2
$\begingroup$

I am working on the class of the Fulton-Macpherson compactification of configuration space in the Grothendieck ring of varieties, over the field of complex numbers. As a first step, I am wondering about the case of 2 points, i.e. the motive of $X[2]/S_2$ for a nonsingular variety $X$. For the case the two points coincide, one was lead naturally to the space $(\mathbb{P}(T_x^2/T_x))/S_2\cong\mathbb{P}((\mathbb{A}^n)^2/\mathbb{A}^n)$. My question is: (1) does this space admit an expression in terms of the Lefschetz motive $\mathbb{L}$? (2) does this space admit a cell decomposition? (i.e. is it a disjoint union of locally closed subvarieties isomorphic to affine spaces?)

$\endgroup$

1 Answer 1

0
$\begingroup$

If we identify $T_x^2 / T_x$ with $\mathbb A^n$, then the action of $S_2$ is multiplication by $-1$. So $S_2$ acts trivially on the projectivization and you are just asking about the existence of a cell decomposition of projective space.


A useful reference might be the last parts of Ezra Getzler's old preprint "Mixed Hodge structures of configuration spaces". He derives explicit formulas for the $S_n$-equivariant "motivic" Euler characteristic of $X[n]$ in terms of the motivic Euler characteristic of $X$. Here "motivic" means that the Euler characteristic is taken in a suitable ring of the form $K$ tensored with the representation ring of $S_n$, where $K$ is e.g. the Grothendieck group of $\ell$-adic Galois representations or the Grothendieck group of Hodge structures. So his formulas include in particular the equivariant Hodge-Deligne polynomial. Then taking $S_n$-invariants just corresponds to the substitution $p_n \mapsto x^n$, where $p_n$ is the $n$th power sum and $x$ is an indeterminate.

$\endgroup$
1
  • $\begingroup$ Yes you are absolutely right, but how about the case of at least three points? Does a clean motivic decomposition still exists? Also, thank you for the excellent reference! $\endgroup$
    – xf495
    Commented Feb 6, 2013 at 8:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .