\documentclass{amsart}

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\begin{document}

\title{Syllabus for Qualifying Exam}
\author{Danny Calegari}
\maketitle

{\bf Exam Committee: Andrew Casson, Robion Kirby, Curtis McMullen, 
Steven Evans (Statistics)}

\vskip 12pt

{\bf Monday 26th February, 3:00 pm}

\vskip 24pt

\section{Major Area: Algebraic Topology}

\begin{itemize}
\item{Fundamental group and higher homotopy groups}

For each space $X$ we can define $\pi_n(X)$ to be the space
$[S^n,X]_0$ of homotopy classes of based maps $S^n \to X$. Denoting
the loop space of $X$ by $\Omega X$ and the suspension by $\Sigma X$,
we have the general fact $$[X,\Omega Y]_0 \cong [\Sigma X,Y]_0$$
Moreover, for $n\ge 2$, $\pi_n(X)$ is abelian.

\item{Seifert-VanKampen theorem}

If $X$ is obtained from $X_1$ and $X_2$ by glueing along $A$, then if $A$
is connected, we have the SVK theorem:
$$\pi_1(X) = \pi_1(X_1) *_{\pi_1(A)} \pi_1(X_2)$$
Related to this is the concept of a HNN extension - if $Y$ is obtained from
$X$ by gluing a subspace $A$ to itself, then
$$\pi_1(Y) = \pi_1(X) *_{\pi_1(A)}$$

\item{Subgroups of fundamental group and covering spaces}

For each subgroup $G$ of $\pi_1(X)$ there is a space $X_G$ with 
$\pi_1(X_G) = G$ such that there is a covering map $\pi: X_G \to X$. If
$G$ is a normal subgroup, then the group $\Gamma = \pi_1(X)/G$ acts properly
discontinuously on $X_G$ and we have $X_G/\Gamma = X$.

\item{Homotopy exact sequence}

For any pair of spaces $(X,A)$ there is a long exact sequence
$$ \dots \to \pi_n(A) \to \pi_n(X) \to \pi_n(X,A) \to \pi_{n-1}(A) \to \dots $$
This map is an exact sequence of $\pi_1(A)$-modules. For $n>2$, $\pi_n(X,A)$
is abelian.

\item{Homology groups: Singular, Cellular, Simplicial}

We define the singular homology groups of a space $X$ with coefficients in
an abelian group $G$ as follows. Let $C_i$ be the free $G$-module generated by
maps from $i$-simplices 
$\sigma: \Delta_i \to X$. We define a boundary operator 
$\partial: C_i \to C_{i-1}$ by the formula 
$$\partial \sigma = \Sigma_j (-1)^j \sigma^j$$ where $\sigma^j$ is the $j$'th
$i-1$-face of $\sigma$. Then $\partial \circ \partial = 0$ and we define
$$H_i(X,G) = ker(\partial_i)/ im(\partial_{i+1})$$

For a CW-complex $W$, we define the chain complex
$$\dots \to H_i(W^i,W^{i-1}) \to H_{i-1}(W^{i-1},W^{i-2}) \to \dots$$
Then the homology of this complex (the Cellular homology of $W$) is isomorphic
to the singular homology of $W$.

\item{Mayer-Vietoris sequence}

There is a long exact sequence in homology for a pair $X_1,X_2$ such that
$X_1 \cup X_2 = X$, $X_1 \cap X_2 = A$ obtained by splicing together the
long exact sequences of $(X,X_1)$ and $(X_2,A)$ whose $3i$ terms are
isomorphic, by excision. This sequence
$$\dots \to H_i(A) \to H_i(X_1) \oplus H_i(X_2) \to H_i(X) \to H_{i-1}(A) 
\to \dots$$
is known as the Mayer-Vietoris sequence.

\item{Morse theory (handlebody decomposition)}

On any manifold $M$ there exists a function $f$ (in fact, many functions)
known as a Morse Function, such that the critical points of $f$ are
isolated, and such that for each critical point $p$, 
there are local co-ordinates
on $M$ such that $f$ takes the form $$f(x_1,x_2 \dots x_n) = x_1^2 + \dots +
x_i^2 - x_{i+1}^2 - \dots - x_n^2$$
We say that $p$ is a critical point of index $n-i$. Then $M$ has the
homotopy type of a CW complex with one cell of dimension $p$ for each critical
point of index $p$.

The theory of normal surfaces in three-manifolds can be described with 
respect to how a surface intersects the $0,1,2$ handles of a handle 
decomposition.

\item{Cohomology groups: Singular, Cellular, Simplicial}

We define the $i$-cochains of a space $X$ to be the elements of the dual
to the $i$-chains, namely $C_i^*$. Then there is a map $\delta:
C_i^* \to C_{i+1}^*$ which is simply the adjoint of $\partial$ under the
natural pairing. As above, we define 
$$H^i(X) = ker(\delta_i)/im(\delta_{i-1})$$

Similarly, cellular cohomology is defined to be the cohomology of the cell
complex
$$ \dots \to H^i(W^i,W^{i-1}) \to H^{i+1}(W^{i+1},W^i) \to \dots $$

\item{De Rham cohomology and Cech cohomology}

On a differentiable manifold $M$ we can define the algebra of exterior forms
$\Omega^*(M)$ which is a graded algebra whose $i$'th component is given by
the differential forms of degree $i$: $$fdx_{r_1} \wedge dx_{r_2} \wedge
dx_{r_3} \wedge \dots \wedge dx_{r_i}$$

Multiplication in this algebra is given by the wedge product, and there is
a differential operator $d$ which acts as follows:
$$d:fdx_{r_1}\wedge \dots \wedge dx_{r_i} = \Sigma_j \partial f/ \partial x_j
dx_j \wedge dx_{r_1} \wedge \dots \wedge dx_{r_i}$$
The closed forms are those whose image under $d$ is zero. The De Rham
cohomology of $M$ is precisely the cohomology of the differential complex
$$\dots \to \Omega^n(M) \to \Omega^{n+1}(M) \to \dots$$
with differential operator $d$.

The Cech cohomology of a space with coefficients in $\Bbb(R)$ is the
cohomology of the complex 
$$ \dots \to C^i(U,{\Bbb R}) \to C^{i+1}(U,{\Bbb R}) \to \dots$$
where $U$ is a good open covering of $M$ (i.e. every open set and 
intersection of open sets is contractible), and the elements of
$C^{i}(U,{\Bbb R})$ are the locally constant functions on the
$i+1$-fold intersections of open sets. More generally, for some sheaf
$\Omega$, the cohomology of the complex
$$\dots \to C^i(U,\Omega) \to C^{i+1}(U,\Omega) \to \dots$$
computes the Cech cohomology of $M$ with coefficients in $\Omega$, where
$C^i(U,\Omega)$ is the sections of the sheaf $\Omega$ over the 
$i+1$-fold intersections. Note the following important theorem:
a sheaf $\mathcal{F}$ on a topological space $X$ is {\it flasque} if
for every inclusion of open sets $V \subset U$ the restriction map
$\mathcal{F}(U) \to \mathcal{F}(V)$ is surjective. Then the Cech cohomology of a
space with coefficients in a flasque sheaf are zero. Moreover, if we have
a short exact sequence of sheaves
$$0 \to \mathcal{F} \to \mathcal{G} \to \mathcal{H} \to 0$$ then this induces a
long exact sequence of cohomology:
$$ \dots \to H^i(X,\mathcal{F}) \to H^i(X,\mathcal{G}) \to H^i(X,\mathcal{H}) 
\to H^{i+1}(X,
\mathcal{F}) \to \dots $$

Generally, we have the following construction. To define $H^*(X,\mathcal{S})$
with $\mathcal{S}$ a sheaf of $K$-modules, let $\mathcal{K}$ be the
trivial $K$-sheaf on $X$, and let
$$0 \to \mathcal{K} \to \mathcal{C}_0 \to \mathcal{C}_1 \to \dots$$
be a fine resolution of $\mathcal{K}$. Then $H^*(X,\mathcal{S})$ is the
cohomology of the complex
$$0 \to \Gamma(\mathcal{C}_0 \otimes \mathcal{S}) \to \Gamma(\mathcal{C}_1 
\otimes \mathcal{S}) \to \dots$$


\item{Hurewicz theorem (statement)}

$$H_1(X,{\Bbb Z}) \cong \pi_1(X) / [\pi_1(X),\pi_1(X)]$$
Moreover, in general if the first $n-1$ homotopy groups of $X$ vanish,
then the first $n-1$ homology groups of $X$ with coefficients in ${\Bbb Z}$
vanish, and the natural map $f: \pi_n(X) \to H_n(X,{\Bbb Z})$ is an
isomorphism.

We also have the Relative Hurewicz theorem; namely, if $\pi_1(A) = 1$ and
$\pi_i(X,A) = 1$ for $i <n$ then if $\pi_n(X,A) \ne 1$ for $n \ge 2$, we have
$\pi_i(X,A) \cong H_i(X,A)$ for $i \le n$.

\item{Whitehead theorem (statement)}

If there is a map $f:X \to Y$ between CW complexes such that
$f_*$ induces an isomorphism on all the homotopy groups of $X$ and $Y$, then
$f$ is a homotopy equivalence.

\item{Classifying spaces (definition and some facts about)}

For any abelian group $\Pi$ and any $n$ there is a space $K(\Pi,n)$ unique
up to homotopy equivalence such that $\pi_i(K(\Pi,n)) = 0$ if $i\ne n$ and
$\pi_i(K(\Pi,n)) = \Pi$ if $i=n$. Moreover, homotopy classes of maps
$[X,K(\Pi,n)]_0$ are in 1-1 correspondence with elements of
$H^n(X;\Pi)$. Note if $n=1$, then there is a $K(\Pi,n)$ for any $\Pi$.

For any group $G$ there is a contractible space $EG$ such that $EG$ is a 
principle $G$-bundle over some classifying space $BG$ such that 
principle $G$-bundles over any space $X$ are in 1-1 correspondence with
homotopy classes of maps $[X,BG]_0$, and the correspondence is precisely by
pulling back the universal bundle $EG$ by this map.

These are special cases of the following {\it Brown Representability Theorem}.
Specifically, this states that if a homotopy functor $H$ satisfies the
following three axioms -      

$H$ is a contravariant functor from based CW complexes and based maps
to based sets such that
\begin{enumerate}
\item{If $f$ is homotopic to $g$ rel. basepoints then $f^* = g^*$}
\item{Inclusions $i_\alpha:X_\alpha \to \bigcup_\alpha X_\alpha$ (one-point
union) induces an isomorphism $H(\bigcup_\alpha X_\alpha) \to \Pi_\alpha
H(X_\alpha)$}
\item{If $e:Y \to Z$ is an inclusion in the reduced double mapping cylinder
of $f,g:X \to Y$ then $\eta \in e^*H(Z)$ iff $f^* \eta = g^* \eta$}
\end{enumerate}

The theorem then states that for every homotopy functor $H$ with
$H(S^0) = pt$, there exists a CW complex $Y_H$ and an element $\eta_H \in
H(Y_H)$ such that for all CW complexes $X$ there is an isomorphism
$[X,Y_H]_0 \cong H(X)$ by sending $[f]$ to $f^* \eta_H$.

\item{Duality theorems: Poincar\' e-Lefschetz duality, Alexander duality}

For an orientable $n$-manifold $M$ with orientable boundary $\partial M$,
there is an isomorphism between the two sequences
$$\dots \to H^i(M,\partial M) \to H^i(M) \to H^i(\partial M) \to \dots$$
and 
$$\dots \to H_{n-i}(M) \to H_{n-i}(M,\partial M) \to H_{n-i-1}(\partial M) \to
\dots$$
given by cap product with the fundamental homology class of $(M,\partial M)$ 
and $\partial M$ as
appropriate. This is known as Poincar\' e-Lefschetz duality.

For $A$ a compact submanifold of ${\Bbb R}^n$ there is an isomorphism
$$H^q(A) \cong H_{n-q-1}({\Bbb R}^n - A)$$
(reduced homology).
More generally, there is an isomorphism for any closed subset $A$ of an
oriented $n$-dimensional manifold $M$
$$H^q(A) \cong H_{n-q}(M,M-A)$$
where $A,M$ are absolute neighborhood retracts, given by cup product with the
fundamental class of $H_n(M, M-A)$.

\item{Universal Coefficient theorems}       

There are isomorphisms
$$H^n(X,G) = Hom(H_n(X;{\Bbb Z}),G)\oplus Ext(H_{n-1}(X,{\Bbb Z}),G)$$
and
$$H_n(X,G) = H_n(X;{\Bbb Z})\otimes G \oplus Tor(H_{n-1}(X;{\Bbb Z}),G)$$

Moreover, there is a short exact sequence
$$0 \to H^*(X) \otimes H^*(Y) \to H^*(X\times Y) \to \oplus_{p+q=n+1}
Tor(H^p(X),H^q(Y)) \to 0$$

In particular, if $H^*(X)$ is free, then $$H^*(X\times Y) \cong
H^*(X) \otimes H^*(Y)$$

\item{Lefschetz fixed point theorem (statement)}

For any map $f:M \to M$ be any self-map of a compact $R$-oriented manifold
where $R$ is a field. Then $\Sigma_q (-1)^q Trace H^q(f)$, the Lefschetz
number of $f$, is given by the intersection number of the graph $\Gamma$
of $f$ with the diagonal $\Delta$ in $M \times M$. In particular, if this
number is non-zero, $f$ has a fixed point.

\end{itemize}

\section{Special Topic: 3-Manifolds and Geometrization}

\begin{itemize}
\item{Moise's theorem - that any three-manifold has a unique (up to
refinement) triangulation (statement)}

We add the assumption that the three-manifold be compact. Moreover, if it
has boundary, then it can be triangulated in a way that extends to the
standard triangulation of the boundary.

Note that this theorem is not true in higher dimensions. There are $n$ 
manifolds $n \ge 4$ which admit many different triangulations, and even some
which admit no triangulations.

\item{Heegard Splitting}

Every closed, connected $3$-manifold admits a Heegard Splitting. That is, it
can be written as a union $V_1 \cup_f V_2$ where $V_1, V_2$ are genus $g$
handlebodies attached along their boundaries by a homeomorphism.
For, triangulate the manifold. Then we can take as $V_1$ a regular neighborhood
of its $1$-skeleton.

\item{Reidemeister-Singer theorem - that any two Heegard splittings are
stably equivalent}

Moreover, any handlebody in a manifold can be stabilized to give a Heegard
splitting. For, we can include the core of the handlebody in the $1$-skeleton
of $M$, then stabilize the splitting by adding the remainder of the 
$1$-skeleton. We then apply Moise's theorem that any two triangulations
have a common refinement, and we are done.

Facts:
\begin{enumerate}
\item{$S^3$ is the unique $3$-manifold of genus $0$.}
\item{A genus $1$ splitting gives a Lens space or $S^2 \times S^1$.}
\item{Any two Heegard splittings of $S^3$ of the same genus are isotopic}
\item{If $\pi_1(M)$ has no presentation with fewer than $n$ generators, then
the genus of $M$ is at least $n$. e.g. $T^3$ has genus $3$}
\item{A handlebody is the unique irreducible manifold with free fundamental
group. This can be proved by showing, by the Whitehead theorem and the
Hurewicz theorem, that it is homotopy equivalent to a wedge of circles}
\end{enumerate}

\item{Scott's Theorem}

If a $3$-manifold group is finitely generated, then it is finitely presented.


\item{Haken's lemma - an essential sphere in a 3-manifold can be ``perturbed" to
meet a splitting surface in a loop (sketch)}

The idea is to look at the sphere which realizes the least number of
intersections with the splitting surface $F$, and then to show it is $1$.
By assumption of minimality, the intersections $\Sigma \cap V_i$ are
incompressible and boundary incompressible. Then by intersecting $\Sigma$
with a set of separating disks for $V_1$, say, we can perform compressions
and boundary-compressions along the circles and arcs of the intersections
to achieve fewer intersections with $F$ unless $\Sigma \cap V_i$ lies in a
collection of disjoint balls. $\Sigma \cap V_i$ therefore consists of a 
finite collection of disks, and therefore it consists of one disk and we are
done.

\item{Loop theorem and Dehn's Lemma}

Loop Theorem:
Let $M$ be any $3$-manifold and $F$ a connected $2$-manifold in $\partial M$.
If $N$ is a normal subgroup of $\pi_1(F)$ and if 
$(ker \pi_1(F) \to \pi_1(M)) - N \ne 0$ then there is a proper embedding
$g:(B^2,\partial B^2) \to (M,F)$ such that $g|\partial B^2]$ is not in $N$.

Dehn's Lemma:
Let $C$ be a circle in $M$. If there is an immersed disk with boundary $C$
such that the disk intersects $C$ only in its boundary, then there is an
embedded disk with boundary $C$.

Proof of Loop Theorem: We consider some map $(B^2,\partial B^2) \to (M,F)$
which kills a non-trivial loop in $F$. Let $V_0$ be a regular neighborhood of
its image. We lift $V_0$ to a double cover, if one exists. Then we look at
a regular neighborhood of a lift of the map into $V_0$ in the cover. Call this
$V_1$. We lift this to a double cover etc. until we can lift no more. This
process must terminate, since all lifts can be made simplicial with respect
to a fixed triangulation of $B$, and each lift identifies fewer simplices than
the lift before. By doing equivariant surgery, we may move ``down" the tower
to arrive at the required map.

\item{Sphere theorem (statement)}

If $M$ is any orientable $3$-manifold and if $N$ is any $\pi_1(M)$-invariant
subgroup of $\pi_2(M)$ with $\pi_2(M) - N \ne 0$ then there is an
embedding $g:S^2 \to M$ with $[g]$ not in $N$. If $M$ is not orientable, then
$g$ may by taken to be a covering map. (It is possible that it must be 
non-trivial. For example, $P^2 \times S^1$ contains no embedded $S^2$ but 
has $\pi_2 = {\Bbb Z}$)

\item{Prime decomposition theorem}

A manifold $M$ is {\it prime} if it cannot be written as a connect sum
$M = M_1 + M_2$ with neither of $M_1, M_2 = S^3$.

A manifold is {\it irreducible} if every embedded $S^2$ bounds a ball.
Note that any orientable prime $3$-manifold is irreducible unless it is
a twisted $S^2$ bundle over $S^1$.

Note that an orientable irreducible manifold with infinite $\pi_1$ is a
$K(\Pi,1)$.

We have the prime decomposition theorem: Any compact $3$-manifold can be
expressed as a connected sum of a finite number of prime factors. Moreover,
this expression is unique (up to order) up to the following ambiguity:
$$S^2 \times_\varphi S^1 + M = S^2 \times S^1 + M$$
for any nonorientable $M$

That is, if a twisted $S^2$ bundle over $S^1$ appears as a summand together with
a nonorientable $M$, we can ``slide" the bundle over an orientation-reversing
loop to make it orientable.

The existence follows from Haken's lemma. For, by Haken's lemma, Heegard
genus is additive under connect sum. Uniqueness follows after splitting off
the twisted $S^2$ bundles by comparing ``outermost" factors.

\item{Jaco Shalen Johannsen decomposition theorem (along embedded incompressible tori
into ``geometrizable" pieces) (statement)}

Any Haken $3$-manifold $M$ that is closed or has incompressible boundary
admits a canonical decomposition along incompressible tori and incompressible
and boundary-incompressible annuli into submanifolds that are either        
Seifert-fibered manifolds, $I$-bundles or simple Haken manifolds. In particular,
there is a pair $(\Sigma, \Phi) \subset (M,\partial M)$ known as the
{\it characteristic Seifert pair} such that any nondegenerate map from
a torus or an annulus into $(M, \partial M)$ is homotopic to a map into
$(\Sigma,\Phi)$. If it does not have incompressible boundary, we first
compress along disks until the boundary is incompressible. Note that this
activity reduces the sums of the genus of the boundary components. Since
by hypothesis, $M$ is irreducible, we can never obtain a $S^2$ boundary unless
it bounds a $3$-ball. Hence this process must stop.

Here a {\it simple} manifold $M$ is 
a manifold such that every two-sided
incompressible torus in $M$ is parallel into $\partial M$. All such with
infinite $\pi_1$ (e.g if $\partial$ is nonempty) are conjectured to be
Hyperbolic.

Note that if one of the ``pieces" was a torus bundle over $S^1$, it would
be better not to cut it in the first place, but rather to give it a 
$Sol$ structure.

\item{Lickorish-Wallace theorem (that any 3-manifold can be presented as
Dehn surgery on a link in $S^3$)}

This follows easily by a ``tunneling" argument if we can prove the
Lickorish Twist theorem; i.e. that the homeomorphisms of a closed surface
are generated by Dehn twists. 

Alternatively, we can use Rourke's argument: given a Heegard splitting
$S({\bf x},{\bf y})$, we can find a ${\bf z}$ with $min({\bf z}\cap {\bf x})$
and $min({\bf z} \cap {\bf y})$ strictly less than
$min({\bf x} \cap {\bf y})$, if this latter is not $0$ or $1$. (Here
$min({\bf x} \cap {\bf y})$ refers to the minimum intersection number of    
some components of ${\bf x}$ and ${\bf y}$.)

After performing integral Dehn surgeries on the bands of $z$ given by the
natural annular neighborhood on the splitting surface, we obtain
$$S({\bf x},{\bf y}) \to S({\bf x},{\bf z}) + S({\bf z},{\bf y})$$

Then inducting on the genus of $S$ and cancelling any trivial pair, to reduce
genus, we are done.


Note in fact that by the construction, the $3$-manifold can be obtained
by integral Dehn surgeries on $S^3$, and even integral surgeries with
$(1,1)$ or $(-1,1)$ framings. Hence $M$ bounds a $4$-manifold, and even
a simply-connected one.

\item{Kirby calculus (statement)}

Any two presentations of a manifold as an (integral) Dehn surgery on a
link in $S^3$ are related by a sequence of moves:
\begin{enumerate}
\item{adding or removing an unknot with a $\pm 1$ framing}
\item{handle sliding a $2$-handle over another $2$-handle - corresponds to
performing a framed band connect sum on two knots of the presentation}
\end{enumerate}

\item{Haken manifolds: incompressible surfaces, existence of hierarchies
(sketch), statement of Waldhausen's theorem}

A compact, orientable, irreducible $3$-manifold is called a {\it Haken
manifold} if it contains a two-sided incompressible surface. This is equivalent
to the statement that $M$ is irreducible and $\pi_1(M)$ splits nontrivially
as an amalgamated free product or a HNN extension. Note that sometimes the
condition of ``orientability" is dropped. Such manifolds are also called
``sufficiently large".

Note if $M$ is irreducible and orientable and $\partial M \ne 0$ then $M^3$ is 
Haken. For, the boundary components are not spheres (for otherwise $M$ would
be reducible) and therefore half of their $H_1$ is killed in $M$ by the
long exact homology sequence of $(M,\partial M)$. In 
particular, half of it is left, so $H_1(M)$ is infinite, and there exists a
map from $M$ to $S^1$ which is epic on $H_1$. Pulling back the preimage of
a point, and making this map transverse, we get a 2-sided surface which can
be surgered equivariantly to give an incompressible surface.

Given any Haken manifold $M$, there exists a hierarchy for $M$. That is, there
is a finite sequence $(M_1,F_1) \dots (M_n,F_n)$ where $M_i$ is Haken,
$F_i$ is a two-sided incompressible surface in $M_i$ which is not
boundary-parallel, and $M_{i+1}$ is obtained from $M_i$ by splitting along
$F_i$. Furthermore, $M_1=M$ and each component of $M_{n+1}$ is a $3$-cell.
This follows by normal surface theory, whereby one can show that there are
a bounded number of incompressible closed non-parallel $2$-sided 
non-intersecting surfaces in $M$.

We have the following important result for Haken manifolds:

Let $M$ and $M'$ be Haken manifolds. Suppose that $\pi_1(M) \ne 1$ and
$f:(M,\partial M) \to (M',\partial M')$ is a map such that
$f_*: \pi_1(M) \to \pi_1(M')$ is an injection. Then there is a homotopy
$f_t$ such that $f_0=f$ and either
\begin{enumerate}
\item{$f_1:M \to M'$ is a covering map}
\item{$M$ is a product $I$ bundle over a closed surface and $f_1(M) \subset 
\partial M'$, or}
\item{$M'$ (and hence also $M$) is a solid torus and $f_1$ is a branch covering
with branching locus a circle.}
\end{enumerate}

This is known as {\it Waldhausen's theorem}.

There is an analogous statement for surfaces 
(all of which admit a ``hierarchy") which is the first step in the proof of 
this theorem.

\item{Statement of Geometrization Conjecture}

Each of the pieces of the Jaco-Shalen-Johannsen decomposition theorem admits
exactly one complete geometric structure based on one of the following eight
geometries: 
If the piece is Seifert-fibered, then if the Euler characteristic
of the base orbifold and the Euler class of the bundle are
$$(>0,\ne 0),(>0,=0),(=0,\ne 0),(=0,=0),(>0,\ne 0),(>0,=0)$$
then the geometric structure will be
$${\Bbb S}^3, {\Bbb S}^2 \times {\Bbb R}, Nil, {\Bbb E}^3, 
{\widetilde SL}_2({\Bbb R}), {\Bbb H}^2 \times {\Bbb R}$$
If the piece is covered by a torus-bundle over $S^1$ with Anosov monodromy,
it has a $Sol$ structure.
Finally, if it is none of these things, it admits a hyperbolic structure.
Note also that every $M$ with $|\pi_1(M)| < \infty$ should be covered by
${\Bbb S}^3$ as $M = {\Bbb S}^3/\Gamma$ with $\Gamma \subset SO(4)$

Note: this theorem has been proved for Haken manifolds, for surface bundles
over $S^1$ and for manifolds admitting a group action which is free on the
complement of a codimension 2 set. It is a theorem of Thurston that the
geometries indicated above are the only maximal 
simply connected manifolds admitting a compact manifold quotient for which
the group of isometries acts transitively on the points of the space.

Note that the manifold $P^3 + P^3$ though not prime, admits a $S^2 \times R$
structure.

\item{Geometrization of Seifert-fibered pieces (i.e.
classification by Euler characteristic of base orbifold and Euler number of
circle bundle) (statement)}

As above. Note that for Seifert-fibered pieces the Geometrization Conjecture
holds. Every Seifert-fibered manifold is Haken unless it has base space a 
triangle orbifold and has $H_1$ finite.

Also, we have the formula for the Euler number $e(E) = -(b + \Sigma_i \beta_i
/\alpha_i)$ and the base orbifold has marked points with angles 
$2 \pi/\alpha_i$, for a standard surgery presentation of the manifold.

\item{Mostow's rigidity theorem}

For $n\ge 3$ every finite volume hyperbolic manifold $M$ admits a unique
hyperbolic structure. In particular, if $M$, $N$ are two finite volume
hyperbolic manifolds and their fundamental groups are isomorphic, then there 
is an isometry between them. Note that because $M$,$N$ are $K(\pi,1)$'s,
they are homotopy equivalent by the hypothesis. The proof proceeds as follows
in the case $M,N$ are compact:

\begin{enumerate}
\item{First we lift a homotopy equivalence $f:M \to N$ to the universal
covers. This map is then a quasi-isometry, since in one direction, it
magnifies lengths by a bounded amount, and then in the other direction, its
inverse magnifies lengths by a bounded amount and moreover is, up to some
constant, an inverse to $f$. It can be shown that a quasi-isometry extends
continuously to a map from $\bar {\Bbb H}^3$ to itself (i.e. it extends
continuously to the boundary)}
\item{Next we show that the following equality holds:
$$vol(M) = vol(N) vol(\sigma)/average(vol(f(\sigma)))$$
Where $v$ is the volume of a regular ideal hyperbolic simplex, and the
average is with respect to some set of coset representatives for $f$ under the
action of $\Gamma_1$. In particular, it is the average volume of the
{\it straightened} simplices with the same endpoints as $f(\sigma)$ where
$\sigma$ ranges over the regular ideal simplices of ${\Bbb H}^3$. To see this,
observe that we can define a cycle $\alpha$ by
$$\alpha = \int_{\Gamma/G} sign(g) \pi(g\sigma) dg$$
Then this is clearly closed and we have a homology class $[\alpha]$.
Now we can evaluate this against the volume form $\Omega$ to obtain
$$<[\alpha],\Omega> = \int sign(g) \int_{\pi(g\sigma)} sign(g)
\Omega dg$$
$$= \int volume(g\sigma) dg$$
$$= volume(\sigma)$$

Now, evaluating its (straightened) image (which defines the same
homology class) against the volume form $\Omega'$ of $N$ we get
$$<h_*[\alpha],\Omega'> = average(volume(f\sigma))$$
However, we have
$$h^* \Omega' = \Omega vol(N)/vol(M)$$
So that we have
$$<h_*[\alpha],\Omega'> = <[\alpha],h^*\Omega'>$$
$$= (vol(N)/vol(M))<[\alpha],\Omega>$$
$$= (vol(N)/vol(M))(vol(\sigma))$$
Therefore the estimate
$$vol(M) = vol(N) volume(\sigma)/average(volume(f\sigma))$$
Since $\sigma$ can have volume as close to the maximum possible as we like,
we can see that $vol(N)\le vol(M)$. Similarly, $vol(M) \le vol(N)$. Therefore
$vol(M)=vol(N)$ and $f$ takes ideal simplices to ideal simplices.}
\item{Finally, any map which preserves ideal simplices is an isometry and we
are done, by extending the action of $\partial {\Bbb H}^3$ to all of
${\Bbb H}^3$.}

Note since $M,N$ are $K(\Pi,1)$'s, any isomorphism of their fundamental
groups implies they are homotopy equivalent, and therefore isometric.
\end{enumerate}

\item{Margulis' lemma}

This theorem states that there is a universal constant $\epsilon$ such that
for $\Gamma$ a discrete group of isometries of ${\Bbb H}^3$, the subgroup
$\Gamma_\epsilon$ which is generated by elements which 
translate points a distance $\le \epsilon$ is
virtually nilpotent. (That is, it contains a subgroup of finite index which is
nilpotent). A brief sketch of a proof is given as follows:

\begin{enumerate}
\item{For any Lie group $G$, the differential of the lie bracket as a smooth
function $[,]:G\times G \to G$ at the identity is zero. For, $[g,id] = id$
and $[id,g] = id$ for all $g$, and the tangents to these subspaces span
the tangent space to $id \times id$. Hence for elements $g,h$ 
sufficiently close
to $0$, $d([g,h],id) \le 1/2 \max(d(g,id),d(h,id))$. }
\item{By proper discontinuity of $\Gamma$, there is an $\alpha$ such that
every element of $\Gamma$ is at least $\alpha$ from the identity. Then choosing
$m$ so that $\epsilon/2^m < \alpha$, the $m$-commutators of the generators of
the group in question are trivial. Since any $m$-commutator in this
group can be written as a product of $m$-commutators of
generators, this group is nilpotent.}
\item{Finally, we show that the group generated by small translations contains
a subgroup of this with finite index. This follows from the compactness of
the point stabilizers of $Isom^+{\Bbb H}^3$ and the fact that parallel
transport around the boundary of a geodesic triangle in $X$ can be controlled
in a universal way by the diameter of the triangle.}
\end{enumerate}

Consequently, we have a classification of the $\epsilon$-thin pieces of
the manifold as product neighborhoods of short geodesics, or $N \times
{\Bbb R}$ for $N$ a Euclidean manifold.

\item{Ideal triangulations}

By a theorem of Thurston, any hyperbolic $3$-manifold has a degree one
decomposition into ideal tetrahedra. If the manifold is non-compact, this
can be taken to be an ordinary decomposition.

Given a collection of ideal tetrahedra and some glueing maps on their
boundaries, we can build a hyperbolic manifold. The condition that this
will be a manifold is firstly that the pieces fit nicely around each edge
(edge consistency conditions) and secondly that the holonomy of
the boundary elements be
parabolic (cusp conditions). Each of these conditions gives an integral
linear combination in $log(z_i), log(1-z_i)$ for the simplex parameters $z_i$
as a multiple of $2\pi i$. 

The set of ideal simplex parameters for a triangulation of a hyperbolic manifold
satisfy the following identity:
$$\Sigma_i (1-z_i) \wedge z_i = 0$$
in $\Lambda^2\bar {\Bbb Q}^*$
This formula captures the fact that the Dehn invariant of any constant
curvature manifold vanishes.

It is a fact that any two ideal triangulations of a manifold determine the
same element of the Bloch group - that is, they are related by cocycle moves,
up to homology. These cocycle moves cannot necessarily be performed in the
manifold, however.

\item{Geometric Dehn surgery theorem (statement)}

For any hyperbolic link complement $M$ in $S^3$, with components $L_1 \dots L_n$
there is an open neighborhood $\Omega$ of $\infty,\infty \dots \infty$ in
$S^2 \times S^2 \times \dots \times S^2$ such that any generalized
geometric Dehn surgery in $\Omega$ gives a complete hyperbolic structure 
on $M \cup_i \gamma_i$ with singular set either geodesics or points as
appropriate.

Moreover, $vol(M_\infty) > vol(M_{p,q})$ for any $(p,q) \ne \infty$.

\end{itemize}

\section{Minor Topic: Complex Analysis}

\begin{itemize}
\item{A function satisfies the Cauchy-Riemann equations iff it has a local
power series expansion (statement)}

This theorem is self explanatory. However, it is proved by using the
Weierstrass theorem, which states that if $f_n(z)$ is analytic in a region
$\Omega$ and the sequence $f_n$ converges uniformly on compact sets to
$f$, then $f$ is analytic in $\Omega$, and moreover $f_n'$ converges uniformly
to $f'$ on every compact subset of $\Omega$.

Note also that having the real and imaginary parts of $f$ harmonic implies
that $f$ is $C^\infty$ by elliptic regularity.

\item{Schwarz-Pick lemma}

An analytic map between Riemann surfaces decreases the hyperbolic metric
unless it is a covering. This follows in a straightforward way from the
maximum modulus priciple and the classification of automorphism of the disk
as functions satisfying the formula
$$\frac {w - w_0} {1 - \bar w w_0} = e^{i \theta} \frac {z - z_0} {1- \bar z z_0}$$
which preserves the infinitesimal form
$$\frac {2dz} {1 - |z|^2}$$ which is the Poincar\' e metric.

\item{Koebe $1/4$ theorem}

A univalent function from $D$ to ${\Bbb C}$ with $|f'(0)|=1$ fixing $0$
contains
the disk of radius $1/4$ about the origin.

Together with the last theorem, this implies that the Poincar\' e metric
on a domain $\Omega$ is quasi-isometric to $\frac {dz} {d(z, \partial \Omega)}$


\item{Cauchy integral formulae}

Generally, for a simple loop $\gamma$ in a region $\Omega$ in which
$$f(z) = a_{-n}(z-w)^{-n} + \dots + a_{-1}(z-w)^{-1} + a_0 + \dots$$
we have the formula
$$a_m = \int_\gamma \frac {f(\zeta)} {(\zeta-w)^{m+1}} d\zeta$$

\item{Rouch\' e's theorem}

If on some curve $\gamma$ we have the estimate that $|f| > |f-g|$ then
the holomorphic function $f$ has the same number of zeroes enclosed by
$\gamma$ as $g$ does. This is otherwise known as the ``dog on a leash"
theorem.

\item{Argument principle}

$$1/2\pi i \int_\gamma \frac {f'(z)} {f(z)} dz$$ is
equal to the number of zeroes minus the number of poles of $f$ in the
region enclosed by $\gamma$.

For, suppose $f$ has no poles. Then in the region in question, write it
as $(z-\alpha_1)(z-\alpha_2) \dots (z-\alpha_n)k(z)$ where $k$ does not
vanish. Then
$$f'(z)/f(z) = \sum_i \frac 1 {(z-\alpha_i)}$$ which integrates to give
the number $n$.

\item{modulus of a quadrilateral; modulus of an annulus}

For any quasicircle $Q$ there is a unique quadrilateral $M$ and a conformal
map $Q \to M$ extending to a continuous map of the boundaries which takes
any four points on $Q$ to the corners of $M$. The modulus of $Q$ is then the
ratio of the two sides of $M$.

Similarly, for any annulus $A$ there is a unique standard annulus $N$ and
a conformal map $A \to N$ extending to a continuous map of the boundaries.
This follows from Andre\' ev's theorem.
The log of the ratio of the inner and outer circles of $N$ is the modulus
of $A$.

\item{Riemann mapping theorem}

Any simply connected domain $\Omega$ in ${\Bbb C}$ which omits at least two
points is conformally equivalent to the unit disk $D$. The proof
consists in first constructing some conformal map into the unit disk 
and then by means
of the Schwartz lemma, and simple facts about normal families, constructing
a maximal such map which must therefore be onto. To
construct some conformal map: first take a single-valued square root of
$(x-p)(x-q)$ where $p,q$ are the two omitted values. The image of this
omits some disk, so by means of a suitable Mobius transformation, we can
put it into the unit disk.

\item{Reflection principle}

If an analytic map $f:A \to B$ maps some circular 
arc $a$ of $\partial A$ to some circular arc $b$ of $\partial B$ we can analytically
continue $f$ from $A \cup \bar A$ to $B \cup \bar B$ where $\bar A$ is the
reflection of $A$ in this arc, and $\bar B$ the reflection of $B$ is the other.
If, more generally, we merely have that for any $z_i \to a$ that $f(z_i) \to b$,
then by a suitable estimate, and use of Morera's theorem, we can make the
same claim.

\item{Morera's theorem}

If a function satisfies Cauchy's integral formula in some domain, and is
continuous there, then it is analytic. Here we use the fact that
$$g(\zeta) = \frac 1 {2 \pi i} \int_\gamma \frac {\varphi(z)}
{(z - \zeta)}$$ is analytic for any continuous $\varphi$. 

\item{Definition of a normal family - condition for a family to be normal}

A family is {\it normal} on some domain $\Omega$ if each sequence of functions
contains a convergent subsequence (i.e. convergent uniformly on
compact subsets). By Arzela-Ascoli, it is sufficient that
the family be equicontinuous and locally bounded. In fact, it is sufficient
to assume the family is locally bounded. In fact, it is enough to assume the
sequence omits two values in ${\Bbb C}$.

\item{Liouville's theorem}

A bounded entire holomorphic function is constant. This follows from the
estimate for $f'$ obtained from
$$f'(\zeta) = \frac 1 {2 \pi i} \int_\gamma \frac {f(z)} {(z-\zeta)^2}$$

\item{Picard's theorem}

An entire holomorphic function which omits two values is constant. This
follows by a developing argument. (Use the modular invariant $\lambda$ to
uniformize the triply punctured sphere)

\item{Maximum modulus principle}

A function holomorphic in an open set $\Omega$ does not attain a maximum
on $\Omega$ unless it is constant. Equivalently, non-constant 
holomorphic maps are open.

\item{The real and complex parts of an analytic function are harmonic}

Follows immediately from the Cauchy-Riemann equations:
if $z=x+iy$ and $f(z) = u+iv$ then we have for an analytic function,
$$\frac {\partial u} {\partial x} = \frac {\partial v} {\partial y}$$
and
$$\frac {\partial u} {\partial y} = -\frac {\partial v} {\partial x}$$

\item{Gauss' lemma (the value of a harmonic function at the center of a
circle is the average of its value on the circle)}

This just follows from the above remark and the Cauchy integral formula for
a circle. In particular, any function $\varphi$ on $\partial D$ can be  
extended to a harmonic function on $D$ by the ``visual extension".

\item{Definition of Weierstrass $\wp$ function}

For any lattice $\Lambda = {\Bbb Z} \oplus \tau {\Bbb Z}$ in ${\Bbb C}$,
we define the function
$$\wp(z) = \frac 1 {z^2} +  
\sum_{\omega \in \Lambda} (\frac 1 {(z-\omega)^2}
- \frac 1 {\omega^2}) $$

This satisfies a functional equation
$$\wp'(z)^2 = 4\wp(z)^3 - 60g_2 \wp(z) - 140 g_3$$
where $g_2$ is the sum of the reciprocals of the fourth powers of elements in
$\Lambda$, and $g_3$ is the sum of the reciprocals of the sixth powers.
The cross product of the roots of this cubic and $\infty$ is the modular
invariant $\lambda(\tau)$.

\end{itemize}

\end{document}