5
$\begingroup$

The (model-theoretic) ultrapowers had been used for studying elementary equivalnce of first-order structures. Then, they have been adapted to Banach spaces, which are, let me say, second-order creatures. Of course, one cannot expect that all the properties (propositions of the 'language of Banach spaces') can be persevered by ultrapowers. Indeed, an ultrapower of a reflexive space need not be reflexive.

But reflexivity is a somewhat complicated issue. What if we ask easier questions? What properties are preserved by ultrapowers? For instance, suppose that we are given a Banach space $X$ and each its subspace isomorphic to $X$ is complemented. Is such property preserved under ultrapowers?

Improve this question
$\endgroup$
7
  • $\begingroup$ The question "What properties are preserved by ultrapowers?" seems overly broad. I don't know the answer to the particular question you give at the end, though. Are you familiar with Heinrich's article Ultraproducts in Banach space theory, J. Reine Angew. Math. 313 (1980), 72--104? $\endgroup$
    – Yemon Choi
    Commented Apr 16, 2012 at 22:40
  • $\begingroup$ Yes, I am. Of course, I am asking about reasonable but non-trivial properties. $\endgroup$
    – Olaf Kummers
    Commented Apr 16, 2012 at 22:44
  • $\begingroup$ In keeping with one of MO's guidelines, I really think it would be helpful if you could suggest some of these reasonable but non-trivial properties. "Is this true?" is an easier question to answer well than "What kinds of thing might be true?" $\endgroup$
    – Yemon Choi
    Commented Apr 16, 2012 at 23:51
  • $\begingroup$ Juris gives good advice in his answer. As for your specific question, "suppose that we are given a Banach space X and each its subspace isomorphic to X is complemented. Is such property preserved under ultrapowers?", the answer is yes because Lindenstrauss and Tzafriri proved that such an X is isomorphic to a Hilbert space. $\endgroup$
    – Bill Johnson
    Commented Apr 16, 2012 at 23:58
  • $\begingroup$ @Bill: I thought that the LT result referred to spaces for which every subspace is complemented, and not those for which only the subspaces isomorphic to the original are complemented? $\endgroup$
    – Yemon Choi
    Commented Apr 17, 2012 at 1:38

2 Answers 2

Reset to default
6
$\begingroup$

Say that a Banach space $X$ has property $K$ (for Kummers) provided every subspace of $X$ that is isomorphic to $X$ is complemented. The classical separable, infinite dimensional spaces that have property $K$ include $\ell_2$, $c_0$, and $\ell_2 \oplus c_0$; the first obviously, the second because $c_0$ is separably injective, and the last is not hard to show. It is known that $\ell_p$ for $1\le p \not= 2 < \infty$ does not have property $K$, and it is not hard to show that other possible examples, such as $\ell_2(c_0)$ and $c_0(\ell_2)$, fail property $K$. Ultrapowers of $\ell_2$ of course have property $K$. I don't know about ultrapowers of $c_0$. Are they all injective (which clearly implies property $K$)? Probably not, but I did not try to check the literature.

However, every (complex) HI space [Gowers-Maurey] is not isomorphic to any proper subspace and hence has property $K$. Some real HI spaces have the same property. Now there are HI spaces that contain uniformly complemented copies of $\ell_1^n$ for all $n$. Probably the original Gowers-Maurey space has this property, but, as Thomas Schlumprecht pointed out to me, it is clear that the HI space $AD$ of Argyros-Delyiani does, and $AD$ has property $K$. Since $\ell_1$ is a dual space, there is an ultrapower $Y$ of $AD$ that contains complemented copies of $\ell_1$ and hence $Y$ is isomorphic to $Y \oplus \ell_1$. But $\ell_1$ contains an uncomplemented copy of $\ell_1$ [Bourgain] and hence $Y$ contains an uncomplemented copy of itself.

Therefore property $K$ is not preserved under ultrapowers.

Improve this answer
$\endgroup$
4
  • $\begingroup$ A wonderful answer. Unfortunately, the above-mentioned property for $\ell_2\oplus c_0$ is not clear to me... $\endgroup$
    – Olaf Kummers
    Commented Apr 20, 2012 at 11:27
  • 1
    $\begingroup$ If $X$ is a subspace of $\ell_2 \oplus c_0$ that is isomorphic to $\ell_2$, then, since the projection onto $c_0$ is strictly singular when restricted to $X$, the projection onto $\ell_2$ is an isomorphism on some finite codimensional subspace of $X$. On the other hand, if $Y$ is subspace of $\ell_2 \oplus c_0$ that is isomorphic to $c_0$, then the projection onto $\ell_2$ is compact on $Y$. Play with these two facts to see that $X + Y$ is complemented (note that it is automatically closed by the two facts since e.g. the intersection of $X$ and $Y$ must be finite dimensional; in fact,...). $\endgroup$
    – Bill Johnson
    Commented Apr 20, 2012 at 15:33
  • 1
    $\begingroup$ ...the sum of any two totally incomparable closed subspaces of any Banach space is closed. $\endgroup$
    – Bill Johnson
    Commented Apr 20, 2012 at 15:34
  • $\begingroup$ Let me point out that ultraproducts of Banach spaces are never injective (unless finite-dimensional); see matematicas.unex.es/~fcabello/files/printable/66sub.pdf Theorem 6.1 $\endgroup$
    – Tomasz Kania
    Commented Nov 14, 2013 at 13:47
6
$\begingroup$

There is currently a great deal of activity in answering your question. The subject is called "continuous logic" and you can learn quite a bit more about it by searching through the work of Ward Henson or Itai Ben Yaacov. The article [Model theory for metric structures, based on continuous logic][1]http://www.math.uiuc.edu/~henson/cfo/UCLA09text.pdf might be good place to start. But if you are specifically interested in Banach spaces, perhaps the work of David Sherman is a better starting point.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .