# Decomposition of manifolds with toroidal boundary

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. )

• 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. Dec 9, 2021 at 14:21

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.

• 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. Jan 7 at 21:26
• @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. Jan 7 at 21:35
• Okay I see. But this does not change your proposed trichotomy, right? Jan 7 at 21:38
• @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. Jan 7 at 21:39
• @G.Blaickner: yes, it is a contradiction, proving that $M=\hat{T}$. Feb 22 at 16:48