TITLE: A question about Weil algebra
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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.