No lattice in space wins at every width

Put a Gaussian of width set by a parameter $t$ on every point of a lattice scaled to unit volume, and add them up:

$$\theta_L(t) = \sum_{v \in L} e^{-\pi t |v|^2}.$$

The sum is the lattice’s energy if its points repel one another through a Gaussian, it is the lattice’s theta series seen as a function rather than as a list of counts, and every inverse-power sum over the lattice is an average of it over widths. So asking which lattice has the smallest sum at a given width is asking which arrangement of points is most spread out, measured at that scale.

In the plane the question has one answer for every width. Among lattices of unit area the hexagonal one has the smallest sum at every $t$ — Montgomery’s theorem of 1988, which the lattice that minimises a sum measured through its consequence for inverse powers. The natural guess is that space behaves the same way, with some lattice — the face-centred cubic, most likely — best at every width.

It does not, and the reason is not a computation. No lattice in three dimensions has the smallest Gaussian sum at every width, and the proof uses two facts that are both on this page already: which lattice is densest, and the identity that relates a lattice’s sum to its dual’s.

The two ends of the width axis #

At large widths the sum is its shortest vectors. Each point contributes $e^{-\pi t |v|^2}$, so as $t$ grows every term except the origin’s is swamped by the terms from the shortest non-zero vectors: $\theta_L(t) - 1$ behaves like the number of shortest vectors times $e^{-\pi t \lambda^2}$, where $\lambda$ is their length. The lattice with the smallest sum at large widths is therefore the one whose shortest vector is longest at unit volume — the densest lattice packing of spheres. In space that is the face-centred cubic lattice and no other, which is Gauss’s theorem of 1831, the three-dimensional counterpart of the densest lattice in the plane.

At small widths the identity turns the question round. For a lattice of unit volume,

$$\theta_L(t) = t^{-3/2}\,\theta_{L^*}(1/t),$$

where $L^*$ is the dual lattice, also of unit volume. As $t$ shrinks, $1/t$ grows, and the sum over $L$ at a small width is the sum over $L^*$ at a large one, times a factor that is the same for every lattice. So the lattice with the smallest sum at small widths is the one whose dual is the densest — the dual of the face-centred cubic lattice, which is the body-centred cubic.

So a lattice with the smallest sum at every width would have to be both. It would have to be the densest lattice, to win as $t$ grows, and the dual of the densest lattice, to win as $t$ shrinks. That is possible only if the densest lattice is its own dual, and in space it is not: the face-centred and body-centred cubic lattices are different lattices, each the other’s dual up to scale.

That is the whole proof. It does not say who wins at any particular width in between; it says only that nobody wins everywhere, and it says it with no numerical input beyond Gauss’s theorem. The computation below is not needed for the conclusion — it is there to find where the change happens and to test what the proof leaves unsaid.

Why the plane escapes #

The same argument applied to the plane has nothing to push against. The densest lattice of the plane is the hexagonal one, and the dual of the hexagonal lattice is the hexagonal lattice turned through a right angle — the same lattice. So the large-width winner and the small-width winner coincide, and a single lattice is free to win everywhere. Montgomery’s theorem says it does.

In fact every plane lattice is its own dual up to rotation and scale, so in the plane the argument never bites for any lattice. Space is the first dimension where it can, and it does at once, because the one lattice everybody would nominate is the one that is not self-dual.

Where the two cubic lattices change places #

The proof says fcc wins at large widths and bcc at small ones. It does not say where they change places, and the identity answers that too. Applied to fcc, whose dual is bcc,

$$\theta_{\text{fcc}}(t) = t^{-3/2}\,\theta_{\text{bcc}}(1/t),$$

and at $t = 1$ the factor is one: the two sums are equal at the self-dual width, exactly. Computed directly, at unit volume, they agree to fifteen significant figures there. Below $t = 1$ the body-centred lattice’s sum is smaller; above, the face-centred lattice’s is. On a grid of widths from 0.35 to 2.8 the sign of the difference never changes except at one.

The crossing is one of the few exact facts about which lattice is best at a finite width. Everything else — that fcc stays best for every width above one, and bcc for every width below — is believed and not proved. The computation checks it on the grid, along the path between the two lattices, and against random lattices near each; none of that is a proof, and it is said to be none.

The path between them #

The face- and body-centred cubic lattices lie on one family: the body-centred tetragonal lattices, a square base with a point in the middle of each cell, and a height that can vary. At a height ratio of one it is body-centred cubic; at $\sqrt2$ it is face-centred cubic. Squashing one into the other along this family is the Bain path, the route iron takes when it transforms between its two cubic structures — body-centred at room temperature, face-centred above 912 °C. Nothing here explains iron, whose preference is set by its electrons and not by a Gaussian repulsion; the path is borrowed because it is the one continuous family on which both cubic lattices sit, and so the natural place to watch them trade places.

At width 0.8 the lowest point of the path is the body-centred end, and at 1.25 the face-centred end. At the self-dual width the two ends tie exactly, and the curve between them rises to a maximum near a height ratio of $2^{1/4} \approx 1.19$. That point is not arbitrary: the body-centred tetragonal lattice at ratio $r$ has as its dual the one at ratio $\sqrt2/r$, up to scale, so the path is carried onto itself reversed by duality, and the lattice at $2^{1/4}$ — the geometric mean of the two ends — is its own dual. At the self-dual width, the self-dual lattice is the worst lattice on the path. It was the one candidate the proof could not rule out, and it loses to both cubic lattices at the only width where it could have been expected to shine.

The ends, and what the sum is made of there #

At width three the sums are already their shortest vectors in disguise. The face-centred lattice has twelve shortest vectors of length 1.1225 at unit volume, the body-centred eight of length 1.0911, the simple cubic six of length one, and the self-dual body-centred tetragonal lattice eight of 1.0987. The face-centred lattice has more of them than bcc and they are longer, and the exponential makes length decisive: its sum minus one is $8.36 \times 10^{-5}$, the smallest of the four. The same table read at a small width, through the identity, is a table of duals, and there the body-centred lattice leads.

This is also where the problem touches packing and covering. The best packer in space is fcc and the best coverer is bcc, and that essay explained the split by duality while saying carefully that duality does not literally exchange a packing radius and a covering radius. For the Gaussian sum it does exchange something exactly: the large-width end of one lattice’s sum and the small-width end of its dual’s. The Gaussian sum is the version of the packing-against-covering story in which the duality argument is a theorem.

A soft crystal that changes its mind #

The Gaussian sum is not only a measure of spreading. It is the exact energy of a physical model, and the model shows the crossing.

Give particles a pair repulsion that is itself a Gaussian, $u(r) = \varepsilon\, e^{-r^2/\sigma^2}$ — a soft, bounded repulsion, the shape of the effective interaction between two polymer coils whose centres can pass through each other. This is Stillinger’s Gaussian-core model of 1976. Put the particles on a lattice at number density $\rho$ and the energy per particle is half the sum over all the others, which is

$$\frac{E}{N} = \frac{\varepsilon}{2}\left(\theta_L(t) - 1\right), \qquad t = \frac{1}{\pi \sigma^2 \rho^{2/3}},$$

the lattice’s Gaussian sum at unit volume, at a width set by the density. Dilute means large $t$, and there the face-centred crystal has the lower energy; dense means small $t$, and there the body-centred crystal does. So the zero-temperature crystal of the Gaussian-core model is face-centred at low density and body-centred at high density, and the two energies per particle are equal at exactly $\rho\sigma^3 = \pi^{-3/2} \approx 0.180$ — the density at which the width is one, where each lattice is the other’s dual.

That is an unusual thing for a crystal to do. Most simple repulsions make the densest packing more favourable the more the particles are squeezed; a bounded one lets them overlap, and once they overlap the question is no longer who has the most room but whose points are most evenly spread at the scale of the overlap — which is the small-width end, where the dual of the densest packing wins. The transition in the real model is not quite a single density: two phases of slightly different density coexist over a narrow range that straddles the crossing, because the energies have to be compared at equal pressure rather than at equal density. But the range is centred where the duality identity puts it, and the identity is the reason there is a transition at all.

A local check on each side #

Sixty random nearby lattices were tried on each side — each basis perturbed by a small random matrix and rescaled to unit volume, the draws seeded so that the numbers are reproducible. At width two every one of the sixty perturbations of fcc has a larger sum than fcc, and at width one half every perturbation of bcc has a larger sum than bcc. Each lattice is a local minimum on its own side of the crossing. That is evidence, and it is the kind of evidence that has been wrong before about lattice problems — a local minimum is not a global one, and sixty directions is a small sample of a five-dimensional space of shapes.

What the proof does not reach: inverse powers #

A lattice sum of an inverse power, $\sum' |v|^{-2s}$ — the Epstein zeta function, which is what an electrostatic or van der Waals energy over a lattice is made of — is an average of the Gaussian sum over every width at once, weighted by $t^{s-1}$. The average draws on both sides of the crossing, so the proof above says nothing about which lattice gives the smallest inverse-power sum: the face-centred lattice’s advantage at large widths and the body-centred lattice’s at small ones are weighed against each other, and the weights depend on the exponent.

The duality survives the averaging in a different form. The Epstein zeta function of a lattice at exponent $s$ is tied by its functional equation to that of the dual at exponent $\tfrac32 - s$, just as the Gaussian sum at width $t$ is tied to the dual’s at $1/t$ — so the self-dual exponent is $\tfrac34$, where the fcc and bcc values coincide exactly for the same reason they coincide at $t = 1$. In the range where the sums converge without any continuation, exponents above $\tfrac32$, which is where the physical ones lie, the face-centred lattice is believed to be the minimiser; a proof of that is also open, and it is a different open question from the one about widths, because it is about an average and not about every point on the axis.

What the argument rests on, and what it does not say #

The proof rests on uniqueness at the densest end. That the face-centred cubic lattice is the densest lattice packing in space, and the only one, is Gauss’s theorem; without the uniqueness the large-width winner could be another lattice of the same density. With it, the large-width winner is determined, and so by duality is the small-width one.

It is about lattices, not arrangements. The hexagonal close-packed arrangement of spheres is as dense as fcc and is not a lattice — it has two points in its cell — so it is outside the question, and the Gaussian sum of a non-lattice arrangement has no dual to compare against in the same way.

The widths are measured at unit volume. The sum depends on the lattice’s scale, and every comparison here is between lattices of the same volume per point; “large width” means large $t$ at that normalisation. The self-dual width is $t = 1$ only because the lattices are scaled to unit volume, where a lattice and its dual have the same volume.

No figure shows a Gaussian. The curves are sums of millions of Gaussians evaluated at single widths, and the lattices are named rather than drawn. The one picture that would make the crossing visible — two clouds of blurred points becoming equally uniform at one blur — is a statement about sums, which the sums say more precisely.

Where the length problem goes from here #

The Gaussian sum is where the question of how many vectors of each length a lattice has becomes a question about the whole list at once: the counts are the coefficients, and the sum weights them by distance. The sum whose answer depends on the shape used the same identity to give a value to a sum that had none, and here it gives an impossibility instead. And two lattices with the same lengths have the same Gaussian sum at every width, so none of these comparisons can tell them apart — in sixteen dimensions, where such pairs exist, the question of a best lattice is a question about classes of lattices rather than about lattices.

The dimensions where the argument gives nothing are exactly the ones where the densest lattice is its own dual. Eight and twenty-four are the famous cases — the $E_8$ lattice and the Leech lattice are self-dual and densest — and there the conclusion is the plane’s rather than space’s: a single lattice wins at every width, which Cohn, Kumar, Miller, Radchenko and Viazovska proved in 2022. The plane, eight and twenty-four are where a best lattice exists at every scale; three is where it provably does not.

Still open: whether the crossing is the only one #

The proof establishes that the winner changes somewhere; the identity says the cubic lattices exchange places at exactly $t = 1$. What neither establishes is that nothing else ever wins — that for every width above one no lattice beats fcc, and for every width below none beats bcc. The conjecture is that the cubic pair are the whole story, with a single change of winner at the self-dual width, and the computation here is consistent with it and cannot prove it.

A proof would have to handle the widths near one, where the two cubic lattices are nearly tied and the self-dual lattice between them is nearly as good, and it would have to rule out every shape in the five-dimensional space of lattices rather than a path through it. The computation that would test it more severely than this page does is a descent from random starting lattices at each width, recording where every descent ends — if any ended somewhere other than fcc or bcc, the conjecture would be false, and the place it ended would be the new lattice.