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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?

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

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  • $\begingroup$ 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, 2016 at 15:26
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    $\begingroup$ That expression will not necessarily give rise to a globally defined vector field. $\endgroup$
    – Pedro
    Nov 17, 2016 at 15:49
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    $\begingroup$ 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, 2016 at 15:50
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    $\begingroup$ 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, 2016 at 15:52
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    $\begingroup$ @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, 2016 at 10:53

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