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This question is inspired by another question of mine, a few years ago. In that question, I asked whether we could "fill a Klein bottle with four-dimensional water". Ultimately, what I think I wanted to ask is whether a Klein bottle embedded into $\mathbb R^4$ (by Whitney) defines a four-dimensional "volume" to be occupied by four-dimensional "water droplets".

My intuition says that this is not possible, since the analogy in three dimensions (a curve, codimension = 2) do not define a volume "to be filled". However, a surface embedded into three-dimensional space can define a volume (codimension = 1). Likewise, in the plane, we have Jordan's curve theorem that says that a sufficiently nice curve embedded into the plane (i.e. a simple closed curve) defines a bounded region called the interior of the curve (codimension = 1).

So it seems that we have a similar result if the codimension is 1. I am curious about higher-dimensional results, and I think I have arrived at the proper-ish formalization of this problem:

Suppose $\mathcal M$ is a closed connected $m$-dimensional manifold. Is there a closed $(m+1)$-dimensional (connected) manifold $\mathcal N$ such that $\partial \mathcal N = \mathcal M$?

In Jordan's curve theorem, the curve is the boundary of the interior. In three-dimensional space, it is intuitive to think about a nice enough surface defining an interior volume. What is lacking from my formalization is how the notion of a simple curve generalizes in the manifold setting. How do I modify my statement to include this property? Is this generalization even something that holds true in higher dimensions?

I vaguely remember theorems about this when I studied differential topology and bordisms for my master's thesis, but I could not find a theorem that answered my question in the positive.

Markus Klyver
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  • Your proper-ish formalization is just fine, if you keep in mind that manifolds can exist in the abstract, without being embedded in any other space such as a Euclidean space. That way, for the purposes of your problem, the whole idea of "simple curve" just... goes away. – Lee Mosher Sep 17 '25 at 20:09
  • @LeeMosher Is there an intrinsic definition of a manifold "not crossing itself"? – Markus Klyver Sep 17 '25 at 20:24
  • Your formalization isn't quite right as stated, at least if you follow the usual convention that "closed manifold" means "compact manifold without boundary". You don't want the manifold $N$ to be closed, otherwise the question is trivial. You also don't want to allow $N$ to be non-compact, since the question is also trivial in that case. – Derived Cats Sep 17 '25 at 20:42
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    "Crossing itself" is an informal feature of an immersion from one manifold to another; immersions are a standard concept of different topology. – Lee Mosher Sep 17 '25 at 20:52
  • As said, your question is best formalized without any map, embedding or immersion or any other kind of map, from one manifold to another. – Lee Mosher Sep 17 '25 at 20:54
  • A similar question. – Michael Albanese Sep 18 '25 at 13:11
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    The Klein bottle is the boundary of a 3-manifold! There are two ways to make a 3-manifold by gluing together the ends of a solid cylinder $D^2\times I$. One way gives a solid torus. The other gives a non-orientable 3-manifold whose boundary is the Klein bottle. – Cheerful Parsnip Sep 18 '25 at 19:22

1 Answers1

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Cobordism is the right thing to review. If $M = \partial\mathcal N$, then in particular all the Pontryagin numbers of $M$ must vanish. Thus, the simplest example of a compact, orientable manifold that cannot be a boundary is $\Bbb CP^2$. See Milnor and Stasheff, among other sources.

EDIT: See @MichaelAlbanese's linked answer for a discussion of the analogous fact that $\Bbb RP^2$ cannot be a boundary because the Stiefel-Whitney numbers of a boundary must all vanish.

Ted Shifrin
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  • Is "orientable" the right generalization for "simple curve"? – Markus Klyver Sep 17 '25 at 19:56
  • There's no need to restrict to orientable manifolds. $\mathbb{R}P^2$ doesn't bound, and this is detected by the Stiefel-Whitney numbers. – Tyrone Sep 17 '25 at 20:30
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    No, “embedded” is perhaps an appropriate generalization of “simple.” I restricted to orientable for no particular reason, just to be proactive if you’d asked “what if we impose orientability?” when given non-orientable examples. – Ted Shifrin Sep 17 '25 at 21:46
  • Perhaps it's worth editing to give a non-orientable example as well. – FD_bfa Sep 18 '25 at 16:13
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    There's a "more elementary" argument that $\mathbb{C}P^2$ does not bound a manifold (in the sense that very few undergraduates learn anything about cobordism).

    If $N$ is compact, one can always form the closed manifold $N' = N \amalg_{\partial N} N$. If $\dim N$ is odd, $N'$ has Euler characteristic $0$ (by mod-2 Poincaré duality). Also, $\chi(N') = \chi(N) + \chi(N) - \chi(\partial N)$.

    So, if $\dim(\partial N)$ is even, $\chi(\partial N)$ must be even.

    However, the Euler characteristic of $\mathbb{C}P^2$ is $3$. So $\mathbb{C}P^2$ cannot be the boundary of a compact manifold.

     – diracdeltafunk Sep 18 '25 at 20:51