A Zoo of Three-Manifolds

Book 7A stated the Poincaré conjecture: a closed, simply connected 3-manifold is the 3-sphere. Book 10A asks the broader question that Thurston's geometrization conjecture answers: what are all the closed 3-manifolds, and what do they look like? The Ricci flow with surgery is a machine that takes any closed 3-manifold and returns pieces. This book describes the pieces, how they fit together and the geometries they carry, so that when the machine runs in Books 11A–12C you know what it must produce.

This first chapter is a zoo: the 3-sphere and its Hopf fibration, the 3-torus, $S^2\times S^1$, lens spaces and the other spherical space forms, the Poincaré homology sphere in three guises, and the Seifert–Weber space. Each will reappear as an example of a piece, a geometry or a fate under the flow.

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

• describe $S^3$ as the unit quaternions and as two solid tori, and compute the fibres of the Hopf fibration;
• build the 3-torus, $\mathbb{RP}^3$ and the lens spaces as quotients, and compute their fundamental groups;
• state which groups act freely on $S^3$, and recognise spherical space forms;
• construct the Poincaré homology sphere and the Seifert–Weber space by gluing the faces of a dodecahedron;
• explain how one could test whether the universe is a nontrivial 3-manifold.

1.1The shape of space

1.2The 3-sphere and the Hopf fibration

The unit sphere $S^3 = \{(z, w) \in \mathbb{C}^2 : |z|^2 + |w|^2 = 1\}$ has several faces (7A.2 Compactness and Compactification, 8A.5 Lie Groups and Group Actions): the one-point compactification of $\mathbb{R}^3$, via stereographic projection; the group $SU(2)$ of unit quaternions; and the union of two solid tori $\{|z| \leq |w|\}$ and $\{|z| \geq |w|\}$ glued along the torus $|z| = |w| = \frac{1}{\sqrt2}$, with the meridian of each glued to the longitude of the other.

The Hopf fibration is the map

Its fibres are the circles $\{(e^{i\theta}z, e^{i\theta}w)\}$: each is the intersection of $S^3$ with a complex line through the origin, hence a great circle (Exercise 1.1). So $S^3$ is a union of disjoint great circles, one for each point of $S^2$, and any two of them are linked once. Under stereographic projection to $\mathbb{R}^3$ the picture is famous (Figure 1.1): one fibre is a vertical line through infinity, one is the unit circle around it, and the rest fill nested tori, each torus made of circles that wind once around each way. The Berger spheres of 9A.5 Computing Curvature and 9B.3 Collapsing and Noncollapsing are $S^3$ with these circles shrunk.

1.3Flat, product and quotient examples

The 3-torus $T^3 = \mathbb{R}^3/\mathbb{Z}^3$, a cube with opposite faces glued. It has a flat metric and $\pi_1 = \mathbb{Z}^3$. Someone living inside a small 3-torus would see copies of themselves repeated in a cubic lattice in every direction (Figure 1.2), which is what the matched-circles test looks for on a cosmic scale.

$S^2\times S^1$, with $\pi_1 = \mathbb{Z}$. It is the prime manifold that is not irreducible (10A.3 The Prime Decomposition): the sphere $S^2\times\{\ast\}$ does not bound a ball. Its product metric is the neck geometry $S^2\times\mathbb{R}$ wrapped up.

Lens spaces. For coprime $p > q \geq 1$, the cyclic group generated by $(z, w) \mapsto (\zeta z, \zeta^qw)$, $\zeta = e^{2\pi i/p}$, acts freely on $S^3$ by isometries, and the quotient is the lens space $L(p, q)$, with $\pi_1 = \mathbb{Z}/p$ (7A.6 Covering Spaces). $L(2, 1) = \mathbb{RP}^3$. Lens spaces are classified by Reidemeister's theorem (1935, extended from piecewise-linear equivalence to homeomorphism by Brody in 1960): $L(p, q)$ and $L(p, q')$ are homeomorphic if and only if $q' \equiv \pm q^{\pm1} \pmod p$. Homotopy equivalence is weaker, $qq' \equiv \pm n^2 \pmod p$ for some $n$, so $L(7, 1)$ and $L(7, 2)$ are homotopy equivalent but not homeomorphic (Exercise 1.3): the fundamental group, even with homotopy type, does not determine a 3-manifold in general. The Poincaré conjecture is the claim that it does when the group is trivial.

Spherical space forms. More generally, a spherical space form is a quotient $S^3/\Gamma$ by a finite group $\Gamma \subset SO(4)$ acting freely. The possible groups were classified by Hopf (1926) and later work: cyclic groups, as for lens spaces, and products of cyclic groups with the binary dihedral, binary tetrahedral, binary octahedral and binary icosahedral groups (the preimages in $SU(2)$ of the rotation groups of regular solids), in suitable combinations. These are exactly the closed 3-manifolds with a metric of constant positive curvature (10A.5 Thurston’s Eight Geometries).

Mapping tori and knot complements. Gluing the ends of $\Sigma\times[0, 1]$ by a diffeomorphism $\phi$ of a surface gives the mapping torus of $\phi$, a 3-manifold fibring over the circle. Removing an open tubular neighbourhood of a knot in $S^3$ gives a knot complement, a compact 3-manifold with torus boundary; the trefoil and figure-eight complements are the first examples of 10A.4 Seifert Spaces and the JSJ Decomposition and 10A.6 Hyperbolic Three-Manifolds.

1.4Dodecahedral spaces

Take a solid regular dodecahedron and glue each face to the opposite face. A face is a pentagon, and the opposite face is rotated relative to it, so a twist is needed.

• Gluing with a $\frac{1}{10}$ turn ($36°$) gives the Poincaré homology sphere, also described as $S^3/I^*$, the quotient by the binary icosahedral group of order $120$ (7A.6 Covering Spaces), and as the link of the singularity $x^2 + y^3 + z^5 = 0$ (the Brieskorn sphere $\Sigma(2, 3, 5)$). Its fundamental group is perfect and nontrivial, so it has the homology of $S^3$ but is not $S^3$: the example that made Poincaré state his conjecture with $\pi_1$ instead of homology.
• Gluing with a $\frac{3}{10}$ turn ($108°$) gives the Seifert–Weber space (Weber and Seifert, 1933), which carries a hyperbolic metric.

The edges of the dodecahedron fall into classes under the gluing, and the angles around each glued edge must add up to $2\pi$. In the Poincaré gluing the edges come in classes of $3$, so the dihedral angle must be $120°$, slightly more than the Euclidean $116.57°$: the dodecahedron must be spherical, one of the $120$ cells of the regular 120-cell in $S^3$. In the Seifert–Weber gluing they come in classes of $5$, so the angle must be $72°$: the dodecahedron must be hyperbolic (Exercise 1.4). The gluing determines the geometry. This is the first glimpse of geometrization.

1.5History

Heinz Hopf introduced his fibration in 1931. Heinrich Tietze defined lens spaces in 1908; Kurt Reidemeister classified them in 1935. Poincaré constructed his homology sphere in 1904; its dodecahedral description is due to Weber and Seifert (1933), as is the hyperbolic space with the $\frac{3}{10}$ twist. Cornish, Spergel and Starkman proposed the circles-in-the-sky test in 1998; the Planck collaboration's topology analysis appeared in 2014 and 2016.

1.6Exercises

Solution

(a) $|2z\bar w|^2 + (|z|^2 - |w|^2)^2 = (|z|^2 + |w|^2)^2 = 1$. If $h$ agrees, then $|z| = |z'|$, $|w| = |w'|$ and $z\bar w = z'\bar w'$, which forces a common phase. (b) The fibre is $S^3\cap\{\lambda(z, w) : \lambda \in \mathbb{C}\}$, the unit circle of a real 2-plane. (c) Writing $w = x_3 + ix_4$ and projecting from $x_4 = 1$, $\{w = 0\}$ is the unit circle $x_1^2 + x_2^2 = 1$ in $x_3 = 0$, and $\{z = 0\}$ is the circle $x_3^2 + x_4^2 = 1$, which projects to the $x_3$-axis together with the point at infinity. The axis passes through the disc bounded by the unit circle once.

Solution

$T^3 = \mathbb{R}^3/\mathbb{Z}^3$: $\pi_1 = \mathbb{Z}^3$, cover $\mathbb{R}^3$. $S^2\times S^1$: cover $S^2\times\mathbb{R}$, deck group $\mathbb{Z}$. $L(p, q) = S^3/(\mathbb{Z}/p)$: $\pi_1 = \mathbb{Z}/p$, cover $S^3$. $\mathbb{RP}^3 = L(2, 1)$: $\mathbb{Z}/2$, cover $S^3$. All universal covers are simply connected; only $S^3$ is compact among $\mathbb{R}^3$, $S^2\times\mathbb{R}$, $S^3$.

Solution

For $p = 5$: $\pm1^{\pm1} = \{1, 4\}$, which does not contain $2$; and $3 \equiv -2$, so $L(5, 2) \cong L(5, 3)$. For $p = 7$: $1\cdot2 = 2$, and the squares mod $7$ are $1, 2, 4$, so $2 = 3^2 \bmod 7$; but $\pm1^{\pm1} = \{1, 6\} \not\ni 2$, so they are not homeomorphic.

Solution

(a) Poincaré: $5 - 10 + 6 - 1 = 0$. Seifert–Weber: $1 - 6 + 6 - 1 = 0$. (b) Around a glued edge class the dihedral angles must sum to $2\pi$: with $3$ edges each angle is $120°$, with $5$ edges $72°$. Spherical polyhedra have larger angles than Euclidean ones of the same shape and hyperbolic ones smaller, and the angle of a regular dodecahedron varies continuously with its size, from the Euclidean value at small size: up to $120°$ (and beyond) in $S^3$, down to $72°$ (and below) in $\mathbb{H}^3$.

Solution

$h(1, \varepsilon v) = (2\varepsilon\bar v, 1 - \varepsilon^2|v|^2)$ to first order, so $dh(0, v) = (2\bar v, 0)$, of length $2|v|$; also $dh(i, 0) = 0$, since $h$ is constant on fibres. So $\frac12h$ maps $S^3$ to $S^2(\frac12)$, kills the fibre direction, and is an isometry on horizontal vectors. $U(2)$ acts transitively on $S^3$ preserving the fibration and the metrics, so the same holds at every point.