One perfect form in space

The densest lattice in the plane is settled here by an argument that does not survive a change of dimension. Three circles of the same size touching each other leave a curved triangle between them; the fraction of a triangle they cover is fixed by the angles, and tiling the plane with those triangles gives π/√12. It is a good argument and it is two-dimensional through and through.

The same question in space has no such argument, and the history is a warning about how hard it is. Kepler’s conjecture — that no packing of equal spheres beats the face-centred cubic arrangement — was open from 1611 to 1998, and the proof that closed it runs to three hundred pages with a computer verification attached. The lattice case is a completely different matter. That one was settled by Gauss in 1831, and this essay is about the machinery Voronoi built around it seventy years later, which turns a maximisation over a continuum into two finite tests.

What “densest” means when only lattices are allowed #

A lattice packing places one sphere at each lattice point, all the same size, as large as they can be without overlapping. The radius is then half the length of the shortest non-zero lattice vector, and the density is the sphere volume divided by the volume of one cell.

Writing that in terms of the Gram matrix G of a basis makes it scale-free. Let m be the minimum — the smallest value of vᵀGv over non-zero integer v — and let det G be the determinant, whose square root is the cell volume. Then

γ(G) = m / (det G)^{1/n}

is unchanged by scaling the lattice and by changing its basis, and the density is an increasing function of it. This is Hermite’s invariant, and maximising it over all lattices of a given dimension is the whole problem.

The awkwardness is that m is a minimum over infinitely many integer vectors, so γ is a minimum of infinitely many smooth functions and is therefore not smooth. It has corners exactly where the set of shortest vectors changes, and the maximum sits on one of those corners rather than anywhere calculus can reach. Voronoi’s two conditions are what replace the derivative.

Perfection is a rank #

The first condition asks whether the shortest vectors determine the lattice.

Suppose the lattice is deformed a little — G moves to G + E for a small symmetric E. Each shortest vector v changes length by vᵀEv, and that expression is linear in E. Keeping every shortest vector at its length is therefore a system of linear equations on E, one per vector, and the question is whether that system has a solution other than zero.

Writing vᵀEv as the inner product of E with the rank-one matrix vvᵀ turns the question into a rank. The space of symmetric matrices has dimension n(n+1)/2 — three in the plane, six in space — and the system has only the zero solution exactly when the matrices vvᵀ, taken over the shortest vectors, span the whole of it. A form whose shortest vectors span that space is called perfect.

The counting is unforgiving and that is the point. A rank-one matrix from an opposite pair ±v is one matrix rather than two, so the number of pairs of shortest vectors is the most rank a lattice can hope for. In space that means a lattice with fewer than six pairs cannot possibly be perfect, whatever the pairs are. The primitive cubic lattice has three pairs, along the axes. The body-centred cubic lattice has four, along the body diagonals. Neither can span six dimensions, and the arithmetic stops before it starts.

That is a striking way to lose. The body-centred cubic lattice is not a bad lattice — it is the best covering lattice in space, the one whose spheres fill every gap with the least overlap, which is the other half of covering and packing wanting different lattices. It fails here on a count of vectors before any density is computed.

Which vectors, and how many #

The vector counts are worth seeing rather than taking on trust, and in the plane they can simply be drawn.

A rectangular lattice whose sides differ has exactly two shortest vectors, ±a along the shorter side. One pair, rank one, and the form can be deformed in two independent directions without shortening anything — lengthen the long side, or shear it. A square lattice has four shortest vectors in two pairs, rank two, and one direction of freedom remains: pull the square into a rhombus and the four stay equal while the determinant changes. A hexagonal lattice has six in three pairs and there is nothing left; any deformation shortens one of them.

Rank three is reached by exactly one plane lattice, and it is the dense one. That is not a coincidence — it is Voronoi’s theorem being visible in a case small enough to see whole — but it is not yet a proof either, because perfection alone does not distinguish a maximum from a saddle.

Eutaxy is a sign #

The second condition is where the maximum comes from. A perfect form is a corner of the function γ, and a corner can be a peak, a trough or a col. Which it is depends on whether the direction of increase points out of the region the corner bounds, and Voronoi’s way of asking that is a statement about the inverse.

The derivative of det G with respect to G is det G · G⁻¹. So the direction in which the determinant grows fastest — the direction that would make the lattice sparser without moving anything else — is G⁻¹, and the corner is a maximum exactly when that direction is blocked: when G⁻¹ lies inside the cone spanned by the shortest vectors’ rank-one matrices with positive coefficients. A form with that property is called eutactic.

For a perfect form the rank-one matrices span, so a solution always exists; the content is entirely in the signs. Both of the perfect forms found here come out with every coefficient strictly positive, and the residual of the fit is at the level of rounding, which is the check that the solution is a solution rather than a least-squares consolation.

Voronoi’s theorem, from 1908, is that a form is extreme exactly when it is perfect and eutactic. Nothing here proves it; what is computed is the two conditions, and the theorem is what makes computing them worth doing. That division is the usual one on this site — Chevalley’s theorem was used the same way, with both of its sides run on thirty-two classes and neither of them proved.

The search, and the bound it is inside #

Two conditions on a single lattice do not answer “which lattice”, because they are local. What closes the question is that in low dimensions there are very few perfect forms at all, and that can be established by enumeration.

A perfect form is determined up to scale by its shortest vectors, and scaling it so that the minimum is two makes it integral in the two dimensions at issue here. So the search is over integer Gram matrices: take every reduced one inside a bound, keep the positive definite ones whose minimum is two, and test each for perfection.

The sorting deserves a note, because the first version of it was wrong in a way worth recording. Grouping the perfect forms by the invariants that are cheap — the number of shortest vectors and the determinant — reported them as one class, and grouping by those plus a prefix of the theta series reported two. The theta prefix was the unreliable one: counting vectors up to a norm inside a fixed coordinate box misses vectors that lie outside the box in one basis and inside it in another, so two spellings of one lattice got different counts. The repair was to stop comparing invariants and start exhibiting the basis change — for each perfect form, search for an integer matrix U of determinant ±1 with UᵀG₁U = G₂ — which is a proof rather than an indication.

One class in the plane. One class in space. In the plane it is the hexagonal lattice; in space it is the face-centred cubic one, with twelve shortest vectors in six pairs, Hermite invariant exactly the cube root of two, and packing density π/√18 ≈ 0.7405.

Since each dimension has only one perfect form, and since the maximum of γ is attained and must be at a perfect eutactic form, the local maximum is the global one by elimination. That is the shape of the whole argument, and it is why perfection matters more than it looks: a condition that would otherwise only certify a local peak becomes a complete answer as soon as the peaks can be counted.

The pair that swaps places #

The two lattices that fail the rank test are not equally interesting failures, and the more interesting one is the body-centred cubic.

It is the dual of the face-centred cubic lattice, and the face-centred is the dual of it — a fact this collection checked from the shell counts when it established that every plane lattice is its own dual and almost no lattice in space is. In the plane, duality is invisible: every lattice is similar to its own dual, so a question about a lattice and a question about its dual are the same question. In space the two are genuinely different objects, and here they hold the two prizes.

The face-centred cubic lattice is the best packing. Its dual, the body-centred cubic lattice, is the best covering — the arrangement in which spheres large enough to leave no gap overlap least. That is not a theorem this essay proves; it is Bambah’s, from 1954, and it is the fact that makes the pairing worth remarking on rather than a curiosity.

The two questions pull opposite ways, and the rank test is where the pulling becomes visible. Packing rewards many shortest vectors, because each one is a contact between two spheres and contacts are what a dense arrangement has: twelve for the face-centred cubic lattice, and twelve is the kissing number in three dimensions. Covering rewards the opposite — a cell whose deepest hole is shallow — and the body-centred cubic lattice’s Voronoi cell is the truncated octahedron, the roundest of the five parallelohedra, whose furthest corner is nearer the centre than any other cell’s.

So the eight shortest vectors that make the body-centred lattice a good covering are exactly what makes it fail the rank test for packing: four pairs against the six that are needed. The failure is not a near miss. It is the same property, read as a virtue on one question and a defect on the other.

What perfection is measuring #

It is easy to read perfection as a technicality — a rank condition invented to make a proof work — and it is worth saying what it means, because it means something a crystallographer already knows.

A lattice’s shortest vectors are what a diffraction pattern’s strongest reflections come from, what determines which planes a crystal cleaves on, and what decides the first shell of neighbours around every atom. Perfection asks: do those vectors already determine the lattice, or is there a family of different lattices sharing them?

For a rectangular lattice the answer is that they do not, and dramatically so. The shortest vectors are one opposite pair, and every lattice obtained by stretching or shearing in the other direction has the same first shell. Knowing the shortest vectors of such a lattice says almost nothing about it. For the hexagonal lattice the answer is that they do: the three pairs fix the shape entirely, and any lattice with those six vectors as its shortest is that lattice.

That is the same question the lengths do not name the lattice asks in a harder setting, where two lattices agree at every shell and are still different. Perfection is the easy end of it — agreement at the first shell only — and the answer there is that agreement at the first shell pins the lattice down for very few lattices, and that those very few are the dense ones.

There is a pleasing consequence for the arithmetic of the search. Because a perfect form is determined by its shortest vectors, and because those vectors are integer vectors once a basis is chosen, a perfect form is rational — its Gram matrix is a rational multiple of an integer one. That is what makes an integer search complete rather than a sampling: there is no perfect form hiding between the integer points, because a perfect form cannot have an irrational shape.

What has and has not been established #

The bound is real and is stated in the figure. What the search establishes is that no other reduced integer form of minimum two inside that bound is perfect. Extending the bound is arithmetic rather than insight, and Voronoi supplied an algorithm that removes the bound entirely: from one perfect form, its perfect neighbours can be constructed, and the process closes on a finite list. That algorithm is not implemented here, and the honest statement of what is above is a bounded enumeration together with the two conditions computed exactly.

Three other limits are worth stating plainly.

Nothing here is about non-lattice packings. The face-centred cubic lattice is the densest lattice packing in space, and that has been the theorem since Gauss. Whether some irregular packing beats it is Kepler’s conjecture, settled affirmatively in the negative direction — nothing beats it — by Hales in 1998 with an argument this collection does not carry, and the same gap the plane case records for Thue’s theorem.

Nor is hexagonal close packing a counterexample. It has exactly the same density, and it is not a lattice: it needs two points per cell, so it is a lattice with a basis rather than a lattice, and it is invisible to every computation above. That two arrangements of such different symmetry share a density to the last digit is the subject of two stackings, one density, and the reason is that density is a local quantity — each sphere touches twelve others in both, and the stacking order beyond the first neighbours does not enter.

And the dimension matters more than it looks. Perfect forms are one in dimension two, one in dimension three, two in dimension four, three in dimension five, seven in dimension six, and by dimension eight there are ten thousand nine hundred and sixteen. The method does not degrade; the list does. What makes two and three answerable in an afternoon is that the answer is a list of length one.

Why a crystallographer might care #

The face-centred cubic lattice is the arrangement of atoms in copper, aluminium, gold, silver, nickel and lead, and the reason usually given is that metals with non-directional bonding pack as tightly as possible. That is a physical claim about energies, and this collection does not compute energies — permission is not presence, and a lattice being densest is a permission.

What the argument above supplies is the other half: the geometry really does have a unique answer, and it is not close. There is not a family of near-optimal lattices among which a material might choose on some other ground. There is one perfect form in space, everything else is strictly below it, and the runner-up among the highly symmetric candidates — the body-centred cubic lattice at 0.6802 — is nine per cent behind. When a metal that packs by density picks a structure, it is picking from a list with one item on it.

That is worth having beside the classification results elsewhere in this collection. Fourteen lattices says how many kinds of lattice there are; the holohedry is the ceiling says how much symmetry each can carry. Neither ranks them, because symmetry is not an ordering. Density is an ordering, and it has a maximum, and the maximum is a rank calculation away.