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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.

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3 Answers 3

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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.

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  • $\begingroup$ That's the type of argument we were trying to construct. But, come to think of it, I'm not sure what is the map $S^*_Y|_{i(X)}$. A priori, it is a map $i(X) \to Y^*$. But I don't think it automatically factors through $i(X)$? $\endgroup$
    – Theo Johnson-Freyd
    Commented Apr 8, 2010 at 2:48
  • $\begingroup$ Maybe the point is to do the following. On the space of functions $X \to Y^*$ there is the convolution product, given by $f*g = m_{Y^*} \circ (f\otimes g) \circ \Delta_X$. Then you will argue that both $i\circ S_X$ and $S^*_Y \circ i$ are two-sided inverses for $i: X \to Y^*$, and so must agree. $\endgroup$
    – Theo Johnson-Freyd
    Commented Apr 8, 2010 at 2:50
  • $\begingroup$ Continuing in the train of thought from the previous comment, it's clear that $i\circ S_X$ is a two-sided inverse for $i$ just as soon as $i$ is an algebra homomorphism, using the Hopf identities. As for $S_Y^* \circ i$, we want to check that $m_{Y^*} \circ ((S_Y^* \otimes i) \otimes i) \circ \Delta_X$ is $1_{Y^*} \circ \epsilon_X$. But $m_{Y^*} \circ ((S_Y^* \otimes i) \otimes i) \circ \Delta_X = m_{Y^*} \circ (S_Y^* \otimes \text{id}) \circ (i\otimes i) \circ \Delta_X = \bigl( (S_Y \otimes \text{id}) \circ \Delta_Y \bigr)^* \circ (i\otimes i) \circ \Delta_X$. $\endgroup$
    – Theo Johnson-Freyd
    Commented Apr 8, 2010 at 3:08
  • $\begingroup$ (continued) So if $(i\otimes i) \circ \Delta_X = m_Y^* \circ i$, we'd be done. And I think this is the other axiom to be a bialgebra pairing. (Oh, and I just noticed an error, which is that both times that I wrote "$(S_Y^* \otimes i)$" I meant "$S_Y^* \circ i$". And there's no "edit" for comments until at least the next upgrade.) $\endgroup$
    – Theo Johnson-Freyd
    Commented Apr 8, 2010 at 3:10
  • $\begingroup$ Ok, so I think even without any nondegeneracy assumptions, just that $X \to Y^*$ and $Y \to X^*$ are both unital algebra homomorphisms, we have that $i \circ S_X = S_Y^* \circ i$, which in terms of elements of $X,Y$ is $\langle S(x),y\rangle = \langle x,S(y)\rangle$. So I'm happy, and I think I'll accept your answer. But do let me know if I've made an error. $\endgroup$
    – Theo Johnson-Freyd
    Commented Apr 8, 2010 at 3:13
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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 $h_i=a''_i$.

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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?

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    $\begingroup$ If Y is not finite dimensional, one cannot define a bialgebra structure on $Y^*$ a priori. The coproduct on $Y$ defines a product on $Y^*$; however the product on $Y$ does not yield a coproduct on $Y^*$. This is because $Y^*\otimes Y^*\subset(Y\otimes Y)^*$ but not conversely. $\endgroup$
    – David Jordan
    Commented Apr 8, 2010 at 0:53
  • $\begingroup$ Right. The question is trivial for $Y$ finite-dimensional, but as DJ says, if $Y$ is infinite-dimensional, then $Y^*$ is not a bialgebra, and the map $X \to Y^*$ is only an algebra homomorphism. There is a canonical Hopf dual for any Hopf algebra, defined by $Y^\circ = \{\mu \in Y^* \text{s.t. } m^*(\mu) \in Y^* \otimes Y^*\}$, where $m^*: Y^* \to (Y \otimes Y)^*$ is dual to the multiplication map. Then $Y^\circ$ is Hopf, a result I think is due to Drinfel'd. But we were not sure whether a bialgebra pairing induces a bialg homo $X \to Y^\circ$. If it does, we're done. $\endgroup$
    – Theo Johnson-Freyd
    Commented Apr 8, 2010 at 2:55

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