The quasi-Riemann hypothesis: an exposition

The prime number theorem says that the number of primes up to x, written $\pi(x)$, is approximately $x/\log x$, in the sense that the ratio of the two tends to 1 as x grows. A closer approximation is the logarithmic integral $\operatorname{Li}(x)=\int_2^x dy/\log y$. How close these approximations are depends on the zeros of the Riemann zeta function

Here $s=\sigma+it$ is a complex number with real part $\sigma=\operatorname{Re}s$ and imaginary part $t=\operatorname{Im}s$. The product, with one factor for each prime p, converges absolutely when σ > 1. An absolutely convergent product of nonzero factors is nonzero which rules out zeros with σ > 1.

Riemann extended ζ to every complex s except s = 1, where it has a pole (it behaves like $1/(s-1)$ near that point). This extension is called analytic continuation. There is only one complex-differentiable function on the larger region that agrees with the series where the series converges; this post takes that fact on trust. In the same way, $1/(1-z)$ continues the geometric series $1+z+z^2+\cdots$ from |z| < 1 to every z ≠ 1.

The extended ζ is zero at s = −2, −4, −6 and so on. These are called the trivial zeros, and all the other zeros are called nontrivial. The nontrivial zeros lie in the critical strip $0<\sigma<1$ and are symmetric about the line $\sigma=\tfrac12$: if ρ is a nontrivial zero, so are $1-\rho$ and the complex conjugate $\bar\rho$.

Write β for the real part of a nontrivial zero ρ. Let ψ(x) count each prime power $p^k\le x$ with weight log p, so that ψ(x) is approximately x. In the explicit formula for $\psi(x)-x$, each zero contributes a term of size about $x^{\beta}/|\rho|$. Its effect on $\pi(x)-\operatorname{Li}(x)$ is about $x^{\beta}/(|\rho|\log x)$. The largest β therefore decides how fast the error can grow. The Riemann hypothesis says every β equals 1/2, and it is equivalent to $|\pi(x)-\operatorname{Li}(x)|\ll\sqrt x\,\log x$.

The quasi-Riemann hypothesis asks for less. It asks for some constant $\sigma_0<1$ such that

at every height t. Before this result no such line was known. The zero-free regions of de la Vallée Poussin (1899), $\sigma>1-c/\log t$ for a constant c > 0, and of Vinogradov and Korobov (1958) both get thinner as the height t grows, with width tending to zero.

The theorem also covers Dirichlet L-functions. An L-function is a variant of ζ whose terms $1/n^s$ are multiplied by a periodic pattern of roots of unity and zeros (Chapter 1 gives the definition). To follow the idea of the proof you only need ζ. For an L-function whose pattern has period q, the classical zero-free region also shrinks as q grows. It allows one possible exception, a single real zero very close to 1, called a Landau–Siegel zero.

The companion paper proves the statement with 11/12 in place of 7/8 by a shorter route, and most of this post follows that route.

Notation

$s=\sigma+it$

A complex variable, with real part $\sigma=\operatorname{Re}s$ and imaginary part $t=\operatorname{Im}s$. The symbol σ is used for nothing else.

$f\ll g$ and $f=O(g)$

Both mean that $|f|\le Cg$ for some constant C, which does not depend on the quantity that is growing (such as x or D).

$f\asymp g$

Both $f\ll g$ and $g\ll f$, so f and g agree up to constant factors.

ε

Any small positive number. A bound such as $A\ll D^{11/12+\varepsilon}$ means that for every ε > 0 there is a constant $C_\varepsilon$ with $|A|\le C_\varepsilon D^{11/12+\varepsilon}$. Factors such as log D are absorbed this way, because $\log D\ll D^{\varepsilon}$ for every ε > 0.

Dirichlet characters and L-functions

Start with the function $\chi_4$ whose values at n = 1, 2, 3, 4 are 1, 0, −1, 0, repeating with period 4. It is periodic, and it is completely multiplicative: $\chi_4(mn)=\chi_4(m)\chi_4(n)$ for all m and n.

A Dirichlet character modulo q is any function χ on the integers that has period q, is completely multiplicative, and is nonzero exactly at the n that share no factor with q. Its nonzero values are roots of unity. The principal character mod q equals 1 at every n coprime to q and 0 elsewhere. Modulo 3 there are two characters: the principal one, and the character $\chi_{-3}$ with $\chi_{-3}(n)=0,\,1,\,-1$ for $n\equiv0,\,1,\,2\pmod 3$. Each character has an L-function. For Re s > 1,

When q = 1 the only character is χ(n) = 1 for all n, and $L(s,\chi)=\zeta(s)$. Each $L(s,\chi)$ extends to the whole complex plane, with a pole at s = 1 only for principal characters, and its nontrivial zeros lie in the critical strip. The generalized Riemann hypothesis (GRH) says that every nontrivial zero of every $L(s,\chi)$ has real part 1/2.

The Eisenstein integers

Think of the Eisenstein integers as a triangular-lattice version of the integers. Let $\omega=e^{2\pi i/3}=-\tfrac12+\tfrac{\sqrt3}{2}i$, a cube root of 1 with $\omega^2+\omega+1=0$. The Eisenstein integers $\mathbb{Z}[\omega]$ are the numbers $m+k\omega$ with m and k ordinary integers, and they form a triangular lattice in the plane. The field $K=\mathbb{Q}(\sqrt{-3})$ is the name of the number system they live in, and one can show that they are exactly the integers of K.

The norm is the square of the absolute value,

It is a positive integer for every nonzero element, and $N(xy)=N(x)N(y)$. The elements of norm 1 are the six units $\pm1,\pm\omega,\pm\omega^2$, the sixth roots of unity. Multiplying a nonzero x by each of the six units gives a set of six associates, and “counted up to units” means taking one representative out of each such set of six. Factorization into Eisenstein primes is unique up to order and units, and each ordinary prime p behaves in one of three ways.

• If $p\equiv1\pmod 3$, p splits into two Eisenstein primes of norm p, complex conjugates of each other. For example $7=(3+\omega)(2-\omega)=6-\omega-\omega^2$, using $\omega+\omega^2=-1$.
• If $p\equiv2\pmod 3$, p stays prime, with norm $p^2$. For example 2 and 5 are Eisenstein primes.
• The prime 3 is the square of a prime of norm 3, up to a unit: $3=-\omega^2(1-\omega)^2$.

An Eisenstein integer $m+k\omega$ coprime to 3 is called primary if it is congruent to 1 mod 3, which means $m\equiv1$ and $k\equiv0\pmod 3$. Exactly one of its six associates is primary, and the proof labels each Eisenstein integer counted up to units by its primary associate.

A finite-order Hecke character ν is the Eisenstein analogue of a Dirichlet character: a completely multiplicative function of Eisenstein integers counted up to units that depends only on the residue class modulo a fixed modulus. Its L-function is $L_K(s,\nu)=\sum_n\nu(n)N(n)^{-s}$, and for ν = 1 it is the Dedekind zeta function $\zeta_K(s)$.

Poisson summation

For a smooth function f on the real line that decays quickly, write its Fourier transform as $\hat f(\xi)=\int_{-\infty}^{\infty}f(x)e^{-2\pi i x\xi}\,dx$. The Poisson summation formula says

It follows from Fourier series. The function $F(x)=\sum_n f(x+n)$ has period 1, and its k-th Fourier coefficient is $\int_0^1F(x)e^{-2\pi ikx}\,dx=\hat f(k)$. Writing F as its Fourier series and putting x = 0 gives the formula. In sampling terms this is aliasing: at the integers the waves $e^{2\pi i n\xi}$ and $e^{2\pi in(\xi+k)}$ agree, and the samples f(n) see every $\hat f(k)$ at once.

The proof uses Poisson summation to turn a long smooth sum into a short one. Let Φ be a fixed smooth bump, S a large length, and χ a character with period q. Splitting n into residue classes mod q and applying the formula in each class gives

Since $\hat \Phi(\xi)$ is negligible once |ξ| is large, only the terms with |k| up to about q/S matter on the right. A sum of length S becomes a sum of length about q/S, whose coefficients are the finite Fourier coefficients of χ. The same formula holds for sums over the lattice $\mathbb{Z}[\omega]$, where the dual sum has about N(q)/S terms.

The classic example is the Jacobi theta function $\theta(x)=\sum_{n\in\mathbb Z}e^{-\pi n^2x}$ for x > 0. The Gaussian $e^{-\pi y^2x}$ has Fourier transform $x^{-1/2}e^{-\pi\xi^2/x}$, and Poisson summation gives

At x = 100 the left side has about two dozen terms of noticeable size, while the right side equals 10 with an error smaller than $10^{-134}$.

Random signs and square-root cancellation

Add up D independent random signs, each +1 or −1 with probability 1/2. The square of the sum is a sum of $D^2$ products; the D squares of the signs equal 1, and the other products average to 0. The mean square is D, and a typical sum has size about $\sqrt D$. A sum of D terms of size at most 1 shows square-root cancellation if it is at most about $D^{1/2+\varepsilon}$. For a family of H such sums, square-root cancellation on average means that the sum of their squares is at most about HD.

Many theorems in number theory were proved assuming the generalized Riemann hypothesis (GRH, Chapter 1). Some of those proofs use only that the zeros of every $L(s,\chi)$ stay a fixed distance to the left of $\operatorname{Re}s=1$, with the same distance for every modulus q. Those results now hold without any assumption.

• Primes in arithmetic progressions. Write $\pi(x;q,a)$ for the number of primes p ≤ x with $p\equiv a\pmod q$, and $\varphi(q)$ for the number of residues mod q that share no factor with q; for example φ(10) = 4. The 11/12 theorem gives, for every modulus $q\le x$ and every a coprime to q,
  $$\Big|\pi(x;q,a)-\frac{\operatorname{Li}(x)}{\varphi(q)}\Big|\ll x^{11/12}\log x$$
  where the implied constant does not depend on q, a or x and can be computed. The main term Li(x)/φ(q) is larger than the error for q up to about $x^{1/12}$, up to logarithmic factors. The Siegel–Walfisz theorem, the previous unconditional result, needed q to be at most a power of log x.
• No Landau–Siegel zeros. For a character χ mod q whose values are only 0 and ±1, the classical zero-free region leaves open one possible real zero $\beta_1$ of $L(s,\chi)$ in the range $1-c/\log q<\beta_1<1$. Once q is large enough that $1-c/\log q>7/8$, the whole range lies to the right of 7/8, and the theorem excludes any zero there.
• Square roots mod p. The theorem implies that the least quadratic nonresidue mod a prime p, the least positive number that is not a square mod p, is at most a constant times a fixed power of log p (for example the 32nd power). Trying 2, 3, 4, … in turn therefore finds one quickly, and the Tonelli–Shanks algorithm uses it to compute square roots mod p deterministically in time polynomial in log p.

Step 1 shows that a bound on a smoothed Möbius sum gives a zero-free half-plane. Step 2 puts that sum in a family twisted by sixth-power residue symbols; for u = p⁶ the twist equals 1 unless p divides n, so the family contains many copies of the original sum. Steps 3 to 5 bound the whole family on average, using Poisson summation, the cubic Gauss sums that appear as coefficients of Kubota’s theta function, and a shrinking recursion. A factorization of L-functions then carries the result to ζ and every Dirichlet L-function.

Chapters 4 and 5 give Steps 1 and 2 in full. Chapters 6 to 8 describe what Steps 3 to 5 achieve and why, and leave the full arguments to the paper.

The Möbius function μ(n) is defined as follows. If n is divisible by the square of a prime, then μ(n) = 0. Otherwise n is a product of k different primes, and $\mu(n)=(-1)^k$. For example μ(1) = 1, μ(2) = μ(3) = −1, μ(4) = 0, μ(6) = 1 and μ(30) = −1. Multiplying out the Euler product $1/\zeta(s)=\prod_p(1-p^{-s})$ produces each product of distinct primes once, with sign $(-1)^k$, and no other n. Therefore

Write $M(x)=\sum_{n\le x}\mu(n)$ for the partial sums, which are called the Mertens function. Suppose that for every ε > 0 there is a constant $C_\varepsilon$ with $|M(x)|\le C_\varepsilon x^{\sigma_1+\varepsilon}$, for some fixed exponent σ₁ < 1. Partial summation, the discrete form of integration by parts, then shows that the series for 1/ζ converges whenever $\operatorname{Re}s>\sigma_1$.

Call the sum of the series F(s). It is a finite, well-behaved function on Re s > σ₁ (holomorphic, which means complex-differentiable). On Re s > 1 we have $\zeta(s)F(s)=1$. Both sides are holomorphic on Re s > σ₁ away from s = 1, so by the uniqueness of analytic continuation, which this post takes on trust, $\zeta(s)F(s)=1$ holds there too. At a zero ρ of ζ with Re ρ > σ₁ the left side would be $0\cdot F(\rho)=0$, which contradicts the identity. There is therefore no zero with Re s > σ₁. The converse also holds: if ζ has no zeros with real part greater than σ₁, then $M(x)\ll x^{\sigma_1+\varepsilon}$. For σ₁ = 1/2 this is Littlewood’s form of the Riemann hypothesis.

The proof works with a smoothed version of this sum over the Eisenstein integers $\mathbb{Z}[\omega]$ (Chapter 1), twisted by a Hecke character ν. From here on p and n denote Eisenstein primes and Eisenstein integers, and N is the norm. Fix a finite-order Hecke character ν and a smooth function W that is zero outside a fixed interval $[y_1,y_2]$ with $0<y_1<y_2$. For each size D put

The sum runs over primary Eisenstein integers n coprime to 6 and to the modulus of ν; leaving out the others removes finitely many Euler factors, which are nonzero for Re s > 0. μ(n) is defined from the factorization of n into Eisenstein primes. Only n with $y_1D\le N(n)\le y_2D$ contribute. There are about D of them and each term has size at most max |W|, which gives $|A_1(D)|\ll D$ with no cancellation at all. The goal is a power saving below that:

The bound becomes a statement about zeros through the Mellin transform, written with a tilde to keep it apart from the Fourier transform:

Substituting $y=e^{x}$ turns it into $\int_{-\infty}^{\infty}W(e^{x})\,e^{\sigma x}e^{itx}\,dx$, a Fourier transform in the variable x = log y. The Mellin transform of W is therefore the Fourier transform of the function $x\mapsto W(e^{x})e^{\sigma x}$, evaluated at $\xi=-t/(2\pi)$ in the convention of Chapter 1. For one term of A₁(D), viewed as a function of D, the substitution y = N(n)/D gives

Summing over n with the weights μ(n)ν(n) gives, for Re s > 1 and up to the omitted Euler factors,

A₁(D) is zero for small D (every N(n)/D is then larger than $y_2$), and for large D the bound applies. The integral defining I(s) therefore converges for $\operatorname{Re}s>11/12$, and I is holomorphic there. The identity $I(s)\,L_K(s,\nu)=\widetilde W(s)$ holds for Re s > 1, and by uniqueness of continuation it holds on Re s > 11/12 too.

Suppose $L_K$ had a zero ρ with $\operatorname{Re}\rho>11/12$. Putting s = ρ into the identity gives $\widetilde W(\rho)=I(\rho)\cdot0=0$, and this would have to hold for every smooth W, because the bound holds for every smooth W. It is enough to find one W with $\widetilde W(\rho)\ne0$. Take a smooth function V ≥ 0 that is zero outside [1, 2] and positive somewhere, and put $W(y)=y^{-\rho}V(y)$. The factor $y^{-\rho}$ cancels the $y^{\rho}$ in the transform:

This contradicts $\widetilde W(\rho)=0$, and $L_K(s,\nu)$ has no zero with real part greater than 11/12. No method is known for proving the bound for one Möbius sum on its own, and the proof instead puts $A_1$ inside a large family of similar sums, where the bound can be proved on average.

The twist is a sixth-power version of the Legendre symbol. For an odd prime p and an integer m, the Legendre symbol $(m/p)$ is 1 if m is a nonzero square mod p, −1 if m is a nonresidue, and 0 if p divides m. Euler’s criterion computes it with one power:

Mod 7, for example, we have $2^3=8\equiv1$ and $3^3=27\equiv-1$, and 2 is a square (it is $3^2=9$ mod 7) while 3 is not.

Over the Eisenstein integers the same construction works with sixth roots, because all six sixth roots of unity $\pm1,\pm\omega,\pm\omega^2$ are Eisenstein integers. Let p be an Eisenstein prime that does not divide 6. The residues mod p form a finite field with N(p) elements. N(p) is either a prime ≡ 1 mod 3 or the square of a prime ≡ 2 mod 3, in both cases ≡ 1 mod 3, and it is odd because p does not divide 2; together these give 6 | N(p) − 1. For u not divisible by p the power $u^{(N(p)-1)/6}$ is a sixth root of 1 mod p, and it is congruent to exactly one of the six units. That unit is the sextic residue symbol $(u/p)_6$. It equals 1 exactly when u is a sixth power mod p, and it is set to 0 when p divides u. For a primary n coprime to 6, define $\chi_n(u)=(u/n)_6$ by multiplying these symbols over the prime factors of n.

The family is indexed by a second Eisenstein integer u:

For u = 1 every $\chi_n(1)$ is 1, and this is the original sum A₁. Think of the family as a table. Each row is one u, each column is one n, and A_u(D) adds up the entries of row u. The rows are all nonzero u with N(u) ≤ H, where $H=D^{1+\vartheta}$ for any small fixed ϑ > 0 (the paper takes ϑ ≤ 1/10). After Poisson summation in Step 3 the rows have norm up to about $D^2/H=D^{1-\vartheta}$, slightly less than the column scale D, and Step 5 uses that gap to shrink the problem.

The central estimate of the proof (Proposition 3.1 of the 11/12 paper) says that these sums cancel as much as random signs would, on average over the rows:

An average bound says nothing about one particular row unless that row appears many times. Here it does. Let p be a primary Eisenstein prime and take $u=p^6$. If p does not divide n, then $\chi_n(p)$ is a sixth root of unity and $\chi_n(p^6)=\chi_n(p)^6=1$. If p divides n, the symbol is 0. Row $p^6$ therefore agrees with row 1 except in the columns n that are multiples of p. If $N(p)\approx Y$, about D/Y of the columns are multiples of p, and each entry has size at most max |W|, which gives

Choose $Y=H^{1/6}$. Then $N(p^6)=N(p)^6\le Y^6=H$, and every $p^6$ is one of the rows. By the prime ideal theorem (the prime number theorem for Eisenstein primes) there are $P\asymp Y/\log Y$ primes p with Y/2 < N(p) ≤ Y, which makes P rows that are all near-copies of $A_1$.

To combine them, use $(x+y)^2\le2x^2+2y^2$, which follows from $(x-y)^2\ge0$. With $x=|A_{p^6}(D)|$, $y=|A_1(D)-A_{p^6}(D)|$ and the triangle inequality it gives $|A_1|^2\le2|A_{p^6}|^2+2|A_1-A_{p^6}|^2$ for each of the P primes. Average over the P primes, and bound the sum over these P rows by the sum over all rows:

In the second line the factor log Y from 1/P is absorbed into $D^{\varepsilon}$. With $Y=H^{1/6}=D^{(1+\vartheta)/6}$ the exponents are

The first exponent is the larger one. Taking square roots gives $|A_1(D)|\ll D^{11/12+5\vartheta/12+\varepsilon}$. Because ϑ can be any small fixed number, choosing it small makes $5\vartheta/12$ as small as we like, which gives $A_1(D)\ll D^{11/12+\varepsilon}$. Step 1 then gives the zero-free half-plane $\operatorname{Re}s>11/12$.

Step 3 trades the Möbius function, which has no structure the proof can use, for Gauss sums, which Step 4 shows are Fourier coefficients of a function with a symmetry. Write $w(n)=\mu(n)\nu(n)W(N(n)/D)$ for the column weights, so that $A_u(D)=\sum_n w(n)\chi_n(u)$. Expanding $|A_u|^2=A_u\overline{A_u}$ and moving the sum over u inside gives

The pairs with n₁ = n₂ are called the diagonal terms. For them $|\chi_n(u)|^2=1$ whenever u and n share no factor, and the inner sum is about H. There are about D such n, each with $|w(n)|^2$ at most a constant, and the diagonal contributes about $DH=D^{2+\vartheta}$, the size the target allows.

For the other pairs the proof uses periodicity. The symbol $(u/p)_6$ is defined by a congruence mod p, and multiplying over the primes of n shows that $\chi_{n_1}(u)\overline{\chi_{n_2}(u)}$ depends only on u mod $n_1n_2$. The paper replaces the sharp row condition N(u) ≤ H by a smooth weight in u, so that Poisson summation (Chapter 1) applies and turns the inner sum into a sum over dual frequencies h. The period has norm about $D^2$, and the dual sum runs over h with N(h) up to about $D^2/H=D^{1-\vartheta}$.

The coefficients of the dual sum are finite Fourier coefficients of characters, called Gauss sums. For a character χ mod an odd prime p that is not principal, and any k not divisible by p, substituting y = xk gives

The constant $g(\chi)$ has absolute value $\sqrt p$. The Fourier transform of a character is therefore the conjugate character at the dual variable, times a constant of size $\sqrt p$. For a real character such as the Legendre symbol the conjugate is the same character, but for the sextic symbol it matters: transforming $\overline{\chi_n}$ produces $\chi_n(h)$ and the Gauss sum of $\overline{\chi_n}$. Write $\gamma_j(n)$ for the Gauss sum of the character $\chi_n^{\,j}$ over $\mathbb{Z}[\omega]$, divided by $\sqrt{N(n)}$. Then $\gamma_{-1}(n)$ belongs to the conjugate character $\overline{\chi_n}$, and $\gamma_2(n)$ to $\chi_n^2$, the cubic residue symbol, whose values are cube roots of unity and 0. For squarefree n coprime to 6 both have absolute value 1.

A Gauss-sum identity, following Hasse and Heath-Brown, then removes the Möbius function. For squarefree primary n away from a fixed finite set of primes,

On the right, $\chi_n(-1)$, the conjugate of $\alpha(n)=n/|n|$, and $G(n)^{-1}$, a phase that depends only on n modulo a fixed modulus, all have absolute value 1. The proof carries them along. The remaining factor is the cubic Gauss sum $\gamma_2(n)$.

The dual terms still involve two columns n₁ and n₂. The paper first removes their common factor. The modulus n₁n₂ then splits into coprime parts, and the Chinese remainder theorem splits each dual coefficient into a factor for one column times the conjugate of the same factor for the other (a finite sum of such products). A Mellin expansion separates the smooth weights into products of a single sum over n₁ and a conjugated single sum over n₂, and Cauchy–Schwarz over the rows then bounds the total by mean squares of one single sum over n, the dual sum

The star means that n runs over squarefree primary Eisenstein integers, U is another smooth cutoff, and ν′ is one of finitely many fixed Hecke characters that collect the remaining phases. From here on R is the norm range of the dual rows: h runs over nonzero Eisenstein integers with N(h) ≤ R, about R of them up to a constant, starting at $R=D^{1-\vartheta}$. The bound needed is

If each $B_h(D)$ behaved like D random signs, the left side would be about $RD=D^{2-\vartheta}$, so the target asks for slightly less than square-root cancellation: it allows a loss of a factor about $D^{\vartheta}$. Schematically, the two families compare as

The first term is the diagonal, and if the dual bound holds the second is also about DH, which gives the mean-square bound of Step 2.

The Jacobi theta function of Chapter 1 can be written with a complex variable τ in the upper half-plane as $\theta(\tau)=\sum_{n\in\mathbb Z}e^{\pi in^2\tau}$. It satisfies

and the second rule is the identity $\theta(1/x)=\sqrt x\,\theta(x)$ of Chapter 1 at τ = ix. A function that changes only by a simple factor under an infinite group of symmetries, as θ does, is called an automorphic form.

Kubota constructed a cubic analogue, which this post writes as Θ and treats as a black box with two properties. First, Θ is a function on three-dimensional hyperbolic space whose Fourier coefficients are cubic Gauss sums up to simple factors. Patterson computed the coefficients $c_\Theta$ of its complex conjugate $\overline\Theta$: for primary n and b coprime to 6, with n squarefree,

Second, Θ has a symmetry of the same kind as the rule for θ(−1/τ), and from the symmetry the proof uses only the summation formula it gives. For a fixed row h and a smooth weight U, the formula has the schematic form

Here $d_\Theta$ are coefficients of $\overline\Theta$ in a related expansion, $U^{\sharp}$ is a transformed weight, (⋯) stands for phases, and ≈ means a finite combination of such sums. A sum over indices of norm about X becomes a sum over indices of norm up to about $N(h)^2/X$.

The formula applies to sums over all indices $nb^3$, including those with a cube factor, while $B_h$ contains only the squarefree indices. The paper therefore completes $B_h$ into a sum $T_h(X)$ that also contains the cube terms, following Dunn and Radziwiłł. The paper normalizes $T_h(X)$ by $1/\sqrt X$, so that $B_h(X)$ equals $\sqrt X$ times the b = 1 part of $T_h(X)$.

The transformation changes the twist by h. At a prime p dividing h, it produces a factor $\chi_p^{-1}$ from the Fourier side and a factor $\chi_p^{-2}$ from the theta coefficients, which combine to $\chi_p^{-3}$. A sextic character takes sixth roots of unity $e^{2\pi ik/6}$ as values, and cubing gives $e^{\pi ik}=\pm1$. The cube of a sextic character is a quadratic character, the analogue of the Legendre symbol for $\mathbb{Z}[\omega]$.

The large sieve is why this matters. Put the values of about R characters in the rows of a matrix and about C column indices in the columns, and write $z_{jk}$ for the entry in row j and column k. In the application the sieve is used after the transformation: the rows are the h with N(h) ≤ R, and the columns are the transformed indices m, whose norm is at most about $R^2/X$, so $C\approx R^2/X$. A large sieve inequality gives a Δ such that, for all coefficients $w_k$,

If the rows were exactly orthogonal, each with squared length C, Bessel’s inequality (the inequality half of Parseval’s identity) would give Δ = C. A matrix and its transpose have the same Δ, and exactly orthogonal columns would give Δ = R. Neither can be beaten when the entries have absolute value 1: putting all of w on one column shows Δ ≥ R, and taking w equal to the conjugate of one row shows Δ ≥ C. The ideal bound is therefore $\Delta\ll(RC)^{\varepsilon}(R+C)$. For quadratic characters, with rows and columns restricted to squarefree indices, it was proved by Heath-Brown over the integers and by Goldmakher and Louvel over number fields such as K. The restriction is needed: a square column index makes every entry of that column equal to 1, and a square row index makes the row a principal character, so many rows or columns would line up. For cubic and sextic characters the known bound has an extra term $(RC)^{2/3}$, and with it this argument would no longer reach the target.

With Cauchy–Schwarz in the cube variable b and the quadratic large sieve in the squarefree part of the transformed index, with C ≈ R²/X, the result is

If the cube terms could be dropped, the normalization above would give

using R ≤ D, which is the target of Step 3. The cube terms cannot be dropped: $T_h$ is its b = 1 part plus the terms with b ≠ 1, and those terms could cancel much of the b = 1 part, so a small $T_h$ does not imply a small b = 1 part. Step 5 removes them.

Step 5 uses the identity

For n = 6 the sum is $\mu(1)+\mu(2)+\mu(3)+\mu(6)=1-1-1+1=0$. In general the signs add up to $(1-1)^k=0$ when n > 1 has k prime factors. The same holds over the Eisenstein integers, and using it to undo a sum over divisors is called Möbius inversion. A function f is completely multiplicative if $f(bc)=f(b)f(c)$ for all b and c, including when they share factors. The character $\chi_4$ is completely multiplicative, and μ is not, since μ(4) = 0 while μ(2)² = 1.

The completed sum $T_h(X)$ is a weighted total of copies of the squarefree sum. The terms with a given cube factor b form the normalized squarefree sum at column scale $X/N(b)^3$, multiplied by a weight of absolute value at most $1/N(b)$, and the weights are completely multiplicative in b. Möbius inversion then expresses the squarefree sum as a combination of $T_h$ at the scales $X/N(d)^3$, over divisors d with signs μ(d) and the same weights.

The proof splits these divisors at a cutoff Q. By Step 4, each term with divisor d costs about $R+R^2N(d)^3/X$, so all d up to Q together cost about $R+R^2Q^3/X$, up to a factor $D^{\varepsilon}$. This stays within the target X when R ≤ X and $R^2Q^3\le X^2$, that is $Q\le(X/R)^{2/3}$. Divisors never need to go beyond $X^{1/3}$, because a cube factor with $N(b)^3>X$ leaves no terms of norm about X. Hence

The divisors d above Q are grouped by the full cube factor b that contains them, and they become squarefree sums $B_h$ at the smaller column scale $X_b=X/N(b)^3$, still with rows up to R. Two more Poisson summations restore the original shape. The first, in h, turns the cubic Gauss sums back into Möbius coefficients, with dual rows up to about $X_b^2/R$. The proof enlarges this range to $XX_b/R$, which can only increase a sum of squares. The second Poisson summation returns to cubic Gauss sums with rows up to $X_b^2/(XX_b/R)=RX_b/X=R/N(b)^3$. The net effect is

Both scales shrink by the same factor, and the ratio stays at $R/X=D^{-\vartheta}$. While R > √X the cutoff is $(X/R)^{2/3}=D^{2\vartheta/3}$, and a cube factor above it has $N(b)^3>D^{2\vartheta}$. Each round therefore divides R by more than $D^{2\vartheta}$.

Every branch ends in one of two ways. When R ≤ √X the cutoff becomes $X^{1/3}$, no divisor lies above it, and the direct bound covers everything. Alternatively a large b can take R to at most 1, where the sum is counted term by term. In the slowest case each round lowers the exponents of R and X by 2ϑ, starting from 1 − ϑ and 1, and R ≤ √X is reached after about $(1-2\vartheta)/(2\vartheta)$ rounds, a number that depends on ϑ alone. Reading the chain backwards proves the dual bound, then the mean-square bound of Step 2.

Call a Hecke character primitive if it does not come from a character with a smaller modulus. Let $\beta^{*}$ be the supremum (least upper bound) of the real parts of all zeros of all primitive finite-order Hecke L-functions over K, with 1/2 included so that the set is not empty. The 7/8 theorem says $\beta^{*}\le7/8$. The proof assumes $\beta^{*}>\sigma_0$ for a proposed boundary σ₀ (here 7/8) and derives a contradiction.

For each character η and each large real number Z the proof uses two quantities. The character sum $J_\eta(Z)$ is built from the same kind of completed cubic-theta sums as Steps 4 and 5 (theta coefficients at indices cn³), twisted by sextic characters and by η, with length set by Z. The signal $f_\eta(Z)$ is an integral of $Z^{\ell(s)}/L_K(s,\eta)$ along a vertical line, times holomorphic factors and with finitely many Euler factors removed, where $\ell(s)=s+c$ for a fixed constant c. It is a Mellin integral like the one in Step 1. The argument runs in five stages.

1. The low estimate bounds the sum directly: $|J_\eta(Z)|\ll Z^{\ell(\sigma_0)+\delta}$, with a small loss $0<\delta<\beta^{*}-\sigma_0$.
2. The high estimate says the sum is close to the signal: $|J_\eta(Z)-f_\eta(Z)|\ll Z^{\ell(\beta^{*})-\kappa}$, with a saving κ > 0. This is where the definition of β* enters. Every L-function in the computation, including the Hecke twists that Poisson summation produces, has no zeros to the right of β*, so the proof can move contours to near Re s = β* and measure the error against $Z^{\ell(\beta^{*})}$.
3. Because ℓ has slope 1, $\ell(\sigma_0)+\delta=\ell(\beta^{*})-(\beta^{*}-\sigma_0-\delta)$, and the triangle inequality gives $|f_\eta(Z)|\ll Z^{\ell(\beta^{*})-\varepsilon^{*}}$ with $\varepsilon^{*}=\min(\beta^{*}-\sigma_0-\delta,\;\kappa)>0$.
4. As in Step 1, a growth bound of this kind makes $1/L_K(s,\eta)$ finite on the half-plane $\operatorname{Re}s>\beta^{*}-\varepsilon^{*}$.
5. Because β* is a supremum, at least one character has a zero with real part greater than $\beta^{*}-\varepsilon^{*}$, and there its $1/L_K$ is infinite. This contradiction gives $\beta^{*}\le\sigma_0$.

The last stage needs δ and κ, and hence ε*, to be the same for every character. Otherwise the zero near β* could belong to a character whose own margin is too small.