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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!

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    $\begingroup$ 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 $\mathfrak g \to \mathcal A(M)$? $\endgroup$
    – Will Sawin
    Mar 16 at 18:34
  • $\begingroup$ @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 ? $\endgroup$
    – asma
    Mar 16 at 19:26
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    $\begingroup$ Sounds right to me. What about a sum of simple tensors? $\endgroup$
    – Will Sawin
    Mar 16 at 19:27
  • $\begingroup$ I think we extend the construction linearly ! $\endgroup$
    – asma
    Mar 16 at 19:28

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