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Saturday, January 5, 2013

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}.

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