The Uncountability of the Harmonic Archipelago Group

In a previous post, we discussed the fundamental group of the harmonic archipelago $\mathbb{HA}\subset\mathbb{R}^3$ in some detail.

Harmonic Archipelago

One item that I skipped earlier is the fact that this group is uncountable. In this post, we’ll see why $\pi_1(\mathbb{HA})$ is uncountable. We’ll use a little bit of work from that original post.

Definition: The Baer-Specker group (or often just Specker group) is the infinite direct product $\prod_{n=1}^{\infty}\mathbb{Z}$ . The elements of this abelian group are infinite sequences $(a_n)$ of integers and the operation is component-wise addition. Let $e_n=(0,0,...,0,1,0,0...)$ denote the sequence which is zero everywhere except the n-th position where it is $1$ .

It is easy to see that the Specker group is torsion free and uncountable (since it projects onto $\prod_{n=1}^{\infty}\mathbb{Z}/2\mathbb{Z}$ which is the power set of $\mathbb{N}=\{1,2,...\}$ ). It is less obvious that the Specker group is not isomorphic to a free abelian group (perhaps a future post!).

The Specker-group does have a natural free subgroup: the infinite direct sum $\bigoplus_{n=1}^{\infty}\mathbb{Z}$ freely generated by the basis $\{e_n|n\geq 1\}$ . This subgroup consists of sequences $(a_n)$ where all but finitely many terms $a_n$ are $0$ . We’re going to make use of subgroups $A\subset\prod_{n=1}^{\infty}\mathbb{Z}$ which are isomorphic to the direct sum $\bigoplus_{n=1}^{\infty}\mathbb{Z}$ .

Lemma 1: If $\displaystyle f:\bigoplus_{n=1}^{\infty}\mathbb{Z}\to\prod_{n=1}^{\infty}\mathbb{Z}$ is any homomorphism, then the quotient group $\displaystyle\prod_{n=1}^{\infty}\mathbb{Z}\big/Im(f)$ is uncountable.

Proof. Notice that the group $\bigoplus_{n=1}^{\infty}\mathbb{Z}$ is countable since it is the countable union of countable free abelian groups $\bigoplus_{k=1}^{n}\mathbb{Z}$ . Thus $Im(f)$ has at most countable cardinality. If the quotient $\displaystyle\prod_{n=1}^{\infty}\mathbb{Z}\big/ Im(f)$ is countable, then $\prod_{n=1}^{\infty}\mathbb{Z}$ is the countable disjoint union of the cosets of $Im(f)$ . But all of these cosets have the same cardinality as $Im(f)$ and must be countable. Since we have written the uncountable Specker group as the countable union of countable sets, we conclude the Specker group is countable, which is a contradiction. $\square$

Assuming the axiom of choice, if $\{P_{\lambda}|\lambda\in\Lambda\}$ is a partition of a set $X$ with cardinality of the continuum (like the Specker group) and each $P_{\lambda}$ is countable, then $\Lambda$ must have cardinality of the continuum. Thus, if we wanted to, we could strengthen the above lemma to say that $\displaystyle\prod_{n=1}^{\infty}\mathbb{Z}\big/Im(f)$ has the cardinality of the continuum.

Now, we’re going to fix a certain homomorphism $f:\bigoplus_{n=1}^{\infty}\mathbb{Z}\to\prod_{n=1}^{\infty}\mathbb{Z}$ to show that $\pi_1(\mathbb{HA})$ naturally surjects onto one of these uncountable quotients.

Corollary 2: The homomorphism $f:\bigoplus_{n=1}^{\infty}\mathbb{Z}\to\prod_{n=1}^{\infty}\mathbb{Z}$ defined as $f(e_n)=e_n-e_{n+1}$ on basis elements is an injection and $\displaystyle\prod_{n=1}^{\infty}\mathbb{Z}\big/Im(f)$ is uncountable.

Proof. The second part of the corollary follows directly from Lemma 1. To see that $f$ is injective, suppose

$f(a_1,a_2,...,a_k,0,0,...)=(a_1-a_2,a_2-a_3,...,a_{k-1}-a_k,a_k,0,0,...)=(0,0,...).$

Then $a_j=a_{j+1}$ for $1\leq j\leq k-1$ and $a_k=0$ . Thus $a_n=0$ for all $n$ . $\square$

It is interesting to notice that

$\displaystyle Im(f)=\left\{(a_n)\in\bigoplus_{n=1}^{\infty}\mathbb{Z}\big|\sum a_n=0\right\}$ .

Proposition 3: $\displaystyle\prod_{n=1}^{\infty}\mathbb{Z}\big/Im(f)$ is torsion free.

Proof. Suppose for some element $(a_n)\in\prod_{n=1}^{\infty}\mathbb{Z}$ that $(ka_n)\in Im(f)$ for some $k>1$ . Then $k\sum a_n=\sum ka_n=0$ and thus $\sum a_n=0$ . Thus $a_n\in Im(f)$ showing that $\displaystyle\prod_{n=1}^{\infty}\mathbb{Z}\big/Im(f)$ is torsion free. $\square$

Now we need to call upon some things we know about the fundamental group of the Hawaiian earring and the harmonic archipelago.

Fact I: The map $q_n:\mathbb{H}\to C_n$ , which collapses all of the circles of the Hawaiian earring except the n-th circle, induces a surjection $(q_n)_{\ast}:\pi_1(\mathbb{H})\to\mathbb{Z}$ which gives the winding number of a loop around the n-th circle. So if $\ell_n$ is the loop going once around $C_n$ in the clockwise direction, then $(q_n)_{\ast}([\ell_n])=1$ , $(q_n)_{\ast}([\ell_n\ell_{n}^{-1}])=1-1=0$ , $(q_n)_{\ast}([\ell_n\ell_n\ell_{n}^{-1}\ell_n])=1+1-1+1=2$ , etc.

Together, these winding number homomorphisms uniquely induce a surjection $\epsilon:\pi_1(\mathbb{H})\to\prod_{n=1}^{\infty}\mathbb{Z}$ to the Specker group, given by $\epsilon([\alpha])=((q_1)_{\ast}([\alpha]),(q_2)_{\ast}([\alpha]),...)$ . Thus $\epsilon$ keeps track of all winding numbers around all circles. See the Hawaiian earring post for more details on $\pi_1(\mathbb{H})$ .

Fact II: The natural inclusion $\mathbb{H}\subset\mathbb{HA}$ of the Hawaiian earring induces a surjection $\phi:\pi_1(\mathbb{H})\to\pi_1(\mathbb{HA})$ where $\ker\phi$ is (freely) generated by the classes $[\ell_{n}\ell_{n+1}^{-1}]$ for each $n\geq 1$ . See the harmonic archipelago post for more details.

Theorem 4: $\pi_1(\mathbb{HA})$ is uncountable.

Proof. Recall the injective homomorphism $f:\bigoplus_{n=1}^{\infty}\mathbb{Z}\to\prod_{n=1}^{\infty}\mathbb{Z}$ from Corollary 2. Let $\pi:\prod_{n=1}^{\infty}\mathbb{Z}\to\prod_{n=1}^{\infty}\mathbb{Z}/Im(f)$ be the canonical projection map.

We are looking to construct the following morphism of short exact sequences of groups. The top row is an exact sequence of non-abelian groups and the bottom row is an exact sequence of abelian groups

Recall that $\ker\phi$ is the conjugate closure of the free subgroup of $\pi_1(\mathbb{H})$ generated by the elements $[\ell_{n}\ell_{n+1}^{-1}]$ . If $g\in\pi_1(\mathbb{H})$ , then

$\epsilon(g^{-1}[\ell_{n}\ell_{n+1}^{-1}]g)=-\epsilon(g)+(e_n-e_{n+1})+\epsilon(g)=e_n-e_{n+1}\in Im(f)=\ker(\pi).$

Consequently, we have an induced homomorphism $\delta:\pi_1(\mathbb{HA})\to\prod_{n=1}^{\infty}/Im(f)$ such that $\delta\circ\phi=\pi\circ\epsilon$ . Moreover, since all other maps in the right square are surjective, $\delta$ is also surjective. Since $\pi(Im(\epsilon\circ i))=\pi(\epsilon(\ker\phi))=\delta\circ\phi(\ker\phi)=0$ , it follows that $Im(\epsilon\circ i)\subseteq\ker(\pi)=Im(f)$ . Since $f$ is injective, we get an induced map $a:\ker\phi\to\bigoplus_{n=1}^{\infty}\mathbb{Z}$ such that $f\circ a=\epsilon\circ i$ . Moreover, for each $n\geq 1$ , we have $f(e_n)=e_n-e_{n+1}=\epsilon([\ell_{n}\ell_{n+1}^{-1}])$ where $[\ell_{n}\ell_{n+1}^{-1}]\in\ker\phi$ . It follows that $Im(f)=Im(\epsilon\circ i)$ and thus $a$ is surjective.

Ok…so we didn’t really need to do this homological algebra, but it personally helps me see what is going on here. By Corollary 2, $\displaystyle\prod_{n=1}^{\infty}\mathbb{Z}/ Im(f)$ is uncountable. Since $\delta$ surjects $\pi_1(\mathbb{HA})$ onto this uncountable group, $\pi_1(\mathbb{HA})$ must also be uncountable. $\square$

Remark 1: If you want to be a little more specific about the cardinality, you could notice this: $\pi_1(\mathbb{HA})$ is the quotient of $\pi_1(\mathbb{H})$ , which has cardinality of the continuum (it projects onto the Specker group and is a subgroup of $\prod_{n=1}^{\infty}F_n$ ). Additionally, $\pi_1(\mathbb{HA})$ surjects onto $\displaystyle\prod_{n=1}^{\infty}\mathbb{Z}\big/Im(f)$ which also has the cardinality of the continuum, so $\pi_1(\mathbb{HA})$ must have the same cardinality as these groups.

Remark 2: All of the vertical maps $a$ , $\epsilon$ , and $\delta$ in the above diagram are surjective. The top row is non-abelian and the bottom row is abelian so it is tempting to conjecture that these three maps are abelianization maps, however, it seems that none of them are!

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