Let $E\longrightarrow B$ be an oriented $G$equivariant vector bundle over $B$, where $G$ is a compact Lie group. Then we have the equivariant Thom form of $E$, denoted by $\Theta_{G}$, which is an equivariant closed form. In further, the equivariant Euler class $e_{G}(E)$ is equivalent to the pullback of $\Theta_{G}$ by the zero section of $E$. In other aspect we have the nonequivariant Thom form $\Theta$ and the associated nonequivariant Euler class $e(E)$. So what is the relation of $\Theta_{G}$ and $\Theta$ ($e_{G}(E)$ and $e(E)$)$?
If $Y$ is any $G$space, then $Y_G$ is defined to be the quotient of $Y\times EG$ by the diagonal $G$action. For purposes of this answer, I will use the definition $$H_G^*(Y):= H^*(Y_G).$$ Now, if $E\rightarrow X$ is a $G$equivariant (say complex) vector bundle , we may endow $E_G$ with the structure of a vector bundle over $X_G$. We then have $e_G(E)=e(E_G)\in H_G^*(X)$.
There is a natural map $f:X\rightarrow X_G$, and it gives morphism $f^{!}:H^*_G(X)\rightarrow H^*(X)$. Note that $$f^!(e_G(E))=f^{!}(e(E_G))=e(f^*(E_G))=e(E),$$ as taking Chern classes commutes with the pullback operations.

$\begingroup$ Thanks for your answer. So if we use the equivariant differential forms to represent the equivariant Thom class and equivariant Euler class, does it means that the top component(or leading component) of the equivariant Thom form (equivariant Euler class), which is an ordinary differential form, is the nonequivariant Thom form (Euler class)? $\endgroup$ – yangyang Oct 27 '13 at 4:48

$\begingroup$ If $E$ has complex rank $n$, then $e_G(E)\in H_G^{2n}(X)$. The equivariant differential forms representing elements of $H_G^{2n}(X)$ are then $d_G$closed forms in $$\bigoplus_{p+2q=2n}\Omega^p(X)\otimes_{\mathbb{R}}S(\frak{g}^*).$$ If $\omega$ represents $e_G(X)$, then are you referring to the component of $\omega$ lying in $\Omega^{2n}(X)$? If so, then I would not be surprised if you were right. $\endgroup$ – Peter Crooks Oct 27 '13 at 13:04

$\begingroup$ Here is something to consider. In the Cartan model of equivariant cohomology, perhaps the map $$f^!:H_G^*(X)\rightarrow H^*(X)$$ is defined by evaluating the polynomials on $\frak{g}$ at $0$. Since $f^!(e_G(E))=e(E)$, you would then be correct. I am not sure about this, however. You might read "The Moment Map and Equivariant Cohomology" by Atiyah and Bott. Another good reference on the Cartan model is "Moment Maps, Cobordisms, and Hamiltonian Group Actions" by Guillemin, Karshon, and Ginzburg. $\endgroup$ – Peter Crooks Oct 27 '13 at 13:11