3
$\begingroup$

In what follows, $\mathfrak{H}$ denotes the upper half plane, and $\Gamma=SL(2,\mathbb{Z})$. On the modular curve $\Gamma\backslash\mathfrak{H}$, consider the non-holomorphic Poincare series, defined by $$P_{m}(z,s)=\sum_{\gamma\in \Gamma_{\infty}\backslash\Gamma}\text{Im}(\gamma z)^s\cdot e^{2\pi im\gamma z}$$ I am interested in finding a reference for the Fourier expansion for this Poincare series. Any references or suggestions would be greatly appreciated. Secondly, I am curious as to whether these Poincare series are in $L^2(\Gamma\backslash\mathfrak{H})$ for $m\neq 0$.

$\endgroup$

2 Answers 2

4
$\begingroup$

Computation of the Fourier expansion of non-holomorphic Poincare series is due to Selberg, in his survey paper "On the estimation of Fourier coefficients of modular forms. 1965 Proc. Sympos. Pure Math., Vol. VIII pp. 1–15 Amer. Math. Soc., Providence, R.I.". Although, I could not find an online version of it.

However, you can find a similar/general result in Yoshida's paper (Theorem B). Also, yes, $P_m$ are indeed $L^2$ functions on the modular surface.

$\endgroup$
0
4
$\begingroup$

This is indeed due to Selberg. We indicate a method of proof in my book with F.~Stromberg: the result is as follows (simplified to your case, but it is easy to generalize to the case of weight $k$ and two characters $\chi_1$ and $\chi_2$): $$P_n(\tau,s)=y^se^{2\pi in\tau}+(4\pi^2y)^s\sum_{m\in\Bbb Z}|m|^{2s-1}b_m^n(y,s)e^{-2\pi|m|y}e^{2\pi imx}\text{ with}$$ $$b_m^n(y,s)=\sum_{j=0}^\infty\dfrac{(-4\pi^2mn)^j}{j!\Gamma(j+s)}U(s,j+2s,4\pi my)Z_{n,m}(2s+2j)\text{ if $m\ge0$},$$ $$b_m^n(y,s)=\sum_{j=0}^\infty\dfrac{(-4\pi^2|m|n)^j}{j!\Gamma(s)}U(j+s,j+2s,4\pi|m|y)Z_{n,-m}(2s+2j)\text{ if $m<0$},$$ with $U$ the standard confluent hypergeometric function and $Z$ is the Selberg $Z$ function defined by $Z_{m,n}(s)=\sum_{c\ge1}K(m,n;c)/c^s$, $K$ Kloosterman sum.

$\endgroup$
1
  • $\begingroup$ many thanks for the helpful response! $\endgroup$ Apr 16, 2018 at 14:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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