cup product of stiefel-whitney class

Question

Let $\xi$ be a vector bundle. Let $w(\xi)$ be the total Stiefel-whitney class. Let $\bar w$ be the dual Stiefel-whitney class. In John Milnor's Characteristic class book, page 40-41 Chap.4, \begin{eqnarray*} \bar w&=&[1+(w_1+w_2+\cdots)]^{-1}\\ &=&1-(w_1+w_2+\cdots)+(w_1+w_2+\cdots)^2\cdots \end{eqnarray*} Why $w_1w_2=w_2w_1$? Does the cup prodcut of $w_i$'s are commutative?

Answer 1

The Stiefel-Whitney classes take values in $\mathbb{F}_2$-cohomology, and the cup product over $\mathbb{F}_2$ is commutative because $-1 = 1$ over $\mathbb{F}_2$. It also happens that the Chern and Pontryagin classes take values in even $\mathbb{Z}$-cohomology, and the cup product over $\mathbb{Z}$ is commutative on even-dimensional classes because $(-1)^2 = 1$. But, for example, once you start writing down Euler classes you need to start keeping track of signs when you commute things.