Is the spectrum of closed geodesics in a closed hyperbolic 3-manifold asymptotically homogeneously dense? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-19T18:12:08Zhttp://mathoverflow.net/feeds/question/41492http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/41492/is-the-spectrum-of-closed-geodesics-in-a-closed-hyperbolic-3-manifold-asymptoticaIs the spectrum of closed geodesics in a closed hyperbolic 3-manifold asymptotically homogeneously dense?Bruno Martelli2010-10-08T08:44:21Z2012-03-07T22:56:53Z
<p>It seems to me that the results in <a href="http://arxiv.org/abs/0910.5501" rel="nofollow">this important paper of Kahn-Markovich</a> 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 $R+\epsilon$. </p>
<p>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 $n$?</p>
http://mathoverflow.net/questions/41492/is-the-spectrum-of-closed-geodesics-in-a-closed-hyperbolic-3-manifold-asymptotica/41521#41521Answer by Richard Kent for Is the spectrum of closed geodesics in a closed hyperbolic 3-manifold asymptotically homogeneously dense?Richard Kent2010-10-08T16:41:04Z2010-10-08T16:50:27Z<p>I think this should just follow from the exponential mixing of the geodesic flow (due to Pollicott).</p>
<p>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 </p>
<p>$\Big| \int_{T^1M} (g_t^*f)h - (\int_{T^1M} f)(\int_{T^1M}h ) \Big | \leq C e^{-qt}$</p>
<p>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.</p>
<p>I played fast and loose there with the constants, but that's the idea, I think.</p>