Elsewhere I asked about ultrapowers of the C*-algebra $A$ of compact operators on separable infinite-dimensional Hilbert space. My question was whether the process of taking ultrapowers of ultrapowers ever stabilizes.

It was pointed out that the definition of an ultrapower of a C*-algebra is slightly different from the usual definition of an ultrapower; that the process never stabilizes in the sense that the canonical embedding is never an isomorphism; but that nevertheless (assuming CH) that if $A^1$ is an ultrapower of $A$ then the ultrapowers of $A^1$ are isomorphic as C*-algebras to $A^1$.

This last statement depends on the fact $A$ is separable. Here I would like to ask about the corresponding question for the C*-algebra $B$ of bounded operators on Hilbert space and also for the Calkin algebra $C=B/A$. Does the process of taking ultrapowers of ultrapowers of $B$ ever stabilize in the sense of producing isomorphic C*-algebras? And the same question for $C$?

  • $\begingroup$ Have you tried to see at the tensor product of ultrafilters? In many cases the iterated ultrapower is the ultrapower with respect to the tensor product of the ultrafilters. Whether or not this is isomorphic to the first ultrapower is most of the times matter of CH. $\endgroup$ – Valerio Capraro May 10 '12 at 20:17

The statement does not depend on separability of A. Actually in the Ge-Hadwin paper they explicitly state the result for C*-algebras of cardinality continuum. So under CH you have that the ultrapowers of C^1 are isomorphic to C^1. It is not completely trivial that C is not isomorphic to C^1. This was proved in arXiv:1112.3898 Countable saturation of corona algebras. Ilijas Farah, Bradd Hart. Also, if CH fails then every C*-algebra has 2^c nonisomorphic ultrapowers, so the sequence does not stabilize. This is in arXiv:0908.2790 Model theory of operator algebras I: Stability. Ilijas Farah, Bradd Hart, David Sherman.

| cite | improve this answer | |
  • 4
    $\begingroup$ Welcome to mathoverflow, Ilijas! $\endgroup$ – Joel David Hamkins May 15 '12 at 21:03

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.