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# Possible Symmetries of Spherical Patterns

All the symmetries of discrete patterns on spheres are described by symbols in the orbifold notation that cost less than \$2. What can we buy? To get the complete list of notational symbols we can afford, we'll first figure out which expensive items we can buy, then see what we can do with the change.

While reading the discussion below, keep in mind that every gyroscopic point costs less than \$1, and every kaleidoscopic point costs less than \$.50. We must spend less than \$2 total.

We can't afford to buy a handle (o) -- that costs \$2 by itself.

We can buy a cross-cap (x) for \$1. Once we've bought the cross-cap, we haven't got enough money left to buy a mirror boundary (*) without going broke, but we can still buy gyration points if we want. So, we can buy the group x and any group nx. Once we've bought one gyration point, we can't afford a second, so those are all the groups we can buy with cross-caps in them. They are called the "polydromic" groups, and are the only sphere groups whose orbifolds are non-orientable. (If you're4 not sure what non-orientable means, don't worry about it.)

We can create an example of a pattern with symmetry group x by painting a spiral starting at the north pole of the sphere and ending at the south pole. Opposite points of the sphere look alike, which is typical of sphere patterns with x's in their symbols. To make a pattern with symmetry group nx, we would use an n-armed spiral, and so add n-fold rotational symmetry.

We could buy a mirror boundary (*) for \$1. If we wish to add a gyration point once we've done that, we may. The symbols n* denote the "polygyros" groups, whose orbifolds consist of one mirror boundary with a single cone point.

We see an object whose symmetry group is represented by n* when we set an object with n-fold rotational symmetry down on a mirror.

If we buy a mirror boundary and an order two gyration point, we will have spent \$1.50. We then have enough change to buy any kaleidoscopic (corner) point! The spherical symmetry groups with orbifold notation 2*n are called polydigyros.

A sphere decorated with n S- and Z- shaped curlicues about its equator might have this symmetry group.

If we buy a mirror boundary and an order three gyration point, we will have less than 33 cents change. There is only one thing we can buy with that -- an order 2 kaleidoscopic point. The group 3*2 is called pyritohedral (don't ask me why).

If we decorate each face of an octahedron with a three-legged pinwheel, we get a pattern with pyritohedral symmetry.

If we buy a mirror boundary and no gyration points, we'll have a whole dollar left to buy kaleidoscopic points! Alas there are no symmetry groups corresponding to the symbols *mn for unequal m and n. The ideal fabric from which we build orbifolds will not alow itself to be stretched into a two-gon with unequal angles. We also cannot form a "one-gon" -- an orbifold with only one corner point.

However, we can buy any orbifold with the symbol *nn -- these groups are called polyscopic. If n is 2, we will have \$.50 change left to buy any other corner point; the groups *22n are called polydiscopic. A sphere with n plusses equally spaced on its equator has a discopic symmetry group.

If we buy an order five kaleidoscopic point, we will have \$.60 change. If we don't want to buy another order five corner point to match the first, we must buy two other corner points. We've already discussed what happens when we add two order two corner points; our only other option is to buy one of order three and one of order two, for approximately \$.25 + \$.33 = \$.58. The achiral icosahedral symmetry group is that of the icosahedron, and is denoted *532. The base triangle of the icosahedral group in Kali has angles of 180/5, 180/3, and 180/2.

The situation for order four and order three kaleidoscopic points is similar to that for order five. Once we've bought one, we can only afford to buy two order two corners or an order three and an order two. The symbols *432 and *332 denote the achiral octahedral and achiral tetrahedral groups, respectively. These are the symmetry groups of the octahedron and tetrahedron. The base triangles shown by KaleidoTile for these groups do have angles of 45, 60, 90 and 60, 60, 90!

If we buy no mirror boundary at all, we can afford at least two gyration points. Once again, we're prohibited from making an orbifold with one lone or two unmatched gyration points. We can, however, form any of the groups nn, called polytropic groups. These describe the symmetries of a sphere with an n legged pinwheel drawn on it.

If we buy two order two gyration points, we will have \$1 left over to buy any other gyration point. The groups 22n are called polyditropic. A sphere with n S's on its equator has this symmetry.

If we buy an order five gyration point, we'll have \$1.20 left to spend on others. The only thing we can afford that we haven't discussed already is 532, the chiral icosahedral group. This symmetry can be illustrated by putting three-legged pinwheels on the faces of an icosahedron, or five-legged pinwheels on the faces of a dodecahedron.

Perhaps you've noticed a correspondence between symbols made up of kaleidoscopic points and those made of gyroscopic points. This is not a coincidence -- kaleidoscopic points cost half what gyroscopic points cost, and once we've bought our mirror boundary we have half as much to spend! We expect that the chiral octahedral and chiral tetrahedral groups (432 and 332, illustrated by drawing on the faces of the octahedron and tetrahedron) will cost less than \$2, and in fact they do.

After a moment's thought, you will find that we cannot possibly buy any other symbol in the orbifold notation for less than \$2. Hence, we have completely classified the spherical symmetry groups!

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