Suppose $X$ is a compact metric space. Is there a good candidate for a dense subalgebra $A\subseteq C(X)$, such that the inclusion induces an isomorphism in $K$-theory? For example, if $X$ was a manifold, I could take $A$ to be the smooth functions. Would it help if $X$ was a finite CW-complex?

P.S.: I know that $A=C(X)$ is a good candidate :)

EDIT: In response to a comment below, let me add a few comments. The problem with $C(X)$ is that, as any other nuclear $C^*$-algebra, it has trivial higher (continuous) cyclic cohomology, therefore we don't get interesting pairings with the corresponding homology. Ideally, I would like $A$ to be such that

  • It has interesting cyclic cohomology
  • The (even, odd) periodic cyclic homology of A coincides with (even, odd) Cech, or Alexander-Spanier, cohomology of $X$.
  • 1
    $\begingroup$ Could you say something more about what nice properties you want $A$ to have, which make it better than $C(X)$ itself? E.g. finiteness/Noetherianity properties, or good cohomological behaviour? $\endgroup$
    – Yemon Choi
    Sep 6, 2017 at 10:47
  • $\begingroup$ I modified my post in response to your comment. $\endgroup$
    – vap
    Sep 6, 2017 at 20:09
  • 4
    $\begingroup$ There is an old paper by Bost (not at my fingertips) which discusses this. If $A$ is a dense $*$-subalgebra of a unital C$^*$-algebra such for all $a \in A$, $1 +a*a$ is invertible in $A$ [obviously it is invertible in the C$^*$-algebra], then the inclusion induces isomorphisms on K$_0$ and on K$_1$ (and the isomorphism on K$_0$ is an isomorphism with respect to the pre-ordering). Also there is an earlier paper by Swan (which deals with commutative C$*$-algebras, that is, $C(X)$). $\endgroup$ Sep 6, 2017 at 20:23
  • $\begingroup$ Thanks for the pointer! I am still interested in the question concerning cyclic homology and cohomology of the underlying space. $\endgroup$
    – vap
    Sep 7, 2017 at 2:08
  • 1
    $\begingroup$ The case of manifolds already says much. The algebra of smooth functions is ok but it is difficult to say it can be easily deduced in a canonical way. Many other dense subalgebras of continuous function would do the game (like smooth everywhere but at a point, and continuous at that point with maybe some additional conditions). In a sens you're looking for some kind of deRham's theorem out of the differentiable context. $\endgroup$ Oct 16, 2017 at 7:52


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.