Finite Rotation Groups and Spherical Space Forms

The Ricci flow ends, for the manifolds in the Poincaré conjecture, at quotients of the round 3-sphere. Hamilton's theorem (11A.11 Hamilton’s 1982 Theorem) turns positive Ricci curvature into constant positive curvature, and a closed manifold of constant curvature $1$ is $S^3/\Gamma$ for a finite group $\Gamma$ of rotations acting freely (10A.1 A Zoo of Three-Manifolds). In Perelman's argument the components that disappear during the flow are recognised as such quotients, or as $S^2\times S^1$ and its relatives, by their geometry just before they disappear (12B.13 Ricci Flow with Surgery for All Time, 12C.1 Reading Off the Topology). And the elliptization theorem, which Perelman's work proves, says that every closed 3-manifold with finite fundamental group is one of them.

So these quotients are the endpoints of the whole story, and this chapter determines them. The route is classical and every step can be done by hand. First, the finite groups of rotations of ordinary space: Felix Klein's list of cyclic, dihedral, tetrahedral, octahedral and icosahedral groups, proved by counting the poles of rotations. Second, their preimages in the unit quaternions, the binary groups, which are all the finite subgroups of $S^3$. Third, the quotients of $S^3$ by those groups and by the more general groups acting freely, with the complete list of the resulting manifolds. Along the way the orbifold Euler characteristic of 10A.7 Seifert Spaces and the JSJ Decomposition reappears as an accounting identity, and the uniqueness of an involution from 10A.1 A Zoo of Three-Manifolds becomes a constraint on groups.

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

• prove that every finite group of rotations of the plane is cyclic;
• prove Klein's classification of the finite groups of rotations of space by counting poles, and identify the groups with the symmetry groups of regular solids;
• show that the finite subgroups of the unit quaternions are the cyclic groups and the binary dihedral, tetrahedral, octahedral and icosahedral groups;
• state the classification of spherical 3-manifolds, recognise their fundamental groups, and know which are determined by their fundamental group;
• compute volumes, diameters and the Ricci flow of the constant-curvature metrics on $S^3/\Gamma$;
• explain why these manifolds are the endpoints of the Ricci flow in Hamilton's and Perelman's theorems.

2.1What we are looking for

Recall from 10A.1 A Zoo of Three-Manifolds the description of the rotations of the 3-sphere. Points of $S^3$ are unit quaternions, and every element of $SO(4)$ is a map $x \mapsto ax\bar b$ for a pair of unit quaternions $(a, b)$, determined up to the simultaneous sign change $(a, b) \mapsto (-a, -b)$. Such a rotation has a fixed point on $S^3$ exactly when $\operatorname{Re}a = \operatorname{Re}b$. A spherical space form is a quotient $S^3/\Gamma$ with $\Gamma \subset SO(4)$ finite and acting freely; it is a closed orientable 3-manifold with a metric of constant curvature $1$, and its fundamental group is $\Gamma$ (a free action of a finite group on the simply connected $S^3$ is a covering, 7A.7 Covering Spaces).

Two questions organise the chapter.

1. Which finite groups $\Gamma$ occur?
2. For a given group, how many different manifolds arise?

The simplest case already contains most of the ideas: take $b = 1$. Then $\Gamma$ is a finite group $G$ of unit quaternions acting by left multiplication, which is always free, since $\operatorname{Re}g < 1 = \operatorname{Re}1$ for $g \neq 1$. So the first task is to find all finite subgroups of $S^3$. Because $S^3$ double covers $SO(3)$, that reduces to finding the finite subgroups of $SO(3)$, and that is a problem about rotations of ordinary space that Klein solved in 1884.

2.2Finite rotation groups of the plane

The same argument shows that a finite group of rotations of space about a fixed axis is cyclic, since such rotations form a copy of $SO(2)$. That is the local ingredient in the next section.

2.3Klein's classification

Every rotation $g \neq \operatorname{id}$ of $\mathbb{R}^3$ is a rotation about an axis through the origin by an angle in $(0, 2\pi)$ (Euler's theorem: an element of $SO(3)$ has eigenvalue $1$, since its characteristic polynomial is real of odd degree and the product of its eigenvalues is $1$). So $g$ fixes exactly two points of the unit sphere $S^2$, where the axis meets it. Call them the poles of $g$.

Let $G \subset SO(3)$ be a finite group of order $N \geq 2$, and let $P \subset S^2$ be the set of all poles of nonidentity elements of $G$. Two facts get the argument started.

• $G$ permutes $P$. If $p$ is a pole of $g$ and $h \in G$, then $hgh^{-1}$ fixes $hp$, and $hgh^{-1} \neq \operatorname{id}$, so $hp$ is a pole.
• Stabilisers are cyclic. For $p \in P$, the stabiliser $G_p = \{g \in G : gp = p\}$ consists of rotations about the axis through $p$, so it is cyclic by Proposition 2.1, of some order $n_p \geq 2$. The orbit of $p$ has $\frac{N}{n_p}$ points (orbit–stabiliser), and all points of one orbit have stabilisers of the same order (they are conjugate).

An accounting identity. The equation $2 - \frac2N = \sum(1 - \frac{1}{n_i})$ is the orbifold Euler characteristic of 10A.7 Seifert Spaces and the JSJ Decomposition in disguise. The quotient $S^2/G$ is a sphere with cone points of orders $n_i$ at the images of the pole orbits, and its orbifold Euler characteristic is $\chi^{\mathrm{orb}} = 2 - \sum(1 - \frac{1}{n_i})$. The identity says

which is the multiplicativity of the orbifold Euler characteristic under the "covering" $S^2 \to S^2/G$. The classification is thus the list of spherical 2-orbifolds that are quotients of the sphere: $S^2(N, N)$, $S^2(2, 2, m)$, $S^2(2, 3, 3)$, $S^2(2, 3, 4)$, $S^2(2, 3, 5)$. These five bases will reappear as the bases of the Seifert fibrations of spherical 3-manifolds.

2.4The finite subgroups of the 3-sphere

The double cover $\rho : S^3 \to SO(3)$, $\rho(q)(v) = qv\bar q$ on pure quaternions, has kernel $\{\pm1\}$ (7A.7 Covering Spaces). The preimage of a finite group $G \subset SO(3)$ of order $N$ is a group $G^* = \rho^{-1}(G)$ of order $2N$, the binary version of $G$. In this way the cyclic groups give cyclic groups, $D_m$ gives the binary dihedral group $D^*_{4m}$ of order $4m$, and $T$, $O$, $I$ give the binary tetrahedral, octahedral and icosahedral groups $T^*$, $O^*$, $I^*$ of orders $24$, $48$ and $120$.

Note that the binary groups are genuinely different from the rotation groups they cover. $D^*_8$, for $m = 2$, is the quaternion group $Q_8 = \{\pm1, \pm i, \pm j, \pm k\}$, not the Klein four-group times $\mathbb{Z}/2$: it has a unique element of order $2$, as the lemma requires, while $\mathbb{Z}/2\times\mathbb{Z}/2\times\mathbb{Z}/2$ has seven. In general $D^*_{4m}$ is generated by $x = e^{i\pi/m}$ (of order $2m$) and $y = j$, with $y^2 = x^m = -1$ and $yxy^{-1} = \bar x = x^{-1}$; the binary groups are central extensions of the rotation groups by $\mathbb{Z}/2$ that do not split.

The first family of space forms. Every finite $H \subset S^3$, acting by left multiplication, gives a spherical space form $S^3/H$ with fundamental group $H$:

$H$	order	$S^3/H$
cyclic $\mathbb{Z}/p$	$p$	the lens space $L(p, 1)$
$D^*_{4m}$	$4m$	a prism manifold
$T^*$	$24$	the binary tetrahedral space
$O^*$	$48$	the binary octahedral space
$I^*$	$120$	the Poincaré homology sphere

For the cyclic case, writing $x = z + wj$ with $z, w \in \mathbb{C}$, left multiplication by $e^{2\pi i/p}$ is $(z, w) \mapsto (e^{2\pi i/p}z, e^{2\pi i/p}w)$, which is the generator of $L(p, 1)$ in 10A.1 A Zoo of Three-Manifolds. That $S^3/I^*$ is the Poincaré homology sphere, the dodecahedral space of 10A.1 A Zoo of Three-Manifolds, was shown in the early 1930s (Threlfall and Seifert; Weber and Seifert): the $120$ elements of $I^*$ move a point of $S^3$ to the centres of the $120$ dodecahedral cells of the regular $120$-cell, and the fundamental domain of the action is one of those dodecahedra.

2.5All spherical space forms

General groups $\Gamma \subset SO(4)$ use both factors of $S^3\times S^3$. The criterion of 10A.1 A Zoo of Three-Manifolds says which ones act freely: no nonidentity element $\pm(a, b)$ may have $\operatorname{Re}a = \operatorname{Re}b$. The lens spaces $L(p, q)$ with $q \neq \pm1$ are the first examples: they come from the cyclic group generated by $(a, b) = (e^{i\alpha}, e^{i\beta})$ with $\alpha - \beta = \frac{2\pi}{p}$ and $\alpha + \beta = \frac{2\pi q}{p}$ (10A.1 A Zoo of Three-Manifolds). The full classification was completed by Hopf (1925–26), Seifert and Threlfall (1930s), and later work on free actions; the result is as follows (statement; Scott's survey gives it in the language used here).

Two remarks put the list in perspective.

• Why the extensions are central, and by even cyclic groups. Every noncyclic group in the list has even order, so it contains an involution, which must be $-\operatorname{id}$ (the element $(1, -1)$, acting as $x \mapsto -x$), and $-\operatorname{id}$ is central. The coprimality conditions are exactly what the fixed-point criterion $\operatorname{Re}a \neq \operatorname{Re}b$ demands of products of elements from the two factors. For example, in $I^*\times\mathbb{Z}/m$, realised by left multiplication by $I^*$ and right multiplication by $m$-th roots of unity in a circle subgroup, an element $(a, \zeta)$ with $a \in I^*$ of order $k$ and $\zeta$ of order $m$ could have a fixed point only if $a$ and $\zeta$ were conjugate, hence of the same order, which coprimality of $m$ with $|I^*|$'s element orders $1, 2, 3, 4, 5, 6, 10$ excludes except in trivial cases.
• The constraint from involutions. By 10A.1 A Zoo of Three-Manifolds, a group acting freely and orthogonally on $S^3$ has at most one element of order $2$, which is central. So, for example, the dihedral groups $D_m$ themselves (with $m$ involutions), the symmetric group $S_3$, and $\mathbb{Z}/2\times\mathbb{Z}/2$ never occur. John Milnor (1957) proved that in any free action of a finite group on any sphere, by homeomorphisms, every element of order $2$ is central; his condition is the topological shadow of the lemma above.

2.6Geometry of the space forms

The round metric of $S^3$ descends to $S^3/\Gamma$, with curvature $1$. Its basic invariants are determined by $\Gamma$.

• Volume. $\operatorname{Vol}(S^3/\Gamma) = \frac{2\pi^2}{|\Gamma|}$, since $S^3 \to S^3/\Gamma$ is a $|\Gamma|$-sheeted local isometry and $\operatorname{Vol}S^3 = 2\pi^2$. The Poincaré sphere has volume $\frac{2\pi^2}{120} = \frac{\pi^2}{60}$.
• Injectivity radius. At a point $x$, the distance to the nearest translate $\gamma x$, $\gamma \neq 1$, is the length of the shortest geodesic loop at $x$, and $\operatorname{inj}(x)$ is half of it or $\pi$, whichever is smaller (on the sphere itself the injectivity radius is $\pi$, 9A.3 Geodesics and the Exponential Map). For left multiplication by $H$, $d(x, gx) = d(1, x^{-1}gx)$ is the angle $\theta$ with $\cos\theta = \operatorname{Re}g$, independent of $x$: the injectivity radius is constant, equal to half the smallest angle of a nonidentity element. For $L(p, 1)$ it is $\frac{\pi}{p}$.
• Diameter. Bounded below in terms of the volume: a manifold of diameter $D$ is a ball of radius $D$ around any of its points, whose volume in curvature $1$ is at most $\operatorname{Vol}B_{S^3}(D) = \pi(2D - \sin2D)$ for $D \leq \pi$ (Bishop's inequality, 9B.2 Volume Comparison). But small volume does not force small diameter. The lens spaces $L(p, 1)$ have volume $\frac{2\pi^2}{p} \to 0$, yet their diameter is at least $\frac\pi2$: the group $\mathbb{Z}/p$ acts along the Hopf circles, so the Hopf map descends to a Riemannian submersion $L(p, 1) \to S^2(\frac12)$, which does not increase distances and is onto. These lens spaces are collapsed: curvature $1$, bounded diameter, volume tending to zero, exactly as for the Berger spheres of 9B.3 Collapsing and Noncollapsing.

The last point matters for the flow. The lens spaces $L(p, 1)$ with $p$ large are thin in the Hopf direction: their injectivity radius $\frac\pi p$ is small while their curvature is $1$. They are $\kappa$-noncollapsed only with $\kappa$ of order $\frac1p$, and this is consistent with Perelman's theorem, whose $\kappa$ depends on the initial metric (12A.4 κ-Noncollapsing).

2.7The flow on a space form

The Ricci flow of the round metric on $S^3/\Gamma$ is the quotient of the flow on $S^3$, because the flow on the cover is invariant under the deck transformations (isometries are preserved, 11A.13 Homogeneous Flows): $g(t) = (1 - 4t)g_{\mathrm{round}}$, extinct at $t = \frac14$, uniformly round all the way. More is true. Every metric on $S^3/\Gamma$ with positive Ricci curvature flows, after normalisation, to a round metric (Hamilton's theorem, 11A.11 Hamilton’s 1982 Theorem). And in Perelman's flow with surgery, a component that becomes extinct with positive curvature is recognised, at a time just before extinction, as close to round after rescaling (an $\varepsilon$-round component) or as a compact component of bounded positive curvature diffeomorphic to $S^3$ or $\mathbb{RP}^3$ (a $C$-component), and so as a spherical space form (12B.5 The Canonical Neighbourhood Theorem).

For the Poincaré conjecture only one entry of the classification is needed, $\Gamma = 1$. The rest of the list is what the flow produces in general, and it is the content of the elliptization theorem.

Combined with Theorem 2.5, elliptization answers the first question of this chapter for all closed 3-manifolds: the finite groups that occur as fundamental groups of closed 3-manifolds are exactly the groups in the list. A purely algebraic consequence: a finite group with two commuting elements of order $2$ generating a Klein four-group, or with a noncentral involution, is never the fundamental group of a closed 3-manifold.

2.8History

Felix Klein classified the finite rotation groups in his Lectures on the Icosahedron (1884), building on the study of regular solids that goes back to Euclid. Heinz Hopf studied the spherical space forms in 1925–26, and Herbert Seifert and William Threlfall completed their classification in dimension three in the early 1930s, identifying the Poincaré sphere with $S^3/I^*$. Georges de Rham, Reidemeister and Brody contributed to the classification of lens spaces (1935–60). Milnor's paper on free actions of finite groups on spheres appeared in 1957. Perelman's work proved elliptization in 2003.

2.9Exercises

Solution

Face centres: $6$ points, stabiliser of order $4$ (quarter turns about the axis through opposite faces). Edge midpoints: $12$ points, stabiliser of order $2$ (half-turns). Vertices: $8$ points, stabiliser of order $3$ (third-turns about a long diagonal). Each orbit has $\frac{24}{n_i}$ points, as it should: $\frac{24}{4} = 6$, $\frac{24}{2} = 12$, $\frac{24}{3} = 8$. The identity: $2 - \frac{2}{24} = \frac{23}{12}$, and $(1 - \frac14) + (1 - \frac12) + (1 - \frac13) = \frac34 + \frac12 + \frac23 = \frac{9 + 6 + 8}{12} = \frac{23}{12}$. As a count of nonidentity elements: $3$ axes through faces with $3$ nonidentity rotations each ($9$), $6$ edge axes with $1$ each ($6$), $4$ diagonals with $2$ each ($8$): $9 + 6 + 8 = 23 = N - 1$.

Solution

One orbit would give $2 - \frac2N = 1 - \frac1n < 1$, but $2 - \frac2N \geq 1$ for $N \geq 2$: there are too few fixed-point incidences for the number of elements. Four orbits would give a right side at least $4\cdot\frac12 = 2$, more than the left side, which is less than $2$: too many incidences. For $SO(2)$ the count $2(N - 1)$ is infinite, and the poles are just two points with infinite stabilisers; the identity, divided by $N$, has no meaning. The finiteness of $G$ is used to count.

Solution

$\rho(i)$ is the rotation $v \mapsto iv\bar i$, which fixes $i$ and sends $j \mapsto iji^{-1} = -j$, $k \mapsto -k$: the half-turn about the $i$-axis. Similarly for $j$ and $k$, and $\rho(\pm1) = \operatorname{id}$. So $\rho(Q_8) = V$ and $Q_8 = \rho^{-1}(V)$ (both have order $8$). $Q_8$ has six elements of order $4$ ($\pm i, \pm j, \pm k$) and one of order $2$; $(\mathbb{Z}/2)^3$ has seven of order $2$, and $\mathbb{Z}/4\times\mathbb{Z}/2$ has three of order $2$; and $Q_8$ is not abelian ($ij = k = -ji$). Its commutator subgroup is $\{\pm1\}$ (generated by $iji^{-1}j^{-1} = -1$), so $Q_8^{\mathrm{ab}} = Q_8/\{\pm1\} \cong V \cong \mathbb{Z}/2\times\mathbb{Z}/2$, and $H_1(S^3/Q_8) = \pi_1^{\mathrm{ab}} = \mathbb{Z}/2\times\mathbb{Z}/2$ (7A.9 Homology in Brief).

Solution

$\{\pm1\}$ acts by $x \mapsto -x$, the antipodal map, whose quotient is $\mathbb{RP}^3$; as $L(p, q)$ with $p = 2$ the generator is $(z, w) \mapsto (-z, -w)$, the same map. The map $\rho : S^3 \to SO(3)$ is a surjective homomorphism with kernel $\{\pm1\}$ and a local diffeomorphism (a 2-sheeted covering, 7A.7 Covering Spaces), so it descends to a bijective local diffeomorphism $S^3/\{\pm1\} \to SO(3)$, a diffeomorphism. As $S^3$ is simply connected and the covering has two sheets, $\pi_1(SO(3)) \cong \{\pm1\} \cong \mathbb{Z}/2$.

Solution

$\mathbb{Z}/2\times\mathbb{Z}/2$ has three elements of order $2$, but a group acting freely by orthogonal maps of $S^3$ has at most one: an orthogonal involution without fixed points has no eigenvalue $1$, so it is $-\operatorname{id}$ (10A.1 A Zoo of Three-Manifolds). By elliptization, a closed 3-manifold with finite fundamental group is a spherical space form, so no closed 3-manifold has fundamental group $\mathbb{Z}/2\times\mathbb{Z}/2$.

Solution

(a) $\operatorname{Ric}(g_0) = 2g_0$, so $g(t) = (1 - 4t)g_0$ (10A.10 Geometrization and Ricci Flow), extinct at $t = \frac14$. (b) Volumes scale like $(\text{length})^3$, and lengths by $\sqrt{1 - 4t}$, so $\operatorname{Vol} = \frac{2\pi^2}{|\Gamma|}(1 - 4t)^{3/2}$. Its derivative is $-6(1 - 4t)^{1/2}\frac{2\pi^2}{|\Gamma|}$. On the other hand $R(g(t)) = \frac{6}{1 - 4t}$, constant in space, so $-\int R\,dV = -\frac{6}{1 - 4t}\cdot\frac{2\pi^2}{|\Gamma|}(1 - 4t)^{3/2}$, the same. (c) The injectivity radius of $g_0$ on $L(p, 1)$ is $\frac\pi p$, so that of $g(t)$ is $\frac\pi p\sqrt{1 - 4t}$, while the rescaled metric $\frac{g(t)}{1 - 4t} = g_0$ has injectivity radius $\frac\pi p$ for all $t$. Rescaling by the curvature keeps the manifold the same size, so the blow-up limit (in the sense of 11B.3 Compactness of Ricci Flows) is the round $L(p, 1)$ itself, compact: the singularity is a round point, and the flow does not see $\mathbb{R}^3$ unless one zooms in further than the curvature scale.