Let’s work out how symmetric groups act on themselves by conjugation. As I’m writing I notice that what I said before about composition of permutations is sort of backwards. It’s one of those annoying conventions that doesn’t really change anything, but can still be a bit confusing. From here on when we write permutations in cycle notation we compose by reading the cycles from right to left. That is, . Before I was reading them left to right. The new way behaves more like group actions. The exposition comes after the jump.
So the second Carnival of Mathematics is up at Good Math, Bad Math. My first post on Rubik’s Cube is up there, as well as one about algebraic topology at UM regular Mikael Johansson’s place, and one on permutation cycle types at The Universe of Discourse.
That last topic is actually really monumentally important, and I’ll be getting back to it after class and the graduate student seminar today.
This just popped up on the arXiv, so I thought I should mention it: Richard Schwartz has a paper up showing that there is a shape for an “outer billiards” table and a starting point whose path gets as far away from the shape as you want. Even better, it’s one of the shapes from the Penrose tiling. Curiouser and curiouser. The first section or so of the paper are very readable, and gives a much better explanation (with pictures!) of outer billiards than I could manage. The proof itself is heavily aided by computer calculations, but seems to be tightly reasoned apart from needing help to handle a lot of cases.
I’m not quite sure how billiards and outer billiards are related. My intuition is that there’s some sort of duality going on, which would exchange lengths of segments in outer billiards with angles in billiards. On the other hand, if there were such a straightforward translation, couldn’t the enormous machinery of billiards have been brought to bear on this problem before now? Do any billiard theorists in the audience know anything about outer billiards?