The lattice that minimises a sum

The densest lattice in the plane is the hexagonal one, and the argument for it turns on a single quantity: the shortest vector, whose length is the diameter of the largest disc that fits. Nothing beyond the first shell of neighbours enters the question at all.

That makes the answer a little suspicious as a general statement about which lattice is best. Change the question so that every distance matters — sum an inverse power of the length over all the non-zero vectors, at fixed cell area — and there is no obvious reason for the same lattice to win. A lattice with an excellent shortest vector might be paying for it further out.

It does not. The hexagonal lattice minimises that sum too, at every exponent this computes, and the computation is worth doing carefully because it is the kind whose answer is decided by a truncation if nobody looks.

The sum, and the scaling that makes it a question #

For a lattice L and an exponent s, the quantity is

$$E_s(L) = \sum_{v \ne 0} |v|^{-2s}$$

with the sum over every non-zero lattice vector. It converges for s > 1 in the plane, since the number of vectors within a radius grows as the area — which is the same counting the theta series makes shell by shell.

The scaling is not a detail. An unscaled sum is made small by making the lattice sparse, so without a constraint the answer is “spread it out” and the question is empty. Fixing the cell area makes the comparison one between shapes, and that is what the moduli region parameterises: every plane lattice is one point of it, and a fixed area leaves the shape as the only variable.

The quantity has physical readings — an interaction energy summed over pairs, for an inverse-power interaction — and this essay does not develop them. What is computed is a sum over a lattice, and its minimiser — the physics of whether any material minimises such a sum is a question this collection does not enter, and permission is not presence applies to a minimiser as much as to a property.

What is being asked that packing does not ask #

Packing asks a question about the first shell: which lattice, at fixed area, has the longest shortest vector. Every lattice with the same shortest vector is equally good, and the further shells are irrelevant.

The sum asks about all of them, with a weighting: at large s the near vectors dominate and the sum is nearly a question about the first shell again; at s close to one the far vectors matter as much, and the answer could in principle be different.

So the interesting range is small s, and the interesting fact is that the answer does not change there. The hexagonal lattice minimises at s = 6, where the question is nearly the packing question, and at s = 3/2, where it is not.

That is not obvious and it is not this collection’s theorem. Rankin proved it for s > 1 in 1953, and Montgomery proved the much stronger statement in 1988: the hexagonal lattice minimises every completely monotone function of the squared distances at once, which contains all these exponents as special cases and a great deal more. What is done here is a measurement, over the same region.

The truncation, and the one row it decides #

A sum over infinitely many vectors is computed over finitely many, and the difference is the tail. Reporting the tail is the difference between a measurement and an assertion.

The vectors beyond a radius R contribute about $\pi R^{2-2s}/(s-1)$, which falls quickly for large s and slowly for s near one. At s = 2 the tail is about 5 × 10⁻³ against a gap between the two lattices of 2.4 × 10⁻¹ — two orders of magnitude of margin, so the comparison stands. At s = 6 the tail is 10⁻¹⁴.

At s = 3/2 it is 2.6 × 10⁻¹ against a gap of 1.5 × 10⁻¹. The tail is larger than the difference being measured, so this computation does not decide that case, and it says so rather than reporting the winner.

That row is the reason the tail is computed at all. A routine reporting five wins out of five would be reporting a fact about its cutoff in one of them, and the honest output is four decided rows and one undecided.

The scan, and where its answer sits #

Beside the two named lattices, the whole region is scanned: every shape, at equal area, evaluated and sorted.

The minimum sits at the corner of the region, which is the hexagonal point — and it sits at the corner for every exponent tried. That is a weaker statement than comparing two lattices, because a grid scan can only locate a minimum to its own spacing, but it is a different statement: it says nothing else in the region beats the hexagonal lattice, which the two-lattice comparison does not.

One feature of the region needs saying or the scan reads wrong. The hexagonal lattice appears at both lower corners — (±½, √3/2) are the same lattice on two bases — so a scan reporting a minimum at the left corner has found the hexagonal lattice, not something else. A check comparing the found minimum to one corner only would report the right answer as wrong, and did, until the region’s own identifications were taken into account.

A worked comparison, in numbers #

At s = 2 the two sums come out 5.7787 for the hexagonal lattice and 6.0225 for the square one, both at unit cell area and both truncated at the same radius. Cut at a different radius the totals move in the third decimal — 5.7606 and 6.0004 out to squared length a hundred and thirty-five — and the margin between them does not: four per cent either way, which is exactly what a truncated comparison has to demonstrate rather than assume.

Where the difference comes from is worth following, because it is not where a reader expects. The hexagonal lattice has six shortest vectors and the square has four, so the first shell contributes more to the hexagonal sum than to the square one — 4.5 against 4.0 at equal cell area. The hexagonal lattice is losing at the first shell.

It does not lose for long, and the usual explanation of how it recovers is wrong. The story told is that the square lattice’s shells are the more crowded further out, so its later contributions pile up. They are not. Both lattices have about πR² vectors inside a radius R at equal cell area — that is what a lattice is — so neither can be the sparser one for long, and the running counts cross back and forth for ever. Counted over ninety-two radii out to squared length a hundred and thirty-five, the square lattice has more vectors inside the radius at thirty-four of them and the hexagonal lattice has more at forty-nine, with nine ties.

What decides the sum is not how many vectors there are but where they sit. The square lattice’s second shell arrives at twice its first squared length; the hexagonal one’s at three times. So the square lattice’s second shell contributes 1.00 where the hexagonal one’s contributes 0.50 — and after the square’s second shell the running totals are 5.00 against 4.50, with the hexagonal lattice already ahead and never behind again. The four per cent it wins by is settled in the first three shells and confirmed by all the rest.

That is why the question is not the packing question in disguise. Packing counts only the first shell, where the hexagonal lattice is behind on the number of shortest vectors and ahead on their length; the sum counts all of them; and the two agreeing is a fact rather than a restatement.

The region, and why a scan over it is enough #

Scanning “every plane lattice” sounds like an infinite task and is not, and the reason is the moduli region this collection built earlier.

Every lattice, at fixed area, is one point of a region bounded by |x| ≤ ½ and x² + y² ≥ 1 — and exactly one point, since the identifications on the boundary are known. So a scan over that region visits every lattice once, and no lattice is missed or counted twice.

Without the region the scan would have to range over all bases, which is an infinite set with each lattice appearing infinitely often, and a minimum found in it would say nothing about lattices. The reduction that brings a basis into the region is the same reduction as everywhere else in this collection, and it is what makes a search over shapes finite.

That is a general remark about this site’s method rather than about this essay: a great many questions about lattices become searches once the space of lattices is parameterised, and parameterising it is the work.

Below one, where the sum is not a sum #

At s ≤ 1 the sum does not converge, and a routine that returned a number for it would be reporting its own cutoff.

The check is direct: quadruple the cutoff radius and see what happens to the answer. At s = 2 the sum changes by three parts in a thousand — it has converged. At s = 0.9 it changes by a factor of 1.69 and keeps growing, because the number of vectors within a radius grows faster than their contributions fall.

So the machinery reports the growth rather than a value, and no lattice is named below s = 1. That is the correct behaviour and it is worth having as a refusal: a question about which lattice minimises a divergent sum is not a question, and the arithmetic should say so instead of comparing two infinities that happen to have been cut off at the same place.

The same lattice, and different questions #

The hexagonal lattice is this collection’s recurring answer, and the questions it answers are not all the same question. Setting them side by side is the useful summary.

Packing — the largest discs that do not overlap. Hexagonal, and the argument needs only the shortest vector.

Covering — the smallest discs that leave no gap. Also hexagonal, and this collection has an essay on the fact that packing and covering do want different lattices in higher dimensions, where the plane’s coincidence stops.

Summing — the least total of an inverse power over every vector. Hexagonal again, at every exponent, and now the whole lattice is involved.

Three questions with three different arguments, and the same winner in the plane. In three dimensions they come apart: the densest packing is not the best covering, and which lattice minimises an energy sum depends on the exponent. The plane’s habit of giving one answer to every question is a fact about the plane and not a general principle, and this collection has been careful to say so each time.

What “universally optimal” means, and what it does not #

Montgomery’s theorem is the strong statement, and it is worth quoting precisely because it is easy to over-read.

It says the hexagonal lattice minimises Σ f(|v|²) over lattices of fixed area, for every completely monotone f — a function whose derivatives alternate in sign, which includes e⁻ᵗ, t⁻ˢ for s > 0, and their positive combinations. So one lattice minimises an infinite family of energies simultaneously, which is a much stronger statement than minimising any one of them.

What it does not say is that the hexagonal lattice minimises every energy. A function that is not completely monotone — one with a well at some distance, which is what a real interatomic potential has — can be minimised by something else entirely, and real materials crystallise in structures that are not close-packed for exactly that reason.

Nor does it say anything about three dimensions, where the corresponding statement is a much harder theorem, proved for dimensions 8 and 24 in 2019 and open in 3. The plane’s case is the easy one and it is still not elementary.

Why the answer is stable, and where it would not be #

The measurement’s most striking feature is how little the minimiser moves: the same lattice at s = 3/2 and at s = 6, exponents differing by a factor of four in how sharply they weight the near vectors.

The reason, in the plane, is that the hexagonal lattice is simultaneously the best at every scale — its shells are as evenly spread as a lattice’s can be, and any deformation towards the square lattice both shortens the first shell relative to the area and crowds the later ones. There is nothing to trade, so no exponent finds a better compromise.

Where that reasoning fails is in higher dimensions, and it fails for a reason worth stating. In eight dimensions and in twenty-four the analogous lattices are again optimal for every such energy, and those cases were settled in 2019 by a proof of a quite different character; in most other dimensions the minimiser is unknown, and in some it is expected to depend on the exponent. The plane’s stability is a small-dimension phenomenon.

So the right reading of the measurement here is not that a lattice minimising one energy minimises all of them. It is that this particular lattice, in this particular dimension, happens to leave nothing to trade — and the theorem saying so is harder than the measurement showing it.

What is measured and what is imported #

Measured. The sums themselves, at five exponents, over a fixed radius, with the tail estimate beside each. The scan over the region and where its minimum sits. The divergence below s = 1, by quadrupling the cutoff.

Imported. Rankin’s theorem, and Montgomery’s. Neither is proved here, and the measurements are consistent with both.

Left open. The row at s = 3/2, where the truncation is too coarse to decide. Extending the sum would settle it, and the honest report is that this computation does not.

The sum is a modular form, which is why the region is the right domain #

The scan runs over the moduli region because every lattice appears there exactly once. There is a second and stronger reason the region is the natural home for these sums, and it explains why the answer is so stable.

Sum $e^{-\pi t |v|^2}$ over a lattice and the result is its theta series. The Epstein zeta function this essay computes is the same object seen through a Mellin transform — an integral over t of the theta series against a power — so the two carry the same information and the sums at every exponent are read off one function.

A theta series of a lattice is a modular form. As a function of the point τ in the upper half-plane, it is invariant under the changes of basis — the same maps τ ↦ τ + 1 and τ ↦ −1/τ that generate the identifications this collection’s moduli region is a fundamental domain for. So the sum is not merely a function that happens to be constant on lattice classes; it is a modular function, and the region is its natural domain rather than a convenience.

That is where the stability comes from. A modular form on this region has its critical points constrained by the region’s own symmetry, and the two corners are the points fixed by something — so the minimiser being at a corner is what one should expect, and which corner is the only thing at issue. A minimiser drifting to a generic point as the exponent changed would be the surprising outcome, and it is the one the measurement rules out.

The problem this is not solving #

There is a physical question in the neighbourhood, it is the one a reader is likely to think is being answered, and it is open.

The sum here is over the vectors of a lattice, minimised over lattices. The physical question — the crystallisation problem — asks something harder: given particles interacting through a pair potential, and no assumption of periodicity at all, do their lowest-energy arrangements form a lattice?

That is not what has been shown. Restricting to lattices assumes the answer to the interesting half of the question, exactly as restricting a packing question to lattice packings assumes away the hard part that Hales’s proof settled.

Some cases are known. For particles in the plane with certain short-ranged potentials, the lowest-energy arrangements have been proved to be the triangular lattice, by arguments about the local geometry rather than about sums. In three dimensions essentially nothing is proved, and the statement that atoms of a single element arrange themselves periodically — the observation the whole of this collection is built on — has no proof from a potential.

That is worth stating plainly at the end of an essay whose figures show a clean minimum. The minimum is over a one-parameter family of lattices and is exact; the claim a reader might take from it is over all arrangements and is unproved.

What the pictures cannot show #

A convergent sum is not visible. The figures shade the region by the value of a sum, and every shade is a number computed to a finite radius. What a picture cannot show is that the shading would be the same at twice the radius, which is what the tail estimate is for.

The region’s identifications are not drawn. The two lower corners are the same lattice, and the vertical edges are identified with one another; the figure draws a region rather than the surface it becomes when those identifications are made. This collection’s moduli essays draw the identifications, and reading this figure without them makes the hexagonal lattice look like two lattices.

And no material appears anywhere. The sum is a sum over a lattice. Reading it as an energy needs an interaction, and reading a minimiser as a predicted structure needs a great deal more than that — which is why this essay’s claims stop at the arithmetic.