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 $\langle X, g \rangle = \langle g, X \rangle = 0$