4
$\begingroup$

There have been a couple of posts and questions on MathOverflow about the proofs of the following two facts:

Fact 1: if $X$ is a topological space, then $\pi_k(X,x)$ is abelian for $k\ge 2$.

Fact 2: if $G$ is a topological group, then $\pi_1(G,e)$ is abelian.

Both facts can be proven using the Eckmann-Hilton argument, which is a cool algebraisation of two more topological proofs that one can get by staring for sufficiently long to a couple of pictures* (after drawing them): while, up until two hours ago, I thought that the two proofs were actually distinct, Ryan Budney proved me wrong and showed me the connection (see comments below)

Now, to the question:

Is there any other example of an incursion of the Eckmann-Hilton argument into the realm of topology? Is there any other such application outside category theory/algebra?

I would like to see some results for which no proof is known that doesn't make use of the E-H argument, or such that any proof avoiding the argument (even somehow disguised) is significantly longer/harder.

EDIT: Thanks to Ryan Budney for pointing out how the two kind of proofs are actually the same proof, and to Tom Goodwillie for making me realise that the question was a bit too rough in an earlier version.

* One such picture is linked above. The other one is just a square with the diagonal, representing the map $(s,t)\mapsto (f(s),g(t))$: this picture gives homotopies between $f\cdot g$ and $f*g$, so they recover as much information as the E-H argument produces, see the question linked above.

Improve this question
$\endgroup$
6
  • 4
    $\begingroup$ I don't see much of a question here -- the Eckmann-Hilton argument is equivalent to the argument with the pictures you suggest. One is couched in more verbal/algebraic language, the other in more direct geometric language. But it's essentially the same argument in that the same homotopies are ultimately used in the Eckmann-Hilton argument, they're just decorated in a different way. $\endgroup$
    – Ryan Budney
    Commented Nov 16, 2011 at 17:54
  • 4
    $\begingroup$ "Intrusion", huh? I could just as well say that the topological proof conceals the fact that it's really an instance of Eckmann-Hilton. But I'd rather say that the choice is a matter of taste, and that for me the two points of view are nicely complementary. $\endgroup$
    – Tom Goodwillie
    Commented Nov 16, 2011 at 17:56
  • $\begingroup$ @Ryan Budney: I struggle to see homotopies in the proof of the E-H argument (that doesn't mean that they're not there). Could you expand a bit? Also, I'm editing the question, trying to make it a bit softer. -- @Tom Goodwillie: I just wanted to make clear "which side I'm on". I agree that it's nice to have both viewpoints, and I quite like the tricky proof. Nevertheless, I feel like the "topological" proof is more instructive for -say- an undergraduate student learning algebraic topology. $\endgroup$
    – Marco Golla
    Commented Nov 16, 2011 at 18:49
  • 1
    $\begingroup$ When you say something like $\pi_k(X,x)$ is abelian, that's saying there's a homotopy between the representatives for $ab$ and $ba$. The Wikipedia article you cite describes how to go from $ab$ to $ba$, and there are various ingredient steps like $a=a.1$ (this is itself an argument that there is a homotopy of representatives) and such. Assemble all these ingredient homotopies into the argument that $ab=ba$ and you get the same picture. The Wikipedia page strangely says this picture is more satisfying, yet it does not give it. $\endgroup$
    – Ryan Budney
    Commented Nov 16, 2011 at 19:10
  • 1
    $\begingroup$ If $E$ is an $E_\infinity$ ring spectrum (actually $E_2$ is enough) then the $E$-homology of a 2-fold loop space is a commutative $E_*$-algebra. This is merely a stable version of the facts you mention so this is probably not an acceptable answer. $\endgroup$
    – Geoffroy Horel
    Commented Nov 16, 2011 at 21:23

1 Answer 1

Reset to default
4
$\begingroup$

I can't resist pointing out that while the EH-argument shows that a group object in the category of groups is an abelian group, this does not apply to a group object in the category of groupoids, which is equivalent instead to a crossed module, which represents a pointed, connected homotopy 2-type.

Higher groupoids are in some sense ``more nonabelian'' than groups, and provide a route to some nonabelian calculations in higher homotopy theory.

So circumstances in which the EH-argument fails (for example a group object in the category of semigroups) are maybe more interesting.

Improve this answer
$\endgroup$
4
  • $\begingroup$ I should modify that comment. The EH-argument can also be phrased as using the interchange law''. This is fundamental in its general form for multiple groupoids, and their applications in homotopy theory, and higher dimensional algebra; it allows what we can call algebraic inverses to subdivision'', that is multiple subdivisions and compositions. Here the cubical apparatus is easiest to apply. $\endgroup$
    – Ronnie Brown
    Commented Dec 22, 2011 at 18:53
  • $\begingroup$ Thanks for the answer. Do you have any specific example of such "nonabelian calculations in higher homotopy theory"? $\endgroup$
    – Marco Golla
    Commented Dec 22, 2011 at 21:52
  • 1
    $\begingroup$ See Part I of our new book downloadable from bangor.ac.uk/r.brown/nonab-a-t.html which calculates homotopy 2-types as crossed modules, using a 2-d van Kampen theorem; most of the 114 papers on the nonabelian tensor product listed on ...../r.brown/nonabtens.html and for example a recent paper Ellis, G.J. and Mikhailov, R. A colimit of classifying spaces. Advances in Math. (2010) arXiv: [math.GR] 0804.3581v1 1--16. uses these nonabelian techniques, with references to the background. Some calculations are in terms of precise nonabelian colimits. (!!!) $\endgroup$
    – Ronnie Brown
    Commented Dec 22, 2011 at 23:29
  • 1
    $\begingroup$ Let me give a simple example. Let $f: G \to H$ be a morphism of groups. Then we have a map of classifying spaces $Bf: BG \to BH$ and we would like to calculate the 2-type of the mapping cone $C(Bf)$. The 2d-van Kampen theorem implies this 2-type is described by an "induced crossed module" $\partial: f_*(G) \to H$ and there are methods of calculating this explicitly, sometimes needing computational group theory. From this (nonabelian) information you can calculate $\pi_2(C(Bf),x)$ as Ker $\partial$ as a module over $\pi_1$. There seems no other way to obtain this information. $\endgroup$
    – Ronnie Brown
    Commented Dec 23, 2011 at 21:54

You must log in to answer this question.

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