# A question on Hawaiian earring group

I have asked this question in MSE but have not got any satisfactory answer, so I am asking it here. Any idea on how to approach this problem will be highly appreciated.

Consider the Hawaiian earring. Suppose $f_n$'s are the loops representing the circles of radius $1/n$ centered at $(1/n, 0)$ for $n=1,2\cdots$. Suppose the following holds in the fundamental group

$$\langle [f_1, f_2][f_3, f_4]\cdots\rangle = \langle [g_1, g_2]\cdots [g_{2k-1},g_{2k}]\rangle$$ for some loops $g_1,g_2,\ldots , g_{2k}$, where $\langle\cdot\rangle$ denotes the homotopy class and $[a,b]=aba^{-1}b^{-1}$ denotes the commutator. Then can we say that $$\langle [f_1, f_2]\cdots [f_{2n-1}, f_{2n}]\rangle = \langle [g_1, g_2]\cdots [g_{2k-1},g_{2k}]\rangle$$ for all sufficiently large $n$, or for some $n>k$ ?

I had this question while I was reading the paper here (see page 76, last paragraph), where the above has been mentioned (in some other equivalent form) without proof.

This question seems to assume that the infinite product loop $$[f_1,f_2][f_3,f_4][f_5,f_6]\dots$$ is null-homologous in the Hawaiian earring, which is false. In fact, it is false by Katusya Eda's $$0$$-form Lemma, which says that a loop $$\alpha$$ in the Hawaiian earring is null-homologous if and only if the reduced representative of $$\alpha$$ factors into a finite concatenation $$\prod_{i=1}^{2n}\alpha_i$$ where there is an inverse pairing among the factors. Formally, this means $$\{1,2,\dots ,2n\}$$ splits into the disjoint union of two $$n$$-element sets $$A,B$$ and there is a bijection $$\phi:A\to B$$ such that for each $$i\in A$$, $$\alpha_i$$ is a reparameterization of the reverse of $$\alpha_{\phi(i)}$$.

The $$0$$-form Lemma can be found as Lemma 3.6 in Singular homology groups of one-dimensional Peano continua. Another nice generalization for the Hawaiian earring is Lemma 4.3 in Cotorsion-free groups from a topological point of view by Eda and Fischer. The double induction of this proof is rather subtle and is a thing of beauty.

I think your question is interesting and deserves to be solved. But, I cannot find the corresponding part in Higman's paper. For $$c$$ in $$G'$$, let $$r(c)$$ be the minimal number $$n$$ such that $$c = [u_1,v_1]\dots[u_n,v_n]$$. Let $$e_i$$ be the generator of the $$i$$-th factor of $$F$$. Then, $$r([e_1,e_2]...[e_{2n-1},e_{2n}]) = n$$, which is known but is not a straightforward fact. Therefore, $$[e_1,e_2]\dots[e_{2n-1},e_{2n}]\dots$$ does not belong to the commutator subgroup.

• Welcome to MathOverflow! Commented Jan 26, 2020 at 22:34
• Indeed by this argument it's not even in the commutator subgroup of the larger group, projective limit of free groups. Hence, the argument uses 2 ingredients (a) the canonical map from $\pi_1$*(Hawaiian earring)* to $\mathrm{limproj}F_n$ is injective (b) in $F_{2n}$ the product $\prod_{1}^n[x_{2i-1},x_{2i}]$ has commutator length $c_n$ tending to infinity.
– YCor
Commented Jan 27, 2020 at 3:57

I believe this sort of question is addressed in the beautiful series of papers by Cannon and Conner (e.g., the first one:

MR1775709 (2001g:20020) Reviewed Cannon, J. W.(1-BYU); Conner, G. R.(1-BYU) The combinatorial structure of the Hawaiian earring group. (English summary) Topology Appl. 106 (2000), no. 3, 225–271.

I am pretty sure your question or some variant is covered in Section 4 of this.

• Commented Jan 27, 2020 at 15:17