Prime manifold#Irreducible manifold

Consider specifically 3-manifolds.

A 3-manifold is {{visible anchor|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, $S=\backslash partial\; D$) which is homeomorphic to the closed ball

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

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 {{visible anchor|reducible}}.

A connected 3-manifold $M$ is prime if it cannot be expressed as a connected sum $N\_1\backslash \#\; N\_2$ of two manifolds neither of which is the 3-sphere $S^3$ (or, equivalently, neither of which is homeomorphic to $M$).

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

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

The 3-sphere $S^3$ is irreducible. The product space $S^2\; \backslash times\; S^1$ is not irreducible, since any 2-sphere $S^2\; \backslash times\; \backslash \{pt\backslash \}$ (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 $p\backslash neq\; 0$ (and thus not the same as $S^2\; \backslash times\; S^1$) is irreducible.

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

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

$$M=N\_1\backslash \#N\_2,$$

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.

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

$$M\; =\; N\_1\backslash \#N\_2.$$

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 $\backslash gamma$ in $M$ intersecting $S$ at a single point. Let $R$ be the union of the two tubular neighborhoods of $S$ and $\backslash gamma.$ The boundary $\backslash partial\; R$ 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 $S^2\; \backslash times\; S^1$ or else the other, non-orientable, fiber bundle of $S^2$ over $S^1.$

{{reflist|group=note}}

• {{cite book|author=William Jaco|author-link=William Jaco|title=Lectures on 3-manifold topology|isbn=0-8218-1693-4}}

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

{{Manifolds}}

Category:Manifolds