Let $\mathcal{M}$ be a compact, connected, oriented 3-manifolds with non-empty connected boundary $\partial\mathcal{M}$. Then, following this article, it is stated that $\mathcal{M}$ can be written as


where $\#_{\partial}$ denotes the boundary connected sum and where $P_{i}$ are $\partial$-prime manifolds, i.e. 3-manifolds with non-empty connected boundary, which are not homeomorphic to a 3-ball and for which a decomposition like $P=Q_{1} \#_{\partial}Q_{1}$ implies that either $Q_{1}$ or $Q_{2}$ is a closed 3-ball. So this is basically a generalization of the famous prime decomposition ("Kneser-Milnor theorem") to the case of manifolds with boundary.

Now, lets say I only consider manifolds $\mathcal{M}$ with the property that $\partial\mathcal{M}$ is homeomorphic to the 2-torus $T^{2}=S^{1}\times S^{1}$. Then, if I understand the theorem above correctly, $\mathcal{M}$ has to be a prime-manifold: Suppose that $\mathcal{M}$ can be decomposed as $\mathcal{M}=P_{1}\#_{\partial}P_{2}$ for two prime manifolds. However, the boundary connected sum has the property that $\partial\mathcal{M}=(\partial P_{1})\# (\partial P_{2})$, which is not possible in our case, since $\partial\mathcal{M}=T^{2}$ and the 2-torus cannot be obtained as the connected sum of two other manifolds.

What I am wondering is the following:

Is every compact, connected, oriented 3-manifold $\mathcal{M}$ with non-empty connected boundary $\partial\mathcal{M}\cong_{\mathrm{homeo.}} T^{2}$ of the form $$\mathcal{M}\cong_{\mathrm{homeo.}}\overline{T^{2}}\#\mathcal{N},$$ where $\overline{T^{2}}=S^{1}\times D^{2}$ denotes the solid torus and where $\mathcal{N}$ is a closed, orientable and connected 3-manifold. The connected sum here is the internal one.

(This is a follow-up question to this MathOverflow post. )

  • 2
    $\begingroup$ Of course not, take the exterior of any nontrivial knot in $S^3$. What is true is that either your manifold admits a nontrivial boundary connected sum decomposition, or it is the solid torus connected sum with a closed manifold, or it has incompressible boundary. $\endgroup$ Dec 9, 2021 at 14:21

1 Answer 1


The situation is similar to the one of closed manifolds. One defines "boundary-prime" manifolds as those that cannot be decomposed nontrivially in a boundary-connected sum.

Note that if $M$ is connected, has nonempty boundary and is not prime, then $M$ is never boundary-prime. Namely, take a 2-sphere $S\subset M$ separating $M$ into two components none of which is a ball. Connect $S$ to $\partial M$ by a 1-handle $D^2\times [0,1]$ ($D^2\times \{0\}\subset S$, $D^2\times \{1\}\subset \partial M$). Now, remove from $S$ the open disk equal to $int(D^2)\times \{0\}$ and add to $S$ the annulus $\partial D^2\times [0,1]$. The resulting surface $S'$ is a 2-dimensional disk with $\partial S'\subset \partial M$. This disk will cut $M$ in two components, none of which is a ball.

This works no matter what $\partial M$ is, in particular, if $\partial M=T^2$.

Thus, in what follows (until the concluding paragraph), I will assume that $M$ is prime.

Lemma. If $\partial M$ is a torus, then $M$ is necessarily boundary-prime.

Proof. Let $D\subset M$ be a properly embedded disk splitting $M$ in two components, none of which is a 3-ball. (Splitting means that you remove from $M$ an open tubular neighborhood of $D$.) Since $D$ separates $M$, $\partial D$ separates $T^2=\partial M$, hence, bounds a disk $D'\subset T^2$. Taking the union $D\cup D'$ we obtain a non-properly embedded 2-sphere in $M$. Pushing the disk $D'$ slightly into $M$, we obtain a 2-sphere $S\subset M$ disjoint from the boundary. Since $D$ was splitting $M$ into two submanifolds none of which is a 3-ball, the same holds for $S$. Hence, $M$ is not prime, contradicting the standing assumption. qed

We continue the discussion of prime manifolds $M$ with toral boundary. Such a manifold still can have compressible boundary. However, if this is the case, a boundary-compressing $D$ disk in $M$ is necessarily nonseparating. (Since every separating loop in the torus $T^2$ bounds a disk in $T^2$.) Cutting $M$ open along $D$ results in a manifold $M'$ with spherical boundary. If $M'$ is homeomorphic to the 3-ball, then $M$ itself is a solid torus, $\hat{T}=D^2\times S^1$. Otherwise, attaching $B^3$ along $\partial M'$ results in a closed 3-manifold $N$ which is not $S^3$. Then $M= N\# \hat{T}$. But there is one more possibility, namely, $\partial M$ is incompressible. There are many manifolds like that, for instance, the exterior of any nontrivial knot in $S^3$.

Hence, we obtain the trichotomy for 3-manifolds $M$ which need not be $\partial$-prime.

To conclude: Suppose that $M$ is a connected (not necessarily prime) 3-manifold and $\partial M$ is homeomorphic to $T^2$. Then one of the following mutually exclusive properties holds:

  1. $M$ is not prime, equivalently, is not $\partial$-prime.

  2. $M=\hat{T}$.

  3. $M$ is prime and $\partial M$ is incompressible.

  • 1
    $\begingroup$ Sorry, I am late to this question, but I was looking for a similar problem. Just a short question: Why is a $3$-manifolds with $\partial\mathcal{M}\cong T^{2}$ necessarily a $\partial$-prime? Because, it could maybe still be possible to write $\mathcal{M}$ as $\mathcal{M}=Q_{1}\#_{\partial} Q_{2}$ where $Q_{1}$ is some other $3$-manifolds with torus boundary and where $Q_{2}$ is some $3$-manifolds with spherical boundary, which is different from the closed $3$-ball. $\endgroup$
    – B.Hueber
    Jan 7, 2022 at 21:26
  • $\begingroup$ @B.Hueber: You are right, I was sloppy and assumed implicitly that $M$ is prime to begin with. Then it will have to be boundary-prime as well. $\endgroup$ Jan 7, 2022 at 21:35
  • $\begingroup$ Okay I see. But this does not change your proposed trichotomy, right? $\endgroup$
    – B.Hueber
    Jan 7, 2022 at 21:38
  • $\begingroup$ @B.Hueber: Item (2) should be read as "$M$ is not prime." (One of the factors need not be a solid torus.) Thank you for noticing. $\endgroup$ Jan 7, 2022 at 21:39
  • 1
    $\begingroup$ @G.Blaickner: yes, it is a contradiction, proving that $M=\hat{T}$. $\endgroup$ Feb 22, 2022 at 16:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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