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We want to calculate $H^{k}_{S^1}(S^2)$. We can choose two open sets $U=S^{2} \setminus p_{+}$ and $V = S^{2} \setminus p_{-}$, where $p_{+}$ and $p_{-}$ are the north and south pole of $S^{2}$. They are fixed points of the action ($S^{1}$-invariant).
So $U \cap V$ is homotopic to $S^{1} \hookrightarrow S^{2}$ on which $S^{1}$ acts freely. So $$ H^{*}_{S^{1}}(U \cap V) \simeq H^{*}_{S^{1}}(S^{1}) \simeq H^{*}\left(\frac{S^{1}}{S^{1}} \right) \simeq H^{*}(pt) .$$ Moreover $$ H^{*}_{S^{1}}(U) \simeq H^{*}_{S^{1}}(p_{-}) \simeq H^{*}(BS^{1}) = \mathbb{R}[x_{-}] $$ where $x_{-}$ has degree $2$ end $H^{*}_{S^{1}}(V)= \mathbb{R}[x_{+}]$.

So when $k>1$ $$ H^{*}_{S^{1}}(S^{2}) \simeq \mathbb{R}[x_{+}] \oplus \mathbb{R}[x_{+}] $$ and if $k=1$ or $k=0$?

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Presumably the action is rotation on the north-south axis. – Mariano Suárez-Alvarez Jan 20 '13 at 0:59
Up to conjugacy the unique maximal compact subgroup of $Diff(S^2)$ is $SO(3)$, within which $S^1$ is a maximal torus, so yes! – Allen Knutson Jan 20 '13 at 21:41
Well, up to multiple. – Allen Knutson Jan 20 '13 at 21:41

This is fine for $\mathbb Z$-coefficients; you don't need to go to $\mathbb R$.

The part of the sequence you need is $$0\to H_{S^1}^0(S^2) \to H_{S^1}^0(U) \oplus H_{S^1}^0(V) \to H_{S^1}^0(U\cap V) \to H_{S^1}^1(S^2) \to H_{S^1}^1(U) \oplus H_{S^1}^1(V) $$ which you worked out to $$ 0\to H_{S^1}^0(S^2) \to {\mathbb Z}^2 \to {\mathbb Z} \to H_{S^1}^1(S^2) \to 0 \oplus 0 $$ so it is enough to notice that the restriction map $H_{S^1}^0(V) \to H_{S^1}^0(U\cap V)$, aka $H^0({\mathbb CP}^{\infty}) \to H^0(pt)$, is onto. Then $H^0_{S^1}(S^2) = \mathbb Z$ and $H^1 = 0$.

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But if I take $\mathbb{R}$ it is still true... Why $\mathbb{Z}^{2}$? – MiliskWall Jan 20 '13 at 1:07

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