we get exactly $F^{-1}_p = G^{-1}_{-p}$.

Going back to the check of axioms we get that Axiom 5 is exactly equivalent to $F_p = G_{-p}$.

As for Axiom 6, if we pose $i(p,x) = (-p,\nabla_{p_1}S_h(-p,p,s(p,x))$, then $$\begin{aligned}
i(p,x)(p,x) & = & (\nabla_xS_h(-p,p,s(p,x),s(p,x))\\
            & = & (0,s(p,x))\\
            & = &\epsilon(s(p,x)),\end{aligned}$$ provided that $x = \nabla_{p_2}S_h(-p,p,\nabla_{p_2}S_h(p,0,x))$, which is guaranteed by Lemma \[formaldiffeo\].

Similarly for Axiom 7, if we put $\tilde i(p,x)= (-p,\nabla_{p_2}(p,-p,t(p,x))$ we get that $(p,x)\tilde i(p,x) = \epsilon(t(p,x))$. Now by Lemma \[relationformdiffeo\] we get $$\nabla_{p_2}S_h(p,-p,t(p,x)) = F_{-p}\circ G_p(x) = x$$ $$\nabla_{p_1}S_h(-p,p,s(p,x)) = G_{-p}\circ F_p(x) = x.$$ Thus $i(p,x) = \tilde i(p,x) = (-p,x)$.

Now using the canonical symplectic bracket on $G_h$, i.e., $$\{F,G\}_{G_h}(p,x)= \langle\nabla_x F(p,x),\nabla_p G(p,x)\rangle-\langle\nabla_p F(p,x),\nabla_x G(p,x)\rangle$$ we can consider the problem of finding a Poisson bracket on $\mathbb R^d$ such that $s_h$ is Poisson algebra homomorphism, i.e., $$s^*_h\{f,g\}_{\mathbb R^d}(p,x) = \{s_h^*f,s_h^*g\}_{G_h}(p,x).$$ The following Proposition answers this question.

There is a unique Poisson structure on $\mathbb R^d$ such that $s^*$ is a (formal) Poisson map. Moreover this Poisson structure is given by $$\{f,g\}_{\mathbb R^d}(x) = \{s_h^*f,s_h^*g\}_{G_h}(0,x)= 2hS_1(df,dg,x).$$

Suppose there exists a Poisson structure $\{,\}_{\mathbb R^d}$ such that $s^*$ is Poisson. This means that $\{f,g\}_{\mathbb R^d}(s_h(p,x)) = \{s_h^*f,s_h^*g\}_{G_h}(p,x)$. In particular if 