Katok's entropy rigidity conjecture

The explanation

Picture a closed three-dimensional space with constant curvature −1, a hyperbolic three-manifold. It is negatively curved, so nearby geodesics, the straightest paths, spread apart exponentially. One way to see this is to count closed geodesics. On this space the number with length at most L grows roughly like e^(2L)/(2L). For L = 10 that is about 24 million. The 2 in the exponent is the topological entropy, the growth rate of the motion's complexity. In dimension n with curvature −1 it equals n − 1. If we double all lengths, the curvature becomes −1/4 and the entropy halves to 1.

There is a second number. Liouville measure is the natural way to spread mass evenly over all positions and directions. Its entropy measures the complexity a typical path sees. A general principle says it is never larger than the topological entropySource 1. For the hyperbolic space the two are equal. Katok's conjecture asks whether equality, in negative curvature, picks out exactly the locally symmetric metricsSource 2. A metric is locally symmetric if near each point it looks the same in every reflected direction. In negative curvature these are the real, complex, quaternionic and Cayley hyperbolic spacesSource 3. So local symmetry need not mean constant curvature.

Katok proved the surface case in 1982Source 4. In higher dimensions, Flaminio showed that a key surface step failsSource 5. Humbert later proved the result for metrics very close to a real or complex hyperbolic oneSource 6. The paper claims the full result. For every closed negatively curved manifold of dimension at least three, the two entropies are equal if and only if the metric is locally symmetricSource 7. When the metric is not locally symmetric, the two natural measures are then mutually singular, meaning they live on separate setsSource 8.

The main obstacle is regularity across leavesSource 9. Negative curvature splits each direction of motion into stable and unstable parts. These form smooth surfaces called leaves. But the way the leaves sit next to each other need not be smooth. The paper shows that equal entropies force this missing smoothness. It works with finite Taylor expansions, then uses algebra to sort the possible symmetries into two cases. One case is ruled out by the geometry at infinity. The other gives a smooth splitting. A theorem of Benoist, Foulon and Labourie then yields the symmetric modelSource 10.

How sure is this? There is no Lean check. A computer proof checker has not looked at any part of it. The proof is long and combines dynamics, normal forms and Lie algebra theory. No specialist review of the paper has been published.

What it doesn't show. This does not cover surfaces, which Katok already settled, or manifolds with some zero or positive curvature. It has no Lean check.