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($(H, \Delta, \epsilon)$ 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.

2 Answers 2

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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).

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    That is indeed what I mean by $H$. The question is how to get to the first notation stated. In our course, we defined counity has the second equality I wrote and the teacher just writes that it can be rewritten as the first one I put. But how? My teacher then states that from that first notation, we can get to the last relation I wrote using the fact that elements of K and of H commute (I don't see why this is true either) – 2017-01-16
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    Your teacher is right. I guess what he has in mind, is that for any vector space $V$ over a field $k$, the notations $kv$ and $vk$ are the same thing. Remember that a vector space over a field can be seen as a module over a commutative ring (here the field is the commutative ring). But generally, the notions of left and right module coincide over a commutative ring. – 2017-01-16
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    Ok. It seems that my teacher is working the other way round then me and I was wondering how he did it. I go from A to B to C and He goes from A to C to B. (A = formal definition, B = last equality I write, C = first Equality I write). But at least there seems to be no errors in my reasoning above – 2017-01-16
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    If you feel there is smt else that could be added to help just let me know. – 2017-01-16
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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

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