Just parallelling this question, that seemed not to admit an easy answer at all, let's "soft down" the category and ask the same thing in the case of $\mathcal{C}^{\infty}$-differentiable manifolds [Edit: we consider only manifolds without boundary].

Well, so:

Is every differentiable manifold diffeomorphic to an open submanifold of a compact one?

Edit: As some comments have pointed out, there are manifolds for which the compactification theorem fails, so someone has suggested to change the question to the more meaningful:

Which differentiable manifolds are diffeomorphic to an open submanifold of a compact one?

  • 3
    $\begingroup$ very nice question. perhaps better: which manifolds embed into a compact one? $\endgroup$ Apr 24, 2010 at 18:58
  • 1
    $\begingroup$ The Whitney embedding theorem implies that any compact smooth manifold can be embedded in to a compact manifold with a high enough dimension. So the only interesting case is non-compact manifolds. $\endgroup$ Apr 24, 2010 at 19:40
  • $\begingroup$ @Kelly: I don't understand your comment. I was asking for an open embedding into a compact one. I presume it's easy to embed any manifold as a locally closed submanifold of a sphere: just Whitney-embed as a closed submanifold of $\mathbb{R}^N$, send $\mathbb{R}^N$ diffeomerphically onto a ball, which sits inside a larger ball, and one-point-compactify the larger ball. $\endgroup$
    – Qfwfq
    Apr 25, 2010 at 4:16
  • $\begingroup$ @unknown I was referring to Martin's comment "very nice question. perhaps better: which manifolds embed into a compact one." which to me seems not very interesting for compact manifolds. Martin did not specify an open embedding just an embedding. $\endgroup$ Apr 25, 2010 at 9:08
  • 1
    $\begingroup$ @Kelly: Ok, I think maybe M. left the word "open" as understood, just referring to the context of the question. $\endgroup$
    – Qfwfq
    Apr 25, 2010 at 17:39

6 Answers 6


No. A surface of infinite genus is not a submanifold of a compact surface.

  • $\begingroup$ ...That's true!... $\endgroup$
    – Qfwfq
    Apr 24, 2010 at 18:09
  • 6
    $\begingroup$ The complement of the Cantor set in $S^2$ is still genus 0, although it has countably many ends. $\endgroup$ Apr 25, 2010 at 15:23
  • 4
    $\begingroup$ To see that Richard's answer is correct, note that the genus is the rank of the intersection form on $H_1$, which cannot drop in any embedding. This generalizes to even-dimensional manifolds, e.g., an infinite connect sum of $\mathbb{CP}^2$'s. Are there any odd-dimensional examples? $\endgroup$ Apr 25, 2010 at 15:41
  • 1
    $\begingroup$ I think the same intersection theory argument applies to the product of an infinite genus surface $S$ with a torus $T^k$ (of any dimension). If the surface handles are represented as $a_i, b_i$, then the intersection number of $a_i$ and $b_j\times T^k$ is $1$ if $i=j$ and $0$ else. It follows that no two of $a_i$'s are homologous in the image of $S\times T^k$ under any codimension zero embedding. $\endgroup$ Apr 25, 2010 at 16:23
  • 2
    $\begingroup$ Why can't one look at the rank of $H_i(M)$ divided by the annihilator of the pairing with $H_{n-i}(M)$ in the odd-dimensional case? That would seem to make the connected sum of an infinite number of $n$-dimensional tori an example in any dimension. $\endgroup$ Apr 25, 2010 at 16:36

There are contractible 3-manifolds which cannot be embedded in any compact 3-manifold. Kister and McMillan constructed a variant of the Whitehead manifold $M'$ which is contractible but which cannot embed into $S^3$. From the Geometrization theorem, the universal cover of any compact 3-manifold embeds into $S^3$. So if $M'$ embedded into a compact 3-manifold $M'\subset M$, its lift $M'\subset \widetilde{M}\subset S^3$ to the universal cover would give a contradiction.


As everyone has said, the answer is "no". You have to make assumptions to ensure that the "ends" of your manifold are sufficiently simple. It appears hard to find using the refs Paul posted, but the key result about this is Larry Siebenmann's thesis. I don't think this was ever published, but it is available on Andrew Ranicki's webpage here. Another source (also on Ranicki's webpage) for this is some lecture notes of Kervaire, available here.

By the way, one obvious necessary condition is for your manifold to have a finitely presentable fundamental group (this is one of the problems with Richard's example). A classic example to show that this is still not enough (even in dimension 3) is the Whitehead manifold.

EDIT : I should also point out one beautiful recent about this. Marden's Tameness Conjecture (recently proved independently by Agol and Calegari-Gabai) says that if M is a hyperbolic 3-manifold with finitely generated fundamental group, then M is homeomorphic to the interior of a compact 3-manifold. The Whitehead manifold mentioned above shows that the assumption that M is hyperbolic is necessary.

  • 2
    $\begingroup$ The Whitehead manifold embeds in the 3-sphere, though. It's the complement of the Whitehead continuum. $\endgroup$ Apr 24, 2010 at 19:53
  • $\begingroup$ Good point! I didn't read the question carefully enough and assumed that he was asking for the interior of a manifold with boundary. $\endgroup$ Apr 24, 2010 at 20:01
  • $\begingroup$ Yes, the problem as stated seems tricky: how to show that a given manifold cannot be the complement of any closed set in any compact manifold? $\endgroup$
    – Paul
    Apr 24, 2010 at 22:49

There is a long history on this problem, starting, in dimensions>4 with Browder-Levine-Livesay:



Follow MR to get to results in dimensions 3,4, etc. You have to first eliminate issues like Richard mentions using finiteness obstructions.

  • 1
    $\begingroup$ You are answering the question "is every differentiable manifold diffeomorphic to the interior of a compact one?". $\endgroup$ Apr 25, 2010 at 13:29

I think none of the above posts answer the question "is every differentiable manifold diffeomorphic to an open submanifold of a compact one?". Rather they answer "is every differentiable manifold diffeomorphic to the interior of a compact one?" The reason for the confusion could be the latter question is fundamental in geometric topology, while the former one has little significance. Anyway,

The connected sum $V$ of infinitely many copies of $CP^3$'s is not diffeomorphic to an open subset of a compact manifold.

EDIT: Hats off to Torsten Ekedahl who pointed out in comments that my argument below is incorrect (thus I don't know whether the above statement about $V$ is true). I decided not to delete it because it illuminates some subtleties of the original question.

The point is that any diffeomorphism onto an open subset pulls back the tangent bundle, and in particular, pulls back the first Pontryagin class $p_1$. Thus if $V$ is an open subset of a compact manifold $M$, then its first Pontryagin class $p_1(V)$ lies in the image of $H^4(M)\to H^4(V)$, which is a finitely generated subgroup of $H^4(V)$, which is the infinite product of $\mathbb Z$'s corresponding to generators of $H^4(CP^3)$. The first Pontryagin class of $CP^3$ is a multiple of a generator of $H^4(CP^3)\cong\mathbb Z$, and removing a finite set of points from $CP^3$ does not affect the $4$th skeleton, so $p_1(V)$ does not lie in a finitely generated subgroup of $H^4(V)$.

I am curious to see low-dimensional answers to the question "is every differentiable manifold diffeomorphic to an open submanifold of a compact one?"

  • 1
    $\begingroup$ I don't understand, any element of an abelian group lies in a finitely generated subgroup. Also $H^4(V)$ is the product not the sum of copies of $\mathbb Z$. If one among the compact manifolds include those with boundary it may not even be true that $H^4(M)$ is finitely generated. $\endgroup$ Apr 25, 2010 at 13:55
  • $\begingroup$ @Torsten, thanks for catching the typo. I edited to replace "sum" by "product". On the other hand, any compact manifold (with or without boundary) has finitely generated homology and cohomology. In the smooth case this is easy to see because by Morse theory they are homotopy equivalent to a finite cell complexes. Also this is a non-issue for if $V$ embeds to a manifold $M$ with boundary, $V$ also embeds into the double of $M$ along the boundary, which is closed. $\endgroup$ Apr 25, 2010 at 14:03
  • 1
    $\begingroup$ You are right that there are no problem for f.g. of $H^4(M)$, my bad. However, I still do not understand your contradiction. $p_1(V)$ lies in the finitely generated subgroup $\mathbb Z p_1(V)$ so where is the contradiction? $\endgroup$ Apr 25, 2010 at 14:12
  • $\begingroup$ By the way there is a beautiful argument in Spanier: Algebraic topology for the finite generation of homology for a compact manifold (with boundary which I should have remembered). Finite generation is algebraically derived from Poincaré duality and the universal coefficient formula. $\endgroup$ Apr 25, 2010 at 14:17
  • $\begingroup$ Let me make an attempt. Pick a sequence of points on $V$ without limit point in $V$ and let $x$ be a limit point in $M$. A coordinate open neighbourhood of $x$ in $M$ must contain an (infinite number of) copies of $CP^3$ so that the tangent bundle of $CP^3$ minus the attaching points is trivial which is not possible as non-triviality is detected by $p_1$. $\endgroup$ Apr 25, 2010 at 14:30

I'm trying to answer the second question. Most of the results can be found in the joint paper with Guilbault https://arxiv.org/abs/1712.05995. Here I just wanted to make a quick summary. An $m$-manifold $M^m$ with (possibly empty) boundary is completable if there exists a compact manifold $\widehat{M^m}$ and a compactum $C \subseteq \partial \widehat{M^m}$ such that $\widehat{M^m}-C$ is homeomorphic to $M^m$. In this case $\widehat{M^m}$ is called a (manifold) completion of $M^m$. One can change the homeomorphism to diffeomorphism or PL homeo. for other categories.

Dim = 0,1 are obvious.

Dim = 2, we can't find a complete classification in the literature, so we provided a theorem in that paper. That is, a connected 2-manifold is completable iff it has finitely generated first homology. This is mainly based on classical work of Kerekjarto and Richards.

Dim = 3, it's mainly due to Tucker, where he showed that a 3-manifold can be completed if and only if each component of each clean neighborhood of infinity has finitely generated fundamental group.

Let me talk about dimensions $\geq 6$ first. Then I'll go back to dimensions 4 and 5. The first breakthrough regarding this problem was due to Siebenmann in 1965. In his PhD thesis, he proved that an open n-manifold $M^n$ is completable (it was called collarable by that time) iff (1) M is inward tame, (2) the end is pro-$\pi_1$ stable and (3) the Wall finiteness obstruction of the end vanishes. In 1983, O'Brien generalized the theorem to one-ended manifold with possibly non-empty boundary.

In our paper, we dropped the O'Brien's assumption on that manifolds are one-ended. By properly generalizing Siebenmann's conditions, we proved that manifolds of dimension at least 6 are completable iff they are inward tame, peripherally $\pi_1$-stable at infinity, of zero Wall and Whitehead torsion. Our proof is based on PL manifolds, but one can employ standard techniques such as "rounding off corners" to handle the other catergories.

Our theorem is still true in dimension = 5 provided that the fundamental groups are good in the sense of Freedman and Quinn.

The theorem fails in dimension = 4. Kwasik-Schultz and (independently) Weinberger discovered that there are open 4-manifold satisfying Siebenmann's condition but fail to be collarable.

Just a quick comment on Ian's reference to a contractible open manifold $M'$ constructed by Kister-McMillan which doesn't embed in $S^3$. The example was first proposed by R. H. Bing. Haken further proved that $M'$ doesn't embed in any 3-manifold using his finiteness theorem. Recently, I showed that $M'$ can't be embedded in any compact, locally connected and locally 1-connected metric space. https://arxiv.org/abs/1809.02628


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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