Zero-free half-planes and exceptional real zeros

Source and verification

This article explains family 003 of the October 6, 2026 OpenAI collection. Research conclusions are attributed to its manuscripts. The review covers statements, introductions, and selected proof mechanisms; it is not an independent verification of the complete proofs.

🏷️ Dirichlet series and zero-free regions

A Dirichlet character modulo $q$ is a periodic multiplicative function $\chi$, zero on integers not coprime to $q$, whose nonzero values lie on the unit circle. For $\mathrm{Re}s>1$,

$$\begin{array}{rl}L(s,\chi )& =\sum _{n\ge 1}\frac{\chi (n)}{n^s}\\ & =\prod _p(1-\chi (p)p^{-s}{)}^{-1}.\end{array}$$

Absolute convergence makes this product nonzero. The difficult question concerns the continuation into the strip $0<\mathrm{Re}s<1$. The principal character gives a pole at $s=1$; for modulus one the function is the Riemann zeta function. A primitive character is one not induced from a smaller modulus; its smallest modulus is its conductor.

A zero-free half-plane has a boundary independent of height. This is stronger than a region whose width decreases as $|\mathrm{Im}s|$ grows. The quasi-Riemann hypothesis asks for some fixed boundary below one. The Riemann hypothesis asks that every nontrivial zeta zero have real part exactly $1/2$.

The family has three manuscripts. Their conclusions and mechanisms should be read separately.

Manuscript	Claimed conclusion	Scope
September 30	No zeros for $\mathrm{Re}s>7/8$	All Dirichlet characters; also the stated Hecke family
October 5	No zeros for $\mathrm{Re}s>11/12$	A separate exposition of the first analytic stage
October 1	$(1-\beta )\log q\ge c>0$	Real zeros of primitive nonprincipal real characters

The first two statements include finite-order Hecke $L$-functions over $\mathbb{Q}(\sqrt{-3})$. These are analogous Dirichlet series indexed by ideals, with ideal norms replacing positive integers. Their principal poles remain allowed. The open boundaries do not exclude zeros on the boundary lines. (OpenAI, 2026) (OpenAI, 2026a)

🏷️ Cancellation and continuation

The Möbius function $\mu (n)$ is zero when a square divides $n$, and otherwise equals $(-1{)}^r$, where $r$ is the number of its prime factors. Euler products give

$$\begin{array}{c}\frac{1}{L(s,\chi )}=\sum _{n\ge 1}\frac{\mu (n)\chi (n)}{n^s},\\ \mathrm{Re}s>1.\end{array}$$

This identity explains why cancellation in Möbius sums can imply zero-free regions. For example, if

$$M_{\chi }(x)=\sum _{n\le x}\mu (n)\chi (n)=O_{\chi }(x^{\theta }),$$

partial summation represents the reciprocal by

$$\frac{1}{L(s,\chi )}=s{\int }_1^{\infty }{M}_{\chi }(x)x^{-s-1}dx.$$

The integral is holomorphic for $\mathrm{Re}s>\theta$. Analytic continuation then prevents a zero of $L$ there, since a zero would produce a pole of its reciprocal. This is a model implication, not an assertion that the required cancellation is elementary.

The October 5 manuscript organizes its argument through mean-square estimates for twisted Möbius sums. The September 30 manuscript compares two exact representations of a completed cubic-theta average. Reflection controls an energy norm. Poisson summation isolates a principal frequency containing the target reciprocal $L$-function, multiplied by a holomorphic nonvanishing factor. The remaining frequencies must be bounded with a power saving. (OpenAI, 2026b) (OpenAI, 2026a)

The completion retains indices of the form $c{n}^3$, with $c$ squarefree but without requiring $c$ and $n$ to be coprime. Removing shared factors would alter the local identities that expose the reciprocal. The principal frequency is a signal to retain, not an error to discard.

🏷️ The two analytic stages

Balanced averaging scales produce the $11/12$ stage. Quadratic and planar large-sieve estimates control the reflected side; zero detection and a sextic large sieve organize the nonprincipal rows on the Poisson side.

The $7/8$ stage changes the average. Selected primes compensate for an unwanted Euler contribution, and the two scales become unequal. A detected zero now supplies both an inverse polynomial, with Möbius coefficients, and a plain character polynomial. Their simultaneous largeness at the same twist height permits stronger row counts. Separate recursive moment estimates are needed for these two polynomials.

Two source distinctions matter. The affine Mellin normalizations $s-2/3$ and $s-11/16$ belong to different averages. Also, a uniform half-plane boundary does not require every intermediate estimate to have a character-independent constant. The sources fix the target character while maintaining the exponent margins needed for continuation.

The final transfer from the quadratic field to Dirichlet functions uses a factorization into $L(s,\chi )L(s,\chi {\chi }_{-3})$, apart from finitely many nonvanishing Euler factors. The pole case is handled separately. The stronger boundary is still well to the right of $1/2$.

🏷️ A separate real-zero argument

The October 1 manuscript claims an absolute $c>0$ such that

$$(1-\beta )\log q\ge c$$

for every real zero $0<\beta <1$ of every primitive nonprincipal real character of conductor $q\ge 3$. This excludes zeros approaching one faster than the logarithmic scale; it does not exclude every real zero in the critical interval. (OpenAI, 2026c)

Its proof is a determinant comparison. A hypothetical very close zero creates a shortage of primes with $\chi (p)=1$. Primes with $\chi (p)=-1$ then supply divisibility of a nonzero interpolation determinant. The construction has $N^4$ columns and natural degree scale $U=N^{4/3}$. Prime divisibility contributes a leading $\log U$, while evaluation sizes contribute $\log N=(3/4)\log U$. Weighted interpolation must establish nonvanishing and preserve this strict advantage.

These notes explain that comparison without developing its algebraic interpolation theory. The fixed-half-plane papers also discuss consequences for least quadratic nonresidues and deterministic square-root extraction; these are consequences of their claimed analytic theorem, not evidence verifying it.

🏷️ Key steps and proof outlines

Sixth powers amplify one Möbius sum into many rows

The October 5 paper’s mean-square proposition concerns a family

$$A_u(D)=\sum _{(n,S)=1}\mu (n)\nu (n){\chi }_n(u)W\left(\frac{\mathrm{N}n}{D}\right).$$

Here $n$ ranges over ideals in the fixed quadratic field, $\mathrm{N}n$ is the size of the quotient by the ideal, $\nu$ is the fixed finite-order character, $S$ removes finitely many bad primes, and $W$ is a smooth compactly supported weight. The residue characters ${\chi }_n$ have order dividing six. The target row is $u=1$. The source proves, for fixed small $\vartheta >0$,

$$\sum _{0<\mathrm{N}u\le D^{1+\vartheta }}|A_u(D){|}^2{\ll }_W{D}^{2+\vartheta +\epsilon }.$$

Proof outline for extracting the exponent. Set $Y=D^{(1+\vartheta )/6}$. For a prime ideal of norm between $Y/2$ and $Y$, with chosen generator $p$, the exact identity ${\chi }_n(p^6)=1_{p\nmid n}$ gives $A_{p^6}(D)=A_1(D)+O_W(D/Y)$. Thus there are $J\asymp Y/\log Y$ different rows that nearly reproduce the target row. Averaging their squares and applying the mean-square proposition with a smaller exponent loss, so that $\log Y$ is absorbed into $D^{\epsilon }$, yields

$$\begin{array}{rl}|A_1(D){|}^2& {\ll }_W{D}^{2+\vartheta +\epsilon }/Y+D^2/Y^2\\ & =D^{11/6+5\vartheta /6+\epsilon }+D^{5/3-\vartheta /3}.\end{array}$$

Taking square roots and then choosing $\vartheta$ and the exponent loss sufficiently small gives $A_1(D){\ll }_W{D}^{11/12+\epsilon }$. The exponent $11/12$ therefore comes from the number of available sixth-power copies and the mean-square cost, rather than from a formal substitution into the Euler product. (OpenAI, 2026a)

A decreasing parameter makes the mean-square recursion finite

The mean-square proposition is the difficult input to this deduction. Poisson summation changes the original character sums into auxiliary energies $\mathcal{E}(\mathcal{H},X,F)$: $\mathcal{H}$ measures the row range, $X$ the column range, and $F$ an additional factor range. Put $\Sigma =XF$. The proof needs the normalized energy $\mathcal{E}/\Sigma$ bounded by an arbitrarily small power of the original scale $D$.

Proof outline. Separate indices according to their cube factors. Small cube factors are treated directly; the others permit two reciprocity/Poisson transfers. Crucially, the output is again an energy of the same allowed form, with

$$\begin{array}{c}{\mathcal{H}}^{\prime }\le D^{-2\kappa }\mathcal{H},{\mathcal{H}}^{\prime }/{\Sigma }^{\prime }\le D^{-\kappa },\\ {\Sigma }^{\prime }\le \Sigma \le D^{C_0}.\end{array}$$

The positive gap $\mathcal{H}/\Sigma \le D^{-\kappa }$ survives the transfer and forces the row parameter to decrease. Induct on the integer $j$ in $\mathcal{H}\le D^{j\kappa }$. At $j=0$, counting and the triangle inequality suffice. At the next stage, every transferred energy falls within the preceding stage. The smoothness requirement increases from derivative order $m$ to at least $4m+12$, but only finitely many stages are needed because $\mathcal{H}\le D^{C_0}$. Splitting the allowable exponent loss between the reduction, transfer, and induction closes the estimate. This identifies both the decreasing quantity and the uniformity that a recursive argument must preserve.

Mellin continuation must retain a nonzero numerator

If $\rho$ were a zero with $\mathrm{Re}\rho >11/12$, choose a nonzero nonnegative smooth $\varphi$ supported in $(1,2)$ and set $W(y)=y^{-\rho }\varphi (y)$. Its Mellin transform satisfies $\hat{W}(\rho )=\int \varphi (y)dy/y>0$. The bound just obtained makes

$${\mathcal{M}}_W(s)={\int }_0^{\infty }{A}_1(D)D^{-s}\frac{dD}{D}$$

holomorphic for $\mathrm{Re}s>11/12$: compact support handles small $D$, and the power bound handles large $D$. For $\mathrm{Re}s>1$, absolute convergence allows termwise integration, giving ${\mathcal{M}}_W(s)=\hat{W}(s)/L_K^S(s,\nu )$. Continue the identity after multiplying by $L_K^S$. Evaluation at $\rho$ would say $0=\hat{W}(\rho )>0$. This is the final contradiction, once the paper’s energy estimates are available.

The September 30 approach uses a more general common signal $f(Z)$. Its two representations give a low bound for an average $J(Z)$ and a high-side approximation to that average. If ${\beta }_{\ast }$ is the supremum of the real parts of the target zeros and $C(s)=s+c$, their bounds are

$$\begin{array}{rl}|J(Z)|& \ll Z^{C({\sigma }_0)+\omega },\\ |J(Z)-f(Z)|& \ll Z^{C({\beta }_{\ast })-\sigma }.\end{array}$$

Here $0<\omega <{\beta }_{\ast }-{\sigma }_0$ and $\sigma >0$. The triangle inequality gives a saving ${ϵ}_{\ast }=\min ({\beta }_{\ast }-{\sigma }_0-\omega ,\sigma )>0$ relative to $C({\beta }_{\ast })$. Rapid decay of $f$ at zero and this saving at infinity continue its Mellin transform to $\mathrm{Re}s>{\beta }_{\ast }-{ϵ}_{\ast }$. The numerator is a Gaussian factor times $H_{\nu }(s)$, with $|H_{\nu }-1|\le 1/2$. Thus it cannot cancel a pole of $1/L$. Only the positive exponent margin must be uniform in $\nu$; the implied constants may depend on the fixed character. (OpenAI, 2026b)

flowchart TD
  A[Completed theta average] --> B[Reflection and energy bound]
  A --> C[Poisson representation]
  C --> D[Principal frequency: reciprocal L-function]
  C --> E[Power saving for other frequencies]
  B --> F[Bound for the common signal]
  E --> F
  D --> G[Identify its Mellin transform]
  F --> G
  G --> H[Holomorphic reciprocal beyond the hypothetical zero boundary]

The two branches concern the same average. An estimate for a different normalization would not close this diagram.

The stronger endpoint requires a strict exponent budget

The $7/8$ argument changes the average and compensates unwanted Euler factors. Detecting a zero yields both an inverse polynomial and an ordinary character polynomial that are large at the same height. Moment bounds for these two objects give complementary restrictions on the number of contributing rows.

Proof outline. Bin the rows by their scale and detector size, apply the two restrictions, and choose the cutoff where they balance. The resulting exponent inequality is checked on a compact parameter region. The source’s “Compensated endpoint certificate” rewrites its polynomial numerator as a square plus nonnegative terms and obtains the explicit margin $49/440640$. With $\Delta ={\beta }_{\ast }-7/8$, the inequality used in the continuation argument is

$$E_{\mathrm{actual}}-\Delta \le -\frac{49}{440640}-\frac{51}{64}\Delta +{ϵ}_{\mathrm{total}}<0.$$

Here $E_{\mathrm{actual}}$ measures the high-side error relative to the scale at $7/8$, and ${ϵ}_{\mathrm{total}}$ collects adjustable losses. Subtracting $\Delta$ converts to the scale at the hypothetical zero boundary. Choose the losses below the strict margin. This explains why an error need not be small relative to the low-side bound to be sufficient for continuation.

The real-zero companion uses divisibility without changing the determinant

The real-zero paper selects independent evaluation rows greedily in increasing weighted degree. For an admissible prime, its evaluation points satisfy a congruence ${\theta }^p\equiv g_p(\theta )(\mathrm{mod}p)$, where $g_p$ is one of the fixed conjugations. Write a row exponent as ${\alpha }_1=pk+r$, with $0\le r<p$.

Proof outline. Replace the factor ${\theta }^{{\alpha }_1}$ by ${\theta }^r({\theta }^p-g_p(\theta ){)}^k$. Every new entry is divisible by $p^k$. Expanding the replacement gives the original row plus rows of strictly smaller weighted degree, since $p$ exceeds the fixed weight parameter. The greedy rule expresses these in earlier retained rows, so the change of rows is triangular with diagonal one and preserves the determinant exactly. Multiplying the extracted prime powers, then taking the field norm, gives a lower bound on a nonzero integer. A size estimate on all conjugates gives the upper bound. Choose the degree weight first, the exponent in $N=\lceil q^{\gamma }\rceil$ second, and only then let the conductor grow under the hypothetical condition $(1-\beta )\log q\to 0$. In that order the error terms disappear while the strict $\log U$ versus $(3/4)\log U$ advantage survives. (OpenAI, 2026c)

🏷️ Connections

Cubic Gauss sums explains a related theta and Poisson mechanism. Its first-moment conclusion is not used as an input to the $7/8$ paper. Primitive roots requires a different Hecke family over varying cyclotomic fields; the Dirichlet half-plane alone does not supply that statement.

🏷️ Manuscripts and verification scope

• The Quasi-Riemann Hypothesis: A Zero-Free Half-Plane Re(s)>7/8 — pinned manuscript and companion source.
• The Quasi-Riemann Hypothesis — pinned manuscript and companion source.
• Uniform exclusion of Landau—Siegel zeros — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. The page selects the 7/8 zeta, Dirichlet, and Hecke statements and the logarithmic real-zero gap; later applications are outside its stated scope. No Lean build, axiom audit, or manuscript-to-formal-statement equivalence check was performed for these notes.

OpenAI collection provides prerequisites and the reading sequence; coverage register records completed and pending families.

🐻  OpenAI 2026a. The Quasi-Riemann Hypothesis.

🐻  OpenAI 2026b. The Quasi-Riemann Hypothesis: A Zero-Free Half-Plane Re(s)>7/8.

🐻  OpenAI 2026c. Uniform exclusion of Landau–Siegel zeros.