The thirties share a lattice

No lattice carries all three twelves settled a question about four dimensions by computing whole symmetry groups. A four-dimensional lattice can have an operation of order twelve in three ways, which three rotations of order twelve distinguished by the characteristic polynomial: a single irreducible block $\Phi_{12}$, a third-turn beside a quarter-turn $\Phi_3\Phi_4$, and a quarter-turn beside a sixth-turn $\Phi_4\Phi_6$. The census found the single block on $D_4$ and on two hexagonal planes, the other two only where a hexagonal plane meets a square one, and never all three, or even the single block with either of the others, in one group.

It ended by looking up a dimension, the same move the restriction in three dimensions made from the plane. Five dimensions adds no new order, since no order needs exactly five. Six adds eight: 7, 9, 14, 15, 18, 20, 24 and 30. Order thirty again has three kinds, and the obvious guess is that six dimensions repeats the pattern of four, three kinds held apart.

The guess is wrong, and the way it is wrong explains the four-dimensional result better than that essay could. All three kinds of order thirty live on one lattice, $A_4 \oplus A_2$, forty-eight symmetries of each kind. What kept the twelves apart was never that there were three of them. It was that one of them is a single block that cannot be broken into pieces, and in six dimensions no thirty is.

What a kind is, and which kinds six dimensions adds #

The convention first. A symmetry of a lattice is an integer matrix, and its characteristic polynomial is a product of cyclotomic polynomials $\Phi_m$, one factor for each block the operation can be broken into over the rationals. The block $\Phi_m$ has dimension $\varphi(m)$, Euler’s totient, and turns its planes by multiples of $1/m$ of a turn. The order of the operation is the least common multiple of the $m$ in its blocks. Two operations are of the same kind when their characteristic polynomials agree, which is the same as being conjugate by a rational change of basis. That is the classification the degrees that name the restriction set up, and a kind is a rational class, not an integral one.

An order first appears in the least dimension that a sum of blocks reaching it fits into. For thirty, a single block $\Phi_{30}$ has dimension $\varphi(30) = 8$, which is too large for six. So every thirty in six dimensions is a sum: a four-dimensional block from $\Phi_5$ or $\Phi_{10}$, which carries the five, beside a two-dimensional block from $\Phi_3$ or $\Phi_6$, which carries the three. The two blocks must reach thirty together, so at least one of them must be even, and the possibilities are exactly three: $\Phi_5\Phi_6$, $\Phi_3\Phi_{10}$ and $\Phi_6\Phi_{10}$. The fourth pair, $\Phi_3\Phi_5$, has order fifteen.

The question the four-dimensional essay closed on listed a single block $\Phi_{30}$ among the candidates. It does not fit. The enumeration of rational types used here lists every multiset of blocks whose dimensions sum to six, and $\Phi_{30}$ is not among them.

The other new orders have kinds of their own. Seven, nine, fourteen and eighteen are single blocks, since $\varphi$ of each is six: $\Phi_7$, $\Phi_9$, $\Phi_{14}$, $\Phi_{18}$. Fifteen is $\Phi_3\Phi_5$. Twenty is a five-fold or ten-fold four-space beside a square plane, $\Phi_4\Phi_5$ or $\Phi_4\Phi_{10}$. Twenty-four is an eight-fold four-space beside a hexagonal plane, $\Phi_3\Phi_8$ or $\Phi_6\Phi_8$. Twelve kinds in all, over eight orders.

Six lattices, counted element by element #

The census uses the same method as the four-dimensional one, run on larger matrices. A lattice is given by the Gram matrix of a basis. A symmetry is found by choosing where each basis vector goes: to a lattice vector of the same length, whose inner products with the images already chosen match the Gram matrix. A search that backtracks over those choices finds every symmetry, and nothing else. Each group’s order is then checked against the order its structure predicts, which is the test a missed or false symmetry would fail first.

The six lattices are chosen so that each of the twelve kinds has an obvious candidate home. $A_6$ is the root lattice of the six-dimensional simplex, whose symmetries permute seven points, so it should carry the sevens. $E_6$ is the exceptional root lattice with a symmetry group of order 103,680, the one the four-dimensional essay named and had not computed. Three hexagonal planes, $A_2 \oplus A_2 \oplus A_2$, can be permuted cyclically while one is turned, which should give a nine. And three sums of a four-dimensional lattice with a plane: $A_4 \oplus A_2$, the five-fold four-space beside a hexagonal plane; $A_4 \oplus \mathbb{Z}^2$, the same four-space beside a square one; and $D_4 \oplus A_2$, the four-dimensional checkerboard lattice, which has eight-fold symmetries, beside a hexagonal plane.

The groups come out at 10,080 for $A_6$, which is $2 \times 7!$, the permutations of seven points with a sign; 103,680 for $E_6$, its Weyl group of order 51,840 with the negative of the identity added; 10,368 for the three hexagonal planes, $12^3 \times 3!$; 2,880 for $A_4 \oplus A_2$, which is $240 \times 12$; 1,920 for $A_4 \oplus \mathbb{Z}^2$; and 13,824 for $D_4 \oplus A_2$. Each matches its prediction exactly, and $E_6$’s took well under a second. That is 142,752 symmetries, and every one is classified by its characteristic polynomial.

Every kind is carried by at least one of the six, so every one of them exists, as an operation on an actual lattice rather than only as a block matrix. The sevens and fourteens are on $A_6$ alone, 720 of each, which are the seven-cycles of seven points and their negatives. The nines and eighteens are on $E_6$, 5,760 of each, and on the three hexagonal planes, 576 of each. Fifteen and all three thirties are on $A_4 \oplus A_2$, forty-eight of each. The twenties are on $A_4 \oplus \mathbb{Z}^2$ and the twenty-fours on $D_4 \oplus A_2$, 288 of each of the latter.

Every permitted order, and no other #

The crystallographic restriction in six dimensions permits seventeen orders: every order whose least dimension is at most six. Those are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 18, 20, 24 and 30.

Together the six groups hold exactly those seventeen. That is a test the census could fail in two directions: a search that admitted a matrix that is not a symmetry could produce an order the restriction forbids, and a search that missed symmetries could leave a permitted order unrealised. Neither happens. The census therefore gives each of the seventeen orders a lattice, which the restriction alone does not do. The restriction says an operation of order thirty is not forbidden; the census exhibits one.

It is worth noticing what the large groups lack. $E_6$ has the largest group, more than seven times the next, and holds only eleven orders: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 and 18. It has no seven, no fifteen, no twenty, no twenty-four and no thirty. $D_4 \oplus A_2$ has twenty-four but neither five nor nine. A large group holds the orders of its own structure, and the new orders of six dimensions are spread across lattices with nothing in common. Four root systems made the same point about $D_4$ in four dimensions: a group can be large because it turns many planes by the same few angles.

The thirties together #

Every symmetry of $A_4 \oplus A_2$ is a symmetry of $A_4$ beside a symmetry of $A_2$, because the two parts are different dimensions and no symmetry can exchange them. The symmetries of $A_4$ are the permutations of five points with a sign, 240 of them, and among them are the twenty-four five-cycles, whose blocks are $\Phi_5$, and their twenty-four negatives, whose blocks are $\Phi_{10}$. The symmetries of $A_2$ are the twelve of a hexagon, among them two third-turns, $\Phi_3$, and two sixth-turns, $\Phi_6$.

So the group contains every pairing: a five-cycle beside a sixth-turn is $\Phi_5\Phi_6$, a negated five-cycle beside a third-turn is $\Phi_3\Phi_{10}$, and a negated five-cycle beside a sixth-turn is $\Phi_6\Phi_{10}$. Each pairing comes twenty-four times two ways, forty-eight times. The census finds exactly those numbers by classifying the 2,880 matrices without being told the structure.

The picture is the explanation. In four dimensions the single-block twelve $\Phi_{12}$ is irreducible over the rationals: it leaves no rational plane alone, so a lattice that carries it cannot be split into a hexagonal plane and a square one, and a lattice that is so split cannot carry it. The other two twelves are exactly such splits. The twelves were kept apart by irreducibility. In six dimensions every thirty splits, into the same kind of four-space and the same kind of plane, and the sum of the two lattices those pieces live on carries every pairing at once. Had $\Phi_{30}$ fitted in six dimensions it would have played the part $\Phi_{12}$ played in four, and the answer would have been the same as there. It does not fit, so the answer is the opposite.

The other orders that split #

The thirties are not special in being split. Of the twelve new kinds, eight are sums of a four-dimensional block and a two-dimensional one, and only four, the sevens, nines, fourteens and eighteens, are single blocks. The eight split kinds behave the same way the thirties do, and the census shows it for each order.

Fifteen has one kind, $\Phi_3\Phi_5$, a five-cycle beside a third-turn, and it lives on $A_4 \oplus A_2$ beside the thirties, forty-eight times. Twenty has two kinds, $\Phi_4\Phi_5$ and $\Phi_4\Phi_{10}$, a five-fold or ten-fold four-space beside a quarter-turn of a square plane, and both live on $A_4 \oplus \mathbb{Z}^2$, forty-eight times each. Twenty-four has two kinds, $\Phi_3\Phi_8$ and $\Phi_6\Phi_8$, an eight-fold four-space beside a third-turn or a sixth-turn of a hexagonal plane, and both live on $D_4 \oplus A_2$, 288 times each. In every case the kinds of one order share one lattice, and the lattice is the sum of the lattice its four-dimensional block needs and the lattice its plane needs.

The four-dimensional pieces are themselves the lattices that where five-fold becomes legal found for the five-fold and ten-fold rotations, which first appear in four dimensions, and the checkerboard lattice whose eight-fold rotations turn two planes by an eighth and three-eighths of a turn. The planes are the two that the crystallographic restriction allows in two dimensions, the square and the hexagonal. Six dimensions does not invent new geometry for its split orders. It places an existing four-dimensional rotation beside an existing plane rotation and lets them run at once, and an order that is the least common multiple of the two appears.

That is why the split kinds were never in doubt, and why the census is a confirmation rather than a discovery for them. The interesting cases are the ones where a single block cannot be assembled from smaller pieces, and those are exactly the cases where two kinds of the same order, or two orders, could be kept apart. In four dimensions there was one such block among the new orders, $\Phi_{12}$, and it kept itself apart. In six there are four, and they come in two pairs, each pair a block and its negative. A block and its negative always share a lattice, since the negative of the identity is a symmetry of every lattice; so the only separation six dimensions can show among its single blocks is between the sevens and the nines.

How small a group can hold all three #

A lattice that carries all three thirties has a symmetry group containing all three, and the group of $A_4 \oplus A_2$ is far larger than it needs to be. The smallest subgroup holding all three kinds can be searched for directly: every subgroup generated by one element of kind $\Phi_5\Phi_6$ and one of either other kind, kept if it contains all three, and the smallest recorded.

It has sixty elements. It is a cyclic group of order thirty, generated by a five-cycle beside a sixth-turn, with one more element added: the negative of the identity on the four-dimensional part, which turns every $\Phi_5$ block into a $\Phi_{10}$ block and leaves the plane alone. Inside it the thirties come eight of each kind, which is the number of generators of a cyclic group of order thirty, once for each way of choosing a sign on the four-space and a power on the plane.

The four-dimensional counterpart has no answer, since no group holds all three twelves. The nearest comparison is that the two two-plane twelves of four dimensions did share a lattice, $A_2 \oplus \mathbb{Z}^2$, for the reason the thirties do here: they are the same two planes, turned by different amounts.

A seven and a nine #

The four new single-block kinds behave as $\Phi_{12}$ did. $\Phi_7$ and $\Phi_{14}$ live on $A_6$, which is the lattice a seven-fold lattice lives on a folded torus reached from another direction. $\Phi_9$ and $\Phi_{18}$ live on $E_6$ and on three hexagonal planes.

The hexagonal planes carry their nines in a way worth seeing once. Permute the three planes cyclically and turn one of them by a third as it moves. Three applications carry each plane round once and turn every plane by a third, so the cube of the operation is a third-turn on all three planes at once, and the operation itself has order nine. Its eigenvalues are cube roots of a primitive cube root of unity, which are primitive ninth roots of unity, and its characteristic polynomial is $\Phi_9$, a single block that leaves none of the three planes alone. $E_6$ contains the three hexagonal planes as a sublattice of index three, and carries ten times as many nines.

No computed group holds a seven and a nine. The reason in each case is visible in the arithmetic. A group with an element of order seven has an order divisible by seven, and only $A_6$’s is. $A_6$’s symmetries are permutations of seven points with a sign, and a permutation of seven points has order nine only if it contains a nine-cycle, which seven points cannot hold. $E_6$’s order, $2^8 \cdot 3^4 \cdot 5$, has no seven.

That is a statement about six lattices, not about all of them, and the step from one to the other is not taken here. The four-dimensional essay took it by quoting Dade’s classification of the maximal finite groups of four-by-four integer matrices, nine groups in six rational types, and checking that its lattices covered all six types. Six dimensions has a corresponding classification, by Plesken and Pohst, with many more maximal groups, and this census has not been checked against it. Whether some six-dimensional lattice carries both a seven and a nine is therefore unproved here. The thirties need no such step, because a positive answer needs only one lattice, and $A_4 \oplus A_2$ is it.

What the matrices cannot say #

The census decides kinds, not integral classes. Two operations of the same kind can sit on lattices that no integral change of basis relates, as $A_4$ and its dual carry the same five-fold symmetries in two inequivalent ways. The question asked here, which kinds share a lattice, depends only on the rational kind, because a rational change of basis preserves characteristic polynomials. A finer question, which integral forms of each kind share a lattice, would need the integral classification as well, the step that in the plane turns ten point groups into thirteen ways to hold a lattice.

Nothing here is a picture of six dimensions. The figures are tables and block diagrams. A six-dimensional operation of order thirty turns three planes, by a fifth and two-fifths of a turn in the four-space and a sixth of a turn in the plane, and the closest a drawing comes is three dials side by side. The argument does not use the dials; it uses the factorisation of a polynomial.

The lattices were chosen, not derived. Each of the six was picked because its structure suggested which kinds it would carry. That choice cannot make the census wrong about what the six carry, since every symmetry is found by search, but it means the census answers existence questions and nothing more. It says that every new kind exists and that the thirties share a lattice. It does not say which lattices are the only homes of each kind.

Who found the pieces #

The least dimension of an order, the sum of totients that decides when an order first appears, was worked out by Hiller in 1985 for all dimensions, after earlier partial results. The symmetry groups of the root lattices are their Weyl groups extended by diagram symmetries and the negative of the identity, from the theory of reflection groups that Coxeter and Witt built in the 1930s. The maximal finite groups of integer matrices in six dimensions were classified by Plesken and Pohst in the 1970s, in a series of papers on the finite subgroups of the general linear groups over the integers in dimensions five to nine. The census here computes six groups from Gram matrices and classifies every element; it quotes no list of elements from those sources and uses none of their classifications.

What the census has to refuse #

The refused claim is the conclusion of the previous essay carried up two dimensions by analogy, and analogy is exactly what fails. The four-dimensional result looked like a fact about three kinds of one order. It was a fact about one irreducible block among them. The first test, the group orders against their structures, is again the one everything else rests on. A search that admitted one false symmetry would change an order, and a search that missed one would change it the other way, before either changed a count of thirties.

Still open: eight dimensions, where thirty becomes a single block #

The pattern makes a definite prediction one step further on. In eight dimensions $\Phi_{30}$ fits, as a single irreducible block of dimension eight, and it is the Coxeter element of $E_8$, whose order thirty is the Coxeter number of that root system. Eight dimensions also holds every six-dimensional thirty with a plane added. So eight dimensions should reproduce the four-dimensional situation one level up: an irreducible thirty living on $E_8$, and split thirties living on sums. Whether the irreducible one shares a lattice with any of the split ones, as $\Phi_{12}$ shared none, would be the test. $E_8$’s symmetry group has 696,729,600 elements, which the backtracking search used here could not enumerate one by one; the question would have to be asked through the group’s structure instead, and it has not been asked.