Algebraic Methods for Solution of Polyhedra

Russian Math. Surveys 66:3 445–505 ⃝c 2011 RAS(DoM) and LMS Uspekhi Mat. Nauk 66:3 3–66 DOI 10.1070/RM2011v066n03ABEH004748

Algebraic methods for solution of polyhedra

I. Kh. Sabitov

Abstract. By analogy with the solution of triangles, the solution of polyhedra means a theory and methods for calculating some geometric parameters of polyhedra in terms of other parameters of them. The main content of this paper is a survey of results on calculating the volumes of polyhedra in terms of their metrics and combinatorial structures. It turns out that a far-reaching generalization of Heron’s formula for the area of a triangle to the volumes of polyhedra is possible, and it underlies the proof of the conjecture that the volume of a deformed flexible polyhedron remains constant. Bibliography: 110 titles.

Keywords: polyhedra, combinatorial structure, metric, volume, bending, bellows conjecture, volume polynomials, generalization of Heron’s formula.

Contents

1. Some history 446 2. Polyhedra in R3 and their generalized volume 448 3. Motivations for choosing the path to the solution of the problem 449 4. The central theorem and its consequences 452 5. The algebraic meaning of the central theorem 453 6. Main lemma 453 7. The Cayley–Menger determinant 454 8. The idea of the proof of the main lemma 455 9. Proof of the central Theorem1 458 10. Another proof of the central Theorem1 459 11. Several conclusions 463 12. Canonical volume polynomials 464 13. Calculating the volume polynomial for some polyhedra 467 14. Calculating the lengths of diagonals of polyhedra 472 15. Generalizing the equations for volumes and diagonals 478 16. Polyhedra with diagonals whose lengths are known 479 17. Isometric immersion problem for a polyhedral metric 480

This work was partially supported by the Russian Foundation for Basic Research (grants no. 09-01-00179 and 10-01-91000АНФ). AMS 2010 Mathematics Subject Classification. Primary 51M20, 52C25; Secondary 51M10, 52B11. 446 I. Kh. Sabitov

18. Some algebro-geometric properties of volume polynomials 482 18.1. Roots of a polynomial as the volumes of polyhedra 482 18.2. Roots of a volume polynomial for a flexible polyhedron 484 19. The strong bellows conjecture 486 20. Another invariant of flexible polyhedra 487 21. Polyhedra in the spherical and the hyperbolic spaces 491 22. Polytopes in multidimensional spaces 493 23. Some unsolved problems 495 Bibliography 498

1. Some history By analogy with the solution of triangles, the solution of polyhedra refers to a recently emergent area in the metric theory of polyhedra which investigates the possibilities for calculating some parameters of a polyhedron (its volume, the lengths of the edges and diagonals, dihedral angles, and so on) it terms of other metric characteristics of it, with account taken of the combinatorial structure of the polyhedron, of course. For the present the metric of the polyhedron and its combinatorial structure are usually assumed to be known, and such questions as the existence of a polyhedron with prescribed metric and combinatorial structure, calculation of its volume and the lengths of its diagonals, establishment of its flexibility or inflexibility, and determination of its isometric deformations are solved on this basis. Historically, the initial question here was the uniqueness of a polyhedron with given metric and combinatorial structure, first investigated by Legendre. In an appendix (Note XII) of the first edition of his well-known textbook [1] he analyzes Euclid’s definition of equality of polyhedra1 and concludes that, first, this definition must in fact be regarded as a theorem to be proved, and second, we must assume in advance that the polyhedra under consideration are convex, otherwise Definition 10 contradicts an earlier example (1756) due to the English mathematician R. Simson: a cube with a quadrangular pyramid with apex outside the cube and base coinciding with a face of the cube (removed beforehand) and an equal cube with a similar pyramid but with apex inside the cube, have faces “equal in multitude and in magnitude” but obviously are not equal. Further, in the same Note XII Legendre proved for certain classes of polyhedra that two convex polyhedra with the same combinatorial structure and equal corresponding faces can be obtained one from the other by a motion, and he conjectured that this might be true for any convex polyhedra. For reasons unknown to us, Legendre left out this part of Note XII in subsequent editions (2nd to 9th) of his book, and restored it only after Cauchy in 1813 proved the equality for all convex polyhedra [3], essentially using Legendre’s ideas in the proof. Hence it would be fair historically to call this result of Cauchy the Legendre–Cauchy theorem, and that is what we shall call it in what follows.2 At the same time, the Legendre–Cauchy theorem asserts that a (strictly) convex polyhedron is inflexible, because in a flex (that is, a continuous deformation of

1From Book XI of Euclid’s Elements [2]: Definition 10. Equal and similar solid figures are those contained by similar planes equal in multitude and in magnitude. 2Several authors have devoted papers to a thorough analysis of Definition 10 and its relations to the Legendre–Cauchy theorem (see, for instance, [4]–[6]). Algebraic methods for solution of polyhedra 447 the polyhedron during which it has absolutely rigid faces) the polyhedron remains convex at the initial instants of the deformation, so that by the Legendre–Cauchy theorem it keeps its original shape and the flex reduces to a simple motion of itas a solid body. Once the Legendre–Cauchy theorem was established, the following question naturally arose: does this inflexibility hold for all polyhedra, not necessarily convex, or do there exist flexible polyhedra?Equality ( in the class of all, not necessarily convex, polyhedra is refuted by Simson’s example.) This question remained open for quite a long period of time until at the end of the 19th century the Belgian mathematician and engineer Bricard proved [7] the existence of flexible octahedra, providing at the same time a complete classification of them. However, all these lie in R3 but not as embedded or even immersed polyhedra, so the question of the existence of ‘proper’ (non-self-intersecting) flexible polyhedra still remained open. Moreover, in 1975 Gluck proved [8] that almost all polyhedra homeomorphic to a sphere are inflexible, and thus the chances for the existence of embedded flexible polyhedra were considered negligible. Nevertheless, the American mathematician Connelly managed to construct examples of flexible polyhedra, first immersed ones [9] (1976) and then embedded ones [10] (1977). Some new examples of flexible polyhedra have been constructed since then (the reader can find details in [11]), and for all of them it was discovered that flexes do not change their volume. This observation provided some grounds for the conjecture that all the flexible polyhedra have the property of volume invariance (see [12] (February 1978), where the author of this conjecture was not mentioned, and [13] (August 1978), where a reference to D. Sullivan was made; however, even today we do not know3 who actually was the first to put forward the conjecture). In view of the well-known physics law pV = Const, where p is the pressure and V the gas volume, Connelly [13] called this the ‘bellows conjecture’: a flexible polyhedron of mathematically ideal design (with a hole in one of the faces, of course) cannot be used as a bellows. This beautiful conjecture was apparently a topic of study for many geometers for almost two decades, but this remained their personal secret, since only in the two publications [16] and [17] by V. A. Alexandrov (Novosibirsk) did the author state explicitly that his ultimate aim was to prove or refute the conjecture. More- over, [18] presents examples of polyhedra which can easily be deformed but whose volume changes during the deformation — so one could assume that they also were constructed with an aim to disprove the conjecture (however, these deformations turned out not to be isometries). Finally, in 1996, a proof of the conjecture was given in [19] for flexible polyhedra of any topological genus (the first advances date back to [20] and more detailed proofs can be found in [21] and [22]). The method of the proof turned out to be very useful also in other questions of the metric theory of polyhedra, and our goal in this paper is to present a short proof of the bellows conjecture4 and a survey of other results in this area of geometry that have appeared during the last 10–15 years. We also state many unsolved problems. The surveys [24], [14], and [25] are closest to our topic. The reader can find a more popular discussion of these equations in our pamphlet [11], and [26] treats similar questions for planar polygons and also gives a large list of unsolved

3The authors of [14] and [15] do not know this either. 4Another proof appeared in [23] and we also present it in what follows. 448 I. Kh. Sabitov problems. The reader who is interested in a wider range of problems in the metric theory of polyhedra and polygons can be referred to the extensive text [15] (to our knowledge there are plans to publish it as a book, but it is presently available as a preprint). We leave unmentioned many papers relating to flexes and infinitesimal flexes of polyhedra.

2. Polyhedra in R3 and their generalized volume We start by introducing the main notions relating to polyhedra in three dimensions. A polyhedron in R3 is characterized by its combinatorial structure or scheme and the geometric realization of this scheme in space. We limit ourselves to polyhedra with triangular faces, so the combinatorial model of a polyhedron will essentially be the combinatorial scheme of a 2-dimensional simplicial complex K. A polyhedron with combinatorial structure K, or a geometric realization of the simplicial complex K in R3 can be obtained in two ways. First let K be given as an abstract 2-dimensional simplicial complex with n 0-dimensional simplexes, which form a set A = {a1, . . . , an}.

Definition 1a. An arbitrary map P : A → R3 is called a polyhedron in R3 with combinatorial structure K. In this case 1) the images of the 0-dimensional simplexes of K are called vertices of the polyhedron P ; 2) 1-dimensional simplexes realized as straight line segments joining the images of the corresponding pairs of 0-dimensional simplexes are called edges of the polyhedron; 3) by definition the faces of the polyhedron are the flat triangles naturally associated by the map P with 2-simplexes of the complex K. In the second method for defining a polyhedron we assume that K is given as a geometric simplicial complex, that is, its simplexes are points, line segments, and (flat) 2-dimensional triangles, and a corresponding rule for identification ofpoints and line segments is described. Let |K| be the body of the complex K, that is, the set-theoretic union of the geometric simplexes in K with account taken of their identifications (sometimes |K| is called the polyhedron of K), endowed with the corresponding topology.

Definition 1b. By definition a polyhedron in R3 with combinatorial structure K is an arbitrary map P : |K| → R3 that is continuous everywhere and linear on each simplex. The image P (|K|) is called a geometric polyhedron with combinatorial structure K. Since a linear map of a simplex is completely determined by its values on the 0-dimensional simplexes, it is obvious that the two definitions are equivalent. On a personal computer (PC) the first definition of a polyhedron (when its edges and faces are given by corresponding lists) is more convenient, while the second definition is better suited for generalizations to polyhedra with non-triangular faces defined as images of cells. To avoid repeated discussions of whether we treat the map defining a polyhedron under consideration in the first or the second way, we shall always denote Algebraic methods for solution of polyhedra 449 it by P : K → R3. This notation shows that we mean a polyhedron in R3 with combinatorial structure K. In the two (equivalent) definitions of a polyhedron P its geometric image in R3 can have various singularities: degenerate edges or faces, self-intersections, and so on. Therefore, for this polyhedral surface we must give a special definition of the volume of the spatial region ‘bounded’ by it. To do this, we assume additionally in our general definition of a polyhedron that the original geometrically defined simplicial complex K is homeomorphic to an orientable manifold M of topological genus g > 0, so that K defines a triangulation of M. To avoid new notation we shall assume that K is itself represented as a manifold triangulated by rectilinear simplexes (that is, we have what topologists call a polyhedral or piecewise linear manifold). Assume that we have an orientation on K. Then we push it forward to all the faces of P (K) by indicating the corresponding order in which the edges follow one another. Definition 2. The generalized volume V (P ) of the oriented geometric polyhedron P (K) is the sum of the algebraic volumes of the compatibly oriented tetrahedra with a common vertex at some point O ∈ R3 and with bases coinciding with the oriented faces of the geometric polyhedron P (K). We recall that if we write out the vertices of an oriented face G ⊂ P (K) in accordance with the order on the boundary, say, as pi, pj, pk, then the algebraic volume V (T ) of the oriented tetrahedron T with vertex at O and with base G is one sixth of the scalar triple product of the vectors going from O to the vertices of G taken in the order of the vertices on the oriented face. Hence, it is easy to see that the value of the generalized volume of a polyhedron is independent of the choice of O, and if a polyhedron is embedded in R3, then its generalized volume is equal to the usual oriented volume.5 Introducing a Cartesian system of coordinates in R3 with origin at some point O and taking this origin for the point O in the definition of the generalized volume, we arrive at the following obvious but important conclusion: the generalized volume of a polyhedron is a homogeneous polynomial of degree three in the coordinates of the vertices of the polyhedron.

3. Motivations for choosing the path to the solution of the problem For a polyhedron P with n vertices suppose that all its vertices are numbered in some order and let (xi, yi, zi), 1 6 i 6 n, be their coordinates. Then we can 3n associate with this polyhedron a point M(x1, y1, z1, . . . , xn, yn, zn) in the space R . The other way round, each point M ∈ R3n is associated in an obvious way with the 3 set of points Mi(xi, yi, zi), 1 6 i 6 n, in R , and from these points, knowing the combinatorial structure of K, we can construct a polyhedron P : K → R3. Hence

5If P (K) has a face degenerating into a line segment or even a vertex, then we assign to its edges the orientation inherited from that of the corresponding 1-simplexes in the inverse image of the face in K. Tetrahedra with bases on degenerate faces have volume zero no matter what orientation might be assigned to the base, and therefore we indicate the orientation of such faces only for a formally correct expression for the corresponding scalar triple product in the definition of the volume, so that the algebraic expression for the sum of the volumes displays its independence from the choice of the common vertex of the tetrahedra. 450 I. Kh. Sabitov there exists a bijective correspondence P ↔ M between the set of polyhedra of given combinatorial type with n vertices and the set of points in R3n, and we can assume it to be a homeomorphism with respect to the suitable topology. Let e be the number of edges of the complex K, and assume that we have numbered all the edges so that to each pair of vertices i and j connected by an edge there corresponds an edge with an index k = k(i, j) = k(j, i) with 1 6 k 6 e. Let 3 lk be the lengths of these edges of the polyhedron P : K → R . Then we have the equations

2 2 2 2 (xi − xj) + (yi − yj) + (zi − zj) = lk, 1 6 k = k(i, j) 6 e. (1)

For given edge lengths lk the solutions of the system (1) determine via the map M → P a set of polyhedra in R3 which have the same combinatorial structure and are mutually isometric. Some of these polyhedra can be obtained from others by rigid body motions in R3. To eliminate such ‘spurious’ solutions, we add to (1) the equations X X X xi = 0, yi = 0, zi = 0, (2) i i i which correspond to putting the centre of mass of the system of vertices of the polyhedron at the origin by a motion in R3, so that further parallel translations of it are ruled out. It is more difficult to eliminate rotations: there exists (toour knowledge) no equation which can be used to eliminate rotations of all the polyhedra of a fixed combinatorial type in all the cases, regardless of the configuration of the polyhedron and the edge lengths.6 For instance, the equations X X X xiyi = 0, yizi = 0, xizi = 0, (3) i i i which correspond to the reduction to principal axes of the ellipsoid of inertia of P 2 P 2 the vertex set of the polyhedron, eliminate rotations only if i xi ̸= i yi ̸= P 2 i zi (that is, only if the ellipsoid of inertia of the vertex set is not an ellipsoid of revolution). Apart from considering the equations (3), we can suppose that one of the faces takes some fixed position, which makes rotating the polyhedron impossible: for example, if there is a non-degenerate face ⟨1, 2, 3⟩, then we can add the three equations y2 = y1, z1 = z2 = z3 to the system (1), (2). However, these equations do not rule out rotations if for some edge lengths the face in question is degenerate: that is, we cannot point out one particular face such that by fixing it we eliminate rotations of the polyhedron for all values of the edge lengths (allowed by the triangle inequality).7 Nevertheless,

6As concerns eliminating continuous rotations of some given polyhedron, such equations do exist (see [12]). 7Thus, we can formulate the problem of eliminating continuous rotations as the problem of replacing the abstract operation of taking the quotient of the solution set of (1) (with respect to the condition that solutions superposable one on the other by a continuous rotation are equivalent) by the addition of three equations, as done when we replaced taking the quotient with respect to parallel translations by adding the three equations (2). In other words, we must find an algebraic surface meeting each trajectory of the action of the rotation group on the solution space of the system (1) at finitely many points; for details see[27], Problem Section. Algebraic methods for solution of polyhedra 451 for given edge lengths we can assume that, one way or another, we have three equations forbidding continuous rotations of the polyhedron in the case when the metric complex K has at least some non-degenerate faces. We recall that the set of isometric polyhedra of the same combinatorial structure in R3 is called the configuration space of any of these polyhedra. Since a polyhedron in R3 can be interpreted as a point in R3n, we see that the configuration space of an arbitrary polyhedron P in R3 is homeomorphic to the real algebraic variety A in R3n defined by the system (1), (2) combined with equations (3) or some analogue of them. Already for the system (1), (2) alone we can show that for fixed values 2 2 2 of the edge lengths all its solutions (x1, . . . , zn) lie in a ball Br : x1 + ··· + zn < r in R3n, where the radius r = r(K, l) depends on the combinatorial structure K and the metric l of the polyhedron. Hence, A is a compact variety. Since an algebraic variety can have only finitely many connected components, the configuration space of a polyhedron consists of finitely many compact connected components. Each of these corresponds to polyhedra that can be superposed one on another by a non-trivial flex. In particular, 0-dimensional (single-point) components correspond to inflexible polyhedra, while points in components of dimension > 1 correspond to isometric polyhedra that can be flexed into one another. Thus, ifthe bellows conjecture holds, then all the polyhedra in the same connected component of A have the same volume, and since there are only finitely many components, we see that if the conjecture holds, then the set of volumes of all possible polyhedra of fixed combinatorial structure with a fixed set of edge lengths must befinite.The converse is also true: if the volumes of isometric polyhedra can assume only finitely many values, then the bellows conjecture holds. This observation dates back to the author’s paper of 1989 ([28], p. 263). We now ask about possible arguments for the conclusion that the volumes of isometric polyhedra with the same combinatorial structure can take only finitely many values. Of course, one starts from the simplest cases, that is, polyhedra with just a few vertices. We shall consider the cases of a tetrahedron and a quadrangular pyramid. The volume of a tetrahedron. We recall that if a tetrahedron (Fig.1) has edge 8 lengths l1, l2, l3, l4, l5, l6, then its volume V satisfies the formula

1 h V 2 = l2l2(l2 + l2 + l2 + l2 − l2 − l2) + l2l2(l2 + l2 + l2 + l2 − l2 − l2) 144 1 5 2 3 4 6 1 5 2 6 1 3 4 5 2 6 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2i + l3l4(l1 + l2 + l5 + l6 − l3 − l4) − l1l2l3 − l2l4l5 − l1l4l6 − l3l5l6 . (4)

The generalized volume V of a quadrangular pyramid (Fig.2) with triangulated base can be calculated as the algebraic sum of the volume V1 of the tetrahedron ⟨Ap1p2p3⟩ and the volume V2 of the tetrahedron ⟨Ap1p3p4⟩: V = V1 ± V2 ⇒ 4 2 2 2 2 2 2 2 2 V − 2(V1 + V2 )V + (V1 − V2 ) = 0, and since V1 and V2 can be expressed in terms of the squares of the lengths of the corresponding edges by the previous formula (4), for the generalized volume V of a quadrangular pyramid we have

8Unfortunately, in our papers [21] and [25] formulae (1) and (3), respectively, for the volume of a tetrahedron contains errors in the indices of the edges in the last four terms, which we put right here (each of the four terms involves the indices of the three edges incident to a face). 452 I. Kh. Sabitov

Figure 1 Figure 2 a polynomial expression of the form

4 2 Q(V ) = V + a1(l)V + a2(l) = 0. (5)

Thus, we conclude that in the simplest cases the volume of a polyhedron is a root of a certain polynomial with leading coefficient 1, the other coefficients of which are polynomials in the squares of the edge lengths of the polyhedron, with numerical coefficients determined by its combinatorial structure. Hence, we could expect that the required finiteness of the set of volumes of polyhedra with given metric and given combinatorial structure would follow from the fact that the volume is a zero of a polynomial independent of the particular configuration of the polyhedron in space. This conjecture was stated for the first time in [29], in § 8 written by the present author.

4. The central theorem and its consequences The conjectural situation does indeed hold: we have the following theorem.

Theorem 1. Let [P ] be the set of polyhedra in R3 with triangular faces, with the same combinatorial type K, and with corresponding edges of the same lengths 9 (l1, . . . , le), where e is the number of edges. Then there exists a monic polynomial

2N 2N−2 2 Q(l, V ) = V + a1(l)V + ··· + aN−1(l)V + aN (l) (6) such that the generalized volume of each polyhedron in [P ] is a zero of this polynomial, whose coefficients ai(l), 1 6 i 6 N, are themselves polynomials in the squares of the edge lengths of the polyhedron, with numerical coefficients depending on the combinatorial structure of the polyhedron.10 This theorem has two obvious consequences. Corollary 1. The set of values of the generalized volumes of isometric polyhedra with the same combinatorial structure is finite. Corollary 2. Flexes preserve the generalized volume of a flexible polyhedron.

9This is how we refer to a polynomial with leading coefficient 1. There are also other terms for such polynomials: for example, some authors call them normalized. 10We let l denote the system of squares of the edge lengths of the polyhedron. Algebraic methods for solution of polyhedra 453

In fact, a flex must change the volume of a polyhedron continuously, but a continuous function taking finitely many values is constant. Thus, the bellows conjecture appears as a simple consequence of our central theorem, which can be naturally regarded as a generalization of Heron’s formula to polyhedra.

5. The algebraic meaning of the central theorem As noted at the end of § 2, the volume V of a polyhedron is a homogeneous polynomial of degree three in the coordinates of the vertices of the polyhedron. We write this as V = V (x), where x is the set of coordinates x1, y1, z1, . . . , xn, yn, zn of all the n vertices of the polyhedron. Equations (1) also express the dependence of the squares of the edge lengths of the polyhedron on the coordinates of the vertices. We express this as l = l(x), where l is the set of squares of the edge lengths. Thus if the central theorem holds for each configuration and each position of the polyhedron in space, then this must be reflected by the following property: substituting for V and l in Q(V ) their expressions in terms of the coordinates of the vertices of the polyhedron, we obtain an identity Ql(x),V (x) ≡ 0 in the variables x1, y1, z1, . . . , xn, yn, zn. This indicates an algebraic relation between the second-degree polynomials in x expressing the squares of the edge lengths and the third-degree polynomial expressing the volume. Thus if a monic polynomial in 2 2 2 2 2 l1, . . . , le,V vanishes identically after replacing V and the quantities li (1 6 i 6 e) by their expressions in terms of the vertices of a polyhedron of given combinatorial type K, then we call it a volume polynomial for polyhedra with this combinatorial structure K.

6. Main lemma For the statement of the main lemma we must agree about terminology. Definition 3. If Theorem1 is known to hold for some polyhedron, then we say that this polyhedron has property Q. Definition 4. If the cycle formed by three edges of a polyhedron bounds no face, then we call it a null-cycle.

The cycle ⟨Ap1p3⟩ in Fig.2 is an example of a null-cycle.

Definition 5. We say that a polyhedron has combinatorial structure of type K0 if at least one vertex of it is not incident to a null-cycle, that is, no null-cycle passes through this vertex (for brevity we say that such vertices are ‘nice’). A tetrahedron and an octahedron give us examples of polyhedra with combinatorial structure of type K0 which have only ‘nice’ vertices. The polyhedron in Fig.2 also has combinatorial type K0: besides ‘non-nice’ vertices it also has some ‘nice’ ones. The Cs´asz´arpolyhedron [30] of torus type with seven vertices provides a counterexample: it has no diagonals and any two vertices on the boundary of the star of an arbitrary vertex are connected by an edge, so that each vertex of the polyhedron is incident to 9 null-cycles (for a triangulation and a picture of this polyhedron see, for instance, [31]). It can be shown that each polyhedron of genus g = 0 with more than four vertices has at least two nice vertices not connected by an edge [32]. 454 I. Kh. Sabitov

Note that if a polyhedron has a vertex of degree 3, then it must have combinatorial type K0. We can now state the main lemma.

Lemma 1. If all polyhedra of fixed topological genus g > 0 with n vertices have property Q, then the polyhedra with n + 1 vertices of the same topological genus and combinatorial type K0 also have property Q.

7. The Cayley–Menger determinant In a series of papers from the 1850s–60s Cayley investigated the geometry of point sets in the plane. In particular, he considered the question of when 6 positive numbers can be pointwise distances between 4 points in the plane. Later on, in the 1920s Menger investigated the analogous question for a general dimension n. Its solution is discussed in detail in [33], and we cite only the result we need: the

Figure 3

3 10 distances between 5 points A, p1, p2, p3, p4 in R (Fig.3) must satisfy the Cayley–Menger equation

0 1 1 1 1 1

2 2 2 2 1 0 lA1 lA2 lA3 lA4

1 l2 0 l2 l2 l2 1A 12 13 14 det(CM) = 2 2 2 2 = 0 (7) 1 l2A l21 0 l23 l24

1 l2 l2 l2 0 l2 3A 31 32 34 2 2 2 2 1 l4A l41 l42 l43 0

(here lAi = liA, 1 6 i 6 4, denotes the distance between the points A and pi, and lij is the distance between pi and pj). The Cayley–Menger equation has the following 4 geometric meaning: the volume VT of a tetrahedron with five vertices in R is 1 expressed by the formula V 2 = − det(CM). The Cayley–Menger equation T 24(4!)2 can be obtained from the observation that the five vertices of a 4-dimensional tetrahedron lie in R3 if and only if its 4-dimensional volume is zero. We shall require the Cayley–Menger equation in the case when we do not know 2 all the distances, so we use special notation for some of them. Namely, let x = l13 Algebraic methods for solution of polyhedra 455

2 and y = l24. Then cumbersome but simple calculations yield an equality of the form

2 2 −8 det(CM) = x [y + a1(l)y + b1(l)] 2 2 + x[a2(l)y + b2(l)y + b3(l)] + b4(l)y + b5(l)y + b6(l) = 0, (8) where the coefficients a1(l) and a2(l) do not vanish if no three points are incident to the same straight line.

8. The idea of the proof of the main lemma For the proof of Lemma1 we assume that the polyhedron P of combinatorial type K0 with n + 1 vertices is in general position: the coordinates of its vertices are algebraically independent, that is, do not satisfy an algebraic equation. We carry out the proof by induction on the degrees of nice vertices (recall that the degree of a vertex is the number of edges incident to it). Assume that P contains a vertex A of degree 3. Then we delete its open star, which leaves us with a polyhedron with triangular boundary. Gluing it up by a triangular face, we obtain an n-vertex polyhedron P1. We have added no new edges by this operation, so we know the lengths of all the edges of P1. By the induction hypothesis P1 has property Q, and therefore its volume V1(P1) satisfies a polynomial equation of the form

Q (V ) = V 2N1 + a(1)V 2N1−2 + ··· + a(1) V 2 + a(1) = 0, 1 1 1 1 1 N1−1 1 N1

(1) with coefficients ai depending on the squares of the edge lengths of P1, that is, on the squares of the lengths of all edges of P with the exception of the three edges incident to A. The volumes V and V1 of P and P1, respectively, are related by V1 = V + εVT , where VT is the volume of the tetrahedron T with vertex A and ε = ±1 (the volume of the removed tetrahedron must be added or subtracted depending on the position of the vertex A). Setting V1 = V + εVT in the equation Q1(V1) = 0 and bearing in mind formula (4), which expresses the square of the volume of T in terms of the squares of its edge lengths, we see that V 2(P ) is a root 4N1 4N1−2 of an equation of the form Q(V ) = V + a1(l)V + ··· = 0. We now use induction on the degree of a nice vertex of the polyhedron. We consider more closely the case when the polyhedron has a nice vertex of degree 4. Thus, let A be a nice vertex of degree 4 not incident to a null-cycle (of three edges). Let St A be the star of A. We number the vertices on the boundary of St A in a cyclic order, starting with any one of them: p1, p2, p3, p4. We now construct from P new polyhedra P1 and P2 with vertices of degree 3. Namely, to construct Pi we remove the edge ⟨Api⟩, i = 1, 2, together with the two incident faces ⟨pi−1Api⟩ and ⟨piApi+1⟩ (we set p0 = p4) and replace them by the new faces ⟨pi−1pipi+1⟩ and ⟨pi−1Api+1⟩ (see Fig.4). This is possible because the diagonals ⟨p1p3⟩ and ⟨p2p4⟩ are not edges of P , so each of them is incident to precisely two new faces in P1 and P2 alike. In the polyhedra P1 and P2, A is a vertex of degree 3. Let D1 and D2 be the lengths of the diagonals ⟨p4p2⟩ and ⟨p1p3⟩, respectively (so that the subscript indicates the index of the point lying ‘between’ the endpoints of the diagonal in accordance with the orientation of the quadrangle). By the induction 456 I. Kh. Sabitov

Figure 4

basis the volumes V1 = vol(P1) and V2 = vol(P2) are roots of polynomial equations of the form

2Ni (i) 2 2Ni−2 Qi = Vi + a1 (l, Di )Vi + ··· = 0, i = 1, 2. (9)

Let V0i be the volume of the tetrahedron ⟨Api−1pipi+1⟩ complementing Pi in P . Then for the volume V (P ) we have

Vi = V − εiV0i, εi = ±1, i = 1, 2. (10)

Substituting this expression for Vi into (9) and using the equalities m m−1 X 2m 2(m−k) X 2m 2(m−k−1) V 2m = V 2kV − ε VV V 2kV , i 2k 0i i 0i 2k + 1 0i k=0 k=0 we use obvious transformations to obtain equations of the form

(i) (i) V 4Ni + A (l, D2,V 2 )V 4Ni−2 + ··· + A (l, D2,V 2 )V 0 = 0, (11) 1 i 0i 2Ni i 0i this time without εi. This means that the resulting equations are independent of the position of St A relative to the rest of P . Substituting in (10) the expression (4) 2 2 for V0i as a function of (l, Di ), we now get that

(i) (i) V 4Ni + B (l, D2)V 4Ni−2 + ··· + B (l, D2)V 0 = 0, i = 1, 2. (12) 1 i 2Ni i

2 We can regard (12) as polynomial equations with respect to Di . To this end we represent (12) as

C(i)(l, V 2)D2Ki + ··· + C(i)(l, V 2) = 0, i = 1, 2, (13) 0 i Ki Algebraic methods for solution of polyhedra 457 where 2Ki 6 12Ni is the highest power of Di in the polynomial equation (12) with 2 2 the l-quantities l1, . . . , le regarded as independent variables (‘symbolic quantities’). If Ki = 0 for i = 1 or 2, then the corresponding equation (13) takes the form (i) C (l, V 2) = V 4Ni + ··· = 0. There are now two possibilities. We substitute in Ki the last equation the expressions for the volume V and the squares of the lengths of the sides as functions of the coordinates of vertices. If this gives us identically zero, then we have obtained a required polynomial of the form (6) for V : Q(V ) = (i) C (l, V 2) = V 4Ni + ··· . But if we obtain a polynomial not vanishing identically, Ki then we see that the coordinates of the vertices of our polyhedron are algebraically dependent, which is impossible for a polyhedron in general position. Hence we either have a required polynomial of the form (6) or Ki > 0. Assume that Ki > 0, (i) 2 2 2 i = 1, 2. We substitute in Cj (l, V ) (0 6 j < Ki) the expressions for l = (l1, . . . , le) and V as polynomial functions of (x1, . . . , zn). If all the resulting functions vanish identically, then again we arrive at a required equation Q(V ) = C(i)(l, V 2) = Ki V 4Ni + ··· = 0. But if at least one of these functions does not vanish identically, then for the P under consideration the coefficient of the corresponding power of Di is non-zero, since P is in general position by assumption. For simplicity of notation (i) 2 (i) 2 let C0 (l, V ) be the first such coefficient, so that(12) holds with C0 (l, V ) ̸= 0. Now recall that the 10 distances between the 5 points A, p0, p1, p2, p3 satisfy the 2 2 Cayley–Menger equation (8). In our case x = D1, y = D2, and the other distances are the known lengths of the corresponding edges of P , so we can denote them by l. Then the corresponding Cayley–Menger equation can be written as

2 2 2 2 D1 (D2 + aD2 + b) + D1(aD2 + bD2 + b) + (bD2 + bD2 + b) = 0, (14)

2 2 where D1 = D1 and D2 = D2 (here we do not indicate the variables p and q since their values are known: they are the lengths of ⟨p0p2⟩ and ⟨p0p2⟩). Moreover, for conciseness we write the coefficients without indices. Since the leading coefficients in (12) for i = 1 and in (14) are non-zero, we can eliminate D1 using the resultant of (13) and (14). This gives us the equation

2 2 2 (D + ··· )(aD + ··· )(bD + ··· ) 0 0 0 2 2 2 2 2 0 (D2 + ··· )(aD2 + ··· ) 0 0 0

...... = 0 2 2 2 0 0 ... (D2 + ··· )(aD2 + ··· )(bD2 + ··· ) (1) (1) 2 2 4N1 C0 (l, V ) C1 (l, V ) ... ∗ (V + ··· ) 0 (1) 2 4N1 0 C0 (l, V ) ... ∗ ∗∗ (V + ··· )

0 (here ∗ and ∗∗ stand for the coefficients of D1 in equations (14) and (13) for i = 1). (1) We recall that V occurs to powers less than 4N1 in the coefficients Cj , j < K1, so after expanding this (K1 + 2) × (K1 + 2) determinant we obtain an equation of the form 2K 2K −1 1 8N1 1 8N1 D2 (V + ··· ) + D2 (aV + ··· ) 2K −2 0 1 8N1 8N1 + D2 (bV + ··· ) + ··· + D2 (bV + ··· ) = 0. (15) 458 I. Kh. Sabitov

2 Next we eliminate D2 in (15) and equation (13) for i = 2 and D2 = D2. It is easy to see that the corresponding resultant yields the equation

(V 8N1 + ··· )(aV 8N1 + ··· ) ... ∗ ... 0

0 (V 8N1 + ··· ) ......

......

0 ... (V 8N1 + ··· ) ...... ∗ = 0. (2) (2) 2 2 4N2 C0 (l, V ) C1 (l, V ) ... (V + ··· ) ... 0 ...... (2) 2 4N2 0 ...C0 (l, V ) ...... (V + ··· )

Hence we see that the highest power of V is 2N = 8K2N1 + 8K1N2 (since V occurs (2) in the coefficients Ci to powers less than 4N2), with leading coefficient 1, that is, we have an equation of the required form, Q(V ) = V 2N + ··· = 0, of degree 2N 6 96N1N2 (this very crude estimate can be improved substantially, but this is not very important for the proof of the lemma and, accordingly, of the theorem). Thus, we have made the induction step in the case when there is a vertex A of degree 4. We do not verify here the induction step in the case of a nice vertex of degree m > 5: this has been done, for instance, in [19], [21], and [22]. To complete the proof of the lemma we observe the following. Our argument holds if the polyhedron P under consideration is in general position. Otherwise we put P in general position by a small perturbation and write equations Qε(Vε) = 0 of the form (6) for nearby polyhedra Pε in which the polynomial functions ai(lε) have numerical coefficients independent of ε (they depend only on the combinatorial structure of P and the choice of A). Passing to the limit as ε → 0 while bearing in mind that the lengths lε occurring in the coefficients of Qε depend continuously on ε → 0, we see that V (P ) satisfies the same equation (6) with coefficients ai(l). This completes the proof of Lemma1. Remark 1. We see from the above arguments that the required polynomial Q(V ) can be found effectively, but the algorithm is not without ambiguity: eachstep depends on the choice of a nice vertex and on the method taken for reducing the problem to a polyhedron with a nice vertex of smaller degree.

9. Proof of the central Theorem1 We shall use induction on the number n of vertices and the genus g. The induction basis is the formula for the volume of a tetrahedron. First assume that the polyhedra under consideration have genus g = 0 and that the theorem has been proved for such polyhedra with n vertices. Let P be a polyhedron of genus 0 with n + 1 vertices. It can be shown that each polyhedron of genus g = 0 has a nice vertex (in fact, there are at least 2; see [32]), and thus all such polyhedra have combinatorial type K0. Hence we can use the main lemma, and the required polynomial equation Q(l, V ) = 0 exists.11

11[21] gives an argument not relying on the existence of a nice vertex. Algebraic methods for solution of polyhedra 459

Assume that we have proved the theorem for all polyhedra of genus g. Consider polyhedra of genus g + 1. Let P0 be a polyhedron of genus g + 1 with the minimum number of vertices nmin (a formula for nmin in terms of the genus g can be found in [34]). Then it must contain a 3-edged cycle not bounding a face. Indeed, let A be a vertex in P0. If there is no such cycle through A, then we can remove the star of A and glue up the hole by triangular faces, some sides of which are diagonals of ∂(St A) going out from some vertex (because no diagonal of the boundary of the star is an edge). Then we obtain a new polyhedron of the same genus g but which has fewer vertices than P0, in contradiction to the definition of P0. Hence P0 contains a 3-edged cycle not bounding a face. We cut P0 along this cycle and glue up the resulting triangular holes by two faces. This gives us one or two polyhedra of lower genus, which are known to have property Q. If P0 does not fall into two pieces after this operation (and then all the isometric realizations of P0 also have this property, since it depends only on the combinatorial structure of the polyhedron), then its volume is a root of the polynomial Q corresponding to the polyhedron of lower genus obtained after gluing. Assume that we get two polyhedra P1 and P2 of lower genus. By the induction hypothesis both have property Q. The volume V0(P0) is related to the volumes V1(P1) and V2(P2) by

V0(P0) = ε1V1(P1) + ε2V2(P2), εi = ±1, i = 1, 2. (16)

Since for V1 and V2 we have polynomial equations of the form Q1(l, V1) = 0 and Q2(l, V2) = 0 with coefficients involving only the edge lengths of P0, we can easily eliminate V1 and V2 in these equations and (16), thereby obtaining the required equation of the form Q(l, V0) = 0. Thus, we have the induction basis for genus-g polyhedra: we have shown that the polyhedron with fewest possible vertices in this class has property Q. Next we resort to induction with the use of the main lemma, and when it cannot be used (that is, when the polyhedron does not have combinatorial type K0), then we cut the polyhedron along the existing null-cycle and reduce the question to a polyhedron of the same topological genus with fewer vertices or to polyhedra which have more vertices but a lower genus, for which we have already established the property Q. The proof of the theorem is complete.

10. Another proof of the central Theorem1 We can show that if we set Ve = 12V for polyhedra, then the monic equation 2N 2N−2 Q(V ) = V + a1(l)V + ··· = 0 in which the polynomials ai(l) have rational coefficients is transformed into a monic equation 2N 2N−2 Qe(Ve) = Ve +a ˜1(l)Ve + ··· = 0 (17) where thea ˜i(l) are polynomials in the squares of the edge lengths as before, but now have integer coefficients. Recall that given afield L and a ring R ⊂ L, an element x ∈ L is said to be integral over R if there exist a positive integer N and elements ai ∈ R (1 6 i 6 N) such that x satisfies the equation N N−1 x + a1x + ··· + aN = 0. (18) Thus, the algebraic meaning of equation (6) explained in § 5 and extended to equation (17) can also be expressed as follows: for a polyhedron P with known combinatorial structure and known coordinates of the vertices, the element Ve of R is integral 460 I. Kh. Sabitov over the ring generated by the squares of its edge lengths (this ring is precisely the set of numbers representable as values of polynomials in the squares of the edge lengths with integer coefficients). A test for elements of a field to be integral over a fixed subring is well known in algebra. Use of it yieldsa second proof of the central theorem as presented in [23]. We shall sketch this proof, following mostly the presentation in [14] and adding some necessary details. To do this we recall several concepts in the theory of places. Let L and F be fields. An F -valued place for L is a map ϕ: L → F ∪ {∞} such that for all x, y ∈ L

1) ϕ(x + y) = ϕ(x) + ϕ(y), 2) ϕ(xy) = ϕ(x)ϕ(y), 3) ϕ(1) = 1.

Elements of F are said to be finite. As regards operations with the symbol ∞, the conventions are as follows: ∞ · ∞ = ∞, and if a ∈ F , then a ± ∞ = ∞, a/∞ = 0, and a/0 = ∞ and a · ∞ = ∞ for a ̸= 0. The expressions ∞/∞, ∞ ± ∞, 0/0, and 0 · ∞ are not defined, and the above equalities 1) and 2) make sense only when their right-hand sides are well defined. Using places for L, we obtain the following test for integral elements of L. Statement 1. For a field L let R be a subring of it. Then an element x ∈ L is integral over R if and only if each place for L which is finite at elements of R is finite at x. The proof can be found, for example, in [35] or [36]. Hence we obtain a simple consequence, which is nevertheless important for what follows. Corollary 3. If x and y are elements of L which are integral over a ring R, then x + y and x − y are also integral over R. Formula (4) for the volume V of a tetrahedron can be expressed as

2 2 2 2 2 2 2 (12V ) + a1(l1, l2, l3, l4, l5, l6) = 0, where a1(l) is a polynomial in the squares of the edge lengths of the tetrahedron, and is explicitly written in (4). Hence, in the terminology introduced above, the 2 2 2 element (12V ) ∈ R is integral over the ring generated by l1, . . . , l6. The proof is now similar to the first proof, proceeding by induction on the number of vertices n and the topological genus g of the polyhedron. The induction step will be based on an analogue of the main Lemma1, whose proof uses the following result. 3 Lemma 2. Let q, p1, p2, . . . , pn be n + 1 distinct points in R , n > 4, with algebraically independent coordinates x0, y0, z0, . . . , zn. Let li be the length of |qpi| and li,j the length of |pipj|, where the indices are identified modulo n. Let ϕ be a place for the field L = Q(x0, . . . , zn) of rational functions of x1, . . . , zn with integer coefficients (L includes the function identically equal to 1), which contains the subring R generated by the squares of the lengths indicated above, regarded as elements of Q in 2 accordance with formula (1). Assume that this place is finite at all the elements li 2 2 and li, i+1. Then there exists i with 1 6 i 6 n − 2 such that ϕ(li,i+2) is finite. Proof of Lemma 2. Assume the contrary:

2 ϕ(li,i+2) = ∞, 1 6 i 6 n − 2. (19) Algebraic methods for solution of polyhedra 461

Then we assert that for all i with 3 6 i 6 n,

2 2 l1,i ϕ(l1,i) = ∞ and ϕ 2 = ∞. (20) l1, i−1

Indeed, these equalities hold for i = 3 by the hypothesis of the lemma and our assumption. If, assuming that they hold for some i > 3, we can also derive them 2 for i+1, then the last of these equalities ϕ(l1,n) = ∞ will contradict the hypothesis of the lemma. Suppose that (20) holds for some i > 3. Then we shall show that

2 l1, i+1 ϕ 2 2 ̸= 0. l1,ili−1, i+1

Again, we assume the opposite:

2 l1, i+1 ϕ 2 2 = 0. (21) l1,ili−1, i+1

For the 10 distances between the 5 points p1, pi−1, pi, pi+1, q the Cayley–Menger equation (7) is

0 1 1 1 1 1 2 2 2 2 1 0 l1, i−1 l1,i l1, i+1 l1 2 2 2 2 1 l1, i−1 0 li−1, i li−1, i+1 li−1 det(CM) = 2 2 2 2 = 0. 1 l1,i li−1, i 0 li, i+1 li 2 2 2 2 1 l1, i+1 li−1, i+1 li, i+1 0 li+1 2 2 2 2 1 l1 li−1 li li+1 0

2 We divide the entries in the 2nd column and 2nd row by l1,i ̸= 0, and the entries 2 in the 5th column and the 5th row by li−1, i+1 ̸= 0, which yields

1 1 0 1 1 1 2 2 l1,i li−1, i+1 2 2 2 1 l1, i−1 l1, i+1 l1 0 1 l2 l2 l2 l2 l2 1,i 1,i 1,i i−1, i+1 1,i l2 1 1, i−1 0 l2 1 l2 2 i−1, i i−1 l1,i 2 = 0. li, i+1 1 1 l2 0 l2 i−1, i l2 i i−1, i+1 1 l2 l2 l2 1, i+1 1 i, i+1 0 i+1 2 2 2 2 2 li−1, i+1 l1,ili−1, i+1 li−1, i+1 li−1, i+1 2 2 l1 2 2 li+1 1 2 li−1 li 2 0 l1,i li−1, i+1 462 I. Kh. Sabitov

We now apply the place ϕ to this determinant. In view of the assumptions (19) and (21) and the equalities (20) for i > 3, we have

0 0 1 1 0 1

0 0 0 1 0 0 2 2 1 0 0 ϕ(li−1, i) 1 ϕ(li−1) ϕ(0) = = 0. 1 1 ϕ(l2 ) 0 0 ϕ(l2) i−1, i i 0 0 1 0 0 0 2 2 1 0 ϕ(li−1) ϕ(li ) 0 0

However, calculating this determinant directly yields ϕ(0) = −1, so the assumption (21) fails and 2 l1, i+1 ϕ 2 2 ̸= 0. l1,ili−1, i+1 Hence 2 2 2 l1, i+1 ϕ(l1, i+1/l1,i) ϕ 2 2 = 2 ̸= 0. l1,ili−1, i+1 ϕ(li−1, i+1)

2 By assumption (19), the value of ϕ(li−1, i+1) involved in this inequality is ∞, so 2 2 ϕ(l1, i+1/l1,i) in the numerator cannot be finite, which verifies the induction step in the proof of (20). However, as already noted, for i = n we then arrive at a contradiction to the assumptions of the lemma. This demonstrates the failure of the assumption (19) and thus proves Lemma2.

Now we can prove an analogue of the main Lemma1, which was used in the first proof.

Lemma 3. If a simplicial polyhedron P has a nice vertex, then verifying the assertion that the volume of P is integral over the ring generated by the squares of the edge lengths reduces to verifying the same assertion for the volume of a certain polyhedron with a nice vertex of smaller degree.

Proof. We start by assuming that P is in general position, that is, the coordinates x1, . . . , zn of all the n vertices of the polyhedron are algebraically independent. Let ϕ be a place for the field Q(x1, . . . , zn) which is finite at the elements of the ring R generated by the squares of the edge lengths of P , and let V be the volume of P . Let q be a nice vertex of P with degree m and let T be its star. Let p1, p2, . . . , pm be the vertices of the boundary of T numbered in a cyclic order. By Lemma2 there exist points p and p such that ϕ(l2 ) is finite. Removing from i0−1 i0+1 i0−1,i0+1

P the faces ⟨qpi0−1pi0 ⟩ and ⟨qpi0 pi0+1⟩ together with the edge qpi0 , we glue up the resulting quadrangular hole by the two flat triangles qpi0−1pi0+1 and pi0−1pi0 pi0+1. This gives us a new simplicial polyhedron P ′ with the same vertices such that q remains a nice vertex, but with degree one less. Let V ′ be the volume of P ′ and assume that we know that 12V ′ ∈ Q is integral over the ring generated by the squares of the edge lengths of P ′. Then any place for Q which is finite over the corresponding ring R′ is also finite atV 12 ′. In particular, ϕ is finite atV 12 ′. ′ ′ Now, the volumes V and V are related by 12V = 12V ± 12V0, where V0 is the Algebraic methods for solution of polyhedra 463

volume of the tetrahedron qpi0−1pi0 pi0+1, at which ϕ also takes a finite value. Therefore, by Corollary3 the place ϕ is also finite atV 12 . That is, to obtain from the test in terms of places that 12V is integral, it is sufficient that the volume of a certain polyhedron P ′ with a nice vertex q of smaller degree be integral, which is the required result. Assume now that the coordinates of the vertices of the polyhedron P under consideration do not have the property of algebraic independence. First of all, we observe that, in view of the representation (1), the squares of the lengths of corresponding edges of different polyhedra with the same combinatorial structure are equal elements of the field Q(x1, . . . , zn) whatever the numerical values of the lengths may be, so for all such polyhedra (with algebraically independent coordinates of the vertices) the polynomial of form (18) for the integral element 12V ∈ Q has the same degree N and the same coefficients ai (1 6 i 6 N) as some polynomials in the squares of the edge lengths. Hence there are two options: 1) either (as in the proof of the main lemma) we construct approximations of P by polyhedra with the same combinatorial structure and with algebraically independent coordinates of vertices and find the polynomial (6) as the limit of polynomials with coefficients of the same form depending on the squares of the edge lengths, or2)we consider the map (˜xi, y˜i, z˜i) → (xi, yi, zi), 1 6 i 6 n, where (˜xi, y˜i, z˜i) are the algebraically independent coordinates of a polyhedron Pe with the same combinatorial structure as P , and since this map is a homomorphism of the corresponding rings, it takes the known polynomial equation for the integral element 12 vol(Pe) to the same equation for 12 vol(P ). The proof is complete. Once we have established this lemma we can continue the proof of the existence of a polynomial expressing the volume, following the same scheme as in the first proof.

11. Several conclusions We summarize by observing that each proof has its advantages and deficiencies. The second proof is much shorter and is more often presented in public, but it only establishes the existence of a polynomial expressing the volume, saying nothing about methods for finding it. The first proof is longer, but it is effective andbesides the existence, it also provides a method for constructing the required polynomial. Moreover, to find a polynomial for the volume we require no information about the position of the polyhedron in space. We need not even know that this polyhedron exists: its combinatorial structure and its edge lengths are all we require. This is equivalent to knowing an abstractly defined development into triangles that must be faces of the polyhedron, that is, we assume that we know a natural development of the polyhedron. Thus, given a triangular development from which the polyhedron must be glued together in accordance with the prescribed combinatorial structure, but with no actual polyhedron at hand and not even knowing if it can be glued together from this natural development, we already know a finite list of possible values of its volume. This observation gives us the following necessary condition for the isometric realization of a given polyhedral metric on a polyhedron in R3. 464 I. Kh. Sabitov

Theorem 2. Assume that a polyhedral metric is given as a triangular development homeomorphic to an orientable manifold. Then for this metric to be realized on a polyhedron in R3 with faces equal to the triangles in the development it is necessary that the polynomial for the squared volume which is calculated for the development have at least one non-negative zero.

Clearly, if we wish to have an isometric embedding of a development in R3, with faces equal to the triangles in the development, then the volume polynomial must have at least one positive zero V 2. Corollary 4. If a polynomial for the squared volume of a polyhedron constructed for a given development has only negative or complex-valued zeros, then this development cannot be formed by the faces of a polyhedron in R3.

Example. Let l1, . . . , l6 be positive numbers such that the right-hand side of (4) is negative. Then these quantities cannot be the edge lengths of a tetrahedron with edges corresponding to particular lengths arranged as in Fig.1, and thus there exists no tetrahedron with faces as in Fig.1 (of course, this is interesting only provided that the corresponding triangle inequalities are satisfied: otherwise the faces will not even exist). Note that Corollary4 does not overlap with Aleksandrov’s well-known result [37] on the realization of a convex polyhedral metric as a metric of a convex polyhedron in R3, since Aleksandrov’s theorem contains no a priori information and imposes no constraints on the combinatorial structure of the required polyhedron. For the same reason there is no overlap with the Burago–Zalgaller theorem [38] on the realization of any polyhedral metric in R3. Incidentally, Corollary4 provides the first test for non-realizability of a polyhedral metric in the class of polyhedra with a fixed combinatorial structure, and this test is valid for polyhedra of any topological genus.

12. Canonical volume polynomials As already mentioned, finding a volume polynomial by the method used in the proof of the existence theorem gives us polynomials of high degree, which definitely have more zeros than there can be isometric realizations of the given polyhedral metric by polyhedra with different values of the volume. Moreover, multiplying a volume polynomial by an arbitrary monic polynomial also produces a volume polynomial of higher degree. It is therefore natural to look for a volume polynomial of smallest degree possible which still has zeros at all the possible values of the volume. Recall (see the end of § 5) that by a volume polynomial for a polyhedron with given combinatorial structure K we mean an arbitrary monic polynomial Q ∈ Q[l][V ] 2 2 with coefficients depending polynomially on the variables l1, . . . , le and vanishing 2 identically after replacing V and the quantities li (1 6 i 6 e) (here and below we denote the set of squares of the lengths by l) by their expressions in terms of the coordinates of the vertices of the polyhedron.12 Let S be the set of all such polynomials. We even extend this class by dropping the a priori assumption that

12We assumed earlier that this must be a polynomial in V 2, but in fact this assumption is not necessary, since this form of the dependence of the volume polynomial on V follows automatically from its definition. Algebraic methods for solution of polyhedra 465 the leading coefficient lc is equal to 1. Assume that we are considering polyhedra of genus 0, that is, homeomorphic to a sphere. It turns out that among the volume polynomials for such polyhedra we can distinguish ones of the smallest degree, which we shall call canonical volume polynomials. A precise statement of this result is given in the following theorem [39]. Theorem 3. Let d be the smallest degree of non-trivial polynomials in S. Then the set of polynomials of degree d in S contains a polynomial Q0 such that 1) its leading coefficient lc(Q0) is equal to 1; 2) it divides all the polynomials in S; 3) such a polynomial Q0 is unique. We start with several lemmas.

Lemma 4. If a polynomial w(l) ∈ Q[l] is not identically equal to zero, then replacing the squares of the lengths by their expressions in terms of the coordinates of the vertices using formula (1) yields a polynomial q(x) = w(l(x)) ̸≡ 0. In fact, a simplicial polyhedron homeomorphic to a sphere and with n vertices has e = 3n − 6 edges, and the number of coordinates of its vertices which uniquely determine the position of the polyhedron up to a space motion is also 3n − 6. By Steinitz’ theorem, for a polyhedron with any combinatorial structure homeomorphic to a sphere there exists an isomorphic (having the same combinatorial structure) strictly convex polyhedron in space. Let x0 be the set of coordinates of vertices of some strictly convex polyhedron with one fixed face and let l0 be the set of squares of the edge lengths of this polyhedron. As shown in [8], for strictly convex polyhedra the map x → l by formula (1) is a diffeomorphism. Hence, in a neighbourhood of x0 and l0 there exists an inverse diffeomorphism l → x, so that if q(x) = w(l(x)) ≡ 0, then also w(l) = q(x(l)) ≡ 0, contradicting the condition w(l) ̸≡ 0. The lemma just proved can be formulated differently as the algebraic independence of the polynomials

2 2 2 lij = (xi − xj) + (yi − yj) + (zi − zj) , defined in terms of the coordinates of the endpoints of the edges of a polyhedron homeomorphic to a sphere.

Lemma 5. Let P1,P2 ∈ S and let lc(P1) = 1. Let R = GCD(P1,P2) ∈ Q(l)[V ] with lc(R) = 1. Then a) deg R > 0; b) R ∈ Q[l][V ]. Proof. a) We shall seek the greatest common divisor R using Euclid’s algorithm, that is, we consider the sequence of polynomials Tn+1 = Tn−1 −UnTn, n > 1, where T1 = P1 and T2 = P2, in which we have deg Tn+1 < deg Tn and Tn,Un ∈ Q(l)[V ]. At the same time we consider the sequence of polynomials Te1 = P1 l(x),V , Te2 = P2 l(x),V , Ten l(x),V ∈ Q(x)[V ] such that Ten+1 = Ten−1 − UenTen. By Lemma4, after going over to the ( x)-variables the non-trivial coefficients of the polynomials Tn(l, V ) will correspond to non-trivial rational functions of (x) multiplying the same powers of V , and therefore the polynomial Re(x, V ) = R l(x),V is 466 I. Kh. Sabitov the GCD for Pe1(x, V ) and Pe2(x, V ). However, the function V (x) provides a common zero for Pe1(x, V ) and Pe2(x, V ), so both polynomials are multiples of V − V (x) in the ring Q(x)[V ], and thus Re(x, V ) is also a multiple of V − V (x), whence deg R = deg Re > 0. b) By what we have proved, R divides P1, so that P1 = RW , where P1 ∈ Q[l][V ], R,W ∈ Q(l)[V ], and lc(P1), lc(R), lc(W ) = 1. We assert that in fact R ∈ Q[l][V ]. Assume not. Then some coefficients in both R(l, V ) and W (l, V ) are rational fractions with non-constant denominators. Consider a polynomial f(l) which is an irreducible factor in the denominator of some coefficient in R. Let ra and wb be the coefficients of the ath and bth monomials in R and W , respectively (where the leading monomial has index 0), such that f occurs in these monomials to the lowest powers ta and ub. As regards R, this means that a > 1 and ta < 0, while for ub we have ub 6 0, for otherwise after multiplying R by W we obtain N−a a fractional coefficient of V in P1 = RW . If there are several coefficients with smallest ta and ub, then we choose the ones multiplying the lowest power of V . Then R and W will be written as

t1 M−1 ta M−a tM R = VM + f A1(l)V + ··· + f Aa(l)V + ··· + f AM (l),

N−M u1 N−M−1 ub N−M−b uN−M W = V + f B1(l)V + ··· + f Bb(l)V + ··· + f BN−M (l), where ti > ta for i < a and ti > ta for i > a. Similarly, uj > ub for j < b and uj > ub for j > b. We get that the coefficient of the polynomial P1 = RW N−a−b multiplying V contains f raised to the power ta + ub < 0, contradicting the assumption that P1 ∈ Q[l][V ]. Proof of Theorem 3. 1) The set S is non-empty. For example, it contains the monic polynomial Q(l, V ) given by formula (6) in the first proof of the central theorem. If this is the unique polynomial in S, then it is the required canonical polynomial Q0 with the properties stated in Theorem3. Suppose that S contains some other polynomials and consider the (one or several) non-trivial polynomials of lowest degree d. Let P , deg P = d, be one of them and consider the two polynomials Q(l, V ) in (6) and P . Applying Lemma5 to them, we see that R = GCD(Q, P ) ∈ Q[l][V ] and lc(R) = 1. On the other hand, in the proof of part a) of Lemma5 we showed that the polynomial R l(x),V is a multiple of V − V (x) (in the ring Q(x)[V ]), so R l(x),V (x) ≡ 0, that is, R(l, V ) ∈ S and therefore deg R > d. Furthermore, deg R 6 deg P = d, so that deg R = d and R is the required polynomial Q0. 2) Consider an arbitrary T ∈ S. By Lemma5 and by what we proved in part 1) of the theorem,

R0 = GCD(Q0,T ) ∈ Q[l][V ], lc(R0) = 1,R0 ∈ S.

Hence deg R0 > d. On the other hand, deg R0 6 deg Q0 = d, therefore R0 = Q0 and T is a multiple of Q0 = R0. 3) Assume that S contains two monic polynomials R1 and R2 of the same degree d which divide all the polynomials in S. Then they divide each other, which is possible only when they coincide. The proof of the theorem is complete. Algebraic methods for solution of polyhedra 467

Thus, we have an algorithm for finding a canonical volume polynomial: we can take an arbitrary polynomial in the set S (for instance, the polynomial (6) constructed in the first proof of the central theorem) and factor it over the field R. Then one of the factors with leading coefficient 1 is the required canonical volume polynomial. Now it is natural to ask about the existence of and methods for finding a canonical volume polynomial for polyhedra of arbitrary topological genus g > 0. Unfortu- nately, the above argument cannot be used, since Lemma4 fails. In fact, simplicial polyhedra of genus g > 1 with n vertices have e = 3n + 6g − 6 edges and 3n − 6 free coordinates of vertices, so the edge lengths of such polyhedra cannot be arbitrary but must satisfy 6g > 6 conditions. This means that in the e-dimensional space of non-negative numbers l1, . . . , le the points corresponding to the lengths in actually existing genus-g polyhedra with given combinatorial structure must lie on some algebraic surface of codimension k = 6g. In other words, to have a solvable system of equations (1) for all the edges of the polyhedron, the right-hand sides must satisfy 6g constraints. If these constraints are Fi(l1, . . . , le) = 0, 1 6 i 6 k, then replacing the lj by their values according to (1), we obtain the identities Fi(l(x)) ≡ 0. Hence there exist non-trivial equations w(l) = 0 which correspond to the identities w(l(x)) ≡ 0, and therefore Lemma4 fails. Thus, the following unsettled question arises.

Question 1. For polyhedra of genus g > 1 prove the existence of a canonical volume polynomial and find a method for constructing it.

The same question was asked in [27], Problem Section.

Remark 2. Expressed in terms of the lengths, a canonical volume polynomial cannot be unique: to any (non-leading) coefficient of such a conjectural polynomial P we can add a non-trivial polynomial in the squares of the lengths which vanishes identically upon substituting the expressions (1) for the lengths, leaving P a volume polynomial (by definition), but changing it as a polynomial in theedge lengths. Perhaps the existence of or the method for finding a canonical polynomial depend on whether the polyhedron can be represented as a triangulated topological manifold of genus g > 1 in the so-called normal form by cutting it along some edges (in other words, whether we can glue the triangles in the development together into a 4g-gon with pairwise identification of the corresponding sides — which isnot always possible).

13. Calculating the volume polynomial for some polyhedra We consider the question of the actual calculation of volume polynomials. In § 3 we found such polynomials for the volumes of polyhedra with n = 4 and n = 5 vertices directly, without appealing to a ‘grand’ theory. Both polynomial equations (4) and (5) have the least possible degree: this is obvious for tetrahedra, and as for polyhedra with five vertices, there are such having two isometric realizations with different values of V 2, so the equation for the volume of a polyhedron with five vertices cannot have degree less than four. Hence (4) and (5) are canonical volume polynomials for the corresponding polyhedra. 468 I. Kh. Sabitov

We now look at polyhedra with n = 6 vertices. First consider a pentagonal pyramid with triangulated base. Without adding new edges we can cut it into 2 pieces, a tetrahedron with volume V1 and a quadrangular pyramid with volume V2. The total volume V is V2 + εV1, ε = ±1. We know equations for the squares of the volumes of the tetrahedron and the pyramid, and it is easy to see that the possible values of the volume of the pentagonal pyramid are zeros of a monic polynomial of degree eight, with number of monomials of order 106, and this degree cannot be reduced. A similar argument shows that the square of the volume of a solid polyhedron bounded by an (n − 1)-gonal pyramid homeomorphic to a sphere and with triangulated base can take 2n−4 values which are zeros of a polynomial of the same degree 2n−4 in V 2 or of degree 2n−3 in V . This is also the degree of a volume polynomial for so-called 3-decomposable polyhedra constructed by successively attaching new tetrahedra to an original tetrahedron or to the polyhedra obtained in the construction process (before gluing, we remove two congruent faces and then glue along the boundaries of these removed faces; see [32] and [40] for details). All these are canonical volume polynomials.

For polyhedra with n = 6 vertices we still have the non-trivial case of an octahedron. If as in [41] we use here the construction developed in the first proof of the central theorem, then we obtain a polynomial of degree 210 = 1024, which does not appear to be minimal. Another method for calculating the volume of an octahedron using only its metric was presented in [42]. Here we sketch it.

For the combinatorial scheme of an octahedron in Fig.5 let X, Y, and Z be the vectors corresponding to the diagonals A1A2, B1B2, and C1C2, respectively. Let x, y, and z be the squares of the lengths of these vectors, that is, the squares of the lengths of the corresponding diagonals. We also consider the vectors a, b,

Figure 5

and c going along the edges B1C1, C1A1, and A1B1. We now calculate the scalar products (aX), (aY), (aZ), (bX), (bY), (bZ) and (XY), (XZ), (YZ). From Algebraic methods for solution of polyhedra 469

Fig.5 we get that −−−→ −−−→ −−−→ X = b + C1A2 = a + c + C1A2 = c + B1A2, −−−→ −−−→ −−−→ −−−→ Y = a + C1B2 = b − c + C1B2 = B1A2 + A2B2, −−−→ −−−→ −−−→ −−−→ Z = B1C2 − a = A1C2 − b = C1A2 + A2C2.

We see from these formulae that the above scalar products are as follows:

(aX) = (l), (bY) = (l), (XY) = (l), (XZ) = (l), (YZ) = (l), 2(bX) = −x + (l), 2(aY) = y + (l), 2(aZ) = −z + (l), 2(bZ) = z + (l), (22) where (l) denotes terms depending only on the squares of the edge lengths of the octahedron. We consider the new vectors va = −Y + 2a + Z and vb = −Z + 2b + X. For the system {vb, x, z, y, va} of five vectors, taking (22) into account, we look at the Gram matrix   a11 − x − z a12 a13 a14 a15 + z  a12 x a23 a24 a25    R =  a13 a23 z a34 a35  ,    a14 a24 a34 y a45  a15 a25 a35 a45 a55 − y − z where the aij are polynomials in the squares of the edge lengths of the octahedron. Since all of these are 3-dimensional vectors, all 4th-order minors of R vanish. This gives us the polynomial equations R11 = ··· = R55 = 0 for the variables x, y, and z, where the Rij are the complementary minors of the corresponding entries. Let v be the volume of the octahedron. Then it is easy to show that the scalar triple product (X, Y, Z) is ±6v, that is,   x a23 a24 2 V = V (x, y, z) = (6v) = det a23 y a34 , a24 a34 z where the aij are equal to the entries of R with the same indices. Multiplying out the determinant, we get that

V = xyz + αx + βy + γz + δ, (23) where the coefficients α, β, γ, δ are polynomials in the squares of the edge lengths of the octahedron. For a given set of lengths assume that there exists a particular octahedron with √ √ √ diagonals x0 , y0 , and z0 . We calculate V0 = V (x0, y0, z0), and, using the polynomials Rij(x, y, z) and V (x, y, z), we produce 27 polynomials in (x, y, z):

V − V0, (V − V0)x, (V − V0)y, (V − V0)z, (V − V0)xy, (V − V0)xz, (V − V0)yz,

(V − V0)xyz, R12,R13,R14,R23,R24,R34,R25,R35,R45,R15,R11,

R55,R22,R33R44,R15x, R15y, R33z, R15xy. 470 I. Kh. Sabitov

We set all of them equal to zero. This system has a solution (x0, y0, z0). We regard it as a homogeneous linear system with respect to the 27 variables

1, x, y, z, xy, xz, yz, xyz, . . . , x2y2z2

i j k (all these variables of the form x y z , 0 6 i, j, k 6 2, occur in the first eight equations, when V (x, y, z) is multiplied in succession by 1, x, y, z, xy, xz, yz, and xyz). With the numerical quantity V0 regarded as a parameter, the system 2 2 2 can have a non-trivial solution (1, x0, . . . , x0y0z0 ) only if Q(V0) = det P (V0) = 0, where P is the matrix of the system for the unknowns taken in the order specified above. It can be represented as

P P P = 1 2 , P3 P4 where P1 is an upper triangular 8 × 8 matrix and P4 has size 19 × 19. The parameter V0 occurs only in the matrix P1, whose diagonal elements have the form δ − V0, so Qe(V0) = det P (V0) is a polynomial in V0 of degree eight with leading coefficient det P4. The first key point of our further arguments is as follows: as a polynomial in the squares of the edge lengths of the octahedron, det P4 does −1 not vanish identically, so the inverse matrix P4 exists at almost every point of the 12-dimensional space l1, l2, . . . , l12. Then we assert that for almost all values of −1 the edge lengths det P = det P4 · det(P1 − P2P4 P3). Indeed,

−1 P1 P2 1 0 P1 − P2P4 P3 P2 −1 = . P3 P4 −P4 P3 1 0 P4

−1 The second key point is that the entries of P1 − P2P4 P3 are polynomials in the squares of the edge lengths of the octahedron. This convenient property was in −1 fact completely unexpected, for the entries of P4 are rational functions (of course, both these key observations were obtained by numerical calculations using a PC). On this basis we can assert that we can replace the equality Qe(V0) = det P (V0) = −1 −1 det(P1 −P2P4 P3) det P4 = 0 by Q(V0) = det(P1 −P2P4 P3) = 0. In the 8th-order −1 determinant det(P1 − P2P4 P3) the variable V0 does not occur below the main diagonal, and the diagonal entries pii have the form qii − V0 with qii, 1 6 i 6 8, independent of V0. Furthermore, in entries over the main diagonal only the first power of V0 occurs. Hence we obtain the highest power of V0 by multiplying the diagonal entries, so that the polynomial equation Q(V ) = 0 has the form V 8 + 7 2 a1(l)V + ··· = 0, where we recall that V = 36v and v is the algebraic volume of the octahedron. We easily produce an example of an octahedron possessing 8 isometric configurations with 8 different values of v2 (you can take the so-called cross-polytope, an octahedron with vertices positioned symmetrically on the coordinate axes, and reflect it in 8 distinct ways in the coordinate planes; see the figuresin[11]). Hence we have found a canonical volume polynomial for the octahedron, since no polynomial of smaller degree can have 8 different roots V 2. In fact, we cannot explicitly write the volume polynomial for octahedra, since in the general case it contains millions (if not billions) of monomials. However, for the Algebraic methods for solution of polyhedra 471

Bricard octahedra of types 1 and 2, whose edges are pairwise equal, calculations produce polynomials with just a few terms (because many similar terms can be combined). For instance, consider the model of an octahedron in Fig.5 with the edge lengths expressed by the following relations:

2 2 2 2 2 2 |A1B1| = |A2B2| = a, |A1B2| = |A2B1| = b, |B1C1| = |B2C2| = c, (24) 2 2 2 2 2 2 |B1C2| = |B2C1| = d, |A1C1| = |A2C2| = e, |A1C2| = |A2C1| = f. (25)

Then

Q(V ) = V 8 − 4ab(c + d + e + f − a − b) + cd(a + b + e + f − c − d) + ef(a + b + c + d − e − f) − (ace + adf + bcf + bde)V 7 = 0, (26) where V = 36v2 and v is the volume of the octahedron. The root V = 0 corresponds to the volume of a flexible Bricard octahedron of type 1, while the second root, distinct from zero, gives us the volume of another, inflexible, realization of an octahedron with the same edge lengths. In particular, it gives the volume of a convex octahedron with these edge lengths. In [39] we present explicitly a volume polynomial for a Bricard octahedron of type 2 which contains only the powers V 8, V 7, and V 6. Besides Bricard octahedra there are many other cases when some of the edges of an octahedron are equal and instead of 12 arbitrary lengths, the formulae for the volume involve fewer free lengths. For instance, this is the case when the structure of the octahedron has some symmetry. In [43] we introduced the notions of combinatorial, metric, and spatial symmetry of polyhedra, and after obtaining a complete classification of the octahedra with metric symmetry (there are 25 main typesof these), we showed that in almost all cases the corresponding volume polynomials contain relatively few monomials, and we wrote out all such polynomials explicitly (with the exception of two types with three subclasses, when there were 7 or 8 free edge lengths). All these polynomials had a common form:

V 8 + aV 7 + bV 6 = 0, so the possible values of the volumes can be calculated explicitly (for some types we even have b = 0). It should be noted that by Theorem1 the zeros of these polynomials contain all possible values of the volumes of octahedra with prescribed metric symmetry, including ones whose configuration displays no spatial symmetry. In the general case the symmetry of a polyhedron or a small number of distinct edge lengths should also make calculations of its volume polynomial much easier. In particular, it would be interesting to find a canonical volume polynomial for the Steffen polyhedron, the simplest flexible polyhedron without self-intersections. It has 9 vertices, 21 edges, and 5 different values of the edge lengths. (A description can be found in many papers; see, for example, [11].) Calculating the volume of polyhedra without knowing their spatial shapes but only the combinatorics and their edge lengths can be useful not only in mathematics, but also in some areas of the natural sciences and technology. For example, in the theory of mixtures and solutions it is important for the investigation of the thermodynamical properties of a solution to know the total volume of the molecules 472 I. Kh. Sabitov of the dissolved salt, which consist of some standard combinations of atoms with constant distances because of some stable physico-chemical bonds between the atoms. The case of molecules having the combinatorial structure of an octahedron and displaying a certain metric symmetry with 2 or 3 distinct edge lengths was considered in [44]. There we could calculate the volumes of the molecules for various versions of the mutual position of the atoms in them. Remark 3. In connection with the determination of possible values of the volume of a polyhedron with a known metric the following observation can be of use: there exist isometric polyhedra with the same combinatorial structure, both convex and non-convex, such that the non-convex polyhedron has a greater volume. This phenomenon was discovered in [38] with the use of complicated construc- tions involving polyhedra with very many vertices. However, this turned out to be possible even for polyhedra with the combinatorial structure of an octahedron. The corresponding example was constructed in [45], and the corresponding octahedra with precise coordinates of their vertices can be seen at the internet site http://www.etudes.ru/ru/mov/mov002/index.php.

14. Calculating the lengths of diagonals of polyhedra Once we know how to calculate the volume of a polyhedron from its combinatorial structure and metric we can find many other geometric parameters of it. In this section we use results from [46] to show how to find the diagonals of a polyhedron. Consider two faces of a polyhedron sharing a common edge. If two vertices of these faces which are not incident to the common edge of the faces are not connected by an edge, then we can join them by a diagonal of the polyhedron, which we call a minor diagonal. For brevity we shall often use the same term for the length of the line segment connecting these vertices in our imagination (the distance between them). We have the following result. Theorem 4. Each minor diagonal satisfies a polynomial equation with coefficients depending only on the choice of this diagonal, the metric, and the combinatorial structure of the polyhedron. For polyhedra in general position not all of these coefficients vanish. We do not give here a complete proof of this theorem (it was thoroughly proved in [46]), but start with a key lemma and then just sketch the proof.

Lemma 6. Let lk be the length of some arbitrarily chosen edge in a polyhedron with a given combinatorial structure K. Then no relation of the following form is possible between the edge lengths of such polyhedra:13

ˆ ˆ S ˆ ˆ L(lk, l) = A0(lk)lk + ··· + AS−1(lk)lk + AS(lk) = 0 (27)

2 ˆ 2 ˆ 2 2 2 2 with A0(lk) + ··· + AS−1(lk) ̸= 0 for l1 + ··· + lk−1 + lk+1 + ··· + le ̸= 0, where e ˆ is the number of edges and the notation lk means that the argument of the function involves all the lengths l1, . . . , le except lk. 13The condition that (27) holds for the edge lengths of polyhedra rather than for the lengths of 1-simplexes in the complex K means that substituting the values l = l(x) from (1) into (4) ˆ transforms this equation into an identity L lk(x), l(x) ≡ 0. Algebraic methods for solution of polyhedra 473

Proof of Lemma 6. In fact, assume that the kth edge in K has endpoints at the vertices with indices i and j and that the faces F and G incident to this edge have vertices i, j, p and i, j, q. We construct a polyhedron P (K) in R3 with the help of the following map P : K → R3: we map the vertices i, j, p, and q to the points Mi(0, yi, zi), Mj(0, yj, zj), Mp(a, 0, 0), and Mq(−a, 0, 0), respectively, a ̸= 0, and map all the other vertices to some points on the Ox axis. Next we rotate Mi and Mj about the Ox axis, with all other vertices fixed. This deformation gives us new polyhedra Pt with the same combinatorial structure K. Then the lengths of all the edges except for the kth remain the same, so if we had equation (27) with the properties described in the hypotheses, then for given lengths l1, . . . , lk−1, lk+1, . . . , le, not all of them zero, the length of lk could take only finitely many values, which is impossible since the value of lk changes continuously in the rotation. The proof is complete.

the polyhedron P the polyhedron P1

Figure 6

Next we sketch the proof of Theorem4. In a polyhedron P let ⟨CD⟩ be a minor diagonal for the dihedral angle between the faces ⟨ABC⟩ and ⟨ABD⟩ sharing the edge ⟨AB⟩. We remove these faces and glue up the resulting hole by the two new faces ⟨ADC⟩ and ⟨BDC⟩, obtaining a new polyhedron P1 such that the minor diagonal ⟨CD⟩ of P coincides with one of its edges (Fig.6). While the volume V = vol(P ) is a root of equation (6), the volume V1 = vol(P1) is a root of another equation:14 2N ˆ 2N−2 ˆ Q1 = V1 + a1(l, d)V1 + ··· + aN (l, d) = 0, (28) with coefficients ai (1 6 i 6 N) depending on the edge lengths of P (with the exception of ⟨AB⟩: this is reflected in our notation ˆl ) and the length d of the diagonal ⟨CD⟩. The volumes V = vol(P ) and V1 = vol(P1) differ by the geometric volume VT > 0 of the tetrahedron ⟨ABCD⟩:

V1 = V + εVT , ε = ±1.

In the general case we do not know whether the volume of this tetrahedron must be added or subtracted, therefore we must find a relation between V and V1 which

14 Of course, in the general case Q1 has degree different from that of(6), say 2N1, but we keep the notation 2N for simplicity. 474 I. Kh. Sabitov is valid for any orientation and any mutual arrangement of these polyhedra. To do this we eliminate ε, obtaining

4 2 2 2 2 2 2 V1 − 2(V + VT )V1 + (V − VT ) = 0. (29)

2 Regarding (28) and (29) as equations for the same unknown v = V1 , we see that their resultant R must vanish:

1 a1 a2 a3 . . . aN 0

0 1 a1 a2 . . . aN−1 aN 2 2 2 2 2 1 −2(V + VT )(V − VT ) 0 ... 0 0

...... R = ...... = 0.

...... 2 2 2 0 0 0 ...... (V − VT ) 0 2 2 2 2 2 0 0 0 ...... −2(V + VT )(V − VT ) (30) We calculate the volume of the tetrahedron (see Fig.7) using the already known formula15

2 2 2 2 2 2 2 2 2 144VT = L d (l1 + l3 + l2 + l4 − L − d ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 + l1l4(L + d + l2 + l3 − l1 − l4) + l3l2(L + d + l1 + l4 − l3 − l2) 2 2 2 2 2 2 2 2 2 2 − L (l1l3 + l2l4) − d (l1l2 + l3l4) (31)

(our notation for the lengths is clear from the picture). Multiplying out the determinant in (30) and collecting the terms containing the same powers of d, we obtain by (31) a polynomial in d of the form

ˆ 2K ˆ 2K−2 ˆ A0(l, L, V )d + A1(l, L, V )d + ··· + AK (l, L, V ) = 0. (32)

Our further analysis aims to demonstrate that at least one coefficient of a non-trivial power of d in (32) does not vanish identically after substituting the expressions for l and V as functions of the coordinates of the vertices of the polyhedron. To do

Figure 7

15Formula (8) in [46] contains a deplorable error: the coefficients of L2 and d2 in the last two (j) (j) terms must be interchanged. Also, in (11) the exponent ν0 must be replaced by ν1 , and in the (j) (j) last equation in (12) µ2 must be replaced by ν2 . Algebraic methods for solution of polyhedra 475 this we use, first, the explicit dependence of the coefficients in(32) on L, taking (30) and (31) into account, and second, our knowledge about the structure of the general term of the resultant. We recall that by the weight theorem the general jth monomial of the resultant (30) has the form

µ(j) µ(j) µ(j) ν(j) (j) (j) (j) 0 1 N 0 ν1 2 2 ν1 2 2 2ν2 cja0 a1 ··· aN b0 (−2) (V + VT ) (V − VT ) (a0 = b0 = 1) (33)

(j) (j) (j) where cj is a non-zero integer and the non-negative integers µ0 , µ1 , . . . , ν2 satisfy µ(j) + µ(j) + ··· + µ(j) = 2, ν(j) + ν(j) + ν(j) = N, 0 1 N 0 1 2 (34) (j) (j) (j) (j) (j) µ1 + 2µ2 + ··· + NµN + ν1 + 2ν2 = 2N. As a result, after fairly meticulous arguments (see [46]) we find that if all the coefficients of powers of d in (32) vanished identically, then setting L equal to the length of the edge AB would transform an equation of the form

16N 2 ˆ 16N 2−2 ˆ c0L + c1(l )L + ··· + c8N 2 (l ) = 0, c0 ̸= 0, into an identity, so that in polyhedra with the same combinatorial structure K as P , one edge length would be determined by the others, which is impossible by Lemma6. Thus, the polynomial (32) in powers of the diagonal length d cannot be trivial. This completes the proof of Theorem4 in [46]. Remark 4. In calculating minor diagonals we do not actually require the volumes of polyhedra. Instead, we could eliminate V in (6) and (32), which would give us a polynomial equation of degree 2KN with respect to d2, with coefficients depending only on the edge lengths. However, there will then be more excessive computa- tions, because when we use volumes we take account only of non-negative roots V 2 of equation (6), so the number of possible values of d2 is in general less than 2KN. We underscore that, in contrast to Theorem1, the leading coefficient of the polynomial (32) for the minor diagonal length d in Theorem4 is not a constant but a polynomial in the squares of the edge lengths of the polyhedron, as is the case for all the other coefficients. Also, there are cases when all the coefficients of this polynomial vanish for particular values of the lengths and volumes. For instance, this happens if changing the minor diagonal in question results in flexing the polyhedron. From Theorem4 and our comments to it we immediately deduce several consequences. Corollary 5. For general-position polyhedra with a given combinatorial structure and given values of the edge lengths, minor diagonals can take only finitely many values defined as roots of equations of the form (32), in each of which V 2 is successively set equal to the possible values of the square of the volume of the polyhedron, which are non-negative roots of equation (6) with respect to V 2. Corollary 6. For a polyhedron to be flexible it is necessary that in at least one equation (32) for minor diagonals all the coefficient functions Ai(l, V ), 0 6 i 6 K, of the edge lengths and the volume of the polyhedron are equal to zero. 476 I. Kh. Sabitov

Corollary 7. Almost all the polyhedra in R3 are inflexible. For polyhedra of genus g = 0 the assertion of Corollary7 was proved in [8], and for g = 1 and higher it is known only from preprints: from [47] and [48] for g = 1 and from [49] in the general case, with a very complicated proof (but valid for any dimension). Question 2. It could be interesting to prove Corollary7 by the same method as in [8], using the first-order rigidity of almost every polyhedron of genus 0 (because of the rigidity of all strictly convex polyhedra). In our case the first-order rigidity of almost every polyhedron of arbitrary genus must be proved for this. It would be sufficient for the same result to prove the second-order rigidity of almost every polyhedron of arbitrary genus. Maybe this approach can also lead to a proof of the inflexibility of almost every polytope in several dimensions. We now describe a procedure for finding the lengths of other diagonals ofa polyhedron if we already know its minor diagonals. This is based on the formula for the distance between two points with known distances from three base points not lying on the same line [50]. We denote the length of an edge ⟨pipj⟩ by lij (see Fig.8). Now let S123 be the area of the triangle ⟨p1p2p3⟩ and let V1234 and V1235 be

Figure 8 the volumes of the corresponding tetrahedra. Then the distance d45 between the vertices p4 and p5 can be calculated by the formula

2 2 2 D + 36εV1234V1235 d45 = l34 + l35 + 2 , (35) S123 where

2l2 l2 − l2 − l2 l2 − l2 − l2 1 23 12 23 13 25 35 23 D = l2 − l2 − l2 2l2 l2 + l2 − l2 , 8 12 23 13 13 35 13 15 2 2 2 2 2 2 l24 − l23 − l34 l13 − l34 − l14 0 and ε = ±1, depending on whether the vertices p4 and p5 lie on different sides or the same side of the plane of the triangle ⟨p1p2p3⟩ (if one of p4 and p5 or both of them lie in the plane of the base, then the corresponding volumes vanish and the choice of ε is irrelevant). Assume that we must find the length of the diagonal connecting two vertices on the boundary of the same star (Fig.9), for instance, the distance between Algebraic methods for solution of polyhedra 477

M1 and M5. The distances M1M3, M2M4, and M3M5, which are the lengths of minor diagonals, are known. We now find the distance between M1 and M4 by formula (35), as the distance between two points whose distances from the vertices of the base triangle ⟨OM2M3⟩ are known, after which we use the same formula to find the distance between M1 and M5 as points with known distances from the base points O, M3, and M4. In this way we can obviously find the lengths of all the diagonals in a star, one after another, discarding at the same time some bad choices of the lengths of minor diagonals, because we can calculate each ‘non-minor’ diagonal in two ways, by approaching the vertex in question from two sides, and requiring that the results of the calculations coincide.

Figure 9 Figure 10

Consider now a diagonal connecting vertices of different stars. We describe an algorithm for calculating its length, taking the diagonal MN in Fig. 10 for example. We find a shortest path (in terms of the number of edges) between M and N (in the picture this is the polygonal line MA1A2A3N). We know how to calculate the distances between vertices of a star, and we shall show how to find the distances between M and vertices of the star of A2. The stars of A1 and A2 share the edge ⟨A1A2⟩ and the face ⟨A1A2M3⟩, which is the base triangle for finding the distance between M and A3 by formula (35), since we know all the distances from these points to the vertices of the face ⟨A1A2M3⟩. It can easily be seen that ⟨A2A3M3⟩ is a base triangle for calculating the distance between M and N1. Once we have found it, M3A3N1 becomes a base triangle for calculating the distance MA4. Finally, knowing the distances from M and N to the vertices of the face ⟨A3A4N1⟩, we can find the distance between M and N. Of course, it should not be overlooked that we find two possible values of the distances under consideration at each step, so as a result we do not get the precise values of the distances, but rather a finite set of their possible values, which is quite sufficient for a finite algorithm. We seethat the algorithm works provided that no three vertices of the polyhedron lie on one line, for otherwise some base triangle can be degenerate and we will not be able to use (35). This procedure for finding the diagonals of a polyhedron leads to the following problem. As we know, there are pairs of isometric polyhedra with one flexible and the other not flexible (you can take a flexible Bricard octahedron P of type 1 or 2 with flat equator and an embedded octahedron isometric to it obtained byamirror reflection of half of P with respect to the equatorial plane). Using the method 478 I. Kh. Sabitov of diagonals described above, we end up with a positive result if the polyhedron does not flex, but this same method does not work for the flexible polyhedron, because one of the polynomials for a minor or ‘non-minor’ diagonal will vanish identically. Now it is natural to ask: how many and what particular diagonals must be known (more precisely, given) to be able to find all the remaining diagonals locally uniquely, and therefore to fix locally the spatial position of the polyhedron in a sufficiently small neighbourhood (in other words, to prove its local unique determinacy or inflexibility in the case of fixed lengths of the diagonals in question)? If the minimum sufficient number of fixed (lengths of) diagonals is p, then we say that the polyhedron is combinatorially p-parametric. This notion and some modifications of it have been studied in[51]–[53] and other papers mentioned there. Among the central unsolved problems in this direction is the problem of finding an algorithm for calculating the lengths of diagonals of inflexible polyhedra: since the polyhedron is inflexible, it is locally uniquely determined by its combinatorial structure and metric, so the lengths of the diagonals are uniquely determined — but how can we find them? For example, by Cauchy’s theorem acon- vex polyhedron is uniquely determined by its combinatorial structure and metric, so it is 0-parametric in the class of convex polyhedra, and therefore the lengths of its diagonals are uniquely determined — but what is an algorithm for computing them?

15. Generalizing the equations for volumes and diagonals Once we have established that we can calculate the lengths of all the diagonals as roots of certain polynomial equations determined by the combinatorial structure of the polyhedron and the squares of the edge lengths, the following result from [54] appears as a natural generalization of these equations.

Theorem 5. Let f(x1, . . . , zn) be a polynomial in the coordinates of the n vertices of a simplicial polyhedron homeomorphic to a sphere and let f be invariant with respect to space motions. Then there exists a polynomial q ∈ Q[l, t] such that f is a root of this polynomial. For instance, the algebraic volume V of the polyhedron defined in § 2 is a homogeneous polynomial of degree three in the coordinates x1, . . . , zn. It is invariant with respect to motions of the first kind and its square is invariant with respect to all motions, so Theorem5 is a new demonstration of the already known existence of a polynomial N N−1 c0(l)t + c1(l)t + ··· + cN (l) with V 2 as a zero. The same holds for the square of the length of any diagonal in the polyhedron. However, Theorem5 says nothing about the form of the leading coefficient, and proving that for f = V 2 this coefficient is equal to 1 was the crucial problem. Proof of Theorem 5. This is a consequence of the algebraic independence (already used in § 12) of the squares of the edge lengths of a polyhedron homeomorphic to a sphere in the ring of polynomials in the coordinates of the vertices, which squares are invariant with respect to motions. Since f is invariant under motions of the polyhedron, we position the polyhedron so that the vertices of one of its Algebraic methods for solution of polyhedra 479 faces (for instance, the ones with indices 1, 2, 3) have coordinates x1 = y1 = z1 = y2 = z2 = z3 = 0. Then f becomes a polynomial F in the n − 6 free variables x2, x3, y3, x4, . . . , zn. The polynomials (1) are algebraically independent and their number is also 3n − 6, so they form a transcendence basis of the algebra of polynomials in the variables indicated above [55]. Hence F is algebraically dependent on the polynomials (1), as required. A more interesting and more difficult result concerns the degree of the polynomial q for given f. In the same paper [54] it is shown that whatever form f may have, the class of polynomial equations for f

N q = c0(l)f + ··· + cN−1(l)f + cN (l) = 0,

e contains a polynomial of degree N 6 2 , where e is the number of the edges of the polyhedron.

16. Polyhedra with diagonals whose lengths are known If we know not just the edge lengths of an oriented simplicial polyhedron but also the lengths of its diagonals, then we can propose an explicit formula for its volume. Indeed, we put the origin at one of the vertices of the polyhedron and give it the index 0: x0 = y0 = z0 = 0. Let (xi, yi, zi), 1 6 i 6 n, be the coordinates of the remaining n vertices. Assume that some orientation of the polyhedron is fixed. We construct the tetrahedra with apexes at the origin and with bases coinciding with faces of the polyhedron. Then the generalized volume V of the polyhedron can be calculated by the formula

x y z  X i i i 6V = det xj yj zj  , (36) F xk yk zk where the sum is taken over all the faces and in each term the vertices of the face are put in the order prescribed by the orientation of that face. Taking the square of both sides of (36), we write the determinants on the right-hand side one time in the original form and a second time as the determinants of the transposed matrices. Then we have

rirp rirq rirs 2 X X 36V = rjrp rjrq rjrs (37)

Fijk Fpqs rkrp rkrq rkrs (the order of summation is obvious). Each scalar product in the matrix entries on the right-hand side can be expressed in terms of the squares of the lengths of the edges and diagonals. For example,

r2 + r2 − d2 r r = i p ip , i p 2

2 2 where ri and rp are the squares of the lengths of the diagonals connecting vertices 2 with indices i and p with the 0th vertex, and dip is the square of the length of an edge or diagonal (depending on whether i and p are the endpoints of an edge). 480 I. Kh. Sabitov

Formula (36) gives us the precise value of the volume as a function of the edge lengths in the case when the polyhedron has no diagonals. The simplest example is formula (4) for the volume of a tetrahedron. Another example is a torus with fewest possible vertices. Such a torus has seven vertices, all of which are connected by edges with one another. A formula for the volume V of such a polyhedron was found (of course, with the use of PC calculations) by O. Davydov, a student at Moscow 1 State University, in his diploma thesis of 1999; it takes 3 2 standard (A4) pages, and being a precise formula of the form V 2 = ··· cannot be simplified!16 (The following observation is important: given the ‘frame’, the 1-skeleton of 21 edges, there are 120 ways to ‘glue up’ this skeleton by triangular faces, thus obtaining 120 toroidal polyhedra with different combinatorial structures, and hence 120 (!) formulae for their volumes; see, for instance, [56].) In the general case we easily find an upper bound for the number of monomials on the right-hand side of (37): it does not exceed 243(2n − 2 + 4g − k)2, where g is the genus of the polyhedron and k is the degree of the vertex taken for the origin, so that it is substantially less than the number of monomials in the volume polynomial not involving diagonals. A formula similar to (37) was known already to Staudt [57]. Sylvester [58] commented on it17 and generalized it.

17. Isometric immersion problem for a polyhedral metric One of the central problems of the metric theory of polyhedra is the isometric realization of a given polyhedral metric. This problem can be put into many different forms. If we are looking for an immersion of the metric in the form of a polyhedral surface, with no a priori assumptions about its combinatorial structure or shape, then the problem has a solution, which is presented in [38]. If we assume in advance that the metric is given on a topological sphere and is convex, then it can be realized on a convex polyhedron, but the combinatorial structure of this polyhedron may be completely different from the combinatorial structure of the development (this is Aleksandrov’s classical result [37]; modern algorithms describing the edges and faces of the resulting shape were proposed in [59] and [60]). Theorem4 and Corollary5 to it allow us to propose the following algorithm for reconstructing a polyhedron from a given natural development: find all the non-negative roots V 2 of equation (6), substitute them in all the equations (32), one after another (in the general case the number of these equations is equal to the number e of edges of the polyhedron), find the non-negative roots d2 of these equations, thereby making a list of all possible values of the lengths of all the minor diagonals, and then on the basis of this list draw up a new list of the possible systems of lengths of minor diagonals for each of the conjectural polyhedra (that is, for each minor diagonal we take one of the possible values of its length),18 and finally, by conducting an exhaustive

16Nevertheless, its physical length can be slightly reduced by replacing the symbols of the form dij for the edge lengths by single letters: you will require 21 letters to do this. 17Staudt proved that the products of the areas of two polygons or of the volumes of two polyhedra are entire algebraic functions (that is, polynomials) of the squares of the various distances between their vertices; Sylvester called these results beautiful and important geometric theorems. 18By the way, we see that in the general case the number of possible realizations of a given development as a polyhedron of topological genus g with n vertices has the bound N · 2K1 · 2K2 ··· 2Ke, where 2N is the degree of the volume polynomial, the 2Ki, 1 6 i 6 e, are the Algebraic methods for solution of polyhedra 481 search of the list, find compatible systems of lengths of diagonals or show that there are no such systems, that is, that the development in question cannot be realized isometrically in R3. In [46] we presented a numerical example due to S. N. Mikhalev in which this scheme is implemented. It involves the realization of a development of an octahedron with prescribed lengths of several edges. Its isometric immersion in accordance with the above scheme results in eight non-congruent isometric configurations of an octahedron. Our method has a drawback: it cannot be used when the polyhedron to be constructed is flexible. An algorithm free from this deficiency was proposed in[61], but so far it has been realized only for genus 0 polyhedra. We fix a development consist- ing of triangles which must be the faces of the desired polyhedron homeomorphic to a sphere. First we make an assumption about the combinatorial structure of the development: it contains no 3-edged null-cycles (as defined in § 6). Then by The- orem 1 in [62] the development contains a Hamilton cycle, that is, a simple cycle formed by edges and passing through all the vertices of the development. It parti- tions the development into two simply connected domains such that the boundary of either domain contains all n vertices of the development. We realize one of these domains in R3 as a polyhedron with boundary: this will be some spatial n-gon M triangulated without adding new vertices. It contains n − 2 faces and n − 3 interior edges which are common to pairs of faces. It is shown in [61] that we can construct M so that the n − 3 dihedral angles between consis- tently oriented faces take prescribed values ϕk ∈ [0, 2π), 1 6 k 6 n − 3. Since all the vertices of the polyhedron are already vertices of M, the position of M uniquely determines the position of the polyhedron with the same vertices. It will have some new edges, whose lengths (the distances between the corresponding vertices) must be equal to the lengths of the remaining n − 3 edges of the development, which we know but have not yet used. To satisfy this condition certain equations are derived, and each solution of these equations is proved to correspond to a unique polyhedron realizing the given development isometrically in R3 (so that if this system is unsolvable, then the isometric realization is impossible, while if the solutions form a continuous family, then we obtain a flexible polyhedron). The construction of the initial polygon M is effective, the equations for the angles ϕk are linear with respect to cos ϕk and sin ϕk, 1 6 k 6 n − 3, and can be written out explicitly, and thus the entire algorithm can be realized on a computer even in the case of developments of rather complex combinatorial structure. On the other hand, if the development contains a 3-edged null-cycle, then we can cut the polyhedron19 along this cycle and partition the development into two parts, with triangular boundaries. Each boundary may be declared to bound a 2-simplex, which makes the new developments homeomorphic to a sphere. If at least one of degrees of the polynomials (32) for the minor diagonals, and e = 3n − 6 + 6g is the number of edges of the polyhedron; here we bear in mind that given the length of a minor diagonal there exist two ways to glue two faces. Of course, the upper bound must in fact be considerably lower, since once we have found the values of n − 3 minor diagonals for the boundary of the star of a vertex of degree n > 4, we have at most two possible values for the length of each of the three remaining diagonals. 19In fact, we cannot cut the polyhedron itself, which is not yet available, and we should rather talk about cutting a sphere homeomorphic to the body of the development. 482 I. Kh. Sabitov them cannot be realized isometrically in R3, then neither can the original development. If both of them can be realized as polyhedra in R3, then they have congruent faces corresponding to the additional 2-simplexes. Removing the interiors of these faces, we glue the resulting polyhedra along the congruent triangular boundaries, and this gives us an isometric realization of the original development in R3. If a development of a polyhedron of genus g > 1 with n vertices contains a Hamil- ton cycle bounding a disc, then in realizing this disc in R3 as a triangulated spatial polygon with the vertices at the required distances from one another we can again form a system of linear equations for the cosines and sines of the n − 3 dihedral angles between the n − 3 triangular faces of the polygon. This time the system will be overdetermined, and we have as yet no software implementation of the algorithm. Remark 5. By the way, the method and result of [61] answer Legendre’s question on the number of parameters uniquely determining the position of a polyhedron. Legendre himself answered the question in the convex case (see [1] or any later edition) by showing that for convex polyhedra the number of such parameters is equal to the number of edges of the polyhedron, but the edge lengths cannot be taken for such parameters. In the case of a general inflexible polyhedron homeomorphic to a sphere it follows from [61] that the number of parameters is also equal to the number of edges: 2n − 3 parameters are the edge lengths of the triangulated polygon M and n − 3 parameters are the dihedral angles ϕk between faces in M.

18. Some algebro-geometric properties of volume polynomials In our opinion the most intriguing feature of a volume polynomial for a polyhedron is that its origin (its definition or the proof of its existence) is based onthe exterior shape of the polyhedron as an object in 3-dimensional space, while it is completely determined by intrinsic characteristics of the polyhedron, namely, its combinatorial structure and metric, and hence we can write out the polynomial given a natural development, before constructing the polyhedron in space. This is similar to the Gaussian curvature of a surface.20 Historically, it was first defined ‘extrinsically’ as the product of the principal curvatures of the surface, and then its intrinsic definition in terms of the metric on the surface was discovered andagreat theory on relations between the intrinsic and extrinsic geometry of surfaces was developed. However, volume polynomials have not yet lived through the second part of the history of Gaussian curvature, in that no intrinsic interpretation of the origins of volume polynomials is known. Perhaps some property of a class of systems of algebraic equations connected by some combinatorial relations even more general than (1) will be discovered here. 18.1. Roots of a polynomial as the volumes of polyhedra. In this subsec- tion we discuss the question of whether each zero of the polynomial Q(V ) in (6) is the volume of some actually existing polyhedron having the same combinatorial structure and metric as the ones for which the polynomial was constructed.21 If there exists such a polynomial, then we say that the root is realized.

20Note, however, a significant distinction from the Gaussian curvature, which has a local definition, while the volume is a global characteristic of the polyhedron. 21By analogy with the determination of a metric from its Gaussian curvature, after which the question of an isometric immersion of this metric can be investigated, we can pose the broader Algebraic methods for solution of polyhedra 483

In fact, this is a very difficult question. Only partial answers to it are known, which are, however, of crucial importance. The starting point of investigations here must be to define a metric simplicial complex, so first of all we define metric Bricard octahedra as combinatorial octahedra with edge lengths satisfying the same conditions as the edges of flexible Bricard octahedra of types 1 and 2. Namely, if in the model of an octahedron given in Fig.5 the edge lengths are related by (24) and (25), then we have a metric Bricard octahedron of type 1. But if these lengths are related by the equalities

2 2 2 2 2 2 |A1B1| = |B1A2| = f, |A2B2| = |A2A1| = e, |C1B2| = |C2B2| = c, 2 2 2 2 2 2 |C1B1| = |C2B1| = d, |A2C1| = |A1C2| = a, |A1C1| = |A2C2| = b, then we have a metric Bricard octahedron of type 2. For these abstractly defined metric octahedra we can write the volume polynomial explicitly. For type 1 it is 8 7 6 given in (26) and for type 2 it has the form V + a1(l)V + a2(l)V . Hence, it has two different zeros for type 1 and three zeros for type 2. In either case oneofits zeros is equal to 0. The following results were established in [45]. Theorem 6. Assume that a given metric Bricard octahedron of type 1 has a canon- 22 ical volume polynomial with a positive zero V0. Then this metric Bricard octahedron has an isometric realization in R3 as an octahedron with squared volume equal to V0. For octahedra of type 2 more cases are possible. Theorem 7. Assume that a given metric Bricard octahedron of type 2 has a canonical volume polynomial with a positive zero V1. Then this metric Bricard octahedron has an isometric realization in R3 as an octahedron with squared volume equal to V1 if one of the following conditions holds: 1) the zero V1 has multiplicity 1; 2) V1 has multiplicity 2 and

(f − d)(e − c) ̸= ab. (38)

Note that if V1 is a multiple zero and (38) holds, then a certain combination of the edge lengths vanishes, which in turn ensures the existence of the required octahedron. In the case when (f − d)(e − c) = ab and the polynomial has a non-zero root V1, this root must have multiplicity 2, but its positivity turns out not to be sufficient for the realization of the metric octahedron. More precisely, there exist examples of realizability and non-realizability in [45]. Hence there are cases when not every root of the polynomial can be realized as the volume of an actually existing polyhedron. question of whether there exists a metric simplicial complex with prescribed combinatorial structure such that the corresponding polynomial Q(V ) has the root in question. Then we can ask whether this root is the volume of some isometric immersion of the simplicial complex obtained. 22To avoid explaining in each case the relation between the variable V in the formula for the polynomial and the volume v, we shall assume that the polynomial has been multiplied by a numerical coefficient such that V = v2. 484 I. Kh. Sabitov

Remark 6. The above theorems do not exhaust the question of necessary and sufficient conditions for the isometric realization in R3 of metric Bricard octahedra of types 1 and 2, since the question of the realization of the root V = 0 remains open. Moreover, in the case of a multiple positive root no sufficient conditions for its realization are known. The question of the possible realization of roots of a canonical volume polynomial for other polyhedra, even for octahedra other than metric Bricard octahedra, is still open. 18.2. Roots of a volume polynomial for a flexible polyhedron. Another problem on roots of volume polynomials is to discover special properties of roots in the case of a flexible polyhedron. We proved the existence of a volume polynomial by constructing the polynomial (6) explicitly with the use of induction. This construction is not unambiguous: it depends on the choice of a nice vertex and the method by which we reduce its degree, and therefore there exist many volume polynomials for each polyhedron. We agree to mean by a standard volume polynomial any polynomial constructed following the proof of the central Theorem1. Then we can state the following result [63]. Theorem 8. The root V 2 of a standard polynomial Q(V ) corresponding to a flexible polyhedron is multiple.

Proof. Let P be a flexible polyhedron with volume V = V0. Hence, it has at least one variable dihedral angle, or in other words, there exists a minor diagonal of variable length. Let D be such a diagonal, of length d. In [22] and [21] it was shown that before the last step in finding the required polynomial (6) we obtain two equations of the following form for the volume V and the length d of the minor diagonal D:

2K 2N 2K−2 2N 0 2N d (V + ··· ) + d b1(l)V + ··· + ··· + d bK (l)V + ··· = 0, (39) where the bi(l) (1 6 i 6 K) are some known polynomials in the squares of the edge lengths of the polyhedron, and

2 2L 2 0 C0(l, V )d + ··· + CL(l, V )d = 0, (40)

2 2M 2M−2 with CL(l, V ) = V + c1(l)V + ··· and the degree in V of the polynomials 2 Ci(l, V ) (0 6 i 6 L−1) less than 2M. The polynomial Q(V ) in (6) is the resultant of the polynomials on the left-hand sides of equations (39) and (40) corresponding to the elimination of d in these equations. However, for a flexible polyhedron P all the coefficients of powers of d on the left-hand sides of (39) and (40) vanish, because otherwise these equations would allow only finitely many values of d, contradicting the assumption that the length of the diagonal D changes continuously. Hence the resultant R of the polynomials on the left-hand sides of (39) and (40) is the 2 2 determinant of a matrix whose entries all vanish at the point V = V0 . Taking ′ 2 the derivative QV 2 of Q(V ), regarded as the resultant R, at the point V0 gives us a sum of the determinants of matrices in each of which all but possibly one of the rows (namely, the row currently differentiated) consist of zero entries, so that ′ 2 QV 2 (V0) = 0. Hence V0 is a multiple root of Q(V ). Algebraic methods for solution of polyhedra 485

However, the above argument is valid only in the ‘regular’ case of finding the polynomial Q(V ), when we use the main Lemma1 directly. When we cannot use this lemma, that is, when the polyhedron has no nice vertices, we must cut the polyhedron along some null-cycle, find volume polynomials for the two pieces of the polyhedron that are possibly obtained as a result, and combine them into a volume polynomial for the entire polyhedron. In this case it can happen that one of the two pieces does not flex and we cannot apply the above argument to its volume polynomial.23 Still, at least one piece of the flexible polyhedron must be flexible, and its volume polynomial has a multiple root corresponding to the volume of this piece. As concerns the proof that the total volume polynomial has a multiple root, we can use the following algebraic lemma. Lemma 7. Consider a system of three polynomial equations

n n−1 m m−1 a0x + a1x + ··· + an = 0, b0y + b1y + ··· + bm = 0, z = x + εy, ε = ±1.

Let x0 be a multiple root of the first equation and y0 a not necessarily multiple root of the second. Then z0 = x0 + εy0 is a multiple root of the equation q(z) = 0 obtained by eliminating x and y in the system, and the multiplicity of this root is no less than that of x0. We shall not prove this lemma since in fact it will only give us that the root 2 V = V0 is multiple, not V0 as we require. Thus, we prove another version of this lemma. Lemma 8. Consider a system of three polynomial equations  Xn + a Xn−1 + ··· + a = 0, where X = x2,  1 n m m−1 2 Y + b1Y + ··· + bm = 0, where Y = y , Z = z2, z = x + εy, ε = ±1.

2 2 Let X0 = x0 be a multiple root of the first equation and Y0 = y0 a not necessarily 2 2 multiple root of the second. Then Z0 = z0 = (x0 + εy0) is a multiple root of the equation q(Z) = 0, Z = z2, obtained by eliminating X and Y in the system, and its multiplicity is no less than that of X0. Proof. We represent the first two equations in the form

2 2 2 2 2 2 2 2 (X − X0) f(X) = (x − x0) f(x ) = 0, (Y − Y0)g(Y ) = (y − y0)g(y ) = 0 and set x = z + εy in the first equation. Then we get that

2 2 2 2 2 2 2 2 2 2 2 [(z + y − x0) + 4z y ]f(t) = −4εzy(z + y − x0)f(t), t = z + y + 2εzy. (41)

2 We can assume that f(t0) ̸= 0 for t0 = (z0+εy0) , since otherwise the corresponding root X0 has a larger multiplicity and we could factor out (X−X0) to a power greater

23We must admit that in [63] we overlooked this case. 486 I. Kh. Sabitov

2 2 2 2 than 2. Taking the squares of both sides of (41), we get that [(z + y − x0) − 2 2 2 2 4z y ] f (t) = 0. Setting x0 = z0 + εy0, we can represent this equation as

2 2 2 2 [(Z − Z0) − 4εz0y0(Z − Z0) + 2(y − y0)(Z − Z0) − 4y (Z − Z0) 2 2 2 2 2 2 2 2 2 − 4Z0(y − y0) − 4ε(y − y0)z0y0 + (y − y0) ] f (t) = 0, which shows that upon eliminating y in this equation and the equation with respect 2 to y in the original system we obtain an equation in which (Z − Z0) is a common factor in all the terms. This completes the proof of Lemma8, which also gives us the theorem, because the method used to derive the equation for V 2 from the equations for the volumes V1 and V2 of the pieces of the polyhedron is just the same as the one we used above for Z = (x + εy)2 (for example, see the beginning of the proof of the main lemma in § 8).

19. The strong bellows conjecture It was conjectured in [64] that flexible polyhedra do not just preserve their volume in flexes, but even remain equicomposed, that is, in any pair of these polyhedra one of them can be partitioned into finitely many polyhedra that can be rearranged by motions without overlapping so as to produce the other polyhedron. This conjecture is usually called the strong bellows conjecture. The question of whether polyhedra with the same volume are equicomposed is Hilbert’s third problem. We briefly recall that in the plane the answer is affirmative: two polygons ofthesame area are equicomposed, while in space this fails in the general case: there are examples of a cube and a tetrahedron with the same volume that are nevertheless not equicomposed. The classical results are thoroughly presented in [65], while what concerns the contemporary approaches to the statement of the problem and newer results can be found in [66], [67], and [68]. The well-known equicomposition test for polyhedra of equal volume is the equality of their Dehn invariants (see [65] about these invariants).24 However, in the case of general flexible polyhedra it is not clear what we should mean by the equicomposition of such polyhedra as space bodies: there exist non-embedded and even non-immersed flexible polyhedral surfaces, so that we cannot speak of the bodies bounded by them. On the other hand, for the definition of the Dehn invariants we do not require a body bounded by the polyhedral surface, so the problem is posed as follows: do polyhedra (treated as polyhedral surfaces) that can be flexed one into the other have the same Dehn invariants? Since the equality of the Dehn invariants for each pair of polyhedra means that each polyhedron in a family of flexes must have the same Dehn invariants as the polyhedron taken as the initial one, flexes must preserve the Dehn invariants if the strong bellows conjecture holds. It is clear

24As regards the necessity part of this test, it is most geometrically expressed by the following Bricard condition: if two polyhedra A and B are equicomposed, then there exist positive integers ′ nj and ni and an integer p such that ′ ′ n1α1 + ··· + nqαq = n1β1 + ··· + nrβr + pπ, where α1, . . . , αq and β1, . . . , βr are the values of the dihedral angles in the polyhedra A and B, respectively. Algebraic methods for solution of polyhedra 487 that if the polyhedra in a family of flexes are embedded, then the constancy of the Dehn invariants means that the polyhedra are equicomposed, because a polyhedron does not change volume during a flex. In [69] the question of the constancy of the Dehn invariants was answered in the affirmative for each of the three types of flexible octahedra and for flexes oftheStef- fen polyhedron. However, already for flexible suspensions with a 6-gonal equator it was shown in [70] that a flex can produce polyhedra some of whose Dehn invariants do not coincide with the corresponding invariants of the original polyhedron. But it should be noted that the flexible suspensions are not embedded polyhedra, so the question of the equicomposition of embedded flexible polyhedra remains open.

20. Another invariant of flexible polyhedra There exists another geometric characteristic of polyhedra which is preserved by flexes, namely, the integral mean curvature H of an orientable embedded polyhedron, which is defined as 1 X H = l (π − α ), (42) 2 i i where the sum is taken over all the edges, li is the length of the ith edge, αi is the dihedral angle at the ith edge, 1 6 i 6 e, and e is the number of edges. We assume that the polyhedron under consideration is oriented by a normal to some face whose direction is extended to the other faces in accordance with the standard rule for matching the orientation of faces. Then the dihedral angle αi at the ith edge is measured as the geometric internal angle of the polyhedron. It is proved in [71] that if an embedded oriented polyhedron is flexible, then it preserves its integral mean curvature during a flex. Moreover, this property of the invariance of n H also holds in several dimensions (in R , n > 4, we must replace the edge length in (42) by the (n−2)-volume of the common (n−2)-face of two (n−1)-dimensional facets, with the appropriate understanding of the orientation and the angle αi). The proof of this result is simple, and we present it here. Let K(t) be a smooth (with respect to the parameter t) family of 3-simplexes in E3. Let L(t) be an edge of K(t) and let F1(t) and F2(t) be the incident faces. We take a point M(t) on L(t) and four vectors at this point: the outward normals N1 and N2 to the faces F1 and F2, respectively, and the outward normals v1 and v2 to L(t) in the planes of F1 and F2, in accordance with the orientation of these faces (Fig. 11). The normals N1 and N2 are taken to have unit length, and the normals v1 and v2 have length l(t) equal to the length of L(t). In the plane T orthogonal to L(t) we introduce a Cartesian system of coordinates with unit basis vectors n1 and n2. Then

N1(t) = cos θ1 n1 + sin θ1 n2,

N2(t) = cos θ2 n1 + sin θ2 n2,

v1(t) = l(t)(− sin θ1 n1 + cos θ1 n2),

v2(t) = l(t)(sin θ2 n1 − cos θ2 n2), where θi is the angle between n1 and Ni, measured in the standard way consistent with the orientation, i = 1, 2. It is easy to see that the internal dihedral angle 488 I. Kh. Sabitov

Figure 11

α(t) between the planes of the faces F1 and F2 is equal to π − θ2(t) − θ1(t) . We want to know how the smooth deformation parametrized by t changes α(t).25 By performing a suitable motion, we can assume that the deformation stabilizes the plane T and the vectors n1 and n2 in it. Then dN 1 = −(sin θ n − cos θ n )θ′ (t), dt 1 1 1 2 1 dN 2 = −(sin θ n − cos θ n )θ′ (t), dt 2 1 2 2 2 which gives us the required equation

dα(t) dN dN dv dv l(t) = 1 , v + 2 , v = − N , 1 − N , 2 . (43) dt dt 1 dt 2 1 dt 2 dt

Now consider a polyhedral surface P with non-degenerate triangular faces (we do not require P to be embedded or even immersed). Supposing it to be flexible, we apply formula (43) to the dihedral angles at all the edges. Then we get that X dαk(t) X X dvij l = − N , , (44) k dt i dt k i j where k is the index of an edge, i is the index of a face, and j ranges between 1 and 3 on each face. The vectors vij are the outward normals to the sides of the ith face, of lengths equal to the lengths of the corresponding sides. By Minkowski’s P3 well-known formula, j=1 vij = 0 for each i, so the sum on the left-hand side of the equation vanishes, which means that when the polyhedron flexes, its mean curvature H has derivative zero. Moreover, this property of the invariance of H persists in the multidimensional case.

25In fact, in place of a tetrahedron we could take two triangles sharing an edge and the dihedral angle at this edge. Algebraic methods for solution of polyhedra 489

Remark 7. It follows from the above formula and the definition (42) of the mean curvature that in deformations of a polyhedron the differential of its mean curvature depends linearly on the differentials of the edge lengths:

1 X dH = − α dl (i ranges over the index set of the edges). 2 i i i An interesting comparison of the methods of the invariance proofs for the volume of a polyhedron and its integral mean curvature was carried out in [72]. The author points out that the invariance of the volume has no infinitesimal analogue: if a polyhedron admits a non-trivial infinitesimal flex of the first order, then the corresponding first variation of the volume is not necessarily zero (however, if the infinitesimal flex is the velocity field of a flex, then of course the variation of the volume vanishes). This indicates in an indirect fashion that we cannot link the differential of the volume with the differentials of the edge lengths, explaining why both known proofsof the invariance of the volume of a flexible polyhedron rely on algebraic methods.26 Still, we see from the above observation that the differential of the mean curvature is linearly expressed in terms of the differentials of the edge lengths. Following [72], we now show that the mean curvature and the edge lengths are not related algebraically. Consider the tetrahedron in Fig.1. By elementary calculations we find the dihedral angle ϕ at the edge l1:

4S2 + 4S2 − l2l2 + [(l2 + l2) − (l2 + l2)]2 cos ϕ = 123 146 1 5 3 4 2 6 , (45) 8S123S146 where S with subscripts denotes the areas of the faces adjoining l1, and the other terms involve the squares of the lengths of pairwise skew edges. Assume that the tetrahedron is based on an equilateral triangle with sides 1 and that the lateral edges have the same length l. Then (45) yields the following values for the cosines of the dihedral angles:

1 2l2 − 1 cos ϕ = √ √ , cos ψ = 3 4l2 − 1 4l2 − 1 (ϕ is the value of the dihedral angles at the base and ψ is the value of the dihedral angles at the lateral edges). Hence the mean curvature of the tetrahedron is

3 1 3 2l2 − 1 H = π − arccos √ √ + l π − arccos . 2 3 4l2 − 1 2 4l2 − 1 We see from this that H cannot be an algebraic function of the edge lengths, because it would then have only finite-order branch points, while the extensions ofthe inverse trigonometric functions into the complex plane branch to infinite order. As

26On the other hand, one of the prospective approaches to the investigation of the bellows conjecture at the time of attempts to prove or disprove it involved analyzing the behaviour of the volume of a non-rigid polyhedron under an infinitesimal flex of the first order (see[16]). Although this approach does not lead to a refutation of the conjecture, in the class of infinitesimal flexes of a non-rigid polyhedron it helps to select the ones extending to actual flexes. 490 I. Kh. Sabitov for the proof that the mean curvature of polyhedra with an arbitrary combinatorial structure is non-algebraic, see [72], which contains the following result: for any combinatorial structure we can find a configuration of polyhedra such that their mean curvature is not an algebraic function of the edge lengths. We now return to the general formula (42) for the mean curvature of a polyhedron. If we know the lengths of the minor diagonals, then we can find the values of the cosines of the dihedral angles by (45) (in (45) replace l5 by the corresponding minor diagonal length d; the other quantities are calculated from the edge lengths), so that we find two possible values of the angle, of the form ϕ and 2π − ϕ, from which we must pick the required value of the internal angle. As a result, we obtain a precise formula for the mean curvature in terms of the lengths of the edges and minor diagonals: 1 X H = l π − arccos f (l, d) (the sum is taken over the set of edges), (46) 2 i i i where l and d are the sets of squares of the lengths of the edges and the lengths of the minor diagonals of the polyhedron, respectively, and the fi are defined by formula (45) with the corresponding values of the lengths of the edges and minor diagonals (each function fi involves five edges and one minor diagonal). Formula (46) is valid for all configurations of polyhedra with given combinatorial structure andedge lengths. We know equations for each minor diagonal, but first, we do not know consistent solutions to the entire set of these equations, and second, in some of these equations (maybe even in all of them) the coefficients may vanish, so we will not be able to find the corresponding minor diagonals. Hence there is a problem in finding an equation which could produce all a priori possible values of the mean curvature. The result in [72] says that such an equation cannot be algebraic, and thus we must seek it among analytic or differential equations. As regards finding the latter, we can present the following observations. The function H is homogeneous of degree 1 (all the minor diagonals have dimension 1, so the arguments of arccos are homogeneous of degree 0). Then X ∂H l = H i ∂l i i and furthermore, ∂H 1 1 ∂fi X 1 ∂fj = π − arccos fi(l, d) + li p + lj q . ∂li 2 1 − f 2 ∂li 2 ∂li i j̸=i 1 − fj

Hence we see that in general position27 all the second-order derivatives of H are algebraic functions of the edge lengths, which gives us some hope of finding a direct method28 for computing a priori values of the mean curvature before constructing the polyhedron from its metric and combinatorial structure, as can be done for the volume. 27 Here we bear in mind that the derivatives of the minor diagonals with respect to the lengths li certainly cannot be defined, for example, in the case of flexible polyhedra. 28That is, a method not requiring the calculation of minor diagonals, which we know to be complicated and, more importantly, not always possible. Algebraic methods for solution of polyhedra 491

21. Polyhedra in the spherical and the hyperbolic spaces Among papers concerned directly with the central subject of our survey, we men- tion first of all [73]. It contains an example of a flexible polyhedron (an octahedron) in the interior of a 3-dimensional hemisphere such that a flex changes its volume. The differentials of the volume and mean curvature in spaces Mn of non-zero constant curvature K are related by Schl¨afli’sformula (see, for instance, [71])

1 X (j) K dV = − V dβ , (47) n 2(n − 1) n−2 j j where n is the dimension of the space, Vn is the n-volume of the body in Mn bounded by the polyhedron P undergoing deformation, and βj is the angle between the two facets adjoining the (n − 2)-face with index j in some numbering and with (j) (n − 2)-volume Vn−2. Therefore, the flex of a polyhedron with variable volume also cannot preserve its mean curvature. Thus, in the spherical space neither of the two quantities, the volume and the mean curvature, is invariant, although they are invariant in R3. Nothing is known about the behaviour of these characteristics for flexible polyhedra in the Lobachevskii space. We can only say that, regarding the volume, its differential is not a linear function of the differentials oftheedge lengths, as we can see from an example in [74]. Another example when problems and results obtained in the metric theory of polyhedra in R3 are compared and extended to the spherical or hyperbolic space is given in [75], where by analogy with ideas in [43] and under the assumption of a certain metric symmetry of an octahedron the authors succeed in finding formulae for the volume as a function of the edge lengths. However, the crucial difference between these two papers is as follows: while the formula for the volume in [43] holds for all octahedra isometric to one with the prescribed symmetry, [75] presents a formula only for the octahedron displaying the symmetry, and the question of the volume of an octahedron isometric to it but lacking the spatial symmetry in its configuration remains open. The starting point in our proof of Theorem1 was formula (4) for the volume of a tetrahedron as a function of the edge lengths. There are no such simple formulae for the volumes of tetrahedra in the spherical and hyperbolic spaces.29 Many authors have proposed analytic formulae (see, for instance, [78]–[81]), however, there is no widely accepted simple formula. But in fact, the problem is not just the simplicity or complexity of the formulae: the point is that they cannot be used for an inductive transition from more complex polyhedra to less complex ones.

29Even an analogue of Heron’s formula for the area of a triangle in the Lobachevskii plane of curvature K = −1 is frighteningly complex ([76], Chap. V, problem 31.1, and [77]):

∆ [(e2σ− 1)(e2σ−2α− 1)(e2σ−2β − 1)(e2σ−2γ −1)]2 sin = 2 (eα + 1)(eβ + 1)(eγ + 1) psinh σ sinh(σ − α) sinh(σ − β) sinh(σ − γ) = , 2 cosh(α/2) cosh(β/2) cosh(γ/2) where ∆ is the area of a triangle with sides of lengths α, β, and γ and 2σ is the perimeter of this triangle. 492 I. Kh. Sabitov

Furthermore, there is no phenomenon of similarity in the spherical and hyperbolic spaces, so that homogeneity in the edge lengths disappears in the volume formulae. As a result, the bulk of papers aim at obtaining formulae for the volumes of particular polyhedra by combining various algebraic, analytic, synthetic, and combinatorial topological methods (see, for example, [82]–[84]). A nice concise survey of such results can be found in [85], while general questions of combinatorial metric geometry of polyhedra in the Lobachevskii space, as well as results on the volumes of some concrete polyhedra were considered in [86]. On the other hand, the behaviour of the volume of a flex of a polyhedron in the spherical or hyperbolic space is a meaningful question, since flexible polyhedra exist in both spaces: for the spherical space this was proved in [73] and for the hyperbolic space in [87], where it was shown that the class of flexible octahedra in the hyperbolic space contains at least the same three types as the ones discovered by Bricard in the Euclidean case. However, we see no approaches to a general volume formula for a family of isometric polyhedra. Moreover, there are some indications arguing against the existence of such a formula. Namely, it is known that a tetrahedron with dihedral angles A, B, C, D, E, F is uniquely determined by its Gram matrix  1 − cos A − cos B − cos F  − cos A 1 − cos C − cos E G =   − cos B − cos C 1 − cos D − cos F − cos E − cos D 1

(each row corresponds to the angles at some vertex). Seidel’s conjecture (1986) asserted that the volume of an ideal hyperbolic tetrahedron is a function of the determinant and permanent of its Gram matrix, while the extended Seidel conjecture posed by I. Rivin and F. Luo asserted that the volume of an arbitrary non-Euclidean tetrahedron is a function of the determinant of its Gram matrix alone. The failure of both conjectures was demonstrated in [88] and [89] (however, it was also shown that the original Seidel conjecture holds for the classes of acute-angled and obtuse-angled ideal tetrahedra). We see that there is little hope of finding a volume polynomial or some other volume function for isometric polyhedra in the hyperbolic space, so could one perhaps prove the bellows conjecture there not as a consequence of an analogue of Theorem1, but rather as an independent result? As concerns the spherical space, V. A. Alexandrov, the author of an example of a flexible octahedron with variable volume, has suggested that his example is accidental in a certain sense, and that ‘most’ flexible polyhedra in the spherical space must preserve their volume. In recent years some authors have associated with knots and links the volumes of certain artificial objects, conditionally called virtual polyhedra (see [90]–[92]), although many of these results have not been published yet. Such objects can be constructed in any geometry of constant curvature, and as bodies they have boundaries obtained by identifying certain surfaces generated by some pieces of these knots and links. If we are in Euclidean geometry, then it can be proved that the volumes of such virtual polyhedra are roots of certain polynomials, as in the case of Theorem1. In connection with these papers we can pose the problem of finding the volumes of polyhedra given in locally Euclidean or locally hyperbolic spaces in which the isometric inverse images of these polyhedra are closed and Algebraic methods for solution of polyhedra 493 without identification of faces. We partially illustrate this idea by the following example. In the Euclidean plane with variables (x, y) we consider a rectangle Π: 0 6 x 6 a, 0 6 y 6 b. Identifying its sides x = 0 and x = a, we obtain an annulus P , a manifold with boundary and locally Euclidean metric, and we can calculate its area. On the other hand, in another plane T with coordinates (u, v) we can take an annulus K : 0 < r2 < u2 + v2 < R2 with locally Euclidean metric ds2 = (u2 + v2)(du2 + dv2), and (for suitably selected r and R) obtain a domain isometric to P without identification. Thus, after having introduced the plane T with locally Euclidean metric we no longer need such an abstract operation as identification of sides (or faces) and can work with an object whose visual image is in one-to-one correspondence with the object itself. Therefore, we can pose the question of the volume of polyhedra in spaces that are locally spaces of constant curvature30 and thereby extend the class of polyhedra to which we can carry over results proved for spaces with a global standard metric of constant curvature. Schl¨afli’sformula (47), which is very useful in non-Euclidean constant curvature spaces, can also be of use in the Euclidean case. To see this, we need to investigate its consequences in Euclidean space, treated as the limit of spaces of curvature K < 0 and K > 0 as |K| approaches zero. This was partially done in [71], p. 676, where it was shown that the formula for the differential of the mean curvature of a polyhedron in Euclidean space can be obtained as the limiting case of Schl¨afli’s formula in the spherical space as the radius R of the sphere increases to infinity. This result can be obtained by considering the leading terms of the first order of smallness with respect to 1/R2 → 0, but there is an open question of the influence of the terms of orders 3, 4, and greater with respect to 1/R → 0.

22. Polytopes in multidimensional spaces Little is known about the flexibility of polyhedra in spaces of dimension greater than 3, and even less is known about their volumes. Flexible cross-polytopes in E4 were described in [93]. These polyhedra are analogues of octahedra defined in n i i any E , n > 3: they have 2n vertices grouped into pairs (p1, p2), 1 6 i 6 n, their n = n(n−1) edges connect pi with pk for i ̸= k and j , j ∈ {1, 2}, and 2 2 j1 j2 1 2 the 2n facets are simplexes with vertices p1 , p2 , . . . , pn , j , j , . . . , j ∈ {1, 2}. j1 j2 jn 1 2 n A combinatorial model of a cross-polytope can be visualized as follows: consider n a system of coordinates Ox1 . . . xn in E , put a pair of points symmetric with respect to the origin on each of the coordinate axes, and in each hyperoctant consider the (n − 1)-simplex spanned by the vertices lying there. Spatial realizations of this model can have various shapes, not all of them convex. Among such non-convex and even self-intersecting polyhedra A. Walz found examples of flexible cross-polytopes, which underlie a more general class of flexible cross-polytopes in [93]. The method of their construction can also be used to construct flexible cross-polytopes in the 4-dimensional elliptic and hyperbolic spaces. As for higher dimensions, we only know some characterizations of non-rigid cross-polytopes (that is, of ones admitting non-trivial infinitesimal flexes of the first order) (see[94]).

30Here we mean that, similarly to the above example, the metric on the space can only locally be reduced to the standard form of metric of a constant curvature space. 494 I. Kh. Sabitov

For the polyhedra in En with n > 3 we can resort to the method in § 16 for calculating their volumes using the lengths of diagonals, and thereby establish an algebraic connection between the volume and the lengths of the edges and diagonals, after which we can try to eliminate the lengths of the diagonals using various relations for the edge lengths in a complete graph. However, this way has not yet been realized (results on the ranks of the distance matrix [95] and the Cayley–Menger matrix [33], [96] can be useful here). There is yet another method for establishing relations between the volume and the lengths of the edges and diagonals. According to the definition of the generalized volume of a simplicial polyhedron in § 2, it is equal to the sum of the volumes of the tetrahedra with bases on the compatibly oriented facets of the polyhedron and with a common vertex at an arbitrarily chosen point O. If we take one of the vertices of the polyhedron as O, then the square of the volume of each tetrahedron is equal to the Cayley–Menger determinant of a matrix with entries equal to the squares of the edge lengths at the base of the tetrahedron (which is a facet of the polyhedron) and the squares of the lengths of the diagonals from the selected vertex of the polyhedron to the vertices on the corresponding facet (with coefficient depending only on n and with sign determined by the orientation of the face). Hence, for the volume V of the polyhedron we have q X 2 V = εk Vk , εk = ±1, k

2 where k ranges over the set of indices of the facets, and the Vk are polynomials in the squares of the lengths of the edges and the diagonals passing from the selected vertex to the other vertices of the polyhedron. Taking repeatedly the square of both sides and grouping the terms appropriately, we obtain an equation of the form

2N 2 2 2N−2 2 2 V + a1(l , d )V + ··· + aN (l , d ) = 0, (48) where the coefficients aj, 1 6 j 6 N, are polynomials in the squares of the edge lengths (the system of them is denoted by l2) and the squares of the lengths of the diagonals (the system of them is denoted by d2). In this way we can obtain an analogue of Theorem1, and as a consequence prove the bellows conjecture in En for one class of polytopes which we call pyramids. Apparently these polytopes were considered for the first time in [97]. We define them as polytopes having at least one vertex connected with all the other vertices by edges. Such simplicial polyhedra in E3 can have arbitrary topological genus and can be orientable or non-orientable [29], and in any En they can be constructed over an arbitrary polyhedral hypersurface with boundary such that all its vertices lie on the boundary: we fix an arbitrary point in the space, connect it to the vertices on the boundary by edges, and ‘glue up’ the boundary by adding the (n − 1)-simplexes spanned by these edges and the (n−2)-simplexes on the boundary. For an orientable simplicial pyramidal polytope we can calculate its volume by taking the vertex of the polytope connected with all the others by edges as the common apex of the n-dimensional tetrahedra. Then the coefficients aj in (48) depend only on the edge lengths, and we obtain an analogue of Theorem1 in any dimension. This construction leads to the following corollary. Algebraic methods for solution of polyhedra 495

Corollary 8. If a polytope with N vertices in En, n > 3, has fewer than N/2 diagonals, then there exists a volume polynomial for the volumes of such polytopes and the bellows conjecture holds for them.

23. Some unsolved problems In conclusion we present a list of problems waiting to be solved, together with some comments to them. This list is far from complete. We note that in the text we have already stated ten or so problems, which we usually (though not always) distinguished using italics or boldface. 1. Prove or disprove the bellows conjecture for polyhedra in the Lobachevskii space. A polynomial analogue of Heron’s formula can hardly be expected, but maybe it can be proved that the volume is a zero of some analytic function (explicitly defined or given as a solution of some differential equation), which wouldalso be sufficient for the proof of the bellows conjecture. 2. Prove the bellows conjecture in a multidimensional Euclidean space. The difficult point here is the lack of integer-valued parameters for classifying thecom- binatorial structure of polyhedra, like the number of vertices and the topological genus, and therefore it is unclear how induction can be used. Moreover, the simplest polytopes, such as suspensions in R4 over a 2-dimensional toroidal polyhedron in R3, are not manifolds, but only pseudomanifolds. For starters, we could at least try to demonstrate the existence of algebraic relations between the volume and the edge lengths of the polytope, following the idea outlined in the second paragraph of § 22. 3. Investigating polytopes, as in §§ 16 and 22, under the additional assumption that the lengths of all their diagonals are known leads to the general problem of describing the structure of an N-point set with all the N(N − 1)/2 distances between pairs of points known, or describing the configuration of a complete graph with straight line edges of known lengths in the Euclidean space Rn, n < N. In fact, even in the simplest problem of describing tetrahedra (for n = 3 and N = 4) for a fixed set of 6 lengths not distributed in advance among the edges, whichthe author posed some time ago to N. Ermilov, a student at Moscow State University, it turns out that in the general case there exist 30 (!) non-congruent configurations of a 4-point set with the same system of distances between pairs of points [98], [99] (these configurations can be identified with the corresponding tetrahedra). And the problem of describing point configurations with known distances between all the pairs of points is very important in many areas of the natural sciences and engineering. For instance, in many physics problems one looks for a minimum of a functional (usually called the potential or energy) depending on the distances between the points in some system (of atoms, molecules, or bodies). Mathemati- cally, these functionals (such as the Lennard-Jones potential) involve only distances, and after finding the values of the distances supplying the minimum, we havethe further problem of determining the geometric structure of a point set with these distances. Thus, we obtain the problem of describing the N-point configurations with prescribed distances between the points and also of distinguishing possible polytopes with prescribed edge lengths among such configurations. Among such problems we can single out problems with just a few different distances (for instance, the same 496 I. Kh. Sabitov problem with four vertices is uniquely solvable if all the distances are equal, has at most 3 solutions if there are only 2 distinct distances, and so on). For several results obtained in this setting see [100]. As concerns estimates for the number of inflexible (locally uniquely defined) configurations of graphs with given edge lengths in constant curvature spaces of arbitrary dimension, see [101]. 4. The problem described above can be linked with the following one. We mentioned in § 16 that, given a 1-dimensional skeleton of a torus with seven vertices, we can glue it up by 14 triangles in 120 ways, thereby obtaining 120 non-congruent (and non-isometric) polyhedra of genus 1. We can pose a similar question in the general case: remove (the interiors of) all the faces of a polyhedron while keeping its edges. If we now try to restore the faces (bearing in mind the standard condition that each edge is incident to precisely two faces and two faces can meet only in a vertex or a full edge), will we necessarily obtain the original polyhedron? Cases when the answer is affirmative were considered in[102] and [103], but there also are non-uniqueness cases (see, for example, [56], [104], or [105]). What is an algorithm for recovering a polyhedron from its 1-dimensional skeleton and how can we obtain a polyhedron with extremal values of the volume or the total area of the faces? 5. What can be called a ‘canonical’ volume polynomial for polyhedra of genus g > 0? If a volume polynomial (see §§ 5 and 12 for the general definition of a volume polynomial) has the lowest degree possible, then is it unique and can there be different volume polynomials for the same combinatorial structure, depending on the edge lengths? Give a consistent definition of a volume polynomial and describe an algorithm for calculating it. 6. Prove that the volume of a flexible polyhedron is a multiple zero of any volume polynomial. The same holds perhaps for the volume of any non-rigid polyhedron (admitting non-trivial infinitesimal flexes of the first order). The converse assertion fails: as mentioned in § 18.1, there exist metric Bricard octahedra of type 2 for which a volume polynomial has a positive zero of multiplicity 2 real- izable as the volume of some actual octahedron, but this octahedron is inflexible, since all the flexible octahedra have volume 0 (however, we have not verified whether this octahedron is rigid, so the question of the validity of the converse assertion for non-rigid polyhedra remains open). 2 7. Assume that the square V0 of the volume of a polyhedron is a multiple zero of a canonical volume polynomial (we know what these polynomials are for polyhedra of genus 0; for genus g > 0 assume that there is some definition of a canonical polynomial). Then what is the geometric meaning of the multiplicity of a zero? Here we must distinguish two cases: the polyhedron is flexible or inflexible. As shown by A. Proskuryakov, a student at Moscow State University (in his unpub- lished diploma thesis of 2005), for flexible octahedra if a root is the square ofthe volume, then its multiplicity (which is 7 for Bricard octahedra of type 1 and 6 for octahedra of type 2) is related to the number of different configurations of the polyhedron that cannot be superposed one onto another by a flex. Maybe it is not the flexibility that is crucial for the answer, but the order of the existing infinitesimal flexes of higher orders. 2 8. Let V0 > 0 be a root of the volume polynomial constructed for a given natural development of a polyhedron. When is this root equal to the square of the volume of an actually existing polyhedron isometric to the development? As we Algebraic methods for solution of polyhedra 497 know from § 18, in the case of octahedra not every root of the volume polynomial is the volume of an actually existing octahedron with this metric.

9. Assume that a root V0 of a canonical volume polynomial is not equal to the volume of an actual polyhedron in R3. Does there exist a polyhedron with complex-valued coordinates of vertices such that its formally defined generalized volume coincides with this root? 10. Design shorter notation for volume polynomials (for instance, by grouping terms or introducing some new variables, such as areas, and so on). In this connection we observe that a simplex in the 4-dimensional space has the same number of 5 5 1- and 2-dimensional faces (= 2 = 3 = 10). Its 1-dimensional faces are edges, and the 4-volume of a simplex can be expressed by the Cayley–Menger determinant involving the squares of the edge lengths. The question is whether we can express the edge lengths in terms of the areas of 2-faces or at least express the volume in terms of the areas of 2-faces. This problem has long been ‘folklore’, and the discussion of a similar question in [106] gives hope that the answer may be affirmative. A similar question can be asked in the general case of n > 4: an n-simplex in En contains the same number of k-simplexes and (n−1−k)-simplexes, for instance, the same number of 1-dimensional simplexes (edges) and (n − 2)-simplexes. Do there exist formulae relating the volumes (of the corresponding dimensions) of these simplexes? In particular, can we express the n-volume of a simplex in terms of the volumes of its faces of dimension greater than 1? 11. For finding volume polynomials for polyhedra with few vertices, design software which could be used in practice at least when the number of vertices does not exceed 10. Maybe this cannot be done in the general case (with symbolic notation for the edges), in which case two directions of further work are possible: assume that some of the lengths coincide, so that there are fewer different lengths than edges (for example, this is the case for the Steffen polyhedron) or assume that the numerical values of the edge lengths are known (as was done, for instance, in [11], p. 18). 12. Design an algorithm for finding a canonical volume polynomial, not necessarily running in polynomial time. There is a ‘naive’ algorithm for genus-0 polyhedra: find some polynomial using the first method of the proof ofTheorem1 and then pick the required polynomial among its divisors. 13. Find a formula or a polynomial for the volume of simplicial polyhedra of given genus g > 0 with the smallest number of vertices. This number nmin(g) is defined by the following formula (see [34]):

1 n (g) = 7 + p1 + 48g , g 0, g ̸= 2, min 2 > where ]a[ is the minimum positive integer that is no smaller than a (the case g = 2 is exceptional: then nmin(2) = 10). If there are no diagonals (for polyhedra of genus g = m(m + 1)/2, m, g ∈ N0), then the volume can be expressed in terms of the edge lengths by an exact formula (see § 16), but there exist sequences of values of g for which polyhedra with fewest vertices have an unboundedly increasing sequence of numbers of diagonals [107]. 498 I. Kh. Sabitov

14. Characterize the non-rigid polyhedra for which the variation of the volume corresponding to an infinitesimal flex is zero. The paper[108] contains an algorithm for verifying whether an infinitesimal flex can be extended to a genuine flex. However, the invariance of volumes during flexes was not known at that time, so now this problem can be considered again, taking account of the volume invariance. This problem is probably not that difficult to solve on a theoretical level, and a software realization of such an algorithm would be of interest. Perhaps the ideas used in [109] and [110] will be useful in this study. 15. One could propose a broad range of investigations aiming to carry over all the above problems, starting from their formulations, to polyhedra in spaces with the local structure of spaces of constant curvature. There is no doubt that some objects and problems in algebra are counterparts of the geometric objects and problems considered here. We hope that algebra and the geometry of polygons and polyhedra are as intimately connected as analysis and the differential geometry of curves and surfaces. In conclusion I use this occasion to thank L. L. Bezkorovainaya, J. Wallner, A. Yu. Vesnin, E.` B. Vinberg, A. D. Mednykh, H. Stachel, and in particular, V. A. Alexandrov for their attention to my numerous calls for help with literature references and for useful consultations.

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I. Kh. Sabitov Received 08/JUL/10 Moscow State University Translated by N. KRUZHILIN E-mail: [email protected]