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.
Definitions and statement of the conjecture
Every smooth surface or curve in Euclidean space
is a metric space
, in which the (intrinsic) distance
between two points of is defined as the infimum of the lengths of the curves that go from to ''along'' . For example, on a closed curve
of length , for each point of the curve there is a unique other point of the curve (called the antipodal of ) at distance from .
surface fills a closed curve if its border (also called boundary
, denoted ) is the curve . The filling is said isometric if for any two points of the boundary curve , the distance between them along is the same (not less) than the distance along the boundary. In other words, to fill a curve isometrically is to fill it without introducing shortcuts.
Question: ''How small can be the area of a surface that isometrically fills its boundary curve, of given length?''
For example, in three-dimensional Euclidean space, the circle
(of length 2) is filled by the flat disk
which is not an isometric filling, because any straight chord along it is a shortcut. In contrast, the hemisphere
is an isometric filling of the same circle , which has twice the area of the flat disk
. Is this the minimum possible area?
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. The surface can be removed from Euclidean space altogether, obtaining a Riemannian surface
, which is an abstract smooth surface
with a Riemannian metric
that encodes the lengths and area. Reciprocally, according to the Nash-Kuiper theorem
, any Riemannian surface with boundary can be embedded in Euclidean space preserving the lengths and area specified by the Riemannian metric. 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.
:Conjecture (Gromov's filling area conjecture, 1983): ''The hemisphere has minimum area among the orientable
compact Riemannian surfaces that fill isometrically their boundary curve, of given length.''
Gromov's proof for the case of Riemannian disks
In the same paper where Gromov stated the conjecture, he proved that
:''the hemisphere has least area among the Riemannian surfaces that isometrically fill a circle of given length, and are homeomorphic
to a disk
Proof: Let be a Riemannian disk that isometrically fills its boundary of length . Glue each point with its antipodal point , defined as the unique point of that is at the maximum possible distance from . Gluing in this way we obtain a closed Riemannian surface 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 , we obtain a disk that fills isometrically its boundary of length .) Thus the minimum area that the isometric filling can have is equal to the minimum area that a Riemannian projective plane of systole can have. 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 area of this round projective plane equals the area of the hemisphere (because each of them has half the area of the sphere).
The proof of Pu's inequality relies, in turn, on the uniformization theorem.
Fillings with Finsler metrics
In 2001, Sergei Ivanov presented another way to prove that the hemisphere has smallest area among isometric fillings homeomorphic to a disk. 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. Moreover, Ivanov's proof applies more generally to disks with Finsler metrics, which differ from Riemannian metrics in that they need not satisfy the Pythagorean equation at the infinitesimal level. The area of a Finsler surface can be defined in various inequivalent ways, and the one employed here is the Holmes–Thompson area, which coincides with the usual area when the metric is Riemannian. What Ivanov proved is that
:''The hemisphere has minimum Holmes–Thompson area among Finsler disks that isometrically fill a closed curve of given length.''
Let be a Finsler disk that isometrically fills its boundary of length . We may assume that is the standard round disk in , and the Finsler metric is smooth and strongly convex. The Holmes–Thompson area of the filling can be computed by the formula
where for each point , the set is the dual unit ball of the norm (the unit ball of the dual norm ), and is its usual area as a subset of .
Choose a collection of boundary points, listed in counterclockwise order. For each point , we define on ''M'' the scalar function . These functions have the following properties:
* Each function is Lipschitz on ''M'' and therefore (by Rademacher's theorem) differentiable at almost every point .
* If is differentiable at an interior point , then there is a unique shortest curve from to ''x'' (parametrized with unit speed), that arrives at ''x'' with a speed . The differential has norm 1 and is the unique covector such that .
* In each point where all the functions are differentiable, the covectors are distinct and placed in counterclockwise order on the dual unit sphere . Indeed, they must be distinct because different geodesics cannot arrive at with the same speed. Also, if three of these covectors (for some