7
$\begingroup$

If we consider a curve with infinitely many distinct lines of reflectional symmetry, we generally think of a circle. However, can we prove that the circle is the only finite plane curve that has infinitely many lines of symmetry, or do more such curves exist? (I am not exactly certain of a formal term to describe this, but by "finite," I mean that the curve does not go to infinity at any point, i.e. for all $(x,y)$ on the curve, there exist some real numbers $a$, $b$, $c$, and $d$ such that $x\in[a,b]$ and $y\in[c,d]$; thus lines do not count as "finite" curves.)


So far, I have considered two possibilities for showing that circles alone have this property, but neither formally proves that circles are the only ones. The first is that finite plane curves with infinite reflectional symmetry will have a locus of points that are equidistant from some "center" (intersection point of two or more lines of symmetry), and I think that only a circle satisfies this property. The second is that a finite curve with any lines of reflectional symmetry can be divided into congruent parts by lines of symmetry:

Two curves initially satisfying the condition of being able to be divided into congruent parts

However, the pieces must be congruent no matter how many cuts are made in order for the curve to have reflectional symmetry. Therefore, the curve must possess infinite rotational symmetry too, because any "piece" (formed by the lines of symmetry) rotated onto any other piece must match. Otherwise, the pieces would intersect at a finite number of points. However, intersecting at a finite amount of points implies that another line of symmetry will cut the curve into non-congruent pieces, and thus it cannot have infinite reflectional symmetry. For example, we see curve A from earlier intersecting itself, showing that it does not have infinite rotational, and consequently reflectional, symmetry:

As smaller and smaller pieces are rotated, the curve fails to pass the procedure

Only a circle, such as curve B, avoids such finite intersections.


Can either of these two demonstrations be formalized, or is there some other way to prove or disprove it? Any college-level answers are fine, but I would appreciate simpler answers.

$\endgroup$
  • $\begingroup$ The circle should not be called a finite curve, but a bounde curve. $\endgroup$ – Jean Marie Jun 3 '17 at 22:08
  • $\begingroup$ @JeanMarie- this is what I initially thought about saying too, but it might be confused with bounded in only one variable (e.g. the curve (0,t) is "bounded", but not completely), as "bounded" is used more often with functions. $\endgroup$ – Robert Zhang Jun 3 '17 at 22:20
  • 2
    $\begingroup$ You can ask that the curve be contained in a compact set. That will force it to be bounded in all directions. $\endgroup$ – Ross Millikan Jun 3 '17 at 22:29
12
$\begingroup$

Claim 1. Let $\ell_1,\ell_2,\ell_3$ be three line in the plane that do not intersect in a single point. Then the only bounded subset of the plane that is symmetric to all three lines is the empty set.

Proof. If two of the lines are parallel, then the composition of the reflections at them is a translation. A non-empty bounded set cannot be translation-invariant.

So we may assume that the three lines surround a triangle $\Delta$. Let $O$ be an interior point of $\Delta$. Let $S$ be a non-empty and bounded and symmetric with respect to the $\ell_i$. Then the closure $\overline S$ also has these properties and additionally is compact. Let $A\in S$ be a point at maximal distance from $O$. Consider the line $OA$. The ray starting at $O$ and not containing $A$ is not contained in in $\Delta$, hence intersects some $\ell_i$. Then $O$ and $A$ are in the same half plane determined by $\ell_i$, which implies that the reflection of $A$ at $\ell_i$ is further away from $O$ than $A$ - contradiction. $\square$

Claim 2. Any non-empty compact set that is symmetric with respect to reflections at infinitely many lines is the union of concentric circles (with radius $0$ allowed).

Proof. As a consequence of claim 1, we need only consider the case of infinitely many axes of symmetry passing through a common point $O$. Let $C$ be our symmetric compact set, $A\in C$ a point, and $S$ the circle around $O$ through $A$. We need to show $S\subseteq C$. Assume $X\in S\setminus C$. As $C$ is compact, there is $r>0$ such that the $r$-ball around $X$ is disjoint from $C$. The intersection of this ball with $S$ is an arc $\widehat{X_1X_2}$ subtending some angle $\angle X_1OX_2=\alpha$. Pick a natural number $n>\frac{4\pi }{\alpha}$ and consider a regular $n$-gon inscribed to $S$. Among our infinitely many symmetry axes, we find two that intersect the same edge of this $n$-gon. Then the angle between these axes is $\le \frac{2\pi}n<\frac\alpha2$. The composition of their reflection is then a rotation by an angle $<\alpha$. After finitely many applications of this rotation, $A$ is transported into our arc $\widehat{X_1X_2}$, contradiction. $\square$

Corollary. The only bounded curve with infinitely many refection symmetries is the circle.

Proof. By claim 2, (the closure of) the curve must be the union of concentric circles. This is a curve only if it is actually a single circle. $\square$

$\endgroup$
  • $\begingroup$ Nice. Your Claim 1 neatly shortcuts the tricky compactness argument I used. $\endgroup$ – Eric Wofsey Jun 3 '17 at 22:51
  • $\begingroup$ I like your proof, but what is the purpose of adding that you need some $r$-ball around X for some arbitrary r? $\endgroup$ – Robert Zhang Jun 4 '17 at 1:45
4
$\begingroup$

Here is a theorem along the lines you are asking for. To avoid worrying about what the definition of "finite curve" should be, I will simply consider arbitary compact subsets of $\mathbb{R}^2$.

Theorem: Let $K\subset\mathbb{R}^2$ be compact and suppose it has infinitely many lines of symmetry (i.e., lines $\ell$ such that reflection across $\ell$ maps $K$ to itself). Then there exists a point $p\in\mathbb{R}^2$ such that $K$ is a union of circles with center $p$ (possibly including the degenerate circle $\{p\}$).

Proof: We may assume $K$ is nonempty. First, $K$ cannot have two parallel lines as lines of symmetry, since the composition of their reflections will give a nontrivial translation and any nonempty set invariant under a nontrivial translation is unbounded.

Given a point $p\in\mathbb{R}^2$, let $G_p$ be the group of rotations centered at $p$ which fix $K$. Note that if two lines of symmetry of $K$ pass through $p$, then $G_p$ is nontrivial, since composition of the reflections across these lines gives a rotation around $p$. If $G_p$ is infinite, then it contains rotations by arbitrarily small angles. It follows that for any $q\in K$, the set of points obtained by applying elements of $G_p$ to $q$ is dense in the circle centered at $p$ passing through $q$. Since $K$ is closed, it thus contains this entire circle. It follows that $K$ is a union of circles centered at $p$.

So we may assume that $G_p$ is finite for all $p$. Let $n_p$ be the order of the group $G_p$. It follows that any two lines of symmetry of $K$ passing through $p$ meet at an angle which is an integer multiple of $\pi/n_p$. In particular, there can only be finitely many such lines. Since $K$ has infinitely many lines of symmetry, there must be infinitely many different points $p$ such that $G_p$ is nontrivial.

Moreover, the numbers $n_p$ must be relatively prime for different points $p$. Indeed, if $n_p$ and $n_q$ are not relatively prime, then there is a nontrivial angle $\theta$ such that both a rotation around $p$ by angle $\theta$ and a rotation around $q$ by angle $-\theta$ fixes $K$. The composition of these two rotations is then a translation (which is nontrivial if $p\neq q$), but no translation can fix $K$.

Finally, the set of points $p$ such that $G_p$ is nontrivial is bounded. Indeed, if $n_p\neq 1$, then $K$ is fixed by a nontrivial rotation around $p$. Since $K$ is nonempty, this means that $K$ must contain a point which is at least as far from the origin as $p$. Since $K$ is bounded, the set of $p$ for which this is possible is bounded.

So the set of points $p$ such that $G_p$ is nontrivial is infinite and bounded. We can thus find a sequence of distinct points $(p_i)$ such that $G_{p_i}$ is nontrivial for all $i$ and $p_i$ converges to some point $p\in\mathbb{R}^2$. Since the numbers $n_{p_i}$ must all be relatively prime, they must go to infinity. This means that for any angle $\theta$, we can find a rotation around a point arbitrarily close to $p$ by an angle arbitrarily close to $\theta$ which fixes $K$: namely, pick $p_i$ for $i$ large enough so that $p_i$ is close enough to $p$ and $n_{p_i}$ is large enough that $\theta$ is close enough to a multiple of $2\pi/n_{p_i}$. Such rotations will then converge pointwise to a rotation by an angle of $\theta$ centered at $p$. Since $K$ is closed, it follows that rotation by an angle of $\theta$ centered at $p$ maps $K$ to itself. But $\theta$ was arbitrary, so this means $G_p$ is infinite, which is a contradiction.

$\endgroup$

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.