notation of counitality using Sweedler notation

Question

($(H, \Delta, \epsilon)$ a $K$-coalgebra)

I was wondering how one could justify the following notation of counitality using Sweedler notation $$ \epsilon(a_{(1)})a_{(2)} = a = a_{(1)}\epsilon(a_{(2)})$$ Normal counitality is written as follows $$ (H\otimes \epsilon) (\Delta(a)) = a \otimes 1_K$$ We develop the left side$$\begin{align} (H\otimes \epsilon) (\Delta(a)) &= (H\otimes \epsilon) \left(\sum a_i^{(1)}\otimes a_i^{(2)}\right)\\ &= \sum a_i^{(1)}\otimes \epsilon(a_i^{(2)})\\ &= \sum \epsilon(a_i^{(2)})(a_i^{(1)}\otimes 1_K)\\ &= \sum (\epsilon(a_i^{(2)})a_i^{(1)})\otimes 1_K)\end{align}$$ The last line I could understand as being written in Sweedler notation as $$ \epsilon(a_{(2)})a_{(1)} \otimes 1_K$$ by applying, the same reasoning and switching $H$ and $\epsilon$ we would find $$ 1_K\otimes\epsilon(a_{(1)})a_{(2)}$$ and so we would have $$ \epsilon(a_{(1)})a_{(2)} = a = \epsilon(a_{(2)}) a_{(1)}$$ But this is not the first equality we have stated.

Answer 1

First of all, I guess the notation $H$ in $(H\otimes \epsilon) (\Delta(a))$ simply implies the identity map $H\equiv Id:H\to H$. If this is the case, then your last relation is exactly the same as the first one, since $\epsilon:C\to k$, thus $\epsilon(a_{(2)})$ is just a scalar (an element of the field), so: $$ \epsilon(a_{(2)}) a_{(1)}=a_{(1)}\epsilon(a_{(2)})=a $$ (This property you are discussing is just the definition of the counity map i.e. part of the definition of a coalgebra).

Answer 2

As I know, we have the definition of counit in the coalgebra structure:

• $(Id_H \otimes \epsilon)\circ \Delta = Id_H\otimes 1_K$
• $(\epsilon \otimes Id_H)\circ \Delta = 1_K \otimes Id_H$

So, if we use Sweedler's notation, i.e. $\Delta(a) = \sum_i a_i^{(1)}\otimes a_i^{(2)}$, then

$$a = Id_H(a)\otimes 1_K = (Id_H\otimes \epsilon)\circ \Delta(a) = \sum_i a_i^{(1)} \otimes \epsilon(a_i^{(2)}).$$

Note that, we silently identify $H\otimes K \cong H \cong K \otimes H$ (see more here).

And remember that $\epsilon: H \rightarrow K$. Then, one can write as

$$(Id_H\otimes \epsilon)\circ \Delta(a) = \sum_i a_i^{(1)} \epsilon(a_i^{(2)})$$