Relation between cohomology of ordered and unordered configuration spaces? For any manifold $M$, the unordered configuration space of $k$ points is obtained as a quotient of ordered configuration space of $k$ points by the group action of symmetric group on $k$ letters. Does it induce some relation between the cohomology algebras of the two spaces?
 A: Oscar Randal-Williams already mentioned the transfer isomorphism $H^*(C_n(M);\mathbb{Q}) \approx H^*(F_n(M);\mathbb{Q})^{\Sigma_n}$. At the risk of self-promotion, one place you can see this transfer in action is in my paper "Homological stability for configuration spaces of manifolds", arXiv:1103.2441. In that paper I used an analysis of $H^*(F_n(M);\mathbb{Q})$ and the action of $\Sigma_n$ on it to conclude that the cohomology of the unordered configuration space $C_n(M)$ is eventually independent of the number of points $n$: $$H^k(C_n(M);\mathbb{Q})\approx H^k(C_{n+1}(M);\mathbb{Q})\text{ for }n\gg k.$$
The key idea is that we can relate $F_{n+1}(M)$ with $F_n(M)$, even when we cannot relate $C_{n+1}(M)$ with $C_n(M)$ directly, and then the transfer map lets us push information from $F_n(M)$ down to $C_n(M)$. The basic framework of this approach is explained in the introduction.
There are also some explicit computations (some going back to Bödigheimer–Cohen–Taylor) of $H^*(C_n(M);\mathbb{Q})$ for various manifolds $M$ in Section 4.2 that might be of interest to you.
Also, you should check out the papers on the cohomology of configuration spaces written by the other participants in this discussion, Oscar Randal-Williams and Giacomo d'Antonio, which should provide a somewhat different perspective.
A: This is an expansion to Oscar's answer. The cited result on the cohomology with rational coefficients is an application of the Transfer Theorem:

Let $G$ be a finite group, $X$ a topological manifold and $F$ a field with 
  $\mbox{char}\,F = 0$ or $\mbox{char}\,F \nmid o(G)$, then
  $$H^*(X/G; F) \cong H^*(X;F)^G$$

There is a nice proof of this theorem on Bredon's book "Introduction to compact transformation groups."
For the case of $\mathbb R^k$ you can compute the cohomology (with complex coefficients) explicitly. The answer depends on the parity of $k$. For odd $k$, $H^*(F_n(\mathbb{R}^k); \mathbb{C})$ is one dimensional in degree $0$ and trivial otherwise, for even $k$ it is one-dimensional in degrees $0$ and $k-1$ and trivial otherwise.
A: If $F_n(M)$ denotes the ordered configuration space and $C_n(M)$ the unordered configuration space, the quotient map gives a map
$$H^*(C_n(M)) \longrightarrow H^*(F_n(M)).$$
If one takes rational coefficients, then this induces an isomorphism
$$H^*(C_n(M);\mathbb{Q}) \longrightarrow H^*(F_n(M);\mathbb{Q})^{\Sigma_n}$$
onto the symmetric group invariant subspace.
In positive characteristic I don't think there is any pleasant relation between them.
