# Cohomology of an orbit space

Let $T^n = (S^1)^n$ be the $n$-torus and $G = (\mathbb{Z}_2)^n$ be product of $n$ copies of the group of order $2$. Consider the action of $G$ on $T^n$ induced by the antipodal action componentwise.

Computing the cohomology $H^*(T^n/G;\mathbb{Z}_2)$, I am tempted to say that $T^n/G \cong (S^1/\mathbb{Z}_2)^n$ since the action happens componentwise, but I am not sure if that is correct. Also, I am trying to avoid a Spectral sequences argument, but if there is a solution in that way will be useful for future computations.

## 1 Answer

Your temptation is correct. If two elements of $T^n$ are related by an element of $G$, that just means that on each coordinate they are related by an element of $\mathbb{Z}_2$. So an orbit of $G$ consists of just a product of orbits of $\mathbb{Z}_2$ on each component. This gives a bijection $f:T^n/G\to (S^1/\mathbb{Z}_2)^n$ which is easily seen to be continuous using the universal property of $T^n/G$. Since $T^n/G$ is compact and $(S^1/\mathbb{Z}_2)^n$ is Hausdorff, you can conclude that $f$ is a homeomorphism.

Since $S^1/\mathbb{Z}_2\cong S^1$, this means that $T^n/G\cong T^n$, so you are just taking the cohomology of $T^n$.

• Thanks @Eric Wofsey for your answer. – C. Zhihao Oct 12 '17 at 13:56