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Sam Nead
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YesEDIT: Here is a substantial rewrite of my previous (incomplete) answer. Suppose 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 \subset \partial M$$S$.

Suppose that $S$ is compressible. Let $C$ be the surface we want to attach alongcharacteristic compression body with positive boundary $S$. Hatcher's theorem(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, correctly generalisedsay, says thatHempel's paper 3-manifolds as viewed from the curve complex. If the set of "boundary slopes"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. (curves which are partSee Section 3.4 of an "essential" surface inBonahon's article.) Let $M$ with boundary only$C$ be the component of the characteristic $I$-bundle meeting $S$. So the "corners of $C$" (the boundaries of its vertical boundary) isgive a "thin" subsetcollection of all curves in $S$. So Similar to Hempel's approach, we can attach a two-handlehandlebody $V$ to $M$ along $S$ whose core is not equal to anyso that all boundaries of these boundary slopesdisks (in $V$) are "sufficiently far" from the corners of $C$ (in the curve complex of $S$). After attachment Again, the$N = M \cup_S V$ will have no new manifoldtwo-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 irreduciblebe Seifert fibered. Induct)

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 gives. His results give a closely related resulthandlebody attachment which again makes $N$ irreducible. (but with stronger hypothesesIn fact, $N$ will be irreducible, atoroidal, and stronger conclusions)will have relatively hyperbolic fundamental group.)

Yes. Suppose that $S \subset \partial M$ is the surface we want to attach along. Hatcher's theorem, correctly generalised, says that the set of "boundary slopes" in $S$ (curves which are part of an "essential" surface in $M$ with boundary only meeting $S$) is a "thin" subset of all curves in $S$. So we can attach a two-handle to $S$ whose core is not equal to any of these boundary slopes. After attachment, the new manifold is again irreducible. Induct.

Lackenby's paper Attaching handlebodies to 3-manifolds gives a closely related result (but with stronger hypotheses and stronger conclusions).

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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Sam Nead
  • 28.2k
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  • 131

Yes. Suppose that $S \subset \partial M$ is the surface we want to attach along. Hatcher's theorem, correctly generalised, says that the set of "boundary slopes" in $S$ (curves which are part of an "essential" surface in $M$ with boundary only meeting $S$) is a "thin" subset of all curves in $S$. So we can attach a two-handle to $S$ whose core is not equal to any of these boundary slopes. After attachment, the new manifold is again irreducible. Induct.

Lackenby's paper Attaching handlebodies to 3-manifolds gives a closely related result (but with stronger hypotheses and stronger conclusions).