5
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ Do you mean the cohomology of the Borel construction? $\endgroup$ Dec 4, 2022 at 8:28
  • $\begingroup$ Yes (what other options are there?) $\endgroup$ 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) $\endgroup$ Dec 4, 2022 at 11:14

1 Answer 1

11
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.