The densest lattice in the plane

Put a disc at every point of a lattice, as large as fits without overlap. What fraction of the plane is covered?

The question has an obvious guess and the guess is right: the hexagonal lattice, at π/√12 ≈ 0.9069. What is interesting is that the argument finishes — there are infinitely many lattices and the search is over a bounded region with an answer at its corner.

Turning infinitely many lattices into a region #

A lattice is described by a positive definite quadratic form a x² + b xy + c y², whose value at (x, y) is the squared length of the corresponding lattice vector. The disc’s diameter is the lattice’s shortest vector and the cell’s area is √(4ac − b²)/2, so the fraction covered is

$$\text{density} = \frac{\pi\,a}{2\sqrt{4ac-b^2}}$$

provided a is the minimum, which is exactly what reduction guarantees. Every lattice has one and only one reduced form, with 0 ≤ b ≤ a ≤ c.

Two more observations close the region. Scaling changes nothing — enlarging a lattice enlarges the discs to match — so a may be fixed at one. And the reduction conditions then say 0 ≤ b ≤ 1 ≤ c.

That is the whole space of plane lattices, and it is two parameters wide. A question about infinitely many objects has become a question about a region — which is what reduction is for, and the same move that makes a crystallographic database able to ask whether two reported cells are the same cell.

The square lattice is the left edge of that region and the hexagonal one is its corner, which is worth locating before the argument begins. A square lattice has $b = 0$: its two basis vectors are perpendicular, so their dot product vanishes, and the reduced form is $x^2 + y^2$. Sliding $b$ from zero up to one shears the square towards the hexagon without changing either basis vector’s length, and at $b = 1$ the two vectors and their difference are all the same length. Everything in the domain is a stage of that shear crossed with a stretch, and the two named lattices a crystallographer would draw are its two extreme positions along one edge.

Why the corner #

The maximum is at b = 1, c = 1, and the reason is monotonicity in each variable separately.

Raising b raises the density. The shortest vector is unchanged — it is still the one of squared length a — and the cell’s area, √(4ac − b²)/2, falls. Same disc, smaller cell, more coverage.

Raising c lowers the density. The shortest vector is again unchanged and the area rises.

So no interior point can beat the corner, and the corner is a = b = c = 1 — the form x² + xy + y², which is the hexagonal lattice. Its density is π/(2√3) = π/√12, and the square lattice’s is π/4.

Two separate monotonicities are doing the work of one maximisation, and that is the whole economy of the argument. A function of two variables has no reason to attain its maximum at a corner of its domain; it might have an interior critical point, or a maximum on an edge, and finding out normally means differentiating and solving. Here neither variable ever wants to move inwards — the density rises with $b$ everywhere and falls with $c$ everywhere — so the search collapses to reading off the extreme value of each, and the two extremes are compatible because the domain is a rectangle in those coordinates rather than some awkward shape. Reduction is what made it a rectangle. Without the conditions $0 \le b \le a \le c$ the same density function ranges over every positive definite form, takes any value at all, and has no maximum to find.

Both monotonicities are checked at every sampled point rather than differentiated. That is a weaker argument than calculus and a much stronger one than looking at the picture, and it is the form this collection prefers: a claim that could fail and is tested where it might.

What the corner has that the others do not #

The winning lattice is the one where a = b = c, and in geometric terms that means the three shortest vectors are all the same length: the two basis vectors and their difference. Equivalently the lattice has six shortest vectors rather than four or two, so each disc touches six others.

That number is the kissing number, and its appearance here is not a coincidence. A packing gets dense by having each disc touch as many as possible, and the maximum in the plane is six — a seventh neighbour would need an angle under sixty degrees between two contacts, which two touching discs of equal size cannot have.

So the densest lattice is the one with the largest kissing number, and in the plane that implication runs both ways. In three dimensions it does not, which is the first sign that the higher-dimensional problem is harder than it looks.

Reading the numbers #

The five densities are worth having in front of one, because the spread is larger than the pictures suggest.

Hexagonal, 0.9069. Six touching neighbours, and the gaps are the small curved triangles between three mutually touching discs.

Square, 0.7854 = π/4. Four touching neighbours, and the gaps are considerably larger — the square hole between four discs has room in it that the hexagonal arrangement does not waste.

Centred rectangular at c = 2a, 0.5937. The cell is stretched, the discs are still limited by the short vector, and a sixth of the plane has gone.

Rectangular at c = 2a, 0.5554, and a general oblique lattice lower still.

The last two are worth noting for what they are not. They are not bad packings of the plane — nobody would arrange discs that way — they are what happens when a lattice is forced to have a particular shape by something other than the packing, which is exactly the situation in a crystal. A structure’s lattice is decided by its chemistry, and the packing fraction is then a consequence rather than a target. Which faces a crystal shows and how densely it packs are both read off a lattice that was chosen by neither.

Lattices are not all packings #

The result above is about lattice packings: arrangements where the discs sit at the points of a lattice. That is a real restriction, and the question of whether some irregular arrangement does better is a completely different problem.

In the plane the answer is that nothing does better. Thue proved it in 1910 and Fejes Tóth gave the first complete proof in 1943: no packing of equal discs, lattice or otherwise, exceeds π/√12. The argument is not the one above — it works by assigning each disc a region of the plane and showing no region can be smaller than the hexagonal one — and it is substantially harder than the lattice case.

The lattice case is easy because reduction makes it finite. The general case has no such reduction: an arbitrary packing has no basis, no cell and no form, so there is nothing to be bounded.

Where the discs actually touch #

The formula above hides one step that is worth doing explicitly, because it is where the geometry enters.

The disc’s radius is half the shortest vector, because two discs at neighbouring lattice points touch when their radii sum to the distance between them. Making the radius any larger makes the two nearest discs overlap; making it smaller wastes room. So the radius is fixed by the lattice, and the only quantities in the problem are the shortest vector and the cell area.

That is why reduction is the right tool. Reduction is exactly the algorithm that finds the shortest vector, and its output is a basis whose first vector is that shortest one. A method for the packing question and a method for the basis question are the same method, which is not obvious in advance — the packing question is about area and the basis question is about arithmetic.

The consequence is a sharper statement of the result. Writing μ for the shortest vector’s length and A for the cell’s area, the density is πμ²/(4A), so the densest lattice is the one maximising μ²/A — the ratio of what a lattice keeps apart to how much room it takes. That quantity is called the Hermite invariant, and its maximum over all lattices of a given dimension is the Hermite constant. In the plane it is 2/√3, and the essay above is the computation of it.

What happens in space #

The same distinction, one dimension up, is one of the famous problems of mathematics.

The densest lattice packing of spheres is the face-centred cubic, at π/√18 ≈ 0.7405. Gauss proved it in 1831 by the same kind of argument as above — reduction of ternary forms, a bounded region of reduced forms, and a maximum found in it. The region is larger and the case analysis longer, and the argument still finishes.

The densest packing of any kind is the Kepler conjecture, stated in 1611 and open until Hales in 1998. His proof is a reduction to a large but finite optimisation, checked by computer, and its formal verification took a further decade. That the general case took four hundred years while the lattice case took Gauss an afternoon is the strongest possible statement of what a reduction theory is worth.

This collection has already met the answer from the other side: cubic and hexagonal close packing both achieve π/√18, they differ in stacking rather than in density, and the hexagonal one is not a lattice packing at all — its spheres do not sit at the points of a lattice. The optimum is achieved by infinitely many arrangements and by exactly one lattice.

The same argument, three times, in this collection #

The move that makes this essay work has been made twice before here and is worth naming as a pattern.

A classification of infinitely many objects becomes finite when the right equivalence is imposed and a canonical representative is available.

For lattices the equivalence is congruence, the representative is the reduced form, and the region is the one drawn above. That is this essay.

For arithmetic classes the equivalence is conjugacy in GL(2,ℤ), the representative is found by a search over conjugating matrices, and the answer is thirteen.

For the seventeen the equivalence is affine conjugacy and the representative is the standard setting, and the finiteness comes from the average that bounds a finite group.

In each case the reduction is the theorem and the enumeration is the bookkeeping. What makes a classification hard is never counting the answers; it is proving that the list of candidates is complete, and a canonical form is what does that.

What the round trip checked, and how #

An unreduced form must be refused. For a form that is not reduced, the leading coefficient is not the lattice’s minimum — 5x² + y² has minimum one and leading coefficient five — and using it as the disc’s diameter reports a density of 1.76, which is not a fraction of anything. The function refuses the form rather than returning the number.

Reducing it must give the right answer. The same lattice, reduced, has the density its shape deserves; and x² + 2xy + 2y², which is the square lattice written in another basis, must come out at π/4.

A form that is not positive definite is not a lattice, and must throw.

The sampled maximum must be at the corner, not merely near it.

And the density must be monotone, which is the difference between “the corner is the best point sampled” and “the corner is the maximum”.

Higher dimensions, where the intuition fails #

The Hermite constant is known in dimensions one to eight and in twenty-four, and nowhere else. That list is worth staring at.

Up to eight the densest lattices are the ones a reader would guess: the hexagonal in two, the face-centred cubic in three, and a family culminating in E₈, the lattice this collection has already met as half of a pair with the same lengths. Viazovska proved E₈ optimal — among all packings, not merely lattice ones — in 2016.

In twenty-four the answer is the Leech lattice, proved optimal the same year by Cohn, Kumar, Miller, Radchenko and Viazovska.

In between and beyond, nothing is known, including in dimension nine. The reason is not that the problem changes character; it is that the reduced domain grows and the corner stops being obvious. In high dimensions the densest known packings are not lattices at all, and there is reason to believe the optimum is not a lattice — which inverts the plane’s situation completely.

A crystallographer meets none of this, and the reason the numbers are here is that they measure how special the plane and space are. Two dimensions have one answer, three have one answer with two arrangements achieving it, and past eight the subject stops giving answers.

Where the exactness stops #

The proof is of the lattice case only, and the general case is quoted. Nothing here rules out an irregular packing beating π/√12 — Thue and Fejes Tóth do that, by an argument this collection does not carry.

The monotonicity is sampled rather than differentiated. The derivatives are elementary and could be written down; testing on a grid catches any failure the grid is fine enough to see and would miss a failure confined to a narrower region. Given that the function is a smooth explicit expression, that is a small risk taken knowingly.

Reduction is the whole argument and reduction is two-dimensional here. In three dimensions the reduced domain is five parameters wide with several inequalities, and the same essay would be a much longer one; nothing above indicates how it goes.

The five drawn lattices are five points of a continuum, not five types. The oblique row of the figure is one oblique lattice among infinitely many, and its density is a value rather than the value. Only the hexagonal and square entries are canonical, because those are the two lattices the reduced domain has corners and edges at.

And density is one measure among several. The hexagonal lattice is densest, and it is not the best for every question a materials scientist asks — the coordination number, the thinnest covering and the best lattice for transmitting a signal are different optimisations with different winners even in the plane.

What a crystal does with any of this #

The plane’s answer is visible in real materials constantly, and the reason is not that atoms are trying to pack densely.

A close-packed layer is the hexagonal lattice, and it is what a layer of equal spheres does when nothing else constrains it — which describes the close-packed metals, the noble-gas solids and the layers that stack into cubic and hexagonal close packing. The density there is a consequence of the atoms being nearly spherical and nearly equal, and it fails as soon as they are not.

An ionic crystal is not close packed and is not trying to be. Its structure is decided by charge balance and by the ratio of the two ions’ radii, and the resulting lattices — rock salt, caesium chloride, fluorite — have packing fractions well below π/√18 with no waste involved. What is being minimised is energy, and density is only one term in it.

And a molecular crystal packs the way its molecules can. Molecules are not spheres, and the relevant statement is Kitaigorodskii’s: molecular crystals adopt the space groups that let awkward shapes interlock, which is why P2₁/c is so common. Density matters there too, and what is being packed is not discs.

So the theorem above is exact, general and applies to almost nothing directly. It is the baseline that says what the geometry permits, and every real structure’s deviation from it is a measurement of what the chemistry demanded instead.

Where the ladder goes next #

The packing thread has now measured the densities in space, counted the polytypes between them and decided the plane’s optimum. What runs through all three is that a question about infinitely many arrangements becomes finite once the right equivalence is imposed — which is the move this whole collection is built on, and which its aperiodic essays are the standing exception to.

Lattices are not all the packings #

The argument here is complete and it answers a narrower question than the one usually asked, so the boundary is worth drawing.

Nothing required a packing to be a lattice. Discs may be placed anywhere at all, with no repetition and no rule, and such an arrangement has a density too — defined as a limit over large regions rather than as a ratio inside a cell.

The space of those arrangements is not two-dimensional. It is not finite-dimensional in any useful sense, so the argument on this page — reduce, bound the region, check monotonicity in each variable, take the corner — has nothing to run on.

The answer is the same and the proof is different. No packing of equal discs in the plane, periodic or not, exceeds the hexagonal density. Thue claimed it in the 1890s and the first proof accepted as complete is Fejes Tóth’s, four decades later — and it works by an averaging argument over the Voronoi cells of the packing rather than by optimising over a region.

So this essay’s result is the lattice case, which is worth having on its own: it is exact, short, and checkable, and it is the case a crystal actually realises, since a crystal is periodic by definition.

Three dimensions, and two dimensions nobody expected #

The same pair of questions in higher dimensions has a famously uneven history.

In three dimensions the lattice case was settled by Gauss. The densest lattice packing of spheres is the face-centred cubic, at a density of about 0.7405, and the argument is a higher-dimensional version of the one here.

The general case took another hundred and seventy years. That no packing of equal spheres beats it is Kepler’s conjecture of 1611. Hales announced a proof in 1998 that ran to hundreds of pages with substantial computation; it was published after years of review with the referees unable to certify it completely, and a machine-checked formal proof was completed in 2014.

And then two dimensions fell out of nowhere. In 2016 Viazovska proved that the $E_8$ lattice is the densest packing in eight dimensions, by constructing a single auxiliary function with prescribed properties — a few pages rather than a few hundred — and the method extended within weeks to twenty-four dimensions and the Leech lattice.

Which leaves the sequence in a strange state. Dimensions one, two, three, eight and twenty-four are settled; four through seven are not, and nor is anything above twenty-four. The two solved exceptional dimensions are the ones with an exceptionally symmetric lattice available, and the pattern that makes them tractable is the same one that makes them interesting.