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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 S^{1} and the twisted 2-sphere bundle over S^{1}.

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.

Consider specifically 3-manifolds.

A 3-manifold is **irreducible** if any smooth sphere bounds a ball. More rigorously, a differentiable connected 3-manifold is irreducible if every differentiable submanifold homeomorphic to a sphere bounds a subset (that is, ) which is homeomorphic to the closed ball

The assumption of differentiability of is not important, because every topological 3-manifold has a unique differentiable structure. The assumption that the sphere is

A 3-manifold that is not irreducible is called **reducible**.

A connected 3-manifold is **prime** if it cannot be expressed as a connected sum of two manifolds neither of which is the 3-sphere (or, equivalently, neither of which is homeomorphic to ).

Three-dimensional Euclidean space is irreducible: all smooth 2-spheres in it bound balls.

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

The 3-sphere is irreducible. The product space is not irreducible, since any 2-sphere (where is some point of ) has a connected complement which is not a ball (it is the product of the 2-sphere and a line).

A lens space with (and thus not the same as ) is irreducible.

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

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

then is obtained by removing a ball each from and from and then gluing the two resulting 2-spheres together. These two (now united) 2-spheres form a 2-sphere in The fact that is irreducible means that this 2-sphere must bound a ball. Undoing the gluing operation, either or 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 or was in fact a (trivial) 3-sphere, and is thus prime.

Let be a prime 3-manifold, and let be a 2-sphere embedded in it. Cutting on one may obtain just one manifold or perhaps one can only obtain two manifolds and In the latter case, gluing balls onto the newly created spherical boundaries of these two manifolds gives two manifolds and such that

Since is prime, one of these two, say is This means is minus a ball, and is therefore a ball itself. The sphere 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 is irreducible.

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

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