Almost All Simply Connected Closed Surfaces Are Rigid

ALMOST ALL SIMPLY CONNECTED CLOSED SURFACES ARE RIGID

HERMAN GLUCK

1, INTRODUCTION, Are closed surfaces rigid? Euler thought so, and conjec- tured in 1766, "A closed spacial figure allows no changes, as long as it is not ripped apart" [6], and expanded on this in letters to Lagrange in 1770. But the conjecture has not yet yielded, and is surely one of the oldest and most beautiful unsolved problems in geometry in the large. What is the evidence in its favor? The experimental evidence is of two sorts. Most cardboard models of closed surfaces, such as the boundaries of the regular solids (but also nonconvex surfaces and those of different topological type) seem to be rigid and not flex. More inter- esting, some models do seem to flex, but in each case the apparent flexing has been traceable to slight distortions, such as bending of the faces or separation of the vertices, due to the nature of construction. The first mathematical advance was made by Cauchy [2] in 1813, who proved that two closed convex polyhedra, constructed from pairwise congruent faces assembled in the same order, were in fact congruent themselves. Hence a closed strictly convex polyhedron must be rigid because any slight flexing of it would still be convex and hence congruent to it. Similar results were obtained by Liebmann [8] in 1899 for analytic surfaces, and by Cohn-Vossen [3] in 1936 for the smooth case. I offer here a simple argument that a closed simply-connected surface in three space is almost always rigid; Euler's conjecture in this case is therefore "statis- tically" true. I think the same should be provable for any closed surface in three space, regardless of topological type, but I have been unable to do this. In order to provide a self-contained introduction to the rigidity problem, I have included Alexandrov's proof of the infinitesimal rigidity of strictly convex polyhedra in section 5. The reader wishing a guide to the history and literature of this problem should first consult Efimov's introduction and appendix in [5], and then the standard bible [i] by Alexandrov. 226

This paper is organized as follows: 2. Rigidity - competing definitions and their equivalence, 3. Infinitesimal rigidity - competing definitions and their equivalence, 4. Infinitesimal rigidity implies rigidity, S. Strictly convex closed surfaces are infinitesimally rigid, hence rigid, 6. Almost all simply connected closed surfaces are infinitesimally rigid, hence rigid. I have profited from reading Robin Langer's honors thesis [7], in which several of the ideas to be discussed below were recast in more elegant form, and I have bor- rowed from his presentation. I have also benefited from many discussions with David Singer. Some new and exciting work on the rigidity conjecture is being done by Robert Connelly, using methods of complex analysis, but this is not yet in print.

2, RIGIDITY, Let K be a simplicial complex whose underlying space is homeomorphic to the two-sphere. K is otherwise arbitrary and will remain fixed for the entire story. A polyhedron in three-space, combinatorially the same as K, is realized by a simplexwise linear map P : K + R 3 . For simplicity, we do not exclude the degenerate maps at this point. Such maps are determined by their values on the vertices vl, v2,..., v V of K, and hence correspond to V-tuples (PI' P2' .... PV ) of points of R 3, The set of 3V all such polyhedra is therefore parametrized by R , and by abuse we allow our- selves to write

P = (PI' P2 .... ' PV ) @ R3V

Our goal is to show that almost all such polyhedra are rigid, in that the set of rigid ones contains an open and dense subset of R 3V. Two polyhedra P = (Pl .... 'Pv ) and Q = (ql' .... qv ) are CONGRUENT, written P ~ Q, if there is a rigid motion h : R 3 + R 3 such that hP = Q : K + R 3. Equiva- lently,

II Pi - Pj II = H qi - qj II for 1 ! i, j iV , where II Jl is the Euclidean norm in R 3. The congruence class of P in R 3V will be denoted by [[P]]. By contrast, P and Q are I$OM£YRIC, written P ~ Q, if each face of P is congruent to the corresponding face of Q, equivalently if P and Q both pull the Riemannian metric on R 3 back to the same (possibly degenerate) metric on K. Let E be the set of pairs (i,j), 1 ~ i, j ~ V, for which an edge of K connects the vertices v. and v.. Then P is isometric to Q if and only if l ] 227

[I Pi - Pj II = II qi - qj II for (i,j) c E .

Note that this capitalizes on the fact that the faces of K are triangular. The isometry class of P in R ~V will be denoted by [P]. Since congruence implies isometry, [[P]] c [p]. Though they are equal for a tetrahedron, they may easily be unequal. For example, if P is a house whose roof slopes on all four sides, and Q is reconstructed from the same pieces but with its roof dipping into the body so as to form a large trough, then P and Q are isometric but not congruent, so that [[P]] is a proper subset of [P]. What do [[P]] and [P] look like as subsets of R3V? Since the equation LI Pi - Pj ;I = II qi - qj II can be written as a quadratic polynomial in the coordinates of the vertices of Q, both [[P]] and [P] are algebraic varieties in R ~V. If P does not lie in any plane in R 3, then the group of all rigid motions of R 3 acts effectively on [[P]], which is therefore homeomorphic to two disjoint copies of R 3 x p~ (where p3 denotes projective 3-space), embedded as an algebraic submanifold of R 3V. The appearance of [P] is less clear. If the rigidity conjecture is correct, then any embedded P is rigid and therefore (as we will see shortly) [P] is a finite disjoint union of congruence classes and hence an algebraic submanifold of R 3V. But a priori we only know that [P] is an algebraic variety. Indeed, if we consider instead the simplexwise linear maps of a quadrilateral into the plane R z, and let P denote a square, then [P] is homeomorphic to R z x S 1 x (S 1 V SI), where S 1 V S 1 denotes the wedge of two circles (a figure eight). Hence in this case, [P] is a four dimensional algebraic variety with singularities along a three dimensional subvariety.

We now crystallize the notion of rigidity by first offering some competing definitions and then observing that they are equivalent. Let H Ii also denote the Euclidean norm in R 3V, so that we may speak of the distance between polyhedra (each a linear map of K into R3).

DEFINITION 2,1. A polyhedron P is RIGID if there exists an E > 0 such that any polyhedron Q, isometric to p and within g of P, is actually congruent to P. In symbols,

Q ~ p and II Q - P II < c ~ Q ~ P .

For the purposes of the following discussion, we also say Q is ~-RIGID.

DEFINITION 2.2, P is RIGID if any path in [P] beginning at P actually lies entirely in [[P]]. 228

DEFINITION 2,3, p is RIGID if any analytic path in [P] beginning at P lies entirely in [[P]].

The first definition calls P rigid if a dissection and reassembly of P into a polyhedron approximating it can only yield a bodily displacement of P. The second calls P rigid if no continuous "flexing" of P is possible, and the third is just a technical modification of this.

REMARK (2,4), If P is g-rigid and Q is congruent to P, then Q is also g-rigid, for the same c. Hence [[P]] has distance > g from the rest of [P].

REMARK (2,5), ~or any P, [P] has only finitely many topological components, since as a real algebraic variety it can be written as a finite disjoint union of connected real algebraic manifolds by [12].

REMARK (2,6), Any collection of polyhedra which are isometric to one another, pairwise noncongruent and g-rigid (for varying g) must be finite. For by (2.4), each P in the collection yields two components of [P], namely the two pieces of [[P]], while by (2.5) there can only be finitely many components.

THEOREM 2,7, The three definitions of rigidity are equivalent.

If P is e-rigid, then a continuous path in [P] beginning at P would, by (2.4), have to remain in [[P]]. Hence the first definition implies the second, which obviously implies the third. Suppose the first definition fails. Then any neighborhood of P in [P] contains points of [P] - [[P]]. Since [P] is a real algebraic variety and [[P]] a subvariety, Lemma 18.3 of [i0] can be applied to yield an analytic path ~(t) in [P], with ~(0) = P and ~(t) E [p] - [[p]] for t > 0. Thus the third definition also fails, so all three are equivalent.

3, INFINITESIMAL RIGIDITY, In this section we give two definitions of infinitesimal rigidity, provide algebraic formulations and geometric motivations for each, and then prove they are equivalent. The notion of infinitesimal rigidity is stricter than that of rigidity, and corresponds to the conditions an engineer would require to be satisfied by a collection of rods, joined but freely pivoting at their ends, before certifying rigidity. Infinitesimal rigidity is a linearized version of rigidity, with a subsequent loss of "precision".

DEFINITION 3,1, Ca) A family (61 ..... 6V) of vectors in R 3 will be called an INFINITESIMAL ISOMETRIC PERTURBATION of the polyhedron P : K ÷ R 3 if 229

(3,2) (Pi - Pj) " (6i " 6j) = 0 for all (i,j) E E , where the dot indicates the inner product in R 3 (b) Suppose that t and r are any two vectors in R 3, and let 6 = i t + (Pi x r) for 1 < i < V, where the cross denotes the vector cross product in R ~ . Then

(Pi - Pj) " (6i - 6j) = (Pi - Pj) ' ((Pi - Pj) × r) = 0 , for all i,j, whether or not corresponding to an edge of K. Such a choice of (61,...,6V) is therefore certainly an infinitesimal isometric perturbation, and is called an INFINITESIMAL CONGRUENCE. (c) An infinitesimal isometric perturbation which is not an infinitesimal congruence is called an INFINITESIf4AL FLEXING. If P admits an infinitesimal flexing, it is INFINITESIMALLY FLEXIBLE; otherwise it is INFINITESIVALLY RIGID.

ALGEBRAIC FORMULATION 3,3, Let K have V vertices, E edges and F faces. Then V - E + F = 2 and 3F = 2E; hence E = 3V - 6. Technically there are 2E equations in system (3.2), but the equations for (i,j) and (j,i) ~ E are really the same. Thus we may regard (3.2) as a system of E = 3V - 6 equations in the 3V scalar unknowns (6 I, ... ,6V). An infinitesimal isometric perturbation of P = (PI'''"Pv) is therefore an element in the kernel of the linear map

L : R 3V ÷ R ~V-6

L(61 ..... 6V) = ( .... (Pi - Pj) " (6i 6j) .... ) , where the component displayed on the right corresponds to the edge v.v. of K, I j with i < j say. If P does not degenerate to a subset of a line in R 3, then it is easy to see that the infinitesimal congruences form a 6-dimensional subspace of the kernel of L. In this case, P is infinitesimally rigid if and only if dim(ker L) = 6. If P does degenerate to a subset of a line, it is easy to see that P must be infinitesimally flexible.

GEOMETRIC MOTIVATION 34 Suppose we have a smooth isometric deformation of P, that is, a smooth one-parameter family P(t) = (p1(t) .... ~Pv(t)) satisfying

(Pi(t) - pj(t)) • (Pi(t) - pj(t)) = constant , for (i,j) ¢ E.

Differentiating with respect to time t,

(Pi(t) - pj(t)) • (pi'(t) - pj'(t)) = 0 for (i,j) E E. 230

Setting Pi = Pi (0) and 6 i = pi'(O), we get system (3.2). Thus the time derivative of a smooth isometric deformation of P is an infinitesimal isometric perturbation of P. If P is moved as a "rigid body" in R a, then it is easy to see that the corresponding time derivative has the form 6 i = t + (Pi x r), where t corresponds to translations and r to pure rotations with angular velocity r. In this case (61 ..... dV) is an infinitesimal congruence. For a different geometric point of view, suppose that (~i,...,@) is an infinitesimal isometric perturbation of P = (PI' .... PV )" Then consider the one-parameter family of polyhedra

P(t) = (Pl + t61 .... 'Pv + t6v) "

The square of the length of its ij edge is

[(Pi - Pj) + t(6i - 6j)] • [(Pi - Pj) + t(di - 6j)] , and the time derivative of this is

2[(Pi - pj) + t(6 i - 6j)] " (@i - 6j) .

Evaluated at t = 0, we get

2(Pi - Pj ) ° (6i - ~j ) = 0 .

This works both ways; that is, (~l,...,@) is an infinitesimal isometric perturbation of P if and only if the time derivative at t = 0 of the squares of the lengths of the edges of P(t) is O. In other words, the edge lengths of P(t) may be changing at t = 0, but this is not detectable in first order terms.

EXAMPLES 3,5, Infinitesimal congruences are easy to visualize, but what do infinitesimal flexings look like? The simplest example: suppose P has a FLAT VERTEX Pi' meaning the dihedral angles at the edges of P containing Pi are all equal to ~. Then (0 .... ,0,6i,0 ..... 0), where ~i is a vector normal to the plane through Pi which contains all the faces of P at Pi' is an infinitesimal flexing of P. This is illustrated in the first figure below.

A more subtle example is illustrated in the second figure. In the octahedron shown there, the quadrilateral ABCD is planar but nonconvex. Edge BC when prolonged meets edge AD at the point B'; similarly, DC when prolonged meets AB at D'. The vertex E is chosen to lie somewhere directly over the line through B' and D'; the vertex F is chosen to lie somewhere directly below this line. Con- structed this way, it turns out that the octahedron is infinitesimally flexible, and 231

in fact in such a way as to "preserve" the planarity of ABCD. The reader may find it instructive to compute the actual infinitesimal flexing given this information.

FIGURE 1

A s

FIGURE 2 232

Now we consider the second version of infinitesimal rigidity.

DEFINITION 3,6, (a) Let {mij : (i,j) • E} be a set of real numbers such that

(i) m.. : m.. lJ Jl Pi - P~ (ii) ~ ~.. = 0 for I < i < V . j : (i,j) • E 13 Ifpi - pjt[

(Notice that for this to make sense, we must have Pi¢ Pj for i # j.) Then {mij} will be called an w-BENDING of P = (pl,...,pv). W.. = !I (b) If we replace ~ij by ~ij Ilpi ~ p]il • then the two conditions become

(i) K. = ~.. zJ 3i

(ii) ~ ~ij (Pi - Pj) = 0 for l

Such a set of numbers will also define an m-bending of P, and permit Pi = Pj for i # j in the definition.

ALGEBRAIC FORMULATION 3,7, Since we require ~ij = %1' let us regard {~ij} as a set of E real numbers, for example by demanding i < j. Then (ii) above may be regarded as a system of 3V equations in E : 3V - 6 unknowns. An e-bending of P is therefore an element in the kernel of the linear map

M : R3V -6 * R aV

M( .... mij .... ) = (~ $1j(Pl - Pj) ..... ~ ~vj(Pv - Pj)) "

The polyhedron P is u-rigid if and only if dim(ker M) = O.

GEOMETRIC MOTIVATION 5,8, SuppOse we have a smooth isometric deformation P(t) = (p~(t),...,Pv(t)) of P = P(O). Let ~ij(t) be the dihedral angle at the edge Pi(t)pj(t) of P(t), measured from the inside• say. If aij'(t) is its time derivative, then

Pi(t) - p~(t) ~ij' (t) i@,i(t) PJ (t)ll

is the angular velocity vector of a face of P(t) with respect to its neighbor across the edge Pi(t)pj(t)" 233

The motion of this face with respect to its neighbor is a rotation of R 3, and hence a path in SO(3) if we choose the origin at Pi(t). If we view things so that at t = 0 we start with the identity, then the tangent vector to this path at t = 0 is an element of the Lie algebra so(3), and is precisely the angular velocity at that instant. Now hold the vertex Pi(t) fixed in mind and consider the succession of faces of P(t) around it. The motion of a face with respect to its neighbor once removed is the composition of two motions with respect to immediate neighbors. Since composition in S0(3) corresponds to addition in so(3), the tangent vector at t = 0 to this composition is the sum of the two corresponding tangent vectors, that is, the sum of the two angular velocity vectors. Composing fully around the vertex Pi(t), we come back to the first face, which does not move with respect to itself. Hence the sum of the angular velocity vectors is 0. Put

~ij = ~ij '(0) and Pi = Pi (0) '

and we get

Pi - Pj = 0 for i < i < V . j : (i,j) 6 E ~ij llpi - pjl]

THEOREM 3,9, Infinitesimal rigidity and m-rigidity are equivalent.

Recall the linear maps

L : R 3V + R 3V-6 (i,j) E E L(61 ..... ~V ) = ( .... (Pi - Pj) " (~i - 6j) .... ) i< j

and

M : R3V -6 + R 3v

M( .... m13 .... ) = (~ ~lj(pl - pj) ..... ~ mvj(Pv - pj)) R 3V used in the algebraic formulation of the competing definitions. Let and R 3V-~ have their usual inner products.

CLAIM, The maps L and M are dual, that is

6 • MCg) = LC¢) • g,

where 6 = (61,...,6V) and ~ = {~ij : (i,j) E E, i < j} .

To see this, 234

L(6} • ~ = [ (Pi - Pj) " (6i - 6j) ~'ij i

Interchange the indices i and j in the second summation, getting

• Z. CPi - Pj) " 6i $ij I,]

: f 6i " (f o~-~jcPi - pj)) i j

= 6 • M(~) , and verify the claim. To complete the proof of the theorem, suppose first that P does not degenerate to a subset of s line. Then

is infinitesimally rigid ~=* dim(ker L) = 6 L is onto *=*M is one-one P is w-rigid .

Finally, suppose P does degenerate to a subset of a line. Then any (61 .... ,~V) with each 6i perpendicular to this line gives an infinitesimal isometric deformation of P. Since P has at least four vertices, the dimension of these defor- mations is greater than 6, hence P is infinitesimally flexible. Furthermore, by duality there exists a nonzero m-bending, so P is also m-flexible. Thus the two definitions are equivalent.

4, INFINITESIMAL RIGIDITY IMPLIES RIGIDITY,

THEOREM 4,1, Infinitesimal rigidity implies rigidity.

This is just the implicit function theorem. Consider the smooth map f : R 3V + R 3V-6 defined by (i,j) 6 E f(P: ..... PV ) = ( .... (Pi - Pj) " (Pi - Pj) .... ) i < j

This map assigns to each polyhedron P the set of squares of the lengths of its edges. Its differential at P is a linear map dfp : R 3V + R 3V-6, and one easily sees that

dfP (61 ..... 6V) = ( .... 2(Pi - Pj) " (~i - 6j) .... ) . 235

Hence dfp = 2L. Therefore

is infinitesimally rigid ~=~ dim(ker L) = 6 ~=~ L is onto ~=~ dfp is onto ~=~ P is a regular point of f .

Now by the implicit function theorem, if P is a regular point of f, then f-lf(p) is a 6-dimensional manifold near P. But f-lf(p) = [p]. Since P is infinitesimally rigid, it cannot degenerate to a subset of a line and therefore, as seen before, [[P]] is also a 6-dimensional manifold near P. Hence [P] and [[P]] coincide near P. It follows that any polyhedron, near P and isometric to it, is actually congruent to it. But this means that P is rigid.

5, STRICTLYCONVEX POLYHEDRA ARE INFINITESIMALLY RIGID, The polyhedron P is STRICTLY C~VEX if through each vertex Pi of P there may be drawn a plane meeting P only at Pi"

THEOREM 5,1, (Cauchy-Dehn-Weyl-Alexandrov) Strictly convex polyhe~a are infinitesimally rigid.

This was first proved by Dehn [4] in 1916 by a clever, somewhat involved, argument, and again by Weyl [ii] the following year. The best proof, from which we extract, was given by A. D. Alexandrov in [i], and this in turn is really an adapta- tion of Cauchy's original argument for the rigidity of convex polyhedra.

LEMMA 5,2, (Cauchy) Let L be a finite graph on the two-sphere $2~ with no circular edges and no region of S 2 - L bounded by just two edges of L. Mark the edges of L randomly with + and -, and let N v be the number of sign changes as one circles round the vertex v. Let N = ~ N v be the total number v of sign changes. Let V be the number of vertices of L. Then

N<4V-8.

In particular, it is IMPOSSIBLEfor each N to be at least 4. v

The reader may care to quickly check the lemma on the one-skeleton of an octahedron, and also to observe the necessity of outlawing regions bounded by just two edges, for otherwise an arbitrary even number of great semicircles can be drawn from the north to south pole and alternately marked + and - Let N G denote the number of sign changes as one traverses the boundary of a region G of S 2 - L. Then N = ~G NG also. Let F n denote the number of regions 236

of S 2 - L having n boundary edges, where an edge is counted twice for G if G lies on both sides of this edge. Note that if G has n edges, then N G is an even number < n. Hence

N = ~G NG <- 2 F 3 + 4 F~ + 4 F s + 6 F 6 + 6 F 7 + ....

Let V, E and F denote the number of vertices, edges and faces of L, as usual, and C the number of components of L. Then

V - E + F = 1 + C > 2

2E = ~ n Fn , F = Z F n

Hence

4V - 8 > 4E - 4F ~(2n- 4)F n

= 2 F 3 + 4 F~ + 6 F s + 8 F 6 + i0 F 7 + ... > 2 F 3 + 4 F + 4 F s + 6 F 6 + 6 F 7 + ...

> N , proving Cauchy's lemma.

L£MMA 5,3, (Cauchy - Alexandrov) Let Po be a strictly convex vertex of P (obvious definition), and pl,...,pn those vertices of P joined to Po by edges. Let ~i ..... ~n be real n~mbers satisfying

n % (Pi - Po ) = 0 • i=l

Then either (1) all ~. = O, or else (2) the sign of ~i changes at least four times as one circles around Po" ignoring any zeros.

Let the number of sign changes, as counted above, be called the II~DEX at Po" It is an even number > 0. We must show two things: (a) If the index is 0, then all ~. must be 0. 1 (b) The index can't be 2. (a) Suppose the index is 0. The vectors Pi - Po all lie in some open half space by strict convexity of P at Po" But then Zi ~--i (Pi - Po ) = 0 with all ~i of the same sign implies all ~i = 0. (b) Suppose the index were 2. Then the vertices about Po can be renumbered so that in circling PO' ~i ..... ~k ~ 0 while ~X+I ..... m--n are L O, By strict convexity, there is a plane through Po separating pl,...,p k from Pk+1 .... 'Pn" The 237

vectors ~i (Pi - Po ) which are nonzero therefore all lie in the same corresponding open half space. Since [i ~i (Pi - Po ) = 0, all ~i = 0, so in fact the index can't equal 2, proving the lemma.

PROOF OF THEOREM 5,1, We want to show that a strictly convex polyhedron P is infinitesimally rigid, and since the two versions of infinitesimal rigidity are equivalent, we will show that P is u-rigid. Suppose {~..} corresponds to an u-bending of P. Mark each edge of P with ij a + sign if the corresponding ~.. > 0, with a sign if ~.. < 0, and leave i] ij it unmarked if ~-.. = 0. Delete the unmarked edges of P and let the remaining z3 edges form the graph L. By Lemma 5.3, the index of each vertex of L is >_ 4. Since L is part of a simplicial complex, the hypotheses imposed on it by Lemma 5.2 are satisfied, while the conclusion is contradicted...unless there is no L at all because all the ~.. = 0. Thus the u-bending is trivial, so P is infinitesimally z] rigid.

COROLLARY 5,4, (Cauchy) Strictly convex polyhedra are rigid.

Strictly convex polyhedra are infinitesimally rigid by Theorem 5.1, and hence rigid by Theorem 4.1.

6. ALMOST ALL SIMPLY CONNECTED SURFACES ARE RIGID, THEOREM 6,1, Let K be a simplicial complex whose underlying space is homeomorphic to the two-sphere, and as before, identify the space of all simplexwise linear maps of K into R 3 with R 3V. Then the infinitesimally rigid polyhedra correspond to an open and dense subset of R 3V, and the rigid ones contain these.

Let IF denote the subset of R 3V corresponding to the infinitesimally flexible polyhedra. A polyhedron p E IF if and only if the corresponding linear map L : R 3V + R 3V-6 has more than a 6-dimensional kernel. The entries in the matrix of L are either 0, x i - xj, Yi - Yj' z i - zj, where Pi = (xi'Yi'Zi)' The condi- tion that L have more than a 6-dimensional kernel can be expressed by taking the sum of the squares of the determinants of all 3V-6 by 3V-6 submatrices of L, and setting this equal to 0. Since this is a polynomial in the coordinates of the 3V vertices of P, IF is a real algebraic variety in R . Is it proper, or could it possibly be all of R3V? We pause to state

STEINITZ' THEOREM. Any simplicial complex K with underlying space homeomorphic to a two-sphere admits a simplexwise linear embedding into R s whose image is strictly convex. (See [9] for a proof.) 238

Therefore no matter which K we start with, some p E R 3V is strictly convex, hence infinitesimally rigid by Theorem 5.i. So IF is a PROPER algebraic subvariety of R 3V, and its complement (the infinitesimally rigid polyhedra) is open and dense in R 3V, The rigid ones contain these by Theorem 4.1, so the argument is complete.

The author thanks the National Science Foundation for its support.

UNIVERSITY OF PENNSYLVANIA PHILADELPHIA, PENNSYLVANIA 239

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