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Timeline for Witten's QFT and Jones Poly paper

Current License: CC BY-SA 3.0

10 events
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S Aug 10, 2013 at 16:34 history suggested Michael Albanese CC BY-SA 3.0
Replaced \\# with \#.
Aug 10, 2013 at 16:27 review Suggested edits
S Aug 10, 2013 at 16:34
Feb 22, 2011 at 9:08 history edited Kelly Davis CC BY-SA 2.5
Expanded on the answer
Feb 22, 2011 at 9:02 history undeleted Kelly Davis
Feb 21, 2011 at 21:02 history deleted Kelly Davis
Feb 21, 2011 at 10:58 comment added Daniel Pomerleano the weird tilde thing is supposed to be the universal cover but it's late and i'm to lazy to fix it...
Feb 21, 2011 at 10:57 comment added Daniel Pomerleano Yeah at least in the simply connected case, that's probably what she means. We can just collapse the two skeleton of M and look at the map S^3 \to G given by a generator of pi_3. By Hurewicz and the fact that the collapsing map induces an iso on H_3, the map must be non-trivial on homology. In the non-simply connected case, pass to the universal cover and note that the map $H_3(\tilde(G))\to H_3(G)$ is multiplication by some non-zero number.
Feb 21, 2011 at 9:00 comment added Aaron Mazel-Gee Presumably you should be able to take a map which is constant on $M\backslash D^3$ (and on the connecting tube) and essential on $(S^3\backslash D^3,\partial D^3) \simeq (S^3,pt)$? Then we could prove this is essential too by looking at the induced map on $H_3$ or something like that. (Or maybe there's an easier direct argument proving that such a construction can't be nullhomotopic.)
Feb 21, 2011 at 7:23 comment added Kevin Wray Yes, I understand the $M=S^3$ case, but not following what you are implying by considering the connected sum $M\# S^3$
Feb 21, 2011 at 7:11 history answered Kelly Davis CC BY-SA 2.5