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Let $M$ be a compact manifold (without boundary) and let $f:M\to \mathbb{R}$ be a fixed Morse-function. My goal is to better understand gradient-like vector fields for $f$.

Question: Do any two gradient-like vector fields necessarily coincide on a sufficiently small neighborhood of any critical point of $f$ ?

This seems to be a natural question to ask. So I'm quite surprised that it's not addressed in any book about Morse theory that I've seen so far.
To eliminate any misunderstanding I will recall the definition of gradient-like vector fields:

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 $\mathbb{R^n}$).

Some motivation: I'm interested whether the space of all gradient-like vector fields for $f$ is contractible. If the answer to my question is affirmative then clearly any convex combination of a gradient-like vector field is still gradient-like and hence the space of gradient-like vector fields is contractible.

Any relevant references are also much appreciated.
Thanks for any contribution.

Edit: I've posted a related question on MathOverflow.

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1 Answer

up vote 3 down vote accepted

No. Since your question is about a neighborhood of a critical point, we can work over $\mathbb{R}^n$ instead of the compact manifold $M$.

Consider $\mathbb{R}^2$ with the following two coordinate charts in a neighborhood of 0. First we have the standard $x,y$ coordinates. Next we have the coordinates

$$ z = x \cos r^2 + y \sin r^2 \qquad w = y \cos r^2 - x \sin r^2 $$

where $r^2 = x^2 + y^2$. We easily verify that $z^2 + w^2 = x^2 + y^2 = r^2$. So that both $(x,y)$ and $(z,w)$ are Morse charts for $f = r^2$.

Let the vector field $X$ be $- x\partial_x - y\partial_y$ in the $(x,y)$ coordinates, and $X'$ be $- z\partial_z - w\partial_w$ in the $(z,w)$ coordinates. You can compute the change of variables explicitly and see that $X \neq X'$ except at the origin.

(It may be easier to see in standard polar coordinates, where $X = r\partial_r$ and $X' = r\partial_r + 2r^2\partial_\theta$. With this you also see that by adding a cut-off at finite $r$ for the perturbation, we can also directly extend this example to any two dimensional manifold. Higher dimensional analogues are also immediate.)

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Thanks a lot for this example @Willie Wong ! It's a shame that the answer is no even though I expected it to be. Do you know by any chance if the space of gradient-like vector fields is convex/connected anyway? –  Dave Hartman Mar 7 '13 at 21:02
    
Note that your statement is actually only dependent on the local structure of Morse charts near a critical point. So the issue of convexity/connectedness can be reduced to studying diffeomorphisms of $\mathbb{R}^{p+q}$ to itself that preserves the level sets of a nondegenerate bilinear form $B^{p,q}$. I think it is true that restricted to each level set the diffeomorphism must coincide with the action by an element of $SO(p,q)$. Overall smoothness will require the dependence on the level to be continuous. This should then imply connectedness. But I haven't tried to work out the details. –  Willie Wong Mar 8 '13 at 9:06
    
BTW: this MO question is related to my comment above. –  Willie Wong Mar 8 '13 at 9:09
    
Thanks for the clarification & the link. Correct me if I'm wrong: but contractibility follows if each diffeomorphism of the form you describe is isotopic to the identity where the isotopy should only go through diffeomorphisms of the same form? It seems rather hard to show that such an isotopy exists... Maybe I'll open a new question concerning convexity/contractibility. Do you think that it's appropriate to ask about this on MO? –  Dave Hartman Mar 8 '13 at 9:26
    
@Dave: if the intuition behind what I wrote is true, then we have a smooth map $\rho: \mathbb{R} \to SO(p,q)$ such that the action of the diffeo is identical to $\rho(l)$ for the level set $f^{-1}(l)$ where $f(x) = B^{p,q}(x,x)$. Then considering $\rho_\lambda = \rho(\lambda\cdot)$ should connect the diffeo to an element of $SO(p,q)$, which leaves $X$ invariant. –  Willie Wong Mar 8 '13 at 9:36
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