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 submanifold of dimension of is a subset that is locally -diffeomorphic to open balls of . Similarily, a manifold with boundary is locally diffeomorphic to open balls of the half space .
If is a differentiable map between manifolds, we denote by the differential of at . Each tangent space has an orthogonal complement in ; the normal bundle of is . In the following, we assume is compact.
Given , we let and , where is the euclidean distance. The set is an -neighborhood of the cross-section in , and is a tubular neighborhood of in . We will prove the following theorem.
Tubular neighborhood theorem. Let be a compact submanifold of , without boundary. For sufficiently small, and are manifolds, diffeomorphic under the map .
Corollary 1. If is sufficiently small, then for each there exists an unique closest point to on .
Note, however, that this corollary may not hold when is only a manifold. We will require to be at least . The proof will make clear why this is necessary, but I also present a counter-example.
Counterexample ( manifolds). Let be the graph of in . It is indeed a compact manifold since is . However, the points have, for all , two closest points on . To see this, first note that if had an unique closest point to , then that point would be , by the parity of . Now, consider the function , its graph being the lower half of a circle centered at and crossing . We find , meaning that the graph of is under near . This is a contradiction, as me may thus shrink the circle to find two intersection points on closer to than .
Proof of the theorem.
In the following, is a compact submanifold of of dimension .
Lemma 1. The normal bundle is a submanifold of and .
Proof. Let and consider a neighborhood of in . It may be chosen so that , for some with surjective. Restricting some more, we can find a diffeomorphism . Using and and , we construct a map having as a regular value and such that . It will follow from the preimage theorem that is a submanifold of dimension . Furthermore, is an open neighborhood of for all and we will have found that is a manifold.
The map is defined as , , where and is a basis of . Because the vectors form a basis of , the zero set is precisely and we find that . To differentiate , we use the fact that is . In its matrix form,
where both and are surjective whenever . Thus is indeed surjective for all . The assertion follows from . QED.
Lemma 2. For all , is a submanifold with boundary.
Proof. Let . For any , we have is surjective. By the preimage theorem, we find that is a submanifold of with boundary . QED.
Lemma 3. The map is a local diffeomorphism onto its image .
Proof. The differential of is simply , an isomorphism since is the direct sum of and . By the inverse function theorem, it follows that is a local diffeomorphism onto its image.
Now, it is clear that . If , then by compacity of we can find a closest point . It is straightforward to verify that , and thus , . QED.
Lemma 4 (See Spivak, 1970). If is taken sufficiently small, then is a diffeomorphism.
Proof. It suffices to show that is bijective, whenever is sufficiently small. The local diffeomorphism will then be a global diffeomorphism. Note that is injective on . Let be the set of points showcasing the non-injectivity of . This set is disjoint from the compact set . Therefore, if we can show that is closed, we will find and taking will suffice.
Let be a sequence of points in converging to some . By continuity of , we must have . We cannot have , as this would contradict the fact that is a local diffeomorphism. Therefore and . QED.
Milnor, J.W. (1965) Topology from a differentiable point of view.
Spivak, M. (1970) A comprehensive introduction to differential geometry.