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.



Compatibility of adjoint action with comultiplication in a Hopf algebra
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?


