# fundamental groups of complements to countable subsets of the plane

This question is a follow-up of this MSE post and a comment by Henno Brandsma:

Question 1. Let $$S$$ be the set of isomorphism classes of fundamental groups $$\pi_1(E^2 - C)$$, where $$C$$ ranges over all countably infinite subsets of the Euclidean plane $$E^2$$. What is the cardinality of $$S$$?

All what I can say is that $$S$$ contains at least two non-isomorphic groups:

• One is the free group $$F_\omega$$ of countably infinite rank, the fundamental group of the complement to a closed discrete countably infinite subset of $$E^2$$, does not matter which one, say, $$C={\mathbb Z}\subset {\mathbb R}\subset {\mathbb R}^2$$.

• The other is $$G_{{\mathbb Q}^2}=\pi_1(E^2-C)$$, where $$C$$ is a dense countable subset of $$E^2$$, again, does not matter which one, for instance, $$C={\mathbb Q}^2$$. This group is not free since it contains, for instance, the fundamental group of the Hawaiian Earrings. (Actually, it contains $$\pi_1$$ of every nowhere dense planar Peano continuum.)

The natural expectation is that $$S$$ is uncountable (more precisely, has the cardinality of continuum: It is clear that the cardinality of $$S$$ cannot be higher than that).

Edit. Following Yves' suggestion:

Question 2. Let $${\mathbb H}$$ denote the Hawaiian Earrings. Is the fundamental group $$\pi_1({\mathbb H})$$ essentially freely indecomposable? (Here a group $$G$$ is essentially freely indecomposable if in every free product decomposition $$G\cong G_1\star G_2$$, one of the free factors $$G_1, G_2$$ is free of finite rank.) One can also ask for the weaker property of $$G=\pi_1({\mathbb H})$$, namely, that $$G$$ does not admit free product decompositions $$G\cong G_1\star G_2$$ with two uncountable factors.

The only relevant result I could find in the literature is the theorem (due to Higman) which (according to "The combinatorial structure of the Hawaiian earring group" by Cannon and Conner) implies that that every freely indecomposable free factor of $$G=\pi_1({\mathbb H})$$ is either trivial or infinite cyclic. Maybe Higman's methods prove more, but his paper ("Unrestricted Free Products, and Varieties of Topological Groups", Journal of LMS, 1952) is behind the paywall.

If Q2 (even in the weaker form) has positive answer, then in Q1 at least one can say that $$S$$ is infinite.

• Note that is $C$ is closed then the $\pi_1$ is free and countable (better denoted $F_\omega$ than $F_\infty$, since uncountable free groups such as $F_{\mathfrak{c}}$ come into the game, at least as subgroup of some of these groups). Also if $C=\{1/n:n\ge 1\}$, then $\mathbf{C}-C$ is maybe homotopy equivalent to the Hawaiian earring space. – YCor Feb 8 '20 at 15:38
• If $D$ is a nonempty compact totally disconnected subset of the sphere $S$, one can produce a closed subset $W_D$ containing $D$, such that $D$ precisely consists of the set of accumulation points in $W_D$. Of course $W_D$ is not unique, but it might be unique up to homeomorphism of $S$ (I'm also only interested in $D$ up to homeomorphism, or [nontrivially] equivalently homeomorphism of $S$).Define $C_D=W_D\smallsetminus D$. Then one could wonder whether the isomorphism class of $\pi_1(\mathbf{R}^2-C_D)$ determines $D$ up to homeomorphism (there are continuum such $D$ up to homeomorphism). – YCor Feb 8 '20 at 15:48
• @YCor Yes, in the first example $E^2 -C$ is h.e. to the H.E. The difficulty with the question, as I currently see it, is that I see very few invariants to distinguish isomorphism types of groups appearing as $\pi_1(E^2- C)$. For instance, I cannot even tell is $G_{{\mathbb Q}^2}$ splits nontrivially as a free product (I am quite sure, it does not) or is not isomorphic to the $\pi_1$ of the H.E (I am sure, it is not). – Moishe Kohan Feb 8 '20 at 21:52
• One can also wonder whether the HE-group can be split as free product of two uncountable groups. If the answer is no, then it's not isomorphic to the $\pi_1$ of the bouquet of two HE along a non-singular point. (This question is maybe more accessible, because this group is well-studied and quite well-understood.) – YCor Feb 9 '20 at 0:40
• To me, Question 2 is a completely different question from Question 1 that should appear as a separate MO question. – Jeremy Brazas Feb 20 '20 at 23:04

## 1 Answer

Thanks to the comments, my original cardinality bound $$\aleph_1\leq |S|\leq \mathfrak{c}$$ has been refined to the equality $$|S|=\mathfrak{c}$$ that I originally suspected.

For Question 1: $$S$$ has the cardinality of the continuum. It's clear that $$|S|\leq \mathfrak{c}$$. Below, I'll argue that $$|S|$$ is at least the cardinality of the set of homeomorphism types of closed nowhere dense subsets of $$[0,1]$$. Since this set has cardinality $$\mathfrak{c}$$ (using Pierre PC's comment below), we have $$|S|\geq \mathfrak{c}$$. For the lower bound, I'll use a construction from this paper.

Consider any infinite closed nowhere dense subset $$A\subseteq[0,1]$$ containing $$\{0,1\}$$. Let $$\mathcal{I}(A)$$ denote the ordered set of components of $$[0,1]\backslash A$$. For each $$I=(a,b)\in\mathcal{I}(A)$$, let $$C_I=\left\{(x,y)\in\mathbb{R}^2\mid y\geq 0,\left(x-\frac{a+b}{2}\right)^2+y^2=\left(\frac{b-a}{2}\right)^2\right\}$$ be the semicircle whose boundary is $$\{(a,0),(b,0)\}$$. Let $$\mathbb{W}_{A}=([0,1]\times\{0\})\cup \bigcup_{I\in\mathcal{I}(A)}C_I$$ with basepoint $$(0,0)$$. Here is an example where $$\mathcal{I}(A)$$ has the order type of the linear order sum $$\omega^{\ast}+\omega+1+\omega^{\ast}$$ where $$\ast$$ denotes reverse order. First, notice that $$\mathbb{W}_A$$ is a one-dimensional Peano continuum (connected, locally path-connected, compact metric space). By picking a single point in the interior of each simple closed curve $$C_I\cup (\overline{I}\times \{0\})$$, we can see that $$\mathbb{W}_A$$ is homotopy equivalent to $$E^2\backslash C$$ for some countably infinite set $$C$$. The fundamental groups $$\pi_1(\mathbb{W}_A)$$ ranging over all such $$A$$ realize continuum-many non-isomorphic groups. Here are the heavy hitting theorems that get the job done.

1. Eda's homotopy classification of 1-dimensional Peano continuua: two one-dimensional Peano continua are homotopy equivalent if and only if they have isomorphic fundamental groups. I would just like to pause and emphasize how incredible and powerful this result is. When you first hear about it and realize the kind of groups and spaces it applies to, you might get the impression that you are being scammed.
2. Let $$\mathbf{w}(X)$$ denote the subspace of $$X$$ consisting of the points at which $$X$$ is not semilocally simply connected, i.e. the 1-wild set of $$X$$. For general spaces, the homotopy type of $$\mathbf{w}(X)$$ is a homotopy invariant of $$X$$, but monodromy actions in one-dimensional spaces have discrete graphs (see 9.13 of this paper of mine with H. Fischer). Among spaces whose monodromy actions have discrete graphs, the homeomorphism type of $$\mathbf{w}(X)$$ becomes a homotopy invariant of $$X$$ (see 9.15 of the same paper). Thus, among 1-dimensional spaces, $$\mathbf{w}(X)$$ is a homotopy invariant. This result is kind-of embedded in Eda's work leading up to 1. but the core idea behind what is really going on is fleshed out in Section 9 of the linked paper.

By combining 1. and 2. we have:

Corollary: If one-dimensional Peano continua $$X$$ and $$Y$$ have isomorphic fundamental groups, then $$\mathbf{w}(X)\cong \mathbf{w}(Y)$$. A direct consequence is that the Hawaiian earring group and the free product of the Hawaiian earring group with itself are not isomorphic because two copies of $$\mathbb{H}$$ adjoined by an arc has two 1-wild points.

Returning back to the spaces $$\mathbb{W}_A$$, notice that $$\mathbf{w}(\mathbb{W}_A)$$ is homeomorphic to the Cantor Bendixsion derivative of $$A$$, i.e. the subspace of non-isolated points of $$A$$. Every closed nowhere dense set $$B\subseteq [0,1]$$ is the Cantor Bendixsion derivative of some other $$A$$. Hence, the 1-wild sets of the $$\mathbb{W}_A$$ realize all closed nowhere dense subsets of $$B\subseteq [0,1]$$. By the corollary, each homeomorphism class of a nowhere dense closed subset of $$[0,1]$$, gives a unique isomorphism class of fundamental group $$\pi_1(\mathbb{W}_A)$$ and hence a unique isomorphism class of a fundamental group $$\pi_1(E^2\backslash C)$$ for some countably infinite set $$C$$. In the comments below, Pierre PC gives a construction of $$\mathfrak{c}$$-many non-homeomorphic closed nowhere dense subsets of $$[0,1]$$ (one can confirm this by analyzing the Cantor Bendixson derivatives of the neighborhoods of the described "super limit points"). Hence, $$|S|=\mathfrak{c}$$.

• In the last paragraph, you say that there are continuum many $B$'s, but then you give as examples the countable compact ordinals. Since there are only $\aleph_1$ countable ordinals, you seem to be assuming the continuum hypothesis here to say that this is the cardinal of the continuum. – Andreas Blass Feb 20 '20 at 14:53
• Right, I don't usually worry about this sort of thing and implicitly applied CH (I apologize to the logicians) but I guess I should either clarify or decide if there really are continuum many non-homeomorphic closed nowhere dense subsets in $[0,1]$. – Jeremy Brazas Feb 20 '20 at 15:04
• Here is, I think, a construction of a different homeomorphism type of some subset $A$ of $[0,1]$ for every subset $S$ of the integers. Define a compact set $K_k$ as the gluing of (first) some copy of $\omega^k+1$, seen as a compact space with "super limit point" $x$, corresponding to the +1, and (second) a Cantor set, glued at respectively $x$ and an extremity (the maximum, say). Then fit a copy of $K_k$ in $(2^{-k-1},2^{-k})$ if and only if $k\in S$. – Pierre PC Feb 20 '20 at 16:49
• @JeremyBrazas: This is quite amazing. I think, you should add to your answer references to Proposition 9.13 (discreteness of monodromy) and 9.15 (discrete monodromy implies that topology of $w(X)$ is a homotopy-invariant of $X$) of your paper with Fischer. – Moishe Kohan Feb 20 '20 at 19:08
• @PierrePC nice construction! Actually this precisely answers this question (which already has an accepted answer, but the latter gives a reference and no explicit construction). – YCor Feb 22 '20 at 1:36