3.1: Space Groups

Space groups identify the possible ways to describe the rotational and translational symmetry of crystalline structures in real space. As we have seen, these aspects of 3-d crystalline symmetry are separately described by 32 crystallographic point groups and 14 Bravais lattices. For any space group, these two types of symmetry must be compatible with each other. For example, consider a 2-d tetragonal lattice, which demands a unit cell with a = b and γ = 90°, decorated with different arrangements of 4-atom molecules (A = red circles; B = yellow circles).

Figure 3.1: Five different 2-d tetragonal cells with different patterns of square-shaped molecules.

If the atomic arrangement has four-fold symmetry at each lattice point, then the crystalline symmetry will retain tetragonal characteristics, i.e., four-fold rotational symmetry at the corners and center of each unit cell. Among these five structures, (a), (c), and (e) are tetragonal. Cases (a) and (c) are constructed of square molecules A₄, whereas case (e) involves a specific packing of A₂B₂ molecules that creates four-fold symmetry between four distinct molecules. Combining the 32 crystallographic point groups and 14 Bravais lattices for 3-d real space in all compatible ways generates 230 space groups. For 2-d space, there are 17 plane groups generated from 10 point groups and 5 Bravais lattices.

Rotation-Translation Operations

Every member of a space group is a rotation-translation operation, which can be represented by augmented matrices and symbolized by the Seitz notation, which facilitates evaluating combinations of these operations. Accordingly, every rotation-translation operation is symbolized by \(\left( R \middle| \boldsymbol{\tau} \right)\), such that R is a proper or improper rotation and τ is a displacement. The effect of this operation on a position vector r is

\[\left( R \middle| \boldsymbol{\tau} \right)\boldsymbol{\ r} = R\boldsymbol{r} + \boldsymbol{\tau}= \begin{pmatrix} R_{11} & R_{12} & R_{13} \\ R_{21} & R_{22} & R_{23} \\ R_{31} & R_{32} & R_{33} \\ \end{pmatrix}\boldsymbol{r} + \begin{pmatrix} \tau_{1} \\ \tau_{2} \\ \tau_{3} \\ \end{pmatrix} = \begin{pmatrix} \begin{matrix} R_{11} & R_{12} \\ R_{21} & R_{22} \\ \end{matrix} & \begin{matrix} R_{13} & \tau_{1} \\ R_{23} & \tau_{2} \\ \end{matrix} \\ \begin{matrix} R_{31} & R_{32} \\ 0 & 0 \\ \end{matrix} & \begin{matrix} R_{33} & \tau_{3} \\ 0 & 1 \\ \end{matrix} \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ 1 \\ \end{pmatrix} \nonumber \]

If \(\boldsymbol{\tau}\) is a lattice translation \(\boldsymbol{T}\), then \(\left( E \middle| \boldsymbol{T} \right) = \left( 1 \middle| \boldsymbol{T} \right)\) symbolizes a lattice translation:

\[\small{\left( 1 \middle| \boldsymbol{T} \right)\boldsymbol{r} = \boldsymbol{r} + \boldsymbol{T}} \nonumber \]

Also, the sequential product of two operations is:

\[\left( R_{2} \middle| \boldsymbol{\tau}_{2} \right)\left( R_{1} \middle| \boldsymbol{\tau}_{1} \right)\boldsymbol{\ r} = \left( R_{2} \middle| \boldsymbol{\tau}_{2} \right)\left\lbrack R_{1}\boldsymbol{r}+\boldsymbol{\tau}_{1} \right\rbrack = R_{2}\left\lbrack R_{1}\boldsymbol{r} +\boldsymbol{\tau}_{1} \right\rbrack + \boldsymbol{\tau}_{2} = R_{2}R_{1}\boldsymbol{r} +\left( R_{2}\boldsymbol{\tau}_{1} +\boldsymbol{\tau}_{2} \right) \nonumber \]

Note that the sequence in the product proceeds right-to-left. Therefore,

\[\left( R_{2} \middle| \boldsymbol{\tau}_{2} \right)\left( R_{1} \middle| \boldsymbol{\tau}_{1} \right) = \left( R_{2}R_{1} \middle| R_{2}\boldsymbol{\tau}_{1} + \boldsymbol{\tau}_{2} \right) \nonumber \]

defines the product of two rotation-translation operations. Let \(\mathcal{G =}\left\{ \left( R_{1} \middle| \boldsymbol{\tau}_{1} \right),\left( R_{2} \middle| \boldsymbol{\tau}_{2} \right),\ldots \right\}\) a set of these operations. For \(\mathcal{G}\) to be a space group, this set must satisfy the definitions of a group:

1. Closure under multiplication: \(\left( R_{2} \middle| \boldsymbol{\tau}_{2} \right)\left( R_{1} \middle| \boldsymbol{\tau}_{1} \right) = \left( R_{2}R_{1} \middle| R_{2}\boldsymbol{\tau}_{1} + \boldsymbol{\tau}_{2} \right)\) must also be a member of \(\mathcal{G}\). The product of two rotations \(R_{2}R_{1}\) is another rotation (Euler’s theorem) and the vector \(R_{2}\boldsymbol{\tau}_{1} + \boldsymbol{\tau}_{2}\) is a displacement, so the new operation is another rotation-translation operation. There may be certain constraints placed on the allowed displacements depending on the nature of the rotational symmetry. We will examine these constraints shortly.
2. Existence of an identity operation: \(\left( E \middle| \mathbf{0} \right) = \left( 1 \middle| \mathbf{0} \right)\) is a member of \(\mathcal{G}\) such that

   \[\left( E \middle| \boldsymbol{0} \right)\ \boldsymbol{r} = E\boldsymbol{r} + \boldsymbol{0} = \boldsymbol{r} \nonumber \]

3. Multiplication is associative: \(\left( R_{3} \middle| \boldsymbol{\tau}_{3} \right)\left\lbrack \left( R_{2} \middle| \boldsymbol{\tau}_{2} \right)\left( R_{1} \middle| \boldsymbol{\tau}_{1} \right) \right\rbrack = \left\lbrack \left( R_{3} \middle| \boldsymbol{\tau}_{3} \right)\left( R_{2} \middle| \boldsymbol{\tau}_{2} \right) \right\rbrack\left( R_{1} \middle| \boldsymbol{\tau}_{1} \right)\). We show this by considering the two different products of the three sequential operations:

   \[\small{\left( R_{3} \middle| \boldsymbol{\tau}_{3} \right)\left\lbrack \left( R_{2} \middle| \boldsymbol{\tau}_{2} \right)\left( R_{1} \middle| \boldsymbol{\tau}_{1} \right) \right\rbrack = \left( R_{3} \middle| \boldsymbol{\tau}_{3} \right)\left( R_{2}R_{1} \middle| R_{2}\boldsymbol{\tau}_{1} + \boldsymbol{\tau}_{2} \right) = \left( R_{3}R_{2}R_{1} \middle| R_{3}{R_{2}\boldsymbol{\tau}_{1} + R_{3}\boldsymbol{\tau}}_{2} +\boldsymbol{\tau}_{3} \right)} \nonumber \]

   \[\small{\left\lbrack \left( R_{3} \middle| \boldsymbol{\tau}_{3} \right)\left( R_{2} \middle| \boldsymbol{\tau}_{2} \right) \right\rbrack\left( R_{1} \middle| \boldsymbol{\tau}_{1} \right) = \left( R_{3}R_{2} \middle| R_{3}\boldsymbol{\tau}_{2} + \boldsymbol{\tau}_{3} \right)\left( R_{1} \middle| \boldsymbol{\tau}_{1} \right) = \left( R_{3}R_{2}R_{1} \middle| R_{3}R_{2}\boldsymbol{\tau}_{1} + {R_{3}\boldsymbol{\tau}}_{2} +\boldsymbol{\tau}_{3} \right)} \nonumber \]

   To verify this assertion, it is very important to maintain the correct order of operations.

4. Existence of an inverse for every operation: Let

   \(\left( R_{1} \middle| \boldsymbol{\tau}_{1} \right) = \left( R \middle| \boldsymbol{\tau} \right)^{- 1}\). Then

   \(\left( R \middle| \boldsymbol{\tau} \right)^{- 1}\left( R \middle| \boldsymbol{\tau} \right) = \left( R_{1} \middle| \boldsymbol{\tau}_{1} \right)\left( R \middle| \boldsymbol{\tau} \right) = \left( R_{1}R \middle| R_{1}\boldsymbol{\tau} + \mathbf{\tau}_{1} \right) = \left( E \middle| \boldsymbol{0} \right).\)

   As a result,

   \(R_{1} = R^{- 1}\) and \(\boldsymbol{\tau}_{1} = -R^{-1}\boldsymbol{\tau}\). Therefore,

   \(\left( R \middle| \boldsymbol{\tau} \right)^{- 1} = \left( R^{-1} \middle| -R^{-1}\boldsymbol{\tau} \right)\).

What specific rotations \(R\) and displacements \(\mathbf{\tau}\) are possible for any general rotation-translation operation \(\left( R \middle| \mathbf{\tau} \right)\) of a space group \(\mathcal{G}\)?

Allowed Rotations (\(R\))

The set of Bravais lattice vectors describes the translational periodicity of the space group. Each Bravais lattice vector corresponds to an operation \(\left( 1 \middle| \boldsymbol{T} \right)\) in \(\mathcal{G}\). Therefore, if \(\left( R \middle| \boldsymbol{\tau} \right)\) is a member of space group\(\mathcal{\ G}\), then so is the similarity transformation

\[\small{\left( R \middle| \boldsymbol{\tau} \right)\left( 1 \middle| \boldsymbol{T} \right)\left( R \middle| \boldsymbol{\tau} \right)^{- 1} = \left( R \middle| \boldsymbol{\tau} \right)\left( R^{- 1} \middle| \boldsymbol{T} - R^{- 1}\boldsymbol{\tau} \right) = \left( RR^{- 1} \middle| R\boldsymbol{T} - RR^{- 1}\boldsymbol{\tau} + \boldsymbol{\tau} \right) = \left( 1 \middle| R\boldsymbol{T} \right)} \nonumber \]

As a result, \(R\boldsymbol{T}\) is a Bravais lattice vector, so that \(R\) must be among the allowed rotations of 3-d lattices, i.e., proper rotations \(1,\ 2,\ 3,\ 4,\ 6\) and improper rotations \(\overline{1},\overline{2} = m,\overline{3},\ \overline{4},\overline{6}\).

This result also points out that the Bravais lattice operations \(\left( 1 \middle| R\boldsymbol{T} \right)\) form collections of operations that make up complete classes of the space group \(\mathcal{G}\). A subset of group operations that form complete classes and is a subgroup is called an invariant subgroup, a result that has profound impacts on the irreducible representations of the group. As an example, consider any centrosymmetric point group. Then, the subgroup \(\mathcal{C}_{i} = \left\{ E,i \right\}\) is an invariant subgroup because both the identity \(E\) and inversion \(i\) commute with every rotation operation. As a result, \(\left\{ E,i \right\}\) is formed by two complete classes for every centrosymmetric point group. To see the impact on irreducible representations (ir’s), the ir’s of every centrosymmetric point group are labeled as either symmetric (gerade) or antisymmetric (ungerade) with respect to inversion \(i\).

Allowed Translations (\(\boldsymbol{\tau}\))

For every rotation \(R = n\) or \(\overline{n}\), there is an integer N such that

\(R^{N} = \text{identity}\). For the proper rotation \(n\), N = n; for the improper rotation \(\overline{n}\), N = n (for n even) and N = 2n (for n odd). By closure, \(\left( R \middle| \boldsymbol{\tau} \right)^{N}\) is also a member of \(\mathcal{G}\). Expanding this expression

\[\small{\left( R \middle| \boldsymbol{\tau} \right)^{N} = \left( R \middle| \boldsymbol{\tau} \right)\left( R \middle| \boldsymbol{\tau} \right)\cdots\left( R \middle| \boldsymbol{\tau} \right) = \left( R \cdot R\cdots R \middle| \boldsymbol{\tau} + R\boldsymbol{\tau} + \cdots + R^{N - 1}\boldsymbol{\tau} \right) = \left( R^{N} \middle| \boldsymbol{\tau} + R\boldsymbol{\tau} + \cdots + R^{N - 1}\boldsymbol{\tau} \right)} \nonumber \]

\(\left( R \middle| \boldsymbol{\tau} \right)^{N} = \left( 1 \middle| \sum_{j = 1}^{N}{R^{j - 1}\boldsymbol{\tau}} \right)\).

Therefore, allowed translations τ must satisfy the condition that \(\sum_{j = 1}^{N}{R^{j - 1}\boldsymbol{\tau}}\) is a Bravais lattice vector. If \(R\) is the identity, then τ must be a Bravais lattice vector. Also, if τ is already a Bravais lattice vector, then this condition is satisfied. So, are there any other vectors τ for an allowed rotation \(R\) that satisfy this equation? Because \(R\) is a linear operator, we can rewrite \(\sum_{j = 1}^{N}{R^{j - 1}\boldsymbol{\tau}}\) as \(\left( \sum_{j = 1}^{N}R^{j - 1} \right)\)τ. The orientation of the rotation \(R\) is specified by the direction of its axis. To evaluate the operation in parentheses and without loss of generality, consider the rotation axis to be parallel to the c-axis of the unit cell and let \(\boldsymbol{\tau} = \tau_{1}\boldsymbol{a} + \tau_{2}\boldsymbol{b} + \tau_{3}\boldsymbol{c}\), such that vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are perpendicular to \(\boldsymbol{c}\). Now, evaluate this condition systematically for each allowed rotation:

For proper rotations \(n\) (matrices are written using the lattice vectors as the basis):

\[2_{\boldsymbol{c}} = \small{\begin{pmatrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}}: \nonumber \]

\[\left( \sum_{j = 1}^{2}2_{\boldsymbol{c}}^{j - 1} \right)\boldsymbol{\tau =}\left( 1 + 2_{\boldsymbol{c}} \right)\boldsymbol{\tau} = \small{\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 2 \\ \end{pmatrix}\begin{pmatrix} \tau_{1} \\ \tau_{2} \\ \tau_{3} \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 2\tau_{3} \\ \end{pmatrix}} = p\boldsymbol{c} = \left( 2\tau_{3} \right)\boldsymbol{c} \nonumber \]

• Distinct solutions: \(\tau_{3}=0,½\) (all other solutions are related by the lattice vector \(\mathbf{c}\));

\[3_{\mathbf{c}} = \small{\begin{pmatrix} 0 & - 1 & 0 \\ 1 & - 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}}:\)

\(\left( \sum_{j = 1}^{3}3_{\boldsymbol{c}}^{j - 1} \right)\boldsymbol{\tau =}\left( 1 + 3_{\boldsymbol{c}} + 3_{\boldsymbol{c}}^{2} \right)\boldsymbol{\tau} = \small{\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 3 \\ \end{pmatrix}\begin{pmatrix} \tau_{1} \\ \tau_{2} \\ \tau_{3} \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 3\tau_{3} \\ \end{pmatrix}} = p\boldsymbol{c} = \left( 3\tau_{3} \right)\boldsymbol{c} \nonumber \]

• Distinct solutions: \(\tau_{3} = 0,\ ⅓,\ ⅔\);

\[4_{\boldsymbol{c}} = \small{\begin{pmatrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}}: \nonumber \]

\[\left( \sum_{j = 1}^{4}4_{\boldsymbol{c}}^{j - 1} \right)\boldsymbol{\tau} = \small{\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 4 \\ \end{pmatrix}\begin{pmatrix} \tau_{1} \\ \tau_{2} \\ \tau_{3} \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 4\tau_{3} \\ \end{pmatrix}} = p\boldsymbol{c} = \left( 4\tau_{3} \right)\boldsymbol{c} \nonumber \]

• Distinct solutions: \(\tau_{3} = 0,\ ¼,\ ½,\ ¾\);

\[6_{\boldsymbol{c}} = \small{\begin{pmatrix} 1 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}}: \nonumber \]

\(\left( \sum_{j = 1}^{6}6_{\boldsymbol{c}}^{j - 1} \right)\boldsymbol{\tau} = \small{\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 6 \\ \end{pmatrix}\begin{pmatrix} \tau_{1} \\ \tau_{2} \\ \tau_{3} \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 6\tau_{3} \\ \end{pmatrix}} = p\boldsymbol{c} = \left( 6\tau_{3} \right)\boldsymbol{c}\)

• Distinct solutions: \(\tau_{3} = 0,\ ⅙,\ ⅓,\ ½,\ ⅔,\ ⅚\).

Therefore, for a proper rotation \(n\), the allowed translations that are not lattice vectors are all rational fractions \(\frac{j}{n}\ (j = 1,\ldots,n - 1)\) of the smallest lattice vector parallel to the rotation axis. These operations are called screw rotations. Translations perpendicular to the rotation axis simply set the rotation away from the origin.

For improper rotations \(\overline{n}\) (matrices written using the lattice vectors as the basis):

\[\overline{1} = \small{\begin{pmatrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \\ \end{pmatrix}}\): \(\left( \sum_{j = 1}^{2}{\overline{1}}^{j - 1} \right)\boldsymbol{\tau} = \left( 1 + \overline{1} \right)\boldsymbol{\tau} = \small{\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}\begin{pmatrix} \tau_{1} \\ \tau_{2} \\ \tau_{3} \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix}} = \boldsymbol{0}\

• (No solutions)

\[{\overline{2}}_{\boldsymbol{c}} = \small{\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \\ \end{pmatrix}}\): \(\left( \sum_{j = 1}^{2}{\overline{2}}_{\boldsymbol{c}}^{j - 1} \right)\boldsymbol{\tau} = \small{\begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}\begin{pmatrix} \tau_{1} \\ \tau_{2} \\ \tau_{3} \\ \end{pmatrix} = \begin{pmatrix} 2\tau_{1} \\ 2\tau_{2} \\ 0 \\ \end{pmatrix}} = m\boldsymbol{a} + n\boldsymbol{b} = \left( 2\tau_{1} \right)\boldsymbol{a +}\left( 2\tau_{2} \right)\boldsymbol{b} \nonumber \]

• Distinct solutions: \(\tau_{1} = 0, ½\) and \(\tau_{2} = 0,\ ½\).

\[{\overline{3}}_{\boldsymbol{c}} = \small{\begin{pmatrix} 0 & 1 & 0 \\ - 1 & 1 & 0 \\ 0 & 0 & - 1 \\ \end{pmatrix}}\): \(\left( \sum_{j = 1}^{6}{\overline{3}}_{\boldsymbol{c}}^{j - 1} \right)\boldsymbol{\tau} = \small{\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}\begin{pmatrix} \tau_{1} \\ \tau_{2} \\ \tau_{3} \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix}} = \boldsymbol{0} \nonumber \]

• (No solutions)

\[{\overline{4}}_{\boldsymbol{c}} = \small{\begin{pmatrix} 0 & 1 & 0 \\ - 1 & 1 & 0 \\ 0 & 0 & - 1 \\ \end{pmatrix}}\): \(\left( \sum_{j = 1}^{4}{\overline{4}}_{\boldsymbol{c}}^{j - 1} \right)\boldsymbol{\tau} = \small{\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}\begin{pmatrix} \tau_{1} \\ \tau_{2} \\ \tau_{3} \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix}} = \boldsymbol{0} \nonumber \]

• (No solutions)

\[{\overline{6}}_{\boldsymbol{c}} = \small{\begin{pmatrix} - 1 & 1 & 0 \\ - 1 & 0 & 0 \\ 0 & 0 & - 1 \\ \end{pmatrix}}\): \(\left( \sum_{j = 1}^{6}{\overline{6}}_{\boldsymbol{c}}^{j - 1} \right)\boldsymbol{\tau} = \small{\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}\begin{pmatrix} \tau_{1} \\ \tau_{2} \\ \tau_{3} \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix}} = \boldsymbol{0} \nonumber \]

• (No solutions)

Only reflections \(\overline{2} = m\) allow translations that are not lattice vectors. These displacements are parallel to the reflection plane, i.e., perpendicular to the improper \(\overline{2}\) axis, and one-half the length of a lattice vector. These operations are called glide reflections. Translations perpendicular to the reflection plane set the reflection away from the origin. Since the other improper rotations occur with respect to a single point in space, any translation will set this point away from the origin.