Talk:Brouwer fixed-point theorem
Latest comment: 19 years ago by Chan-Ho Suh in topic Hex in multiple dimensions?
Courant and Robbins provide an accessible proof.
According to Lyusternik Convex Figures and Polyhedra, the theorem was first proved by a Lettish mathematician named Bol. No references are provided. Anyone know what this is about?--192.35.35.36 00:08, 18 Feb 2005 (UTC)
Proofs
Why is wikipedia not the place to reroduce a long proof? --anon
Hex in multiple dimensions?
The article says:
- A quite different proof can be given based on the game of Hex. The basic theorem about Hex is that no game can end in a draw. This is equivalent to the Brouwer fixed point theorem for dimension 2. By considering n-dimensional versions of Hex, one can prove that in general that Brouwer's theorem is equivalent to the "no draw" theorem for Hex.
But how does one play an "n-dimensional version of Hex"?
- Consider the original 2D hexboard as being made from a lattice, where one connects the lower left corner of a square of a lattice to the upper right corner with an edge. These added diagonals make it so you can have 6 neighbors instead of 4 (on the original lattice). It's easy to see how to cut out an n by n Hex board from this modified lattice. In general, to create an n x n .... x n board, consider a lattice in R^m (where m is the number of dimensions of the board) and then in each m-dimensional cube add a diagonal. Then cut out a board as in 2D.
- Each player has an opposite pair of sides as before, but now some sides belong to neither player. The game is played the same way as before. It doesn't appear to be so interesting to play, but people have devised other higher dimensional versions which are probably more fun and interesting mathematically. In any case, in the version I described, there can never be a draw, and this no-draw result is equivalent to the Brouwer fixed point theorem. --Chan-Ho (Talk) 07:03, 15 January 2006 (UTC)