Let $G$ ba a compact Lie group with Lie algebra $\mathfrak{g}$, $\mathfrak{g}^{\*}$ be the dual of $\mathfrak{g}$. We known the Weil algebra is


$$W(\mathfrak{g})=\wedge(\mathfrak{g}^{\*})\otimes S(\mathfrak{g}^{\*}).$$


Choose a basis $e\_{1},\cdots,e\_{n}$ for $\mathfrak{g}$ and let 
$e\_{1}^{\*},\cdots,e\_{n}^{\*}$ be the dual basis of $\mathfrak{g}^{\*}$. 
Let $\theta^{1},\cdots,\theta^{n}$ be the dual basis of $\mathfrak{g}^{\*}$ generating the exterior algebra $\wedge(\mathfrak{g}^{\*})$ and let $\phi^{1},\cdots,\phi^{n}$ be the dual basis of $\mathfrak{g}^{\*}$ generating the symmetric algebra $S(\mathfrak{g}^{\*})$.


On $(S(\mathfrak{g}^{\*})\otimes\Omega^{\*}(M))^{G}$, there is the Cartan model. 


The exterior differential in Cartan model of equivariant cohomology is defined by 
$$D=1\otimes d-\sum\_{j=1}^{n}\phi^{j}\otimes\iota\_{j}$$ 
here, $\iota\_{j}$ is the contraction operator.


If we choose $\widetilde{e}\_{1},\cdots,\widetilde{e}\_{n}$ as the another basis of 
$\mathfrak{g}$, then the dual basis of $\mathfrak{g}^{\*}$ generating the symmetric algebra 
$S(\mathfrak{g}^{\*})$ is $\widetilde{\phi}^{1},\cdots,\widetilde{\phi}^{n}$.


How to express the operator $\sum\_{j=1}^{n}\phi^{j}\otimes\iota\_{j}$
in the basis $\widetilde{e}\_{1}^{\*},\cdots,\widetilde{e}\_{n}^{\*}$
and Why is it?



A question about Weil algebra
If $\tilde e\_i = \sum\_j A\_{ij} e\_j$ then for the dual basis we have 
$\tilde e\_k^\ast = \sum\_\ell ((A^t)^{-1})\_{kl} e\_l^\ast$, so the the base change matrix cancels out of the formula.


