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.

No comments:

Post a Comment