PDF《arXiv:math.MG/0111003 v1 1 Nov 2001》

arXiv:math.

MG/0111003 v1

1 Nov

2001 Flexible polyhedra in the Minkowski 3-space Victor Alexandrov Sobolev Institute of Mathematics, Novosibirsk-90, 630090, Russia. E-mail: [email protected] Abstract We prove that ?exible polyhedra do exist in the Minkowski 3-space and each of them preserves the (generalized) volume and the (total) mean curvature during a ?ex. To prove the latter result, we introduce the notion of the angle between two arbitrary non-null nonzero vectors in the Minkowski plane.

2000 Mathematics Subject Classi?cation: 52C25, 51B20, 52B70, 52B11, 51M25 Key words: ?exible polyhedron;

Minkowski space;

Minkowski plane;

volume;

total mean curvature;

angle

1 Introduction Recall that the Minkowski n-space Rn

1 is the linear space which consists of all ordered n-tuples of reals x = (x1, x2,xn) and is endowed with the following scalar product: (x, y) = x1y1 + x2y2 xn?1yn?1 ? xnyn. The length |x| = (x, x) of a vector x is either a positive number or the product of a positive number by the imaginary unity i, or zero (see, for example, [15]). Let Σ be a connected (n ? 1)-dimensional simplicial complex which is a manifold. A continuous map P : Σ → Rn

1 is said to be a polyhedron in Rn

1 if it is a?ne and injective on each simplex of Σ. However

1 2 Existence We use standard notations from the theory of the Minkowski spaces freely. The reader can ?nd them in [14] or [15]. Let eu : Rn

1 → Rn be the identity mapping of the Minkowski n-space Rn

1 to the Euclidean n-space Rn . Note that a set D ? Rn

1 is convex if and only if the set eu(D) ? Rn is convex. A convex polyhedron is said to be strictly convex if there is no straight angle among its dihedral angles. It is crucial for the following Lemma that, according to our de?nition, all faces of a polyhedron are triangular. Lemma 1. Let P : Σ → R3

1 be a sphere-type strictly convex polyhedron and let Q be a disk-type polyhedron which is obtained from P by removing two adjanced faces. Then Q is ?exible. Proof. Choose a simplex ? ? Σ such that P(?) ? Q. Let Σ have v vertices (0-simplices) V1, V2,Vv and let Vv?2, Vv?1, Vv ∈ ?. Suppose Σ has e edges (1-simplices) and the edges indexed by e ? 2, e ? 1, and e belong to ?. Consider the family of all polyhedra P : Σ → E3 2,1 such that P|? = P|?. This family depends on 3v ?

9 parameters, x1, y1, z1,xv?3, yv?3, zv?3, where (xj, yj, zj) = P(Vj) (j = 1, 2,v ? 3) are the coordinates of the jth vertex of P. Suppose vertices P(Vj) and P(Vk) (j, k = 1, 2,v ? 3) are joined by an edge and suppose that the edge is indexed by m (m = 1, 2,e ? 3). De?ne an auxiliary real-valued function fm by the formula fm(x1, y1, z1,xv?3, yv?3, zv?3) = [(xj ? xk)2 + (yj ? yk)2 ? (zj ? zk)2 ]/2. The components of the vector-valued function f = (f1, f2,fe?3) represent the half of the squared edge lengths of all edges of P which do not belong to P(?). Euler'

s formula yields 3v ? e =

6 and, thus, e ?

3 = 3(v ? 3). This means that the Jacobian matrix of f is a square matrix. Obviously, its mth row is as follows: (0 . . .

0 xj ? xk yj ? yk zk ? zj

0 . . . . . .

0 xk ? xj yk ? yj zj ? zk

0 . . . 0) . The determinant of this Jacobian matrix does not vanish at the point corresponding to P : Σ → R3 1. To prove this statement, we have to repeat the above constructions for the convex polyhedron eu(P) ? R3 . As a result, we obtain a vector-valued function g = (g1, g2,ge?3) whose components gm(x1, y1, z1,xv?3, yv?3, zv?3) = [(xj ? xk)2 + (yj ? yk)2 + (zj ? zk)2 ]/2 (m = 1, 2,e ? 3) represent the half of the squared (Euclidean) lengths of those edges of eu(P) which do not belong to eu(P)(?). The mth row of the Jacobian matrix of g is as follows: (0 . . .

0 xj ? xk yj ? yk zj ? zk

0 . . . . . .

0 xk ? xj yk ? yj zk ? zj

0 . . . 0) . This means that the determinant of the Jacobian matrix of f is the product of ±1 by the determinant of the Jacobian matrix of g, while the latter is known to be nonzero at the point corresponding to the convex polyhedron eu(P) of the Euclidean 3-space. This statement was obtained in [10] by direct calculations and is known to be equivalent to the ?rst-order rigidity of a strictly convex polyhedron in the Euclidean 3-space [1]. Since the determinant of the Jacobian matrix of f does not vanish at the point corresponding to P : Σ → R3 1, it follows that f maps a neighborhood U of that point homeomorphically onto its image f(U). For t close enough to zero, there exists a point ft = (ft 1, ft 2,ft e?3) ∈ f(U) such that α) ft m equals the sum of t and the value of fm at the point corresponding to P, if m corresponds to the edge of Σ shared by those two simplices which should be removed from P to obtain Q, and

2 β) ft m equals the value of fm at the point corresponding to P, otherwise. Since f is a local homeomorphism, it follows that there exists a polyhedron P? t such that f maps the point of U corresponding to P? t into ft . This means that, arbitrarily close to P, there exists a polyhedron P? t which is not congruent to P and shares with P the lengths of all edges but the edge of Σ shared by those two simplices which are removed from P to obtain Q. Removing from P? t those two simplices, we obtain a polyhedron Q? t which is arbitrary close to Q, is isometric to Q, and is not congruent to Q. In [12] it is shown that existence of such a family of polyhedra Q? t implies existence of a family of polyhe........