Prime manifold

A 3-manifold is irreducible if any smooth sphere bounds a ball. More rigorously, a differentiable connected 3-manifold ${\displaystyle M}$ is irreducible if every differentiable submanifold ${\displaystyle S}$ homeomorphic to a sphere bounds a subset ${\displaystyle D}$ (that is, ${\displaystyle S=\partial D}$) which is homeomorphic to the closed ball

$${\displaystyle D^{3}=\{x\in \mathbb {R} ^{3}\ |\ |x|\leq 1\}.}$$

The assumption of differentiability of ${\displaystyle M}$ is not important, because every topological 3-manifold has a unique differentiable structure. The assumption that the sphere is smooth (that is, that it is a differentiable submanifold) is however important: indeed the sphere must have a tubular neighborhood.

A 3-manifold that is not irreducible is called reducible.

A connected 3-manifold ${\displaystyle M}$ is prime if it cannot be expressed as a connected sum ${\displaystyle N_{1}\#N_{2}}$ of two manifolds neither of which is the 3-sphere ${\displaystyle S^{3}}$ (or, equivalently, neither of which is homeomorphic to ${\displaystyle M}$).

Three-dimensional Euclidean space ${\displaystyle \mathbb {R} ^{3}}$ is irreducible: all smooth 2-spheres in it bound balls.

On the other hand, Alexander's horned sphere is a non-smooth sphere in ${\displaystyle \mathbb {R} ^{3}}$ that does not bound a ball. Thus the stipulation that the sphere be smooth is necessary.

The 3-sphere ${\displaystyle S^{3}}$ is irreducible. The product space ${\displaystyle S^{2}\times S^{1}}$ is not irreducible, since any 2-sphere ${\displaystyle S^{2}\times \{pt\}}$ (where ${\displaystyle pt}$ is some point of ${\displaystyle S^{1}}$) has a connected complement which is not a ball (it is the product of the 2-sphere and a line).

A lens space ${\displaystyle L(p,q)}$ with ${\displaystyle p\neq 0}$ (and thus not the same as ${\displaystyle S^{2}\times S^{1}}$) is irreducible.

An irreducible manifold ${\displaystyle M}$ is prime. Indeed, if we express ${\displaystyle M}$ as a connected sum

$${\displaystyle M=N_{1}\#N_{2},}$$

then ${\displaystyle M}$ is obtained by removing a ball each from ${\displaystyle N_{1}}$ and from ${\displaystyle N_{2},}$ and then gluing the two resulting 2-spheres together. These two (now united) 2-spheres form a 2-sphere in ${\displaystyle M.}$ The fact that ${\displaystyle M}$ is irreducible means that this 2-sphere must bound a ball. Undoing the gluing operation, either ${\displaystyle N_{1}}$ or ${\displaystyle N_{2}}$ is obtained by gluing that ball to the previously removed ball on their borders. This operation though simply gives a 3-sphere. This means that one of the two factors ${\displaystyle N_{1}}$ or ${\displaystyle N_{2}}$ was in fact a (trivial) 3-sphere, and ${\displaystyle M}$ is thus prime.

Let ${\displaystyle M}$ be a prime 3-manifold, and let ${\displaystyle S}$ be a 2-sphere embedded in it. Cutting on ${\displaystyle S}$ one may obtain just one manifold ${\displaystyle N}$ or perhaps one can only obtain two manifolds ${\displaystyle M_{1}}$ and ${\displaystyle M_{2}.}$ In the latter case, gluing balls onto the newly created spherical boundaries of these two manifolds gives two manifolds ${\displaystyle N_{1}}$ and ${\displaystyle N_{2}}$ such that

$${\displaystyle M=N_{1}\#N_{2}.}$$

Since ${\displaystyle M}$ is prime, one of these two, say ${\displaystyle N_{1},}$ is ${\displaystyle S^{3}.}$ This means ${\displaystyle M_{1}}$ is ${\displaystyle S^{3}}$ minus a ball, and is therefore a ball itself. The sphere ${\displaystyle S}$ is thus the border of a ball, and since we are looking at the case where only this possibility exists (two manifolds created) the manifold ${\displaystyle M}$ is irreducible.

It remains to consider the case where it is possible to cut ${\displaystyle M}$ along ${\displaystyle S}$ and obtain just one piece, ${\displaystyle N.}$ In that case there exists a closed simple curve ${\displaystyle \gamma }$ in ${\displaystyle M}$ intersecting ${\displaystyle S}$ at a single point. Let ${\displaystyle R}$ be the union of the two tubular neighborhoods of ${\displaystyle S}$ and ${\displaystyle \gamma .}$ The boundary ${\displaystyle \partial R}$ turns out to be a 2-sphere that cuts ${\displaystyle M}$ into two pieces, ${\displaystyle R}$ and the complement of ${\displaystyle R.}$ Since ${\displaystyle M}$ is prime and ${\displaystyle R}$ is not a ball, the complement must be a ball. The manifold ${\displaystyle M}$ that results from this fact is almost determined, and a careful analysis shows that it is either ${\displaystyle S^{2}\times S^{1}}$ or else the other, non-orientable, fiber bundle of ${\displaystyle S^{2}}$ over ${\displaystyle S^{1}.}$