15
$\begingroup$

In this recent question, I learned that any two separable Banach spaces are homeomorphic. Based on some readings, I'm guessing that $L^2(\mathbb R)$ is homeomorphic to $\prod_{n=1}^{\infty} (0,1)$ (infinite product of the open interval, with the product topology).

Warm-up question: Is it true that $L^2(\mathbb R)$ is homeomorphic to $\prod_{n=1}^{\infty} (0,1)$?

(The space $\prod_{n=1}^{\infty} (0,1)$ is known as the pseudo-interior of the Hilbert cube $I^\infty:=\prod_{n=1}^{\infty} [0,1]$.)

Spaces locally homeomorphic to $I^\infty$ are known as Hilbert cube manifolds; spaces locally homeomorphic to $L^2(\mathbb R)$ are known as $\ell_2$-manifolds.

I seem to recall that the classification of compact Hilbert cube manifolds is very interesting: compact Hilbert cube manifolds are in bijective correspondence with compact CW-complexes, up to simple homotopy equivalence. But I'm not completely sure of the above statement, so I'll repeat it in my question below:

Questions:
• What is the classification of compact Hilbert cube manifolds?
• What is the classification of non-compact Hilbert cube manifolds?
• What is the classification of $\ell_2$-manifolds?

PS: Let's assume that all spaces are separable (or whatever implies Polish), to avoid pathologies.

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ The topological classification of $\ell_2$ coincides with their homotopical classification -- this is a well-known result of classical infinite-dimensional topology (see the book of Bessaga and Pelczynski). The classification of Hilber cube manifolds is more complicated but is very well written in the old book of Chapman. $\endgroup$
    – Taras Banakh
    Commented May 10, 2017 at 18:02
  • 1
    $\begingroup$ Although the classification of infinite dimensional Banach spaces as topological spaces is complete, their classification as uniform spaces is not and is a fairly hot topic currently. Consult the book of Benyamini and Lindenstrauss for an introduction into this topic. $\endgroup$
    – Bill Johnson
    Commented May 10, 2017 at 19:05
  • $\begingroup$ Related to the warm-up question, to quote from the MR review of this article "The author proves that the sequential Hilbert space $l_2$ is homeomorphic to the countable infinite product $s$ of real lines." So one gets $L^2[0,1]$, at least. $\endgroup$
    – David Roberts ♦
    Commented Jul 23, 2017 at 8:18

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .