Let $M$ be Poisson manifold with Poisson bracket $\{\cdot,\cdot \}.$ Let $f,g \in C^{\infty} (M).$ Let $(U,x^i)$ be a local coordinate of $M.$ Then it is claimed in my book that $$\{f,g\} \rvert_{U} = \sum\limits_{i,j} \pi_{ij} \frac {\partial f} {\partial x^i} \frac {\partial g} {\partial x^j}$$where $\pi_{ij} = \{x^i, x^j\}.$
Could anyone please explain to me as to where does the expression come from? Any help would be warmly appreciated.
Thanks for your time.
For some point $p = (p_1,\dots,p_n)$, choose coordinates so that $p = (0,0, \dots, 0)$. Now approximate $f$ and $g$ by Taylor series: $$ f(x) = f(0) + \left( \sum_i \frac{\partial f}{\partial x_i}(0) \cdot x_i \right) + F(x) $$ $$ g(x) = g(0) + \left( \sum_j \frac{\partial g}{\partial x_j}(0) \cdot x_j \right) + G(x) $$ where $F(x)$ and $G(x)$ are the higher order terms (of degree $\geq 2$).
Now use these expressions to compute $\{f,g\}$. Since $f(0)$ and $g(0)$ are constants, they can be ignored (since constants are Casimirs), and since $\frac{\partial f}{\partial x_i}(0)$ and $\frac{\partial g}{\partial x_j}(0)$ are constants, they can be factored out: $$ \begin {align*} \{f,g\}(x) &= \sum_{i,j} \frac{\partial f}{\partial x}_i(0) \frac{\partial g}{\partial x_j}(0) \{x_i, x_j\} \\ &\phantom{=} + \sum_i \frac{\partial f}{\partial x_i}(0) \{x_i, G\} + \sum_j \frac{\partial g}{\partial x_j}(0) \{F, x_j\} \\ &\phantom{=} + \{F,G\} \end {align*} $$
Notice that the first line of this equation is exactly what you want. So you need to check that the terms on the second and third line vanish when $(x_1,x_2, \dots, x_n) = (0,0,\dots,0)$. Remember that $F(x)$ and $G(x)$ are power series with all terms of degree $\geq 2$. So when you use the Leibniz rule, every term will have a factor of some $x_k$, so when $x \to 0$, all these terms will be zero.
