The Prime Decomposition

Integers factor uniquely into primes. Closed orientable 3-manifolds do too, with the connected sum in place of multiplication and the 3-sphere in place of $1$. This is the first of the two canonical ways of cutting a 3-manifold into simpler pieces. It cuts along 2-spheres, and its pieces are the prime manifolds.

The prime decomposition matters directly for the proof. A Ricci flow on a 3-manifold develops necks, regions that look like $S^2\times$ an interval, and Perelman's surgery cuts along the central 2-sphere of each neck and caps the two sides with balls (12B.4 Surgery). Topologically, that is undoing a connected sum. So the flow with surgery carries out a prime decomposition by geometry. If the manifold is simply connected, every prime piece is simply connected, which is where the Poincaré conjecture enters (12C.1 Reading Off the Topology).

By the end of this chapter you will be able to:

• define the connected sum and explain why $S^3$ is its identity;
• define prime and irreducible manifolds, and show that $S^2\times S^1$ is the only prime orientable manifold that is not irreducible;
• state Alexander's theorem and the Kneser–Milnor prime decomposition theorem, and sketch why it holds;
• compute the fundamental group of a connected sum;
• state the sphere theorem, Dehn's lemma and the loop theorem, and use the sphere theorem to recognise aspherical manifolds.

3.1Connected sums

Given closed connected oriented 3-manifolds $M_1, M_2$, remove an open ball from each and glue the two boundary spheres by an orientation-reversing homeomorphism. The result is the connected sum $M_1\#M_2$. It does not depend on the choices of balls or gluing map, because any two orientation-preserving embeddings of a ball are isotopic and every orientation-preserving homeomorphism of $S^2$ is isotopic to the identity. Orientation matters: in general $M_1\#M_2$ and $M_1\#\overline{M_2}$ (with $M_2$'s orientation reversed) can be different manifolds. The operation is commutative and associative, and $M\#S^3 = M$, since removing a ball from $S^3$ leaves a ball.

Geometrically, $M_1\#M_2$ is $M_1$ and $M_2$ joined by a thin tube, a neck $S^2\times[0, 1]$, and the 2-sphere in the middle of the neck separates the two summands (Figure 3.1).

3.2Prime and irreducible manifolds

A closed orientable 3-manifold $M$ is prime if $M = P\#Q$ implies $P = S^3$ or $Q = S^3$. It is irreducible if every embedded 2-sphere in $M$ bounds a ball. (Spheres here are smooth or piecewise-linear; wild topological spheres are excluded.)

An irreducible manifold is prime: a decomposition $M = P\#Q$ produces a separating sphere, which bounds a ball on one side, so that side's summand is $S^3$. The converse almost holds.

$S^2\times S^1$ is prime: in any decomposition one summand would have to be simply connected and contain the non-separating sphere, which leads to a contradiction (Exercise 3.6). It is not irreducible: $S^2\times\{\ast\}$ does not bound a ball, since it does not even separate (Figure 3.2).

James Waddell Alexander proved this in 1924, and in the same year constructed the horned sphere, a topological embedding of $S^2$ whose outside is not simply connected, showing that the smoothness hypothesis is needed. The proof (Hatcher, chapter 1) puts the sphere in general position with respect to horizontal planes and cuts it along the circles of intersection, reducing to simpler spheres by induction.

3.3The prime decomposition theorem

Existence (Hellmuth Kneser, 1929). Splitting off summands one at a time could a priori go on forever, so the point is a finiteness theorem: in a fixed triangulation of $M$, a family of disjoint spheres, no two cobounding a product region $S^2\times I$ and none bounding a ball, has a number of members bounded in terms of the number of tetrahedra. Kneser proved this by putting the spheres in normal position with respect to the triangulation, so that each meets each tetrahedron in triangles and quadrilaterals, of which there are only a few types.

Uniqueness (John Milnor, 1962). Given two systems of spheres defining two decompositions, one makes them disjoint by cutting and pasting along circles of intersection, using irreducibility of the summands and Alexander's theorem; the remaining bookkeeping is where the special role of $S^2\times S^1$ appears.

Two consequences used later:

• $\pi_1(M_1\#M_2) = \pi_1(M_1) * \pi_1(M_2)$, the free product (Exercise 3.4). So $\pi_1$ of a connected sum is trivial only if both summands' groups are trivial.
• If $M$ is simply connected, every prime summand is simply connected. Given the Poincaré conjecture each summand is $S^3$, so $M = S^3$. Conversely, the Poincaré conjecture reduces to prime (and so irreducible) simply connected manifolds.

3.4Spheres, discs and loops

Three theorems from the 1950s let one find embedded surfaces from algebraic information. They were proved by Christos Papakyriakopoulos in 1957.

• Dehn's lemma. If a loop on the boundary of $M$ bounds a singular disc in $M$ (a continuous map of a disc, possibly with self-intersections, but embedded near the boundary), then it bounds an embedded disc.
• The loop theorem. If the map $\pi_1(\partial M) \to \pi_1(M)$ has a nontrivial kernel, then some essential simple closed curve on $\partial M$ bounds an embedded disc in $M$.
• The sphere theorem. If $\pi_2(M) \neq 0$ (for $M$ orientable), then $M$ contains an embedded 2-sphere that is essential: it is nonzero in $\pi_2(M)$.

The sphere theorem gives a clean criterion: an irreducible $M$ has $\pi_2(M) = 0$. If in addition $\pi_1(M)$ is infinite, then $M$ is aspherical: its universal cover is contractible (Exercise 3.8). Every closed hyperbolic manifold is of this kind (Cartan–Hadamard, 9A.7 Jacobi Fields and Curvature versus Topology), and so are most of the pieces of 10A.4 Seifert Spaces and the JSJ Decomposition.

3.5History

Alexander's theorem and the horned sphere date from 1924. Kneser proved the existence of prime decompositions in 1929; Milnor proved uniqueness, and isolated the role of $S^2\times S^1$, in 1962. Max Dehn stated his lemma in 1910, with a gap found by Kneser in 1929; Papakyriakopoulos proved it, with the loop theorem and the sphere theorem, in 1957. Wolfgang Haken developed normal surface theory into an algorithmic tool in the 1960s.

3.6Exercises

Solution

The two pieces overlap in $S^2\times(-\varepsilon, \varepsilon)$, which is simply connected, so van Kampen gives the free product of $\pi_1(M_i\setminus B)$. Removing a ball doesn't change $\pi_1$: by van Kampen again, $M_i = (M_i\setminus B)\cup B$ with overlap $S^2\times(-\varepsilon, \varepsilon)$, simply connected, so $\pi_1(M_i) = \pi_1(M_i\setminus B) * \pi_1(B) = \pi_1(M_i\setminus B)$.

Solution

The natural maps $A \to A * B$ and $B \to A * B$ are injective (normal form for free products), so if $A * B = 1$ then $A = B = 1$. By induction over the summands, every prime summand of a simply connected $M$ is simply connected. Each is then $S^3$ by the conjecture, and $S^3\#\cdots\#S^3 = S^3$.

Solution

$\mathbb{Z} = \pi_1(P) * \pi_1(Q)$. A free product of two nontrivial groups is nonabelian (nontrivial elements $a$, $b$ of the two factors satisfy $ab \neq ba$ by the normal form), and $\mathbb{Z}$ is abelian, so one factor is trivial. A simply connected $Q$ need not be $S^3$ without the conjecture. Milnor's argument: isotope the separating sphere $\Sigma$ of the sum off the non-separating sphere $S = S^2\times\{\ast\}$ by cutting along circles of intersection. Then $\Sigma$ lies in the complement $S^2\times(0, 1)$, a shell in $\mathbb{R}^3$, so by Alexander's theorem it bounds a ball in $\mathbb{R}^3$. The ball cannot contain the inner boundary sphere, since then $\Sigma$ would be parallel to $S$ and would not separate $S^2\times S^1$; so the ball lies in the shell, and the summand it represents is $S^3$.

Solution

$\pi_1 = \mathbb{Z}/2 * \mathbb{Z}/2$, which contains the infinite-order product of the two generators. Both maps are isometric involutions of $S^2\times\mathbb{R}$ without fixed points ($-x \neq x$ on $S^2$). Their composition $(x, t) \mapsto (x, t + 2)$ generates an index-2 subgroup $\mathbb{Z}$, so the quotient is double covered by $S^2\times S^1$. Each involution, restricted to a neighbourhood of its fixed slice $t = 0$ or $t = 1$, gives a twisted $I$-bundle over $\mathbb{RP}^2$, which is $\mathbb{RP}^3$ minus a ball; two of them glued along their boundary spheres form $\mathbb{RP}^3\#\mathbb{RP}^3$.

Solution

(a) If $\pi_2(M) \neq 0$, the sphere theorem gives an essential embedded sphere; by irreducibility it bounds a ball, hence is null-homotopic, a contradiction. Covering maps induce isomorphisms on $\pi_2$. (b) $\tilde M$ is simply connected with $H_2 = \pi_2 = 0$ (Hurewicz), $H_3 = 0$ (noncompact), and no higher homology; by Hurewicz inductively $\pi_k = H_k = 0$ for all $k$, and a CW complex with all homotopy groups trivial is contractible. (c) A closed hyperbolic manifold is $\mathbb{H}^3/\Gamma$; a sphere in it lifts to a sphere in $\mathbb{H}^3 \cong \mathbb{R}^3$, which bounds a ball by Alexander's theorem, and that ball projects to a ball (a standard argument with the covering action), so $M$ is irreducible; its $\pi_1 = \Gamma$ is infinite.