A seven-fold lattice lives on a folded torus

A five-fold lattice has one free number found every four-dimensional lattice with a five-fold rotation. They are all one ring, the integers of the fifth roots of unity, with one number deciding the shape. Folded onto an interval, that number runs from $A_4^*$ at one end to $A_4$ at the other, and centring by fifths reflects the interval end to end. The essay closed by asking the same of seven. In six dimensions the ring $\mathbb{Z}[\zeta_7]$ again has class number one, its real subfield is cubic rather than quadratic, and the shapes should have two free numbers rather than one. It asked two things: whether centring by sevenths is again a reflection, and whether the family has special points other than the two that $A_6$ and $A_6^*$ supply.

The answers are no and yes, and they come from the same fact. The shapes form a torus rather than an interval, and the symmetry that folds it is a third-turn rather than a reflection.

A seven-fold rotation leaves three weights free #

A rotation of order seven in six dimensions that fixes no vector has characteristic polynomial $\Phi_7(x) = x^6 + x^5 + \dots + x + 1$, the only polynomial of degree six whose roots are primitive seventh roots of unity. No other kind of order-seven operation fits in six dimensions, since a seven-fold block needs all six of them, as three rotations of order twelve worked out for the orders four dimensions admit. So the rotation makes the lattice a module over $\mathbb{Z}[\zeta]$, $\zeta = e^{2\pi i/7}$, of rank one: an ideal of that ring. Because the ring has class number one, every ideal is principal, and the lattice is $\mathbb{Z}[\zeta]$ itself, with the rotation acting as multiplication by $\zeta$.

What is left to choose is the inner product. Over the real numbers $\mathbb{Z}[\zeta]$ spans three complex lines, one for each pair of embeddings $\zeta \mapsto \zeta^k, \zeta^{-k}$ with $k = 1, 2, 3$. Multiplication by $\zeta$ turns the $k$-th line by $2\pi k/7$. An inner product that the rotation keeps can weight the three lines differently but cannot mix them, so it is three positive numbers $A_1, A_2, A_3$:

$$Q(x) = A_1 |x_1|^2 + A_2 |x_2|^2 + A_3 |x_3|^2 .$$

On the basis $1, \zeta, \dots, \zeta^5$ the Gram matrix has entries $\sum_k 2A_k \cos(2\pi k(i - j)/7)$. With all three weights equal to one, every diagonal entry is six and every other entry is minus one, which is $A_6^*$ scaled by seven. The forms with integer entries are those whose weights are the three real embeddings of a totally positive element of the cubic field generated by $\eta = \zeta + \zeta^{-1}$, and that is the trace form $\mathrm{Tr}(a\,x\bar{x})$ of the five-fold essay with one more dimension of choice in $a$. Up to scale, two free numbers remain.

Why the shapes form a torus #

Two different weightings can give the same lattice, and the ways they can are what turn the plane of weights into a closed surface.

The first comes from the units. Multiplying every element of $\mathbb{Z}[\zeta]$ by a unit $u$ is a bijection of the ring onto itself, and it carries the form with weights $A_k$ to the form with weights $A_k |u_k|^2$. Every unit of this ring is a power of $\zeta$ times a real unit $\varepsilon$, so $|u_k|^2 = \varepsilon_k^2$. In logarithms the weights live in a plane once the overall scale is removed, and the squares of the real units form a lattice of translations of that plane. Two real units generate them all, $\eta$ and $1 + \eta$: each has three embeddings whose product is $\pm 1$, and the area they span, the regulator, comes out at 0.525455. That is the field’s own regulator, the smallest any two independent units can span, so nothing is missing. This is the same arithmetic as how many lattices share a determinant, where a class number counted the plane lattices of one determinant; here the class number is one, and the units do the counting instead. The plane modulo these translations is a torus.

The second comes from the field’s automorphisms. Sending $\zeta$ to $\zeta^2$ is a ring automorphism of $\mathbb{Z}[\zeta]$, so it maps the lattice onto itself, and it carries the weights $(A_1, A_2, A_3)$ to $(A_2, A_3, A_1)$. In the plane of logarithms that cyclic shift is a rotation by a third of a turn about the point of equal weights. The translations must be invariant under it, so they form a hexagonal lattice, and they do: the two generators have equal length, 2.0503, at sixty degrees. The shapes of seven-fold lattices are a hexagonal torus folded by a third-turn.

That fold is the picture at the head of this essay. A third-turn of a hexagonal torus has exactly three fixed points, and those are where the special lattices of the family must sit, because a lattice at a fixed point is carried onto itself by the Galois map composed with a unit, which is a symmetry that a generic member does not have.

The three fixed points #

The three fixed points can be named in advance. The ideals $(1 - \zeta)^k$ of the ring, given 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 their shapes are fixed points. Seven is a unit times $(1 - \zeta)^6$, so the step from one power to the next has order dividing six on the torus. Lying among the fixed points, which form a group of three, it has order three. The fixed points are therefore $(1 - \zeta)^0$, $(1 - \zeta)^1$ and $(1 - \zeta)^2$, at $(0, 0)$, $(\tfrac13, \tfrac13)$ and $(\tfrac23, \tfrac23)$ in the torus’s own coordinates. That is what the computation finds.

The first is $A_6^*$, as the all-ones weighting already said, and the second is $A_6$, the root lattice. Each is identified by an explicit isometry to a Gram matrix built from scratch, $A_6$’s from its Cartan matrix and $A_6^*$'s from that matrix’s inverse. Both have 10,080 symmetries, the order of the symmetric group on seven letters times two, and both contain 120 different subgroups of order seven, since the 720 seven-cycles of seven letters fall six to a subgroup.

The third is new. It is the ideal $(1 - \zeta)^2$ with the trace form, one of the lattices Craig introduced in the 1970s, and it is neither of the root lattices nor a dual of one. It has 672 symmetries and contains 8 subgroups of order seven. Its shortest vectors number 42, as $A_6$’s do. And it packs spheres better than either root lattice: its centre density at determinant one is $7^{-3/2} \approx 0.05399$, against $1/(8\sqrt{7}) \approx 0.04725$ for $A_6$ and $0.02603$ for $A_6^*$.

The factorisation in the table is the part worth reading slowly. In all three lattices the subgroups of order seven are conjugate, and the normaliser of one of them has 84 elements. Those 84 are accounted for exactly. Multiplication by $\pm\zeta^k$ gives fourteen, complex conjugation doubles them to twenty-eight, and the Galois third-turn that fixes the point triples that to 84. What separates the three is how many other seven-fold rotations the lattice has, rotations not conjugate to $\zeta$ by anything in the family’s own symmetry: 120 conjugates in the root lattices and 8 in the third. The same split holds one prime down. Run on $A_4$ and $A_4^*$, the same count finds 240 symmetries in each as 6 subgroups of order five times a normaliser of 40, which is the twenty symmetries of a generic five-fold member times the fold of order two.

Everything else has twenty-eight #

A member off the three points keeps the fourteen rotations and negatives and the conjugation, and nothing else. That is twenty-eight symmetries, and the census finds no member with more. There are 96 integral members with small coefficients. Of these, 77 lie off the three points and every one has exactly 28, while the other 19 land on the special points and are identified there. A grid of 324 real weightings, chosen with no regard to integrality, gives 28 at each of the 321 points that are not fixed points.

That is the six-dimensional form of the four-dimensional result that every lattice strictly between $A_4^*$ and $A_4$ has twenty symmetries. There the generic count was ten rotations and negatives times two. Here it is fourteen times two. In both cases the extra symmetry appears only at the points the Galois fold fixes. There is no line of extra symmetry on the torus. The grid’s points on the diagonal through all three special points have 28 like everything else.

Centring by sevenths #

A centring by sevenths that the rotation keeps is a subspace of the lattice modulo seven that multiplication by $\zeta$ maps into itself. Those subspaces are the ideals of $\mathbb{Z}[\zeta]/7$, and because seven is a unit times $(1 - \zeta)^6$ they are the five ideals $(1 - \zeta)^k/7$ for $k$ from one to five. There is one in each dimension from five down to one. Four dimensions had three, at fifths.

Centring by the $k$-th of them makes the lattice $(1 - \zeta)^{k - 6}\mathbb{Z}[\zeta]$ with the same form. That is again a principal ideal, so again the ring itself with its weights multiplied by $|1 - \zeta^j|^{2(k-6)}$. On the torus that is a translation by $k - 6$ steps of the order-three step: a third of the long diagonal when $k$ is one or four, two thirds when $k$ is two or five, and nothing at all when $k$ is three. The last case has index $7^3 = 343$, and the centred lattice is the original scaled by $\sqrt{7}$, which is the six-dimensional cousin of the sublattices that are the same shape. In space, the shapes a lattice in space can thin to found that a similar sublattice must have a cube for its index; here the index is a cube of seven for the same reason, a determinant taken six ways. Every member of the family has a similar superlattice of index 343, and therefore a similar sublattice of the same index, whatever its weights.

None of this is taken on trust from the algebra. The centred lattice is built as an explicit superlattice of $\mathbb{Z}[\zeta]$, its basis is found by Hermite reduction and its Gram matrix by direct computation, and an isometry to the predicted member is found by search. That was done at all five indices, for a member chosen at a generic point of the torus.

So the answer to the first question is no. Only two and three can centre a lattice derived the centring letters as subspaces modulo a prime that a holohedry keeps, and the subspaces here are the same objects for a group that is not a holohedry. Centring by fifths in four dimensions looked like a reflection, but it was really a translation of order two on a circle, folded by the Galois map of order two into an interval and seen there as $t \mapsto 1 - t$. Centring by sevenths is the same construction with three in place of two: a translation of order three on a torus that commutes with the third-turn. It fixes no member. It carries the three special points round in a cycle, and at index seven the cycle runs $A_6 \to A_6^* \to (1 - \zeta)^2 \to A_6$. That every lattice in the family is three centrings of index seven away from a scaled copy of itself is the same statement as $7 =$ unit $\times (1 - \zeta)^6$ read halfway.

Duality is a half-turn #

The dual of a member is another member. It is the dual ideal with the reciprocal weights, multiplied back into the ring by the different, $(1 - \zeta)^5$. On the torus the dual of the point $p$ is the point $p_0 - p$, where $p_0$ is the step. That is a half-turn about the point where $2p = p_0$, which is $(1 - \zeta)^2$.

So duality exchanges $A_6$ and $A_6^*$, as it must, and fixes the third lattice. $(1 - \zeta)^2$ is similar to its own dual, and the search confirms it by finding an explicit isometry from its dual’s Gram matrix, rescaled to determinant one, to its own. In four dimensions the corresponding point of the circle was the middle of the interval, $t = \tfrac12$, where the earlier essay found no integral member at all. In six dimensions the point exists, it is integral, and it is the best lattice in the family.

The dual lattice builds the dual as one point per family of rows. A lattice that is similar to its own dual, an isodual lattice, is one where the rows and the points have the same arrangement up to scale. Craig’s lattice has that property for a reason the torus makes visible: it sits at the centre of the half-turn.

Which member packs best #

The shading in the head figure already suggests the answer, and the line through all three special points makes it plain.

The three lattices order themselves by packing in the opposite order to their symmetry: $A_6^*$ with 10,080 symmetries is the worst packer in the family, $A_6$ with the same number is a local peak, and $(1 - \zeta)^2$ with 672 is the best. On a grid of 3,600 points over the whole torus no member does better than the third lattice. That is a grid, not a proof. The maximum sits at the third fixed point, where the third-turn and the half-turn both act, and a smooth function invariant under both has a critical point there.

This is the surprising part. In four dimensions the two ends of the interval were the two root-lattice relatives, and the better packer, $A_4$, was at one end, with the measure rising steadily towards it. In six dimensions a reader would expect $A_6$ to play the same part, and it does not. The best seven-fold lattice is a point the four-dimensional picture had no room for, because a fold of order two on a circle has only two fixed points. And it is not the best lattice in six dimensions. That is $E_6$, with centre density $1/(8\sqrt{3}) \approx 0.0722$, and $E_6$ has no seven-fold rotation at all: its symmetry group has 103,680 elements, and seven does not divide that number. The densest packing in six dimensions has no rotation of order seven, and among the lattices that do have one, the densest is the lattice at the centre of the duality half-turn.

What the census does and does not show #

The symmetry orders are found by a search that assigns images to six basis vectors from the list of vectors no longer than the longest basis vector, checking inner products as it goes. For the integral members every inner product is an integer and the comparisons are exact. For the real weightings on the grid they are made to a tolerance of $10^{-7}$, so a symmetry that held to one part in ten million and not exactly would be reported as a symmetry. The claim that every non-special member has 28 symmetries is checked on 77 integral members and 321 real ones. The argument for it, that any extra symmetry would have to normalise the rotation or bring in a new one and that both happen only at fixed points, is sketched here and not proved.

The claim about the best packer is also a measurement. The grid has 3,600 points, and the true maximum could lie between grid points in a place the grid misses. It is the symmetry of the fixed point that makes the third lattice a natural candidate for the maximum. The grid supports that, and a proof would need the local behaviour of the minimum function, which has corners, near that point.

The picture cannot show the lattices themselves. A six-dimensional lattice is drawn here as a single point on a surface, and everything that makes one lattice different from another — which vectors are shortest, how the symmetries act on them — lives in the tables and the prose, not in the shading.

The convention the counts depend on #

“A seven-fold rotation” means an integer matrix of order seven with no fixed vector. In six dimensions that is every order-seven operation. In seven dimensions it would not be, because a lattice could then have a seven-fold rotation fixing one direction. The weights are normalised to product one, so that every member is measured at determinant one, and the packing measure is the cube of the shortest squared length, proportional to the density of the sphere packing on the lattice. The torus coordinates are those of the two units $\eta$ and $1 + \eta$; another choice of generators would shear the rhombus without moving the three fixed points or the cycle through them.

Where the facts come from #

That $\mathbb{Z}[\zeta_7]$ has class number one is classical, and true for every prime up to 19. The units, the regulator and the identification of the three fixed points are computed here. The lattices $(1 - \zeta)^k\mathbb{Z}[\zeta_p]$ with the trace form are Craig’s, from his construction of dense packings in high dimensions in the 1970s, and the densities quoted for $A_6$ and $E_6$ are the standard ones, reproduced here for $A_6$ and quoted for $E_6$. The census itself, the torus, the third-turn and the cycle, is assembled for this essay. The order of $E_6$’s symmetry group is quoted rather than computed; the argument uses only that seven does not divide it.

The refused claims are the two the earlier essay made natural. Its interval had two ends and a reflection, so a reader extrapolating would expect two special lattices and a reflection again. The first refusal records that the centring moves every point by a third of the diagonal and needs three steps to return, so it is not an involution. The second records the 672 symmetries of the lattice that sits where the extrapolation put nothing. The body-centred hypercube is the checkerboard lattice is where the census of centrings by fifths began, and the same caution applied there: a pattern seen in one dimension is a prediction in the next, not a result.

Still open: eleven, in ten dimensions #

The same construction runs for every prime whose cyclotomic ring has class number one, which is every prime up to nineteen. For eleven, in ten dimensions, the real subfield has degree five. The shapes would then form a four-dimensional torus folded by a rotation of order five, the ideals $(1 - \zeta)^k$ for $k$ from zero to four would give five fixed points, and centring by elevenths would be a translation of order five cycling them. That is the prediction the two computed cases make. It assumes the step has order $(p - 1)/2$ for eleven as it did for five and seven, which nothing here checks.

The packing question is the interesting one. In four dimensions the best five-fold lattice was a root lattice; in six the best seven-fold lattice is the isodual point between them. With five fixed points in ten dimensions, which of the five packs best, and whether it is again the one fixed by duality, is a census of the same kind with larger matrices. It would need a shortest-vector search in ten dimensions, which the backtracking here has not been asked to do. No lattice carries all three twelves and the space every lattice lives in are the nearest relatives: one counts which rotations a dimension admits, the other draws the space of all shapes in the plane. This essay draws the shapes that carry one particular rotation, and in six dimensions those shapes form a folded torus.