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I'm reading the article of equivariant cohomology with generalized coefficients by Kumar and Vergne and I have this question from that article

Let G be a compact 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).

We denote by $C^{-\infty}(\mathfrak{g},A(M) )$ the space of generalized functions on $\mathfrak{g}$ with values in the space A(M). This is by definition, the space of continuous $\mathbb{R}$- linear maps $Hom(\mathfrak{D}(\mathfrak{g}), A(M))$ from the space of smooth compactly supported densities $\mathfrak{D}(\mathfrak{g})$ on $\mathfrak{g}$ to the space A(M). That if $\alpha $ is an element of $C^{-\infty}(\mathfrak{g},A(M) )$ and if $\phi$ is a smooth compactly supported density on $\mathfrak{g}$, then $(\alpha, \phi)$ is a differential form on M , s.t $(\alpha, \phi):= \int_\mathfrak{g} \alpha(X)\phi(X)dX.$

Why does the equation (7) in the following paragraph hold:

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I guess the point is that $$ \int_{\mathfrak g} \alpha(X) \Phi(X) dX ,$$ by definition, is $$(\alpha, \Phi)$$ since the idea of generalized functions is that the linear form represents the integral. Using this, (7) reads

$$ | \det_{\mathfrak g}(g)| ( \alpha, \Phi^g) = g^{-1} \cdot (\alpha, \Phi). $$

However, $| \det_{\mathfrak g}(g)| $ since $G$ is compact so its determinant acting on any representation has unit norm. Furthermore, $\Phi^g = g^{-1} \cdot \Phi$ because these have the same definition. So we can express the identity as

$$(\alpha, g^{-1} \cdot \Phi) =g^{-1} \cdot (\alpha ,\Phi)$$ which follows from the definition $g \alpha= \alpha$ of equivariance.

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  • $\begingroup$ thank you so much @Will Sawin for your answer ! I just still have a question , why the determinant appear in this formula, does it come from a change of variable ? $\endgroup$
    – asma
    Feb 4 at 8:20
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    $\begingroup$ @asma Essentially, yes. What's strange is the integral is defined in a purely formal way for generalized functions, so the usual change of variable formula doesn't exist, but I think the author has chosen to write this to look like the usual change of variable formula. It's somewhat irrelevant here because that term is always 1. I think for a noncompact group where the determinant was not always 1, you would need to put a determinant term in the first formula as well, to match this one. $\endgroup$
    – Will Sawin
    Feb 4 at 16:59

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