Chern Class

Universal rank $n$ bundle

For $p:EU(n)\to BU(n)$ the universal principal $U(n)$-bundle, consider the associate bundle with standard representation $\rho :U(n)\to {\mathbb{C}}^n$, we define ${\gamma }_{\mathbb{C}}^n:=EU(n){\times }_{U(n)}{\mathbb{C}}^n$ Similarly, we can define for real: ${\gamma }_{\mathbb{R}}^n:=EO(n){\times }_{O(n)}{\mathbb{R}}^n$

The idea of universal rank $n$ bundle comes from we want to construct a vector bundle which is comparable to universal principal bundle, thus we would be able to find similar correspondence in theorem, representing vector bundle as pull back of a “classifying space”.

Property

For $p:E\to B$ a complex vector bundle, $E\cong Fr(E{)}_{U(n)}\times {\mathbb{C}}^n$. Let $f:B\to BU(n)$, then by theorem, we get $Fr(E)\cong f^{\ast }(EU(n))$, which means following diagram commute: $\begin{array}{ccc}Fr(E)& ⟶& EU(n)\\ \downarrow & & \downarrow \\ B& \stackrel{f}{\to }& BU(n)\end{array}$. So together we get $E\cong f^{\ast }(EU(n)){\times }_{U(n)}{\mathbb{C}}^n\cong f^{\ast }(EU(n){\times }_{U(n)}{\mathbb{C}}^n)\cong f^{\ast }({\gamma }_{\mathbb{C}}^n)$

Remark

The isomorphism $E\cong Fr(E{)}_{U(n)}\times {\mathbb{C}}^n$ can be understand as $Fr(E{)}_x$ represents all well-ordered basis for $E_x$ at $x$ and $U(n)$ act on ${\mathbb{C}}^n$ gives a transition between basis, which means if we mod out the action of $U(n)$, we’ll get a class of basis for $E_x$ and $\times {\mathbb{C}}^n$ gives us $E_x$.

First Chern Class for complex line bundle

$\begin{array}{ccc}L& ⟶& EU(1)\cong S^{\infty }\\ \downarrow & & \downarrow \\ B& \stackrel{f}{\to }& BU(n)\cong \mathbb{C}{P}^{\infty }\end{array}$.