Let $M$ be a manifold – by which I mean a second countable Hausdorff smooth manifold.

Here's an "obviously" correct definition of (embedded) submanifold:

Definition A. A subspace $S$ of $M$ is a submanifold if there is an atlas $\{ \phi_\alpha : U_\alpha \to U'_\alpha \}$ for $S$ such that each $\phi^{-1}_\alpha : U'_\alpha \to U_\alpha$ is a smooth immersion (or, equivalently, embedding) considered as a map to $M$.

But what if we relax these conditions slightly?

Definition B. A subspace $S$ of $M$ is a submanifold if there is some smooth structure on $S$ making the inclusion $S \hookrightarrow M$ into a smooth immersion.

This definition makes it clear that the image of any smooth embedding is automatically a submanifold, but it is not clear how the smooth structure of $S$ is induced from $M$, nor whether such a smooth structure is unique. A little thought shows that definitions A and B are equivalent.

Definition C. A subspace $S$ of $M$ is a submanifold if it has the following property: for all points $p$ in $S$, there exist an open neighbourhood $U$ of $p$ in $M$ and a chart $\phi : U \to U'$ such that $S \cap U$ is the inverse image of a linear subspace of $U'$.

This definition leaves no room for doubt about the existence and uniqueness of a smooth structure of $S$ and gives a clear, concrete picture of how it is induced from the smooth structure of $M$. But on the other hand, one worries that this definition is too restrictive: it seems to imply that every submanifold can be locally thickened without self-intersection, which sounds like a non-trivial condition to me.

Question. Are definitions A, B, and C equivalent? If they are, a proof-sketch or reference would be much appreciated.

  • $\begingroup$ What is $U^\prime$? $\endgroup$ – Daniel Moskovich May 7 '12 at 1:05
  • 1
    $\begingroup$ @DanielMoskovich, it has got to be an open set in $\mathbb R^n$ for the appropriate $n.$ So the "linear subspace" has got to be the intersection of that with a lower dimensional plane in $\mathbb R^n.$ $\endgroup$ – Will Jagy May 7 '12 at 1:29

For compact manifolds with boundary and $S$ a neat submanifold, or for $S$ closed, this is Theorem II.2.3 in Kosinski's Differential Manifolds. It is indeed a nontrivial fact, and it indeed implies that a neat submanifold has a collar. If $S$ is allowed to have corners, it needs also to be "regular" for the definitions to coincide- the reference is Proposition A.3 in Farber's Topology of Closed One-Forms.

  • $\begingroup$ I'm confused. From what I've been able to glean from the Google Books previews, a neat submanifold is my definition C, and the theorem in [Kosinski] talks about submersions instead of immersions... For the purposes of this question, I am thinking only of manifolds without boundary. Are all three definitions equivalent then? $\endgroup$ – Zhen Lin May 7 '12 at 21:31
  • $\begingroup$ If you don't care about boundaries, ignore (b). Kosinski's definition of submanifold is "the image of an imbedding" (immersion + topological imbedding. essentially your definition B). Theorem II.2.3 then says that this is equivalent to your definition C. Its proof explains how (page 29 paragraph 3 to the end). This is true mutatis mutandis even if S has a boundary, as long as a transversality condition is satisfied. $\endgroup$ – Daniel Moskovich May 8 '12 at 1:19

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