The polyhedra that can flex

The count that promises a mechanism sets up Maxwell’s arithmetic: a framework of V joints and E bars in three dimensions has 3V − 6 freedoms to remove and E bars to remove them with, so a shortfall promises a motion. The mechanisms a count cannot see is where the promise fails, because bars can be redundant and the count cannot tell.

Both are about frameworks of rods. This is about the same shapes made of plates, and the two give different answers on the same solid.

The distinction is worth drawing carefully because the two objects share a picture and share almost nothing else. A bar framework fixes distances between joints; a plate framework fixes whole faces and lets them turn about shared edges. A cube has the same eight corners and twelve edges in both readings, and the answers differ by six degrees of freedom. So “is a cube rigid” has no answer until it is said what is being held fixed, and half the confusion about rigidity comes from not saying.

Rods and plates are different objects #

A cube of rods hinged at the corners folds flat. Everybody who has assembled shelving knows it, and the count says why: eight joints give eighteen freedoms, twelve bars remove at most twelve, so six remain.

A cube of cardboard does not fold, because its faces are rigid. Cauchy proved the general statement in 1813: a convex polyhedron with rigid faces, hinged along its edges, is rigid — and more, that two convex polyhedra assembled from the same faces in the same pattern are congruent. It is one of the oldest theorems in the subject and it is usually proved by a sign-counting argument on the edges.

A rank computation sees it without the argument.

The rigidity matrix is the whole instrument and it is worth stating once. It has one row per bar and three columns per joint; the row for a bar carries the bar’s direction at one end and minus it at the other. A vector in its kernel is an assignment of velocities to the joints that keeps every bar’s length constant to first order — and six of those are always there, being the translations and rotations of the whole thing. So the framework cannot move if and only if the rank is 3V − 6, and the shortfall counts the motions.

That is an infinitesimal statement, and the distinction matters: a framework can be infinitesimally rigid and still not rigid, or the reverse. For everything in this essay the two coincide, and where they might not the flex is followed explicitly rather than inferred from a rank.

The pattern in the first half of that table has a one-word explanation.

Twelve diagonals on the cube and sixty on the dodecahedron, and both become rigid. That is Cauchy’s theorem in the only form a computation can hold: making the faces rigid is adding their diagonals, and the rank goes to 3V − 6.

The dodecahedron is the extreme case in the table and it repays a look. Twenty vertices give fifty-four freedoms and thirty edges remove thirty, so a dodecahedron of rods has twenty-four ways to move — more freedoms than bars. It is barely a structure at all, and it collapses in the hand. Adding sixty diagonals brings the rank to fifty-four exactly, with no redundancy anywhere, which is a small coincidence worth noticing: the plate framework of every one of the five is exactly isostatic.

There is a second reading of the diagonals which is the useful one for a structure rather than a solid. A square face of a framework contributes two freedoms, and one diagonal removes one of them; the second diagonal is redundant for rigidity but not for the plate, since a plate is rigid in both directions at once. So the count of diagonals added above is larger than the count strictly needed — the cube needs six diagonals rather than twelve to be rigid, one per face — and taking twelve is the honest translation of “the faces are plates”. The rank does not distinguish, because rank ignores redundancy, which is exactly the blind spot the mechanisms a count cannot see is about.

Cauchy’s own proof is worth a paragraph even though it is not what is computed here, because its shape explains why convexity is unavoidable. Suppose two convex polyhedra have the same faces assembled the same way but different dihedral angles. Mark each edge with a sign according to whether its angle grew or shrank. A lemma about spherical polygons says that around any vertex the sign pattern must change at least four times; counting sign changes over the whole surface against Euler’s formula then gives a contradiction. Every step of that uses convexity — the vertex figures are convex spherical polygons — and there is no version of the lemma for a vertex that folds inward.

Where the hypothesis earns its place #

Cauchy’s theorem says convex, and a theorem’s hypotheses are worth testing rather than reading past. Bricard found in 1897 that non-convex octahedra can flex, and the easiest of his three families to build needs no cleverness at all.

Take any three points. Let the other three be their images under a half turn about a line. That is a line-symmetric octahedron, and it has one internal degree of freedom for a reason that is a count: nine free coordinates, two isometries that respect the symmetry, six independent edge lengths, and 9 − 2 − 6 = 1.

The count deserves unpacking because it explains why the construction is so easy. Three free points is nine coordinates. The isometries that respect a half turn about a fixed line are rotations about that line and translations along it — two of them, not six, because everything else would move the line. And the twelve edge lengths reduce to six independent ones, since the half turn pairs them up. Nine minus two minus six leaves one, and a one-dimensional configuration space is a flex. Nothing about the three points was special, so almost every line-symmetric octahedron flexes.

One detail of the numerical method is worth naming because it was the bug. The kernel of the rigidity matrix here is one-dimensional, so its basis vector has an arbitrary sign — and computing it fresh at each step gives a sign chosen at random, which makes the configuration random-walk on the constraint surface instead of travelling along it. Carrying the sign forward from the previous step fixes it, and the difference is visible in the first column of the table: with the sign carried the vertices travel four tenths of a unit in forty steps, and without it they end up almost where they started.

One more property of the flex is worth recording because it explains why the family is easy to find and hard to see. The half turn is preserved throughout: the octahedron stays line-symmetric at every step, because the motion was found inside the symmetric family and nothing takes it out. So the flex is a symmetry-preserving mechanism, which is the same kind of object a fold that keeps its symmetry is about — a motion that exists because a symmetry reduces the count of independent constraints, rather than in spite of the symmetry.

A conserved quantity that was never there #

Sabitov proved in 1996 what had been conjectured for decades: a flexing polyhedron keeps its volume. The bellows theorem — a bellows works by changing shape, and the theorem says a polyhedral one could not pump air.

The octahedron above keeps its volume through the whole flex. That is not evidence for anything.

The check is cheap and it is the sort that is easy not to run. Computing the volume at forty configurations and finding it constant is a result; computing it at the first configuration and finding it zero is the thing that decides whether the result means anything. The habit generalises past this case: a conserved quantity should be evaluated once before the motion starts, because a quantity that is identically zero on the whole family is conserved for a reason that has nothing to do with the theorem being illustrated.

The reason is two lines. A half turn is a rotation, so it preserves determinants. It carries each face of the octahedron to the opposite face, and opposite faces have opposite orientations in the surface. So the two contributions to the algebraic volume cancel, in pairs, for every line-symmetric octahedron whatever its three free vertices are — and the sum is zero before anything has moved.

This is worth stating because the alternative is a figure that looks like a demonstration and is not. A quantity that stays at zero throughout a motion tells nothing about whether it would have stayed constant at some other value, and a reader shown “the volume does not change” beside a flexing polyhedron would reasonably conclude that Sabitov’s theorem had been checked. It has not been, here or anywhere in this collection.

There is a way to read the whole episode that is more useful than the counterexample itself. The flex was found by asking for the kernel of a matrix, the volume was computed because the bellows theorem said it should be constant, and the constancy turned out to be an artefact of the symmetry that produced the flex. Every step was mechanical and the last one was a trap — the kind that is only visible if the quantity is checked at the start as well as along the way. Checking a conserved quantity at the initial configuration, before anything moves, costs nothing and is the whole of the defence.

It is worth being clear about what Cauchy’s theorem does not say, since the counterexample can be over-read. It does not say convex polyhedra are the only rigid ones — most non-convex polyhedra are perfectly rigid, and Bricard’s are a thin family. It says convexity is sufficient, so a convex polyhedron never needs checking. Dropping the hypothesis does not make a shape flexible; it makes the question open, and then it has to be computed.

What this has to do with crystals #

Three things, and the third is the reason the essay is in this collection rather than in a book about polyhedra.

The first is that a corner-sharing framework of rigid tetrahedra — a silicate, a zeolite, a perovskite — is a plate-hinged structure and not a bar-jointed one. The tetrahedra are the rigid faces and the shared corners are the hinges, and asking whether such a framework has a rigid unit mode is asking Cauchy’s question about a non-convex, infinite, periodic body. Convexity is unavailable, so the theorem gives nothing, and the answer has to be computed — which is what the mechanisms a count cannot see does for the periodic case.

The second is that the rank is the same instrument in both. Maxwell’s count is the difference between two integers and the rank is what actually happens; the gap between them is redundancy, and redundancy is why a count promises more motions than a structure has. Every number in the tables above is a rank rather than a count, and the two agree here only because these frameworks have no redundant bars.

The third is about hypotheses. Convexity is doing all the work in Cauchy’s theorem and nothing in the statement suggests how much: the theorem reads as though it were about polyhedra, and it is about convex ones, and the difference between the two is a shape that visibly moves. Every parallelohedron is a shadow of a cube meets the same pattern from another direction — Minkowski’s condition on a tiling body also assumes convexity, and dropping it admits shapes that tile in ways the classification does not describe.

The third of those is the one that keeps the counterexample honest. A rank computation that reported eleven for every octahedron would prove nothing about symmetry — it would prove the code had a bug — so a deliberately lopsided octahedron is run through the same routine and comes back at twelve. One shape moves and the other does not, from the same twenty lines.

One more comparison is worth drawing, because this collection has met the same question in a completely different language. Twelve pentagons and no way round them counts the faces of a closed surface by Euler’s formula, and the same formula is what makes a triangulated polyhedron isostatic: a triangulation on V vertices has 3V − 6 edges exactly, which is exactly the number of constraints a rigid body in three dimensions needs. So Euler’s formula and Maxwell’s count are the same arithmetic, and the fact that a triangulated sphere is exactly isostatic is a topological accident rather than a mechanical one.

There is a last observation about the two solids that fail as rods, and it is a fact about materials rather than about mathematics. A cube of rods has six freedoms and a dodecahedron of rods twenty-four, and in both cases the freedoms are exactly the face diagonals not present. So a structure built from polygons larger than triangles is soft in proportion to how many sides its polygons have, which is why every space frame anybody builds is triangulated and why a geodesic dome is made of triangles rather than of the hexagons its pattern suggests.

There is one more crystallographic reading, and it concerns what a framework analysis can and cannot promise. A mechanism that is a wave treats a periodic framework’s motions as modes with a wavevector, and the rank computation there is the same one done Fourier component by Fourier component. What that machinery inherits from this essay is the caution: a rank is an infinitesimal statement, and an infinitesimal mode need not extend to a finite motion. Silicates that are predicted to have rigid unit modes and turn out to be stiff are usually stiff for exactly that reason.

A last note on what the tables would look like for a solid that is not Platonic. Nothing in the computation uses regularity: the vertices go in as coordinates, the edges are the shortest pairs, the faces are the planes with everything on one side, and the rank is the rank. Running it on an irregular convex polyhedron gives the same two answers — flexible as rods unless triangulated, rigid as plates always — because Cauchy’s theorem does not mention regularity either. The five are here because they are familiar, not because they are special.

One number in the first table is worth a second glance because it is the only surprise in it. The octahedron is rigid as rods with rank exactly twelve against twelve — no redundancy, no slack — so removing any single bar from it leaves a mechanism. The icosahedron is the same: thirty of thirty. A triangulated sphere is exactly isostatic rather than comfortably rigid, which means every bar in it is load-bearing and none is spare. That is an unusual state for a structure to be in and it is the state every triangulated closed surface is in.

Where this stops #

The flex here is followed numerically rather than solved. Bricard’s octahedra have closed-form parameterisations and this collection does not carry them; what is computed is a path on the constraint surface, stepped along the kernel and projected back, and the evidence that it is a genuine flex is that the edge lengths hold to the arithmetic’s own precision over forty steps. That is strong evidence and it is not a proof, and the difference is worth keeping.

And a self-intersecting polyhedron is not a polyhedron in the sense a reader may have in mind. Bricard’s octahedra pass through themselves. Steffen built an embedded flexible polyhedron in 1978 — nine vertices, fourteen faces, no self-intersection — and it is the object the bellows theorem is really about. Building it needs a specific set of edge lengths rather than three arbitrary points, and it is not here.