The Harmonic Archipelago Group is not Free

Posted on May 18, 2014 by Jeremy Brazas

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

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

Posted in earring group, earring space, Free groups, Fundamental group, Group homomorphisms, harmonic archipelago | Tagged , , , , , , , , | 2 Comments

The earring group is not free (Part II)

Posted on May 16, 2014 by Jeremy Brazas

This post is Part II in an explanation of why the fundamental group of the earring space is not a free group. I’ll be referencing the notation and results that we worked through in Part I.

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Recall that the earring group \pi_1(\mathbb{E})  is an uncountable group isomorphic to the free \sigma-product of \mathbb{Z}, denoted \#_{\mathbb{N}}\mathbb{Z}.

Theorem 1 (de Smit [1]): \pi_1(\mathbb{E})=\#_{\mathbb{N}}\mathbb{Z} is not isomorphic to a free group.

Proof. Since \#_{\mathbb{N}}\mathbb{Z} is uncountable (we proved this fact in this post), if it were free, then it would be isomorphic to a free group F(X) on an uncountable set X. But the set of group homomorphisms Hom(F(X),\mathbb{Z}) to the additive group of integers is in bijective correspondence with the (obviously uncountable) set \mathbb{Z}^{X} of functions X\to\mathbb{Z}. However, according to Theorem 7 (which we prove below), Hom(\#_{\mathbb{N}}\mathbb{Z},\mathbb{Z}) is isomorphic to the direct sum \bigoplus_{j=1}^{\infty}\mathbb{Z}, which is countable. Therefore, we can conclude that \#_{\mathbb{N}}\mathbb{Z} is not a free group. \square.

To fill in the remaining gaps (namely a proof of Theorem 7), we’ll need a couple of lemmas to break it up into more manageable bits.

Let  G_{\geq j} be the subgroup of \#_{\mathbb{N}}\mathbb{Z} consisting of sequences (w_n) in which the letters g_1,...,g_{j-1} do not appear. Formally,

G_{\geq j}=\{(w_n)|\#_{k}(w_n)=0\text{ for all }n\geq 1\text{ and }k<j\}

The elements of G_{\geq j} were also used in Lemma 3 in Part I.

Lemma 5: For each j>1, the group \#_{\mathbb{N}}\mathbb{Z} is the free product of the subgroup F_{j-1} and G_{\geq j}

 Proof. Since the sets of letters \{g_1,...,g_{j-1}\} and \{g_j,g_{j+1},...\} used in the two groups are mutually exclusive we can use the uniqueness of reduced words in free groups to prove this lemma. In particular, we need to show that any element of \#_{\mathbb{N}}\mathbb{Z} can be written uniquely as a product a_1b_1a_2...a_pb_p where a_i\in F_{j-1} and b_i\in G_{\geq j} (where at most one of a_1 or b_p is allowed to be the identity).

Fix an element (w_n)\in\#_{\mathbb{N}}\mathbb{Z}. By definition, \lim_{n\to\infty}\#_{k}(w_n)=M_k<\infty for all k\geq 1. In particular, for k=1,...,j-1, there is a n_{k} such that \#_{k}(w_{n})=M_k for all n\geq n_j. Let N=\max\{n_1,...,n_{j-1}\}. Thus the appearances of the letters g_1,...,g_{j-1} do not change in the words w_{N},w_{N+1},.... Rather, we can only get w_{N+q+1} from w_{N+q} by adding letters g_{k} (where k>j) in between the already present elements of F_{j-1}. So by the uniqueness of reduced words in free groups, there are elements

a_{1},...,a_{p}\in F_{j-1}  and  b_{1}^{(N+q)},...,b_{p}^{(N+q)}\in F(g_{j},...,g_{N+q})

such that w_{N+q}=a_1b_{1}^{(N+q)}...a_{p}b_{p}^{(N+q)}. We can demand that at most one of a_1 or b_{p}^{(N+q)} may be an identity element. Let r_n:F_{N}\to F_n be the projection when n<N. Then if

b_i=(r_1(b_{i}^{(N)}),r_2(b_{i}^{(N)}),...,r_{N-1}(b_{i}^{(N)}),b_{i}^{(N)},b_{i}^{(N+1)},...)\in\#_{\mathbb{N}}\mathbb{Z}

we get the desired, unique representation (w_n)=a_1b_1...a_pb_p. \square

Lemma 6: Every homomorphism f:\#_{\mathbb{N}}\mathbb{Z}\to\mathbb{Z} such that f(g_j)=0 for all j\geq 1 is the trivial homomorphism.

 Proof. Suppose f(g_j)=0 for all j\geq 1 and that f(x)\neq 0 for some x\in\#_{\mathbb{N}}\mathbb{Z}. By Lemma 5, we can (for each j>1) write x=a_1b_1\cdots a_pb_p where a_i\in F_{j-1} and b_i\in G_{\geq j}. Let s_j=b_1b_2\cdots b_p. Basically, s_j is the unique word obtained from x by deleting all appearances of the letters g_1,...,g_{j-1}. Notice that since f(g_j)=0, we have f(a_i)=0 and thus f(s_j)=f(x) for each j\geq 1.

The sequence s_j meets the two conditions necessary to apply Lemma 3 from the previous post. This means there is a self-homomorphism h:\#_{\mathbb{N}}\mathbb{Z}\to\#_{\mathbb{N}}\mathbb{Z} such that h(g_j)=s_j. But now f\circ h:\#_{\mathbb{N}}\mathbb{Z}\to\mathbb{Z} is a homomorphism such that f\circ h(g_j)=f(s_j)=f(x)\neq 0 for all j> 1 and this contradicts Lemma 4 (again from the previous post) which says that all but finitely many g_j must be killed by every homomorphism \#_{\mathbb{N}}\mathbb{Z}\to\mathbb{Z}! \square

Remark about the proof of Lemma 6: One might be tempted to skip Lemma 5 altogether and try to construct s_j using the self-homomorphism q_j:\#_{\mathbb{N}}\mathbb{Z}\to\#_{\mathbb{N}}\mathbb{Z} induced by collapsing the first j-1 circles of \mathbb{E} to the identity and leaving the other circles untouched. Indeed, it is true that s_j=q_j(x), but a priori it is not clear that you get the next step f(x)=f(s_j) unless you have the free product decomposition!

Theorem 7: The natural homomorphism Hom(\#_{\mathbb{N}}\mathbb{Z},\mathbb{Z})\to\prod_{j=1}^{\infty}\mathbb{Z} given by f\mapsto (f(g_1),f(g_2),f(g_3),...) is an injection onto the direct sum \bigoplus_{j=1}^{\infty}\mathbb{Z}.

 Proof. Lemma 6 says precisely that Hom(\#_{\mathbb{N}}\mathbb{Z},\mathbb{Z})\to\prod_{j=1}^{\infty}\mathbb{Z} is injective so we are left to identify the image. By Lemma 4 from the previous post, for every homomorphism f:\#_{\mathbb{N}}\mathbb{Z}\to \mathbb{Z}, there is an N\geq 1 such that f(g_n)=0 for all n\geq N. Thus the image of the homomorphism Hom(\#_{\mathbb{N}}\mathbb{Z},\mathbb{Z})\to\prod_{j=1}^{\infty}\mathbb{Z} lies in the direct sum \bigoplus_{j=1}^{\infty}\mathbb{Z}. To see that the image is equal to the direct sum take any element (z_1,z_2,...,z_n,0,0,...)\in\bigoplus_{j=1}^{\infty}\mathbb{Z}. The natural map \mathbb{E}\to \bigcup_{j=1}^{n}C_j collapsing the j-th circle, j>n induces a surjection \#_{\mathbb{N}}\mathbb{Z}\to F_n. Now define a homomorphism F_n\to\mathbb{Z} by g_j\mapsto z_j. The composition of these to homomorphisms yields the homomorphism f:\#_{\mathbb{N}}\mathbb{Z}\to \mathbb{Z} for which we have f(g_j)=z_j when j\leq n and f(g_j)=0 otherwise. \square

Here is a strengthened version of Lemma 6 that relates this stuff to a non-commutative version of slender groups and suggests that \#_{\mathbb{N}}\mathbb{Z} behaves in many ways like the non-commutative analogue of the Specker group \prod_{n=1}^{\infty}\mathbb{Z}.

Theorem 8: Suppose S is any group which has \mathbb{Z} as a quotient group. For every homomorphism f:\#_{\mathbb{N}}\mathbb{Z}\to S, there is a n\geq 1 such that f(G_{\geq n})=1.

Proof. Since we are assuming the existence of a surjective homomorphism S\to\mathbb{Z}, it suffices to prove the theorem for S=\mathbb{Z}. Again, we’ll use the fact that there is an N\geq 1 such that f(g_n)=0 for all n\geq N. But notice that the retract \mathbb{E}_{\geq N}=\bigcup_{n\geq N}C_n of \mathbb{E} consisting of all but the first N-1 circles is homeomorphic to the earring space. Moreover, notice that the image of the inclusion \mathbb{E}_{\geq N}\to\mathbb{E} is precisely G_{\geq N}. Since the restriction of f to G_{\geq N}=\pi_1(\mathbb{E}_{\geq N})\cong\pi_1(\mathbb{E}) takes g_k to 0 for all k\geq N. So by Lemma 6, f(G_{\geq N})=f(\pi_1(\mathbb{E}_{\geq N}))=0. \square

I think it is interesting that Theorem 8 applies to all free groups S=F(X) but does not apply when S is any finite group.

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.

Posted in earring group, earring space, Free groups, Fundamental group, Group homomorphisms | Tagged , , , , , , | 5 Comments

The earring group is not free (Part I)

Posted on May 9, 2014 by Jeremy Brazas

The main goal of this two-part post will be to study the homomorphisms out of the 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{E}),\mathbb{Z}) to the additive group of integers is countable.  This may seem a bit strange considering \pi_1(\mathbb{E}) is an uncountable group. As a direct consequence, we can see that \pi_1(\mathbb{E}) is not isomorphic to any free group F(X) on a set X.

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

The 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 earring with the CW-topology) is isomorphic to the free group F(g_1,g_2,...) on a countably infinite set of generators.

The earring group

In a previous post, I discussed how to begin understanding the algebraic structure of the fundamental group \pi_1(\mathbb{E}) of the earring space \mathbb{E}=\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.

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We decided that the fundamental group \pi_1(\mathbb{E}) 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{E}) 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{E})\cong\#_{\mathbb{N}}\mathbb{Z}.

So to study the properties of the 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<j
  • For all n\geq 1 and k<j, 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<n\\ 1 &\text{ if }j=n \end{cases}

and

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

the following diagram commutes.

selfhomomorphism

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<j 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<n_2<n_3<\cdots 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<c_j

Proof: Since a_j\geq 3, we have a_1<a_1a_2 and inductively if b_{j-1}<c_{j-1}, b_{j}=b_{j-1}+c_{j-1}<c_{j-1}+c_{j-1}=2c_{j-1}<a_{j}c_{j-1}=c_{j}. \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<c. 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<c_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.

Posted in earring group, earring space, Free groups, Fundamental group, Group homomorphisms | Tagged , , , , , , | 7 Comments