In eight dimensions one lattice wins at every width

No lattice in space wins at every width compared lattices of unit volume by a single sum, the Gaussian sum

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

and found that no lattice in three dimensions has the smallest one at every width $t$. The reason was the identity from the sum that turns a lattice into its dual, $\theta_L(t) = t^{-d/2}\,\theta_{L^*}(1/t)$ for a lattice of unit covolume. It says the sum at a wide Gaussian is the dual’s sum at a narrow one. At large $t$ the sum is decided by the shortest vectors, so the densest lattice wins there, which in space is face-centred cubic. At small $t$ the identity hands the question to the duals, so the dual of the densest wins, which is body-centred cubic. A lattice best at every width would have to be both, and fcc is not bcc.

That essay ended by naming the dimensions where the argument gives nothing: those whose densest lattice is its own dual. There the lattice that wins at large widths also wins at small ones, and nothing stops it winning in between. In eight dimensions the densest lattice is $E_8$, and $E_8$ is its own dual. This essay computes the eight-dimensional comparison, which the space essay could only point at, and places it among the dimensions where the argument is decisive and the ones where it is silent.

The sums, in closed form #

Eight-dimensional lattices are too large to handle by listing vectors across the widths that matter, but the ones a reader would put up against $E_8$ all have Gaussian sums in closed form. With $q = e^{-\pi t}$, Jacobi’s three theta functions $\theta_2$, $\theta_3$ and $\theta_4$ are the sums of $q^{n^2}$ over the integers, over the half-integers, and over the integers with alternating signs. The cubic lattice $\mathbb{Z}^8$ has sum $\theta_3^8$. The lattice $D_8$ of integer vectors whose coordinates add to an even number keeps half the terms, $(\theta_3^8 + \theta_4^8)/2$. $E_8$ is $D_8$ together with the coset of vectors whose coordinates are all half-odd-integers adding to an even number, and its sum is $(\theta_3^8 + \theta_4^8 + \theta_2^8)/2$. The dual of $D_8$ is the cubic lattice together with the all-halves coset, with sum $\theta_3^8 + \theta_2^8$. A fifth rival, two copies of the four-dimensional $D_4$, has the square of $D_4$’s sum.

Each has a covolume, the volume of its unit cell: one for $\mathbb{Z}^8$ and $E_8$, two for $D_8$, a half for its dual, four for the pair of copies of $D_4$. A lattice scaled by $s$ has sum $\theta_L(s^2 t)$, so every lattice is scaled to unit covolume before any comparison, which is the only fair comparison, since a sparser lattice trivially has a smaller sum.

The closed forms are checked against a direct count. Every integer and half-integer vector in a box of eight dimensions is listed, tested for membership, and counted by its squared length. $E_8$ has 240 vectors of squared length two, 2160 of four and 6720 of six, and none of any odd squared length. Those are $240\,\sigma_3(n)$ for $n = 1, 2, 3$, with $\sigma_3$ the sum of the cubes of the divisors, which are the coefficients of the Eisenstein series $E_4$ and the classical statement that $E_8$’s sum is that modular form. $D_8$ has 112 vectors of squared length two. At unit covolume $D_8$’s shortest vectors are shorter than $E_8$’s by a factor of $2^{1/8}$, and it has fewer of them. $E_8$’s 240 shortest vectors, the root system four root systems met in the plane’s version, are both longer and more numerous at equal density than any rival’s.

Why $E_8$ is its own dual #

Self-duality is the property the whole argument turns on, and for $E_8$ it follows from two facts that can each be checked in a line. The first is that every vector of $E_8$ has an even squared length: the counts above find nothing at any odd norm, and the reason is that the coordinates of a vector either are all integers adding to an even number or are all half-odd-integers adding to an even number, and in both cases the sum of the squares comes out even. The second is that $E_8$ has covolume one. It is twice as dense as $D_8$, which has covolume two, because it adds one coset to $D_8$.

Even squared lengths make every inner product between two vectors of $E_8$ a whole number, by expanding the squared length of their sum. So every vector of $E_8$ has whole-number inner products with every vector of $E_8$, which is the definition of lying in the dual lattice: $E_8$ is contained in its dual. A lattice and its dual have reciprocal covolumes, and $E_8$’s is one, so its dual has covolume one too. A lattice contained in another lattice of the same covolume is that lattice. $E_8$ is its own dual because it is even and has the smallest covolume an integral lattice can have. The same two facts make the Leech lattice self-dual in twenty-four dimensions, and a theorem about even unimodular lattices says they exist only in dimensions that are multiples of eight. That is part of why the dimensions where the duality argument goes quiet are so few.

Every rival loses everywhere #

The picture at the head of this essay is the comparison, and it has one message. At every width from 0.2 to 5, every rival’s sum exceeds $E_8$’s. The ratio is taken after subtracting the constant term one, which every lattice shares and which says nothing. The margin is smallest at $t = 1$, where $D_8$ and its dual have exactly the same sum and both are 1.31 times $E_8$’s. Away from $t = 1$ the margins grow rapidly. At $t = 5$, where only the shortest vectors matter, the cubic lattice’s excess is more than four hundred thousand times $E_8$’s, because its shortest vectors at unit covolume have squared length one against $E_8$’s two.

At narrow widths the comparison needs care, and it is made through the identity. As $t$ falls, every sum at unit covolume is dominated by the same term, $t^{-4}$, the volume of the Gaussian, and the differences between lattices are hidden many digits down, where floating-point arithmetic loses them. The identity recovers them exactly: $t^4 \theta_L(t) - 1 = \theta_{L^*}(1/t) - 1$, a comparison of the duals at the reciprocal width, where their short vectors decide it. That is also why the ratios on the left of the figure are the ratios on the right with the lattices exchanged for their duals: $D_8$’s excess at $t = 0.2$ is $D_8^*$'s at $t = 5$.

The dual pair shows exactly what $E_8$ escapes. $D_8$ is the better of the two at wide widths, where its shortest vectors, of squared length 1.68 at unit covolume, beat its dual’s 1.19. $D_8^*$ is the better at narrow widths, by the same margins reflected. They cross at $t = 1$. Each is the other’s mirror image in the width, and neither can win at both ends. $E_8$ has no partner to cross, so the argument that stops every three-dimensional lattice does not stop it.

Duality is a mirror in the width #

The identity has a clean geometric reading. Multiply the sum at unit covolume by $t^{d/4}$, here $t^2$, and plot it against the logarithm of $t$. The identity then says that a lattice’s curve reflected in the line $t = 1$ is its dual’s curve. A self-dual lattice’s curve is its own reflection, symmetric about $t = 1$, and $E_8$’s is symmetric to one part in $10^{13}$ across every width computed. $D_8$’s and $D_8^*$'s curves are reflections of each other and meet on the line.

That picture makes the space argument visible in one glance. If the winning lattice at wide widths is not self-dual, its reflection, the dual, is a different curve, and at narrow widths the reflection is the lower of the two. A single curve lowest everywhere must be its own reflection. Self-duality is necessary for one lattice to win at every width, and it is a condition on the lattice, checkable by comparing shortest vectors, before any sum is computed.

It is not sufficient, and the rivals show that too. The cubic lattice $\mathbb{Z}^8$ and the pair $D_4 \oplus D_4$ are both similar to their own duals, so their curves are symmetric as well. They lose to $E_8$ at every width anyway, the cubic lattice by at least a factor of two and the pair of copies of $D_4$ by at least 1.43. Symmetry leaves room for a lattice to win everywhere, and it takes the right short vectors to use the room. The pair of copies of $D_4$ is the instructive loser. Each copy is the best-known lattice of its own four dimensions, and stacking two of them side by side gives an eight-dimensional lattice that is self-dual up to scale and made entirely of good pieces. It still loses by at least 1.43 at every width, because its short vectors lie in two separate four-dimensional planes and never mix the two. $E_8$’s vectors use all eight directions at once, and no arrangement built from lower-dimensional pieces matches it.

What the width means #

The Gaussian sum has a physical reading that makes “at every width” concrete. Place a particle at every lattice point and let each pair repel with an energy $e^{-\pi t r^2}$ that falls off with their distance $r$. The sum over the lattice, less the constant term, is then the energy of one particle in the field of all the others. A large $t$ is a short-range repulsion that only nearest neighbours feel, and the lattice with the fewest close neighbours at the longest distance has the lowest energy. That is the densest packing. A small $t$ is a soft, long-range repulsion felt across many cells, and there the energy depends on the arrangement at large scales, which the identity turns into a question about the dual’s short vectors.

So the width is the range of a force, and a lattice that wins at every width is one that is the lowest-energy arrangement for every Gaussian repulsion, short-ranged or long-ranged. Every repulsion that falls with distance as a completely monotone function, including every inverse power, is a mixture of Gaussians of different widths, so a lattice that wins at every width wins for all of them at once. That is the sense in which $E_8$ is called universally optimal, and why the comparison is made at every width rather than at one.

Which dimensions leave room #

Running the argument across dimensions gives a short table. In one dimension there is only one lattice up to scale. In the plane the hexagonal lattice is its own dual up to a rotation, and among lattices it wins at every width, which is Montgomery’s theorem and the result the lattice that minimises a sum measured. In space fcc’s dual is bcc, with twelve shortest vectors against eight, and there is no single winner. In four dimensions $D_4$ is similar to its dual, both with twenty-four shortest vectors, so the argument is silent. In five, six and seven the densest known lattices, $D_5$, $E_6$ and $E_7$, have duals with ten, fifty-four and fifty-six shortest vectors against their own forty, seventy-two and a hundred and twenty-six, so no lattice wins at every width there. In eight dimensions $E_8$ is its own dual, and in twenty-four the Leech lattice is.

The rows for dimensions one to four and eight are computed or counted here. The rows for five, six, seven and twenty-four are quoted from the known lattices. The pattern they make is the one worth remembering: the argument that forbids a universal winner in space forbids one in five, six and seven dimensions too, and leaves exactly one, two, four, eight and twenty-four open. In eight and twenty-four, Cohn, Kumar, Miller, Radchenko and Viazovska proved in 2022 that $E_8$ and the Leech lattice win not only against lattices but against every arrangement of points of the same density, at every width. Their proof built special functions whose Fourier transforms vanish at exactly the lengths these lattices’ vectors take. The comparison here, against four rivals at seventeen widths, is not a proof of anything. It is what the theorem looks like when measured.

The Gaussian result also sits inside a longer story about $E_8$ that is quoted here rather than computed. $E_8$’s 240 shortest vectors are the most spheres that can touch one sphere in eight dimensions, the kissing number, proved optimal in 1979. Its packing fills $\pi^4/384$, about a quarter, of eight-dimensional space, and Viazovska proved in 2016 that no packing of equal spheres does better, lattice or not. The universal optimality of 2022 extended that proof from one energy, the hard-sphere packing, to every completely monotone energy. The comparison measured above is the weakest member of the family: a handful of lattices at a handful of widths, where the theorem covers every configuration at every width.

Four dimensions, where the question stays open #

Four dimensions is the one open case the table leaves, and it is instructive because the measurement says less than it seems to. $D_4$ at unit covolume has exactly the same sum as its dual at unit covolume, to fifteen digits at every width, which is the numerical statement that $D_4$ and its dual are the same lattice turned and rescaled. Against the cubic lattice $\mathbb{Z}^4$, $D_4$ wins at every width, and the ratio is symmetric about $t = 1$ because both lattices are self-dual up to similarity. So $D_4$ passes every test that $E_8$ passed in its own dimension.

That does not make it the winner. $E_8$ was compared with four rivals and backed by a theorem; $D_4$ is compared here with one, and the space of four-dimensional lattices is a nine-parameter family in which a better one could hide. Whether $D_4$ has the smallest Gaussian sum at every width among all four-dimensional lattices is an open problem, and it is conjectured that it does. What the duality argument establishes is only that nothing rules it out. The measurement here establishes that one obvious rival loses.

What the comparison has to refuse #

The refused claim is the natural mistake: that the denser lattice of a dual pair is simply the better lattice. $D_8$ is denser than its dual, with longer shortest vectors at equal volume, and it does win at wide widths. At narrow widths the sum is decided by the dual’s short vectors, $D_8$’s dual is $D_8^*$, and $D_8$ loses. Density is a statement about one end of the width axis. Only a lattice that is the same at both ends can be judged by its density alone, and in eight dimensions $E_8$ is that lattice.

Still open: twenty-four, and the energies that are not Gaussians #

The comparison here stops at eight dimensions because the closed forms stop there. The Leech lattice’s sum is a modular form of weight twelve, with 196,560 vectors in its first shell, and the natural rivals in twenty-four dimensions are less obvious than $D_8$ is in eight. Repeating the measurement there would need the sums of the rivals as modular forms too, and it would show the same symmetric curve with a much wider margin.

The other direction is the one the theorem addresses and a Gaussian comparison does not. A sum of Gaussians at every width is enough to decide every energy that is a mixture of Gaussians, and that includes every inverse power of the distance, which is how the lattice that minimises a sum turned a Gaussian result into one about inverse powers. It does not include energies with a minimum, such as the attraction and repulsion of real atoms. For those, $E_8$ need not win, and which eight-dimensional lattice minimises such an energy at a given density is a question the duality argument does not touch.