The Hurewicz theorem tells us that if $X$ is a path-connected space then $H_1(X, \, \mathbb{Z})$ is isomorphic to the abelianisation of $\pi_1(X)$. This gives a potential method for computing the abelianisation of a (sufficiently nice) group $G$: realise it as the fundamental group of a space $X$ and then compute $H_1(X, \, \mathbb{Z})$ by your favourite means.

Is this method ever used in practice? Are there nice examples of abelianisations which are easily (best?) computed in this way?

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    $\begingroup$ Take $G$ to be the fundamental group of the complement $X$ of a link of $n$ circles in $S^3$, or $n$ complex hyperplanes in some complex, finite-dimensional vector space. Then $G$ itself can be quite hard to compute, but its abelianization is isomorphic to $H_1(X)=\mathbb{Z}^n$. $\endgroup$ – Alex Suciu Jan 26 '17 at 15:11
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    $\begingroup$ If the group G is given by a finite presentation, or if a finite presentation is easy to compute, then Smith normal form is the easiest way to compute the abelianization so somehow G should either not be finitely presented or we should not have any access to a presentation if it might be. $\endgroup$ – Benjamin Steinberg Jan 26 '17 at 15:44
  • $\begingroup$ Among the answers I could distinguish (a) cases where the group is explicit and however the abelianization is not so clear to describe (b) cases where the abelianization is easier to describe directly without using the group explicitly. I'm not sure what the OP had in mind. Cases with finitely presented groups fall in (b) (since when a presentation is explicit the abelianization is trivial to compute). Of course, I find (a) more surprising than (b). $\endgroup$ – YCor Jan 27 '17 at 21:04

Yes, the fundamental group of the Hawaiian earring $\pi_1(\mathbb{H},b_0)$ is an important group which is sometimes called the free sigma product $\#_{\mathbb{N}}\mathbb{Z}$. Its is often defined in purely algebraic terms as a group of "transfinite words" in countably many letters. In many ways this group behaves like the non-abelian version of the Specker group $\prod_{\mathbb{N}}\mathbb{Z}$ and it is the key to the homotopy classification of 1-dimensional Peano continua. However, it's abelianization is not $\prod_{\mathbb{N}}\mathbb{Z}$; it is a good deal more complicated. The abelianization was first described by Eda and Kawamura in the following paper:

The Singular Homology Group of the Hawaiian Earring, Journal of the London Mathematical Society, 62 no. 1 (2000) 305–310.

The proof explicitly uses the Hawaiian earring and the authors mention in Remark 2.7 that

Though the question itself is formulated algebraically, we have not succeeded in finding a purely algebraic proof of it.

One way to represent the isomorphism class of this abelianization is as $$Ab(\#_{\mathbb{N}}\mathbb{Z})\cong \prod_{\mathbb{N}}\mathbb{Z}\oplus \prod_{\mathbb{N}}\mathbb{Z}/\bigoplus_{\mathbb{N}}\mathbb{Z}$$

Similar methods have been used to identify abelienizations of fundamental groups of other 1-dimensional spaces, e.g. the Menger curve.

  • $\begingroup$ FWIW, Wikipedia claims that the abelianization of the fundamental group of the Hawaiian earring has "no known simple description": en.m.wikipedia.org/wiki/Hawaiian_earring $\endgroup$ – Sam Hopkins Jan 26 '17 at 23:10
  • $\begingroup$ @SamHopkins I've actually thought of editing that for some time but "simple" is a pretty relative word. If you think the group I describe is simple, then wikipedia is wrong. If not, then there is no problem with it. To be fair, the isomorphism is not in any way natural; it relies heavily on classification results in infinite abelian group theory. $\endgroup$ – Jeremy Brazas Jan 26 '17 at 23:20
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    $\begingroup$ When we have a big group such as this abelianization, "identify" the abelianization has possible distinct meanings. One is "provide a reasonably defined group isomorphic to it". One is the same, but also specify the quotient map. $\endgroup$ – YCor Jan 27 '17 at 19:04

A usually hard problem in Algebraic Geometry is computing the fundamental group of the complement $U=\mathbb{P}^n - V$, where $V$ is a reduced hypersurface.

At least in principle, the computation of $\pi_1(U)$ can be carried out by using Seifert-Van Kampen theorem; the tricky part is that the relations in the presentation depend not only on the singularities of $V$, but also on their mutual position.

A classical example due to Zariski is when $V \subset \mathbb{P}^2$ is a curve of degree $6$ having six ordinary cusps and no other singularities. Then $\pi_1(U)$ is the free product $\mathbb{Z}/2 \mathbb{Z} * \mathbb{Z}/3 \mathbb{Z}$ if the six cusps lie on the same conic, and the direct product $\mathbb{Z}/ 2 \mathbb{Z} \times \mathbb{Z}/3 \mathbb{Z}$ otherwise.

However, by using Lefschetz duality we can compute in a straightforward way $H_1(U, \, \mathbb{Z})$ for all hypersurfaces, so that the abelianization of $\pi_1(U)$ is always known even if a presentation for $\pi_1(U)$ is not available. In fact, there is the following result that can be found in A. Dimca's book Singularities and topology of hypersurfaces, Chapter 4.

Proposition. Assume that the hypersurface $V \subset \mathbb{P}^n$ has $k$ irreducible components $V_1, \ldots, V_k$ of degrees $d_1, \ldots, d_k$. Let $d$ be the greatest common divisor of the integers $d_1, \ldots, d_k$. Then $$H_1(U, \, \mathbb{Z}) = \mathbb{Z}^{k-1} \oplus \mathbb{Z}/ d \mathbb{Z}.$$

In particular, $H_1(U)$ does not depend on the singularities of $V$, but only on the number and degrees of its irreducible components. As a consequence, we recover the fact that in both cases of Zariski's example quoted above (where $k=1, \, d_1=6$) the abelianization of $\pi_1(U)$ is the cyclic group of order $6$.


The mapping class group of a smooth manifold $M$ is the group of all its self diffeomorphisms up to isotopy, i.e. $\pi_0(\operatorname{Diff}(M))\cong \pi_1(B\operatorname{Diff}(M))$.

A large portion of geometric topology is concerned with gaining a better understanding of the homotopy type of $B\operatorname{Diff}(M)$, in particular of its cohomology as the latter is the ring of characteristic class of smooth $M$ bundles.

In the last 15 years, initiated by Madsen and Weiss' solution of the Mumford conjecture, the understanding of the homology of $B\operatorname{Diff}(M)$ has fundamentally improved which led, among others, to calculations of $H_1(B\operatorname{Diff}(M))\cong \pi_0(\operatorname{Diff}(M))^{ab}$ for some $M$, e.g. by Galatius and Randal-Williams in the following paper.

Abelian quotients of mapping class groups of highly connected manifolds, Mathematische Annalen June 2016, Volume 365, Issue 1, pp 857–879

  • $\begingroup$ In the third paragraph, I think you mean that $H_1(B\mathrm{Diff}(M))$ is isomorphic to the abelianization of $\pi_0(\mathrm{Diff}(M))$. $\endgroup$ – HJRW Jan 26 '17 at 20:21

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