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Saturday, January 19, 2013

Category Theoretic Methods and Drinfeld's Quantum Double

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:

Theorem Let 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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