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Let $H$ be a Hopf algebra with a bijective antipode $S$. Does the equality $\sum\limits_{(h)} h_2 \otimes S^{-1}(h_1) = \sum\limits_{(h)} h_1 \otimes S(h_2)$ hold for any $h \in H$, where $\Delta(h)=\sum h_1 \otimes h_2$? Thank you.

1985
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  • Welcome to MSE! It is more likely that you'll receive an answer if you showed us that you've made an effort. – Vee Hua Zhi Dec 19 '18 at 09:47
  • I could prove the equality for cocommutative Hopf algebras. As in that case $\sum h_2 \otimes S^{-1}(h_1) =\sum h_1 \otimes S^{-1}(h_2)=\sum h_1 \otimes S(h_2)$. But I do not know in general. – 1985 Dec 20 '18 at 05:24

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Observe that your requirement is equivalent to the following $$\Delta= (S^2\otimes \mathrm{id})\circ\tau\circ \Delta,$$ where $\tau$ is the flip map.

For a cocommutative Hopf algebra it is known that $S^2=\mathrm{id}$, therefore in that case it is true.

In general, consider the Sweedler's Hopf algebra as an counterexample.

mikis
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