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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?

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What is the tag gt? –  Anweshi Jan 24 '10 at 22:46
    
Of course, smoothness has nothing to do with it. There are immersions of schemes as well. I believe that immersions make sense as long as we have a locally ringed space. –  Harry Gindi Jan 24 '10 at 23:31
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algori: figure eight decomposes as the union of a circle and an embedded interval. As for "disjoint union" your are formally correct because I never specified that the domain of the immersion is connected but this is a natural assumption. –  Igor Belegradek Jan 25 '10 at 0:30
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Are there any immersed submanifolds that cannot be decomposed as the union of countably many embedded submanifolds? Can't you just cover the manifold that you are immersing by a countable number of balls so that the restriction of the immersion to each ball is an embedding? –  Andrey Gogolev Jan 25 '10 at 0:52
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@Harry: you can define closed immersions for any locally ringed space, but if your locally ringed space is a smooth manifold, then the two notions of immersion do not coincide. The lrs version yields a closed embedding on the level of topological spaces. –  Pete L. Clark Jan 25 '10 at 2:33

4 Answers 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)

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(The content of this answer is the same as that of Andrey Gogolev's simultaneous comment.) –  macbeth Jan 25 '10 at 0:54
    
Thanks, HRM and Andrey, this is a good point. –  Igor Belegradek Jan 25 '10 at 2:14
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When I looked again at Igor's questions, I realised that what he calls an "immersed submanifold" is what I call an "immersion". So the above is actually complete answer: "yes"; "no". The issue raised by my misreading still remains, though: are there images-of-immersions ("immersed submanifolds", in Igor/Wikipedia's terminology) which aren't images-of-injective-immersions? The spectre of a union of two crazy torus-maps, say $\{(t, \sqrt{2}t)\}$ and $\{(t, \sqrt{3}t)\}$, makes me suspect the answer is yes. –  macbeth Jan 25 '10 at 2:24

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.

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Regarding your last paragraph: an immersion of what? I imagine of something more interesting than itself :) –  Mariano Suárez-Alvarez Jan 25 '10 at 1:53
    
I meant the drawing of the disk with two handles that is usually represented in the plane. Early work of George Francis and others studied immersed surfaces in the plane that had given immersed curves as boundaries. The example that I indicated is used in the Torus trick (Kirby Siebenmann), but don't ask me how, I don't really remember right now. –  Scott Carter Jan 25 '10 at 15:11

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...

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The wikipedia definition is the one I think of when I hear "immersion," and it's the one I really use. Allowing anything that's a local embedding allows too many maps. It seems to me that it's more common to stay in the smooth category. –  Richard Kent Jan 25 '10 at 1:16
    
Oh, I guess you mean "smooth local embedding." I don't get your objection. They're the same thing. –  Richard Kent Jan 25 '10 at 1:21
    
I think Anton does not like the term "immersed submanifold" and in general does not like the kind of objects as in the bottom picture of wikipedia page. –  Igor Belegradek Jan 25 '10 at 2:06
    
I myself probably wouldn't say "immersed submanifold" either. I don't object to images of immersions, but I don't think of them as submanifolds. –  Richard Kent Jan 25 '10 at 2:14
    
Searching in mathscinet on "immersed submanifold" gives 206 articles, so the term is not completely obscure. –  Igor Belegradek Jan 25 '10 at 2:21

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 taling about is a tangent vecotr 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 submanifolds 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.

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