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The space $\ell_1$ has loads of (isomorphic) predulas. They can be as weird as possible but I am interested in Banach lattices.

Question: Let $X$ be a Banach lattice with dual isomorphic to $\ell_1$. Must $X$ be isomorphic to $C(K)$ for some countable $K$?

Well, the classical Bourgain-Delbaen spaces are not good candidates because they have no copies of $c_0$, hence they are not isomorphic to a Banach lattice (a Banach lattice without a copy of $c_0$ is weakly sequentially complete and weakly sequentially Banach lattices are complemented in their biduals).

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  • $\begingroup$ My guess is "yes". I think it is known that $\ell_1$ has a unique lattice structure, which means that the lattice structure on $X^*$ as a dual to the lattice $X$ is just the usual lattice structure on $\ell_1$. This means that disjoint sequences in $X$ add in an $\ell_\infty$ way, which says that $X$ is (isomorphically) an abstract $M$-space. An abstract $M$-space whose dual is $\ell_1$ should be (isomorphic to) $C(K)$ for some countable $K$, yes? $\endgroup$
    – Bill Johnson
    Commented Jun 24, 2012 at 18:12
  • $\begingroup$ Gasparis has several strong results on $\ell_1 $ preduals. His papers have very well-written and lengthy introductions. It may be worth taking a look: arxiv.org/find/math/1/au:+Gasparis_I/0/1/0/all/0/1 $\endgroup$
    – Kevin Beanland
    Commented Jun 25, 2012 at 13:33
  • $\begingroup$ I think that the lattice structure on $\ell_1$ need not be unique (cf. p. 84, par. 3 of Lacey's and Wojtaszczyk's paper: jstor.org/stable/10.2307/2040755 ) $\endgroup$
    – Tomasz Kania
    Commented Jun 26, 2012 at 12:57
  • $\begingroup$ I was thinking isomorphically, Tomek. The example in Lacey-Wojtaszczyk is lattice isomorphic to $\ell_1$. $\endgroup$
    – Bill Johnson
    Commented Jun 29, 2012 at 16:14
  • $\begingroup$ Benyamini proved that a separable $M$ space is isomorphic to a $C(K)$ space, and $K$ must then be a space of ordinals if the dual is separable. So I think the outline I gave above really does yield a proof. $\endgroup$
    – Bill Johnson
    Commented Jun 29, 2012 at 16:18

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