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Let $M$ be a compact, orientable, irreducible 3-manifold with boundary (possibly more than one component). Let $S\subseteq\partial M$ be one of its boundary components, which is an orientable surface of genus $g$. Can we glue a genus $g$ handlebody $H_g$ along $S$ in an appropriate way so that $M'=M\cup_{S} H_g$ is still irreducible?

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    $\begingroup$ Pretty please, don't "break" the notation $\partial M$. You could call your special surface $S$, or $F$, or something similar. $\endgroup$
    – Sam Nead
    Commented Jul 31 at 17:14

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EDIT: Here is a substantial rewrite of my previous (incomplete) answer. I think that this proof is a bit "heavy", but I haven't yet thought of a better approach.


The answer is "yes". We split into cases, depending on the quality of the boundary component $S$.

Suppose that $S$ is compressible. Let $C$ be the characteristic compression body with positive boundary $S$. (See Section 3.3 of Bonahon's article Geometric structures on 3–manifolds for definitions and theorems.) We now attach a handlebody $V$ to the positive boundary of $C$ following, say, Hempel's paper 3-manifolds as viewed from the curve complex. If the attaching map has sufficiently high "distance" in Hempel's sense, then $N = M \cup_S V$ will have no new two-spheres, and so will be irreducible.

So we now may suppose that $S$ is incompressible. Suppose that $(M, S)$ admits essential annuli. (See Section 3.4 of Bonahon's article.) Let $C$ be the component of the characteristic $I$-bundle meeting $S$. So the "corners of $C$" (the boundaries of its vertical boundary) give a collection of curves in $S$. Similar to Hempel's approach, we attach a handlebody $V$ to $M$ along $S$ so that all boundaries of disks (in $V$) are "sufficiently far" from the corners of $C$ (in the curve complex of $S$). Again, $N = M \cup_S V$ will have no new two-spheres.

So we now may suppose that $S$ is "an-annular": that is, incompressible and without essential annuli. Suppose that $M$ is Seifert fibered. So $S$ is a torus. In this case can we use Hatcher's theorem to obtain a non-boundary slope to attach along. (And we can arrange that the result $N = M \cup_S V$ will again be Seifert fibered.)

Suppose that $M$ is toroidal but not Seifert fibered. Then we can cut along the JSJ decomposition of $M$ and deal with the component of the decomposition containing $S$.

Suppose, finally, that $M$ is atoroidal and $S$ is an-annular. I will also assume that all boundary components of $M$ are an-annular (dealing with the other possibility is an exercise). So $M$ is "simple" in sense given by Lackenby's paper Attaching handlebodies to 3-manifolds. His results give a handlebody attachment which again makes $N$ irreducible. (In fact, $N$ will be irreducible, atoroidal, and will have relatively hyperbolic fundamental group.)

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    $\begingroup$ By attaching a 2-handle onto $S$ along a curve $\alpha$, suppose we obtain a reducible manifold $N$. I found in many articles (mainly about toral boundary) the following fact which I'm a little confused. Suppose $F=S^2$ is a reducing sphere in $N$, and up to isotopy we can assume $F\cap \partial M$ is some parallel copies of $\alpha$. If $F$ is chosen such that the number of components of $F\cap \partial M$ is minimal, then $F\cap M\subseteq M$ is incompressible and $\partial$-incompressible. I understand why it is incompressible, but I don't get the point why it is $\partial$-incompressible. $\endgroup$
    – YC Su
    Commented Aug 2 at 17:37
  • $\begingroup$ @YCSu - I've answered (elsewhere) your new question, and I've rewritten (here) my answer to this question. Please let me know if you have questions (or comments!) about either. $\endgroup$
    – Sam Nead
    Commented Aug 4 at 7:58

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