Affine Bernstein rigidity through dimension nine and a smooth dimension-ten counterexample

The explanation

Start with a convex function of three variables, u = (x² + 4y² + 9z²)/2. Its graph is a bowl called an elliptic paraboloid. The Hessian is the table of second derivatives. Here it is the diagonal table 1, 4, 9, and its determinant is 36 everywhere. The affine maximal equation is a fourth-order equation built from that determinantSource 1. When the determinant is constant, every term vanishes, so this bowl solves it. Now try u = cosh(x) + (y² + z²)/2. Its Hessian determinant is cosh(x), which changes. At x = 0 the equation gives −0.8, not 0, so this graph is not a solution.

Solutions are balanced for affine area, a notion of area that stays the same under stretching and shearingSource 1. The affine Bernstein problem asks whether a complete solution must be a paraboloid. Complete means every path of finite length stays on the surface, with no missing edge. Chern posed the two-dimensional caseSource 2. Trudinger and Wang proved it for surfaces with Euclidean completenessSource 3. They reduced higher dimensions to one estimate: a uniform control of how the graph separates from its tangent planesSource 4. Getting that estimate without extra assumptions was the obstacleSource 5.

The second paper proves the answer is yes in dimensions three through nineSource 6. A smooth graph with positive-definite Hessian, solving the equation and complete for ordinary distance, is a paraboloidSource 6. No growth bound is assumedSource 7. The proof rescales the graph and studies convex limits. It counts the directions in which a limit opens up, and shows only one is possible. The dimension limit enters through one quadratic estimate that stays positive only up to dimension nineSource 8.

Trudinger and Wang had found a dimension-ten solution, √(|x|⁹ + t²), but it is not smooth at the originSource 9. The first paper claims a smooth solution over all of ten-dimensional space that is not quadraticSource 10. It keeps the shape of their example but replaces |x|^(9/2) with a smooth convex functionSource 11. The main obstacle is to show an auxiliary function never reaches zero at a finite radiusSource 12. So the range three through nine cannot be extended to ten.

How sure is this? The Lean scope note says the formalization proves the Euclidean-complete graph theorem for dimensions three through nine. The extension to hypersurfaces that are not graphs is outside that note. The dimension-ten counterexample has no Lean check. No specialist review has been published.

What it doesn't show. This does not settle the problem when only the affine metric is complete in dimension ten. The Lean proof covers only the dimension three to nine graph theorem, not the dimension-ten example.