# Residue of Rankin-Selberg L-function for non-trivial nebentypus

Let $f\in S_k(\Gamma_0(N),\chi)$ be a normalized holomorphic newform (i.e. weight $k$, level $N$, nebentypus $\chi$) and write its Fourier expansion as $$f(z)=\sum_{n\ge 1} \lambda_f(n)n^{(k-1)/2}e^{2\pi i n z}$$ for $\Im z >0$ where $\lambda_f(1)=1$ and, by Deligne's bound, $|\lambda_f(n)|\le d(n)$. Here $d(n)$ denotes the number of positive divisors of $n$. Define the Rankin-Selberg convolution $L$-function as $$L(s,f\times \bar{f}) = \sum_{n=1}^\infty \frac{|\lambda_f(n)|^2}{n^s}$$ for $\Re s >1$. This $L$-function has a simple pole at $s=1$.

Questions:

(1) What is this residue?

(2) How does one compute this residue?

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I think the residue is $\langle f,\overline{f}\rangle$ (the Petersson inner product). Henri Darmon thought a course on L-functions last semester and there was a portion of the course devoted to the Rankin-Selberg method. Have a look at this: math.mcgill.ca/darmon/courses/nt/notes/lecture22.pdf (although it seems you don't have the same normalisation). – M Turgeon Jul 18 '12 at 20:41
I think the residue is $\langle f,\bar{f}\rangle$ for the completed $L$-function, though obviously the two residues are related. Also, it seems that Darmon is handling the case where the character is trivial. – Jim Jul 19 '12 at 0:00
Are you sure that it has a pole at $s = 1$? I think the pole is actually at $s = k$. – David Loeffler Jul 19 '12 at 7:08
@David: He is using the "analytic" normalization so that the critical line is $\Re s =1/2$. – Jim Jul 19 '12 at 18:04
@MTurgeon, Could you send to me this notes, the link doesn't work with me – Med Feb 17 at 18:11

$$\operatorname{res}_{s=1} L(s,f\times\overline f) =\lim_{x\rightarrow\infty}\frac1x\sum_{n\leq x}\left|\lambda_f(n)\right|^2.$$

This fact may be found in a paper by Micah Milinovich and Nathan Ng on the arXiv at arXiv:1306.0854 [math.NT]. (See the remark on page 4.)

In the simple case that $f$ is a normalized primitive holomorphic cusp form on the full modular group, then $f$ is self dual and $$L(s,f\times\overline f)=L(s,f\times f) = \sum_1^\infty \lambda_f^2(n)n^{-s}.$$ Then $$\operatorname{res}_{s=1} L(s,f\times f)=Z(1,f)M/\varphi(M)$$ (equation (2.35) in `Low-lying zeros of families of $L$-functions' by Iwaniec-Luo-Sarnak [note they use the notation $L(s,f\otimes f)$ for what we describe as $L(s,f\times f)$).

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As you have said, let $f$ we a weight $k$ form of nebentypus $\chi$. Then in particular, $f(\gamma z) = (cz + d)^k \chi(d) f(z)$, where $\gamma = \left(\begin{smallmatrix} a&b\\c&d \end{smallmatrix}\right)$ is a matrix in our congruence subgroup du jour.

Notice that $\lvert f(z) \rvert^2 y^k$ is invariant under the slash operator, as \begin{align} \lvert f(\gamma z) \rvert^2 (\gamma y)^k &= f(\gamma z) \overline{f(\gamma z)} \frac{y^k}{\lvert cz + d \rvert^{2k}} \\ &= (cz+d)^k \overline{(cz + d)^k} \chi(d) \overline{\chi(d)} f(z) \overline{f(z)} \frac{y^k}{\lvert cz + d\rvert^{2k}} \\ &= \lvert f(z) \rvert^2 y^k. \end{align}

This means that it is meaningful to take the inner product against the normal Eisenstein series $$E(z,s) = \sum_{\gamma \in \Gamma_\infty \backslash \Gamma_0(N)} \text{Im}(\gamma z)^s.$$

Performing an unfolding of the integral along the critical strip, we can see that $$\langle \lvert f \rvert^2 y^k, E(z,s) \rangle = \frac{\Gamma(s + k - 1)}{(4 \pi)^{s + k - 1}} \sum_n \frac{a(n)^2}{n^{s + k - 1}},$$ where $a(n) = \lambda(n)n^{(k-1)/2}$ are the Fourier coefficients of $f$. Notice the sum on the right is the Rankin-Selberg $L$-function, and I deliberately avoid indicating any normalization anywhere.

As we understand the analytic behaviour of the Eisenstein series and the $\Gamma$ function, we can understand the analytic behavior of the $L$-function. Most importantly, since the Eisenstein series has a single pole at $s = 1$ of known residue, we understand the pole of the $L$-function, and it has residue $$\langle fy^{k/2}, fy^{k/2} \rangle R \frac{(4\pi)^k}{\Gamma(k)},$$ where $R$ is the residue of the Eisenstein series attached to the congruence subgroup (and often looks something like $\frac{3}{\pi}$).

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Ah, I see suddenly that this is really old, and the other answerer edited his question for... some reason. So this might never be read or help anyone. Go figure. – mixedmath Nov 20 '14 at 4:08