5
$\begingroup$

Gowers' dichotomy theorem asserts that every Banach space either contains an HI subspace or a subspace having an unconditional basis. There are examples of HI spaces without quotients having unconditional bases (was Argyros the first who proved that?). This strange phenomenon tempts me to ask

whether every reflexive non-HI space has a quotient with an unconditional basis?

My apologies if this is well-known.

Improve this question
$\endgroup$

3 Answers 3

Reset to default
5
$\begingroup$

Every quotient of the original Gowers-Maurey space is HI; Ferenczi proved this. Argyros-Felouzis produced examples of reflexive spaces that are not HI (e.g. contain some $\ell_p$) and have HI duals. Any of these spaces give an example of a space such that no quotient has an unconditional basis. I think it's safe to say that very little in the theory of HI spaces can be classified as 'well-known'.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks Kevin for these examples. I guess that in this theory everything can go bad, so perhaps there are even superreflexive counterexamples to my question? $\endgroup$
    – MeasureConcentrator
    Commented Feb 20, 2013 at 23:44
  • $\begingroup$ Oh, yes; the Argyros-Felouzis examples are better than what I mentioned. I'll delete my answer. $\endgroup$
    – Bill Johnson
    Commented Feb 20, 2013 at 23:45
  • $\begingroup$ No, please, let me see... $\endgroup$
    – MeasureConcentrator
    Commented Feb 20, 2013 at 23:46
  • $\begingroup$ I'm not sure about superreflexive examples. One probably exists though.. $\endgroup$
    – Kevin Beanland
    Commented Feb 20, 2013 at 23:51
6
$\begingroup$

The space of Gowers and Maurey is reflexive and H.I. , and as proved by Ferenczi every of its quotients and its dual is also H.I. (as mentioned in the answers of Bill Johnson (giving the reference) and Kevin Beanland). The results of Argyros and Felouzis are answering to the question.

Moreover the remark of Bill Johnson is to the point. S. Argyros and myself provided a reflexive and unconditionally saturated Banach space $X$ with an H.I. dual $X^*$. Thus every quotient of $X$ is indecomposable. Additionally we show that every quotient of $X$ has a further quotient which is H.I.

S. Argyros, A. Tolias, Indecomposability and unconditionality in duality. Geom. Funct. Anal. 14 (2004), no. 2, 247–282.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks for the addition, Andreas. I thought someone had done this but did not remember who; sorry about that. $\endgroup$
    – Bill Johnson
    Commented Mar 5, 2013 at 16:12
3
$\begingroup$

The original Gowers-Maurey HI space GM is reflexive. Ferenczi proved that the dual (and also every quotient) of GM is HI. Then GM $\oplus$ GM is not HI but cannot have a quotient with unconditional basis, for then its dual would have a subspace with an unconditional basis. But if $X\oplus X$ has a subspace with an unconditional basis, then also $X$ has a subspace with an unconditional basis (the two complementary projections onto the copies of $X$ in the direct sum $X \oplus X$ cannot both be strictly singular on the same subspace).

I would guess that the experts even know that there is an unconditionally saturated reflexive space $X$ whose dual is HI and thus every quotient of $X$ cannot have an unconditional basis.

Ferenczi, V. Quotient hereditarily indecomposable Banach spaces. Canad. J. Math. 51 (1999), no. 3, 566–584.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you for this. This is a very good (and pedagogical) example! $\endgroup$
    – MeasureConcentrator
    Commented Feb 20, 2013 at 23:59

You must log in to answer this question.

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