What I learned today: Braided (or quasi-triangular) Hopf Algebras

This is a post in my ``psuedo-daily'' series ``What I learned Today''

A Hopf algebra $H$ is endowed with both a multiplication map $m: H \otimes H \rightarrow H$ and a comultiplication map $\Delta: H \rightarrow H \otimes H$. Many of our ``quantum'' objects lack commutivity or co-commutitivity, but not completely. In cases of co-commutitivity, the entities that lack it often have the lack of co-commutivity controlled by an invertible element $\mathcal{R} \in H \otimes H$ in the following manner:

$\displaystyle \tau \circ \Delta h = \mathcal{R}(\Delta h)\mathcal{R}^{-1}$

Where $\tau(x\otimes y) = y \otimes x$ is the flip operator. If we also put some restrictions on what element $\mathcal{R}$ we can chose, we'll arrive at Drinfeld's theory of Quasi-tranigular, or Braided, Hopf Algebras. But first, let's look at an example: Let $G$ be the Klein four-group, id est, $G = \langle x, y : x^2 = y^2 = (xy)^2 = 1 \rangle$ and let $kG$ be the group Hopf algebra. For $q \in k$ define

$\displaystyle \mathcal{R}_q = \frac{1}{2} (1\otimes 1 + 1\otimes x + x\otimes 1 - x\otimes x) + \frac{q}{2}(y \otimes y + \otimes xy +xy \otimes xy - xy \otimes y)$

Let's change the coproduct on $kG$ to be $\Delta x = x\otimes x$ and $\Delta y = 1 \otimes y + y\otimes x$. Also change the antipode to $Sx = x$ and $Sy = xy$ and the counit $\epsilon (x) = 1$ and $\epsilon (y) = 0$. This new Hopf algebra indeed obeys the $\tau \circ \Delta h = \mathcal{R}_q (\Delta h) \mathcal{R}_q^{-1}$. It also obeys the additional relations

$\displaystyle (\Delta \otimes I) \mathcal{R} = \mathcal{R}_{13}\mathcal{R}_{23}$

$\displaystyle (I \otimes \Delta) \mathcal{R} = \mathcal{R}_{13}\mathcal{R}_{12}$

Where $\mathcal{R}_{23} = 1 \otimes \mathcal{R}$, for instance. These three conditions are what defines a Braided Hopf algebra (we call them braided because their category of modules generates a braided category. More on that in a later post.) These Hopf algebras generate solutions to the \href{http://en.wikipedia.org/wiki/Yang

$\displaystyle (c \otimes I)(I \otimes c)(c \otimes I) = (I \otimes c)(c \otimes I)(I \otimes c)$

for an automorphism $c$ of $V \otimes V$, where $V$ is an $H$-module. To see this, let $V$ and $W$ be $H$ modules and define an isomorphism of $V \otimes W$ and $W \otimes V$ by

$\displaystyle c_{V,W} (v\otimes w) = \tau_{V,W} (\mathcal{R} (v\otimes w))$

From here, one can show that for $H$-modules $U$, $V$, and $W$ we have

$\displaystyle c_{U\otimes V, W} = (c_{U,W} \otimes I_V)(I_U \otimes c_{V,W}) $

$\displaystyle c_{U, V\otimes W} = (I_V \otimes c_{U,W})(c_{U,V} \otimes I_W)$

and finally

$\displaystyle (c_{V,W} \otimes I_U) (I_V \otimes c_{U,W})(c_{U,V} \otimes I_W) = (I_W \otimes c_{U,V})(c_{U,W} \otimes I_V)(I_U \otimes c_{V,W})$

Letting $V = W = U$, we see we have a solution of the Yang-Baxter equation. It's also worth noting that the element $\mathcal{R}$ itself satisfies a version of the Yang-Baxter equation: $\mathcal{R}_{12}\mathcal{R}_{13}\mathcal{R}_{23} = \mathcal{R}_{23}\mathcal{R}_{13}\mathcal{R}_{12}$.