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?
 A: 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.
A: 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. 
