# Filling area conjecture

In mathematics, in Riemannian geometry, Mikhail Gromov's filling area conjecture asserts that among all possible fillings of the Riemannian circle of length 2π by a surface with the strongly isometric property, the round hemisphere has the least area. Here the Riemannian circle refers to the unique closed 1-dimensional Riemannian manifold of total 1-volume 2π and Riemannian diameter π.

## Explanation

To explain the conjecture, we start with the observation that the equatorial circle of the unit 2-sphere

$S^2 \subset \R^3 \,\!$

is a Riemannian circle S1 of length 2π and diameter π. More precisely, the Riemannian distance function of S1 is the restriction of the ambient Riemannian distance on the sphere. This property is not satisfied by the standard imbedding of the unit circle in the Euclidean plane, where a pair of opposite points are at distance 2, not π.

We consider all fillings of S1 by a surface, such that the restricted metric defined by the inclusion of the circle as the boundary of the surface is the Riemannian metric of a circle of length 2π. The inclusion of the circle as the boundary is then called a strongly isometric imbedding of the circle. In 1983 Gromov conjectured that the round hemisphere gives the "best" way of filling the circle among all filling surfaces.

## Relation to Pu's inequality

An animation of the Roman surface representing RP2 in R3

The case of simply-connected fillings is equivalent to Pu's inequality for the real projective plane RP2. Recently the case of genus-1 fillings was settled affirmatively, as well (see Bangert et al). Namely, it turns out that one can exploit a half-century old formula by J. Hersch from integral geometry. Namely, consider the family of figure-8 loops on a football, with the self-intersection point at the equator (see figure at the beginning of the article). Hersch's formula expresses the area of a metric in the conformal class of the football, as an average of the energies of the figure-8 loops from the family. An application of Hersch's formula to the hyperelliptic quotient of the Riemann surface proves the filling area conjecture in this case.