Prime manifold

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). == Examples ==

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. == References ==