Formula for nth Pontryagin class

82 Views Asked by At

For the first Pontryagin classes we have: $$ p_0=1 \\ p_1 = -\tfrac{1}{8\pi^2}\text{Tr}(F^2)\\ p_2=\tfrac{1}{128 \pi^4}(\text{Tr}(F^2)^2-2\text{Tr}(F^4))\\ p_3 = -\tfrac{1}{3072 \pi^6} (\text{Tr}(F^2)^3 - 6\text{Tr}(F^2)\text{Tr}(F^4) + 8 \text{Tr}(F^6))$$

I want a formula for the $p_n$ Pontryagin class. I have tried using the Newton-Girard formulas, specifically: $$ p_k = (-1)^{k-1}ke_k+\Sigma_{i=1}^{k-1} (-1)^{k-1+i}e_{k-i}p_i$$

I had assumed that $e_n=(\tfrac{1}{2\pi})^{2n}\text{Tr}(F^{2n})$ but this does not give the correct answer.

1

There are 1 best solutions below

1
On

I personally define the Pontryagin classes in terms of the Chern classes:

Let $\pi : E \to X$ be a real vector bundle. Then

$$p_i(E)= (-1)^i c_{2i}(E\otimes_{\mathbb R} \mathbb C) \in H^{4i}(X;\mathbb R)$$

p19 of these (excellent) course notes on characteristic classes and topological K-theory is where this comes from.