Why does the formula (7) in the article equivariant cohomology with generalized coefficients hold

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:

1 Answer

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.

• 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 ?
– asma
Feb 4 at 8:20
• @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. Feb 4 at 16:59