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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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    $\begingroup$ If an integral curve was not a geodesic the gradient vector would have trouble being tangent to the integral curve, no? $\endgroup$ – Ryan Budney Jan 9 '11 at 18:57
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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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    $\begingroup$ 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. $\endgroup$ – Sak Jan 11 '11 at 2:44
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    $\begingroup$ Right, $|\nabla f|$ constant is all we need. Different magnitudes of $|\nabla f|$ just correspond to different scalings of $f$. $\endgroup$ – Jason DeVito Jan 11 '11 at 3:32
  • $\begingroup$ Dear Jason: this is great. It is easy to think about the second step, but could you explain a bit how one can think of the first step? It seems to be coming from nowhere to me! Thanks a lot! $\endgroup$ – Wu Yaochen May 10 '17 at 13:08
  • $\begingroup$ I'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. $\endgroup$ – Jason DeVito May 11 '17 at 15:59
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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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  • $\begingroup$ Why does it have to happen that $|f(x(t1))−f(x(t0))|=|t1−t1|$ ? $\endgroup$ – Sak Jan 11 '11 at 3:20
  • $\begingroup$ Nevermind, I see it now :) (there's a little typo it shoud be $|t1−t0|$) $\endgroup$ – Sak Jan 11 '11 at 3:32
  • $\begingroup$ Why would $\vert f(x(t_1))-f(x(t_0))\vert=\vert t_1-t_0\vert$ or $\vert f(y(t_1))-f(y(t_0))\vert\leq\vert t_1-t_0\vert$ for that matter be true?? The function $f:M \to \mathbb R$ is an arbitrary differentiable function, so the function which maps everything to zero is also allowed. This would however mean that $0 = \vert t_1-t_0 \vert $ which is not correct unless $t_1=t_0$. $\endgroup$ – Gonenc Mogol Jul 4 '17 at 6:19
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    $\begingroup$ @GonencMogol Rather late to the party but: $df(x(t))/dt = $ derivative of $f$ in the direction of $gamma'(t) = \langle grad(f),dx/dt\rangle = ||grad(f)||^2 = 1$ by our assumption. Integrating both sides with respect to $t$ we get what we want. $\endgroup$ – Asvin Jun 10 '18 at 2:03
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    $\begingroup$ @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. $\endgroup$ – Asvin Jun 10 '18 at 2:17
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Another strategy: define coordinates $(x^1, \ldots, x^{n-1}, y)$ at points $p \in M$ so that $(x^1, \ldots, x^{n-1})$ are local slice coordinates for the level sets $\{q \in M : f(q) = f(p)\}$, and $y \in \mathbb R$ satisfies $f(\gamma_p(y)) = f(q)$, where $\gamma_p$ is the integral curve of $\mathrm{grad} f$ starting at $p$. In these coordinates, $\mathrm{grad} f = \partial_y$, and $g(\partial_{i}, \partial_y) \equiv 0$, where $\partial_i := \partial_{x^i}$. Furthermore, $g(\partial_y, \partial_y) = |\mathrm{grad}f|^2 \equiv 1$. We'll prove $D_t \dot\gamma(t) \equiv 0$ in these coordinates for every integral curve $\gamma$ of $\mathrm{grad}f$.

By symmetry of the Levi-Civita connection $\nabla$ and commutativity of coordinate vector fields, $\nabla_{\partial_i} \partial_y = \nabla_{\partial_y} \partial_i$. So, since $\nabla$ is compatible with the Riemannian metric, for $i = 1, \ldots, n-1$, \begin{align*} \left\langle D_t \dot\gamma(t), \partial_i\right\rangle &= \frac d{dt} \left\langle \dot\gamma(t), \partial_i\right\rangle - \left\langle \dot\gamma(t), D_t \partial_i \right\rangle = \frac{d}{dt} \langle \partial_y, \partial_i \rangle - \left\langle \partial_y, \nabla_{\partial_y} \partial_i \right\rangle = -\left\langle \partial_y, \nabla_{\partial_i} \partial_y \right\rangle \\ &= \left\langle \nabla_{\partial_i} \partial_y, \partial_y \right\rangle - \partial_i \left\langle \partial_n, \partial_n \right\rangle = \left\langle \nabla_{\partial_i} \partial_y, \partial_y \right\rangle = -\left\langle D_t \dot\gamma(t), \partial_i\right\rangle. \end{align*} So $\left\langle D_t \dot\gamma(t), \partial_i\right\rangle = 0$ for $i = 1, \ldots, n-1$. Furthermore, $$ \left\langle D_t \dot\gamma(t), \partial_y\right\rangle = \left\langle D_t \partial_y, \partial_y\right\rangle=\frac d{dt} \left\langle \partial_y, \partial_y \right\rangle - \left\langle \partial_y, D_t \partial_y \right\rangle = - \left\langle \partial_y, D_t \partial_y \right\rangle. $$ So, $\left\langle D_t \dot\gamma(t), \partial_y\right\rangle = 0$, whence $\left\langle D_t \dot\gamma(t), X\right\rangle \equiv 0$ for every vector field $X$ over $M$. So $D_t \dot\gamma(t) \equiv 0$, so $\gamma(t)$ is geodesic.

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