# 3-manifolds

## 1 Introduction

In the 3-dimensional setting there is no distinction between smooth, PL and topological manifolds neccesary; the categories of smooth, PL and topological manifolds are equivalent (TODO ref). A lot of techniques have been developed in the last century to study 3-manifolds but most of them are very special and don't generalise to higher dimensions. One key idea is to decompose manifolds along incompressible surfaces into smaller pieces, to which certain geometric models apply. A great progress was made in with the proof of the Poincaré conjecture and Thurton's geometrization conjecture by Perelman in 2003.


## 2 Construction and examples

Basic examples are $\mathbb{R}^3, S^3, S^1 \times S$$\mathbb{R}^3, S^3, S^1 \times S$ with $S$$S$ any surface. Important types of 3-manifolds are Haken-Manifolds, Seifert-Manifolds, 3-dimensional lens spaces, Torus-bundles and Torus semi-bundles.

There are two topological processes to join 3-manifolds to get a new one. The first is the connected sum of two manifolds $M_1$$M_1$ and $M_2$$M_2$. Choose embeddings $f_1:D^3\rightarrow M_1$$f_1:D^3\rightarrow M_1$ and $f_2:D^3\rightarrow M_2$$f_2:D^3\rightarrow M_2$, remove the interior of $f_1(D^3)$$f_1(D^3)$ and $f_2(D^3)$$f_2(D^3)$ and glue $M_1$$M_1$ and $M_2$$M_2$ together along the boundaries $f_1(S^3)$$f_1(S^3)$ and $f_2(S^3)$$f_2(S^3)$. The second uses incompressible surfaces. Let $M$$M$ be manifold and $S\subset M$$S\subset M$ a surface. $S$$S$ is incompressible, if there is no disk $D$$D$ in $M$$M$ with $D\cap S=\partial D$$D\cap S=\partial D$. The torus sum is the process which glues incompressible tori boundary components together.

(TODO What is incompressibility needed? / What is is good for/ What happen if one takes a compressible surface ?)

## 3 Invariants

In the 3-dimensional world the fundamental group is a powerful invariant to distiguish manifolds. It determines already all homology groups:

• $H_1(M)$$H_1(M)$ = abelization of $\pi_1(M)$$\pi_1(M)$.
• $H_2(M) = H^1(M) = H_1(M)/$$H_2(M) = H^1(M) = H_1(M)/$torsion
• $H_3(M) = \Zz$$H_3(M) = \Zz$
• $H_n(M) = 0$$H_n(M) = 0$ for $n > 3$$n > 3$

## 4 Classification/Characterization

By reversing the process of connected and torus sum every 3-manifold can be decomposed into pieces which admit a geometric structure. We describe the details in the following.

### 4.1 Prime decomposition

Definition 4.1. A manifold $M$$M$ is called prime, if it can't be written as a non-trivial connected sum, i.e. $M=M_1 \# M_2$$M=M_1 \# M_2$ implies $M_1 = S^3$$M_1 = S^3$ or $M_2 = S^3$$M_2 = S^3$. A manifold $M$$M$ is called irreducible if every embedded $S^2$$S^2$ bounds a ball, i.e. the embedding extends to an embedding of $D^3$$D^3$

Irreducibility is only slightly stronger than being prime. A orientable prime 3-manifold is either $S^2 \times S^1$$S^2 \times S^1$ or every embedded 2-sphere bounds a ball.

Theorem 4.2 Kneser. Every orientable, compact 3-manifold $M$$M$ has a decomposition $M=P_1 \# \ldots \# P_n$$M=P_1 \# \ldots \# P_n$ into prime manifolds $P_i$$P_i$ unique up to ordering and $S^3$$S^3$ summands.

A orientable prime 3-manifold is either $S^2 \times S^1$$S^2 \times S^1$ or every embedded 2-sphere bounds a ball, in which case the manifold is called irreducible.

Van Kampen's theorem tells you, that $\pi_1(M \# N)=\pi_1(M)*\pi_1(N)$$\pi_1(M \# N)=\pi_1(M)*\pi_1(N)$. Hence any 3-manifold, whose fundamental group cannot be written as a free product of two nontrivial subgroups, can only be written as the connected sum of another 3-manifold with a simply connected 3-manifold. By the Poincaré conjecture a simply connected 3-manifold is already homeomorphic to $S^3$$S^3$. Hence each such manifold is prime.

Prime 3-manifolds can be distinguished by their fundamental groups into the following 3 types:

#### 4.1.1 Type I: finite fundamental group

The universal cover $\tilde{M}$$\tilde{M}$ is a simply-connected 3-manifold. As the fundamental group already determines the homology of a oriented, closed compact 3-manifold, it has to be a homology sphere. Using the Hurewicz-theorem, its fundamental class is represented by a degree 1 map $S^3 \rightarrow \tilde{M}$$S^3 \rightarrow \tilde{M}$. This map induces isomorphisms on the homology and on the fundamental group. Hence it is a weak homotopy equivalence, and hence a homotopy equivalence by Whitehead's theorem (ref?). Hence every prime $3$$3$-manifold with finite fundamental group arises as the quotient of a homotopy sphere by a free action of a finite group. With the use of the Poincaré conjecture every homotopy 3-sphere is homeomorphic to $S^3$$S^3$ and we can write $M=S^3/\Gamma$$M=S^3/\Gamma$. If $\Gamma$$\Gamma$ is cyclic $M$$M$ is known as lens space (ref).

#### 4.1.2 Type II: infinite cyclic fundamental group

$S^1\times S^2$$S^1\times S^2$ is the only orientable closed prime 3-manifold of this type. Futhermore it is the only not irreducible prime manifold. (TODO: proof/ref)

#### 4.1.3 Type III: infinite non-cyclic fundamental group

Such a manifold $M$$M$ is always aspherical (TODO ref). The sphere theorem states, that every map $S^2\rightarrow M$$S^2\rightarrow M$ is homotopic to an embedding; and - as $M$$M$ is irreducible - it is nullhomotopic. Hence $\pi_2(M)=0$$\pi_2(M)=0$. Consider the universal covering $\tilde{M}$$\tilde{M}$ of $M$$M$. Its first homology vanishes as it is simply connected. The long exact sequence of homotopy groups of the fibration $\pi_1(M)\rightarrow \tilde{M}\rightarrow M$$\pi_1(M)\rightarrow \tilde{M}\rightarrow M$ gives a isomorphism $\pi_2(M)\cong \pi_2(\tilde{M})$$\pi_2(M)\cong \pi_2(\tilde{M})$. Hence by Hurewicz' theorem $H_2(\tilde{M})=0$$H_2(\tilde{M})=0$. Furthermore $H_3(\tilde{M})=0$$H_3(\tilde{M})=0$, as $M$$M$ is noncompact. Applying Hurewicz theorem again we get that all homotopy groups of $\tilde{M}$$\tilde{M}$ vanish and hence by Whitehead's theorem $\tilde{M}$$\tilde{M}$ is contractible. This means that $M$$M$ is apherical. Hence the homotopy type of a prime 3-manifold with infinite non-cyclic fundamental group is uniquely determined by its fundamental group. Furthermore not every group can occur as a fundamental group of a prime 3-manifold. The equivariant cellular chain complex of $\tilde{M}$$\tilde{M}$ is a projective resolution of the trivial $\Zz[\pi_1(M)]$$\Zz[\pi_1(M)]$-module $\Zz$$\Zz$. Hence .... For any subgroup $F\le \pi_1(M)$$F\le \pi_1(M)$ the space $\tilde{M}/F$$\tilde{M}/F$ is a finite-dimensional model for $K(F,1)$$K(F,1)$. For example a finite group cannot have such a model (by group homology ref) and hence $\pi_1(M)$$\pi_1(M)$ must be torsionfree. Furthermore it is a Poincaré duality group (link).

### 4.2 Torus decomposition

According to the previous section it remains to classify irreducible prime 3-manifolds. After cutting along spheres which don't bound balls as far as possible the next canonical step is to consider incompressible tori which are disjoint from the boundary.

Theorem 4.3 Jacob-Shalen, Johannson. If $M$$M$ is an irreducible compact orientable manifold, then there is a collection of disjoint incompressible tori $T_1, \ldots ,T_n$$T_1, \ldots ,T_n$ in $M$$M$ such that splitting $M$$M$ along the union of these tori produces manifolds $M_i$$M_i$ which are either Seifert-fibered or atoroidal, i.e. every incompressible torus in $M_i$$M_i$ is isotopic to a torus component of $\partial M_i$$\partial M_i$. Furthermore, a minimal such collection of tori $T_j$$T_j$ is unique up to isotopy in $M$$M$.

Thurston's geometrization conjectures states that all the pieces we get by this JSJ-decomposition admit one of eight possible geometric structures: There is a list of eight simply connected Riemannian manifolds - the so called model geometries. A geometric structure on $M$$M$ is the choice of a Riemannian metric on $M$$M$, with the property that its universal covering $\tilde{M}$$\tilde{M}$ equipped with the pull-back metric is isometric to one of the eight model geometries. It might a priori be easier to classify all cocompact actions on the several model geometries.

The Seifert-fibered pieces are well understood since the work of Seifert in the 30s (TODO: mention classification theorem). The atoroidal pieces are described by the following Hyperbolization theorem which was stated by Thurston (ref) and proven by Perelman. \begin{thm} Every irreducible atoroidal closed 3-manifold that is not Seifert-fibred is hyperbolic. \end{thm}

### 4.3 Dehn surgery

Dehn surgery is a way of constructing (TODO oriented ? neccesary) 3-manifolds. Given a link in a $3$$3$-manifold $N$$N$

$\displaystyle L: \coprod_{i=1}^n S^1\rightarrow N,$

and a choice of a tubular neighborhood of $L$$L$

$\displaystyle L': \coprod_{i=1}^n S^1\times D^2\rightarrow N\mbox{ with }L'(x,0)=L(x)$
.

(This choice essentially is the choice of a trivialization of the normal bundle; TODO find a correct formulation for this). This gives us a family of embedded, disjoint, full tori. The idea of Dehn surgery is to remove these Tori and glue them back in using a twist.\\ Let us restrict to the case with only one solid torus $L':S^1\times D^2\rightarrow N$$L':S^1\times D^2\rightarrow N$. Choose any self-homeomorphism $f$$f$ of the torus $S^1\times S^1$$S^1\times S^1$. The result of the Dehn surgery at $L$$L$ with the twist $f$$f$ is defined as

$\displaystyle N_{f,L'}:=N\setminus L'(S^1\times \mathring{D}^2) \cup_f S^1\times D^2=N\setminus L'(S^1\times \mathring{D}^2) \amalg S^1\times D^2/\sim,$

where the equivalence relation identifies for $(x,y)\in S^1\times S^1$$(x,y)\in S^1\times S^1$ the points $L'(x,y)$$L'(x,y)$ in the left component and $f(x,y)$$f(x,y)$ in the right component. If $f$$f$ is the coordinate flipping, Dehn surgery is nothing but usual codimension $2$$2$ surgery.

Lemma 4.4. Suppose $f,f'\in \Homeo(T^2)$$f,f'\in \Homeo(T^2)$ are isotopic and let $L':S^1\times D^2 \rightarrow N$$L':S^1\times D^2 \rightarrow N$ be any embedding of the full Torus in a $3$$3$-Manifold $N$$N$.Then $N_{f,L'}$$N_{f,L'}$ and $N_{f',L'}$$N_{f',L'}$ are homeomorphic.

Proof. Let $j:T^2\times [0;1] \rightarrow T^2$$j:T^2\times [0;1] \rightarrow T^2$ be an isotopy from $f$$f$ to $f'$$f'$. This gives a homeomorphism:

$\displaystyle J:T^2\times [0;1]\rightarrow T^2\times [0;1] \qquad (x,y,t)\mapsto (j(x,y,t),t).$

(TODO: is its inverse $(x,y,t)\mapsto (j(-,-,t)^{-1}(x,y),t)$$(x,y,t)\mapsto (j(-,-,t)^{-1}(x,y),t)$ continuous ?). The idea is to grab some additional space, where one can use the map $J$$J$. TODO

$\square$$\square$

TODO formulate a lemma, that M_{f,L'} also only depends on the isotopy class of $L'$$L'$ (which is hopefully true). Hence, we have to classify all self-homeomorphisms of $T^2$$T^2$ up to isotopy.

Lemma 4.5. Every self-homeomorphism of $T^2$$T^2$ is isotopic to exactly one homeomorphism of the shape

$\displaystyle f_A:\Rr^2/\Zz^2 \rightarrow \Rr^2 /\Zz^2 \qquad \left(\begin{array}{c}x\\y\end{array}\right)\mapsto A\cdot\left(\begin{array}{c}x\\y\end{array}\right),$

where $A\in GL_2(\Zz)$$A\in GL_2(\Zz)$ (reference of proof).

Proof. Since the torus is a $K\left(\Zz^2,1\right)$$K\left(\Zz^2,1\right)$-space, we have that $\pi_0 Map\left(T^2,T^2\right)\rightarrow Hom\left(\Zz^2,\Zz^2\right)$$\pi_0 Map\left(T^2,T^2\right)\rightarrow Hom\left(\Zz^2,\Zz^2\right)$ is an isomorphism. Homotopic surface homeomorphisms are isotopic (Reference?). Thus the restriction $\pi_0Homeo\left(T^2\right)\rightarrow Hom\left(\Zz^2,\Zz^2\right)$$\pi_0Homeo\left(T^2\right)\rightarrow Hom\left(\Zz^2,\Zz^2\right)$ is injective. Moreover, each $A\in Hom\left(\Zz^2,\Zz^2\right)$$A\in Hom\left(\Zz^2,\Zz^2\right)$ is realised by $f_A$$f_A$, therefore $\pi_0Homeo\left(T^2\right)\rightarrow Hom\left(\Zz^2,\Zz^2\right)$$\pi_0Homeo\left(T^2\right)\rightarrow Hom\left(\Zz^2,\Zz^2\right)$ is also surjective. TODO find a reference in ANY source about the mapping class group

$\square$$\square$

The next lemma tells us, that composition of self-homeomorphisms corresponds to two successive Dehn surgeries.

Lemma 4.6. Let $f,g\in \Homeo(T^2)$$f,g\in \Homeo(T^2)$ be given and let $L':S^1\times D^2\rightarrow N$$L':S^1\times D^2\rightarrow N$ is an embedding of the full torus in a 3-manifold. Then we have map

$\displaystyle L'':S^1 \times D^2 \rightarrow N\setminus L'(S^1\times \mathring{D}^2) \cup_f S^1\times D^2=N_{f,L'}$

given by the map $S^1\times D^2\rightarrow S^1 \times D^2 \quad (x,y)(x,y/2)$$S^1\times D^2\rightarrow S^1 \times D^2 \quad (x,y)(x,y/2)$ postcomposed with the canonical inclusion in the second coordinate. Then $(N_{f,L'})_{g,L''} \cong N_{f\circ g,L'}$$(N_{f,L'})_{g,L''} \cong N_{f\circ g,L'}$. TODO right order of composition ? We will see in the proof.

Proof.

$\square$$\square$

We have to find out, which self-homeomorphisms of the torus don't change the homeomorphism type of the manifold.

Lemma 4.7. Consider a matrix of the form $\left( \begin{array}{cc} 1 & 0 \\ k & 1 \end{array}\right)$$\left( \begin{array}{cc} 1 & 0 \\ k & 1 \end{array}\right)$ and let $L':S^1\times D^2 \rightarrow N$$L':S^1\times D^2 \rightarrow N$ be any eembedding. Then $N_{f_A,L'}\cong N$$N_{f_A,L'}\cong N$.

TODO are there any orientation reversing homeos, that also extend ? Think so. Also add them here.

Proof. The homeomorphism $f_A \in \Homeo(T^2)$$f_A \in \Homeo(T^2)$ extends to a homeomorphism of $\bar{f_A}\in \Homeo(S^1\times D^2)$$\bar{f_A}\in \Homeo(S^1\times D^2)$:

$\displaystyle \bar{f_A}(x,y):=(x,x^ky),$

where $x\in S^1 = \{z\in \Cc| |z|=1\}, y\in D^2=\{y\in\Cc||y|\le 1\}$$x\in S^1 = \{z\in \Cc| |z|=1\}, y\in D^2=\{y\in\Cc||y|\le 1\}$. Using this homeomorphism one can define a homeomorphism from $N_f$$N_f$ to $N$$N$:

$\displaystyle N = N\setminus L(S^1\times \mathring{D}^2)\cup_1 S^1\times D^2 \rightarrow N_{f_A}=N\setminus L(S^1 \times \mathring{D}^2)\cup_{f_A} S^1\times D^2$

given by the identity on the left component and $\bar{f_A}$$\bar{f_A}$ on the right component.

$\square$$\square$

Together with (link to comment about composition), this tells us, that $N_{f_A}$$N_{f_A}$ really only depends on the coset $A\cdot \left(\begin{array}{cc}1&*\\0&1\end{array}\right)$$A\cdot \left(\begin{array}{cc}1&*\\0&1\end{array}\right)$ (TODO check right or left coset). This coset is uniquely determined by the image $(p,q)$$(p,q)$ of $(1,0)$$(1,0)$ with $p$$p$ and $q$$q$ coprime.

The ratio $p/q$$p/q$ is called the surgery coefficient. (TODO what is the quotient good for ?)<++>

TODO does the result give different manifolds.

TODO does the result only depend on the isotopy class of the link.

Every compact (oriented /able, neccesary ?) 3-manifold might be obtained from $S^3$$S^3$ by a Dehn surgery along a link (TODO ref). Of course this does not satisfy to classify 3-manifolds without having a good classification of links in $S^3$$S^3$.

## 5 References

[Scott1983], [Thurston1997], [Hatcher2000], [Hempel1976]