Functional Equations for L-Functions arising from Modular Forms

In this note, I remind myself of the functional equations for the $latex {L}$-functions $latex {\displaystyle \sum_{n\geq 0} \frac{a(n)}{n^s}}$ and $latex {\displaystyle \sum_{n\geq 0} \frac{a(n)}{n^s}e(\frac{n\overline{r}}{c})}$, where $latex {\overline{r}}$ is the multiplicative inverse of $latex {r \bmod c}$.

Let $latex {f}$ be a weight $latex {k}$ modular cusp form, so that $latex {f}$ satisfies $latex {f(\gamma z) = (cz + d)^kf(z)}$, where $latex {\gamma = \left( \begin{smallmatrix} a&b \ c&d \end{smallmatrix} \right)}$ (and $latex {\gamma}$ will always be this), and where the Fourier expansion of $latex {f}$ is $latex {f(z) = \displaystyle \sum_{n \geq 1} a(n) e^{2\pi i n z} = \sum_{n \geq 1}a(n)e(nz)}$. Firstly, we note the relationship between the first L-function $latex {\displaystyle \sum \frac{a(n)}{n^s}}$ and the integral $latex {\displaystyle \int_0^\infty f(iy)y^s \frac{\mathrm{d}y}{y}}$.

\begin{align} \int_0^\infty f(iy)y^{s + \frac{k-1}{2}} \frac{\mathrm{d}y}{y} &= \int_0^\infty \sum_{n \geq 1} a(n)e(iny)y^{s +\frac{k-1}{2}}\frac{\mathrm{d}y}{y} \
&= \sum_{n \geq 1} \int_0^\infty a(n)e^{-2\pi n y}y^{s + \frac{k-1}{2}} \frac{\mathrm{d}y}{y} &\left(y \mapsto \frac{y}{2\pi n}\right) \
&= \frac{1}{(2 \pi n)^{s + \frac{k-1}{2}}} \sum_{n \geq 1} \int_0^\infty a(n)e^{-y}y^{s + \frac{k+1}{2} – 1}\frac{\mathrm{d}y}{y} \
&= \sum_{n \geq 1} \frac{a(n)}{(2 \pi n)^{s + \frac{k-1}{2}}}\Gamma(s + \frac{k+1}{2}) \
&= \frac{\Gamma(s + \frac{k+1}{2})}{(2 \pi)^{s + \frac{k-1}{2}}} \sum_{n \geq 1} \frac{a(n)}{n^{s+\frac{k-1}{2}}} := \Lambda(f, s), \end{align
} where we might notice that the notation has the $latex {\frac{k-1}{2}}$ built into the $latex {\Lambda}$ expression.

As $latex {f}$ is modular, $latex {f\left( \left(\begin{smallmatrix} 0&-1\1&0 \end{smallmatrix}\right)iy\right) = (iy)^kf(iy) = f(\frac{-1}{iy}) = f(\frac{i}{y})}$, so we also have that

\begin{align} \int_0^\infty f(iy) y^{s +\frac{k-1}{2}} \frac{\mathrm{d}y}{y} &= \int_0^\infty f\left( \begin{pmatrix} 0&-1\1&0 \end{pmatrix} iy\right) y^{s + \frac{k-1}{2}} \frac{\mathrm{d}y}{y} \
&= i^{-k} \int_0^\infty f\left( \frac{i}{y} \right) y^{s + \frac{k-1}{2} – k} \frac{\mathrm{d}y}{y} \
&= i^{-k} \int_0^\infty f\left( \frac{i}{y} \right) y^{s + \frac{-k-1}{2}} \frac{\mathrm{d}y}{y} &\left( y \mapsto \frac{1}{y} \right) \
&= i^{-k} \int_0^\infty f(iy) y^{-s + \frac{k+1}{2}} \frac{\mathrm{d}y}{y} \
&= i^{-k} \sum_{n \geq 1} \int_0^\infty a(n) e^{-2\pi n y}y^{\frac{k+1}{2} – s} \frac{\mathrm{d}y}{y} &\left( y \mapsto \frac{y}{2\pi n} \right) \
&= \sum_{n \geq 1} \frac{i^{-k}a(n)}{(2\pi n)^{\frac{k+1}{2} – s}} \int_0^\infty e^{-y} y^{(1-s) + \frac{k+1}{2} – 1} \frac{\mathrm{d}y}{y} \
&= \frac{i^{-k}\Gamma(1 – s + \frac{k+1}{2})}{(2 \pi)^{(1 – s) + \frac{k-1}{2}}}\sum_{n \geq 1} \frac{a(n)}{n^{(1 – s) +\frac{k-1}{2}}} = i^{-k}\Lambda(f, 1-s),
\end{align
}
giving us the following proposition.

Proposition 1. $latex {\Lambda(f, s) = i^{-k}\Lambda(f, 1-s)}$, where

$latex \displaystyle \Lambda(f,s) = \frac{\Gamma(s + \frac{k+1}{2})}{(2 \pi)^{s + \frac{k-1}{2}}} \sum_{n \geq 1} \frac{a(n)}{n^{s+\frac{k-1}{2}}}. \ \ \ \ \ (1)$

Now we do the same sort of idea for the twisted function cusp form $latex {\sum_{n \geq 1} a(n) e\left( \frac{n\overline{r}}{c} \right)e(nz)}$. To examine this, we will look at the integral $latex {\displaystyle \int_0^\infty f\left( \frac{\overline{r}}{ c} + \frac{iy}{ c} \right) y^{s + \frac{k-1}{ 2}} \frac{\mathrm{d}y}{y}}$.

\begin{align} \int_0^\infty f\left( \frac{\overline{r}}{ c} + \frac{iy}{ c} \right) y^{s + \frac{k-1}{ 2}} \frac{\mathrm{d}y}{y} &= \sum_{n \geq 1}\int_0^\infty a(n) e\left( \frac{n\overline{r}}{ c} + \frac{iny}{ c} \right)y^{s + \frac{k-1}{ 2}} \frac{\mathrm{d}y}{y} \
& \qquad \qquad \left( y \mapsto \frac{cy}{ 2\pi n} \right) \
&= \sum_{n \geq 1} \left( \frac{c}{ 2\pi n} \right)^{s + \frac{k-1}{ 2}} \int_0^\infty a(n) e^{\frac{2\pi i n \overline{r}}{ c}} e^{-y}y^{s + \frac{k-1}{ 2}} \frac{\mathrm{d}y}{y} \
&= \left( \frac{c}{ 2\pi} \right) ^{s + \frac{k-1}{ 2}}\sum_{n \geq 1} \frac{a(n)}{n^{s + \frac{k-1}{ 2} } } e\left(\frac{n\overline{r}}{ c} \right) \int_0^\infty e^{-y}y^{s + \frac{k+1}{ 2} -1} \frac{\mathrm{d}y}{y} \
&= \left( \frac{c}{ 2\pi} \right) ^{s + \frac{k-1}{ 2} } \Gamma \left(s + \frac{k+1}{ 2} \right) \sum_{n \geq 1} \frac{a(n)}{n^{s + \frac{k-1}{ 2} } } e\left(\frac{n\overline{r}}{ c} \right) \
&:= \Lambda \left(f, s, \frac{\overline{r}}{c}\right), \end{align
} and where we see that my notation suggests that

$latex \displaystyle L\left(f, s, \frac{\overline{r}}{c}\right) = \sum_{n \geq 1} \frac{a(n)}{n^{s + \frac{k-1}{ 2} } } e\left(\frac{n\overline{r}}{ c} \right). $

Now, as $latex {f}$ is modular, it is invariant under the action of $latex {\left( \begin{smallmatrix} r&\alpha \ c&\overline{r} \end{smallmatrix} \right)}$. So on the one hand,

$latex \displaystyle f\left(\begin{pmatrix} r&\alpha \\ -c&\overline{r} \end{pmatrix}\left( \frac{\overline{r}}{ c} + \frac{iy}{ c} \right) \right) = f\left( -\frac{r}{c} + \frac{1}{cy}i\right), $

while on the other hand

$latex \displaystyle f\left(\begin{pmatrix} r&\alpha \\ -c&\overline{r} \end{pmatrix}\left( \frac{\overline{r}}{ c} + \frac{iy}{ c} \right) \right) = \left(-c\left( \frac{\overline{r}}{ c} + \frac{iy}{ c} \right) + \overline{r} \right)^kf\left( \frac{\overline{r}}{ c} + \frac{iy}{ c} \right) = (-iy)^kf\left( \frac{\overline{r}}{ c} + \frac{iy}{ c} \right). $

So, now that we know that $latex {\displaystyle (-iy)^{-k} f\left( -\frac{r}{c} + \frac{1}{cy}i\right) = f\left( \frac{\overline{r}}{ c} + \frac{iy}{ c} \right)}$, we can say that

\begin{align} \int_0^\infty f\left( \frac{\overline{r}}{ c} + \frac{iy}{ c} \right) y^{s + \frac{k-1}{ 2} } \frac{\mathrm{d}y}{y} &= \int_0^\infty (-iy)^{-k} f\left( -\frac{r}{ c} + \frac{1}{ cy } i \right) y^{s + \frac{k-1}{ 2}} \frac{\mathrm{d}y}{y} \
&= (-i)^{-k}\int_0^\infty f\left( -\frac{r}{ c} + \frac{1}{ cy } i \right) y^{s + \frac{-k – 1}{ 2} } \frac{\mathrm{d}y}{y} \
&= (-i)^{-k} \sum_{n \geq 0} a(n) e\left( \frac{-rn}{ c} \right) \int_0^\infty e\left( \frac{in}{ cy} \right) y^{s + \frac{-k-1}{ 2} } \frac{\mathrm{d}y}{y} \
& \qquad \qquad \left( y \mapsto \frac{2\pi n}{ cy} \right) \
&= (-i)^{-k}\left( \frac{c}{ 2 \pi} \right) ^{\frac{k + 1}{ 2} -s} \sum_{n \geq 1} \frac{a(n)}{n^{\frac{k+1}{ 2} -s}} e\left( \frac{-rn}{ c} \right) \int_0^\infty e^{-y} y^{\frac{k + 1}{ 2} -s} \frac{\mathrm{d}y}{y} \
&= (-i)^{-k} \left( \frac{c}{ 2 \pi} \right) ^{(1-s) + \frac{k-1}{ 2}}\Gamma\left( (1-s) + \frac{k+1}{ 2} \right) \sum_{n \geq 1}\frac{a(n)}{ n^{(1-s) + \frac{k-1}{ 2} }} e\left( \frac{-rn}{ c} \right) \
&= (-i)^{-k} \Lambda\left(f, 1-s, \frac{-r}{ c} \right). \end{align
}

Proposition 2. $latex \displaystyle \Lambda\left(f, s, \frac{\overline{r}}{c} \right) = (-i)^{-k} \Lambda\left( f, 1-s, \frac{-r}{ c} \right), \ \ \ \ \ (2)$

where

$latex \displaystyle \left( \frac{c}{ 2\pi} \right) ^{s + \frac{k-1}{ 2} } \Gamma \left(s + \frac{k+1}{ 2} \right) \sum_{n \geq 1} \frac{a(n)}{n^{s + \frac{k-1}{ 2} } } e\left(\frac{n\overline{r}}{ c} \right) := \Lambda \left(f, s, \frac{\overline{r}}{c}\right)$

So what, we might ask? This will allow us to convert the $latex {\overline{r}}$ part of an annoying Kloosterman sum into a much nicer $latex {r}$, giving us something that we can handle, in a forthcoming application. (And is something that we’ve done enough that I want to have the computation written down completely and nicely).

This note can be found online at davidlowryduda.com, and in note form here.

This entry was posted in Math.NT, Mathematics and tagged , , , , . Bookmark the permalink.

Leave a Reply