Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

$\newcommand{\RR}{\mathbb{R}}$The present question arises from some confusion on my part regarding the precise statement of the strong Whitney embedding theorem for non-compact manifolds.

The strong Whitney embedding theorem is usually stated as follows.

Theorem: If $M$ is a smooth $n$-dimensional manifold, then $M$ admits a smooth embedding into $\RR^{2n}$.

In fact, the theorem is stated in essentially this form in Whitney's original article "The self-intersections of a smooth $n$-manifold in $2n$-space". For definiteness, I will assume that all manifolds are Hausdorff, second countable, and smooth.

Question 1: Can we always take the embedding in the above theorem to be closed? If so, is there a reference for such a statement of the theorem?

It seems that Whitney's original proof produces an embedding whose image is not closed when $M$ is open. In fact, immediately after the construction, Whitney explicitly poses the following problem: "Does there exist an imbedding, for $M$ open, with no limit set?"

Having thought about the matter for a short while, I am inclined to believe that Whitney's trick (introduced in the aforementioned article by Whitney) allows the cancellation of infinitely many double points in a manner that preserves closed immersions. Is this correct? Or is my argument getting trapped in some pitfall?

My second question concerns possible dimensional restrictions in the above embedding theorem, stemming from the failure of Whitney's trick for $n=2$.

Question 2: Does every $2$-dimensional manifold embed in $\RR^4$? If so, can we also take the embedding to be closed in this case?

Here is the suggested proof in Whitney's article: "For $n=2$, we imbed the sphere, projective plane, or Klein bottle in $E^4$, and add the necessary number of handles to obtain the given manifold." I can see that this procedure should work for compact surfaces, but I am unable to carry it out in the non-compact case.

Finally, I would also be interested to hear about more recent, good references concerning Whitney's strong embedding theorem.

share|improve this question
    
Have a look at Whitney's "Geometric integration theory". It contains a proof of his embedding theorem which shows that the image of the embedding is closed. –  Liviu Nicolaescu May 24 '13 at 11:52
    
@Liviu: The only related result I could quickly find in Whitney's "Geometric integration theory" is his weak embedding theorem concerning embeddings of a $n$-dimensional manifold into $\mathbb{R}^{2n+1}$. This is much more standard than the strong embedding theorem I am asking about, concerning embeddings into $\mathbb{R}^{2n}$. Did I happen to miss this result in the book you mentioned? –  Ricardo Andrade May 24 '13 at 22:54
    
Whitney defines an embedding is a 1-1 proper map. –  Liviu Nicolaescu May 25 '13 at 11:01
    
The image of a proper embedding is a closed subset. –  Liviu Nicolaescu May 25 '13 at 11:03
1  
Also, let me make a curious note on evolution of terminology. When Whitney says that $f:X\to Y$ is "proper", he does not mean that: (#) the inverse image of a compact subspace of $Y$ by $f$ are compact (the current usual meaning of proper map). In fact, if $f$ is injective, Whitney's notion of "proper" just means that $f$ is a homeomorphism onto its image. A map $f$ verifying (#) would actually be called by Whitney a "mapping without limit set". See, for example, the discussion preceding the statement of theorem IV.1A in "Geometric integration theory". –  Ricardo Andrade May 25 '13 at 11:57

1 Answer 1

up vote 9 down vote accepted

Regarding question 1, yes you can always ensure the image is closed. You prove the strong Whitney by perturbing a generic map $M \to \mathbb R^{2m}$ to an immersion, and then doing a local double-point creation/destruction technique called the Whitney trick. So instead of using any smooth map $M \to \mathbb R^{2m}$, start with a proper map -- one where the pre-image of compact sets is compact. You can then inductively perturb the map on an exhausting collection of compact submanifolds of $M$, making the map into an immersion that is also proper.

Regarding question 2, generally speaking if a manifold is not compact the embedding problem is easier, not harder. Think of how your manifold is built via handle attachments. You can construct the embedding in $\mathbb R^4$ quite directly. Think of $\mathbb R^4$ with its standard height function $x \longmapsto |x|^2$, and assume the Morse function on $M$ is proper and takes values in $\{ x \in \mathbb R : x > 0 \}$. Then I claim you can embed $M$ in $\mathbb R^4$ so that the Morse function is the restriction of the standard Morse function. The idea is every $0$-handle corresponds to creating an split unknot component in the level-sets, etc.

edit: The level sets of the standard morse function on $\mathbb R^4$ consists of spheres of various radius. So when you pass through a critical point (as the radius increases) either you are creating an split unknot component, doing a connect-sum operation between components (or the reverse, or a self-connect-sum), or you are deleting a split unknot component. By a split unknot component, I'm referring to the situation where you have a link in the $3$-sphere. A component is split if there is an embedded 2-sphere that contains only that component, and no other components of the link. So a split unknot component means that component bounds an embedded disc that's disjoint from the other components.

Regarding your last question, the Whitney embedding theorem isn't written up in many places since all the key ideas appear in the proof of the h-cobordism theorem. So Milnor's notes are an archetypal source. But Adachi's Embeddings and Immersions in the Translations of the AMS series is one of the few places where it occurs in its original context. You can find the book on Ranicki's webpage.

share|improve this answer
    
@Ryan: Thank you very much for your answer. I did not remember to look in Adachi's book. Thank you for mentioning that reference and Milnor's notes. Finally, I am afraid I do not quite understand what you mean by "every 0-handle corresponds to creating a split unknot component in the level-sets, etc". Can you explain what you mean by a split unknot? –  Ricardo Andrade May 24 '13 at 10:45
    
@Ryan: Thank you very much for the clarification on the knot-theoretic terminology. I will think about the details of your suggestion for embedding surfaces in $\mathbb{R}^4$. –  Ricardo Andrade May 24 '13 at 11:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.