TITLE: Equidistibution of horocycles through Hecke eigenvalues of Maass cusp forms
QUESTION [5 upvotes]: At the end of this very nice post:
http://blogs.ethz.ch/kowalski/2012/05/21/who-needled-buffon/
E. Kowalski talks about the equidistribution of the points $\frac{j+i}{N}$ when $j=1,\dots,N$ and $N$ tends to $\infty$ in the fundamental domain of $SL_2(R)/SL_2(Z)$. 
He says that it "follows here most easily from non-trivial bounds on Hecke eigenvalues of Maass cusp forms".
Can anyone fill in the details of give a good reference? any related thoughts will also be appreciated!

REPLY [4 votes]: Weyl's criterion is that to check equidistribution it is enough to check against a basis of test functions.  Ignoring the non-compactness of the space for the moment, a natural basis is given by the eigenfunctions of the Laplacian, that is by the Maass eigenforms.  These have a Fourier expansion:
$$f(z) = \sum_{n\neq 0} a_n \sqrt{|n|y} K_{ir}(2\pi|n|y)e(nx)\,.$$
where $K_{ir}$ is the $K$-Bessel function and $e(w)=\exp(2\pi iw)$.  It is now easy to compute the average above in those terms: set $y=1/N$ and average $x$ over $\frac{1}{N}\mathbb{Z}/\mathbb{Z}$ -- the average of $e(nx)$ being zero unless $n$ is divisible by $N$, when it is one. Letting $\mu_N(f)$ denote the average of $f$ over the given set, and changing variables via $n=Nm$, we have
$$ \mu_N(f) = \sum_{m\neq 0} a_{Nm} \sqrt{m} K_{ir}(2\pi|m|)\,.$$
The goal is to show that this tends to zero with N.  Since the $K$-Bessel function decays exponentially, by the Bounded Convergence Theorem it is enough to know that the $a_{n}$ decay with $n$.
It now happens that if $f$ is also a Hecke eigenform then the $a_n$ are basically the corresponding Hecke eigenvalues, and then bounds toward the Ramanujan Conjecture show that the Hecke eigenvalues, hence the Fourier coefficients, decay.  It is believed that $a_n$ decay roughly like $n^{-1/2}$, and bounds of the form $n^{-\theta}$ are known.
Returning to the non-compactness, there is also continuous spectrum, but it is explicit (as is its Fourier expansion) so the analogous calculation for it is not hard.