2
$\begingroup$

Let $A$ be a $\mathbb{Z}_2$-graded $C^*$-Algebra. Then we can take the direct limit $$ colim_{A_\sigma} KK_*(\mathbb{C}, A_\sigma) $$ over all $\sigma$-unital graded $C^*$-sub-algebras $A_\sigma \subseteq A$. This gives me the $K$-theory of graded $C^*$-algebras. Now, this is supposed to be tha same as taking the direct limit over all separable, graded $C^*$-subalgebras. For that to be true for any $\sigma$-unital, graded $C^*$ subalgebra $A_\sigma$ should be contained in a separable, graded $C^*$-sub-algebra $A_s$. But I don't quite see why.

Thank you

$\endgroup$
2
  • 1
    $\begingroup$ Any subset of a separable metric space is separable. $\endgroup$
    – Nik Weaver
    Feb 4, 2017 at 20:36
  • 1
    $\begingroup$ Every C*-algebra is the directed union of its separable sub-C*-algebras, and K-theory is a continuous functor, that is, it commutes with direct limits (=colimits). Therefore, $K_0(A)$ is the colimit of $K_0(A')$ over all separable sub-C*-algebras $A'$, ordered by inclusion. I think the grading will not cause any problems. One only needs to use that every graded C*-algebra is also the directed union of its graded, separable sub-C*-algebras. $\endgroup$ Feb 5, 2017 at 9:26

0

Your Answer

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