Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 24 / 2

Work in the smooth or piecewise-linear category; embedded spheres and disks are locally flat. A closed connected orientable three-manifold is an irreducible three-manifold when every embedded $S^2$ bounds an embedded three-ball. It is a prime three-manifold when $M\cong P\#Q$ forces $P\cong S^3$ or $Q\cong S^3$. A separating sphere gives such a connected sum after capping its two sides with balls. Irreducibility therefore implies primeness, but a prime manifold can still have a nonseparating sphere.

A geometrizable three-manifold is one whose prime summands, after sphere splitting and capping, can be cut along disjoint essential tori into pieces whose interiors admit complete geometric structures modeled on the eight Thurston geometries:Here a geometric structure is a locally homogeneous quotient structure modeled on the stated simply connected space. The original manifold need not itself carry one geometry before decomposition.

A Heegaard splitting is $M=H_1{\cup }_{\varphi }{H}_2$, where $H_1,H_2$ are handlebodies with their boundaries identified. Its genus is the common boundary genus. The Heegaard genus $g_H(M)$ is the least possible genus of such a splitting. Thus “genus $g$” as a minimum requires both a genus-$g$ splitting and the nonexistence of one of smaller genus; an arbitrary genus-$g$ splitting only proves $g_H(M)\le g$.

The sphere $\{p\}\times S^2$ in $S^1\times S^2$ is nonseparating: its complement is $(S^1\setminus \{p\})\times S^2$, which is connected. A sphere bounding a ball separates, so this product is not an irreducible three-manifold.

To show it is prime, suppose a separating sphere cuts it into compact pieces $V,W$ with one spherical boundary each. The Seifert-van Kampen theorem givesA free product of two nontrivial groups is nonabelian: choose nonidentity elements in the two factors; their products in the two orders are distinct reduced words. Hence one factor, say ${\pi }_1(V)$, is trivial. The inclusion of this simply connected compact piece lifts to the universal cover $\mathbb{R}\times S^2\cong {\mathbb{R}}^3\setminus \{0\}$. The lift is an embedding: equality of lifted points would imply equality of their projections in $V$. Its image $\tilde{V}$ is compact with boundary the lifted sphere.

Use the three-dimensional Schoenflies theorem: a locally flat embedded sphere in ${\mathbb{R}}^3$ bounds a ball, and its compact side is that ball. The interior of $\tilde{V}$ is both open and closed in the corresponding complement component, and compactness rules out the unbounded component. Thus $\tilde{V}$ is the bounded ball, and $V$ itself is a ball. Capping $V$ gives $S^3$, so every connected-sum decomposition is trivial. Therefore $S^1\times S^2$ is prime but not irreducible.

Now let a prime closed connected orientable $M$ fail irreducibility. A sphere not bounding a ball cannot separate: if it did, primeness would make one capped side $S^3$, and the three-dimensional Schoenflies theorem would make its uncapped side a ball. Thus there is a nonseparating sphere $S$. Its two-sided neighborhood is $S^2\times [-1,1]$. Since the complement is connected, join its two boundary spheres by an embedded arc in that complement and thicken the arc. The union is a punctured $S^2\times S^1$: the thickened arc joins the two ends of the sphere-cylinder, and the remaining boundary is one sphere. This is the nonseparating sphere gives a sphere-circle summand construction and yieldsThe product is not $S^3$, since its fundamental group is $\mathbb{Z}$. Primeness forces $N\cong S^3$, soThe background results used are van Kampen, the explicit universal cover of the product, and the locally flat three-dimensional Schoenflies theorem; no Poincaré-conjecture argument is needed.

Let $D_{\alpha },D_{\beta }$ be the disjoint complete meridian-disk system of $H_1$, and let $D_{\gamma },D_{\delta }$ be that of $H_2$. The identification $\varphi (\alpha )=\gamma$ joins $D_{\alpha }$ and $D_{\gamma }$ into an embedded sphere $S$ in $M$. Each disk is nonseparating in its handlebody: cutting along it leaves a genus-one handlebody, and cutting along the other then gives a ball. After cutting $M$ along $S$, the two cut handlebodies are connected and are still joined along the complement of the common boundary circle. Hence $S$ is nonseparating.

Cap the two new sphere boundaries with balls. The resulting closed manifold $N$ inherits a genus-one Heegaard splitting: its two sides are the solid tori obtained by cutting the original handlebodies along their common disks. Their meridians are the surviving curves $\beta$ and $\delta$. The hypothesis that $\varphi (\beta )$ avoids $\gamma$ makes it descend to the compressed torus, and its single intersection with $\delta$ remains a single intersection there.

A genus-one splitting whose two meridians intersect once is $S^3$. To see this directly, choose the target torus basis $(\mu ,\lambda )$ with $\mu$ its meridian. The other meridian is $a\mu \pm \lambda$. A meridional Dehn twist, extending across the target solid torus, removes the integer $a$, leaving the longitude. Gluing a second solid torus with meridian equal to that longitude is the standard splitting of $S^3\subset {\mathbb{C}}^2$ into the neighborhoods of its two coordinate circles. Therefore $N\cong S^3$.

By the common nonseparating disks in a Heegaard splitting construction and part (b), reversing the cut and caps givesThe product metric on $S^2\times \mathbb{R}$ is homogeneous, and quotienting by an integer translation in the second factor gives $S^2\times S^1$. Thus $M$ carries this Thurston geometry and is a geometrizable three-manifold.

There is a wording qualification: if “Heegaard genus two” means the minimum defined in part (a), the disk data contradict it. The product splits as $S^1\times D^2$ on each hemisphere of $S^2$, giving a genus-one splitting, while genus zero gives only $S^3$, so $g_H(S^1\times S^2)=1$. The genuine construction is a genus-two splitting of a genus-one manifold; the geometrizability conclusion remains valid. The proof used the complete disk systems, the common-sphere cut-and-cap construction, and the explicitly justified genus-one meridian-intersection criterion.