# Tubular neighborhoods

Complete proof of the tubular neighborhood theorem for submanifolds of euclidean space. I was unable to find an elementary version in the litterature.

Olivier Binette
12-02-2016

Let me introduce some notations. A $$\mathcal{C}^l$$ submanifold of dimension $$m$$ of $$\mathbb{R}^k$$ is a subset $$M$$ that is locally $$\mathcal{C}^l$$-diffeomorphic to open balls of $$\mathbb{R}^m$$. Similarily, a $$\mathcal{C}^l$$ manifold with boundary is locally diffeomorphic to open balls of the half space $$\mathbb{H}^k = \{(x_1, \dots, x_m)\in \mathbb{R}^m | x_m \geq 0\}$$. If $$f : M \rightarrow N$$ is a differentiable map between manifolds, we denote by $$df_x: T_x M \rightarrow T_{f(x)} N$$ the differential of $$f$$ at $$x$$. Each tangent space $$T_x M$$ has an orthogonal complement $$N_x M$$ in $$\mathbb{R}^k$$; the normal bundle of $$M$$ is $$N(M) = \{(x, v) \in \mathbb{R}^{2k} | x \in M,\, v \in N_x M\}$$. In the following, we assume $$M$$ is compact.

Given $$\varepsilon \gt; 0$$, we let $$V_\varepsilon = \{(x, v) \in N(M) \,|\, |v| \le \varepsilon\}$$ and $$P_\varepsilon = \{y \in \mathbb{R}^k | d(y, M) \le \varepsilon \}$$, where $$d$$ is the euclidean distance. The set $$V_\varepsilon$$ is an $$\varepsilon$$-neighborhood of the cross-section $$M \times \{0\}$$ in $$N(M)$$, and $$P_\varepsilon$$ is a tubular neighborhood of $$M$$ in $$\mathbb{R}^k$$. We will prove the following theorem.

Tubular neighborhood theorem. Let $$M$$ be a compact submanifold of $$\mathbb{R}^k$$, without boundary. For $$\varepsilon \gt; 0$$ sufficiently small, $$V_\varepsilon$$ and $$P_\varepsilon$$ are manifolds, diffeomorphic under the map $$F : V_\varepsilon \rightarrow P_\varepsilon : (x, v) \mapsto x+v$$.

Corollary 1. If $$\varepsilon \gt; 0$$ is sufficiently small, then for each $$w \in P_\varepsilon$$ there exists an unique closest point to $$w$$ on $$M$$.

Note, however, that this corollary may not hold when $$M$$ is only a $$\mathcal{C}^1$$ manifold. We will require $$M$$ to be at least $$\mathcal{C}^2$$. The proof will make clear why this is necessary, but I also present a counter-example.

Counterexample ($$\mathcal{C}^1$$ manifolds). Let $$M$$ be the graph of $$f: [-1,1] \rightarrow \mathbb{R} : t \mapsto t^{4/3}$$ in $$\mathbb{R}^2$$. It is indeed a compact $$\mathcal{C}^1$$ manifold since $$f$$ is $$\mathcal{C}^1$$. However, the points $$w_\varepsilon = (0, \varepsilon)$$ have, for all $$\varepsilon \gt; 0$$, two closest points on $$M$$. To see this, first note that if $$M$$ had an unique closest point to $$w_\varepsilon$$, then that point would be $$(0,0)$$, by the parity of $$f$$. Now, consider the function $$g_\varepsilon :[-\varepsilon, \varepsilon] \rightarrow \mathbb{R}: t \mapsto \varepsilon - \sqrt{\varepsilon^2 - t^2 }$$, its graph being the lower half of a circle centered at $$w_\varepsilon$$ and crossing $$(0,0)$$. We find $\lim_{t \rightarrow 0} g_\varepsilon'(t)/f'(t) = \lim_{t \rightarrow 0} \frac{3}{4}\frac{t^{2/3}}{\sqrt{\varepsilon^2 - t^2}} = 0,$ meaning that the graph of $$g_\varepsilon$$ is under $$M$$ near $$(0,0)$$. This is a contradiction, as me may thus shrink the circle to find two intersection points on $$M$$ closer to $$w_\varepsilon$$ than $$(0,0)$$.

## Proof of the theorem.

In the following, $$M$$ is a compact $$\mathcal{C}^2$$ submanifold of $$\mathbb{R}^k$$ of dimension $$m$$.

Lemma 1. The normal bundle $$N(M)$$ is a $$\mathcal{C}^1$$ submanifold of $$\mathbb{R}^{2k}$$ and $$T_{(x,v)} N(M) = T_xM \times N_x M$$.

Proof. Let $$(x_0, 0) \in N(M)$$ and consider a neighborhood $$\mathcal{U}$$ of $$x_0$$ in $$\mathbb{R}^k$$. It may be chosen so that $$M\cap \mathcal{U} = \phi^{-1}(0)$$, for some $$\phi : \mathcal{U} \rightarrow \mathbb{R}^{k-m}$$ with $$d\phi_x$$ surjective. Restricting $$\mathcal{U}$$ some more, we can find a $$\mathcal{C}^2$$ diffeomorphism $$\psi : \mathbb{R}^m \rightarrow M\cap \mathcal{U}$$. Using $$\phi$$ and and $$\psi$$, we construct a $$\mathcal{C}^1$$ map $$f : \mathbb{R}^k \times \mathbb{R}^k \rightarrow \mathbb{R}^k$$ having $$0$$ as a regular value and such that $$N(M\cap \mathcal{U}) = f^{-1}(0)$$. It will follow from the preimage theorem that $$N(M \cap \mathcal{U})$$ is a $$\mathcal{C}^1$$ submanifold of dimension $$k$$. Furthermore, $$N(M \cap \mathcal{U})$$ is an open neighborhood of $$(x_0, v)$$ for all $$v \in T_{x_0}M$$ and we will have found that $$N(M)$$ is a $$\mathcal{C}^1$$ manifold.

The map $$f$$ is defined as $$f: \mathbb{R}^k \times \mathbb{R}^k \rightarrow \mathbb{R}^k$$, $$f(x, v) = \left(\phi(x), u(x, v)\right)$$, where $$u : \mathbb{R}^{2k} \rightarrow \mathbb{R}^m : (x,v) \mapsto (\langle v, d\psi_{\psi^{-1}(x)}(e_1) \rangle, \dots, \langle v, d\psi_{\psi^{-1}(x)}(e_m) \rangle)$$ and $$(e_i)$$ is a basis of $$\mathbb{R}^m$$. Because the vectors $$d\psi_{\psi^{-1}(x)}(e_i)$$ form a basis of $$T_x M$$, the zero set $$u^{-1}(0)$$ is precisely $$N_x M$$ and we find that $$f^{-1}(0) = M \cap \mathcal{U}$$. To differentiate $$f$$, we use the fact that $$\psi$$ is $$\mathcal{C}^2$$. In its matrix form,

$df_{(x,v)} = \left[\begin{array}{cc}d\phi_x& 0\\ *& \partial_2 u_{(x,v)}\end{array}\right]$

where both $$d\phi_x$$ and $$\partial_2 u_(x,v) = u(x, \cdot)$$ are surjective whenever $$x \in M$$. Thus $$df_{(x,v)}$$ is indeed surjective for all $$(x,v) \in f^{-1}(0)$$. The assertion $$T_{(x,v)} N(M) = T_xM \times N_x M$$ follows from $$T_{(x,v)} N(M) = ker(df_{(x,v)}$$.  QED.

Lemma 2. For all $$\varepsilon \gt; 0$$, $$V_\varepsilon \subset N(M)$$ is a submanifold with boundary.

Proof. Let $$f : N(M) \rightarrow \mathbb{R} : (x,v) \mapsto ||v||^2$$. For any $$(x, v) \in f^{-1}(\varepsilon^2)$$, we have $$df_{(x,v)}: T_x M \times N_x M \rightarrow \mathbb{R} : (y, u) \mapsto 2\langle u, v \rangle$$ is surjective. By the preimage theorem, we find that $$f^{-1}((-\infty, \varepsilon^2])$$ is a submanifold of $$N(M)$$ with boundary $$f^{-1}(\varepsilon^2)$$. QED.

Lemma 3. The map $$F: V_\varepsilon \rightarrow \mathbb{R}^k$$ is a local diffeomorphism onto its image $$N_\varepsilon$$.

Proof. The differential of $$F$$ is simply $$dF_{(x,v)} : T_xM \times N_xM \rightarrow \mathbb{R}^k : (a, b) \mapsto a+b$$, an isomorphism since $$\mathbb{R}^k$$ is the direct sum of $$T_xM$$ and $$N_xM$$. By the inverse function theorem, it follows that $$F$$ is a local diffeomorphism onto its image. Now, it is clear that $$F(V_\varepsilon) \subset N_\varepsilon$$. If $$w \in N_\varepsilon$$, then by compacity of $$M$$ we can find a closest point $$x \in M$$. It is straightforward to verify that $$w-x \in N_xM$$, and thus $$(x, w-x)\in V_\varepsilon$$, $$F(x, w-x) = w$$. QED.

Lemma 4 (See Spivak, 1970). If $$\varepsilon \gt; 0$$ is taken sufficiently small, then $$F : V_\varepsilon \rightarrow N_\varepsilon : (x, v) \mapsto x+v$$ is a diffeomorphism.

Proof. It suffices to show that $$F$$ is bijective, whenever $$\varepsilon \gt; 0$$ is sufficiently small. The local diffeomorphism will then be a global diffeomorphism. Note that $$F$$ is injective on $$M\times \{0\}$$. Let $$A = \{(a, b) \in N(M)^2 | a \not = b,\, F(a) = F(b)\}$$ be the set of points showcasing the non-injectivity of $$F$$. This set is disjoint from the compact set $$(M \times \{0\})^2$$. Therefore, if we can show that $$A$$ is closed, we will find $$d(A, (M \times \{0\})^2) \gt; 0$$ and taking $$\varepsilon \lt; d(A, (M \times \{0\})^2)$$ will suffice. Let $$\{(a_n, b_n)\}$$ be a sequence of points in $$A$$ converging to some $$(a,b)$$. By continuity of $$F$$, we must have $$F(a) = F(b)$$. We cannot have $$a = b$$, as this would contradict the fact that $$F$$ is a local diffeomorphism. Therefore $$a \not = b$$ and $$(a,b) \in A$$. QED.

References:

Milnor, J.W. (1965) Topology from a differentiable point of view. Spivak, M. (1970) A comprehensive introduction to differential geometry.