Notes for SwissKnots-1105/0:19:28: Difference between revisions
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A Leibniz algbera is a Lie algebra minus the anti-symmetry of the bracket; I have previously erroneously asserted that here {\mathcal A}(K) is Lie; however, |
A Leibniz algbera is a Lie algebra minus the anti-symmetry of the bracket; I have previously erroneously asserted that here <math>{\mathcal A}(K)</math> is Lie; however, |
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Latest revision as of 11:04, 1 June 2011
A Leibniz algbera is a Lie algebra minus the anti-symmetry of the bracket; I have previously erroneously asserted that here is Lie; however,
Jim, Ooops, you are probably right and I should retract my claim and revert to the older version, which just said that gr is a Leibniz algebra. Do you allow me to post this conversation as is (minus your email address) as a reference? Where did you find the Lie claim? I just made it now at Swiss Knots 2011, but I have the feeling I made it elsewhere too. Best, Dror. On Wed, 1 Jun 2011, James Conant wrote: > Hi Dror, > > I know you must be busy, but I have a quick question about a claim on one of > your slides that the quadratic approximation to the associated graded object > for a quandle (with unit) is a (graded) Lie algebra. This caught my eye since > I am on the look-out for interesting constructions of Lie algebras associated > to knots and links. In any event, I haven't been able to prove antisymmetry. > It's pretty obvious that on the basis {(v-1)} for the augmentation ideal that > (v-1)^(v-1)=0. If we also knew that x^x=0 for arbitrary linear combinations > of these basic v-1s, we'd be done, but I don't see how to show that. Perhaps > I'm missing something obvious. > > Thanks, > > -Jim