On $n$-th cohomology of $M_{0,n+3}$ Can anyone please provide me with a reference on $H^n(M_{0,n+3},{\mathbb C})$ where $M_{0,n+3}$ is the (affine) scheme parametrizing $n+3$ labeled distinct points on ${\mathbb C\mathbb P}^1$? I am looking for combinatorial description, dimension, weight filtration, etc.
 A: The action of the symmetric group can be described in a compact way by means of the isomorphisms
$$ \newcommand{\Lie}{\mathsf{Lie}}\Lie(\!(n+3)\!) \cong H^n(M_{0,n+3}) \otimes \mathrm{sgn}_{n+3}$$ 
for all $n$. These are due to Getzler: see his paper "Operads and moduli spaces of genus 0 Riemann surfaces". Here we denote by $\Lie(\!(N)\!)$ the component in arity $N$ of the cyclic Lie operad; if $\Lie(n)$ denotes the ordinary Lie operad then
$$ \Lie(n) = \mathrm{Res}^{S_{n+1}}_{S_n}\Lie(\!(n+1)\!).$$
There is a very simple combinatorial description of $\Lie(n)$: it is the vector space spanned by Lie words in the symbols $a_1,\ldots, a_n$ such that each symbol appears exactly once. The module $\Lie(\!(n)\!)$ is subtler but can be computed from $\Lie(n)$ via
$$ \Lie(\!(n)\!) \oplus \Lie(n) \cong \mathrm{Ind}_{S_{n-1}}^{S_n} \Lie(n-1).$$
See e.g. these notes of Stanley.
A: $M_{0,n+3}$ can be identified with the set $(z_1, z_2, \ldots, z_n) \in \mathbb{C}^n$ so that $z_i \neq 0$, $z_i \neq 1$ and $z_i \neq z_j$ for every $i \neq j$. In other words, it is the complement of $2n+\binom{n}{2}$ hyperplanes in $\mathbb{C}^n$. The cohomology of the complement of a hyperplane arrangement was computed by Orlik and Solomon; here is an expository paper which covers the full result in great detail. 
In summary, consider the complement in $\mathbb{C}^n$ of the hyperplanes given by linear equations $\lambda_1(z)=0$, $\lambda_2(z)=0$, ..., $\lambda_N(z)=0$. The cohomology ring is generated in $H^1$ by the forms $d \lambda_i/\lambda_i$; these are Hodge-Tate of type $(1,1)$, so $H^n$ is entirely of type $(n,n)$. There are explicit combinatorial formulas for the dimensions in terms of matroid theory. In your case, I get that $\dim H^n(M_{n+3}) = (n+1)!$. 
