Concept
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. Every smooth surface M or curve in Euclidean space is a metric space, in which the (intrinsic) distance dM(x,y) between two points x, y of M is defined as the infimum of the lengths of the curves that go from x to y along M. For example, on a closed curve of length 2L, for each point x of the curve there is a unique other point of the curve (called the antipodal of x) at distance L from x. A compact surface M fills a closed curve C if its border (also called boundary, denoted ∂M) is the curve C. The filling M is said to be isometric if for any two points x,y of the boundary curve C, the distance dM(x,y) between them along M is the same (not less) than the distance dC(x,y) 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 2pi) 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 C, 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.