## Section12.2Symmetry

An isometry or rigid motion in $${\mathbb R}^n$$ is a distance-preserving function $$f$$ from $${\mathbb R}^n$$ to $${\mathbb R}^n\text{.}$$ This means that $$f$$ must satisfy

\begin{equation*} \| f({\mathbf x}) - f({\mathbf y}) \| =\|{\mathbf x} - {\mathbf y} \| \end{equation*}

for all $${\mathbf x}, {\mathbf y} \in {\mathbb R}^n\text{.}$$ It is not difficult to show that $$f$$ must be a one-to-one map. By Theorem 12.8, any element in $$O(n)$$ is an isometry on $${\mathbb R}^n\text{;}$$ however, $$O(n)$$ does not include all possible isometries on $${\mathbb R}^n\text{.}$$ Translation by a vector $${\mathbf x}\text{,}$$ $$T_{\mathbf y}({\mathbf x}) = {\mathbf x} + {\mathbf y}$$ is also an isometry (Figure 12.11); however, $$T$$ cannot be in $$O(n)$$ since it is not a linear map.

We are mostly interested in isometries in $${\mathbb R}^2\text{.}$$ In fact, the only isometries in $${\mathbb R}^2$$ are rotations and reflections about the origin, translations, and combinations of the two. For example, a glide reflection is a translation followed by a reflection (Figure 12.12). In $${\mathbb R}^n$$ all isometries are given in the same manner. The proof is very easy to generalize.

Let $$f$$ be an isometry in $${\mathbb R}^2$$ fixing the origin. We will first show that $$f$$ preserves inner products. Since $$f(0) = 0\text{,}$$ $$\| f({\mathbf x})\| = \| {\mathbf x} \|\text{;}$$ therefore,

\begin{align*} \| {\mathbf x} \|^2 - 2 \langle f({\mathbf x}), f({\mathbf y}) \rangle + \| {\mathbf y} \|^2 & = \| f({\mathbf x}) \|^2 - 2 \langle f({\mathbf x}), f({\mathbf y}) \rangle + \| f({\mathbf y}) \|^2\\ & = \langle f({\mathbf x}) - f({\mathbf y}), f({\mathbf x}) - f({\mathbf y}) \rangle\\ & = \| f({\mathbf x}) - f({\mathbf y}) \|^2\\ & = \| {\mathbf x} - {\mathbf y} \|^2\\ & = \langle {\mathbf x} - {\mathbf y}, {\mathbf x} - {\mathbf y} \rangle\\ & = \| {\mathbf x} \|^2 - 2 \langle {\mathbf x}, {\mathbf y} \rangle + \| {\mathbf y} \|^2\text{.} \end{align*}

Consequently,

\begin{equation*} \langle f({\mathbf x}), f({\mathbf y}) \rangle = \langle {\mathbf x}, {\mathbf y} \rangle\text{.} \end{equation*}

Now let $${\mathbf e}_1$$ and $${\mathbf e_2}$$ be $$(1, 0)^\transpose$$ and $$(0, 1)^\transpose\text{,}$$ respectively. If

\begin{equation*} {\mathbf x} = (x_1, x_2) = x_1 {\mathbf e}_1 + x_2 {\mathbf e}_2\text{,} \end{equation*}

then

\begin{equation*} f({\mathbf x}) = \langle f({\mathbf x}), f({\mathbf e}_1) \rangle f({\mathbf e}_1) + \langle f({\mathbf x}), f({\mathbf e}_2) \rangle f({\mathbf e}_2) = x_1 f({\mathbf e}_1)+x_2 f({\mathbf e}_2)\text{.} \end{equation*}

The linearity of $$f$$ easily follows.

For any arbitrary isometry, $$f\text{,}$$ $$T_{\mathbf x} f$$ will fix the origin for some vector $${\mathbf x}$$ in $${\mathbb R}^2\text{;}$$ hence, $$T_{\mathbf x} f({\mathbf y}) = A {\mathbf y}$$ for some matrix $$A \in O(2)\text{.}$$ Consequently, $$f({\mathbf y}) = A {\mathbf y} + {\mathbf x}\text{.}$$ Given the isometries

\begin{align*} f({\mathbf y}) & = A {\mathbf y} + {\mathbf x}_1\\ g({\mathbf y}) & = B {\mathbf y} + {\mathbf x}_2\text{,} \end{align*}

their composition is

\begin{equation*} f(g({\mathbf y})) = f(B {\mathbf y} + {\mathbf x}_2) = AB {\mathbf y} + A{\mathbf x}_2 + {\mathbf x}_1\text{.} \end{equation*}

This last computation allows us to identify the group of isometries on $${\mathbb R}^2$$ with $$E(2)\text{.}$$

A symmetry group in $${\mathbb R}^n$$ is a subgroup of the group of isometries on $${\mathbb R}^n$$ that fixes a set of points $$X \subset {\mathbb R}^n\text{.}$$ It is important to realize that the symmetry group of $$X$$ depends both on $${\mathbb R}^n$$ and on $$X\text{.}$$ For example, the symmetry group of the origin in $${\mathbb R}^1$$ is $${\mathbb Z}_2\text{,}$$ but the symmetry group of the origin in $${\mathbb R}^2$$ is $$O(2)\text{.}$$

We simply need to find all of the finite subgroups $$G$$ of $$E(2)\text{.}$$ Any finite symmetry group $$G$$ in $${\mathbb R}^2$$ must fix the origin and must be a finite subgroup of $$O(2)\text{,}$$ since translations and glide reflections have infinite order. By Example 12.10, elements in $$O(2)$$ are either rotations of the form

\begin{equation*} R_{\theta} = \begin{pmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \end{equation*}

or reflections of the form

\begin{equation*} T_{\phi} = \begin{pmatrix} \cos \phi & - \sin \phi \\ \sin \phi & \cos \phi \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} \cos \phi & \sin \phi \\ \sin \phi & - \cos \phi \end{pmatrix}\text{.} \end{equation*}

Notice that $$\det(R_{\theta})=1\text{,}$$ $$\det(T_{\phi})=-1\text{,}$$ and $$T_{\phi}^2=I\text{.}$$ We can divide the proof up into two cases. In the first case, all of the elements in $$G$$ have determinant one. In the second case, there exists at least one element in $$G$$ with determinant $$-1\text{.}$$

###### Case 1.

The determinant of every element in $$G$$ is one. In this case every element in $$G$$ must be a rotation. Since $$G$$ is finite, there is a smallest angle, say $$\theta_0\text{,}$$ such that the corresponding element $$R_{\theta_0}$$ is the smallest rotation in the positive direction. We claim that $$R_{\theta_0}$$ generates $$G\text{.}$$ If not, then for some positive integer $$n$$ there is an angle $$\theta_1$$ between $$n \theta_0$$ and $$(n+1) \theta_0\text{.}$$ If so, then $$(n+1) \theta_0 - \theta_1$$ corresponds to a rotation smaller than $$\theta_0\text{,}$$ which contradicts the minimality of $$\theta_0\text{.}$$

###### Case 2.

The group $$G$$ contains a reflection $$T\text{.}$$ The kernel of the homomorphism $$\phi : G \rightarrow \{-1, 1\}$$ given by $$A \mapsto \det(A)$$ consists of elements whose determinant is 1. Therefore, $$|G/ \ker \phi|=2\text{.}$$ We know that the kernel is cyclic by the first case and is a subgroup of $$G$$ of, say, order $$n\text{.}$$ Hence, $$|G| = 2n\text{.}$$ The elements of $$G$$ are

\begin{equation*} R_{\theta}, \ldots, R_{\theta}^{n-1}, TR_{\theta}, \ldots, TR_{\theta}^{n-1}\text{.} \end{equation*}

These elements satisfy the relation

\begin{equation*} TR_{\theta}T = R_{\theta}^{-1}\text{.} \end{equation*}

Consequently, $$G$$ must be isomorphic to $$D_n$$ in this case.

### SubsectionThe Wallpaper Groups

Suppose that we wish to study wallpaper patterns in the plane or crystals in three dimensions. Wallpaper patterns are simply repeating patterns in the plane (Figure 12.16). The analogs of wallpaper patterns in $${\mathbb R}^3$$ are crystals, which we can think of as repeating patterns of molecules in three dimensions (Figure 12.17). The mathematical equivalent of a wallpaper or crystal pattern is called a lattice.

Let us examine wallpaper patterns in the plane a little more closely. Suppose that $${\mathbf x}$$ and $${\mathbf y}$$ are linearly independent vectors in $${\mathbb R}^2\text{;}$$ that is, one vector cannot be a scalar multiple of the other. A lattice of $${\mathbf x}$$ and $${\mathbf y}$$ is the set of all linear combinations $$m {\mathbf x} + n {\mathbf y}\text{,}$$ where $$m$$ and $$n$$ are integers. The vectors $${\mathbf x}$$ and $${\mathbf y}$$ are said to be a basis for the lattice.

Notice that a lattice can have several bases. For example, the vectors $$(1,1)^\transpose$$ and $$(2,0)^\transpose$$ have the same lattice as the vectors $$(-1, 1)^\transpose$$ and $$(-1, -1)^\transpose$$ (Figure 12.18). However, any lattice is completely determined by a basis. Given two bases for the same lattice, say $$\{ {\mathbf x}_1, {\mathbf x}_2 \}$$ and $$\{ {\mathbf y}_1, {\mathbf y}_2 \}\text{,}$$ we can write

\begin{align*} {\mathbf y}_1 & = \alpha_1 {\mathbf x}_1 + \alpha_2 {\mathbf x}_2\\ {\mathbf y}_2 & = \beta_1 {\mathbf x}_1 + \beta_2 {\mathbf x}_2\text{,} \end{align*}

where $$\alpha_1\text{,}$$ $$\alpha_2\text{,}$$ $$\beta_1\text{,}$$ and $$\beta_2$$ are integers. The matrix corresponding to this transformation is

\begin{equation*} U = \begin{pmatrix} \alpha_1 & \alpha_2 \\ \beta_1 & \beta_2 \end{pmatrix}\text{.} \end{equation*}

If we wish to give $${\mathbf x}_1$$ and $${\mathbf x}_2$$ in terms of $${\mathbf y}_1$$ and $${\mathbf y}_2\text{,}$$ we need only calculate $$U^{-1}\text{;}$$ that is,

\begin{equation*} U^{-1} \begin{pmatrix} {\mathbf y}_1 \\ {\mathbf y}_2 \end{pmatrix} = \begin{pmatrix} {\mathbf x}_1 \\ {\mathbf x}_2 \end{pmatrix}\text{.} \end{equation*}

Since $$U$$ has integer entries, $$U^{-1}$$ must also have integer entries; hence the determinants of both $$U$$ and $$U^{-1}$$ must be integers. Because $$U U^{-1} = I\text{,}$$

\begin{equation*} \det(U U^{-1}) =\det(U) \det( U^{-1}) = 1; \end{equation*}

consequently, $$\det(U) = \pm 1\text{.}$$ A matrix with determinant $$\pm 1$$ and integer entries is called unimodular. For example, the matrix

\begin{equation*} \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} \end{equation*}

is unimodular. It should be clear that there is a minimum length for vectors in a lattice.

We can classify lattices by studying their symmetry groups. The symmetry group of a lattice is the subgroup of $$E(2)$$ that maps the lattice to itself. We consider two lattices in $${\mathbb R}^2$$ to be equivalent if they have the same symmetry group. Similarly, classification of crystals in $${\mathbb R}^3$$ is accomplished by associating a symmetry group, called a space group, with each type of crystal. Two lattices are considered different if their space groups are not the same. The natural question that now arises is how many space groups exist.

A space group is composed of two parts: a translation subgroup and a point. The translation subgroup is an infinite abelian subgroup of the space group made up of the translational symmetries of the crystal; the point group is a finite group consisting of rotations and reflections of the crystal about a point. More specifically, a space group is a subgroup of $$G \subset E(2)$$ whose translations are a set of the form $$\{ (I, t) : t \in L \}\text{,}$$ where $$L$$ is a lattice. Space groups are, of course, infinite. Using geometric arguments, we can prove the following theorem (see [5] or [6]).

The point group of $$G$$ is $$G_0 = \{A : (A,b) \in G \text{ for some } b \}\text{.}$$ In particular, $$G_0$$ must be a subgroup of $$O(2)\text{.}$$ Suppose that $${\mathbf x}$$ is a vector in a lattice $$L$$ with space group $$G\text{,}$$ translation group $$H\text{,}$$ and point group $$G_0\text{.}$$ For any element $$(A, {\mathbf y})$$ in $$G\text{,}$$

\begin{align*} (A, {\mathbf y}) (I, {\mathbf x}) (A, {\mathbf y})^{-1} & = (A,A {\mathbf x} + {\mathbf y}) (A^{-1},-A^{-1} {\mathbf y})\\ & = (A A^{-1},-A A^{-1} {\mathbf y} + A {\mathbf x} + {\mathbf y})\\ & = (I, A {\mathbf x}); \end{align*}

hence, $$(I, A {\mathbf x})$$ is in the translation group of $$G\text{.}$$ More specifically, $$A {\mathbf x}$$ must be in the lattice $$L\text{.}$$ It is important to note that $$G_0$$ is not usually a subgroup of the space group $$G\text{;}$$ however, if $$T$$ is the translation subgroup of $$G\text{,}$$ then $$G/T \cong G_0\text{.}$$ The proof of the following theorem can be found in [2], [5], or [6].

To answer the question of how the point groups and the translation groups can be combined, we must look at the different types of lattices. Lattices can be classified by the structure of a single lattice cell. The possible cell shapes are parallelogram, rectangular, square, rhombic, and hexagonal (Figure 12.21). The wallpaper groups can now be classified according to the types of reflections that occur in each group: these are ordinarily reflections, glide reflections, both, or none.

The 17 wallpaper groups are listed in Table 12.22. The groups p3m1 and p31m can be distinguished by whether or not all of their threefold centers lie on the reflection axes: those of p3m1 must, whereas those of p31m may not. Similarly, the fourfold centers of p4m must lie on the reflection axes whereas those of p4g need not (Figure 12.24). The complete proof of this theorem can be found in several of the references at the end of this chapter, including [5], [6], [10], and [11].

### SubsectionHistorical Note

Symmetry groups have intrigued mathematicians for a long time. Leonardo da Vinci was probably the first person to know all of the point groups. At the International Congress of Mathematicians in 1900, David Hilbert gave a now-famous address outlining 23 problems to guide mathematics in the twentieth century. Hilbert's eighteenth problem asked whether or not crystallographic groups in $$n$$ dimensions were always finite. In 1910, L. Bieberbach proved that crystallographic groups are finite in every dimension. Finding out how many of these groups there are in each dimension is another matter. In $${\mathbb R}^3$$ there are 230 different space groups; in $${\mathbb R}^4$$ there are 4783. No one has been able to compute the number of space groups for $${\mathbb R}^5$$ and beyond. It is interesting to note that the crystallographic groups were found mathematically for $${\mathbb R}^3$$ before the 230 different types of crystals were actually discovered in nature.