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)})

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6 comments

  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.

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