# equivariant whitney product formula

Let $G$ be a compact Lie group and $B$ be a closed smooth $G$-manifold. If we have two smooth oriented $G$-vector bundles: $$E\longrightarrow B$$ and $$F\longrightarrow B$$ then is there exists an equivariant version of Whitney product formula $$e_{G}(E\oplus F)=e_{G}(E)\wedge e_{G}(F)?$$ Where $e_{G}(E)$ represents the equivariant Euler class.

I believe the answer is yes. For a $G$-manifold $X$, let $X_G$ denote its Borel mixing space. Recall that $$H_G^*(X)=H^*(X_G).$$
Now, note that $E_G$, $F_G$, and $(E\oplus F)_G$ are the total spaces of vector bundles over $B_G$. Also, $$(E\oplus F)_G\cong E_G\oplus F_G$$ as vector bundles over $B_G$. (I am confident this is true, but it is worth checking.) Computing equivariant Euler classes, we obtain $$e_G(E\oplus F)=e((E\oplus F)_G)$$ $$=e(E_G\oplus F_G)$$ $$=e(E_G)e(F_G)$$ $$=e_G(E)e_G(F).$$
• Could you elaborate on the step $e(E_G\oplus F_G)=e(E_g)e(F_G)$? – Michael Dec 22 '13 at 16:21
• Sure. I used the non-equivariant version of the Whitney product formula. I am taking the ordinary Euler class of a Whitney sum of vector bundles over $B_G$. This equals the product of the Euler classes of the bundles in $H^*(B_G)=H_G^*(B)$. – Peter Crooks Dec 22 '13 at 17:49
• That's what confused me. I thought Whitney product formula worked only mod $Z_2$, for Stieffel-Whitney classes, not for Euler class. – Michael Dec 22 '13 at 20:05