$\mathbb Z_2$-equivariant cohomology of tori

Question

I consider the tori $X=(\mathbb R/ 2 \pi \mathbb Z)^d$ on which the group $\mathbb Z_2$ acts by $x \mapsto -x$ for the nontrivial element of $\mathbb Z_2$. This action has $2^d$ fixed points, which are of the form $x=(x_1,...,x_d)$ with each $x_i$ equal to $0$ or $\pi$. I would like to know what are the equivariant cohomology groups $H^{\bullet}_{\mathbb Z_2}(X, \mathbb Z)$ (although I suspect $\mathbb Z_2$ coefficients might be easier and would be happy to learn about these cohomology groups as well). I know that there is a spectral sequence with $E_2^{p,q}=(\mathbb{RP}^{\infty},H^q(X,\mathbb Z))$, but it is unclear to me how to extract useful information from that.

Answer 1

Since in the fibration $X \hookrightarrow X \times_\mathbb{Z_2}E{\mathbb{Z}_2} \to B\mathbb{Z}_2 \simeq \mathbb{RP}^{\infty}$ the fiber's cohomology groups are freely generated, your situation calls for the Leray–Hirsch theorem (in Hatcher it is p. 432). You want then the inclusion $X \hookrightarrow X \times_\mathbb{Z_2}E{\mathbb{Z}_2}$ to induce a surjection on cohomologies. For that, it is enough to show the surjectivity of $H_\mathbb{Z_2}^1(X) \to H^1(X)$ (use multiplicativity). I can't now think of a clever way to show that but to investigate 1- and 2-dimensional cells seems to be enough. (Another way which might work is to replace absolute circles with pairs of intervals and their boundaries: in the relative case you won't have the first row in the spectral sequence, so the differential $d_2^{0,1}$ would vanish, which is equivalent to surjectivity.)

UPD: The spectral sequence reference was unnecessary, deleted.