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.


1 Answer 1


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.

Edit: Let me plug a recent paper (appeared December 2014, but is dated 2013) which has lots of examples, where the split cases typically have to be separated from the generic nonsplit ones,

Realizing dimension groups, good measures, and Toeplitz factors, Ill J Math Volume 57, Number 4 (2013), 1057-1109.

  • $\begingroup$ 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? $\endgroup$
    – Min Ro
    Commented Jan 1, 2015 at 20:21
  • 1
    $\begingroup$ Yes; K$_0$ just picks out the projections. $\endgroup$ Commented Jan 1, 2015 at 21:36

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