Talk:Kervaire-Milnor Braid (Ex)
For starters the braid looks as follows:
Filling in the -groups (which were computed by Kervaire and Milnor), the homotopy groups of the orthogonal group (which are given by Bott periodicity), the framed bordism groups (which were shown to be the stable stems by Pontryagin and in low dimensions computed by Serre), and the -homomorphism (which was computed by Adams and Quillen, but can be computed by hand in low dimensions) we arrive at
where we have used to denote the map induced by the Kervaire invariant . Using that this is surjective ( with its Lie-group framing has Kervaire-invariant ) and the fact that the signature of an almost framed -manifold is divisible by (and is actually the signature of ???), we obtain the following maps:
The extension at is split since it surjects onto (right lower map) which is free. Clearing this and the obvious 's, we find:
which is only possible for , since the upper map cannot possibly be surjective in the other cases ( having a common factor with both and ). With this as input we claim that there is a unique isomorphism class of braids left and it looks as follows (proof given below):