Is a bialgebra pairing of Hopf algebras automatically a Hopf pairing? The following question came up in the course on Quantum Groups here at UC Berkeley.  (If you care, I have been TeXing uneditted lecture notes.)
Let $X,Y$ be (infinite-dimensional) Hopf algebras over a ground field $\mathbb F$.  A linear map $\langle,\rangle : X\otimes Y \to \mathbb F$ is a bialgebra pairing if $\langle x,y_1y_2 \rangle = \langle \Delta x,y_1\otimes y_2\rangle$ and $\langle x_1x_2,y\rangle = \langle x_1\otimes x_2,\Delta y\rangle$ for all $x,x_1,x_2 \in X$ and $y,y_1,y_2 \in Y$.  (You must pick a convention of how to define the pairing $\langle,\rangle : X^{\otimes 2} \otimes Y^{\otimes 2} \to \mathbb F$.)  And we also demand that $\langle 1,- \rangle = \epsilon_Y$ and $\langle -,1\rangle = \epsilon_X$, but this might follow from the previous conditions.  (See edit.)
A bialgebra pairing is Hopf if it also respects the antipode: $\langle S(x),y \rangle = \langle x,S(y)\rangle$.  A pairing $\langle,\rangle : X\otimes Y \to \mathbb F$ is nondegenerate if each of the the induced maps $X \to Y^*$ and $Y \to X^*$ has trivial kernel.

Question: Is a (nondegenerate) bialgebra pairing of Hopf algebras necessarily Hopf?  (Does it depend on whether the pairing is nondegenerate?)

My intuition is that regardless of the nondegeneracy, the answer is "Yes": my motivation is that a bialgebra homomorphism between Hopf algebras automatically respects the antipode.  But we were unable to make this into a proof in the infinite-dimensional case.
Edit: If $\langle,\rangle: X\otimes Y \to \mathbb F$ is nondegenerate, then it is true that as soon as it satisfies $\langle x,y_1y_2 \rangle = \langle \Delta x,y_1\otimes y_2\rangle$ and $\langle x_1x_2,y\rangle = \langle x_1\otimes x_2,\Delta y\rangle$, so that the induced maps $X \to Y^*$ and $Y \to X^*$ are (possibly non-unital) algebra homomorphisms, then it also satisfies $\langle 1,- \rangle = \epsilon_Y$ and $\langle -,1\rangle = \epsilon_X$, so that the algebra homomorphism are actually unital.  But I think that this does require that the pairing be nondegenerate.  At least, I don't see how to prove it without the nondegeneracy assumption.  So probably the nondegeneracy is required for the statement about antipodes as well.
 A: Hi
I was in class when the question came up, and I remember the discussion on whether a bialgebra pairing of $A$ and $B$ always induces a bialgebra map $A\to B^\circ$. You guys tricked me back then :): it is, in fact, true. In other words, a bialgebra pairing is always (regardless of (non)degeneracy) exactly the same thing as a bialgebra map from one of the two bialgebras to the finite dual of the other. 
I'm going to view the elements of $a$ as linear maps on $B$ via the pairing. We want to show that the map from $A$ to the dual of $B$ induced by the pairing actually lands in $B^\circ$. 
This is pretty clear using the following characterization of the finite dual: $B^\circ$ is precisely the set of $f\in B^*$ for which one can find finitely many $g_i,h_i\in B^*$ such that $f(xy)=\sum_ig_i(x)h_i(y)$ (for all $x,y\in B$). It follows immediately from the bialgebra pairing conditions that any $a\in A$ satisfies this property: if, say $\Delta(a)=\sum_i a'_i\otimes a''_i$, then take $g_i=a'_i$ and $h_i=a''_i$.
A: Theo,
I think one can argue like this in the case of a non-degenerate pairing.  I didn't check everything here carefully, so don't believe it unless you confirm it yourself.
One has from the pairing an inclusion of $i:X\hookrightarrow Y^*$. One has two maps on $i(X)$, $S_X$, and $S_Y^*|_{i(X)}$.  Both of these satisfy the axioms of an antipode on $i(X)$ (which we can check by pairing with elements of $Y$), so they must agree, as desired, since being a Hopf algebra is a property, not a structure.
I'll think about the degenerate situation.  My guess is that it's not true, but I don't know.  
A: The induced map $X \to Y^\ast$ is a map of bialgebras, and therefore a map of Hopf algebras. So it takes $S(x)$ to $\langle x, S(-) \rangle$ as desired. This seems too simple, though, so something might be wrong. Is the antipode of the dual Hopf algebra not what I think it is?
