The normaliser only acts modulo two

Two hundred and nineteen over seventy-three counted the space groups the way the theory says to count them. Each arithmetic class has a finite group of translation classes, its first cohomology $H^1$. The changes of basis that carry the class’s point group to itself, its normaliser, move those classes around, and each orbit is one space-group type. The census came out at 219 types, and 230 when mirror-image pairs are kept apart, which are the Tables’ numbers. It reached them with a shortcut, and said so. It used only the normaliser elements whose entries are $-1$, $0$ or $1$, a finite set. In the monoclinic and triclinic classes the normaliser is infinite, and a finite subset of an infinite group could in principle miss an identification. The census therefore called its own agreement “evidence rather than proof”. It named what a proof would need: the full normaliser, or an argument that the normaliser’s action on the cohomology factors through a finite group that the small elements already generate.

This essay supplies both, and a third argument that was already available. It changes the status of the census class by class.

The count can only ever err upward. Each element the census used was checked to normalise its class. So every identification it made is genuine, and a missing element can only leave two classes apart that should have been joined. Each class’s count is therefore at least its true count. The census totals equal the Tables’ totals, and a sum of terms each at least as large as its partner, with the same total, has every term equal. On the strength of the two published numbers alone, the census was already exact in every class.

In the monoclinic classes it is exact without them. There the infinite normaliser acts on the cohomology only through its reduction modulo two. In the finite group that reduction lands in, the elements with entries $-1$, $0$ and $1$ already produce every matrix the rest of the normaliser could. The infinite part of the normaliser is real, and it is invisible to the count.

A count that can only be too high #

The census joins two cohomology classes when some normaliser element carries one to the other. It joins them transitively, so the orbits are the connected pieces of a graph whose edges are those carryings. Adding an element to the set can only add edges, and edges can only merge pieces. So an orbit count computed with a subset of the normaliser is never smaller than the count computed with all of it. The one way for the census to undercount is to use a matrix that is not a normaliser element at all. The census cannot do that: the action is refused outright on a matrix that fails to carry the point group to itself, and that refusal is tested below.

So each of the seventy-three counts is an upper bound on the true number of space-group types in its class. The earlier essay compared the sum of the upper bounds with the Tables’ total and found them equal, 219 affine types and 230 proper ones. If any class were overcounted, the census sum would exceed 219. The earlier essay read the agreement as a “severe test”, in which “a missed identification would make a total too large, an extra one would make it too small”. Only the first half of that is possible, and it is that asymmetry that turns the test into a proof. The proof is conditional on one fact from outside the computation, that the true totals are 219 and 230. The rest of this essay removes that condition where the normaliser is infinite.

What the cohomology of one class looks like #

Before the argument, the object. In the class of P2/m, the point group has four operations: the identity, a two-fold rotation, the mirror perpendicular to it, and the inversion. A space group over the class attaches a translation to each, and up to the lattice and a change of origin there are eight ways to do it. They form the group $H^1 = (\mathbb{Z}/2)^3$, which Seventy-three Smith forms computed class by class.

Each class can be read by what it does to the two operations that matter. The intrinsic translation of the two-fold, the part along its own axis, is either zero or half a lattice step: a rotation or a screw. The intrinsic translation of the mirror, the part within its own plane, is either zero or half of one of three lattice vectors in that plane: a mirror, or a glide along $a$, along $c$, or along the diagonal. The class with neither is the symmorphic group P2/m, and the class with only the screw is P2₁/m. The three glides are one space group, P2/c, because a change of basis in the mirror plane carries any of the three glide directions to any other. The three classes with a screw and a glide are likewise one group, P2₁/c. The normaliser’s job in the census is exactly this identification of the three glide directions, and it is where an infinite normaliser might in principle do something a finite subset cannot.

Where the normaliser is infinite #

The normaliser of an arithmetic class is infinite only in the triclinic and monoclinic systems. In the triclinic classes the cohomology is trivial, so there is nothing to move. That leaves the six monoclinic classes, five of which have non-trivial cohomology. C2 has none.

A monoclinic point group has at most one operation of each kind: one two-fold, one mirror, one inversion. A change of basis that carries the group to itself must carry each operation to one of the same kind, because it preserves determinant and trace. So it carries each operation to itself, and the normaliser is the centraliser: the integer matrices that commute with the two-fold, or with the mirror. Such a matrix must preserve the two-fold’s axis and the plane perpendicular to it. On the lattice those are two sublattices, a line $L_+$ and a plane $L_-$. In the plane it can do anything that preserves the lattice. That freedom is the whole of $GL(2, \mathbb{Z})$ on a primitive lattice, and a subgroup of it of index three on a centred one, and it is why the normaliser is infinite.

Halves, and therefore modulo two #

The cohomology of every monoclinic class has exponent two: each class, added to itself, gives zero. So every class has a representative whose translations are halves of lattice vectors. A normaliser element $n$ that centralises the point group acts on a representative $t$ by $(n \cdot t)(g) = n\, t(g)$. It multiplies each half-translation by an integer matrix, and the result modulo the lattice depends only on the matrix modulo two. Two normaliser elements that agree modulo two act on the cohomology identically.

So the action of the whole infinite normaliser passes through a finite group: its image in $GL(3, \mathbb{F}_2)$, the invertible $3 \times 3$ matrices over the field with two elements, of which there are 168. That is the finite group the earlier essay’s proof needed. The remaining question is whether the elements with entries $-1$, $0$ and $1$ reach every element of that image.

Two bounds that meet #

The image can be bounded from both sides without knowing the normaliser completely.

From below, it contains the reductions of the elements the census used. Those elements, and the products of their reductions, form a subgroup of $GL(3, \mathbb{F}_2)$ that can simply be listed.

From above, every normaliser element preserves the two sublattices $L_+$ and $L_-$. So its reduction preserves their reductions: the images of $L_+$ and $L_-$ in the eight-element space $\mathbb{F}_2^3$. The matrices of $GL(3, \mathbb{F}_2)$ that preserve two given subspaces can also simply be listed.

The two lattices behave differently here, and the difference is the centring. On a primitive lattice the axis and the plane are complementary. Their reductions are a line and a plane of $\mathbb{F}_2^3$ that meet only at zero, and the matrices preserving both are those acting as $GL(2, \mathbb{F}_2)$ on the plane and fixing the line: six of them. On a centred lattice the axis and the plane together span only half the lattice. The index is two, because the centring vector is half an axis vector plus half a plane vector. Reduced modulo two, the axis vector falls inside the plane’s reduction, and the matrices preserving a line inside a plane number eight.

The picture at the head of this essay is that comparison. The subgroup generated by the reductions of the census’s elements has six matrices in each primitive class and eight in each centred one, exactly the upper bound in all six classes. The image of the infinite normaliser is squeezed between two sets that turn out to be the same set, so it equals both. Every action the whole normaliser has on the cohomology is already the action of an element with entries $-1$, $0$ and $1$. The orbits the census found are the orbits of the full normaliser, and the monoclinic counts, two types over P2, Pm, C2/m and Cm, four over P2/m and one over C2, are proved without reference to any table.

The argument is not a statement about generators, and the centred classes show why that matters. Their normaliser contains matrices such as a shear by two in the plane, with an entry of $2$. In the basis the census works in, a search finds eighty normaliser elements of C2/m with entries up to two that the census did not use, sixteen of them reducing to the identity modulo two. They are genuine normaliser elements, and they reach nothing new: their reductions are already reached, and the ones that reduce to the identity act as the identity. Whether the census’s elements generate the whole normaliser is a question the argument never has to answer. They generate its image modulo two, and that image is all the count can see.

Why three glides are one, and never two #

The reduction also explains the orbit sizes of the P2/m figure, without the census. A glide in the mirror plane is a half-translation along a vector of the plane’s lattice $L_-$, and what matters about it is that vector modulo two: an element of $L_-/2L_-$, a plane with four points over $\mathbb{F}_2$. The zero vector is no glide at all. The three non-zero vectors are the glides along $a$, along $c$ and along $a + c$, the three the figure shows joined into one orbit.

On a primitive lattice the normaliser acts on that plane through all of $GL(2, \mathbb{F}_2)$, the six matrices of the upper bound. $GL(2, \mathbb{F}_2)$ is the symmetric group on the three non-zero vectors of the plane: it can send any one of them to any other. So the three glide directions must form a single orbit. Its size is three, and no class with a glide can stand alone. The screw on the axis sits in the other factor, $L_+/2L_+$, which the normaliser fixes. So the screw’s presence is invariant, and it splits each orbit into a with-screw and a without-screw version. That is four types, of sizes one, three, one and three. The arithmetic of $\mathbb{F}_2$ alone predicts the census’s orbit sizes for P2/m, and the same count gives Pm’s two types and the single orbit of three glides among them.

The centred lattice changes the count because it changes the plane. There the axis vector reduces into $L_-$'s image, and the eight matrices of the upper bound must fix that one non-zero vector of the plane while moving the other two. The three glide directions are no longer equivalent. One of the three non-zero vectors is the reduction of the axis itself. A glide by half of a plane vector with that reduction differs from the mirror by a centring translation, which only moves the mirror, so it is no glide at all. The other two are exchanged by the normaliser, which is why C2/m has exactly one space group with a glide, C2/c, and not two.

A million in the entries #

The argument predicts something testable at a scale the census never touched. Take a long product of normaliser elements, with entries growing into the thousands or the millions. Its action on the cohomology should be identical to that of a small element with the same reduction modulo two.

For each of the five monoclinic classes with non-trivial cohomology, forty products of fourteen random normaliser elements were formed. Their entries reach 1,026,641 in C2/m. Each product’s action on every cohomology class was computed directly, by conjugating the point group and transforming the translations, and compared with the action of the census element that has the same reduction. All two hundred agree. The action of a matrix with a million in it is decided by nine bits.

The other sixty-one, searched #

Past the monoclinic system the normaliser of every class is finite. The argument above does not apply to those classes, because there is no infinite part to factor away, and the question becomes whether the census’s elements are the whole normaliser. That is checked by search rather than proved.

The search is complete for entries up to two in absolute value. On every lattice past monoclinic, primitive or centred, it finds exactly the normaliser elements the census already had. It finds none with a $2$ in it, for any of the fifty-six classes with non-trivial cohomology there. On the two monoclinic lattices it finds the extra elements described above, and the counts do not move. That is evidence of the same kind as the totals and weaker than the monoclinic argument, since a normaliser element with an entry of three would escape it. Combined with the one-sided argument, the counts in those classes stand on the published totals and on this search, and nothing found contradicts either.

Two reductions, and two different jobs #

This is the second time a reduction has settled a question about the space groups, and the two uses are opposite. Reduction modulo three proved that the list of finite integer groups is finite. It works because reduction modulo three is faithful on finite groups: no two operations of a finite group collide, so the group survives the reduction intact and can be counted in a finite place. Reduction modulo two here works for the opposite reason. On the monoclinic normaliser it is deliberately unfaithful: an infinite kernel of matrices congruent to the identity is thrown away, and the argument is that the cohomology never sees that kernel.

The two primes cannot be swapped. Modulo two, a finite integer group can lose elements: $-1$ and $1$ collide. That is why the finiteness argument needs three. Modulo three, the normaliser’s action is not the right quotient, because the cohomology here consists of halves, and a half multiplied by a matrix congruent to the identity modulo three is not the same half. The prime in each argument is chosen by what the argument has to preserve.

What these pictures cannot show #

The cube of eight points and the tiles of nine squares are finite pictures of a finite quotient. They show that the image of the normaliser is small and that it is reached. They do not show the normaliser itself, an infinite group with elements of every size. The lifts figure samples that group and cannot display it. The argument that the whole group reduces into the drawn set is the centraliser argument in the prose, not anything drawn.

Nor do the pictures show the proper count. The 230 proper types come from orbits under the normaliser elements of determinant one. For the monoclinic classes that changes nothing, because the inversion lies in every monoclinic centraliser, has determinant $-1$ and reduces to the identity modulo two. So restricting to determinant one removes no action, and the monoclinic classes contain no mirror-image pairs, as the Tables confirm. The argument above covers both counts, and no figure displays the determinants.

What the argument has to refuse #

The first refused claim is the worry the census began with: that the census’s small set had missed an identification, so a larger set would join more classes. The search found every monoclinic normaliser element with entries up to two, 208 on each primitive class and 136 on each centred one, and added them. No orbit merged. The second refusal guards the one-sided argument. That argument needs every element the census uses to be a genuine normaliser element, and when the action is applied to a matrix with entries $-1$, $0$ and $1$ that fails to normalise P2/m, it is refused rather than computed.

The test that most deserves the word “test” is the one on the upper bound. It could have come out larger than the lower bound, and then the argument would have needed a finer bound or a new element. On the primitive lattices the bound is six, the full $GL(2, \mathbb{F}_2)$ acting on the plane. On the centred ones it is eight, which is a larger group than on the primitive lattices, not a smaller one, because a line inside a plane is preserved by more matrices than a line beside one. Both were reached.

Still open: a proof for the finite normalisers #

The fifty-six classes with finite normalisers and non-trivial cohomology are covered by the published totals and by the search to entries of two, not by an argument. A proof would show that every element of those normalisers has entries $-1$, $0$ or $1$ in the census’s basis. The route is visible. A class’s normaliser lies in the normaliser of its Bravais group, which is finite for every lattice past monoclinic, so it preserves a positive definite form in that Bravais group’s space of forms. Every automorphism of such a form is an integer matrix of bounded size. Whether the bound, worked out lattice by lattice, comes down to one is a finite computation about fourteen lattices, not seventy-three classes, and it has not been done here.