What are some examples of manifolds that do not have boundaries and are not boundaries of higher dimensional manifolds?

Is any $n$-dimensional closed manifold a boundary of some $(n+1)$-dimensional manifold?

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    $\begingroup$ It sounds like you're confusing the idea of a boundary in the sense of chain complexes (i.e., an element of the image of $d$ or $\partial$) with a boundary in the sense of a manifold (i.e., the topological boundary of a manifold). Are you asking about manifolds $M$ that are not of the form $\del N$ for any (say, compact) $N$, manifolds that carry no cohomology, or (co)homological elements that are closed but not exact? $\endgroup$
    – anomaly
    Aug 5, 2015 at 20:17
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    $\begingroup$ Hi Anomaly. I realized that my real question indeed does not match with the original title of my question. Yes you are right I asked a wrong question. I intended to ask a question about chain complexes. Gary answered my question. Thank you very much for the editions. My major is not mathematics. I am a physics student who studies homology by himself. It is difficult for me to ask mathematical questions in correct ways. I am sorry for that. $\endgroup$ Aug 5, 2015 at 23:21
  • $\begingroup$ After reading Gary's answer, I realized that my first question was really silly. I forgot the Torus totally. $\endgroup$ Aug 5, 2015 at 23:23

3 Answers 3


Two closed $n$-dimensional manifolds $M$, $N$ are said to be cobordant if there is a compact $(n+1)$-dimensional manifold $W$ such that $\partial W = M\sqcup N$. If $N = \emptyset$, then $\partial W = M$ and $M$ is said to be null-cobordant. In this terminology, your question becomes

Is every closed manifold null-cobordant?

The answer to this question is no. For example, a single point is not the boundary of any compact one-dimensional manifold (the only such manifolds are disjoint unions of finitely many copies of $S^1$ and finitely many intervals, all of which have an even number of boundary points).

More generally, we have the following:

Proposition: If a closed manifold $M$ is null-cobordant, then $\chi(M)$ is even.

Proof: If $M$ has odd dimension, then $\chi(M) = 0$ so there is nothing to prove. Suppose then that $M$ has even dimension and that $W$ is a compact manifold with $\partial W = M$. The double of $W$, denoted by $D(W)$, is a closed odd-dimensional manifold and hence $\chi(D(W)) = 0$. But $\chi(D(W)) = \chi(W) + \chi(W) - \chi(\partial W) = 2\chi(W) - \chi(M)$, so $\chi(M) = 2\chi(W)$ which is even.

So for example, we also see that $\mathbb{RP}^2$ is not null-cobordant as $\chi(\mathbb{RP}^2) = 1$.

The converse of the above result does not hold; that is, there are closed manifolds $M$ with $\chi(M)$ even which are not null-cobordant. The simplest example is the four-dimensional manifold $M =(\mathbb{RP}^2\times\mathbb{RP}^2)\#\mathbb{RP}^4$ which has $\chi(M) = 0$, but is nonetheless not null-cobordant.

Pontryagin & Thom proved that a closed smooth manifold is null-cobordant if and only if all of its Stiefel-Whitney numbers are zero. The top Stiefel-Whitney number is the mod $2$ reduction of the Euler characteristic, so we recover the condition that $\chi(M)$ must be even. This result also explains why $M = (\mathbb{RP}^2\times\mathbb{RP}^2)\#\mathbb{RP}^4$ is not null-cobordant. On a closed four-manifold, there are three possibly non-zero Stiefel-Whitney numbers: $\underline{w}_1^4$, $\underline{w}_2^2$, and $\underline{w}_4$. Although $\underline{w}_4(M) = 0$ (i.e. $\chi(M)$ is even), $\underline{w}_1^4(M) = \underline{w}_2^2(M) = 1 \neq 0$.

You can ask the same question for oriented manifolds: is every oriented closed $n$-dimensional manifold the boundary of an oriented $(n+1)$-dimensional compact manifold? In this case, we say two oriented closed $n$-dimensional manifolds $M$ and $N$ are said to be orientedly cobordant if there is a compact oriented $(n+1)$-dimensional manifold $W$ such that $\partial W = M\sqcup\overline{N}$ where $\overline{N}$ denotes $N$ with the opposite orientation. If $N = \emptyset$, then $\partial W = M$ and $M$ is said to be orientedly null-cobordant. In this terminology, the question becomes

Is every oriented closed manifold orientedly null-cobordant?

Again, the answer is no (again, consider a single point). More generally, if a closed oriented manifold is orientedly null-cobordant, then it is in particular null-cobordant. So for example, $\mathbb{CP}^2$ is not orientedly null-cobordant because it is not null-cobordant ($\chi(\mathbb{CP}^2) = 3$).

Note however that an oriented closed manifold could be null-cobordant without being orientedly null-cobordant. The simplest example is the four-dimensional manifold $M = \mathbb{CP}^2\#\mathbb{CP}^2$.

It was shown by Wall that a smooth oriented closed manifold is orientedly null-cobordant if and only if all of its Stiefel-Whitney numbers and Pontryagin numbers vanish. In particular, we again see that a smooth orientedly null-cobordant manifold is also null-cobordant. This result also explains why $M = \mathbb{CP}^2\#\mathbb{CP}^2$ is not orientedly null-cobordant. On a smooth oriented closed four-manifold, there is only one Pontryagin number, $\underline{p}_1$, which satisfies $\underline{p}_1 = 3\tau$ where $\tau$ is the signature. Although all the Stiefel-Whitney numbers of $M$ are zero, $\underline{p}_1(M) = 6 \neq 0$.

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    $\begingroup$ Thank you so much Michael. I am a physics student who is studying cohomology by himself. I was confused by the simplicial complex because all the closed shapes that came into my mind look like boundary of something. I still do not have any intuitive visualization of projective plane. $\endgroup$ Aug 5, 2015 at 18:30

The circle $x^2+y^2=1$ in the punctured $(x,y)$-plane $\mathbb R^2\setminus \{(0,0)\}$ is a cycle but not a boundary .

  • $\begingroup$ Thanks Georges. My question wasn't precise. I should have mentioned that it is a boundary of a compact manifold. $\endgroup$ Aug 5, 2015 at 23:25
  • $\begingroup$ Why? The unit disk $x^2+y^2\leq 1$ is a $2$-dim manifold with boundary. So the unit circle is the boundary of it. $\endgroup$
    – user685167
    Dec 14, 2019 at 16:29
  • $\begingroup$ @Sssss The unit "disk" in the punctured plane $\mathbb R^2\setminus \{0\}$ described by your equation $x^2+y^2\leq1$ is not compact,while Xiaoyi Jing asks about cycles which are boundaries of compact manifolds, as explained in his comment just above yours. $\endgroup$ Dec 14, 2019 at 19:50

1) Consider a latitude or longitude of a torus $S^1 \times S^1$, neither of those cycles bounds in the torus . There can be such a cycle only in any manifold with non-trivial homology (although the cycle may not be a (sub) manifold) and no such cycle otherwise: trivial homology means that every cycle bounds.

2) No; Real projective plain is not the boundary of any manifold, because of its Euler characteristic. The Euler characteristic is an obstruction. See obstruction theory

The general area of study is Cobordism theory

  • $\begingroup$ Thank you so much Gary. Your example is really simply but brilliant. I am wondering why I didn't find such a simply example on Torus. Thanks a lot. $\endgroup$ Aug 5, 2015 at 18:32
  • $\begingroup$ @XiaoyiJing: No problem, glad it helped. $\endgroup$
    – Gary.
    Aug 5, 2015 at 18:37

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