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As far as I understand it the closing lemma implies that closed geodesics on surfaces of negative curvature are dense. So: how can they be constructed in general?

A concrete answer that dovetails with the construction of such surfaces with constant negative curvature and genus $g$ from regular hyperbolic $(8g-4)$-gons along lines indicated by Adler and Flatto and gives the endpoints of the geodesics in the Poincaré disk model would be ideal. More useful still would be a way to construct all the closed geodesics that cross the boundaries of translates of the fundamental $(8g-4)$-gon some specified number of times (I am pretty sure this ought to be a finite set, but I couldn't say why off the top of my head).

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What do you mean by "construct"? One can show that there is one in each homotopy type of closed curves in the surface (which is a component of the loop space), but I guess that is not a construction in your sense, no? –  Mariano Suárez-Alvarez Feb 1 '10 at 22:24
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(To be precise, each non-trivial free homotopy class of closed curves contains a geodesic) –  Mariano Suárez-Alvarez Feb 1 '10 at 22:49
    
Presumably your surface isn't given as $H^2 / G$ where $G$ is a group of covering tranformations? Because then your question boils down to finding the axis of these transformations, which is straight-forward. How are you specifying your surface? –  Ryan Budney Feb 1 '10 at 23:09
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There is one closed geodesic per conjugacy class in the fundamental group of the surface. That is the statement I commented on above. –  Mariano Suárez-Alvarez Feb 2 '10 at 1:28
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The crossing pattern with the boundary indicates an element of the fundamental group. So, you can list the possible crossing patterns, and compute the element of the fundamental group, and then use Ryan Budney's answer to compute the geodesic corresponding to that conjugacy class. Do you mind that some of these closed geodesics will cross themselves, i.e., they are immersed instead of embedded? –  Douglas Zare Feb 2 '10 at 3:56

5 Answers 5

up vote 7 down vote accepted

If you think of your surface as the upper half plane modulo a group of Moebius transformations $G$, start by representing each of your Moebius transformations $ z \longmapsto \frac{az+b}{cz+d}$ by a Matrix.

$$A = \pmatrix{ a & b \\\ c & d}$$

And since only the representative in $PGL_2(\mathbb R)$ matters, people usually normalize to have $Det(A) = \pm 1$.

The standard classification of Moebius transformations as elliptic / parabolic / hyperbolic (loxodromic) is in terms of the determinant and trace squared. You're hyperbolic if and only if the trace squared is larger than $4$. Hyperbolic transformations are the ones with no fixed points in the interior of the Poincare disc, and two fixed points on the boundary, and they are rather explicitly "translation along a geodesic".

Elliptic transformations fix a point in the interior of the disc so they can't be covering transformations. Parabolics you only get as covering transformations if the surface is non-compact, because parabolics have one fixed point and its on the boundary -- if you had such a covering transformation it would tell you your surface has non-trivial closed curves such that the length functional has no lower bound in its homotopy class.

So your covering tranformations are only hyperbolic. That happens only when $tr(A)^2 > 4$. So how do you find your axis? It's the geodesic between the two fixed points on the boundary, so you're looking for solutions to the equation:

$$ t = \frac{at+b}{ct+d}$$

for $t$ real, this is a quadratic equation in the real variable $t$. If I remember the quadratic equation those two points are:

$$ \frac{tr(A) \pm \sqrt{tr(A)^2 - 4Det(A)}}{2c}$$

Is this what you're after?

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There's countably infinite many closed geodesics, one for every conjugacy class of elements in $\pi_1$ of the surface. It's a theorem in hyperbolic geometry that in every homotopy-class $S^1 \to M$ for $M$ hyperbolic there is a unique geodesic representative. –  Ryan Budney Feb 2 '10 at 2:20
    
OK, I'll be able to make sense of that. Thanks. –  Steve Huntsman Feb 2 '10 at 2:50

Are you willing to buy that the set of non-closed geodesics are dense? If so, here is an argument that goes back to Birkhoff and Hadamard. Take the surface's universal cover -- which is to say the upper half plane. Tile it with fundamental domains for said surface's fundamental group (relative to a fixed const neg curv metric). These have some number of edges, a, b, c, ... . Now count how a geodesic crosses the edges: acbaf... thus getting an (infinite) word -- or symbol sequence-- in the edges. Theorem: the symbol sequence is periodic if and only if the geodesic is. Theorem: if we are given a (variable) negatively curved metric on the surface then the symbol sequence uniquely determines the geodesic. Theorem: if a sequence $s_N$ of symbol sequences converges to a symbol sequence $s$, in the sense that for any sufficiently large `window' L of word length centered around 0, the finite word of length L of arbitrary length eventually agree, then the corresponding geodesics also converge. Now, approximate your given geodesic -- ie symbol sequence -- by a periodic sequence. symbol sequence by longer and longer periodic sequences.

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There is a simple, geométrical construction of the closed geodesic freely homotopic to a given closed curve on the surface in John Stillwell's Geometry of surfaces. I'll not sketch the argument because his explanation even has pictures :P

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I imagine this is the construction where you lift the curve to the universal cover, take the two limiting points on the boundary of the Poincare disc, then take "staight-line" (geodesic) homotopy between your original curve and the geodesic with the same endpoints? –  Ryan Budney Feb 2 '10 at 1:43
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Yes, it is. I might add that this idea goes back to Poincar\'e in 1904. –  John Stillwell Feb 4 '10 at 1:11

Of course closed geodesics can be explicitly constructed in the fundamental polygon by making sure that the angles at the end points match so that we get a closed geodesic rather than a geodesic loop.

The question of which immersed curves can be isotoped to a closed geodesic on a hyperbolic surface is quite subtle. I do not know the complete answer but the following paper may help: [Angenent, Sigurd B. Curve shortening and the topology of closed geodesics on surfaces. Ann. of Math. (2) 162 (2005), no. 3, 1187--1241.] The point is to start with a configuration of curves and use curve shortening flow to flow the configuration to a closed geodesic.

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Fix your favorite point in the surface. Look at your favorite preimage $x$ of that point in the universal cover. Now take all the lines going through that point and any other preimage (for each other preimage there is a unique such line). Remember that all preimages are just the orbits of $x$ under the covering group $\Gamma$. Those are all the closed geodesics passing through your favorite point.

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How do you know that the geodesic returns to the point at the correct angle? I don't believe there is a closed geodesic passing through each point on a hyperbolic surface. –  Deane Yang Feb 2 '10 at 1:25
    
Yup, there's only countably many closed geodesics, and a surface isn't a union of countably many 1-dimensional submanifolds by basic measure theory. –  Ryan Budney Feb 2 '10 at 1:32
    
In fact, there is a unique closed geodesic in each non-trivial free homotopy class of closed curves on the surface, so there are only countably many closed geodesics. –  Mariano Suárez-Alvarez Feb 2 '10 at 1:35
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Ben is probably thinking about a closed geodesic arc where one corner is allowed. –  Ryan Budney Feb 2 '10 at 1:39
    
On the other hand the set of unit tangents to closed geodesics is dense in the unit tangent bundle to the surface. –  Sam Nead Feb 2 '10 at 15:36

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