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' $d\Omega=\omega \wedge \Omega$

  • $\begingroup$ Surely you mean $\omega\in \Omega^{1}(E, \mathfrak{g}).$ $\endgroup$ – Andy Sanders Jan 4 at 0:23
  • $\begingroup$ @AndySanders Thanks, you are right. I had made a typo. Have corrected it now. $\endgroup$ – Ishan Mata Jan 4 at 13:46

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