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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Feb 1, 2013 at 6:00 vote accept David Roberts
Aug 21, 2011 at 21:59 answer added Ryan Budney timeline score: 4
Jan 1, 2011 at 21:24 answer added Jeffrey Giansiracusa timeline score: 3
Dec 31, 2010 at 22:51 comment added Ryan Budney The complement of an $n$-knot for $n>1$ is aspherical if and only if the knot complement has the homotopy-type of $S^1$. This is an old result of Dyer and Vasquez, 1973. The reference is in Hillman's book "2-knots and their groups" but Google books isn't bringing up that part of the book, and my actual copy is in my office... Google books does bring up Eckmann's 1976 proof, though.
Dec 31, 2010 at 22:34 answer added Peter Teichner timeline score: 13
Dec 31, 2010 at 22:28 comment added Peter Teichner Ryan, are you sure that a 2-knot complement isn't a 1-type? I remember that Whitehead's conjecture implies that complements of ribbon disks in the 4-ball are aspherical.
Aug 19, 2010 at 0:29 comment added David Roberts "...2-knot complements in S^4..." - cool. This is the sort of thing I was after. If you want to put this as an answer I'll accept it.
Aug 18, 2010 at 4:38 comment added Ryan Budney The wedge $S^1 \vee S^2$ is interesting. $2$-knot complements in $S^4$ have an interesting $2$-type, as well.
Aug 18, 2010 at 4:35 comment added David Roberts I don't need a 2-type per se, but a space with an interesting 2-type. But this is a good point you raise +1.
Aug 18, 2010 at 4:14 comment added Ryan Budney It's been a while but aren't there theorems to this effect: there are no finite-dimensional $K(\pi,n)$ spaces for $n\geq 2$. So if you had a finite-dimensional 2-type, its universal cover would be a finite-dimensional $K(\pi,2)$. If I recall, it's some kind of Serre spectral sequence argument, showing that $H_i(K(\pi,n))$ is non-trivial for infinitely-many $i$. I may be mis-remembering?
Aug 18, 2010 at 3:41 history asked David Roberts CC BY-SA 2.5