gpt 6 astra pro attempt on Math-77

Siegel Zeros

Definitions and mathematical statement

A primitive quadratic character $\chi$ has minimal modulus, its conductor $q$, and values in $\{0,\pm1\}$. Using its Dirichlet $L$-function, the uniform zero-free assertion is \[ \exists\delta>0\ \forall q\ge3\ \forall\chi\text{ primitive quadratic of conductor }q, \quad L(\beta,\chi)\ne0\quad\left(1-\delta/\log q<\beta<1\right). \] The same $\delta$ must work for all conductors, and only real $\beta$ are quantified here. A separate positive zero-free radius for each individual character would not establish the target.

Formulation and concept references: [S1].

Short English statement

Can one uniformly exclude exceptionally close real zeros of quadratic Dirichlet $L$-functions near one?

Research attempt

Current frontier.

Lu--Zaman--Zhao report a rigorous computational exclusion of real zeros in $\sigma\ge1-1/(5\log q)$ for quadratic characters with $q\le10^{10}$ [S1], Theorem 1.1. Their verified finite range is not an all-conductor theorem. Their algorithm's asymptotic termination analysis uses GRH, separately from the finite verifications. The established zero-free-region theorem allows at most one exceptional real zero rather than excluding that zero. The distinction between lower bounds for $L(1,\chi)$ and a zero-free interval of width $1/\log q$ is analyzed by Friedlander--Iwaniec [E1], Sections 1--2. These are the relevant inputs; a general proof claim retained in [S2] is not used as a theorem.

Attack route.

Three possible routes are: control the variation of $L$ relative to its value at one; exploit positivity of the coefficients of $\zeta L$; or obtain a uniform lower bound for the positive prime sums used by the numerical exclusion method. The first requires an estimate proportional to $L(1,\chi)$, not just an absolute derivative estimate. The second must continue through a pole without illegally continuing a positive series. The third requires uniform information on splitting primes in a range where exceptional zeros are precisely an obstruction. We carry out the first route explicitly and test the positivity shortcut against a model.

Attempt: an unconditional derivative calculation.

Let $q\ge3$, $\ell=\log q>1$, and let $\chi$ be primitive quadratic. It is nonprincipal. Set $A(t)=\sum_{n\le t}\chi(n)$. Periodicity and the zero sum of a nonprincipal character over one full period give $A(q)=0$ and $|A(t)|\le q$. Partial summation yields, for real $s>0$, \[ L(s,\chi)=\sum_{n\le q}\frac{\chi(n)}{n^s} +s\int_q^\infty A(t)t^{-s-1}\,dt. \tag{43.1} \] The integral converges absolutely and agrees with the continuation from $s>1$. On any compact subset of $s>0$, its derivative is dominated by an integrable multiple of $(1+\log t)t^{-1-\sigma_0}$, where $\sigma_0>0$. Differentiation is therefore justified, giving \begin{align*} L'(s,\chi)={}&-\sum_{n\le q}\frac{\chi(n)\log n}{n^s} +\int_q^\infty A(t)t^{-s-1}\,dt\\ &-s\int_q^\infty A(t)(\log t)t^{-s-1}\,dt. \tag{43.2} \end{align*} Take $1-1/(4\ell)\le s\le1$. Then $s\ge3/4$ and $q^{1-s}\le e^{1/4}<2$. The finite sum has absolute value at most \[ e^{1/4}\sum_{n\le q}\frac{\log n}{n} \le e^{1/4}\ell(1+\ell)<4\ell^2. \] The two integral terms in $43.2$ together have absolute value at most \[ q^{1-s}\left(\frac1s+\log q+\frac1s\right) \le2\left(\ell+\frac83\right)<8\ell^2. \] In particular the deliberately nonoptimal bound \[ \sup_{1-1/(4\ell)\le s\le1}|L'(s,\chi)|\le16\ell^2 \tag{43.3} \] is uniform in $q$ and $\chi$. The order of this estimate is familiar from [E1], equation $1.4$; the calculation above supplies an elementary proof, not an improvement on that literature.

If a real zero $\beta$ lies in this interval, the fundamental theorem of calculus and the classical fact $L(1,\chi)>0$ imply \[ 0<L(1,\chi)=\int_\beta^1 L'(s,\chi)\,ds \le16(1-\beta)\ell^2. \tag{43.4} \] Here positivity at one is the usual nonvanishing and positivity theorem for real nonprincipal Dirichlet characters; it is also implicit in the special-value bounds discussed in [S1] and [E1]. Thus even a hypothetical uniform estimate $L(1,\chi)\ge c/\ell$ would yield from this route only $1-\beta\ge c/(16\ell^3)$ inside the working interval. That is two powers of $\log q$ short of the required scale. Increasing the finite verification range does not repair this asymptotic loss.

Attempt: a relative-variation criterion.

Here is a sufficient hypothesis that identifies the missing cancellation without assuming a value for $L(1,\chi)$: \[ \exists C>0\ \ \forall\chi,q:\quad \sup_{1-1/(4\log q)\le s\le1}|L'(s,\chi)| \le C(\log q)L(1,\chi). \tag{43.5} \] If $43.5$ holds, the target follows with $\delta=\min\{1/4,1/(2C)\}$. Indeed, a zero with $0<1-\beta<\delta/\log q$ would give \[ L(1,\chi)\le C(1-\beta)(\log q)L(1,\chi) <\tfrac12 L(1,\chi), \] a contradiction. This is a complete proof of the conditional implication, not a proof of $43.5$. A one-sided integral estimate could be enough; absolute control over the whole interval is stronger than what the contradiction uses. In particular $43.5$ should not be described as an established equivalence or as an independently easier conjecture.

To seek $43.5$, separate the principal part of $43.2$ into sign classes. Its finite sum is \[ -\sum_{\substack{n\le q\\\chi(n)=1}}\frac{\log n}{n^s} +\sum_{\substack{n\le q\\\chi(n)=-1}}\frac{\log n}{n^s}. \] Bounding both sums separately destroys the cancellation that $43.5$ requires when $L(1,\chi)$ is small. Character periodicity supplies $43.3$ but does not relate this cancellation to the small special value. Friedlander--Iwaniec's more refined relation to a smoothed positive coefficient sum explains why that link is substantive [E1], Proposition 2.1. No such uniform relative estimate has been proved in this attempt.

Attempt: testing positivity across the pole.

For $s>1$ the product has the absolutely convergent expansion \[ Z(s)=\zeta(s)L(s,\chi)=\sum_{n\ge1}\frac{a(n)}{n^s}, \qquad a(n)=\sum_{d\mid n}\chi(d)\ge0. \tag{43.6} \] The last assertion follows prime by prime. If $\chi(p)=1$, then $a(p^j)=j+1$; if $\chi(p)=-1$, it equals one for even $j$ and zero for odd $j$; if $\chi(p)=0$, it equals one. Multiplicativity proves the claim, including $a(m^2)\ge1$. Hence $Z(s)>0$ for $s>1$ and its alternating real derivatives have the signs supplied by a positive Laplace transform. This calculation is checked on small conductors in the accompanying exact script.

An attempted continuation of these signs to $s<1$ is invalid. To expose the exact logical failure, put $t=s-1$ and, for any $\epsilon,C>0$, define \[ Z_\epsilon(t)=C\left(1+\frac{\epsilon}{t}\right) =C\int_{[0,\infty)}e^{-tu} (\delta_0+\epsilon\,du),\qquad t>0. \tag{43.7} \] It is positive and completely monotone for $t>0$, has a positive-residue simple pole at zero, and nevertheless its meromorphic continuation has a zero at $t=-\epsilon$. After removing the pole, $L_\epsilon(t)=tZ_\epsilon(t)=C(t+\epsilon)$ has $L_\epsilon'(0)/L_\epsilon(0)=1/\epsilon$. Thus positivity of the transform alone cannot control the relative variation. This model is neither a Dirichlet $L$-function nor a counterexample to the problem: it lacks the arithmetic Euler factors and functional equation. It falsifies only the proposed inference from positivity and a simple pole.

Stress test.

The finite computations in the cited paper prove a finite statement, not termination of a conductor-by-conductor argument for all $q$. Applying its GRH-conditional termination estimate here would assume a substantially stronger unresolved zero-location statement. Likewise, excluding all but one exceptional character in a range does not exclude that last character. The derivative calculation covers the stipulated real interval and uses uniform constants; it does not claim complex zero-freeness. Every exchange of derivative and integral in $43.2$ was made in $s>0$ with an explicit integrable majorant, whereas no use of the series $43.6$ was made left of its convergence domain.

For any fixed finite collection of characters, positivity and continuity at one give a positive minimum zero-free radius after rescaling by $\log q$. That elementary compactness argument applies to a finite set, not to an unbounded sequence of conductors. The potential counterexample regime is precisely a sequence for which the rescaled radii tend to zero. None of the inequalities derived above excludes that regime.

Outcome.

Outcome: Conditional reduction. The target follows from the explicit relative-variation estimate $43.5$. The supplied proof of the implication handles all conductors uniformly. What was proved unconditionally is a coarse derivative bound and the resulting weaker special-value-to-zero-distance estimate, together with a countermodel to a positivity-only shortcut. These auxiliary calculations are not claimed to be historically new. The missing cancellation in $43.5$, or an adequate one-sided substitute, has not been established; there is no unconditional exclusion of all Landau--Siegel zeros here.

Sources

• S1 Rick F. Lu, Asif Zaman and Haonan Zhao, Numerical Computations Concerning Landau–Siegel Zeros; arXiv2602.03626v1 February3,2026. https://arxiv.org/pdf/2602.03626v1. Catalogue formulation source.
• S2 On the Extended, Generalized, and Grand Riemann Hypotheses: A Unified Approach via General Properties of L-Functions, v18. https://www.preprints.org/manuscript/202506.0481/v18. Catalogue research/status reference; not a proof endorsement.
• S3 Numerical Computations concerning Landau–Siegel Zeros. https://arxiv.org/html/2602.03626v1. Catalogue research/status reference; not a proof endorsement.
• S4 Landau-Siegel zeros of Rankin-Selberg L-functions. https://arxiv.org/abs/2601.04189. Catalogue research/status reference; not a proof endorsement.
• E1 John B. Friedlander and Henryk Iwaniec, A note on Dirichlet $L$-functions, Expositiones Mathematicae 36 $2018$, 343--350; arXiv:1701.03771v1 $2017$, Section 1 and Proposition 2.1. https://arxiv.org/html/1701.03771v1. The distinction between special-value estimates and real zero-free intervals. Consulted 27 September 2026.