One of the best ways to get a handle on a group is to recognize it as isomorphic to a set of symmetries of some structure. The dihedral group of order $2n$ is easily recognized as the set of symmetries of a regular $n$-gon, the symmetric group as the permutations of a set, the Klein-four group as the symmetries of the 'cross'---one pair of opposing prongs are longer than the other pair.

The previous descriptions are also in a sense 'exhaustive': every 'sensible' symmetry you could conceive of those structures is represented by some element of the group. An example where the group is not exhaustive over its structure is $C_n$ over the regular $n$-gon (reflection is missing), or $A_n$ over the set of $n$ elements.

I wish to know of a structure that a free group---specifically $F_2$---naturally acts on which it is also in a sense 'exhaustive'. I'm aware of $F_2$'s Cayley graph, but I want other examples.


The question may have been a bit unfair and imprecise, but I can't help that it is really. One of the reasons it may be unfair is that 'exhaustive' is a bit subjective. For example, I myself cannot point out a structure that $C_n$ naturally acts on that $D_{2n}$ could not be conceived of as acting on also; however, both of these groups also sit in $S_n$ in a more-or-less canonical fashion. But there are structures, such as polygons, that we think of as $D_{2n}$ acting on naturally rather than $S_n$.

I'm trying to find structures such that when we think of their 'natural' set of symmetries, the group of these symmetries would be isomorphic to a free group. I want a couple of examples, as I'm slightly disappointed in having only the Cayley graph or finite tuples of words as a reference. Other groups have many examples of acting on a structure. For example the Klein four group could also be realized as the symmetries of a proper rectangle, $A_4$ can be realized as the rigid motions of the tetrahedron rather than the even permutations of a 4-element set.

I also don't want these examples to be 'cheap'. What I mean by this is that the action is not arrived at by quotienting by a normal subgroup to get an action that is more properly realized by another group.

  • 1
    $\begingroup$ Just an idea: Take the set $X = \operatorname{conv}(\{e^{k\frac{2\pi i}{\sqrt{2}}}|k\in\mathbb{Z}\})$, and let $a^n\in F_1$ act on $X$ by multiplication with $e^{n\frac{2\pi i}{\sqrt{2}}}$. I doubt $X$ has any symmetries other than by rotation, though I'm not sure how to prove it (which is why I made this a comment). You can generalize this to $F_k$ by taking the $k$th cartesian product, and let the generators of $F_k$ act componentwise. $\endgroup$ – roman Feb 7 '14 at 9:04
  • 2
    $\begingroup$ The free group on $n$ generators acts on anything that has $n$ specified automorphisms. $\endgroup$ – Zhen Lin Feb 7 '14 at 10:07
  • 3
    $\begingroup$ @commentors: Have you read the question carefully? $\endgroup$ – Martin Brandenburg Feb 7 '14 at 10:33
  • 1
    $\begingroup$ Are fundamental groups too far from "symmetries"? Because one can obtain free groups as fundamental groups of wedge sums of circles: en.wikipedia.org/wiki/… $\endgroup$ – Manny Reyes Feb 7 '14 at 16:43
  • 1
    $\begingroup$ Free groups can be viewed as subgroups of $GL_2(\mathbb{Z})$. This means that they act on the abelian group $\mathbb{Z}\times \mathbb{Z}$, as well as on the plane $\mathbb{R}\times\mathbb{R}$. However, in neither case is the free group the whole automorphism group. But what we can learn from this action is that free groups are residually finite: automorphism groups of finitely generated, residually finite groups are residually finite (this is a $10$-line proof), and then $\mathbb{Z}\times\mathbb{Z}$ is f.g. and r.f. so it automorphism group, $GL_2(\mathbb{Z})$, is r.f. Thus, $F_2$ is r.f. QED. $\endgroup$ – user1729 Feb 8 '14 at 21:13

$F_2$ acts on a certain tiling of the hyperbolic plane. It looks sort of like this:


The above tiling is acted on by the modular group $\Gamma \cong \text{PSL}_2(\mathbb{Z})$, which naturally sits as a subgroup inside of the full group $\text{PSL}_2(\mathbb{R})$ of isometries of the hyperbolic plane. Abstractly, this group is the free product $C_2 \ast C_3$. It has congruence subgroups $\Gamma(N)$ given by the image of the kernel of the quotient maps $\text{SL}_2(\mathbb{Z}) \to \text{SL}_2(\mathbb{Z}/N\mathbb{Z})$ in $\Gamma$. For $N \ge 2$ these groups act freely on the hyperbolic plane, and the topological structure of their quotients are known (they are uncompactified modular curves, all of which will be compact Riemann surfaces minus finitely many points, and both the genus and the number of points removed are known). In particular, $\mathbb{H}/\Gamma(2)$ is known to be $\mathbb{R}^2$ minus two points, which has fundamental group $F_2$. Hence

$$\Gamma(2) \cong F_2$$

acts on the hyperbolic plane $\mathbb{H}$. You can find a fundamental domain for this action by taking a union of finitely many fundamental domains for the action of $\Gamma$ (pictured above), and this gives a tiling of the hyperbolic plane whose automorphism group should be exactly $\Gamma(2)$ (I think).

Groups that arise in this way can be thought of as hyperbolic analogues of wallpaper groups.

In general, an interesting way to find an interesting space on which a group $G$ acts is to find a space $X$ with $\pi_1(X) \cong G$ and look at the action of $G$ on the universal cover $\tilde{X}$, especially if $X$ has extra structure for its universal cover to inherit. In the case of $F_2$ we can pick $X$ to be the wedge of two circles and this reproduces the action of $F_2$ on its Cayley graph, but for other kinds of spaces $X$ we get more interesting spaces. Above we instead pick $X$ to be $\mathbb{R}^2$ minus two points and get an action on the hyperbolic plane; this is a special case of uniformization.

Edit: Perhaps an easier description of what I'm talking about is in terms of the von Dyck groups. In the Poincaré disk model the relevant tiling looks like this:

enter image description here

$F_2$ is (Edit:) the group of orientation-preserving symmetries of this tiling that can be obtained by rotating around the vertices of any of the triangles in it. (The original statement that was here was wrong; the tiling above clearly admits a rotational symmetry of order $3$, but $F_2$ has no torsion. The resolution is that this rotational symmetry fixes the origin, which is not a vertex.)

  • 2
    $\begingroup$ Regarding a side comment in the OP, $C_n$ acts on an oriented $n$-gon. There are various equivalent ways to describe what an orientation is here; one way is to think of it as a consistent orientation of the edges so that they all point clockwise or all point counterclockwise. $\endgroup$ – Qiaochu Yuan Feb 10 '14 at 7:39
  • $\begingroup$ @Alex: it's from Wikipedia. I added links to make this clearer. $\endgroup$ – Qiaochu Yuan Feb 10 '14 at 8:54
  • $\begingroup$ It wouldn't have mattered if it were from a calculus textbook--it's still a nice picture of a disk :) $\endgroup$ – Alex Youcis Feb 10 '14 at 8:56
  • 2
    $\begingroup$ It should be pointed out that these groups are called Fuchsian groups. They play an important role in, for example, studying the structure of hyperbolic groups (which are groups whose coarse structure "looks" hyperbolic), as hyperbolic groups can be split as a graph of groups in a canonical way over virtually-cyclic subgroups, and then vertex groups are either (bounded) Fuchsian or have finite outer automorphism group. This yields information about the outer automorphism groups and the model theory of hyperbolic groups, and has led to the resolution of the isomorphism problem for them. $\endgroup$ – user1729 Feb 11 '14 at 10:56
  • 2
    $\begingroup$ @QiaochuYuan: To picture $F_2$ acting here you need to take the union of two adjacent (white and back) triangles; let's call it a "quadrilateral" $Q$. Call the sides of $Q$: $A, B, C, D$ (in cyclic order); where $A, B$ are on "white" side and $C, D$ are on the "black" side. One generator $g$ of $F_2$ will "rotate" $D$ to $A$, while the other $h$ will "rotate" $C$ to $B$. When you glue the corresponding sides of $Q$ via these "rotations", you will get a triply punctured sphere (with the fundamental group $F_2$, of course). Hence, $g, h$ indeed generate a free subgroup of rank 2. $\endgroup$ – Moishe Kohan Feb 15 '14 at 14:15

Your Answer

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