Tubular neighborhoods

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

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 > 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 > 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 > 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 > 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,


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 > 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 > 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 > 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) > 0 and taking \varepsilon < 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.


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

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