What is an immersed submanifold?

Question

An immersed submanifold is by definition the image of a smooth immersion. I know some examples but I lack general understanding of what immersed submanifolds look like. For example, can one characterize subsets of a manifold $M$ that are immersed submanifolds of a given dimension?

For example, subsets of $M^n$ that are embedded smooth $k$-dimensional submanifolds $S$ are those for which $(M, S)$ is locally diffeomorphic to $(\mathbb R^n, \mathbb R^k)$, so by looking at $S$ one can instantly tell if it is a smooth submanifold. How does one do the same for immersed submanifolds? Is the union of countably many embedded submanifolds an immersed one? Are there any immersed submanifolds that cannot be decomposed as the union of countably many embedded submanifolds?

Answer 1

Bing's house with two rooms is the image of an immersed sphere that is not in general position.

General position immersions are easy to build out of local pictures --- well sort of easy to build. Consider inside an $n$-ball

$D^n=$ $\{ (x_1,\ldots, x_n) : \sum x_j^2 \le 1 \}$ all of the $k$-dimensional sub-spaces that have $(n-k)$ of the $x_j =0$. This is the local picture for a minimal dimension multiple point. (The greater the multiplicity, the smaller the dimension of the intersection). You don't have to choose all such subspaces, but only some of them. In this way you have local pictures to patch together. Now if you know how to attach handles to spaces, you can attach handles that have immersed pieces together. One can construct Boy's surface from this point of view.

Sometimes you get stuck. For example, a figure 8 has one double point. Boy's surface has one triple point. Capping of a generic sphere eversion gives a $3$-manifold in $4$-space with one quadruple point. But if you start from the intersection $(a,b,c,d,0)$ $\cap (a,b,c,0,e)$ $\cap (a,b,0,d,e)$ $\cap (a,0,c,d,e)$ $\cap(0,b,c,d,e)$ in the $5$-ball, there is no way to close this off to get a $4$-manifold with one quintuple point. There are plenty of codimension $1$ immersions in $5$-space, but they all have an even number of quintuple points.

You should also consider equatorial spheres in a large dimensional sphere. This is the boundary of the second example I gave. You can connect these with handles to get connected immersions.

A very cool example in 3-space (beyond Boy's surface and an acme Klein bottle) is obtained by twisting a figure 8 a full rotation. A half a twist gives a Klein bottle, a full-twist gives an immersed torus whose stable framing is induced by the Lie group structure.

Codimension $0$ examples are also very important. The standard $2$-disk with two handles that represents a punctured torus is the image of an immersion into the plane.

Answer 2

I think Anton said basically the same thing, but I'll expand a bit. When I think of an immersed submanifolds, two reasonable definitions come to my mind:

1. A map $f:N \to M$ such that $N$, $M$ are both differential manifolds, $\dim M >\dim N$, and the map is locally an embedding, i.e. the derivative matrix at each point has no kernel.

2. The same as above, but with the additional requirement that the map be transverse to itself.

(In fact, for me an immersion is almost always number 2, but 1 might make more sense sometimes. In general, all books I've seen say that there is no universally agreed upon definition of immersed/embedded submanifolds.)

I do not think it makes sense to think of the submanifold as just the image of that map. In particular, the main reason to have submanifolds is to talk about tangent vectors to the submanifolds, and this makes no sense unless you have the map. (When you imagine a tangent vector to the image, what you are actually talking about is a tangent vector to $N$.)

If you accept 2 as the definition, it's an interesting question of whether you can reconstruct the map $f$ from just the image in any reasonable unique way. I think the answer should be yes for reasonable examples, but there might be a weird counterexample. If 1 is the definition, the answer is certainly "no" (just imagine a figure eight where the self-intersection is a small interval rather than just a point). In any case, I don't think you'll be able to do anything with your immersed submanifold unless you have the map.

---

My answers to the specific questions of the original poster:

1. Union of countably many submanifolds is an immersed submanifold iff you consider a disjoint union of countably many abstract manifolds a manifold. Note that for embedded submanifolds, it's always possible to construct a map from the corresponding abstract manifold to $M$.

2. This depends on whether you require the embedded submanifolds to be closed. A figure eight cannot be decomposed into a union of embedded closed differentiable manifolds. If they don't have to be closed, as Andrey said in the comments, you can cover $N$ by open sets small enough that the map is an embedding on each.

Answer 3

I looked in wikipedia definition and I do not like it. I do not see what this def is it good for and I'm sure no one really use it.

For me immersed submanifold is a slang for local embedding.

Nevertheless, the answer to your last questions: YES and NO, and both follow directly from definition...

Answer 4

I think the answer to your final question is no, and more generally: countable unions of embedded submanifolds are precisely the images of (not-necessarily-injective) immersions.

Sketch proof: A countable union of manifolds is a manifold, so a countable union of embeddings is an immersion. Conversely, by the Inverse Function Theorem, an immersion $f: M\to N$ is locally-in-$M$ an embedding; we thus obtain a "cover of the immersion by embeddings", and since manifolds are Lindelöf there's a countable subcover.

To answer your second-last question, we then need to analyse whether there are images-of-immersions that aren't immersed submanifolds (= images-of-injective-immersions).

(edit: fixed typo)