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're 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. We will learn in the next section that only orbifold symbols that cost less than $2 have positive "orbifold euler characteristic". In order for an orbifold symbol to describe a spherical symmetry group, its euler characteristic must be strictly positive, and all finite symmetry groups of the sphere are described by some orbifold symbol. So, by finding all the orbifold symbols that cost less than $2 we have classified all possible (finite) symmetries of spherical patterns!

Author: Heidi Burgiel

Comments to:
webmaster@geom.umn.edu

Created: Dec 7 1995 ---
Last modified: Wed May 28 11:33:56 1997

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