Glueing together functions defined on the leaves of a foliation

Question

Even though my question can be asked for very general types of foliations, I am interesetd only in its answer for Poisson manifolds, which are what I am currently studying.

Consider a Poisson manifold $(M, P)$ and let its symplectic leaves be $(M_a) _{a \in A}$, with $i_a : M_a \hookrightarrow M$ the usual injective Poisson immersions. Let $f_a \in C^\infty (M_a)$. Is it possible to test whether there exist a "simultaneous extension" $f \in C^\infty (M)$ such that $\left. f \right| _{M_a} := i_a ^* (f) = f_a$?

Answer 1

Suppose that the foliation ${\cal F}$ is regular, You can cover $M$ by an atlas $(U_i)_{i\in I}$ such that the restriction of ${\cal F}$ to $U_i$ is simple. This is equivalent that there exists a transversal $T_i\subset U_i$ such that $U_i=T_i\times V_i$ and the leaves of ${\cal F}_{\mid U_i}$ are $(x,V_i), x\in T_i$. Let $f^i_x$ be restriction of the function defined on the leaf containing $(x,V_i)$ to $(x,V_i)$. You may define $F_i:U_i\longrightarrow R$ by $F_i(x,y)=f^i_x(x,y)$. Suppose that $F_i$ is differentiable, you can define $F:M\rightarrow R$ by $F_{\mid U_i}=F_i$. So it is enough to check that you can extend the $f_a$ locally.