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Prime manifold - Wikipedia

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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.

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. This is somewhat analogous to the notion in algebraic number theory of prime ideals generalizing Irreducible elements.

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.

Irreducible manifold

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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.

A connected 3-manifold {\displaystyle M} is prime 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}).

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

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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).

A lens 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

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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

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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

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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 neighborhoods of {\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 bundle of {\displaystyle S^{2}} over {\displaystyle S^{1}.}