What six lengths decide and nine do not

The polyhedra that can flex builds an octahedron that moves while every edge keeps its length, and then finds that its volume stays constant throughout — and that the constancy is vacuous, because the algebraic volume of that particular octahedron is zero for a reason having nothing to do with the motion.

This essay asks the question that failure left open. If the edges cannot change, what about the volume is decided and what is not?

Six lengths and a determinant #

For four points the answer is complete, and it is a determinant.

Write dᵢⱼ for the squared distance between points i and j, and form the matrix with a border of ones and a zero in the corner. Its determinant — the Cayley–Menger determinant — is 288 V² for a tetrahedron. Nothing about the points appears except their mutual distances, so the volume of a tetrahedron is a function of its six edge lengths and of nothing else.

The alternating sign is worth pausing on, because it is not bookkeeping. For an n-simplex the determinant equals (−1)^(n+1) · 2ⁿ · (n!)² · V², so a triangle’s is positive and a tetrahedron’s is negative, and the squared content that comes out is a genuine square only when the sign is right. A value with the wrong sign is not a small volume or a numerical problem. It is a proof that the distances belong to no set of points at all.

The formula deserves a check against something independent, because a determinant is exactly the kind of object that can be transcribed slightly wrong and still produce plausible numbers.

The scaling check is the more informative of the two, because it can fail in a way no single example reveals. Multiply every squared length by a factor and the determinant must multiply by that factor’s cube, since a volume is a length cubed and the determinant is a squared volume in quantities that are themselves squared lengths. A formula with a wrong power would agree with the coordinates on nothing, but a formula with a wrong constant would agree on nothing either — and it is the two checks together, one about a value and one about a scaling law, that pin it.

There is a structural fact behind the scaling that is worth naming, because it is what Sabitov’s theorem generalises. The determinant is a polynomial in the squared edge lengths with integer coefficients — expanded, it is a sum of products of six of them at a time with coefficients from the set {−1, 1, 2} — so 288 V² is not merely computable from the lengths, it is a polynomial in them. That is a much stronger statement than “determined by”. It says the volume satisfies an algebraic equation over the lengths, and the whole content of the theorem discussed below is that some such equation survives when the shape stops being a simplex.

There is a second way to write the same object that makes the connection to the rest of this collection visible. Fixing one point as an origin, the squared distances determine the Gram matrix of the three edge vectors from it, and the volume is the square root of that matrix’s determinant divided by six. So the Cayley–Menger determinant is a Gram determinant with the choice of origin removed — which is why a lattice cannot have all its vectors long can speak of a lattice’s covolume as the square root of a Gram determinant and mean the same arithmetic. A lattice’s fundamental cell is a parallelepiped rather than a simplex, and the two differ by exactly the factorial that appears in the sign formula above.

What the inequalities miss #

The determinant does something the obvious test does not, and this is where it starts to earn its place.

The obvious test for whether six lengths are a tetrahedron’s is the triangle inequality on each of the four faces. It is necessary, and it is not sufficient, and the gap between the two is easy to exhibit.

Take a unit equilateral triangle and ask for a fourth point at distance r from each of its corners. The three side faces are then triangles with sides 1, r, r, legal as soon as 2r > 1. But such a point must lie on the perpendicular through the base triangle’s circumcentre, and its distance to a corner is the hypotenuse of the circumradius and the height. So it exists only when r exceeds the circumradius 1/√3 ≈ 0.5774, and for r between 0.5 and 0.5774 every face is legal and nothing is there.

That is a small gap and it is not a small point. It says that the local conditions — each face is a triangle — do not add up to the global one, and that assembling a solid out of legal pieces can fail for reasons no piece knows about. The determinant knows, because it is a statement about all four points at once.

There is a family resemblance here to a polyhedron is two properties of a graph, where a graph passes a local test on every vertex and fails a global one about connectivity. In both cases the local conditions are cheap and necessary, the global one is the theorem, and the counterexamples that separate them are small enough to draw.

The same shape of failure appears one level up, in the subject this ladder is about. The mechanisms a count cannot see finds frameworks whose freedoms a Maxwell count gets wrong, because the count is a statement about how many constraints there are and the truth is a statement about whether those constraints are independent. Here the triangle inequalities are the count and the determinant is the rank: four legal faces are four satisfied constraints, and whether they can be satisfied simultaneously is a different question with a different answer. In both cases the cheap test is a sum and the correct test is a determinant, and the gap between them is where the interesting examples live.

Where the lengths stop deciding #

Four points are the last case where the lengths settle it, and the failure at five is not subtle.

Glue two tetrahedra on a common triangle, with apexes at heights h above and k below its plane. Now reflect the lower apex through that plane. Its three distances to the base corners are unchanged, because reflection is an isometry and the base triangle is fixed by it; the base’s own edges are untouched; the combinatorics — which triangles meet along which edges — is the same list it was. So the two shapes have the same edge lengths and the same faces, and their volumes are (h + k) and (h − k) times a third of the base area. With h = 0.9 and k = 0.4 that is 0.1876 against 0.0722, a ratio of 2.6.

One honest caveat belongs immediately with the example. Only the first arrangement is an embedded solid: with both apexes on the same side, the smaller tetrahedron sits inside the larger and the surface passes through itself. That is what makes the counterexample cost nothing to construct, and it is exactly why the harder examples matter historically — an embedded polyhedron that flexes was not found until Steffen’s in 1978, sixty years after the flexible immersed ones.

The reflection generalises, and knowing how far tells how large the finite set can be. Any vertex of a triangulated surface that has exactly three neighbours can be reflected through the plane those three neighbours span: its three edge lengths are distances to points of that plane and reflection preserves every one of them. So a surface with several degree-three vertices has at least 2ᵏ arrangements with the same edge lengths, one for each subset of them reflected, and the volumes need not coincide. The glued tetrahedra have two such vertices, which is where the four arrangements came from. That construction gives a lower bound on the size of the finite set and says nothing about an upper one — the upper bound is the degree of Sabitov’s polynomial, and it is far larger.

The sweep is the more useful picture, because it shows the shape of the answer rather than one instance of it. As the lower apex moves, the two volumes move apart continuously and never become three or become an interval. That is the pattern, and it is a theorem: Sabitov proved in 1996 that the volume of any closed triangulated surface in space is a root of a monic polynomial whose coefficients are polynomials in the squared edge lengths. So for fixed edge lengths there are finitely many possible volumes — a root set, not a range.

Why that settles the bellows #

The bellows theorem falls straight out of the finiteness, and the argument is two sentences.

A flexing polyhedron moves continuously while every edge keeps its length. Its volume is therefore a continuous function of time taking values in a finite set. A continuous function into a finite set is constant.

That is the whole proof, given Sabitov’s theorem, and it explains why the result was hard: the difficulty is entirely in showing the polynomial exists, and none of it is in the deduction. It also explains why the polyhedra that can flex could not demonstrate the theorem by watching a volume stay constant. Watching a constant is compatible with the constancy being an accident of the example, which is what happened there — the algebraic volume of a line-symmetric octahedron is identically zero, before any motion is considered.

The right demonstration is the one here, and it is a demonstration of the other half. Constancy along a flex is the easy consequence; the content is that the possibilities form a finite set at all, and the smallest case where that set has more than one element is the pair of glued tetrahedra above. A single example with two volumes does more for the theorem than a hundred frames of a flex with one.

There is a converse worth stating because it bounds what the theorem claims. Finiteness does not say the volume is determined, and it does not say a polyhedron cannot be deformed at all. It says that a deformation which preserves every edge length either keeps the volume exactly or jumps — and jumping is not something a continuous motion does. So the theorem forbids gradual change and permits discrete change, which is exactly what the reflection above performs: the two arrangements are both there, and no motion through legal shapes connects them.

It is worth setting the theorem beside the other invariant this collection computes for polyhedra, because the two behave oppositely. The angle that is not a fraction of a turn computes the Dehn invariant, which is what volume alone fails to capture: two polyhedra of equal volume are cut-and-reassemble equivalent only when their Dehn invariants also agree. There, volume is the weak invariant and something else is needed. Here, volume is the thing that will not move, and the edge lengths are what fail to determine it. The two statements are about different operations — cutting, and flexing — and a quantity that is too coarse for one can be exactly the quantity the other conserves.

What this says about a structure #

Two things, and the second is about instruments rather than about crystals.

The first is that a structure determination which fixes bond lengths has not thereby fixed a volume. A framework of rigid units with known connectivity and known link lengths — a silicate, a metal–organic framework, any of the objects the count that promises a mechanism counts freedoms for — can have several distinct configurations with the same lengths and different cell volumes, and the number of them is finite but is not one. Reporting a volume therefore requires more than a length list, and the extra thing required is a choice among finitely many, which is a discrete piece of information a refinement can get wrong without any residual noticing.

There is a version of this that a reader of five solids from one inequality will recognise. That essay settles which regular solids exist by turning a geometric question into an integer inequality with finitely many solutions, and the answer is a list. The finiteness here has the same character and a different source: the shapes with given edge lengths form a finite list not because a count runs out but because a polynomial has finitely many roots. Both are the good case, where a continuum of possibilities collapses to something enumerable, and both are worth more than a bound would be.

The second is that the negative determinant is a usable test. Given a list of interatomic distances — from a refinement, from a model, from an assumed coordination polyhedron — the Cayley–Menger determinant of any four of them decides whether those four distances are geometrically possible, in one evaluation, before any structure is built. A set that fails is inconsistent, and the failure is exact rather than statistical. That test costs nothing and is not usually run.

The second of those is the one that keeps the instrument honest. Four coplanar points give a determinant of exactly zero, not a small number, and a formula returning 10⁻¹⁷ for a flat configuration would be a formula whose zeros could not be trusted. Since the whole realisability test is a question about a sign, a zero that is not exactly zero would put every boundary case in doubt.

One implementation note belongs with the refusals, because it is what makes the sign test usable. The determinant is evaluated by Gaussian elimination with partial pivoting rather than by expanding the permutation sum, which for a five-by-five matrix would be a hundred and twenty products and is where a hand-transcribed formula goes wrong. Pivoting matters for the boundary cases in particular: near the circumradius above, the determinant passes through zero, and an elimination that divided by a small pivot would report a large number of the wrong sign exactly where the answer is most delicate. The coplanar test in the refusals is the check on that — four points in a plane must give zero and not a small number, and they do.

Where this stops #

The polynomial is not constructed here. Sabitov’s proof produces one of degree 2⁴ = 16 in V² for the octahedron and much larger degrees for larger surfaces, and computing its coefficients is a substantial piece of elimination theory. What is computed here is the finite set in one case, by enumerating the arrangements directly, which is possible only because that case has four of them.

The counterexample is also immersed rather than embedded, as noted, and the embedded question is genuinely harder. Steffen’s polyhedron has nine vertices and twenty-one edges and flexes while remaining a surface in space with no self-intersection, and constructing it is a separate piece of work that this collection has not done.

One more limitation is worth stating precisely, because the essay’s headline could be read as stronger than it is. “Nine lengths do not decide” is a statement about nine lengths and a combinatorial type: it is not that the shape is undetermined up to rigid motion — the two arrangements really are different shapes — but that knowing every length and every face does not pick one out. A stronger set of data would: adding the three distances across the shared triangle to the far apex, which are not edges of this surface, distinguishes the two immediately. So the failure is a failure of a particular kind of data and not of geometry, and what Sabitov’s theorem adds is that the data still constrains the volume to finitely many values rather than leaving it free.

And nothing here touches dimensions above three. The Cayley–Menger determinant generalises immediately — it computes the content of an n-simplex from its n(n+1)/2 edge lengths at every n — but the bellows theorem is known in three and four dimensions and open in higher ones, which is the usual pattern: the formula extends and the theorem does not.