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I'm not especially well-versed in Hopf algebra theory, so apologies in advance if the following question has a very easy answer. Given a Hopf algebra $H$, let $\Delta$ denote the comultiplication, $\sigma$ the coinverse, and $*$ the adjoint action of $H$ on itself. Then, explicitly, the adjoint action is given by $X*Y = \sum X' Y \sigma(X'')$, where $\Delta X = \sum X' \otimes X''$. We can extend the adjoint action to an action of $H$ on $H \otimes H$ in the standard way that one usually extends a left Hopf algebra action to a tensor product; I'll also call this action $*$. In the case that $H$ is the hyperalgebra (or enveloping algebra) of the unipotent radical of a Borel of a semisimple algebraic group, it appears to be the case that $\Delta(X*Y) = X*\Delta(Y)$ for all $X$ and $Y$.

My question is: Are there simple conditions on a Hopf algebra, eg cocommutativity, that imply $\Delta(X*Y) = X*\Delta(Y)$ for all $X$ and $Y$ in $H$? Or is this always true? I have a somewhat ugly ad-hoc proof for the case mentioned above, but this seems like the kind of thing that should follow easily from simple facts in Hopf algebra theory.

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By "coinverse" you mean "antipode"? –  darij grinberg Jun 17 '11 at 20:10
    
If yes, cocommutativity is indeed enough. I'll post a proof in a few minutes. –  darij grinberg Jun 17 '11 at 20:12
    
Yes, "coinverse" does mean "antipode." –  Chuck Hague Jun 18 '11 at 18:47
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Yes, this is true for any cocommutative Hopf algebra.

Let me rewrite your statement using my (or actually my professor's) brand of Sweedler notation: I write $x_{(1)}\otimes x_{(2)}$ for $\Delta\left(x\right)$, without sum sign. And I denote the antipode of the Hopf algebra $H$ by $S$. Also I will denote your action by $\rightharpoonup $ rather than by $*$ (so I will write $X\rightharpoonup Y$ for your $X*Y$) because I prefer to use $*$ for convolution. So we have $X\rightharpoonup Y = X_{(1)} Y S\left(X_{(2)}\right)$ for all $X,Y\in H$.

Now, any $X,Y\in H$ satisfy

$\Delta\left(X\rightharpoonup Y\right) = \Delta\left(X_{(1)} Y S\left(X_{(2)}\right)\right) = \Delta\left(X_{(1)}\right) \Delta\left(Y\right) \Delta\left(S\left(X_{(2)}\right)\right)$ (since $\Delta$ is an algebra homomorphism),

whereas

$X\rightharpoonup \Delta\left(Y\right) = X\rightharpoonup \left(Y_{(1)} \otimes Y_{(2)}\right) = \left(X_{(1)} \rightharpoonup Y_{(1)}\right) \otimes \left(X_{(2)} \rightharpoonup Y_{(2)}\right)$

$= \left(X_{(1)}\right)_{(1)} Y_{(1)} S\left(\left(X_{(1)}\right)_{(2)}\right) \otimes \left(X_{(2)}\right)_{(1)} Y_{(2)} S\left(\left(X_{(2)}\right)_{(2)}\right) $

$= X_{(1)} Y_{(1)} S\left(X_{(2)}\right) \otimes X_{(3)} Y_{(2)} S\left(X_{(4)}\right) $

$= X_{(1)} Y_{(1)} S\left(X_{(4)}\right) \otimes X_{(2)} Y_{(2)} S\left(X_{(3)}\right) $

(since $X_{(1)} \otimes X_{(2)} \otimes X_{(3)} \otimes X_{(4)} = X_{(1)} \otimes X_{(4)} \otimes X_{(2)} \otimes X_{(3)}$, which is because $H$ is cocommutative)

$= \underbrace{\displaystyle \left(X_{(1)} \otimes X_{(2)}\right)}_{\displaystyle =\Delta\left(X_{(1)}\right)} \underbrace{\displaystyle \left(Y_{(1)} \otimes Y_{(2)}\right)}_{\displaystyle = \Delta\left(Y\right)} \underbrace{\displaystyle \left(S\left(X_{(4)}\right) \otimes S\left(X_{(3)}\right)\right)}_{\displaystyle = \Delta\left(S_{(3)}\right)\text{ (since }S\text{ is a anti-coalgebra homomorphism)}}$

$= \Delta\left(X_{(1)}\right) \Delta\left(Y\right) \Delta\left(S\left(X_{(2)}\right)\right) = \Delta\left(X\rightharpoonup Y\right)$,

qed.

This can be cast in diagram-chasing form, but that will make it totally unreadable.

What I used are the facts that the antipode of a Hopf algebra is an anti-coalgebra homomorphism, and that if a coalgebra is cocommutative, then tensors of the form $X_{(1)} \otimes X_{(2)} \otimes ... \otimes X_{(n)}$ are symmetric. Should I prove any of these?

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For a cocommutative algebra, you can take it further away and simply write $x\otimes x$ :) –  Mariano Suárez-Alvarez Jun 17 '11 at 20:50
    
Great, thanks! The main fact I was missing is that tensors of that form are symmetric in a cocommutative Hopf algebra. Is that easy to see? –  Chuck Hague Jun 18 '11 at 18:45
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Yes. It is enough to show that they are invariant under any transposition, i. e. that $X_{(1)}\otimes X_{(2)}\otimes ...\otimes X_{(n)} = X_{(1)}\otimes X_{(2)}\otimes ...\otimes X_{(i+1)} \otimes X_{(i)} \otimes ... \otimes X_{(n)}$ (on the right hand side, we have transposed $X_{(i)}$ and $X_{(i+1)}$ while leaving the rest of the tensor invariant). But the left hand side of this is equal to $X_{(1)}\otimes X_{(2)}\otimes ...\otimes X_{(i-1)} \otimes \Delta\left(X_{(i)}\right) \otimes X_{(i+1)} \otimes X_{(i+2)} \otimes ...\otimes X_{(n-1)}$, while the right hand side ... –  darij grinberg Jun 18 '11 at 18:49
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... is equal to $X_{(1)}\otimes X_{(2)}\otimes ...\otimes X_{(i-1)} \otimes \left(\sigma\circ \Delta\right)\left(X_{(i)}\right) \otimes X_{(i+1)} \otimes X_{(i+2)} \otimes ...\otimes X_{(n-1)}$. (In fact, here we are using that $X_{(1)}\otimes X_{(2)}\otimes ...\otimes X_{(n)} = X_{(1)}\otimes X_{(2)}\otimes ...\otimes X_{(i-1)} \otimes \Delta\left(X_{(i)}\right) \otimes X_{(i+1)} \otimes X_{(i+2)} \otimes ...\otimes X_{(n-1)}$. This is a consequence of coassociativity. (Its proof depends on how exactly we define $X_{(1)}\otimes X_{(2)}\otimes ...\otimes X_{(n)}$...) –  darij grinberg Jun 18 '11 at 19:15
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