# Existence of a Tubular neighborhood of a hypersurface

Suppose $$H$$ is a co-dimension 1 embedded submanifold of $$M$$. Let $$X$$ be a vector field on $$M$$ such that $$\forall x \in H$$, $$T_xM = T_xH \oplus X_x$$. Now I want to show that there exixts an open set $$U$$ such that $$N \subset U$$, $$U$$ diffeo to $$H \times (-\epsilon, \epsilon)$$.

The proof is supposed to go as follows:

Let $$\phi(x,t): M \times (-a,a) \to M$$ be an integral flow of $$X$$, such that $$\phi(x,0) = id_M$$, $$\phi(\cdot,t)$$ is a diffeo for each $$t$$, and $$\frac{\partial}{\partial t} \phi = X$$. Then by the inverse function theorem, $$\phi$$ is a diffeo on a neighborhood of $$U \times V$$ of each $$(x,0)$$.

How does the inverse function theorem work? I understand that the construction is trying to make that the differential of the map land in TH and X separately. But there is no guarantee that the first part is in TH. Why can the local diffeo be elevated to a neighborhood containing the entire H?

• What is the derivative of $\phi$ at $(x,0)$ for any $x\in H$? Commented Dec 9, 2018 at 0:09
• Must it be decomposed into a vector tangent to $H$ and a vector in $X$? Commented Dec 9, 2018 at 0:19
• Yes, best to think of this as a matrix with respect to (some) basis built on that decomposition. Commented Dec 9, 2018 at 0:20
• I see, may I ask whether you have any idea why does a local diffeo solve the question? Commented Dec 9, 2018 at 0:21
• Aha ... The problem, as stated, is wrong unless $H$ is compact. If $H$ is non-compact, the $\epsilon$ may have to shrink as you "go to infinity" in $H$. So, see if you can do it when $H$ is compact. Commented Dec 9, 2018 at 0:22

Here's a sketch of the argument that when $$H$$ is compact, we can find $$\epsilon>0$$ so that $$\phi$$ is injective on $$X\times (-\epsilon,\epsilon)$$. Suppose not. Then for each $$n\in\Bbb N$$ we have $$x_n,x'_n\in H$$ and $$|t_n|,|t'_n|<1/n$$ so that $$\phi(x_n,t_n)=\phi(x'_n,t'_n)$$. By compactness of $$H$$, we can find convergent subsequences $$x_{n_k}\to x$$ and $$x'_{n_k}\to x'$$. Then from $$\phi(x_{n_k},t_{n_k})=\phi(x'_{n_k},t'_{n_k})$$ we infer that $$\phi(x,0) = \phi(x',0)$$, and so $$x=x'$$. But, by the inverse function theorem, $$\phi$$ is a bijection on a neighborhood of $$(x,0)$$. This contradiction completes the proof.
• When $H$ is not compact, this may be of some interest mathoverflow.net/questions/54799/…