4
$\begingroup$

If $A$ is a Hopf algebra, $\varepsilon$ is the co-unit and $S$ is the antipode then $\varepsilon(S(a))=\varepsilon(a)$. I am not sure how to show this, however...

$\endgroup$

1 Answer 1

4
$\begingroup$

If for any $h\in H$, we use the notation $\Delta(h)=\sum h_{(1)}\otimes h_{(2)}$ for the comultiplication $\Delta:H\rightarrow H\otimes H$, then by definition the antipode satisfies: $$ \sum h_{(1)}S(h_{(2)}) = \varepsilon(h)1_{H} $$ Applying $\epsilon:H\rightarrow k$, ($k$ is the field), to both sides of the above relation, we get $$ \varepsilon\Big(\sum h_{(1)}S(h_{(2)})\Big) = \sum \varepsilon\big(h_{(1)}\big)\varepsilon\big(S(h_{(2)})\big)=\varepsilon(h)\varepsilon(1_{H})=\varepsilon(h) \Leftrightarrow \\ \\ \\ \sum \varepsilon\Big(\varepsilon(h_{(1)})S(h_{(2)})\Big)=\varepsilon(h) \Leftrightarrow \\ \\ \sum \varepsilon\bigg(S\big(\varepsilon(h_{(1)})h_{(2)}\big)\bigg)=\varepsilon(h) \Leftrightarrow \\ \\ \varepsilon\circ S\bigg(\sum \varepsilon(h_{(1)})h_{(2)}\bigg)=\varepsilon(h) \Leftrightarrow \\ \\ \varepsilon\circ S(h)=\varepsilon\big(S(h)\big)=\varepsilon(h) $$ for any $h\in H$, because $\varepsilon:H\rightarrow k$ is an algebra homomorphism and at the same time both $\varepsilon$, $S$ are $k$-linear maps, with $\varepsilon(1_H)=1\in k \ $ and: $\sum \varepsilon(h_{(1)})h_{(2)}=h$ (by the definition of the counity map $\varepsilon:H\rightarrow k$).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .