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Let P denote the real projective plane. It has an action of the circle group S1. (e.g. Let S1 act on the 2-sphere by rotations about an axis, then this action descends to the quotient P). I have a simple question.

What is the equivariant cohomology $H^\ast_{S^1}(P;\mathbb{F}_2)$?

Ideally I want a description as a module over $H^\ast_{S^1}(\mathrm{pt};\mathbb{F}_2)$, but I can't even manage to compute the structure as an abelian group myself.

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    $\begingroup$ Do you mean the cohomology of the Borel construction? $\endgroup$
    – Fernando Muro
    Commented Dec 4, 2022 at 8:28
  • $\begingroup$ Yes (what other options are there?) $\endgroup$
    – Peter McNamara
    Commented Dec 4, 2022 at 9:35
  • $\begingroup$ Have you tried writing down an $S^1$-cell structure for $P$ ? There is a very simple one for $S^2$, and you can modify it appropriately to get one on $P$, which should give you a description over the cohomology of the point (regardless of whether you mean Borel or Bredon cohomology) $\endgroup$
    – Maxime Ramzi
    Commented Dec 4, 2022 at 11:14

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I think one gets $$H^*_{S^1}(\mathbb{RP}^2; \mathbb{F}_2) = \mathbb{F}_2[x, y]/(xy) $$ where $|x|=1$ and $|y|=2$. The module structure over $H^*_{S^1}(pt; \mathbb{F}_2) = \mathbb{F}_2[t]$ is given by $t \mapsto x^2 + y$.

I got this by writing $\mathbb{RP}^2$ as the union of $\mathbb{RP}^1$ equipped with the "double speed" action of $S^1$, and $D^2$ equipped with the natural action of $S^1$. The corresponding pushout of Borel constructions takes the form $$BC_2 \leftarrow pt \to BS^1$$ as spaces over $BS^1$, which gives the claimed formula.

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