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.

  • $\begingroup$ ,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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.