For a coassociative coalgebra $A$, we have a comultiplication map $\Delta \colon A \to A \otimes A$. An element $c \in A$ is sent to a sum of simple tensors, which can be a mess of indices, so we can use Sweedler's notation to try to write down the comultiplication more cleanly: $$ \Delta(c) = \sum_i c_{{(1)}_i} \otimes c_{{(2)}_i} = \sum_{(c)} c_{{(1)}} \otimes c_{{(2)}} $$

Sometimes even the sigma is omitted in the notation, and we simply write $\Delta(c) = c_{(1)} \otimes c_{(2)}$. Working with this sumless Sweedler notation takes some getting used to. What are some good introductory exercises to give a student practice working with Sweedler notation? I assume that there must be some canonical exercises like this one that only use the very basic properties of bialgebras or Hopf algebras.

  • 1
    $\begingroup$ There are a few in Chapter 1 of Darij Grinberg and Victor Reiner, Hopf algebras in combinatorics. See the First solution to Exercise 1.3.4 (b) and the First solution to Exercise 1.4.4 (a) and the Solution to Exercise 1.4.5 and the First solution to Exercise 1.4.26 (a) for examples of proofs using Sweedler notation (I show alternative proofs avoiding it as well). $\endgroup$ – darij grinberg Jan 23 '18 at 2:47

I like this question because it reminded me of my postgrad school days, when I was striving to get accustomed to Sweedler's notation of the comultiplication. I remembered that there were a few exercises in Sweedler's book on Hopf algebras, pages 11-12, and a couple of them in Abe's book, page 57. So I made little search in my library and here they are.

Sweedler's Book

  1. Prove that this is an equivalent expression of coassociativity: $$\sum_{(c)} \Delta(c_{(1)}) \otimes c_{(2)} = \sum_{(c)} c_{(1)} \otimes \Delta(c_{(2)})$$

  2. Similarly for generalized coassociativity.

  3. Show using this notation that $$\sum_{(c)} c_{(1)} \otimes \dotsb \otimes \epsilon(c_{(i)}) \otimes \dotsb \otimes c_{(n+1)} = \sum_{(c)} c_{(1)} \otimes \dotsb \otimes c_{(n)}\,.$$

  4. Let $T \colon C \otimes C \to C \otimes C$ be the map such that $T \colon c \otimes d \mapsto d \otimes c$. So $T$ is the twist map. Show that $T\Delta(c) = \sum_{(c)} c_{(2)} \otimes c_{(1)}$ (find the bilinear map from which $T$ arises).

  5. Verify the following identities:

    • $\displaystyle \sum_{(c)} \epsilon(c_{(2)}) \otimes \Delta(c_{(1)}) = \Delta(c)$
    • $\displaystyle \sum_{(c)} \Delta(c_{(2)}) \otimes \epsilon(c_{(1)}) = \Delta(c)$
    • $\displaystyle \sum_{(c)} c_{(1)} \otimes \epsilon(c_{(3)}) \otimes c_{(2)} = \Delta(c)$
    • $\displaystyle \sum_{(c)} c_{(1)} \otimes c_{(3)} \otimes \epsilon(c_{(2)}) = \Delta(c)$
    • $\displaystyle \sum_{(c)} \epsilon(c_{(1)}) \otimes c_{(3)} \otimes c_{(2)} = \sum_{(c)} c_{(2)} \otimes c_{(1)}$
    • $\displaystyle \sum_{(c)} \epsilon(c_{(1)}) \otimes \epsilon(c_{(3)}) \otimes c_{(2)} = c$

Notice that the first bullet of exercise 5 is identical to the one in your linked post.

Abe's Book

Prove the following equalities:

$$\Delta(c) = \sum_{(c)} \epsilon(c_{(2)}) \otimes \Delta(c_{(1)}) = \sum_{(c)} \Delta(c_{(2)}) \otimes \epsilon(c_{(1)})\,,$$

$$\Delta(c) = \sum_{(c)} c_{(1)} \otimes \epsilon(c_{(2)}) c_{(3)} = \sum_{(c)} c_{(1)} \otimes \epsilon(c_{(3)}) c_{(2)}\,,$$

$$c = \sum_{(c)} \epsilon(c_{(1)}) \epsilon(c_{(3)}) c_{(2)}\,.$$

It might also be instructive and interesting to have a look at pages 5-8 of Dascalescu's book on Hopf algebras, which contains some lemmas, properties, and examples on Sweedler's notation, coassociativity, and generalized coassociativity. Finally, you can try and search around Math.SE and MathOverflow. There are various posts which present detailed computations using Sweedler's notation (though not necessarily at an introductory level). Here are some links, but I am sure you can find more:

Good luck ;)


Prove that if a bialgebra $H$ admits a $k$-linear antipode map $S\colon H \to H$, then that map must be unique.

This is a good exercise to work out in Sweedler notation, and uses only basic axioms of a Hopf algebra $$ S(h_{(1)})h_{(2)} = \epsilon(h) = h_{(1)}S(h_{(2)}) \quad\text{and}\quad \epsilon(h_{(1)})h_{(2)} = h = h_{(1)}\epsilon(h_{(2)}) \,. $$


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