12
$\begingroup$

It's easy to derive a presentation of the fundamental group of a 4-manifold if you have a Kirby diagram: The 1-handles are generators and the 2-handles are the relations. The 3- and 4-handles, which are invisible in the Kirby diagrams, don't contribute.

Is there anything like this for the homotopy 2-type?

I'm imagining an algorithm that takes a Kirby diagram and spits out a groupoid, or a crossed module. Probably, the 1-handles generate the 1-morphisms, the 2-handles the 2-morphisms, and the higher handles are relations? How can the relations be visualised, though?

$\endgroup$
  • 3
    $\begingroup$ I thought about this at some point but didn't really get anywhere. The problem is that the 3-handles actually do matter, already when you want to get a description of $H_2(X)$, for example. The best I could do is to describe $H_2(X)/torsion$ and the intersection form on this group for closed $X$. (This is written down in an appendix to my thesis, but probably also in other places.) [to be continued...] $\endgroup$ – Stefan Behrens Feb 9 '17 at 14:33
  • 3
    $\begingroup$ To get $\pi_2(X)$ with all its structrure ($\pi_1$-module, equivariant intersection form) the obvious thing to try is to produce an infinite Kirby diagram for the universal cover and take it from there. But this looked like a combinatorial nightmare and I soon stopped trying. Of course, this doesn't mean that it's impossible to get information in this way, only that I didn't manage to do so. I'd still be interested in an answer to the question, though. $\endgroup$ – Stefan Behrens Feb 9 '17 at 14:33
7
$\begingroup$

Yes: given a CW-complex $X$ the fundamental crossed module $\Pi_2(X,X^1)=(\partial \colon \pi_2(X,X^1) \to \pi_1(X^1))$, where $X^1$ is the 1-skeleton, represents the homotopy 2-type of $X$ (this can be stated in several ways). Moreover $\Pi_2(X,X^1)$ can be calculated combinatorially: $(\partial \colon \pi_2(X^2,X^1) \to \pi_1(X^1))$ is a totally free crossed module, and when you attach three handles you solely need to impose relations on $\pi_2(X^2,X^1)$ in order to get to $\pi_2(X,X^1)$

Now if you have a Kirby diagram (i.e. a handlebody decomposition), then squashing the handles along their core yields a CW-complex. So in theory the fundamental crossed module of the associated CW-complex can be calculated combinatorially. Except that it might be a non-trivial exercise to determine the attaching maps of the 3-handles in such that way that the relations on $\pi_2(X^2,X^1)$ become transparent. (For complements of knotted surfaces this is quite a doable thing.)

Some discussion is in J. Faria Martins, The fundamental crossed module of the complement of a knotted surface, Trans. Amer. Math. Soc. 361 (2009), no. 9, 4593–4630 https://arxiv.org/abs/0801.3921

and also in: Higher lattices, discrete two-dimensional holonomy and topological phases in (3+1) D with higher gauge symmetry (https://arxiv.org/abs/1702.00868) section 3.4 and On 2-Dimensional Homotopy Invariants of Complements of Knotted Surfaces (https://arxiv.org/abs/math/0507239)

$\endgroup$
5
$\begingroup$

A step towards what you seem to be imagining may be in the following paper (available here):

J. Faria Martins, The fundamental crossed module of the complement of a knotted surface, Trans. Amer. Math. Soc. 361 (2009), no. 9, 4593–4630

As the title states this looks at the case of the complement of a knotted surface. I do not know if that helps.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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