# Are separable F-spaces (completely metrizable topological vector space) homeomorphic to $l_2$?

An F-space is a completely metrizable topological vector space, i.e. the vector topology is induced by a complete metric. A Fréchet space is, by definition, a locally convex F-space.

It is known that all (infinite dimensional) separable Fréchet spaces are homeomorphic to $l_2$, the space of square summable sequences of real numbers. (See e.g. Anderson, R. D.; Bing, R. H., A complete elementary proof that Hilbert space is homeomorphic to the countable infinite product of lines. Bull. Amer. Math. Soc. 74, 1968, 771–792.)

More precisely any such space, including $l_2$ is homeomorphic to a countable infinite product of copies of real lines.

Can this result be extended to (any) non-locally convex F-spaces?

I am interested in any reference or recent literature on the topology (homeomorphism class) of F-spaces.

More precisely, I am interested in the following examples: $L_p([0,1])$ or $l_p$ with $0<p<1$, which are in fact quasi-Banach spaces. Meaning, the p-sum/integral defines a quasi-norm only, that is $||f+g||_p\leq K(||f||_p+||g||_p)$, holds only for a uniform constant $K>1$. Nevertheless $d(f,g)=||f-g||_p^p$ is a complete metric inducing the vector topology.

This question arose when wondering if $l_p$ has a structure of a Hilbert manifold.

• The Mazur mappings shows that every $\ell_p$, respectively, $L_p$ is homeomorphic to $\ell_2$, respectively, $L_2$, and hence all are homeomorphic. – Bill Johnson Aug 24 '17 at 23:49
• Thank you for this comment. The extension to $0<p<1$ of Mazur's proof was done by A. Weston in his thesis. The following paper by him on uniform homeomorphisms of $L^p(\mu)$ spaces contains the needed inequalities. ( maths-proceedings.anu.edu.au/CMAProcVol29/… ) – Thibaut Dumont Aug 25 '17 at 11:52

There is a famous linear metric space constructed by R. Cauty [Un espace métrique linéaire qui n'est pas un rétracte absolu, Fund. Math. 146 (1994)] whose completion is a separable $F$-space which is not an AR. I do not have Cauty's paper handy but the latter fact is stated on the first page of Cauty's space enhanced in [Topology Appl. 159 (2012), no. 1, 28–33].
Since $\ell_2$ is an AR, the above $F$-space is not homeomorphic to $\ell_2$.
Incidentally, a separable $F$-space is homeomorphic to $\ell_2$ if and only if it is a non-locally-compact AR, see Corollary 5.2.2 in [Absorbing sets in Infinite-Dimensional Manifolds, T. Banakh, T. Radul, M. Zarichnyi].
• I think if we add the requirement that the separable, completetely metrisable $F$-space is an AR, we do have that is homeomorphic to $\ell_2$. – Henno Brandsma Aug 25 '17 at 21:40
• @HennoBrandsma: $\mathbb R^n$ is a counterexample to what you claim. – Igor Belegradek Aug 25 '17 at 21:58