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Thursday, December 6, 2012

What I learned today: Dually Paired Hopf Algebras

To help keep me motivated and mathematically active, I will be blogging about "what I learned today" in my various projects. This is the second post in this "psuedo-daily" series.

I'm still trying to catch up on Quantum Groups. Any reader who is familiar with the field will be able to tell from these posts that I'm a long way away from doing research. But we're slowly getting somewhere!
Let H = kG be the group Hopf algebra for a finite group G. In particular, the product map m: H \otimes H \rightarrow H is just the group operation g \otimes h \rightarrow gh for g, h\in G, the coproduct is g \mapsto g\otimes g, the antipode is group inversion x \mapsto x^{-1}, the group idenity is the unit and g \mapsto 1 is the counit. We want to talk about its dual H^\ast = \text{Hom}(H,k).

Kassel's Quantum Groups text gives us several propositions to prove that the dual of a finite dimensional Hopf algebra is also a Hopf Algebra. Since G is a finite group, H is finite dimensional, and we'll follow Kassel in proving that H^\ast in particular is a Hopf Algebra.

Now, H^\ast is the space of linear functions from kG to k. Any such linear functional will be determined by basis elements that themselves are determined by their action on G, hence H^\ast is the space k(G) of functions on G with values in k. Clearly this space has pointwise multiplication to make it into an algebra - but we want to derive this from the dual of the coproduct map \Delta: H \rightarrow H \otimes H.

So let's talk about \Delta^\ast: (H\otimes H)^\ast \rightarrow H^\ast. This is, by definition, the map that takes each \alpha\otimes\beta \in (H\otimes H)^\ast to another linear functional \gamma = \Delta^\ast(\alpha\otimes\beta) on H such that \gamma(h) = \alpha\otimes\beta(\Delta(h)). But before we can work out how this becomes a product map, we need to work out some technicalities:

Can we say that (H\otimes H)^\ast \cong H^\ast \otimes H^\ast so that \Delta^\ast is actually a product? Yes! this is provided by a theorem in Kassel's book which says that the map \lambda: \text{Hom}(U, U') \otimes \text{Hom}(V,V') \rightarrow \text{Hom}(V \otimes U, U' \otimes V') by (f\otimes g)(v \otimes u) = f(u)\otimes g(v) is an isomorphism when one of the pairs (U, U'), (V, V'), or (U,V) consists of only finite dimensional spaces. We won't discuss the proof here, but I do want to point that the theorem does require finite dimensionality .

Back to the dual of comultiplication on H: We have that \Delta(h) = h \otimes h. So \alpha \otimes \beta (h\otimes h) = \alpha(h) \beta(h) = \gamma(h). Hence \Delta^\ast(\alpha \otimes \beta)(h) = \alpha(h)\beta(h) for \alpha, \beta \in H^\ast and h\in H. Hence the dual of comultiplication of a group Hopf algebra is just pointwise multiplication, as we expected.

Now what about the dual of multiplication? Same definition as above, m^\ast: H^\ast \rightarrow H^\ast \otimes H^\ast must be the map taking \alpha \in H^\ast to an m^\ast(\alpha)=\beta \otimes \gamma such that \beta\otimes\gamma(h_1\otimes h_2) = \alpha(m(h_1,h_2)) = \alpha(h_1 h_2). Hence we let m^\ast be the function m^\ast(\alpha)(x\otimes y) = \alpha(xy)

Continuing in this fashion, you'll see that the antipode is the map S(\alpha(x)) = \alpha(x^{-1}), the unit is the identity function, and the counit the map \alpha \mapsto \alpha(1).

Hence we have a Hopf algebra H^\ast = k(G) that is dual to H = kG. If we let \langle, \rangle: H^\ast \otimes H be the evaluation map \langle \alpha, x \rangle = \alpha(x), we can see write down the behavior of this map and see if we can come up with a more general situation to more varied sets of Hopf algebras. For instance, extending the map to tensor products pairwise, note that

\displaystyle \langle \alpha\beta, h \rangle = \langle \alpha\otimes\beta, \Delta(h) \rangle
\displaystyle \langle \Delta (\alpha), h\otimes g \rangle = \langle \alpha, hg\rangle
\displaystyle \langle S(\alpha), h \rangle = \langle \alpha, S(h) \rangle
\displaystyle \langle 1, h \rangle = 1
\displaystyle \langle \alpha, 1 \rangle = \alpha(1)
Replacing those last two conditions with the same thing expressed in units and counits, we have
\displaystyle \langle 1, h \rangle = \epsilon(h)
\displaystyle \langle \alpha, 1 \rangle = \epsilon(\alpha)

Thus now we have a definition: Let H_1, H_2 be Hopf algebras. We say that they are dually paired if there is a linear map \langle, \rangle : H_2 \otimes H_1 \rightarrow k that satisfies the above 5 conditions.

To make things explicit, k(G) and kG are dually paired by the evaluation map. Majid's book gives a more exotic example of a ``quantum group'' that is paired with itself:

Let q \in k be nonzero and let U_q(b+) be the k-algebra generated by elements g, g^{-1}, and X with the relation gX = qgX. Majid's book assures me we'll see how to find this ``quantum group'' in the wild later on, but for now he gives it a Hopf algebra structure with the coproduct \Delta X = X\otimes 1 + g\otimes X, \Delta g = g \otimes g, \epsilon X = 0, \epsilon g = 1, SX = -g^{-1} X, and Sg = g^{-1}

This Hopf algebra is dually paired with itself by the map \langle g, g \rangle = q, \langle X, X \rangle = 1, and \langle X, g \rangle = \langle g, X \rangle = 0

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