Symmetry Breaking: One-Point Theorem

Abstract

Symmetry breaking is crucial in many areas of physics, mathematics, biology, and engineering. We investigate the symmetry of regular convex polygons, non-convex regular polygons (stars), and symmetric Jordan curves/domains. We demonstrate that removing a single point from the boundary of regular convex and non-convex polygons and symmetrical Jordan curves reduces the symmetry group of the polygon to the trivial $C_1$ group when the point does not belong to the axis of symmetry of the polygon. The same is true for solid and open 2D regular convex polygons and symmetric Jordan curves. The only exception is a circle. Removing a single point from the boundary of a circle creates a curve characterized by the $C_2$ group. The symmetry of circles is reduced to the trivial $C_1$ group by removing a triad of non-symmetrical points. The same is true for a solid circle. The “effort” necessary to break the symmetry of a circle is maximal. A 3D generalization of the theorem is exemplified. Thus, the classification of symmetrical curves following the minimal number of points necessary to break their symmetry becomes possible. The demonstrated theorem shows that the symmetry group action on curves and domains becomes trivial when an asymmetric perturbation is introduced, when the curve is not a circle. An informational interpretation of the demonstrated theorem, which is related to the Landauer principle, is provided.

1. Introduction

Symmetry breaking is crucial in many areas of physics, mathematics, and even biology and engineering. It often reveals deeper structures, initiates critical transitions, or explains how diversity and complexity emerge from simple rules [1,2]. In the Higgs mechanism, the underlying laws remain symmetric, but the vacuum state itself breaks that symmetry [3]. Rigorously speaking, Lagrangian has a symmetry, but the lowest energy state (vacuum) of the system does not respect it [3,4]. This gives particles mass in the Standard Model. Without symmetry breaking, all particles would remain massless—contradicting reality.

Symmetry breaking plays a central role in phase transitions [5,6,7]. In particular, spontaneous symmetry breaking plays a crucial role in second-order (continuous) phase transitions [8,9,10,11]. Phase transitions associated with spontaneously broken global symmetries can have important cosmological implications [12]. Today, symmetry breaking is effectively exploited in materials engineering; it plays a significant role in 2D layered materials, defining their macroscopic electrical, optical, magnetic, and topological properties [13].

Ideas of symmetry breaking are intensively discussed in modern chemistry; a catalyst design based on symmetry-breaking sites activating nonpolar CO₂ molecules has been reported [14]. Furthermore, symmetry breaking in living matter has been addressed [15]. It has also been suggested that symmetry breaking is closely related to esthetics [16]. Our study is devoted to the mathematical aspects of symmetry breaking. We address the following fundamental question: what minimal effort is necessary to break the symmetry of symmetric curves and shapes? This paper is structured as follows: (i) the symmetry breaking of polygons is addressed, and the minimal effort necessary to break their symmetry is established; (ii) the symmetry breaking of Jordan curves is investigated; (iii) applications of the introduced approach are discussed; (iv) the symmetry breaking of 3D objects is considered; and (v) the minimal effort necessary for symmetry breaking is discussed in the context of the Landauer principle.

2. Results

2.1. Breaking the Symmetry of Regular Polygons

Consider a symmetrical polygon. We pose the following fundamental question: how many points should be removed from the polygon in order to break its symmetry? Let us start from the simplest example of an equilateral triangle. The symmetry group of an equilateral triangle is the dihedral group $D_3.$ This is a group with an of order six, and it comprises three rotations and three reflections; namely, the elements of this group are

$$D_3=\left\{r_0,r_1,r_2, s_1,s_2,s_3\right\},$$

(1)

where $r_0$ is the identity (rotation by 0 rad), $r_1$ is rotation by ${\displaystyle \frac{2}{3}}\pi$, $r_2$ is rotation by ${\displaystyle \frac{4}{3}}\pi$, $s_1$ is reflection across the axis through vertex “1” and the midpoint of the opposite side, and $s_2, s_3$ are reflection axes through vertices “2” and “3”, as depicted in Figure 1.

We remove a single point, denoted by A (see Figure 1), located on the boundary of the equilateral triangle. This point does not lay on the symmetry axes $s_i, i=1,\dots ,3.$ Let us analyze this procedure from a topological point of view. The initial boundary of the triangle is homeomorphic to a circle $S^1$, which is compact, connected, and without a boundary. After removing point A, the boundary becomes homeomorphic to an open interval $\left(\mathrm{0,1}\right)$, which is non-compact and connected and possesses two “ends” (though no actual boundary points because the endpoints are missing).

After removing point A, the boundary loses all elements of symmetry, with a single exception of the identity element, i.e., rotation by 0 rad. This is easily checked by sequentially examining the symmetry of the triangle with the removed point A relative to the $D_3$ group elements. Thus, as a result of the suggested procedure, the dihedral group $D_3$ is reduced to the trivial symmetry group, which is usually labeled $C_1$ or $\left\{e\right\}$ within graph-theoretic notation. We denote the removal of a single point from the boundary $\left(1pr\right).$ Thus, the entire process of symmetry breaking is briefly expressed in Equation (2) as follows:

$$D_3\to \left(1pr\right)\to C_1$$

(2)

We conclude that removing a single point from the equilateral triangle boundary, which does not lay on the axes of symmetry of the triangle, completely destroys its symmetry. The order of the symmetry group is reduced from six to unity. Removing a point that is located on one of the axes of symmetry of the equilateral triangle reduces the symmetry to the $C_2$ group.

The same is true for an isosceles triangle, as shown in Figure 2. The dihedral group of an isosceles triangle is usually labeled $D_1$, and it contains two elements, i.e., an identity element $r_0$ and a symmetry axis $s_1$ (see Figure 2).

We select point A, which does not belong to the symmetry axis $s_1$, and remove it from the boundary. After removing point A, the boundary becomes homeomorphic to an open interval $\left(\mathrm{0,1}\right)$, and the symmetry of the boundary is reduced to the $C_1$ group, including the identity element only. In other words, Equation (3) is true, as follows:

$$D_1\to \left(1pr\right)\to C_1$$

(3)

The order of the symmetry group is reduced from two to unity. Removing a point that is located on one of the axes of symmetry of the isosceles triangle reduces the symmetry to the $C_2$ group.

Now, we consider a solid 2D triangle T, which is homeomorphic to a closed 2-disk $D^2$, as depicted in Figure 3. The symmetry group of this shape is $D_3=\left\{r_0,r_1,r_2, s_1,s_2,s_3\right\}$ (see Figure 3). We propose removing a single point A from the triangle (see Figure 3). Point A does not belong to the axes of symmetry of the triangle. If the point is removed from the interior of the triangle, then the shape will be homeomorphic to a punctured disk $D^2\setminus \left\{A\right\}.$

If point A is removed from the boundary, then the shape will be equivalent to a closed disk with a point removed from its boundary, which is topologically homeomorphic to the original triangle. Removing point A, which does not lay on the axis of symmetry $s_1$, again destroys the symmetry of the isosceles triangle. If the removed point is at the center (the intersection of the three symmetry axes), then every symmetry of the equilateral triangle fixes it. Thus, the symmetry group remains as $D_3$. If the removed point is any other point on the symmetry axis (e.g., a vertex, an edge midpoint, or any interior point on that axis but not at the center), then the only nontrivial symmetry that fixes it is the reflection across that axis. Thus, the symmetry group drops to $C_2$ due to that reflection.

The same approach is applicable to an open isosceles/equilateral triangle, which is homeomorphic to an open 2-disk. In this case, point A, which is to be removed from the triangle, belongs to the interior of the triangle and does not lay on its axis of symmetry.

Thus, the following lemma is demonstrated:

Lemma 1.

Consider the boundary of an isosceles/equilateral triangle. Removing a single point from the boundary (a point that does not belong to the axis of symmetry of the triangle) reduces the symmetry group of the triangle to the trivial $C_1$group. The same is true for a solid 2D isosceles/equilateral triangle and an open isosceles/equilateral triangle.

Again, as a result of the suggested puncturing of the triangle, the dihedral group $D_3$ is reduced to the trivial symmetry group, labeled $C_1$.

Now, we consider regular polygons. Consider a regular pentagon, as depicted in Figure 4. The circumcenter of the pentagon is denoted by O. The symmetry group of the regular pentagon is the dihedral group $D_5$, and it includes five rotations (including the identity) and five reflections (through lines that pass through a circumcenter and the midpoint of the opposite side, denoted by $s_i, i=1,\dots ,5$).

We remove a single point, denoted by A (see Figure 4), located on the boundary of the regular pentagon. This point does not lay on the symmetry axes $s_i, i=1,\dots ,5.$ The initial boundary of the pentagon is homeomorphic to a circle $S^1$, which is compact, connected, and without a boundary. After removing point A, the boundary becomes homeomorphic to an open interval $\left(\mathrm{0,1}\right)$, which is non-compact and connected and possesses two “ends”.

After removing point A, the boundary becomes homeomorphic to an open interval $\left(\mathrm{0,1}\right)$, and the symmetry of the boundary is reduced to the trivial group $C_1$, including the identity element only. This is easily demonstrated with the triangulation procedure, shown with dashed lines in Figure 4. Removing point A from the boundary breaks the symmetry of the isosceles triangle $\left(1O2\right)$ (see Figure 4) according to Lemma 1. Breaking the symmetry of the isosceles triangle $\left(1O2\right)$ breaks the symmetry of the entire pentagon. In other words, for a regular pentagon, Equation (4) is true, as follows:

$$D_5\to \left(1pr\right)\to C_1$$

(4)

The order of the symmetry group of the pentagon is reduced from five to unity. This approach is easily extended to any regular n-gone. Thus, Theorem 1 is demonstrated:

Theorem 1.

Let $\left\{B_1, B_2\dots B_n\right\}$ be a regular polygon/n-gon (for $n\ge 3$). Removing a single point from $\left\{B_1, B_2\dots B_n\right\}$ that is not fixed under any nontrivial symmetry operation of $\left\{B_1, B_2\dots B_n\right\}$ reduces its symmetry group to the trivial group $C_1$. The same is true for a solid 2D regular n-gon and an open regular n-gon.

Removing point A from the boundary of the pentagon, which is located on one of its axes of symmetry, reduces the symmetry group from $D_5$ to $C_2$.

Theorem 1 is also true for a non-convex pentagon, such as a five-pointed star, as shown in Figure 5.

The symmetry group of the five-point star is the same dihedral group $D_5$. We remove a single point, denoted by A (see Figure 5), located on the boundary of the regular pentagon. This point does not lay on the symmetry axes $s_i, i=1,\dots ,5.$ Removing point A from the boundary breaks the symmetry of the star, and Equation (5) is true. This is trivially demonstrated by dividing the star into five isosceles triangles and a pentagon and considering Lemma 1. Removing point A from the boundary of the star, which is located on one of its axes of symmetry, reduces the symmetry group from $D_5$ to $C_2$. Note that the five-point star depicted in Figure 5 is not a convex polygon but still a Jordan curve.

It is also noteworthy that the demonstrated Lemma does not work for the set of isolated vertices of regular polygons themselves. Removing one of the vertices does not reduce the symmetry of the set to the trivial $C_1$ group. Indeed, the vertices of regular polygons are located on their axes of symmetry.

2.2. Breaking the Symmetry of Curves

Now, consider the ellipse depicted in Figure 6. The ellipse is centered at the origin (or, more generally, with its axes aligned with coordinate axes).

The group of symmetry of an ellipse is the dihedral group $D_2$. An ellipse is centered at the origin. The ellipse has the following symmetries: reflection across the major axis $E{E}^{\prime }$; reflection across the minor axis $D{D}^{\prime }$ (see Figure 6); rotation by 180° around the center; and identity transformation. The order of the symmetry group is four. Removing a single point A from the ellipse breaks the symmetry of the curve, and Equation (5) is true:

$$D_2\to \left(1pr\right)\to C_1$$

(5)

We discuss this result below in the context of billiards.

Removing point A from the boundary of the ellipse, which is located on one of its axes of symmetry, reduces the symmetry group from $D_2$ to $C_2$.

It seems that the behavior of symmetrical curves resembles that of regular polygons; namely, removing a single point that does not belong to the axes of symmetry destroys the symmetry of the shape and reduces it to the trivial symmetry group $C_1.$ Thus, the suggested approach may be easily extended to symmetric Jordan curves. However, a remarkable exception exists, and this exception is a circle. The full symmetry group of a circle in the plane is called $O\left(2\right)$, an orthogonal group in 2D. It includes $SO\left(2\right),$ i.e., all rotations about the center (this is a Lie group, topologically a circle), and reflections, i.e., an infinite number of reflections across lines (diameters) through the center. This is an infinite, continuous group unlike the finite symmetry groups of regular polygons. Any point on a circle belongs to one of its axes of symmetry. Thus, two very different cases, as depicted in Figure 7, are distinguished. Inset A of Figure 7 illustrates a situation where A is removed from the circle. Thus, an open circle is formed. The symmetry group of the open circle is $C_2$ and not $C_1$, which is inherent for an open regular polygon and ellipse. Axes $D{D}^{\prime }$, as shown in Figure 7A, are the axes of symmetry of the open circle. It can be easily observed that even removing an arbitrary pair of points again reduces the $O\left(2\right)$ symmetry of a circle to $C_2$ symmetry; $C_2=\left\{e, R_{\pi }\right\}$. It is necessary to remove at least three non-symmetrically located points from the circle in order to reduce the $O\left(2\right)$ symmetry group to the trivial symmetry group $C_1$, as depicted in inset C of Figure 7. Additionally, it should be emphasized that, if a simple closed curve in the plane has the same symmetry group as the circle (i.e., $O\left(2\right)$), then it must be a circle. Thus, all symmetrical curves may be classified as follows according to the number of points that must be removed from them in order to reduce their symmetry to the $C_1$ group:

(i)

Symmetrical curves: Symmetry is reduced to the trivial $C_1$ group by removing a single point that does not belong to one of the axes of symmetry.

(ii)

Circles: Symmetry is reduced to the trivial $C_1$ group by removing a triad of non-symmetrical points. The same is true for solid and open circles.

Inset B of Figure 7 illustrates a situation where the axis of symmetry $D{D}^{\prime }$ is prefixed for a circle (thus, the $O\left(2\right)$ initial symmetry of the circle is reduced to $C_2$), and, afterwards, point A, which does not belong to the fixed axis $D{D}^{\prime },$ is removed. In this case, for a circle with axis $D{D}^{\prime }$ attached to the curve, we obtain an open circle characterized by the trivial symmetry group $C_1$.

2.3. Extension to Symmetrical Jordan Curves

It is possible to extend the suggested approach to symmetrical Jordan curves that are not circles. A formal proof is provided below:

Let $\mathsf{\gamma }\subset {\mathrm{R}}^2$ be a Jordan curve that is not a circle. Let $\mathrm{G}\subset \mathrm{I}\mathrm{s}\mathrm{o}\mathrm{m}{\mathrm{R}}^2$ be the symmetry group of $\mathsf{\gamma }$ assumed to be non-trivial. $A\in \mathsf{\gamma }$, and suppose that $A\notin$ Fix(g) for any nontrivial $\mathrm{g}\in \mathrm{G}$; i.e., A is not fixed under any symmetry. Consider ${\mathsf{\gamma }}^{\prime }=\mathsf{\gamma }\setminus \mathrm{A}$. We demonstrate that $\mathrm{S}\mathrm{y}\mathrm{m}\left({\mathsf{\gamma }}^{\prime }\right)={\mathrm{C}}_1$.

A Jordan curve $\mathsf{\gamma }$ can have (i) reflection symmetry with respect to a line l such that reflection ${\mathsf{\sigma }}_{\mathrm{l}}$ maps $\mathsf{\gamma }$ to itself, and it can have (ii) rotational symmetry about a center O such that rotation ${\mathrm{R}}_{\mathsf{\theta }}$ by angle $\mathsf{\theta }<2\mathsf{\pi }$ maps $\mathsf{\gamma }$ to itself. The group G of symmetries of the curve is finite (since a Jordan curve is compact) and acts as a finite subgroup of the Euclidean group ${\mathrm{E}}_2$. Thus $\mathrm{G}\subset {\mathrm{D}}_{\mathrm{n}}$ or $\mathrm{G}\subset {\mathrm{C}}_{\mathrm{n}}$ for some $\mathrm{n}\ge 2$.

Let $\mathrm{g}\in \mathrm{G}\setminus \left\{\mathrm{e}\right\}$. Then, g is a nontrivial isometry such that $\mathrm{g}\left(\mathsf{\gamma }\right)=\mathsf{\gamma }$. As $A\in \mathsf{\gamma }$, then $\mathrm{g}\left(A\right)\in \mathsf{\gamma }$, and as $A\notin \mathrm{F}\mathrm{i}\mathrm{x}\left(\mathrm{g}\right)$, it follows that $\mathrm{g}\left(A\right)\ne \mathrm{A}.$ Then,

$${\mathit{\gamma }}^{\prime }=\mathit{\gamma }\setminus \mathit{A}$$

$$\mathit{g}\left({\mathit{\gamma }}^{\prime }\right)=\mathit{g}\left(\mathit{\gamma }\setminus \mathit{A}\right)=\mathit{\gamma }\setminus \left\{\mathit{g}\left(\mathit{A}\right)\right\}$$

Therefore, $g\left({\gamma }^{\prime }\right)\ne {\gamma }^{\prime }$.

Because $g\left(A\right)\ne A$, ${\gamma }^{\prime }$ is missing point A, while $g\left({\gamma }^{\prime }\right)$ is missing $g\left(A\right)$. Hence,

$$\mathit{g}\left({\mathit{\gamma }}^{\prime }\right)\ne {\mathit{\gamma }}^{\prime }\to \mathit{g}\notin \mathit{S}\mathit{y}\mathit{m}\left({\mathit{\gamma }}^{\prime }\right)$$

(6)

This is true for all nontrivial $g\in G$. Thus, we conclude that

$$\mathit{S}\mathit{y}\mathit{m}\left({\mathit{\gamma }}^{\prime }\right)=\left\{\mathit{e}\right\}={\mathit{C}}_1$$

(7)

Thus, Theorem 2 is demonstrated.

Theorem 2.

Consider a symmetrical Jordan curve $\gamma$ that is not a circle. By removing point A from the symmetric Jordan curve $\gamma$ such that A is not fixed under any symmetry of the curve, the set $\gamma \setminus P$ loses all its symmetries, except for its identity symmetry. Hence, its symmetry group is the trivial group $C_1$.

The proof for an open or closed Jordan domain is trivial.

2.4. A 3D Generalization of Symmetry Breaking

The suggested approach may be generalized to 3D bodies. Consider a cube. If the orientation of the cube is preserved, then the symmetry group of the cube is $S_4$. Its order is 24, and it contains 24 distinct rotations. If we consider the full list of symmetries of a non-oriented cube, which implies all rigid motions (rotations + reflections + inversion), then the symmetry group is $S_4\times C_2$. It is usually denoted by $O_h$, and it contains 48 distinct elements. Removing one point anywhere on the surface or inside of the cube but not lying on any axis of rotation, mirror plane, or special point (face center, edge midpoint, vertex, or center) reduces the symmetry group to $C_1,$ regardless of the initial symmetry group $S_4$ (the orientation is preserved) or $O_h$ (all rigid motions of the cube).

Consider an oriented cube. If we remove the point on a face center (the center of one of the six faces), then the symmetry group is reduced to the group $C_4$ (the order of the group is four). Consider a non-oriented cube. If we remove the point on a face center, then the symmetry group will be reduced to the group $D_8$ (the order of the group is 8). We again consider an oriented cube. If we remove a single point on an edge midpoint, then the symmetry group will be reduced to the group $C_2$. If we perform the same procedure for a non-oriented group, then the symmetry group will be $D_2$. If we remove a single point at the cube center, then the initial symmetry of the group will not change.

We generalize the following: removing a single point from anywhere on the surface or inside of an oriented or non-oriented cube but not lying on any element of its symmetry reduces the symmetry group to $C_1$, regardless of the initial symmetry group. Thus, we extend the suggested approach to 3D objects.

2.5. Informational Interpretation of the Suggested Approach: Erasure of the Single Bit Enables Symmetry Breaking of the Entire String Program

Let us imagine a computer program represented by a symmetrical string of “zeros” and “ones”, such as that depicted in Figure 8.

It is sufficient to delete a single bit in order to break the symmetry of the entire symmetric program, as shown in Figure 8. This re-shaping of the suggested approach is important in view of its possible physical applications. According to the Landauer principle, the erasure of one bit of information in a computing device demands the minimal energy $W_{min}$, given by Equation (8), as follows:

$${\mathit{W}}_{\mathit{m}\mathit{i}\mathit{n}}={\mathit{k}}_{\mathit{B}}\mathit{T}\mathit{l}\mathit{n}\mathbf{2},$$

(8)

where ${\mathrm{k}}_{\mathrm{B}}$ is the Boltzmann constant [17,18,19,20,21,22,23,24,25,26,27,28]. Thus, minimal symmetry breaking effort is established with the Landauer limit and supplied with Equation (8). Moreover, it has recently been demonstrated that the Landauer principle emerges from the underlying second law bound, as formulated by Kelvin, as a consequence of the spontaneous symmetry breaking that accompanies logical irreversibility [28].

3. Discussion

Let us put the suggested approach in the context of modern mathematics. Symmetry may be seen as an automorphism, i.e., a transformation of an object that preserves its structure. Symmetry breaking leads to the reduction in automorphisms. Thus, the demonstrated theorem is similar to distinguishing numbers in graph theory or geometric structures, i.e., the minimum number of labels (or modifications) needed to destroy all nontrivial automorphisms. Removing a point is akin to labeling it differently. In graph theory, the distinguishing number of a graph is a concept related to breaking the symmetries (automorphisms) of the graph using vertex labels [29]. The distinguishing number $D\left(G\right)$ of a graph G is the smallest number of labels (colors) needed to label its vertices such that the only automorphism that preserves the labeling is the identity automorphism (i.e., does nothing) [29]. We, in turn, introduce the distinguishing number of the curve/domain $\mathcal{L}$, i.e., the smallest number of points to be removed from the curve/domain $\mathcal{L}$ such that the only automorphism that preserves its initial shape is the identity automorphism. The distinguishing number of the curve/domain captures how symmetric a curve/domain is: a lower distinguishing number means that it is easier to break its symmetry. We demonstrated that the curve/domain whose symmetry is the hardest to break is a circle, possessing an infinite set of symmetry axes. It is necessary to remove three asymmetric points in order to break the symmetry of a circle. Within the class of regular polygons and symmetric Jordan curves examined in this work, we did not identify any curve for which at least two removed boundary points broke all symmetries while one point did not. This remains an open issue for more general curves. In a purely mathematical sense, the demonstrated theorem is important for understanding moduli spaces, where small defects change equivalence classes [30].

Now, let us discuss the physical applications of the theorem. The demonstrated theorem has a direct relation to the concept of modern physics, which is called spontaneous symmetry breaking, where a tiny local change (the removal of a point) destroys the global symmetry [31]. In dynamical systems, even small asymmetries (such as removing a point) can destroy the integrals of motion related to symmetries [32]. Removing a point from a symmetric geometric object can be regarded as introducing a localized perturbation that explicitly breaks the symmetry of the system. In a previous study, a particle moving inside an elliptical billiard was addressed [33]. Within elliptical billiard trajectories, conserving the sum of distances to the two foci (related to the separability of the Hamilton–Jacobi equation in elliptical coordinates) yields a second integral of motion besides energy. Removing a single point from the elliptical boundary (introducing a small scatterer or indentation) destroys the symmetry. Thus, we destroy separability in elliptical coordinates, and the second integral of motion is no longer conserved.

Perhaps the most important aspect of symmetry in theories of physics is the idea that the states of a system do not need to have the same symmetries as the theory that describes them [31]. A system usually contains domains in which symmetry is broken. Sometimes, it is broken spontaneously. Such a spontaneous breakdown of symmetries governs the dynamics of phase transitions, the emergence of new particles and excitations, and the rigidity of the collective states of matter, and it is one of the main ways classical physics emerges in a quantum world [31]. It was noted that the limit of an infinite system size, often called “the thermodynamic limit” in the literature, is not a mandatory condition for spontaneous symmetry breaking to occur in practice. Additionally, an even stronger principle should be clearly understood: for almost all realistic applications of symmetry breaking theory, it is a rather useless limit in the sense that it is never exactly realized in nature [31]. Even in situations where the object of interest can be considered large, the coherence length of ordered phases is generically small, and a single domain (say “phase”) cannot, in good faith, be considered to approximate any sort of infinite size [31]. Thus, a reasonable question is as follows: what is the minimal effort necessary to break the symmetry of the phase structural unit? The present study addresses this question and demonstrates that eliminating a single point that is not located on the axis of symmetry of a Jordan curve/domain reduces its symmetry to the trivial $C_1$ group when the curve is not a circle.

In real crystals, a defect (a vacancy or a single point) in the lattice breaks the symmetry and alters the physical behavior [34]. In this case, we deal with a set of vertices of regular polygons, and the symmetry is decreased, but it is not always reduced to the trivial group $C_1$, as discussed above. A vacancy defect (an atom missing from a lattice site) typically reduces the space group to a subgroup that preserves the defect. If the vacancy is at a generic position (not on any symmetry axis or plane), then the symmetry is fully broken to the trivial group $C_1$. In condensed matter physics, removing a lattice site can change the conductivity or magnetic response [35]. Removing symmetry (even at a small scale) introduces mode coupling or splitting in vibrational modes [36]. The fact that the absence of a single feature (point) can reduce the symmetry classification drastically is of primary importance in image analysis, computer vision, and shape recognition [37].

The challenging goal to be addressed in future investigations is the extension of the introduced approach to non-Jordan curves. For a non-Jordan “figure-eight curve” (also called a lemniscate or self-intersecting loop), the extension is trivial. However, the general case of symmetric non-Jordan curves/domains looks challenging.

4. Conclusions

This study estimates the minimal “effort” necessary to break the symmetry of geometrical shapes. The concept of symmetry breaking is one of the most profound and far-reaching ideas in both mathematics and physics. Symmetry breaking is foundational; it explains how complex structures and phenomena emerge from simple laws. Spontaneous symmetry breaking is at the heart of the Standard Model. Without symmetry breaking, mass would not exist in the way that we observe it [38]. Thermodynamic phase transitions are classic cases of symmetry breaking [7]. During the Big Bang, the Universe likely underwent a series of symmetry-breaking events, with these transitions shaping its structure and content [39]. Symmetry breaking bridges perfect physical laws and the imperfect, observable world. Symmetry breaking is a mechanism of emergence, and it explains how simpler principles give rise to rich, varied phenomena. This study focuses on the mathematical aspects of symmetry breaking, and it poses the following fundamental question: how many points need to be removed from a symmetrical shape (a curve or domain) to break its symmetry? In other words, what is the minimal “effort” necessary to break the symmetry of curves and shapes (open or closed)? It is demonstrated that removing a single point from the boundary of regular convex and non-convex polygons and symmetrical Jordan curves reduces the symmetry group of the polygon to the trivial $C_1$ group when the eliminated point does not belong to the axis of symmetry of the polygon. The same is true for solid and open 2D regular convex polygons. The only and remarkable exception is a circle. Removing a single point from the boundary of a circle creates a curve characterized by the $C_2$ group. The symmetry of circles is reduced to the trivial $C_1$ group by removing a triad of non-symmetrical points. Thus, the “effort” necessary to break the symmetry of a circle is maximal. The same is true for solid and open circles. A 3D generalization of the theorem is demonstrated. Thus, the classification of symmetrical curves/domains following the minimal number of points necessary to break their symmetry becomes possible. The demonstrated theorem shows that the symmetry group action on curves and shapes becomes trivial when a point asymmetric perturbation is introduced. The introduction of symmetry breaking is accompanied by topological metamorphosis.

The demonstrated theorem is applicable to analyses of the modes of vibration in crystals (the removed point corresponds to a vacancy in a crystal). In dynamical systems, even small asymmetries (such as removing a single point) can destroy the integrals of motion tied to symmetries. In this study, an informational interpretation of the demonstrated theorem, which is related to the Landauer principle, enabling the estimation of the minimal energy necessary for symmetry breaking within computations is provided. Future research directions are envisaged. Symmetric Jordan curves, for which the minimal number of points that must be eliminated to achieve symmetry breaking is two, remain to be further explored.