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.
Best Answer
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.