# "Adding one dimension" to an infinite dimensional topological vector space

Let $$V$$ be an infinite-dimensional Hausdorff topological vector space over $$\mathbb{R}$$. Is it true that $$V\oplus\mathbb{R}$$ is isomorphic (as a TVS) to $$V$$ itself? (How about $$V\oplus V$$?) If yes, I also wonder whether for every $$v \ne 0$$ in $$V$$ there exists an isomorphism $$f : V \cong V \oplus \mathbb{R}$$ such that the projection of $$f(v)$$ to $$\mathbb{R}$$ is nonzero.

If no:

1. What is the simplest counterexample?

2. What are some minimal conditions (complete/locally convex/normed/inner product etc.) that guarantee the existence of such isomorphisms? (If $$V$$ is a Hilbert space, desired isomorphisms exist in all three cases above.)

I looked briefly into Schaefer and Wolff's Topological Vector Spaces but did not find any clue.

• I will say, though, that in any Locally convex space such that $V\cong V\oplus \mathbb R$, the proof of this result (math.stackexchange.com/a/1169419/1210477), along with Hahn Banach for locally convex spaces, seems to indicate that your followup question is true, since by Hahn Banach any $v\in V$ is complementary to a codimension one closed subspace (namely, kernel under Hahn banach functional).
– M W
Sep 21 at 2:09
• [Reposting my deleted first comment which was kind of fubar] For $V\oplus \mathbb R \cong V$, I suspect the answer is no in general, based on the wikipedia claim (en.wikipedia.org/wiki/…) that the answer is yes specifically for Banach spaces. For the general $V\oplus V\cong V$ case, the answer certainly seems to be no from same wiki article (en.wikipedia.org/wiki/…) as $V\oplus V\cong V$ is listed as a nontrivial requirement for a decomposition result.
– M W
Sep 21 at 2:18
• @MW Thanks but Gowers's counterexample shows the Wikipedia claim is problematic; for details see my answer. Sep 22 at 5:33

Since this is a question in Functional Analysis, I will assume that isomorphism means "bicontinuous bijective linear map".

The answer is no in both cases. Not sure what the "simplest example is". The category "topological vector space", without further qualifications, allows for fairly nasty things to happen.

Let $$V=L^p[0,1]$$, with $$0. The "$$p$$-norm" in this case is not a norm, but $$V$$ is a complete metric topological vector space with the metric $$d(f,g)=\int_0^1 |f-g|^p.$$ It turns out that $$V^*=\{0\}$$, that is $$V$$ has no nonzero continuous linear functionals. But $$V\oplus\mathbb R$$ (you are not saying what topology you consider there, but I will assume that it is the product topology) clearly has the continuous functional $$\varphi(f,t)=t$$, so its dual is nontrivial. So $$V$$ and $$V\oplus\mathbb R$$ cannot be isomorphic as topological vector spaces.

As for Banach spaces $$V$$ such that $$V\not\simeq V\oplus V$$, these exist but you will not find them among the classical examples of Banach spaces. Easiest examples (bear in mind that I'm not an expert) seem to be of the form infinite $$\ell^2$$-sums of spaces $$\ell^{n_j}_{p_j}$$ (this would be $$\mathbb R^{n_j}$$ with the $$p_j$$-norm). There even exists a Banach space that is not isomorphic to any square of any Banach space. The keyword to search for all this stuff is "Banach spaces nonisomorphic to their squares" (this is almost verbatim the title of a paper that has several references).

As far as I can tell, there is no characterization of Banach spaces isomorphic to their squares.

• Thanks for the answer! The trivial dual examples are pretty easy to find, but evaded me when I asked the question. You almost fully answer my questions, but did not yet eliminate the hope that $V \cong V \otimes \mathbb{R}$ might hold for all locally convex $V$, since they have enough continuous linear functionals, and the question is whether the (some?) kernels are isomorphic to $V$. Sep 21 at 4:15
• This (and the followup question) was motivated by the question whether the projectivization (the space of lines through the origin) of a TVS $V$ is a "manifold" modelled by $V$ itself, and there still seems to be hope that it's true for locally convex $V$. (We're trying to formalize the manifold structure in Lean: leanprover.zulipchat.com/#narrow/stream/287929-mathlib4/topic/… ) Sep 21 at 4:15
• Any Banach space with a Schauder basis (which are all that you will find in "real life") have a Schauder basis, and then they trivially satisfy $V\simeq V\oplus \mathbb R$. A counterexample can only occur if $V$ does not have a Schauder basis. These spaces are rare and none of them is "natural", if I'm not wrong. Sep 21 at 22:24
• Hmm I don't see why the existence of a Schauder basis implies $V \cong V \oplus \mathbb{R}$ and apparently Gowers's counterexample admits a Schauder basis outofthenormmaths.files.wordpress.com/2012/04/… You may be proposing that the map $\sum_{i=1}^\infty a_i e_i \mapsto \sum_{i=1}^\infty a_{i+1} e_i$ is continuous, but it's not even clear to me that the RHS always converges if the LHS converges. Sep 22 at 5:32
• Yes, you are right. I'm too biased by Hilbert spaces. My bad. Sep 22 at 8:10

Sophie Morel on Zulip found this counterexample (by Gowers) to $$V \cong V \oplus \mathbb{R}$$ with $$V$$ a Banach space: A Solution to Banach's Hyperplane Problem, and there's a nice blog post about it; it turns out Banach asked both questions (about $$V \oplus \mathbb{R}$$ and $$V \oplus V$$) for Banach spaces.

The claim on Wikipedia that "a finite-codimensional subspace of a Banach space $$X$$ is always isomorphic to $$X$$" is apparently problematic, and Antoine Chambert-Loir offered an explanation how the mistake could have been made.

For the followup question, the answer should be true if $$V^*$$ (the continuous linear functionals) separates points in $$V$$ (which is true in any locally convex space by Hahn-Banach): from $$V \cong V \oplus \mathbb{R}$$ we know at least one closed hyperplane in $$V$$ is isomorphic to $$V$$, but all closed hyperplanes are isomorphic (I believe the proof works in any Hausdorff TVS, and so does this stronger result), so for any nonzero continuous linear functional $$f \in V^*$$ we get a decomposition $$V \cong \text{ker }f \oplus \mathbb{R} \cong V \oplus \mathbb{R}$$.