VibeMathedMath problems solved with AI

The symmetric Mahler conjecture

For an origin-symmetric convex body K⊂RnK\subset\mathbb R^n let K∘={y:⟨x,y⟩≤1 for all x∈K}K^\circ=\{y:\langle x,y\rangle\le1\ \text{for all}\ x\in K\} be its polar and P(K)=∣K∣ ∣K∘∣P(K)=|K|\,|K^\circ| its volume product, which is invariant under invertible linear maps. Mahler, in his work on transference in the geometry of numbers, asked how small P(K)P(K) can be. The cube and its polar cross-polytope give 4n/n!4^n/n!, as do all Hanner polytopes (built from intervals by products and convex-hull joins). Known cases before this work: the plane (Mahler), unconditional bodies (Saint-Raymond), zonoids (Reisner), dimension three (Iriyeh-Shibata), and the bound up to a factor cnc^n (Bourgain-Milman). Is ∣K∣ ∣K∘∣≥4n/n!|K|\,|K^\circ|\ge4^n/n! for every origin-symmetric convex body K⊂RnK\subset\mathbb R^n in every dimension, with equality only for linear images of Hanner polytopes?

Result
Proved(see note)
Status
Candidate (review pending)
AI contribution
AI-discovered
Method
Argument
Field
Convex geometry: volume products and polarity
Posed by
Kurt Mahler, Ein Ubertragungsprinzip fur konvexe Korper, Casopis pro pestovani matematiky a fysiky 68 (1939)
Year posed
1939
Years open
87y
Solved
2026-09-22
Model
Unreleased internal OpenAI model
Vendor
OpenAI
Collaborators
—
Verification
Lean-checked, statement unaudited
Publication
Announced
Collection
OpenAI math release (October 2026), version adc7f12
Significance
60 / 100
Disclosed cost
—
Wikipedia
No dedicated article

What was actually shown

Theorem 1.1: for every n≥1n\ge1 and every origin-symmetric convex body K⊂RnK\subset\mathbb R^n, ∣K∣ ∣K∘∣≥4n/n!|K|\,|K^\circ|\ge4^n/n!, with equality iff KK is an invertible linear image of a Hanner polytope. The proof uses a planar conformal lens (a rotation of Gross's uniform-distribution map), a holomorphic mass estimate proved by Stokes' theorem, and, for equality, a metric-median property of the norm and Hansen-Lima's classification. The companion Symplectic Balls in Symmetric Polar Products gives an independent proof of the inequality (not of the equality cases) through the Gromov width of int K×int K∘\mathrm{int}\,K\times\mathrm{int}\,K^\circ. It does not treat the non-symmetric problem (a separate entry) and gives no stability estimate.

What the AI did

The OpenAI math release (github.com/openai/math, commit adc7f12) states that its results were produced by an unreleased internal OpenAI model under one fixed procedure, averaging about three hours of ChatGPT Pro thinking compute per result, across roughly 4,000 posed problems; outputs were then grouped into families and filtered for significance. This result is not among the README's stated exceptions (the Riemann zeta zero-free region work and the Hodge conjecture for CM abelian varieties). The manuscript is credited to OpenAI alone and names no human author. The README also cautions that unformalized results could have issues. OpenAI also released an abridged summary of the model's reasoning for this family (reasoning_traces/symmetric-and-general-mahler-conjectures.pdf).

Verification

No independent mathematician has checked this yet. Checked here: the abstract, introduction and Theorem 1.1 of the TeX source, read against Mahler's symmetric problem as the manuscript cites it; the proof was not refereed. Lean-checked on two Comparator challenges listed in the release's formalization catalogue. MahlerConjecture (OAI.SymmetricMahler.symmetric_mahler, OAI/Analysis/Mahler/MainTheorem.lean) states 4n/n!≤∣K∣ ∣K∘∣4^n/n!\le|K|\,|K^\circ| for every compact convex origin-symmetric K⊆RnK\subseteq\mathbb R^n with nonempty interior, n≥1n\ge1, the polar taken with the coordinate inner product. SymmetricMahlerEquality (OAI.SymmetricMahler.symmetric_mahler_equality, OAI/Analysis/Mahler/Classification.lean) states that equality holds iff KK is a linear image of a Hanner body, defined inductively from centered intervals by products and convex-hull joins. Together they state the headline claim in full. Permitted axioms: propext, Quot.sound, Classical.choice. The statements were read here but not independently audited, and the development was not rebuilt here.

Sources

Changelog1 change

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