3
$\begingroup$

Let $M^m$ be a smooth manifold and $f:M\to\mathbb{R}$ a smooth, non singular function. Prove that there exists a vector field $X\in\frak{X}$$(M)$ such that $X(f)\equiv 1$.

Take a chart $(U, \phi)$ at $p\in M$ with $U$ small enough so that we can assume, with no loss in generality, that $\frac{\partial f}{\partial \phi^{n}}\neq 0$ in $U$ (that's possible since $f$ is non singular). That way, take any $g_1, ...,g_{n-1}\in C^{\infty}(M)$ and define

$$g_n:=\left(\frac{\partial f}{\partial \phi^{n}}\right)^{-1}\left(1-\sum_{i=1}^{n-1}g_i\frac{\partial f}{\partial \phi^{i}}\right)$$

which is obviously in $C^{\infty}(M)$. Now, defining $X_{U}:=\sum_{i=1}^n g_i\frac{\partial }{\partial \phi^{i}}$, we have a local smooth vector field with $X_U(f)\equiv 1$ in $U$.

I've tried to use a partition of unity to define some $X:=\sum X_{U}$, but I could not guarantee that $X(f)\equiv 1$ and I don't know how to work this out. Any ideas?

$\endgroup$
5
$\begingroup$

Fix a Riemannian metric on $M$ and set $X = \frac{\nabla f}{||\nabla f||^2}$.

$\endgroup$
  • $\begingroup$ Thanks!, @Pedro! But couldn't we simply define $X:=\frac{1}{\sum_{i=1}^{n}\left(\frac{\partial f}{\partial x_i}\right)^2}\sum_{i=1}^{n}\left(\frac{\partial f}{\partial x_i}\right)\frac{\partial }{\partial x_i}$? Why is it necessary to talk about a Riemannian metric? $\endgroup$ – rmdmc89 Nov 17 '16 at 15:26
  • 1
    $\begingroup$ That expression will not necessarily give rise to a globally defined vector field. $\endgroup$ – Pedro Nov 17 '16 at 15:49
  • 1
    $\begingroup$ To be able to define the gradient of a function in an arbitrary manifold, you need a Riemannian connection. The gradient is uniquely determined by the identity $\langle \nabla f, Y \rangle = df(Y)$. $\endgroup$ – Pedro Nov 17 '16 at 15:50
  • 1
    $\begingroup$ Furthermore, the expression of $\nabla f$ in coordinates will in general NOT be $\sum \frac{\partial f}{\partial x_i} \frac{\partial}{\partial x_i}$. There are terms depending on the metric. You may consult any Riemannian geometry book to learn about this. $\endgroup$ – Pedro Nov 17 '16 at 15:52
  • 1
    $\begingroup$ @AguirreK: You do not need to know Riemannian geometry, instead you can consider your $M$ as a submanifold of $R^N$, extend $f$ smoothly to $R^N$ (do you know how to do so?) and then use the Euclidean gradient (the one from vector calculus). $\endgroup$ – Moishe Kohan Dec 1 '16 at 10:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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