Dirichlet Nonvanishing

Chris Birkbeck David Loeffler Michael Stoll

0.1 What this project is about

We’re trying to prove the following theorem:

Let $N \ge 1$ and $\chi $ a complex-valued Dirichlet character mod $N$. Then $L(\chi , 1 + it) \ne 0$ for all real $t$.

See Theorem 12 below.

0.2 High-level outline

If $t \ne 0$ or if $\chi ^2 \ne 1$ then this follows from the product for $\zeta (s)^3 L(\chi , s)^{-4} L(\chi ^2, s)$ which is in EulerProducts. So we may assume that $t = 0$ and $\chi $ is quadratic.
Assume for contradiction $L(\chi , 1) = 0$. Then the function

\[ F(s) = L(\chi , s) \zeta (s) \]

is entire.
For $\Re (s) {\gt} 1$, $F(s)$ is given by a convergent Euler product with local factors XXX. Hence its Dirichlet series coefficients are positive real numbers.
Hence the iterated derivatives of $F(s)$ on $(0, \infty )$ alternate in sign.
By an analytic result from EulerProducts, this implies that $F(s)$ is real and non-vanishing for all real $s$.
However, $F(-2) = 0$. This gives the desired contradiction.

0.3 A more detailed plan

Lemma 1

✓

Let $\chi $ be a Dirichlet character modulo $N$. Then for all $\varepsilon {\gt} 0$, we have

\begin{equation} \label{eqn:product} |L(1, 1 + \varepsilon )^3 L(\chi , 1 + \varepsilon + it)^4 L(\chi ^2, 1 + \varepsilon + 2it)| \ge 1 \, . \end{equation}

Proof ▶

This follows from a trigonometric inequality.

Lemma 2

✓

Let $t \in \mathbb {R}$ and let $\chi $ be a Dirichlet character. If $t \ne 0$ or $\chi ^2 \ne 1$, then

\[ L(\chi , 1 + it) \ne 0 \, . \]

Proof ▶

Assume that $L(\chi , 1 + it) = 0$. Then the (at least) quadruple zero of $L(\chi , s)^4$ at $1 + it$ will more than compensate for the triple pole of $L(1, s)^3$ at $1$, so the product of the first two factors in 1 will tend to zero as $\varepsilon \searrow 0$. If $t \ne 0$ or $\chi ^2 \ne 1$, then the last factor will have a finite limit, and so the full product will converge to zero, contradicting Lemma 1.

So it suffices to prove that if $\chi $ is a quadratic character we have $L(\chi , 1) \ne 0$.

Definition 3

✓

A bad character is an $\mathbb {R}$-valued (hence quadratic) Dirichlet character such that $L(\chi , 1) = 0$.

Definition 4

✓

Define $F \colon \mathbb {C} \to \mathbb {C}$ by

\[ F(s) = \begin{cases} \zeta (s) L(\chi , s) & \text{if $s \ne 1$} \\ L’(\chi , 1) & \text{if $s = 1$} \end{cases} \]

Lemma 5

✓

If $\chi $ is a bad character, then $F$ is an entire function.

Proof ▶

This is easy for $s \ne 1$ since we know analyticity of both factors. To prove analyticity at $s = 1$, it suffices to show continuity (Riemann criterion) and that should follow easily since we know that $\lim _{s \to 1} (s - 1) \zeta (s)$ exists.

Lemma 6

✓

We have $F(-2) = 0$.

Proof ▶

Follows from the trivial zeroes of Riemann zeta.

Lemma 7

✓

For $\Re (s) {\gt} 1$, $F(s)$ is equal to the $L$-series of a real-valued arithmetic function $e$ defined as the convolution of constant 1 and $\chi $.

Proof ▶

We have Euler products for both $L(\chi , s)$ and $\zeta (s)$.

Lemma 8

✓

The weakly multiplicative function $e(n)$ whose Euler product is $\mathcal{E}$ takes non-negative real values.

Proof ▶

It suffices to show this for prime powers. We have $e(p^k) = (k + 1)$ if $\chi (p) = 1$, $e(p^k) = 1$ if $\chi (p) = 0$, and if $\chi (p) = -1$ then $e(p^k) = 1$ if $k$ even, $0$ if $k$ odd.

Lemma 9

✓

An entire function $f$ whose iterated derivatives at $s$ are all real with alternating signs (except possibly the value itself) has values of the form $f(s) + \text{nonneg. real}$ along the nonpositive real axis shifted by $s$.

Proof ▶

This follows by considering the power series expansion at zero.

Lemma 10

✓

If $a \colon \mathbb {N}\to \mathbb {C}$ is an arithmetic function with $a(1) {\gt} 0$ and $a(n) \ge 0$ for all $n \ge 2$ and the associated $L$-series $f(s) = \sum _{n \ge 1} a(n) n^{-s}$ converges at $x \in \mathbb {R}$, then the iterated derivative $f^{(m)}(x)$ is nonnegative for $m$ even and nonpositive for $m$ odd.

Proof ▶

The $m$th derivative of $f$ at $x$ is given by

\[ f^{(m)}(x) = \sum _{n=1}^\infty (-\log n)^m a(n) n^{-x} = (-1)^m \sum _{n=1}^\infty (\log n)^m a(n) n^{-x} \, , \]

and the last sum has only nonnegative summands.

Lemma 11

✓

If $\chi $ is a nontrivial quadratic Dirichlet character, then $L(\chi , 1) \ne 0$.

Proof ▶

Assume that $L(\chi , 1) = 0$, so $\chi $ is a bad character. By Lemma 5, we then know that $F$ is an entire function. From Lemmas 7 and 8 we see that $F$ agrees on $\Re s {\gt} 1$ with the $L$-series of an arithmetic function with nonnegative real values (and positive value at $1$). Lemma 10 now shows that $(-1)^m F^{(m)}(2) \ge 0$ for all $m \ge 1$. Then Lemma 9 (applied to $f(s) = F(2+s)$) implies that $F(x) {\gt} 0$ for all $x \le 2$. This now contradicts Lemma 6, which says that $F(-2) = 0$. So the initial assumption must be false, showing that $L(\chi , 1) \ne 0$.

Theorem 12

✓

If $\chi $ is a Dirichlet character and $t$ is a real number such that $t \ne 0$ or $\chi $ is nontrivial, then $L(\chi , 1 + it) \ne 0$.

Proof ▶

If $\chi $ is not a quadratic character or $t \ne 0$, then the claim is Lemma 2. If $\chi $ is a nontrivial quadratic characters and $t = 0$, then the claim is Lemma 11.