Given a bivector, determine the decomposition of $\mathbb{R}^3$ into symplectic leaves

Question

Say I have a bivector, given by $f(z) \partial_x \wedge \partial_y$, where $f(z)$ is a smooth function of $z$. How do I determine the decomposition of $\mathbb{R}^3$ into symplectic leaves from this?

Answer 1

Let us denote the bivector field by $$\Pi=f(z)\partial_{x}\wedge\partial_{y}.$$ There is an associated vector bundle map $$\Pi^{\sharp}:T^{*}\mathbb{R}^{3}\rightarrow T\mathbb{R}^{3}:\alpha\mapsto\iota_{\alpha}\Pi,$$ which takes a one-form $\alpha$ and contracts $\Pi$ with it. The symplectic foliation integrates the (singular) distribution $\text{Im}(\Pi^{\sharp})$. In this case, we have \begin{align} \text{Im}(\Pi^{\sharp})&=\text{Span}\{\Pi^{\sharp}(dx),\Pi^{\sharp}(dy),\Pi^{\sharp}(dz)\}\\ &=\text{Span}\{f(z)\partial_{y},f(z)\partial_{x}\}. \end{align} Note that this gives at each point $(x,y,z)$ either a zero-dimensional or a two-dimensional subspace, depending on whether or not $z$ is a zero of $f$. It follows that the symplectic leaves are the following:

1) Each horizontal plane $z=c$ is a leaf whenever $f(c)\neq 0$.

2) If $f(c)=0$, then all points $(x,y,c)$ are symplectic leaves.