The Hawaiian Earring Group is not Free (Part II)

This post is Part II in an explanation of why the fundamental group of the Hawaiian earring 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 Hawaiian earring group \pi_1(\mathbb{H})  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{H})=\#_{\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{H} 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{H}\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{H}_{\geq N}=\bigcup_{n\geq N}C_n of \mathbb{H} consisting of all but the first N-1 circles is homeomorphic to the Hawaiian earring. Moreover, notice that the image of the inclusion \mathbb{H}_{\geq N}\to\mathbb{H} is precisely G_{\geq N}. Since the restriction of f to G_{\geq N}=\pi_1(\mathbb{H}_{\geq N})\cong\pi_1(\mathbb{H}) takes g_k to 0 for all k\geq N. So by Lemma 6, f(G_{\geq N})=f(\pi_1(\mathbb{H}_{\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.

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4 Responses to The Hawaiian Earring Group is not Free (Part II)

  1. Pingback: The harmonic archipelago group is not free | Wild Topology

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  3. Pingback: The Baer-Specker Group | Wild Topology

  4. Pingback: The Hawaiian Earring Group is not Free (Part I) | Wild Topology

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