Skip to main content

Analytic Number Theory Course Notes: Math 507, Spring 2024, Boise State University

Zach Teitler

Section 7.5 Dirichlet L-functions

Definition 7.5.1.

Let \(q\) be a positive integer and \(\chi\) a Dirichlet character modulo \(q\text{.}\) We define
\begin{equation} L(\chi,s) = \sum_{n=1}^{\infty} \frac{\chi(n)}{n^s}\text{,}\tag{7.5.1} \end{equation}
called the Dirichlet \(L\)-function associated to \(\chi\text{.}\) Here \(s\) is a complex variable; \(L(\chi,s)\) is a function defined on the subset of \(\bbC\) where this series converges, with values in \(\bbC\text{.}\)
When we think of \(s\) as a complex number, we can write \(s = \sigma + it\text{,}\) where \(\sigma\) is the real part of \(s\) and \(t\) is the imaginary part. However for the proof of Dirichlet’s theorem we will only deal with real values of \(s\text{,}\) so we will mostly just use the letter \(s\text{.}\) (In fact we will mostly just deal with \(s=1\text{.}\))
The series converges absolutely if \(s\) has real part strictly greater than \(1\text{,}\) i.e., if \(\sigma \gt 1\text{.}\) To see this, recall that every value \(\chi(n)\) is either a root of unity, or else zero. Then,
\begin{equation} \sum_{n=1}^{\infty} \left| \frac{\chi(n)}{n^s} \right| \leq \sum_{n=1}^{\infty} \frac{1}{n^\sigma}\text{.}\tag{7.5.2} \end{equation}
We know that this series converges if \(\sigma \gt 1\text{,}\) and diverges if \(\sigma \leq 1\text{,}\) by the theory of \(p\)-series.
However we will also be interested in Dirichlet \(L\)-functions when \(0 \lt \sigma \leq 1\text{.}\) Now the series fails to converge absolutely, but it might still converge conditionally.
Having said that, if the character \(\chi=\chi_0\) is the principal character, then it’s easy to see that \(\sum \chi_0(n)/n^s\) diverges to \(\infty\) if \(\sigma \leq 1\text{.}\) It is not exactly a \(p\)-series because \(\chi_0(n) = 0\) when \((n,q) \gt 1\text{.}\) However, for each arithmetic sequence corresponding to a residue class modulo \(q\text{,}\) we can show that the sum diverges, via a comparison with a \(p\)-series. For example,
\begin{equation*} \sum_{n \equiv 2 \pmod{5}} \left|\frac{1}{n^s}\right| = \frac{1}{2^\sigma} + \frac{1}{7^\sigma} + \frac{1}{12^\sigma} + \dotsb \gt \frac{1}{5^\sigma} + \frac{1}{10^\sigma} + \frac{1}{15^\sigma} + \dotsb = \frac{1}{5^\sigma} \left( \frac{1}{1^\sigma} + \frac{1}{2^\sigma} + \frac{1}{3^\sigma} + \dotsb \right)\text{,} \end{equation*}
where the last sum is a normal \(p\)-series, so it diverges if \(\sigma \leq 1\text{.}\)
So, the Dirichlet \(L\)-series for the principal character diverges if \(\sigma \leq 1\text{,}\) and for nonprincipal characters, it does not converge absolutely if \(\sigma \leq 1\text{.}\) The only thing left to settle is whether the series for a nonprincipal character converges conditionally. And it turns out that it does.

Proof.

We will use Riemann-Stieltjes integration and integration by parts, as we have done before (but it would be equivalent to use Abel summation).
Let \(A(x) = \sum_{n \leq x} \chi(n)\text{.}\) Since \(\chi\) is a nonprincipal character, we get that \(\sum_{n=1}^q \chi(n) = 0\) and in general \(\sum_{n=1}^{kq} \chi(n) = 0\) for every positive integer \(k\text{.}\) Therefore,
\begin{align} |A(x)| \amp \leq \left| \sum_{n=1}^{kq} \chi(n) \right| + \sum_{kq+1 \leq n \lt x} |\chi(n)| \tag{7.5.5}\\ \amp = \sum_{kq+1 \leq n \lt x, (n,q)=1} 1 \tag{7.5.6}\\ \amp \leq \phi(q) \tag{7.5.7}\\ \amp = O(1), \tag{7.5.8} \end{align}
and then by Riemann-Stieltjes integration and integration by parts (or by Abel summation),
\begin{align} \sum_{x \lt n \leq y} \chi(n) f(n) \amp = \int_x^y f(t) dA(t) \tag{7.5.9}\\ \amp = A(y)f(y) - A(x)f(x) - \int_x^y A(t) f'(t) dt \tag{7.5.10}\\ \amp \ll f(y) + f(x) + \int_x^y |f'(t)| dt \tag{7.5.11}\\ \amp = f(y) + f(x) - \int_x^y f'(t) dt \tag{7.5.12}\\ \amp = f(y) + f(x) - (f(y)-f(x)) \tag{7.5.13}\\ \amp = 2f(x) \tag{7.5.14}\\ \amp \ll f(x) \tag{7.5.15} \end{align}
This is the first claim.
(A remark: Above, we wrote \(dA(t)\) even though \(A\) is a complex-valued function. However, our previous discussion of Riemann-Stieltjes integration was only for real-valued functions. To deal with a complex-valued function, we could handle the real and imaginary parts of \(A\) separately, and just recombine them at the end.)
Next, assume that \(f(x) \to 0\) as \(x \to \infty\text{.}\) Since
\begin{equation*} \sum_{x \lt n \leq y} \chi(n) f(n) \ll f(x)\text{,} \end{equation*}
or
\begin{equation*} \sum_{N \lt n \leq M} \chi(n) f(n) \ll f(N)\text{,} \end{equation*}
then these partial sums for \(x \lt n \leq y\) or \(N \lt n \leq M\) go to zero as \(x,N \to \infty\text{.}\) Therefore the infinite series \(\sum_{n=1}^{\infty} \chi(n)f(n)\) converges by the Cauchy criterion.
Finally, the partial sum for \(n \leq x\) is equal to to the sum of the whole series minus the sum of the tail \(n \gt x\text{.}\) And the sum of that tail is equal to the limit of the sum for \(x \lt n \leq y\) as \(y \to \infty\text{.}\) By the first claim, the tail is \(O(f(x))\) regardless of the value of \(y\text{.}\)

Proof.

The previous theorem with functions \(\frac{1}{x}\text{,}\) \(\frac{\log x}{x}\text{,}\) and \(\frac{1}{\sqrt{x}}\) respectively.
A crucial step in the proof of Dirichlet’s theorem is to prove that \(L(1,\chi) \neq 0\text{.}\) Very simply, the idea is to define a certain function \(A(x)\text{,}\) and to show that
  1. \(A(x) \to \infty\) as \(x \to \infty\) (\(A(x)\) is unbounded), but
  2. \(A(x)\) is equal to \(L(1,\chi)\) times some unbounded function, plus \(O(1)\text{.}\)
The point is that if \(L(1,\chi)=0\) then \(A(x)=O(1)\text{,}\) a contradiction. Specifically, this argument works for real-valued nonprincipal characters; we will give a separate argument for non-real nonprincipal characters. We postpone the details.
Very, very simply put, the reason we care that \(L(1,\chi)\) is nonzero is that we will show that \(L(1,\chi)\) times a certain sum is bounded, and then, as long as \(L(1,\chi) \neq 0\text{,}\) we can deduce that the sum itself is bounded.
PrevTopNext