4
$\begingroup$

Let $M$ be a manifold.

Let $F(M,n)$ be the configuration space of $n$-tuples on $M$.

Let $B(M,n)=F(M,n)/S_n$, where $S_n$ is the symmetric group of order $n$, be the corresponding unordered configuration space.

If $M$ is compact, then the graded vector space structure of $H_*(B(M,n);F)$, where $F=\mathbb{Q}$ or $\mathbb{Z}/p\mathbb{Z}$ is a field, is given in the paper C.-F. BODIGHEIMER, F. COHEN, L. TAYLOR, On the homology of configuration spaces, Topology 1989.

If $M$ is non-compact, for example, $M=S^2\times \mathbb{R}$, how to understand the graded vector space structure of $H_*(B(M,n);F)$?

Is there any reference or procedure to follow?

$\endgroup$
2
  • 4
    $\begingroup$ Over the rationals, you can take a look at Ben Knudsen's recent work: arxiv.org/abs/1405.6696. Alternatively, in a range depending on the number of particles you can use homological stability and scanning. This works particularly well over the rationals, where rational homotopy theory is available. $\endgroup$
    – skupers
    Feb 1, 2015 at 3:41
  • $\begingroup$ How about Z/2Z? I mostly want to know Z/2Z. $\endgroup$ Feb 1, 2015 at 5:58

1 Answer 1

5
$\begingroup$

I believe that, although the results of Bodigheimer-Cohen-Taylor are stated for compact manifolds (potentially with boundary), they hold for noncompact manifolds which are homeomorphic to the interior of a compact manifold. See, for instance, the last sentence of 2.1 of that article.

If $M$ is a compact manifold with boundary, and $N = M \setminus \partial M$ its interior, then the natural inclusion $B(N, n) \to B(M, n)$ is a homotopy equivalence (with inverse induced by an injective map $M \to N$ which ``pushes $M$ in from its boundary;" this is isotopic to the identity). So to compute the cohomology of $B(N, n)$, it suffices to compute that of $B(M, n)$. If mod 2 coefficients are what you're after, Bodigheimer-Cohen-Taylor is precisely what you need.

Regarding your specific question of $S^2 \times \mathbb{R}$, this is homeomorphic to the interior of the compact manifold $S^2 \times [0, 1]$, so BCT applies.

$\endgroup$

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.