Timeline for Complete knot invariant?
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12 events
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| Mar 28 at 20:12 | comment | added | Ian Agol | Yes, I think that’s right. In principle one could incorporate the orientation into the group theory, by taking an orientation of the relative fundamental class of the fundamental group (since the knot complement is a $K(\pi,1)$). This orientation of the peripheral subgroup is more-or-less equivalent since it orients the peripheral torus, and hence the 3-manifold by the long exact sequence in homology. | |
| Mar 28 at 16:34 | comment | added | Minkowski | Alright, but then it doesn't follow from Waldhausen theorem that the homeomorphism between the knot exteriors is orientation-preserving, right? In other words, if I am understanding correctly, the peripheral group system is a complete invariant of unoriented knots up to mirror images, is that right? And if the knots are oriented, then the peripheral group systems of two knots $K,K'$ are isomorphic if and only if $K'=K$ or $K' = rmK$ the reverse of the mirror image of $K$. Is this correct? | |
| Mar 28 at 2:31 | comment | added | Ian Agol | @Minkowski: if the knots admit an orientation, then the orientation of $S^3$ induces an orientation of the meridians. If the map on peripheral elements preserves the orientation of meridian and longitude, then the map of knots will also be orientation preserving. And orientation preserving homeomorphisms of $S^3$ are connected, so the knots are isotopic. If the knots are unoriented, just try both orientations to compare them. So eg one can also tell if a knot is invertible or amphichiral. | |
| Mar 27 at 20:41 | comment | added | Minkowski | Quite late here but how does it follow from the Waldhausen theorem that the homeomorphism between the knot exteriors is orientation-preserving (which is needed to conclude that the knots are isotopic by the Gordon-Luecke theorem)? | |
| Aug 20, 2010 at 14:22 | comment | added | Dave Futer | @Ryan: your summary of what's known about link complements is spot on. | |
| Aug 17, 2010 at 3:30 | vote | accept | Peter Samuelson | ||
| Aug 16, 2010 at 16:01 | history | edited | Ian Agol | CC BY-SA 2.5 |
correction
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| Aug 15, 2010 at 21:02 | comment | added | Ryan Budney | Since knots are essentially classified perhaps I should have phrased that as, it's an open problem to find an efficient procedure to go from one link and construct all the links whose complements are homeo/diffeomorphic to your original link complement. | |
| Aug 15, 2010 at 20:53 | comment | added | Ryan Budney | I think it's still an open problem as which links have the same link complement. I believe Gordon has some results on this but as far as I know these results are not known to be complete? | |
| Aug 15, 2010 at 20:47 | comment | added | Ryan Budney | @Samuelson: the "filling slope" construction I gave is basically the way of avoiding dealing with the Gordon-Luecke knot complement problem. In particular, you can't use Gordon-Luecke for link complements, but the filling-slope technique does generalize. | |
| Aug 15, 2010 at 20:45 | history | edited | Ian Agol | CC BY-SA 2.5 |
added 134 characters in body
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| Aug 15, 2010 at 20:40 | history | answered | Ian Agol | CC BY-SA 2.5 |