# Strong Whitney embedding theorem for non-compact manifolds

$\newcommand{\RR}{\mathbb{R}}$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.

• 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
• 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
• @Liviu: I strongly believe that the passage you quote refers to the function $f^{-1}$, not to the operation of taking inverse image of subsets of the target. In fact, your quote from section IV.A.1 of Whitney's book skips/omits an important part which clarifies what Whitney means. I will now give the full quote: "It is easy to see that a one-one mapping $f$ is proper if and only if the inverse $f^{-1}$ is continuous in $f(M)$, or, if and only if $f^{-1}$ carries compact sets into compact sets." That sentence is valid only when the second $f^{-1}$ also denotes the inverse function. – Ricardo Andrade May 26 '13 at 20:13
• I thus reiterate my previous statement: an injective continuous map $f$ between two manifolds is proper in Whitney's sense if and only if $f$ is a homeomorphism onto its image. – Ricardo Andrade May 26 '13 at 20:22

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.
• @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