What is the natural coaction $\delta: V \to V \otimes V^* \otimes V$? Let $V$ be a vector space. A professor said that there is a natural coaction $\delta: V \to V \otimes V^* \otimes V$. What is this natural coaction? Thank you very much.
 A: There is a natural counit $K\to {\rm End}(V)$ defined by $\lambda\mapsto\lambda\cdot{\rm Id}$.
There is a natural isomorphism $V\otimes V^*\xrightarrow{\sim}{\rm End}(V)$ defined by $v\otimes w\mapsto w(\cdot)v$.
Mix these ingredients to get a map $K\otimes V\to (V\otimes V^*)\otimes V$. Check it's a coaction.
There is also a natural isomorphism $K\otimes V\xrightarrow{\sim}V$ defined by $\lambda\otimes v\mapsto \lambda v$.
A: Another way to see the coaction, which uses the natural isomorphisms $V\otimes V^*\cong\mathrm{End}(V)$ and $V^{**}\cong V$ ($V$ finite dimensional) is to dualize the (right) action of $\mathrm{End}(V^*)$ on $V^*$. That is, given the multiplication map
$$
m:V^*\otimes\mathrm{End}(V^*)\longrightarrow V^*,
$$
one has
$$
\delta=m^*:V\cong V^{**}\longrightarrow(V^*\otimes\mathrm{End}(V^*))^*\cong V\otimes V^*\otimes V
$$
A: Let $e_1, \ldots, e_n$ be a basis of $V$. The natural coaction $\delta: V \to V \otimes V^* \otimes V$ is given by the following formula
$$
e_i \mapsto \sum_j e_i \otimes e_j^* \otimes e_j. 
$$
Let $c_{ij} = e_i \otimes e_j^*$. The comultiplication on $V \otimes V^*$ is given by 
$$
\Delta(c_{ij}) = \sum_k c_{ik} \otimes c_{kj} = \sum_k e_i \otimes e_k^* \otimes e_k \otimes e_j^*.
$$
We have 
\begin{align}
(1 \otimes \delta) \delta(e_i) & = (1 \otimes \delta) (\sum_j e_i \otimes e_j^* \otimes e_j) \\
& = \sum_{j} \sum_{k} e_i \otimes e_j^* \otimes e_j \otimes e_k^* \otimes e_k \\
& = \sum_{k} \sum_{j} e_i \otimes e_k^* \otimes e_k \otimes e_j^* \otimes e_j.
\end{align}
Here $1$ is the identity map on $V \otimes V^*$.
\begin{align}
(\Delta \otimes 1) \delta(e_i) & = (\Delta \otimes 1) (\sum_j e_i \otimes e_j^* \otimes e_j) \\
& = \sum_{j} \sum_{k} e_i \otimes e_k^* \otimes e_k \otimes e_j^* \otimes e_j \\
& = \sum_{k} \sum_{j} e_i \otimes e_k^* \otimes e_k \otimes e_j^* \otimes e_j.
\end{align}
Here $1$ is the identity map on $V$.
Therefore $\delta$ is a coaction.
