No lift at all?

0:00Cones from polynomials

A homogeneous polynomial is hyperbolic in a direction e if it is nonzero at e and, along every line parallel to e, all its roots are real. Precisely, for every point x, the polynomial in t given by p(te − x) has only real roots, and where all of them are nonnegative, x lies in the hyperbolicity cone. Take x₀² − x₁² − x₂², with e pointing along the x₀ axis. The roots are x₀ ± the distance from that axis, and the cone is the ice-cream cone. For the determinant of a symmetric matrix, with e the identity, the roots are the eigenvalues, and the cone is the positive semidefinite matrices. Gårding proved in 1959 that every hyperbolicity cone is convex, and Güler showed that these cones carry the barrier functions of interior-point methods, so they support optimization beyond semidefinite programming. A spectrahedron is a slice of the positive semidefinite cone: the points x where a matrix depending affinely on x is positive semidefinite. For a cone the dependence is linear, and when the matrix is positive definite somewhere, its determinant is hyperbolic, so spectrahedral cones are hyperbolicity cones. Is the converse true? In 3 variables, yes: the theorem of Helton and Vinnikov, with an equivalence of Lewis, Parrilo and Ramana, settled Lax’s original question. The generalized Lax conjecture says that every hyperbolicity cone is a spectrahedron. A weaker version, the projected Lax conjecture of Netzer and Sanyal, allows extra variables: every hyperbolicity cone should be a spectrahedral shadow, the projection of a spectrahedron. Netzer and Sanyal proved this when every nonzero boundary point is a smooth point of the polynomial. Scheiderer found other convex sets that are not shadows, and asked whether his methods could settle the question for hyperbolicity cones. The difference between a slice and a shadow already shows in the plane. The region where x⁴ + y⁴ ≤ 1 is a shadow: add variables u and v with u ≥ x², v ≥ y², and u² + v² ≤ 1, each a 2 × 2 matrix condition. In our check this matches the region at 20,000 random points. But the region is not a spectrahedron. Along a line through an interior point, its quartic boundary polynomial has four roots, and on every such line exactly two are real. Helton and Vinnikov showed that for a spectrahedron, the lowest-degree polynomial vanishing on its boundary has only real roots along every line through an interior point.

2:48The claims

Result 095 is three OpenAI preprints. The first, dated September 24, 14 pages, writes down an explicit polynomial of degree 20 in 23 variables, a 4 × 4 determinant built from two symmetric matrices X and Z and a vector y in 3 dimensions. It claims the polynomial is hyperbolic but its cone is not a spectrahedron, in any matrix size: a counterexample to the generalized Lax conjecture. Dividing out one factor lowers the degree to 16. A companion, dated October 5, 10 pages, shows that this cone is nonetheless a shadow: a 100 × 100 matrix with 307 extra variables describes it exactly. The third preprint, also dated October 5, 29 pages, claims a hyperbolic polynomial whose cone has no semidefinite lift at all, with any number of extra variables and any real coefficients. That disproves the projected Lax conjecture. The third paper’s polynomial is not written down. It is an existence argument in very high dimension, not a bound on the size of representations: with the parameters the paper states, the degree comes to about 67 billion. The sizes in the first two papers are not claimed minimal. The first paper also gives a counterexample to a step in an earlier preprint by González Nevado that claimed the conjecture was true. OpenAI published a Lean formalization of the first paper’s theorem, 101 files and about 6,600 lines, matched against a separate statement. The other two papers are not formalized. We have not run the check.

4:32How the proofs go

The first paper starts from the Choi–Lam form, a polynomial b in two triples of variables, z and y, quadratic in each. It is never negative, yet the only bilinear form whose square fits under it is zero: b vanishes at certain pairs, and along certain curves it grows only like the fourth power of the parameter. We checked these identities exactly. Writing b as zᵀQ(y)z gives a matrix Q(y) that is positive for every y, and from it a positive map Φ. The polynomial is the determinant of (det X) Z − Φ(adj X). As the later papers show, along every line parallel to e its 20 roots are the eigenvalues of one symmetric 20 × 20 matrix, so they are real. On 20 random integer lines, our exact root count found all 20 real each time. Suppose some matrix pencil described the cone. Where X is positive definite, membership means that Z dominates Φ(X⁻¹). The paper rescales the pencil, pushes X⁻¹ and Z⁻¹ toward rank-one projections, and gets an exact identity for kernel projections of the pencil’s blocks. Differentiating one of them produces bilinear forms whose squares fit under b. So they all vanish, which would make b zero everywhere: a contradiction, so no pencil exists. The companion lifts the same cone. It covers the y-space by planes through the vertical axis; on each, Q has a Gram factor linear in the plane’s coordinates, giving a 20 × 20 pencil. A linear functional on trigonometric polynomials replaces the angle. Tested against 5 basis functions, it gives one matrix of size 20 × 5 = 100, and of its 330 coordinates, 23 are the original variables, leaving 307 extra ones. The third paper keeps the determinant construction but lets Q be any quadratic matrix map that is positive at every input. It reduces the question to one convex set: positive matrices A and numbers t with t at least the trace of A†Q(y). To build Q, it uses moment matrices on quadratic forms in 8 variables, a space of dimension 36, starting from a rank-20 seed matrix, with its rank conditions certified by explicit determinants over the field with 7 elements. More than 12,960 copies, and a quadratic form that separates one vector from every limit of the moment images, give Q. Finally, a positive moment functional produces points that violate the defining inequality by order ε⁴, while any finite lift would force feasible points within order ε¹⁶. That contradiction is the theorem.

7:28Where it stands

The three preprints are not peer reviewed and have not been revised. Lean covers the first paper’s theorem: its degree-20 polynomial is hyperbolic and its cone is not a spectrahedron; we have not run it. The exact lift and the headline result, no lift at all, are checked only on paper. The headline proof is closely related to Scheiderer’s local sum-of-squares obstructions. None of the papers is among those OSVerify reviewed, and we have seen no expert review. If it holds, hyperbolic programming truly reaches beyond semidefinite programming: some hyperbolicity cones cannot be described exactly by semidefinite constraints, even with extra variables. No explicit cone without a lift is given, and no small one.