Four groups are made of mirrors

Closing the plane from two centres took two rotation centres of given orders at a given distance, closed the set of motions they generate, and watched the shortest translation. It settled when the least common multiple of the two orders was one a lattice permits, and it fell without end when it was not — so the crystallographic restriction arrived as a condition on a closure, with the lattice as its output. That closure reached the five plane groups made of rotations and nothing else. The eleven groups with a reflection or a glide were out of its reach, because two rotations never compose to a reflection.

This is the other half: close the plane from located mirrors. Two mirrors meeting at an angle compose to a rotation by twice the angle, and two parallel mirrors to a translation by twice their distance — the doubling law — so mirrors can reach everything rotations reached and more. The question is which configurations of mirrors close to a plane group, which plane groups their own mirrors close to, and what the closure produces that nobody put in.

The short answer has three parts. Four triangles of mirrors close discretely, and one of them is not what it appears to be. Four of the seventeen plane groups are generated by their mirrors, three more contain a kaleidoscope at a finite index that their orbifold symbols name, and three have mirrors that can never generate them. And the closure of mirrors produces glides, which have no fixed point and so no location of the kind the rotation closure computed.

Three mirrors, and the corners they make #

Put three mirrors along the sides of a triangle. At each corner two of them meet, at the corner’s angle, and their product is a rotation about the corner by twice that angle. If the angle is $\pi p/q$ in lowest terms, the rotation turns by $2\pi p/q$ and has order $q$. So every triangle of mirrors with rational angles carries three rotation centres of orders equal to the denominators of its angles, and its closure contains rotations of every order the least common multiple of those denominators allows.

The crystallographic restriction then decides the matter, through the same mechanism the rotation closure found. A group holding a rotation of order $L$ and a translation also holds the translation turned and subtracted from itself, and for $L = 5$ and for every $L$ of seven or more that difference, or the corresponding sum, is shorter than the translation it came from. The shortest translation then falls without limit and the closure is dense. For $L$ of 1, 2, 3, 4 or 6 no such fall is forced, and a lattice can form.

The census checks that prediction against the closure itself. There are forty-four triangles with rational angles and denominators up to twelve. Each is closed under its three mirrors, word by word, and the shortest translation among the products is recorded at every word length. For forty of them it steps down: a longer word finds a shorter translation, and then a still longer one a shorter one again. For four it is flat from the first translation found. The four flat ones are exactly the four whose corner orders have a least common multiple of three, four or six: all corners at 60°; 45°, 45°, 90°; 30°, 60°, 90°; and 30°, 30°, 120°.

The staircases are the forbidden rotations at work. The triangle with angles of 36°, 72° and 72° has five-fold corners; its shortest translation drops from 1.18 to 0.73 to 0.45 as the words lengthen, each step a factor of 0.618 — which is $2\cos(2\pi/5)$, the factor the rotation closure found the five-fold rotation forces. The eight-fold triangle drops by a factor of 0.54, the seven-fold one by 0.55 and then 0.45, and none of them stops within the words tried. That is the restriction arriving a second time, now from reflections, and arriving in the same form: as a factor below one in a contraction the group cannot avoid.

The triangle that is two triangles #

Three of the four flat triangles are the familiar ones. Their angles are $\pi/3, \pi/3, \pi/3$; $\pi/4, \pi/4, \pi/2$; and $\pi/6, \pi/3, \pi/2$, each of the form $\pi/p, \pi/q, \pi/r$ with $1/p + 1/q + 1/r = 1$, and those are the only flat triangle groups there are. Their closures are p3m1, p4m and p6m, and in each the triangle is a fundamental domain: its images tile the plane, one image for each element of the group.

The fourth is not of that form. Its angles are 30°, 30° and 120°, and 120° is $2\pi/3$, not $\pi$ over anything. Its closure is discrete all the same — and it is p6m, the same group the 30–60–90 triangle generates. The reason is visible in the figure at the head of the page: the 30–30–120 triangle is two 30–60–90 triangles glued along the altitude from its obtuse corner. The closure contains the mirror along that altitude, which was none of the three mirrors it was built from. It is the product of the three in a particular order, and it cuts the triangle into two fundamental domains.

So the closure does something the generators did not specify: it can find a mirror inside the region the given mirrors bound. That is the general reason a triangle with an angle of the form $2\pi/q$ can still close — the extra mirror bisects it — and why the census has exactly one such triangle among the discrete ones. An angle of $2\pi/3$ bisects into $\pi/3$; an angle of $2\pi/5$ would bisect into $\pi/5$, and the five-fold corner is forbidden anyway.

Why triangles, and one rectangle #

The census looked only at triangles, and there is a reason it could afford to. A polygon of mirrors that closes to a plane group with the polygon as its fundamental domain must have every angle of the form $\pi/n$, since the mirrors at a corner generate a dihedral group whose chamber is that angle. And the angles of a flat polygon with $k$ sides add to $(k-2)\pi$. So

$$\frac{1}{n_1} + \frac{1}{n_2} + \cdots + \frac{1}{n_k} = k - 2,$$

with every $n_i$ at least two. For a triangle the right side is one, and the solutions are $(3,3,3)$, $(2,4,4)$ and $(2,3,6)$. For a quadrilateral it is two, and since no term exceeds a half, all four must be exactly a half: the rectangle, and pmm. For five sides or more the right side is at least three while the left is at most $k/2$, which is smaller. So the kaleidoscopes of the plane are three triangles and one rectangle, and there is nothing else to look for. That is the same count fundamental domains arrives at by cutting, and the same equation every wall names a generator reads off the corners of a domain; here it is the condition under which a closure of mirrors reproduces the polygon it started from.

The 30–30–120 triangle is outside that argument because it is not a fundamental domain, and the census is what finds it: an argument about domains cannot see a region that is two domains, and a closure can.

Which plane groups their own mirrors make #

Turn the question round. Take a plane group, take every mirror it has — every lattice translate of every mirror line — and close. What comes out is the reflection subgroup, the part of the group its mirrors account for, and its index says how much is left over.

Of the ten plane groups with a reflection, four are generated by their mirrors outright: pmm, p3m1, p4m and p6m. Their orbifold symbols are $*2222$, $*333$, $*442$ and $*632$ — a star and nothing before it — and their fundamental domains are polygons bounded by mirrors on every side: a rectangle and the three triangles.

Three more contain a kaleidoscope and are larger than it. cmm holds a pmm at index two, p31m holds a p3m1 at index three, and p4g holds a pmm at index four. Their orbifold symbols are $2{*}22$, $3{*}3$ and $4{*}2$, and the number before the star is in each case the index. That is not a coincidence of three rows. A number before the star is a rotation centre that lies on no mirror, and the group acts on the kaleidoscope’s chamber by rotating it about that centre; the order of the rotation is how many chambers make one fundamental domain’s worth of the larger group. The census computes the index by closing mirrors and counting translations and point operations, the orbifold symbol comes from a separate computation of rotation centres and mirror boundaries, and the two are required to agree for all ten groups.

The last three have mirrors in one direction only. pm, cm and pmg — $**$, $*{\times}$ and $22*$ — have parallel mirrors, and parallel mirrors compose only to translations perpendicular to themselves. Their reflection subgroup has translations in one direction and none in the other; it is a strip, not a plane group, and its index in the group is infinite. The strip is itself a familiar object: a row of parallel mirrors with the translations between them is a frieze group — the one whose motif is reflected across lines perpendicular to the strip — laid across the plane, and a plane group with only parallel mirrors is a frieze of mirrors with a second translation added from outside. That is where an older essay went wrong: three reflections, and never four listed pm among the groups generated by their reflections, and it is not. Its mirrors are symmetries, but no product of them slides along them, so the translation along the mirrors has to be supplied separately. The essay has been corrected.

Glides nobody asked for #

The closure of mirrors does not only produce rotations and translations. It produces glides.

In p4m, generated by its mirrors, a mirror composed with a translation that the mirrors generate — one not perpendicular to it — is a reflection followed by a slide along the line. When the slide is half a lattice vector along the line, no mirror is there: the product is a glide, along a line halfway between two diagonal mirrors. The closure of mirrors made it, and it is in every p4m pattern for that reason rather than because anyone put a glide into the group.

The census finds this in four places: p4m, p3m1 and p6m, and p31m’s kaleidoscope. It finds it nowhere in pmm, whose mirrors meet only at right angles and whose translations are all perpendicular to one family or the other, so a mirror is never composed with a slide along itself by less than a whole cell. And the glides of cmm and p4g are not products of their mirrors: those groups’ glides come from the generator their mirrors do not supply, the rotation before the star.

This matters for the question these closures have been asking, which is where things are. A rotation centre is a point; a mirror is a line; both are located. A glide has no fixed point at all — it moves every point of the plane — so the closure that located every rotation centre it produced cannot locate a glide in the same sense. What it can locate is the glide’s line: the one line the glide carries to itself. So the located-elements picture survives the glides, with the location of an element downgraded from a point to the set it preserves, which for a glide is a line and for a translation is nothing at all.

What the closure depends on, and cannot show #

The triangle census is bounded twice. The angles have denominators at most twelve, and the closure is run to words of thirteen mirrors within a window of the plane. A dense closure is recognised by its shortest translation stepping down within that bound. All forty dense triangles do step down, but four of them only at the twelfth word or later; a triangle whose first step came past the thirteenth would be misread as discrete, and nothing but the restriction’s contraction argument rules that out. The census is evidence for the rule, and the rule is the argument.

The reflection subgroups are computed on a finite piece of each group. Every mirror with a translation part in a small box, closed within a larger box, with the translation lattice read off from the determinants of the translations found. That finds the lattice’s index exactly when the box holds a basis of it, and the agreement with the orbifold symbols for all ten groups is the check that it did.

And no figure here shows a group. The tiling shows the images of a triangle, and the dashed line in it is drawn where the computation found a mirror; that the closure contains that mirror is a statement about a set of motions, which a picture of their images can only illustrate. The same holds for the glides: a dashed line is where a glide acts, and the slide it performs is not drawn.

Mirrors against rotations #

The rotation closure and the mirror closure end in the same place for the same reason. Both find a lattice when the rotations they produce have orders one, two, three, four or six, and both find a dense set otherwise, because both contain rotations, and a rotation of any other order contracts a translation. What differs is what the closure can build. Two rotation centres reach the five rotation groups; mirrors reach the kaleidoscopes directly and, with one extra rotation, the others that contain them. Neither alone reaches pg or pgg, whose operations are glides and rotations with no mirror anywhere.

That division of the seventeen is the orbifold’s, read off a closure. A group is generated by its mirrors when its orbifold symbol is a star and corners; it holds a kaleidoscope at a finite index when a rotation stands before the star; it holds only a strip of mirrors when the star has no corners; and it has no mirrors when there is no star. The seventeen are, in that sense, four kaleidoscopes, three kaleidoscopes turned, three strips and seven groups that mirrors cannot build at all.

Whose the pieces are #

The kaleidoscopes are old. Groups generated by reflections were classified in every dimension by Coxeter in the 1930s, and the plane’s three triangles and one rectangle are the smallest case of his list, with the equation above as their condition. The reading of a plane group’s symbol as a kaleidoscope with some rotations and some twists added is Conway’s orbifold notation, from the 1990s, in which the star marks the mirror boundary and the digits before it the rotation centres that lie on no mirror. That the digits before the star multiply to the index of the mirrors’ subgroup follows from the orbifold picture and is not new; what the census adds is that the index, computed by closing the mirrors of each group and counting what they generate, agrees with it group by group, and that the one error on the subject in this collection — pm listed as a reflection group — was exactly a case the orbifold symbol $**$ would have flagged, since a star with no corners is a strip.

The 30–30–120 triangle is a standard exercise rather than a discovery. Its interest here is only that a closure finds it without being told, which is the property a derivation by closure has to have.

Still open: the closure from mirrors and glides together #

The one kind of generator left is the glide itself. A closure from located glides — two glides whose lines cross at an angle, or a glide and a mirror — reaches pg, pgg, pmg and cm directly, and it raises the question this essay could not: a glide’s line is its location, but its slide is a second datum, and two glides on the same line with different slides are different motions. Whether the classification of the seventeen falls out of closing located glide lines with their slides, as it falls out of closing located centres and mirrors, is the question that would complete the derivation. The arithmetic is the doubling law again — two glides compose to a rotation about a point determined by both lines and both slides — and the census that would settle it is the same one run here, with a slide attached to each line.