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Suppose $X$ is a Banach space with an unconditional basis $(e_n)$. Then, one may easily define a Boolean algebra of projections in $\mathcal{L}(E)$ which is isomorphic to the power-set of $\mathbb{N}$ (by a Boolean algebra of projections I understand a family of bounded idempotents on a Banach space which is a Boolean algebra under operations $P\wedge Q = PQ$ and $P\vee Q = P+Q-PQ$, zero-element equal to zero operator and unit equal to the identity on $X$).

What are sufficient conditions for a Banach space to have a complete Boolean algebra of projections? Is there a separable Banach space $X$ without unconditional basis with some Boolean algebra of projections isomorphic to the power-set of $\mathbb{N}$?

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Please do not delete questions after they received extensive answers. –  t.b. Apr 12 '12 at 15:34
    
@t.b. To be fair, that answer was extensive largely because it answered a different question to the one asked. The original question had a two-dimensional counterexample –  user16299 Apr 12 '12 at 17:37
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@Yemon: the present question was deleted, too. See here. –  t.b. Apr 12 '12 at 19:38
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1 Answer 1

A fairly trivial sufficient condition for the existence of a complete Boolean algebra of projections is for the space $X$ to be (up to isomorphism) of the form $(\sum_{n\in\mathbb{N}} \oplus E_n)_{\ell_p}$, $1\leq p < \infty$, or $(\sum_{n\in\mathbb{N}} \oplus E_n)_{c_0}$ (where the spaces $E_n$ are, of course, nonzero).

So, an example of a space with the property requested in the OP's second question is an space of the form $c_0(E)$ or $\ell_p(E)$, $1\leq p < \infty$, where $E$ is a separable Banach space that does not embed in any Banach space having an unconditional basis; examples of such spaces $E$ include:

  • The James space $J$.

  • The James tree space $JT$.

  • $L_1[0,1]$.

  • Let $K$ be either an uncountable compact metric space (e.g., $[0,1]$), or equal to a compact ordinal interval $[0,\alpha]$, where $\alpha \geq \omega^\omega$ and $[0,\alpha]$ is equipped with its natural order topology. Then one can take $E=C(K)$.

  • The universal basis space $U$ of Pelczynski, which has (a basis and) the property that every Banach space with a basis is isomorphic to a complemented subspace of $U$. In particular, $U$ does not embed in a space with an unconditional basis since it contains (complemented) copies of $L_1[0,1]$ and the $C(K)$ spaces mentioned above (since these spaces all have a basis).

With the exception of the James space, all of the examples $E$ given above have the property that $E$ is isomorphic to $c_0(E)$ or $\ell_p(E)$ for some $1\leq p < \infty$ (in the case of Pelczynski's space, we actually have that $U$ is isomorphic to all of the spaces $c_0(U)$ and $\ell_p(U)$, $1\leq p < \infty$); so, with the exception of the James space, we can in fact take any of the spaces $E$ given as examples above as an answer to the second question.

Finally, I mention that one could also consider, in a similar way, direct sums with respect to any unconditional basis.

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