3
$\begingroup$

The homology cylinder group is the monoid of homology cylinders modulo homology cobordism. I wonder whether there is any known finite order element in this group.

$\endgroup$
3
  • $\begingroup$ Could you explain which dimension(s) you're interested in? Since there's a 3-manifold tag, presumably the cylinders or their boundaries have dimension 3? $\endgroup$
    – Ian Agol
    Commented May 11, 2012 at 14:18
  • $\begingroup$ @Agol: I think he is referring to 3-dimensional cylinders with surface boundary. $\endgroup$
    – Jim Conant
    Commented May 11, 2012 at 14:42
  • $\begingroup$ @Agol: yes, Conant is right. $\endgroup$
    – hopflink
    Commented May 11, 2012 at 14:56

1 Answer 1

7
$\begingroup$

Certainly there is a lot of $2$-torsion. One easy way to construct such an example is to look at the embedding of the string link cobordism group on $n$ strands into the homology cobordism group of genus $n$. (Levine explains this embedding in his paper "Homology cylinders: an enlargment of the mapping class group." See Theorem 4 of that paper.) The string link group has lots of $2$-torsion. For example just tie an amphicheiral knot (which is not slice) into one of the strands. In fact the knot concordance group has infinitely many $\mathbb Z_2$ summands which inject into the homology cylinder group.

$\endgroup$
4
  • $\begingroup$ Thank you! You helps me a lot. Is there any other example, for surfaces with boundary ? $\endgroup$
    – hopflink
    Commented May 11, 2012 at 14:16
  • 1
    $\begingroup$ @HJ: this construction is for surfaces with one boundary component. I am sure there are other examples too. This paper arxiv.org/abs/0909.5580 shows there are infinitely many $\mathbb Z_2$ invariants, which should be realizable by actual homology cylinders, though I don't know off the top of my head. $\endgroup$
    – Jim Conant
    Commented May 11, 2012 at 14:42
  • $\begingroup$ @Conant: Would there be more example for more boundary componant? $\endgroup$
    – hopflink
    Commented May 12, 2012 at 7:18
  • $\begingroup$ @HJ: Sure. The map from string links to homology cobordisms extends to arbitrary numbers of components. Also, a string link is a homology cobordism of a planar surface so you get lots of direct examples that way. $\endgroup$
    – Jim Conant
    Commented May 12, 2012 at 12:15

You must log in to answer this question.

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