Is there a Whitney Embedding Theorem for non-smooth manifolds? - MathOverflow

Is there a Whitney Embedding Theorem for non-smooth manifolds?

2010-08-05T17:20:50Z

For smooth $n$-manifolds, we know that they can always be embedded in $\mathbb R^{2n}$ via a differentiable map. However, is there any corresponding theorem for the topological category? (i.e. Can every topological manifold embed continuously into some $\mathbb R^N$, and do we get the same bound for $N$?)

Answer by Andy Putman for Is there a Whitney Embedding Theorem for non-smooth manifolds?

2010-08-05T17:26:33Z

I'm not sure about $\mathbb{R}^{2n}$, but you can embed them in $\mathbb{R}^{2n+1}$ using dimension theory. The theorem is that every compact metric space whose covering dimension is $n$ can be embedded in $\mathbb{R}^{2n+1}$. The example of non-planar graphs (which are $1$-dimensional) shows that this is the best you can do in general.

The classic source for this is Hurewitz-Wallman's beautiful book "Dimension Theory", which I recall being pretty readable to me when I was an undergraduate, though I haven't looked at it in a while. There is also a nice discussion of this in Munkres's book on point-set topology -- when I last taught a point-set topology class, I used this as one of the capstone theorems in the course.

Answer by Andrea Ferretti for Is there a Whitney Embedding Theorem for non-smooth manifolds?

2010-08-05T17:28:23Z

I don't know if you can get the same bound, but you have the embedding in some big $\mathbb{R}^N$. The proof is the same as in the smooth case, even simpler, Let me show how it works assuming $M$ is compact, say of dimension $n$.

Cover $M$ by finitely many charts $U_1, \dots, U_k$ homeomorphic to $\mathbb{R}^n$. For every $i$ consider the map $f_i \colon M \to S^n$ which collapse the complement of $U_i$ to a point. Of course you can see $f_i$ as a map to $\mathbb{R}^{n+1}$. Then $f = (f_1, \dots, f_k)$ is the desired embedding.

Answer by José Figueroa-O'Farrill for Is there a Whitney Embedding Theorem for non-smooth manifolds?

2010-08-05T17:28:32Z

Yes. For compact manifolds, the proof can be found in Munkres Topology, a first course. In the 1975 edition which I used it's in §4.5, but not sure in newer editions.

According to Munkres, the theorem also follows without the compactness assumption but "the proof is a good deal harder." Others can surely point to the original literature.

Answer by Jack Lee for Is there a Whitney Embedding Theorem for non-smooth manifolds?

2010-08-05T17:31:28Z

In Munkres's Topology, 2nd edition, Corollary 50.8 says "Every compact $m$-manifold can be imbedded in $R^{2m+1}$." Then Exercise 6 on page 315 shows (with hints) how to extend it to noncompact manifolds. I don't know if the dimension can be lowered to 2m, though.

Answer by Robin Chapman for Is there a Whitney Embedding Theorem for non-smooth manifolds?

2010-08-05T17:36:08Z

As Andy says, each compact metric space of dimension $n$ embeds in $\mathbb{R}^{2n+1}$ (but some don't in $\mathbb{R}^{2n}$). It is the case that this extends to second countable locally compact Hausdorff spaces (including second countable Hausdorff manifolds).

Answer by Tom Goodwillie for Is there a Whitney Embedding Theorem for non-smooth manifolds?

2010-08-05T22:41:23Z

A bit off-topic, but I'd like to mention some big differences between Whitney's $2n$ theorem and his $2n+1$ theorem.

The idea of the $2n+1$ theorem is that "most" smooth maps from a compact smooth $n$-manifold to a $2n+1$-manifold are embeddings -- in particular every map is smoothly homotopic to an embedding, by an arbitrarily short homotopy. This, coupled with the relatively easy result that every continuous map between smooth manifolds is homotopic to a smooth map, implies that every map is homotopic to a smooth embedding. In particular every $n$-manifold embeds in $\mathbb R^{2n+1}$.

The $2n$ theorem has a trickier proof. Step 1, most smooth maps from an $n$-manifold to a $2n$-manifold are immersions (locally embeddings) without triple points or non-transverse double points. Step 2, there is a procedure for eliminating a transverse double point by a homotopy under certain broad hypotheses. Differences: (1) The homotopy is not short. This is not about "most maps" being embeddings. (2) Step 2 fails if $n=2$. That's because you use an embedded $2$-disk in constructing the homotopy, but the construction of an embedded $2$-disk in a $2n$-manifold won't be had for free as in the $2n+1$ theorem if $2n=4$. (3) Step 2 also requires the choice of some path in the domain and a nullhomotopy of some loop in the codomain, which means that it fails if the given map of manifolds is non-injective on $\pi_0$ or non-surjective on $\pi_1$. It's OK for embedding in $\mathbb R^{2n}$, but there are simple counterexamples in general. Also, for $2$-manifolds in $\mathbb R^4$ you cheat and use the classification of surfaces. For $2$-manifolds in $4$-manifolds there are interesting surprises: not every map $S^2\to \mathbb CP^2$ or $S^2\to S^2\times S^2$ is homotopic to an embedding.

Answer by Alfredo Hubard for Is there a Whitney Embedding Theorem for non-smooth manifolds?

2011-08-11T15:31:33Z

It Is easy to see that every simplcial complex embeds in R^{2n+1}, just take generic points on the moment curve as vertices and extend linearly.

Stallings proved that every n dimensional complex embeds upto homotopy into R^{2n}. Not sure if adding the manifold hypothesis lets you get rid of the homotopy.