A five-fold lattice has one free number

The body-centred hypercube is the checkerboard lattice ran a centring census in four dimensions. For six lattices whose symmetry groups are maximal, it took every subspace modulo a prime that the group keeps and built the lattice centred by it. Only two and three can centre a lattice had shown that in three dimensions only halves and thirds make new lattices, because every holohedry’s order there is a product of twos and threes. In four dimensions the census found a prime three dimensions never could. The 240 symmetries of $A_4$ include rotations of order five, and centring $A_4$ by fifths makes $A_4^*$, its dual.

That essay ended on the subgroups. Five also divides the orders of smaller groups inside $A_4$’s, the cyclic group of one five-fold rotation and the dihedral groups containing it. A lattice whose symmetry group is one of those would have centrings by fifths of its own, and nothing said whether any of them is new. That is the question of four-dimensional lattices with a five-fold rotation that are not $A_4$ or $A_4^*$.

The answer needs almost no search, because the lattices with a five-fold rotation turn out to have almost no freedom. Every one of them is the same ring with one number attached, and centring by fifths changes only the number. No new Bravais type comes out, and the reason is a reflection of an interval.

A five-fold rotation leaves one number free #

A rotation of order five in four dimensions, written as an integer matrix on the lattice’s basis, has a characteristic polynomial of degree four whose roots are fifth roots of unity. It cannot fix a vector, because the other three dimensions would then need an integer rotation of order five of their own, and an integer rotation of order five needs four. So its roots are the four primitive fifth roots of unity, and the polynomial is $x^4 + x^3 + x^2 + x + 1$.

That makes the lattice something more structured than a set of vectors. Call the rotation $\zeta$. Every vector can be multiplied by $\zeta$, and so by any integer combination of its powers, and the lattice becomes a module over the ring $\mathbb{Z}[\zeta]$ of integers of the fifth roots of unity. It has rank one over that ring, and a module of rank one without torsion is isomorphic to an ideal. Ideals of $\mathbb{Z}[\zeta_5]$ are all principal, the classical fact that its class number is one. So every four-dimensional lattice with a five-fold rotation is, as a group with that rotation acting, the ring $\mathbb{Z}[\zeta]$ itself.

What distinguishes one such lattice from another is only its inner product, and the inner products $\zeta$ preserves are exactly $\mathrm{Tr}(a\,x\,\bar y)$ for a number $a$ in the real field $\mathbb{Q}(\sqrt5)$, positive under both its embeddings. The trace adds the four conjugates of a number of the field of fifth roots of unity. Two values of $a$ give similar lattices when they differ by a positive rational factor, which is a change of scale, or by the square of a unit of the ring. The units of the ring include the golden ratio $\varphi$, so $a$ and $a\varphi^2$ describe the same lattice. And $a$ and its conjugate $\sigma(a)$, with $\sqrt5$ replaced by $-\sqrt5$, give isometric lattices, through the automorphism of the ring that sends $\zeta$ to $\zeta^2$.

So the whole family has one real parameter. The natural coordinate is

$$t = \frac{\log\big(a/\sigma(a)\big)}{2\log\varphi},$$

which multiplying $a$ by a rational leaves alone, multiplying by $\varphi^2$ shifts by 2, and conjugating changes in sign. Taken modulo 2 and folded by $t \mapsto -t$, it places every lattice of the family at one point of the interval from 0 to 1.

The family computed #

The Gram matrix of $\mathrm{Tr}(a\,x\,\bar x)$ on the basis $1, \zeta, \zeta^2, \zeta^3$ has entries $\mathrm{Tr}(a\,\zeta^{i-j})$. With $a = u + v\sqrt5$ they are integers, since $\mathrm{Tr}(\zeta^m)$ is 4 when five divides $m$ and $-1$ otherwise, and $\sqrt5$ is the Gauss sum $\zeta - \zeta^2 - \zeta^3 + \zeta^4$. At $a = 1$ the Gram matrix is 4 on the diagonal and $-1$ everywhere else.

The census takes every totally positive $a = u + v\sqrt5$ with $u$ from 1 to 9, twenty-seven of them, reduces each Gram matrix, and finds its symmetry group by the same backtracking over short vectors the four-dimensional census used. Sixteen have 20 symmetries and eleven have 240. Every one of the eleven is shown, by a map carrying one Gram matrix onto the other, to be $A_4$ or $A_4^*$. The trace form itself, $a = 1$, is $A_4^*$, and so are $3 \pm \sqrt5$, $7 \pm 3\sqrt5$ and $9 \pm 4\sqrt5$, all of them a rational times a power of $\varphi^2$. $A_4$ appears at $5 \pm \sqrt5$ and $5 \pm 2\sqrt5$, which are $\sqrt5$ times powers of $\varphi$.

The twenty symmetries of the others are the least a member can have. There are the ten multiplications by $\pm\zeta^k$, and the ten maps $x \mapsto \pm\zeta^k\bar x$, since complex conjugation preserves $\mathrm{Tr}(a\,x\,\bar x)$ whenever $a$ is real. That is a dihedral group of order twenty, the symmetries of a regular decagon, and the lattices that have exactly those are the decagonal lattices of four dimensions.

The picture at the head of this essay places the twenty-seven on the interval. They fall at eight distinct points, since members that differ by a rational or by $\varphi^2$ coincide. $A_4^*$ is at 0, $A_4$ at 1, and everything strictly between is decagonal. Nothing in the scan has any other number of symmetries, and nothing has 240 symmetries away from the ends.

Which end packs better #

The interval is not only a list of shapes. Each member also packs spheres, one centred at each lattice point, and how well it does so can be measured without drawing anything: the square of the lattice’s shortest length divided by the square root of its determinant, which is proportional to the square of the fraction of space the spheres fill. The scan computes it for every member.

At the $A_4^*$ end it is 1.431, and at the $A_4$ end 1.789. In between it climbs steadily with $t$: 1.493 at $t = 0.43$, 1.525 at 0.53, 1.552 at 0.60, 1.594 at 0.69, 1.662 at 0.81 and 1.695 at 0.87. The interval is ordered by packing as well as by shape, and $A_4$ is its best packing. The number of shortest vectors does not climb with it. Every decagonal member has ten, the orbit of one shortest vector under the ten rotations $\pm\zeta^k$, and so does $A_4^*$. Only $A_4$ has twenty, because at that end conjugation stops merely preserving the form and starts producing new shortest vectors.

That $A_4$ packs better than $A_4^*$ is known, and on this family’s evidence it is the best packing a lattice with a five-fold rotation can make. The two lattices that stop a descent short of D4 found both as places where a search for better packings is held up. What the family adds is the path between them. A lattice moving along the interval from $A_4^*$ gains packing density all the way to $A_4$ and never gains a symmetry until it arrives. The lattice that minimises a sum found, in the plane, that two different measurements of how well a lattice fills space pick out the same lattice. Here one measurement ranks a whole family, and the ranking agrees with the symmetry only at the ends.

Centring by fifths #

A centring by fifths adds to the lattice a fifth of some vectors, and for the result to keep the five-fold rotation, the subspace modulo five it adds must be one the rotation keeps. In the ring those are exactly the ideals containing five. Five factors in $\mathbb{Z}[\zeta]$ as a unit times $(1-\zeta)^4$, so the ideals between $5\,\mathbb{Z}[\zeta]$ and $\mathbb{Z}[\zeta]$ are $(1-\zeta)^k$ for $k$ from 0 to 4. That is three proper subspaces modulo five, of dimensions three, two and one.

The census confirms this by brute force. Of every line, plane and three-space of $\mathbb{F}_5^4$, exactly three are kept by both $\zeta$ and conjugation, one of each dimension. The line is spanned by $(1, 2, 3, 4)$ in the basis $1, \zeta, \zeta^2, \zeta^3$, which by a coincidence of bases has the same coordinates the earlier census reported for the line $A_4$’s symmetries keep in the basis of its simple roots.

Centring by the ideal $(1-\zeta)^k/5$ gives the lattice $(1-\zeta)^{k-4}\,\mathbb{Z}[\zeta]$. That is again a principal ideal, so again the ring $\mathbb{Z}[\zeta]$, with its form’s number multiplied by $|1-\zeta|^{2(k-4)}$. Now $|1-\zeta|^2 = (5-\sqrt5)/2$, whose conjugate is $(5+\sqrt5)/2$, and the ratio of the two is $(5 - \sqrt5)/(5 + \sqrt5) = \varphi^{-2}$. So multiplying $a$ by $|1-\zeta|^{-2}$ shifts $t$ by exactly one. The centring of index 25 multiplies $a$ by $|1-\zeta|^{-4}$, shifts $t$ by two and returns the same lattice, rescaled. The centrings of index 5 and 125 shift $t$ by one, which on the folded interval is the reflection $t \mapsto 1 - t$.

The prediction is checked the direct way. For six members, the centred lattice is built from the Gram matrix, reduced, and its symmetry group computed. The member the ring predicts is built independently, with $a$ multiplied by $(5+\sqrt5)/10$, and a map between the two Gram matrices is searched for. It is found every time. $A_4^*$ centres to $A_4$, and $A_4$ to $A_4^*$, which is the earlier census’s finding seen from the family. Each decagonal member centres to a decagonal member, the one at the reflected position.

So centring by fifths never leaves the family. It exchanges the two ends and pairs every decagonal lattice with the one reflected through the middle of the interval. The middle itself, $t = 1/2$, would be a lattice centred into itself, but it would need $a/\sigma(a) = \varphi$, a number of norm $-1$, and $a/\sigma(a)$ always has norm one. No lattice sits there, and every decagonal lattice has a partner different from itself.

What this says about the earlier question #

The earlier census asked whether the subgroups of $A_4$’s group containing a five-fold rotation give lattices with new centrings. They give no lattices of their own at all. A lattice with a five-fold rotation has at least the twenty decagonal symmetries, because conjugation and $-1$ come free with the ring. It has exactly twenty unless it is $A_4$ or $A_4^*$. So the only symmetry groups a five-fold lattice can have in four dimensions are the decagonal group of order twenty and the group of order 240 shared by $A_4$ and $A_4^*$. The cyclic group of order five, the dihedral group of order ten and the other subgroups between them are never the whole symmetry group of a lattice. They are not holohedries, and the question of their centrings does not arise.

The decagonal lattices are one Bravais type, not a continuum of them. Every member of the family has the same symmetry matrices, $\zeta$, conjugation and their products, on the same basis, and what varies is only the form those matrices keep. Four dimensions therefore have three Bravais types with a five-fold rotation, the decagonal type and $A_4$ and $A_4^*$, and centring by fifths maps the three onto themselves. The decagonal type goes to itself, and the other two are exchanged.

That is the same kind of statement as the dual lattice made in the plane, where every lattice is similar to its dual, and as the three-dimensional pair of body-centred and face-centred cubic, which are each other’s duals. There the involution is duality. Here it is a centring, and it pairs a whole family rather than two lattices. The two lattices that stop a descent short of D4 found $A_4$ and $A_4^*$ as the two places a search for dense packings gets stuck, and they are the two ends of this interval, the only members with more symmetry than their neighbours.

The ring, seen in the plane #

The ring has a second life. Its four coordinates can be split into two complex embeddings, $\zeta \mapsto e^{2\pi i/5}$ and $\zeta \mapsto e^{4\pi i/5}$, and the lattice sits in four dimensions as the pairs of images. Place each element at its image under the first, and keep it only when its image under the second is small. What remains is a set of points in the plane with five-fold symmetry and no period, the construction cut and project describes, and with a window of the right shape it gives the vertices of a Penrose tiling. Five-fold centres stay few, and mirrors do not used exactly this lift to explain which patches of that tiling are symmetric. So the lattice under a Penrose tiling is a member of this family. In the form $\mathrm{Tr}(x\,\bar x)$ it is $A_4^*$, and the family’s parameter is a change of shape that the projection does not see.

What the census does and does not show #

The classification rests on three classical facts, not on the computation: that a four-dimensional rotation of order five has the polynomial $\Phi_5$, that $\mathbb{Z}[\zeta_5]$ has class number one, and that the forms a five-fold rotation keeps are the traces $\mathrm{Tr}(a\,x\,\bar y)$. The computation confirms their consequences on twenty-seven members and six centrings. It does not replace them.

The scan is finite. That no member between the ends has more than twenty symmetries is shown for the eight distinct lattices the scan reaches. The argument for all of them uses the known list of finite groups of four-by-four integer matrices: a larger group would lie inside the 240 symmetries of $A_4$ or of $A_4^*$, and every group there that properly contains the decagonal group contains either the automorphism $\zeta \mapsto \zeta^2$ or a group acting irreducibly, and either one keeps only the form of an end.

A centring here means one the five-fold rotation keeps. A lattice in the family also has centrings by fifths that break the rotation, and centrings by other primes. Those leave the family, and the question they answer, which lattices without a five-fold rotation are reached, is a different one.

Where the facts come from #

That the integers of the fifth roots of unity have unique factorisation was known to Kummer, whose work on cyclotomic fields in the 1840s rests on exactly such class-number facts. Dade listed the maximal finite groups of four-by-four integer matrices in 1965, and the classification of four-dimensional lattices and their symmetry groups, with its Bravais types, was completed by Brown, Bülow, Neubüser, Wondratschek and Zassenhaus in 1978. Lattices built from rings of integers with a trace form are a standard construction, and the description of the Penrose tiling as a projection from them goes back to de Bruijn in 1981. The interval, its reflection and the census above are worked out on this page.

What the census must refuse #

The refused claim is the question the earlier census left open, in the form that would have made it interesting. A decagonal lattice centred into a lattice with a different symmetry group, or with the same group acting a different way, would have been a new Bravais type. The census finds every centring of every scanned member landing back in the family at the predicted place. The test that makes this more than a coincidence of symmetry orders is the isometry. Two lattices with twenty symmetries each could still be different lattices, and a map carrying one Gram matrix onto the other is found for every pair.

Seven, and the next ring #

The same argument runs for any prime $p$ whose cyclotomic ring has class number one, in dimension $p - 1$. For $p = 7$, in six dimensions, the ring $\mathbb{Z}[\zeta_7]$ again has class number one, the real field is cubic rather than quadratic, and the family of lattices with a seven-fold rotation has two free parameters instead of one. The special members would include $A_6$ and its dual, and centring by sevenths would act on a two-dimensional space of shapes rather than an interval. No lattice carries all three twelves found which orders of rotation six dimensions admit. Whether centring by sevenths is again a reflection, and whether the family has other special points than the two ends that $A_6$ and $A_6^*$ supply, is a computation of the same kind in two more dimensions. A seven-fold lattice lives on a folded torus makes it: the shapes form a torus folded by a third-turn, centring by sevenths is a translation of order three rather than a reflection, and the fold has a third special point that packs better than either root lattice.