The Hawaiian Earring Group is not Free (Part I)

The main goal of this two-part post will be to study the homomorphisms out of the Hawaiian earring group. Click here to get to Part II. In particular, we’ll end up concluding that the set of homomorphisms $Hom(\pi_1(\mathbb{H}),\mathbb{Z})$ to the additive group of integers is countable.  This may seem a bit strange considering $\pi_1(\mathbb{H})$ is an uncountable group. As a direct consequence, we can see that $\pi_1(\mathbb{H})$ is not isomorphic to any free group $F(X)$ on a set $X$.

Theorem 1 (de Smit [1]): $\pi_1(\mathbb{H})$ is not isomorphic to a free group.

The Hawaiian earring group is complicated enough that this should not be completely obvious. In this post I’ll fully hash out the details of Bart de Smits proof in [1]. Working through all these details has certainly helped me understand it better.

Theorem 1 is in contrast with the fact that the fundamental group $\pi_1\left(\bigvee_{n=1}^{\infty}S^1\right)$ on a countable wedge of circles (the Hawaiian earring with the CW-topology) is isomorphic to the free group $F(g_1,g_2,...)$ on a countably infinite set of generators.

The Hawaiian earring group

In a previous post, I discussed how to begin understanding the algebraic structure of the fundamental group $\pi_1(\mathbb{H})$ of the Hawaiian earring $\mathbb{H}=\bigcup_{n=1}^{\infty}C_n\subset \mathbb{R}^2$, where $C_n$ is the circle of radius $\frac{1}{n}$ centered at $\left(\frac{1}{n},0\right)$. The basepoint is the origin $x_0=(0,0)$, the one point where the shrinking circles meet.

We decided that the fundamental group $\pi_1(\mathbb{H})$ is an uncountable group that could be understood a subgroup of the inverse limit

$\varprojlim\left(\cdots\to F_{n+1}\to F_n\to \cdots \to F_2 \to F_1\right)$

where $F_n$ is the free group on the generators $g_1,...,g_n$ and the map $p_{n+1}:F_{n+1}\to F_n$ identifies the letter $g_{n+1}$ to the identity element (or empty word). The inverse limit $\varprojlim_{n}F_n$ consists of sequences $(w_n)$ where $w_n \in F_n$ is obtained from $w_{n+1}$ by removing all occurrences of the letter $g_{n+1}$. The fundamental group $\pi_1(\mathbb{H})$ corresponds to a certain subgroup of $\varprojlim_{n}F_n$.

Definition: Suppose $w=g_{k_1}^{\epsilon_1}g_{k_2}^{\epsilon_2}...g_{k_m}^{\epsilon_m}\in F_n$ is a reduced word in the letters $g_1,...,g_n$. Reduced means that $k_i\neq k_{i+1}$ and $\epsilon_i\neq 0$ for each $i$. The k-weight of $w$ is

$\#_{k}(w)=\sum_{k_m=k}|\epsilon_{k_m}|$.

Essentially, $\#_{k}(w)$ is the number of times $g_{k}$ or $g_{k}^{-1}$ appears in $w$.

If $(w_n)\in\varprojlim_{n}F_n$, then the sequence $\#_{k}(w_n)$ (for fixed $k$) is non-decreasing since the projections $p_n$ only delete letters. We wish to consider the elements of the inverse limit where each such sequence is also bounded (and thus eventually constant). Let

$\#_{\mathbb{N}}\mathbb{Z}=\left\{(w_n)\in\varprojlim_{n}F_n\Big|\lim_{n\to\infty}\#_{k}(w_n)<\infty\text{ for all }k\geq 1\right\}$

be the subgroup of $\varprojlim_{n}F_n$ consisting of sequences where every k-weight is eventually constant. These sequences are usually called locally eventually constant sequences and the group $\#_{\mathbb{N}}\mathbb{Z}$ is often called the free $\sigma$-product of $\mathbb{Z}$.

In a previous post, we found the following canonical group isomorphism.

Theorem 2: $\pi_1(\mathbb{H})\cong\#_{\mathbb{N}}\mathbb{Z}$.

So to study the properties of the Hawaiian earring group, we can focus our attention on the purely algebraic structure of $\#_{\mathbb{N}}\mathbb{Z}$. For instance, $\#_{\mathbb{N}}\mathbb{Z}$ is certainly uncountable and also must be torsion free since it is a subgroup of the torsion free group $\varprojlim_{n}F_n$.

Some homomorphisms out of $\#_{\mathbb{N}}\mathbb{Z}$

First let’s exploit the inverse limit structure of $\varprojlim_{n}F_n$ to construct some interesting self-homomorphisms of $\#_{\mathbb{N}}\mathbb{Z}$. For simplicity of notation, we’ll just write $g_k$ for the element $(1,1,...,1,g_k,g_k,...)\in\varprojlim_{n}F_n$ where the first non-trivial term is in the k-th position. More generally, we could also identify $w\in F_n$ with it’s image under the canonical embedding $F(g_1,g_2,...)\to\#_{\mathbb{N}}\mathbb{Z}$ of the infinite free group.

Consider a sequence $s_j=(w_{n}^{(j)})\in\#_{\mathbb{N}}\mathbb{Z}$ of locally eventually constant sequences. The $s_j$ should satisfy the follow two properties:

• $w_{n}^{(j)}=1$ for all $n
• For all $n\geq 1$ and $k, we have $\#_{k}(w_{n}^{(j)})=0$

For instance, we could have something like:

$s_1=(g_1, g_1g_2, g_1g_2g_3,...)$   which corresponds to an infinite word $g_1g_2g_3g_4...$

$s_2=(1, g_2, g_2g_3, g_2g_3g_4, ...)$   which corresponds to an infinite word $g_2g_3g_4g_5...$

$s_3=(1, 1, g_3, g_3g_4, g_3g_4g_5, ...)$   which corresponds to an infinite word $g_3g_4g_5g_6...$

and so on…

What is important is that  1) the first $j-1$ terms of $s_j$ are trivial and that 2) the letters $g_1,...,g_{j-1}$ don’t show up in $s_j$.

Condition 1) means precisely that the sequence $s_j$ must converge to the identity element $(1,1,1,...)$ when $\varprojlim_{n}F_n$ has the inverse limit topology, i.e. as a subspace of $\prod_{n}F_n$.

Lemma 3: There is a self-homomorphism $h:\#_{\mathbb{N}}\mathbb{Z}\to\#_{\mathbb{N}}\mathbb{Z}$ where $h(g_j)=s_j$.

Proof. Define a homomorphism $h_n:F_n\to F_n$ on the free group by $h_n(g_j)=w_{n}^{(j)}$ (obviously for $j\leq n$). Since

$p_n(h_n(g_j))=p_n(w_{n}^{(j)})=\begin{cases} w_{n-1}^{(j-1)} & \text{ if }j

and

$h_{n-1}(p_n(g_j))=\begin{cases} h_{n-1}(g_j)=w_{n}^{(j-1)} & \text{ if }j

the following diagram commutes.

Consequently, we get a self-homomorphism $h:\varprojlim_{n}F_n\to\varprojlim_{n}F_n$ on the inverse limit such that $h(g_j)=s_j$. We check that $h(\#_{\mathbb{N}}\mathbb{Z})\subseteq\#_{\mathbb{N}}\mathbb{Z}$ and then use the restriction of $h$ to prove the lemma. If $v=(v_n)\in\#_{\mathbb{N}}\mathbb{Z}$ then for each $k\geq 1$, we have $\lim_{n\to\infty}\#_{k}(v_n)=M_k<\infty$. Now for fixed $k\geq 1$:

$\#_{k}(h(v)_n)=\#_{k}(h_n(v_n))\leq\sum_{j=1}^{n}\left(\#_{j}(v_n)\cdot\#_{k}(h_n(g_j))\right)=\sum_{j=1}^{n}\left(\#_{j}(v_n)\cdot\#_{k}(w_{n}^{(j)})\right)$

The inequality must be there since when we replace each letter $g_j$ of $v_n$ with $h_n(g_j)$, we may have some word reduction to do. Our restriction that $\#_{k}(w_{n}^{(j)})=0$ for all $n\geq 1$ and $k means that

$\#_{k}(h(v)_n)\leq\sum_{j=1}^{k}\left(\#_{j}(v_n)\cdot\#_{k}(w_{n}^{(j)})\right)$

Since, by assumption,  $w_{n}^{(j)}\in\#_{\mathbb{N}}\mathbb{Z}$, we have $\lim_{n\to\infty}\#_{k}(w_{n}^{(j)})=N_k<\infty$. Thus

$\lim_{n\to\infty}\#_{k}(h(v)_n)\leq\sum_{j=1}^{k}M_k\cdot N_k<\infty$

showing that $h(v)\in\#_{\mathbb{N}}\mathbb{Z}$ is locally eventually constant. $\square$

Now let’s see what happens when we map $\#_{\mathbb{N}}\mathbb{Z}$ to the additive group of integers $\mathbb{Z}$.

Lemma 4: If $f:\#_{\mathbb{N}}\mathbb{Z}\to\mathbb{Z}$ is a homomorphism, then there is an $N\geq 1$ such that $f(g_n)=0$ for all $n\geq N$.

Proof. Suppose $f(g_{n_j})\neq 0$ for $n_1 and let

$s_j=\begin{cases} g_{n_j}^{3} & \text{ if }f(g_{n_j})>0\\ g_{n_j}^{-3} & \text{ if }f(g_{n_j})<0 \end{cases}$

By Lemma 3, there is a homomorphism $h:\#_{\mathbb{N}}\mathbb{Z}\to\mathbb{Z}$ such that $h(g_j)=s_j$ for all $j\geq 1$. Thus $f\circ h(g_j)=3|f(g_{n_j})|\geq 3$ for each $j\geq 1$. We might as well now replace $f$ with $f\circ h$ so from now on let’s assume that

$a_j=f(g_j)\geq 3$.

Ok, now let’s define a special sequence $z^{(j)}\in\#_{\mathbb{N}}\mathbb{Z}$ that will help us arrive at a contradiction. We define the n-th term $z^{(j)}_{n}\in F_n$ to be

\begin{cases} 1 & \text{ if }n<j\\ g_j & \text{ if }n=j \\ g_jg_{j+1}^{a_j} & \text{ if }n=j+1 \\ g_j(g_{j+1}g_{j+2}^{a_{j+1}})^{a_j} & \text{ if }n=j+2 \\ g_j(g_{j+1}(g_{j+2}g_{j+3}^{a_{j+2}})^{a_{j+1}})^{a_j} & \text{ if }n=j+3 \\ … & … \end{cases}

So the general form for $n>j$ is

$z^{(j)}_{n}=g_j(g_{j+1}(g_{j+2}...(g_{n-2}(g_{n-1}g_{n}^{a_{n-1}})^{a_{n-2}})^{a_{n-3}}...)^{a_{j+1}})^{a_{j}}$.

Notice that removing $g_k$ from $z^{(j)}_{k}$ gives $z^{(j)}_{k-1}$ and that $\lim_{n\to\infty}\#_{j+k}(z^{(j)}_{n})=a_{j+k-1}<\infty$ for $k\geq 1$ (i.e. the letter $g_{j+k}$ never appears more than $a_{j+k-1}$ times despite the fact that the appearances get further and further apart). Thus $z^{(j)}_{n}$ is a well defined element of $\#_{\mathbb{N}}\mathbb{Z}$.

The main feature of this sequence is that $z^{(j)}=g_j(z^{(j+1)})^{a_j}$ so that when we apply $f$, we get

$f(z^{(j)})=f(g_j)+f\left((z^{(j)})^{a_j}\right)=a_j+a_jf(z^{(j+1)})$

Iterating this formula $j-1$ times for $f(z^{(1)})$ gives

$f(z^{(1)})=a_1+a_1a_2+a_1a_2a_3+\cdots+a_1a_2...a_{j-1}+a_1a_2...a_{j-1}a_j+a_1a_2...a_{j-1}a_jf(z^{(j)}).$

So if we let

$b_j=a_1+a_1a_2+a_1a_2a_3+\cdots+a_1a_2...a_{j-1}$    and    $c_j=a_1a_2...a_j$,

then we see that $f(z^{(1)})=b_j+c_j+c_{j}f(z^{(1)})$ and thus

$f(z^{(1)})=b_j\text{ mod}(c_j).$

Let’s make a few more observations about $b_j$ and $c_j$:

• $b_j\to\infty$

Proof: this one is pretty obvious since $a_j\geq 3$. $\square$

• $b_j

Proof: Since $a_j\geq 3$, we have $a_1 and inductively if $b_{j-1}, $b_{j}=b_{j-1}+c_{j-1}$\square$

• $c_j-b_j\to\infty$

Proof: Here we also use the fact that $a_j\geq 3$ for each $j$. Notice

$c_2-b_2=a_1a_2-a_1=a_1(a_2-1)\geq 2a_1$,

$c_3-b_3=a_1(a_2(a_3-1)-1)\geq a_1(2a_2-1)\geq 5a_1$,

$c_4-b_4=a_1(a_2(a_3(a_4-1)-1)-1)\geq 14a_1$, and so on.

If we recursively define the increasing sequence $p_2=2$, $p_{j+1}=3p_{j-1}-1$, we get $c_j-b_j\geq p_ja_1$, which does the trick. $\square$

Added Remark: de Smit’s construction uses $a_j\geq 3$, however, the inductive proofs of these three bullet points also seem to work if you only assume $a_j\geq 2$. So actually, I believe this slight simplification can be made.

Now, let’s finally finish the proof of Lemma 4 by showing that $f(z^{(1)})$ satisfies way to many modular equations.

In general, suppose $y=b\text{ mod}(c)$ where $b. If $y\geq 0$, then we must have $y\geq b$ and if $y<0$, then we must have $y\leq b-c$.

Consequently, if $f(z^{(1)})\geq 0$, then we must have $f(z^{(1)})\geq b_j$ for all $j\geq 1$ since $f(z^{(1)})=b_j\text{ mod}(c_j)$ and $b_j. But this contradicts the fact that $b_j\to \infty$. On the other hand, if $f(z^{(1)})<0$, then we must have $f(z^{(1)})\leq b_j-c_j$ for all $j \geq 1$ but this contradicts the fact that $b_j-c_j\to-\infty$. $\square$

References.

[1] B. de Smith, The fundamental group of the Hawaiian earring is not free, International Journal of Algebra and Computation Vol. 2, No. 1 (1992), 33-37.