The Harmonic Archipelago

Another fascinating space that receives a lot of attention is the so-called harmonic archipelago \mathbb{HA} which is the following subspace of \mathbb{R}^3.

Harmonic Archipelago

Harmonic Archipelago

You can describe the construction like this: Start by drawing the usual earring space \mathbb{E} onto a solid disk D in the xy-plane. Now between the 1st and 2nd hoops, draw a small disk and push it up so that it becomes a smooth hill with unit height. Do the same thing between the 2nd and 3rd hoops of \mathbb{E}  and then the 3rd and 4th hoops and so on. Notice that the earring \mathbb{E} is naturally a subspace of \mathbb{HA} and each hill is hollow underneath. Although the height of the hills always remains at 1, the projections of the hills to the xy-plane have diameters that tend to 0.

The Harmonic Archipelago Space

If you can believe it, the fundamental group \pi_1(\mathbb{HA},x_0), which we might refer to as the archipelago group, is even crazier than the earring group \pi_1(\mathbb{E})!

First, let’s get some notation down:

  • Let C_n=\left\{(x,y,0)|\left(x-\frac{1}{n}\right)^2+y^2=\frac{1}{n^2}\right\} be the n-th circle so that \mathbb{E}=\bigcup_{n\geq 1}C_n with basepoint x_0=(0,0,0).
  • Let \mathbb{E}_{\geq m}=\bigcup_{n\geq m}C_n be the smaller copies of the earring space.
  • Let B_n\subset \mathbb{HA} be the open disk between C_n and C_{n+1} which contains the n-th hill in the archipelago.
  • Let D_n=\{(x,y,z)\in B_n \mid \, z>\frac{1}{2} be the open disk which is the upper half of the hill B_n.

Now some observations:

1) \mathbb{HA} is path-connected and locally path connected.

2) \mathbb{HA} is non-compact since it does not include limit points on the z-axis. For instance, the sequence given by the top of the hills converges to (0,0,1) which is not included. This means that if X is compact and f:X\to \mathbb{HA} is continuous, then the image f(X) can only hit finitely many of the open disks D_n.

3) If \ell_n:S^1\to C_n is the loop that traverses C_n once counterclockwise in the xy-plane, then \ell_{n} can be deformed over finitely many hills (but not infinitely many!). So the homotopy classes [\ell_{n}]=[\ell_{n+1}] in \pi_1(\mathbb{HA},x_0) are the same for all n\geq 1 but yet are non-trivial since deforming \ell_1 over every hill should violate continuity.

Lemma 1: Every based loop \alpha:S^1\to \mathbb{HA} is homotopic to a loop \beta_{m}\colon S^1 \to\mathbb{E}_{\geq m} for each m\geq 1.

Proof: As mentioned above, the image of any based loop \alpha:S^1\to \mathbb{HA} can intersect at most finitely many of the disks D_n. Thus \alpha must have image in one of the spaces that looks like this:

Mostly chopped Harmonic Archipelago

In such a space you can deformation retract each of the infinitely many chopped hills $B_n\backslash D_n$, which are now cylinders down to height z=0.  Therefore,  \alpha is homotopic rel. basepoint to a loop in a space that looks like this:

Notice the holes get smaller and smaller so this subspace is a deformation retract of one of the form

B_1\cup B_2\cup\cdots\cup B_{m-1}\cup\mathbb{E}_{\geq m+1}.

The retraction is given by expanding each little circle in the xy-plane to the entire hole usually present in the earring space. At this point, we can choose m to be as large as we want (by adding some hills back in). The resulting space looks like the one-point union of a smaller copy of the earring and a bumpy region that is homotopy equivalent to a circle.

Now deform the bumpy region B_1\cup B_2\cup\cdots\cup B_{m-1} onto the smallest circle which bounds the hills, namely C_m. The composition of these deformation retracts provides a homotopy of \alpha to a loop in \mathbb{E}_{\geq m}. Since we could choose m to be arbitrarily large, the lemma is proven. \square

Lemma 1 basically says that every based loop is homotopic to an arbitrarily small loop.

Corollary 2: The homomorphism on fundamental groups \phi :\pi_1(\mathbb{E},x_0)\to\pi_1(\mathbb{HA},x_0) induced by inclusion is surjective and \phi([\ell_m])=\phi([\ell_n]) for all n,m\geq 1.

More generally, if g_n=[\ell_n]\in\pi_1(\mathbb{E},x_0), then g_{k_1}^{\epsilon_1}g_{k_2}^{\epsilon_2} \cdots g_{k_p}^{\epsilon_p}\in\ker\phi whenever \sum_{j}\epsilon_{j}=0. However, the fundamental group \pi_1(\mathbb{E},x_0) is way bigger than the free subgroup F_{\infty}=<g_n|n\geq 1> so we should not expect that these are the only elements of \ker\phi.

Let’s make sure no other surprising homotopies of loops can show up.

Let s_{n}:\mathbb{E}_{\geq n}\to \mathbb{E}_{\geq n+1} be the natural retraction which collapses C_n homeomorphically onto C_{n+1}. These maps induce retractions c_n:\pi_1(\mathbb{E}_{\geq n},x_0)\to\pi_1(\mathbb{E}_{\geq n},x_0) which together form a directed system:

Notice that if \phi_n:\pi_1(\mathbb{E}_{\geq n},x_0)\to\pi_1(\mathbb{HA},x_0) is the homomorphism induced by inclusion, then we have, by Corollary 2, that \phi_{n+1}\circ c_n=\phi_n for each n\geq 1. Consequently, we get a canonical homomorphism \Phi from the direct limit:

Theorem 3: \Phi:\varinjlim_{n}\pi_1(\mathbb{E}_{\geq n},x_0)\to\pi_1(\mathbb{HA},x_0) is an isomorphism of groups.

Proof: Since \phi is surjective (Corollary 2), so is \Phi. Since each c_n is a retraction it suffices to show that if [\alpha]\in\pi_1(\mathbb{E},x_0)=\pi_1(\mathbb{E}_{\geq 1},x_0) and \phi([\alpha])=1, then c_{n-1}\circ c_{n-2}\circ\dots\circ c_1([\alpha])=1 for some n>1. Since \phi([\alpha])=1, there is a homotopy H:[0,1]\times[0,1]\to\mathbb{HA} contracting \alpha to the constant loop at {x_0}. By compactness, the image of H can intersect only finitely many hills. Apply the composition of deformation retracts from the proof of Lemma 1 to obtain an n and a homotopy G:[0,1]\times[0,1]\to\mathbb{E}_{\geq n} which contracts s_{n-1}\circ s_{n-2}\circ\dots\circ s_1\circ\alpha to the constant loop {x_0}. Thus c_{n-1}\circ c_{n-2}\circ\dots\circ c_1([\alpha])=1 in \pi_1(\mathbb{E}_{\geq n},x_0). \square

Identifying  \pi_1(\mathbb{HA},x_0) as a direct limit illustrates a kind of “universal property.”

Corollary 4: Suppose Y is a space which is first countable at it’s basepoint y_0. For every shrinking sequence of based loops \beta_n\to y_0 such that \beta_n\simeq\beta_{n+1} for all n\geq 1, there is a unique induced homomorphism f:\pi_1(\mathbb{HA},x_0)\to \pi_1(Y,y_0) such that f(g_n)=[\beta_n].

Here is one last interpretation of Theorem 4: Recall that we can represent a homotopy class [\alpha]\in \pi_1(\mathbb{E},x_0) as a sequence (w_1,w_2,...)\in\varprojlim_{n}F_n, i.e. where w_n is the word in the free group F_n on letters g_1,...,g_n obtained by removing all appearances of the letter g_{n+1} from w_{n+1}. Also, the number of times a given letter g_k can appear in w_n stabilizes as n\to \infty (in other words, (w_1,w_2,...) is locally eventually constant).

If n<m, let \sigma_m(w_n)=1  and if n\geq m, let \sigma_m(w_n) be the reduced word in F_n obtained after each letter g_1,\dots,g_{m-1} is replaced by g_m. Now let


where the first possible non-trivial word appears in the m-th position. It is pretty straightforward to check that \rho_m(w_1,w_2,w_3,\dots) is still a locally eventually constant element of \varprojlim_{n}F_n.

Corollary 5: If [\alpha]\in\pi_1(\mathbb{E},x_0) corresponds to the sequence (w_1,w_2,...)\in\varprojlim_{n}F_n, then [\alpha]\in \ker\phi if and only if there is an m\geq 1 such that \rho_m(w_1,w_2,w_3,\dots)=(1,1,1,\dots,1,\sigma_m(w_m),\sigma_m(w_{m+1}),\dots) is the trivial element of \varprojlim_{n}F_n.

I’m going to call the next statement a Corollary because technically it does follow from Corollary 5 if you’re comfortable with which elements of the earring group are trivial. However, it does follow more simply from observing that \mathbb{HA} is obtained by attaching a sequence of 2-cells to the subspace A=\{(x,y,z)\in \mathbb{HA}\mid z\leq \frac{1}{2}\}, which deformation retracts on to \mathbb{E}. Recall that when you attach 2-cells to a space X along a set of attaching loops \alpha_j, j\in J, then an application of the van Kampen theorem is that the resulting space has fundamental group \pi_1(X)/N where N is the conjugate closure of \{[\alpha_j]\mid j\in J\}. Combining this with some of the above ideas gives:

Corollary 6: \ker\phi is the conjugate closure of the free subgroup of \pi_1(\mathbb{E},x_0) generated by the elements g_{n}g_{n+1}^{-1}, n\in\mathbb{N}.

There’s also an important algebraic consequence of Theorem 3. A group G is locally free if every finitely generated subgroup of G is a free group. Even though the earring group \pi_1(\mathbb{E},x_0 is not free, it is locally free. It’s a nice exercise to show that any direct limit of locally free groups is a locally free group. Hence, we have the following.

Theorem 7: The archipelago group \pi_1(\mathbb{HA},x_0) is locally free.



Apparently the first appearance of the harmonic archipelago (where it was also named) was in the following unpublished note:

[1] W.A. Bogley, A.J. Sieradski, Universal Path Spaces, Unpublished preprint.

Some unpublished notes on understanding the fundamental group of the harmonic archipelago:

[2] P. Fabel, The fundamental group of the harmonic archipelago, preprint.


Thanks to Moaaz AlQady for sending me some corrections (11/29/20).

This entry was posted in Algebraic Topology, earring space, Fundamental group, harmonic archipelago and tagged , , , , , . Bookmark the permalink.

10 Responses to The Harmonic Archipelago

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

  2. Pingback: The Uncountability of Harmonic Archipelago Group | Wild Topology

  3. Pingback: Homomorphisms from the harmonic archipelago group to finite groups | Wild Topology

  4. Pingback: The Griffiths Twin Cone | Wild Topology

  5. Pingback: Homotopically Hausdorff Spaces I | Wild Topology

  6. Pingback: Homotopically Hausdorff Spaces II | Wild Topology

  7. Pingback: What is an infinite word? (Part I) | Wild Topology

  8. Pingback: Homotopically Reduced Paths: Part I | Wild Topology

  9. Pingback: The Griffiths twin cone and the harmonic archipelago have isomorphic fundamental group (Part 1) | Wild Topology

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s