# longest simple closed geodesic

I am obviously not familiar with differential geometry. But some times I do want to know detailed answers to the following questions? May someone help?

When will there be a longest simple closed geodesic on a metric space? Of course, this is too general a question. To be more touchable, what is the case for Riemannian manifolds with non-positive curvature?

Or more generally is there any reference for the the relationship between longest/shortest simple closed curve, diameter, area and curvature?

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You'll need some kind of assumption of finiteness of topology at least. For example, by stitching together a sequence of pairs of pants properly, you'll be able to construct a complete, connected surface with curvature $K=-1$ that has a sequence of simple closed geodesics whose lengths go off to infinity. – Robert Bryant Jan 22 '13 at 12:38
It seems that the question is ill-posed. On the 2-dimensional torus, there is in each homology class a simple closed geodesic of shortest length, and this length goes to infinity with the homology class. Maybe you want to restrict to manifolds of positive curvature? (Their fundamental group is finite.) – ACL Jan 22 '13 at 13:03
Thanks Professor Bryant! You mean if we consider a hyperbolic surface which is topologically infinite (the sum of genus, punctures and ideal boundaries is infinite), then it is possible that the lengths tend to infinity? But what about the case for a closed hyperbolic surface of finite genus $g \geq 2$? Is it follows from the compactness that the lengths of simple closed curves are upper bounded? – silktomath Jan 22 '13 at 13:12
@silktomath: The answer to your first question is 'yes'. For the answer to your second (and third) question, which is 'no', see Maryam Mirzakhani's paper Growth of the number of simple closed geodesics on hyperbolic surfaces, Annals of Mathematics, 168 (2008), 97–125, which gives an estimate for how fast the number of such simple closed geodesics grows with given length $L$. Boundedness not only fails, but fails spectacularly on a hyperbolic surface of genus $g>1$. – Robert Bryant Jan 22 '13 at 17:27
@Robert Bryant: Thanks for all comments! I will check the paper you mentioned which might be of further interest. – silktomath Jan 23 '13 at 4:38

The following paper proves that, on any surface homeomorphic to a sphere of unit area, the shortest closed geodesic cannot be longer than $8$: