The Bellows conjecture.

TLDR: In this article, it was shown that for any oriented polyhedron P, the volume of P is a root of a polynomial depending only on the combinatorial structure and edge lengths of P.

Abstract: Cauchy’s rigidity theorem states: If P and P’ are combinatorially equivalent convex polyhedra such that the corresponding facets of P and P’ are congruent, then P and P’ are congruent polyhedra. For many years it was unknown whether the same theorem was true in general for non-convex polyhedron. In 1977, more than 160 years after the work of Cauchy, Robert Connelly discovered a polyhedron P (without selfintersections) that allowed for a continuous deformation keeping the facets of P flat and congruent. It was soon noticed that the volume of Connelly’s polyhedron remained constant under the flexing motion, and the same fact was found to be true of all later-discovered flexible polyhedra. It was conjectured that this fact would hold in general, and it came to be known as the bellows conjecture. In 1995, Idjad Sabitov proved the bellows conjecture by showing that for any (oriented) polyhedron P, the volume of P is a root of a polynomial depending only on the combinatorial structure and edge lengths of P. Moreover, the coefficients of the polynomial are themselves polynomials of the squares of the lengths of the edges of P with rational coefficients, with the coefficient polynomials depending only on the combinatorial structure of P. Hence, the volume of a polyhedron P with combinatorial structure K is a finitely valued function of the edge lengths of P, and in this way the theorem may be viewed as a generalization of Heron’s formula for the area of a triangle. 1 Formulas from Antiquity Often attributed to Heron of Alexandria (though perhaps known already to Archimedes), Heron’s formula expresses the area of a triangle as a function of the side lengths. Heron’s Formula. Let T be a triangle with side lengths (a, b, c). Area(T ) = 1 4 √ (a+ b+ c)(a+ b− c)(a− b+ c)(−a+ b+ c) We all know from elementary geometry that a triangle is determined by its side lengths. With this knowledge, Heron’s formula is, perhaps, unsurprising. For quadrilaterals, we quickly observe that the side lengths do not uniquely determine the area, so there is no hope of finding a similar formula that holds in the same generality. However, if we restrict to the case of cylic quadrilaterals, i.e., quadrilaterals whose vertices all lie on a common circle, then an analogous formula does exist. Brahmagupta’s Formula. Let Q be a cyclic quadrilateral with side lengths (a, b, c, d). Area(Q) = 1 4 √ (a+ b+ c− d)(a+ b− c+ d)(a− b+ c+ d)(−a+ b+ c+ d) If we let an edge of Q have length 0, we see that Heron’s Formula appears as a special case of Brahmahgupta. It is natural to ask whether or not a similar formula exists for cyclic pentagons, hexagons, etc. Now Heron’s formula may be easily derived using nothing more than the Pythagorean theorem, and in [3], we see that Brahmagupta’s formula follows easily from Heron. Despite the relative ease of these first two cases, the formula for the cylic pentagon went unknown for over 1300 years after Brahmagupta recorded his formula for the quadrilateral. In 1994 D. P. Robbins discovered the formulas for both the cyclic pentagon and the cyclic hexagon and proved the following more general result. Theorem (Robbins [6]). For any natural number n, there exists a unique (up to sign) irreducible homogenous polynomial f (where we regard the first argument as degree 4, the rest as degree 2) with integer coefficients such that for any cyclic n-gon P , the edge lengths (a1, a2, . . . , an) and area K of P satisfy f(16K, a1, . . . , a 2 n) = 0.