# Filling area conjecture

In differential geometry, Mikhail Gromov's filling area conjecture asserts that among all possible Riemannian surfaces (not to be confused with Riemann surfaces) that isometrically fill a Riemannian circle of given length, the hemisphere has the least area.

## Mathematical statement

Every Riemannian manifold M (including any curve or surface in Euclidean space) is a metric space, in which the (intrinsic) distance dM(x,y) between two points x and y of M is the infimum of the lengths of the curves that go along M from x to y. In particular, any Riemannian circle of total length 2π is a metric space with diameter π, in fact any point x of the circle has a unique opposite point y in the circle, at the maximum distance π from x.

A compact Riemannian surface is said to fill a Riemannian circle if the boundary of the surface is the circle, and the filling is said isometric if for any pair of points on the circle, the intrinsic distance between them along the surface is the same, and not less, than the intrinsic distance along the circle. In other words, the filling is isometric if it does not introduce shortcuts between the points of the circle.

For example, in three-dimensional Euclidean space, the circle

${\displaystyle S^{1}=\{(x,y,0):\ x^{2}+y^{2}=1\}}$

is filled both by the flat disk

${\displaystyle D^{2}=\{(x,y,0):\ x^{2}+y^{2}\leq 1\}}$

and by the hemisphere

${\displaystyle S_{+}^{2}=\{(x,y,z):\ x^{2}+y^{2}+z^{2}=1{\text{ and }}z\geq 0\}.}$

Of these two fillings, only the hemisphere is isometric. The flat disk is not an isometric filling because, in it, the intrinsic distance between any two opposite points is 2, which is less than π.

In 1983, Mikhail Gromov conjectured[1]:13 that the hemisphere has least area among the orientable compact Riemannian surfaces that fill isometrically a circle of given length. In fact he conjectured an analogous statement for higher dimensions as well.

## Known cases and partial results

In 1983, Gromov proved his conjecture restricted to fillings homeomorphic to a disk.[1] He did so by applying Pu's systolic inequality to the closed surface, homeomorphic to the real projective plane, obtained from the filling surface by identifying opposite points of its boundary.

In 2004, the case of fillings of genus one was settled affirmatively, as well.[2] In this case, the non-orientable surface obtained by identifying opposite points of the boundary has an orientable double cover of genus two, and every Riemannian surface of genus two is hyperelliptic. The proof then exploits a 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. 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.

In 2001, Sergei Ivanov generalized the conjecture to allow Finsler metrics, conjecturing that the Euclidean hemisphere has least Holmes–Thompson area among the Finsler surfaces that isometrically fill the circle. He proved his conjecture in the case that the surface is homeomorphic to a disk.[3][4][5] His proof used a calibration and relied on the topological fact that two curves on a disk cross if their endpoints are interlaced on the boundary. The calibration comes from a finite-sum approximation of Santalo’s formula[6] for the area of a metric with no conjugate points in terms of boundary distances, which in turn uses Hamiltonian dynamics of the geodesic flow to adapt a formula of integral geometry to curved surfaces.

In 2002, Burago and Ivanov showed that there exists a Finsler disk that isometrically fills the circle N ≤ 10 times (in the sense that its boundary wraps around the circle N times), but whose Holmes–Thompson area is less than N times the area of the hemisphere.[7] In other words, the Finsler filling area conjecture cannot hold if 2-chains with rational coefficients are allowed as fillings, instead of orientable surfaces (which are equivalent to 2-chains with integer coefficients). This rules out the possibility of proving the Finsler version of the conjecture using projections or calibrations.

In 2010, the same authors showed that if Riemannian manifold that is almost flat cannot be replaced by an orientable Riemannian manifold with the same boundary and less volume without reducing the distance between the boundary points.[8] This implies that if a piece of sphere is sufficiently small (and therefore, nearly flat), then it cannot be replaced by an orientable surface of smaller area without reducing the distances between boundary points. If this result can be extended to larger regions (namely, to an hemisphere), then the filling area conjecture is true. Their proof involved embedding the manifold in ${\displaystyle L^{\infty }}$, and then showing that any isometric replacement of the manifold can also be mapped into the same space, and projected onto the embedded manifold, without increasing its volume. This implies that the replacement has not less volume than the original manifold.