No lattice carries all three twelves

Three rotations of order twelve sorted the operations of a four-dimensional lattice by their characteristic polynomials, and found three kinds of order twelve. The first, $\Phi_{12}$, is a single irreducible block: it turns one plane by a twelfth of a turn and the perpendicular plane by five-twelfths, and no plane of the lattice is carried to itself. The second, $\Phi_3\Phi_4$, turns one lattice plane by a third and another by a quarter. The third, $\Phi_4\Phi_6$, turns one by a quarter and the other by a sixth. Each exists, and each was built from blocks of companion matrices. That essay ended by asking which lattices carry which of them, and whether any carries all three.

Answering needs more than blocks. An operation exists in some lattice as soon as its characteristic polynomial is a product of cyclotomic ones, but two operations live in the same lattice only if some lattice’s symmetry group contains both. That is a question about whole groups of integer matrices, and there is a finite list of the groups that matter. This essay computes each group element by element. The answer is that no lattice carries all three, and more: the single-block twelve never shares a lattice with either of the others.

Six lattices, and their symmetries counted one by one #

Every finite group of integer matrices preserves some lattice, and it sits inside the full symmetry group of that lattice. So the groups worth computing are the largest ones, the symmetry groups of the most symmetric lattices. Six lattices cover every kind, for a reason given below. They are the hypercubic lattice $\mathbb{Z}^4$, the checkerboard lattice $D_4$ of points with an even coordinate sum, the root lattice $A_4$ of the four-dimensional simplex, the sum $A_2 \oplus A_2$ of two hexagonal planes at right angles, the sum $A_2 \oplus \mathbb{Z}^2$ of a hexagonal plane and a square one, and the cubic lattice with a perpendicular line, $\mathbb{Z}^3 \oplus \mathbb{Z}$.

Each lattice is given by the Gram matrix of a basis, the table of inner products of its basis vectors. A symmetry is an integer matrix $M$ with $M^{\mathsf T} G M = G$, carrying the basis to four lattice vectors with exactly the same lengths and mutual inner products. The search for them is a backtrack. Choose an image for the first basis vector among the lattice vectors of its length, then for the second among those of its length with the right inner product with the first, and so on. The candidates come from a box of coefficients checked to be wide enough, by confirming that it holds every shortest vector: the box finds 8 in $\mathbb{Z}^4$, 24 in $D_4$ and 20 in $A_4$, the known kissing numbers.

The six groups have 384, 1,152, 240, 288, 96 and 96 elements. These are the orders their root systems predict, which is the first check. $\mathbb{Z}^4$’s group is every signed permutation of the four axes, $2^4 \cdot 4! = 384$. $D_4$’s is the Weyl group of $F_4$, 1,152. $A_4$’s is the permutations of five things, with a sign, 240. The two hexagonal planes give twelve symmetries each and a swap, $12 \cdot 12 \cdot 2 = 288$. The hexagonal and square planes give $12 \cdot 8 = 96$, and the cube with a line $48 \cdot 2 = 96$.

The picture at the head of this essay is then read off by computing every element’s characteristic polynomial and factoring it into cyclotomic polynomials. $D_4$ holds 96 single-block twelves and nothing of the other two kinds. $A_2 \oplus A_2$ holds 24 single-block twelves and nothing else. $A_2 \oplus \mathbb{Z}^2$ holds four of each two-plane kind and no single-block twelve. $\mathbb{Z}^4$, $A_4$ and $\mathbb{Z}^3 \oplus \mathbb{Z}$ hold no element of order twelve at all.

The orders, all nine and no more #

The restriction in four dimensions, which the crystallographic restriction begins in the plane and three rotations of order twelve corrected for orders rather than angles, permits exactly the orders $n$ whose least dimension $\psi(n)$ is at most four: 1, 2, 3, 4, 5, 6, 8, 10 and 12. The six groups together contain exactly those orders, and each is found somewhere. Order five and ten appear only in $A_4$, whose symmetries permute five points, the case where five-fold becomes legal described. Order eight appears in $\mathbb{Z}^4$ and $D_4$, as a signed four-cycle of the axes. Order twelve appears in $D_4$, $A_2 \oplus A_2$ and $A_2 \oplus \mathbb{Z}^2$. That the union is exactly the permitted list is a check on the census from the opposite direction, since the permitted list was computed from characteristic polynomials alone with no group in sight.

Why the kinds do not mix #

The three twelves differ in how they sit in space, and the difference decides where they can live.

A two-plane twelve is a pair of plane rotations of the lattice. $\Phi_3\Phi_4$ turns one rational plane by a third, which only a hexagonal plane allows, and another by a quarter, which only a square plane allows. A lattice with that symmetry contains a hexagonal plane and a square plane, each carried to itself, and up to a sublattice of finite index it is $A_2 \oplus \mathbb{Z}^2$. $\Phi_4\Phi_6$ needs the same two planes, turned by a quarter and a sixth, and the same lattice has it.

The single-block twelve is different in kind. Its characteristic polynomial $\Phi_{12}(x) = x^4 - x^2 + 1$ is irreducible over the rationals, so it leaves no rational plane alone, and neither of its real invariant planes contains a single lattice point other than the origin. A lattice point in one of them would be carried round that plane by the operation, and the orbit would span a rational plane left alone, which the irreducibility forbids. It is a genuinely four-dimensional operation, and it lives in lattices with no hexagonal-and-square splitting. The census finds it in two: $D_4$ and the two hexagonal planes. In $A_2 \oplus A_2$ it exchanges the two hexagonal planes while turning each, so that its square turns both by a sixth. The two planes are there, but the operation itself carries each onto the other, and only its even powers leave them in place. Any symmetry of $A_2 \oplus A_2$ that keeps both planes is a pair of plane rotations or reflections, with a characteristic polynomial that factors, and cannot be a single block.

$D_4$ fails to hold the two-plane twelves for a reason that shows in its whole list of characteristic polynomials. Its 1,152 symmetries fall into seventeen of them. Among the ones that turn two planes without fixing a line are $\Phi_3^2$, sixteen symmetries that turn two planes by a third each, $\Phi_4^2$, twelve that turn two by a quarter each, and $\Phi_6^2$, sixteen that turn two by a sixth each. There are also the 144 single-block operations of order eight and the 96 of order twelve. What the list never contains is a pair of different turns: not $\Phi_3\Phi_4$, not $\Phi_4\Phi_6$, and not $\Phi_3\Phi_6$ either. Whenever $D_4$ turns two planes, it turns them alike. The quaternion description below explains why, since multiplying by a unit on one side turns two planes by the same angle. The two hexagonal planes of $A_2 \oplus A_2$, by contrast, do turn two planes differently, by a third and a sixth, in 32 of their 288 symmetries, but a quarter-turn and a third-turn needs a square plane they do not have.

The census says more than the geometry forces. A single group could in principle contain an operation that splits into a hexagonal and a square plane and another that splits the space no way at all. None of the six does, and no finite group of integer matrices does, by the argument below.

One twelfth of the group, every time #

The counts 96 and 24 are no accident. In both lattices the single-block twelve commutes with nothing but its own powers: its centraliser has order twelve, and all the single-block twelves form one conjugacy class of size $|G|/12$. In $D_4$ this is a known object. $D_4$’s symmetry group is the Weyl group of $F_4$, whose Coxeter number is twelve, and a Coxeter element of a Weyl group always has a centraliser that is the cyclic group it generates. The single-block twelve of $D_4$ is the Coxeter element of $F_4$. In four root systems the product of a rank-two system’s two simple reflections is a rotation whose order is that system’s Coxeter number. Here the product of the four simple reflections of $F_4$, taken one after another, has order twelve, and it is the single-block twelve.

That identification explains why $D_4$ has only this kind. The elements of a Weyl group are classified by their characteristic polynomials into a known list, and for $F_4$ the list’s order-twelve entries are the Coxeter element’s class alone. The census rediscovered that from the matrices. The degrees that name the restriction reads the same number twelve from the other side, as the largest of $F_4$’s invariant degrees 2, 6, 8 and 12.

The two-plane twelves of $A_2 \oplus \mathbb{Z}^2$ commute with 24 symmetries each rather than twelve. They are products of a rotation in each plane, and they commute with every product of a rotation of the hexagonal plane and a rotation of the square plane, six times four of them. So their centralisers have order 24, and their classes are four elements each, a twenty-fourth of 96.

A third-turn on one side and a quarter-turn on the other #

$D_4$ has another description in which its twelves become products of simpler things. Up to scale, it is the ring of Hurwitz quaternions, the numbers $a + bi + cj + dk$ with $a, b, c, d$ all integers or all halves of odd integers. Its lattice points are the quaternions themselves, and multiplying the whole ring on the left by a unit $u$ and on the right by a unit $v$ carries it to itself without changing any length. The ring has 24 units: $\pm 1$, $\pm i$, $\pm j$, $\pm k$ and the sixteen $\tfrac12(\pm 1 \pm i \pm j \pm k)$. They form the binary tetrahedral group, with one unit of order one, one of order two, six of order four, eight of order three and eight of order six.

The maps $x \mapsto uxv$ and $x \mapsto u\bar{x}v$ give 576 distinct symmetries, exactly half of $D_4$’s 1,152. The other half needs units from a larger ring that normalises the Hurwitz quaternions without belonging to them. All 96 single-block twelves lie in the Hurwitz half, and each is a product of a unit of order four on one side with a unit of order three or six on the other, 48 each way round. The eigenvalues show why. Left multiplication by a unit turns two perpendicular planes by the unit’s angle in the same sense, and right multiplication turns the same two planes by its own angle, in the same sense in one plane and the opposite sense in the other. A quarter-turn and a sixth-turn so combined give $90 + 60 = 150$ degrees in one plane and $90 - 60 = 30$ in the other. The two plane rotations of the two-plane twelve $\Phi_4\Phi_6$ are there in the angles, but distributed across both planes at once. In quaternion form the difference between the kinds is the difference between turning two planes separately and turning the whole space from both sides.

The dodecagonal lattice is two hexagonal planes #

The single-block twelve’s other home, $A_2 \oplus A_2$, also has a second description, and it is the one a quasicrystal needs. Take the ring $\mathbb{Z}[\zeta_{12}]$ of whole-number combinations of twelfth roots of unity, with basis $1, \zeta, \zeta^2, \zeta^3$, and measure lengths by the trace form, the sum of $|x|^2$ over the four ways of embedding the ring in the complex numbers. Its Gram matrix comes out as

$$\begin{pmatrix} 4 & 0 & 2 & 0 \\ 0 & 4 & 0 & 2 \\ 2 & 0 & 4 & 0 \\ 0 & 2 & 0 & 4 \end{pmatrix},$$

which, with the basis reordered as $1, \zeta^2$ and then $\zeta, \zeta^3$, is two copies of twice the hexagonal Gram matrix. So the ring of twelfth roots of unity is two hexagonal lattices at right angles, and multiplication by $\zeta$, which preserves the trace form, is a single-block twelve with characteristic polynomial $x^4 - x^2 + 1$. That is the lattice whose projection onto a plane, rotating one of its two invariant planes by $30$ degrees, gives the dodecagonal quasicrystals, just as where five-fold becomes legal projects a five-fold pattern from the ring of fifth roots of unity. A dodecagonal quasicrystal is cut from a lattice with no other kind of twelve in its symmetry group.

Why six lattices are enough #

The census covers six lattices, and the claim is about every four-dimensional lattice. What bridges the two is a classification of the maximal finite groups of four-by-four integer matrices, the groups contained in no larger finite group of integer matrices. Every finite group of integer matrices lies in one of them. Everett Dade found them in 1965, and there are nine.

The nine fall into the six types computed here, once a change of basis with rational entries is allowed. Two of the nine are the symmetry groups of $A_4$ and of its dual, which are the same group written in two bases. Three are the symmetry groups of the primitive, face-centred and body-centred cubic lattices, each with a perpendicular line, and they too are one group written in three bases. The other four are the symmetry groups of $\mathbb{Z}^4$, $D_4$, $A_2 \oplus A_2$ and $A_2 \oplus \mathbb{Z}^2$. A rational change of basis does not change a characteristic polynomial, so the twelves a group can contain depend only on its rational type. Every finite group of four-by-four integer matrices lies in one of the nine, so every such group lies, up to a rational change of basis, in one of the six groups computed here, and inherits their verdict. No four-dimensional lattice carries all three twelves, and none carries the single-block twelve together with either of the others.

The classification is quoted here, not derived. The census computes the six groups completely and checks them against their predicted orders. The step from six lattices to every lattice rests on Dade’s theorem, and a reader who wanted it checked would need the classification of four-dimensional Bravais groups. That classification was later recomputed as part of the complete enumeration of the 4,783 four-dimensional space-group types by Brown, Bülow, Neubüser, Wondratschek and Zassenhaus in 1978.

What the matrices cannot say #

Nothing here is a picture of four dimensions. The figures show tables, and the one that shows planes shows two circles side by side, because a four-dimensional rotation is two plane rotations at once and nothing more can be drawn. That the single-block twelve’s planes contain no lattice point is an argument about irreducibility. The drawing only illustrates the two angles.

The convention is the integral one. Two groups are counted as one type when a rational change of basis relates them, and as different forms when only an integral one would do. The question about twelves depends only on the rational type, which is why the nine maximal groups reduce to six. A question about which lattices a group can act on, a Bravais-type question, would need all nine.

The census is about point groups, not space groups. In the plane the corresponding step, from a group of integer matrices to the ways it can act on a lattice, turns ten point groups into thirteen ways to hold a lattice, and the space groups are a step further again. A twelve that a lattice carries may still combine with fractional translations in a space group, as a screw does in three dimensions. Which four-dimensional space groups contain each kind of twelve is a larger count, and it is not made here.

Who found the pieces #

Hurwitz introduced his quaternion integers in 1896, as the ring in which quaternion arithmetic has unique factorisation. The identification of $D_4$’s symmetry group with the Weyl group of $F_4$, and of the Coxeter element’s centraliser with the cyclic group it generates, belong to the theory of reflection groups that Coxeter built from the 1930s and that Kostant and Springer completed for Coxeter elements in the 1950s and 1970s. Dade’s classification of the maximal finite groups of four-by-four integer matrices appeared in 1965. The census here computes the six groups from Gram matrices, rather than taking their elements from those sources, and uses only Dade’s list from outside.

What the census has to refuse #

The refused claim is the natural first guess, and it fails in the most instructive way. $D_4$ is the densest lattice in four dimensions and has the largest symmetry group of any, and its symmetries include operations of every order four dimensions permit except five and ten. Yet its twelves are all of one kind. Symmetry in bulk is not the same as symmetry of every kind: a group can be large because it turns planes in many directions by the same angles, and $D_4$’s group is large in exactly that way. The test the rest depends on is the first, the six group orders against their root systems. A backtracking search that missed a symmetry or admitted a false one would change an order before it changed a count of twelves.

Still open: five and six dimensions #

In five dimensions the permitted orders gain nothing new, since no order has a least dimension of exactly five. In six dimensions they gain orders 7, 9, 14, 15, 18, 20, 24 and 30, and the operations of each order split into kinds by characteristic polynomial just as the twelves did. The maximal finite groups of integer matrices have been classified well beyond four dimensions, by Plesken and Pohst among others. Which kinds of order thirty, a block $\Phi_{30}$ or a $\Phi_5\Phi_6$ or a $\Phi_3\Phi_{10}$, share a lattice would be the same census with larger matrices. The backtracking search that computed $D_4$’s 1,152 symmetries in under a second would need to find the symmetries of $E_6$, 103,680 of them, and it has not been run.