In differential geometry, Mikhail Gromov's filling area conjecture asserts that the hemisphere has minimum area among the orientable surfaces that fill a closed curve of given length without introducing shortcuts between its points.
Question: How small can be the area of a surface that isometrically fills its boundary curve, of given length?
In contrast, the hemisphere is an isometric filling of the same circle C, which has twice the area of the flat disk.
The surface can be imagined as made of a flexible but non-stretchable material, that allows it to be moved around and bended in Euclidean space.
None of these transformations modifies the area of the surface nor the length of the curves drawn on it, which are the magnitudes relevant to the problem.
Thus the filling problem can be stated equivalently as a question about Riemannian surfaces, that are not placed in Euclidean space in any particular way.
that is homeomorphic to the real projective plane and whose systole (the length of the shortest non-contractible curve) is equal to
(And reciprocally, if we cut open a projective plane along a shortest noncontractible loop of length
can have is equal to the minimum area that a Riemannian projective plane of systole
But then Pu's systolic inequality asserts precisely that a Riemannian projective plane of given systole has minimum area if and only if it is round (that is, obtained from a Euclidean sphere by identifying each point with its opposite).
The proof of Pu's inequality relies, in turn, on the uniformization theorem.
In 2001, Sergei Ivanov presented another way to prove that the hemisphere has smallest area among isometric fillings homeomorphic to a disk.
[2][3][4] His argument does not employ the uniformization theorem and is based instead on the topological fact that two curves on a disk must cross if their four endpoints are on the boundary and interlaced.
What Ivanov proved is that Let (M,F) be a Finsler disk that isometrically fills its boundary of length 2L.
are vertices, listed in counterclockwise order, of a convex polygon inscribed in the dual unit ball
, then we can rewrite this lower bound using the Stokes formula as The boundary integral that appears here is defined in terms of the distance functions
In summary, our lower bound for the area of the Finsler isometric filling converges to
Unlike the Riemannian case, there is a great variety of Finsler disks that isometrically fill a closed curve and have the same Holmes–Thompson area as the hemisphere.
A Euclidean disk that fills a circle can be replaced, without decreasing the distances between boundary points, by a Finsler disk that fills the same 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 disk.
An orientable Riemannian surface of genus one that isometrically fills the circle cannot have less area than the hemisphere.
[7] The proof in this case again starts by gluing antipodal points of the boundary.
The non-orientable closed surface obtained in this way has an orientable double cover of genus two, and is therefore hyperelliptic.
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.
-near the standard Euclidean metric), then M is a volume minimizer: it cannot be replaced by an orientable Riemannian manifold that fills the same boundary and has less volume without reducing the distance between some boundary points.
[8] This implies that if a piece of sphere is sufficiently small (and therefore, nearly flat), then it is a volume minimizer.
If this theorem can be extended to large regions (namely, to the whole hemisphere), then the filling area conjecture is true.
It has been conjectured that all simple Riemannian manifolds (those that are convex at their boundary, and where every two points are joined by a unique geodesic) are volume minimizers.
[8] The proof that each almost flat manifold M is a volume minimizer involves embedding M in