# configuration spaces of real projective space

Let $F(\mathbb{R}P^n,k)$ be the $k$-th ordered configuration space on $\mathbb{R}P^n$. In http://arxiv.org/abs/1502.04258, the cohomology ring $$H^*(F(\mathbb{R}P^n,k);R)$$ is obtained for any commutative ring $R$ with unit and $2$ invertible. I want to find $$H^*(F(\mathbb{R}P^n,2);\mathbb{Z}).$$

When I use the Serre spectral sequence for the fibration $$\mathbb{R}P^n\setminus *\simeq \mathbb{R}P^{n-1}\to F(\mathbb{R}P^n,2)\to \mathbb{R}P^n,$$ I do not know how to determine the differentials. How can I get $$H^*(F(\mathbb{R}P^n,2);\mathbb{Z})?$$

Are there more general results for $$H^*(F(\mathbb{R}P^n,k);\mathbb{Z})?$$

$F(\mathbb RP^n,2)$ is the space of ordered pairs of distinct points in real projective space. If you pass to the cover $S^n$ that means you're studying ordered pairs of distinct 1-dimensional subspaces of $\mathbb R^{n+1}$.

So there is a fibre bundle $F(\mathbb RP^n) \to Gr_{n+1, 2}$ (the Grassmannian) and the fibre is the space of ways of writing $\mathbb R^2$ as a direct sum of two 1-dimensional subspaces. The fibre has the homotopy-type of a circle.

Thus, $F(\mathbb RP^n,2)$ is a fibre bundle over the Grassmannian $Gr_{n+1,2}$ of 2-dimensional subspaces of $\mathbb R^{n+1}$ with fibre a circle. I suspect it's a non-trivial bundle provided $n \geq 2$.

As a space, I suppose an efficient way to describe it would be $$O_{n+1} / ( O_{n-1} \times D_4 )$$

And the bundle to the Grassmannian would be the projection

$$O_{n+1} / ( O_{n-1} \times D_4 ) \to O_{n+1} / ( O_{n-1} \times O_2 )$$

This is a moderately efficient approach but it has less fuel when there's more than two points in your configuration, then this approach looks more like your original approach.

Using the approach you suggest is pretty productive. But it becomes a long inductive process. For example, $F(\mathbb RP^2,2)$ is a bundle over $\mathbb RP^2$ with fibre a circle. The $\pi_1$ monodromy is a non-trivial (orientation reversing) automorphism of the circle. This is just a linear algebra argument. All the monodromies are linear algebra constructions. So in principle you can climb up the inductive ladder and compute the homology and cohomology of $F(\mathbb RP^n,k)$.

So this isn't a complete answer to your question but it gives you some more things to work with.