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2
votes
Example equivariant Mayer-Vietoris for $H^{*}_{S^{1}}(S^{2})$
This is fine for $\mathbb Z$-coefficients; you don't need to go to $\mathbb R$.
The part of the sequence you need
is $$0\to H_{S^1}^0(S^2) \to H_{S^1}^0(U) \oplus H_{S^1}^0(V) \to H_{S^1}^0(U\cap V …
5
votes
Accepted
Why is the equivariant Euler class a character ?
Well, there's going to be some map $T^* \to H^2_T(pt)$ taking a weight $\lambda$ to the equivariant Euler class of the corresponding line bundle over a point. If you add weights, that tensors the line …
1
vote
Accepted
Elementary question: Intuition for equivariant cohomology
I'm guessing that, unstated, $M,G$ are finite-dimensional and $G$ is connected Lie.
Then $H^*(M/G)$ vanishes for $* \gg 0$, but $H^*_G$ is positively graded, so $H^*(M/G)$ must be a torsion module. …