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We want calculate $H^{*}_{S^1}(S^2)$. We can choose two open sets $U= S^{2} \setminus p_{+}$ and $V = S^{2} \setminus p_{-}$, where $p_{+}$ and $p_{-}$ are the north pole and south pole of $S^{2}$. There are fixed points of action and they are $S^{1}$-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 $2$ and $H^{*}_{S^{1}}(V)= \mathbb{R}[x_{+}]$.

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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up vote 2 down vote accepted

Assuming $k$ to be the degree of the equivariant cohomology (and that you are trying to apply MV sequence), I try to answer (by using Cartan model).

First of all we do not need MV sequence to find $H^0_{S^1}(S^2)$. A cohomology class of degree $0$ is represented by an $S^1$-invariant function $f$ with $d_Gf=0$. Since $d_Gf=df=0$, $f$ is a locally constant function. Because $S^2$ is connected, we conclude $H^0_{S^1}(S^2)=\mathbb{R}$.

Next, here is part of MV sequence. $$H_{S^1}^0(U) \oplus H_{S^1}^0(V) \xrightarrow{\phi} H_{S^1}^0(U \cap V) \to H_{S^1}^1(S^2) \to 0.$$ If we recall the definition of $\phi$ and the above explanation for $H^0_{S^1}$, we can easily find that $\phi$ is surjective. So $H_{S^1}^1(S^2)=0$.

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