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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?

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  • $\begingroup$ What is $M$? I'd guess that $\iota_j$ is the insertion operator of $e_i$ in the former case and of $\tilde{e}_i$ in the latter. $\endgroup$ Commented Jun 22, 2013 at 18:46
  • $\begingroup$ @robot: $M$ is the closed manifold with $G$-action. $\iota_{j}$ is the operator as you guess, here it is a contraction operator on $\Omega^{*}(M)$. $\endgroup$
    – Chen
    Commented Jun 24, 2013 at 11:19

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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.

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