## Monday, January 14, 2013

### Bosonisation of Quantum Groups, part II (Braided Categories)

Let's talk about the category of modules ${_H\mathcal{M}}$ for a braided Hopf algebra ${H}$ for a minute. If ${V}$ and ${W}$ are ${H}$-modules, then we have an easy action of ${H}$ on ${V \otimes W}$ as well, by:

$\displaystyle h \triangleright(v\otimes w) = \sum h_1\triangleright v \otimes h_2\triangleright w$

Hence we can think of ${\otimes}$ as a functor ${_H\mathcal{M} \times _H\mathcal{M} \rightarrow _H\mathcal{M}}$. This sort of functor is what turns ${_H\mathcal{M}}$ into a monoidal category. We also have the opposite function ${\otimes^\text{op}}$, which does what one would expect: ${V \otimes^\text{op} W = W \otimes V}$.

We have another operation on this category, the flip map ${\tau}$, which carries the action of one functor ${\otimes}$ to another ${\otimes^\text{op}}$. We'd like it to be a natural transformation, actually a natural isomorphism (one can see already that it's invertible) from ${\otimes \rightarrow \otimes^\text{op}}$. For this to work we must have, for each ${f: V \rightarrow V'}$ and ${g: W \rightarrow W'}$ in the category, a morphism ${\tau_{V \otimes W}: V \otimes W \rightarrow W \otimes V}$ such that the square

commutes. But before we check that, let's first verify that ${\tau_{V \otimes W}}$ is even a morphism in the category (to take a step back from category-theoretic language, this means that we need to check that ${\tau_{V \otimes W}(v\otimes w) = w \otimes v}$ is an ${H}$-linear map). But it's not necessarily true that ${h \triangleright \tau_{V \otimes W}(v\otimes w) = \tau_{V \otimes W}(\sum h_{(1)}\triangleright v \otimes h_{(2)}\triangleright w)}$. In particular, this is only true when ${H}$ is co-commutative.

But ${H}$ is a braided Hopf algebra, meaning that its failure to be co-commutative is controlled by an invertible element ${\mathcal{R} \in H \otimes H}$. Can we use this to construct an natural transformation ${\otimes \rightarrow \otimes^{op}}$? Yes, we can, and this is exactly the structure that gives us a braided category.

To be brief, we can define a braided category to be a monoidal category together with a natural isomorphism ${\Psi : \otimes \rightarrow \otimes^\text{op}}$ that obeys the following conditions: ${ \Psi_{V \otimes W, Z} = \Psi_{V,Z} \circ \Psi_{W,Z}}$ and ${\Psi_{V,W\otimes Z} = \Psi_{V,Z} \circ \Psi_{V,W}}$. (These are the hexagon conditions''. I'm not writing the structure maps of the monodial category, but when you include them, you express these as two hexagonal commutative diagrams.)

In the case of a braided, or quasitriangular Hopf algebra, our ${\Psi_{V,W} = \tau(\mathcal{R} \triangleright v\otimes w) = \sum R_2 \triangleright w \otimes R_1 \triangleright v}$. The reader can verify that this meets the axioms of a braided category.

There's more we can say about the braiding ${\Psi}$ and it's connection to braid groups, et cetera, but I don't want to go down that route this evening. Rather, I want to discuss the dual version of this theory. First, I need find the dual notion of a braided Hopf algebra. We'll call this a co-braided Hopf algebra. By this I mean that we want to control how far ${H}$ is from being commutative, instead of co-commutative.

A co-braided Hopf algebra is a Hopf algebra ${H}$ together with a linear functional ${R: H\otimes H \rightarrow k}$. We require ${R}$ to be convolution invertible, id est, that there exists another linear functional ${R^{-1}}$ such that ${\sum R^{-1}(h_1 \otimes g_1) R(h_2 \otimes g_2) = \epsilon(h)\epsilon(g) = \sum R(h_1 \otimes g_1) R^{-1}(h_2\otimes g_2)}$. We also require it to obey three other relations:

$\displaystyle \sum g_1h_1 R(h_2\otimes g_2) = \sum R(h_1 \otimes g_1) h_2g_2$

$\displaystyle R (hf \otimes f) = \sum R(h\otimes f_1)R(g\otimes f_2)$

$\displaystyle R(h\otimes gf) = \sum R(h_1 \otimes f)R(h_2 \otimes g)$

for ${g,h,f \in H}$.

Let ${V}$ and ${W}$ be ${H}$-co-modules with structure maps ${\beta_V: V \rightarrow H \otimes V}$ by ${v \mapsto \sum g_1 \otimes v_1}$ and ${\beta_W: W \rightarrow H \otimes W}$ by ${w \mapsto \sum h_1 \otimes w_1}$ The tensor functor works like such:

$\displaystyle \beta_{V\otimes W}(v\otimes w) = \sum g_1 h_1 \otimes v_1 \otimes w_1$

What does the braiding on the category of co-modules ${^H\mathcal{M}}$ look like? I propose that the braiding ${\Psi}$ is then:

$\displaystyle \Psi_{V,W}(v\otimes w) = \left(\sum R(g_1 \otimes h_1) \right) w_1 \otimes v_1$

I've not proved it yet.