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.
1 Answer
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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 $$e_G(E\oplus F)=e((E\oplus F)_G)$$ $$=e(E_G\oplus F_G)$$ $$=e(E_G)e(F_G)$$ $$=e_G(E)e_G(F).$$
I hope this helps.

$\begingroup$ Could you elaborate on the step $e(E_G\oplus F_G)=e(E_g)e(F_G)$? $\endgroup$– MichaelCommented Dec 22, 2013 at 16:21

$\begingroup$ Sure. I used the nonequivariant 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 $H^*(B_G)=H_G^*(B)$. $\endgroup$ Commented Dec 22, 2013 at 17:49

$\begingroup$ That's what confused me. I thought Whitney product formula worked only mod $Z_2$, for StieffelWhitney classes, not for Euler class. $\endgroup$– MichaelCommented Dec 22, 2013 at 20:05

$\begingroup$ It also works for Euler classes. A reference is Theorem 6.6 of Basic Bundle Theory and KCohomology Invariants, by Husemoller. $\endgroup$ Commented Dec 22, 2013 at 21:09

1$\begingroup$ @Zhang Yang, have I answered your question, or were you seeking a different kind of answer? $\endgroup$ Commented Dec 24, 2013 at 16:05