## The Griffiths twin cone and the harmonic archipelago have isomorphic fundamental group (Part 1)

This is a guest post by Sam Corson, who is a Heibronn Fellow at the University of Bristol.

This first post will provide background on the infinite word combinatorics which are used in the description of the fundamental group of each of the spaces in question. The Griffiths twin cone space $\textbf{TC}$ first appeared in print in H. B. Griffith’s paper The fundamental group of two spaces with a common point, Quart. J. Math. Oxford 2 (1954), 175-190. The first appearance of the harmonic archipelago $\textbf{HA}$ seems to be in the work of W. A. Bogley and A. J. Sieradski Weighted combinatorial group theory and wild metric complexes, Groups-Korea ’98 (Pusan), de Gruyter, Berlin, 2000, 53-80. For more background into these two spaces, you can consult some of Brazas’ old blog posts: harmonic archipelago and Griffiths twin cone. The conjecture that $\pi_1(\textbf{TC}) \simeq \pi_1(\textbf{HA})$ originated with James W. Cannon and Gregory R. Conner.

Recall that the earring space $\textbf{E}$ is the shrinking wedge of countably infinitely many circles. More formally if $p \in \mathbb{R}^2$ we let $C(p, r)$ denote the circle centered at $p$ of radius $r$. The subspace $\textbf{E} \subseteq \mathbb{R}^2$ is given by $\textbf{E} = \bigcup_{n\in \mathbb{N}} C((0, \frac{1}{n + 1}), \frac{1}{n + 1})$ (this post of Brazas gives some nice background). It is well-known that the fundamental group of a wedge of circles is a free group (with each circle corresponding to a free generator), and so one would expect that the the fundamental group $\pi_1(\textbf{E})$ will be like a free group. While this is true, it is emphasized that $\pi_1(\textbf{E})$ is not a free group. This is best illustrated by the curious fact that $| \pi_1(\textbf{E})| =2^{\aleph_0}$ and $\pi_1(\textbf{E})$ cannot homomorphically surject onto a free group of infinite rank (for this latter result, see Theorem 1 of G. Higman, Unrestricted free products and topological varieties, J. London Math. Soc. 27 (1952), 73-81.)

Let $A = \{a_n^{\pm 1}\}_{n \in \mathbb{N}}$ be a countably infinite collection of symbols, which we will call letters, which is equipped with formal inverses. Usually the superscript $1$ is not written. A word $W$ is a finite-to-one function $W: \overline{W} \rightarrow A$ where the domain $\overline{W}$ is a totally ordered set (finite-to-one means in this case that for each $n \in \mathbb{N}$ and $\epsilon \in \{\pm 1\}$ the set $\{t \in \overline{W} : W(t) = a_n^{\epsilon}\}$ is finite). It follows that the domain $\overline{W}$ of a word $W$ must be countable (possibly finite or empty). As an example the infinite string

$a_0^{-1}a_1a_2^{-1}a_3a_4^{-1}a_5 \cdots$

is a word; more formally it is the word $W: \mathbb{N} \rightarrow A$ given by $W(n) = a_n^{{(-1)}^{n + 1}}$ (notice that each element of the alphabet $A$ is utilized at most once in the word). The infinite string

$a_0a_1a_0a_3a_0a_5 \cdots$

given by the rule $n\mapsto \begin{cases}a_0 \text{ if }n\text{ is even}\\a_n\text{ if }n\text{ is odd } \end{cases}$ is not a word since the letter $a_0$ is used infinitely often. Let $E$ denote the empty word, i.e. the word with empty domain. A word can have more exotic domain than $\mathbb{N}$: any finite-to-one function $W: \mathbb{Q} \rightarrow A$ is a word. As a technical aside, we consider two words $W_0$ and $W_1$ to be equivalent, and write $W_0 \equiv W_1$, provided there exists an order isomorphism $\iota: \overline{W_0} \rightarrow \overline{W_1}$ such that $W_0(t) = W_1(\iota(t))$ for all $t\in \overline{W_0}$. We form the concatenation of two words $W_0$ and $W_1$, denoted $W_0W_1$, by declaring that $W_0W_1$ has domain which is the disjoint union $\overline{W_0} \sqcup \overline{W_1}$ with the elements in $\overline{W_0}$ being ordered below those in $\overline{W_1}$ and having

$W_0W_1(t) = \begin{cases}W_0(t)\text{ if }t\in \overline{W_0}\\W_1(t)\text{ if }t\in \overline{W_1}\end{cases}$

Analogously, given a totally ordered set $\Lambda$ and collection of words $\{W_{\lambda}\}_{\lambda \in \Lambda}$ indexed by $\Lambda$ we can form a function whose domain is the disjoint union $\bigsqcup_{\lambda \in \Lambda}\overline{W_{\lambda}}$, ordered in the natural way, and defined by $t \mapsto W_{\lambda}(t)$ where $t \in \overline{W_{\lambda}}$. This function we denote $\prod_{\lambda} W_{\lambda}$ and it is a word provided it is finite-to-one.

A word $W$ has an inverse, which is denoted $W^{-1}$, given by letting $\overline{W^{-1}}$ be the set $\overline{W}$ under the reverse order and $W^{-1}(t) = (W(t))^{-1}$. For example the inverse of the word

$a_0^{-1}a_1a_2^{-1}a_3a_4^{-1}a_5 \cdots$

will be the word

$\cdots a_5^{-1}a_4a_3^{-1}a_2a_1^{-1}a_0$

Given $N \in\mathbb{N}$ and word $W$ we let $p_N(W)$ be the finite word given by the restriction $W \upharpoonright\{t\in \overline{W}: W(t) \in \{a_0^{\pm 1}, \ldots, a_N^{\pm 1}\}\}$. Given words $W_0, W_1$ we write $W_0 \sim W_1$ if for each $N \in \mathbb{N}$ the words $p_N(W_0)$ and $p_N(W_1)$ are equal as elements in the free group. For example, the word $W$

$a_0a_1^2a_4a_5a_6a_7a_8a_9 \cdots \cdots a_9^{-1}a_8^{-1}a_7^{-1}a_6^{-1}a_5^{-1}a_4^{-1}a_1^{-3}a_0$

has $p_0(W) \equiv a_0^2$, $p_1(W) \equiv a_0a_1^2a_1^{-3}a_0 \equiv p_2(W) \equiv p_3(W)$ and for $N \geq 4$ we get

$p_N(W) \equiv a_0a_1^2a_4 \cdots a_N a_N^{-1}\cdots a_4^{-1}a_1^{-3}a_0$

It is easy to see that $a_0a_1^{-1}a_0 \sim W$.

The group $\pi_1(\textbf{E})$ is isomorphic to the collection of equivalence classes over $\sim$. The binary operation is given by concatenation: $(W_0/\sim) * (W_1/\sim) = (W_0W_1)/\sim$ and the $\sim$ class of the empty word $E$ plays the role of the group identity. Inverses in the group are predictably defined by $(W/\sim)^{-1} = W^{-1}/\sim$.

Analogously to a free group, there are specific words with which we prefer to work. Given a word $W$ we say that $W_1$ is a subword of $W$ if there exist words $W_0, W_2$ (either or both of which may be empty) such that $W \equiv W_0W_1W_2$. Moreover $W_1$ is an initial (respectively terminal) subword provided $W_0$ (resp. $W_2$) in the above writing is empty. Finally a word $W$ is reduced if for every subword $W_1$ we have $W_1\sim E$ implies $W_1 \equiv E$. Clearly every subword of a reduced word is itself reduced. The proof of the following result is far more difficult than that of the free group analogue:

Lemma. Every $\sim$ class contains a reduced word which is unique up to $\equiv$. Letting $\textbf{Red}(W)$ denote the reduced representative of the $\sim$ class of word $W$ we have for all words $W_0, W_1, W_2$ that $\textbf{Red}(W_0 \textbf{Red}(W_1W_2)) \equiv \textbf{Red}(\textbf{Red}(W_0W_1)W_2)$. Moreover, given reduced words $W, W'$ there exist words $W_0, W_1, W_0', W_1'$ such that

(1) $W \equiv W_0W_1$;

(2) $W' \equiv W_0'W_1'$;

(3) $(W_1)^{-1} \equiv W_0'$;

(4) $W_0W_1'$ is reduced.

For further reading on (reduced) words see Section 1 of K. Eda, Free $\sigma$-products and noncommutatively slender groups, J. Algebra 148 (1992), 243-263.

The nice qualities of reduced words motivate one to consider the earring group as the set $\textbf{Red}$ of reduced words with binary operation $W_0*W_1 \equiv \textbf{Red}(W_0W_1)$. We introduce two alphabets with formal inverses:

$H = \{h_n^{\pm 1}\}_{n \in \mathbb{N}}$ (with $H$ for “h”armonic archipelago); and

$T = \{t_{i, n}^{\pm 1}\}_{i \in \{0, 1\}, n \in \mathbb{N}}$ (with $T$ for “t”win cone).

Define words, concatenation, $\sim$, reduced word, etc. just as before for each of these new alphabets and let $\textbf{Red}_H$ and $\textbf{Red}_T$ denote the respective sets of reduced words. These two sets are each groups under the binary operation $W_0*W_1 \equiv \textbf{Red}(W_0W_1)$ and both are isomorphic to $\textbf{Red}$ (the isomorphism with $\textbf{Red}_H$ is given by the word mapping which extends $a_n^{\pm 1} \mapsto h_n^{\pm 1}$ and the isomorphism with $\textbf{Red}_T$ is given by $a_n^{\pm 1} \mapsto t_{i, m}^{\pm 1}$ where $n = 2m + i$).

A word $W \in \textbf{Red}_T$ is $(0, T)-$pure if the first subscript in each of the letters appearing in $W$ is $0$, and $(1, T)$pure is defined analogously. A word is $T$pure provided it is either $(0, T)$-pure or $(1, T)$-pure. For $i \in \{0, 1\}$ every subword of a $(i, T)$-pure word is again $(i, T)$-pure, and the only word which is both $(0, T)$-pure and $(1, T)$-pure is $E$. Let $\textbf{Pure}_T$ denote the set of $T$-pure words. The group $\pi_1(\textbf{TC})$ is isomorphic to $\textbf{Red}_T/\langle\langle \textbf{Pure}_T \rangle\rangle$, where the notation $\langle\langle \cdot \rangle\rangle$ denotes the smallest normal subgroup which includes the input. This isomorphism can be seen by two applications of van Kampen’s Theorem (see e.g. Section 4 in K. Eda, H. Fischer, Cotorsion-free groups from a topological viewpoint, Topology Appl. 214 (2016), 21-34.)

A word $W \in \textbf{Red}_H$ is $(n, H)$pure, where $n \in \mathbb{N}$, provided all subscripts of letters appearing in $W$ are $n$ (i.e. $W$ is of form $h_n^j$ where $j \in \mathbb{Z}$). A word is $H$pure provided it is $(n, H)$-pure for some $n \in \mathbb{N}$ and we let $\textbf{Pure}_H$ denote the set of $H$-pure words. The group $\pi_1(\textbf{HA})$ is isomorphic to $\textbf{Red}_H/\langle\langle \textbf{Pure}_H\rangle\rangle$ (see Theorem 5 of G. R. Conner, W. Hojka, M. Meilstrup, Archipelago groups, Proc. Amer. Math. Soc. 143 (2015), 4973-4988.)

Now the task of establishing the isomorphism $\pi_1(\textbf{TC}) \simeq \pi_1(\textbf{HA})$ is reduced to producing an isomorphism between $\textbf{Red}_T/\langle\langle \textbf{Pure}_T \rangle\rangle$ and $\textbf{Red}_H/\langle\langle \textbf{Pure}_H\rangle\rangle$. This is not an easy task. It’s a nice exercise to check that any continuous function $f: \textbf{TC} \rightarrow \textbf{HA}$ induces a trivial homomorphism $f_*: \pi_1(\textbf{TC}) \rightarrow \pi_1(\textbf{HA})$ (using the fact that $\textbf{TC}$ is a Peano continuum and any continuous Hausdorff image of a Peano continuum is again a Peano continuum). While it is possible to give a continuous function $f: \textbf{HA} \rightarrow \textbf{TC}$ so that $f_*$ is surjective, it is not possible to make such an $f_*$ injective as well. Thus, the natural (spacial) homomorphisms are ruled out. The fact that each element of $\textbf{Red}_T$ is a (possibly infinitary) concatenation of $T$-pure words and similarly each element of $\textbf{Red}_H$ is a (possibly infinitary) concatenation of $H$-pure words should be used in some way. A confounding issue is that $|\textbf{Pure}_T| = 2^{\aleph_0}$ and $|\textbf{Pure}_H| = \aleph_0$. We will continue in Part 2.

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