Why should we define the differential in Weil model as follows? I could understand $\sum\_{j,k} c\_{jk}^i \theta\_j \wedge \theta\_k$ plays a role in the formula because it is the dual of the structure map of $\mathfrak{g}$. The rest of formula looks mysterious to me.


[Cartan-Weil model for Equivariant Cohomology](https://mathoverflow.net/questions/79041/cartan-weil-model-for-equivariant-cohomology)



> 
> We define its Weil algebra by $W^\*(\mathfrak{g}^\*)=S^\*(\mathfrak{g}^\*) \otimes \wedge^\*(\mathfrak{g}^\*)$ there is also a natural differential operator $d\_W$ which makes $W\*(\mathfrak{g}^\*)$ into a complex. We define $d\_W$ as follows:
> 
> 
> Choose a basis $e\_1,...,e\_n$ for $\mathfrak{g}$ and let $e^\*\_1,...e^\*\_n$ its dual basis in $\mathfrak{g}^\*$. Let $\theta\_1,...,\theta\_n$ be the image of $e^\*\_1,...e^\*\_n$ in $\wedge(\mathfrak{g}^\*)$ and let $\Omega\_1,...,\Omega\_n$ be the image of $e^\*\_1,...e^\*\_n$ in $S(\mathfrak{g}^\*)$. Let $c\_{jk}^i$ be the structure constants of $\mathfrak{g}$, that is $[e\_j,e\_k]=\sum\_{i=1}^nc\_{jk}^ie\_i$. Define $d\_W$ by
> \begin{eqnarray}
> d\_W\theta\_i=\Omega\_i- \frac{1}{2}\sum\_{j,k} c\_{jk}^i \theta\_j \wedge \theta\_k
> \end{eqnarray}
> and
> \begin{eqnarray}
> d\_W\Omega\_i=\sum\_{j,k}c\_{jk}^i\theta\_j \Omega\_k
> \end{eqnarray}
> and extending $d\_W$ to $W(\mathfrak{g})$ as a derivation.
> 
> 
> 



Differentials in Weil model for equivariant cohomology
Suppose $E \to B $ is a principal $G$-bundle with connection $\omega \in \Omega^1(E,\mathfrak{g})$, and corresponding curvature $\Omega \in \Omega^2(E,\mathfrak{g})$. Then $\Omega^\*(E)$ is a differential graded algebra, with the exterior derivative as the differential, and wedge product of forms being the multiplication. 


Let $\omega\_i$'s (resp. $\Omega\_i$'s) be the connection 1-forms (resp. curvature 2-forms) on $E$ w.r.t. the chosen basis of $\mathfrak{g}$. Thus $\omega=\sum\_{i=1}^n \omega\_i e\_i $ and $\Omega=\sum\_{i=1}^n \Omega\_i e\_i $


There is an algebra homomorphism $f: W(\mathfrak{g}^\*) \to \Omega^\*(E)$ given by $\theta\_i \mapsto \omega\_i$, and $ u\_i \mapsto \Omega\_i$ (for clarity of notation, I am using $u\_i$'s as generators of $S(\mathfrak{g}^\*)$ instead of your notation $\Omega\_i$).


Defining the Weil-differential in the manner it is defined makes sure that this algebra homomorphism is a differential graded algebra homomorphism (i.e. is compatible w.r.t. the derivation operation) by the virtue of the 'structure equation' :
$\Omega=d\omega + \frac{1}{2}[\omega,\omega]$ and the 'Bianchi identity' $d\Omega=\omega \wedge \Omega$


