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,


<pre>
<pre>

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