Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For a topological space $X$ and a positive integer $k\in \mathbb{N}_{>0}$ let $F_k(X):= \{ (x_1,\ldots,x_k)\in X^k |x_i\neq x_j \text{ for } i\neq j \}$ be its $k$-configuration space. Let $f:M\to \mathbb{R}$ be a Morse-function on a compact manifold $M$. The space $GVect(f)$ of all gradient-like vector fields for $f$ is convex (if a suitable definition of gradient-like is used) and hence contractible.

Question: Is the $k$-configuration space $F_k(GVect(f))$ contractible?


My approach so far:

Let $Q_m\subset GVect(f)$ denote a set of $m\geq 0$ distinct points.

Claim:

  1. $GVect(f)$ is a manifold without boundary
  2. $GVect(f)\setminus Q_m$ has trivial homotopy groups
  3. $F_k(GVect(f))$ is a $CW$-complex.

Using parts 1 & 2 of the claim, Theorem 2.5 and the proof of Theorem 2.7 of this paper, one can show that the homotopy groups of $F_k(GVect(f))$ are trivial. Whitehead's-theorem and part 3 of the claim now imply that $F_k(GVect(f))$ is in fact contractible.

However, I'm not sure if the claim is true (or if it's even sensible) and how one may prove it.

Thank you for any contribution.

share|improve this question
    
In what sense is it a manifold? –  Dan Petersen Apr 17 '13 at 13:38
    
Is GVect(f) infinite dimensional ? If this is the case, you are right. If not, I think claim 2 is not true. –  Geoffroy Horel Apr 17 '13 at 13:44
    
@Geoffroy: For a small enough bump function $\rho\in C^\infty(M)$ and $X\in Gvect(f)$ the vector field $\rho\cdot X$ is still gradient like. Since $C^\infty(M)$ is an infinite dimensional vector space this suggests that the dimension of $Gvect(f)$ is not finite. I'd really appreciate if you could elaborate on your comment. What do you think I'm right/wrong about and why? –  Dave Hartman Apr 17 '13 at 15:19
    
@Dan Petersen: I'm sorry but I don't know if or in what sense $Gvect(f)$ is a manifold... If you have an idea please let me know, thank you. –  Dave Hartman Apr 17 '13 at 15:22
1  
If $E$ is an infinite dimensional vector space, a finite configuration of points $X$ of $E$ is contained in a finite dimensional subspace $V$. You can apply the Mayer Vietoris long exact sequence for the inclusion $V-X\to E$. It tells you that $E-X$ has the same connectivity as $V-X$ i.e. $dim(V)-1$. Since you can do that for arbitrarily large $V$'s you can show that $E-X$ is contractible. If $Gvect(f)$ is homeomorphic to an infinite dimensional vector space then you are fine but I'm not sure that it is the case. –  Geoffroy Horel Apr 17 '13 at 18:11
show 1 more comment

1 Answer

up vote 2 down vote accepted

It seems that $GVect(f)$ is a manifold without boundary. Building on my answer to your last question (did you prove all of it?) let us argue as follows: A vector field $X$ is in $GVect(f)$ if:

  • $X(p)=0$ for each critical point $p$ of $f$. This describes a closed linear subspace $G_1$ of the Frechet space $\mathfrak X(M)$.

  • Near each critical point $p$ the function $df(X)$ is Morse with a maximum at $p$. This is a $C^2$-open condition in $G_1$: The differential must be transversal to the zero section, at $p$.

  • Off the critical points we have $df(X)<0$. This would be a $C^1$-open condition in $G_1$ if it holds on a closed subset of $M$. We can take the closed subset as the complement of the union of small open neighborhoods of the critical points of $f$. But these neighborhoods depend on $X$. So we have to look at all of them and take the union. Since the union of open sets is open, we are done.

So $GVect(f)$ is open in the Frechet space $G_1$.

The rest seems to be done by the comments to your question, by Geoffroy.

share|improve this answer
    
Thank you very much for your answer! Do you know whether $F_k(GVect(f))$ is a $CW$-complex? I might have another question about your answer once I've thought about it a bit more. –  Dave Hartman Apr 17 '13 at 22:15
    
since GVect(f) is a manifold, F_k(GVect(f)) is a submanifold of (GVect(f))^k - and manifolds are particular instances of CW-complexes. –  Tarje Bargheer Apr 17 '13 at 22:36
    
@Tarje Bargheer: Thanks. –  Dave Hartman Apr 18 '13 at 7:10
add comment

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.