## § John Conway: The symmetries of things

Original way to classify wallpaper groups: think of geometric transforms that fix the pattern. Thurston's orbifold solution: think of quotients of $\mathbb R^2$ by groups --- this gives you an orbifold (orbit manifold). Take a chair, surround it around by a sphere. The symmetries of a physical object fixes the center of gravity. So we pick the center of the sphere to be the center of gravity. The "celestial sphere" (the sphere around the chair) is a nice manifold (We only have the surface of the sphere). The vertical line that divides the chair also divides the sphere into two parts.
• The points of the orbifold are orbits of the group.
• So now the orbifold gives us a hemisphere in this case.
• The topology of the orbifold determines the group.
• This is astonishing, because the group is a metrical object: elements of the grouppreserve the inner product of the space.
• And yet, geometrical groups are determined by the topology of their orbifolds!
• Thurston's metrization conjecture: certain topological problems reduce to geometrical ones.
Conway came up with his notation for wallpaper groups/orbifolds. There are only four types of features.
• The hemisphere orbifold is *. (group of order 2). * denotes the effect onthe orbifold. * really means: what is left out of a sphere when I cut out a hemispherical hole. * is the name for a disk, because a hemisphere is a disktopologically. It has metrical information as well, but we're not going tospeak about it, because all we need is the topological information.
• One-fourth of a sphere (symmetry group of rectangular table)is denoted by * 2 2. The * for the hemisphere, and 2, 2for the angles of pi/2.
• If the table is a sphere, then we have diagonal symmetry as well. In this case,the orbifold has angle pi/4. So the table is * 4 4.
• If we take a cube, then we have an even more complicated orbifold. The "fundamental region"of the cube has 2, 3, and 4 mirrors going through them. So in the orbifold, we gettriangles of angles pi/2, pi/3, pi/4. This would be * 4 3 2.
• Draw a swastika. This has no reflectionsymmetry. This has a gyration: a point about which the figure can be rotated,but the point is NOT on a line of reflection. We can tear the paper and makeit into a cone. This gives us a cone point. The angle around the cone pointis 2pi/4. This is the orbifold of the original square with a swastika on it.
An orbifol can be made to carry some amount of metrical information. The cone point only has 90 degrees, so it is in some sense, "a quarter of a point".
• Draw a cube with swastikas marked on each face. This has no reflectionsymmetry. Once again, we have a gyration, and again, only the gyration/singularitiesmatter. This group is again 4, 3, 2 , but in blue. In this notation,red is reflection, blue is "true motion" (?).
Let us try to work out the euler characteristic of the rectangular table orbifold by using $V - E + F$. The orbifold as one face. The wrong thing to say is that the orbifold has two edges and two vertices. It is untrue because the edge of the orbifold is only half an edge --- let's say that lines have thickness. In this case, we will have $V = 2/4$, $E = 2/2$, and $F = 1$. The euler characteristic works out to be a half. This is appropriate, because the orbifold is a type of divided manifold.
• If we work this out for a cube, we get $2/48$. This is because the sphere getsdivided into 48 pieces, and the sphere has an euler characteristic of 2!
• Alternatively, we can think that we started out with 2 dollars, and we are thenbuying the various features of our orbifold. * costs 1\$, a blue numberafter a star, for example: 2 costs 1/2 a dollar. 3 costs 2/3 of a dollar,4 costs 3/4 of a dollar. In general, n costs 1 - 1/n.The red numbers are children, so they cost half an much: n consts 1/2(1 - 1/n) = (n-1)/2n.
Now, see that we started with positive euler characteristic (2), and we divide it by some n (the order of the group). So we end up with a positive euler characteric. By a sort of limiting argument, the euler characteristic of the wallpaper groups, which are infinite, is zero. However, see that we must get to the zero by starting with two dollars and buying things off the menu! If we try and figure out what all the possible ways are to start with 2 dollars and buy things till we are left with exactly 0 dollars, we get that there are 17 possible ways of buying things on the menu! Thus, this the reason for there being 17 wallpaper groups.
• To buy more than two dollars, you are buying symmetries from the hyperbolicplane!
Because we can completely enumerate 2-manifolds, we can completely enumerate 2-orbifolds, which are essentially the same thing as symmetry groups. The real power is in the 3D case. We don't have a full classification of 3-manifolds. But we maybe able to go the other way. This is the metrization theorem.