# Problems with Sweedler notation HW3, question 3

I’m utterly confused by this one:

$\sum_{c} h_{(1)} S(h_{(2)}) \otimes h_{(3)} = h$

It looks like the left-hand side lives in $H \otimes H$ and the rhs lives in H.  I’m assuming this:  first, $h_{(1)} S(h_{2})$ is a multiplication operation, and since S sends H to H, this would live in H.

I’ve tried mapping the elements, and that hasn’t helped clarify this.

What am I doing wrong here?   I assume it has something to do with the relationship between $h_{(1)}$ and  $S(h_{(2)})$

1. yimp34

what I belive is,

$h_{(1)}S(h_{(2)})$ lives in $H$ right. But it is an special element from $H$ because it lives (recall the definition of the antipode) in the image of the unit map $u$. So the left hand really lives in $u(\mathbb{K})\otimes H\cong \mathbb{K}\otimes H\cong H$ so we can stablish the equality as an equivalence with respect to this congruence.

2. LaTeX corrected. I hate WordPress for not having a preview button…

I agree with Lisa: the formula $\sum_h h_{(1)} S\left(h_{(2)}\right) \otimes h_{(3)} = h$ makes no sense. The left tensorand rather clearly is considered an element living in $H$ rather than in the ground field (although it does live in the ground field as yimp34 correctly noticed), so we can’t “cancel” it.

I am not sure whether the problem was trying to say $\sum_h h_{(1)} S\left(h_{(2)}\right) \otimes h_{(3)} = 1\otimes h$ or $\sum_h h_{(1)} S\left(h_{(2)}\right) h_{(3)} = h$, but I am pretty positive it is one of these.

3. I am totally agree with Yimp34’s response here,Lisa. From the commutative diagram,we see that $\sum h_{(1)}S(h_{(2)}) \otimes h_{(3)}$ can be viewed as $\sum (I*S \otimes I) (h_{(1)}h_{(2)} \otimes h_{(3)}) = \sum (\epsilon u \otimes I) (h_{(1)}h_{(2)} \otimes h_{(3)}) = ...$. It is mention on Page 37 of the lecture note. Hope it helps. =)

4. I take back what I said above–

5. Sebastián O.

Well, in the last HW, in the “(Practice with Sweedler notation.)” I had a rough time trying to understand the same issue, so punctualizing the problem is simple, indeed the expression on the left lives in $H \otimes H$, and the right one is from $H$, but we can use the product to take out the tensor expression and symplify it. Remember that Sweedler’s notation it’s only a repesentation like, fon example, $c=\sum_{c}c_{1}\otimes\epsilon(c_{2})$ even if they don’t come from the same space.

I hope this can clarify the situation a little bit.

6. Lisa, it’s a fair objection. Sweedler’s notation is hiding the canonical identification of $H$ with $\mathbf{K} \otimes H$, which is itself identified with $u(\mathbf{K}) \otimes H \subset H \otimes H$ — precisely as yimp34 suggested.

In fact Sweedler’s book states this question as:
$\sum_{c} h_{(1)} S(h_{(2)}) \otimes h_{(3)} = h \quad ({\textrm or } \quad 1 \otimes h)$
and in the next questions it just writes $h$ for $1 \otimes h$. I probably should have done that too, sorry for the confusion.