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\define\GL{\operatorname{GL}}
\define\ORTH{\operatorname{O}}
\define\Hom{\operatorname{Hom}}
\define\End{\operatorname{End}}
\define\Aut{\operatorname{Aut}}
\define\Mat{\operatorname{Mat}}
\define\Ker{\operatorname{Ker}}
\define\Coker{\operatorname{Coker}}
\redefine\Im{\operatorname{Im}}
\define\Rank{\operatorname{Rank}}
\define\R{\Bbb R}
\define\C{\Bbb C}
\define\rp{\Bbb{RP}}
\define\cp{\Bbb{CP}}

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\title       The Geometry of Vector Bundles and an Introduction to Gauge
Theory \\
Lecture 29
\endtitle
\author      Professor Steven Bradlow \\ Class Notes From Math 433
\endauthor
\affil       University of Illinois at Urbana-Champaign
\endaffil
\address     273 Altgeld Hall, 1409 W. Green Street, Urbana, IL  61801
\endaddress
\email       bradlow\@math.uiuc.edu
\endemail
\date        April 8,1998
\enddate

\endtopmatter

\document
Recall that given a vector bundle, $E \to B$ and any $\GL_n$ invariant polynomial $P$, we can define cohomology classes, $[P(E)] \in H^*(B
;\Bbb F)$, where $\Bbb F=\R$ or $\Bbb F=\C$. If $I_{\GL_n}$ is the set of 
all $\GL_n$ invariant polynomials on $\Mat_n$, then we can think of $I_{\GL_n}$ as a ring under functional addition and multiplication and hence get a map,
 $I_{\GL_n} \to H^*(B; \Bbb F)$ of rings, $P \mapsto [P(E)]$.  This is called 
the {\it Chern-Weil} homomorphism.
\vskip.25in

\remark{Note}  If $E$ admits a flat connection, i.e, $D$ such that $F_D=$, 
then $[P(E)]=0$ for all $P$.  Since every trivial bundle has $[P(E)]=0$ and 
there are flat bundles which are not trivial, there exist bundles with the 
same characteristic classes but which are not isomorphic.
\endremark
\vskip.25in

If $E' \to B'$ is a bundle and $f:B \to B'$ is a map with pullback bundle 
$f^*(E')=E \to B$, i.e, we have the following set up,
$$\sarrowlength=.42\harrowlength
\commdiag{E&&E' \cr \mapdown\ && \mapdown \cr
B &\mapright^{f}&B'}$$

For any $\GL_n$ invariant polynomial $P$, we have $[P(E)] \in H^*(B;\Bbb F),
[P(E')] \in H^*(B';\Bbb F)$, and we would like for these classes to match up; that is, for $[P(E)]=f^*[P(E')]$.  Given a connection $D'$ on $E'$ we can 
define the pullback connection on $E$, $D=f^*(D')$.  Since $F=f^*(F')$, we 
do indeed get $[P(E)]=f^*[P(E')]$ by definition.
\vskip.25in

\leftline{{\bf The Complex Bundle Case}}
\vskip.25in
{\bf Question:}  What are all of the $\GL(n,\C)$ invariant polynomials?  That
is, can we classify all of the polynomials, $P:\Mat(n,\C) \to \C$ which are 
invariant under conjugation by elements of $\GL(n,\C)$?
\vskip.25in

Consider the case of $$\aligned P \in \text{DIAG}&=\{A \in \Mat(n,\C):A\text{ is diagonalizable}\} \cr
&=\{gAg^{-1}=\displaystyle{
\pmatrix \lambda_1 & 0 &\dots & \dots &0\\
0 &\lambda_2 &0 &\dots &0\\
\dots & \dots &\dots &\dots &\dots\\
0 & 0 & \dots & \dots &\lambda_n
\endpmatrix}:g \in \GL(n,\C)\}\endaligned
$$
If $P$ is a $\GL(n,\C)$ invariant polynomial, then by definition, 
$$P(A)=P(\displaystyle{
\pmatrix \lambda_1 & 0 &\dots & \dots &0\\
0 &\lambda_2 &0 &\dots &0\\
\dots & \dots &\dots &\dots &\dots\\
0 & 0 & \dots & \dots &\lambda_n
\endpmatrix})$$ where the $\lambda_i$ are the Eigenvaluues.  Thus, $P$ depends
only on the $\lambda_i$.

\vskip.25in
\proclaim{Claim}  $P(\lambda_1,\dots,\lambda_n)$ must actually be a symmetric function of $\lambda_i$.
\endproclaim
\vskip.25in
{\bf Proof:}  By appropriate conjugation in $\GL(n,\C)$, we can permute the 
$\lambda_i$.  Since $P$ is invariant under such conjugation, $P$ must be 
invariant under the permutation of the $\lambda_i$.  That is, $P$ must 
be a symmetric function of the $\lambda_i$.\qed
\vskip.25in
\example{Example}  $$\displaystyle{
\pmatrix \lambda_1 & 0 \\
0 &\lambda_2\\
\endpmatrix
=\pmatrix 0 & 1 \\
1 &0\\
\endpmatrix
\pmatrix \lambda_2 & 0 \\
0 &\lambda_1\\
\endpmatrix
\pmatrix 0 & 1 \\
1 &0\\
\endpmatrix}$$
\endexample
\vskip.25in

{\bf Fact:}  All such $P$ are generated by the elementary symmetric functions, 
$$\aligned \sigma_0(\lambda_1,\dots,\lambda_n)&=1 \cr
\sigma_1(\lambda_1,\dots,\lambda_n)&=\sum_{i=1}^n \lambda_i \cr
\sigma_2(\lambda_1,\dots,\lambda_n)&=\sum_{i,j} \lambda_i \lambda_j \quad i\neq j \cr
&\dots \cr
\sigma_n(\lambda_1,\dots,\lambda_n)&=\lambda_1 \lambda_2 \dots \lambda_n \endaligned$$
\vskip.25in

\example{Example}  $$\sum_{i=1}^n \lambda_i^2=(\sum_{i=1}^n \lambda_i)^2-2(\sum_{i \neq j} \lambda_i)=\sigma_1^2-2\sigma_2$$
\endexample
\vskip.25in

This is a basic fact from algebra and essentially says that any symmetric
polynomial can be written as $$Q(\sigma_0(\lambda_1,\dots,\lambda_n),\dots,
\sigma_n(\lambda_1,\dots,\lambda_n))$$ where $Q$ is a polynomial.
\vskip.25in

\remark{Note} $$\aligned P(A)&=\text{det}(I+A)=\displaystyle{
\text{det}(\pmatrix 1+\lambda_1 & 0 &\dots & \dots &0\\
0 &1+\lambda_2 &0 &\dots &0\\
\dots & \dots &\dots &\dots &\dots\\
0 & 0 & \dots & \dots &1+\lambda_n
\endpmatrix}) \cr
&=\prod_{i=1}^n(1+\lambda_i)=1+\sigma_1 +\dots+\sigma_n \endaligned$$
So, the $\sigma_i$ arrise as the homogeneous parts of $P(A)$.
\endremark
\vskip.25in

\remark{Note}  So far, all of this applies equally to the real case.
\endremark
\vskip.25in

We want not quite $\sigma_j(F)$, but rather $\sigma_j(\frac{i}{2 \pi} F)$.
\vskip.25in

\definition{Definition 1}  The {\it Chern classes} of $E$ are defined by 
$c_k(E)=[\sigma_k(\frac{i}{2 \pi} F)]$.  $c_k$ is called the {\it k-th Chern
class of E}.  These are the homogeneous parts of the {\it total Chern class}, 
$c(E)=[\text{det}(I+\frac{i}{2 \pi}F)]$.
\enddefinition
\vskip.25in

We have that $$\text{det}(I+\frac{i}{2 \pi})=1+\text{Tr}(\frac{i}{2 \pi}F)+\dots+\text{det}(\frac{i}{2 \pi} F)$$
\vskip.25in

\remark{Note}  $c_k(E) \in H^{2k}(B; \C)$.  We will see, in fact, that $c_k(E) \in H^{2k}(B;\R)$.
\endremark
\vskip.25in

We still need to answer the question as to why it is enough only to look at 
diagonalizable matrices, DIAG.


\remark{Note}  This is the point where the real case differs from the complex case.
\endremark

\itemitem{(1)}  $\text{DIAG} \subset \Mat_n$ is dense.  If $A \in \Mat_n$, then 
there is a sequence of $A_i$ with each $A_i \in \text{DIAG}$ with $$A=\lim_{i \to \infty} A_i$$  Since the invariant polynomials are continuous functions, we can evaluate $$P(A)=\lim_{i \to \infty} P(A_i)$$

\itemitem{(2)}  Any complex bundle, $E \to B$, can be given a Hermitian metric.
So, we can take a compatible connection, $D$ and with respect to a unitary 
frame, $F$ is skew Hermitian; i.e, $F+F^*=0$.  In general, $AA^*=A^*A$ implies 
that $A$ is diagonalizable.  So, if $F+F^*=0$, then $F$ is diagonalizable.
So, for purposes of evaluating $c_k$, we can restrict to diagonalizable 
matrices.
\vskip.25in

\remark{Note}  If $(\frac{i}{2 \pi}F)^*=\frac{i}{2 \pi}F$, then $(I+\frac{i}{2 \pi}F)^*=I+\frac{i}{2 \pi}F$.  So, $$\overline{\text{det}(I+\frac{i}{2 \pi}F)}=\text{det}(I+\frac{i}{2 \pi}F)^*=\text{det}(I+\frac{i}{2 \pi}F)$$ This 
says that $c(E)=\overline{c(E)}$ and hence that $c_n(E) \in H^{2k}(B;\R)$ for 
all $k$.
\endremark
\vskip.25in

\remark{Remark}  The factors $\frac{1}{2 \pi}$ are inserted so that the classes
$c_k(E)$ are integer valued; i.e, $c_k(E) \in H^{2 k}(B;\Bbb Z)$.
\endremark

\enddocument
\end
