Spacetime Penrose inequalities: enclosing area, charge, rotation, and anti-de Sitter extensions

The explanation

Take the simplest black hole, the Schwarzschild solution of Einstein's equations. Use units where the speed of light and the gravity constant are 1. A black hole of mass m = 1 then has a horizon of radius 2m = 2. Its area is 4π × 2², which is 16π, about 50.27. In everyday units, a black hole as heavy as the Sun has a horizon radius near 2.95 km. Its area is near 110 square km. Turn this around. If a horizon has area A, the mass should be at least the square root of A/(16π). For A = 16π, that gives 1, so Schwarzschild sits exactly on the line.

Penrose's null-shell argument made this comparison concreteSource 1. It came from a test of cosmic censorship, the idea that collapse ends in a black hole hidden by a horizon. To state it as a theorem, you describe one moment of time by initial data. This is a space, its metric (the rule for distances), and its second fundamental form (how the slice bends inside spacetime). The mass is the ADM mass, read off far away. The horizon is replaced by a trapped boundary, a surface where outgoing light cannot spread out.

The case with zero bending was settled long ago. Bray's conformal flow proves the sharp bound for the total area of a possibly disconnected horizonSource 2. With general bending, nonzero second fundamental form, the method breaks. Neither scalar curvature nor the boundary mean curvature has the required Riemannian signSource 3. That sign problem is the main obstacle.

The papers claim that the enclosing-area form of the spacetime Penrose conjecture holds in their class of dataSource 4. The area used is the smallest area of any surface that encloses the boundary. They do not assume the boundary is outermost or area-minimizingSource 5. With electric and magnetic charge Q, they claim m ≥ Q and r_A ≤ m + √(m² − Q²), where r_A is the enclosing-area radiusSource 6. For m = 1 and Q = 0.6, the radius is at most 1.8, so the area is at most about 40.7. A rotating version, the Kerr–Newman Penrose inequality, is also claimedSource 7. One paper builds examples that break that rotating version when the angular momentum omits its electromagnetic partSource 8. Another set claims versions for a negative cosmological constant, the anti-de Sitter settingSource 9. For equality cases, the papers claim the data must be a slice of a Schwarzschild–Tangherlini exteriorSource 10.

How sure is this? Lean, a computer proof checker, covers only one supporting step. It formalizes replacing one type of curved end by a flat end while keeping energy and area under control. It also checks that Schwarzschild data gives equality at every positive mass. The main inequality, the charged, rotating and anti-de Sitter versions, and the equality theorems rest on thirteen long papers. No specialist review of them has been published.

What it doesn't show. This does not prove cosmic censorship, and Lean does not check the main inequality, the charged or rotating versions, or the rigidity theorems.