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Prime manifold

According to a theoremofHellmuth 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[edit]

Consider specifically 3-manifolds.

Irreducible manifold[edit]

A 3-manifold is irreducible if every 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. 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 reducible.

Prime manifolds[edit]

A connected 3-manifold ${\displaystyle M}$isprime 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 ${\displaystyle M}$).

Examples[edit]

Euclidean space[edit]

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.

Sphere, lens spaces[edit]

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).

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

Prime manifolds and irreducible manifolds[edit]

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

From irreducible to prime[edit]

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.

From prime to irreducible[edit]

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 neighborhoodsof${\displaystyle S}$ 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 bundleof${\displaystyle S^{2}}$ over ${\displaystyle S^{1}.}$

References[edit]

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

See also[edit]

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