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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.

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If an integral curve was not a geodesic the gradient vector would have trouble being tangent to the integral curve, no? – Ryan Budney Jan 9 '11 at 18:57
up vote 5 down vote accepted

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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Thanks! I've alredy done the calculations and it works wonders :). I have one question though: It seems to me that this works if we let that $|\mathrm{grad}(f)|$ is constant. – Sak Jan 11 '11 at 2:44
Right, $|\nabla f|$ constant is all we need. Different magnitudes of $|\nabla f|$ just correspond to different scalings of $f$. – Jason DeVito Jan 11 '11 at 3:32

An intuitive method of approaching this is as follows. If $t\mapsto x(t)$ is an integral curve then $\vert f(x(t_1))-f(x(t_0))\vert=\vert t_1-t_0\vert$. However, any other curve $t\mapsto y(t)$ joining points $y(s_0)=x(t_0)$ and $y(s_1)=x(t_1)$ satisfies $\vert f(y(s_1))-f(y(s_0))\vert\le\vert s_1-s_0\vert$ when parameterized by arc length. So $x$ is the shortest curve between the points.

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Why does it have to happen that $|f(x(t1))−f(x(t0))|=|t1−t1|$ ? – Sak Jan 11 '11 at 3:20
Nevermind, I see it now :) (there's a little typo it shoud be $|t1−t0|$) – Sak Jan 11 '11 at 3:32

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