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.

The notions of irreducibility in algebra and manifold theory are related. An irreducible manifold is prime, although the converse does not hold. From an algebraist's perspective, prime manifolds should be called "irreducible"; however the topologist (in particular the 3-manifold topologist) finds the definition above more useful. The only compact, connected 3-manifolds that are prime but not irreducible are the trivial 2-sphere bundle over the circle S¹ and the twisted 2-sphere bundle over S¹.

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 irreducible 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

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

The assumption of differentiability of $M$ 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.

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 $M$).

Examples

Euclidean space

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

On the other hand, Alexander's horned sphere is a non-smooth sphere in $\mathbb {R} ^{3}$ 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 $S^{1}.$

References

• William Jaco. Lectures on 3-manifold topology. ISBN 0-8218-1693-4.

See also

• 3-manifold
• Connected sum
• Prime decomposition (3-manifold)