iterating ultrapowers of C*-algebras

Let $A$ be something interesting like the C*-algebra of compact operators on a separable infinite-dimensional Hilbert space and let $A^1$ be an ultrapower of $A$. Then $A^1$ is a primitive C*-algebra strictly containing $A$ (Ge and Hadwin). Now let $A^2$ be an ultrapower of $A^1$. Then $A^2$ is a primitive C*-algebra containing $A^1$, presumably strictly. In the same way we may define $A^3$, $A^4$, etc. Since $A^n$ contains $A^{n-1}$ we may define an inductive limit $A^{\omega}$, and then continue transfinitely with $A^{{\omega} +1}$ etc.

My question is, Does the process ever stabilise? If so does it stabilise at some finite step, or at $\omega$ or at the first uncountable ordinal?

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This might, perhaps, depend on the ultrafiltera used at each atep. –  Juris Steprans May 1 '12 at 23:08
In light of my answer, could you clarify what you mean by stabilize? The successive ultrapower maps are never isomorphisms, but perhaps you just want the models to be externally isomorphic. –  Joel David Hamkins May 2 '12 at 0:43
@ Joel. Yes, I was just thinking of external isomorphism. So in my question, I am interested in whether $A^1$ and $A^2$ are isomorphic as C*-algebras. –  Douglas Somerset May 2 '12 at 7:51
@Joel. Does "externally isomorphic" just mean "isomorphic"? I'm not familiar with this term. –  Nik Weaver May 2 '12 at 13:03
By "externally" isomorphic, I had just meant isomorphic, but not necessarily by the ultapower map under consideration. –  Joel David Hamkins May 2 '12 at 19:44

In the C*-algebra ultrapower construction, instead of identifying two sequences if they agree on a set in the ultrafilter, you identify two sequences if their difference goes to zero along the ultrafilter. You also throw out those sequences whose norm goes to infinity along the ultrafilter.

I think the simple answer to the question is that any infinite dimensional C*-algebra is properly contained in any ultrapower coming from a free (nonprincipal) ultrafilter. In fact this will be true of any infinite dimensional Banach space, as a simple consequence of the fact that its unit ball is not precompact. We can find a sequence of vectors in the unit ball such that the distance between any two of them is at least 1/2, and this will always give rise to a new element in the ultrapower (it is not identified with any constant sequence). So no, the process never stabilizes.

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The argument of your second paragaph is showing that the ultrapower map itself is not an isomorphism. But this is not the same as showing that the ultrapower algebra is not isomorphic to the original algebra. –  Joel David Hamkins May 2 '12 at 10:36
For example, in analogy with the usual first-order ultrapower construction, it seems likely that if the CH holds, then the double ultrapower of a $C^*$ algebra of size continuum will be externally isomorphic to any single ultrapower. –  Joel David Hamkins May 2 '12 at 12:09
@Joel. But in the original post, he is unambiguously asking whether you reach a point where the natural containment is not strict. Otherwise the notation $A^\omega$ doesn't even make sense. –  Nik Weaver May 2 '12 at 13:09
Joel --- Yes, Ge and Hadwin did this. (Ultraproducts of C∗-algebras, in Recent advances in operator theory and related topics (Szeged, 1999), 305–326, Oper. Theory Adv. Appl., 127, Birkhäuser, Basel, 2001.) –  Nik Weaver May 2 '12 at 22:53
Great! Then it seems to be a proof that the iterated ultrapowers (for seperable $C^*$ algebras) stabilize after one step, assuming the continuum hypothesis. –  Joel David Hamkins May 3 '12 at 0:53

Iterated ultrapowers are intensely studied in set theory, and this topic is fundamental to the theory of large cardinal inner model theory.

The usual set-theoretic perspective on ultrapowers is broad-minded about what it is that we are actually taking the ultrapower of: why bother taking the ultrapower only of the integers, or only of the reals, or only of a particular $C^*$-algebra, when one can take the ultrapower of the entire set-theoretic universe? For a given ultrafilter $U$, if you are entertaining the ultrapower of a specific mathematical structure by $U$, then why not add all possible relations on that structure as well, for this will give nonstandard analogues of those relations in the ultrapower; indeed, why not add all possible sets of relations and so on, iterating transfinitely? The point is that all ultrapowers of any structure by the same ultrafilter fit together coherently into a grand ultrapower of the entire universe.

Meanwhile, for a given ultrafilter $U$, there are several ways to proceed with iterated ultrapowers of the universe $V$, and from this perspective the question is ambiguous about what it means to iterate the ultrapower transfinitely.

Let me explain. On the one hand, the most typical way to proceed is by means of what it called the internal ultrapower. Having started with $V=M_0$ and formed the $\alpha$-th ultrapower $M_\alpha$, one has the canonical map $j_{0,\alpha}:V\to M_\alpha$. The internal perspective is to view $j_{0,\alpha}(U)$ as an ultrafilter $U_\alpha$ inside the model $M_\alpha$, and then form the next structure $M_{\alpha+1}$ as the internal ultrapower of $M_\alpha$ by the ultrafilter $U_\alpha$. At limit stages, one takes the direct limit and continues iterating.

Alternatively, one may form the external ultrapower $M_{\alpha+1}$ as the ultrapower of $M_\alpha$ via the original ultrafilter $U$. It seems that this is what you had in mind.

For the finite ultrapowers, the two ways of proceeding, either internally or externally, actually give rise to isomorphic models, even though this is not obvious. Nevertheless, the resulting system of maps $j_{n,m}:M_n\to M_m$ are definitely not the same, and this difference applies even when taking the ultrapower of the integers or of your $C^*$-algebra. Furthermore, when one takes the limit model $M_\omega$, then the internal iteration can be very different from the external iteration, and no longer isomorphic. For example, in the large cardinal context of iterating the ultrapower by a normal measure on a measurable cardinal, Kunen proved famously that all the internal iterated ultrapowers are well-founded, but the external iterations are definitely ill-founded at $\omega$. Similar and analogous differences arise even for ultrafilters on $\omega$.

Set theorists often also commonly consider more complicated systems of iterating ultrapowers, where one uses different ultrafilters at each step, often on different underlying sets, leading to the concept of extender embeddings. The highly developed subject known as large cardinal inner model theory is about the fundamental nature of these extender embeddings and the universes of set theory that one can build in this way. For example, the concept of iterating a measure "out of the universe" is an elementary part of this theory: one has an ultrafilter and considers the proper-class-length iterated internal ultrapower by that ultrafilter, looking at the residue of what is left behind below the successive critical points of the embeddings.

In addition, Haim Gaifman showed how to form the iteration of an ultrafilter along any linear order, not just the well-founded iterations. Thus, one can form a system of models that is the iteration of an ultrafilter along the integers, or the rationals or even the surreal numbers.

There is another ambiguity in your question, concerning what it means to reach a fixed point. On the one hand, neither the internal nor the external iteration ever reaches a fixed point, if what is meant is that the ultrapower map $j_{\alpha,\alpha+1}:M_\alpha\to M_{\alpha+1}$ is an isomorphism, since this map is never an isomorphism unless the structure is smaller than the degree of completeness of the ultrafilter. In particular, if you are using a nonprincipal ultrafilter on $\mathbb{N}$, then no ultrapower map of an infinite structure, such as your $C^*$-algebra, is an isomorphism. So in this strong sense, the process never terminates, no matter how long you proceed, and your question has a negative answer.

But if you seek a weaker fixed-point concept, where you have only that $M_\alpha$ is isomorphic to $M_{\alpha+1}$, but not necessarily by the ultrapower map connecting these structures, then there can be a positive answer already at the second step. Specifically, if the CH holds, then the ultrapower of any structure of size continuum by a nonprincipal ultrafilter on $\mathbb{N}$ is isomorphic to the two-step-iterated-ultrapower, because these are both saturated structures of size $\omega_1$ and with the same theory, so they are isomorphic by a back-and-forth argument. In this sense, your question has a positive answer.

If you don't have CH, then I'd have to think a bit harder about where one reaches a fixed point.

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I seem to recall that in the Ge-Hadwin paper to which Douglas refers, and indeed in other papers on operator algebras, the definition of a $C^\ast$-algebraic ultrapower is not quite the same as the ultrapower in the set-theoretic sense. However (a) I could be misremembering (b) your comments might still apply. –  Yemon Choi May 2 '12 at 0:47
On further reflection, I guess that the ultraproducts being used by Ge and Hadwin would be along the lines of Henson's constructions/set-up. Do your comments still apply there? –  Yemon Choi May 2 '12 at 0:51
Yemon, yes, that could be right. The OP did seem, however, to suggest a certain flexibility with what his structure was---"Let $A$ be something interesting...."---so I was imagining $A$ as a first-order structure, whose ultrapowers I understand better than $C^*$-algebras. But I think that even for $C^*$-ultrapowers, the ultrapower map itself will never be an isomorphism, so my first answer still applies, and for the second answer, if the CH holds then I find it likely that the saturation arguments still apply to get the isomorphism for the weak fixed point concept. –  Joel David Hamkins May 2 '12 at 0:55
I'm sorry that I am just not sufficiently familiar with the modified construction for ultrapowers of $C^*$-algebras. I am imagining that as in Hans Schoutens cataproduct work they take a standard part at some point to turn hyperreals back into reals? I find it likely that this wouldn't affect either of my answers: the ultrapower map itself will never be an isomorphism, and if CH holds, then the resulting models will be isomorphic; but I'd have to know the exact construction to be sure. –  Joel David Hamkins May 2 '12 at 1:01