OpenAI claim summary for Math-2776
mathematicsTop Open Problemscontributionliterature_reportClaim: literature report
Summary of a claim published in openai/math; not a claim by this platform.
This question uses the source's determinant-one $\mathrm{SL}_r$ character variety and ambient integrality convention. Its theorem gives one finite number-field extension with dense integral points for every smooth complex curve and rank, for exact quasi-unipotent boundary classes, for their characteristic polynomials, and without boundary conditions. It imposes no genus or Jordan-form restriction. The theorem does not assert the corresponding statement for unrestricted $\mathrm{GL}_r$ representations.
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Manuscript
Contents
Integral-point density on curve character varieties
OpenAI's claim
OpenAI presents the following claims relevant to this question, with manuscript numbers following paper order in Sources. In manuscript 1 (family 027; 2026-09-25), the asserted result is that every component of the $\mathrm{SL}_r$ character variety of a smooth complex curve has Zariski-dense integral points after one finite field extension, including prescribed exact quasi-unipotent boundary classes.Mathematical statement of the claimed result
Manuscript 1.
Take $r\ge1$ and an algebraic curve $X/\mathbb C$ that is connected and smooth. Let $M_r(X)$ be the natural integral $\operatorname{SL}_r$ character scheme. At the punctures of a smooth projective completion, prescribe quasi-unipotent conjugacy classes $C_i\subset\operatorname{SL}_r(\mathbb C)$ and let $Y$ be their reduced relative locus of semisimple representations. Choose a number field $K\subset\mathbb C$ containing all boundary eigenvalues. The claim gives one finite number-field extension $L/K$ such that \[ \overline{M_r(X)(\mathcal O_L)\cap Y(\mathbb C)}^{\rm Zar}=Y. \] Integrality means extension to the ambient character scheme over the full integer ring, not avoidance of smaller Jordan strata at all reductions. Prescribing only boundary characteristic polynomials with root-of-unity roots, or no boundary data, also works. For the relative cases and for projective $X$, distinct primes $p,q>r$ permit \[ F=K(\sqrt2,\zeta_{pq}),\qquad L=F(\zeta_r,\{u^{1/r}:u\in\mathcal O_F^{\times}\}). \] Without boundary data one may start with $K=\mathbb Q$.Outline of the argument
For manuscript 1 (family 027; 2026-09-25), filtered local systems are cut along seams and glued with integral splittings. Rank induction handles closed and nonclosed facets together, while flag comparisons recover exact nonsemisimple boundary data rather than only its semisimplification.Formal verification
No Lean formalization is listed at this snapshot for manuscript 1 (family 027; 2026-09-25).This site has not executed the Lean code. Described statement coverage is not an independent proof check.
Status caveat
The manuscripts were produced by an internal OpenAI model. OpenAI cautions that results without formalization can contain mistakes. This site has not independently checked the proofs; the label records the scope of OpenAI's claim.Sources
- OpenAI, Integral points on character varieties of curves, 2026-09-25; citation key
OAI:Integral-points-on-character-varieties-of-curves-September-25-2026. PDF (GitHub); PDF (direct); Manuscript page and citation; Latest version; Statement source: thm:main, eq:main-field. - OpenAI catalogue entry.