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My question is about computing the distance between two points in a Riemannian manifold.

Suppose that $(M,g)$ is compact so that it is geodesically complete and geodesically convex.

Let $X\in\Gamma(TM)$ be a vector field. Fix a point $p\in M$. Let $\gamma:\mathbb{R}\to M$ denote the unique geodesic with initial velocity $X_p$. That is, $$(\exp)_p(X)=\gamma(1) \ \ \text{ and } \ \ (\exp)_p(tX)=\gamma(t).$$

My question is, why does $$d\left((\exp)_p(X),(\exp)_p(tX)\right)=|1-t||X_p| \ ?$$

Here $d:M\times M\to \mathbb{R}$ denotes the Riemann distance function. That is

$$d(p,q)=\inf\left\{\int|\rho^\prime(t)|dt \ ; \rho \text{ is an admissible curve between $p$ and $q$}\right\}$$

Since geodesics are length minimizing wouldn't $\gamma$ be the unique curve between $(\exp)_p(X)$ and $(\exp)_p(tX)$ which minimizes the distance function? But the integral of $|\gamma^\prime(t)|$ isn't $|1-t||X_p|$ ?

Any help is very much appreciated.

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  • $\begingroup$ I don't understand what bothers you. Yes, $\gamma$ is length minimizing, and $|\dot{\gamma}|=|V|$. What's the problem? $\endgroup$ – Amitai Yuval Aug 30 '14 at 22:53
  • $\begingroup$ It is. takes $1-t$ time to move to the end-point $\exp_p(V)$ $\endgroup$ – Troy Woo Aug 30 '14 at 22:55
  • $\begingroup$ So where is the $1-t$ coming in? The distance is $\int_0^t|\gamma^\prime(s)|ds$? $\endgroup$ – user162520 Aug 30 '14 at 22:59
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    $\begingroup$ No, you're not starting at $p$. You're starting at $\exp_p(1X)$. $\endgroup$ – Ted Shifrin Aug 30 '14 at 23:41
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    $\begingroup$ Isn't the distance in question at most |1-t||X|? $\gamma$ is a geodesic joining the points $\gamma(1)$ and $\gamma(t)$, but it may not be the minimal length geodesic joining those points. Maybe for $t$ sufficiently close to $1$, that would be the case. $\endgroup$ – Phillip Andreae Aug 31 '14 at 0:18
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Geodesics are only locally length-minimizing, i.e., only for $t$ sufficiently close to $1$ will $\gamma$ necessarily be the geodesic of shortest length joining $\gamma(1)$ and $\gamma(t)$. All that can be said in general is that the distance in question is at most $\left|1-t \right| \, |X_p|$.

To elaborate, there may be multiple geodesics joining $\gamma(1)$ and $\gamma(t)$; $\gamma$ is one such geodesic, but it may not be the geodesic of shortest length. A good example to keep in mind is the round sphere $S^2$: between any two distinct (non-antipodal) points $p$ and $q$ on the sphere, there is a unique great circle passing through $p$ and $q$, but there are two geodesics joining $p$ and $q$, each corresponding to one of the two ways to go around the great circle.

Edit (to explain $|1-t| \, |X_p|$): The point is that the velocity vector of a geodesic defined via the exponential map has constant length along the geodesic, i.e., for all $s$, $|\gamma'(s)| = |\gamma'(0)| = |X_p|$. This is a consequence of Gauss's lemma, which says that the exponential map is a "radial isometry". Assuming that $t<1$, the length of $\gamma(s)$ between $s=1$ and $s=t$ is $$ \int_t^1 |\gamma'(s)| \, ds = \int_t^1 |X_p| \, ds = (1-t)|X_p|$$ (and similarly if $t > 1$).

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  • $\begingroup$ Could you please explain to me how you are even getting $|1-t||V|,$ or really $|1-t||X_p|$? $\endgroup$ – user162520 Aug 31 '14 at 21:10
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    $\begingroup$ Oh! Geodesics have constant speed! Thank you very much! $\endgroup$ – user162520 Sep 1 '14 at 2:21

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