Earlier this month I discussed braided Hopf algebras, but I neglected to give an example of one. Today I'll describe how we can find a large collection of braided Hopf algebras via a construction by Drinfeld called the Quantum Double.

Let $H$ be a **finite dimensional** Hopf algebra with invertible antipode. Then $H^* = \text{Hom}(H, k)$ is also a Hopf algebra: it's a basic result to show that the dual vector space of a coalgebra is an algebra with the product map $\Delta^*$, the transpose of the original coproduct, moreover, in the finite dimensional case, $m^*$ becomes a coproduct, making the dual of a finite dimensional algebra a coalgebra. Since $H$ is finite dimensional by assumption, and the product and coproduct maps are [co]algebra morphisms, we see we have a Hopf algebra structure on $H^*$, too. That the transpose of the antipode is an antipode and that we have the necessary co/units is a more straightfoward result. The finite dimensional requirement is needed because $H^* \otimes H^*$ is not necessarily isomorphic to $(H\otimes H)^*$ without it.

To be explicit, we have the product

$ \displaystyle \phi \psi (h) = \sum \phi(h_1) \psi(h_2)$

and the coproduct

$ \displaystyle \Delta \phi (h \otimes g) = \phi(hg)$

along with the unit $1(h) = \epsilon(h)$, counit $\epsilon(\phi) = \phi(1)$ and antipode $S(\phi)(h) = \phi(h)(Sh)$. Of course, $H^*$ and $H$ are dually paired by the evaluation map

$ \displaystyle \langle \phi, h \rangle = \text{ev}_\phi(h) = \phi(h)$

we also have the reverse dual pairing $\langle , \rangle : H\otimes H^{\text{op}*}$ for $H$ and $H^*$ with the opposite multiplication. From this, we can define the Quantum Double $D(H)$ as the vector space $H^* \otimes H$ with the product:

$ \displaystyle (\phi \otimes h) (\psi \otimes g) = \sum \left( \langle Sh_1, \psi_2 \rangle \langle h_3, \psi_3\rangle \right) \psi_2 \phi \otimes h_2 g$

the coproduct:

$ \displaystyle \Delta(\phi, h) = \sum (\phi_1 \otimes h_1) \otimes (\phi_2 \otimes h_2)$

the antipode:

$ \displaystyle S(\phi\otimes h) = (1\otimes Sh) (S^{-1}\phi \otimes 1)$

and finally with the unit $1\otimes1$ and counit $\epsilon(\phi \otimes h) = \epsilon(\phi)\epsilon(h)$.

We can also describe the quantum double as the bicrossed product of $H$ with $H^{\text{op}*}$, but we aren't focusing on that here. Rather, let me point out that $D(H)$ is a braided Hopf algebra with

$ \displaystyle \mathcal{R} = \sum_{k=1}^n (f^k \otimes 1) \otimes (1 \otimes e_k)$

Where $\langle e_1, \ldots e_n \rangle$ is a basis for $H$ and $\langle f^1,\ldots f^n \rangle$ is a dual basis.

We can also define the Quantum Double for an infinite dimensional Hopf algebra so long as we can find another dually paired Hopf algebra $H'$, which means finding the map $\langle, \rangle: H \otimes H' \rightarrow k$. But I want to describe it in a better (in my opinion) way. First, I need to quote some results:

Let $_H^H \mathcal{M}$ denote the category of **crossed $H$-modules**, that is, vector spaces $V$ that are compatibly both $H$-modules and $H$-comodules in the following sense: $\triangleright: H \otimes V \rightarrow V$ is the module action and $v \mapsto \sum h^{(1)} \otimes v^{(1)}$ is the coaction, then

$ \displaystyle \sum h_1 h^{(1)} \otimes h_2 \triangleright v^{(1)} = \sum (h_1 \triangleright v)^{(1,H)} h_2 \otimes (h_1 \triangleright v)^{(1,V)}$

for all $h \in H$ and $v \in V$ (the pair $(1,V)$, et al, is to denote which is the $H$ component and which is the $V$ component of the coaction). The morphisms in this category are linear maps that commute with both the action and the coaction. This category is braided with

$ \displaystyle \Psi_{V,W}(v \otimes w) = \sum h^{(1,v)} \triangleright w \otimes v^{(1)}.$

Here the pair $(1,v)$ denotes that this is the coaction on $V$ and not on $W$. Now, we have the following result:

TheoremLet $H$ be a finite dimensional Hopf algebra with invertible antipode. Then $_H^H \mathcal{M} = _{D(H)}\mathcal{M}$, that is, the category of crossed $H$-modules is category of modules of the quantum double of $H$.

I won't (and can't yet) prove this, but you can see an argument for it by noting that $D(H)$ contains both $H$ and $H^{\text{op}*}$ as Hopf sub-algebras; a $D(H)$ module is then an $H$ module and an $H^{\text{op}*}$ left module, but that's the same as a $H^*$ right module, which is the same as a left $H$-comodule.

Now, if $H$ is not finite dimensional, then can we find a Hopf algebra $\mathcal{H}$ such that $_H^H\mathcal{M} = _\mathcal{H}\mathcal{M}$? We again invoke those Tannakian reconstruction theorems and thus have a suitable definition for $D(H) = \mathcal{H}$ for $H$ of any dimension, and without having to specify a dually paired Hopf algebra $H'$.

The obvious question is, what does $D(H)$ look like for an infinite dimensional $H$? Perhaps I'll try answering this later.

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