It may well be a trivial question but I was wondering if there is any relation between $K$-groups and ultrapowers of $C^*$-algebras. For instance, if $A$ is a $C^*$-algebra does $K_0(A^U)$ depend on the choice of a free ultrafilter $U$? What if $A$ is a von Neumann algebra with a trace and $A^U$ is the tracial ultrapower? For instance, what is $K_0(R^U)$, where $R$ is the hyperfinite $II_1$ factor?


If $R$ is type II finite AW* or W* factor, then $K_0(R^U) $ is naturally order isomorphic to the reals, as $R^U$ is again a type II finite AW* or W* factor. More generally, if $A$ is a C* algebra with stable range 1, then $l^{\infty}(A)$ (the algebra of bounded sequences of elements of $A$) has the interesting property that all of its simple images whose kernels contain $c_0(A)$ (the ideal consisting of sequences that converge to zero) are at least finite AW*-factors. A reference for the latter is

D Handelman [me], Homomorphisms from C${^*}$-algebras to AW$^*$-algebras, Michigan Math. J, 1981, 229--241. Here is the link to the paper.

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  • $\begingroup$ ,If $A=\prod_n M_n(\Bbb C)$,what is $K_0(A)$? $\endgroup$ – math112358 Dec 30 '18 at 19:51
  • $\begingroup$ Subgroup of $\prod {\bf Z}$ consisting of sequences $(a_n)$ for which there exists a constant $c$ with $|a_n| \leq c n$; the ordering is given as the coordinatewise ordering, and the sequence $(n)$ is the canonical order unit determined by the free module on one generator. $\endgroup$ – David Handelman Dec 30 '18 at 22:25

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