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.
Integral curves of the gradient
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1If an integral curve was not a geodesic the gradient vector would have trouble being tangent to the integral curve, no? – 2011-01-09
3 Answers
Since this seems to be homework, here is just an outline of the proof.
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.
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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0I'm not sure what I was thinking 6 years ago. Probably I had already seen that outline before. A guess is that Petersen's Riemannian geoemtry text book has it, either worked out or as an exercise. But, as you said, step 2 is relatively clear (how do you show a vector is 0? Inner product it with everything!), and then step 1 follows as something necessary for step 2 to work. So, perhaps, attempting Step 1 is somewhat natural. – 2017-05-11
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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1@GonencMogol As for the second part, if we parametrize by arc length, then $||y'(s)|| = 1$ and as in the first part, $f(s_2) - f(s_1)$ = \integral_{s} df(y(s))/ds ds = \integral_s \langle grad(f) , y'(s) \rangle ds . Now, this inner product is between two vectors of length one and therefore has magnitude $\leq 1$ as required. – 2018-06-10