Antipode of a Hopf algebra being an antihomomorphism: unable to follow the proof

Question

A PhD thesis contains the following proof that antipode of a Hopf algebra is algebra antihomomorphism (page 22): Here $\nu = \eta \circ \varepsilon$, where $\eta$ is the unit map and $\varepsilon$ is the counit map, 2.1.4 refers to $$\nu(h)=\sum h_{1}S(h_2)=\sum S(h_1) h_2 \ \ \ \ (2.1.4)$$ by definition of the antipode as the convolutional inverse of $\text{id}_{H}$, Sweedler notation for comultiplication $\Delta$ $$\Delta h = \sum h_{1}\otimes h_{2}$$ is used, 2.1.5 refers to $$h = (\text{id} * \nu) h = \sum h_{1}\nu(h_{2}) \ \ \ \ (2.1.5)$$ which follows from the fact that $\nu$ is the identity in the convolution algebra of $\text{End}(H)$, 2.1.6 refers to $$S(h)=\sum S(h_{1})\nu(h_{2}) \ \ \ \ (2.1.6)$$ and 2.1.7 is $$\nu(hl)=\nu(h)\nu(l) \ \ \ \ (2.1.7)$$ The notation $v(h)$ is not used in the text above, at least I wasn't able to find any definition for it. I assumed it is a typo, and $\nu(h)=v(h)$. Even so, I wasn't able to follow through the second line. The author seems to denote repeated summation through a single sum sign, and assume that $$\nu(h_2)=\sum h_{2}S(h_3)$$ but this is not true, the correct way to write this down should be $$\sum (h_{2})_{1} S((h_2)_2)$$ and if you redefine $h_{2}$ as $(h_{2})_{1}$, you can't use it later as a synonym for the old $h_{2}$. Also it is not clear how $S((hl)_{1})$ becomes $S(h_{1}l_{1})$, and how the second summation is dealt with.

Answer 1

Here are the few tricks of the trade:

By coassociativity we have $$(id\otimes\Delta)(\Delta(h)) = (\Delta\otimes id)(\Delta(h))=:\sum h_1\otimes h_2\otimes h_3\qquad(*)$$ (more generally: if you use $\Delta$ several (say $k-1$) times, you will always get the same result, which is denoted by $\sum h_1\otimes h_2\otimes\dots\otimes h_k$). We can rewrite $(*)$, using $\Delta(h) = \sum h_1\otimes h_2$, as $$\sum \Delta(h_1)\otimes h_2 = \sum h_1\otimes \Delta(h_2) = \sum h_1\otimes h_2\otimes h_3.$$ This hopefully settles your first problem.

As for your 2nd problem, why $(hl)_1=h_1l_1$ - I would say that the notation in the proof got a bit unclear, but it refers to the compatibility of the product and of the coproduct, which is in Sweedler's notation $$\sum(hl)_1\otimes(hl)_2 = \sum h_1\,l_1\otimes h_2\,l_2$$