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

Why is $H^*_G(M)$ a torsion $H^*_G$-module, where $H^*_G=H^*_G(pt)=H^*(BG)$?

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 $0$?

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?

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    $\begingroup$ Crossposted on MSE. Please don't do that. $\endgroup$
    – Michael Albanese
    Commented Aug 2, 2016 at 1:40
  • $\begingroup$ @MichaelAlbanese I have edited my post to mention that. If this is not ok, I can delete the MSE post. $\endgroup$
    – HLC
    Commented Aug 2, 2016 at 1:46
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    $\begingroup$ Don't ask the question on both sites at the same time. Pick one and ask there. $\endgroup$
    – Michael Albanese
    Commented Aug 2, 2016 at 1:47
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    $\begingroup$ The module structure comes from the map $X_G = (X \times EG)/G \to EG/G = BG$. When $G$ acts freely on $X$, $X_G = X/G \times EG$. When $X$ is a finite-dimensional manifold, $X/G$ has finite-dimensional cohomology, so the $t_i$ must be torsion in its cohomology ring. It is not strictly true that the non-torsion part of $H^*_G(M)$ is contributed by the $G$-fixed part, but rather by non-free $G$-orbits; though in the case of $T^n$, the non-free $G$-orbits have trivial rational cohomology ring if they're not fixed; for a more complicated case see $SU(2)$ acting on $S^2$. Keyword: localization. $\endgroup$
    – mme
    Commented Aug 2, 2016 at 2:09

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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 $0$.

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  • $\begingroup$ What does it mean $\deg t_i=2$? Why is it not $1$? (I'm really sorry for burdening MO like this.) $\endgroup$
    – HLC
    Commented Aug 2, 2016 at 17:12
  • $\begingroup$ $H^*_{S^1}(pt) := H^*(ES^1/S^1) = H^*(\mathbb{CP}^\infty) = \mathbb Z[t]$ where $\deg t = 2$. This is very, very basic to the study of equivariant cohomology. Note as a mnemonic that since these are cohomology rings, they are supercommutative not commutative, so will only give you polynomial rings if the generators are in even degree. $\endgroup$
    – Allen Knutson
    Commented Aug 4, 2016 at 12:18

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