Geometrisation of 3-manifolds

In this post we discuss the geometrisation of three dimensional manifolds, whose proof was completed by Perelman in 2002 using Ricci flow. Due to the proof's length and technical complexity we will not say anything about this, but there are accounts of this online. We also won't prove very many other things, so this is more of an introductory sightseeing tour using Martelli's book as our guide. 

In what follows a closed manifold is a manifold which is compact and without boundary. The Uniformization theorem implies that every closed 2-manifold admits a metric of constant curvature, among other things. In particular, all points have isometric neighbourhoods, and the geometry looks the same. It is natural to wonder what happens in higher dimensions, and we do have the following theorem:

Theorem 1 Let $M_n$ be a complete Riemannian manifold with constant sectional curvature $K$. Up to rescaling, its universal cover is (isometric to) one of: 

1.  The sphere $S^n$ with the round metric if K >0; 
2.  Euclidean $n$-space $E^n$ if $K =0$; 
3.  Hyperbolic $n$-space $H^n$ if $K =−1$.

We will capture the notion of all points being locally isometric as follows:

A pseudogroup $G$ on a topological space $X$ is a set of homeomorphisms between open sets of $X$ such that:

1. The domains of the elements of $G$ cover $X$.
2. The restriction of an element of $G$ to an open set contained in its domain is also in $G$.
3. When the composition of two elements of $G$ exists, it is in $G$.
4. The inverse of an element of $G$ lies in $G$.
5. If $g: U \to V$ is such that each $x \in U$ has an open neighbourhood $U_x \subseteq U$ with $g|_{U_x} \in G$, then $g \in G$.

Examples:

    1.  $C^0$, the pseudogroup of homeomorphisms between open subsets of $\mathbb{R}^n$.

    2. $C^\infty$, the pseudogroup of diffeomorphisms between open subsets of $\mathbb{R}^n$.

    3. $H$, the pseudogroup of holomorphic bijections between open subsets of $\mathbb{C}^n$.

    4. The restrictions of the elements of any Lie Group $G$ of homeomorphisms of $X$ to all the open subsets generate a pseudogroup, denoted $(G, X)$.

 An atlas $A$ for a topological space $M$ is a collection of homeomorphisms from open sets in $M$ to open sets in $X$, called charts, whose domains cover $M$. Two such charts, $\varphi$ and $\psi$, are $G$-compatible if their transition function $\vartheta = \psi \circ \varphi^{-1}$ lies in $G$. A $G$-atlas is an atlas for which all pairs of charts are $G$-compatible, and a $G$-manifold is a Hausdorff second countable topological space with a maximal $G$-atlas $A$.

For example, a topological manifold is a $C^0$-manifold, a differential manifold is a $C^\infty$-manifold, and a complex manifold is an $H$-manifold. However, we will be most interested in $(G, X)$-manifolds, whose geometry is locally like that of $X$ with isometry group $G$.

    A Euclidean manifold is an $(\text{Isom}(E^n), E^n)$-manifold.

    A spherical manifold is an $(\text{Isom}(S^n), S^n)$-manifold.

    A hyperbolic manifold is an $(\text{Isom}(H^n), H^n)$-manifold.

A model geometry $(G, X)$ is a manifold $X$ with a Lie Group $G$ of diffeomorphisms of $X$ such that:

    1. $X$ is connected and simply connected.

    2. $G$ acts transitively on $X$ with compact point stabilizers.

    3. $G$ is not contained in any larger groups of diffeomorphisms of $X$ with compact point stabilizers.

    4. There exists at least one compact $(G, X)$-manifold.

When we say a geometric structure exists on a manifold $M$ we mean a discrete subgroup $\Gamma<G$ such that $M$ is diffeomorphic to $X/\Gamma$ for some model geometry $(G, X)$. 

In dimension 2 uniformisation implies that there are exactly 3 model geometries in dimension 2. However, it also should not surprising that there are more geometries in higher dimensions, and it turns out not every manifold can be given a geometric structure. It turns out that in dimension 3 there are 8 geometric structures, which we discuss below, and every manifold can be cut into pieces which each admit a geometric structure, but in dimension 4 there is a countably infinite family of geometries plus 18 geometries which don't fit into that family and not every manifold can even be cut into geometric pieces. Hence we focus on dimension 3. Note that condition 4, that there exists at least one compact $(G, X)$-manifold, is important. Even in dimension 2 there are lots of weird surfaces, and if we drop condition 4 then in dimension 3 there are uncountably many model geometries.

Of course, when we talk about geometry we usually think of something with a metric, so we should check that manifolds modelled on this notion of model geometry admit sensible Riemannian metrics. 

Proposition 2 Let $G$ be a Lie Group acting transitively on a smooth manifold $X$. Then there exists a Riemannian metric on $X$ invariant under the action of $G$ if and only if for some (and hence any) $x \in X$ the image of the stabilizer $G_x$ in $GL(T_xX)$ has compact closure.

Proof: If such a metric exists, then the image is conjugate to a subgroup of $O(T_xX)$, which is compact. Conversely, if the image has compact closure $H_x$, let $\langle \cdot, \cdot \rangle$ be any positive definite symmetric bilinear form on $T_xX$, and define

\[(u, v) = \int_{H_x} \langle d g u, d g v \rangle d g\]

for $u, v \in T_xX$, where $H_x$ is Haar measure on $H_x$. This gives an inner product on $T_xX$ which is invariant under the action of $G_x$, and is finite because $H_x$ is compact. This can now be propagated to the whole of $X$ using the action of $G$. $\blacksquare$

Hence condition 2 in the definition of model geometries guarantees the existence of an invariant Riemannian metric. One can further show that such a metric is complete, but we move on to discussing the 3-dimensional geometries.

Three of them are the sphere $S^3$, Euclidean space $E^3$, and hyperbolic space $H^3$.  Two of them are built up from the product of lower dimensional spaces, with the corresponding product metric: $S^2 \times \mathbb{R}$ and $\mathbb{H}^2 \times \mathbb{R}$. Exhibiting compact manifolds with these geometric structures is left to the reader. Another geometry is simply known as $\tilde{SL_2}$, the universal cover of $SL_2(\mathbb{R})$, which is also the unit tangent bundle of the hyperbolic plane. For examples of manifolds modelled on this, take the unit tangent bundle of a compact hyperbolic surface. This leaves two more geometries, named after group theoretic properties.

One of them is called Nil, modelled on the three dimensional Heisenberg group, i.e. the group of upper triangular 3 by 3 matrices with ones on the diagonal. This fits into a short exact sequence of Lie groups:

$$ 0 \longrightarrow \mathbb{R} \longrightarrow Nil \longrightarrow \mathbb{R}^2 \longrightarrow 0$$

where $\mathbb{R} = [Nil, Nil] $ is the centre of $Nil$. Therefore Nil is naturally a line bundle over $\mathbb{R}^2$. One can further show that the group of orientation preserving isometries of Nil is nilpotent, so any manifold modelled on Nil has nilpotent fundamental group. An example of compact manifold with Nil structure is given by the quotient of Nil by the integer Heisenberg group.

The last geometry is called Sol, and is a bit of an oddball. The Lie group $Sol$ is the space $\mathbb{R}^3$ equipped with the following operation

$$(x,y,z)\cdot(x',y',z') = (x+ e^{-z}x', y+e^zy', z+z').$$

One checks that

\[(x,y,z)^{-1} = (-xe^z, -ye^{-z}, -z)\]

\[[(x,y,z),(x',y',z')] = \big(x(1\!-\!e^{-z'}) - x'(1\!-\!e^{-z}), y(1\!-\!e^{z'}) - y'(1\!-\!e^z), 0\big), \]

\[[(x,y,z), (x',y',0)] = \big(- x'(1-e^{-z}), - y'(1 - e^z), 0\big). \]

The subgroup $\mathbb{R}^2$ consisting of all elements $(x,y,0)$ is the centre of $Sol$ and by setting $p(x,y,z)=z$ we get an exact sequence

$$0 \longrightarrow \mathbb{R}^2 \longrightarrow Sol \stackrel p \longrightarrow \mathbb{R} \longrightarrow 0.$$

Therefore $Sol$ is a plane bundle over $\mathbb{R}$. We have $[Sol, Sol] = \mathbb{R}^2$ and hence $Sol$ is solvable. However $[Sol, \mathbb{R}^2] =\mathbb{R}^2$ and hence $Sol$ is not nilpotent. With a bit more work one can show that manifolds modelled on Sol have fundamental groups which are virtually solvable but not virtually nilpotent.

In dimension 3 the group theory (of the fundamental group) and the manifold geometry/topology are particularly intricately linked. We give the following as one of many possible pieces of supporting evidence:

Theorem 3: 

Let $M$ be a closed manifold modelled on one of the eight geometries $\mathbb{X}$.

1. if $\pi_1(M)$ is finite, then $\mathbb{X} = S^3$; otherwise
2. if $\pi_1(M)$ is virtually cyclic, then $\mathbb{X} = S^2\times \mathbb{R}$; otherwise
3. if $\pi_1(M)$ is virtually abelian, then $\mathbb{X} = \mathbb{R}^3$; otherwise
4. if $\pi_1(M)$ is virtually nilpotent, then $\mathbb{X} = Nil$; otherwise
5. if $\pi_1(M)$ is virtually solvable, then $\mathbb{X} = Sol$; otherwise
6. if $\pi_1(M)$ contains a normal cyclic group $K$, then:
• if a finite-index subgroup of the quotient lifts, then $\mathbb{X} = \mathbb{H}^2 \times \mathbb{R}$, otherwise $\mathbb{X} = \widetilde{SL_2}$; 

If none of the above are satisfied, then the manifold is hyperbolic.

In general, one approach to describing the topology of a manifold is asking what sort of metrics they admit, and if there are lots of them one asks for a 'best' metric in some sense. 2- and 3-dimensional manifolds admit very symmetric metrics, as borne out by the various conditions we imposed in the definition of model geometry. Being able to show that manifolds with a geometric structure have special topological properties and then being able to cut up general (compact) 3-manifolds into geometric pieces, as we discuss below, is a powerful tool for proving theorems about 3-manifolds in general.

Cutting up manifolds 

Geometrisation talks about cutting up manifolds, so here we discuss a number of ways of decomposing general 3-manifolds that are fundamental to the theory. First we cut along spheres:

Definition A connected, oriented 3-manifold $M$ is prime if every connected sum $M=M_1\# M_2$ is trivial. It is irreducible if every (smoothly embedded) sphere $S\subset int(M)$ bounds a ball.

Being prime is equivalent to be irreducible, with a single exception.

Proposition 4 Every oriented 3-manifold $M\neq S^2 \times S^1$ is prime if and only if it is irreducible.

Proof: The inverse operation of a connected sum $M=M_1\# M_2$ consists of cutting along a separating sphere $S\subset M$ and then capping off the two resulting manifolds $N_1, N_2$ with balls. The capped $N_i$ is $S^3$ if and only if $N_i$ is a ball. Therefore the connected sum is trivial if and only if $S$ bounds a ball on one side. Therefore $M$ is prime if and only if every separating sphere $S\subset M$ bounds a ball.

If $M$ is irreducible, then it is clearly prime. If $M$ is prime and not irreducible, there is a non-separating sphere $S\subset M$. There is a simple closed curve $\alpha\subset M$ intersecting $S$ transversely in one point. Pick two tubular neighbourhoods of $S$ and $\alpha$: their union is a manifold $N$ with a boundary sphere $\partial N = S'$. The sphere $S'$ is separating and $M$ is prime, hence $S'$ bounds a ball $B$ on the other side and $M=N\cup B$. 

We now prove that $M=S^2\times S^1$. We embed $S\cup \alpha$ naturally in $S^2\times S^1$ as $S = S^2\times y$ and $\alpha = x \times S^1$. Decompose $S^2= D\cup D'$ in two discs and $S^1 = I\cup I'$ in two intervals. The manifold

$N$ also embeds as $S^2\times I \cup D \times S^1$ and its complement $B=D'\times I'$ is a ball. Therefore $M = S^2\times S^1$. $\blacksquare$

Theorem 5 (Prime decomposition theorem) Every compact oriented 3-manifold $M$ with (possibly empty) boundary decomposes into prime manifolds:

$$M = M_1\# \ldots \# M_k$$

This list of prime factors is unique up to permutations and adding/removing copies of $S^3$.

This was proved by Kneser and Milnor.

The next two decompositions will be along tori. To state what exactly what we mean by this, we need to introduce some terminology:

Dehn Filling and Seifert manifolds

If a 3-manifold $M$ has a spherical boundary component, we can cap it off with a ball. If $M$ has a toric boundary component, there is no canonical way to cap it off: the simplest object that we can attach to it is a solid torus $D\times S^1$, but the resulting manifold depends on the gluing map. This operation is called a Dehn filling.

Let $M$ be a 3-manifold and $T\subset \partial M$ be a boundary torus component.

Definition A Dehn filling of $M$ along $T$ is the operation of gluing a solid torus $D\times S^1$ to $M$ via a diffeomorphism $\varphi\colon \partial D\times S^1 \to T$.

The closed curve $\partial D \times \{x\}$ is glued to some simple closed curve $\gamma\subset T$. The result of this operation is a new manifold $M^{\rm fill}$, which has one boundary component less than $M$.

Lemma 6 The manifold $M^{\rm fill}$ depends only on the isotopy class of the unoriented curve $\gamma$.

We hope the reader finds this intuitive and skip the proof. We say that the Dehn filling kills the curve $\gamma$, which is what it does to the fundamental group by Seifert- van Kampen. We now give the definition of Seifert manifolds.

Let $M$ be the (unique) oriented bundle over a compact connected (possibly non-orientable) surface $S$ with boundary. We denote by $S$ the zero-section.

Let $T_1,\ldots, T_k$ be the boundary tori of $M$. On each $T_i$ we choose an orientation for the meridian $m_i=T_i\cap \partial S$ and for the fibre $l_i$ of the bundle so that the basis $(m_i,l_i)$ for $H_1(T_i,\mathbb{Z})$ be positively oriented.

A $(p_i,q_i)$-Dehn filling on $T_i$ kills the slope $p_im_i + q_il_i$. We say that the Dehn filling is fibre-parallel if $p_i=0$, i.e. if it kills a fibre.

Definition A Seifert manifold is any 3-manifold $N$ obtained from $M$ by Dehn filling some $h\leqslant k$ boundary tori in a non-fibre-parallel way, that is with $p_i\neq 0$ for all $i$.

Out of the geometric manifolds we have discussed so far, all but the Sol and hyperbolic ones are Seifert manifolds. One can write down invariants of Seifert manifolds and show that the geometric structures are completely classified by their Seifert invariants.

Torus decompositions

Definition      

• An embedded torus \( T \to M \) is compressible if there exists a disc \( D \subseteq M \) with  \( D \cap T = \partial D \) and \( \partial D \) not contractible in \( T \). Otherwise, \( T \) is incompressible.  
• An irreducible 3-manifold \( M \) is atoroidal if any \( \mathbb{Z} \oplus \mathbb{Z} \) subgroup of \( \pi_1(M) \) is conjugate to the inclusion of the fundamental group of a toric boundary component of \( M \). This is equivalent to requiring  that any incompressible torus is boundary-parallel, in the sense that it separates \( M \) into a copy of \( M \) and \( \partial M \times I \).  

Definition A torus decomposition of an irreducible manifold \( M \) is a collection of embedded incompressible tori  \( T_1, \dots, T_n \) such that the components of \( M - \bigcup_i T_i \) are all either atoroidal or Seifert fibred

A torus decomposition is minimal if no proper subset of $S$ is a torus decomposition.

Theorem 7 (JSJ decomposition) Let $M$ be an orientable irreducible and $\partial$-irreducible compact 3-manifold with (possibly empty) boundary consisting of tori. A minimal torus decomposition for $M$ exists and is unique up to isotopy.

The last decomposition that we state is the geometric decomposition:

The geometric decomposition is a slight variation of the canonical torus decomposition that is more suited to the geometrisation perspective that we will encounter in the next chapters. It is constructed from the canonical torus decomposition $S$ for $M$ as follows. Whenever a block $N$ of the torus decomposition is diffeomorphic to $K\tilde{\times} I$,  

we substitute the torus $\partial N$ in $S$ with the core Klein bottle $K$ of $K\tilde{\times} I$. This substitution has the effect of deleting $N$ from the list of blocks of the decomposition.

The geometric decomposition consists of incompressible tori and Klein bottles. 

Proposition 8 The geometric decomposition of a finite cover $\tilde M$ is the preimage of that of $M$.

Geometrisation

 

We can finally state the geometrisation conjecture and its consequences for the structure theory of 3-manifolds. 

Theorem 9 (Geometrisation Conjecture) Let $M$ be an irreducible orientable compact 3-manifold with (possibly empty) boundary consisting of tori. Then every block of the geometric decomposition of $M$ is geometric.

As said before, the proof is hard. Its consequences are very deep, and we can state and prove them.

Theorem 10 (Poincare conjecture) Every simply connected closed 3-manifold $M$ is diffeomorphic to $S^3$.

Proof Via the prime decomposition we may restrict to the case $M$ is prime, hence irreducible. The group $\pi_1(M)$ is trivial and hence does not contain $\mathbb{Z} \times \mathbb{Z}$: every torus in $M$ is thus compressible and the geometric decomposition is trivial. By geometrisation $M$ is itself geometric. The only geometry with finite fundamental groups is $S^3$, and hence $M= S^3/_\Gamma$ is elliptic. Since $M$ is simply connected, the group $\Gamma = \pi_1(M)$ is trivial and hence $M=S^3$. $\blacksquare$

Theorem 11 (Elliptisation) Every closed 3-manifold $M$ with finite $\pi_1(M)$ is elliptic.

Same proof as above. 

Theorem 12 (Hyperbolisation) Every closed irreducible 3-manifold $M$ with infinite $\pi_1(M)$ not containing $\mathbb{Z} \times \mathbb{Z}$ is hyperbolic.

Proof Since $\pi_1(M)$ does not contain $\mathbb{Z}\times \mathbb{Z}$ every torus is compressible and the geometric decomposition of $M$ is trivial. By geometrisation $M$ is geometric. Its geometry is not $S^3$ since $\pi_1(M)$ is infinite, and is not $S^2\times \mathbb{R}$ since $M$ is irreducible. In the other Seifert geometries and in $Sol$ the fundamental group $\pi_1(M)$ always contain a $\mathbb{Z}\times \mathbb{Z}$ (there is always a finite covering containing an incompressible torus). $\blacksquare$

Theorem 13 Let $\tilde M \to M$ be a finite covering. If $\tilde M$ is geometric, then $M$ also is (with the same geometry).

Proof Since $\tilde M$ is geometric, it is irreducible and hence also $M$ is. Proposition 8 implies that the geometric decomposition of $M$ is trivial, and by geometrisation $M$ is geometric. $\blacksquare$

Of course, all this talk of geometrisation might make you wonder what it really feels like to be in one of these manifolds, and there are increasing numbers of visualisation tools. For starters, see here.