Chapter 7 Functional equations for Dirichlet \(L\)-functions
Section 7.1 Even characters
For \(\chi\) even, we can derive a functional equation for \(L(s,\chi)\) by imitating the argument we used for \(\zeta\text{.}\) Start with
\begin{equation*}
\chi(n) \pi^{-s/2} N^{s/2}
\Gamma(s/2) n^{-s} = \int_0^\infty \chi(n) e^{-\pi n^2 x/N}
x^{s/2-1}\,dx
\end{equation*}
and sum over \(n\) to obtain
\begin{equation}
\pi^{-s/2} N^{s/2}
\Gamma(s/2) L(s,\chi) = \frac{1}{2}
\int_0^\infty x^{s/2-1} \theta(x,\chi)\,dx\tag{7.1.1}
\end{equation}
for
\begin{equation*}
\theta(x,\chi) := \sum_{n=-\infty}^\infty \chi(n) e^{-\pi n^2 x/N}.
\end{equation*}
(Notice there is no additive constant because \(\chi(0) = 0\text{.}\))
Applying the Poisson summation formula (Theorem 6.6) to \(\theta(x, \chi)\) looks problematic, because \(\chi(n)\) does not extend nicely to a function on all of \(\RR\text{.}\) Fortunately we can avoid this by doing a bit more Fourier analysis, but this time on the additive group \(\ZZ/N\ZZ\text{:}\) write
\begin{equation*}
\chi(n) = \sum_{m=1}^N c_{\chi,m} e^{2\pi i mn/N}
\end{equation*}
with
\begin{equation*}
c_{\chi,m} := \frac{1}{N} \sum_{l=1}^N \chi(l) e^{-2 \pi i lm/N}.
\end{equation*}
With a bit of fiddling (Exercise 7.3.6), we obtain
\begin{equation}
\theta(x, \chi) = (N/x)^{1/2} \sum_{m=1}^N c_{\chi,m} \sum_{n=-\infty}^\infty
e^{-\pi (nN + m)^2/(xN)}.\tag{7.1.2}
\end{equation}
To get further, we need a description of the \(c_{\chi,m}\) which is uniform in \(m\text{:}\) if \(m\) is coprime to \(N\text{,}\) then
\begin{align*}
c_{\chi,m} \amp= \frac{1}{N} \sum_{l=1}^N \chi(l) e^{-2\pi i lm/N}\\
\amp= \frac{1}{N} \sum_{l=1}^N \overline{\chi(m)} \chi(lm) e^{-2\pi i lm/N}\\
\amp= \overline{\chi}(m) c_{\chi,1}.
\end{align*}
For \(m\) not coprime to \(N\text{,}\) we must assume \(\chi\) is primitive, and then again
\begin{equation}
c_{\chi,m} = \overline{\chi}(m) c_{\chi,1}\tag{7.1.3}
\end{equation}
but this is not so obvious; see exercises.
This gives us
\begin{equation*}
\theta(x,\chi) = (N/x)^{1/2} c_{\chi,1} \theta(x^{-1}, \overline{\chi}),
\end{equation*}
and now we are home free: again split the integral (7.1.1) at 1 and substitute \(x \mapsto x^{-1}\) in one term, to obtain
\begin{align*}
\pi^{-s/2} N^{s/2} \Gamma(s/2) L(s,\chi) \amp= \frac{1}{2} \int_1^\infty x^{s/2-1} \theta(x,\chi)\,dx
+ \frac{1}{2} \int_1^\infty x^{-s/2-1} \theta(x^{-1},\chi)\,dx\\
\amp= \frac{1}{2} \int_1^\infty x^{s/2-1}\theta(x,\chi)\,dx + \frac{1}{2} N^{1/2} c_{\chi,1} \int_1^\infty x^{(1-s)/2 - 1}
\theta(x, \overline{\chi})\,dx.
\end{align*}
Similarly,
\begin{equation*}
\pi^{-s/2} N^{s/2} \Gamma(s/2) L(s,\overline{\chi}) =
\frac{1}{2} \int_1^\infty x^{s/2-1}\theta(x,\overline{\chi})\,dx
+ \frac{1}{2} N^{1/2} c_{\overline{\chi},1} \int_1^\infty x^{(1-s)/2 - 1}
\theta(x, \chi)\,dx.
\end{equation*}
It is elementary to check that \(c_{\chi,1} c_{\overline{\chi},1} = N^{-1}\) (see exercises); we thus obtain
\begin{equation}
\pi^{-(1-s)/2} N^{(1-s)/2} \Gamma((1-s)/2) L(1-s, \overline{\chi})
= N^{1/2} c_{\overline{\chi},1} \pi^{-s} N^{s/2} \Gamma(s/2) L(s,\chi).\tag{7.1.4}
\end{equation}
Again, the extra factors of \(\pi, N, \Gamma\) should be thought of as an “extra Euler factor” coming from the “prime at infinity”.
Pay close attention to the fact that unless \(\chi = \overline{\chi}\text{,}\) the functional equation (7.1.4) relates two different \(L\)-functions. In a few circumstances, this makes it less useful than if it related a single \(L(s,\chi)\) to itself, but so be it.
Also note that quantity \(c_{\chi,1}\) is related to the more commonly introduced Gauss sum associated to \(\chi\text{:}\)
\begin{equation*}
\tau(\chi) = N c_{\overline{\chi},1} = \sum_{l=1}^N
\chi(l) e^{2\pi i l/N}.
\end{equation*}
For more about Gauss sums, see the exercises.
Section 7.2 Odd characters
We have to do something different if \(\chi(-1) = -1\text{,}\) as then the function \(\theta(x,\chi)\) as defined above is identically zero. Instead we use
\begin{equation*}
\theta_1(x, \chi) = \sum_{n=\infty}^\infty n \chi(n) e^{-n^2 \pi x/N}
\end{equation*}
and shift \(s\) around a bit. Namely,
\begin{equation*}
\pi^{-(s+1)/2} N^{(s+1)/2} \Gamma((s+1)/2) L(s,\chi) =
\frac{1}{2} \int_0^\infty \theta_1(x,\chi) x^{(s+1)/2 - 1}\,dx.
\end{equation*}
Again you split the integral at \(x=1\) and use an inversion formula; this time the right identity is
\begin{equation}
\sum_{n=-\infty}^\infty n e^{-n^2\pi x/N + 2\pi i mn/N} =
i (N/x)^{3/2} \sum_{n=-\infty}^\infty \left( n + \frac{m}{N} \right) e^{-\pi (n+m/N)^2 N/x}.\tag{7.2.1}
\end{equation}
You should end up with the functional equation
\begin{equation}
\pi^{-(2-s)/2} N^{(2-s)/2} \Gamma((2-s)/2) L(1-s, \overline{\chi})
= -i \tau(\overline{\chi})N^{-1/2} \pi^{-(1+s)/2} N^{(1+s)/2} \Gamma((1+s)/2) L(s,\chi).\tag{7.2.2}
\end{equation}
Since this is now the third time through this manner of argument, I leave further details to the exercises (Exercise 7.3.7).
Exercises 7.3 Exercises
1.
Prove the following functional equations for \(\Gamma\text{:}\)
\begin{align*}
\Gamma(s) \Gamma(1-s) \amp= \frac{\pi}{\sin \pi s}\\
\Gamma(s) \Gamma(s + 1/2) \amp= 2^{1-2s} \pi^{1/2} \Gamma(2s).
\end{align*}
Then use these to give a simplified functional equation for \(\zeta\) of the form “\(\zeta(1-s)\) equals \(\zeta(s)\) times some explicit function.”2.
Prove that (7.1.3) holds for \(\chi\) primitive whether or not \(m\) is coprime to \(N\text{.}\)3.
Prove that for \(\chi\) primitive, \(\tau(\chi) \overline{\tau(\chi)} = N\text{.}\) (Warning: the value of \(\tau(\chi) \tau(\overline{\chi})\) depends on whether \(\chi\) is even or odd.) Then exhibit an example where this fails if \(\chi\) is imprimitive.4.
For \(\chi\) a Dirichlet character of level \(N\text{,}\) based on the functional equation, where does \(L(s,\chi)\) have zeroes and poles in the region \(\Real(s) \leq 0\text{?}\)5.
Use Poisson summation to prove that
\begin{equation*}
\sum_{n=-\infty}^\infty e^{-(n+\alpha)^2 \pi/x} = x^{1/2} \sum_{n=-\infty}^\infty
e^{-n^2 \pi x + 2\pi i n \alpha} \qquad (\alpha \in \RR, x > 0),
\end{equation*}
then prove (7.2.1) by the same method.6.
Use the previous exercise to deduce (7.1.2).7.
Prove the functional equation (7.2.2).8.
Pick an example of a nonprincipal nonprimitive character \(\chi\text{,}\) and write out the functional equation for \(L(s,\chi)\text{.}\)9. (Dirichlet).
For \(a,b \in \ZZ\) and \(f\colon \RR \to \CC\) a function obtained by taking a continuous function on \([a,b]\) and setting its other values to 0, the Poisson summation formula still holds if interpreted as
\begin{equation}
\frac{1}{2} f(a) + f(a+1) + \cdots + f(b-1) + \frac{1}{2} f(b) =
\sum_{n=-\infty}^\infty \hat{f}(n).\tag{7.3.1}
\end{equation}
Taking (7.3.1) as a given, apply it to the function
\begin{equation*}
f(t) = \begin{cases} e^{2\pi i t^2/N} \amp t \in [0,N] \\
0 \amp \mbox{otherwise}
\end{cases}
\end{equation*}
in order to evaluate \(\sum_{n=1}^N e^{2 \pi i n^2/N}\) for \(N\) a positive integer. Then use this to compute \(G(\chi)\) for \(\chi\) the quadratic character \(\chi(m) = \left( \frac{m}{p} \right)\text{.}\) (You may wish to check your work by computing \(G(\chi)^2\) in a more elementary way.)
