Stress–flex orthogonality on spheres and a torus obstruction

Caius Folkerts

A stress balances forces along the edges of a framework; an infinitesimal flex moves its vertices without changing edge lengths to first order. The strong stress–flex conjecture asks whether these two kinds of data always satisfy a further orthogonality identity when a closed polyhedral surface is coned from a point. This note proves the identity for every polyhedral sphere in three dimensions, without convexity or embeddedness, and disproves the general conjecture with an explicit nine-vertex embedded torus. The distinction is topological: a closed vector-valued cochain has global face potentials on a sphere, whereas the torus supports a nonzero period.

Let be a finite cell decomposition of a closed oriented surface with simple graph . Realize its vertices as distinct points , so that the vertices of each face lie in a plane and include three noncollinear points. The realization need not be embedded. Choose a point outside every face plane, and join to every vertex. An equilibrium stress assigns scalars to the edges of and to the cone edges; at each vertex, the weighted sum of its incident edge vectors is zero. An infinitesimal motion with the cone point fixed assigns velocities satisfying

Theorem 1. If is a sphere, every equilibrium stress and every infinitesimal motion with the cone point fixed satisfy

There is an embedded polyhedral torus with nine vertices, and a cone point outside all its face planes, for which this identity fails.

The torus disproves the strong stress–flex conjecture of Connelly, Gortler, Theran, and Winter [1], whose formulation allows closed surfaces with flat faces. The spherical result rests on the classical relation between stresses and polyhedral liftings [2]; the argument below gives the required correspondence directly. It identifies the topological obstruction.

1. Moving the cone point

Let have rows . A coned stress is encoded by a symmetric matrix , supported off the diagonal on , satisfying . Its outer-edge stresses are , and its cone-edge stresses are . The equilibrium equation at the cone follows by summing the outer-vertex equations.

The following observation applies in every dimension. Moving the cone point in an arbitrary direction tests the corresponding component of the stress-weighted velocity.

Lemma 2. Suppose an outer stress matrix of a framework coned from the origin extends continuously to outer stress matrices for all cone points near , while the outer vertices remain fixed. Then for every infinitesimal motion with the cone point fixed.

Proof. Write and , again with row vectors, and put . The infinitesimal edge equations give

On the other hand, , so the same trace equals . Set , divide by , and let tend to zero. Continuity gives for every . □

The required continuation comes from a vector space that does not depend on the cone point. Let

The affine functions may differ from face to face; their vertex values must agree along common edges.

Proposition 3. If is a sphere and lies outside its face planes, the space of coned stresses is naturally isomorphic to

In particular, its dimension is , and every stress extends real analytically under small motions of .

Proof. Fix and abbreviate . If an oriented edge has face on its left and face on its right, assign to the dual edge the vector

This assignment is antisymmetric. Around the dual cell corresponding to vertex , its sum is, with a consistent orientation,

Thus is closed. The first cohomology of the sphere vanishes, so there are vectors , unique up to a common additive vector, such that .

The difference across an edge annihilates both endpoint vectors. Consequently is independent of the incident face , since those faces form a cycle around . On these values are given by the affine function , and hence .

Conversely, fix . For each face there is a unique vector such that

Indeed, linear functions on restrict to all affine functions on a plane missing the origin, and three noncollinear vertices of that plane give three linearly independent vectors . Across an edge , the difference is perpendicular to and , so it has a unique expression . The cross product is nonzero because is on no incident face plane. Telescoping these differences around shows that is parallel to ; there is therefore a unique diagonal entry making .

This linear construction is onto by the first part of the proof. Its kernel consists of common face vectors , since the dual graph is connected, and thus has dimension three. Finally, keep fixed while moving . Solving (3) and recovering the entries of uses rational functions with locally nonzero denominators, giving the asserted analytic continuation. □

After translating the cone point to the origin, Proposition 3 and Lemma 2 prove the spherical assertion of Theorem 1. More generally, the same argument proves (1) for any stress on a closed oriented surface whose cochain (2) has zero periods. Only the passage from a closed cochain to face potentials used the topology of the sphere.

2. A torus with a nonzero period

Let

and, with indices read modulo three, set

Take the nine quadrilateral faces

They are the boundary of an embedded solid torus. To see this without a projection argument, put and . The map

is a homeomorphism onto its image, with an inverse that recovers from the gauge of and then recovers . Its boundary is precisely the stated nine-face surface. If ranges over the three side-defining linear forms of , the face planes are and . None contains the cone point .

Let and assign velocities

These define an infinitesimal flex of the coned framework. On the lower triangle, each relative velocity is perpendicular to its edge. Every other edge incident to has horizontal component parallel to , so its scalar product with also vanishes; this includes the cone edge. All remaining endpoints are stationary.

The four vertices span a complete subgraph and satisfy

The affine dependence with coefficients gives an equilibrium stress by assigning to each of its six edges and zero elsewhere. Its cone-edge coefficients are at , respectively. Therefore

which proves the second assertion of Theorem 1.

The obstruction is visible on the dual surface. The cycle through the three lower trapezoids crosses the edges , of which only is stressed. Its period is consequently, up to orientation,

Thus the stress cannot be represented by global face potentials, precisely at the step that is automatic on a sphere.

References

[1]: R. Connelly, S. J. Gortler, L. Theran, and M. Winter, The Stress-Flex Conjecture, preprint, 2024. arXiv 2404.15590.
[2]: W. Whiteley, Motions and stresses of projected polyhedra, Structural Topology 7 (1982), 13–38. HDL 2099/989.