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 $N$?)
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19$\begingroup$ Small point, but when $n=0$ you need the manifold to be connected for the above statement to be true. $\endgroup$– Ryan BudneyCommented Aug 5, 2010 at 17:45
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4$\begingroup$ In Munkres's Topology, 2nd edition, pg 316 exercise 11: A space $X$ can be imbeded as a closed subspace of $\mathbb{R}^N$ for some $N$ if and only if $X$ is locally compact and Hausdorff with a countable basis, and has finite topological dimension. So any finite dimensional topological manifold (i.e. finite dimensional, Hausdorff, secound countable, locally Euclidean space) can embed into some $\mathbb{R}^N$. (Note that locally Euclidean space is locally compact, see tag [topological manifold] in nLab) $\endgroup$– Z WuCommented Mar 22, 2021 at 15:58
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2$\begingroup$ The case $n=4$ follows from: (a) The embedding theorem holds for compact topological manifolds; (b) every open 4-manifold is smoothable (Quinn). $\endgroup$– Moishe KohanCommented Apr 17, 2021 at 3:05
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$\begingroup$ @MoisheKohan If you post it as an answer (with slight details) I'll give it the bounty! Just @ me $\endgroup$– John SamplesCommented Apr 21, 2021 at 21:31
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$\begingroup$ Or at least, I'll try! Might not let me. $\endgroup$– John SamplesCommented Apr 21, 2021 at 21:39
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8 Answers
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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.
answered Aug 5, 2010 at 17:26
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5$\begingroup$ To get the dimension down to $\mathbb R^{2n}$ you'd have to follow Whitney's proof in the tame topological category. My understanding is transversality was worked out in that setting by Kirby, Siebenmann, and perhaps others. But I don't know enough of the details to know whether or not the machine can be pushed through to the extent that you can prove Whitney's embedding theorem in the tame topological category. $\endgroup$ Commented Aug 5, 2010 at 17:53
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10$\begingroup$ There may be an indirect way to get the embedding in $\Bbb R^{2n}$: in one of my papers, I showed that there is a Poincare embedding in $S^{2n}$ when $M^n$ is a connected finite Poincare duality space. My understanding from talking in the past to Andrew Ranicki is that the Browder-Casson-Sullivan-Wall theorem holds in the topological category (cf. Wall's book, Chapter 11). If that is indeed the case, then you can use my work coupled with the TOP case of BCSW to get the embedding in the topological category. $\endgroup$ Commented Jan 24, 2011 at 3:13
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10$\begingroup$ The earliest references for the embedding in $\Bbb R^{2n}$ that I'm aware of are Bryant-Mio and Johnston's 1999 papers in Topology, available at Ranicki's website (maths.ed.ac.uk/~aar/homology). I guess their approaches are roughly along the lines of John Klein's comment. A more elementary proof (more in the spirit of Ryan Budney's suggestion with the Whitney trick, but avoiding any use of Kirby-Siebenmann or topological transversality) is here: front.math.ucdavis.edu/0612.5085 $\endgroup$ Commented Aug 27, 2011 at 15:36
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2$\begingroup$ @SergeyMelikhov Your second link seems to be broken. Do you happen to have an updated reference? (There's an ongoing discussion about this at Math Stack Exchange.) $\endgroup$– Jim BelkCommented Mar 13, 2021 at 11:08
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3$\begingroup$ @JimBelk It may be arxiv.org/pdf/math/0612085.pdf (note the similarity of the identifiers) $\endgroup$ Commented Apr 16, 2021 at 10:58
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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.
answered Aug 5, 2010 at 22:41
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14$\begingroup$ A note for readers who have not seen this stuff before. The "not short" homotopy in Step 2 is obtained by what has become known as the Whitney trick. This trick is the key to proving the s-cobordism theorem, and thus lies at the foundation of an enormous amount of high-dimensional manifold topology. For a beautiful account of it (with pictures!), I suggest looking at the first chapter of Scorpan's book "The Wild World of 4-Manifolds". $\endgroup$ Commented Aug 6, 2010 at 6:45
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5$\begingroup$ It's also a key ingredient in surgery theory (in the sense of Kervaire-Milnor, Browder, Wall ...), which is just as crucial as the $s$-cobordism theorem to our understanding of high-dimensional manifolds. $\endgroup$ Commented Aug 6, 2010 at 16:12
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$\begingroup$ I think we mean the same thing -- I was thinking of the s-corbordism theorem as the foundation of surgery theory, but the Whitney trick is used elsewhere in surgery theory as well, so maybe I should have just said that... $\endgroup$ Commented Aug 6, 2010 at 19:02
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1$\begingroup$ @Nikita: Maps $S^2 \to \mathbb CP^2$ not homotopic to an embedding, is that what you mean? Every map $S^2 \to \mathbb CP^2$ induces a map on $\pi_2$. If this induced map is injective but not an isomorphism I believe your map can't be homotopic to an embedding. I think the argument roughly goes like this: the boundary of a tubular neighbourhood of such an embedding would have to be a lens space, the type that has torsion in $H_1$. Then apply Poincare duality to the complement to derive a contradiction. $\endgroup$ Commented Aug 11, 2011 at 21:29
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2$\begingroup$ @Ryan: No, a complex conic in $\mathbb CP^2$ is an embedded $S^2$. $\endgroup$ Commented Oct 12, 2011 at 16:27
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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.
answered Aug 5, 2010 at 17:28
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$\begingroup$ Why is $f_i$ continuous? I presume your definition of $f_i$ is $h \circ g_i$ where: $g_i : M \rightarrow \mathbb{R}^n \cup \{\infty\}$, the latter codomain being the one-point compactification of $\mathbb{R}^n$; $g_i = f_i$ on $U_i$ and $g_i = \infty$ on $M \setminus U_i$; and $h: \mathbb{R}^n \cup \{\infty\} \rightarrow S^n$ is the usual homeomorphism. So my question is why is $g_i$ continuous? It seems to me some modfication is needed to ensure that, possibly involving a "shrinking" of the original cover. $\endgroup$– murrayCommented Aug 4, 2023 at 14:53
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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.
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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.
answered Aug 5, 2010 at 17:28
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3$\begingroup$ Whitney theorem for non-compact manifolds: math.toronto.edu/mgualt/MAT1300/week7.pdf $\endgroup$ Commented Sep 18, 2010 at 7:33
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3$\begingroup$ But these notes appear to consider smooth case, no? $\endgroup$ Commented May 31, 2012 at 19:48
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My contribution here is very limited:
Every topological 4-manifold admits an embedding in $R^8$.
Proof. It suffices to consider the case of connected manifolds. If a connected 4-manifold is noncompact then it admits a smooth structure (equivalently, a triangulation). A proof can be found in
Quinn, Frank, Ends of maps. III: Dimensions 4 and 5, J. Differ. Geom. 17, 503-521 (1982). ZBL0533.57009.
Hence, such manifold admits an embedding in $R^8$ by Whitney's theorem.
It is a general result that every compact $n$-dimensional manifold admits an embedding in $R^{2n}$. This observation is due to Sergey Melikhov (in comments to Andy Putman's answer). All the available references prove a stronger result, namely, an embedding theorem for compact generalized manifolds:
Bryant, J. L.; Mio, W., Embeddings in generalized manifolds, Trans. Am. Math. Soc. 352, No. 3, 1131-1147 (2000). ZBL0934.57023.
and for certain classes of compact ANRs (which include all compact manifolds):
S. Melikhov, E.Shchepin, The telescope approach to embeddability of compacta, unpublished, 2006.
Combining the two, one gets the embedding theorem for 4-manifolds. qed
My sense is that similar arguments to the ones by Bryant-Mio and Melikhov-Shchepin will go through in the noncompact case (I looked only at the paper by Melikhov and Shchepin) and can be used to prove an embedding theorem for noncompact manifolds but somebody would have to check the details...
answered Apr 21, 2021 at 21:47
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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).
answered Aug 5, 2010 at 17:36
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It Is easy to see that every simplcial complex embeds in $\Bbb R^{2n+1}$, just take generic points on the moment curve as vertices and extend linearly.
Stallings proved that every n dimensional complex embeds upto homotopy into $\Bbb R^{2n}$. Not sure if adding the manifold hypothesis lets you get rid of the homotopy.
