Differential Geometry – Integral Curves of the Gradient

differential-geometry

Let $f : M \rightarrow \mathbb{R}$ be a differentiable function defined on a riemannian manifold. Assume that $| \mathrm{grad}f | = 1$ over all $M$. Show that the integral curves of $\mathrm{grad}f$ are geodesics.

Best Answer

Since this seems to be homework, here is just an outline of the proof.

  1. Show that the map $X\rightarrow \nabla_X \nabla f$ is self adjoint, that is, that $g(\nabla_X \nabla f, Y) = g(\nabla_Y \nabla f, X)$ for any vector fields $X$ and $Y$. You'll need to use the fact that $\nabla f$ is a gradient field, but you won't need the fact that it has norm 1.

  2. Show that $g(\nabla_{\nabla f} \nabla f, X) = 0$ for all $X$ by using 1. to write it as $g(\nabla_X \nabla f, \nabla f)$ and expanding. Here, you'll need to use the fact that $\nabla f$ has norm 1. Once you show this, conclude that $\nabla_{\nabla f} \nabla f = 0$, i.e., that the integral curves are geodesics.

Assuming I remember, or that you send a comment, I can update this in a few days with full solutions to either 1 or 2.

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