Field

What a lattice forbids

Only two-, three-, four- and six-fold rotations are compatible with periodicity. The proof takes one line and the exception took a Nobel Prize.

The five rotations a lattice will carry. One motif and every rotation a plane lattice permits: orders 1, 2, 3, 4, 6, and nothing else up to 12. Each panel turns the motif by its own operation as many times as the order allows, on the lattice that operation requires — oblique for the identity and the half turn, hexagonal for the third and the sixth of a turn, square for the quarter. The trace printed under each is the sum of the diagonal of the operation's matrix written in the lattice's own basis, and it is a whole number in every panel, which is the entire content of the crystallographic restriction. The list of orders is produced twice, once from that trace condition and once from the degree of a cyclotomic polynomial, and the figure refuses to draw if the two disagree.

The crystallographic restriction

A repeating pattern may have rotations of order two, three, four or six, and nothing else whatever. The proof is one line of arithmetic, and everything finite in the subject descends from it.

Assuming a 5-fold rotation. The shortest lattice vector, its rotated copies, and the combination of them that is itself a lattice vector. Where that combination comes out shorter than the vector assumed shortest, the assumed rotation cannot exist.

Why five-fold is impossible

A second proof, geometric rather than algebraic: assume a five-fold centre, and out of it construct a lattice vector shorter than the shortest one there is.

The 48 point symmetries of a cubic lattice. Every operation that maps a cubic lattice onto itself, built as the integer matrices preserving that system's metric: 48 of them, of which 24 are proper rotations and 24 reverse handedness. The orders occurring among the rotations are 1, 2, 3, 4 — the same list the plane gives, so the crystallographic restriction does not change in three dimensions, and there is no six anywhere. The 13 rotation axes are counted from the rotations they carry rather than drawn from memory, and every rotation but the identity is checked to belong to exactly one of them.

The restriction in three dimensions

Space is roomier than the plane in every other respect, so the natural expectation is that it permits more rotation orders. It permits exactly the same five, and seeing why is more interesting than the result.

An integer matrix of order 5. The companion matrix of the 5th cyclotomic polynomial has whole-number entries and order exactly 5, so it is a genuine 5-fold symmetry of a 4-dimensional lattice. The plane it rotates sits at an irrational angle to that lattice, and the lattice's shadow on it is dense — which is why a projection needs a window before it becomes a pattern.

Where five-fold becomes legal

A five-fold rotation with whole-number entries exists — in four dimensions, as a four-by-four matrix that can be written down. The plane forbids it because the plane is too small, and knowing which dimension is large enough changes what a quasicrystal is.

Every solution of the axis equation. The integer solutions of 2 − 2/N = Σ(1 − 1/nᵢ), which is what counting the pairs (rotation, fixed pole) two ways gives. Two classes of axis force n₁ = n₂ = N and give the cyclic groups; three classes give the dihedral family and exactly three sporadic answers — (2, 3, 3), (2, 3, 4) and (2, 3, 5), of orders 12, 24 and 60, which are the rotation groups of the tetrahedron, the octahedron and the icosahedron. Four classes are impossible, because four terms of at least a half already exceed the left-hand side. Nothing about crystals has been used.

Before the lattice has a say

Every finite group of motions of the plane is a Cₙ or a Dₙ, and every finite group of rotations of space is one of five families. Both lists come out of counting rather than out of crystallography — and then the crystallographic restriction deletes almost all of them, leaving eleven.

A 5-fold cluster in a crystal that has no 5-fold axis. A cluster of 10 points with an exact 5-fold axis at the centre of each cell, repeated by the lattice. Two measurements, on the same points. The cluster is carried onto itself by a turn of 72° to within 2e-16 of a cell — exact, as far as the arithmetic goes. The pattern is not: applying the same turn about a lattice point sends some atoms 1.19 of a cell from the nearest atom, which is most of the way across it. Both are true at once. The axis is a symmetry of the contents of one cell and not of the crystal, which is what non-crystallographic symmetry means and why a virus with a sixty-fold capsid can crystallise in an ordinary space group.

A fivefold axis in an ordinary crystal

A virus with sixty-fold symmetry crystallises in a space group that has none of it. The restriction forbids a fivefold axis to the lattice and says nothing about what sits inside one cell — so the axis is exact, the crystal genuinely lacks it, and both statements are measurable on the same set of atoms.

$p2 in one cell, p4 on average. On the left, a molecule in one orientation at a site whose symmetry is larger than its own: the arrangement has 2 operations and the detector says p2. On the right, the average over the 2 orientations the site offers, which is what a diffraction experiment measures because different cells choose differently and nothing prefers one choice. The average has 4 operations — it is p4 — and every atom in it is present in half of the cells. Both groups are detected from the point sets rather than assumed, and the difference between them is the reason a refined structure can have symmetry no molecule in the crystal has.$

The symmetry of an average

A diffraction experiment measures an average over some 10²⁰ unit cells, and the average of several orientations is more symmetric than any of them. So a refined structure can carry symmetry that no molecule in the crystal has — including, in the worst case, a centre of inversion in a crystal built entirely of one hand.

Thirteen ways to hold a lattice

The crystallographic restriction is about one matrix. A crystal has a whole group of them acting on one lattice at once, and asking how many such groups there are gives thirteen — not the ten of the plane point groups, and not the seventeen of the plane groups.

The most of an icosahedron a crystal can keep

C₆₀ sits in crystals and virus capsids sit in crystals, and neither of them stops being icosahedral. What a lattice can fix is a subgroup — and the largest crystallographic subgroup of the sixty rotations has order twelve, at index five. The five are Kepler's five cubes.

Why the seventeen is a number at all. The classification is finite because three counts in a row are finite, and the first two are where the work is. Finitely many lattice types, because a lattice's symmetry group is a finite group of integer matrices; finitely many such groups, by Minkowski's lemma and his bound; and finitely many ways to attach translations to each, which is the extension problem. Every step is a count this site makes elsewhere — five, thirteen, seventeen — and this is the reason each of those searches was allowed to stop.

Why there is a list at all

Five lattices, seventeen groups, thirty-two classes, two hundred and thirty. Every one of those counts came out of a search that had to know when to stop, and the reason it could stop is a divisibility Minkowski proved in 1887.

Modulo 3 injective on all thirteen, modulo 2 on 5. Minkowski's lemma says the kernel of reduction modulo an integer of at least three is torsion-free, so a finite group of integer matrices is carried faithfully into a finite group of matrices over ℤ/3 — which is why the classification is finite, before any bound is computed. The middle column checks it on every finite subgroup of GL(2,ℤ) there is: thirteen classes, no collapses. The right column is the case the lemma has to exclude. Modulo 2, minus the identity is the identity, and 8 classes lose operations.

Reduction modulo three

A finite group of integer matrices survives being reduced modulo three: no two of its operations collide. That single fact proves the classification finite without computing any bound — and modulo two it is false, refuted by the inversion centre.

Five solids from one inequality

Five families of rotation group in space, five regular solids, three regular tilings of the plane and an endless supply of hyperbolic ones — all of it is 1/p + 1/q compared with a half, read at its three signs.

Averaging a metric over the group. The 3 pale ellipses are the unit circle carried by each element of a finite group of rational matrices — none of them a rotation, because the group has been skewed out of the orthogonal ones on purpose. Their average is the heavy ellipse, and it is invariant: MᵀAM = A for every element, exactly, in rational arithmetic. So a finite group of matrices is always a group of isometries of some inner product, and every question about how large such a group can be becomes a question about the symmetries of an ellipse. The space of invariant forms here is 1-dimensional, so up to scale the average is the only one.

The average that makes it finite

Two arguments every classification leans on are usually assumed rather than made: that a finite group of motions fixes a point, and that a finite group of integer matrices preserves a metric. They are the same trick — average over the group — and the trick fails exactly where it should.

60 vertices, 12 pentagons. A closed net with three edges at every vertex: 60 vertices, 90 edges and 32 faces, of which 12 are pentagons and 20 are hexagons. The pentagons are picked out in the second colour. Their number is not a property of this cage — it is twelve for every closed trivalent net of pentagons and hexagons, at any size, and the hexagon count is free.

Twelve pentagons, and no way round them

The crystallographic restriction forbids a five-fold face in a flat repeating net. Curve the net into a closed cage and the same three lines of arithmetic require exactly twelve of them — at any size, with the hexagon count free. What a lattice forbids, closing up compels.

The same group means the same pattern

Seventeen patterns is not the same statement as seventeen groups. Two patterns that look nothing alike could in principle have symmetry groups that are abstractly the same, and then the classification would be a classification of drawings. Bieberbach's theorem says they cannot — and the affine map that proves it can be recovered from the two operation sets alone.

5 units of 70.53°: 7.36° left. 5 tetrahedral units of face-centred cubic metal, each the mirror image of its neighbour in a {111} plane, arranged about a common ⟨110⟩ edge. The angle between two such planes is arccos(1/3) = 70.53°, computed from the plane normals rather than quoted, and 5 of them come to 352.64°. The shaded sector is what is left over: 7.36°, or 2.04 per cent of a full turn, which must be taken up by strain, by a gap, or by a defect along the axis.

Five copies, and the gap they leave

Gold, silver and silicon grow particles with a five-fold axis down the middle, out of a lattice that forbids one. Nothing is violated: five tetrahedral pieces of ordinary face-centred metal, each the mirror image of its neighbour, come to three hundred and fifty-two and a half degrees rather than three hundred and sixty — and the seven degrees left over have to go somewhere.

orders 5 and 7 reach a site of symmetry 1 and no more. A molecule whose only symmetry is one n-fold axis, and the highest site symmetry it may occupy in any of the 45 space groups this site builds. The site's symmetry has to be a subgroup of the molecule's, so the site's order must divide n and the site group must be cyclic. Orders 1, 2, 3, 4 and 6 reach a site of their own order. Orders 5 and 7 reach one, because no site symmetry in any space group contains an operation of order five or seven — the orders available are 1, 2, 3, 4, 6, computed by asking every operation of every group whether it moves a point. A five-fold molecule keeps its axis; the crystal simply has no use for it.

What a molecule gives up to sit in a crystal

A molecule brings its own symmetry. A crystal offers sites with symmetries of their own, and the two have to be compatible — the site's symmetry must be a subgroup of the molecule's. So a molecule may always keep more than its site offers, and a molecule with a five-fold axis may sit only where the crystal offers nothing at all.

Rotation orders 1, 2, 3, 4, 6 and no others. Every net in this collection, with the orders of the rotations its own symmetry group has, and the degrees of its vertices beside them. The orders are 1, 2, 3, 4, 6 — the crystallographic restriction, arrived at with no length anywhere in the argument: the translations of a net are ℤ² by construction, an automorphism carries translations to translations, so it acts on ℤ² by an integer matrix, and an integer trace in the interval from minus two to two is one of five numbers. The degree column is there because the two are constantly confused: a net may perfectly well have vertices of degree five, and one here does.

The restriction, with no lattice assumed

The proof that only two-, three-, four- and six-fold rotations are possible is usually stated about a lattice, and every step of it turns out to need no lengths at all. A periodic graph has the same theorem, proved the same way — and a graph may have a five-fold symmetry the plane cannot receive.

The invariant degrees exist for every n; the lattice permits five of them. The reflection group with an n-fold rotation has an invariant ring generated in degrees 2 and n, for every n whatever — the dimensions on the right are counted by pairing monomials in complex coordinates, which needs no matrix and therefore no lattice. Five of these groups can be written in integer matrices, and those five are named in the middle column; the rest cannot, because a lattice has no five-fold or seven-fold rotation. The crystallographic restriction is usually a statement about traces of matrices. Here it is the statement that only five of these invariant rings belong to a crystal, and the two arguments have nothing in common but their answer.

The degrees that name the restriction

The reflection group with an n-fold rotation has invariants of degrees 2 and n — for every n, with no lattice anywhere in the argument. Which of those groups a crystal may have is then the only question left, and its answer is the crystallographic restriction arriving from a direction nobody points it from.

The sphere fixes a count; the torus fixes only a difference. Euler's relation for a trivalent net gives Σ (6 − n) pₙ = 6χ, so the surface fixes one linear combination of the face counts and nothing else. On a sphere that combination is twelve, which with no face smaller than a pentagon forces exactly twelve pentagons. On a torus it is zero, which permits any number of pentagons provided as many heptagons pay for them — and permits none at all, which is the plain hexagonal net. On a surface of two holes it is minus twelve, so heptagons become compulsory instead.

As many heptagons as pentagons

A trivalent net on a sphere must have exactly twelve pentagons. The same three lines of arithmetic on a torus give zero — which does not forbid pentagons, it makes them pay: every pentagon has to be balanced by a heptagon, and the counts are otherwise free. One rotated bond in a wrapped honeycomb makes two of each and changes nothing else.

Four angles, and the integer that picks them. Two roots at angle θ have Cartan integers whose product is 4cos²θ. Both are whole numbers and the product is below four, so it is nought, one, two or three — and each value fixes the angle between the two roots, and with it the angle between the mirrors perpendicular to them. The shaded wedge is the region the pair of mirrors folds the plane onto; the smaller it is, the larger the group they generate.

Four root systems, and the same four rotations

Two mirrors meeting at an angle generate a group. Ask that the group be finite and that a certain pairing between the mirrors come out a whole number, and the angle has only four possible values — from which the rotations that survive are of order two, three, four and six. The crystallographic restriction arrives with no lattice anywhere in the argument.

The same accounting, at every coordination number. One row per number of edges at a vertex. The bill a sphere charges is 2dχ; the face worth nothing is 2d/(d − 2), which is a whole number at three, four and six and is 10/3 at five; the faces that can pay are those with fewer sides than that; and the last column is every way of paying the whole bill with faces of a single size. At three edges a vertex there are three such ways and twelve pentagons is one of them. At six there are none, which is the statement that six-fold coordination belongs to the plane and to no closed surface at all.

The twelve belongs to the vertex

Twelve pentagons is read as a fact about closing a surface. It is not: it is a fact about three edges meeting at a point. Let four edges meet instead and the sphere charges eight triangles; let five meet and it charges twenty; let six meet and it cannot be paid at all.

Every closed surface, and the two that charge nothing. The same accounting indexed by Euler characteristic rather than by genus. An orientable surface has χ = 2 − 2g, so it only ever occupies an even row; a non-orientable one has χ = 2 − k and occupies every row from one downwards. The odd rows therefore belong to surfaces that cannot be oriented and to nothing else — and the first of them, the projective plane, charges six. Six pentagons is a bill no orientable surface presents.

The surfaces a count by genus skips

A count indexed by genus steps in twelves and lands only on even numbers. A closed surface can have any characteristic at or below two, and the odd ones belong to the surfaces that cannot be oriented — where the projective plane charges six pentagons, a bill no orientable surface ever presents.

19 site symmetries, and the counts each can impose. Every distinct site symmetry across the space groups this site builds, named by the multiset of its operation types, with the orientation counts a disordered molecule there may take. The counts are the indices of the site group's subgroups, computed by closing subsets under multiplication rather than looked up. Nearly every row offers every divisor of its order. One does not: a site of order twelve whose group is the tetrahedral rotation group refuses an orientation count of two, because that group has no subgroup of order six.

How many orientations a disorder needs

A molecule at a site with more symmetry than it has resolves the contradiction by occupying several orientations at once. How many is not fitted: it is the index of the molecule's symmetry in the site's, so the occupancy is the reciprocal of a whole number — and one site in the census refuses a divisor of its own order.

Eleven, eleven and ten

Twenty-one of the thirty-two crystal classes contain a mirror, a centre or a rotoinversion, and not one of them is a new group. Each is a group of rotations with the inversion added, or a group of rotations with half of itself negated — and which half is left alone is the whole of the choice.

The seven friezes rolled into cylinders are the seven axial families. Each of the seven frieze groups drawn on a strip 3 cells long, beside the same strip rolled into a cylinder so that its ends meet. A translation by one cell becomes a rotation by a 3th of a turn about the axis, a mirror across the strip a mirror containing the axis, the centre line a mirror perpendicular to it, a half-turn in the strip a half-turn about a horizontal axis, and a glide a rotation by half a cell's angle combined with that perpendicular mirror. Each cylinder's symmetry group was built from the rolled strip and again from the family's own generators, and the two agree. At n = 3 the orders are 3, 6, 6, 6, 6, 12, 12, and the last column names the crystal class each member is, coloured by whether it is proper, contains the centre, or is neither.

Seven friezes round a cylinder

A point group with one principal axis belongs to one of seven infinite families, and there are seven frieze groups. They are the same seven. Draw a frieze on a strip, roll the strip into a cylinder, and every translation becomes a turn about the axis and every glide a rotoreflection.

Where the sphere and the projective plane have no net. The number of different closed nets with three bonds at every atom and faces that are pentagons and hexagons only. On the sphere, with twelve pentagons and k hexagons for k up to 12, every count has at least one net except k = 1. On the projective plane, with six pentagons and h hexagons, each count sits under the sphere count it lifts to, since every hexagon of a projective net becomes two on the sphere. The projective counts for h = 0 to 6 are 1, 0, 0, 1, 1, 3, 3, so the projective plane has no net at h = 1 or 2: two gaps where the sphere has one. Every sphere count was found by enumeration and agrees with the published one.

A gap the sphere does not have

A net of pentagons and hexagons on the projective plane must have six pentagons, and the count permits any number of hexagons. Not every number happens. Lifting each net to the sphere turns the question into one about which cages have a centre — and the answer leaves two gaps where the sphere has one.

The occupancy does not name the disorder

A molecule disordered on a special position takes a number of orientations fixed by a group index, and its occupancy is the reciprocal. Many different disorders share one occupancy — eleven at a single kind of tetragonal site — and what separates them is which of the site's operations the molecule keeps, which the averaged structure records and the occupancy does not.

Two turns and their undoing leave a slide. A turn g by 90° about the point c and a turn h by 60° about d. The marked point p is carried back 60° about d, back 90° about c, forward 60° about d and forward 90° about c, and does not return: it arrives displaced by a vector of length 2.371, which is 4·sin 45°·sin 30°·|c − d|. Two other points put through the same four motions move by the same vector, drawn beside them, because the commutator g h g⁻¹ h⁻¹ of two rotations of the plane is a translation — (I − A)(I − B)(c − d) exactly — whatever the angles and the centres.

What forces a lattice

Every enumeration here starts from a lattice of translations, and the lattice is usually taken as given. It need not be. A group of motions that is discrete, and leaves no point far from an orbit, has to contain one — in the plane by an argument four lines long, each line a picture, and in space by an inequality whose threshold turns out to be the six-fold rotation.

The rectangle a (4, 2) tube is rolled from. A patch of honeycomb turned so that the rolling vector C = 4a₁ + 2a₂ lies along the page. C has length √28 ≈ 5.292; the shortest lattice vector perpendicular to it, T, has length 4.583; and the rectangle on the two holds 28 hexagons and 56 atoms. Rolling the rectangle so that its left and right edges meet makes one repeat of the tube. The two lines through the corner are the sheet's mirror directions nearest C: the zigzag direction along a₁ and the armchair direction thirty degrees from it. C makes an angle of 19.11° with the first and lies on neither.

The tube has a screw no lattice allows

Roll a honeycomb along one of its lattice vectors and the tube turns and climbs with a screw of order 14, 98 or 794 — orders the flat sheet could never have. The rolling keeps the sheet's translations and spends them on turns, and it keeps the sheet's mirrors only along two directions, which is why almost every carbon nanotube comes in two hands.

Whether a rolled sheet ever comes back round. Three plane lattices, each with the same rolling vector C = 3a₁ + a₂ drawn from the origin and the line through the origin perpendicular to it. A translation of the rolled pattern straight up the tube, with no turn, is a lattice vector on that line. The square lattice has one, marked T, and the tube repeats every 10 turns. The general rectangular lattice has none in this direction — only along its cell edges — and the general oblique lattice has none in any direction at all, so its rolled pattern climbs forever without returning to the same angle.

Most sheets roll into a tube that never repeats

Rolling the honeycomb along a lattice vector always gives a tube with a repeat, and that is a property of the honeycomb rather than of rolling. Over the seventeen plane groups, 567 of 1,008 rolling directions give a tube with no translation along its axis at all — and every direction of an oblique pattern is one of them.

The average is the site's orbit, with occupancies. A molecule at a site of symmetry mmm keeping a subgroup of order two takes four orientations, and the average over them is the site group's orbit of each of the molecule's atoms, every image at one over the length of its own orbit. Atoms in general positions give eight images at an eighth each, and give the same eight whichever subgroup the molecule keeps. Atoms on a locus the model keeps give a shorter orbit at a higher occupancy, drawn larger and darker, and those are the only atoms that differ between models. The total scattering is the same for every model, so all of them agree exactly at zero scattering angle.

The molecule size that hides a disorder

Two disorder models with the same occupancy leave averaged structures that differ only in a handful of partial atoms. The difference is 20% in structure factors for a ten-atom molecule and 3% for a sixty-atom one — so the data choose between the models for a small molecule and stop choosing for a large one, and seven pairs are identical at any size.

Fifty-four of the seventy-five rod groups are a rolled plane pattern. Every rod group, one dot each, grouped by its crystal class. A dot is filled when some plane pattern rolled along some lattice vector has exactly that group, and the 106 crystallographic rollings of the seventeen plane groups fill 54 of them. The eighteen improper classes are full: every one of the 43 achiral rod groups is reached. The nine proper classes are not, and the twenty-one groups named on the right are what is missing — the sixteen whose screw is one of a left- and right-handed pair, and the five bare axes with no climb at all, p1, p112, p3, p4 and p6.

A rolled sheet is never one of a pair

Rolled up along every lattice vector that gives a crystallographic tube, the seventeen plane groups reach fifty-four of the seventy-five rod groups: every achiral one and eleven of the chiral. Not one of the sixteen screws that come in left- and right-handed pairs is among them, and the reason is a single fact about how far a rolled lattice can climb.

A centre at every other ring

A census cannot settle an infinite row, and the construction proposed to settle it was a tube capped at both ends, lengthened a ring at a time. Carried out, it alternates: a centre appears at every other ring and never between, the two families it permits reach two arithmetic progressions rather than a row, and the first of them opens with exactly the cage the census found could not halve.

Everything except the hexagons

Three counts of what a closed net must carry end on the same admission: an arithmetic saying what a net must charge does not say that a net exists. Eberhard's theorem says how close the charge comes to being enough, and the answer has a shape nobody would guess — it fixes every face count except the hexagons, and the hexagons are exactly the entry it cannot see.

The fewest contacts twelve pentagons can have, by size. For every cage of pentagons and hexagons up to forty-four atoms, the number of pairs of pentagons sharing a bond. The lower line is the fewest any cage of that size achieves — 30, 24, 21, 18, 17, 15, 14, 12, 11, 10, 9, 8 — the upper line the most, and the dashed line the bound that counting edges gives: the twelve pentagons carry sixty edges between them, a contact uses two and an edge to a hexagon uses one, so the contacts cannot fall below 30 − 3h with h hexagons. The bound is attained while the hexagons are few and goes loose at five, after which each extra hexagon removes about one contact rather than three. The number of cages at each size is printed beneath, and it is the least rather than the average that the bound is about.

How close the twelve must be

The charge fixes twelve pentagons and says nothing about where they go, because it is a sum over faces and cannot see which face touches which. What it cannot see is a graph on twelve points, and the fewest edges that graph can have falls from thirty to eight over the cages a census reaches — then keeps falling at a rate that puts its first zero exactly where the truncated icosahedron is.

Three conditions, and a near-miss for each. Zassenhaus's characterisation asks a group for a normal subgroup that is free abelian of finite rank, of finite index, and maximal among the group's abelian subgroups. Four groups against those three clauses. The free group on two letters has no non-trivial abelian normal subgroup at all; the discrete Heisenberg group has one that is free abelian of rank two and maximal abelian, and its index is infinite; ℤ² × ℤ/2 has a free abelian normal subgroup of index two, and the maximal one has torsion in it. Each fails a different clause, which is what shows no clause is redundant. The infinite dihedral group passes and is crystallographic in one dimension.

Which groups a crystal could have

Bieberbach's theorem is a statement about a group acting: discrete, no point far from an orbit. Zassenhaus turned it round into a statement a group can satisfy on its own — a maximal abelian normal subgroup, free of finite rank, of finite index — and each of those three clauses is kept out of redundancy by a group that fails it and nothing else.

An orbit on a parabola, discrete and cocompact. The images of the origin under the group generated by two commuting affine maps of the plane: A slides one step along x and lifts y by the x it started at plus a half, and B is the translation by one in y. The images are the points with whole-number first coordinate and second coordinate a whole number above half the square of it, so the large dots lie on the dashed parabola and the small ones are the rest of the orbit. No two distinct images come closer than 1.000, and no point of the square between the axes lies farther than 0.610 from one — so the action is discrete and its quotient is compact, which is exactly what Bieberbach's first theorem asks for.

Straight lines, and no distances

Every finiteness met so far rests on the motions preserving a metric, because the trick that produces one is an average and an average needs something to average over. Keep the straight lines and drop the distances, and Bieberbach's first theorem is false in the plane — by an example two lines long, whose group is the plane's own translations and whose translations have rank one.

Two, seventeen, two hundred and thirty, and then. The number of arithmetic crystal classes and the number of crystallographic groups in each of the first six dimensions, with the second divided by the first. The classes multiply by between five and fourteen a dimension; the groups multiply by much more, and the quotient — how many groups an average class carries — goes 1.00, 1.31, 3.15, 6.74, 36.5 and 339. The last column says what is derived on this page and what is quoted: the plane in full, six of the seventy-three classes in space, and nothing at all above three dimensions, where the counts come from machine enumerations of the 1970s onwards.

Finitely many is not few

Bieberbach's third theorem says each dimension holds finitely many crystallographic groups and gives no idea how many. The counts are 2, 17, 230, 4783, 222018 and 28927922, and dividing them by the number of arithmetic classes says which of the classification's three steps supplies the explosion — the step that attaches translations, not the one that finds the matrix groups.

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