smooth structure from submanifolds


The purpose of this page is to investigate the question:

How much of a smooth manifold’s structure can be determined by knowing its submanifolds?

This can be interpreted in several ways mathematically. We start with two smooth manifolds, MM and NN, and a bijection Φ:MN\Phi \colon M \to N. We want to learn something about what Φ\Phi must or can be by testing it on submanifolds. Ideally, we would like a condition equivalent to Φ\Phi being a diffeomorphism. Some conditions to consider are the following.

  1. SMS \subseteq M is a submanifold if and only if Φ(S)N\Phi(S) \subseteq N is a submanifold.
  2. SMS \subseteq M is a submanifold of dimension mm if and only if Φ(S)N\Phi(S) \subseteq N is a submanifold of dimension mm.
  3. For some fixed mm and for all kmk \le m, SMS \subseteq M is a submanifold of dimension kk if and only if Φ(S)N\Phi(S) \subseteq N is a submanifold of dimension kk.
  4. For some fixed mm, SMS \subseteq M is a submanifold of dimension mm if and only if Φ(S)N\Phi(S) \subseteq N is a submanifold of dimension mm.


One reason that I am interested in this is that it is a variation of the idea of generalised smooth spaces. In a generalised smooth space, one defines “smoothness” to be “that which is detectable when tested on known spaces”. The point is that the known spaces are fixed and independent of the object being tested. Here, the known spaces can (and, indeed, do) vary with the object being tested. So each object has its own family of test objects. Closely related to this is the question of automorphisms of a category. If we have a category with test objects inside it then applying an automorphism will change the test objects so to study this situation we have to find some other way of characterising the test objects. In comparative smootheology?, I showed that there were no (essentially) non-trivial automorphisms of the category of Frölicher spaces by showing that one could identify the real line by its endomorphism monoid.

Preliminary Thoughts

For \mathbb{R}, it is the case that preserving submanifolds is equivalent to being a homeomorphism so long as we are considering submanifolds of dimension 11. By cardinality, we can separate submanifolds of dimension 11 and 00 and so from any of the statements we see that Φ\Phi takes submanifolds of dimension 11 to submanifolds of dimension 11. But submanifolds of \mathbb{R} of dimension 11 are precisely the open sets, whence Φ\Phi is a homeomorphism.

This extends more generally. If Φ: n m\Phi \colon \mathbb{R}^n \to \mathbb{R}^m is a bijection with the property that Φ(S)\Phi(S) is a submanifold of m\mathbb{R}^m if and only if SS is a submanifold of n\mathbb{R}^n then Φ\Phi is a homeomorphism. The proof is slightly different depending on whether we allow the dimension of a submanifold to be locally constant? or insist that it be constant?.

We start with an open set U nU \subseteq \mathbb{R}^n. As this is a submanifold, Φ(U)\Phi(U) is a submanifold of m\mathbb{R}^m. For cardinality reasons, Φ(U)\Phi(U) cannot be zero dimensional everywhere. Indeed, it can be zero dimensional at at most a countable number of points. Choose a point pUp \in U such that dim Φ(p)Φ(U)0\dim_{\Phi(p)} \Phi(U) \ne 0.

We choose an open set VV with pVV¯Up \in V \subseteq \overline{V} \subseteq U. By assumption, Φ(V)\Phi(V) is a submanifold of m\mathbb{R}^m and Φ(p)Φ(V)\Phi(p) \in \Phi(V). Suppose that dim Φ(p)Φ(U)m\dim_{\Phi(p)} \Phi(U) \ne m. Then there is a connected submanifold TT of m\mathbb{R}^m of dimension at least 11 which meets Φ(U)\Phi(U) transversally at Φ(p)\Phi(p). In particular TΦ(U)={Φ(p)}T \cap \Phi(U) = \{\Phi(p)\}. Applying Φ 1\Phi^{-1}, we find that Φ 1(T)\Phi^{-1}(T) is a submanifold of n\mathbb{R}^n which meets UU only at pp. Thus pp must be an isolated point of Φ 1(T)\Phi^{-1}(T). Since the closure of VV is contained within UU, the union VΦ 1(T)V \cup \Phi^{-1}(T) is therefore a submanifold of n\mathbb{R}^n. Applying Φ\Phi, we see that Φ(V)T\Phi(V) \cup T is a submanifold of m\mathbb{R}^m. But Φ(V)\Phi(V) and TT both have non-zero dimension at pp and meet transversally there, so their union cannot be a submanifold of m\mathbb{R}^m.

Thus at those points where dimΦ(U)0\dim \Phi(U) \ne 0, it must be of dimension mm. To conclude the proof, we just need to show that there are no points where dimΦ(U)=0\dim \Phi(U) = 0. Suppose that there is such a point, say pp. Then Φ(p)\Phi(p) is an isolated point from Φ(U)\Phi(U). Choose an open set W mW \subseteq \mathbb{R}^m such that Φ(p)W\Phi(p) \in \partial W and WΦ(U)=W \cap \Phi(U) = \emptyset. Then W{Φ(p)}W \cup \{\Phi(p)\} is not a submanifold of m\mathbb{R}^m, but Φ(W)U=\Phi(W) \cap U = \emptyset so Φ(W){p}\Phi(W) \cap \{p\} is a submanifold of n\mathbb{R}^n. Hence there cannot be any such points.

In conclusion, Φ(U)\Phi(U) is a submanifold of m\mathbb{R}^m of dimension mm, whence an open subset. As the same holds for Φ 1\Phi^{-1}, we conclude that Φ\Phi is a homeomorphism.

Initial Motivation

The initial motivation for this page came from this question on

Created on April 13, 2011 11:40:33 by Andrew Stacey (