Probing a manifold with geodesics - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T02:16:37Z http://mathoverflow.net/feeds/question/81622 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81622/probing-a-manifold-with-geodesics Probing a manifold with geodesics Joseph O'Rourke 2011-11-22T16:07:56Z 2011-12-19T14:54:50Z <p>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: <br /> &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <img src="http://cs.smith.edu/~orourke/MathOverflow/TorusGeodesic.jpg" alt="Toru s Geodesic" /> <br /> &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <sub>(Based on an image created by <a href="http://www.rdrop.com/~half/math/torus/illustrating.html" rel="nofollow">Mark Irons</a>.) </sub></p> <p>I would like to know what information one could learn about $M$ from such <em>geodesic probes</em>. 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,</p> <blockquote> <p>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 $p$?</p> </blockquote> <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., "<a href="http://www.emis.de/journals/ERA-AMS/2000-01-013/2000-01-013.html" rel="nofollow">Metric with Ergodic Geodesic Flow is Completely Determined by Unparameterized Geodesics</a>." Vladimir Matveev and Petar Topalov. <em>Electronic Research Announcements of the AMS</em>. 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!</p> <p>(Tangentially related MO question: <a href="http://mathoverflow.net/questions/26081/" rel="nofollow">Shortest-path Distances Determining the Metric?</a>.)</p> http://mathoverflow.net/questions/81622/probing-a-manifold-with-geodesics/82186#82186 Answer by Richard Montgomery for Probing a manifold with geodesics Richard Montgomery 2011-11-29T16:04:57Z 2011-11-29T16:04:57Z <p>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.</p> <p>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. </p> <p>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 $p$.</p> http://mathoverflow.net/questions/81622/probing-a-manifold-with-geodesics/83726#83726 Answer by Robert Bryant for Probing a manifold with geodesics Robert Bryant 2011-12-17T17:28:54Z 2011-12-19T14:54:50Z <p>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. </p> <p>The question, then, is "Can you recover the metric on the surface $M$ from the data $\kappa$ and $\tau$?" </p> <p>Not surprisingly, the answer is <em>yes</em> 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 $$f(t,\theta) = \frac{\sqrt{|\tau(t,\theta)|}}{2\ \tau(t,\theta)} \int_0^t\frac{\kappa_\theta(\rho,\theta)}{\sqrt{|\tau(\rho,\theta)|}}\ d\rho\ .$$</p> <p>(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.)</p> <p>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. </p> <p>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 $0\le t &lt; T(\theta)$ 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 $$\chi(M) = \frac{-1}{2\pi}\int_0^{2\pi} f_t\bigl(T(\theta),\theta\bigr)\ d\theta.$$</p> <p>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.</p>