Field

The classification

Exactly seventeen ways to fill the plane with a repeating pattern. Not a fact to memorise: a theorem, with a finite argument and no eighteenth case.

The seventeen

Every way of repeating a pattern across a flat surface, exhibited rather than tabulated. The number is a theorem with a finite proof, and the proof is walkable in an afternoon.

The wallpaper group p4m. A pattern with the symmetry of p4m, generated by applying the group's 8 operations to an asymmetric motif and repeating across the lattice. The symmetries of the result were then found independently and match the group exactly.

Reading Hermann–Mauguin

p4g looks like a licence plate and is in fact a set of instructions. Half an hour with the rules turns the seventeen from a list to be memorised into a notation that can be read.

The wallpaper group p3m1. A pattern with the symmetry of p3m1, generated by applying the group's 6 operations to an asymmetric motif and repeating across the lattice. The symmetries of the result were then found independently and match the group exactly.

Two groups with the same lattice, the same point group and the same number of operations, differing only in where the mirrors sit. The pair is the clearest evidence that position is as much a part of a symmetry as presence.

Why p4 cannot be drawn with dots. The same group applied to a single dot and to a motif with no symmetry of its own. The dot's orbit turns out to have more symmetries than the group it was made with, so a figure drawn that way illustrates a different group from the one in its caption.

The motif must be a comma

A dot is too symmetric to illustrate most wallpaper groups. Its orbit acquires mirrors nobody asked for, and the resulting figure is quietly of a different group from the one in its caption.

The seven frieze groups. Every way of repeating a motif along a strip. Seven, and no more: the only ingredients are a translation, a mirror across the strip, a mirror along it, a half-turn and a glide, and most combinations of those turn out to generate one another.

Seven friezes

The same classification argument on a strip instead of a plane, where it is short enough to check by hand. Seven ways to repeat a motif along a line, with names like hop, step and sidle.

The seventeen sorted by lattice: 2, 5, 2, 3, 5. The five plane lattices, each drawn from the basis every other figure here uses, with the wallpaper groups that sit on it and the order of each against its lattice's holohedry. The counts are 2, 5, 2, 3, 5, which is seventeen again, arrived at by a different route from the case analysis on rotation order. Two relations hold and both are checked. Every group's order divides its lattice's holohedry, because an operation has to map the lattice onto itself before it can map the pattern onto itself — which is why a quarter turn has nowhere to live but a square lattice. And the converse fails on every one of the five: each lattice carries at least one group whose order falls short of what the lattice offers, so knowing the lattice narrows the group to a handful of candidates and never to one. The pairs printed in the accent colour are the groups that take everything their lattice permits.

The classification proof, one branch at a time

Seventeen is a theorem, and the argument that establishes it is a finite case analysis that fits on a few pages. Working through it is the difference between knowing the number and knowing why there is no eighteenth.

A fundamental domain for p6m. One representative from every orbit of p6m, shaded, with the images that tile the rest of the cell. The domain was found by computing orbits rather than by drawing a region, and every sample's orbit was checked to meet it exactly once — so the region has neither a gap nor an overlap.

Orbifold notation, the shorter language

Fold a pattern up along its own symmetries and what remains is a small surface with marked points. Its shape is a complete name for the group, and reading the name off costs an arithmetic sum that has to come to two.

How often a dot gives the wrong group. Every position on a grid inside the cell, tried as a single-dot motif for each of the seventeen groups. The bar is how often the resulting pattern turned out to have more symmetry than the group it was made with — so the caption would have been wrong and nothing about the picture would have shown it.

The groups ornament actually uses

Seventeen exist and decoration does not use them evenly. Which are common is an empirical question that published surveys answer differently, and part of the reason is a hazard this site can measure exactly.

Why seven — all 16 candidates. Every subset of the 4 extras available on a strip, closed under composition and named from the operations that come out. 16 candidates give 7 distinct groups: 9 of them generate operations they were not given and land on a group already listed.

Why sixteen become seven

Four extra operations give sixteen combinations and seven groups. The nine that vanish are not cases anybody forgot — each one comes back from the closure holding something it was never given, and one of them changes the lattice underneath it.

The friezes inside the seventeen

Take one lattice row of a wallpaper pattern and keep only the symmetries that leave that row where it is. What survives is a frieze group — and which of the seven it turns out to be is a fact about the plane group that its symbol does not state.

Two-colourings of the seventeen. How many ways each of these 17 plane groups can be two-coloured so that every symmetry either preserves the colours or exchanges them. 74 in all, each one a subgroup of index two enumerated by trying every assignment of colours to a generating set and keeping the assignments that turn out to be consistent. p3 admits none: a homomorphism onto a group of order two has nothing to send a three-fold rotation to but the identity, and once the rotation and its conjugates are killed nothing is left to reverse the colours. pmm admits the most, with 15. Every count is one less than a power of two because the homomorphisms of a group onto the two-element group are the non-zero elements of a vector space over that field.

Two colours, and a symmetry that swaps them

A chessboard and a grid of identical squares have the same group, which is plainly not what anybody sees. Admitting the colour swap as an operation gives a finer classification — and one of the seventeen turns out to admit no two-colouring at all.

The Alhambra question

Textbooks say the Alhambra contains all seventeen wallpaper groups. Careful analysts of the same building have counted eleven, thirteen, fourteen and seventeen — and the disagreement is not about the mathematics but about what counts as an instance.

Six classifications, and which are enumerated here. The families of symmetry groups by how many directions they repeat in and how many they live in. The thirty-two crystal classes, the seven friezes and the seventeen plane groups are each built from their own operations and counted. The seventy-five rod groups, the eighty layer groups and the two hundred and thirty space groups are numbers from the literature, marked as such wherever they appear: reaching them needs the translation extensions and their equivalences in full, which is the content of the classification rather than an application of it. The subperiodic cases sit exactly between the two halves, which is why they are so easy to assume are already known.

A layer is not a wallpaper

A sheet repeats in two directions and lives in three, and its symmetry group is not one of the seventeen. There are eighty of them, the difference between one and another is a single sign per operation, and the arithmetic that supplies those signs is the arithmetic of a two-coloured pattern.

Subgroups of index 3, across the seventeen. Every subgroup of index 3 with cyclic quotient in each of the seventeen plane groups, sorted into the two kinds: 4 keep all the translations and lose operations, 22 keep all the operations and lose translations, and the total is 26. The split is decided by whether the homomorphism onto ℤ3 kills the two lattice translations, which is a property of the kernel and not a judgement. Every one of them is found by enumeration inside the finite quotient by 3Λ, and the count for the whole classification is a measurement.

Three colours, and why most patterns cannot have them

Seventy-four of the seventeen plane groups' subgroups have index two, and every group but one has at least one. At index three there are twenty-six, and ten of the seventeen have none at all — because a symmetry of order two cannot survive being asked to permute three colours.

The seventeen signatures, and the seventeen groups. Every combination of features costing exactly two, beside the plane group each one names. The left column is produced by an accounting identity that has never heard of a lattice; the right by reading seventeen groups' own operations — their rotation centres and orders, which of those lie on mirrors, and how many closed curves the mirror lines make once equivalent lines are identified. The map between the two lists is a bijection, and the figure does not appear unless it is one — in both directions. A signature with no group and a group whose signature is not on the list are both refused, and so is the failure that actually happens: two groups deriving one signature, which costs exactly two and passes every check but injectivity.

Seventeen dollars

Conway's magic theorem prices the features a folded-up pattern can have — a handle costs two, a mirror boundary one, a cone point of order n almost one — and requires the total to come to exactly two. There are seventeen ways to pay, and the classification falls out of an accounting identity that never mentions a lattice.

The symmetry of a cut through Pnma. Every height in one cell, and the number of operations of Pnma that map the plane at that height to itself. The answer is 2 almost everywhere and rises to 4 at the special heights, where the plane group named above the spike is what a reader looking down at that surface would see. The rule is two conditions and no more: the operation must not tilt the plane, and the plane must come back to its own height — so a twofold axis lying in the plane survives at two heights per cell and a screw axis along the normal survives nowhere.

What a cleave leaves

A surface is a crystal that has been cut, and the symmetry it presents is what the space group leaves of itself on that plane. Two conditions decide it — the plane must not tilt, and it must come back to its own height — and the answer changes with where the cut was made.

How much a count of descriptions over-counts. For each plane group that has any two-colouring at all: how many colourings it has, how many designs those come to, and the ratio between them. Over the seventeen the ratio is 1.61, and group by group it runs from 1.00 — where nothing is identified — to 3.50 at p2, whose seven colourings fall into one class of six and one of one. The tick on each row is that row's largest single class, and it is at least the bar and usually more. The largest class anywhere is p2's 6, and that same group over-counts by only 3.50, because a factor is a mean over the group's classes and a mean reaches its largest term only when every term equals it. Reading the largest class as the over-count is therefore an over-statement, always. And the factor varies from group to group, which is why no single correction turns a count of descriptions into a count of designs after the fact.

Seventy-four colourings, forty-six groups

This site counts the two-colourings of the seventeen and gets seventy-four. The literature says there are forty-six two-colour wallpaper groups. Both numbers are right, and the gap between them is a disagreement about when two coloured patterns are the same pattern.

The (2, 3, 7) group, in the Poincaré disk. A triangle with angles π/2, π/3 and π/7, reflected in its own three sides until depth 12: 380 triangles, alternating in handedness because every generator is a reflection. The sum 1/2 + 1/3 + 1/7 is less than one, so the triangle does not fit in the flat plane and the drawing is of the hyperbolic one, with the whole plane squeezed inside a disk. Every triangle has the same hyperbolic area; the ones near the edge look small because the model shrinks distances there, and the tiling stops at the edge of the drawing rather than at the edge of anything.

Past two, the list does not stop

Conway's accounting says a wallpaper group costs exactly two dollars, and there are seventeen ways to spend it. Spend less and the answer is a finite group. Spend more and the list is infinite — but the cheapest thing past two costs two and one eighty-fourth, and nothing at all lies in between.

Every group decided by a window of radius 1. For each of the seventeen, the radius at which a round window on the pattern admits exactly the group's own operations and no others — with the numbers it admits at each smaller radius beside it. Two opposite failures are visible. Most groups under-report at small radii, because an operation carrying points out of the window cannot be tested at all; cm over-reports, admitting operations the pattern does not have. The groups that take longest to settle are the ones distinguished by a glide, which moves a point half a cell before anything can be compared.

How much pattern is enough

Every claim here about a pattern's group is a claim about an infinite pattern. A reader sees a patch. Measuring what a finite window can decide gives a number — about one cell's radius — and two opposite ways of being wrong on the way there.

Twenty-one vertices, eleven tilings

Regular polygons meeting at a point must fill exactly a turn, which is a Diophantine equation with seventeen answers and twenty-one cyclic arrangements. Ten of the twenty-one tile nothing at all — and the argument that kills them counts places round a polygon rather than measuring anything.

$The cell of 3.4.6.4, and the vertices in it. 3.4.6.4 drawn with the cell its own translations define. The lattice is hexagonal and the cell holds 6 vertexes, marked. Neither was chosen: the translations are the vertex-to-vertex vectors that carry every polygon of the patch onto a polygon of the patch, and the cell is the shortest independent pair of them. Expressed in that basis the vertices have coordinates that are exact and are not fractions — a vertex of this tiling sits at 1/(1 + √3) of a cell — which is why the detector that decides its group works in ℚ(√3) rather than in the rationals.$

Eleven tilings, five groups

Hand each of the eleven uniform tilings to a detector that has never heard of tilings and ask what its symmetry is. Six of them answer p6m. Twelve of the seventeen wallpaper groups never appear at all — and the coordinates the question has to be asked in are not fractions.

3.4.6.4 and its dual. The tiling in pale outline with its dual drawn over it: one dual vertex at the centre of every tile, one dual edge across every shared edge, and one dual tile round every vertex. 3.4.6.4 has 3 kinds of tile and one kind of vertex; its dual has one kind of tile and 3 kinds of vertex, and the congruence of those tiles is checked rather than eyeballed — every dual face presents the same cyclic sequence of squared edge lengths, compared exactly. That swap is what the eleven duals are for: read one way the list classifies tilings with all vertices alike, read the other it classifies tilings with all tiles alike.

Eleven duals, one tile each

Swap the vertices of a uniform tiling for its tiles and the eleven come back as eleven tilings by a single repeated shape. Three of those shapes are pentagons — which is worth pausing over on a site whose other essays prove that five-fold symmetry cannot exist.

p1 folds into a torus. The cell of p1 with its edges marked as the group joins them: both pairs by a plain translation, both arrows the same way round. Gluing top to bottom gives a tube and gluing its ends gives a torus. Nothing in p1 holds a point still, so the surface has no marked points and its first homology is two copies of the integers.

The two that fold into a surface

Fold a wallpaper pattern along its own symmetries and what is left is usually a shape with corners and edges nobody drew. For two of the seventeen it is a plain surface with no marks on it at all — a torus and a Klein bottle — and which two is decided by a single question asked of every operation.

a general quadrilateral tiles. A general quadrilateral — convex, with no equal sides and no parallel edges — with copies placed by half-turns about edge midpoints. The patch was checked by sampling 2000 points inside a disc: every one of them lies in exactly one tile, so there is no gap and no overlap anywhere in the region tested.

Which shapes tile by themselves

Every triangle tiles the plane. So does every quadrilateral, convex or not. Six sides admits three families, seven sides admits nothing at all — and the five-sided case took a hundred years and finished with a computer search. The bound at seven needs no search: it is Euler's relation with the curvature set to zero.

75 rod groups over 27 axial classes. Each axial crystal class, with the number of rod groups it carries: every consistent choice of translation along the axis, in every way the class can sit on the rod, with two groups counted as one when a shift of the origin along the rod or a turn about it carries one onto the other. The total is 75, and every row as well as the total agrees with the International Tables, which are compared with this enumeration rather than used to produce it.

Seventy-five ways to be a thread

Eighty layer groups and seventy-five rod groups are usually quoted, both as numbers from the literature. The second is derived here, class by class — and the total alone turned out to be no check at all, because two errors of six groups each give seventy-five as well.

18 extension classes, 17 groups. Each of the thirteen arithmetic classes with the number of ways translations may be attached to it — its cohomology — the shape of that group, and how many distinct plane groups the classes come to once the changes of basis that are mere relabellings are quotiented out. The two columns differ in exactly one row, 2mmp, where four extension classes are three groups because two of them are the same group with the axes swapped. No lattice is drawn anywhere in this computation.

Seventeen, without a picture

Every other count of the plane groups has a plane in it — a pattern generated, a domain folded, an orbifold's curvature spent. The same seventeen come out of pure algebra: attach translations to a point group, keep the assignments that close, throw away the ones that differ only by where the origin was put, and add up over the thirteen arithmetic classes.

8 tiles over 5 colours. Wang tiles: unit squares with a colour on each edge, which may be laid side by side only where the touching edges agree, and which may never be turned or reflected. That last restriction is what makes them a computational object rather than a jigsaw — an edge colour is a symbol passed from one tile to its neighbour, and turning a tile would let a symbol change direction. The set here was generated from a stated seed.

Nothing decides whether a set of tiles tiles the plane

This collection rests on decidability — generate a pattern, forget the group, rediscover it, compare. One question in the same subject has no procedure at all: given a finite set of tiles, whether they cover the plane cannot be decided by any algorithm whatever. What can be done is two half-searches, and measuring what they leave behind.

Y-pentomino: A B C D E F, with 6 arcs. The boundary of the Y-pentomino cut into six arcs. A runs from one corner to another and D is the same arc traversed backwards, so D is a translate of A and the translation is (3, 1) cells. Each of B, C, E and F is carried onto itself by the half turn about its own midpoint, and those midpoints are the four marked dots — That is Conway's criterion, and a shape meeting it tiles the plane by translations and half turns.

A tiling of the whole plane, decided on one tile's edge

Whether a shape tiles the plane is a question about an infinite object, and there is no procedure that answers it. There is a procedure that answers it *sometimes*, and it reads nothing but the shape's own boundary — a closed path of a few dozen steps, cut into six arcs. When the cut exists the tiling exists, and the cut names the group that makes it.

anisohedral: 2 orbits of congruent tiles. A tiling of the plane by 8 copies of one shape per cell of a lattice of index 64, drawn 1 cell across and 8 up, and coloured by which orbit of the tiling's own symmetry group each tile belongs to. The group has 4 operations per cell and 2 orbits: every tile is congruent to every other, and no motion of the whole pattern carries a tile of one colour to a tile of another. Congruence is a fact about the shapes; an orbit is a fact about the pattern, and they are different facts.

One shape, two kinds of tile

A tiling by copies of a single shape looks as though it must be homogeneous — every tile is congruent to every other, so what could distinguish them? The symmetry group can. There are shapes that tile the plane and admit no tiling whose group carries any tile to any other, and the smallest of them has eight cells.

Surrounded twice over, and covering nothing

A shape that tiles the plane can be surrounded by copies of itself for ever. A shape that tiles nothing cannot be surrounded for ever — but it can be surrounded once, and sometimes twice, and the number of times is a measurement of how much local success a global impossibility permits.

11 nets, and one accounting. Every plane net folds onto a torus when its own translations are divided out, and a torus has Euler characteristic zero — so the quotient's vertices, edges and faces satisfy n − e + f = 0 and the number of faces is not something to be counted off a drawing but e − n. Dividing through gives one over the mean face size plus one over the mean degree equal to a half, which is the same relation that forbids a plane tiling by pentagons, reached here with no geometry in it at all. It holds for every net in the table.

Every net folds onto a torus

Divide a plane net by its own translations and the quotient is a finite graph drawn on a doughnut. A doughnut has Euler characteristic zero, so the number of faces is not something to count — it is forced, and with it a relation between how many edges meet at a vertex and how many bound a face.

3 whole-number solutions: (6, 3), (4, 4), (3, 6). Every pair of whole numbers from three to 12, with the mean face size across and the mean degree down. A square in the first colour is a pair satisfying one over p plus one over q equals a half exactly — the flat case, where a periodic net is possible — and there are 3 of them: 6 and 3, 4 and 4, 3 and 6. The lighter squares above and to the left have a sum greater than a half, which is a closed polyhedron rather than a plane tiling; the ones below and to the right have a sum less than a half and belong to a surface of negative curvature. The plane is the boundary between them and it is thin.

Three answers in whole numbers

One over the face size plus one over the degree equals a half. Ask for whole numbers and there are exactly three answers, which are the three nets everybody has drawn since childhood — and the pairs on either side of them are a closed polyhedron and a plane the plane has no room for.

$A wavevector of thirds, and the boxes that cannot see it. Which sizes of box can carry the wavevector at the corner of a hexagonal zone. The characters of the box's translation group are its wavevectors, and there are exactly N² of them — the fractions with denominator dividing N. A wavevector of thirds is therefore present in a box of three, six, nine or twelve cells and absent from one of two, four or five: not approximated badly, not resolved coarsely, absent. A mechanism or a level living there is invisible to such a calculation, and that is the practical content of a mechanism count depending on the cell it was looked for in.$

Crystallography in a box

A calculation over a crystal is not performed on a crystal. It is performed on a finite block with its edges glued, and the block has a symmetry group of its own — finite, complete in one direction and missing something decisive in the other.

p4m from 3 orbits of waves — detected p4m. A density built as a sum of 3 symmetry-adapted waves of p4m, each of them the average of a plane wave over the group, shaded from light to dark across one cell. The level set of this density — the darkest points of it — was handed to the same detector the pattern figures use, and it reports p4m, which is exactly the group the waves were built from. The waves are invariant by construction, so the density can never have less symmetry than the group; the interesting direction is the other one.

How many waves a group permits

A pattern can be written as a sum of waves instead of as an orbit of a motif, and then the group ties the coefficients together and forbids some outright. Building a density from the permitted ones and handing it back to the detector closes the same loop through a different door — and at low resolution the density has symmetry the crystal has not.

656 sets, every one decided. Every set of one, two, three and four tiles over two colours — sixteen tiles exist in all, so these are complete lists rather than samples — reduced by relabelling the two colour alphabets, and each set decided by the two half-searches. The last column is the one that matters: it is empty. At these sizes there is no room for a set that tiles the plane and admits no periodic tiling, which is the residue undecidability lives in. The smallest aperiodic set is known to have eleven tiles and four colours.

How much room a hard question needs

No algorithm decides whether a set of tiles covers the plane. Every set of four or fewer tiles over two colours is nevertheless decided here, exhaustively, in under a second — because the sets that defeat the two half-searches have nowhere small to live.

The parity argument loses 36 pairs it had won alone. The argument that refutes ten of the twenty-one species walks round a polygon of odd size: the ring of polygons about it is a closed walk of odd length in a graph the species decides, and a bipartite graph has no such walk. With two species at a vertex the flanking pairs come from the union of two graphs, and a union of bipartite graphs need not be bipartite — so the walk stops being constrained. The fourth row is the cost: pairs whose members the argument kills on their own and which it cannot kill together.

The argument that closes eleven

Twenty-one vertex species satisfy the angle equation; a parity argument kills ten before anything is drawn, and the eleven survivors are all built. Asking the same question of tilings with two kinds of vertex, the parity argument evaporates — it constrains a walk in a graph one species decides, and two species decide the union of two graphs, which need not be bipartite. What is left is a search, and a search cannot close a count.

The cross, from the selection rule alone. The layer lines of a helix with the first maximum of each marked on both sides. Nothing here is a picture of a photograph: each mark is at the radius where the Bessel function of the lowest order the selection rule permits on that layer line first peaks, and that radius is proportional to the order. The order rises by one per layer line until the middle of the repeat, so the maxima lie on two straight lines through the origin — the X — and the larger marks are the layer lines that reach the axis.

What a thread scatters

A helix with ten subunits in a turn is not a screw axis a crystal may have, and nothing about its diffraction pattern is lawless. The pattern lies on layer lines, and on each one only certain angular orders may contribute — a selection rule as hard as any extinction condition. The lowest permitted order rises by one per layer line, a Bessel function of order n does nothing until its argument is about n, and the maxima therefore lie on two straight lines through the origin.

One group refuses two colours and three refuse three. The two counts side by side, with the rows that refuse a number of colours marked. p3 is the only group with no two-colouring; p4, p4m and p4g are the only ones with no three-colouring. Neither list is a subset of the other and both come from the same arithmetic — a rotation order that divides nothing the symmetric group has.

What a half-turn does to three colours

Ten of the seventeen plane groups have no three-colouring, because a half-turn cannot permute three colours cyclically — that is the first rung of this ladder and it is true. Drop the word cyclically and the answer changes completely: a half-turn permutes three colours perfectly well by swapping two and fixing one, and only the three four-fold groups refuse three colours at all.

Why a tetrahedron is not a cube cut up. The two invariants side by side. A cube's twelve right angles are each a rational part of a turn and contribute nothing; a regular tetrahedron's six edges each contribute one α, giving six. Cutting a polyhedron and rearranging the pieces cannot change the invariant, so no dissection takes one to the other however the volumes are matched. That is Hilbert's third problem, and the whole of it is one angle.

The angle that is not a fraction of a turn

Any two polygons of equal area can be cut into pieces that rearrange into each other. In space that fails, and the obstruction is a sum over edges of length against dihedral angle — zero for anything that fills space, and not zero for a regular tetrahedron. The whole argument reduces to one claim about one angle, and that claim is an integer computation: a sequence that is never divisible by three, when it would have to be.

The five, as generator counts. Each of the five convex bodies that tile space by translation, built as the set of combinations of a handful of vectors with coefficients between zero and one. Three generators give a cube, four give either a hexagonal prism or a rhombic dodecahedron depending on whether three of them are coplanar, five give the elongated dodecahedron and six the truncated octahedron. The last column is what the same number of generators would give in general position, and the shortfall is the number of faces lost to coplanarity.

Every parallelohedron is a shadow of a cube

Take a few vectors and form every combination of them with coefficients between zero and one. All five of the convex bodies that tile space by translation come out of that recipe, from three vectors, four, four, five and six — and since the recipe is exactly the image of a cube of that many dimensions, the truncated octahedron is a three-dimensional shadow of a six-dimensional cube. The five are not the generic answers: they are the degenerate ones, and the degeneracy is what the tiling demands.

One group without an axis, and as many as the order with one. For a rotation of each order that an integer matrix can have in a small dimension, the number of space groups its arithmetic class admits — computed from the cohomology rather than enumerated. A rotation acting on the smallest lattice that will hold it fixes no direction and admits exactly one group: the symmorphic one, with no screw. Add a direction it leaves alone and the count becomes the order of the rotation, and the extra groups are its screws. The four-fold with an axis gives four, which are P4, P4₁, P4₂ and P4₃; the five-fold with an axis gives five, in five dimensions, where no published table exists to check it against.

The screw a dimension does not have

The extension count is a machine that runs in any dimension, and the seventeen were the case where every step could be checked against a list arrived at four other ways. Run on a cyclic point group it has a closed form two lines long — and it says a five-fold screw axis does not exist in four dimensions, which is a prediction rather than a check.

$Tight where there is an axis and vacuous where there is not. The order of a point group annihilates its cohomology, so every intrinsic translation is a multiple of one over that order — the bound every account of the subject quotes. What occurs is one over the exponent of the cohomology, which divides the bound. For a rotation with a direction it fixes the two agree exactly: a four-fold screw does need quarters and a six-fold sixths. For a rotation acting with no fixed direction the exponent is one — the cohomology is trivial and no fraction occurs at all — so the bound is slack by the whole order. The same bound is sharp and useless in the same table.$

The denominator a group actually needs

The order of a point group annihilates its cohomology, so every intrinsic translation is a multiple of one over that order — a bound every account of the subject quotes. What occurs is one over the exponent, which divides it. The same bound turns out to be attained exactly and to be slack by its whole size, in two rows of one table, and what decides which is whether the rotation fixes a direction.

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