Three mirrors make every solid

The five solids as three groups counted the rotations that carry a regular solid to itself — twelve for the tetrahedron, twenty-four for the cube and octahedron, sixty for the dodecahedron and icosahedron — and found that there are only three such groups because only three whole-number solutions exist to one inequality. It set one thing aside deliberately: reflections. A cube can also be carried to itself by reflecting it in a plane through its centre, which no rotation achieves, and the full symmetry group of each solid is twice the size of its rotation group.

This essay puts the reflections back, and finds that they are not an afterthought. They are the simplest way to build the whole group. Three mirrors, arranged correctly, generate every symmetry by reflecting in one another; and a single point placed between them generates, by reflection, not only the regular solids but most of their Archimedean relatives as well.

The mirrors of a cube #

Start with the cube, whose mirrors can be listed by hand. There are three planes parallel to its faces, through its centre — each cuts the cube into two slabs that are reflections of each other. And there are six planes through pairs of opposite edges, each cutting the cube along a diagonal rectangle. That is nine mirrors, the same nine that the octahedron has, since the octahedron sits inside the cube with a corner at the centre of each face.

Draw the nine planes where they meet a sphere around the centre, and each becomes a great circle. The nine great circles cut the sphere into pieces, and the pieces are all congruent triangles — 48 of them, as the opening figure shows. Each triangle has one corner at the projection of a vertex of the octahedron, one at the projection of the middle of one of its edges, and one at the projection of the centre of one of its faces. The angles there are $180°/4$, $180°/2$ and $180°/3$: at an octahedron’s vertex four mirrors meet, so eight triangles fit round it, each with angle $45°$; at an edge’s middle two mirrors meet at right angles; at a face’s centre three meet.

The count 48 is not a coincidence. Any symmetry of the octahedron carries the mirror arrangement to itself, and so carries each triangle to a triangle. Conversely, given any two triangles, there is exactly one symmetry carrying the first to the second — and the symmetry is determined by where it sends one triangle. So there are exactly as many symmetries as triangles. The rotations are the symmetries that preserve which way round each triangle’s corners go; the reflections reverse it; and they are 24 each.

A group generated by three reflections #

Take the three mirrors along the sides of one triangle and ask what reflecting in them repeatedly produces. Reflecting a triangle in one of its sides gives the neighbouring triangle across that side. Reflecting that one in its sides gives its neighbours, and so on. Every triangle on the sphere is reached from the first by crossing sides one at a time, so every triangle is the image of the first under some sequence of reflections in the three mirrors — and since a symmetry is determined by where it sends one triangle, every symmetry is a product of the three reflections.

That is exactly what a kaleidoscope does. A child’s kaleidoscope is three strips of mirror arranged in a triangular tube; the objects at the end are reflected in each mirror, and the reflections are reflected again, and the eye sees one triangle of beads repeated round in a pattern. The pattern closes up — the reflections of reflections fit together without overlapping — only when the angles between the mirrors are $180°$ divided by whole numbers. A toy maker uses $60°$, $60°$, $60°$ in a flat pattern; the regular solids use the spherical triangles.

The figures do exactly this computation. Starting from the three reflections of one triangle, they multiply them together in every possible order until no new symmetry appears, and count what they have: 24 for the tetrahedron, 48 for the octahedron, 120 for the icosahedron — twice the rotation counts of the earlier essay, as they must be. Then they reflect the three mirrors themselves by every symmetry found, and count the distinct mirrors that result: 6, 9 and 15.

These groups are not only mathematical objects. A molecule’s shape has a symmetry group, and chemists classify molecules by it because the group decides which vibrations the molecule can have and which of them absorb light. Methane, a carbon atom with four hydrogens at the corners of a tetrahedron, has exactly the 24 symmetries of the tetrahedral kaleidoscope; sulphur hexafluoride, with six fluorines at the corners of an octahedron, has the 48 of the octahedral one; and the football-shaped carbon cage of buckminsterfullerene has all 120 of the icosahedral. The mirrors in the figures are, for those molecules, planes that a spectroscopist can detect.

Why only these three triangles #

The kaleidoscope picture gives the classification of the earlier essay a new and very short form. A kaleidoscope of three mirrors through a point has a spherical triangle as its basic tile, with angles $180°/p$, $180°/q$, $180°/r$ for whole numbers $p, q, r$ — the mirrors must meet at such angles for the reflections to fit. A spherical triangle has angle sum greater than $180°$, so

$$\frac{1}{p} + \frac{1}{q} + \frac{1}{r} > 1.$$

The solutions in whole numbers of at least two are $(2, 2, n)$ for any $n$ — the kaleidoscopes of the prism and the $n$-gon — and exactly three more: $(2, 3, 3)$, $(2, 3, 4)$, $(2, 3, 5)$. Those are the tetrahedral, octahedral and icosahedral kaleidoscopes. It is the same inequality the rotation groups satisfied, arriving from the reflections, and it is also the reason only five regular solids exist: at a corner of a regular solid, $q$ faces with $p$ sides fit only if the angles add to less than a full turn.

The area of a spherical triangle is its angle excess, the amount by which its angles exceed $180°$, so each kaleidoscope’s triangle has a definite area: for $(2, 3, 5)$ it is $180°/2 + 180°/3 + 180°/5 - 180° = 6°$, which is $1/120$ of the sphere’s total excess of $720°$. That is the count 120 again, now as a ratio of areas — and seven hundred and twenty degrees is Descartes’s angle defect for every convex solid, which is the same number doing the same job.

One point, reflected: Wythoff’s construction #

Now put something inside the triangle — a single point — and reflect it in all the mirrors. Its images, one in each of the 48 or 120 triangles, are a set of points on the sphere, arranged with the full symmetry of the kaleidoscope. They are the corners of a solid.

Where the point sits decides which solid. Put it exactly at the triangle’s vertex corner, and it lies on two of the mirrors, so reflecting in those does not move it; its images are the corners of the octahedron itself, six of them. Put it at the face corner, and its images are the face centres — eight points, the corners of the cube. Put it at the edge corner, and its images are the twelve midpoints of the edges, the corners of the cuboctahedron.

Put it elsewhere, and the result is an Archimedean solid. A point on a side of the triangle lies on one mirror and is off the other two; if it is placed at equal distance from those two, every edge of the resulting solid has the same length, because each edge joins the point to its reflection in one mirror and the two distances are equal. That equal-distance condition is the whole of what makes the solid uniform: all edges equal, and every corner surrounded by the same faces in the same order, since the group carries any corner onto any other. The construction is Willem Wythoff’s, from 1918, and Coxeter made it the standard way of organising the uniform solids. What is remarkable about it is how little it asks for: no lengths, no angles of faces, no list of which polygon meets which — only a triangle and a point, and the solid, with every one of its faces, falls out of the reflections.

Seven solids from each kaleidoscope #

A point in a triangle can be at one of three corners, on one of three sides, or inside: seven placements. Each gives a solid.

The table reads as a map of the family. Along the side between the octahedron’s corner and the cube’s corner, the solids pass from one regular solid to the other through the truncations: cutting the corners off an octahedron gives the truncated octahedron, cutting deeper gives the cuboctahedron at the halfway point, and cutting the cube’s corners gives the truncated cube on the other side. That is the parameter the earlier essay on the Archimedean solids varied continuously, here quantised into the special positions where the edges come out equal. The side opposite gives the rhombicuboctahedron, which is not a truncation of anything — its squares come from pulling the faces of the cube outward and filling the gaps — and the inside point gives the solid with the most corners, one in each of the 48 triangles.

The icosahedral table has the same structure with the pentagon in place of the square. Its truncated icosahedron, twelve pentagons and twenty hexagons, is the pattern of a football and the carbon cage of buckminsterfullerene; twelve pentagons, whatever the hexagons explains why the pentagons must number exactly twelve, and the kaleidoscope explains why they sit where they do. The tetrahedral kaleidoscope, with its triangle of angles $60°, 60°, 90°$, gives only one solid not already on the other two lists — the truncated tetrahedron — because its other placements reproduce the octahedron, the cuboctahedron and the truncated octahedron.

Thirteen, and the two that need something else #

Counting the distinct solids across the three kaleidoscopes gives the five regular solids and eleven of the thirteen Archimedean solids: five from the octahedral table, five from the icosahedral, and the truncated tetrahedron. The missing two are the snub cube and the snub dodecahedron, and they are missing for a reason the construction makes visible.

A snub solid has the rotations of the cube or the icosahedron but not the reflections: it comes in a left-handed and a right-handed form, mirror images that cannot be rotated into each other. Wythoff’s construction, which uses the full group of reflections, can only produce solids that are their own mirror images. The snubs come from a modification of it — take the inside point, keep only its images under the rotations, which is every other triangle, and adjust the point’s position so that the resulting edges are equal. That is Coxeter’s alternation, and it is why the snubs had to be pulled apart and twisted by an angle that must be found rather than cut off at a depth that can be chosen.

The kaleidoscope therefore accounts for sixteen of the eighteen convex uniform solids that are not prisms — the five regular and the thirteen Archimedean — directly, and for the other two by a single, principled change — which is as close to a complete explanation of the list as anything in the subject.

When the angles add up to exactly half a turn #

The inequality that limited the kaleidoscopes to three has two neighbours, and they explain what the regular solids are a special case of.

If the three angles $180°/p$, $180°/q$, $180°/r$ add up to exactly $180°$ — that is, $1/p + 1/q + 1/r = 1$ — the triangle is flat. The solutions are $(3, 3, 3)$, $(2, 4, 4)$ and $(2, 3, 6)$: an equilateral triangle, a right isosceles triangle, and half of an equilateral one. Their kaleidoscopes do not close up round a point into a finite group; they tile the whole plane, with infinitely many reflections. Wythoff’s construction still works: put a point in the triangle, reflect it everywhere, and join the images to get a tiling — the square grid, the honeycomb, the pattern of triangles and hexagons on a woven basket, the octagon-and-square floors. The uniform tilings of the plane by regular polygons, apart from a few alternations, come out of these three kaleidoscopes exactly as the solids came out of the spherical ones.

If the angles add up to less than $180°$, the triangle lives in the hyperbolic plane, where triangles have angle sums below half a turn, and there are infinitely many kaleidoscopes — $(2, 3, 7)$, $(2, 4, 5)$, $(3, 3, 4)$ and every other triple with $1/p + 1/q + 1/r < 1$. Each tiles the hyperbolic plane, each gives its own family of uniform tilings by Wythoff’s construction, and they are the patterns Escher drew in his circle limits.

So the regular solids are the positively curved third of a single picture. The same three mirrors, at angles that sum to more than, exactly, or less than half a turn, produce the finite groups of the sphere, the symmetries of the flat tilings, and the infinite groups of hyperbolic geometry. Which one appears is decided by one inequality, and the angle defect — positive on the sphere, zero in the plane, negative in hyperbolic space — is that inequality made into a number.

What the kaleidoscope pictures cannot show #

The mirror figures show great circles on a sphere, drawn from one viewpoint, with the far side faint. They cannot show the whole sphere at once, and the count of triangles is a computation, not a count of what is visible. The solids are drawn with their hidden faces removed, so a reader sees at most half of each; the face counts in the captions include the other half.

More substantially, the figures show the kaleidoscopes that exist, not the proof that there are no others. That proof is the inequality above, which says a spherical triangle tiling the sphere by reflections must have angles $180°/2$, $180°/3$ and one of $180°/3$, $180°/4$, $180°/5$, or be one of the prism kaleidoscopes. The figures also compute rather than prove the group orders: they multiply the three reflections together until nothing new appears, which establishes 24, 48 and 120 for these groups exactly, but a statement about every kaleidoscope needs the argument about triangles and symmetries given earlier, not a search. And the figures do not show the higher-dimensional story, where the same construction with four mirrors through a point of four-dimensional space produces the regular four-dimensional polytopes, six of them, and their uniform relatives.

Still open: uniform solids in higher dimensions #

In three dimensions everything here is settled: the finite reflection groups are the three kaleidoscopes and the prism family, and the convex uniform solids are completely listed. The classification of finite reflection groups in every dimension is also complete — it is Coxeter’s classification of 1934, one of the cleanest results in algebra, with its diagrams of dots and lines.

What is not complete is the list of uniform polytopes in four and more dimensions. Wythoff’s construction produces most of them from the reflection groups, alternation produces more, and in four dimensions the convex uniform polytopes have been enumerated — 64 of them outside two infinite families of prisms — by a combination of the construction and computer search. One of those 64, the grand antiprism found by John Conway and Michael Guy in 1965, comes from neither Wythoff’s construction nor a simple alternation of it — the first sign that the construction does not account for everything. In five and more dimensions no complete list is known, and how many exceptions of that kind exist, or whether they can be organised by any construction as clean as the kaleidoscope, is an open question.