I'm taking a course on differential geometry, and up until now I'd always thought that the definition of a geodesic is (loosely speaking) a curve on a surface with the minimal length between its endpoints. My professor, taking his lead from do Carmo, however, defines it as any curve whose geodesic curvature $\kappa_g=0$. We showed that this is equivalent to satisfying the following pair of nonlinear ordinary differential equations:

$$(\boldsymbol{E}u' + \boldsymbol{F}v')' = \frac12(\boldsymbol{E}_u(u')^2 + 2\boldsymbol{F}_uu'v' + \boldsymbol{G}_u(v')^2)$$

$$(\boldsymbol{F}u' + \boldsymbol{G}v')' = \frac12(\boldsymbol{E}_v(u')^2 + 2\boldsymbol{F}_vu'v' + \boldsymbol{G}_v(v')^2)$$

We then went through an incredibly painful calculation on the length of the family of curves $\gamma_\lambda$ to show that geodesics (i.e, those curves satisfying the geodesic equations above) are critical points of the functional

$$\displaystyle\mathcal{L}(\lambda) = \int_a^b{\left\|\frac{d\gamma_\lambda}{dt}\right\| dt},$$

which is the length of the curve. Therefore, according to my professor's (and the textbook's) definition, geodesics are not necessarily length-minimizing, just critical points of $\mathcal{L}$. Therefore, on a sphere, two non-antipodal points have two geodesics: the obvious length-minimizing one, and the other one going the long way around the sphere (which is, in this case, a saddle point of $\mathcal{L}$). This is not just an oversight on my professor's part, he explicitly brought attention to this fact.

My question is, what are the advantages and disadvantages of these two conflicting definitions? I still see the length-minimizing one almost everywhere.

On a related note, the fact that a geodesic is only a critical point, not necessarily a minimum, leaves open the possibility of a geodesic actually being the longest path between two points. Are there any situations where this is actually possible? It seems you could always perturb a curve slightly to stay within the image of a chart while still increasing its length infinitesimally. Are there some weird spaces where this is not the case?

  • $\begingroup$ I don't think there can ever be a longest path: if $\gamma$ is a path then we can compose $\gamma$ with the path taking us backwards along $\gamma$ and then compose again with $\gamma$ to get a path 3 times as long as $\gamma$. Edit: I'm guessing though you meant local maximum. $\endgroup$ – Eric O. Korman Mar 23 '11 at 18:59
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    $\begingroup$ Geodesics are locally length minimizing, in that any sufficiently small portion is the shortest path between the endpoints of that portion. $\endgroup$ – Joseph O'Rourke Mar 23 '11 at 19:16
  • $\begingroup$ There can be no local maximum for the length functional. For one thing, you can always overshoot your target and double back (or more precisely, you can make a continuous family of paths do this). And if your manifold has dimension more than 1, then you can also wiggle your path side to side in any small neighborhood. $\endgroup$ – Aaron Mazel-Gee Mar 23 '11 at 23:06
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    $\begingroup$ By the way, the calculations are significantly cleaner in the general case. Check out the other do Carmo ("Riemannian Geometry"), it's pretty accessible. $\endgroup$ – Aaron Mazel-Gee Mar 23 '11 at 23:48

I don't really see any advantage to restricting the definition of geodesic to be minimal -- after all, those are just what we call "minimal geodesics"! As you do more geometry (Riemannian and otherwise), you'll encounter many other definitions that are given via differential equations. These all have their local theories -- in this case, we find that every point on a Riemannian manifold has a neighborhood where minimizing geodesics are unique -- and this does not detect the global behavior. But this can be a good thing, because once you've nailed down the local picture then you have firmer footing to ask global questions. Here, we might ask: When exactly does a geodesic stop being a minimizing geodesic?

These words may not mean anything, and they don't really need to, but a similar differential-geometric example that might shed light by analogy is Darboux's theorem, which says that all symplectic manifolds of the same dimension are locally symplectomorphic. That is, as far as the stuff we care about is concerned (namely the "symplectic structure"), neighborhoods of any two points on any two equidimensional symplectic manifolds are indistinguishable. This is true too of smooth manifolds, but not of Riemannian manifolds, since curvature gives us a local invariant with which to distinguish them. (In fact curvature is the only local invariant! But that's another story.) But nevertheless people call themselves symplectic geometers, and indeed there are some very deep global questions in symplectic geometry.

The moral of the story is that in geometry one often starts with an idea (e.g. "the shortest path between two points"), examines the local behavior, and then works to understand how the local story pieces together to form a global picture. This makes it very natural to begin with local definitions such as the one you mention.

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    $\begingroup$ "When exactly does a geodesic stop being a minimizing geodesic?" (I know this was a rhetorical question, but...) The answer is: at the cut locus. The cut locus w.r.t. point $x$ is where the geodesics from $x$ cease to be shortest paths, i.e., shortest paths are cut off there. $\endgroup$ – Joseph O'Rourke Mar 24 '11 at 1:38
  • $\begingroup$ Thank you for this great response, and thanks also to @Joseph for the pointer. Lots to think about. $\endgroup$ – Adrian Petrescu Mar 24 '11 at 1:43
  • $\begingroup$ @ Joseph: Perhaps, but that's pretty tautological if you ask me! $\endgroup$ – Aaron Mazel-Gee Mar 24 '11 at 2:14
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    $\begingroup$ it doesn't hurt to put keywords out there to help people who want to find out more. The cut locus is only a definition, but if you know it then you can more easily look it up and find out about conjugate points and Klingenberg's theorem. $\endgroup$ – yasmar Mar 24 '11 at 7:54
  • $\begingroup$ Sure! I guess the way it was phrased suggested to me that it was meant to give information rather than just mention the name. But you're right, it's nice to have heard the words. $\endgroup$ – Aaron Mazel-Gee Mar 24 '11 at 8:12

A short answer is that the definition via geodesic curvature is much easier to verify, hence much easier to prove things about.


One example of a geodesic being the longest path in some sense is in Einstein's relativity. It is related to the Twin Paradox, where two twins set off from some point in spacetime and then meet again at another point in spacetime, to discover one has aged more than the other.

The geodesic is the path which takes the longest as measured by a clock passing along it. For Special Relativity, this is a straight line and a constant speed, i.e. inertial motion, and any clock going by any other path will measure less time.

  • $\begingroup$ Interesting. It seems like this is a different use of the word "longest", as a measurement in a particular coordinate with no reference to the metric on the 4-manifold. What is the metric, anyways? $\endgroup$ – Aaron Mazel-Gee Mar 24 '11 at 2:22

It seems to me that the only possible situation where a geodesic is the longest path (among the class of injective paths) between two points is where it is the only path; in particular, on some one-dimensional (sub)manifolds (lines but not circles). Here it is also the shortest path.

Keep in mind that all minimisers of the length functional are also critical points, thus every "shortest path" is also a geodesic, but not vice-versa.

  • $\begingroup$ Paths don't need to be injective, which I think invalidates your first statement. $\endgroup$ – Aaron Mazel-Gee Mar 23 '11 at 23:03
  • $\begingroup$ Sure, of course you are right, on one can always continue to loop over the path or go back and forth to generate paths longer than the geodesic. It also isn't true for S^1. Sigh. I'll edit. $\endgroup$ – Glen Wheeler Mar 24 '11 at 0:06

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