# Is the space of gradient-like vector fields contractible?

Let $M$ be a compact manifold (without boundary) and let $f:M\to \mathbb{R}$ be a fixed Morse-function.

Question: Is the space $GVect(M,f)$ of all gradient-like vector-fields for $f$ contractible or even convex?

To eliminate any misunderstanding, this is the definition of gradient-like vector fields that I'm using:

Definition: A vector field $X\in Vect(M)$ is called gradient-like for $f$ if
1. $\forall q\in M\setminus Crit(f):\; df(q)X(q)<0$.
2. For each critical point $p\in Crit(f)$ of $f$, $\exists$ Morse coordinate chart $\phi:U\to\mathbb{R^n}$ (i.e. $\phi(p)=0$ and $f\circ\phi^{-1}(x)=f(p)-\sum_{i=1}^k x_i^2+\sum_{i=k+1}^n x_i^2$ for $x\in\phi(U)$, $k=index(p)$) such that $$\phi_*X(x)=d\phi(\phi^{-1}(x))X(\phi^{-1}(x))=(2x_1,\ldots,2x_k,-2x_{k+1},\ldots,-2x_n)^T$$ for $x\in\phi(U)$ (i.e. in this Morse-chart $X$ coincides with the negative gradient of $f$ w.r.t. the euclidean metric on $\mathbb{R^n}$).

Thank you in advance for any insights.

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I've also posted a related question on StackExchange, see math.stackexchange.com/questions/322330/… –  Dave Mar 8 '13 at 17:19

$GVect(M,f)$ is a convex set in the space of all vector fields. Thus it is contractible.

Edit, since you asked for details: I assume that the Morse charts $\Phi$ are the same for all $X$.
Let $X,Y$ be gradient-like for $f$, then:
1. $df(q)(t.X(q)+(1-t).Y(q)) = t.df(q)(X(q)) + (1-t).df(q)(Y(q)) < 0$ for $0\le t\le 1$.
2. In the Morse charts the two vector fields look the same.

Second edit: Okay, you want the Morse chart to depend on the vector field. Maybe, your definition of a gradient-like vector field is equivalent to:

a) At a critical point $p$ we have $X(p)=0$.
b) Near each critical point $p$, the function $-X(f)= -df(X)$ is Morse with critical point $p$ a minimum. Then you can use $-df(X)$ as the square distance from $p$ for a Riemannian metric, and the Morse chart construction with respect to that Riemannian metric leads to a Morse chart as you require.
c) Off the critical points, we have $df(X)<0$ (i.e. X is transversal to the level sets and points towards lower level sets).

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Could you please elaborate? Why exactly is it convex? –  Dave Mar 8 '13 at 16:15
Why can you assume that the Morse charts are the same for the two gradient-like vector fields? This is not clear to me at all. –  Dave Mar 8 '13 at 17:11
You did not say that they depend on the field. Actually, into the construction of a Morse chart goes a Riemann metric and the neg. gradient of $f$ with respect to $g$. The space of Riemann metrics is also convex. –  Peter Michor Mar 8 '13 at 17:28
How much freedom do you have in choosing a Morse chart for a fixed Morse function? Maybe, this is the core of your problem! –  Peter Michor Mar 8 '13 at 17:34
There are lots of Morse charts around a Morse point. Any diffeomorphism preserving the level sets will give you another Morse chart, so you can slide around'' the level sets. For example, in the case of a positive definite Morse point, you can rotate the spherical level sets, each by a different angle. –  Ben McKay Mar 11 '13 at 17:06