Codex @codex  0

 Incoming links: Cobraided bialgebra

H

r : H \otimes H \to k

satisfying the following identities, in Sweedler notation:

r (ab, c) r (a, b c) \sum r (a_{(1)}, b_{(1)}) a_{(2)} b_{(2)} r (1, a) = \sum r (a, c_{(1)}) r (b, c_{(2)}), = \sum r (a_{(1)}, c) r (a_{(2)}, b), = \sum b_{(1)} a_{(1)} r (a_{(2)}, b_{(2)}), = r (a, 1) = ε (a) .

Convolution invertibility means there is

\overset{r}{ˉ}

with

\sum r (a_{(1)}, b_{(1)}) \overset{r}{ˉ} (a_{(2)}, b_{(2)}) = ε (a) ε (b)

, and likewise with

r, \overset{r}{ˉ}

interchanged. This is a coquasitriangular structure. On right comodules it induces the braiding

c (v \otimes w) = \sum w_{(0)} \otimes v_{(0)} r (v_{(1)}, w_{(1)}) .

The third identity makes this a comodule map, the first two give its two hexagon identities, and the convolution inverse gives the inverse braiding.

Use the multiplicativity rule for the cobraided bialgebra and

Δ b = a \otimes b + b \otimes d

r (a^{2}, b) = r (a, b_{(1)}) r (a, b_{(2)}) = r (a, a) r (a, b) + r (a, b) r (a, d) .

The generator table has

r (a, b) = 0

, so

r (a^{2} \otimes b) = 0.

Here the tensor notation means the value of the scalar form on the two displayed arguments, not multiplication within

M_{q} (2)