Assume that $M$ is a differentiable manifold. Is there a Riemannian metric on $M$ such that the space of all gradient vector fields on $M$ would be closed under the Lie bracket?
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1 Answer
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If $M$ is the circle $S^1$ there is not such metric. Indeed, any Riemannian metric on $S^1$ is isometric to the one induced on a circle of radious $r$ of the plane centered at some point. Then take the angular "coordinate" $\theta$ and the functions $\cos(\theta)$ and $\sin(\theta)$ and check that the Lie bracket of their gradients is not a gradient.
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$\begingroup$ Thank you for your answer. Since the Lie bracket is a non vanishing vector field. yes? $\endgroup$ Commented Apr 1, 2017 at 10:48
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$\begingroup$ yes indeed : functions on compact manifolds have critical points. $\endgroup$ Commented Apr 1, 2017 at 10:50