The Harmonic Archipelago Group is not Free

In recent posts, I’ve been writing about the behavior of fundamental groups of the most fundamental “wild” spaces. We did a decent amount of work in a two-part post to convince ourselves that the fundamental group $\pi_1(\mathbb{E})$ of the earring space $\mathbb{E}$ (called the earring group) is not free (Part I and Part II). Playing a key role is an explicit description of the earring group, which allows us to make arguments without having to discuss loops.

Now I’m going to work through an interesting property of another important group: the fundamental group of the harmonic archipelago $\mathbb{HA}\subset\mathbb{R}^3$

Harmonic Archipelago

In this post, I’m going to try to convince you that every homomorphism $\pi_1(\mathbb{HA})\to\mathbb{Z}$ to the additive group of integers is trivial. In other words $Hom(\pi_1(\mathbb{HA}),\mathbb{Z})=0$ . This may seem a bit strange considering $\pi_1(\mathbb{HA})$ is uncountable and torsion free.

You can construct the harmonic archipelago by drawing a copy of the earring space $\mathbb{E}$ on a solid disk and pushing up a hill of height one in between each of the hoops. Notice that the earring space is naturally a subspace of $\mathbb{HA}$ . So $\mathbb{HA}$ is basically just a really bumpy disk that differs topologically from a disk only at the one wild point of the earring space.

From the inclusion $\mathbb{E}\subset\mathbb{HA}$ , we have an induced homomorphism $\phi:\pi_1(\mathbb{E})\to\pi_1(\mathbb{HA})$ . In my first post on the harmonic archipelago, I argued that $\phi$ is surjective and hence that $\pi_1(\mathbb{HA})$ is a natural quotient of $\pi_1(\mathbb{E})$ . In particular, if $g_n=[\ell_n]$ is the homotopy class of the loop going once around the n-th circle of the earring space, then $\ker\phi$ is the conjugate closure of the set $\{g_{n}g_{n+1}^{-1}\mid n\geq 1\}$ . This is essentially because $\mathbb{HA}$ is homotopy equivalent to the space obtained by attached 2-cells to $\mathbb{E}$ using the loops $\ell_{n}\ell_{n+1}^{-1}$ , $n\geq 1$ as attaching maps. Unfortunately, the characterization as $\pi_1(\mathbb{HA})\cong\pi_1(\mathbb{E})/\ker\phi$ is not helpful for understanding the combinatorial structure of the group so I showed that $\pi_1(\mathbb{HA})$ is a direct limit of copies of the earring group. As a direct limit of locally free groups, $\pi_1(\mathbb{HA})$ is a locally free group (“locally free” means that all finitely generated subgroups are free). However, it’s still not completely obvious that $\pi_1(\mathbb{HA})$ is not an uncountable free group; we’ll rule that out in this post too.

Homomorphisms out of $\pi_1(\mathbb{HA})$

Let’s first recall the two key facts that we needed to show $\pi_1(\mathbb{E})$ is not a free group. These facts are Lemma 4 from Part I and Lemma 6 from Part II respectively.

Fact I: For every homomorphisms $f:\pi_1(\mathbb{E})\to\mathbb{Z}$ , there is an $N\geq 1$ such that $f(g_n)=0$ for all $n\geq N$ .

Fact II: If $f:\pi_1(\mathbb{E})\to\mathbb{Z}$ is a homomorphism such that $f(g_n)=0$ for all $n\geq 1$ , then $f=0.$

Theorem 1: Every homomorphism $h:\pi_1(\mathbb{HA})\to\mathbb{Z}$ is trivial, i.e. $Hom(\pi_1(\mathbb{HA}),\mathbb{Z})=0$ .

Proof. Consider the composition $h\circ\phi:\pi_1(\mathbb{E})\to\mathbb{Z}$ . According to Fact I, there is an $N\geq 1$ such that $h\circ\phi(g_n)=0$ for all $n\geq N$ . But since $\phi(g_n)=\phi(g_m)$ for all $n,m\geq 1$ , we have $h\circ\phi(g_n)=0$ for all $n\geq 1$ . Now according to Fact II, we have $h\circ\phi=0$ . Since $\phi$ is surjective, $h=0$ . $\square$

Corollary 2: $\pi_1(\mathbb{HA})$ is not isomorphic to a free group.

Proof. There is no non-trivial free group $F$ such that every homomorphism $F\to\mathbb{Z}$ is trivial. Therefore $\pi_1(\mathbb{HA})$ is not free. $\square$

One might be tempted to conjecture that the groups $\pi_1(\mathbb{E})$ and $\pi_1(\mathbb{HA})$ are isomorphic since they are both uncountable, torsion free, locally free, and we get $\pi_1(\mathbb{HA})$ by introducing countably many relations in the uncountable group $\pi_1(\mathbb{E})$ . Now it is clear that this cannot be the case.

Corollary 3: $\pi_1(\mathbb{HA})$ is not isomorphic to $\pi_1(\mathbb{E})$ .

Proof. Non-trivial homomorphisms $\pi_1(\mathbb{E})\to\mathbb{Z}$ exist. In fact, $Hom(\pi_1(\mathbb{E}),\mathbb{Z})\cong\bigoplus_{n=1}^{\infty}\mathbb{Z}$ as we saw in my last post. For example, the retraction $\mathbb{E}\to S^1$ to the unit circle which collapses all but the first circle of $\mathbb{E}$ induces a surjective homomorphism $\pi_1(\mathbb{E})\to\pi_1(S^1)\cong\mathbb{Z}$ . $\square$