The densest eleven-fold lattice is not its own dual

A seven-fold lattice lives on a folded torus found every six-dimensional lattice with a rotation of order seven. They are one ring, the integers of the seventh roots of unity, with three positive weights, and their shapes form a hexagonal torus folded a third of a turn by the ring’s own Galois automorphism. The fold has three fixed points, $A_6^*$, $A_6$ and a lattice of Craig’s, the ideal $(1 - \zeta)^2$. Centring by sevenths carries them round in a cycle, and duality is a half-turn that fixes the third. That third lattice was also the densest in the family.

The essay ended with a prediction for eleven, in ten dimensions. The ring $\mathbb{Z}[\zeta_{11}]$ also has class number one, its real subfield has degree five, and the shapes should form a four-dimensional torus folded by a rotation of order five. That would give five fixed points, the ideals $(1 - \zeta)^k$ for $k$ from zero to four, and a centring by elevenths that cycles them. The prediction rested on an assumption the essay named and did not check, that the step from one ideal to the next has order five on the torus, as it had order two and three for five and seven.

All of that holds, and the computation below checks each piece. What does not carry over is the packing. In six dimensions duality fixed one special lattice and that lattice was the best packer. In ten dimensions duality fixes $(1 - \zeta)^3$, which is third best of the five, and the densest member of the family is $(1 - \zeta)^2$, whose dual is $(1 - \zeta)^4$. The densest lattice with a seven-fold rotation sat at the centre of the duality symmetry. The densest with an eleven-fold rotation does not, and the reason is how many special points a cycle of five has room for.

Five weights and a four-torus #

Ten is the fewest dimensions a rotation of order eleven can live in. A lattice rotation’s characteristic polynomial has integer coefficients, so a rotation of order $n$ needs at least $\varphi(n)$ dimensions, the degree of $\Phi_n$, which is how the degrees that name the restriction put a number on every order and why five-fold becomes legal only in four dimensions. For eleven the count is ten, and a rotation of order eleven in ten dimensions that fixes no vector has $\Phi_{11}$ for its characteristic polynomial, so it makes the lattice a module of rank one over $\mathbb{Z}[\zeta]$, $\zeta = e^{2\pi i/11}$. With class number one every such module is the ring itself, and what remains to choose is an inner product the rotation keeps. Over the reals the ring spans five complex lines, one for each pair of embeddings $\zeta \mapsto \zeta^{\pm k}$, and an invariant inner product is a positive weight on each line:

$$Q(x) = \sum_{k=1}^{5} A_k\, |\sigma_k(x)|^2 .$$

Five weights, so four free numbers once the overall scale is removed. The same two identifications as in six dimensions turn the space of weights into a closed space. Multiplying the ring by a unit $u$ is an isometry that multiplies the weights by $|\sigma_k(u)|^2$, and in logarithms the squared units form a lattice of translations of the four-dimensional space of log-weights. The real subfield has four independent units, by Dirichlet’s theorem, and the cyclotomic units $\sin(\pi a/11)/\sin(\pi/11)$ for $a = 2, \dots, 5$ generate all of them up to sign, because the real subfield’s class number is one. So the shapes are a four-dimensional torus. Sending $\zeta$ to $\zeta^2$ is an automorphism of the ring, and it permutes the five lines in a single cycle, $1 \to 2 \to 4 \to 3 \to 5 \to 1$. That cycle rotates the space of log-weights by a fifth of a turn about the line of equal weights.

The first check is that the fold exists at all. A rotation folds a torus only if it maps the torus’s lattice of translations onto itself. The computation expresses the rotated units in terms of the originals, and every coefficient comes out an integer, so the fifth-turn acts on the unit lattice by an integer matrix whose characteristic polynomial is $x^4 + x^3 + x^2 + x + 1$. The shapes of eleven-fold lattices form a four-torus folded by a rotation of order five, as predicted.

Five fixed points, and the step that reaches them #

A rotation of a torus has as many fixed points as the absolute value of $\det(I - R)$ on its lattice. For a rotation whose characteristic polynomial is $\Phi_5$ that determinant is $\Phi_5(1) = 5$, and the computation finds exactly five fixed points. They lie on the torus’s diagonal, at the origin and at one, two, three and four fifths of the way along it.

The ideals $(1 - \zeta)^k$, with the trace form, are members of the family with weights $|1 - \zeta^j|^{2k}$, and the Galois map sends each to itself up to a unit, so each sits at a fixed point. The step $p_0 = \mathrm{Log}\,|1 - \zeta|^2$ moves from one to the next. Eleven is a unit times $(1 - \zeta)^{10}$, so ten steps return to the start and the step’s order divides ten. It lies among the five fixed points, which form a group of order five, so its order is five or one. The computation finds the step at four fifths of the diagonal: order five, exactly as the seven-fold essay assumed, and the five ideals $(1 - \zeta)^0$ to $(1 - \zeta)^4$ are the five fixed points.

The picture at the head of this essay names them. $(1 - \zeta)^0$ with the trace form is $A_{10}^*$ and $(1 - \zeta)^1$ is $A_{10}$, each identified by finding an explicit isometry to a Gram matrix built from the Cartan matrix of the root system and from its inverse. Their symmetry groups are the root system’s, twice the symmetric group on eleven letters, 79,833,600 elements each. The other three are Craig’s lattices, and an exhaustive search for their isometries over their shortest vectors finds exactly 2,640 in each.

That number factors as twelve subgroups of order eleven times 220, and the 220 is accounted for. Multiplication by $\pm\zeta^k$ gives twenty-two symmetries, complex conjugation doubles them, and the Galois fold of order five that fixes the point multiplies by five. The same factorisation fits the root lattices, where the eleven-cycles of eleven letters make $10!/10 = 362{,}880$ subgroups of order eleven, and $362{,}880 \times 220 = 2 \times 11!$. A member of the family off the five points has the forty-four symmetries that every member has, and the search on a member with weights chosen at random finds exactly forty-four.

Centring cycles them, duality reflects the cycle #

A centring by elevenths that the rotation keeps is an ideal of $\mathbb{Z}[\zeta]/11$, and there are nine of them, $(1 - \zeta)^k/11$ for $k$ from one to nine, one in every dimension from nine down to one. Centring by the $k$-th turns the ring into the ideal $(1 - \zeta)^{k - 10}$, which on the torus is a move of $k - 10$ steps. At $k = 5$ that is ten steps, or nothing: a superlattice of index $11^5 = 161{,}051$ similar to the original, scaled by $\sqrt{11}$. Every member of the family therefore has a similar sublattice of that index, whatever its weights, the ten-dimensional relative of the sublattices that are the same shape, where an index has to be a power for the determinant to come out right. The subspaces here are the same objects only two and three can centre a lattice used to derive the centring letters: subspaces of the lattice modulo a prime that the symmetry keeps. For a holohedry in three dimensions only the primes two and three give any. For a single rotation of order eleven, the prime eleven gives nine. Every other index moves every member, and the five special ones are carried round their cycle, $A_{10} \to A_{10}^* \to (1 - \zeta)^4 \to (1 - \zeta)^3 \to (1 - \zeta)^2 \to A_{10}$ at index eleven. Unlike the six-dimensional essay, which built each centred lattice explicitly and found an isometry to the predicted member, this follows from the ideal arithmetic, checked here only as positions of weights on the torus.

Duality is the other symmetry of the family. The dual of a member is the dual module with reciprocal weights, and the dual module is the codifferent, $(1 - \zeta)^{-9}$. So on the torus the dual of the point $x$ is $-x - 9p_0$, which on the five special points is $k \mapsto 1 - k$. That pairs $A_{10}^*$ with $A_{10}$, as it must, and $(1 - \zeta)^2$ with $(1 - \zeta)^4$, and fixes $(1 - \zeta)^3$ alone. In the picture duality is the reflection of the pentagon through $(1 - \zeta)^3$, as in six dimensions it was the reflection of the triangle through $(1 - \zeta)^2$. The dual lattice built the dual as one point for each family of lattice planes. A member at the reflection’s axis is similar to its own dual: its planes and its points have the same arrangement up to scale.

Which one packs best #

In six dimensions the three special lattices ordered themselves by packing as $A_6^*$, $A_6$, $(1 - \zeta)^2$, and the best was the self-dual one. The natural guess for ten dimensions is that $(1 - \zeta)^3$, the self-dual one, is again the best.

The diagonal through all five says otherwise. Each special lattice is a sharp peak, a point where a new set of vectors becomes shortest at once, except $A_{10}^*$, which is the lowest point of the line. The peaks have centre densities 0.00942 for $A_{10}$, 0.01892 for the self-dual $(1 - \zeta)^3$, 0.02212 for $(1 - \zeta)^4$ and 0.02741 for $(1 - \zeta)^2$. The densest is $(1 - \zeta)^2$, with 220 shortest vectors, and it is not its own dual. Its dual, $(1 - \zeta)^4$, is second. The self-dual point is third of five.

A line through the special points is not the whole family, so the computation also samples the torus on a grid of eight points in each of its four directions, 4,096 lattices. Most of the family packs worse than $A_{10}$: the histogram peaks near a centre density of 0.006. The best lattice on the grid reaches 0.0202, below both $(1 - \zeta)^4$ and $(1 - \zeta)^2$. That is a grid, not a proof, but it makes $(1 - \zeta)^2$ the best eleven-fold lattice found, and a grid that came within a quarter of it would have been worth refining. The best known lattice in ten dimensions, $\Lambda_{10}$, has centre density 0.0361; the best eleven-fold lattice reaches three quarters of that.

Kissing numbers that double and halve #

The kissing numbers of the five, 22, 110, 220, 110 and 132, say something the densities alone do not. In six dimensions the three special lattices had 14, 42 and 42 shortest vectors, so the new lattice there matched $A_6$’s kissing number and beat it on density only through its determinant. Here $(1 - \zeta)^2$ has twice $A_{10}$'s kissing number, 220 shortest vectors, which is ten orbits of twenty-two under the rotations and their negatives. Its shortest vectors can be named. The products $(1 - \zeta^a)(1 - \zeta^b)$ with $a \ne \pm b$, times $\pm\zeta^c$, built exactly as polynomials in $\zeta$, are 220 distinct elements of the ideal, all of the same trace norm, and that norm is the lattice’s minimum. So every shortest vector is a product of two of the ten elements $1 - \zeta^a$, each of which generates the ideal $(1 - \zeta)$, taken with $b$ different from $a$ and from $-a$. The squares and the products $(1 - \zeta^a)(1 - \zeta^{-a})$ are longer, by half again.

$(1 - \zeta)^3$ falls back to 110, the same as $A_{10}$, and $(1 - \zeta)^4$ rises again to 132. Kissing number and density do not move together along the cycle: $(1 - \zeta)^3$ has fewer shortest vectors than $(1 - \zeta)^4$ and packs less densely, while $A_{10}$ and $(1 - \zeta)^3$ share a kissing number and differ in density by a factor of two. Like the body-centred hypercube, where a centring turned one familiar lattice into another, the special points of this family are lattices with names, and their names say little about which of them packs best.

Why the self-dual point stopped being best #

The six-dimensional result looked like a principle: the best lattice of the family is the one at the centre of the duality symmetry. The ten-dimensional family shows it was a coincidence of small numbers, and the cycle diagram shows what the coincidence was.

Packing density is not a function duality keeps. The dual of a dense lattice can be a poor packer, and $A_{10}$ and $A_{10}^*$ are the standard example, at 0.00942 and 0.00201. So there is no reason for the self-dual point to be special for packing, and no reason for it not to be. What is forced is only that the two root lattices, $A_n$ and its dual, are always two of the special points, and that in their own dimensions they pack badly. In a cycle of three, the root lattices take two places and leave one, which duality must then fix. The best packer of the three could only be that one. In a cycle of five there are three non-root places, duality fixes one and pairs the other two, and nothing makes the fixed one the best. The computation finds the pair beating the fixed point, and the better of the pair, $(1 - \zeta)^2$, on top.

So the right statement for both dimensions is that the densest member is a Craig lattice, $(1 - \zeta)^2$ in both, and that it was self-dual in six dimensions because seven is small. For thirteen, in twelve dimensions, the prediction this essay makes is a cycle of six fixed points with the root lattices at two of them, duality $k \mapsto 1 - k$ fixing none of them (since $2k \equiv 1$ has no solution modulo six), and the densest member again a Craig lattice that is not self-dual.

What the computation does and does not establish #

The torus, the fold and the fixed points are exact. The integer matrix of the fold, its five fixed points and the step’s order are checked to rounding precision on quantities that must be integers or fractions with denominator five, and the ideal arithmetic behind them is exact.

The symmetry counts are searches. The 2,640 symmetries of each Craig lattice come from an exhaustive backtracking search over shortest vectors, which finds every isometry with the right Gram matrix and keeps those of determinant one. The 44 of a generic member are the same search on one member, not a proof for all.

The packing comparison is a grid. No member beats $(1 - \zeta)^2$ on 4,096 points, and the diagonal shows each special point as a peak. A proof that $(1 - \zeta)^2$ is the global maximum would need a bound on the packing function over the whole torus, and this essay does not attempt one.

The centrings are not built. The six-dimensional essay constructed each centred lattice explicitly by Hermite reduction and found an isometry to the predicted member. Here the centrings are read from the ideal arithmetic, which is exact, but the explicit construction was not repeated in ten dimensions.

Who worked out the pieces #

Kummer studied the arithmetic of the cyclotomic integers in the 1840s, and that $\mathbb{Z}[\zeta_p]$ has class number one for every prime up to nineteen was established in the following century. Dirichlet’s unit theorem fixes the rank of the unit lattice, four here. Craig introduced the lattices $(1 - \zeta)^k$ with the trace form in 1976 as a source of dense packings in dimensions $p - 1$, and Conway and Sloane’s tables give the centre densities of the best known lattices against which they are compared. The torus of shapes and its fold are the space every lattice lives in seen through a single rotation, and a five-fold lattice has one free number is the same construction with one dimension of shape.

What the family must refuse #

The first refusal is the claim the six-dimensional result invited. Had $(1 - \zeta)^3$ come out the densest, the principle would have survived a second case. It came out third, and the test records that the densest special lattice is not the one duality fixes. The second refusal guards the prediction’s own assumption. The step’s order divides ten, and only the fold forces five. A computation that found ten would have meant the torus coordinates were wrong, or the fixed points were not where the ideals sit.

Still open: thirteen, and a fixed point duality misses #

The prediction for thirteen above is the next case to check, and it is the first where duality has no fixed point among the special lattices at all, so the family would have no isodual special member. It needs a twelve-dimensional shortest-vector search and the symmetry groups of four Craig lattices, all within reach of the same searches as here. The question it would settle is whether, with the root lattices at two of six places, the densest member is still $(1 - \zeta)^2$, or whether a higher power of $1 - \zeta$ takes over once the cycle is long enough to leave it room.