I am stuck on this computation of the Fourier coefficients of Eisenstein series. For $\Gamma = SL(2, \mathbb{Z})$ and $\Gamma_\infty = \left\{ \left( \begin{array}{cc} 1 & m \\ 0 & 1 \end{array}\right): m \in \mathbb{Z} \right\}$ we can define an Eisenstein series as the sum over the cosets:

$$ E_\Gamma(z,s) = \sum_{\gamma \in \Gamma_\infty \backslash \Gamma} (y(\gamma z))^s $$

The Fourier coefficients of this Eisenstein series are written in terms of an L-function and Bessel functions:

$$ \int_0^1 E_\Gamma(z,s) e^{2\pi i m x} \, dx= \begin{cases} y^s + \left( \sum_{c \geq 0} \frac{r_{0, \Gamma}(c)}{c^{2s}} \right) \frac{\pi^{1/2}\Gamma(s - \frac{1}{2})}{\Gamma(s)} y^{1-s} & m = 0 \\ \left( \sum_{c \geq 0} \frac{r_{n, \Gamma}(c)}{c^{2s}} \right) \frac{2\pi^s|m|^{s - \frac{1}{2}}K_{s- \frac{1}{2}}(2\pi|m|y)}{\Gamma(s)}& m \neq 0 \end{cases} $$

Not begin an expert on Automorphic forms (in fact a total newbie), I consulted Anton Dietmar's text and found not-quite the same formula:

$$ E(z,s) = \pi^{-s} \Gamma(s) \zeta(2s) \sum_{\Gamma_\infty \backslash \Gamma} \mathrm{Im}(\gamma z)^s $$

This formula looks the same up to some factors of gamma and zeta functions. Then

$$ \int_0^1 E(z,s) e^{2\pi i m x} \, dx= \begin{cases} \pi^{-s}\Gamma(s) \zeta(2s)y^s + \pi^{s-1} \Gamma(1-s)\zeta(2-2s)y^{1-s} & m = 0 \\ 2|m|^{2 - \frac{1}{2}}\sigma_{1-2s}(|m|)\sqrt{y}K_{s - \frac{1}{2}}(2\pi|m|y)& m \neq 0 \end{cases} $$

I don't understand why in the first case I get Ramanujan sums and in the second case, I get the sum-of-divisors function:

$$ r_m(c) = \sum_{(c,d)=1} e^{2\pi i m \frac{d}{c}}$$

Is this consistent? I know the two version of $E$ are written slightly differently, but I have not yet verified that they agree.


1 Answer 1


I think all your questions are answered by the following calculation (assume $m\geq 1$ and $\Re(s)>1$): $$ \sum_{c=1}^\infty\frac{r_m(c)}{c^{2s}} = \sum_{c=1}^\infty\frac{1}{c^{2s}}\sum_{\substack{\text{$d$ mod $c$}\\{(d,c)=1}}}e\left(m\frac{d}{c}\right) = \sum_{c=1}^\infty\frac{1}{c^{2s}}\sum_{\text{$d$ mod $c$}}e\left(m\frac{d}{c}\right)\sum_{n\mid(c,d)}\mu(n) $$ $$ = \sum_{n=1}^\infty\mu(n)\sum_{\substack{c=1\\{n\mid c}}}^\infty\frac{1}{c^{2s}}\sum_{\substack{\text{$d$ mod $c$}\\{n\mid d}}}e\left(m\frac{d}{c}\right) = \sum_{n=1}^\infty\mu(n)\sum_{c=1}^\infty\frac{1}{(nc)^{2s}}\sum_{\text{$nd$ mod $nc$}}e\left(m\frac{nd}{nc}\right)$$ $$ = \sum_{n=1}^\infty\frac{\mu(n)}{n^{2s}}\sum_{c=1}^\infty\frac{1}{c^{2s}}\sum_{\text{$d$ mod $c$}}e\left(m\frac{d}{c}\right) = \sum_{n=1}^\infty\frac{\mu(n)}{n^{2s}}\sum_{c\mid m}\frac{c}{c^{2s}} = \frac{\sigma_{1-2s}(m)}{\zeta(2s)}.$$

P.S. The above calculation is classical, e.g. see (1.5.4) in Titchmarsh: The theory of the Riemann zeta-function.

  • 2
    $\begingroup$ One might add that even these relatively simple classical calculations may be easier if/when one realizes that everything factors over primes, so that a "local" computation suffices. Easier to avoid getting lost, maybe. $\endgroup$ Jul 13, 2015 at 2:09

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