A counterexample to Griffiths’ positivity conjecture

The explanation

Picture the Riemann sphere: all the complex numbers, plus one extra point called infinity. The paper uses two motions of this sphere, called A and B. A multiplies every number by a fixed number whose angle is 120 degrees. So A is a third of a turn about the points 0 and infinity. Three steps bring every point home. B also returns home after three steps, but it turns about two other points. A and B have no fixed point in common.

Now combine them: do B, then A. In suitable coordinates, this combined motion multiplies every number by about 3.55. Each step pushes points away from one of its fixed points and toward the other. That number comes from its matrix, whose two stretch factors are about 1.883 and 0.531. Their ratio is about 3.55.

Try to sprinkle sand on the sphere so that A and B both leave the pattern unchanged. The amount of sand is fixed in advance, and it is not zero. The combined motion would drag any sand that sits away from its two fixed points. So all the sand must sit on those two points. A must also leave the pattern unchanged, so it can at most swap the two piles. But a swap needs two steps to undo, and A returns home after three. So A fixes both piles, and so does B. That gives A and B a common fixed point, which they lack. So no such sand exists. This small clash drives the whole paper.

The sand matters because of a question Phillip Griffiths asked in 1969. Geometers often attach a small flat space of directions to each point of a shape. The whole family is called a vector bundle. Here each of those spaces is a plane, so the bundle has rank two. A Hermitian metric is a smooth rule for measuring lengths in each plane, by a formula like Pythagoras'. Its curvature records how those rulers bend from point to point. The metric is strictly Griffiths-positive when that curvature is positive in every pair of directions. Such a metric forces an algebraic kind of positivity called ampleness. Griffiths' conjecture asks for the reverse: every ample bundle should carry such a metric.

For bundles with one direction at each point, the two notions agree. On curves, which are one-dimensional shapes, Hiroshi Umemura proved the reverse direction. He did it first for elliptic curves and then for compact Riemann surfaces. In 2026, Rei Murakami gave an analytic proof of the full curve case. So a surface is the first place a counterexample could live. Other work found strong curvature only after changing the bundle, by twisting it or taking powers. Bo Berndtsson did this in 2009, and Kefeng Liu, Xiaofeng Sun and Xiaokui Yang in 2013. Those changes give no such metric on the original bundle. Two days before this paper's date, Yun-Heng Du and Song-Yan Xie posted a preprint. It states an ample rank-two bundle on a different surface with no metric even of nonnegative curvature.

This paper works on the quadric surfaceSource 1. Its points are pairs (x, y), each taken from its own Riemann sphere. Inside it sit three curves: the pairs where y equals x, where y equals Ax, and where y equals Bx. The paper glues two simple bundles together along these curves, using A and B to match them. The matching works because the matrices for A, B and one combination of them repeat. They repeat after 6, 6 and 8 steps. So the paper raises the gluing maps to the 24th power, a common multiple of all threeSource 2. The result is a bundle called G. The paper then pulls G back by the map that raises each coordinate to the m-th power. It also adds one fixed twist. Call the result E_m. Every E_m is ample, for every positive whole number mSource 3.

The hard part is ruling out every metric. A metric can be chosen freely at each point, so there are endless candidates. No one can test them one by one. The paper instead supposes that infinitely many E_m carry a strictly positive metricSource 4. It averages each one over the m-th power map and lets m grow. The limit is a nonzero object that still respects the three gluing curves. On one sphere it becomes sand left unchanged by A and B. The opening clash says that sand cannot exist. So only finitely many E_m can carry such a metric.

That gives a threshold m_0. From m_0 on, no E_m has a strictly positive metric. The threshold comes from a limit argument, so the paper cannot say what m_0 isSource 5. Smaller members of the family may or may not carry such metrics. The result also excludes only strictly positive metrics, not those whose curvature is merely nonnegativeSource 6. And the metric must be HermitianSource 7. Shoshichi Kobayashi showed in 1975 that ampleness does match a curvature condition for a looser kind of length rule.

How sure is this? A computer proof checker called Lean has checked a formal version of the main theorem. It covers the bundle G, ampleness of every E_m, and the missing metrics for all large m. The paper's separate, more explicit ampleness certificate is outside that check. No specialist review has been published.

What it doesn't show. This does not say which members of the family first fail, or exclude metrics with merely nonnegative curvature.