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It seems to me that the results in this important paper of Kahn-Markovich imply the following fact. Let $M^3$ be any closed hyperbolic 3-manifold. For every $\epsilon > 0$ there is a natural number $R(\epsilon)$ such that for every $R>R(\epsilon)>0$ there is a closed geodesic whose lenght is between $R-\epsilon$ and $R+\epsilon$.

That is, the more $R$ grows, the more the spectrum of all geodesics having length more than $R$ becomes uniformly crowded, without holes. Am I right that this is a consequence of their results? If so, is there a more direct way to prove this? Does this property generalize to closed hyperbolic manifolds with arbitrary dimension $n$?

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Do you mean a geodesic whose length is between $R \pm R(\epsilon)$? – Kimball Oct 8 '10 at 13:34
No, I really mean between $R \pm \epsilon$. – Bruno Martelli Oct 8 '10 at 13:58
I guess the $M$ should be "any closed hyperbolic $3$-manifold" – Roberto Frigerio Oct 8 '10 at 14:15
Yes Roberto, thanks (I add "hyperbolic") – Bruno Martelli Oct 8 '10 at 15:26
Oh, got it. Sorry, I misread it. – Kimball Oct 8 '10 at 21:38
up vote 12 down vote accepted

I think this should just follow from the exponential mixing of the geodesic flow (due to Pollicott).

Exponential mixing says that there is a constant $q$ such that if you have two smooth functions $f$ and $h$ on the unit tangent bundle, and $g_t$ is the geodesic flow, then there is a constant $C$ depending on $f$ and $g$ (some function of some Sobolev norm) such that

$\Big| \int_{T^1M} (g_t^*f)h - (\int_{T^1M} f)(\int_{T^1M}h ) \Big | \leq C e^{-qt}$

If you take a vector $v$ and then let $f = h$ be a function with integral $1$ supported on an $\epsilon$-neighborhood of $v$, then you find that there is some constant $T(\epsilon)$ depending only on $M$ and $\epsilon$ such that for any $T \geq T(\epsilon)$, there is a closed path whose length is within $\epsilon$ of $T$ and which is a geodesic except at the basepoint, where it is broken at an angle of $\pi - \epsilon$. If $\epsilon$ is small, this path will be close to a bona fide geodesic, which should give you what you want.

I played fast and loose there with the constants, but that's the idea, I think.

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Thank you, the idea is very clear. – Bruno Martelli Oct 8 '10 at 19:39
Glad to be of some help. – Richard Kent Oct 8 '10 at 21:01

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