## The tubular neighborhood theorem November 5, 2009

Posted by Akhil Mathew in differential geometry, MaBloWriMo.
Tags: , , ,

If ${M}$ is a manifold and ${N}$ a compact submanifold, then a tubular neighborhood of ${N}$ consists of an open set ${U \supset N}$ diffeomorphic to a neighborhood of the zero section in some vector bundle ${E}$ over ${N}$, by which $N$ corresponds to the zero section.

Theorem 1 Hypotheses as above, ${N}$ has a tubular neighborhood.

When ${M= \mathbb{R}^n}$, the idea is to take ${E}$ as the normal bundle of ${N}$, whose fiber at ${p}$ consists of tangent vectors in ${\mathbb{R}^n}$ perpendicular to ${T_p(N)}$. There is a map ${f: E \rightarrow \mathbb{R}^n}$ sending ${(v,p)}$ for ${v \in E_p, p \in N}$ to ${v+p \in \mathbb{R}^n}$. When restricted to the zero section, this map is just the identity on ${N}$, and the differential is the natural map

$\displaystyle E_p \oplus T_p(M) \rightarrow \mathbb{R}^n,$

which is an isomorphism, by construction. Using the inverse function theorem, we can find a bounded neighborhood ${V \subset E}$ containing the zero section, such that ${f: V \rightarrow \mathbb{R}^n}$ is locally a homeomorphism.

Now, following Spivak, I quote a lemma:

Lemma 2 If ${X' \subset X}$ is a closed subset of the compact metric space ${X}$, and ${f: X \rightarrow Y}$ is a local homeomorphism that is injective on ${X'}$, then there is a neighborhood ${U \supset X'}$ with ${f: U \rightarrow Y}$ injective.

Suppose this failed; then we can find a sequence ${U_n}$ of open sets containing ${X'}$ (e.g. the ${\frac{1}{n}}$ neighborhood) with ${\bigcap U_n = X'}$ such that ${f}$ isn’t one-to-one on ${U_n}$. Thus one picks ${x_n,y_n \in U_n}$ with ${f(x_n)=f(y_n), x_n \neq y_n}$. By taking subsequences if necessary, assume both ${x_n,y_n}$ converge to ${x,y \in X'}$. Then clearly ${f(x)=f(y)}$. I claim ${x \neq y}$; otherwise for ${n}$ very large there would be ${x_n \neq y_n }$ close to ${x}$ with ${f(x_n)=f(y_n)}$, and this contradicts the local homeomorphism condition. Now ${x \neq y}$ but ${f(x)=f(y)}$, so this contradicts the injectivity on ${X'}$.

So returning to the special case of the theorem, we can find a suitably small neighborhood ${U \subset E}$ such that the map ${f: U \rightarrow \mathbb{R}^n}$ is injective, hence an isomorphism onto the image.

Now for the general case. We can make ${M}$ into a metric space via a Riemannian metric, which can be constructed through a standard partition-of-unity argument. Similarly, we can obtain a connection on ${M}$ (not necessary for it to be compatible with the Riemannian metric), which leads to exponential maps as before.

We can restrict the bundle ${TM}$ to ${N}$ to get a vector bundle ${TM|N}$, which contains as a subbundle ${TN}$. Let ${E}$ be a complement (e.g. obtained by putting a nondegenerate inner product on ${TM|N}$) of ${TN \subset TM|N}$, i.e. such that ${E \oplus TN \simeq TM|N}$.

Now there is an open neighborhood ${U}$ of the zero section in ${E}$ and the exponential map ${U \rightarrow M}$; this is smooth by the ODE theorem, and the differential at ${0_p \in E_p}$ for ${p \in N}$ is seen to be an isomorphism as before (yesterday we computed the differential of the exponential map at ${0}$). This fact, with the lemma above, completes the proof of the tubular neighborhood theorem.

Note in particular that ${N}$ is a smooth deformation retract of the tubular neighborhood ${U}$ (via ${v \rightarrow (1-t)v}$, ${0\leq t \leq 1}$).

Here is another variant of this result:

Theorem 3 (Collar Neighborhood Theorem) Let ${N}$ be a manifold-with-boundary, with compact boundary ${\partial N}$. There is a neighborhood ${U \supset \partial N}$ isomorphic to ${\partial N \times I}$ for ${I \subset \mathbb{R}}$ an open interval.

Choose a Riemannian metric on ${N}$ (which is defined in the same way as for a regular manifold). On each point ${p}$ of the boundary ${\partial N}$, we can assign a normal ${n_p \in T_p(N)}$ perpendicular to ${T_p(\partial N)}$ that points “inward” to ${N}$. We can find a connection ${\nabla}$ on ${N}$, and using the exponential map, we can consider ${F(p,t) = \exp_p( t n_p)}$. Then ${F}$ is defined on ${\partial N \times I}$ for ${I}$ small enough. Now this is a local isomorphism at each point of ${\partial N \times 0}$, and we can apply the same type of argument as before.

1. Gary K - June 13, 2012

I don’t know if I am missing something, but, does the sum v+p in the

middle of the 2nd paragraph always

make sense? What if, say, M=R^7 , N=S^2 . Then E would have

dimension 5, right?

Gary K - June 13, 2012

Nevermind, Sorry.

2. tinyurl.com - March 1, 2013

Where did u actually obtain the points to publish ““The
tubular neighborhood theorem | Delta Epsilons” 37thvannats ?
Regards ,Tressa

3. Dion - July 20, 2013

what would be a reference for the collar neighbourhood theorem?