7
$\begingroup$

What is an example of a $*$-algebra $\cal{A}$, which admits two non-equivalent norms $\| \cdot \|_1$ and $\| \cdot \|_2$, with respect to which we can complete $\cal{A}$ to give two $C^*$-algebras $A_1$ and $A_2$, such that the two associated $K$-theory groups are non-isomorphic, that is: $$ K(A_1) \not\simeq K(A_2). $$

$\endgroup$
2
  • 2
    $\begingroup$ How about $C(\mathbf{F}_2)$ and $C^r(\mathbf{F}_2)$ which are completions of $\ell_1(\mathbf{F}_2)$ under suitable norms? $\endgroup$ Commented Nov 21, 2018 at 19:12
  • 4
    $\begingroup$ @TomekKania Hi Tomek, I think that the canonical map from the full algebra of ${\bf F}_2$ onto the reduced algebra actually induces an isomorphism on K-theory; if my memory is correct this is an Acta paper of E. C. Lance, but I haven't had time to check. (The relevant phrase is "K-amenability") $\endgroup$
    – Yemon Choi
    Commented Nov 21, 2018 at 19:25

1 Answer 1

11
$\begingroup$

Any infinite discrete group $\Gamma$ with Kazhdan's property (T) gives an example. Since it is not amenable, the full and reduced C*-algebras (which are both completions of the group algebra) do not coincide. Moreover, the full C*-algebra contains a projection with non-trivial K-theory class (so-called Kazhdan projection) which is mapped to $0$ in the reduced C*-algebra.

As for free groups, they are $K$-amenable, meaning that the canonical surjection between the full and reduced C*-algebras induces an isomorphism in $K$-theory.

The basic reference for this is J. Cuntz's article K-theoretic amenability for discrete groups, J. reine angew. Math. 1983 (where he also acknowledges overlapping work of E.C. Lance).

$\endgroup$
1
  • 1
    $\begingroup$ Nice answer --- I suppose it is still possible that the $K$ groups are isomorphic, just not naturally ... $\endgroup$
    – Nik Weaver
    Commented Nov 21, 2018 at 21:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .