most recent 30 from http://mathoverflow.net 2013-05-26T02:56:46Z http://mathoverflow.net/feeds/user/3046 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58656/what-is-the-subfactor-planar-algebra-of-type-tildea-n-of-index-4 Stephen Bigelow 2011-03-16T16:18:25Z 2011-03-16T18:05:07Z

<p>As I understand it, there is a subfactor whose principal graph is the affine Dynkin diagram $\tilde{A}_n$. Since every vertex has two neighbors, does that mean the space of 1-boxes is two dimensional? Is that allowed?</p>

http://mathoverflow.net/questions/11305/constructing-the-hecke-algebra-from-the-burau-representation/13461#13461 Stephen Bigelow 2010-01-30T06:48:17Z 2010-01-30T06:48:17Z

<p>There's a more old-fashioned way to see the connection between the Burau representation and the Alexander polynomial. The Burau representation of a braid is the action on the first homology of the punctured disk with local coefficients. The Alexander polynomial of a knot is an invariant of the first homology of the knot complement with local coefficients. When you close up the braid, each element of homology of the punctured disk on the bottom becomes identified with its image in the punctured disk at the top. Thus you're killing all the columns in the matrix $\psi^r_n(\beta) - I$. The actual proof uses some long exact sequences, and I remember it as being a bit fiddly. It's probably in Birman and/or Rolfsen's books.</p> <p>This doesn't (yet?) generalize to other quantum invariants, so it might not be the "right" way to think about it.</p>

http://mathoverflow.net/questions/10974/does-homology-detect-chain-homotopy-equivalence Stephen Bigelow 2010-01-06T22:05:33Z 2010-01-06T23:50:39Z

<p>Is the following true: If two chain complexes of free abelian groups have isomorphic homology modules then they are chain homotopy equivalent.</p>