What I learned today: Weird differences between the action of a Hopf algebra and its dual

This is another post in my ``psuedo-daily'' series ``What I learned Today''

Let $H$ be a Hopf algebra. Today we're going to talk about its actions (and its coactions, time permitting). A Hopf algebra is just an algebra, which is just a ring, and ring's have actions (equiv. modules), and we're comfortable with that already. For this to make sense on a Hopf algebra, we merely require that the action agrees with all the necessary units, products, coproducts, et cetera.

More precisely, we say that $H$ acts on an algebra $A$ (or that $A$ is an $H$-module algebra) if we have a linear map $\triangleright : H \otimes A \rightarrow A$, written $h \triangleright a$ such that

$\displaystyle h \triangleright (ab) = m(\Delta(h) \triangleright (a \otimes b))$

with $\triangleright$ extended to tensor products and $m$ the product of $A$, and

$\displaystyle h\triangleright 1_A = \epsilon(h) 1_A$

Some weirdness emerges here, for instance, if $H=kG$, the group hopf algebra of a finite group $G$ (the vector space having elements of $G$ as a basis, the product being the group operation, and the coproduct defined by $\Delta(g) = g \otimes g$, antipode, units, et cetera are the obvious choice), then a hopf action collapses to a typical group action:

$\displaystyle g\triangleright(ab) = (g\triangleright a) (g\triangleright b)$

But if we forget the Hopf algebra structure for a second, we know that $kG$ is exactly the same as $k(G)$, that is, as functions defined on $G$ with values in $k$, as any ``vector'' in $kG$ defines a function and vice versa. Hence, as vector spaces $kG$ and $k(G)$ are naturally isomorphic. This is not the case with the Hopf algebra structure, however (though we have seen how the two are dually paired already), as the coproduct on $k(G)$ is $\Delta(f)(x,y) = f(xy)$, where we identify $k(G \times G)$ with $k(G) \otimes k(G)$ (this is exactly what a tensor product is for, actually.) In this case, the hopf algebra action on an algebra $A$ is the same as a $G$-grading on $A$. Let me explain this (with help from a kind person on math.stackexchange):

Say $A$ is $G$-graded and for each $a \in A$ let $|a|$ denote the element in $G$ such that $a \in A_{|a|}$. We can get a $k(G)$ action from this by $f \triangleright a = f(|a|) a$. To go the other way, start from a Hopf algebra action $\triangleright: k(G) \otimes A \rightarrow A$. Then for each $g \in G$, let $A_g$ be the set of all elements of $A$ on which every $f\in k(G)$ acts via scalar multiplication by $f(g)$. One then has to prove that these sets are nonempty, that they are subspaces, that $A$ decomposes into direct sums of them, and that they obey the axioms of a grading. I've been assured by various sources that it can be done!