Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

Added later:
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?

share|improve this question
1  
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. –  tharris Apr 19 '13 at 14:34
    
@TomHarris: Thanks. –  Dave Apr 20 '13 at 14:21
1  
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. –  Martin 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
1  
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. –  Martin Apr 22 '13 at 21:29

1 Answer 1

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.

I'd like to thank Martin for his helpful comments.

share|improve this answer
    
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). –  Martin 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. –  Martin 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. –  Dave Apr 24 '13 at 9:55
    
@Martin: Does the argument work in this setting? (For some more background, see my question on MO.) –  Dave Apr 24 '13 at 10:09
    
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. –  Martin Apr 24 '13 at 10:27

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.