Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 39 (1998), No. 2, 367-378.

Sufficient Conditions for the Extendibility of an n-th Order Flex of Polyhedra

Victor Alexandrov

Sobolev Institute of Mathematics Novosibirsk-90, 630090, Russia e-mail: alex@math.nsc.ru

Abstract. A new approach to describing a high order flex of polyhedra is given. Sufficient conditions are derived under which an infinitesimal flex of polyhedra has an extension to an authentic flex. It is shown that one set of these conditions is fulfilled for Bricard octahedra.

1. Introduction

A very long-standing (and still open) problem is to prove that a smooth compact surface in the Euclidean 3-space is rigid, i.e., one cannot find a continuous family of smooth surfaces in the Euclidean 3-space each of which is isometric (in the intrinsic metric) to any other one and is not obtained from the initial surface by a rigid motion [9]. Classical attempts to solve this problem were based on the notion of an infinitesimal flex (see [6] or [7] for further references).

Since a counterexample was constructed in the class of polyhedral surfaces [2], particular attention was given to studying infinitesimal flexes of polyhedra. In particular, the “classical” definition of a high order infinitesimal flex occurs to be improper from the functional point of view [4].

Without revising the “classical” definition of a high order infinitesimal flex, we give a new technique of calculations in Section 2 below. Section 3 contains some auxiliary facts from linear algebra. The main results of this paper are given in Section 4 and provide some sufficient conditions under which an infinitesimal flex of a polyhedron can be extended to an authentic flex. In Section 5, we discuss results of computer experiments demonstrating that one set of our sufficient conditions is fulfilled for Bricard octahedra.

In [8] some formal sufficient conditions were given for extendibility of an infinitesimal flex of a smooth surface vaguely resembling our conditions.

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix}1998 Heldermann Verlag

368 V. Alexandrov: Sufficient Conditions for the Extendibility of an n-th Order Flex …

2. Reduction of the flex problem for polyhedra to an algebraic problem

Definition. A closed polyhedron Q in the Euclidean 3-space is called flexible if there exists an analytic (with respect to a parameter t ∈ [0, 1]) family of polyhedra Q_{t} such that

(i) Q coincides with Q_{0}; (ii) for each t ∈ [0, 1], the polyhedron Q_{t} is isometric to the polyhedron Q_{0} in the intrinsic

metrics; moreover, the restriction of this isometry onto each face of Q_{t} can be extended to an isometry of the whole 3-space;

(iii) the polyhedron Q_{1} cannot be obtained from Q_{0} by a rigid motion.

Remark. (i) Some other natural definitions of flexible polyhedra are known. In [5] it is shown that all of them are equivalent to the above definition. The family of polyhedra Q_{t} is called an authentic flex of Q. (ii) Without loss of generality, we will always assume in this paper that all faces of a flexible polyhedron are triangular. Otherwise it suffices to triangulate all faces in advance and consider the new polyhedron that is obviously flexible.

Let Q be a closed flexible polyhedron. Suppose that Q has v vertices (x_{i}, y_{i}, z_{i}), 1 ≤ i ≤ v, e edges, and f faces. If vertices (x_{i}, y_{i}, z_{i}) and (x_{j}, y_{j}, z_{j}) are joined together by an edge of Q, then we assign a number e(i, j), 1 ≤ e(i, j) ≤ e, to this edge and denote by l_{e(i,j)} the length of the edge.

The coordinates of the i-th vertex of the polyhedron Q_{t} are analytic functions with respect to t, 0 ≤ t ≤ 1. Denote them by x_{i}(t), y_{i}(t), and z_{i}(t).

Since all faces of Q are supposed to be triangular, the statement that the polyhedra Q_{t} and Q_{0} are isometric to each other in intrinsic metrics is equivalent to the statement that relevant edges of these polyhedra have equal lengths. In other words, if vertices i and j are joined by an edge of Q then

\sqrt{[x_{i}(t) − x_{j}(t)]^{2} + [y_{i}(t) − y_{j}(t)]^{2} + [z_{i}(t) − z_{j}(t)]^{2}} =

= \sqrt{[x_{i}(0) − x_{j}(0)]^{2} + [y_{i}(0) − y_{j}(0)]^{2} + [z_{i}(0) − z_{j}(0)]^{2}} = l_{e(i,j)}.

Substituting Maclaurin expansions of the functions x_{j}, y_{j}, and z_{j},

x_{j}(t) = ∑_{n=0}^{∞} \frac{1}{n!}[\frac{d^{n}}{dt^{n}} x_{j}(t)] |_{t=0} t^{n} =^{def} ∑_{n=0}^{∞} x_{j,n}t^{n},

y_{j}(t) = ∑_{n=0}^{∞} \frac{1}{n!}[\frac{d^{n}}{dt^{n}} y_{j}(t)] |_{t=0} t^{n} =^{def} ∑_{n=0}^{∞} y_{j,n}t^{n},

z_{j}(t) = ∑_{n=0}^{∞} \frac{1}{n!}[\frac{d^{n}}{dt^{n}} z_{j}(t)] |_{t=0} t^{n} =^{def} ∑_{n=0}^{∞} z_{j,n}t^{n},

into the expression

[x_{i}(t) − x_{j}(t)]^{2} + [y_{i}(t) − y_{j}(t)]^{2} + [z_{i}(t) − z_{j}(t)]^{2} = l_{e(i,j)}^{2},

we obtain

∑_{n=0}^{∞}{ ∑_{m=0}^{n}[(x_{i,m} − x_{j,m})(x_{i,n−m} − x_{j,n−m}) + (y_{i,m} − y_{j,m})(y_{i,n−m} − y_{j,n−m})+

+(z_{i,m} − z_{j,m})(z_{i,n−m} − z_{j,n−m})]}t^{n} = l_{e(i,j)}^{2}.

Therefore, the equation

∑_{m=0}^{n}[(x_{i,m} − x_{j,m})(x_{i,n−m} − x_{j,n−m})+

+(y_{i,m} − y_{j,m})(y_{i,n−m} − y_{j,n−m}) + (z_{i,m} − z_{j,m})(z_{i,n−m} − z_{j,n−m})] = 0 (1)

holds for all n ≥ 1 and all vertices i, j joined by an edge.

Formula (1) gives a necessary condition for the polyhedron Q to be flexible. Note that this condition is close to a sufficient one. Indeed, if the set of numbers x_{i,m}, y_{i,m}, and z_{i,m} satisfies (1) and leads to a convergent Maclaurin series, then the corresponding functions x_{i}, y_{i}, and z_{i} generate a flex of Q.

Let B : R^{3v} × R^{3v} → R^{e} be the bilinear map that takes each couple of vectors

and  \begin{matrix}X = (x_{1}, y_{1}, z_{1}, x_{2}, y_{2}, z_{2}, & …, x_{v}, y_{v}, z_{v}) \\ U = (u_{1}, v_{1}, w_{1}, u_{2}, v_{2}, w_{2}, & …, u_{v}, v_{v}, w_{v}), \end{matrix}

into the vector B(X, U) ∈ R^{e} whose e(i, j)-th component is equal to

(x_{i} − x_{j})(u_{i} − u_{j}) + (y_{i} − y_{j})(v_{i} − v_{j}) + (z_{i} − z_{j})(w_{i} − w_{j})

(recall that e(i, j) stands for the index of the edge joining vertices i and j).

Using the operator B, we can rewrite (1) in the form

∑_{m=0}^{n} B(X_{m}, X_{n−m}) = 0,  (2)

where the following notation is used:

X_{k} = (x_{1,k}, y_{1,k}, z_{1,k}, x_{2,k}, y_{2,k}, z_{2,k}, …, x_{v,k}, y_{v,k}, z_{v,k}).

Remark. Under the customary presentation of the theory of infinitesimal flexes, the vector X_{0} + tX_{1} + ⋯ + t^{k}X_{k} is called the k-th order infinitesimal flex of the polyhedron X_{0} (see, for example, [6] or [7]).

We now can reformulate the problem of verifying if a given polyhedron X_{0} is flexible. It is sufficient to verify that

a) the system (2) has a solution X_{n} for each n ≥ 1;

b) the series

∑_{n=0}^{∞} X_{n}t^{n}

converges for all numbers t sufficiently close to zero;

c) there exists a number n ≥ 1 such that the vector X_{0} + tX_{1} + ⋯ + X_{n} is not generated by any rigid motion of X_{0}.

3. Bilinear game

Suppose that L and K are vector spaces, B : L × L → K is a bilinear map, and x_{0} is a vector in L.

To make the first step of the bilinear game, one has to find (if possible) a vector x_{1} ∈ L satisfying the equality

B(x_{0}, x_{1}) + B(x_{1}, x_{0}) = 0.

Suppose that vectors x_{0}, x_{1}, …, x_{n−1} are obtained according to the rules of the bilinear game. To make the n-th step of the bilinear game, one has to find (if possible) a vector x_{n} ∈ L satisfying the equality

∑_{j=0}^{n} B(x_{j}, x_{n−j}) = 0.  (3)

Definition. An ordered set of vectors x_{0}, x_{1}, …, x_{n} obtained according to the rules of the bilinear game is called a protocol of the bilinear game; the number n is said to be the length of the protocol.

We say that a protocol x_{0}, x_{1}, …, x_{n} can be extended to a protocol of length n + p if there exist vectors x_{n+1}, x_{n+2}, …, x_{n+p} such that each vector x_{n+k}, k = 1, …, p, can be obtained from the vectors with smaller indexes, x_{0}, x_{1}, x_{2}, …, x_{n+k−1}, according to the rules of the bilinear game, i.e., if, for each k = 1, …, p, the following equality similar to (3) holds:

∑_{j=0}^{n+k} B(x_{j}, x_{n+k−j}) = 0.

We say that a protocol x_{0}, x_{1}, …, x_{n} can be infinitely extended if there exists an infinite sequence of vectors x_{n+1}, x_{n+2}, … such that each vector x_{n+k}, k = 1, 2, …, can be obtained from the vectors with smaller indexes, x_{0}, x_{1}, x_{2}, …, x_{n+k−1}, according to the rules of the bilinear game.

We seek for conditions implying that a protocol x_{0}, x_{1}, …, x_{n} of the bilinear game can be infinitely extended.

Remark. An important point of our solution to this problem is that one has to return from a protocol of length n+N to the protocol of the smaller length n, and only afterwards obtains an infinite extension of the protocol of the smaller length n. We demonstrate this point by the following simple example.

Example. Suppose that L is a vector space and e_{0}, e_{1}, …, e_{n−1} is a basis in L. Let B be the bilinear map from L × L into L such that

B(e_{i}, e_{j}) = \{ \begin{matrix}0, & if i ≠ 1 or j ≠ 2; \\ e_{1}, & if i = 1 and j = 2. \end{matrix}

An easy direct test shows that, for each N ≥ 2, the protocol x_{0} = e_{0}, x_{1} = x_{2} = ⋯ = x_{N−1} = e_{1}, x_{N} = e_{2} of length N cannot be extended to a protocol of length N + 1, while the protocol x_{0} = e_{0}, x_{1} = x_{2} = ⋯ = x_{N−1} = e_{1} can be infinitely extended.

To each vector x_{k} belonging to a protocol x_{0}, x_{1}, …, x_{n} assign a linear operator A_{k} : L → K such that A_{k}x = B(x, x_{k}) + B(x_{k}, x) for all x ∈ L. Denote by ker A_{0} the kernel of A_{0} and by A_{k}(ker A_{0}) the image of ker A_{0} under A_{k}. Denote by im A_{0} the image of L under A_{0}.

Theorem 1. Let x_{0}, x_{1}, …, x_{n} be a protocol of the bilinear game. Suppose that the linear span of the subspaces im A_{0} and A_{1}(ker A_{0}) coincides with K. In other words im A_{0} + A_{1}(ker A_{0}) = K. Then the protocol x_{0}, x_{1}, …, x_{n−1} can be infinitely extended.

Proof. It suffices to prove that, for every N ≥ n and every protocol x_{0}, x_{1}, …, x_{N}, the protocol x_{0}, x_{1}, …, x_{N−1} can be extended to a protocol x_{0}, x_{1}, …, x_{N−1}, x_{N} + ξ_{N}, x_{N+1} of length N + 1.

Let k = N or k = N + 1. According to the rules of the bilinear game, to obtain a recurrent vector x_{k}, one has to solve the equation

∑_{j=0}^{k} B(x_{j}, x_{k−j}) = 0  (4)

with respect to x_{k}. Using the notation

b_{N+1} = − ∑_{j=2}^{N−1} B(x_{j}, x_{N+1−j}), b_{N} = − ∑_{j=1}^{N−1} B(x_{j}, x_{N−j})

and grouping terms in (4) by pairs, we obtain the following equations for the redefined x_{N+1} and x_{N}:

A_{0}x_{N+1} + A_{1}x_{N} = b_{N+1},  (5)
A_{0}x_{N} = b_{N}.  (6)

Note that A_{0}, A_{1}, b_{N}, and b_{N+1} are not functions of x_{N} and the yet to be defined x_{N+1}. By the hypothesis, there are vectors x_{N+1} and ξ_{N} such that

A_{0}x_{N+1} + A_{1}ξ_{N} = b_{N+1} − A_{1}x_{N}

where A_{0}ξ_{N} = 0. Hence the equations

A_{0}x_{N+1} + A_{1}(x_{N} + ξ_{N}) = b_{N+1},
A_{0}(x_{N} + ξ_{N}) = b_{N}

follow. Therefore, the ordered set of the vectors x_{0}, x_{1}, …, x_{N} + ξ_{N} is a protocol of the bilinear game that can be extended to the protocol x_{0}, x_{1}, …, x_{N} + ξ_{N}, x_{N+1} of length N + 1.

Another set of sufficient conditions for a protocol to be infinitely extendable is given by the following result.

Theorem 2. Let x_{0}, x_{1}, …, x_{2n+1} be a protocol of the bilinear game and let L_{n} be the linear span of the vectors x_{1}, …, x_{n}. Suppose that (i) x_{j} ∈ L_{n} for each j = n+1, …, 2n+1; (ii) the equation A_{0}x = −B(x_{i}, x_{j}) − B(x_{j}, x_{i}) has a solution x ∈ L_{n} for each 1 ≤ i ≤ n, 1 ≤ j ≤ n. Then the protocol x_{0}, x_{1}, …, x_{2n+1} can be infinitely extended.

Proof. Suppose that, for some k ≥ 0, there exists an extension x_{0}, x_{1}, …, x_{2n+1+k} of the protocol x_{0}, x_{1}, …, x_{2n+1} such that all the vectors x_{2n+2}, …, x_{2n+1+k} belong to L_{n}. It suffices to prove that there exists a vector x_{2n+k+2} ∈ L_{n} satisfying the equality

A_{0}x_{2n+k+2} = − ∑_{j=1}^{2n+k+1} B(x_{j}, x_{2n+k+2−j}).  (7)

The right-hand side of (7) is a linear combination of the vectors y = B(x_{i}, x_{j}) + B(x_{j}, x_{i}) (1 ≤ i ≤ n, 1 ≤ j ≤ n) in the image of A_{0}. Moreover, for each of these vectors y, a solution to the equation A_{0}x = y can be taken as a linear combination of the vectors x_{1}, …, x_{n}. Therefore, (7) has a solution x_{2n+k+2} ∈ L_{n}.

4. Sufficient conditions for extendibility of an infinitesimal flex of a polyhedron

Let Q be a closed polyhedron with triangular faces in the Euclidean 3-space. Using Q, construct a bilinear mapping B according to Section 2 above. Suppose that the bilinear game associated with B has a protocol X_{0}, X_{1}, …, X_{n}. Assign a linear operator A_{k} to each vector X_{k} according to Section 3 above.

The k-th order infinitesimal flex X_{0}+tX_{1}+⋯+t^{k}X_{k} of the polyhedron Q = X_{0} = Q_{0} is said to be trivial if there exists a one-parameter motion group of the Euclidean 3-space such that (i) the coordinates x_{i}(t), y_{i}(t), and z_{i}(t) of the i-th vertex of the polyhedron Q_{t} are generated by this group from the coordinates x_{i}, y_{i}, and z_{i} of the i-th vertex of the polyhedron Q_{0}; (ii) each vector X_{j} (1 ≤ j ≤ k) is generated by the functions x_{i}(t), y_{i}(t), and z_{i}(t) according to Section 2 above.

Lemma. Let X_{0}, X_{1}, X_{2}, …, X_{n}, … be an infinite sequence of vectors such that (i) for each n, the vector X_{0} + tX_{1} + t^{2}X_{2} + ⋯ + t^{n}X_{n} is an n-th order infinitesimal flex of the polyhedron Q = X_{0} and (ii) the n-th order infinitesimal flex X_{0} + tX_{1} + t^{2}X_{2} + ⋯ + t^{n}X_{n} is not trivial for some n. Then the polyhedron Q is flexible.

Proof. The proof is based on the following algebraic theorem of M. Artin [1]: Given a system of polynomial equations f(x, y) = 0, where f = (f_{1}, …, f_{k}), x = (x_{1}, …, x_{m}), and y = (y_{1}, …, y_{n}), there exists an integer β = β(m, n, d, α) depending on m, n, on the total degree d of the polynomials f, and on a nonnegative integer α such that if f(x, \overline{y}(x)) = 0 (mod x^{β}), β = β(m, n, d, α), for some polynomial \overline{y}(x) (that can be referred to as a β-approximate solution), then the system f(x, y) = 0 has a solution y(x) that can be taken in the form of a convergent power series whose initial coefficients coincide with those of the polynomial \overline{y}(x) up to the term x^{α}.

According to Section 2 above, to prove that Q is flexible, we have to solve a system F (t, Y) = 0 of algebraic equations of the form

[x_{i}(t) − x_{j}(t)]^{2} + [y_{i}(t) − y_{j}(t)]^{2} + [z_{i}(t) − z_{j}(t)]^{2} = l_{e(i,j)}^{2}

with respect to the unknown variables t and Y (t)=(x_{1}(t), y_{1}(t), z_{1}(t), …, x_{v}(t), y_{v}(t), z_{v}(t)) (the first variable is not explicitly used in the system F (t, Y) = 0).

Let X_{0} + tX_{1} + t^{2}X_{2} + ⋯ + t^{α}X_{α} be a nontrivial flex of Q. Choose β as in the Artin theorem. Condition (i) of the lemma implies that there exists a β-approximate solution \overline{Y} (t) = X_{0} + tX_{1} + t^{2}X_{2} + ⋯ + t^{β}X_{β} to the system under consideration. According to the Artin theorem, the system has a solution Ỹ(t) that can be taken in the form of a

convergent power series Ỹ(t) = ∑_{j=0}^{∞} c_{j}t^{j}X_{j} such that c_{0} = c_{1} = ⋯ = c_{α} = 1. Grouping

coordinates of Ỹ by triples, we obtain an authentic flex of Q. Thus, Q is flexible.

Remark. As is mentioned in [3] and [7], the idea of using the Artin theorem to obtain the above lemma was suggested by M. Gromov.

Corollary 1. Let a polyhedron Q = X_{0} have a second-order infinitesimal flex X_{0} + tX_{1} + t^{2}X_{2} such that (i) the first-order infinitesimal flex X_{0} + tX_{1} is not trivial; (ii) the linear span of the subspaces im A_{0} and A_{1}(ker A_{0}) coincides with R^{e}, in other words im A_{0} + A_{1}(ker A_{0}) = R^{e}. Then Q is flexible.

Proof. According to Theorem 1, the protocol X_{0}, X_{1} can be infinitely extended. Then the statement follows from the lemma.

Corollary 2. Let a polyhedron Q = X_{0} have a nontrivial infinitesimal flex X_{0}+tX_{1}+⋯+ t^{2n+1}X_{2n+1} and let L_{n} ⊂ R^{3v} be the linear span of the vectors X_{1}, …, X_{n}. Suppose that (i) X_{j} ∈ L_{n} for each j = n+1, …, 2n+1; (ii) the equation A_{0}X = −B(X_{i}, X_{j})−B(X_{j}, X_{i}) has a solution X ∈ L_{n} for each 1 ≤ i ≤ n, 1 ≤ j ≤ n. Then Q is flexible.

Proof. According to Theorem 2, the protocol X_{0}, X_{1}, … X_{n} can be infinitely extended. Then the statement follows from the lemma.

5. Computer experiments

Consider the Bricard octahedron Q of first type with the following vertex coordinates: A_{1} = (2, 0, 0), A_{2} = (1, \sqrt{3}, 0), A_{3} = (−2, 0, 0), A_{4} = (−1, \sqrt{3}, 0), A_{5} = (0, 0, 1), and A_{6} = (0, 0, −1). It is presupposed that the vertices A_{1}, A_{2}, A_{3}, and A_{4} lie on the equator of Q in the order named, while A_{5} and A_{6} are the north and the south poles of Q (see, e.g., [7] for more details about Bricard octahedra).

Enumerate the edges of Q according to the following table:

1 2 3 4 5 6
1 ∗ 1 ∗ 4 5 9
2 1 ∗ 2 ∗ 6 10
3 ∗ 2 ∗ 3 7 11
4 4 ∗ 3 ∗ 8 12
5 5 6 7 8 ∗ ∗
6 9 10 11 12 ∗ ∗

Here, a number at the intersection of the column with the top entry i and of the row with the left entry j defines the index e = e(i, j) of the edge of Q that joins the i-th and the j-th vertices of Q; the symbol ∗ is used to indicate that the corresponding vertices are not joined by any edge of Q.

The following picture gives an indication of the Bricard octahedron under study.

Numbers given near each edge generate the above introduced edge enumeration.

The bilinear map B : R^{18} × R^{18} → R^{12} associated with Q according Section 2 above has the following components:

B_{1}(X, U) = (x_{1} − x_{2})(u_{1} − u_{2}) + (y_{1} − y_{2})(v_{1} − v_{2}) + (z_{1} − z_{2})(w_{1} − w_{2}),
B_{2}(X, U) = (x_{2} − x_{3})(u_{2} − u_{3}) + (y_{2} − y_{3})(v_{2} − v_{3}) + (z_{2} − z_{3})(w_{2} − w_{3}),
B_{3}(X, U) = (x_{3} − x_{4})(u_{3} − u_{4}) + (y_{3} − y_{4})(v_{3} − v_{4}) + (z_{3} − z_{4})(w_{3} − w_{4}),
B_{4}(X, U) = (x_{4} − x_{1})(u_{4} − u_{1}) + (y_{4} − y_{1})(v_{4} − v_{1}) + (z_{4} − z_{1})(w_{4} − w_{1}),
B_{5}(X, U) = (x_{5} − x_{1})(u_{5} − u_{1}) + (y_{5} − y_{1})(v_{5} − v_{1}) + (z_{5} − z_{1})(w_{5} − w_{1}),
B_{6}(X, U) = (x_{5} − x_{2})(u_{5} − u_{2}) + (y_{5} − y_{2})(v_{5} − v_{2}) + (z_{5} − z_{2})(w_{5} − w_{2}),
B_{7}(X, U) = (x_{5} − x_{3})(u_{5} − u_{3}) + (y_{5} − y_{3})(v_{5} − v_{3}) + (z_{5} − z_{3})(w_{5} − w_{3}),
B_{8}(X, U) = (x_{5} − x_{4})(u_{5} − u_{4}) + (y_{5} − y_{4})(v_{5} − v_{4}) + (z_{5} − z_{4})(w_{5} − w_{4}),
B_{9}(X, U) = (x_{6} − x_{1})(u_{6} − u_{1}) + (y_{6} − y_{1})(v_{6} − v_{1}) + (z_{6} − z_{1})(w_{6} − w_{1}),
B_{10}(X, U) = (x_{6} − x_{2})(u_{6} − u_{2}) + (y_{6} − y_{2})(v_{6} − v_{2}) + (z_{6} − z_{2})(w_{6} − w_{2}),
B_{11}(X, U) = (x_{6} − x_{3})(u_{6} − u_{3}) + (y_{6} − y_{3})(v_{6} − v_{3}) + (z_{6} − z_{3})(w_{6} − w_{3}),
B_{12}(X, U) = (x_{6} − x_{4})(u_{6} − u_{4}) + (y_{6} − y_{4})(v_{6} − v_{4}) + (z_{6} − z_{4})(w_{6} − w_{4}),

where

and  \begin{matrix}X = (x_{1}, y_{1}, z_{1}, x_{2}, y_{2}, z_{2}, x_{3}, y_{3}, z_{3}, & x_{4}, y_{4}, z_{4}, x_{5}, y_{5}, z_{5}, x_{6}, y_{6}, z_{6}) \\ U = (u_{1}, v_{1}, w_{1}, u_{2}, v_{2}, w_{2}, u_{3}, v_{3}, w_{3}, & u_{4}, v_{4}, w_{4}, u_{5}, v_{5}, w_{5}, u_{6}, v_{6}, w_{6}) \end{matrix}

are arbitrary vectors in R^{18}.

It is well-known that the Bricard octahedron is flexible. For the sake of convenience, suppose that the vertices A_{5} and A_{6} have the following coordinates when the absolute value of the time variable t is small enough: A_{5}(t) = (0, 0, 1 − t) and A_{6}(t) = (0, 0, −1 + t). Moreover, suppose that the edges A_{1}A_{2} and A_{3}A_{4} are symmetric to each other with respect to the plane spanned by the vectors (1, 0, 0) and (0, 0, 1). Under these conditions, we have the following equations to determine the coordinates of the points A_{1}(t) = (x_{1}(t), y_{1}(t), 0), A_{2}(t) = (x_{2}(t), y_{2}(t), 0), A_{3}(t) = (x_{1}(t), −y_{1}(t), 0), and A_{4}(t) = (x_{2}(t), −y_{2}(t), 0) as functions of the time variable t:

[x_{1}(t) − x_{2}(t)]^{2} + [y_{1}(t) − y_{2}(t)]^{2} = 4,
[x_{1}(t) + x_{2}(t)]^{2} + [y_{1}(t) − y_{2}(t)]^{2} = 12,
[x_{1}(t)]^{2} + [y_{1}(t)]^{2} + (1 − t)^{2} = 5,
[x_{2}(t)]^{2} + [y_{2}(t)]^{2} + (−1 + t)^{2} = 5.

376 V. Alexandrov: Sufficient Conditions for the Extendibility of an n-th Order Flex … Differentiating these equations with respect to t as it was described in Section 2 above, we obtain the following protocol of the bilinear game associated with the bilinear mapping B:

X_{0} = (2, 0, 0, 1, \sqrt{3}, 0, −2, 0, 0, −1, \sqrt{3}, 0, 0, 0, 1, 0, 0, −1),

X_{1} = (\frac{1}{2}, \frac{2}{\sqrt{3}}, 0, −\frac{1}{4}, \frac{5}{4\sqrt{3}}, 0, −\frac{1}{2}, \frac{2}{\sqrt{3}}, 0, \frac{1}{4}, \frac{5}{4\sqrt{3}}, 0, 0, 0, −1, 0, 0, 1),

X_{2} = (−\frac{31}{48}, −\frac{11}{6\sqrt{3}}, 0, \frac{37}{96}, −\frac{113}{96\sqrt{3}}, 0, \frac{31}{48}, −\frac{11}{6\sqrt{3}}, 0, −\frac{37}{96}, −\frac{113}{96\sqrt{3}}, 0, 0, 0, 0, 0, 0, 0),

X_{3} = (\frac{445}{576}, \frac{223}{144\sqrt{3}}, 0, −\frac{649}{1152}, \frac{1325}{1152\sqrt{3}}, 0, −\frac{445}{576}, \frac{223}{144\sqrt{3}}, 0, \frac{649}{1152}, \frac{1325}{1152\sqrt{3}},

0, 0, 0, 0, 0, 0, 0),

X_{4} = (−\frac{30239}{27648}, −\frac{3683}{1728\sqrt{3}}, 0, \frac{50249}{55296}, −\frac{101413}{55296\sqrt{3}}, 0, \frac{30239}{27648}, −\frac{3683}{1728\sqrt{3}}, 0, −\frac{50249}{55296},

− \frac{101413}{55296\sqrt{3}}, 0, 0, 0, 0, 0, 0, 0),

X_{5} = (\frac{566191}{331776}, \frac{283771}{82944\sqrt{3}}, 0, −\frac{1027159}{663552}, \frac{2128499}{663552\sqrt{3}}, 0, −\frac{566191}{331776}, \frac{283771}{82944\sqrt{3}}, 0, \frac{1027159}{663552},

\frac{2128499}{663552\sqrt{3}}, 0, 0, 0, 0, 0, 0, 0),

X_{6} = (−\frac{7751729}{2654208}, −\frac{225415}{36864\sqrt{3}}, 0, \frac{14770211}{5308416}, −\frac{3512383}{589824\sqrt{3}}, 0, \frac{7751729}{2654208}, −\frac{225415}{36864\sqrt{3}}, 0,

− \frac{14770211}{5308416}, −\frac{3512383}{589824\sqrt{3}}, 0, 0, 0, 0, 0, 0, 0),

X_{7} = (\frac{170000039}{31850496}, \frac{92241497}{7962624\sqrt{3}}, 0, −\frac{332342771}{63700992}, \frac{730291327}{63700992\sqrt{3}}, 0, −\frac{170000039}{31850496},

\frac{92241497}{7962624\sqrt{3}}, 0, \frac{332342771}{63700992}, \frac{730291327}{63700992\sqrt{3}}, 0, 0, 0, 0, 0, 0, 0),

X_{8} = (−\frac{31271000117}{33057647616}, −\frac{2176523761}{95551488\sqrt{3}}, 0, \frac{61898822459}{6115295232}, −\frac{138739916623}{6115295232\sqrt{3}}, 0,

\frac{31271000117}{3057647616}, −\frac{2176523761}{95551488\sqrt{3}}, 0, −\frac{61898822459}{6115295232}, −\frac{138739916623}{6115295232\sqrt{3}}, 0, 0, 0, 0, 0, 0, 0),

X_{9} = (\frac{2230982344027}{110075314176}, \frac{1264655218003}{4194304 · 3^{8}\sqrt{3}}, 0, −\frac{4441640263531}{220150628352}, \frac{10101894456215}{220150628352\sqrt{3}}, 0,

− \frac{2230982344027}{110075314176}, \frac{1264655218003}{4194304 · 3^{8}\sqrt{3}}, 0, \frac{4441640263531}{220150628352}, \frac{10101894456215}{220150628352\sqrt{3}},

0, 0, 0, 0, 0, 0, 0),

X_{10} = (−\frac{108770280601489}{2641807540224}, −\frac{31215406114457}{16777216 · 3^{9}\sqrt{3}}, 0, \frac{217110627870619}{5283615080448},

− \frac{499163140516943}{268435456 · 3^{9}\sqrt{3}}, 0, \frac{108770280601489}{2641807540224}, −\frac{31215406114457}{16777216 · 3^{9}\sqrt{3}}, 0, −\frac{217110627870619}{5283615080448},

− \frac{499163140516943}{268435456 · 3^{9}\sqrt{3}}, 0, 0, 0, 0, 0, 0, 0),

X_{11} = (\frac{2701985126197979}{31701690482688}, \frac{1564814608325921}{134217728 · 3^{10}\sqrt{3}}, 0, −\frac{5399405279552831}{63403380965376},

\frac{12515889452241211}{1073741824 · 3^{10}\sqrt{3}}, 0, −\frac{2701985126197979}{31701690482688}, \frac{1564814608325921}{134217728 · 3^{10}\sqrt{3}}, 0, \frac{5399405279552831}{63403380965376},

\frac{12515889452241211}{1073741824 · 3^{10}\sqrt{3}}, 0, 0, 0, 0, 0, 0, 0).

Denote by L_{n} the linear span of the vectors X_{1}, …, X_{n}. One can verify that L_{1} ⊂ L_{2} ⊂ L_{3} ⊂ L_{4} ⊂ L_{5} = L_{6} = L_{7} = L_{8} = L_{9} = L_{10} = L_{11}. Here, the formula L_{i} ⊂ L_{j} means that L_{i} is a subset of L_{j}, but L_{i} ≠ L_{j}.

Moreover, direct calculations show that B(X_{i}, X_{j}) ∈ A_{0}(L_{5}) for all i, j = 1, …, 5. Hence, Corollary 2 can be applied to the Bricard octahedron under consideration. In other words, we can assert that the Bricard octahedron is flexible according to Corollary 2.

The conditions of Corollary 1 are not fulfilled for the Bricard octahedron under consideration.

6. Final remarks and acknowledgments

Remark. (i) One can see from the proof of Theorem 1 that some new interesting theorems can be obtained if some additional information about the behavior of the sums

∑_{j=k}^{n−k} B(x_{j}, x_{n−j})

is derived. We mean a statement of the following form: if sufficiently many sums of this kind lie in a subspace, then one can choose consequent vectors x_{j} in such a way that all the subsequent sums lie in this subspace too. (ii) One can reduce the problem of bending a smooth surface and a bar framework to a bilinear game by analogy with Section 2.

The author is grateful to I. V. L’vov for a discussion and for the example given in Section 3 above; to E. P. Volokitin for discussions and computer experiments given in Section 5 above and made by use of the definitive system for mathematical computations Mathematica; and to Wolfram Research, Inc. and to the American Mathematical Society for donating Mathematica to the Siberian Society for Promotion Science and Education. The author also thanks the anonymous referee for improvements to the paper arising from his/her comments and suggestions.

References

[1] Artin, M.: Algebraic approximations of structures over complete local rings. Inst. Hautes Études Sci. Publ. Math. 36 (1969), 23–58.

[2] Connelly, R.: The rigidity of polyhedral surfaces. Math. Mag. 52 (5) (1979), 275–283.

[3] Connelly, R.: The rigidity of certain cabled frameworks and the second-order rigidity of arbitrarily triangulated convex surfaces. Adv. Math. 37 (1980), 272–299.

[4] Connelly, R.; Servatius, H.: Higher-order rigidity — what is the proper definition? Discrete Comput. Geom. 11 (1994), 193–200.

[5] Gluck, H.: Almost all simply connected closed surfaces are rigid. Lect. Notes Math. 438 (1975), 225–239.

[6] Ivanova-Karatopraklieva, I.; Sabitov, I. Kh.: Surface deformation I. J. Math. Sci. 70 (2) (1994), 1685–1716 [translation from the Russian original Itogi Nauki Tekh., Ser. Probl. Geom. 23 (1991), 131–184].

[7] Sabitov, I. Kh.: Local theory of bendings of surfaces. In: Geometry III. Theory of surfaces. Encycl. Math. Sci. 48 (1992), 179–250 [translation from the Russian original Itogi Nauki Tekh., Ser. Sovrem. Probl. Mat., Fundam. Napravleniya 48 (1989), 196–270].

[8] Švec, A.: On infinitesimal isometries of a hypersurface. Czech. Math. J. 24 (1974), 150–163.

[9] Yau, S. T.: Problem section. Ann. of Math. Stud. 102 (1982), 669–706.

Received August 13, 1997, revised version January 14, 1998.