# Banach space modulo a one-dimensional subspace =?

My question is the following:

Given an infinite dimensional Banach space $E$ and a one-dimensional linear subspace $F\subset E$. It is well-known that this one-dimensional linear subspace is closed and always complemented. So, we have the following topological isomorphism:

$$E \cong F \times E/F.$$

Now, what is the isomorphism type of $E/F$ ? In particular, is $E/F$ isomorphic to $E$ ?

I know this is true for all Hilbert spaces since they admit orthonormal bases and I would expect that the existence of some sort of "topological basis" could be related to this question but so far I could not find anything. Furthermore, I did not know where to look for an answer to this question.

By the way: Of course, I believe in the axiom of choice.

EDIT: All isomorphisms are meant to be topological isomorphisms, i.e., linear homeomorphisms. Isometric isomorphism are too restrictive, I would assume (although in the Hilbert space case, one can even get isometric isomorphisms, but I do not need that)

• By isomorphism, you mean isometrically isomorph or linear continuous bijection (whose inverse is hence also going to be continuous) ? Dec 1, 2014 at 14:15
• Thanks for your comment. I meant topological isomorphisms and edited the question accordingly.
– Tom
Dec 1, 2014 at 14:41
• $E/F$ is isomorphic to each one-codimensional closed subspace of $E$, and does not need to be isomorphic to $E$. For example, when $E$ is the hereditarily indecomposable space of Gowers and Maurey, $E/F$ is not isomorphic to $E$. Dec 1, 2014 at 15:09

Gowers proved in "A solution to Banach’s hyperplane problem" (1994)

An infinite-dimensional Banach space X is constructed which is not isomorphic to X ⊕ R. Equivalently, X is not isomorphic to any of its closed subspaces of codimension one. This gives a negative answer to a question of Banach. In fact, X has the stronger property that it is not isomorphic to any proper subspace. It also happens to have an unconditional basis.

http://blms.oxfordjournals.org/content/26/6/523.abstract

• It seems user62448 does not exist! Dec 1, 2014 at 15:21
If you consider the spaces constructed by Gowers, Maurey and others exotic, then it turns out that the answer is also negative in the class of $C(K)$-spaces. Indeed, Koszmider constructed a compact, Hausdorff space $K$ so that $C(K)$ is not isomorphic to its hyperplanes:
Of course there is a price you have to pay for this–$C(K)$ is inseparable (i.e. $K$ is non-metrisable) and it is easy to see that a separable $C(K)$-space will never have this property as it contains a complemented copy of $c_0$ which apparently has this property.