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

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

sesheha

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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One Response to The Uncountability of the Harmonic Archipelago Group

  1. Pingback: Homotopically Hausdorff Spaces I | Wild Topology

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