The Hawaiian Earring Group

Here is one of my favorite spaces: The Hawaiian earring, the “shrinking wedge of circles.”

The Hawaiian earring

This space is the first step into the world of “wild” topological spaces. This post is meant to be an introduction into how one can understand the fundamental group of this space, often just called the Hawaiian earring group.

The Hawaiian earring $\mathbb{H}$ is usually defined as the following planar set: let $C_n=\left\{(x,y)\in \mathbb{R}^2\Big|\left(x-\frac{1}{n}\right)^2+y^2=\frac{1}{n^2}\right\}$ be the circle of radius $\frac{1}{n}$ centered at $\left(\frac{1}{n},0\right)$. Now take the union $\mathbb{H}=\bigcup_{n\geq 1}C_n$ with the subspace topology of $\mathbb{R}^2$.

The key feature of this space is that if ${U}$ is any open neighborhood of the “wild” point $x_0=(0,0)$, then there is an ${N}\geq {1}$ such that ${C_n}\subset {U}$ for all ${n}\geq {N}$. Note that the Hawaiian earring has the same underlying set as the infinite wedge $\bigvee_{n=1}^{\infty} S^1$  of circles, however the topology of $\bigvee_{n=1}^{\infty} S^1$ is finer than that of $\mathbb{H}$.  So there is a canonical continuous bijection $\bigvee_{n=1}^{\infty} S^1\to \mathbb{H}$ which is not a homeomorphism.

Topological facts: $\mathbb{H}$ is a one-dimensional, compact, locally path connected, metric space.

Other ways to construct $\mathbb{H}$:

1. As a one-point compatification: ${\mathbb{H}}$ is homeomorphic to the one-point compactification of a countable disjoint union $\coprod_{n=1}^{\infty}(0,1)$ of open intervals.
2. As a subspace of $\prod_{n=1}^{\infty} S^1$: View $\bigvee_{n=1}^{\infty} S^1$ as a subspace of $\prod_{n=1}^{\infty} S^1$ in the obvious way and give it the subspace topology. The resulting space is homeomorphic to $\mathbb{H}$.
3. As an inverse limit: Let $X_n=\bigcup_{k=1}^{n}C_k$. If $n>m$, there is a retraction $r_{n,m}:X_n\to X_m$ which collapses the circles $C_k$, $m to $x_0$. These maps form an inverse system $\cdots \to X_{n+1}\to X_{n} \to X_{n-1}\to \cdots \to X_1$. The inverse limit $\varprojlim_{n}X_n$ of this inverse system is homeormorphic to $\mathbb{H}$.

The really interesting things happen when you start considering loops and their homotopy classes, i.e. the fundamental group $\pi_1(\mathbb{H})$. For each $n\geq 1$ consider the loop $\ell_n:[0,1]\to \mathbb{H}$, where $\ell_n(t)=\left(\frac{1}{n}\cos(2\pi t-\pi)+\frac{1}{n},\frac{1}{n}\sin(2\pi t-\pi)\right)$ which traverses the n-th circle ${C_n}$ once in the counterclockwise direction (and is based at $x_0$). Let’s write $\ell_{n}^{-1}$ for the reverse loop $\ell_{n}^{-1}(t)=\ell_{n}(1-t)$ which goes around in the opposite direction. The loop $\ell_{n}$ is definitely not homotopic to the constant loop (for a proof of this, consider the retraction $q_n:\mathbb{H}\to C_n$ collapsing all other circles to $x_0$). It seems that together, the homotopy classes $g_n=[\ell_n]$ should “generate” $\pi_1(\mathbb{H})$ in some way but these will not be group generators in the usual sense.

A space $X$  is semilocally simply connected at a point ${x}\in {X}$ if there is an open neighborhood ${U}$ of ${x}$ such that every loop in ${U}$ based at ${x}$ is homotopic to the constant loop at ${x}$ in ${X}$ (but not necessarily by a homotopy in ${U}$). This definition is very important in covering space theory. In particular, one must typically require a space to be semilocally simply connected in order to guarantee the existence of a universal covering.

Proposition: ${\mathbb{H}}$ is not semilocally simply connected.

Proof. Every neighborhood of the wild point $x_0$ contains all but finitely many of the circles $C_n$ and therefore the non-trivial loops $\ell_n$. ${\square}$

In fact, the Hawaiian earring does not have a universal covering (though there is a known suitable replacement) and one must attack the fundamental group $\pi_1(\mathbb{H})$ using other methods.

Wild loops: The combinatorial structure of $\pi_1(\mathbb{H})$ is complicated by the fact that we can form “infinite” concatenations of loops. For instance, we can define a loop ${\alpha}{:}{[0,1]}\to \mathbb{H}$ by defining $\alpha$ to be $\ell_n$ on the interval $\left[\frac{n-1}{n},\frac{n}{n+1}\right]$ and $\alpha(1)=x_0$. This loop is continuous because of the topology of $\mathbb{H}$ at ${x_0}$. In this way we obtain an infinite “word”  $g_1 g_2 g_3 ...\in\pi_1(\mathbb{H})$. What is intuitive but (formally) less obvious is that $[\alpha]=g_1 g_2 g_3 ...$ is not in the free subgroup of ${\pi}_{1}(\mathbb{H})$ generated by the set $\{g_1,g_2,g_3,...\}$.

With all these wild loops floating around, we have a pretty big group on our hands.

Proposition: $\pi_1(\mathbb{H})$ is uncountably generated.

Proof. If $\pi_1(\mathbb{H})$ were countably generated, then $\pi_1(\mathbb{H})$ would be countable. Thus it suffices to show $\pi_1(\mathbb{H})$ is uncountable. Recall that the infinite product $\prod_{n=1}^{\infty}\mathbb{Z}/2\mathbb{Z}$ of the cyclic group ${\mathbb{Z}}/{2}\mathbb{Z}=\{0,1\}$ of order ${2}$ is uncountable. For any sequence $s=(a_n)\in\prod_{n=1}^{\infty}\mathbb{Z}/2\mathbb{Z}$, we construct a loop ${\alpha_s}:[0,1]\to\mathbb{H}$ by defining ${\alpha_s}$ to be constant on $\left[\frac{n-1}{n},\frac{n}{n+1}\right]$ if ${a_n}={0}$ and ${\alpha_s}$ to be $\ell_n$ on $\left[\frac{n-1}{n},\frac{n}{n+1}\right]$ if $a_n=1$. We also define $\alpha_s(1)=x_0$. In this way we obtain an uncountable family of homotopy class $[\alpha_s]\in {\pi_1}(\mathbb{H})$. It suffices to show ${[\alpha_s]}\neq{[\alpha_t]}$ whenever ${s}\neq{t}$. Suppose ${s}={(a_n)}\neq {(b_n)}={t}$. Then, without loss of generality, we have ${a_N}={1}$ and ${b_N}={0}$ for some ${N}$.  We again call upon the retraction ${q_N}:\mathbb{H}\to C_N$ which collapses all circles but ${C_N}$. If ${[\alpha_s]}={[\alpha_t]}$, then ${[q_N\circ\alpha_s]}={[q_N\circ\alpha_t]}$ in ${\pi_1}{(C_N)}={\mathbb{Z}}$.  But ${[q_N\circ\alpha_t]}={0}\in{\mathbb{Z}}$ is trivial  and ${[q_N\circ\alpha_s]}={[q_N\circ\ell_N]}={1}\in{\mathbb{Z}}$ is non-trivial, which is a contradiction. Therefore ${[\alpha_s]}\neq{[\alpha_t]}$. ${\square}$

Uncountability and the Specker group: Another way to show that $\pi_1(\mathbb{H})$ is uncountable is to show $\pi_1(\mathbb{H})$ surjects onto the uncountable infinite product $\prod_{n=1}^{\infty}\mathbb{Z}$ which is usually called the Specker group and happens to be the first Cech homology group of $\mathbb{H}$. Each map $q_n:\mathbb{H}\to C_n$ collapsing all but the n-th circles to the basepoint induces a retraction of groups $(q_n)_{\ast}:\pi_1(\mathbb{H})\to\pi_1(C_n)=\mathbb{Z}$ which essentially picks out the “winding number” around the n-th circle. Together, these winding numbers uniquely induce a homomorphism $\epsilon:\pi_1(\mathbb{H})\to\prod_{n=1}^{\infty}\mathbb{Z}$ given by $\epsilon([\alpha])=((q_1)_{\ast}([\alpha]),(q_2)_{\ast}([\alpha]),...)$. To check surjectivity, convince yourself that $\epsilon$ sends the homotopy class of a loop $\alpha$ defined as $(\ell_{n})^{a_n}$ on the interval $\left[\frac{n-1}{n},\frac{n}{n+1}\right]$ and $\alpha(1)=x_0$ to the generic sequence $(a_n)\in\prod_{n=1}^{\infty}\mathbb{Z}$.

One might be tempted to think that all elements of $\pi_1(\mathbb{H})$ can be realized as products of infinite sequences of shrinking loops like $[\alpha]=g_1g_2g_3...$ but alas, this is also too much to hope for. Not only is this too much to hope for, but the combinatorial structure of $\pi_1(\mathbb{H})$ is far from free [3] since we can have “infinite” cancellations of the letters ${g_n}$ when we multiply two elements. As a first example, notice that $[\alpha]^{-1}$ can be thought of as the infinite word ${...}{g}_{3}^{-1} {g_{2}^{-1}}{g_{1}^{-1}}$ and the product $g_1 g_2 g_3......g_{3}^{-1} g_{2}^{-1} g_{1}^{-1}=[\alpha][\alpha]^{-1}=e$ is the identity element. In more geometric terms, this means we can construct a null-homotopy of ${\alpha}{\cdot}{\alpha}^{-1}$ by nesting “small null-homotopies” of the loops $\ell_n\cdot \ell_{n}^{-1}$ inside of each other.

You can take this one-step further by considering the following iterative construction. Start with

$g_{1}g_{1}^{-1}$

Now insert more trivial pairs, but make the index of the $g_i$‘s get larger at each step so the construction is actually represented by a continuous loop.

$g_{1}(g_{2}g_{2}^{-1})(g_{3}g_{3}^{-1})g_{1}^{-1}$

$g_{1}(g_{2}(g_{4}g_{4}^{-1})(g_{5}g_{5}^{-1})g_{2}^{-1})(g_{3}(g_{6}g_{6}^{-1})(g_{7}g_{7}^{-1})g_{3}^{-1})g_{1}^{-1}$

$g_{1}(g_{2}(g_{4}(g_{8}g_{8}^{-1})(g_{9}g_{9}^{-1})g_{4}^{-1})(g_{5}(g_{10}g_{10}^{-1})(g_{11}g_{11}^{-1})g_{5}^{-1})g_{2}^{-1})$              (cont. on next line)

$(g_{3}(g_{6}(g_{12}g_{12}^{-1})(g_{13}g_{13}^{-1})g_{6}^{-1})(g_{7}(g_{14}g_{14}^{-1})(g_{15}g_{15}^{-1})g_{7}^{-1})g_{3}^{-1})g_{1}^{-1}$

$\cdots$

At every stage and in the limit, this construction should represent the identity element of the group, however, in the “transfinite word” which is the limit, there are no straightforward cancellation pairs $g_{k}g_{k}^{-1}$ to be found anywhere! This is because we went on to put new letters in the middle of every such pair. So the cancellations that go on in $\pi_1(\mathbb{H})$ can be quite subtle. How could you possibly define a loop representing the above word? Well, if you look closely at where new pairs are inserted, you can see that it has a “Cantor set-ish” feel to it.

To describe loops representing all elements of ${\pi}_1(\mathbb{H})$, we call upon the middle-third Cantor set ${C}\subset {[0,1]}$. There are countably many open intervals ${[0,1]}{\backslash}{C}=\bigcup_{k\geq 1}(a_k,b_k)$. We can define a loop ${\alpha}{:}{[0,1]}\to \mathbb{H}$ by defining $\alpha(C)=x_0$ and defining ${\alpha}$ on $[a_k,b_k]$ to either be the constant loop or to be one of the loops $\ell_{n_k}^{\pm 1}$ for some ${n_k}\geq {1}$. We have one restriction to ensure that ${\alpha}$ is continuous. We must ensure that for each ${n}\geq {1}$, we only have ${n_k}={n}$ for finitely many ${k}$. This means for fixed $n$, the loops $\ell_{n}$ and $\ell_{n}^{-1}$ can only be used finitely many times. Otherwise, we would admit infinite concatenations like ${\ell_1}\cdot {\ell_1}\cdot {\ell_1}{\cdots}$ which clearly cannot be continuous. It turns out that any element of ${\pi_1}(\mathbb{H})$ is represented by a loop constructed in this way. You can convince yourself of this by first noticing that for any loop $\alpha$, the preimage $\alpha^{-1}(\mathbb{H}\backslash \{x_0\})$ is a countable union of disjoint open intervals.

We’ve yet to really compute $\pi_1(\mathbb{H})$. We could argue exactly what I mean by “compute” here but I really mean “identify the isomorphism class as a reasonably familiar group so that we can make formal algebraic arguments about the group structure without appealing to loops.” This is done using shape theory. Before we do this, I should mention that this shape theoretic approach can fail to provide an explicit characterization of $\pi_1$ when you start considering subsets of $\mathbb{R}^3$.

Recall that one way to construct ${\mathbb{H}}$ is as an inverse limit $\varprojlim_{n}X_n$ where where ${X_n}=\bigvee_{i=1}^{n}S^1$ is the union of the first n-circles. Note that ${\pi}_{1}(X_n)=F_n$ is the free group on the generators ${g_1},{g_2},{...},{g_n}$. If we apply the fundamental group ${\pi_1}$ to the entire inverse system

$\cdots \to X_{n+1}\to X_{n} \to X_{n-1}\to \cdots \to X_1$,

we get an inverse system of free groups

$\cdots \to F_{n+1}\to F_{n} \to F_{n-1}\to \cdots \to F_1$

where the homomorphism $h_n:F_{n+1}\to F_n$ collapses ${g}_{n+1}$ to the identity. The inverse limit $\check{\pi}_1(\mathbb{H})= \varprojlim_{n}F_n$ is the first shape group of $\mathbb{H}$. To be fair, the shape group cannot always be constructed in this way but this is a nice way to understand the one-dimensional case.

We also have projections ${p_n}{\colon}\mathbb{H}=\varprojlim_{n}X_n\to {X_n}$ which collapse ${C_k}$ to the basepoint for ${k}>{n}$. The induced homomorphisms $(p_n)_{\ast}:\pi_1(\mathbb{H})\to \pi_1(X_n)=F_n$ clearly agree with the bonding homomorphisms $h_n:F_{n}\to F_{n-1}$  in the inverse system of free groups so we get an induced homomorphism ${\Psi }:{\pi_1}\left(\mathbb{H}\right)\to\varprojlim_{n}{F_n}$ to the first shape group.

The inverse limit of free groups ${\varprojlim_n}F_n$ is constructed as a subgroup of $\prod_{n=1}^{\infty}F_n$. Specifically, ${\varprojlim_n}F_n$ consists of the sequences $(w_n)$ of words $w_n\in F_n$ such that $h_n(w_n)=w_{n-1}$. This means we can think of elements of ${\varprojlim_n}F_n$ as sequences of words $(w_n)$ where the word $w_{n-1}$ (in letters $g_1,..,g_{n-1}$) is obtained from the word $w_n$ (in letters $g_1,..,g_{n}$) by removing all instances of the letter $g_n$. The homomorphism $\Psi$ is defied as $\Psi([\alpha])=(w_n)$ where $w_n=[p_n\circ\alpha]\in F_n$.

The key to understanding $\pi_1(\mathbb{H})$ is the following theorem which originally appeared in a paper of H.B. Griffiths [1]. Griffiths’ proof apparently had some sort of error in it; a corrected proof was given by Morgan and Morrison [2] and many have since appeared.

Theorem: ${\Psi }:{\pi_1}\left(\mathbb{H}\right)\to\varprojlim_{n}{F_n}$ is injective.

The main idea in the proof of this theorem is to use the data of infinite “word reduction” to construct a null-homotopy of a loop $\alpha$ such that $\Psi([\alpha])=1$ (equivalently  $[p_n\circ \alpha]=1\in \pi_1(X_n)=F_n$ for each $n\geq 1$). It is helpful to imagine doing this for the example above where we kept inserting trivial pairs $g_kg_{k}^{-1}$ between trivial pairs and so on. The details of a full proof are somewhat non-trivial so I’ll skip it for now (but plan to come back to it later). The upshot of the theorem is that we can now understand elements of ${\pi}_{1}\left(\mathbb{H}\right)$ as sequences of words in ${\varprojlim_{n}}{F_n}$.

The question then remains: what is the image of $\Psi$?

Proposition: $\Psi : \pi_{1}\left(\mathbb{H}\right) \to \varprojlim_{n}F_n$ is not surjective.

Consider the sequence $(w_n)\in\varprojlim_{n}F_n$ of commutators $w_n = (g_{1} g_{2} g_{1}^{-1} g_{2}^{-1} )(g_{1} g_{3} g_{1}^{-1} g_{3}^{-1} )(g_{1} g_{4} g_{1}^{-1} g_{4}^{-1}) \cdots (g_{1} g_{n} g_{1}^{-1} g_{n}^{-1})$. Note that as ${n}{\to}{\infty}$ the number of appearances of ${g_1}$ grows without bound. But we can’t have a loop ${\alpha} :{[0,1]}\to\mathbb{H}$ that corresponds to this element  since no continuous loop can traverse ${C_1}$ infinitely many times. This geometric restriction suggests which subgroup we should be looking for.

Definition: If ${1}\leq {k}\leq {n}$ and ${w\in F_n}$, let ${\#}_{k}(w)$ be the number of times $g_{k}^{\pm 1}$ appears in the reduced word $w$. We say an element $(w_n)\in\varprojlim_{n}F_n$ is locally eventually constant if for each ${k} \geq {1}$, the sequence $\#_{k}(w_n)$  is eventually constant (as $n\to\infty$). Let $\#_{\mathbb{N}}\mathbb{Z}\leq\varprojlim_{n}F_n$ be the subgroup of locally eventually constant sequences.

If $\Psi ([\alpha])=(w_n)$ is not locally eventually constant, then we’d have some $k$ where the number of times $g_{k}^{\pm 1}$ appears is unbounded and this contradicts the continuity of $\alpha$. On the other hand, our method of using the Cantor set to construct loops provides a nice way to represent every locally eventually constant sequence by a continuous loop. We conclude that the locally eventually constant sequences are precisely the sequences corresponding to continuous loops.

Theorem [2]: ${\Psi} : {\pi}_{1}\left(\mathbb{H}\right) \to {\varprojlim_{n}}{F_n}$ embeds ${\pi}_{1}\left(\mathbb{H}\right)$ isomorphically onto $\#_{\mathbb{N}}\mathbb{Z}$.

The group ${\pi_1}\left(\mathbb{H}\right){\cong}\#_{\mathbb{N}}\mathbb{Z}$ is sometimes called the free ${\sigma}$-product of $\mathbb{Z}$ in infinite group theory.

Summary

Let’s sum up this combinatorial description of $\pi_1(\mathbb{H})$: The fundamental group of the Hawaiian earring $\pi_1(\mathbb{H})$ is isomorphic to the group of sequences $(w_n)$ where

1. $w_n\in F_n$ is a reduced word in the free group on letters $g_1,...,g_n$,
2. removing the letter $g_n$ from $w_n$ gives the word $w_{n-1}$,
3. for each $k\geq 1$, the number of times the letter $g_k$ appears in $w_n$ stabilizes at $n\to\infty$ (i.e. the sequence $\#_k(w_1),\#_k(w_2),...$ is eventually constant for each $k$).

References.

[1] H.B. Griffiths, Infinite products of semigroups and local connectivity, Proc. London Math. Soc. (3), 6 (1956), 455-485.

[2]  J. Morgan, I. Morrison, A van kampen theorem for weak joins, Proc. London Math. Soc. 53 (1986) 562–576.

[3] B. de Smit, The fundamental group of the Hawaiian earring is not free, Internat. J. Algebra Comput. 2 (1) (1992) 33–37.

Another great reference on the Hawaiian earring group is

[4] J.W. Cannon, G.R. Conner, The combinatorial structure of the Hawaiian earring group, Topology Appl. 106 (2000) 225-271.