Completeness

Let $\vf\:M\to N$ be a Poisson map and $F$ a leaf in $M$. One could ask whether $\vf$ brings symplectic leaves of $M$ into symplectic leaves of $N$. This is easily seen not to be the case. Let us take $\vf\:\bR^2\to\bR$, $\vf(x,y)=x$ is Poisson with respect to the standard Poisson structure in $\bR^2$ and zero structure in $\bR$. But $\vf(\bR^2)$ is a union of leaves. From this example one could guess that in general $\phi(F)$ is a union of leaves. Even this turns out to be wrong, though for a subtler reason. Consider $U\subseteq \bR^{2n}$ open set and $i\:U\to \bR^{2n}$ with the standard Poisson bivector $\Pi$ on $\bR^{2n}$ and $\Pi\vert _U$ on $U$. The image of the leaf $U$ is not a whole leaf but just an open set in the leaf. Why is it so?

Consider now $\vf(F)$ and take $\vf(x)\in S$, where $S$ is a leaf through $\vf(x)$ in $N$. Take $y\in S$ and a piecewise Hamiltonian curve from $y$ to $\vf(x)$. We would like to lift this curve from $N$ to $M$. Say the first Hamiltonian piece is the flow of $X_h$. Even if $X_h$ is complete $X_{\vf^* h}$ is not necessarily complete.

Then we immediately have



Remark that also when we consider algebraic smooth Poisson varieties and alegbraic maps between them, properness, in the algebraic sense, implies completeness. This is often used when dealing with algebraic Poisson groups.