Compatibility of adjoint action with comultiplication in a Hopf algebra 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.
 A: 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?
