A few days ago I sketched out some of the basic category theoretic ideas behind the theory of bosonization of quantum groups. Today I want to talk about the dual version, co-bosonization.

First note that this whole theory is trivial, as dualizing such a theory is a straightforward and obvious process, but I'm working through the proofs of bosonization in a dualized setting by hand, just so I can learn Quantum Groups.

So let $(H,\mathcal{R})$ be a *dually* quasitriangular Hopf algebra (I called this a **co-braided Hopf algebra** in an earlier post on braided categories) and let $\mathcal{C} = ^H\mathcal{M}$ by the braided category of $H$-comodules. If $(V, \beta_V: v \mapsto \sum h^{(1,v)}\otimes v^{(1)})$ and $(W, \beta_W: w \mapsto \sum h^{(1,w)}\otimes w^{(1)})$ are objects in the category, then the braiding is given by

$ \displaystyle \Psi_{V,W}(v\otimes w) = \sum \mathcal{R}\left(h^{(1,v)} \otimes h^{(1,w)}\right) w^{(1)} \otimes v^{(1)}$

I won't give the proof here (I may not give it at all!). As before, a Hopf algebra $B$ in the category $^H\mathcal{M}$ means that $B$, $B\otimes B$ and all the structure maps are morphisms in the category, thus any twist involved is actually an invocation of the braiding $\Psi$.

Also as before, $^B \mathcal{C}$ is the category of $B$-comodules within the category $\mathcal{C}$, and the idea behind co-bosonization is to find a Hopf algebra $\text{cobos}(B)$ such that $^B\mathcal{C} \cong \, ^{\text{cobos}(B)}\mathcal{M}$.

We're going to propose what this new Hopf algebra $\text{cobos}(B)$ should be. I won't give any proofs in this post (because I haven't done them yet), but hopefully I've dualized things correctly and I'm not stating anything that's not true. We'll start with the tensor product $B \otimes H$ and we'll put an coalgebra structure on it by first noting that in the bosonization case, the coalgebra $\Delta B \otimes H \rightarrow (B\otimes H) \otimes (B \otimes H)$ is

$ \displaystyle \left(\text{Id}_B \otimes \Psi_{B,H} \otimes \text{Id}_H\right) \circ \left( \Delta_B \otimes \Delta_H \right) $

This coproduct *almost* makes sense here, we just need to treat $H$ as a co-module of itself (so the braiding makes sense). But this is no difficult matter, because **every Hopf algebra coacts on itself as a coalgebra** by

$ \displaystyle \beta_H (h) = \sum h_{(1)} S(h_{(3)}) \otimes h_{(2)}$

So we just use the dual-braiding $\Psi$ instead of the original one, and write

$ \displaystyle \Delta (b \otimes h) = \sum \mathcal{R}\left( h^{(1,b_{(2)})} \otimes h^{(1,h_{(2)})} \right) b_{(1)} \otimes h_{(1)}^{(1)} \otimes b_{(2)}^{(1)} \otimes h_{(2)}$

Now on to the product for $\text{cobos}(B) = B \otimes H$: in the bosonization case (were $H$ acts, rather than coacts, on $B$), the product is given by

$ \displaystyle (b \otimes h) (c\otimes g) = \sum b(h_{(1)} \triangleright c) \otimes h_{(2)}g$

This doesn't immediately make sense in our case, but we have the following proposition:

**Proposition.** *There exists a monoidal functor (functor that respect the tensor product of objects) $^H \mathcal{M} \rightarrow ^H_H\mathcal{M}$, the crossed modules we mentioned for the quantum double, by $(V, \beta_V) \mapsto (V,\beta_V,\triangleright)$, where $h \triangleright v = \sum \mathcal{R}\left( h \otimes h^{(1,v)} \right)v^{(1)}$*

Hence we do have an action of $H$ on $B$, and we can write the product

$ \displaystyle (b \otimes h) (c \otimes g) = \sum \mathcal{R}\left( h_{(1)} \otimes h^{(1,c)}\right) bc^{(1)} \otimes h_{(2)}g$

The unit and counit are obvious in this case ($1 \otimes 1$ and $\epsilon(b \otimes h) = \epsilon(b)\epsilon(h)$), but we still need an antipode. Again, we refer to bosonization and try to find the correct dual notion. In bosonization, the antipode $S: B \otimes H \rightarrow B \otimes H$ is given by:

$ \displaystyle B\otimes H \xrightarrow{S_B \otimes S_H} B \otimes H \xrightarrow{\Psi_{B,H}} H \otimes B \xrightarrow{\Delta_H \otimes \text{Id}_B} H \otimes H \otimes B \xrightarrow{\text{Id}_H \otimes \Psi_{H,B}} H \otimes B \otimes H \xrightarrow{\triangleright \otimes \text{Id}_H} B \otimes H$

This makes perfect sense for co-bosonization without modification. To summarize, we need proofs for the claims

- That $\Psi$ is the braiding for the category $^H \mathcal{M}$. (I have this.)
- That the product and coproduct as defined above give us a bialgebra $\text{cos}(B)$ for $B$ a Hopf algebra in the category $^H \mathcal{M}$. (Not yet proved.)
- That the above antipode gives us a Hopf algebra $\text{cos}(B)$. (Not yet proved.)
- That $^B\mathcal{C} = ^{\text{cos}(B)}\mathcal{M}$. (Not yet proved).