we put $p=0$ we get exactly that $\{f,g\}_{\mathbb R^d}(x)  = \{s_h^*f,s_h^*g\}_{G_h}(0,x)= 2hS_1(df,dg,x)$ which show uniqueness. Now it remains to prove that $\{f,g\}_{\mathbb R^d}(s_h(p,x)) = \{s_h^*f,s_h^*g\}_{G_h}(p,x)$ which proves as well that the induced bracket is Poisson.

Then we have to check that $$\{s_h^*f,s_h^*g\}_{G_h}(0,s_h(p,x)) = \{s_h^*f,s_h^*g\}_{G_h}(p,x).$$ An easy computation gives us that this equation is equivalent to the following $$\frac{\partial s_h^k}{\partial p^l}(0,s_h(p,x))-\frac{\partial s_h^l}{\partial p^k}(0,s_h(p,x)) =
\sum_{i=1}^d\big\{\frac{\partial s_h^k}{\partial x^i}(p,x) \frac{\partial s_h^l}{\partial p^i}(p,x)-
\frac{\partial s_h^l}{\partial x^i}(p,x) \frac{\partial s_h^k}{\partial p^i}(p,x)\big\}.$$ Differentiating the SGA equation first with respect to $p_3$ and then to $p_2$ and then putting $p_1=p$, $p_2=p_3=0$, we get $$\nabla_{p^1_k}\nabla_{p^2_l}S(0,0,s_h(p,x)) = \sum_{i=1}^d\nabla_{x^i}\nabla_{p^2_k}S(p,0,x)\nabla_{p^1_i}
\nabla_{p^2_l}S(p,0,x)-\nabla_{p^2_k}\nabla_{p^2_l}S(p,0,x).$$ Taking the difference between this equation and the same but with the indices $k$ and $l$ interchanged we finish the proof.

Basic examples {#examples}
==============

Let us see on some examples what are the generating functions and the formal symplectic groupoids. We already know what happens in the case of the trivial Poisson structure over $\mathbb R^d$. The generating function is $S_0(p_1,p_2,x) = (p_1+p_2)x$ and the associated symplectic groupoid is the cotangent bundle $T^*\mathbb R^d$ with structure maps $s(p,x) = x$, $t(p,x)=x$, $\epsilon(x) = (0,x)$ $i(p,x) = (-p,x)$. The composition is the fiberwise addition.

### Constant Poisson structure

Suppose one has 