How many homomorphisms?

In this post I’ll present some results and ideas from the paper The number of homomorphisms from the Hawaiian earring group, 2019.  First I’ll give some motivating background and historical remarks, and then state the main result and concepts used in its proof.

The question “How many?” is an old one, and the ideas advanced by Cantor gave this question meaning in the infinite setting.  Where previously an answer of “Infinitely many” was as descriptive as necessary, one can now ask the followup question “Which infinite?”  When the answer is “as many as is naively allowable” then you get a sense of abundance.  For example one might wonder how many different ultrafilters exist on an infinite set X.  There are |P(P(X))| collections of subsets of X.  Pospisil showed that there are 2^{2^{|X|}} ultrafilters on X (see Remark on bicompact spaces, 1937) and since 2^{2^{|X|}} = |P(P(X))| we have the greatest number imaginable.

When the question is “How many group homomorphisms are there from H to G?”, a very naive upper bound is the total number of functions from H to G, namely |G|^{|H|}.  For example there are 2^{2^{\aleph_0}} group homomorphisms from \mathbb{R} to \mathbb{Q} and this is the same as the number of functions from \mathbb{R} to \mathbb{Q}.  One can see this by picking a basis for \mathbb{R} as a vector space over \mathbb{Q} and considering all possible maps from basis elements to \mathbb{Q}.

When considering other groups the answer can become more subtle.  Recall that the Hawaiian earring is the space E = \bigcup_{n\in \omega} C_{(0, \frac{1}{n+1}), \frac{1}{n+1}} where C_{p, r} \subseteq \mathbb{R}^2 is the circle of radius r centered at the point p.  Its fundamental group, which we denote HEG, is uncountable and not free; in fact any homomorphism to a free group will have finitely generated image.  For each n\in \omega the group HEG has a homomorphic retraction p_n to a free subgroup HEG_n of rank n which is induced by the topological retraction map which takes all points which are not in the outermost n circles of E to the point (0, 0).

A group G is n-slender if for every homomorphism \phi: HEG \rightarrow G there exists some n\in \omega for which \phi = \phi \circ p_n (see posts Automatic continuity parts 2 and 3).  Right-angled Artin groups, Baumslag-Solitar groups, and torsion-free word hyperbolic groups are examples of n-slender groups, and groups containing torsion or a subgroup isomorphic to \mathbb{Q} are not n-slender.  A rough intuitive way of understanding an n-slender group G is that homomorphisms from HEG to G can only be superficial- they cannot involve the infinitely deep parts of HEG.

Curiously there exist models of Zermelo-Fraenkel set theory plus the axiom of dependent choices in which if G is not n-slender we have 2^{\aleph_0} \leq |G| (see Deeply concatenable subgroups might never be free, with Shelah).  In particular, any homomorphism which witnesses that a group of cardinality less than 2^{\aleph_0} is not n-slender is somehow “nonconstructive.”  From this one suspects that such a homomorphism probably relies on performing many choices and is therefore not unique in witnessing n-slenderness (compare the construction of a non-principal ultrafilter).  It turns out that there are as many homomorphisms from HEG to a  small cardinality group which is not n-slender:

Theorem  If G is a group with |G|<2^{\aleph_0} then

|Hom(HEG, G)| = \begin{cases}|G| \text{ if } G \text{ is n-slender} \\2^{2^{\aleph_0}} \text{ if }G \text{ is not n-slender} \end{cases}

The requirement that |G|<2^{\aleph_0} is essential here, since HEG is not n-slender (consider the identity map), |HEG| = 2^{\aleph_0}, and |Hom(HEG, HEG)| = 2^{\aleph_0} (this latter equality follows from Corollary 2.11 of Eda’s Free \sigma-products and fundamental groups of subspaces of the plane, 1998).  It was previously known that if G is a nontrivial finite group then the number of surjections from HEG to G is 2^{2^{\aleph_0}} (see Conner and Spencer, Anomalous behavior of the Hawaiian earring group, 2005).  From more recent results one sees more generally that if G is a compact Hausdorff topological group there exist |G|^{2^{\aleph_0}} many homomorphisms from HEG to G (using theorems from Tlas, Big free groups are almost free, 2015 and Zastrow, The non-abelian Specker group is free, 2000).

It is straightforward to show that if G is n-slender then |Hom(HEG, G)| = |G|.  The challenge lies in producing many homomorphisms from the existence of just one nonconstructive homomorphism.  The trick for this lies in using the Harmonic archipelago HA (this space appeared in Bogley, Sieradski, Universal path spaces).  This space can be imagined as taking the disc D = \{p\in \mathbb{R}^2\mid d(p, (\frac{1}{2}, 0)) \leq \frac{1}{2}\} and from each subdisc D_{n} = \{p\in D\mid d(\frac{1}{2^{n+1}}, 0) \leq \frac{1}{2^{n+3}}\} you form a hill of height 1.  This gives a space whose fundamental group is uncountable and onto which HEG can be naturally surjected.

First, one shows that if |G|<2^{\aleph_0} and G is not n-slender then there exists a nontrivial homomorphism \phi: \pi_1(HA) \rightarrow G (this fact was told to me by Greg Conner).  Next one shows that there exist 2^{2^{\aleph_0}}-many homomorphisms \psi_i: \pi_1(HA) \rightarrow \pi_1(HA) for which \phi \circ \psi_i differs for distinct \psi_i.  This is the most difficult part of the argument and involves manipulation of infinitary words.  Once one has shown this the theorem follows.

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