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I'm reading the paper 'Uniform distribution of eigenfunctions on compact hyperbolic surfaces' by Steven Zelditch (Duke Mathematical Journal, 55, pp. 919-941 (1987), MR916129, Zbl 0643.58029) and am having trouble figuring out one line on page 933, in the proof of lemma 4.1. The line in question is about an integral which comes up when trying to prove a trace formula. But first, some set-up:

We have a compact hyperbolic surface $\Gamma\setminus D$, where $D$ is the open unit disk in $\mathbb{C}$ and $\Gamma$ is a discrete co-compact subgroup of $SU(1,1)$. Given a particular $\gamma\in \Gamma$ let $F_\gamma$ be a fundamental domain of $C_\Gamma(\gamma)$, the centralizer of $\gamma$ in $\Gamma$. Furthermore, given $z\in D$ and $b\in\partial D$, define $\langle z,b\rangle= \frac{1}{2}\log\left(\frac{1-|z|^2}{|b-z|^2}\right)$ (which is also the distance from the origin to the horocycle containing $z$ and $b$).

Now, given a symbol $a\in C^\infty_c(\Gamma\setminus D)$, we have to estimate the integral

$$ I_\gamma(t)=\int_{\partial D}\int_{F_\gamma}\int_{0}^\infty a(z,b)e^{-i\mu t}e^{(i\mu+1)\langle z,b\rangle}e^{(-i\mu+1)\langle \gamma z,b\rangle}\mu\tanh(\frac{\pi}{2}\mu) d\mu dz db. $$

(This looks a little different in Zelditch's paper; he uses notation for $\mu\tanh(\frac{\pi}{2}\mu)d\mu$ and I've changed the order of integration). Zelditch then claims that after a straightforward application of stationary phase we get an expansion

$$ I_\gamma(t)=\left(\int_{\gamma_0}a\right)\cfrac{\delta(t-L_\gamma)}{\sinh\left(\frac{L_\gamma}{2}\right)}+R_\gamma(t), $$

where $R_\gamma$ is in $L^1_{\textrm{loc}}$ and $\gamma_0$ is "the" prime geodesic in $F_\gamma$ (I do not understand why we can say "the" prime geodesic in $F_\gamma$; is there only one?). I believe that $L_\gamma$ is the length of the geodesic connecting $0$ to $\gamma\cdot 0$, but it also might be the length of $\gamma_0$.

Basically, I do not see how to go from the integral to the given expansion. Applying stationary phase in $t$ doesn't work as the phase function is nowhere stationary. I also thought of integrating $z,b$ first and treating $\mu$ as the parameter for stationary phase, but I could not get out the details. I would basically like to see why the integral of $a$ is concentrated on $\gamma_0$, where the $\delta(t-L_\gamma)$ term comes from and how to show the remainder is locally summable.

Thanks for any help. I have only ever used stationary phase to do things like finding asymptotic expansions of bessel functions or proving general estimates in semi-classical quantization. I would really like to understand this particular application in detail.

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    $\begingroup$ >"the" prime geodesic< Since the surface is compact, every element of $\Gamma$ is hyperbolic. Also, any centralizer is a cyclic subgroup. From it it is easy to see in the upper half-plane model, after transforming the axis of the centralizer into a vertical line, that $F_\gamma$ is a domain between two concentric semi-circles and the prime geodesic is indeed unique [namely the part of the vertical line in the domain], any other geodesic segment will not glue up in the quotient. $\endgroup$ Commented Aug 6 at 2:58
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    $\begingroup$ I would say it is more of "non stationary phase"/orthogonality. One clearly needs to integrate $d\mu$ (as the result is $t$ dependent). Now we see that when one fixes the $t$ value to the be value of Busemann cocycle (which will give you the length of the geodesic), it fixes $\mu$ also by orthogonality. The rest of $dzdb$ computation is basically verbatim as the regular computation of the orbital integral $\endgroup$
    – Asaf
    Commented Nov 25 at 18:52

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