One ring is E8 and a lattice with no mirror

Thirty as one block lives on two lattices found that an operation of order thirty, acting as a single irreducible block in eight dimensions, does not pick out a lattice at all. It picks out a family of forms, a whole continuum of inner products that it preserves on one fixed module. That essay read two points of the family, $E_8$ and $A_2 \otimes A_4$, and asked about the rest: which forms have any symmetry beyond the block, and whether any form but $E_8$ has a reflection.

The prime cases have been walked before. A five-fold lattice has one free number folded the four-dimensional family onto a segment with $A_4^*$ and $A_4$ at its ends and twenty symmetries everywhere between, and a seven-fold lattice lives on a folded torus found three fixed points on a torus with nothing between them and the generic count. In both, a member is either generic or one of a very few special lattices. Thirty is different in kind, and the difference can be stated before computing anything, because thirty is composite. Its field contains smaller fields, so a number that determines a form can lie in a subfield, and that should do something.

It does. And the walk finds something it was not looking for, in the ring the whole subject is most famous for. $E_8$ turns out not to be an isolated point of the family at all. It is the end of a line, and that line is $E_8$’s own ring under a sliding choice of form. All along it the group stays the same size, without one reflection, until the last point, where it jumps by a factor of nearly fifty thousand.

The family, in one line #

The setup follows the earlier essays, and the restriction with no lattice assumed is why a block of this order needs eight dimensions at all. A block $\Phi_{30}$ makes a lattice into a module over $\mathbb{Z}[\zeta_{30}]$, the integers of the thirtieth roots of unity. Since $-\zeta_{15}$ is a primitive thirtieth root of unity, that is the same ring as $\mathbb{Z}[\zeta_{15}]$, and it has class number one, so every such module is the ring itself. All the freedom is in the form, and the forms it preserves are

$$\langle x, y \rangle = \mathrm{Tr}(\alpha\, x\, \bar{y}),$$

for a number $\alpha$ in the real subfield, positive at every real embedding. The real subfield has degree four, with integer basis $1$, $\eta_1$, $\eta_2$, $\eta_3$, where $\eta_k = \zeta^k + \zeta^{-k}$. So up to scale $\alpha$ has three free numbers, and the earlier essay called this a four-parameter family counting the scale. Units act on it and fold it, as in the prime cases, into a compact three-dimensional region.

Any form in the family has at least the block’s own symmetries, the thirty powers of $\zeta_{30}$, and complex conjugation $x \mapsto \bar x$, since $\alpha$ is real. That makes sixty, the minimum.

The sampling here takes every primitive totally positive $\alpha$ with coefficients at most four in size and leading coefficient at most sixteen, 5,902 forms, reduces each Gram matrix, and counts its symmetries by backtracking on short vectors found by the Fincke–Pohst method. It stops counting at 3,000, since $E_8$’s 696,729,600 cannot be listed and need not be. The sample is a box and not the whole of a region that folds in a complicated way. That caveat matters at the end.

The result has several levels, and the middle ones are what the prime cases do not have. Sixty is everywhere. A hundred and twenty lies on a set of forms of lower dimension, with two lattices at 1,440 among them. Above that are the few forms the count could not finish, and those need a section of their own.

A line with a known name #

The level at 120 is explained by the arithmetic of the field, and this is where thirty’s being composite comes in.

The real subfield contains one quadratic field, $\mathbb{Q}(\sqrt{5})$, and the number $\eta_3 = \zeta^3 + \zeta^{-3}$ lies in it. Suppose $\alpha$ is a combination of $1$ and $\eta_3$ alone. Then the ring splits as $\mathbb{Z}[\zeta_{15}] = \mathbb{Z}[\zeta_3] \otimes \mathbb{Z}[\zeta_5]$, the trace splits, and $\alpha$ belongs entirely to the second factor. So the form is a Kronecker product:

$$\mathrm{Tr}_{15}(\alpha\, x \bar{y}) \;\cong\; \mathrm{Tr}_{3}(u \bar{v}) \otimes \mathrm{Tr}_{5}(\alpha\, s \bar{t}).$$

The first factor is the trace form of the cube roots of unity, which is $A_2$. The second is a member $L$ of the family in the five-fold essay, the segment with $A_4^*$ at one end and $A_4$ at the other. So the whole of that four-dimensional family sits inside this eight-dimensional one, as a line of forms $A_2 \otimes L$. The Kronecker product was built directly and compared with the form on the fifteenth roots of unity, and the invariants agree at every point tested.

Now the counts have an explanation. $A_2$ has twelve symmetries and $L$ has twenty, and a tensor product takes the product of the groups, except that $-1 \otimes -1$ is the identity, so the count is $12 \times 20 / 2 = 120$. At the ends, $L$ has 240, and $12 \times 240 / 2 = 1{,}440$. That is exactly the order of $A_2 \otimes A_4$ that the earlier essay found by backtracking. It now has a place: it is one end of this line, and $A_2 \otimes A_4^*$ is the other. The trace form on the fifteenth roots of unity, $\alpha = 1$, is that second one.

So the earlier essay’s second lattice, which looked as though it might stand beside $E_8$ as another isolated point, is in fact the corner of a smaller family. The block’s being composite gives it the line. Neither prime essay has anything like it, since a field of prime degree has no subfield of the kind to stop in.

The second line #

Off the tensor line there is almost nothing but the generic sixty. In this sample three sets of invariants break that, and all three are beyond the count’s limit of 3,000. Raising the limit, one of them stops at 14,400, one at 28,800, and the third does not stop.

The third is $E_8$. It has 240 short vectors, every one a root, and determinant one when rescaled, and its Hermite number is exactly two, the highest in eight dimensions. In eight dimensions one lattice wins at every width showed how far this lattice stands from anything else of its kind.

The other two have 120 short vectors, with Hermite numbers $1.7889$, which is $4/\sqrt{5}$, and $1.8091$. One hundred and twenty looks like something. It is the number of icosians, the unit quaternions forming the binary icosahedral group, whose points are the vertices of the 600-cell. The icosian ring is the set of integer combinations of those units, a rank-four module over $\mathbb{Z}[\tau]$ and so of rank eight over the integers. It is the ring from which $E_8$ is classically constructed. The norm of an icosian has the form $a + b\sqrt{5}$, and the construction takes the rational number $a + b$, a twist of the obvious choice, and gets $E_8$.

The obvious choice would be the plain trace, $2a$. And there are others in between, since any totally positive $\beta = u + v\sqrt{5}$ gives a rational form by taking the trace of $\beta$ times the norm.

The computation builds the 120 units as explicit quaternions, reduces them by integer elimination to a basis of rank eight, and writes down these forms. The twisted one has the invariants of $E_8$, and an isometry carries it onto the form $\alpha = 2 + 2\eta_1 - \eta_2 - 3\eta_3$ from the sample. The plain one has exactly the numbers of the lattice with 28,800, and a search finds an isometry onto $\alpha = 6 + 3\eta_1 - \eta_2 - 2\eta_3$, scaled. The third, with 14,400, is the same ring with $\beta = 10 - 3\sqrt{5}$, and it also has an isometry onto a form of the sample, $\alpha = 10 + \eta_1 + 4\eta_2$. Nothing asked for any of this. The block’s family was sampled by coefficient, and the icosians came back.

So $E_8$ is the end of a line, running from the plain-trace lattice, and the line runs through the family as the tensor line does. The counts have a clean reason. A form on the icosian ring is preserved by left and right multiplication by the 120 units, $120 \times 120$ pairs of which each map is given twice, for 7,200. Conjugating the quaternion doubles it to 14,400, and that holds for every $\beta$, which is the count along the line. At the plain trace something further appears and doubles it again. At the $E_8$ end, the group is 48,384 times larger than it is one step away. This has the same shape as the five-fold segment and the tensor line: a one-parameter family with a constant group, special at two ends. Only the size of the jump is unlike anything else on this page.

A lattice with no mirror #

The earlier essay’s direct question was whether any form but $E_8$ holds a reflection. The test that first suggests itself is to ask, of each short vector $v$, whether $2\langle v, x\rangle/\langle v, v\rangle$ is an integer for all $x$, since that is what it means for the reflection in $v$ to be an integer map. It finds 120 roots in $E_8$ and none in anything else in the sample. A test on short vectors alone, however, cannot rule out a longer root.

So for the lattices whose groups can be listed, the group is listed. A reflection in eight dimensions is an involution of trace six. The plain-trace icosian lattice has 28,800 symmetries, each written as a matrix, and none of them is an involution of trace six. The same goes for the 14,400 inside the line, and for $A_2 \otimes A_4^*$ and its 1,440. That is a statement about the complete group, not only about short vectors.

That is a striking contrast to carry away. It is the same ring with the same 120 units as vectors throughout, and the same left and right multiplications. At one end of the line those units are half of the 240 roots of the largest exceptional root system, and the group is generated by reflections, with 696,729,600 elements. A small step along the line, the same 120 vectors are all the shortest vectors, the group is less than a forty-thousandth the size, and it contains no reflection at all. Four root systems, and the same four rotations has shown how much of what a lattice can do is tied up with its reflections. The difference between them is a difference in the single linear functional that takes $a + b\sqrt{5}$ to a rational number, and every value of that functional but one gives no reflection.

There is a similar statement in a different field. The seven-fold essay found its special lattices at the fixed points of the Galois action, lattices whose number $\alpha$ is taken to a unit multiple of itself by the field’s automorphism. That suggests the same here. The family’s ring has a Galois group of order four on its real subfield. The two lines would then be the two components of the fixed set of its element of order two, and the four end points the fixed points of the elements of order four. Showing that means finding those maps acting on each form, and that has not been done, so this paragraph records a pattern and not a result.

Symmetry against packing #

The picture at the head of this essay has a second axis. Packing and symmetry are usually discussed as though they go together: the densest lattice in a dimension is generally a very symmetric one. In the family here that is true at the top, where $E_8$ sits, and it is not true elsewhere, which can be seen from the shape of the two lines.

Packing changes continuously along a line and symmetry does not. On the icosian line the Hermite number rises steadily from $1.7889$ to $2$, and the forms just short of the end pack almost as well as $E_8$, with fourteen thousand symmetries rather than seven hundred million. In the same box of coefficients there are forms with only sixty symmetries, the fewest possible, that reach $1.8765$. Those pack better than the plain-trace lattice with 480 times as many symmetries, and the most symmetric forms on the tensor line reach only $1.5444$. That is the sharpest disagreement on this page with the idea, tested in the densest lattice in the plane, that the best packing is the most symmetric one.

The earlier essay asked whether any other form in this family comes near $E_8$ on packing. The answer is yes and it is less interesting than it first sounds. The family is continuous, so forms arbitrarily close to $E_8$ exist with exactly sixty symmetries, and the Hermite number is continuous across them. What is exceptional about $E_8$ is that it is a local maximum, a perfect form, not that everything else is far away. The numbers quoted depend on the box, and $1.8765$ would rise with a larger one.

What the walk does not reach #

The sample is a box of coefficients, and it can be misleading. Each lattice in the family recurs many times over in any sufficiently large box, because units carry the coefficients around, and the box is not a fundamental region. So the count of forms in each row is not a measure of how common each kind is, and the number of distinct sets of invariants, 1,544, keeps growing with the box. A form with a group larger than 60 that needs larger coefficients, off the tensor line, would be missed. The statement that nothing else lies above sixty off the two lines means in this box. The icosian line itself shows the danger: its interior needed coefficients of ten before one member turned up at all. So a third line that begins at larger coefficients would not have been seen here.

Invariants are matched to the tensor product along that line at a sample of points, and on the icosian line an isometry has been found only for three members. The Kronecker statement itself is a piece of algebra about the ring splitting, and so it holds on the whole tensor line regardless. For the icosian line the corresponding algebra is not written out, and it rests on these three isometries and on the counts.

The $E_8$ symmetry count is known, not computed: the count stops at 3,000, and 696,729,600 is the classical order. Nothing on this page lists it.

And the other rings of degree eight have not been treated here. $\mathbb{Z}[\zeta_{16}]$, $\mathbb{Z}[\zeta_{20}]$ and $\mathbb{Z}[\zeta_{24}]$ all have class number one and all have families of the same dimension, each with subfields of its own, and they contain other familiar lattices. A complete account of eight dimensions would compare all four.

What the walk had to show #

The isometries are the strongest of these checks. Matching invariants can be a coincidence. An explicit integer map carrying one Gram matrix onto another cannot. The icosian ring is built here by a route that has no connection with cyclotomic fields, starting from quaternion coordinates, so its appearing inside the family at all is an independent test of both constructions. A mistake in the family’s Gram matrices or in the icosian’s norm would have ended the search with no map, and the three maps found are three chances for that to have happened.

Still open: the automorphism behind the two lines #

The natural explanation is one automorphism of order four. The Galois group of the real subfield is cyclic of order four, and its element of order two acts on the three-dimensional folded family as an involution. If its fixed set is the two lines found here, then the tensor line and the icosian line are one phenomenon, the two components of one set, and they should meet nothing else. Its elements of order four would have isolated fixed points, and they would act at such a point as extra symmetries, at least quadrupling sixty. Four such points are known, the ends of the two lines. Whether a form exists anywhere in the family with exactly 240 symmetries, or any count not on this page, would test that account. If the fixed sets are exactly these, the census is complete and the box does not matter. Showing it needs the fixed points of a map on a three-dimensional torus, in the unit group of a field of degree four, which is a small and finite calculation. It has not been done here.