# Extending function using a vector field

Let $M$ be a smooth manifold with a smooth submanifold $N$ and vector field $v$ that's never parallel to $N$. Suppose we want to extend some function $f: N \to \mathbb{R}$ to a function $F$ on all of $M$, subject to the constraint that $dF(v)>-1$ everywhere. If $\gamma:[t_1,t_2]\to M$ is an integral curve of $v$ beginning and ending on $N$, then we must have $$f(\gamma(t_2))-f(\gamma(t_1))=\int_{t_1}^{t_2} dF(\dot \gamma(t)) \, dt>\int_{t_1}^{t_2}\dot (-1)\, dt=t_1-t_2.$$ Is this the only form of obstruction to finding an extension $F$?

Note: I'm particularly interested in the case where $M$ is a closed 3-manifold with contact structure $\ker \alpha$, $N$ is a knot, and $v=R_\alpha$ is the (nonvanishing) Reeb vector field associated to $\alpha$.

• So I assume that $dF(v)>0$ is only assumed at $N$? If $M$ is compact, $dF = 0$ somewhere. – user99914 Nov 5 '15 at 7:12
• @JohnMa Tori can have no singularities :) I think that this is not the only obstruction. As far as I see, you want to construct a function that is increasing along trajectories. You won't be able to construct such function if your vector field has periodic orbits. – Evgeny Nov 5 '15 at 11:25
• @Evgeny : all functions $F : M \to \mathbb R$ has a maximum if $M$ is compact. So $dF = 0$ at that point. – user99914 Nov 5 '15 at 11:35
• @JohnMa Yes, you are right, my bad. I was thinking about vector fields singularities before functions singularities. – Evgeny Nov 5 '15 at 11:38
• @JohnMa: Ah, you're right. I asked for $dF(v)>0$ when I should have asked for $dF(v)>c$ for some fixed constant $c$. Editing the question now. – Kyle Nov 5 '15 at 12:36