Fifteen halves, and a three that fixes too much

No cell joins the seven halves ended on a coincidence that it suspected was only a coincidence. A lattice in space can be halved in seven ways: there are seven sublattices of index two, one for each non-zero way of reading a lattice vector’s coordinates modulo two. The group of all changes of basis permutes the seven transitively, so as descriptions they are all one thing. But no single cell does it. For any choice of lengths and angles, the isometries of that cell split the seven into at least two classes, and the reason is short. A group that moves seven things around in one orbit has order divisible by seven, so it contains an element of order seven, and the crystallographic restriction says that no three-by-three integer matrix has order seven.

Seven is $2^3 - 1$, and that it is also an order the restriction forbids looked like an accident of the number three. The essay said where to test it. In four dimensions a lattice has $2^4 - 1 = 15$ halves, and fifteen is three times five. Three-fold rotations act on four-dimensional lattices, and so do five-fold ones: a five-fold lattice has one free number describes every lattice that carries one. So the obstacle is gone, and the question becomes a finite one about four-by-four integer matrices.

The answer is that the fifteen are still never joined, and the reason is no longer an order. It is a trace.

The arithmetic first #

Before computing anything in four dimensions it is worth knowing which dimensions the arithmetic leaves open at all, because the argument from the seven halves generalises at once. In $n$ dimensions a lattice has $2^n - 1$ halves. A group that permutes them in one orbit has order divisible by each prime factor $q$ of that number, so by Cauchy’s theorem it contains an element of order $q$. An integer matrix of prime order $q$ has the cyclotomic polynomial of degree $q - 1$ among the factors of its characteristic polynomial, and so it needs at least $q - 1$ dimensions to act at all. That is the degrees that name the restriction read backwards. So a cell can join every half in $n$ dimensions only if every prime factor of $2^n - 1$ is at most $n + 1$.

That condition is rare. The table runs to sixteen; a computation in exact integers runs to four hundred, and the dimensions that pass are one, two, four, six and twelve, and nothing else in that range. In three dimensions the obstacle is seven, which needs six dimensions. In five it is thirty-one, which needs thirty. In eight, the number of halves is 255, which is three times five times seventeen, and seventeen needs sixteen dimensions. In ten the culprit is thirty-one again, and in sixteen it is 257.

The short list has a reason, and it is a theorem from 1892. Zsigmondy proved that $2^n - 1$ almost always has a primitive prime factor, one that divides no $2^m - 1$ for a smaller $m$. Such a prime $q$ has the property that two has order exactly $n$ modulo $q$, so $n$ divides $q - 1$ and $q$ is at least $n + 1$. The condition above needs $q \le n + 1$, so the primitive factor must be exactly $n + 1$, which must therefore be a prime with two as a primitive root. Two, four and twelve are of that kind, with primitive factors three, five and thirteen. Six is the one exception Zsigmondy’s theorem allows. The number $63 = 3^2 \cdot 7$ has no primitive prime factor at all, because seven already divides $2^3 - 1$ and three divides $2^2 - 1$. So the dimension six survives the arithmetic for a reason that has nothing to do with lattices.

Dimension one is trivial, a lattice with one half. Dimension two is the plane, where a hexagonal cell’s three-fold rotation permutes the three halves cyclically, and the earlier essays on index-two subgroups of the plane groups found it doing so. Dimension four is the first case where the arithmetic permits a joining and something else has to decide.

Half a lattice is a functional #

One convention has to be fixed before any counting, and getting it wrong changes the answer. A half of a lattice $L$ is a sublattice of index two. Every such sublattice contains $2L$, the lattice doubled, so it is determined by what it does to the cosets of $2L$, a vector space of dimension $n$ over the field of two elements. A half is the set of vectors on which some non-zero linear functional on that space vanishes. So the halves are not the non-zero vectors of $L/2L$. They are the non-zero vectors of its dual, the space of functionals. A symmetry $M$ of the lattice acts on a half by acting on its functional, which means by the transpose, $f \mapsto f M$.

The distinction looks pedantic, because the two actions always have the same number of orbits. A matrix and its transpose fix subspaces of the same dimension, so every element fixes as many functionals as vectors, and by Burnside’s lemma the number of orbits is the average number of fixed points. But the sizes of the orbits can differ, and in the earlier essay they did.

The body-centred cubic lattice has an index-two sublattice that every symmetry keeps: the simple cubic lattice of its corners, without the body centres. So its halves include an orbit of size one, and the other six form the second orbit. The face-centred cubic lattice has no half that every symmetry keeps; its halves fall as three, the tetragonal sublattices with one even coordinate, and four. The earlier census acted on vectors rather than on functionals, and reported the two the other way round, body-centred as three and four and face-centred as one and six. The counts were right and the conclusion, at least two orbits on any cell, stands. The sizes were exchanged, and they are corrected in that essay now.

In four dimensions the same thing happens without any pair of dual lattices to explain it. The lattice $A_3 \oplus \mathbb{Z}$, the face-centred cubic lattice with an orthogonal line beside it, has five orbits on its fifteen vectors modulo two, of sizes six, six, one, one and one. On its fifteen halves it also has five orbits, of sizes four, four, three, three and one. Equal counts, unequal sizes, and every number below is about halves.

Seven lattices and their fifteen halves #

Any finite group of four-by-four integer matrices lies inside the full symmetry group of some lattice, since averaging any positive form over the group gives a form the group keeps. And orbits only merge as a group grows. So the question of whether some cell joins the fifteen is answered by the lattices with the largest groups, and those are the lattices the body-centred hypercube and its neighbours have already used.

The hypercubic lattice $\mathbb{Z}^4$, with 384 symmetries, sorts its halves by how many coordinates their functional reads: six that read two, four that read one, four that read three, and the one that reads all four. The checkerboard lattice $D_4$, with 1,152 symmetries, the largest group of any four-dimensional lattice, joins twelve of them and leaves three. The lattice $A_2 \oplus A_2$, two hexagonal planes at right angles, with 288 symmetries, makes nine and six. And $A_4$ and $A_4^*$, each with 240, make ten and five. The lower groups, $A_2 \oplus \mathbb{Z}^2$ and $\mathbb{Z}^3 \oplus \mathbb{Z}$, leave five and seven classes.

So the floor is two, as in space, but it is reached three different ways, by three groups of quite different shape. None of them reaches one. That is a result about seven lattices. To make it a result about every four-dimensional cell, it is enough to know which lattices could do better, and the arithmetic says exactly which.

The five decides it #

A group that joins the fifteen has order divisible by fifteen, so it contains an element of order five. In four dimensions an integer matrix of order five fixes no vector, and its characteristic polynomial is $x^4 + x^3 + x^2 + x + 1$, since that is the only way to have order five with four dimensions to fill. The earlier essay on the five-fold lattices worked out what that forces. The lattice is a module over the ring of fifth roots of unity, every such module is the ring itself because its class number is one, and the forms the rotation keeps make one family with one free number, $u + v\sqrt5$, totally positive.

The scan takes every primitive pair $u$, $v$ with $u$ up to nine and the number totally positive, twenty-seven forms in all. Each one’s symmetry group is computed by searching for every integer matrix that preserves its Gram matrix. Sixteen have twenty symmetries: the ten powers of the rotation and its negative, and the ten products with the complex conjugation that every member of the family has. Eleven have 240, and every one of the eleven is identified as $A_4$ or $A_4^*$ by an explicit isometry to the reference lattice. No other order appears. Twenty is not divisible by three, so a group of twenty cannot join fifteen halves either.

The scan is a sample of the family, not the whole of it, and the reason it can be trusted is structural rather than statistical. A group of four-by-four integer matrices containing both a five-fold and a three-fold element acts irreducibly, since the five-fold element alone fixes no line or plane over the rationals. Such a group is determined up to finitely many possibilities by the classification of finite subgroups of the four-dimensional integer matrices, which Dade completed in 1965, and among his maximal groups only those of $A_4$ and $A_4^*$ have order divisible by five. The scan finds those two, and nothing between twenty and 240, which is what the classification predicts.

So the whole question comes down to $A_4$. Its group is the symmetric group on five letters, permuting the five coordinates of a space in which $A_4$ is the lattice of integer vectors with zero sum, times the sign that reverses everything. The sign does nothing modulo two. And the census says the symmetric group makes ten and five.

The same group, acting the other way #

That would be the end of it, except that the abstract group is perfectly capable of joining fifteen things. The alternating group on five letters, the even half of the symmetric group, is isomorphic to $SL(2,4)$, the group of two-by-two matrices of determinant one over the field of four elements. That group acts on the plane over the four-element field, which has sixteen vectors, and it moves the fifteen non-zero ones in a single orbit. A plane over the four-element field is a four-dimensional space over the two-element field, written out coordinate by coordinate. So there is an action of exactly the right group, on a space of exactly the right size, that is transitive. Its order is sixty, which is four times fifteen. The lattice does not use it, and the reason is visible in a single element.

Take an element of order three in each. In $SL(2,4)$ it is, after a change of basis, the diagonal matrix with entries $\omega$ and $\omega^2$, the two primitive cube roots of unity in the four-element field. It multiplies one coordinate by $\omega$ and the other by $\omega^2$, so it fixes no non-zero vector at all. In $A_4$ an element of order three is a three-cycle of the five coordinates, and it fixes the vectors whose three permuted coordinates are equal. Its trace is one: of its four eigenvalues, two are one and two are the complex cube roots of unity. So it fixes a whole plane of the lattice, and modulo two it fixes a plane of the four-dimensional space over the two-element field, three non-zero vectors, and three halves. All twenty three-fold rotations of $A_4$ have trace one and fix three halves.

Burnside’s lemma turns that difference into the difference between one orbit and two. The number of orbits of a group is the average over its elements of the number of points each fixes. The sixty elements of the alternating group are the identity, fifteen involutions, twenty elements of order three and twenty-four of order five. The identity fixes all fifteen halves. An element of order five fixes none, in both actions, because $x^4 + x^3 + x^2 + x + 1$ has no root modulo two. The involutions fix three each in both actions. What differs is the twenty three-fold elements. In $SL(2,4)$ they fix nothing, and the average is sixty over sixty, one orbit. In $A_4$ they fix three each, which adds sixty more fixed points, and the average is 120 over sixty, two orbits. The whole difference between joining the fifteen and failing to is those sixty fixed points.

And a lattice cannot avoid them. The trace of a three-fold element is a statement about the rational representation, not about any particular lattice, and the alternating group has only one faithful rational representation in four dimensions: the permutation of five coordinates with the sum removed. Its other small representations, of dimension three, have characters involving $\sqrt5$, and a lattice’s symmetries have integer traces. So in every lattice on which this group acts, a three-fold element has trace one, fixes a plane and fixes three halves. The transitive action has trace minus two on its three-fold elements, two pairs of complex cube roots, and that is a representation no lattice can carry. What a group does to a function found characters deciding what symmetry permits; here a character decides which permutation of halves a group can make.

So the plane was the exception #

Taken together, the four-dimensional result says that the plane’s success is the oddity and space’s failure is the rule, though not for the reason the seven halves suggested. In space, the order was forbidden. In four dimensions the order is allowed and the action is not. The first obstacle is arithmetic and the second is representation theory, and the second is subtler: it needs the group’s elements to fix the wrong number of things, which no counting of orders can see.

Beyond four dimensions the computation is easiest on the root lattices, whose groups are generated by reflections and can be followed modulo two without listing their elements. A reflection in a root of norm two acts on the halves as a transvection, and closing under the simple reflections gives the orbits directly. The plane’s $A_2$ is the only one that joins all its halves. In three dimensions $A_3$, which is the face-centred cubic lattice, makes four and three. $E_6$ makes 36 and 27, a split anyone who has met the 27 lines on a cubic surface will recognise. $E_8$, whose group has nearly 700 million elements, makes 135 and 120. In five and seven dimensions the lattices drawn need three classes, and whether some other lattice there reaches two is not decided here.

Six is the dimension the arithmetic left open by Zsigmondy’s exception, and it is worth saying what is known there. Joining 63 halves needs elements of order seven and order nine. $E_6$ has no seven, and splits them 36 and 27. $A_6$, whose group is the symmetric group on seven letters times a sign, has both, and splits them by weight: 35, 21 and 7. That makes six look like four, with the necessary orders present and the action wrong. But it has not been proved here that no other six-dimensional lattice has a group containing both orders. Twelve is in the same position: $A_{12}$, the only obvious candidate with the thirteen the arithmetic demands, splits its 4,095 halves into six orbits by weight.

What the halves mean for a crystal #

Four dimensions is not only a mathematical extension. Modulated crystals and quasicrystals are described as sections of lattices in four or more dimensions, and their superstructures of index two are halves of those lattices. So the result has a direct reading. In a crystal described by a four-dimensional superspace lattice, the fifteen ways of doubling the superspace cell are never all equivalent by symmetry. However symmetric the parent, at least two different kinds of doubling exist, and they will in general show different satellite reflections in the way the reflections a superlattice adds describes for ordinary superstructures.

It is equally a statement about bookkeeping. The affine normaliser of the four-dimensional translation group, every integer change of basis, still joins all fifteen. Any half can be made “the even last coordinate” by choosing a basis. A table of subgroups that treats descriptions as equivalent lists one index-two sublattice; a table that respects a metric lists at least two. The gap between those two counts is exactly what one cell reaches every Wyckoff set found closed for positions and the earlier essay found open for halves. In four dimensions it stays open.

What the census does not reach #

The four-dimensional result is complete in one sense and borrowed in another. The orbit computations are exact: every group is listed as integer matrices, and every half is followed under every element. The completeness of the list of candidate groups is borrowed. That only $A_4$ and $A_4^*$ have groups with both a three and a five rests on the structure of the five-fold family together with Dade’s classification of the maximal finite subgroups of the four-dimensional integer matrices. The scan in the figure is evidence that agrees with it, not a proof that replaces it.

The picture of the two actions is an illustration of an argument, not the argument itself. Rings and colours cannot show a trace. What the figure shows is the consequence of the trace, three fixed halves against none, and the step from trace to fixed halves is the observation that an integer matrix fixing a plane fixes that plane’s reduction modulo two.

The convention that decides everything here is the choice of action. Halves are functionals, and the group acts on them by the transpose. Acting on vectors gives the same counts, so none of the conclusions about joining would change, but the orbit sizes would, and the earlier essay is the evidence that the two are easy to confuse.

Who worked out the pieces #

Cauchy’s theorem on elements of prime order and Burnside’s counting lemma are nineteenth-century group theory. Zsigmondy’s theorem on primitive prime divisors is from 1892, and its exception at $2^6 - 1$ is the reason six survives. Dade classified the maximal finite subgroups of the four-dimensional integer matrices in 1965. That the alternating group on five letters is the special linear group of two-by-two matrices over the field of four elements is one of the small exceptional isomorphisms, known since the nineteenth century. Hering classified the linear groups that are transitive on the non-zero vectors of a space over a finite field, which is the general setting the transitive action belongs to. The census of halves for each four-dimensional lattice, the reduction to $A_4$, the fixed-point account of why it fails, and the correction to the three-dimensional orbit sizes are what this essay adds.

What the census must refuse #

Each check can fail visibly. The orbit computation reports sizes, so a mistaken action shows up as the wrong partition, which is how the exchange of body- and face-centred cubic was noticed. The transitive action of $SL(2,4)$ is built from three matrices and closed under multiplication, and it must come out at sixty elements, one orbit, and no fixed vector for any three-fold element. The trace of every three-fold rotation of $A_4$ must be one, and the claim that $A_4$’s 240 symmetries join the fifteen halves is put to the census and refused.

Still open: the halves in six dimensions #

Four dimensions is settled and six is not. The arithmetic admits it because 63 has no primitive prime factor, and the two lattices drawn here fail it, $A_6$ with the right orders and the wrong action, $E_6$ with the wrong orders. A proof would follow the four-dimensional pattern. A group joining the 63 halves needs an element of order seven, which in six dimensions makes the lattice a module over the seventh roots of unity, a ring whose class number is one, so the lattices form a family like the five-fold one, with more free numbers. The groups in that family that also contain a nine are the candidates, and the question for each is the same fixed-point count: whether its three-fold elements fix the number of halves a transitive action allows. The seven-fold family was mapped in a seven-fold lattice lives on a folded torus, so the search would start from a known list, and it has not been run.