Zoran Skoda twiststar

Denote $\mathcal{F}^{-1} = \sum_\alpha \bar{\mathbf{f}}^\alpha\otimes \bar{\mathbf{f}}_\alpha$, where we will skip summation sign. Define

$f\star g = m \mathcal{F}^{-1}(\triangleright\otimes\triangleright)(f\otimes g) = (\bar{\mathbf{f}}^\alpha\triangleright f)(\bar{\mathbf{f}}_\alpha\triangleright g)$

Then

$\array{ (f\star g) \star h &=& m \mathcal{F}^{-1} (\triangleright\otimes\triangleright)((f\star g)\otimes h)\\ &=& m\{\bar{\mathbf{f}}^\alpha\triangleright [m\mathcal{F}^{-1}(\triangleright\otimes\triangleright)(f\otimes g)]\otimes (\bar{\mathbf{f}}_\alpha\triangleright h)\}\\ &=& m\{\bar{\mathbf{f}}^\alpha\triangleright [(\bar{\mathbf{f}}^\beta\triangleright f)(\bar{\mathbf{f}}_\beta\triangleright g)]\otimes (\bar{\mathbf{f}}_\alpha\triangleright h)\}\\ & \overset{Leibniz}= & m \{ m [\Delta_0(\bar{\mathbf{f}}^\alpha)(\triangleright\otimes\triangleright)[(\bar{\mathbf{f}}^\beta\triangleright f) \otimes (\bar{\mathbf{f}}_\beta\triangleright g)]]\otimes (\bar{\mathbf{f}}_\alpha\triangleright h)\}\\ &=& m(m\otimes id)[(\Delta_0\otimes id)\mathcal{F}^{-1}](\mathcal{F}^{-1}\otimes id)(\triangleright\otimes\triangleright\otimes\triangleright)(f\otimes g\otimes h) }$

Similarly,

$f\star (g \star h) = m(id\otimes m)[(id\otimes \Delta_0)\mathcal{F}^{-1}](id\otimes \mathcal{F}^{-1})(\triangleright\otimes\triangleright\otimes\triangleright)(f\otimes g\otimes h)$

Created on November 28, 2014 at 20:17:01. See the history of this page for a list of all contributions to it.