Thirteen-fold, and nothing is its own dual

The densest eleven-fold lattice is not its own dual took a pattern found in six dimensions and tested it in ten. Every lattice with a rotation of prime order $p$ fixing no vector is the ring of integers of the $p$-th roots of unity with a positive weight on each pair of complex embeddings, and the shapes of such lattices form a torus folded by the ring’s own Galois automorphism. The fold’s fixed points are the ideals $(1 - \zeta)^k$, centring by $p$-ths carries them round in a cycle, and duality reflects the cycle. In six dimensions the cycle had three places, duality fixed one of them, and that one was the densest. In ten it had five, duality fixed the third, and the densest was the second, $(1 - \zeta)^2$.

The essay ended by naming the next case and what it would settle. For thirteen the cycle has six places, an even number, so a reflection of it need not pass through any of them, and the family would have no special member that is its own dual. And with six places rather than five there is more room along the cycle: perhaps a higher power of $1 - \zeta$ would take over as the densest. This essay runs the computation in twelve dimensions. The first prediction holds exactly. The second fails, and the way it fails is the useful part.

Six weights and a five-torus #

Twelve is the fewest dimensions a rotation of order thirteen can live in, for the reason the degrees that name the restriction gave every order: a lattice rotation’s characteristic polynomial has whole-number coefficients, so a rotation of order thirteen that fixes no vector has the thirteenth cyclotomic polynomial as its characteristic polynomial, of degree twelve. The lattice is then a module over the ring $\mathbb{Z}[\zeta_{13}]$, and because that ring has class number one, every such module is the ring itself. What varies from one lattice to another is only the inner product. An inner product the rotation preserves is a positive weight on each of the six pairs of complex embeddings,

$$Q(x) = \sum_{j=1}^{6} A_j\, |\sigma_j(x)|^2 ,$$

six weights, five free numbers once the overall scale is removed.

Two identifications make that five-dimensional space of weights into a closed space, as in ten dimensions. Multiplying the ring by a unit is an isometry that rescales the weights, so in logarithms the squared units act as translations, and the real subfield of degree six has five independent units, which the cyclotomic units $\sin(\pi a/13)/\sin(\pi/13)$ for $a = 2, \dots, 6$ generate because that field’s class number is also one. The space of shapes is therefore a five-dimensional torus. And the Galois automorphism $\zeta \mapsto \zeta^2$ is an isomorphism between a lattice and one whose weights have been permuted in a single cycle of six, so the torus is folded by a rotation of order six.

The computation checks each of those steps rather than citing it. The rotation, written on the lattice of log-units, comes out an integer matrix, which is what makes the fold well defined, and its sixth power is the identity and no lower power is.

Six fixed points, and the step that reaches them #

A rotation of a torus has as many fixed points as the absolute value of $\det(I - R)$, and here that determinant is the product of the cyclotomic polynomials $\Phi_2$, $\Phi_3$ and $\Phi_6$ evaluated at one, which is six. The computation finds exactly six by solving the congruence on the torus.

The ideals $(1 - \zeta)^k$ are fixed by the fold, since every Galois automorphism carries $1 - \zeta$ to a unit times itself. Moving from one to the next is a translation of the torus by the log-weights of $1 - \zeta$, and that step turns out to sit five sixths of the way along the torus’s diagonal, so it has order six: six steps return to the start, because $13$ is a unit times $(1 - \zeta)^{12}$ and the weights of $(1 - \zeta)^{12}$ are a scale. The six ideals $(1 - \zeta)^0$ to $(1 - \zeta)^5$ are the six fixed points, at the sixths of the diagonal, and the computation matches each to one solution of the congruence.

The first two are familiar lattices. $(1 - \zeta)^1$ with the trace form is the root lattice $A_{12}$, and $(1 - \zeta)^0$ is its dual $A_{12}^*$; both identifications are made by finding an explicit isometry to the standard Gram matrices, not by matching invariants. The other four are lattices of Craig’s, the ideal lattices he introduced in 1976 in a search for dense packings in high dimensions.

The table at the head of this essay gives the numbers. Kissing numbers run 26, 156, 390, 312, 78 and 26. The trace-form minimum climbs by twenty-six at each step from $A_{12}$ to $(1 - \zeta)^4$, from 26 to 104, and then jumps by fifty-two to 156 at the last step, where the kissing number falls back to twenty-six, the size of one orbit of the rotations $\pm\zeta^k$. Centre densities, the number of sphere centres per unit volume when the spheres have unit radius, run from 0.00055 for $A_{12}^*$ to a highest of 0.02133 at $(1 - \zeta)^2$, with $(1 - \zeta)^3$ second at 0.01869.

Duality has no fixed point among them #

The dual of the ring with the trace form is the inverse different, which for $\mathbb{Z}[\zeta_{13}]$ is $(1 - \zeta)^{-11}$. Taking duals therefore sends the ideal $(1 - \zeta)^k$ to a lattice similar to $(1 - \zeta)^{1-k}$, and on the six fixed points, read modulo six, that is the map $k \mapsto 1 - k$. It pairs $A_{12}^*$ with $A_{12}$, as it must, then $(1 - \zeta)^2$ with $(1 - \zeta)^5$ and $(1 - \zeta)^3$ with $(1 - \zeta)^4$. No special lattice is its own dual, because $1 - k \equiv k$ modulo six would need $2k \equiv 1$, and two has no inverse modulo six.

The computation confirms the pairing by locating each dual on the torus and finding which fixed point it lands on. The property depends on nothing more than the parity of $(p - 1)/2$, the length of the cycle. For $p = 7$ the cycle has three places and for $p = 11$ five, odd both times, so a reflection of the cycle passes through one of them. For $p = 13$ it has six and the reflection passes between them.

The parity is that of $p$ modulo four, so the rule is short: a family with a $p$-fold rotation has a self-dual special member exactly when $p$ leaves remainder three on division by four. Thirteen is not the first prime to fail it. Five fails it too, and a five-fold lattice has one free number found the consequence in four dimensions without stating the rule: its family’s two special lattices, $A_4$ and $A_4^*$, are exchanged by duality, and the isodual point in the middle of its interval is not integral at all. Seven, eleven and nineteen have a self-dual special lattice; five, thirteen and seventeen do not. The eleven-fold essay’s description of thirteen as the first case without one was right about the sequence it was extending and wrong about the sequence as a whole.

The figure puts the three cycles side by side. In each, centring by the prime is the arrow from one special lattice to the next: a sublattice or superlattice of index $p$, the step that took $A_6^*$ to $A_6$ in six dimensions and $A_{12}^*$ to $A_{12}$ here. Duality is the mirror. The seven-fold cycle’s mirror passes through $(1 - \zeta)^2$, the eleven-fold cycle’s through $(1 - \zeta)^3$, and the thirteen-fold cycle’s through the gap between $A_{12}^*$ and $A_{12}$ on one side and between $(1 - \zeta)^3$ and $(1 - \zeta)^4$ on the other. That a lattice equal to its own dual should be special is a common heuristic, and every plane lattice is its own dual found how little it means in the plane, where every lattice is. Here the family has no such member to be special.

What centring by thirteenths does #

The arrow round the cycle is a centring. Multiplying a lattice’s points by $(1 - \zeta)^{-1}$ adds new points, and because $1 - \zeta$ has norm thirteen the new lattice contains the old one with index thirteen: twelve new points in each old cell, arranged by the rotation into a single orbit. That is the twelve-dimensional analogue of the centrings the fourteen Bravais lattices are built from, where a body centre or face centres are added to a primitive cell, and it is the operation where five-fold becomes legal used to find the lattices a five-fold rotation allows in four dimensions. Here it carries $A_{12}$ to $A_{12}^*$, adding the holes of the root lattice that the dual’s extra points sit in, and in the other direction it carries $(1 - \zeta)^2$ to $A_{12}$ by keeping one point in thirteen.

The cycle closes after six steps rather than twelve. Twelve steps multiply by thirteen up to a unit, which changes no shape, so twelve was guaranteed; that six already suffice is what the computation of the step’s order on the torus finds, and it is the same halving that gave five places for eleven and three for seven. Each centring changes the lattice; the shape returns after half the steps the norm alone would suggest. A rotation of order thirteen cannot fix any vector in twelve dimensions, since its non-trivial part already needs all twelve, which is why the family is the whole story here; in four dimensions three rotations of order twelve found an order with several ways of splitting space, and thirteen has only one.

The densest stays where it was #

The question the eleven-fold essay posed was whether a longer cycle would let a higher power of $1 - \zeta$ become the densest. The reason to expect it was that a higher power has a larger minimum relative to its determinant, up to a point, and in a longer cycle that point is further round. The reason to doubt it was that in both earlier families the densest member was the same one, two steps from the dual root lattice.

$(1 - \zeta)^2$ is the densest of the six, at 0.02133, and $(1 - \zeta)^3$ comes second at 0.01869. The cycle grew by one place and the peak did not move. In six dimensions $(1 - \zeta)^2$ was densest and self-dual; in ten it was densest and not self-dual; in twelve it is densest in a family where nothing is self-dual. The densest member of these families is not the one duality singles out, and it is not drifting round the cycle as the cycle lengthens. It is pinned to the second power.

The diagonal of the torus passes through all six fixed points, and the figure follows the packing density along it. The special lattices with large kissing numbers are sharp peaks, as they were in ten dimensions, because each is a point where a large set of vectors become shortest together. Between them, every lattice on the diagonal has exactly twenty-six shortest vectors, a single orbit of $\pm\zeta^k$, and a density well below the peaks. Two of the special lattices behave differently from anything in ten dimensions. $(1 - \zeta)^4$, with seventy-eight shortest vectors, is only a slight peak, and lattices a little way from it along the diagonal pack better. $(1 - \zeta)^5$, with twenty-six, is not a peak at all: the diagonal passes through it at a point lower than its neighbours. A fixed point of the fold is a special lattice in the sense of symmetry, and that sense need not coincide with being locally best at anything.

The rest of the family, sampled #

A diagonal is one line through a five-dimensional space, and a peak on it could be a saddle in the space. So the whole torus was sampled twice. On a grid of three points in each direction, 243 lattices including the point of $(1 - \zeta)^2$, nothing packs better than $(1 - \zeta)^2$. On a grid of four, 1,024 lattices, which happens to include $(1 - \zeta)^3$ and not $(1 - \zeta)^2$, the best is $(1 - \zeta)^3$ at 0.01869 and the next best is at 0.01039; nothing reaches $(1 - \zeta)^2$. The histogram shows how unrepresentative the special lattices are. Most of the family packs worse than $A_{12}$, the second-worst special member, and the dense ones are isolated spikes.

A grid is not a proof, and nothing here proves that $(1 - \zeta)^2$ is the densest thirteen-fold lattice. What the grids establish is that no lattice at their points beats it, and that the peaks the diagonal shows are not ridges along which some other direction climbs higher.

How good the best one is #

The densest lattice known in twelve dimensions is the Coxeter–Todd lattice $K_{12}$, with centre density $1/27$, about 0.0370, and kissing number 756. $(1 - \zeta)^2$ reaches 0.0213, fifty-eight per cent of it, with kissing number 390. The thirteen-fold rotation is a strong constraint, and the family it defines does not contain the best twelve-dimensional packing. That is the expected answer: $K_{12}$ has a large symmetry group of its own, built on the Eisenstein integers with a three-fold structure rather than a thirteen-fold one, and nothing requires the densest lattice of a dimension to have any particular rotation in its symmetry. What the census establishes is the best a thirteen-fold rotation allows among the lattices sampled, not the best twelve dimensions allows.

The same comparison was implicit in ten dimensions, where the eleven-fold family’s best is likewise well short of the best known ten-dimensional lattice. Covering and packing want different lattices is the reminder that “best” also depends on the question: a family can be poor at packing and good at something else, and nothing here measures covering.

What the computation does and does not establish #

Established, exactly: the rotation on the unit lattice is an integer matrix of order six; it has six fixed points; the step has order six; the six ideals are the fixed points; duality pairs them as $k \mapsto 1 - k$ and fixes none; the first two are $A_{12}$ and $A_{12}^*$, by explicit isometry.

Established numerically, by shortest-vector enumeration in twelve dimensions with a check that the enumeration finds at least the reduced basis’s own shortest vector: the minima, kissing numbers and centre densities of the six, the densities along the diagonal and on two grids.

Not established. The symmetry groups of the four Craig lattices were not computed. An exhaustive isometry search on $(1 - \zeta)^2$, which took seconds for the ten-dimensional analogues, ran past three hundred million steps without finishing, and was abandoned rather than given a larger cap. What is known by construction is a subgroup of order 312 in each: the rotations $\pm\zeta^c$, twenty-six, and the twelve Galois automorphisms, each of which carries the ideal to itself and only permutes the terms of the trace form. A generic member of the family has 52 symmetries, the rotations and complex conjugation, found by exhaustive search. Whether the special members have more than 312, as the ten-dimensional ones had 2,640 against a guaranteed 220, is open. And that $(1 - \zeta)^2$ is the densest member of the whole family is a statement about the lattices sampled.

Who worked out the pieces #

Kummer’s work on cyclotomic integers in the 1840s supplies the ring and the ideals; the class number of $\mathbb{Z}[\zeta_{13}]$ is one, and the class number of $\mathbb{Z}[\zeta_p]$ is one for every prime up to nineteen and for no prime beyond, so the method here has three more primes to run before every module stops being the ring. Maurice Craig introduced the lattices $(1 - \zeta)^k$ with the trace form in 1976, as a source of dense packings in dimensions where nothing better was known, and for some primes and powers they held records for a while. Coxeter and Todd found $K_{12}$ in 1953. The treatment of lattices with a fixed automorphism as a moduli space folded by Galois is the seven-fold essay’s, carried here to the next prime, where the five-fold lattices had a circle rather than a torus and every question had a one-line answer.

What the family must refuse #

Each test could fail. The fixed points are found by solving a congruence, not by assuming the ideals are them, and the matching is then checked point by point. The duality is computed by locating each dual on the torus. And the two claims the essay set out to test are refused by the numbers: that a higher power of $1 - \zeta$ takes over as the cycle lengthens, which fails because $(1 - \zeta)^2$ remains the densest, and that some special member is self-dual, which fails because six is even.

Still open: seventeen and nineteen, and why the second power #

Two primes remain at which the ring has class number one and the same construction works unchanged: seventeen, in sixteen dimensions, with a cycle of eight, and nineteen, in eighteen, with a cycle of nine. The pattern so far, the densest member at $(1 - \zeta)^2$ in cycles of three, five and six, makes a prediction for both, and there is a reason to doubt it. Craig’s construction reaches its best density at a power of $1 - \zeta$ that grows with the prime, and for large primes the best is well past the second power. So somewhere the peak must move. At seventeen or nineteen the shortest-vector searches are in sixteen and eighteen dimensions, where enumeration is still feasible for lattices this symmetric, and the computation would say whether the second power still wins at a cycle of eight and nine or whether the move has begun. It has not been run here.