Given a Poisson bracket $\{\cdot,\cdot\}:C^{\infty}(M)\times C^{\infty}(M)\to C^{\infty}(M)$, it is well known that there is a correspondence between this bracket and bivector fields $\pi\in\mathfrak{X}^2(M)$ satisfying $[\pi,\pi]=0$ (this is the Schouten bracket). However, I understand that this correspondence is not "complete", because maybe there are $1-$forms $\alpha,\beta\in T^*M$ such that $\pi(\alpha,\beta)$ is well defined, but if these $1-$forms are not exact, we can't find two functions $f,g\in C^\infty(M)$ such that $\{f,g\}\equiv\pi(\alpha,\beta)$. So, there are elements in $T^*M$ (the domain of $\pi$) that do not correspond to any element in $C^\infty(M)$ (the domain of $\{\cdot,\cdot\}$). Is this correct?
Thanks in advance.
It's true that the Poisson bivector can act on "new", objects, but that does not mean that Poisson bivectors "extend" Poisson brackets in any meaningful way. $\pi:\mathfrak{X}^*M\times\mathfrak{X}^*M\to C^\infty M$ is fully determined by its action on exact forms due to $C^\infty M$-linearity, so, given a bracket, it requires no new information to specify the corresponding bivector.