I asked this question in MSE but I received no answer

https://math.stackexchange.com/questions/4009524/why-is-the-following-operator-independent-of-the-choice-of-basis/4013636#4013636 Let $G$ be a lie group with lie algebra $\mathfrak{g}$ and let $M$ be a $G$-manifold. We denote the space of smooth differential forms on $M$ by $A(M)$. Let $E^i$ be a basis of $\mathfrak{g}$ and let $E_i$ $\in \mathfrak{g}^*$ be the dual basis. Let $S(\mathfrak{g}^*)$ be the symmetric algebra of $\mathfrak{g}^*$. Define the operator $\overline{d}$ on $S(\mathfrak{g}^*) \otimes A(M)$ by

$$\overline{d}(P \otimes \alpha)= \sum_i E_iP \otimes \iota (E^i_M) \alpha$$

for $P \in S(\mathfrak{g}^*)$ and $\alpha \in A(M)$ (note that $\iota (E^i_M)$ is the contraction by the vector field $E^i_M$ wich is defined by $$(E^i_M.f)(m)= \frac{d}{dt} \bigg|_{t=0} f(\exp(tE^i)m),$$ for $f \in C^\infty(M)$ and $m \in M$).

My question is why the definition of $\overline{d}$ is independent of the choice of the basis $E^i$ ?

Actually, I get this question by reading section $2$ of the article of equivariant cohomology with generalized coefficients by Kumar and Vergne, and the authors have commented on this by saying the following

This expression is independent of the choice of the basis $E^i$, as the element $\sum_i E_i \otimes E^i \in \mathfrak{g}^* \otimes \mathfrak{g}$ is the canonical element $I \in \operatorname{End}(\mathfrak{g})$, where $I$ is the identity element of $\operatorname{End}(\mathfrak{g})$.

But I didn't understand their comment! Moreover, I tried to prove this independence in the case where $\mathfrak{g}$ is one dimensional without going any where.

Please help with this.

  • 1
    $\begingroup$ The key thing to note is that the map $(X \mapsto X_M) : \mathfrak{g} \to \mathfrak{X}(M)$ is actually $\mathbb{R}$-linear, so all the usual elementary linear algebra that tells you that $\sum_i E_i \otimes E^i = \sum_i \tilde{E}_i \otimes \tilde{E}^i$ for any two bases $\{E^i\}$ and $\{\tilde{E}^i\}$ for $\mathfrak{g}$ will still apply. $\endgroup$ Feb 3 at 15:02
  • $\begingroup$ @Branimir, thank you so much for your comment, could you please tell me where I can find it because this is the first time I see it $\endgroup$
    – asma
    Feb 3 at 19:32
  • 2
    $\begingroup$ Thanks for mentioning that you asked the question already on math.stackexchange. When you do this in the future, could you please make sure to edit each question to include a link to the other? That way if someone makes progress on one, people can easily see it before working on the other. $\endgroup$
    – Will Sawin
    Feb 5 at 2:42
  • $\begingroup$ @Will Sawin, thank you for your remark, I'll add the link of my question in MSE. $\endgroup$
    – asma
    Feb 5 at 11:09

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