2
$\begingroup$

There is a reference or a methode which by it we can calculate the cohomology of a configuration space of a compact manifold simply connected? It is possible to find a spectral sequence converging to this cohomology (but the Cohen-Taylor spectral sequence)?

$\endgroup$
5
  • $\begingroup$ What kind of configuration spaces do you have in mind? For example, the configuration space of k points in the 2-dimensional disc is the classifying space of the k-strand braid group, and there is a rather large literature on its properties including its cohomology. Is that an example of what you are asking about? $\endgroup$
    – Lee Mosher
    Feb 23, 2013 at 20:04
  • $\begingroup$ I mean how to calculate the rational cohomology of the configuration space of a compact manifold simply connected in general, or if it is possible determinate a model fot the configuration space, i know that Kriz and Totaro gave a model for the configuration spaces $F(M,k)$ when $M$ is a complex projective manifold, but in general case, it is possible to use the same technics to determinate it? $\endgroup$ Feb 23, 2013 at 22:05
  • $\begingroup$ Let me rephrase my question. What definition of a configuration space are you using? For example, is the space of k-element subsets of the 2-dimensional disc an example of your type of configuration space? The terminology "configuration space" is not completely standard, hence the need to state what definition you are using. $\endgroup$
    – Lee Mosher
    Feb 24, 2013 at 5:14
  • $\begingroup$ @Lee Mosher: here configuration space means $F(M,k) = M^k \setminus \Delta$ where $\Delta$ is the "big diagonal". $\endgroup$ Feb 24, 2013 at 8:18
  • $\begingroup$ I use the classical the following definition of ordered configuration space: Let $M$ an $m-$dimensional manifold. The space of ordered configurations of $k$ pointsis the space $$F(M,k)=\{(x_1,...,x_k)\in M^k ;x_i\neq x_j for i\neq j \}$$, and we ask to find the rational cohomology of this space. $\endgroup$ Feb 24, 2013 at 9:57

2 Answers 2

4
$\begingroup$

Shameless promotion ahead. If you are interested in cohomology with base field $\mathbb{R}$, then:

  • For simply connected manifolds without boundary you have my paper and a paper of Campos and Willwacher. Both papers give so-called "real models" for the configuration spaces; these models are commutative differential graded algebras, and in particular the cohomology of the model is the cohomology of the configuration space.
  • For simply connected manifolds with boundary (and interiors of such manifolds, it's the same thing), we have a paper with Campos, Lambrechts, and Willwacher, where we also give a real model (several, in fact). If $\dim M \ge 4$ then this model is fairly explicit, otherwise it's complicated.

Besides all the models we give are equipped with a symmetric group action that models the natural symmetric group action on $\operatorname{Conf}_k(M)$, so if you are interested in unordered configuration spaces then you can compute their cohomology by considering invariants.

$\endgroup$
0
2
$\begingroup$

I would be extremely surprised if there was anything other than the Cohen-Taylor spectral sequence that you can do in this generality.

As you know, the first nontrivial page of Cohen-Taylor spectral sequence depends only on the ring $H^\bullet(M)$. I think it's written up somewhere that the higher order differentials are defined by Massey products on $M$. In particular the result you refer to in a comment that $H^\bullet(F(M,n))$ (or its associated graded) can be computed explicitly from $H^\bullet(M)$ extends to arbitrary formal manifolds $M$, not just smooth projective varieties. And if you give yourself a minimal model of $M$ you should be able to compute $H^\bullet(F(M,n))$ for all $n$ as well (but not a minimal model of $F(M,n)$ -- see the paper of Longoni and Salvatore).

$\endgroup$
1
  • $\begingroup$ The paper of Longoni and Salvatore shows that configuration spaces are not homotopy invariant in the non-simply connected case. I believe the question of whether the rational homotopy type of $F(M,n)$ is an invariant of rational homotopy type of simply-connected closed $M$ is still open. See the paper eudml.org/doc/116129 of Lambrechts and Stanley, which treats the $2$-connected case. $\endgroup$
    – Mark Grant
    Feb 25, 2013 at 7:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.