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?



Do ultrapowers of classical Banach spaces have unconditional bases?
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.


