Lagrangian fiber bundles and Poisson commutative subalgebras Let $M$ be a symplectic manifold and $\pi : M \to B$ a fiber bundle (fibers are manifolds). Is it true that $\pi$ is an isotropic fiber bundle i.e. all fibers are isotropic submanifolds if and only if $\pi^*(C^\infty(B))$ is Poisson commutative?
More precisely, the latter condition means that $\{\pi^*(f),\pi^*(g)\}=0$ for all $f,g \in C^\infty(B)$ and we use Poisson bracket produced by the symplectic structure.
How to formulate condition that the fiber bundle is Lagrangian in terms of commutative Poisson subalgebras? Do they correspond to maximal such subalgebras?

A counterexample is the fiber bundle $\mathbb{R}^{3}\times\mathbb{R}\rightarrow\mathbb{R}^{3}$ with projection $\pi:(x_1,x_2,x_3,z)\mapsto(x_1,x_2,x_3)$ and symplectic form $\omega\in\Omega^{2}(\mathbb{R}^{3}\times\mathbb{R})$ given by
$$
\omega=dx_1\wedge dx_2 + dx_3\wedge dz.
$$
The fibers are one-dimensional, hence isotropic. But $\pi^{*}(C^{\infty}(\mathbb{R}^{3}))$ is not Poisson-commutative, since
$\{x_1,x_2\}=\pm 1$ (depending on sign conventions).
