I know the embedding theorems that allow you to embed $n$-manifolds into $\mathbb{R}^k$, provided $k$ is chosen large enough. Here I'm interested in the possibility of taking $k=n$ in the case of compact manifolds.

From the classification of compact surfaces I can see that the closed ones cannot be embedded in the plane, and that the ones that can be embedded are disks and annuli, which have non empty boundary.

I'd like to know if this intuition is still sound in dimension $n\geq 3$, so my question is: if a compact manifold of dimension $n$ embeds in $\mathbb{R}^n$, is it forced to have a non empty boundary?

I've read the wiki article about Whitney embedding and it's section about "sharper results", but there they give general estimates for the whole class of compact $n$-manifolds, whereas I would be interested in just a single example of a compact boundaryless $n$-manifold embedded in $\mathbb{R}^n$ (possibly in low dimensions), or a proof/reference if this can never happen. Notice I'm not interested in distinction between orientable/non-orientable manifolds, but in the presence of a boundary.

Thanks in advance!

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    $\begingroup$ A map between $n$-dimensional manifolds is an open mapping, by the Invariance of Domain theorem. So this image of $M$ under this embedding is open in $\mathbb{R}^n$, but it is also closed (because $M$ is compact). But $\mathbb{R}^n$ is connected... $\endgroup$
    – user641
    Commented Oct 10, 2012 at 12:11
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    $\begingroup$ Dear @Steve, it is false that "A map between n-dimensional manifolds is an open mapping". And you certainly don't need a difficult theorem like invariance of domain for such an easy question. $\endgroup$ Commented Oct 10, 2012 at 18:05
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    $\begingroup$ Well I meant embeddings! And I disagree about this being a difficult theorem, since it is equivalent to the implicit function theorem in this case. $\endgroup$
    – user641
    Commented Oct 10, 2012 at 18:17
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    $\begingroup$ I think it is good to note that the invariance of domain theorem is necessary for the proof of this statement for topological manifolds (the other proofs rely on smoothness). $\endgroup$
    – user17786
    Commented Oct 11, 2012 at 16:38
  • $\begingroup$ @user17786 I am not very familiar with topological manifolds. Is it still true that embeddings are open maps? $\endgroup$
    – Kirby Lee
    Commented Nov 17, 2018 at 7:36

4 Answers 4


Here's an alternate proof which doesn't use invariance of domain. It also gives a slightly stronger result.

Let $M^n$ be compact without boundary. Then there is no immersion $f:M\rightarrow \mathbb{R}^n$.

Proof: (sketch). Assume for a contradiction there is such an $f$. Since $M$ is compact, $f$ is a closed map, that is, it maps closed sets to closed sets. To see this, let $F\subseteq M$ be closed. Then it's compact because $M$ is, so $f(F)$ is compact, hence closed. (Here, we just use the fact that $f$ is continuous).

Further, $f$ is an open map. That it, it maps open sets to open sets. To see this, note that it is enough to map really small open sets to open sets because $f(\bigcup U_i) = \bigcup f(U_i)$.

It's a fact (a consequence of the implicit function theorem, if I recall) that every immersion locally looks like the natural inclusion $\mathbb{R}^k\rightarrow \mathbb{R}^n$ into the first $k$ coordinates. (This uses the boundarylessness of $M$ - if $p$ is on the boundary, this part doesn't work.)

Said more precisely, given our immersion $f$ and a point $p\in M$, there is a chart around $p$ and around $f(p)$ so that in these chart coordinates, $f$ looks like the inclusion.
Now, in our case $k = n$ and then the natural inclusion is a homeomorphism. This implies open sets in these special charts are mapped to open sets, so $f$ is open.

Putting this together, we've now seen that $f$ is an open and closed mapping. Now, what is $f(M)$? It must be compact because $M$ is, but it must also be both open and closed in $R^n$ because $M$ is open and closed in itself. This implies $f(M) = \mathbb{R}^n$ since that's the only nonempty clopen subset of $\mathbb{R}^n$, but $\mathbb{R}^n$ isn't compact, so we've reached a contradiction.


I claim we can work even less hard than the arguments that have been proposed so far! Let $M$ be a closed $n$-manifold and let $f : M \to \mathbb{R}^n$ be a smooth map. Pick a coordinate $x : \mathbb{R}^n \to \mathbb{R}$. The composition $x \circ f : M \to \mathbb{R}$ attains a maximum at some point $m \in M$ (by compactness). At this point $m$, $df$ cannot be injective (because its $x$-component is zero), so $f$ is not an immersion.

  • $\begingroup$ If $M$ has boundary and $f$ is an embedding then the conclusion is that $m$ has to lie on the boundary. By varying $x$ we get that if $f$ is an embedding then the interior of $f(M)$ has to be "inside" of its boundary (e.g. it must be contained in the convex hull of the boundary). $\endgroup$ Commented Mar 8, 2015 at 1:42
  • $\begingroup$ The equivalence of injective differentials and of local immersion relies on the constant rank theorem, right? $\endgroup$ Commented Jun 21, 2018 at 21:48

The embedded submanifolds $\emptyset \neq S\subset M$ of codimension $0$ (i.e. $dim M-dimS=0$) of a manifold $M$ are exactly the open subsets of $M$ : Lee, Introduction to smooth manifolds Proposition 5.1, page 99.
Hence they are not compact if $M$ is connected and not compact, which applies to $M=\mathbb R^n$.

  • $\begingroup$ I have given a reference in order to reassure beginners, but the result I quote is very easy to prove.(By the way, the reference is to the second edition of Lee's book, which is even better than the first one) $\endgroup$ Commented Oct 10, 2012 at 18:08
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    $\begingroup$ Georges--thanks for the kind remark about my book. You need to amend what you wrote slightly: it's not true in general that "they are not compact if $M$ is not compact." For example, the noncompact manifold $\mathbb S^1\times \mathbb Z$ has lots of compact, embedded, open, codimension-0 submanifolds. But you're right that there are no compact codimension-0 manifolds embedded in $\mathbb R^n$. (A simple way to see this is to note that a nonempty compact $n$-submanifold of $\mathbb R^n$ has a point at max distance from the origin, but the inclusion map is open by the inverse function theorem.) $\endgroup$
    – Jack Lee
    Commented Oct 11, 2012 at 0:26
  • $\begingroup$ Dear @Jack, yes you are right: thanks for your comment. I have edited my answer and added that $M$ should be connected, which suffices to make the statement correct and yet completely general (I'd rather not restrict the answer to $\mathbb R^n$) . And let me repeat that I find your new edition really excellent: you must have put a lot of work in the complete reorganization of your book. $\endgroup$ Commented Oct 11, 2012 at 8:23
  • $\begingroup$ Lee's book assumes invariance of domain and implicit function/constant rank/inverse function theorem(s). So it is not independent of those results. $\endgroup$ Commented Jun 21, 2018 at 21:40

Suppose you have an n-Manifold $M$ embedded in $\mathbb{R}^{n}$. Hence you can cover your manifold just with one coordinate chart: for example take the identity $I:M\rightarrow \mathbb{R}^{n}$. Note that this application is an homeomorphism, so by the "invariance of domain" $I(M)$ is a compact set such that $int(M)\neq\emptyset$, where $int$ denotes interior.

Now every open set in $\mathbb{R}^{n}$ must have a boundary unless it is all $\mathbb{R}^{n}$, but this is not our case, so $M$ ha non empty boundary.


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