How does one introduce characteristic classes

Question

How do you introduce or how are you introduced to characteristic classes.

I am assuming the student is comfortable with principal bundles and connections on principal bundles.

Answer 1

I was always confused by characteristic classes until I understood the definition of characteristic classes via the classifying map.

Corresponding to a vector bundle with structure group $G$ there is a unique homotopy classes of maps $M\rightarrow BG$. In cohomology with coefficients in $A$ this gives a map $H^*(BG;A)\rightarrow H^*(M;A)$. We call the image of any class in $H^*(BG;A)$ a characteristic class.

Now it is true that for certain groups $G$ and coefficients $A$ we know the cohomology ring $BG$ very well. For example for $G=O(n)$ with $A=\mathbb{Z}/{2\mathbb{Z}}$ this leads to the Stiefel-Whitney classes or for $G=U(n)$ with $A=\mathbb{Z}$ this leads to the Chern classes.

The relations between these classes directly come from relations between the groups.

EDIT: Let me tell you how to get this map for the tangent bundle of a manifold. Any manifold $M^n$ can be embedded in $\mathbb R^k$ for $k$ large by Whitney's theorem. But then for every $x\in M$ the tangent space of $M$ at $x$ can be seen as an $n$ dimensional subspace of $\mathbb{R}^k$. Thus we obtain a map $M\rightarrow G_n(\mathbb{R}^k)$ into the Grassmannian of $n$ planes in $\mathbb{R}^k$. With some work you can show that if we have two embeddings, and $k$ is sufficiently large any two embeddings are isotopic and the resulting maps are homotopic. In the end one ends up with a homotopy class of mappings into $G_n(\mathbb{R}^\infty)=BGL(n)$.

To work this out for general vector bundles you can work as follows: Given a vector bundle $E$ over a reasonable space $M$ find a vector bundle $F$ such that $E\oplus F$ is the trivial bundle over $M$. Then the map that associates to $x\in M$ the subspace $E_x\subset E_x\oplus F_x\cong \mathbb R^k$ defines the classifying map. Again you will have to work a bit to show that this does not depend on the choices if one works up to homotopy.