Notes for AKT-170113/0:50:48: Difference between revisions
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I put the <math>h^2</math> term to make the inverse <math>R^{-1}</math> be identical but with negative <math>h</math>, the factorial is just a hint of more to come. |
I put the <math>h^2</math> term to make the inverse <math>R^{-1}</math> be identical but with negative <math>h</math>, the factorial is just a hint of more to come. |
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I thought it was fun to have an example of this in <math>U(sl_2)</math> where you can check that <math>r_{ij} = E_iF_j + \frac{1}{4} H_iH_j</math> is a solution to CYBE. |
I thought it was fun to have an example of this in <math>U(sl_2)</math> where you can check that <math>r_{ij} = E_iF_j + \frac{1}{4} H_iH_j</math> is a solution to CYBE. |
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If I got it right the positive Reidemeister 1 curl yields the value <math>1+ h(EF+\frac{1}{4}H^2)+\frac{1}{2}h^2(2E^2F^2 + EH^2F+EFH+\frac{H^4}{8})</math> bad news, we need the element <math>S</math> to |
If I got it right the positive Reidemeister 1 curl yields the value <math>1+ h(EF+\frac{1}{4}H^2)+\frac{1}{2}h^2(2E^2F^2 + EH^2F+EFH+\frac{H^4}{8})</math> bad news, we need the element <math>S</math> to get an invariant in this case. |
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Revision as of 08:09, 14 January 2017
Roland At 38:12 Dror mentions a solution to CYBE already gives a knot invariant by setting [math]\displaystyle{ R = 1 + hr + \frac{1}{2!}h^2r^2 }[/math] and working modulo [math]\displaystyle{ h^3 }[/math]. I put the [math]\displaystyle{ h^2 }[/math] term to make the inverse [math]\displaystyle{ R^{-1} }[/math] be identical but with negative [math]\displaystyle{ h }[/math], the factorial is just a hint of more to come. I thought it was fun to have an example of this in [math]\displaystyle{ U(sl_2) }[/math] where you can check that [math]\displaystyle{ r_{ij} = E_iF_j + \frac{1}{4} H_iH_j }[/math] is a solution to CYBE. If I got it right the positive Reidemeister 1 curl yields the value [math]\displaystyle{ 1+ h(EF+\frac{1}{4}H^2)+\frac{1}{2}h^2(2E^2F^2 + EH^2F+EFH+\frac{H^4}{8}) }[/math] bad news, we need the element [math]\displaystyle{ S }[/math] to get an invariant in this case.
