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Paper 08 · Convex geometry

The largest lattice body with one interior lattice point

How large can a convex body be when its center is its only interior whole-number point?

Abhirup GhoshAugust 202613 min read

Put a convex body on the integer grid. Its balance point is the origin, and the origin is its only interior grid point. In 1964, Eugène Ehrhart predicted the exact largest possible volume. This manuscript claims the sharp bound in every dimension: no such body beats the centered simplex.1manuscript

Volume on an integer grid

A lattice point has integer coordinates. A body is convex when every segment joining two of its points remains inside it; its barycenter is the geometric center of mass.

Assume the barycenter is the origin and that no other lattice point lies strictly inside. Ehrhart asked for the exact volume ceiling in n dimensions. He predicted (n+1)ⁿ/n!, reached by the n-dimensional analogue of a triangle.2manuscript

The hard part is the factorial. Symmetry arguments can control overall width, but they forget how n independent directions combine to make a simplex.

Lattice points and slope bounds
A centered lattice triangle scaled around the originAt the sharp scale the origin is the only interior integer point. Enlarging the triangle admits more interior lattice points.
Scale 1.00×
1 interior lattice point

At the sharp centered simplex, the origin is the only interior lattice point.

The triangle is the two-dimensional sharp example. The theorem concerns every full-dimensional compact convex body, not only polygons or symmetric shapes.

What was known—and what kept failing

Ehrhart proved the planar case and the all-dimensional simplex case. For arbitrary centered bodies, symmetrizing the body and applying Minkowski’s theorem gave vol(K) ≤ 4ⁿ, far above the sharp simplex value.3manuscript

Thin-shell methods later improved the general estimate to 4ⁿe−c√n. This was a meaningful advance, but it did not recover the exact factorial constant.4Huang

Toric geometry had already proved the sharp inequality for broad structured classes, including certain rational polytopes with controlled facet normals. That route did not cover every convex body—especially an arbitrary irrational one.5Berman

Why harmonic symmetrization stalls

A harmonic symmetral turns K into a centrally symmetric lattice-free body, so Minkowski applies. Remarkably, it has exactly the right volume for the extremal simplex. But the needed comparison depends on the full distribution of simplex coordinates; centroid bounds see only coarse averages and lose the factor n!. The method recognized the right extremizer without explaining its volume.6walkthrough

Stop symmetrizing the body. Convert it into a convex potential, convert lattice points into holomorphic monomials, and make the desired volume appear between a lower and an upper slope of one convex function.

Astra’s route: from a real body to a complex slope

1. Give any centered body a toric potential

A theorem of Berman and Berndtsson supplies a smooth strictly convex function φ on ℝⁿ whose gradient maps ℝⁿ onto the interior of K and satisfies det D²φ = e−φ. Thus the density e−φ has total mass vol(K). Crucially, this works directly for an arbitrary centered convex body; rational vertices are unnecessary.7Berman

2. Turn interior lattice points into monomials

Move to the complex torus (ℂ*)ⁿ and read ordinary real coordinates as logarithms of complex radii. At level k, the square-integrable Laurent monomials zm are indexed exactly by integer points m inside kK. Therefore dim Hk = #(int(kK) ∩ ℤⁿ), and the original “one interior lattice point” assumption becomes H1 = ℂ: only the constant monomial survives.8manuscript

3. Jet counting forces a lower slope

At p = (1,…,1), filter Hk by order of vanishing. Vanishing to order j kills every Taylor coefficient of degree below j, at most (n+j−1 choose n) conditions. Summing these dimensions and passing to the lattice-counting limit yields an initial-slope lower bound containing (n! vol(K))1/n. This is where the missing n! finally appears: it is the leading coefficient of n-variable Taylor-jet counting.9manuscript

Vanishing to order j removes one coefficient for every multi-index inside an n-dimensional simplex. Its leading count is jⁿ/n!, which is where the previously missing factorial enters.

4. Level one turns Bergman positivity into ordinary convexity

The filtered bases define a limiting ray of complex potentials ψt and a normalized partition function L(t). Simply integrating pointwise-convex rays would be invalid: a variance term can have the wrong sign. The walkthrough explicitly identifies this failed shortcut.10walkthrough

The rescue is H1 = ℂ. Because the relevant holomorphic space has rank one, its Bergman kernel is exactly the inverse partition function. Berndtsson’s positivity theorem then makes L genuinely convex—not merely pointwise plausible.11Berndtsson12manuscript

5. A shrinking complex ball forces the upper slope

A section that vanishes to high order becomes tiny near p. In the filtered ray, the compensating neighborhood has complex radius proportional to e−t/2. Since ℂⁿ has real dimension 2n, its volume scales like e−nt. This gives L(t) ≤ nt + O(1), hence convexity forces L′+(0) ≤ n.13manuscript

Convexity makes the initial slope obey both estimates. Comparing them leaves only the sharp simplex volume.

The factorial and the exponent meet in one squeeze

Real body. Monge–Ampère transport builds φ and remembers the full volume V.
Integer grid. Interior points of kK become a basis of monomials in Hₖ.
Taylor jets. Vanishing-order counts force L′₊(0) ≥ n/(n+1) · (n!V)¹⁄ⁿ.
Rank one. H₁ = ℂ lets Bergman positivity prove L is convex.
Complex ball. A radius e⁻ᵗ⁄² in real dimension 2n forces L′₊(0) ≤ n.
Compare. The two slopes imply V ≤ (n+1)ⁿ/n!.

The walkthrough’s central insight is that the lower constant comes from an n-dimensional simplex of Taylor multi-indices, while the upper n comes from the 2n real dimensions of a complex ball.14walkthrough

Technical layer · the slope calculation

Set V = vol(K) and cK = (n!V)1/n. The jet filtration Fkj has codimension at most (n+j−1 choose n). After truncating vanishing orders at cKk, a Riemann-sum limit gives ∫g dμ ≥ [n/(n+1)]cK. Because L′+(0) = ∫g dμ, this is the lower slope.15manuscript

The potential obeys ∇φ(ℝⁿ) = int(K) and det D²φ = e−φ. Weighted integrability gives Hk = span{zm : m ∈ int(kK)∩ℤⁿ}, while lattice counting gives dim Hk/kⁿ → V. The rank-one level H1 = ℂ converts Berndtsson positivity into convexity of L; the local ball estimate gives its upper slope. Hence [n/(n+1)](n!V)1/n ≤ n.16manuscript

Exact theorem, equality, and limits

Theorem 1.1. Let K ⊂ ℝⁿ be full-dimensional, compact, and convex, with barycenter 0. If int(K)∩ℤⁿ = {0}, then vol(K) ≤ (n+1)ⁿ/n!. The centered simplex (n+1)Δn − (1,…,1) satisfies the hypotheses and reaches equality.17manuscript

The theorem establishes the best numerical constant. It does not classify all bodies attaining equality. The expected uniqueness statement—that every extremizer is a unimodular image of the centered simplex—remains a separate refinement.18Nill

The public EhrhartVolumeInequality.lean file contains a formal set-level theorem with convexity, compactness, nonempty interior, barycenter, and unique-interior-lattice-point hypotheses leading to the sharp bound. This is substantial machine-checked evidence for the encoded theorem, but still warrants independent review of the definitions and analytic interfaces.19formal artifact

Is the claim overhyped?

If the manuscript and formalization withstand expert scrutiny, this is a major resolution. The manuscript settles the numerical inequality; the equality classification and independent validation remain open.

Supported

“Ehrhart’s volume conjecture is proved.”

At manuscript level, yes: Theorem 1.1 is exactly the sharp all-dimensional numerical inequality with an equality example.20manuscript

Qualified

“The extremal body is completely understood.”

No. The centered simplex is an extremizer, but the manuscript explicitly does not classify every equality case.21manuscript

Qualified

“Astra solved it.”

That is OpenAI’s attribution. Its announcement, manuscript, walkthrough, and Lean file are same-release evidence—not an independent referee report or reproduction.22announcement

Full bibliography

22 fully annotated sources
  1. 01 · primary manuscript

    OpenAI. The Sharp Inequality in Ehrhart’s Volume ConjectureChapter 8, abstract and Theorem 1.1, pp. 217–218. The released manuscript and source of the main theorem; not independent validation.

  2. 02 · primary manuscript

    OpenAI. The Sharp Inequality in Ehrhart’s Volume ConjectureIntroduction, pp. 217–218. Problem statement, sharp simplex, and historical context.

  3. 03 · primary manuscript

    OpenAI. Previous work on Ehrhart’s conjecturep. 218, “Previous work”. Locates Ehrhart’s planar and simplex cases and the general 4^n symmetrization bound.

  4. 04 · peer-reviewed predecessor

    Han Huang, Boaz Slomka, Tomasz Tkocz, and Beatrice-Helen Vritsiou. Improved bounds for Hadwiger’s covering problem via thin-shell estimatesJEMS 24 (2022), Proposition 6.2. A pre-resolution general bound of 4^n exp(−c√n).

  5. 05 · peer-reviewed predecessor

    Robert Berman and Bo Berndtsson. The volume of Kähler–Einstein Fano varieties and convex bodiesCorollary 1.4 and Theorem 1.5. Proved the sharp inequality for broad structured classes of convex bodies and rational polytopes.

  6. 06 · reasoning walkthrough

    OpenAI. Reasoning Walkthroughs: The Sharp Ehrhart InequalityChapter 9, §9.2, pp. 36–37. Retrospective account of harmonic symmetrization and why it lost the factorial; not independent evidence.

  7. 07 · peer-reviewed predecessor

    Robert Berman and Bo Berndtsson. Real Monge–Ampère equations and Kähler–Ricci solitons on toric log Fano varietiesTheorem 1.1. Existence of the convex transport potential for an arbitrary centered body.

  8. 08 · primary manuscript

    OpenAI. The Sharp Inequality in Ehrhart’s Volume ConjectureLemma 2.1, pp. 219–220. Identifies weighted Laurent monomials with interior lattice points of kK.

  9. 09 · primary manuscript

    OpenAI. The Sharp Inequality in Ehrhart’s Volume ConjectureLemma 3.1 and Proposition 3.2, pp. 222–224. Jet counting and the sharp lower bound on the initial slope.

  10. 10 · reasoning walkthrough

    OpenAI. Reasoning Walkthroughs: The Sharp Ehrhart InequalityChapter 9, §§9.5–9.6, pp. 38–39. Explains the limiting Bergman ray and the finite-level convexity pitfall.

  11. 11 · peer-reviewed predecessor

    Bo Berndtsson. Subharmonicity properties of the Bergman kernelTheorem 1.1; Annales de l’Institut Fourier 56 (2006). Positivity theorem used to obtain convexity of the rank-one partition function.

  12. 12 · primary manuscript

    OpenAI. The Sharp Inequality in Ehrhart’s Volume ConjectureLemma 4.1, pp. 224–225. Uses H1 = C to identify the Bergman kernel with the inverse partition and prove convexity.

  13. 13 · primary manuscript

    OpenAI. The Sharp Inequality in Ehrhart’s Volume ConjectureLemma 4.2, p. 225. Shrinking complex-ball estimate giving the upper slope L′+(0) ≤ n.

  14. 14 · reasoning walkthrough

    OpenAI. Reasoning Walkthroughs: The Sharp Ehrhart InequalityChapter 9, §§9.3–9.7, pp. 37–40. Discovery-level proof map from jet counting through the slope squeeze.

  15. 15 · primary manuscript

    OpenAI. The Sharp Inequality in Ehrhart’s Volume ConjectureEquation (2) and §§2–4, pp. 218–225. Technical lower- and upper-slope inequalities.

  16. 16 · primary manuscript

    OpenAI. The Sharp Inequality in Ehrhart’s Volume ConjectureEquations (3)–(8), pp. 219–220. Monge–Ampère normalization and lattice Bergman spaces.

  17. 17 · primary manuscript

    OpenAI. The Sharp Inequality in Ehrhart’s Volume ConjectureTheorem 1.1, p. 218. Exact hypotheses, bound, and sharp centered simplex.

  18. 18 · peer-reviewed predecessor

    Benjamin Nill and Andreas Paffenholz. On the equality case in Ehrhart’s volume conjectureConjecture 1.1 and Theorem 1.4. Equality context for structured classes; the global equality classification remains separate.

  19. 19 · formal certificate

    OpenAI. EhrhartVolumeInequality.leanlines 55737–55755. Lean declaration of the set-level sharp volume inequality under convexity, compactness, full-dimensionality, barycenter, and lattice hypotheses.

  20. 20 · primary manuscript

    OpenAI. The Sharp Inequality in Ehrhart’s Volume ConjectureTheorem 1.1. Supports the numerical inequality and sharpness claim at manuscript level.

  21. 21 · primary manuscript

    OpenAI. The Sharp Inequality in Ehrhart’s Volume ConjectureIntroduction, equality discussion. Explicitly says the manuscript does not classify all equality cases.

  22. 22 · official announcement

    OpenAI. Ten advances in mathematicsitem 8. Evidence for OpenAI’s attribution and framing only; not independent mathematical review.