A normaliser must permute the axes

Two hundred and nineteen over seventy-three counted the space groups as orbits. Each of the seventy-three arithmetic classes has a finite group of translation classes, and the changes of basis that carry the class’s point group to itself, its normaliser, move those classes about. One orbit is one space-group type. The census used only the normaliser elements whose entries are $-1$, $0$ and $1$, and it got 219 types, 230 when mirror images are kept apart, which are the published numbers.

The normaliser only acts modulo two turned that agreement into a proof in two places. A count made with too few normaliser elements can only be too high, never too low, so equal totals force every class. And in the six monoclinic classes, where the normaliser is infinite, its action on the translation classes depends only on the matrix modulo two, where the small elements already reach everything. That left the classes past the monoclinic system resting on the published totals and on a search of every matrix with entries up to two. The essay ended by naming what was missing. It wanted a proof that every element of those finite normalisers has entries $-1$, $0$ or $1$. That needed a bound, worked out lattice by lattice, on the size of a matrix that preserves a form.

The bound turns out to be unnecessary. The finite normalisers can be found whole, by an argument about where a lattice’s rotation axes can go.

The axes are what a normaliser cannot avoid moving #

Three steps, each short.

The first reduces seventy-three classes to fourteen lattices. A class’s point group $P$ keeps some quadratic forms, the metrics a crystal of that class can have. The integer matrices that keep every one of those forms make up the class’s Bravais group $B$, the full symmetry group of its lattice. A matrix $n$ that carries $P$ to itself carries the set of forms $P$ keeps onto itself, and so it carries $B$ to itself. So the normaliser of a class lies inside the normaliser of its lattice’s group. That is one of fourteen groups, eleven of them past the monoclinic system. The census checks the step’s hypothesis directly: every one of the seventy-three classes keeps exactly the forms its lattice’s group keeps.

The second step is the geometric one. Every rotation $g$ in $B$ has an axis, a line of lattice points it leaves fixed, and along that line lie two primitive lattice vectors, $v$ and $-v$. If $n$ normalises $B$, then $n g n^{-1}$ is again a rotation in $B$, and its axis is $n v$. An invertible integer matrix sends primitive vectors to primitive vectors. So $n$ sends the finite set $V(B)$ of primitive axis vectors into itself, and being invertible it permutes them.

The third step is counting. Past the monoclinic system a lattice’s rotation axes point in at least three independent directions, so $V(B)$ contains three linearly independent vectors. A linear map is fixed by what it does to three independent vectors. So $n$ is determined by three choices, each from a finite list. Trying every choice, keeping the results that are integer matrices of determinant $\pm 1$, permute $V(B)$ and carry $B$ to itself, gives every element of the normaliser. Nothing is missed, because nothing outside the list can be a normaliser element.

That is the whole argument. It needs no bound on the entries, because it never searches a box. It enumerates the only places a normaliser element can send three vectors.

What the enumeration finds #

The picture at the head of this essay is the census of all fourteen. On the triclinic lattice the group is the identity and the inversion, and the inversion has no axis, so there are no axis vectors at all. On the two monoclinic lattices there is one two-fold axis and so two axis vectors, $\pm v$, on a single line. Three independent vectors cannot be chosen, the enumeration has nothing to run on, and these are exactly the three lattices whose normaliser is infinite. The method fails at the point where the answer stops being finite, and for the same reason. One axis leaves the plane perpendicular to it free, and in that plane every change of basis is allowed.

The orthorhombic lattices have six axis vectors, $\pm a$, $\pm b$ and $\pm c$ in a primitive basis or their equivalents on a centred one. The tetragonal lattices have ten, the rhombohedral eight, the hexagonal fourteen and the cubic twenty-six. The number of triples tried is the cube of that count: 216 for an orthorhombic lattice, 17,576 for a cubic one. Eleven normalisers come back, and every one of them is exactly the set of matrices with entries $-1$, $0$ and $1$ that carry the lattice’s cone of forms to itself, which is the set the census used. Not one element was missing from that set and not one was added.

The largest entry anywhere in the eleven normalisers is one. That is the statement the earlier essay wanted, and it now follows from the enumeration rather than from a bound.

The axes decide the group #

The enumeration applies its tests in order, and the order shows how much each one does. A triple of images gives a rational matrix, and most triples fail at once, because the matrix is not integral or not unimodular. On the tetragonal P lattice, 1,000 triples leave 240 unimodular integer matrices. Of those, only 16 send every axis vector to an axis vector. On the hexagonal lattice, 2,744 triples leave 432 unimodular matrices and then 24. On the primitive cubic lattice, every triple of vectors with entries $-1$, $0$ and $1$ is a triple of axis vectors, so the 17,576 triples give all 6,960 unimodular matrices with such entries, and only 48 of them permute the twenty-six axis vectors.

The last test, that the survivor carries the group to itself, never rejects anything. On all eleven lattices every matrix that permutes the axis vectors is already a normaliser element. That was not required by the argument, which needs only the other direction, and it says something the argument did not ask: the set of axis vectors determines the group. A change of basis that respects where the axes are respects the rotations about them too. Nothing in the enumeration says why a permutation of the vectors cannot send a three-fold axis to the line of a two-fold one; it simply never does. This is observed on eleven lattices rather than proved, and it is the reason the enumeration’s final filter could be dropped without changing a single answer. It is kept, because a test that is never needed in three dimensions is cheap insurance against a dimension where it would be.

The orbit structure also explains the sizes in the hero figure. A permutation of the axis vectors that comes from a linear map must preserve every linear relation among them. On a cubic lattice the twenty-six vectors satisfy many relations, such as the sum of two face diagonals being twice an edge, and only forty-eight permutations respect them all. On an orthorhombic lattice the six vectors satisfy only the relations $v + (-v) = 0$, so every signed permutation of three independent axes respects them, and the count is $3! \times 2^3 = 48$ again, by a completely different route.

Three orthorhombic lattices with a cubic normaliser #

The sizes are worth a look, because they are not the sizes of the lattices’ own groups.

The primitive orthorhombic lattice has eight symmetries, but its normaliser has forty-eight, six times as many. The three two-fold axes lie along $a$, $b$ and $c$, and relabelling the axes is a change of basis. Swapping $a$ and $b$ does not preserve the lattice’s metric, because $a$ and $b$ have different lengths. But it carries the group of the lattice to itself, since that group is three perpendicular two-fold rotations and the swap merely reorders them. So the normaliser contains all six permutations of the axes, with all signs, which is the group of the cube. The same happens on the body-centred and face-centred orthorhombic lattices, whose centrings are symmetric in the three axes. On the C-centred lattice the centring singles out the $c$ axis and only $a$ and $b$ may be swapped, so the normaliser is sixteen, twice the group.

Every other lattice past the monoclinic system turns out to be its own normaliser. The tetragonal, rhombohedral, hexagonal and cubic groups have nothing more to give. Their axes come in kinds — the four-fold axis and the two-folds of a tetragonal lattice, the three kinds of axis of a cubic one — and a change of basis must keep the kinds apart. The group already does every permutation of each kind that the lattice allows. That self-normalising is what the normaliser is not a function of the group needs in the other direction: there an accidental extra symmetry of the metric enlarges what relabelling can do, and here the full group of the metric already contains all of it.

Coordinates that are not small #

There is a tempting shortcut to the conclusion, and it is wrong. A normaliser element permutes axis vectors, so perhaps its entries are bounded by theirs, and perhaps the axis vectors all have small coordinates.

On the face-centred cubic lattice, written on a primitive basis, the vector along a three-fold axis is $(3, -1, -1)$ or one of its relatives. On the body-centred cubic and tetragonal lattices and on the hexagonal one, axis vectors have a coordinate of two. So the coordinates of the axis vectors do not bound the normaliser by one. A matrix’s columns are the images of the basis vectors, not of the axis vectors, and the basis vectors of a centred lattice are not axis vectors. The entries of the normaliser come out at one for a reason the enumeration finds rather than one it assumes. That is why the earlier essay’s suggested route, a bound computed from a form, was not taken: a bound would have had to come down from three to one, and here it is simply not needed.

Sixty-five classes, and the counts again #

From the eleven lattice normalisers the normaliser of each class follows by one more filter: its elements are the ones that carry the class’s own point group to itself. There are sixty-five classes past the monoclinic system. Fifty-six of them have translation classes to act on and nine have none, so they carry one space group each with nothing to identify.

In all sixty-five classes the whole normaliser is the set the census used, element for element. The type counts are therefore the census’s counts. Recomputed with the whole normaliser, as orbits on the translation classes, they come out the same in every class, and the same again with only the determinant-one elements, which keeps mirror-image pairs apart.

So the totals now stand without anything outside the computation.

The triclinic classes have nothing to act on. In the monoclinic classes the reduction modulo two settles the infinite normaliser. Everywhere else the whole finite normaliser is in hand. The sum is 219 types, and 230 when the eleven pairs of mirror images are counted twice. The earlier proof depended on the published totals to force the counts class by class. This one does not: every class is counted with its whole normaliser, or with the part of it that acts, and the totals are a result rather than an input. The figure lists them by family: 2 triclinic, 13 monoclinic, 59 orthorhombic, 65 tetragonal, 7 rhombohedral, 38 hexagonal and 35 cubic, with the enantiomorphic pairs adding 3, 7 and 1 in the last three of those that have any.

That does not make the 230 a theorem proved from nothing, and it is worth saying exactly what it still stands on. The seventy-three classes are taken from seventy-three, without a search, which derives them from the fourteen lattices. The translation classes of each are the Smith forms of seventy-three Smith forms. And the rule that two space groups are the same type exactly when a normaliser element joins their translation classes is a theorem, Bieberbach’s second and the counting that follows it, which the same group means the same pattern works through in the plane. What this essay removes is the last place a published number entered: the totals that the earlier proof used to close the classes past the monoclinic system.

Why three is the number #

The argument hangs on three independent axes, and the number three has a meaning beyond the dimension. A lattice group with axes in only one direction leaves a plane in which its normaliser can do anything, so the normaliser is infinite. The axes have to pin down the whole space for the normaliser to be finite. Pinning down the space takes as many independent directions as there are dimensions, and with them the normaliser is finite and found by listing.

This is a lattice-group version of a fact the average that makes it finite used about forms. A finite group of integer matrices keeps some positive definite form, which is why it is finite. A normaliser keeps something too, less than a form: a finite set of directions. When those directions span space, the set is a cage, and only finitely many matrices fit inside it. When they do not span, the cage has an open side.

The same reasoning predicts where the method would fail in four dimensions, where finitely many is not few counts 4,783 groups. A four-dimensional lattice group whose rotations fix only a plane’s worth of directions, or none, would leave its normaliser room to be infinite. Its axis vectors would not reach four independent directions, and the enumeration would have nothing to run on. That is a prediction about the method’s reach, not a result about four dimensions, and it has not been tested.

What the pictures cannot show #

The stereograms show where the axes point and the vectors along them. They do not show the normaliser, which is a set of matrices, and the claim that every normaliser element is a permutation of those points is made in the prose, by the argument about $n g n^{-1}$, not by anything drawn. The square-per-class figure shows counts, and the counts are what the census already had. Its content is that they did not change when the whole normaliser replaced the small one, and no picture can show that a set is complete.

There is a convention in all of this, and the result depends on it. Entries $-1$, $0$ and $1$ are a statement about a basis. The census works on a primitive basis of each lattice, chosen once for each of the fourteen. In the conventional centred bases of the International Tables, where an F-centred cell has four lattice points, the same normaliser has entries of a half, and in a badly chosen primitive basis it could have entries in the hundreds. The claim is that in the census’s own bases, the ones every count was made in, the normaliser needed nothing it did not have.

Who did this first #

The space groups were first enumerated in 1891 by Fedorov and by Schoenflies, independently, and their lists were reconciled by correspondence. The method of counting them as orbits of normalisers on cohomology is Zassenhaus’s, from 1948, and it is how computer enumerations in higher dimensions have worked since. The International Tables give the normalisers of all 230 groups, Euclidean and affine, in a volume of their own. What this essay adds is not a new fact about any normaliser but a way of knowing it is complete. The normalisers here were not looked up, and the argument that they are whole does not depend on having looked them up.

What the argument has to refuse #

The first refusal is the shortcut of the section on coordinates, refused by the four lattices whose axis vectors carry a two or a three. The second is the boundary of the method. On a monoclinic lattice the two axis vectors lie on one line, no three are independent, and the enumeration declines rather than returning a wrong finite answer. A version of the method that quietly enumerated permutations of two vectors would have found four matrices and reported the monoclinic normaliser finite, which the reduction modulo two shows it is not. The test that matters most is the second of those that hold. It could have found a two, and then the census’s set would have been incomplete and some count possibly wrong. It found none, on eleven lattices, by a search that cannot miss.

Still open: a lattice group with too few axes #

The argument works because a lattice group past the monoclinic system has rotations about three independent axes. Some finite groups of integer matrices in three dimensions have fewer: the cyclic groups generated by a single rotation, which are point groups of classes rather than lattice groups. The argument never needed them, because it ran on the lattice’s group, which always has more axes than its classes do. In four dimensions there are lattice groups with no rotation axes at all. Their rotations turn two planes at once and fix no line, as three rotations of order twelve found for the operations of order twelve. A normaliser there would have to permute something other than axes: the planes, or the eigenspaces in general. Whether such a finite set always exists for a finite four-dimensional normaliser, and whether it is small enough to enumerate, is the next form of the same question.