf Z}[1/n]$.
Also, you don't mean Aff $S(G,u)$ is isomorphic to $C(X,{\bf Z})$ (it isn't), but K$_0(C(X))$ when $X$ zero-dimensional compact.
• Thanks! Just to make sure I correct that last part, it is true that $\theta(K_0(C(X)) \cong C(X,\mathbb{Z})$ right? Jan 1, 2015 at 20:21
• Yes; K$_0$ just picks out the projections. Jan 1, 2015 at 21:36