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What is $H^*(\mathbb{CP}^{2^N-1}/\Sigma_n;\mathbb{Z})$ when $N=\binom{n}{2}$?
$H^*((S^3)^N/\Sigma_n;\mathbb{Q})$ is computed here.
It makes a little more sense to compute $H^*((S^2)^N/\Sigma_n;\mathbb{Q})$ given that global phase is irrelevant. The proof is exactly the same.
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