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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.


  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.

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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 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 Apr 17 '13 at 15:22
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
up vote 3 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.

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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 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 Apr 18 '13 at 7:10

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