1
$\begingroup$

I'm considering finite index abelian (regular) covering of link complement:

$$ X \rightarrow S^3\setminus L$$

where $L$ is a minimally twisted chain link.

I'm interested in covering space. Can we compute its cohomology (or maybe fundamental group) in terms of $S^3\setminus L$? Is it true that $X$ is link complement? I would appreciate any helpful tip on this.

Improve this question
$\endgroup$
5
  • $\begingroup$ What is a (the?) "finite index universal abelian cover" of a manifold? $\endgroup$
    – Sam Nead
    Commented Mar 15, 2023 at 19:54
  • $\begingroup$ For the fundamental group of a link complement, the relevant Google search terms are "Wirtinger's algorithm" and "Wirtinger presentation." There's a really nice chapter of Rolfsen's "Knots and Links" that talks about coverings of $S^3 \setminus L$ using the fact that every oriented link in $S^3$ bounds a Seifert surface. $\endgroup$
    – gdd
    Commented Mar 16, 2023 at 0:28
  • $\begingroup$ The homology of abelian covers of link exteriors, and of branched covers of links in spheres were two of the first invariants ever considered, in the beginning of knot theory. So yes we can compute these things. Some have taught computers to do it for us. $\endgroup$
    – Ryan Budney
    Commented Mar 16, 2023 at 3:31
  • 1
    $\begingroup$ To expand on Sam Nead's question: the "universal abelian cover" of a manifold $M$ usually means the covering space corresponding to the kernel of the Hurewicz map $\pi_1(M)\to H_1(M,\mathbb{Z})$. In the case of a link complement, this covering spaces never has finite index (because $H_1(M,\mathbb{Z})$ is infinite). So the words "finite index" make the question unclear. Assuming you just want to delete the words "finite index", many tools are available, as the other comments have indicated. But please edit to clarify! $\endgroup$
    – HJRW
    Commented Mar 16, 2023 at 9:44
  • 1
    $\begingroup$ Welcome to MathOverflow! Please use the Contact Us form to have your accounts merged, to regain full control over your posts. $\endgroup$
    – Glorfindel
    Commented Mar 16, 2023 at 11:22

2 Answers 2

Reset to default
1
$\begingroup$

  1. Yes there are (implemented!) algorithms to compute the (co)homology of manifolds and their covers.

  2. In general finite covers of link complements, even abelian ones, need not be link complements themselves. As an example, consider the snappy manifold s776. I believe that this is the minimally twisted three component chain link. The manifold s776 has as a double cover the manifold with triangulation isosig “mLvMALPAQbeffeijjikllldadedadedaaed”. This cover has torsion in its first homology, so it is not a link complement in the three-sphere.

Improve this answer
$\endgroup$
0
$\begingroup$

A nice hands-on method for computation could be using C-complexes. To get started, there is a fantastic paper on them by Cimasoni-Florens.

Improve this answer
$\endgroup$

You must log in to answer this question.

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