A new characterization of an annulus in the plane?

Question

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

Answer 1

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

Answer 2

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.

Answer 3

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.