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Supposed you stand at a point $p \in M$ on a smooth 2-manifold $M$ embedded in $\mathbb{R}^3$. You do not know anything about $M$. You shoot off a geodesic $\gamma$ in some direction $u$, and learn back the shape of the full curve $\gamma$ as it sits in $\mathbb{R}^3$. (One could imagine a vehicle traveling along $\gamma$, sending back $xyz$-coordinates at regular time intervals; assume $t \rightarrow \infty$.) For example, if the geodesic happens to be closed, your probe might return the blue curve left below:
          Toru
s Geodesic
                                                       (Based on an image created by Mark Irons.)

I would like to know what information one could learn about $M$ from such geodesic probes. I am interested in the best case rather than the worst case. For example, you might learn that $M$ is unbounded, if you are lucky enough to shoot a geodesic to infinity. In particular,

Are there circumstances (a manifold $M$, a point $p$, directions $u$) that permit one to definitively conclude that the genus of $M$ is nonzero, by shooting (perhaps many) geodesics from one fixed (well-chosen) point $p$?

I believe that, if one knew all the geodesics through every point, then there are natural circumstances under which the metric is determined [e.g., "Metric with Ergodic Geodesic Flow is Completely Determined by Unparameterized Geodesics." Vladimir Matveev and Petar Topalov. Electronic Research Announcements of the AMS. Volume 6, Pages 98-101, 2000]. But I am more interested what can be determined from a single point $p$ (and many directions $u$). Thanks for thoughts/pointers!

(Tangentially related MO question: Shortest-path Distances Determining the Metric?.)

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I thought that on a compact $M$, other than a sphere, the typical geodesic is dense in $M$. This would imply that if you really have the image of all of $[0,\infty)$ under a geodesic path then you know $M$ exactly and can thus deduce its genus and any other property you care about. –  Noam D. Elkies Nov 22 '11 at 17:35
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Having two orthogonal circles of different radii (like on the torus) already tells you something. You do not even need geodesics; if two closed curves on the manifold have just one intersection and it is transversal, that certainly should ring some bells though I'd better leave it to the many topologists here to tell what and how exactly can be derived from it. :). –  fedja Nov 22 '11 at 18:17
    
@Noam: Perhaps you mean, "other than homeomorphic to a sphere," rather than precisely a sphere. Because Zoll's surface has the property that every geodesic is closed and simple (mathoverflow.net/questions/28622). I am surprised to learn that typical geodesics are dense. I thought there would generally be "unreachable" sections. Glad to have my faulty intuition corrected! –  Joseph O'Rourke Nov 22 '11 at 21:35
    
@fedja: Excellent point (thanks!), but I am uncertain what conclusions could be drawn... –  Joseph O'Rourke Nov 22 '11 at 21:39
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By recording the geodesics intersections you can collect information on the cut locus of the surface, which in turn provide you with insight into the topology of the surface. –  Dror Atariah Nov 23 '11 at 11:52

2 Answers 2

There's a different kind of answer to this that you might be interested in: Suppose that, when you fire off a probe along a unit speed geodesic starting at $p\in M$, you record the direction $\theta$ in which you sent it, and the probe reports the inertial forces it is experiencing, i.e., it sends back a running report on the curvature and torsion of the curve it is traveling along. Thus, you get to record these two data as functions $\kappa(t,\theta)$ and $\tau(t,\theta)$ of $t$, the time since the probe was launched, and $\theta$, the direction in which it was sent.

The question, then, is "Can you recover the metric on the surface $M$ from the data $\kappa$ and $\tau$?"

Not surprisingly, the answer is yes in the generic situation. If, for example, you are in the situation in which $\tau\not=0$, you find (by computing with the structure equations) that the induced metric must be of the form $$ g = dt^2 + f(t,\theta)^2\ d\theta^2, $$ where $$ f(t,\theta) = \frac{\sqrt{|\tau(t,\theta)|}}{2\ \tau(t,\theta)} \int_0^t\frac{\kappa_\theta(\rho,\theta)}{\sqrt{|\tau(\rho,\theta)|}}\ d\rho\ . $$

(Obviously, there will be some singularity issues at places where $\tau$ vanishes, but, generically, this isn't a problem. The case in which $\tau$ vanishes identically, such as when $p$ is a pole of rotational symmetry of the surface $M$, has to be treated separately.)

Since you can recover the curvature from $f$ by the formula $K = -f_{tt}/f$, you could detect when, for example, the surface $M$ is locally convex, and so forth.

As for computing the Euler characteristic, since $K\ dA = -f_{tt}\ dt\wedge d\theta = -d\left(\ f_t\ d\theta\ \right)$, it follows that, if you could figure out the star-shaped domain (in good cases, of the form $0\le t < T(\theta)$ for some piecewise differentiable $2\pi$-periodic function $T$ ) that maps one-to-one and onto the complement of the cut locus of $p$, then you could compute the Euler characteristic as $$ \chi(M) = \frac{-1}{2\pi}\int_0^{2\pi} f_t\bigl(T(\theta),\theta\bigr)\ d\theta. $$

How numerically stable all these calculations are, I don't know. (I also don't know how hard it would be to figure out or approximate $T$.) Since you know that the result is an integer, though, you might be able to tolerate a reasonable amount of numerical error.

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@Robert: What a cool idea! Thanks! –  Joseph O'Rourke Dec 17 '11 at 19:20

Fix one $u$, its resulting geodesic probe, and vary $u$ slightly. Comparing the two geodesics at the same arclength values yields a close approximation to the solution to the Jacobi equation along the first curve. Consequently, you can approximate the curvature $K$ along the first curve. Stepping around the unit circle in the tangent space based at $p$ in this way, I have $N$ geodesic probes, and along each an approximation to the curvature. Taking $N$ large, I get an approximation to the curvature of the whole surface. Now you can also approximate the area element on the surface from your skeleton of curves, so you get an approximation of $\int K dA$ and hence the surface's genus.

This seems a bit of a cheat. Your map is the exponential map from the point $p$. I am coming close to saying: pull back the metric on the surface to the tangent space via the exponential map. Now you have a description of the whole surface.

A comment on the dense vs. Zoll business. If the surface is compact then in all circumstances we have Poinare recurrence: any geodesic through $p$ returns to an arbitrarily small neighborhood of $p$.

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@Richard: Actually, your last remark is not true as you have stated it. It's easy to construct a compact surface with a geodesic that $\alpha$-limits to one closed geodesic and $\omega$-limits to another, and such a geodesic won't pass through any point $p$ such that the geodesic returns arbitrarily close to $p$. Poincaré recurrence does, of course, apply to the (volume preserving) geodesic flow on the unit tangent bundle of a compact Riemannian surface, but its interpretation in terms of return of geodesics on the surface is different from what you have written. –  Robert Bryant Dec 16 '11 at 20:34

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