Siegel–Walfisz vocabulary: $\psi(N;q,a)$, the zero-free region and exceptional sets

Davenport.psiAP

psiAP takes three natural-number arguments $N$, $q$, $a$ and returns the real number

where $\Lambda$ is the von Mangoldt function ($\Lambda(n) = \log p$ when $n$ is a positive power of a prime $p$, and $\Lambda(n)=0$ otherwise, in particular $\Lambda(0)=\Lambda(1)=0$), its value being coerced from the extended-real/real-valued arithmetic function into $\mathbb{R}$. The index $n$ runs over the finite range $\{0,1,\dots,N-1\}$: the lower endpoint $0$ is included and the upper endpoint $N$ is excluded, so the sum is over $n < N$ strictly, and $N = 0$ gives the empty sum $0$. The congruence test is performed by reducing both $n$ and $a$ from $\mathbb{N}$ into $\mathbb{Z}/q\mathbb{Z}$ and asking for equality there; consequently $a$ is not required to lie in $\{0,\dots,q-1\}$ (it is reduced modulo $q$), and no coprimality hypothesis $\gcd(a,q)=1$ is imposed anywhere — $a$ may share a factor with $q$, or be $0$. Two degenerate moduli are silently included. If $q = 0$ then $\mathbb{Z}/0\mathbb{Z}$ is $\mathbb{Z}$ and the test becomes the equality of integers $n = a$, so the sum collapses to the single term $\Lambda(a)$ when $a < N$ and to $0$ when $a \ge N$. If $q = 1$ then $\mathbb{Z}/1\mathbb{Z}$ is trivial, the test always succeeds, and the sum is $\sum_{n<N}\Lambda(n)$, i.e. the full Chebyshev sum over $n < N$ with no congruence restriction at all. There is no hypothesis $N > 0$, $q > 0$, or any relation between $N$, $q$ and $a$.

Davenport.regionBoundary

regionBoundary takes a real number $c$, a natural number $q$ and a complex number $s$, and returns the real number

where $q$ is coerced from $\mathbb{N}$ to $\mathbb{R}$ and the logarithm is the real logarithm. The value depends on $s$ only through $|\operatorname{Im} s|$; the real part of $s$ plays no role in the formula. The argument of the logarithm is $q(|t|+2)$ with $t = \operatorname{Im} s$, which is $\ge 2q$ and hence never negative. Edge cases arising from total-function conventions: if $q = 0$ the argument of the logarithm is $0$, the convention $\log 0 = 0$ applies, and division by $0$ yields $0$, so $\beta(c,0,s) = 1$ for every $c$ and every $s$. If $q \ge 1$ then $q(|t|+2) \ge 2 > 1$, so the logarithm is strictly positive and no division by zero occurs; in that case $\beta < 1$ when $c > 0$, $\beta = 1$ when $c = 0$, and $\beta > 1$ when $c < 0$. No hypothesis constrains $c$ to be positive, small, or nonzero, and no hypothesis constrains $q$.

Davenport.InRegion

InRegion takes a real number $c$, a natural number $q$ and a complex number $s$ and is the proposition

that is, $\beta(c,q,s) \le \operatorname{Re} s$ with the boundary value of the previous definition. The inequality is non-strict, so points exactly on the boundary curve $\operatorname{Re} s = \beta(c,q,s)$ belong to the region, and the region is unbounded to the right: every $s$ with sufficiently large real part qualifies, and there is no upper restriction on $\operatorname{Re} s$ and no restriction on $|\operatorname{Im} s|$. Since $\beta$ ignores $\operatorname{Re} s$, the region is the set of $s = \sigma + it$ with $\sigma \ge 1 - c/\log(q(|t|+2))$. Degenerate readings follow from regionBoundary: when $q = 0$ the condition is exactly $\operatorname{Re} s \ge 1$; when $q \ge 1$ and $c = 0$ the condition is again exactly $\operatorname{Re} s \ge 1$; when $q \ge 1$ and $c < 0$ the condition is strictly stronger than $\operatorname{Re} s \ge 1$, namely $\operatorname{Re} s \ge 1 + |c|/\log(q(|t|+2)) > 1$; only for $c > 0$ and $q \ge 1$ does the region reach to the left of the line $\operatorname{Re} s = 1$. Nothing here requires $c$ to be small, so for large $c$ the boundary may lie far to the left of $0$ and the region may include points with negative real part.

Davenport.IsExceptionalSet

IsExceptionalSet takes an implicit natural number $q$ together with the typeclass assumption that $q$ is nonzero (so $q \ge 1$; the case $q = 0$ is excluded here, though $q = 1$ is not), an explicit real number $c$, a Dirichlet character $\chi$ modulo $q$ with values in $\mathbb{C}$, and a set $E \subseteq \mathbb{C}$; it is the conjunction of the following three statements.

1. $E$ has at most one element (it may be empty; it is not required to be nonempty).

2. Every $z \in E$ satisfies all seven of: $\operatorname{Im} z = 0$ (so $z$ is real); $0 < \operatorname{Re} z$; $\operatorname{Re} z < 1$ (both inequalities strict, so $z$ lies strictly inside the interval $(0,1)$ of the real axis); $z$ lies in the region of the previous definition, i.e. $1 - c/\log(q(|\operatorname{Im} z|+2)) \le \operatorname{Re} z$, which for such a $z$ (having $\operatorname{Im} z = 0$) reads $\operatorname{Re} z \ge 1 - c/\log(2q)$; the analytically continued Dirichlet $L$-function satisfies $L(z,\chi) = 0$; the character $\chi$ is quadratic (each of its values is $0$, $1$ or $-1$); and $\chi$ is not the trivial character modulo $q$. Note that the last two conditions concern $\chi$ alone but sit inside the quantifier over $z \in E$, so they are asserted only when $E$ is nonempty: if $E = \emptyset$ this entire clause holds vacuously and $\chi$ may be trivial or non-quadratic.

3. For every complex $s$ with $s \ne 1$, if $s$ lies in the region ($1 - c/\log(q(|\operatorname{Im} s|+2)) \le \operatorname{Re} s$) and $s \notin E$, then $L(s,\chi) \ne 0$. The point $s = 1$ is exempted unconditionally, whether or not it belongs to $E$; every other point of the region that is not the single possible member of $E$ is asserted to be a non-zero of $L(\cdot,\chi)$, including points with $\operatorname{Re} s \ge 1$ and points arbitrarily far to the right or with arbitrarily large imaginary part.

Several consequences of the literal statement are worth making explicit. There is no hypothesis on $c$ at all: if $c \le 0$ (or, more generally, whenever the region contains no point with real part $< 1$, which by the analysis above happens for every $c \le 0$ since $q \ge 1$ forces $\log(q(|t|+2)) > 0$), clause 2 cannot be satisfied by any $z$, so clause 1–2 force $E = \emptyset$ and the definition reduces to the assertion that $L(s,\chi) \ne 0$ for all $s \ne 1$ with $\operatorname{Re} s \ge 1$. Similarly, when $q = 1$ the only character modulo $1$ is the trivial one, so the requirement $\chi \ne 1$ in clause 2 fails and $E$ is again forced to be empty. The definition does not assert that an exceptional set exists, nor that it is unique, nor that $E$ is nonempty when $\chi$ is quadratic; it is a predicate that a given $E$ may or may not satisfy, and $E = \emptyset$ satisfies clauses 1 and 2 automatically. Nothing requires the element of $E$ (when present) to be a simple zero, nor forbids other zeros of $L(\cdot,\chi)$ outside the region, nor says anything about zeros at $s = 1$. Finally, $\chi$ is only constrained through clause 2 (hence only when $E \ne \emptyset$); no primitivity, no non-triviality and no quadraticity is assumed of $\chi$ as a standing hypothesis of the definition.