How to show that action of an algebra $H$ on a vector space is the same as the coaction of $H^*$ on $V$?

Question

Let $H$ be a Hopf algebra and $V$ a finite dimensional $H$-module. How to show that action of an algebra $H$ on a vector space is the same as the coaction of $H^*$ on $V$? Thank you very much.

Answer 1

First, you need to be careful about left/right here. Left $H$-modules would induce right comodules over the dual. Provided that the dual actually makese sense, which leads us to...

Secondly, if $H$ is infinite dimensional then $H^*$ does not make sense as a Hopf algebra. The multiplication map $m\colon H\otimes H\to H$ would formally dualize to $m^*\colon H^*\to (H\otimes H)^*$, and in infinite dimensions $H^*\otimes H^*$ is a proper subspace of $(H\otimes H)^*$, and the image of $m^*$ is almost never fully contained in this proper subspace. But this is required for $m^*$ to be a comultiplication. Whence the algebra structure generally does not dualize to a coalgebra structure. This problem disappears in finite dimensions, and the dualizing makes sense.

For the infinite dimensional case you need to use something called the Sweedler dual of $H$, and suppose that the orbit of any $v\in V$ under the left $H$-action is finite dimensional. I'll skip over the specific details for that.

Picking $v\in V$ and letting $\{v_1,...,v_n\}$ be a basis of $Hv$, then for any $h\in H$ we can write $hv = \displaystyle{\sum_i^n f_i(a)v_i}$ for some $f_i\in H^*$. One can show that these $f_i$ always lie in the Sweedler dual in the infinite dimensional case, under the finite dimensional orbit assumption

We then make $V$ into a right comodule over the (Sweedler) dual via $\rho(v) = \displaystyle{\sum_i^n v_i\otimes f_i}$.

Note this only requires $H$ to be an algebra.

On the other hand, supposing you have a right $C$-comodule $V$, for $C$ a coalgebra, with structure map $\rho(v)=v_0\otimes v_1$, we can define a left $C^*$-module (note that a coalgebra always dualizes to an algebra) structure on $V$ by $fv=\langle f,m_1\rangle m_0$ for all $f\in C^*$.

It is then easy to see that the right comodule we obtained from the left module will then dualize back to the original left module. It is in this sense that they are "the same".