# What is the circle-equivariant cohomology of the real projective plane

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.

• Do you mean the cohomology of the Borel construction? Dec 4, 2022 at 8:28
• Yes (what other options are there?) Dec 4, 2022 at 9:35
• 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) Dec 4, 2022 at 11:14

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.