# Contractibility of convex set

Suppose that $\Omega$ is a convex open subset of an infinite dimensional vector space $E$ such that $\Omega$ is not contained in any finite dimensional subspace of $E$. Let $Q_m\subset \Omega$ denote a set of $m\in \mathbb{N}_{>0}$ distinct points.

Question: Is the space $\Omega\setminus Q_m$ contractible?

Assume that for any set $Q_m'$ of $m$ distinct points $E\setminus Q'_m$ is contractible (I believe that this is true, see my answer below).
By "blowing up $\Omega$ like a balloon" we get a homeomorphism $\varphi$ from the convex open set $\Omega$ to $E$. More precisely, fix $x_0\in\Omega$ and for any open interval $I\subset \mathbb{R}$ let $\varphi_I:I\to \mathbb{R}$ be a (suitable) homeomorphism. Now, define $\varphi:\Omega \to E$ by $$\varphi(x):=\begin{cases}\varphi_{\Omega\cap \mathbb{R}\cdot (x-x_0)}(x), \text{ if } x\neq x_0 \\ x_0 , \text{ if } x= x_0.\end{cases}$$ (I know that the definition of $\varphi$ is not formally correct but I think that the idea is clear.) Restricting $\varphi$ to $\Omega\setminus Q_m$ we get a homeomorphism between $\Omega\setminus Q_m$ and $E\setminus \varphi(Q_m)$. By assumption $E\setminus \varphi(Q_m)$ is contractible and hence $\Omega\setminus Q_m$ is also contractible.

Is the intuition behind this heuristic argument any good? Can it be made into a complete and rigorous argument?

• Kuiper's Theorem has it's roots in the fact that the unit sphere $S^{\infty}$ in infinite dimensional Hilbert space is contractible. The unit ball is convex and removing a point gives a set homotopic to the unit sphere so maybe that's a good place to start. Apr 19 '13 at 14:34
• @TomHarris: Thanks.
– Dave
Apr 20 '13 at 14:21
• What are you assuming about the surrounding topological vector space? Open convex subsets are homeomorphic to the whole space $X$, and if $X$ admits a continuous norm then $X \setminus K$ is homeomorphic to $X$ whenever $K$ is compact (but there may be much easier proofs of what you want). A reference is Bessaga and Pelczynski, Selected topics in infinite-dimensional topology, chapter III, §5. Apr 22 '13 at 20:52
• @Martin : Thanks a lot for your comment and also for the reference. I'll definitely check it out. I suspect that the surrounding vector space is a Fréchet space. So, do you think that the answer to my original question is affirmative?
– Dave
Apr 22 '13 at 21:11
• Definitely yes if the space is a separable Fréchet space. By a deep theorem of Anderson and Kadec these are all homeomorphic to $\ell_2$. Deleting a compact set from $\ell_2$ does not change the homeomorphism type, as I pointed out. Thus, every open convex set minus finitely many points is contractible in a separable Fréchet space. There must be much better (simpler and more explicit) solutions for your problem, but I hope it helps to know that it is true. Apr 22 '13 at 21:29

Following the suggestion of Martin I'll post an answer to my own question. I would appreciate some feedback. Please let me know if you notice any errors.

Let $Q'_m\subset E$ be a set of $m\in \mathbb{N}$ distinct points.

Claim: $E\setminus Q'_m$ is contractible.

Proof: Write $E$ as $E\cong W\oplus V$ where $W\subset E$ is a finite dimensional subspace containing $Q'_m$. Let $\Phi:W\oplus V \to W\oplus V$ be a linear map such that $\Phi|_W=id_W$ and $\Phi|_V$ has no non-zero eigenvalues. (For example $\Phi|_V$ could be chosen as $(v_1,v_2,v_3,\ldots)\mapsto (0,v_1,v_2,v_3,\ldots)$.) The map $\Phi$ gives rise to an isomorphism from $E$ to a genuine subspace $im(\Phi)\subsetneq E$. It also induces a map $\Phi:E\setminus Q'_m\to E\setminus Q'_m$.
Now, $\Phi_t:E\setminus Q'_m\to E\setminus Q'_m$ defined by $$\Phi_t(x):=(1-t)x+t\Phi(x)\quad t\in [0,1]$$ is a homotopy from $id_{E\setminus Q'_m}$ to $\Phi$.
Pick $x_0\notin im(\Phi)\cup Q'_m$. Then $\Psi_t:E\setminus Q'_m\to E\setminus Q'_m$ given by $$\Psi_t(x):=(1-t)\Phi(x)+tx_0\quad t\in [0,1]$$ is a homotopy from $\Psi_0=\Phi$ to the constant map $\Psi_1\equiv x_0$. Hence, $E\setminus Q'_m$ is contractible.$\qquad\square$

This claim together with the fact that $\Omega\setminus Q_m$ is homeomorphic to $E\setminus Q'_m$ (see the comments by Martin and my argument in the question) shows that $\Omega\setminus Q_m$ is contractible.

• Why is there a continuous linear map $\Phi$ on $E$ having all the properties needed? Note that writing $(v_1,v_2,v_3, \dots) \mapsto (0,v_1,v_2,v_3)$ makes sense in a Hilbert space, but not for a general topological vector space $E$ (without further hypotheses). Apr 24 '13 at 4:49
• I think what I'm really feeling uncomfortable with is that you still didn't say what precise assumptions you make on $E$. You talk about a convex open set $\Omega$, but $E$ is just an infinite-dimensional vector space, no topology specified. // I still think the reference to chapter III, §5 of Bessaga-Pelczynski is quite relevant and will be helpful for getting a precise statement and argument. The technique developed there is nice and of a rather elementary nature. Apr 24 '13 at 6:12
• @Martin: Thank you, I'll definitely look at the reference as soon as I get my hands on a copy. Sorry that I didn't make the assumptions more precise, I'll give some background here: The case I'm interested in is where $E$ is a closed infinite dimensional subspace of the space $Vect(M)$ of all vector fields on a compact, closed manifold $M$. I'm guessing that $Vect(M)$ is a Hilbert space with inner product: $<X,Y>:=\sup_{p\in M} g_p(X(p),Y(p))$, where $g$ is a complete Riemannian-metric on $M$, right? If this is correct, then $E$ should be a Hilbert-space itself.
• Thanks for providing the background. Are you sure that $\operatorname{Vect}(M)$ is complete? I think it isn't: if $M = S^1$ you get the $C^\infty$-functions on $S^1$ and this space is dense in $L^2(S^1)$, which is strictly larger. However, that's actually a plus, because then the rather elementary proposition 5.1 of Bessaga-Pelczynski applies and proves the stronger property that $E \setminus Q_m$ is homeomorphic to $E$. I made a scan for you: page 1 and page 2 and I, Cor. 3.3. Apr 24 '13 at 10:27