# Differentials in Weil model for equivariant cohomology

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

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.

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' $$d\Omega=\omega \wedge \Omega$$
• Surely you mean $\omega\in \Omega^{1}(E, \mathfrak{g}).$ – Andy Sanders Jan 4 at 0:23