# Functional Equations for L-Functions arising from Modular Forms

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

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

\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 ${\frac{k-1}{2}}$ built into the ${\Lambda}$ expression.

As ${f}$ is modular, ${f\left( \left(\begin{smallmatrix} 0&-1\1&0 \end{smallmatrix}\right)iy\right) = (iy)^kf(iy) = f(\frac{-1}{iy}) = f(\frac{i}{y})}$, 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. ${\Lambda(f, s) = i^{-k}\Lambda(f, 1-s)}$, where

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

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

\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} \
&= \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

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

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

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

while on the other hand

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

So, now that we know that ${\displaystyle (-iy)^{-k} f\left( -\frac{r}{c} + \frac{1}{cy}i\right) = f\left( \frac{\overline{r}}{ c} + \frac{iy}{ c} \right)}$, 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} \
&= (-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. $\displaystyle \Lambda\left(f, s, \frac{\overline{r}}{c} \right) = (-i)^{-k} \Lambda\left( f, 1-s, \frac{-r}{ c} \right), \ \ \ \ \ (2)$

where

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

So what, we might ask? This will allow us to convert the ${\overline{r}}$ part of an annoying Kloosterman sum into a much nicer ${r}$, 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.

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