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What, if anything, can be said about a surface whose geodesics are all algebraic? One example is of course the sphere.

flag	asked Jun 3 2010 at 23:26 lhf 1,875●1●8●19

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...and the plane. – Mariano Suárez-Alvarez Jun 3 2010 at 23:27

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If the surface is compact, then all geodesics have to be closed. There is a very nice book by Besse on manifolds whose geodesics are all closed, but I do not recall if he considers those which are also algebraic. – Mariano Suárez-Alvarez Jun 3 2010 at 23:39

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If you allow a Lorentz metric on the ambient space, the hyperboloid model of the hyperbolic plane. – Will Jagy Jun 3 2010 at 23:53

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Out of curiosity, what is the simplest example of a non-algebraic geodesic? – Pete L. Clark Jun 4 2010 at 0:46

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Pete, on a torus of revolution in $R^3$ there is a mixture of positive and negative Gauss curvature. So, fixing a point on the outer rim, by symmetry arguments one can describe three types of geodesics. If the angle with the outer "equator" is small, repetitive but not winding. If the angle is large enough, nearly orthogonal to the equator, repetive and winding. At a certain critical angle in the middle, the geodesic approaches the inner "equator" but never reaches it. This last type is not algebraic, whatever can happen with the others. – Will Jagy Jun 4 2010 at 1:10

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Steve Huntsman · Answer 1 · 2010-06-04 00:33:42Z UTC

A class of examples is provided by projective spaces, in which case prime closed geodesics are great circles on spheres.

See Klingenberg's Lectures on Closed Geodesics, p. 178, Theorem 5.2.1. The relevant page is not available on Google, but is accessible on Amazon.

Surfaces whose geodesics are algebraic

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