A group $G$ acts freely on a manifold $M$, then $H^\*\_G(M)=H^\*(M/G)$.



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> Why is $H^\*\_G(M)$ a torsion $H^\*\_G$-module, where $H^\*\_G=H^\*\_G(pt)=H^\*(BG)$?
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> If $G=T=(S^1)^{n+1}$ is a torus then $H^\*\_G=H^\*\_T=\mathbb{Q}[t\_0,...,t\_n]$. Why does $t\_i$ act on $H^\*\_G(M)$ by multiplication by $0$?
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More importantly, I would like to understand the intuition behind these.


For example: does the answer to the second question above have to do with the infinitesimal action of the Lie algebra of $T$ on $M$? what does it mean to say that the non-torsion part of $H^\*\_G(M)$ is contributed by the $G$-fixed part of M?



Elementary question: Intuition for equivariant cohomology
I'm guessing that, unstated, $M,G$ are finite-dimensional and $G$ is connected Lie.


1. Then $H^\*(M/G)$ vanishes for $\* \gg 0$, but $H^\*\_G$ is positively graded, so $H^\*(M/G)$ must be a torsion module. (Non-example: $M$ is the unit sphere in Hilbert space, $G=U(1)$.)
2. $H^{\*>0}(T/T)=0$, and each $\deg t\_i = 2$, so they must act as $0$.


