16
$\begingroup$

I recently thought of this problem and have made very little progress. I conjecture that $\pi_1(\mathbb{R}^2\setminus \mathbb{Q}^2)$ is the biggest possible such group, in the sense that every other group of this form is isomorphic to a subgroup of it, but I have no idea how to justify this. Is classifying these groups even possible? Do these groups always have to be free?

$\endgroup$
  • 6
    $\begingroup$ I don’t think these groups would have to be free. I imagine you could take the complement of the Hawaiian earring and remove a countable dense subset inside it and the end result would be a space who’s fundamental group contains a copy of the fundamental group of the Hawaiian earring, which implies the fundamental group is not free. $\endgroup$ – Connor Malin Jul 23 '19 at 1:38
  • 1
    $\begingroup$ Notably, if $\pi_1(\mathbb{R} \setminus \mathbb{Q})$ is free, then the fundamental group of the previous space wouldn’t embed in it. $\endgroup$ – Connor Malin Jul 23 '19 at 2:11
  • $\begingroup$ At least your groups are in general uncountable. $\endgroup$ – Paul Frost Jul 23 '19 at 12:27
9
$\begingroup$

Recall that a Peano continuum is a compact connected and locally connected metrizable topological space. Equivalently, it is the image of a continuous map of $[0,1]$ to a metrizable space. For a subset $X\subset S^2$ I will denote the complement $S^2-X$ as $X^c$.

Consider $A$, a Peano continuum in $S^2$. I will denote $U_i$, $i\in I$, the components of $A^c$. Since $A$ is connected, each $U_i$ is simply-connected. In particular, it is conformal to the unit disk $\Delta$. It is known that the Riemann mapping $f_i: \Delta\to U_i$ extends to a continuous map
$$F_i: cl(\Delta)\to cl(U_i)$$ (this is the Caratheodory-Torhorst extension theorem, see my answer here). Therefore, given a point $b_i\in U_i$ and $f_i$ sending $0\in \Delta$ to $b_i$, we define $r_i: U_i-\{b_i\}\to \partial U_i$ by composing $$ F_i\circ R_i \circ f_i^{-1} $$ where $R_i: \Delta- \{0\}\to S^1=\partial \Delta$ is the radial projection. One verifies that the map $r_i$ extends continuously (by the identity) to $\partial U_i$.

Proposition 1. Set $B=\{b_i: i\in I\}$. Then there exists a retraction $r: B^c\to A$.

Proof. Combine the retractions $r_i: cl(U_i)- \{b_i\}\to \partial U_i$ defined above. Extend to the rest of $A$ by the identity map. (Use local connectivity of $A$ to check that the resulting map is continuous along $A$.) $\square$

Remark. If you only want to understand examples such as when $A$ is the Hawaiian Earrings, you can construct a retraction $r$ directly without need for any deep extension theorems.

Corollary 1. The inclusion map $A\to B^c$ induces a monomorphism $\pi_1(A)\to \pi_1(B^c)$.

Corollary 2. For every subset $C\subset A^c$ which has nonempty intersection with each $U_i$, the inclusion map $A\to C^c$ induces a monomorphism $\pi_1(A)\to \pi_1(C^c)$.

Proof. For each $U_i$ pick a point $b_i\in C\cap U_i$. Set $B=\{b_i: i\in I\}$. The homomorphism $\pi_1(A)\to \pi_1(B^c)$ factors as $$ \pi_1(A)\to \pi_1(C^c) \to \pi_1(B^c). $$ Now, the claim follows from Corollary 1. $\square$

Corollary 3. Suppose that $C$ is a dense subset of $S^2$ and $A\subset C^c$ is a Peano continuum. Then the inclusion map $A\to C^c$ induces a monomorphism $\pi_1(A)\to \pi_1(C^c)$.

Proof. Observe that, by density, $C$ has nonempty intersection with each component of $A^c$. Now, the claim follows from Corollary 2. $\square$

Recall that for any two dense countable subsets $X, Y\subset E^n$ there is a homeomorphism (actually, one can even find a diffeomorphism) of pairs $(E^n,X)\to (E^n,Y)$. (This is Brouwer's theorem, see my answer here.) In particular, for $n=2$, we have $\pi_1(E^2-X)\cong \pi_1(E^2-Y)$.

In particular, for every dense countable subset $X\subset E^2$, we have $\pi_1(E^2-X)\cong \pi_1({\mathbb R}^2 - {\mathbb Q}^2)$.

Corollary 4. $\pi_1({\mathbb R}^2 - {\mathbb Q}^2)$ is not free.

Proof. Start with the Hawaiian earrings $A=E\subset E^2$. Take a dense countable subset $C\subset A^c$ containing the point $\infty$. Now, apply Corollary 3 and obtain a monomorphism $\pi_1(E)\to \pi_1(E^2 - C)$. $\square$

What's more, by the same argument we obtain:

Proposition 2. $\pi_1({\mathbb R}^2 - {\mathbb Q}^2)$ contains an isomorphic copy of the fundamental group of each nowhere dense Peano continuum $A\subset E^2$.

Remark. With more work one can remove the nowhere dense assumption.

Now, for an arbitrary countable subset $Y\subset E^2$, its complement $E^2-Y$ can be represented as a union of all Peano continua $A_\alpha, \alpha\in J$, in $E^2-Y$, forming a direct system. Therefore,
$$ \pi_1(E^2-Y)\cong \lim_{\alpha\in J} \pi_1(A_\alpha) $$ (the direct limit). As noted above, every group $\pi_1(A_\alpha)$ embeds in $\pi_1({\mathbb R}^2 - {\mathbb Q}^2)$. Maybe from this one can construct a monomorphism $$ \pi_1(E^2-Y) \to \pi_1({\mathbb R}^2 - {\mathbb Q}^2). $$ I do not see how to do this at the moment. (But maybe I am missing something simple.)

In any case, as far as I am concerned, classical algebro-topological invariants such as homotopy groups and singular homology, were designed for "nice" topological spaces. For spaces such as complements to arbitrary countable subsets of $E^2$ one should use other invariants. In particular, the "classification" question for their fundamental groups is hopeless. It reminds me of the joke that the classification of mathematical problems as linear and nonlinear is like classification of the Universe as bananas and non-bananas.

$\endgroup$
  • $\begingroup$ Sure these groups are uncountable but it is an unfortunately common misconception that there is no already developed technology to deal with them. There are an enormous amount of tools for working with these groups, which have an infinite reduced word calculus analogous to that of free groups. Also, the classification of fundamental groups of one-dimensional (and planar) Peano continua is not hopeless. In a very practical sense, it is already done (work of Eda, Conner, Kent, and others). $\endgroup$ – Jeremy Brazas Feb 20 at 14:58
  • $\begingroup$ There are plenty of open questions about the algebra of these groups (just as there are also open questions about finitely generated groups) but in order not to immediately scared off by these things one needs to realize that the fundamental groups are not just groups but are groups with additional infinite product operations. $\endgroup$ – Jeremy Brazas Feb 20 at 14:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.