Timeline for Preduals of $\ell_1$
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jun 29, 2012 at 16:18 | comment | added | Bill Johnson | 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. | |
| Jun 29, 2012 at 16:14 | comment | added | Bill Johnson |
I was thinking isomorphically, Tomek. The example in Lacey-Wojtaszczyk is lattice isomorphic to $\ell_1$.
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| Jun 26, 2012 at 12:57 | comment | added | Tomasz Kania | 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 ) | |
| Jun 25, 2012 at 13:33 | comment | added | Kevin Beanland | 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 | |
| Jun 24, 2012 at 18:12 | comment | added | Bill Johnson | 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? | |
| Jun 23, 2012 at 23:54 | history | asked | Jan Vardøen | CC BY-SA 3.0 |