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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

$$\mathcal{M}=P_{1}\#_{\partial}\dots\#_{\partial}P_{n}$$

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. )

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    $\begingroup$ 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

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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.

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    $\begingroup$ 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 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 at 21:35
  • $\begingroup$ Okay I see. But this does not change your proposed trichotomy, right? $\endgroup$
    – B.Hueber
    Jan 7 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 at 21:39
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    $\begingroup$ @G.Blaickner: yes, it is a contradiction, proving that $M=\hat{T}$. $\endgroup$ Feb 22 at 16:48

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