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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Aug 18, 2010 at 3:05 comment added Theo Johnson-Freyd I have made some small formatting changes.
Aug 18, 2010 at 3:04 history edited Theo Johnson-Freyd CC BY-SA 2.5
small formatting changes
Aug 17, 2010 at 21:32 answer added DamienC timeline score: 4
Aug 17, 2010 at 21:25 comment added DamienC Applying the flip $\tau_{2,3}$ to $$ R_{23}R_{12}R_{13}-R_{13}R_{12}R_{23} $$ we get $$ R_{32}R_{13}R_{12}-R_{12}R_{13}R_{32} $$ which vanishes.
Aug 17, 2010 at 12:26 comment added Abtan Massini Ok, I agree that $ r(u^k_l \otimes I^{ij}_{mn}) = \sum_{r,s,x} R^{ji}_{rs}R^{kr}_{xn}R^{xs}_{lm} - \sum_{r,s,x} R^{ki}_{xr}R^{xj}_{ls}R^{sr}_{nm} = (R_{23}R_{12}R_{13} - R_{13}R_{12}R_{23})^{kij}_{lnm}$, but I don't see why this vanishes. In particular, I don't see why this resolves the specific problem of $R(u^2_1 \otimes I^{12}_{22})$ seeming to not equal zero..
Aug 17, 2010 at 9:11 comment added DamienC Clearly, we do have $\sum_{r,s,x} R^{ji}_{rs}R^{kr}_{xn}R^{xs}_{lm} - \sum_{r,s,x} R^{ki}_{xr}R^{xj}_{ls}R^{sr}_{nm}= (R_{23}R_{12}R_{13}-R_{13}R_{12}R_{23})^{kij}_{lnm}$ and $R_{23}R_{12}R_{13}-R_{13}R_{12}R_{23}$ is not yang-Baxter (and it does not have any meaning in terms of braids).
Aug 17, 2010 at 3:02 comment added Mariano Suárez-Álvarez Kassel's book, if I recall correctly, has all details.
Aug 17, 2010 at 0:59 history edited Abtan Massini CC BY-SA 2.5
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Aug 17, 2010 at 0:22 history edited Abtan Massini CC BY-SA 2.5
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Aug 17, 2010 at 0:17 history asked Abtan Massini CC BY-SA 2.5