Let $H$ be a Poisson Hopf algebra. Then the antipode $S : H \to H$ is a Poisson algebra antimorphism provided $H$ is commutative.
This fact is proved in Lemma 2.6 of the paper.
An identity about antipode is used: \begin{align} \{a_1, b_1\}S(a_2 b_2) + a_1 b_1 S(\{a_2, b_2\})=0. \end{align} Here $a_1 \otimes a_2 = \Delta(a)$. How to prove this identity? Thank you very much.
Let $a,b\in H$. Then by the first part of the proof of Lemma 2.6 of the paper, we have \begin{align} \epsilon(\{a,b\}) = 0. \end{align} By the axioms of Hopf algebras we have \begin{align} m\circ (1 \otimes S) \circ \Delta = \eta \circ \epsilon. \end{align} Therefore \begin{align} (m\circ (1 \otimes S) \circ \Delta)(\{a,b\}) = (\eta \circ \epsilon)(\{a,b\})=0. \end{align} By the axioms of Poisson Hopf algebras, we have \begin{align} & (m\circ (1 \otimes S))(\Delta \{a,b\}) \\ & = (m\circ (1 \otimes S))(\{\Delta(a),\Delta(b)\}) \\ & = (m\circ (1 \otimes S))(\{a_1 \otimes a_2,b_1 \otimes b_2\}) \\ & = (m\circ (1 \otimes S))(\{a_1,b_1\} \otimes a_2 b_2 + a_1 b_1 \otimes \{a_2, b_2\}) \\ & = \{a_1,b_1\} S( a_2 b_2 ) + a_1 b_1 S(\{a_2, b_2\}). \end{align} Therefore \begin{align} \{a_1,b_1\} S( a_2 b_2 ) + a_1 b_1 S(\{a_2, b_2\}) = 0. \end{align}