Thirty as one block lives on two lattices

The thirties share a lattice explained why six dimensions behaves differently from four. In four dimensions there are three kinds of operation of order twelve, and no lattice carries all three. In six there are three kinds of order thirty, and one lattice carries every one of them. The difference, that essay argued, is that one of the twelves is a single block that cannot be broken into pieces, while no thirty is, since a single block of order thirty needs eight dimensions.

So it made a prediction. Eight dimensions is where the single block arrives, and that should reproduce the four-dimensional situation one level up: an irreducible thirty on the lattice $E_8$, whose Coxeter element it is, and split thirties living on sums of smaller lattices, held apart. The test would be whether the irreducible one shares a lattice with any split one. The essay added that the symmetry group of $E_8$ has 696,729,600 elements, which could not be listed, and that the question would have to be asked through the group’s structure instead.

Asked that way, the question has a short answer and the prediction is wrong. $E_8$ holds the single block and three split thirties together. The surprise is less that they share a lattice than how few of the possibilities get in: seventeen kinds of order thirty fit in eight dimensions and $E_8$ takes four, for a reason that can be read off a table of permutations of five things. And the prediction was not wrong in the way it first appears, because the single block has a second home, a less famous lattice on which it really is alone.

What fits in eight dimensions #

A kind of operation, as in the earlier essays, is a characteristic polynomial. An integer matrix of finite order has one that factors into cyclotomic polynomials, and two matrices of different kinds cannot be the same operation in any basis. The degree of the $n$-th cyclotomic polynomial $\Phi_n$ is the number of dimensions that block occupies, so listing the kinds is a finite exercise: choose blocks whose degrees add up to eight and whose indices have thirty as their least common multiple.

There are seventeen. One is the single block $\Phi_{30}$, of degree eight. The other sixteen all contain a five-fold piece of dimension four, either $\Phi_5$ or $\Phi_{10}$, and a three-fold piece of dimension two, either $\Phi_3$ or $\Phi_6$, and they differ in how they fill the two dimensions that remain and in whether the factor of two has been supplied by one of these or by a separate sign.

The picture at the head of this essay is that list sorted by the answer. Four boxes are filled and thirteen are empty, and nothing about degree or order separates them. $\Phi_5\Phi_6\Phi_1\Phi_2$ has the same degree as $\Phi_3\Phi_5\Phi_1\Phi_2$, and the same order, and only the second is present. The rule that decides which orders a dimension permits, the arithmetic of the totient that the restriction in three dimensions introduced, has nothing further to say here. It has done its work by the time the list is written down, and whatever separates the four from the thirteen is a fact about one lattice.

That is the shape every result on this subject now has. A dimension says which operations could exist, and a lattice says which of them do. The interesting question is always the gap between the two, and here for the first time the gap can be explained rather than just measured, because the lattice is small in the one way that counts.

A group too large to list has a small part that matters #

The obstacle the earlier essay named is real for most questions about this group, and it is not an obstacle to this one. A split thirty contains a five-fold block. Take its sixth power, and the block that had order thirty overall now has order five while everything of order dividing six has become the identity: the result is an element $u$ of order five whose characteristic polynomial is $\Phi_1^4\Phi_5$. It rotates a four-dimensional piece and leaves the other four dimensions fixed.

An element commutes with its own powers, so every split thirty lies inside the centraliser of such a $u$. This is the whole trick. Nobody needs the 696,729,600 elements of the group; they need the elements that commute with one five-fold rotation. And those are few. $E_8$’s 240 roots include twenty orthogonal to the copy of $A_4$ that $u$ rotates, and those twenty form a second $A_4$, whose reflections generate the 120 permutations of five points. Adding the five powers of $u$ and the matrix $-1$ gives a group of exactly 1,200 elements, built by closure and checked to commute with $u$.

Those 1,200 are small enough to read one at a time, and reading them produces three split kinds of order thirty, eighty elements each, and none of the thirteen others.

The table says why it could not be otherwise. A permutation of five points acts on the four-dimensional lattice $A_4$ through its cycle type, and there are seven cycle types. A cycle of length three and a cycle of length two give $\Phi_1\Phi_2\Phi_3$, and set beside a five-fold block that is the first thirty: order lcm(5, 6). Its negative is the second. A three-cycle alone gives $\Phi_1^2\Phi_3$, and set beside the five-fold block that has order only fifteen, but its negative doubles to thirty, and that is the third. Nothing else in the table reaches thirty, because no other cycle type contains a three. The thirteen missing kinds ask for something on the second copy of $A_4$ that no permutation of five things can be. $\Phi_3\Phi_6$ in those four dimensions, for instance, would be a matrix with eigenvalues at both the third and the sixth roots of unity and no eigenvalue of one, and a permutation with a three-cycle always has the eigenvalue one somewhere.

The missing kinds are not forbidden by anything about thirty. They are forbidden because the part of $E_8$ that a five-fold rotation leaves over is so rigid. It is one reason the block structure of a split operation carries more information than its order, and it is the same point the earlier essay made about the twelves, seen from the other side.

One assumption has been used and should be named. The argument needs every element of kind $\Phi_1^4\Phi_5$ to be conjugate to this particular $u$, so that its centraliser is this group up to a change of basis. That is Carter’s classification of the conjugacy classes of Weyl groups, which is quoted here rather than rederived. It can at least be tested, because it predicts a number.

What a random walk through the group says #

A group too large to list can still be sampled. Multiplying long random strings of reflections walks through the group, and in the long run the walk spends time in each element equally, so the share of steps landing on a given kind is the fraction of the group that kind occupies. For one conjugacy class that fraction is one over the size of the centraliser of any member. So the listed group makes a prediction nobody would otherwise think to check.

The kind $\Phi_1^4\Phi_5$ should occupy one part in 1,200 of the group, and the walk agrees. Each split thirty should occupy one part in sixty: a group of 1,200 contains eighty of them, which fall into four classes under the four non-trivial powers of $u$, so each element has a centraliser of sixty inside that group, and everything commuting with it already lies inside. The walk finds all three at one in sixty.

The single blocks come from a different theorem. Springer’s theory of regular elements says that a reflection group’s element of regular order $d$ has a centraliser whose order is the product of the group’s degrees divisible by $d$. The degrees of $W(E_8)$ are 2, 8, 12, 14, 18, 20, 24 and 30, and only one of them is divisible by thirty, so the centraliser of a Coxeter element is just its own thirty powers. The same reasoning gives one part in twenty-four for $\Phi_{24}$ and one in twenty for $\Phi_{20}$, and the walk finds those too. Exactly half of all the elements of order thirty in the group are the single block, one part in thirty against three parts in sixty. A random element of order thirty in $E_8$ is as likely to be irreducible as split.

This walk taught a small lesson of its own. Its first version drew its random reflections from a linear congruence, and its shares were wrong by half: the split thirties came out near one in thirty-nine rather than one in sixty, with nothing failing and every kind present. The low bits of such a sequence cycle, and a walk through a group is exactly the kind of process that exposes them. The prediction from the listed centraliser is what showed it. A sampling figure with nothing to compare to would have been published with the wrong numbers.

The glue that forbids one sign #

The table had a suspicious feature. It offers a permutation beside the five-fold block and the negative of both. Why not $\Phi_5$ beside the negative of a permutation, which would give kinds such as $\Phi_5\Phi_6\Phi_1\Phi_2$ and fill many of the empty boxes? The two copies of $A_4$ are separate lattices, and each has its own minus one. Why can they not be used separately?

Because $E_8$ is not the sum of the two copies. $A_4 \oplus A_4$ sits inside it with index five, and the rest of $E_8$ is made by gluing. Each copy of $A_4$ has a dual lattice slightly finer than itself, and the quotient has five classes. $E_8$ consists of the pairs of dual vectors whose classes match, class $i$ on the first copy going with class $2i$ on the second. That construction is what turns two lattices of determinant five into one of determinant one.

Now the operations can be read off. A permutation of five points fixes each of the five classes, so it leaves the gluing alone, and this is why the two sides may be permuted independently. Minus one sends class $i$ to class $-i$. Apply it to one copy only and the matching breaks, because class $i$ on the left now pairs with what class $-2i$ should have gone to on the right. Apply it to both and the matching survives. The signs are tied together, and that one line removes every missing kind.

There is a check on this reasoning that requires no group at all. A lattice that is glued from two pieces has a discriminant that has to match on both sides, and $\Phi_5$ and $\Phi_6$ have a resultant of one, which means no gluing can join a block of the first to a block of the second directly. They would have to be linked through a third block, and so a sum like $\Phi_5\Phi_6\Phi_3$ with no fixed vector and no reversed one has nothing to glue through and could only live on a decomposable lattice. $E_8$ is not decomposable. That is the same conclusion arrived at from the arithmetic of the polynomials rather than from a list of elements, and for the kinds it covers the two agree.

The same block on another lattice #

The prediction said that the irreducible thirty would stand apart. On $E_8$ it does not, so the natural thing is to conclude that eight dimensions simply differs from four. But the prediction has a second reading, and it turns out to be true.

The lattice $E_8$ is not the only one a $\Phi_{30}$ can act on. There is a simple way to build another. The hexagonal lattice $A_2$ has a rotation of order six, and $A_4$ has a symmetry of order five. Their tensor product $A_2 \otimes A_4$ is a lattice of dimension eight, and the operation that applies one on each factor has eigenvalues that are products of a sixth root and a fifth root, which are exactly the primitive thirtieth roots. So this lattice carries a single block of order thirty.

Its group is small enough to compute directly. Its vectors of smallest norm, sixty of them, are all that an automorphism can send a basis to, and a backtracking search over those finds 1,440 symmetries. That is what the tensor structure predicts: twelve from $A_2$ and 240 from $A_4$, divided by two because a pair of signs on the factors cancels. Classifying them finds forty-eight of $\Phi_{30}$, forty-eight of its negative $\Phi_{15}$, and no split thirty at all.

The reason is structural. An element of a tensor product’s group is a pair, one operation on each factor, and its eigenvalues are all the products of one eigenvalue from each. For the order to reach thirty, the factor on $A_4$ has to supply the five, and a five-fold operation on $A_4$ fills all four of its dimensions with one block. The operation on $A_2$ then multiplies every one of those eigenvalues by the same pair of numbers. There is no spare factor for a three-fold piece to occupy on its own, so the result is either one block of degree eight or a repeated smaller one. So the original prediction describes $A_2 \otimes A_4$ and not $E_8$. The single block does have a lattice where it stands alone, and the famous lattice is the one where it does not. The analogy with four dimensions, where the most symmetric lattice $D_4$ has the single block and no split twelve, was reasonable. What it could not see is that $E_8$ has two copies of $A_4$ inside it, and those are what let the split kinds in.

One module, two ways of measuring #

These two lattices differ a great deal. $E_8$ has determinant one and a group of nearly seven hundred million elements, while $A_2 \otimes A_4$ has determinant 2,025 and a group of 1,440. Yet there is a precise sense in which they are the same object.

An integer matrix with characteristic polynomial $\Phi_{30}$ turns the lattice it acts on into a module over the integers of the thirtieth cyclotomic field, and that ring has class number one, the property how many lattices share a determinant counted in the plane. So there is only one such module up to isomorphism, and only one such matrix up to integer change of basis. Every lattice carrying a $\Phi_{30}$ is that module. All that differs between them is the quadratic form put on it, which is to say how lengths are measured. A seven-fold lattice lives on a folded torus found the same structure in six dimensions, where every lattice with a seven-fold rotation is one ring with three weights on it. Here there are four weights.

That can be shown rather than cited. On each lattice one finds a vector $v$ whose eight images under the operation, $v, wv, \ldots, w^7v$, are a basis. In that basis the operation is the same companion matrix on both lattices, and the form is fixed by eight numbers, the inner products of $v$ with its images.

On $E_8$ the numbers are 2, −1, 0, 1, −1, 1, 0, −1; on $A_2 \otimes A_4$ they are 4, −1, 0, 0, 1, 2, −2, 0. Everything else about the two is decided by that row of small integers. It is a striking way to see a point this subject keeps returning to, which the normaliser made about the plane: a symmetry operation in a basis is a fact about the integers, and which further symmetries come along with it is a fact about the metric. One integer matrix is the same matrix whichever form is chosen. What it has for company depends on the form.

The prediction’s error can be put in those terms. It treated “the lattice with the irreducible thirty” as one thing, because a Coxeter element is what $E_8$ is famous for. The operation does not pick out a lattice. It picks out a four-parameter family of forms, and $E_8$ is the point of that family with the most symmetry.

What this does not show #

This does not list the lattices that carry a $\Phi_{30}$. The invariant forms make a family, of which two points have been examined. There are other forms, and some of them may carry other split kinds, or other groups entirely. The classification of the maximal finite groups of integer matrices in eight dimensions is in the literature and is not reproduced or checked here. “Alone” is a statement about $A_2 \otimes A_4$, not about every other lattice.

The picture of glue is a picture. The diagram of five classes on two circles is true to the arithmetic of the discriminant groups, but it shows nothing about what the lattice looks like, and no drawing of $E_8$ in the plane could. What was computed is the closure of a set of matrices and the characteristic polynomials of its members.

The uniqueness of the five-fold class is quoted. The whole argument about the thirteen rests on Carter’s list, supported by a random walk’s agreement with the 1,200. A sampling check cannot prove an absence, and the kinds marked missing could in principle occupy a class too small for sixty thousand steps to meet; the centraliser argument, not the walk, is what excludes them, and the argument is only as good as that quotation.

Nothing here touches the physics of an eight-dimensional crystal, because there is none, and it would be a mistake to read a statement about $E_8$ as a statement about any material. Where operations like these have a physical use is in the superspace of quasicrystals, as in a five-fold lattice has one free number, whose dimension counts are smaller than this.

How the four were found, and what could have refused them #

The 240 roots are found by closing the simple roots under reflection, and every matrix is an integer matrix on the root basis. The characteristic polynomials are computed exactly and factored by dividing by cyclotomic polynomials, so a kind is never estimated. The centraliser is built by closure from the five-fold element, the reflections in the twenty roots orthogonal to its $A_4$, and minus one, and each of the 1,200 elements is checked to commute with it.

Each number can fail independently. The group of $A_2 \otimes A_4$ has to come out at 12 · 240 / 2, predicted before the search rather than read from it. The share of Coxeter elements has to come out near one in thirty, which is a theorem unrelated to the way the walk works. The eighty elements in each split kind have to be divisible by the four powers of $u$, or the classes could not be what they are said to be. And a claim that the prediction might have wanted is turned down: that a lattice with the single block must have split thirties too, which is what $E_8$ alone would suggest. $A_2 \otimes A_4$ is the counterexample, with none.

Still open: the forms between #

Two points of a four-parameter family have been read, and they answer the question very differently. Between them lie all the other forms that a $\Phi_{30}$ preserves. Each is a positive quadruple of numbers up to scale, and each has a group containing the thirty powers of the operation and possibly much more. A natural measurement is how that group changes across the family: which forms have any symmetry beyond the cyclic group and the negative, which of those are maximal, and whether $E_8$ and the tensor product are the only forms whose groups contain a reflection. In eight dimensions one lattice wins at every width asks a question about $E_8$ with no operation in it at all, and whether any other form in this family comes near it on packing would connect that to this. The same structure exists at every dimension of a cyclotomic field with class number one, and the family has not been walked here.