Seventeen dollars

The classification proof establishes that there are seventeen wallpaper groups by walking a case analysis: which rotations are possible, what can be combined with what, which combinations collapse into others. It is a few pages long, it is finite, and it is a proof.

There is a second route to the same number that fits on a postcard, and it is not a shortcut through the first. It is an accounting identity.

Folding a pattern up #

Orbifold notation is the prerequisite and the idea is worth restating in one paragraph. Take a wallpaper pattern and glue together every pair of points the group identifies. What remains is a small closed surface with marks on it: a cone point where a rotation centre was, of the same order as the rotation; a mirror boundary where a reflection axis was, which becomes an edge rather than a line; a corner where two mirrors crossed; a cross-cap where a glide was; a handle if the surface has one. That object is the orbifold, and its list of features is a complete name for the group.

The magic theorem prices the features:

and says: a pattern of the flat plane is one whose features cost exactly two. Less than two and the pattern lives on a sphere, where the group is finite. More than two and it lives on the hyperbolic plane, where the list of possibilities never ends.

The costs are not arbitrary weights. They are 2 − 2χ, where χ is the orbifold Euler characteristic — a handle costs what a handle costs a surface’s Euler characteristic, a cone point of order n removes all but 1/n of a point, a corner removes half of that again because a boundary point counts half. The theorem is Gauss–Bonnet in a coat, and the flat condition is the total curvature being zero.

Seventeen ways to pay #

The enumeration is then mechanical, and it is short enough to run in a paragraph.

With cone points and nothing else, the cost is Σ(1 − 1/nᵢ) = 2, so Σ 1/nᵢ = k − 2 for k cones. Two cones give 1/p + 1/q = 0, impossible. Three cones give 1/p + 1/q + 1/r = 1, whose solutions are (3, 3, 3), (2, 4, 4) and (2, 3, 6) and no others. Four cones give Σ 1/nᵢ = 2, forcing all four to be 2. So four signatures: 333, 442, 632, 2222 — which are p3, p4, p6 and p2.

With one mirror boundary, the boundary costs 1 and the corners must make up the other 1: Σ(1 − 1/nᵢ)/2 = 1, the same equation halved, giving 333, 442, 632 and 2222 — p3m1, p4m, p6m and pmm. Or the boundary can carry fewer corners and the rest be paid in cones: 222, 22, 33, 42 — cmm, pmg, p31m, p4g.

With no cones and no corners, two dollars must be found among boundaries, cross-caps and handles: ** (two boundaries), *× (a boundary and a cross-cap), ×× (two cross-caps), ○ (a handle) — pm, cm, pg, p1.

Four plus eight plus four is sixteen; the seventeenth is 22×, two cones of order two and a cross-cap, which is pgg.

Why the coins have those values #

The prices look arbitrary on first meeting and they are forced, which is worth showing because it turns a mnemonic into an argument.

The Euler characteristic of a closed surface counts vertices minus edges plus faces, and it is 2 for a sphere, 0 for a torus, and 2 − 2g for a surface with g handles. An orbifold Euler characteristic does the same count with fractional weights: a point fixed by a rotation of order n counts 1/n of a point, because n of them were glued into one; a point on a mirror counts half, because the surface was folded there.

Working that through gives the orbifold characteristic as

$$\chi = 2 - 2h - x - b - \sum_i (1 - 1/n_i) - \tfrac{1}{2}\sum_j (1 - 1/m_j),$$

and the cost in the table above is exactly 2 − χ, feature by feature. A flat orbifold has χ = 0, so its cost is 2. A spherical one has χ > 0 and a hyperbolic one χ < 0, which is why the three cases are less than, exactly and more than two.

The reason χ = 0 means flat is Gauss–Bonnet: the total curvature of a closed surface is 2πχ, so a surface of zero characteristic carries zero total curvature and can be given a flat metric. The plane is flat, so the orbifold a plane pattern folds up into must be. That is the whole theorem, and the table is its bookkeeping.

The half that makes it an argument rather than a coincidence #

Enumerating seventeen signatures proves nothing about these seventeen groups until the two lists are set against each other, and the site’s habit is that a claim of agreement is a computation rather than a remark. So each group’s signature is derived from its own operations, by four measurements:

• The rotation centres, found exactly by solving (I − M)x = t over the rationals, sorted into orbits under the group, with the order of each read off its stabiliser.
• Which of those lie on mirrors, which separates cone points from corners: a centre whose stabiliser contains a reflection is a corner, and one whose stabiliser is pure rotation is a cone.
• How many closed curves the mirror lines make once equivalent lines are identified — the number of boundaries, which is not the number of mirror lines nor the number of their orbits. p4m has six mirror lines, three orbits of them, and one boundary curve; pm has two lines, two orbits and two curves. No two of those columns can stand in for a third.
• What is left over once all of that has been paid for, which is spent on a cross-cap if the group has a glide and on a handle if it has not.

The four together give a signature for each of the seventeen, and the map onto the enumerated list is required to be a bijection. It is.

The fourth measurement is not “does the group have a glide”, and it is worth saying why, because the shorter statement is the one everybody reaches for. Ten of the seventeen have a glide somewhere and only three carry a cross-cap. pmg, cmm, p4m, p4g, p3m1, p31m and p6m all have glides and none of them has a cross-cap, because their boundary curves and cone points have already spent the whole two dollars and there is nothing left to buy one with. The implication runs one way: a cross-cap means a glide, and a glide means a cross-cap only when the accounting leaves something over. The one handle in the list belongs to p1, which has neither a mirror nor a glide and so has two whole dollars and nothing to spend them on.

The mistake this derivation made, and it is not the kind that shows #

The third measurement is where the derivation went wrong, and the interesting thing about the wrong answer is how thoroughly it looks like a right one.

The line of a reflection is available in closed form: writing w for the direction the operation reverses, a fixed point must satisfy ⟨w, x⟩ = ⟨w, t⟩/2, so the candidate line is settled by one division by two. That is necessary and not sufficient, and the gap is exactly the difference between a mirror and a glide: the component of the translation along the line must also vanish, and when it does not the operation fixes nothing whatever.

cm is where that bites first. Modulo the primitive rhombic lattice cm has one reflection coset; adding a lattice translation to it gives a second candidate line half a period along, and that second operation has a translation along its own axis, so it is a glide and there is no second family of mirrors. Ten of the seventeen have at least one glide, and every one of them acquires candidate lines it does not have — cm one, p4g six, p6m six.

Run the whole derivation that way and every group still costs exactly two. That is the part worth stopping at. The miscount does not overshoot and it does not fall short, because a glide read as a mirror pays for a boundary out of exactly the dollar it stops spending on a cross-cap; the two features cost one apiece, and swapping one for the other is free. Nor does it produce a name that is off the list: all seventeen answers are among the seventeen signatures.

What breaks is that the map stops being one-to-one. cm comes back **, and so does pg, and so does pm, which is the group ** actually belongs to. pgg comes back 22*, which is pmg’s. Seventeen groups, fourteen names.

So the check that catches this is not a check on the arithmetic. A test that every derived signature costs two passes on all seventeen; a test that every derived signature is on the list of seventeen passes on all seventeen; and only requiring the seventeen groups to derive seventeen different signatures fails. That is what the bijection is for, and it is why it has to be checked in both directions rather than by counting how many derived names turn up on the enumerated list.

This is the second time cm has caught this site out on exactly this distinction. Centring, and why cm is not pm records the first: an early draft claimed the detector finds a glide in cm’s primitive description and it does not, because there the glide is the mirror composed with a lattice vector. The two errors are mirror images of one another — one added a glide that was not there, one added a mirror that was not there — and both come from forgetting that an operation’s translation splits into a part along its own axis and a part across it.

Below two, where the groups are finite #

A spherical orbifold’s cost is 2 − 2/|G|, so a signature costing less than two names a group of order 2/(2 − cost). Running the same enumeration under two therefore produces the finite groups — and it produces them in a form that can be checked against something this site already has.

Before the lattice has a say solves the axis equation

$$2 - 2/N = \sum_i (1 - 1/n_i)$$

by search, and gets the cyclic families, the dihedral families and exactly three sporadic solutions, of orders 12, 24 and 60. That equation is this sum: for a signature with cone points and nothing else the cost is Σ(1 − 1/nᵢ), and the order is 2/(2 − cost), which is the same statement rearranged. So the two enumerations must agree, and they do — 22 is C₂, 322 is D₃, 332 is the tetrahedral group of order twelve, 432 the octahedral group of order twenty-four, 532 the icosahedral group of order sixty.

And what is left over is exactly the bad orbifolds. The signature 62 costs 4/3 and would name a group of order 3; no such group exists. A sphere with two cone points of different orders is not the quotient of a sphere by anything, and neither is a sphere with one cone point. Fifteen such signatures pass the arithmetic in the box searched here, and every one of them has fewer than three cone points — which is the axis equation’s way of saying that a rotation has two poles, so its cone points cannot fail to come in matched orders.

That is worth being blunt about: integrality is necessary here and it is not sufficient. The cost accounting decides the flat case completely and does not decide the spherical case on its own, and an essay that said otherwise would be claiming more for the theorem than it has.

The seven friezes, for the price of an infinity #

A frieze group repeats along a line and is finite across it. Folded up, its translation becomes a cone point of infinite order — a rotation with no finite order at all — and the limit of (n − 1)/n as n grows is 1, while the limit of a corner’s (n − 1)/2n is ½.

Allow those two coins and re-run the enumeration at a total of exactly two. Seven new signatures appear: ∞∞, ∞∞, ∞, ∞×, 22∞, 22∞ and 2∞. Seven, and no eighth. They are the seven friezes, which this site derived by an entirely different route — sixteen candidate combinations of four extra operations, nine of which collapse into others — and the two arguments have nothing in common but the answer.

Getting the plane and the strip out of one sum with one parameter changed is the strongest thing this notation does. The usual treatment proves the two classifications separately, in different chapters, with different case analyses.

What the accounting does not do #

Three limits, all of them worth naming because the theorem’s compactness invites overclaiming.

It does not prove the theorem. Everything here is an enumeration under the magic theorem, taken as given, together with a check that the enumerated list matches seventeen groups derived independently. The theorem itself is Gauss–Bonnet for orbifolds, and this site has not proved that and does not attempt to.

The bounds are checked rather than argued. The search runs over at most one handle, two cross-caps, two boundaries, four cones, four corners and orders up to twelve. Every one of those bounds is widened — to thirty, three, three, five, five — and the answer is required not to change. That is evidence rather than a proof that nothing lives outside the box, and it is stated as such.

A signature is not a construction. Knowing that p4g is 4*2 says the orbifold has a cone of order four, a mirror boundary, and one corner of order two on it. It does not say where they are in the cell, and p3m1 and p31m is the standing warning that where is as much a part of a symmetry as what: the two groups have the same lattice, the same point group and the same number of operations, and their signatures — 333 and 33 — differ precisely because the arrangement differs. The notation is a complete invariant, which is exactly why it must be read carefully rather than quickly.

Reading the seventeen off the ledger #

The completed table repays being read as a table rather than as a list of names, and three things fall out of it that the conventional ordering hides.

The four groups with no rotation at all sit together. The seventeen exhibits them in the conventional order, where they are not adjacent at all. ○, ××, ** and *× are p1, pg, pm and cm — the whole of the “no rotation” family, distinguished by whether the surface is orientable and how many boundaries it has. In Hermann–Mauguin they are scattered by lattice type; here they are adjacent, and the reason cm is the odd one is visible as a cross-cap rather than as a centring convention.

The three groups with a fourfold rotation are 442, 442 and 42. Same cone, three ways of adding mirrors: none, all, or half. That is p4, p4m and p4g, and the fact that there are exactly three is the fact that a cone of order four leaves 2 − 3/4 = 5/4 to spend and only three ways to spend it.

The awkward pair is separated by where the boundary is. 333 has three corners on one boundary and no cones; 33 has one cone and one corner. Those are p3m1 and p31m, and the notation says the difference in four characters where the Hermann–Mauguin symbols say it in a convention about which direction the numeral refers to.

Where the ladder goes next #

This is the sixth rung of the seventeen and probably the last of the classification proofs: the site now has the case analysis, the notation, the pair that the notation separates, and the accounting. What it has not got is the hyperbolic case, where the same sum above two gives infinitely many groups and the enumeration becomes a subject rather than a paragraph — and where the seventeen turn out to be a boundary case of something far larger.

The other direction is the one Conway’s notation makes natural and this site has not taken: the two-dimensional orbifolds themselves, as surfaces, and the sense in which a wallpaper pattern is a map from an orbifold rather than a set of operations. That is a change of viewpoint rather than another rung, and it would want its own anchor.

Why the search’s bounds are not arbitrary #

The enumeration runs over a stated range of features and orders, and the essay records that as a bound checked rather than argued. The argument is available and is short, and having it turns the search into an enumeration.

Every feature costs at least a half. A handle costs two, a cross-cap or a boundary one, and the cheapest cone or corner is the smallest available: a cone of order two costs 1 − 1/2 = 1/2, and a corner of order two costs half of that, a quarter. So a signature costing exactly two has at most eight features, and at most four if none of them is a corner.

And every order is bounded. A cone of order n costs 1 − 1/n, which rises towards one and never reaches it, so a cone can never cost a whole dollar. With a total of two to spend, at most two cones can be present alongside anything else — and once the number of features is fixed, the orders are constrained by the requirement that the costs sum exactly, which is a Diophantine equation in a bounded number of unknowns.

So the enumeration is finite by an argument rather than by a cap. Widening the search’s bounds cannot produce a new signature, and the widening the essay performs as a check is confirming an argument rather than substituting for one.

Why the same trick does not classify the space groups #

Conway extended the notation to three dimensions, and the accounting does not extend with it — which is worth stating, because the plane’s argument is so compact that a reader expects a three-dimensional version of the same length.

The reason is a fact about Euler characteristics. In odd dimensions the Euler characteristic of a closed manifold is zero, automatically, for every such manifold. So the three-dimensional analogue of cost equals two is satisfied by everything, carries no information, and cannot be the equation the classification falls out of.

That is why the two hundred and thirty were enumerated by extension arithmetic — cocycles over arithmetic classes, a computation with no short form — and why nothing in three dimensions has the character of an accounting identity. The plane’s classification is small enough to be a sum because two is even and small, and the coincidence does not repeat.

What does extend is the notation. Conway’s three-dimensional symbols name the two hundred and thirty and are a genuine improvement on the Hermann–Mauguin ones for some purposes — they say what the quotient looks like — and they are read off a group rather than deriving it. A notation and a classification are different things, and the plane is the case where one supplied the other.