# $*$-algebras, completions, and $K$-theory

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: $$K(A_1) \not\simeq K(A_2).$$

• How about $C(\mathbf{F}_2)$ and $C^r(\mathbf{F}_2)$ which are completions of $\ell_1(\mathbf{F}_2)$ under suitable norms? – Tomasz Kania Nov 21 '18 at 19:12
• @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") – Yemon Choi Nov 21 '18 at 19:25

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.
• Nice answer --- I suppose it is still possible that the $K$ groups are isomorphic, just not naturally ... – Nik Weaver Nov 21 '18 at 21:24