# A new characterization of an annulus in the plane?

Let $K$ a connected compact subset of the Euclidean plane which has an infinite set of reflection symmetries.

Does this imply that $K$ is an annulus ?

Source : les dattes à Dattier

• [connected] compact [subset] of the Euclidean plane? What operations are you proposing? – rschwieb Jan 24 '17 at 20:25
• By axial symmetry do you mean a symmetry about a line passing through the origin? – Daniel Robert-Nicoud Jan 24 '17 at 20:34
• Not necessarily – Dattier Jan 24 '17 at 20:36
• @Dattier: Could you please clarify your definition of "an axial symmetry"? The natural guesses are "reflection about a line through the origin" and "rotation about the origin". If either of those is your definition, then "no, an annulus is not the only compact, connected set with infinitely many axial symmetries". – Andrew D. Hwang Jan 24 '17 at 21:16
• @Dattier It's not clear to me what an "orthogonal symmetry" means? Is it treating the line as a mirror? If so, it's what I meant by "symmetry about a line" in my previous comment. I also think that in order to get a bounded set you will find that all of your lines must have a common point (which we can take to be the origin without loss of generality). – Daniel Robert-Nicoud Jan 24 '17 at 21:34

This is not an answer to the answer, but the proof of the following fact, that I stated in a comment to the OP.

Lemma: Let $X\subseteq\mathbb{R}^2$ be a non-empty compact set, and let $L$ be the set of distinct lines about which $X$ is symmetric. Then there is a point $p\in\mathbb{R}^2$ which is common to every line in $L$.

Proof: We will proceed in two parts, both by contradiction on the fact that $X$ is compact. Suppose there are two lines $\ell_1$ and $\ell_2$ in $L$ that have no points in common, i.e. they are parallel. Let $\ell'$ be the line exactly in the middle of $\ell_1$ and $\ell_2$. Then for every point $x\in X$ there is one of the two lines which is closer than the other to $x$, say $\ell_1$ (except if the point is exactly $\ell'$, in which case choose whichever one, it doesn't matter). Let $x'$ be the point which is mirrored to $x$ by $\ell_2$, then the distance of $x'$ from $\ell'$ is greater than the distance of $x$ from $\ell'$. By assumption, $x'\in X$, so we can repeat this procedure to construct a sequence of points of $X$ going to infinity, which contradicts the fact that $X$ is compact. It follows that every two lines in $L$ must intersect.

Now take three lines $\ell_1,\ell_2$ and $\ell_3$ in $L$ and suppose they don't have a common point. We know by what we have said before that they must bound a triangle in the plane, let $c$ be the geometric center of this triangle. Let $x\in X$ and let $\ell$ be the line passing through $x$ and $c$. Suppose without loss of generality that if you follow $\ell$ starting from $x$ and going towards $c$, the last one of the other lines you cross is $\ell_1$. Let $x'$ be the point mirrored to $x$ by $\ell_1$, then the distance from $x'$ to $c$ is strictly greater than the distance from $x$ to $c$. Again, this leads to a contradiction to the fact that $X$ is compact.

It is straightforward to see that this implies that all lines in $L$ have a common point. QED

• $d(x,c)=\max\{d(z,c)|z\in K\}$ ? – Dattier Jan 24 '17 at 22:19
• @Dattier $\min$ – Daniel Robert-Nicoud Jan 24 '17 at 23:44
• @Daniel I have an argument that $K$ is an annulus starting from your result that the axes of symmetry have a common point. On the one hand, I'd like to append it to your post (because your argument is an essential ingredient, and in my reckoning I owe you 100 rep points over that bounty a couple of years back). On the other hand, I don't want to put you in an awkward spot, and am happy to post the answer as community wiki. – Andrew D. Hwang Jan 24 '17 at 23:54
• @AndrewD.Hwang No, please, do post it a your answer. I think it's only fair, I have no idea how to solve the initial question... – Daniel Robert-Nicoud Jan 25 '17 at 0:42
• Good start. The next step is to show that every infinite compact subgroup $K$ of $SO(2)=S^1$ is $S^1$ itself. Hint: Use the fact that $K$ has to have cardinality of continuum, its complement in $S^1$ is at most countable and $K$ acts transitively on itself. – Moishe Kohan Jan 25 '17 at 2:27

The only compact, connected (non-empty) subsets $K$ of the Euclidean plane posessing an infinite number of axial symmetries are annuli.

By Daniel Robert-Nicoud's argument, we may assume the axes of symmetry all pass through some point, which by translation we may assume is the origin. For convenience below, identify the plane rotation group $SO(2)$ with the unit circle $S^{1} = \{e^{it} : \text{$t$real}\}$.

Let $K$ be a compact, connected, non-empty subset of the plane having infinitely many axes of symmetry, let $G$ be the group of Euclidean symmetries of $K$ (rotations and reflections), and let $G_{0} = G \cap SO(2)$ be the subgroup of rotations. Since $G$ is infinite, $G_{0}$ is infinite. (A composition of reflections in lines about the origin is a rotation about the origin.)

If $x \in K$ is not the origin, and if $S$ denotes the circle of radius $\|x\|$ centered at the origin, then the orbit of $x$ under $G$, the set $Gx = K \cap S$, is homeomorphic to $G_{0}$.

It therefore suffices to show that if $G_{0}$ is closed in $SO(2)$, then $G_{0} = SO(2)$.

Lemma 1: Every finite subgroup $H$ of $SO(2)$ is cyclic.

Sketch of proof: Because $H$ is finite, there exists a smallest positive real number $t_{0}$ such that $e^{it_{0}} \in H$. It's straightforward (essentially the division algorithm) to check that $e^{it_{0}}$ generates $H$.

Lemma 2: An infinite subgroup $H$ of $SO(2)$ is dense.

Sketch of Proof: If $H$ contains an element of infinite order (i.e., an element $e^{it}$ with $t$ not a rational multiple of $\pi$), that element generates a dense subgroup.

If instead every element of $H$ has finite order, then $H$ contains a sequence of "increasingly dense subgroups", hence is itself dense.

Precisely, pick an element $h_{1} \neq 1$ arbitrarily, let $H_{1}$ be the (cyclic) subgroup generated by $h_{1}$, and let $t_{1}$ be the smallest positive real number such that $e^{it_{1}} \in H_{1}$. Assume inductively that a finite (hence cyclic, by Lemma 1) subgroup $H_{m}$ has been constructed, and that $t_{m}$ is the smallest positive real number such that $e^{it_{m}} \in H_{m}$. Because $H$ is infinite, there exists an element $h_{m+1}$ in $H\setminus H_{m}$; let $H_{m+1}$ be the (finite, hence cyclic) subgroup generated by $H_{m}$ and $h_{m+1}$, and let $t_{m+1}$ be the smallest positive real number such that $e^{it_{m+1}} \in H_{m+1}$.

The sequence $(t_{m})_{m=1}^{\infty}$ decreases to $0$. (Indeed, $t_{m}$ is an integer multiple of $t_{m+1}$ for each $m$.) It follows that for every open subset $U$ of $S^{1}$, there exists a positive integer $m$ such that $U \cap H_{m}$ is non-empty. That is, the union $\bigcup_{m} H_{m} \subset H$ is dense in $SO(2)$. Consequently, $H$ itself is dense in $SO(2)$.

Proposition: The only closed infinite subgroup of $SO(2)$ is $SO(2)$.

Proof: An infinite subgroup is dense by Lemma 2, and the only closed dense subset of $SO(2)$ is $SO(2)$.

To summarize, if $K$ is compact and has infinitely many axes of symmetry through the origin, then the intersection of $K$ with every circle centered at the origin is either empty or the entire circle. Since $K$ is connected, $K$ is an annulus.

• Bravo, but I have not understand, the Daniel's proof. – Dattier Jan 25 '17 at 11:57
• Very nice proof, +1. A remark: the sentence "The sequence $(t_m)$ decreases to $0$" in the proof of Lemma 2 is not strictly true as you could have elements of the form $2\pi k + \epsilon$ by your definition of $S^1$. Formally, the correct statement is that the sequence $(e^{it_m})$ has an accumulation point at the origin. – Daniel Robert-Nicoud Jan 25 '17 at 13:07
• @Daniel: The number $t_{m}$ is the smallest positive real for which $e^{it_{m}}$ generates $H_{m}$. :) – Andrew D. Hwang Jan 25 '17 at 23:09
• @AndrewD.Hwang Ah, right. Sorry, I take back my remark. – Daniel Robert-Nicoud Jan 26 '17 at 0:10

Call $\mathcal L$ the set of all lines of symmetry, and $\mathcal{R}$ the set of all reflections w.r.t. these lines. Consider the group $G$ of all isometries that fix $K$. In particular $\langle \mathcal R\rangle<G<\mathrm{Isom}(\mathbb R^2)$ where $\langle \mathcal R\rangle$ is the subgroup generated by $\mathcal R$.

First, there is a fixed point w.r.t. all transformations in $G$ (which implies there is a unique common intersection of all lines in $\mathcal L$, as soon as there are two concurring lines in $\mathcal L$), namely the barycenter $m$ of $K$. If $K$ were of positive measure we could define $$m=\frac{1}{|K|}\int_K x \,dx.$$ Indeed, for every $A\in G$ we have that $AK=K$ and that the Jacobian determinant of the transformation is $1$, therefore by a change of variable $$Am=\frac{1}{|K|}\int_K Ax \,dx=\frac{1}{|K|}\int_{AK} y \,dy=\frac{1}{|K|}\int_K y\,dy=m.$$ If $K$ is not of positive measure we can use Haar probability measure $\mu$ on $G$ (whose compactness follows from the compatness of $K$; $G$ is a topological group viewed for instance as a subgroup of $(n+1)\times(n+1)$ matrices) and similarly define $$m=\int_G g(0)d\mu(g)$$ whose invariance follows from the invariance of $\mu$.

Suppose wlog that the common center is the origin. In particular, $G< O(2)$. By compactness (of $\mathbb S^1$), there are arbitrarily close lines of reflections in $\mathcal L$, and therefore, since the composition of two reflections is a rotation of twice the angle betweeen the lines, there are arbitrarily small rotations in $G$, therefore $G$ is dense in $O(2)$. By compactness of $G$, $G=O(2)$. This means that $K$ is invariant under rotations, and is composed by a union of concentric circumferences. The connectedness assumption implies $K$ is an annulus.

• It is clever (for barycenter). – Dattier Jan 25 '17 at 12:56
• I had seen something similar before. In particular if the set (or the group) is finite there's no need to invoke Haar measure, you just take the average (which can be seen as using a discrete Haar measure) – Del Jan 25 '17 at 13:45
• How does compactness of $G$ follow from compactness of $K$? – tomasz Apr 5 '17 at 17:49