- 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.

- 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, 2`

for 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.

- 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. In this notation,*blue*is reflection,*red*is "true motion" (?).*blue*

- 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`

.

`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!