Prime manifold

In topology, a branch of mathematics, a prime manifold is an n-manifold that cannot be expressed as a non-trivial connected sum of two n-manifolds. Non-trivial means that neither of the two is an n-sphere. A similar notion is that of an irreducible n-manifold, which is one in which any embedded (n − 1)-sphere bounds an embedded n-ball. Implicit in this definition is the use of a suitable category, such as the category of differentiable manifolds or the category of piecewise-linear manifolds.

A 3-manifold is irreducible if and only if it is prime, except for two cases: the product $$S^2 \times S^1$$ and the non-orientable fiber bundle of the 2-sphere over the circle $$S^1$$ are both prime but not irreducible. This is somewhat analogous to the notion in algebraic number theory of prime ideals generalizing Irreducible elements.

According to a theorem of Hellmuth Kneser and John Milnor, every compact, orientable 3-manifold is the connected sum of a unique (up to homeomorphism) collection of prime 3-manifolds.

Definitions
Consider specifically 3-manifolds.

Irreducible manifold
A 3-manifold is  if every smooth sphere bounds a ball. More rigorously, a differentiable connected 3-manifold $$M$$ is irreducible if every differentiable submanifold $$S$$ homeomorphic to a sphere bounds a subset $$D$$ (that is, $$S=\partial D$$) which is homeomorphic to the closed ball $$D^3 = \{x\in\R^3\ |\ |x|\leq 1\}.$$ The assumption of differentiability of $$M$$ is not important, because every topological 3-manifold has a unique differentiable structure. However it is necessary to assume that the sphere is smooth (a differentiable submanifold), even having a tubular neighborhood. The differentiability assumption serves to exclude pathologies like the Alexander's horned sphere (see below).

A 3-manifold that is not irreducible is called .

Prime manifolds
A connected 3-manifold $$M$$ is prime if it cannot be expressed as a connected sum $$N_1\# N_2$$ of two manifolds neither of which is the 3-sphere $$S^3$$ (or, equivalently, neither of which is homeomorphic to $$M$$).

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

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

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

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

Prime manifolds and irreducible manifolds
A 3-manifold is irreducible if and only if it is prime, except for two cases: the product $$S^2 \times S^1$$ and the non-orientable fiber bundle of the 2-sphere over the circle $$S^1$$ are both prime but not irreducible.

From irreducible to prime
An irreducible manifold $$M$$ is prime. Indeed, if we express $$M$$ as a connected sum $$M=N_1\#N_2,$$ then $$M$$ is obtained by removing a ball each from $$N_1$$ and from $$N_2,$$ and then gluing the two resulting 2-spheres together. These two (now united) 2-spheres form a 2-sphere in $$M.$$ The fact that $$M$$ is irreducible means that this 2-sphere must bound a ball. Undoing the gluing operation, either $$N_1$$ or $$N_2$$ 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 $$N_1$$ or $$N_2$$ was in fact a (trivial) 3-sphere, and $$M$$ is thus prime.

From prime to irreducible
Let $$M$$ be a prime 3-manifold, and let $$S$$ be a 2-sphere embedded in it. Cutting on $$S$$ one may obtain just one manifold $$N$$ or perhaps one can only obtain two manifolds $$M_1$$ and $$M_2.$$ In the latter case, gluing balls onto the newly created spherical boundaries of these two manifolds gives two manifolds $$N_1$$ and $$N_2$$ such that $$M = N_1\#N_2.$$ Since $$M$$ is prime, one of these two, say $$N_1,$$ is $$S^3.$$ This means $$M_1$$ is $$S^3$$ minus a ball, and is therefore a ball itself. The sphere $$S$$ 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 $$M$$ is irreducible.

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