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\begin{document}
\title{Four-dimensional Manifolds: An introduction}
\author{Kevin Iga}
\date{August 13, 1995}
\maketitle
\section{Introduction}
\section{The Classical Theory}
Before 1980, relatively little was known about four-dimensional manifolds.
Most of this knowledge concerned the homotopy type of a four-dimensional
manifold.
\subsection{The Homology and Cohomology of a Four-Manifold}
In this section we assume $M$ is a four-dimensional closed (compact and without
boundary) topological manifold, and we wish to study properties of $M$ that
only depend on its homotopy type. So, of course, the results still hold in
the PL (Piecewise Linear) and smooth categories.
First, every closed topological manifold of dimension n has finitely generated
homology groups (Spanier, ch 6 p342), and thus finitely generated cohomology
groups, by the Universal Coefficient Theorem. Furthermore, if $M$ is
orientable, Poincar\'e Duality says that if $G$ is an abelian group, then
\[H^q(M;G)\cong H_{n-q}(M;G).\]
Even if $M$ is not orientable, $M$ has an orientable double cover
which is a closed topological orientable manifold, so Poincar\'e
Duality holds for this double cover. In either case,
$H_q(M;R)\cong H^q(M;R)\cong 0$ for $q>n$.
We will furthermore decompose $M$ into its path components, so from now on we
will assume $M$ is connected. Since if $M$ is not orientable, $M$ has an
orientable double cover which is an orientable closed topological manifold,
we can study $M$ by studying its orientable double-cover, we will assume
$M$ is orientable.
Under these assumptions, $H_0(M)\cong {\bf Z}$ and $H^0(M)\cong {\bf Z}$, and Poincar\'e
Duality holds, so $H_4(M)\cong H^4(M)\cong {\bf Z}$
Now we write
\[H_1(M)\cong {\bf Z}^{b_1}\oplus T\]
where $T$ is the torsion
subgroup and $b_1$ is the rank of $H_1$, called the first betti
number. By the Universal Coefficient Theorem,
\[H^1(M)\cong {\bf Z}^{b_1}.\]
By Poincar\'e Duality, we have
\[H^3(M)\cong {\bf Z}^{b_1}\oplus T\]
\[H_3(M)\cong {\bf Z}^{b_1}.\]
If we write
\[H_2(M)\cong {\bf Z}^{b_2}\oplus T'\]
then by the universal coefficient theorem,
\[H^2(M)\cong {\bf Z}^{b_2}\oplus T\]
(recall that $\Hom(T,Z)\cong 0$ and $\Ext(T,Z)\cong T$ if $T$ is torsion). But by
Poincar\'e Duality, $H_2(M)\cong H^2(M)$, so $T\cong T'$. Summarizing, we have the
following homology and cohomology groups:
\begin{eqnarray}
H_0(M)\cong {\bf Z} && H^0(M)\cong {\bf Z} \\
H_1(M)\cong {\bf Z}^{b_1}\oplus T && H^1(M)\cong {\bf Z}^{b_1}\\
H_2(M)\cong {\bf Z}^{b_2}\oplus T && H^2(M)\cong {\bf Z}^{b_2}\oplus T\\
H_3(M)\cong {\bf Z}^{b_1} && H^3(M)\cong {\bf Z}^{b_1} \oplus T\\
H_4(M)\cong {\bf Z} && H^4(M)\cong {\bf Z}
\end{eqnarray}
\subsection{The Intersection Form}
Poincar\'e Duality is more than a statement about homology and
cohomology groups; it also says that the cup product induces the
isomorphism in the following sense: if $x\in H^q(M;R)$ and $y\in
H^{n-q}(M;R)$ ($R$ a commutative ring), then $x\cup y=\langle
x,PD(y)\rangle =\langle y,PD(x)\rangle$, where $PD$ is the Poincar\'e
Duality isomorphism, and $\langle,\rangle$ is the evaluation map of
cohomology cycles on homology cycles. If $R$ is a field, then this
evaluation map is the duality map between vector duals. So the
bilinear pairing
\[\cup:H^q(M;R)\otimes_R H^{n-q}(M;R) \rightarrow H^n(M;R)\cong R\]
is {\em perfect}, that is, the induced adjoint map
\[\cup*:H^q(M;R)\rightarrow (H^{n-q}(M;R))^* \otimes_R R \cong
(H^{n-q}(M;R))^* \]
is an isomorphism.
So the cup product between $H^q$ and $H^{n-q}$ in differing dimensions
gives the evaluation map, while on $H^2$ is a bilinear map, which over
fields is perfect. The pairing on $H^2$ is called the {\bf intersection
form}, since $x\cup y=PD(x)\cap PD(y)$ where $\cap$ denotes the number
of intersections between two surfaces representing the homology classes
which intersect transversally, counted with signs. (For a full account,
see Fulton, Intersection Theory). Since $a\cup b=(-1)^{|a||b|}b\cup a$,
and on $H^2$, $|b|=|a|=2$, the intersection form is symmetric.
Any homomorphism between groups $G\rightarrow G'$ induces homomorphism
$H_q(M;G)\rightarrow H_q(M,G')$ and $H^q(M;G)\rightarrow H^q(M,G')$ and
if $G$ and $G'$ are rings, and the homomorphism a ring homomorphism, then
the induced homomorphism on cohomology is a ring homomorphism.
So if we take the inclusion ${\bf Z} \subset {\bf Q}$, the induced ring
homomorphism $H^q(M)\rightarrow H^q(M;{\bf Q})$ shows that on the free
part of $H^2$, the intersection form is a perfect pairing. Since
$\Hom(T\otimes T,Z)\cong 0$ and $\Hom(T\otimes Z,Z)\cong 0$, and since the cup product
is bilinear, $x\cup y=0$ if $x$ or $y$ is in $T$.
So the intersection pairing is supported completely on the free part of $H^2$,
where it is a perfect pairing to ${\bf Z}$. Given a basis on the free
part of $H^2$, we get a matrix whose entries are integers and whose determinant
is $\pm 1$. Such a matrix is called {\em unimodular}.
To summarize, we have the following information:
\begin{enumerate}
\item The first betti number $b_1$.
\item The second betti number $b_2$.
\item The torsion group $T$.
\item The intersection form $\cup:{\bf Z}^{b_2}\otimes {\bf Z}^{b_2}
\rightarrow {\bf Z}$, which is a symmetric unimodular bilinear form.
\end{enumerate}
Of course, the last form depends on a chosen basis for $H^2(M)$, and
changing the basis changes the form in the usual way, i.e. $A'=Q^tAQ$.
\subsection{Unimodular Symmetric Bilinear Forms}
To better understand this last invariant, let us study unimodular symmetric
bilinear forms over the integers as algebraic objects. In order for an
invariant to be of any use, we must be able to tell when two invariants
are the same or not. For instance, consider the invariant of a manifold
which is the manifold itself. This is not a useful invariant, since if we
wanted to see if two manifolds were the same or not, applying the invariant
would leave us with the same problem we started with.
Similarly, the intersection form is an invariant of the manifold, but one
which is difficult to use. Given two four-dimensional manifolds, we may be
able to compute their intersection forms, but can we tell if these intersection
forms are the same (up to change of basis)?
Unimodular, symmetric bilinear forms are, in general, not well understood. So
with the current state of knowledge, the answer would be no.
But we can find numerical invariants of such forms which are easy to compute,
and this gives us useful invariants.
The first invariant of a unimodular, symmetric bilinear form is its
rank. That is, the size of the space it acts on (since the associated matrix
has determinant $\pm 1$). In our case, this is simply $b_2$.
The next invariant comes when you consider the form as a form on the reals.
We can do this, of course, since the reals are a module over the integers,
so we can take the tensor product of a ${\bf Z}$ module with ${\bf R}$ over
${\bf Z}$ and get an ${\bf R}$ module, and we can extend the form over this
vector space. Over ${\bf R}$ these forms are well-understood. They are
diagonalizable, and by rescaling basis vectors, we can make the diagonal
entries 1, 0 or -1. Since this matrix must be unimodular, none of the
diagonal entries are 0. So we have two numbers, the dimension of the
positive space $b_2^+$, and the dimension of the negative space $b_2^-$.
Of course these are not independent since $b_2^++b_2^-=b_2$. It is
traditional to use $\sigma=b_2^+-b_2^-$, which is called the {\em signature},
instead of $b_2^+$ and $b_2^-$.
Whenever $b_2^+=0$, we say the form is {\em negative definite}, and
whenever $b_2^-=0$, we say it is {\em positive definite}. If either is
the case, we say it is {\em definite}, and if neither is the case we say
it is {\em indefinite}.
Doing the same trick but with ${\bf Z}_2$ instead of ${\bf R}$ gives
yet another invariant. Over ${\bf Z}_2$ we cannot necessarily
diagonalize a matrix, but we can come close. A matrix consists of 1's
and 0's. If there is a 1 in the diagonal, reorder the basis so that
it is the first. We can use row operations to remove 1's in the first
column except in the first row. Doing so also performs column
operations to remove 1's in the first row except in the first column.
We can continue this process until the form splits into $A\oplus B$
where $A$ is the identity and $B$ has only 0's on the diagonal. Now
$B$ must still be unimodular, so for the first column there is a row
such that there is a 1. Reorder the basis so that this is the second
row. We now perform row operations to remove 1's from the first two
columns except in the first two rows. Doing so performs column
operations to reomve 1's from the first two rows except in the first two
columns.
Continuing this procedure we get a direct sum of spaces where the form is
the identity, and where the form is
\[H=\left(\begin{array}{ll}
0 & 1\\
1 &0
\end{array}\right)\]
If there is at least one $(1)$ and at least one $H$, we can do operations to
turn the $H$ into two $(1)$'s as follows:
\[
\left(\begin{array}{lll}
1&0&0\\
0&0&1\\
0&1&0
\end{array}\right)\rightarrow
\left(\begin{array}{lll}
1& 1& 0\\
1& 1& 1\\
0& 1& 0
\end{array}\right)\rightarrow
\left(\begin{array}{lll}
1& 0& 0\\
0& 1& 1\\
0& 1& 0
\end{array}\right)\rightarrow
\left(\begin{array}{lll}
1& 0& 0\\
0& 1& 0\\
0& 0& 1
\end{array}\right)
\]
So we will either have the identity or a direct sum of spaces with $H$.
If there is at least one 1 on the diagonal, we get the former, otherwise
we get the latter.
At any rate, any transformation preserves the property of having only
zeros on the diagonal, since if $\{e_1,...,e_n\}$ is a basis so that
$\langle e_i,e_i\rangle=0$, then if $v=\sum a_i e_i$, then
\begin{eqnarray}
\langle v,v\rangle &=& \sum_{i,j} a_i a_j \langle e_i,e_j \rangle\\
&=&\sum_i a_i^2 \langle e_i,e_i\rangle + 2\sum_{i0$. Find a maximal linear subspace on which $\rho$ is positive
definite. Let $V^+$ be this space. Let $V^-$ be the $\rho$-orthogonal
complement to $V^+$. Since $\rho(v,v)>0$ for all $v\in V^+$, we have
that $V^+\cap V^-=0$. Furthermore, $V^++V^-=V$, since if $x\in V$, we can
apply Gramm-Schmidt to make $(x,v)=0$ for all basis vectors $v$ in $V^+$.
Now if $w\in V^-$, then $\rho(w,w)\le 0$, since if $\rho(w,w)>0$, then
\[\rho(av+bw,av+bw)=a^2\rho(v,v)+b^2\rho(w,w)>0\]
for all $a,b\in {\bf Z}$ and $v\in V^+$, so the space spanned by $V^+$ and
$w$ is positive definite, contradicting maximality of $V^+$.
Also, if $w\in V^-$, then $\rho(w,w)<0$, since if $\rho(w,w)=0$, then for all
$v\in V^+$, $\rho(v,w)=0$, and for all $u\in V^-$,
\[\rho(au+bw,au+bw)=\rho(u,u)a^2+ 2\rho(u,w)ab\]
and I can clearly choose $b$ so that this quantity is positive, unless
$\rho(u,w)=0$. So since these vectors are supposed to be in $V^-$, we must
have $\rho(u,w)=0$ for $u\in V^-$. From this we conclude that $\rho(x,w)=0$
for all $x\in V$, but this means $\rho$ is degenerate, which contradicts
our assumption.
So $V^+$ is positive-definite, $V^-$ is negative-definite, and $V$ is the
orthogonal direct sum of $V^+$ and $V^-$.
\qed
\begin{lemma}
If $\rho$ is indefinite (symmetric, unimodular), then there exists
$v\in V$ such that $\rho(v,v)=0$.
\end{lemma}
Proof of lemma: [see Milnor and Husemoller, Symmetric Bilinear Forms, Springer-Verlag, 1973]
First, we note that when we tensor with the rationals ${\bf Q}$, we can
diagonalize our form. By unimodularity, the diagonal elements are $\pm 1$.
By the indefiniteness, there is at least one (rational) vector $x$ with
$\rho(x,x)=1$ and at least one (rational) vector $y$ with $\rho(y,y)=-1$,
and such that they are orthogonal to each other.
The linear combination $x+y$ has $\rho(x+y,x+y)=1-1=0$. So there is a rational
vector with the desired property. Multiplying by a suitable integer, we get
an integral vector $v$ with $\rho(v,v)=0$.
\qed
Continuing proof of theorem:
[Kirby]
Choose an indivisible $v$ with $\rho(v,v)=0$. There is a $w$ with
$\rho(v,w)=1$. Let $U$ be the orthogonal complement to the space spanned
by $v$ and $w$. If $\rho(w,w)$ is even, then take a vector $u$ in $U$ with
$\rho(u,u)$ odd. (This is possible since $\rho$ is odd and neither $v$ nor
$w$ have odd norms). Then substitute $w+u$ for $w$. Then $\rho(w+u,w+u)=
\rho(w,w)+\rho(u,u)$ which is odd. So we have a subspace where the
form is the matrix
\[\left(\begin{array}{ll}
0 & 1\\
1 & \makebox{odd}
\end{array}\right).\]
Replacing $w$ with $w-kx$ for some $k$, we get the matrix
\[\left(\begin{array}{ll}
0 & 1\\
1 & 1
\end{array}\right).\]
Then replacing $v$ with $v-w$, we get
\[\left(\begin{array}{ll}
-1 & 0\\
0 & 1
\end{array}\right).\]
Choose either $v$ or $w$ so that its orthogonal complement is indefinite,
and repeat the argument until we have the decomposition.
\qed
\subsection{Characteristic Elements}
In this section we investigate another curious feature of unimodular symmetric
forms over ${\bf Z}_2$ which will be useful later: the existence of a
characteristic element.
Let $V$ be a finitely generated vector space over ${\bf Z}_2$. Let
$\rho(x,y):V\otimes V\rightarrow {\bf Z}_2$ be a unimodular symmetric
form over the integers modulo 2. Then the ``norm'' function
$\mu(x)=\rho(x,x)$ is actually {\em linear}. We can see this by
\begin{eqnarray}
\mu(x+y)&=&\rho(x+y,x+y)\\
&=&\rho(x,x)+\rho(x,y)+\rho(y,x)+\rho(y,y)\\
&=&\rho(x,x)+2\rho(x,y)+\rho(y,y)\\
&=&\rho(x,x)+\rho(y,y)
\end{eqnarray}
since we are working over ${\bf Z}_2$. So this linear function is an element
of $V^*$. Now the adjoint of $\rho$ is $\rho^*:V\rightarrow V^*$, and the
unimodular condition means it is an isomorphism. So we can take
\[w={\rho^*}^{-1}(\mu)\]
so that $\rho(x,x)=\rho(x,w)$ for all $x\in V$. Since $\rho^*$ is
one-to-one, such a $w$ is unique. We say $w$ is the {\em characteristic
element} of $\rho$.
Examples:
{\em The form (1)}
If the vector space is 1-dimensional, and $\rho(x,x)=1$, then the
characteristic element is $x$.
{\em Even intersection forms}
If the intersection form is even, then $\mu(x)=0$ for all $x$. Certainly
this is true for a basis element, for any basis, and $\mu$ is linear. So
the characteristic element is 0.
{\em Direct sums}
If $\rho_1$ is such a bilinear form on $V_1$ and $\rho_2$ such a bilinear
form on $V_2$, then we can consider the bilinear form $\rho_1\oplus\rho_2$
on $V_1\oplus V_2$, defined to be $\rho_1\oplus\rho_2 (v_1\oplus v_2,
w_1\oplus w_2)=\rho_1(v_1,w_1)\oplus\rho_2(v_2,w_2)$. In this case, the
characteristic element of $\rho_1\oplus\rho_2$ is the characteristic
element of $\rho_1$ direct summed with the characteristic element of $\rho_2$.
{\em Negatives}
If $\rho$ is a symmetric unimodular bilinear form on $V$, then so is
$-\rho$. Now if $v$ is a characteristic element of $\rho$, then
$-\rho(x,x)=-\rho(x,v)=\rho(x,-v)$, so $-v$ is a characteristic element
of $-\rho$.
{\em $\oplus(1)^p\oplus\oplus^q(-1)$}
By our calculations on $(1)$, direct sums, and negatives, if we have a basis
$e_1,...,e_k,f_1,...,f_l$, and so that $\rho(e_i,e_j)=\delta^i_j$, and
$\rho(f_i,f_j)=\delta^i_j$, and $\rho(e_i,f_j)=\rho(f_j,e_i)=0$, then
our characteristic element is $\sum e_i-\sum f_i$.
\subsection{Characteristic Elements for intersection forms over ${\bf Z}$}
If we have an intersection form over V, a ${\bf Z}$ module, we can
take the tensor product with ${\bf Z}_2$, and study the characteristic
element $w\in V\otimes {\bf Z}_2$. Now since
\[\pi:V\rightarrow V\otimes {\bf Z}_2\]
is surjective, there are preimages $\pi^{-1}(w)\in V$, which we call
{\em characteristic elements} of V. If we denote one
characterstic element by $W_0$, then the set of characteristic elements is
$W_0+2V$. That is, the difference between any two characteristic elements is
twice an element of $V$. So we have:
\begin{theorem}
$V$ has characteristic elements, and any two characteristic elements in
$V$ differ by twice an element of $V$.
\end{theorem}
From the characteristic property of $w$, we can conclude a similar property for
a $W$:
\begin{theorem}
If $W$ is a characteristic element for $V$, then $\rho(W,X)-\rho(X,X)$ is even
for all $X\in V$.
\end{theorem}
Proof:
Apply $\pi$ to the expression. Since
$\rho(w,\pi(X)))=\rho(\pi(X),\pi(X))$, the theorem follows.
\qed
\begin{theorem}
The norm $\rho(W,W)$ modulo 8 is independent of the choice of characteristic
element $W$.
\end{theorem}
Proof:
Suppose $W_1$ and $W_2$ are two characteristic elements. Then $W_2=W_1+2X$
for some $X\in V$.
\begin{eqnarray}
\rho(W_2,W_2)&=&\rho(W_1+2X,W_1+2X)\\
&=&\rho(W_1,W_1)+4\rho(W_1,X)+4\rho(X,X)\\
&=&\rho(W_1,W_1)+4(\rho(X,X)+2K)+4\rho(X,X)\\
&=&\rho(W_1,W_1)+8\rho(X,X)+8K
\end{eqnarray}
\qed
\begin{theorem}
The signature $\sigma$ of a 4-manifold is equal to the norm of a characteristic
element $\rho(W,W)$, modulo 8.
\end{theorem}
Proof:
If the intersection form is odd and indefinite, then there is a basis in which
the form is $\oplus^p (1) \oplus \oplus^q(-1)$. The signature is $p-q$, and
the sum of the basis elements is a lift of the characteristic element, and
its square is $p-q$. So the theorem is true for odd, indefinite forms.
If the intersection form $\rho$ is even, or if it is definite, then we
take the direct sum $\rho'=\rho\oplus (1)\oplus (-1)$, which is an
odd, indefinite form. The signature of a direct sum is the sum of the
signatures, so the signature of $\rho'$ is the signature of $\rho$. The
lift of the characteristic element of $\rho'$ is the lift of the
characteristic element of $\rho$ direct summed with the characteristic
element for $(1)\oplus (-1)$, and the square gives the square for the
characteristic element of $\rho$. So this case reduces to the odd,
indefinite case.
\qed
\begin{corollary}
The signature of an orientable closed four-manifold with even intersection
form is a multiple of 8.
\end{corollary}
Proof:
We can take 0 as a lift of the characteristic element.
\qed
\subsection{Even intersection forms}
In the last section, we showed that the signature of any even
intersection form is a multiple of 8. Now there is an even symmetric
unimodular bilinear form over {\bf Z} whose signature is 8. It is the
Cartan matrix for the Lie Group $E_8$ and is given by:
\[E_8=\left(\begin{array}{llllllll}
2&1&0&0&0&0&0&0\\
1&2&1&0&0&0&0&0\\
0&1&2&1&0&0&0&0\\
0&0&1&2&1&0&0&0\\
0&0&0&1&2&1&0&1\\
0&0&0&0&1&2&1&0\\
0&0&0&0&0&1&2&0\\
0&0&0&0&1&0&0&2
\end{array}\right)\]
To prove this is unimodular, we compute its determinant. To
facilitate this, we invent another matrix and compute its determinant.
The determinant of the $n\times n$ matrix
\[A_n=\left(\begin{array}{llllllll}
2&1&0&0&\cdots&0\\
1&2&1&0&\cdots&0\\
0&1&2&1&\cdots&0\\
\vdots & & &\ddots & &\vdots \\
0&0&\cdots&1&2&1\\
0&0&\cdots&0&1&2\\
\end{array}\right)\]
can be computed by induction. By minors, we can see that
\[\det(A_n)=2\det(A_{n-1})-\det(A_{n-2})\]
and $\det(A_1)=2, \det(A_2)=3$. From this we can show that $\det(A_n)=n+1$.
Computing the determinant of $E_8$ by minors, using the last column, we see
that
\[\det(E_8)=2\det(A_7)-\det(A_2)\det(A_4)=2(8)-(3)(5)=16-15=1\].
So this is a unimodular matrix. Now we show $E_8$ is positive-definite.
To do this we again resort to $A_n$. We can diagonalize $A_n$ over ${\bf R}$
inductively, by diagonalizing the upper left $A_{n-1}$, then adding a multiple
$k_n$ of the penultimate row to the last row, diagonalizing $A_n$ and leaving
a $s_n$ in the bottom-right corner.
Inductively, we get the relations
\[k_n=-\frac{1}{s_{n-1}}, s_n=2+k_n\]
and initial conditions $k_1=0, s_1=2$. Combining these relations we get
\[s_n=2-\frac{1}{s_{n-1}}, s_1=2\].
We solve this and get:
\[s_n=1+1/n.\]
So applying this to $E_8,$ for the first seven rows, we get
\[E_8\cong \left(\begin{array}{llllllll}
2&0&0&0&0&0&0&0\\
0&\frac{3}{2}&0&0&0&0&0&0\\
0&0&\frac{4}{3}&0&0&0&0&0\\
0&0&0&\frac{5}{4}&0&0&0&0\\
0&0&0&0&\frac{6}{5}&0&0&1\\
0&0&0&0&0&\frac{7}{6}&0&0\\
0&0&0&0&0&0&\frac{8}{7}&0\\
0&0&0&0&1&0&0&2
\end{array}\right)\]
and, doing the same trick between the 5th row and the 8th row, we get:
\[E_8\cong \left(\begin{array}{llllllll}
2&0&0&0&0&0&0&0\\
0&\frac{3}{2}&0&0&0&0&0&0\\
0&0&\frac{4}{3}&0&0&0&0&0\\
0&0&0&\frac{5}{4}&0&0&0&0\\
0&0&0&0&\frac{6}{5}&0&0&0\\
0&0&0&0&0&\frac{7}{6}&0&0\\
0&0&0&0&0&0&\frac{8}{7}&0\\
0&0&0&0&0&0&0&\frac{7}{6}
\end{array}\right)\]
This is clearly positive-definite, so the signature is 8.
\qed
Note: the negative of this matrix is negative definite, and has
signature -8, and using direct sums of these, we can get an even
unimodular symmetric form with signature any multiple of 8. By taking
direct sums of $(1)$ and $(-1)$, we can get an odd unimodular symmetric form
with signature any integer.
\begin{theorem}
Any even indefinite unimodular symmetric bilinear form on ${\bf Z}^n$ is
of the form $\oplus^p H\oplus\oplus^q(E_8)$ or the negative of such.
\end{theorem}
Proof:
[Milnor and Husemoller]
Let $W$ be ${\bf Z}^2$, with the form $(1)\oplus(-1)$. Note that any
unimodular symmetric form direct summed with $W$ is an odd indefinite
unimodular symmetric form. Recall that this can be written in the
form $\oplus^k(1)\oplus\oplus^l(-1)$. So in particular, any even form
is a direct summand of an odd indefinite form.
Now consider such an odd form $\rho$ on some $V$, and let
\[V_0=\{v\in V|\rho(v,v) \equiv 0 \pmod{2}\}.\]
Now $\rho$ restricted to $V_0$ is an even form, and $V_0$ is a
subgroup of $V$ of index 2. Now consider $V$ embedded in the vector
space $V\otimes {\bf R}\cong {\bf R}^n$. Now define
\[V_0^*=\{v\in V\otimes {\bf R} | \rho(v,w)\in {\bf Z}
\makebox{ for all } w\in V_0\}.\]
If $(Y,\phi)$ is an even unimodular symmetric form, then $(V,\rho)=(Y\oplus W,
\phi\oplus(1)\oplus(-1))$ is an odd indefinite unimodular symmetric form.
In this case, $V_0=Y\oplus W_0$, so $V_0^*=Y\oplus W_0^*$. Now $V_0$ is
a subgroup of index 4 in $V_0^*$, with factor group ${\bf Z}_2^2$. So there
are three intermediate groups. One of them is $V$, and the other two are
isomorphic to $Y\oplus H$.
Now suppose $(Y,\phi)$ and $(Y',\phi')$ are two even unimodular symmetric
forms of the same rank and signature. Then $Y\oplus W\cong Y'\oplus W$
by the classification of odd indefinite unimodular symmetric forms. Applying
the above construction to both sides, we get that $Y\oplus H\cong Y'\oplus H$
If we can prove that every indefinite even unimodular symmetric form has
an $H$ as a direct summand, then we will have proven that the rank and
signature determine an even indefinite unimodular symmetric form.
\begin{lemma}
Every indefinite even unimodular symmetric form has an $H$ as a direct summand.
\end{lemma}
Proof:
If $(V,\rho)$ is such a form, then there exists a vector $v\in V$ such
that $\rho(v,v)=0$. Take $v$ so that it is not a multiple of another
vector. Then there is a vector $w$ such that $\rho(v,w)=\pm 1$, since
otherwise all such $\rho(v,w)$ will have a common factor. Then in any
basis where $v$ is one basis element, the column will have a common factor,
so the determinant will not be $\pm 1$. Now by choosing between $w$ and $-w$,
we can make $\rho(v,w)=1$. Then on the space spanned by $v$ and $w$ we get
the matrix:
\[\left(\begin{array}{ll}
0&1\\
1&\makebox{even}
\end{array}\right)\]
By adding a sufficient multiple of $v$ to $w$, we can make $\rho(w,w)=0$.
By adding a sufficient multiple of $v$ and $w$ to the other rows, we can
make the inner product of $v$ or of $w$ with any other basis element 0.
This gives the direct decomposition into an even form and $H$.
\qed
Continuation of the proof of the theorem:
This tells us that any indefinite even unimodular symmetric form is determined
by its rank and signature. Clearly, the rank must be strictly greater than
the signature. Now since the rank $r=b_++b_-$, and the signature $\sigma
=b_+-b_-$, we know that modulo 2, $r\equiv \sigma\bmod{2}$. Since an even
form has signature a multiple of 8, the rank must be even.
If $(V,\rho)$ is an even indefinite unimodular symmetric form, of rank $r$ and
signature $\sigma$, then let $(V',\rho')=\oplus^{\sigma/8}E_8\oplus
\oplus^{(r-\sigma)/2}H$ if $\sigma\ge 0$ and
$(V',\rho')=\oplus^{-\sigma/8}-E_8\oplus \oplus^{(r+\sigma)/2}H$ if
$\sigma< 0$. Then since $(V,\rho)$ and $(V',\rho')$ are indefinite even
unimodular symmetric forms with the same rank and signature, we know
that they are isomorphic.
\qed
\begin{corollary}
Every indefinite unimodular symmetric form is determined by its rank, type,
and signature.
\end{corollary}
\subsection{The Fundamental Group}
We cannot classify four-dimensional manifolds up to homotopy type. The reason
for this is the fundamental group. Given any finite presentation of a group,
we can construct a four-dimensional manifold with that group as a fundamental
group. Any classification of four-dimensional manifolds up to homotopy type
will give a classification of finite group presentations, which is known to
be impossible (Robinson, Matiyasevich?).
The construction goes as follows: for every generator, take $S^1\times B^3$.
Take the connected sum of these pieces along 3-disks on their boundary. The
fundamental group of this a free group on the generators of each $S^1\times
B^3$. For each relation we take an embedded (smooth) curve on the boundary
that represents that relation in the fundamental group. We can assume
these curves are transverse, and since the boundary is 3-dimensional,
transverse curves never intersect. A tubular neighborhood of each in the
boundary is homeomorphic to $S^1\times B^2$. For each relation, we take
a four-ball which we write as $B^2\times B^2$. Its boundary is $S^1\times B^2
\cup B^2\times S^1$. Take the $S^1\times B^2$ part of its boundary and
associate it with the tubular neighborhood corresponding to that relation.
Using the Seifert-Van Kampen theorem, we easily see the resulting quotient
space has as fundamental group the group in question. Take two copies of
this space, and associate each point on the boundary of one copy to the
corresponding point on the boundary of the other copy. The resulting
quotient space is a compact manifold without boundary, and the Seifert-Van
Kampen theorem again says that this has the proper fundamental group.
\qed
So the problem of classifying manifolds is at least as hard as the problem
of classifying groups, which was shown to be noncomputable.
Still, there is hope. Although we cannot tell many groups apart, we can
tell some groups apart. For instance, if we restrict our attention to
abelian groups, or finite groups, or some class like this, we have some
hope of classifying manifolds with fundamental groups in that class.
Even if we restrict our attention to simply-connected manifolds, we
still have a non-trivial problem. In fact, the diffeomorphism problem
for simply-connected four-manifolds is unsolved. So in the next section we
restrict our attention to these manifolds.
\subsection{Simply Connected Four-manifolds}
If $\pi_1(M)\cong 0$, then by the Hurewicz theorem, $H_1(M)\cong 0$, so by our
observations earlier, $H^1\cong H_3\cong H^3\cong 0$. Also $H_2\cong H^2$ is torsion-free.
Note also that every simply connected manifold is already orientable,
since on each path component we can choose an orientation on one point, and
for every point in that component, we choose a path to the point where we have
chosen an orientation, then take the orientation induced by this path. Another
choice of path will give the same orientation since the two paths form a
closed curve which spans a disk, and on this disk we have an induced
choice of orientation.
A more fancy way of doing this is to say that every $O(n)$ bundle on a
simply connected manifold lifts to an $SO(n)$ bundle since the
fibration $SO(n)\rightarrow O(n)\rightarrow {\bf Z}_2$ gives a fibration
of classifying spaces $BSO(n)\rightarrow BO(n)\rightarrow B{\bf Z}_2
\cong K({\bf Z}_2,1)$, and an $O(n)$ bundle on $M$ is a map from $M$ to $BO(n)$.
The induced map from $M$ to $K({\bf Z}_2,1)$ is null-homotopic since such a
map gives an element of $H_1(M;{\bf Z}_2)$ which is zero. So the map from
$M$ to $BO(n)$ lifts to a map from $M$ to $BSO(n)$, which is to say that the
$O(n)$ bundle can be considered an $SO(n)$ bundle, that is, it is orientable.
Applied to the tangent bundle of a manifold, we get that the manifold is
orientable.
Of course, our notion of orientable above assumes we are dealing with
a smooth manifold, but if we have a topological manifold, we can still
define orientation as a choice of isomorphism $H_n(X,X-p)\cong {\bf Z}$ for each
$p$, and apply the first proof.
\begin{theorem}
The homotopy type of a closed, simply-connected manifold
depends only on the intersection form (and hence, $b_2$).
\end{theorem}
Proof:
Pick a point and a coordinate neighborhood around the point in the manifold.
Let $U$ be a ball around the point whose closure is in the coordinate
neighborhood. Then $M-U$ is a CW complex with homology up to dimension
2. So by the cellular approximation theorem, we can deformation retract it
to a two-dimensional CW complex $M_0$. By the Hurewicz theorem,
$\pi_2(M_0)\cong H_2(M_0)$.
Let $X$ be the wedge of $b_2$ copies of $S^2$. Now $\pi_2(M_0)$
induces a map from $X$ to $M_0$, taking each sphere to a generator of
$\pi_2(M_0)$. This induces an isomorphism on $\pi_2(M_0)$ so induces
an isomorphism $H_2(X)\rightarrow H_2(M_0)$. Of course, this also
induces an isomorphism $H_0(X)\rightarrow H_0(M_0)$. Since these are
the only nonzero homology groups, the map induces an isomorphism on
all homology groups, and by Whitehead's theorem, is a homotopy
equivalence. So $M_0$ is homotopy equivalent to a wedge of spheres.
Under the deformation retract of $M-U$ to $M_0$, the attaching map
$\partial U\rightarrow M-U$ induces a map $S^3\rightarrow X$. So the
homotopy type of $M$ is classified by $X$ and a map from $S^3$ to $X$.
Let $K$ be the $b_2$-fold product of $CP^\infty$. Now $S^2=CP^1$
embeds canonically into $CP^\infty$, just taking the first three
projective coordinates. So we have the inclusion
\[X=\wedge_{b_2} S^2 \subset \prod_{b_2} S^2 \subset \prod_{b_2} CP^\infty=K.\]
Now
\[\pi_3 X \cong \pi_4(K,X)\]
by the homotopy sequence of the pair, and the fact that $K$ has zero
homotopy groups except in dimension 2. Likewise,
\[\pi_4(K,X) \cong H_4(K,X)\]
by the relative Hurewicz theorem, using the fact that since $K$ is built
from $X$ with cells of dimension four and higher only, and so
$\pi_i(K,X)=0$ for $i<4$. Also, we have
\[H_4(K,X) \cong H_4(K)$\]
by the homology sequence of the pair, and that $X$ has no homology
above dimension 2. By the K\"unneth formula, if $s_i$ is the set of
generators for $H_2(K)$, then the elements of $H_4(K)$ are given by
finite sums of pairs of $s_i$:
\[H_4(K)=\{\sum k_i s_i\times s_j\}\]
Since $H_*(K)$ has no torsion, we can think of $H_*(K)$ as $Hom(H^*(K),\zz)$,
and as such an element of $H_4(K)$ is a symmetric bilinear form on $H^2(K)$.
Furthermore, all symmetric bilinear forms on $H^2(K)$ are realized by an
element of $H_4(K)$.
Now $H^2(K) \cong H^2(X)$ since $K$ is obtained from $X$ by adding only
cells of dimension 4 and greater.
So elements of $\pi_3 X$ are in one-to-one correspondence with symmetric
forms on $H^2(X)$.
We now show that the map $S^3\to X$ obtained by the attaching map
for a particular four-manifold gives rise to the intersection form
under this correspondence.
Suppose we have two elements $a,b$ of $H_2(M)$. These correspond
uniquely to elements of $H_2(M_0)$, which we still call $a$ and $b$.
Under the homotopy equivalence $M_0 \sim X$ they are represented by
spheres in $X$. continue...
[Milnor and Husemoller]
The preimage of a point is
generically a finite set of nonintersecting closed curves in $S^3$,
that is, a link. The linking matrix for this link is the intersection
form for $M$.
[why does this determine the link?]
\qed
\subsection{Tangent Bundle Data}
Here we will assume $M$ is a {\em smooth} closed connected orientable
four-manifold. With this comes a tangent bundle, and we can examine
its characteristic classes (ref: Milnor and Stasheff. Characteristic Classes).
Since the structure group is $SO(4)$, we have available to us the
characteristic classes $w_2, w_3, w_4, p_1,$ and $e$, where
$w_i\in H^i(M;{\bf Z}_2)$ are the Stieffel-Whitney classes,
$p_1\in H^4(M;{\bf Q})$ is the Pontrjagin class, and $e\in H^4(M;{\bf Z})$ is
the Euler class. Now $e$ is the Euler characteristic times the fundamental
class $PD(1)$, so is $(2-2b_1+b_2)PD(1)$. By the Hirzebruch Signature Formula,
$p_1\cong 3\sigma PD(1)$ where $\sigma$ is the signature of the intersection form.
The Wu formula (ref: Milnor and Stasheff. Characteristic Classes) gives a
formula for the Stiefel-Whitney classes in terms of the cup product and the
Steenrod Square operations. So we see that these characteristic classes
can only detect homotopy type; despite the fact that they were defined in
terms of smooth information, i.e. the tangent bundle, they are independent
of all information except the homotopy type.
This is a bit dissapointing, but we will compute the Stiefel-Whitney classes
anyway. Recall that the Wu formula says that if $v_i\in H^i(M;{\bf Z}_2)$ is
the characteristic element in the sense that $v_i\cup x=x\cup x$ for all
$x\in H^i(M;{\bf Z}_2)$, then actually $v_i\cup x=Sq^i(x)$ for all
$x\in H^j(M;{\bf Z}_2)$, and if we write $v=1+v_1+v_2+\cdots$, then
\[w=Sq(v),\]
that is
\[1+w_1+w_2+\cdots=(Sq^0+Sq^1+Sq^2+\cdots)(1+v_1+v_2+\cdots).\]
Expanding, and using the fact that $Sq^0=Id$ and $Sq^i(x)=x\cup x$ if $x\in
H^i(M;{\bf Z}_2)$, and is 0 if $x\in H^j(M;{\bf Z}_2)$ for $i>j$, we see that
\begin{eqnarray}
w_1&=&v_1\\
w_2&=&v_2+v_1^2\\
w_3&=&v_3+Sq^1(v_2)\\
w_4&=&v_4+Sq^1(v_3)+v_2^2
\end{eqnarray}
Since we are in an orientable manifold, $w_1=0$, so $v_1=0$. This means
$w_2=v_2$, so $w_2$ is the characteristic element of our intersection form.
Since in a four-dimensional space, cup producting elements of dimension
3 or 4 with themselves gives 0, the unique characteristic element for 3 and 4
must be 0, so $v_3=v_4=0$. So we get:
\begin{eqnarray}
w_1&=&0\\
w_2&=&v_2\\
w_3&=&Sq^1(w_2)\\
w_4&=&w_2^2
\end{eqnarray}
Recall that $Sq^1$ is in a sense the B\"ockstein homomorphism mod 2. The
B\"ockstein homomorphism is defined as follows: from the sequence
\[0\rightarrow {\bf Z}\rightarrow {\bf Z}\rightarrow {\bf Z}_2\rightarrow 0\]
we get the long exact sequence
\[\cdots\rightarrow H^i(M)\rightarrow H^i(M)\rightarrow H^i(M;{\bf Z}_2)
\rightarrow H^{i+1}(M)\rightarrow \cdots\]
The last homomorphism, the connecting homomorphism, is called the B\"ockstein
homomorphism. Its kernel consists of exactly the elements which come from
modulo-two reductions of elements of the cohomology with integer coefficients.
It is well-known that the B\"ockstein homomorphism composed with the reduction
to modulo two again is $Sq^1$.
We will now prove that $w_2$ has an integral lift, that is, it comes
from the reduction of a class in $H^2(M;{\bf Z})$. It will then
follow that $w_3=Sq^1(w_2)=0$.
\begin{theorem}
Let $M$ be a closed, orientable four-manifold. Then $w_2(M)$
has an integral lift.
\end{theorem}
Proof:
Let $H^2(M)\cong {\bf Z}^{b_2}\oplus T$ where $T$ is the torsion group. Let
$S\cong T\otimes {\bf Z}_2$, the ${\bf Z}_2$ reduction of $T$.
By the universal coefficient theorem, $H^2(M;{\bf Z}_2)\cong {\bf Z}_2^{b_2}\oplus
S\oplus S$.
The first term comes from the reduction of $H^2$ to ${\bf Z}_2$, the second
comes from the torsion term in $H^1$ in the universal coefficient theorem.
We know that the homomorphism $r:H^*(M)\rightarrow H^*(M;{\bf Z}_2)$ is a
ring homomorphism, so if $x,y\in H^*(M)$, then $r(x)\cup r(y)=x\cup y \pmod{2}$.
Now recall that if $x$ or $y$ was in $T$, then $x\cup y=0$. So in that
case, $r(x)\cup r(y)=0$.
Over ${\bf Z}_2$, then, the annihilator of the second summand includes
the image of $r$, which is the first and second summands. As a matrix, then,
the intersection form is
\[\left(\begin{array}{lll}
A&0&B\\
0&0&C\\
B&C&D
\end{array}\right)\]
where $A$ is a $b_2\times b_2$ matrix, $B$ a $b_2\times \dim(S)$ matrix,
and $C$ and $D$ $\dim(S)\times dim(S)$ matrices. We can change the basis
on the third summand so that $C$ is diagonal. There can be no zeros here
or else a column in the second summand would have all zeros, contradicting
unimodularity. So the annihilator of the second summand is precisely the
image of $r$.
Now $x\cup x=0$ for all $x$ in the second summand, so any characteristic
element annihilates the second summand. So by the previous paragraph,
the characterstic element is in the image of $r$. This means it has an
integral lift.
\qed
\begin{corollary}
$w_3(M)=0$.
\end{corollary}
\begin{theorem}
Any integral lift of the characteristic element $w_2\in H^2(M;{\bf Z}_2)$
is a characteristic element in $H^2(M;{\bf Z})$.
\end{theorem}
Proof:
By the universal coefficient theorem, $H^2(M;{\bf Z}_2)\cong H^2(M)\otimes
{\bf Z}_2 \oplus H^3(M)*{\bf Z}_2$, and
$r:H^2(M;{\bf Z})\rightarrow H^2(M;{\bf Z}_2)$ is the map $H^2(M)\rightarrow
H^2(M)\otimes{\bf Z}_2$ onto the first component of $H^2(M;{\bf Z}_2)$.
\qed
\begin{corollary}
The square of an integral lift of $w_2$ is equal to the signature modulo 8.
\end{corollary}
Proof:
This is true of a characteristic element.
\qed
\begin{claim}
Every unimodular symmetric bilinear form is the intersection form for some
simply connected topological four-manifold. Not every unimodular symmetric
bilinear form is the intersection form for some simply-connected smooth
four-manifold.
\end{claim}
The topological case is due to Freedman, and the smooth case is a corollary
to Rohlin's theorem, which states that a necessary condition for the form
to come from a smooth simply-connected four-manifold is that its signature
be a multiple of 16.
\end{document}