# Prime manifold

In topology (a mathematical discipline) 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

Let us consider specifically 3-manifolds.

### Irreducible manifold

A 3-manifold is irreducible if any 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 reducible.

## 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).

## 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

$M=N_{1}\#N_{2},$ 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

$M=N_{1}\#N_{2}.$ ## Bibliography

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