I have some question about the equivariant differential forms on a smooth manifold: \ The equivariant differential forms over some smooth manifold $M$, on which the compact Lie group $G$ acts, are defined to be $$ \Omega_{G}^q(M)= \oplus_{2i+j=q} (S^{i}(\mathfrak g^*) \otimes \Omega^j(M))^G, $$ where $\mathfrak g^*$ denotes the dual of the Lie algebra $\mathfrak g$ of $G$. Then many authors say that these forms can be considered as polynomial functions on the Lie algebra $\mathfrak g$ of $G$, but I am not sure how this is to be done. For example if we consider the element $(x_1 \otimes... \otimes x_i) \otimes \omega$ where $x_1,..., x_i$ are elements of $\mathfrak g^*$ and $\omega$ is a differential form, what is the evaluation of this element on some $a \in \mathfrak g$ in the Lie algebra of $G$.

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    $\begingroup$ I think you should ask this question in mathoverflow.net. $\endgroup$ – emiliocba Jan 22 '12 at 23:22
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    $\begingroup$ Dear Phil, If $V$ is a vector space, then $S^i(V^*)$ equals the space of homogeneous of degree $i$ polynomial functions on $V$: precisely, the value of $x_1\otimes \cdots \otimes x_i$ on a vector $v$ is equal to the product $x_1(v)\cdot x_i(v)$. Looking at your definition, the equivariant differential forms will then be polynomial functions on $\mathfrak g$ (via the preceding formula), with values in the graded algebra of differential forms on $M$. If this answers your question, tell me, and I'll post this comment as an actual answer. Otherwise, what else do you want to know? Regards, $\endgroup$ – Matt E Jan 23 '12 at 3:27
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    $\begingroup$ Now the question has been posted even on MathOverflow, precisely here: mathoverflow.net/questions/86419/equivariant-differential-forms $\endgroup$ – agtortorella Jan 23 '12 at 8:09
  • $\begingroup$ @MattE I do not understand. Differential forms live in the exterior and not the symmetric algebra or is it that their coefficients are, here, considered as polynomial functions ? $\endgroup$ – Duchamp Gérard H. E. Aug 7 '16 at 1:20

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