6
$\begingroup$

A Banach space $X$ is said to be prime if every infinite dimensional complemented subspace is isomorphic to the space $X$. The space $X$ is primary if it has an infinite dimensional subspace $Y$ such that every complemented subspace is either isomorphic to $X$ or to the subspace $Y$. The space $X$ is quasiprime if it is primary and the only decomposition of $X$, as a direct sum, into two infinite dimensional subspaces is $X +Y$. Prime spaces are quasiprime which are primary. Also there are examples that separate the above classes.

A Banach space $X$ is hereditarily prime (primary, quasiprime) if every infinite dimensional subspace is prime (primary, quasiprime). Hilbert spaces are hereditarily prime. Also there exists a non-Hilbertian space $X$ which is hereditarily quasiprime.

Question I Is every hereditarily prime space isomorphic to a Hilbert space?

Question II Do there exist subspaces of $\ell_p$, which are not primary?

Improve this question
$\endgroup$
11
  • 1
    $\begingroup$ The classical definition of primary is that if $X$ is written as the direct sum of two closed subspaces, then at least one of the subspaces is isomorphic to $X$. The space $L_p$ is primary when $1<p<\infty$ and $p\not= 2$ under the classical definition, but not under yours. You should change your terminology. $\endgroup$
    – Bill Johnson
    Sep 6, 2021 at 17:28
  • $\begingroup$ I guess the simplest criterion for a space $X$ not to be primary under either your definition or the classical definition is to write $X$ as the direct sum of two infinite dimensional closed subspaces $V $ and $W$ s.t. neither embeds complementably into the other. When $1<p<2$ one can take $p < r<s<2$ and set $V=(\sum_{n=1}^\infty \ell_r^n)_p$ and $W = (\sum_{n=1}^\infty \ell_s^n)_p$; then I think $X=V\oplus W$ works (check that neither of the spaces $V^*$ and $W^*$ embeds into the other). $\endgroup$
    – Bill Johnson
    Sep 6, 2021 at 17:47
  • $\begingroup$ For $2<p<\infty$, I think the examples of subspaces of $\ell_p$ given by Szankowski and me (using a variation of Davies' construction) that fail the approximation property give examples. IIRC, the construction leads to many (actually, a continuum) of examples of pairs of subspaces $V$ and $W$ of $\ell_p$ s.t. neither embeds complementably into the other. $\endgroup$
    – Bill Johnson
    Sep 6, 2021 at 17:54
  • $\begingroup$ Thanks Bill! For p < 2 I had figured out what you suggested but for 2< p I had no idea what such an example looks like. Concerning the definition of the primary spaces you already pointed out the classical one. By the way under my definition is there a simple example of a space which is primary but not prime. It follows from a work of Raikoftsalis and me that the there exists a quasiprime and not prime space. $\endgroup$
    – S Argyros
    Sep 6, 2021 at 18:15
  • 1
    $\begingroup$ I'd just like to say, for those who can see the deleted "answer" and AlexM's comments, that people here could afford to be a bit more polite and a bit less aggressive when seeking to enforce the norms of the site $\endgroup$
    – Yemon Choi
    Oct 9, 2021 at 19:03

2 Answers 2

Reset to default
2
$\begingroup$

I think that the following provides a partial positive answer to Question I.

Fact If X is separable, Hereditarily Prime and Decomposable then it is isomorphic to a Hilbert space.

The proof goes as follows. We write X as V + W with both of infinite dimension.Also both are isomorphic to the space X. Let Z be a subspace of V and Y = Z + W. Then W is a complemented subspace of Y hence Y is isomorphic to W. The subspace Z is complemented in Y hence isomorphic to Y which in turn is isomorphic to the subspace V. Hence V is isomorphic to any of its subspaces and from Gowers' theorem it is isomorphic to a Hilbert space.

The remaining case does not seem easy. Notice that any hereditarily prime space must be $l_2 $ saturated. Indeed it does not contain HI subspace hence it is unconditionally saturated and the above yields that all these subspaces are isomorphic to $ l_2 $.

A space satisfying the following properties provides a counterexample.

  1. The space X is indecomposable and $ l_2 $ saturated. (The existence of such a space is still open but I think could exist. )

  2. Every subspace is isomorphic to all further subspaces with finite codimension.

  3. Every subspace is either indecomposable or isomorphic to $l_2$.

Improve this answer
$\endgroup$
0
2
$\begingroup$

Actually the above three conditions yielding a counterexample are satisfied by any counterexample and this is easy to be checked. However such a space has extremely peculiar structure. In particular it does not contain any subspace of the form Y + Z with Y Hilbertian and Z indecomposable. This follows from the previous positive partial answer. That means that for every indecomposable subspace Z and every Hilbertian subspace Y dist( $S_Z$ , $S_Y$ )= 0. Further the same holds for all Y subspaces of X. The last follows from the fact that X is $l_2$ saturated. It is hard to think how such a space is defined.

However if the following has a positive answer then the original question I has also a positive one.

Question III Let X be a separable non Hilbertian and $l_2$ saturated Banach space.Does the space X contain Y +Z with Y Hilbertian and Z non Hilbertian?

Improve this answer
$\endgroup$
5
  • $\begingroup$ AFAIK, there could be a counterexample $X$ to Question III with $X^*$ isomorphic to $\ell_1$. Did you check whether the existing $\ell_2$ saturated isomorphic $\ell_1$ preduals satisfy the conclusion of Question III? $\endgroup$
    – Bill Johnson
    Oct 7, 2021 at 12:53
  • $\begingroup$ @BillJohnson Do you mean the appropriate version of the initial Bourgain - Delbaen construction? It is not clear to me that the space is indecomposable. If it is decomposable then it contains a subspace Y+Z as in question III. If not then we have a space which is indecomposable and đť‘™2 saturated (great!!). $\endgroup$
    – S Argyros
    Oct 9, 2021 at 21:23
  • $\begingroup$ I discussed with Pavlos Motakis the problem of the decomposition of the Bourgain Delbaen spaces ( the initial ones ) He has a proof that all these spaces are decomposable. Anyone interested for the result could contact Pavlos $\endgroup$
    – S Argyros
    Oct 9, 2021 at 21:24
  • $\begingroup$ @SpyridonArgyros : Pavlos' result is interesting. He should include it in some paper, IMO. Failing that, I hope he posts it on the Arxiv. $\endgroup$
    – Bill Johnson
    Oct 10, 2021 at 16:27
  • $\begingroup$ @BillJohnson His proof is not trivial. In general, in my opinion , a space is indecomposable if you take a special care for that. The property does not appear accidentally. I also agree that Pavlos result should appear somewhere. $\endgroup$
    – S Argyros
    Oct 10, 2021 at 21:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.