# Closed, complemented subspaces of $l^1(X)$ when $X$ is uncountable

... are all isomorphic to $l^1$ on some other index set. At least, that much I "know" from 2nd-hand sources, since the original proof is apparently in a paper of Köthe from the 1930s 1960s (in German) that I can't get hold of have had trouble digesting. Since there are some Banach space specialists reading MO, I wondered if someone could sketch how the proof differs from the countable case, or point to a more recent text, preferably in English or French, that gives the proof? I hope this is a well-defined question for MO, since I'm not baiting with something where I know the answer.

(Some background for other readers: the analogous result when $X$ is countably infinite follows from combining two steps: one first uses a block basis argument to show that a closed, complemented subspace $V$ inside $l^1(\bf N)$ must either be finite-dimensional, or contain an infinite-dimensional, closed complemented subspace $W$ that is isomorphic to $\ell^1({\bf N})$. In the former case, $V$ is then obviously isomorphic to some finite-dimensional $\ell^1$. In the latter case, one applies Pelczynski decomposition. My impression is that it's the first step which might prove problematic if attempted for $\ell^1(X)$ when $X$ is uncountably infinite, but I could well be wrong and would welcome corrections.)

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I don't read German, but according to arxiv.org/abs/math/0501048 Köthe did this in springerlink.com/content/u58k505417h14214, published in 1966 and hopefully accessible to you as it is to me via institutional login. I don't know an English or French source. – Jonas Meyer Nov 15 '09 at 2:22
Ah, I misremembered the date, but that's the title of the paper I was thinking of. Hmm, didn't realise it was now on SpringerLink (it's been a while since the original "need" I had for it arose). Thanks, although if anyone reading can give me an explanation of the relevant part that'd be great. – Yemon Choi Nov 15 '09 at 2:59

A proof in English (modulo some details involving the Pelczynski decomposition method) can be found in the article 'On relatively disjoint families of measures' (Studia Math, 37, p.28-29) by Haskell Rosenthal.

Regarding the analogous result for $\ell_p (X)$ ($p\in (1,\infty)$) and $c_0 (X)$, I seem to recall reading somewhere that it was solved by Joram Lindenstrauss, but now I can't seem to find any reference to it. I seem to think that I saw something about it in the Appendix to the English translation of Banach's book on linear operations (the appendix is by Bessaga and Pelczynski), but it would take me a while to sift through it to find it, and family dinner is being dished up very shortly. I wonder anyhow how much can be gleaned from Matthew Daws' classification of the closed, two-sided ideals in $\mathcal{B}(\ell_p (X))$, the Banach algebra of all bounded linear operators on $\ell_p (X)$? The relevant paper can be downloaded at http://www.amsta.leeds.ac.uk/~mdaws/pubs/ideals.pdf . The paper 'The lattice of closed ideals in the Banach algebra of operators on a certain dual Banach space' by Laustsen, Schlumprecht and Zsak illustrates how classification of the complemented subspaces of a Banach space $E$ can follow from the classification of closed, two-sided ideals in $\mathcal{B}(E)$ if all the closed, two-sided ideals are generated by projections onto complemented subspaces having certain nice properties. How much of this can be done using Matt's results I haven't checked, but I think that at the very least some partial results could be obtained. Matt might comment of this if he passes by, or if no one else does I might try to look into it in the next day or so and edit this answer accordingly.

The analogous result for $\ell_\infty (X)$ does not hold for uncountable $X$ in general. Indeed, every $\ell_\infty (X)$ is the dual of $\ell_1 (X)$, every Banach space embeds isomorphically into some $\ell_\infty (X)$, but there are injective Banach spaces that are not isomorphic to any dual Banach space; the first such example seems to have been found by Haskell Rosenthal in his paper 'On injective Banach spaces and the spaces $L^\infty (\mu)$ for finite measure $\mu$' (Acta Mathematica, 124, Corollary 4.4), and the existence of such a space provides the desired counterexample.

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Excellent! Thanks Phil - I didn't think to look in the Rosenthal paper, but on a quick look it seems just what I wanted. (The original motivation, back in 2003-04, was that Kothe's result tells us there are no interesting projective Banach spaces, since they'd all have to be complemented subspaces of some l^1(X). At one point I might have wanted/needed to use the result and was reluctant to do so, since I'd never read the proof.) – Yemon Choi Nov 15 '09 at 8:47

The case of $Z=\ell_p(X)$, 1 < p < infinity, and $Z=c_0(X)$ with $X$ uncountable are a bit easier because if $T$ is a bounded linear operator from $Y$ into $Z$ and the density character of $Y$ is smaller than $|X|$, then $T$ cannot be one to one. So if $Y$ is a subspace of $Z$ with density character $|X|$, then a maximal set of disjointly supported unit vectors in $Y$ has cardinality $|X|$ (use the obvious fact that a subspace $W$ of $Z$ is contained in $\ell_p(Q)$ [or $c_0(Q)$] with $|Q|$ equal to the density character of $W$). So every subspace of $Z$ with density character $|X|$ has a norm one complemented subspace that is isometric to $Z$. This and a second use of the obvious fact" gives the desired result.