The two lattices where a four-dimensional descent stops short

No lattice in space wins at every width asked which lattice of unit covolume makes the Gaussian sum $\theta(t) = \sum_v e^{-\pi t |v|^2}$ smallest, and found that in three dimensions the answer changes with the width $t$. The face-centred cubic lattice wins for narrow Gaussians and its dual, the body-centred cubic, for wide ones. The reason is a single identity, $\theta_L(t) = t^{-d/2}\,\theta_{L^*}(1/t)$, which ties a lattice at one width to its dual at the reciprocal width. If the winner at narrow widths is not its own dual, its dual must win at the wide ones, so no single lattice can win everywhere. In eight dimensions one lattice wins at every width measured the case where that argument fails because the champion is its own dual. There, $E_8$ is proved to win everywhere.

That essay ended with four dimensions, where the argument also fails and nothing is proved. The densest four-dimensional lattice, $D_4$, is similar to its own dual, so the duality identity cannot rule it out as a winner at every width. It was measured against one rival, $\mathbb{Z}^4$, and beat it everywhere. This essay measures it against every lattice a local search can reach.

The answer is almost uniform, and the exception is the interesting part. Descents from random four-dimensional lattices end at $D_4$ from fifty-nine of sixty starts. The sixtieth stops at a different lattice, and that lattice turns out to be one of exactly two in four dimensions where a descent can stop short. $A_4$ traps descents for narrow Gaussians and its dual $A_4^*$ for wide ones, and at the middle width neither does. They are the traps they are for a reason Voronoi’s theory of perfect forms supplies.

Every lattice of four dimensions, parametrised #

A lattice of unit covolume in four dimensions, taken up to rotation, is a positive definite $4 \times 4$ Gram matrix of determinant one: ten entries less one constraint, nine free parameters. Here it is written through the upper-triangular factor of the Gram matrix, three of whose diagonal entries enter through their logarithms so that every choice of nine real numbers is a valid lattice, and the whole matrix is then rescaled to determinant one. The search moves through this nine-dimensional space. Each point of it is a lattice, and each lattice appears many times, once for every basis.

The quantity being minimised is evaluated exactly rather than approximated. For a given lattice and width, every lattice vector with $|v|^2$ below the radius at which the Gaussian falls under $10^{-13}$ is enumerated by the Fincke–Pohst recursion on the Cholesky factor, after the basis has been reduced so that the enumeration stays small. The sum is then carried out. Two cases check it: $\mathbb{Z}^4$, whose sum is Jacobi’s $\vartheta_3^4$, and $D_4$, whose sum is $(\vartheta_3^4 + \vartheta_4^4)/2$. The enumeration reproduces both at three widths to eleven decimal places.

$D_4$ against the named rivals #

Before any search, the obvious comparison is with the four-dimensional lattices that have names: the integer lattice $\mathbb{Z}^4$, the root lattice $A_4$ and its dual $A_4^*$, each scaled to unit covolume.

Every ratio stays above one at every width from 0.3 to 3, so $D_4$ has the smallest sum of the four throughout. $\mathbb{Z}^4$ falls away fastest: at $t = 2$ its sum, less the constant, is four and a half times $D_4$’s. $A_4$ and $A_4^*$ are much closer, and they are closer still to each other. At $t = 1$ they tie exactly, at 1.0718 times $D_4$’s, because each is the other’s dual and the duality identity at $t = 1$ equates their sums. Either side of that width they swap. $A_4^*$ is the nearer rival for wide Gaussians, within 0.15 per cent of $D_4$ at $t = 0.5$, and $A_4$ for narrow ones.

That the rivals come so close at wide Gaussians is not a sign of weakness in $D_4$. As $t$ falls, every lattice of unit covolume has a sum that approaches the same $1/t^2$, because a wide Gaussian samples the lattice so coarsely that only its covolume matters. All the ratios therefore approach one at the wide end. Comparisons are informative in the middle of the range and at the narrow end, where the shortest vectors dominate. There $D_4$’s twenty-four shortest vectors of squared length $\sqrt2$, the most any four-dimensional lattice has, give it the advantage that decides the packing problem.

Sixty descents #

The named lattices are four points in a nine-dimensional space. The search tests the rest. At each of five widths, twelve starting lattices are drawn at random. From each start the sum is minimised by the Nelder–Mead method, then refined by a second pass with a smaller step, and the lattice it ends at is identified by its sum and by counting its shortest vectors.

The picture at the head of this essay is the result. Fifty-nine of the sixty descents end at a lattice whose sum agrees with $D_4$’s to seven figures and which has twenty-four shortest vectors. It is $D_4$, reached from random starts that looked nothing like it. At four of the five widths all twelve starts find it.

The sixtieth descent, at $t = 0.5$, ends at a lattice with ten shortest vectors and a sum of $3.01793$ against $D_4$’s $3.01328$. It is $A_4^*$, to every figure the sum is computed to. The descent did not fail to converge. It converged to a lattice where no small change of shape lowers the sum, and that lattice is not the best.

Two traps, and where they close #

A descent that stops at $A_4^*$ raises the obvious question of which lattices, and at which widths, are local minima of the sum at all. The test is direct. Perturb a lattice by a small random change of shape, descend, and see whether the descent returns to it or leaves. A lattice to which every perturbed descent returns is a local minimum at that width.

$D_4$ is a local minimum at every width tested: all six perturbed descents return, at $t = 0.5$, 1 and 2. $\mathbb{Z}^4$ is a local minimum at none. Every descent from it leaves and, apart from one that stalled on the way, reaches $D_4$. $A_4$ and $A_4^*$ behave in a way that duality alone predicts once one of them is known. $A_4$ is a local minimum at $t = 2$ and not at $t = 0.5$ or 1. $A_4^*$ is a local minimum at $t = 0.5$ and not at $t = 1$ or 2. At the middle width, $t = 1$, neither is: perturbed descents from both leave and reach $D_4$.

The pairing is exactly what the duality identity forces. It carries the sum of $A_4$ at width $t$ to the sum of $A_4^*$ at $1/t$, and a change of shape of $A_4$ to a corresponding change of its dual. So $A_4$ is a local minimum at $t$ exactly when $A_4^*$ is one at $1/t$. The measurement shows the traps open at the two ends of the range and closed at the self-dual width in the middle. The random search met the one at $t = 0.5$ once in twelve tries.

Why $A_4$, of all lattices #

The reason $A_4$ is a trap for narrow Gaussians is a classical theorem, and one perfect form in space described the theory behind it. As $t$ grows, the Gaussian sum is dominated more and more by a lattice’s shortest vectors. Minimising it for very narrow Gaussians is therefore the same as maximising the length of the shortest vector at fixed covolume, which is the lattice packing problem. Voronoi showed that the lattices where the packing density is locally maximal are exactly the perfect and eutactic ones. Perfect means the shortest vectors pin the shape down. Eutactic means the inverse Gram matrix is a positive combination of the shortest vectors’ outer products. In two and three dimensions there is one perfect form, and the earlier essay found it by search. In four dimensions there are two, both found by Korkine and Zolotareff in the nineteenth century: $D_4$ and $A_4$. Both are eutactic, so both are local maxima of the packing density.

So for narrow enough Gaussians the sum has at least two local minima, $D_4$ and $A_4$, because the packing problem does. The descent that starts in $A_4$’s basin stops there. At $t = 2$ the measurement already shows it: $A_4$ is a local minimum. Duality then supplies the other end. $A_4^*$'s status at wide Gaussians is $A_4$’s at narrow ones, so $A_4^*$ is a local minimum at $t = 0.5$ for no reason of its own. It borrows the reason from $A_4$.

That makes the middle width the interesting one. At $t = 1$ neither trap is open, and the single minimum found is $D_4$. The self-dual width is where the four-dimensional problem is simplest. A local search from anywhere finds $D_4$, and the question the duality argument leaves open, whether $D_4$ wins at every width, looks easiest to settle there.

Where the traps open #

A local minimum becomes a saddle when some direction of change stops raising the sum and starts lowering it. That is a question about second derivatives, and it can be answered at a lattice without running any descent. $D_4$, $A_4$, $A_4^*$ and $\mathbb{Z}^4$ are critical points of the sum at every width for a reason of symmetry. The gradient of the sum at a lattice is a quadratic form that the lattice’s own automorphisms must leave unchanged. For each of these four lattices the only such form is a multiple of the Gram matrix itself, and a change along that direction only rescales the lattice, which the fixed covolume forbids. So the gradient vanishes. Whether each is a minimum is then decided by the Hessian, the matrix of second derivatives over the nine directions of change.

For $A_4$ those nine directions fall into two families of four and five, the two irreducible pieces into which its symmetry group, of order two hundred and forty, splits the traceless quadratic forms. Within each family every direction has the same curvature. At $t = 0.9$ the five-dimensional family curves downwards, so $A_4$ is a saddle and a descent leaves it. At $t = 1.1$ it curves upwards, and with the four-dimensional family also upward, $A_4$ is a local minimum. $A_4^*$ shows the same pattern reversed.

Bisection on the width finds the crossing. $A_4$’s five-dimensional curvature changes sign at $t = 0.99998$ and $A_4^*$'s at $t = 1.00002$. Their product is one to within the precision of the second derivatives. The duality identity guarantees that product. It carries $A_4$’s curvature at width $t$ onto $A_4^*$'s at $1/t$ with a positive factor, so wherever one changes sign the other changes sign at the reciprocal width. What duality does not say is where the first crossing falls. The measurement puts it at the self-dual width itself, so that at exactly $t = 1$ both lattices sit on a five-dimensional flat of the sum, degenerate critical points that a descent can drift away from. That fits the perturbed descents at $t = 1$, which all left both lattices. Why the crossing falls at $t = 1$ and nowhere else is measured here, not explained.

The symmetry that leaves the question open #

Dividing each rival’s whole sum by $D_4$’s shows the duality directly. $\mathbb{Z}^4$ and $D_4$ are both their own duals at unit covolume, so $\mathbb{Z}^4$’s ratio is exactly symmetric about $t = 1$. $A_4$’s curve is $A_4^*$'s reflected about the same point. The symmetries are checked to $10^{-9}$ and they are the reason the three-dimensional argument fails here. It needed the winner at one end to differ from its dual. $D_4$ does not, so it can in principle win at both ends and everywhere between. Whether it does is what the search is evidence for.

What the widths mean for particles #

The Gaussian sum is not only a lattice invariant. It is the energy per particle of the Gaussian core model, particles that repel one another with a Gaussian of fixed width, arranged on a lattice. Fixing the covolume and changing $t$ is the same as fixing the width of the repulsion and changing the density. A large $t$ is a Gaussian narrow compared with the spacing, which is a dilute system, and a small $t$ is a dense one. In three dimensions, the lattice that minimises the energy switches from face-centred cubic in the dilute system to body-centred cubic in the dense one, which is the duality argument read as a phase transition. Frank Stillinger’s work on this model in the 1970s made the Gaussian core the standard example of a system that freezes into two different crystals at two ends of its density range.

In four dimensions the search says what the corresponding phase diagram looks like among lattices. There is one crystal, $D_4$, at every density tested. $A_4$ appears as a metastable crystal in the dilute system and $A_4^*$ in the dense one, local minima a system could be trapped in and would leave for $D_4$ if it could. At the self-dual density neither is metastable. The two false minima are, in the language of phases, two metastable crystals whose regions of stability meet at a single density and do not overlap.

What a search cannot show #

A descent that ends at $D_4$ from sixty random starts is evidence that $D_4$ is the smallest sum among all four-dimensional lattices at those widths. It is not a proof. A basin of attraction that no start happened to enter, belonging to some lattice with a smaller sum than $D_4$, is not excluded. The search found $A_4^*$'s basin, which is a small target, in one start of twelve. So a basin as small as that is unlikely to hide at the other widths, but a smaller one could.

The search is also only over lattices. The deeper question, whether $D_4$ minimises Gaussian energy among all periodic arrangements of points in four dimensions, including those with several points in a cell, is where the conjecture that it is universally optimal actually lives. For $E_8$ and the Leech lattice that question was settled in 2022 by Cohn, Kumar, Miller, Radchenko and Viazovska. For $D_4$ it is open, and a lattice search cannot reach it.

What the search has to refuse #

The refused claim is the natural misreading of the three-dimensional result: that the duality argument rules out a single winner in four dimensions as it does in three. It cannot, because $D_4$ at unit covolume is its own dual. The sum at width $0.7$ equals the sum at $1/0.7$ divided by $0.7^2$, to nine decimal places, and an argument that needs a winner and its dual to differ has nothing to separate. The test that makes the traps credible is the pairing. $A_4$’s status at $t = 2$ and $A_4^*$'s at $t = 0.5$ were measured separately, by descents that knew nothing of duality, and they came out the same.

Still open: why the traps open at one, and a proof for lattices #

The crossing at the self-dual width asks for a reason the computation does not give. One candidate is a path of lattices from $A_4$ to $A_4^*$ along which the sum at $t = 1$ stays constant. Duality maps such a path to itself, and it would supply the flat direction, and by symmetry all five. Another is a relation between the fourth moments of $A_4$’s shells and $A_4^*$'s at the same width, which the numbers suggest is nearly, but not exactly, a change of sign. Neither has been checked. Both are finite computations on two explicit lattices.

For lattices alone, the question of whether $D_4$ wins at every width is finite-dimensional, and the traps show where a proof would have to work. At the self-dual width there is apparently one minimum. At the ends there are two, and a proof must show that $D_4$’s is the lower of them, which the ratios above measure and do not prove.