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I am trying to imagine (to some extent, of course) the geometry of ultrapowers of certain 'easy-to-handle' Banach spaces. Let me start with $X = \ell_p$, $p\in (1,\infty)$ or $X=c_0$.

Since the differences between partiuclar choices of ultrafilters are not clear to me enough, let me take any free ultrafilter $\mathcal{U}$ over an arbitrary infinite index set $I$.

So, my beginners question is, does the ultrapower $(X)_\mathcal{U}$ have an unconditional basis?

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  • $\begingroup$ For anything interesting to happen, you want a non-principal ultrafilter. Otherwise the ultrapower will just be $X$. $\endgroup$ – Andrew Parker May 17 '12 at 21:50
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    $\begingroup$ Another issue with the generality in which you have posed the question is that if $I$ is very large and there is a measurable cardinal, then the ultrafilter you use on $I$ might be complete beyond the size of $X$, in which case the ultrapower of $X$ will again just be $X$, even though the ultrafilter is nonprincipal. $\endgroup$ – Joel David Hamkins May 17 '12 at 22:37
  • $\begingroup$ Andrew, in this context, "free" means "non-principal". $\endgroup$ – Bill Johnson May 18 '12 at 6:08
  • $\begingroup$ Bill, since "free" was added to the question in the (currently) latest edit, I conjecture that it was added as a result of Andrew's comment. $\endgroup$ – Andreas Blass May 18 '12 at 12:25
  • $\begingroup$ Yes, apologies for not pointing it out. $\endgroup$ – Alfredo Ortuño May 18 '12 at 15:31
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For $\ell_1$ and $c_0$, the answer to your (implicit) question is easy. If you take a free ultrafilter on the natural numbers (or most any ultrafilter used in Banach space theory) the ultrapower contains $L_1$ (or $C[0,1]$) and hence does not have an unconditional basis.

For $\ell_p$, $1<p\not= 2 < \infty$, the situation might be trickier. Enflo and Rosenthal

Some results concerning Lp(μ)-spaces. J. Functional Analysis 14 (1973), 325–348

proved that if $\mu$ is a finite measure s.t. the density character of $L_p(\mu)$ is at least $\aleph_\omega$, then $L_p(\mu)$ does not embed into a space that has an unconditional basis. Now such an $L_p(\mu)$ embeds into some ultrapower of $\ell_p$ but I don't know off the top of my head that when the density character of $L_p(\mu)$ is equal to $\aleph_\omega$ [EDIT: I meant $\aleph_1$ here] whether (under some or any set theoretic axioms about the size of $\aleph_\omega$ [EDIT: I meant $\aleph_1$ here]) $L_p(\mu)$ embeds into an ultrapower of $\ell_p$ when the ultrafilter is a free ultrafilter on the natural numbers.

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  • $\begingroup$ Thank you. Perhaps I should restrict my attention to ultrafilters over integers only. $\endgroup$ – Alfredo Ortuño May 18 '12 at 9:04
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    $\begingroup$ Bill, I'm confused --- isn't any ultrapower of a Hilbert space also a Hilbert space, and doesn't every Hilbert space have an unconditional basis? $\endgroup$ – Nik Weaver May 18 '12 at 9:25
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    $\begingroup$ Sure, Nik. I should have written $1<p\not= 2 < \infty$. I'll edit. $\endgroup$ – Bill Johnson May 18 '12 at 10:39
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    $\begingroup$ Bill: To embed such $L_p(\mu)$ into an ultrapower of $l_p$ with an ultrafilter on $\omega$ you need (at the very least) that $2^{\aleph_0} > \aleph_\omega$. $\endgroup$ – Ramiro de la Vega May 18 '12 at 12:33
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    $\begingroup$ Sorry, Ramiro; I meant $\aleph_1$ at that point. I'll edit again, but make it clear that it is an edit so that your comment makes sense. $\endgroup$ – Bill Johnson May 18 '12 at 13:53

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