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

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.


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