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Consider a smooth $3$-manifold $M$, a submanifold $P$ of dimension $2$ and a non-vanishing vector field $V$ transverse to $P$. Does there exist a neighborhood of $P$ diffeomorphic to $P\times (-\varepsilon, \varepsilon)$, such that $V$ is pushed forward to $\partial z$?

Here $\partial z$ is vector field corresponding to $(-\varepsilon, \varepsilon)$ direction.

If that is not the case, what would be necessary conditions for that? I am not sure if this helps but this question is motivated by trying to think about convex surfaces for Reeb vector field in contact geometry.

OSBM
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    Consider $\varphi$ the flow of $V$, and $F\colon U = \cup_{p\in P} {p}\times (-\varepsilon(p),\varepsilon(p))\to M$ defined by $F(p,t)=\varphi_t(p)$. Here, $\varepsilon(p)>0$ is some constant such that the flow of $V$ starting at $p$ is defined on $(-\varepsilon(p),\varepsilon(p))$. If $P$ is compact, you can find a uniform $\varepsilon$. If $P$ is not compact, a proper rescaling of $V$ would maybe do the trick – Didier Oct 12 '22 at 22:01
  • Would this construction work if $P$ was $1$-dimensional, e.g. $S^1$? What I am thinking about is vector field "wrapping around" $S^1$ like a mobius band. – OSBM Oct 12 '22 at 22:17
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    This construction works for any codimension $1$ submanifold with a transverse vector field. There is no such vector field transverse to $S^1$ in the mobius band – Didier Oct 12 '22 at 22:20
  • Thank you for your answer! Wouldn't this construction work for any codimension submanifold of $n$-manifold, as long as we have the vector field? If not, what would be the reason? – OSBM Oct 12 '22 at 22:43
  • If $P$ has codimension $\geqslant 2$, $P\times (-\varepsilon,\varepsilon)$ would have codimension $\geqslant 1$ in $M$ and cannot be open. This will not be a tubular neighbourhood. If you find as many vector fields, everywhere on $P$ independent, as the codimension of $P$ and which are transverse, then this construction works. You can also look at the notion of "normal bundle" – Didier Oct 13 '22 at 06:46

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