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I've reading some introductory quantum group material and am trying to understand the algebra-space correspondence in the classical case. One object I'm stuck on is the adjoint coaction $$ Ad_R: a \mapsto a_{(2)} \otimes S(a_{(1)})a_{(3)} $$ of a Hopf algebra $H$, where $a \in H$ and $S$ is the co-inverse of $H$. I assume this is the dual of the standard conjugate representation of a group on itself $$ \hat{g}(h) = g^{-1}hg. $$ But shouldn't this map $$ a \mapsto a_{(1)} \otimes a_{(2)} \otimes S(a_{(3)})? $$ Or why not try $$ Ad_R: a \mapsto a_{(2)} \otimes a_{(3)}S(a_{(1)}) $$

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I suppose my problem arises from the fact that I don't really understand the precise correspondence between actions $(R,G) \to G$ ($R$ a ring, $G$ a group) and coactions $G \to G \otimes R$.

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Your formula cannot work because it gives you three tensor factors, and the coaction has to give you only two... –  Mariano Suárez-Alvarez Mar 9 '10 at 18:04
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(what you call co-inverse is called by most people 'antipode', by the way) –  Mariano Suárez-Alvarez Mar 9 '10 at 18:05
    
... I suppose I thought co-inverse was more natural ... is this not standard terminology? –  Jean Delinez Mar 9 '10 at 18:13
    
The standard terminology is 'antipode'. Your new proposal does not work because it is not coassociative: try to check coassociativity to see why... –  Mariano Suárez-Alvarez Mar 9 '10 at 18:15
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"co-inverse" is a name you might have invented if you thought of a Hopf algebra as a generalization of the coordinate ring of an affine group scheme. (This is not an endorsement of the name.) –  S. Carnahan Mar 10 '10 at 5:11

3 Answers 3

The warm-up for any Hopf algebra construction is to try it on the two Hopf algebras $\mathbb K G$ and $C(G,\mathbb K)$, the group ring and ring of functions respectively for a finite group $G$.

In the group ring $\mathbb K G$, the group conjugation manifests in the forward direction. Recall that the structure maps on $\mathbb K G$ are simply the linearizations of the structure maps on $G$, where the comultiplication $\Delta$ is the duplication map $g \mapsto g\otimes g$ for $g\in G$ and the antipode $S$ is $g \mapsto g^{-1}$. Then the conjugation is: $$ g\otimes h \mapsto S(g_{(1)}) \, h \, g_{(2)} $$ where I have used the Sweedler notation to denote the comultiplication, and the multiplication is simply concatenation. So this is a map $\mathbb K G^{\otimes 2} \to \mathbb K G$ extending $(g,h) \mapsto g^{-1}hg$. Incidentally, it's probably better to use the other conjugation $(g,h) \mapsto ghg^{-1}$, as that gives a left-action of $G$ on $G$, and you are writing the actor on the left. But I'll keep the ordering you gave in the question, for now. At the end, I'll come back to this, and it will be clear where the confusion is.

In $C(G,\mathbb K)$, everything is naturally reversed. In particular, for $G$ finite, $C(G,\mathbb K)$ has a basis given by the delta functions $\delta_g$ for $g\in G$. The multiplication is commutative and the comultiplication is anticommutative. We want the map $\delta_{g^{-1}hg} \mapsto \delta_g \times \delta_h$, or perhaps in the other order, depending on your conventions.

So let's read backwards. We first blow up $a = \delta_{g^{-1}hg}$ to $a_{(1)} \otimes a_{(2)} \otimes a_{(3)}$, where we think of $a_{(1)}$ as $\delta_{g^{-1}}$, etc. (of course, it isn't just that, but we'll pick out that term). Now we need to move the last term past the middle term, and antipode the first term: $S(a_{(1)}) \otimes a_{(3)} \otimes a_{(2)}$. Then the final multiplication makes sure that $S(a_{(1)})$ and $a_{(3)}$ are supported at the same points in $G$. So, all together, I think it's reasonable to call: $$ a \mapsto S(a_{(1)})a_{(3)} \otimes a_{(2)} $$ a "coadjoint coaction", with the caveat as above that it's probably on the wrong side.

In any case, up to left and right, this is your second proposal. And left and right is hard, for the following reason. There seems to be no consensus in the quantum groups literature for whether the dual to $A \otimes B$ is $A^* \otimes B^*$ or $B^* \otimes A^*$. Or rather, it's clear from the representation theory of Hopf algebras that it must be the latter, but many of the early (and later) texts on Hopf algebras use the former dual when working in vector spaces (defining duals to Hopf algebras, etc.).

But also, for $C(G,\mathbb K)$ it doesn't matter: $ab = ba$. So really what you must do is check that your proposal is really a coaction, because in Hopf land this will pick out the difference.

In any case, I think you'll find when you work it out that this is not a coaction, exacly because $(g,h) \mapsto g^{-1}hg$ is no an action of groups (the $g$ is on the wrong side). If you do the correctly-sided action $(g,h) \mapsto ghg^{-1}$, then in $\mathbb K G$ this is $g\otimes h \mapsto g_{(1)}hS(g_{(2)})$, and in $C(G,\mathbb K)$ it is $a \mapsto a_{(1)}S(a_{(3)}) \otimes a_{(2)}$ (or the one you initially start with, if you have the opposite left-right convention when dualizing). Then you can just check directly that this is in fact a coaction of Hopf algebras.

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The most simple explanation that I can think of is to look at the diagram figure 3, page 3 of our paper on Cohomology of the Adjoint Action of Hopf algebras. Turn the diagram upside,read it algebraically, and it becomes the book's formula.

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The coadjoint action should be defined in such way that the quantum Fourier transform becomes an isomorphism between the adjoint and co-adjoint actions.

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