Knot at Lunch, May 24, 2007: Difference between revisions
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* <math>[abw]=[baw]</math> if <math>|w|\geq 2</math> and similar identities. |
* <math>[abw]=[baw]</math> if <math>|w|\geq 2</math> and similar identities. |
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* <math>\phi(a,b,c)=\phi(a,b)</math>, <math>\Phi=\exp\phi</math> and in our case, this is just <math>1+\phi</math>! |
* <math>\phi(a,b,c)=\phi(a,b)</math>, <math>\Phi=\exp\phi</math> and in our case, this is just <math>1+\phi</math>! |
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* <math>[a^nb^mab]=(-1)^{n+m}[ab]a^nb^m</math>. |
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* <math>\phi=[ab]\lambda(a,b)</math>. With this, we have as follows: |
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* Unitarity becomes <math>\lambda(a,b)=\lambda(b,a)</math>. |
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* <math>\Phi^{312}=1+[ca]\lambda(ac)</math> where <math>c=-a-b</math>. |
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* Likewise for all other terms in the hexagon, which becomes |
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{{Equation*|<math>e^{b+c}=(1+[ab]\lambda(a,b))e^b\cdots = e^be^c+[ab]\left(\lambda(a,b)e^{b+c}+\lambda(b,c)e^c+\lambda(a,c)\right)</math>}} |
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* Simplifying <math>e^{b+c}-e^be^c</math> using frozen feet, this becomes |
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{{Equation*|<math>[cb]\left(\frac{e^{b+c}-1-b-c}{b(b+c)}-\frac{e^c-1-c}{c}\right) = [ab]\left(\lambda(a,b)e^{b+c}+\lambda(b,c)e^c+\lambda(a,c)\right)</math>}} |
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* Alternatively, |
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{{Equation*|<math>[ab]\left(\ldots\right) = [ab]\left(\lambda(a,b)e^{b+c}+\lambda(b,c)e^c+\lambda(a,c)\right)</math>}} |
Latest revision as of 12:46, 24 May 2007
First meeting for summer 2007! Peter Lee is telling us about associators with frozen feet. See also his handout from the CMS Winter 2006 Session on Knot Homologies - front: , back: and Dror's very partial paperlet, Associators with Frozen Feet.
- Definition of and its relation with finite type invariants.
- The frozen feet quotient.
- The action of , , and of , etc. (Dror: see also VS, TS and TG Algebras.)
- Generators and relations: , , the hexagons and pentagon, unitarity, non-degeneracy, group-like property.
- if and similar identities.
- , and in our case, this is just !
- .
- . With this, we have as follows:
- Unitarity becomes .
- where .
- Likewise for all other terms in the hexagon, which becomes
- Simplifying using frozen feet, this becomes
- Alternatively,