# G-equivariant differential forms

I posted this question here but I don't get any answer! I hope that someone could help me here.

Let G be a lie group with Lie algebra $$\mathfrak{g}$$. We denote by $$S(\mathfrak{g}^*)$$ the symmetric algebra of $$\mathfrak{g}$$. Let M be a smooth manifold on wich G acts. We denote by $$\mathcal{A}(M)$$ the space of differential forms on M.

There are two definitions of the space of G-equivariant differential forms on M:

1. the space of G-equivariant differential form on M is the space of polynomial maps $$\alpha: \mathfrak{g} \rightarrow \mathcal{A}(M)$$ such that $$\alpha(gX)= g.\alpha(X)$$ for $$g \in G.$$

2. The space of G-equivariant differential forms is $${(S(\mathfrak{g}^*) \otimes \mathcal{A}(M))}^G$$ ,(where the coadjoint action of G on $$\mathfrak{g}^*$$ induced the G action on $$S(\mathfrak{g}^*)$$).

What is the explicit isomorphism between these two spaces ? I know that $$S(\mathfrak{g}^*)$$ can be identified with polynomial functions on $$\mathfrak{g}$$ and that the space $$S(\mathfrak{g}^*) \otimes \mathcal{A}(M)$$ is identified with the space $$Hom_\mathbb{R}{((S(\mathfrak{g}^*))}^*, \mathcal{A}(M))$$, however I couldn't write down precisely the relation between these two spaces!

• Tensor products of infinite-dimensional vector spaces are not really spaces of homomorphisms (they're finite-rank homomorphisms). Given a simple tensor in $S( \mathfrak g^*) \otimes \mathcal A(M)$, do you see how to obtain a polynomial map $\mathfrak g \to \mathcal A(M)$? Mar 16 at 18:34
• @Will Sawin, thank you for your comment! My thought about this is If we choose $p \otimes \alpha \in S(\mathfrak{g}^*) \otimes \mathcal{A}(M)$, we associate to it the polynomial with same monomials as P and with coefficients same as P but all are multiplied by $\alpha$ , Is this true ?
– asma
Mar 16 at 19:26
• Sounds right to me. What about a sum of simple tensors? Mar 16 at 19:27
• I think we extend the construction linearly !
– asma
Mar 16 at 19:28