3.3: Glide Reflections
- Page ID
- 474767
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)A glide reflection, symbolized by a letter other than \(m\) (see below), is a reflection \(m = \overline{2}\) accompanied by a displacement that is one-half a lattice translation parallel to the plane of the reflection. The general Seitz symbol is \(\left( m_{\boldsymbol{T}} \middle| \frac{\boldsymbol{T'}}{2} \right)\), in which \(\boldsymbol{T}\) is the lattice direction perpendicular to the reflection plane and \(\frac{\boldsymbol{T'}}{2}\) is one-half the lattice vector parallel to the plane. Therefore, \(\boldsymbol{T'}\) is perpendicular to \(\boldsymbol{T}\). There are three distinct types of glide reflections depending on the direction of the displacement with respect to the unit cell vectors:
Axial glides
(a) Axial glides, symbolized as \(a,b,\) or \(c\), have displacements parallel to the unit cell side specified by the letter. For example, an \(a\)-glide perpendicular to the c-direction, which can be symbolized as \(a_{\boldsymbol{c}} = \left( m_{\boldsymbol{c}} \middle| \frac{\boldsymbol{a}}{2} \right)\), is a reflection perpendicular to \(\boldsymbol{c}\) along with displacement of \(\frac{\boldsymbol{a}}{2}\), so the lattice vector \(\boldsymbol{a}\) must be parallel to the mirror plane. Similarly, the \(b\)-glide perpendicular to c is \(b_{\boldsymbol{c}} = \left( m_{\boldsymbol{c}} \middle| \frac{\boldsymbol{b}}{2} \right)\).
These two axial glides are illustrated to the right. For the axial glides shown, reflection occurs with respect to the \(\boldsymbol{ab}\)-lattice plane through the origin. A second consecutive axial glide reflection results in a lattice translation:
\({a_{\mathbf{c}}}^{2}:\) \(\left( m_{\boldsymbol{c}} \middle| \frac{\boldsymbol{a}}{2} \right)^{2} = \left( m_{\boldsymbol{c}}^{2} \middle| m_{\boldsymbol{c}}\left( \frac{\boldsymbol{a}}{2} \right) + \frac{\boldsymbol{a}}{2} \right) = \left( 1 \middle| \boldsymbol{a} \right)\).
In space group symbols, there is also the symbol “\(e\)”, which stands for a single plane showing axial glide displacements along two different directions. This type of axial glide occurs only for some centered, non-primitive lattices.
Diagonal glides
(b) Diagonal glides, symbolized as \(n\), involve displacements along a face-diagonal direction of the unit cell, i.e., \(\frac{\boldsymbol{a} + \boldsymbol{b}}{2}\), \(\frac{\boldsymbol{a} + \boldsymbol{c}}{2}\), or \(\frac{\boldsymbol{b} + \boldsymbol{c}}{2}\). As an example, the diagonal glide \(n_{\boldsymbol{a}} = \left( m_{\boldsymbol{a}} \middle| \frac{\boldsymbol{b} + \boldsymbol{c}}{2} \right)\) is a reflection perpendicular to a accompanied by displacement along the diagonal of the bc-face of the unit cell.
Diamond glides
(c) Diamond glides, denoted as d, resemble diagonal glides but occur only for face- (F) and body-centered (I) lattices. For example, consider a diamond glide reflection that is perpendicular to the c-axis of a face-centered orthorhombic lattice. Vectors of this lattice will be integer combinations of \(\frac{\boldsymbol{a} + \boldsymbol{b}}{2}\), \(\frac{\boldsymbol{b} + \boldsymbol{c}}{2}\), and \(\frac{\boldsymbol{a} + \boldsymbol{c}}{2}\). This diamond glide, which is illustrated to the right using two perspectives, is \(d_{\boldsymbol{c}} = \left( m_{\boldsymbol{c}} \middle| \frac{\boldsymbol{a} + \boldsymbol{b}}{4} \right)\).
The orientation of a reflection plane is given by the perpendicular vector to the plane and is determined by the symbol’s position in the point group symbol. For example, the point group \(\mathcal{D}_{2h}\) is \(mmm\) in the International notation. With respect to an orthorhombic unit cell, the reflection planes are oriented perpendicular to a, b, and c, respectively, in the symbol. Then, if an orthorhombic crystal exhibits any glide reflections, then the symbol for the space group may show “\(nma\)”. In this case, a diagonal glide is perpendicular to a, \(n_{\boldsymbol{a}} = \left( m_{\boldsymbol{a}} \middle| \frac{\boldsymbol{b} + \boldsymbol{c}}{2} \right)\), and an \(a\)-axial glide is perpendicular to c, \(a_{\boldsymbol{c}} = \left( m_{\boldsymbol{c}} \middle| \frac{\boldsymbol{a}}{2} \right)\).
The general Seitz symbol for a glide reflection is \(\left( m_{\boldsymbol{T}} \middle| \frac{\boldsymbol{T}_{\bot\boldsymbol{T'}}}{2} + \boldsymbol{\tau}_{||\ \boldsymbol{T}} \right)\) in which \(\boldsymbol{T'}_{\bot\boldsymbol{T}}\) is a Bravais lattice vector parallel to the reflection plane (perpendicular to the axis of the plane) and \(\boldsymbol{\tau}_{||\ \boldsymbol{T}}\) is a displacement perpendicular to plane (parallel to the axis of the plane). The vector \(\boldsymbol{\tau}_{||\ \boldsymbol{T}}\) determines the position of the plane relative to the origin.
In summary, the types of rotation-translation operations allowed for 3-d space groups include:
- Bravais lattice translations \(\left( 1 \middle| \boldsymbol{T}_{mnp} \right)\).
- Proper rotations \(\left( 2_{\boldsymbol{T}} \middle| \boldsymbol{\tau}_{\bot\boldsymbol{T}} \right),\ \left( 3_{\boldsymbol{T}} \middle| \boldsymbol{\tau}_{\bot\boldsymbol{T}} \right),\ \left( 4_{\boldsymbol{T}} \middle| \boldsymbol{\tau}_{\bot\boldsymbol{T}} \right)\), and \(\left( 6_{\mathbf{T}} \middle| \mathbf{\tau}_{\bot\mathbf{T}} \right)\), oriented by their axes \(\mathbf{T}\) and located away from the origin if \(\boldsymbol{\tau}_{\bot\boldsymbol{T}} \neq \boldsymbol{0}\).
- Improper rotations \(\left( \overline{1} \middle| \boldsymbol{\tau} \right)\), \(\left( m_{\boldsymbol{T}} \middle| \boldsymbol{\tau}_{||\ \boldsymbol{T}} \right)\), \(\left( {\overline{3}}_{\boldsymbol{T}} \middle| \boldsymbol{\tau} \right)\), \(\left( {\overline{4}}_{\boldsymbol{T}} \middle| \boldsymbol{\tau} \right)\), and \(\left( {\overline{6}}_{\boldsymbol{T}} \middle| \boldsymbol{\tau} \right)\). Higher order improper rotations are oriented by their axes \(\boldsymbol{T}\); reflection planes are oriented by the direction perpendicular to the plane. If \(\boldsymbol{\tau} \neq \boldsymbol{0}\), then the operation is shifted away from the origin.
- Screw rotations \(2_{1},\ 3_{1},\ 3_{2},\ 4_{1},\ 4_{2},\ 4_{3},\ 6_{1},\ 6_{2},\ 6_{3},\ 6_{4},\ 6_{5}\).
- Glide reflections \(a,\ b\), or \(c\) (axial; also \(e\)), \(n\) (diagonal), and \(d\) (diamond).
Determine the Seitz symbol for the following operations:
- a \(4_{1}\) screw rotation along a and its axis intersecting the bc-plane at (0, ¼, 0) in a cubic cell;
- a \(b\)-glide reflection perpendicular to a and intersecting (½, 0, 0) in an orthorhombic cell;
- a \(\overline{4}\) improper rotation with respect to c and the point (¼, ¼, ¼) in a tetragonal cell;
- a \(3\)-fold proper rotation with respect to c and located at (⅔, ⅓, 0) in a trigonal cell.
Solution
The general strategy follows the procedure discussed in slide (18) for considering an operation when its symmetry element (point, line, or plane) does not intersect the origin. That is,
\(\left( R \middle| \boldsymbol{\tau} \right) = \left( 1 \middle| \boldsymbol{t} \right)\left( R \middle| \boldsymbol{\tau}_{R} \right)\left( 1 \middle| \boldsymbol{-t} \right)\),
in which \(\boldsymbol{t}\) is the displacement specified by the invariant point in space for the symmetry element. If the operation is a proper or improper rotation, then \(\boldsymbol{\tau}_{R} = \boldsymbol{0}\); if it is a screw rotation or glide reflection, then \(\boldsymbol{\tau}_{R} \neq \boldsymbol{0}\) and depends on the type of operation it is as discussed in (27) and (28).
- \(4_{1}\) screw rotation along a through the origin is \(\left( 4_{\boldsymbol{a}} \middle| \frac{\boldsymbol{a}}{4} \right)\) and \(\boldsymbol{t} = \frac{\boldsymbol{b}}{4}\) :
\(\left( R \middle| \boldsymbol{\tau} \right) = \left( 1 \middle| \frac{\boldsymbol{b}}{4} \right)\left( 4_{\boldsymbol{a}} \middle| \frac{\boldsymbol{a}}{4} \right)\left( 1 \middle| \frac{- \boldsymbol{b}}{4} \right) = \left( 1 \middle| \frac{\boldsymbol{b}}{4} \right)\left( 4_{\boldsymbol{a}} \middle| \frac{\boldsymbol{a} - \boldsymbol{c}}{4} \right) = \left( 4_{\boldsymbol{a}} \middle| \frac{\boldsymbol{a} + \boldsymbol{b} - \boldsymbol{c}}{4} \right)\).
Note:\(4_{\boldsymbol{a}}\left(\boldsymbol{-b} \right) = \left( \boldsymbol{-c} \right)\).
- \(b\)-glide reflection perpendicular to a through the origin is \(\left( m_{\boldsymbol{a}} \middle| \frac{\boldsymbol{b}}{2} \right)\) and \(\boldsymbol{t} = \frac{\boldsymbol{a}}{2}\):
\(\left( R \middle| \boldsymbol{\tau} \right) = \left( 1 \middle| \frac{\boldsymbol{a}}{2} \right)\left( m_{\boldsymbol{a}} \middle| \frac{\boldsymbol{b}}{2} \right)\left( 1 \middle| \frac{- \boldsymbol{a}}{2} \right) = \left( 1 \middle| \frac{\boldsymbol{a}}{2} \right)\left( m_{\boldsymbol{a}} \middle| \frac{\boldsymbol{a} + \boldsymbol{b}}{2} \right) = \left( m_{\boldsymbol{a}} \middle| \boldsymbol{a} + \frac{\boldsymbol{b}}{2} \right)\).
- \(\overline{4}\) improper rotation with respect to c at the origin is \(\left( {\overline{4}}_{\boldsymbol{c}} \middle| \boldsymbol{0} \right)\) and \(\boldsymbol{t} = \frac{\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c}}{4}\) :
\(\left( R \middle| \boldsymbol{\tau} \right) = \left( 1 \middle| \frac{\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c}}{4} \right)\left( {\overline{4}}_{\boldsymbol{c}} \middle| \boldsymbol{0} \right)\left( 1 \middle| \frac{- \boldsymbol{a} - \boldsymbol{b} - \boldsymbol{c}}{4} \right) = \left( 1 \middle| \frac{\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c}}{4} \right)\left( {\overline{4}}_{\boldsymbol{c}} \middle| \frac{\boldsymbol{b} - \boldsymbol{a} + \boldsymbol{c}}{4} \right) = \left( {\overline{4}}_{\boldsymbol{c}} \middle| \frac{\boldsymbol{b} + \boldsymbol{c}}{2} \right)\).
Note:\({\overline{4}}_{\boldsymbol{c}}\left( - \boldsymbol{a} \right) = \left( + \boldsymbol{b} \right)\),\({\overline{4}}_{\boldsymbol{c}}\left( - \boldsymbol{b} \right) = \left( - \boldsymbol{a} \right)\), and\({\overline{4}}_{\boldsymbol{c}}\left( - \boldsymbol{c} \right) = \left( + \boldsymbol{c} \right)\).
- \(3\)-fold proper rotation with respect to c through the origin is \(\left( 3_{\boldsymbol{c}} \middle| \boldsymbol{0} \right)\) and \(\boldsymbol{t} = \frac{2\boldsymbol{a} + \boldsymbol{b}}{3}\) :
\(\left( R \middle| \boldsymbol{\tau} \right) = \left( 1 \middle| \frac{2\boldsymbol{a} + \boldsymbol{b}}{3}\ \right)\left( 3_{\boldsymbol{c}} \middle| \boldsymbol{0} \right)\left( 1 \middle| \frac{- 2\boldsymbol{a} - \boldsymbol{b}}{3} \right) = \left( 1 \middle| \frac{2\boldsymbol{a} + \boldsymbol{b}}{3}\ \right)\left( 3_{\boldsymbol{c}} \middle| \frac{- 2\boldsymbol{b} + \boldsymbol{a} + \boldsymbol{b}}{3} \right) = \left( 3_{\boldsymbol{c}} \middle| \boldsymbol{a} \right)\).
Note:\(3_{\boldsymbol{c}}\left( - 2\boldsymbol{a} \right) = \left( - 2\boldsymbol{b} \right)\)and\(3_{\boldsymbol{c}}\left( - \boldsymbol{b} \right) = \left( \boldsymbol{a} + \boldsymbol{b} \right)\)
Describe the following operations for a cubic cell from the Seitz symbol:
- \(\left( 4_{\boldsymbol{c}} \middle| \boldsymbol{a}+\frac{\boldsymbol{c}}{2} \right)\)
- \(\left( m_{\boldsymbol{b}} \middle| \frac{\boldsymbol{a}}{2} + \frac{\boldsymbol{b}}{2} + \frac{\boldsymbol{c}}{2} \right)\)
- \(\left( \overline{1} \middle| \boldsymbol{a} + \boldsymbol{b} \right)\)
- \(\left( 2_{\boldsymbol{a}} \middle| \frac{\boldsymbol{a}}{2} + \frac{\boldsymbol{b}}{2} + \frac{\boldsymbol{c}}{2} \right)\)
Solution
Again, the general strategy follows the procedure discussed in slide (18) for considering an operation when its symmetry element (point, line, or plane) does not intersect the origin:
\(\left( R \middle| \boldsymbol{\tau} \right) = \left( 1 \middle| \boldsymbol{t} \right)\left( R \middle| \boldsymbol{\tau}_{R} \right)\left( 1 \middle| \boldsymbol{-t} \right)\).
For this exercise, we need to identify \(\boldsymbol{t}\), i.e., the location of the invariant point for the symmetry element of the operation, and \(\boldsymbol{\tau}_{R}\).
- \(\left( 4_{\boldsymbol{c}} \middle| \boldsymbol{a} + \frac{\boldsymbol{c}}{2} \right)\): \(R\) is 4-fold proper rotation about \(\boldsymbol{c}\); \(\boldsymbol{\tau}\) is lattice translation by \(\boldsymbol{a}\) and shift by \(\frac{\boldsymbol{c}}{2}\), which is parallel to the rotation axis. This operation is a \(4_{2}\) screw rotation along c. Now, where is the axis located?
\(\left( 4_{\boldsymbol{c}} \middle| \boldsymbol{a} + \frac{\boldsymbol{c}}{2} \right) = \left( 1 \middle| \boldsymbol{t} \right)\left( 4_{\boldsymbol{c}} \middle| \frac{\boldsymbol{c}}{2} \right)\left( 1 \middle| \boldsymbol{-t} \right) = \left( 4_{\boldsymbol{c}} \middle| \frac{\boldsymbol{c}}{2} + \boldsymbol{t}-4_{\boldsymbol{c}}\boldsymbol{t} \right)\).
Therefore, \(\boldsymbol{a} = \small{\begin{pmatrix} 1 \\ 0 \\ 0 \\ \end{pmatrix}} = \left( \boldsymbol{t} - 4_{\boldsymbol{c}}\boldsymbol{t} \right) = \begin{pmatrix} t_{x} \\ t_{y} \\ t_{z} \\ \end{pmatrix} - \begin{pmatrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} t_{x} \\ t_{y} \\ t_{z} \\ \end{pmatrix} = \begin{pmatrix} - t_{x} + t_{y} \\ - t_{x} + t_{y} \\ 0 \\ \end{pmatrix} \rightarrow \boldsymbol{t} = \frac{\boldsymbol{a} + \boldsymbol{b}}{2}\).
\(\left( 4_{\boldsymbol{c}} \middle| \boldsymbol{a} + \frac{\boldsymbol{c}}{2} \right)\) = \(4_{2}\) screw rotation along c with its axis intersecting the ab-plane at (½, ½, 0).
- \(\left( m_{\boldsymbol{b}} \middle| \frac{\boldsymbol{a}}{2} + \frac{\boldsymbol{b}}{2} + \frac{\boldsymbol{c}}{2} \right)\): \(R\) is reflection perpendicular to \(\boldsymbol{b}\); \(\boldsymbol{\tau}\) is glide by \(\frac{\boldsymbol{a}}{2} + \frac{\boldsymbol{c}}{2}\) and shift by \(\frac{\boldsymbol{b}}{2}\). This operation is a diagonal glide reflection. Now, where is the plane located?
\(\left( m_{\boldsymbol{b}} \middle| \frac{\boldsymbol{a}}{2} + \frac{\boldsymbol{b}}{2} + \frac{\boldsymbol{c}}{2} \right) = \left( 1 \middle| \boldsymbol{t} \right)\left( m_{\boldsymbol{b}} \middle| \frac{\boldsymbol{a}}{2} + \frac{\boldsymbol{c}}{2} \right)\left( 1 \middle| - \boldsymbol{t} \right) = \left( m_{\boldsymbol{b}} \middle| \frac{\boldsymbol{a}}{2} + \frac{\boldsymbol{c}}{2} + \boldsymbol{t} -m_{\boldsymbol{b}}\boldsymbol{t} \right)\).
Therefore, \(\frac{\boldsymbol{b}}{2} = \small{\begin{pmatrix} 0 \\ ½ \\ 0 \\ \end{pmatrix}} = \left( \boldsymbol{t} - m_{\boldsymbol{b}}\boldsymbol{t} \right) = \begin{pmatrix} t_{x} \\ t_{y} \\ t_{z} \\ \end{pmatrix} - \begin{pmatrix} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} t_{x} \\ t_{y} \\ t_{z} \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 2t_{y} \\ 0 \\ \end{pmatrix} \rightarrow \boldsymbol{t} = \frac{\boldsymbol{b}}{4}\).
\(\left( m_{\boldsymbol{b}} \middle| \frac{\boldsymbol{a}}{2} + \frac{\boldsymbol{b}}{2} + \frac{\boldsymbol{c}}{2} \right)\) = \(n\)-glide reflection perpendicular to b and intersecting (0, ¼, 0).
- \(\left( \overline{1} \middle| \boldsymbol{a} + \boldsymbol{b} \right)\): \(R\) is inversion; \(\boldsymbol{\tau}\) is lattice translation by \(\boldsymbol{a} + \boldsymbol{b}\). This operation is inversion located away from the origin:
\(\left( \overline{1} \middle| \boldsymbol{a} + \boldsymbol{b} \right) = \left( 1 \middle| \boldsymbol{t} \right)\left( \overline{1} \middle| \boldsymbol{0} \right)\left( 1 \middle| - \boldsymbol{t} \right) = \left( \overline{1} \middle| \boldsymbol{t}-\overline{1}\boldsymbol{t} \right) = \left( \overline{1} \middle| \boldsymbol{2}\boldsymbol{t} \right)\). Therefore, \(\boldsymbol{t} = \frac{\boldsymbol{a} + \boldsymbol{b}}{2}\).
\(\left( \overline{1} \middle| \boldsymbol{a} + \boldsymbol{b} \right)\) = an inversion center located at (½, ½, 0).
- \(\left( 2_{\boldsymbol{a}} \middle| \frac{\boldsymbol{a}}{2} + \frac{\boldsymbol{b}}{2} + \frac{\boldsymbol{c}}{2} \right)\): \(R\) is 2-fold proper rotation about \(\boldsymbol{a}\); \(\boldsymbol{\tau}\) is shift by \(\frac{\boldsymbol{a}}{2}\), which is parallel to the rotation axis, as well as a shift by \(\frac{\boldsymbol{b}}{2} + \frac{\boldsymbol{c}}{2}\). This operation is a \(2_{1}\) screw rotation along a. Now, where is the axis located?
\(\left( 2_{\boldsymbol{a}} \middle| \frac{\boldsymbol{a}}{2} + \frac{\boldsymbol{b}}{2} + \frac{\boldsymbol{c}}{2} \right) = \left( 1 \middle| \boldsymbol{t} \right)\left( 2_{\boldsymbol{a}} \middle| \frac{\boldsymbol{a}}{2} \right)\left( 1 \middle| - \boldsymbol{t} \right) = \left( 2_{\boldsymbol{a}} \middle| \frac{\boldsymbol{a}}{2} + \boldsymbol{t}-2_{\boldsymbol{a}}\boldsymbol{t} \right)\).
Therefore, \(\frac{\boldsymbol{b}}{2} + \frac{\boldsymbol{c}}{2} = \small{\begin{pmatrix} 0 \\ ½ \\ ½ \\ \end{pmatrix}} = \left( \boldsymbol{t} - 2_{\boldsymbol{a}}\boldsymbol{t} \right) = \begin{pmatrix} t_{x} \\ t_{y} \\ t_{z} \\ \end{pmatrix} - \begin{pmatrix} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \\ \end{pmatrix}\begin{pmatrix} t_{x} \\ t_{y} \\ t_{z} \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 2t_{y} \\ 2t_{z} \\ \end{pmatrix} \rightarrow \boldsymbol{t} = \frac{\boldsymbol{b} + \boldsymbol{c}}{4}\).
\(\left( 2_{\boldsymbol{a}} \middle| \frac{\boldsymbol{a}}{2} + \frac{\boldsymbol{b}}{2} + \frac{\boldsymbol{c}}{2} \right)\) = \(2_{1}\) screw rotation along a with its axis intersecting the bc-plane at (0, ¼, ¼).