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Suppose that $G=S^1$ acts on a smooth, connected, compact manifold with discrete fixed points, additionally assume that there is at least one fixed point.

Let $\alpha \in H^{2}_{S^1}(M)$ be such that $\alpha|_{p} = 0$ for any fixed point $p$. The space of such $\alpha$ forms an additive sub-group $A \subset H^{2}_{S^{1}}(M)$.

Question: is there an example where $A \neq \{0\}$?

How about if we assume that $M$ is a compact symplectic manifold with a Hamiltonian $S^1$-action with discrete fixed points?

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This subgroup $A$ is precisely the torsion subgroup in $H^2_{S^1}(M)$ (since $H^2_{S^1}(M^{S^1})$ is obviously free, and the relative $H^2_{S^1}(M,M^{S^1})$ is torsion). So, it will be trivial if and only if the equivariant cohomology in degree 2 is free. This is true for compact $M$ with Hamiltonian $S^1$-action by Kirwan (see, for example, Theorem 14.1 of http://www.math.ias.edu/~goresky/pdf/equivariant.jour.pdf), but I think this is too much to hope for in general (I don't know a connected counterxample off-hand; obviously, you can take a free action on a compact manifold, and take disjoint union with a fixed point).

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  • $\begingroup$ Thanks for the answer! it helped a lot. I have one slight issue, It would seem to make more sense to me in terms of the exact sequence if you had written that $H^{2}_{S^1}(M,M^{S^1})$ is torsion, since that maps into $H^{2}_{S^1}(M)$? perhaps I am misunderstanding $\endgroup$
    – Nick L
    Commented May 22, 2017 at 11:01
  • $\begingroup$ No, you're right; I should be able to use exact sequences correctly by now, but apparently not. $\endgroup$
    – Ben Webster ♦
    Commented May 22, 2017 at 13:12

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