Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

It is known (thanks to Hingston, Bangert, Franks, Birckhoff, etc) that $(S^2, g)$ has lots of primitive closed geodesics for any Riemannian metric $g$ (Riemannian is crucial here, this is not true for Finsler metrics).The question is: does the length spectrum determine the metric? This is obviously a completely different question from the similar question for hyperbolic surfaces, where spectral methods are available. In particular, it is not even obvious that if closed geodesics are the same length, then $g$ is the (appropriately scaled) round metric.

share|improve this question
    
This sounds to me more like an open-ended research question than anything anyone can answer definitively. –  Deane Yang Feb 29 '12 at 17:07
    
Well, I find it hard to believe that people haven't thought about this at all, so any pointers would be appreciated. –  Igor Rivin Feb 29 '12 at 17:09
    
See also the MO question, "Surfaces all of whose geodesics are both closed and simple," mathoverflow.net/questions/28622 , which cites some of the same references provided by BS & alvarezpaiva. –  Joseph O'Rourke Feb 29 '12 at 18:25
    
I stand corrected. The answer is definitely no. –  Deane Yang Feb 29 '12 at 19:05
add comment

2 Answers 2

up vote 8 down vote accepted

Dear Igor,

The answer is that the length spectrum cannot determine the metric, not even the round metric : there are metrics on the sphere all of whose geodesics are closed and of the same length that are not isometric to the round metric. In fact, by a theorem of Guillemin (The Radon transform on Zoll surfaces. Advances in Mathematics 22 (1976), 85–119) there are lots and lots of Zoll surfaces.

If you pass to Finsler, even in the reversible case where there are lots of closed geodesics, there are more examples. For instance: take a three-dimensional normed space where the unit sphere $S$ is smooth and of positive Gaussian curvature (for any auxiliary Euclidean metric), then the length spectrum of $S$ equals the length spectrum of the sphere in the dual normed space (both with the induced length metric). This is in Dual spheres have the same girth, American journal of mathematics 128 (2), 361-371.

share|improve this answer
    
I just wanted to add that I don't know of any examples of Riemannian metrics on the projective plane that have the same length spectrum. For reversible Finsler metrics, the example above with the unit spheres in a normed space and its dual will pass to ${\mathbb RP}^2$. –  alvarezpaiva Feb 29 '12 at 21:54
add comment

As to the question of closed geodesics all of the same length, there are Zoll metrics, so the length spectrum doesn't characterize even the round metric. See e.g. Besse's "Manifolds all of whose geodesics are closed" Springer 1978, or recent work by LeBrun and Mason on twistor methods applied to Zoll metrics.

On the other hand (and extremally opposite case), for a suitably generic metric, there is the Duistermaat-Guillemin trace formula which implies that the length and Laplace spectra determine each other. But there are isospectral non isometric surfaces. I don't know if there are spherical examples (this seems plausible via Sunada -- or rather Buser -- construction), and even if Zoll metrics are isospectral (which I doubt).

Hope this helps.

EDIT : in fact Berger proved that the round 2-sphere is determined by its laplace spectrum inside smooth metrics, so Zoll's aren't all isospectral.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.