International
Tables for
Crystallography
Volume A1
Symmetry relations between space groups
Edited by Hans Wondratschek and Ulrich Müller

International Tables for Crystallography (2011). Vol. A1, ch. 1.6, pp. 52-54

Section 1.6.5. Handling cell transformations

Ulrich Müllera*

aFachbereich Chemie, Philipps-Universität, D-35032 Marburg, Germany
Correspondence e-mail: mueller@chemie.uni-marburg.de

1.6.5. Handling cell transformations

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It is important to keep track of the coordinate transformations in a sequence of group–subgroup relations. A Bärnighausen tree can only be correct if every atomic position of every hettotype can be derived from the corresponding positions of the aristo­type. The mathematical tools are described in Sections 1.2.2.3[link] , 1.2.2.4[link] and 1.2.2.7[link] of this volume and in Part 5[link] of Volume A.

A basis transformation and an origin shift are mentioned in the middle of a group–subgroup arrow. This is a shorthand notation for the matrix–column pair (Seitz symbol; Section 1.2.2.3[link] ) or the 4 × 4 matrix (augmented matrix, Section 1.2.2.4[link] ) to be used to calculate the basis vectors, origin shift and coordinates of a maximal subgroup from those of the preceding space group. If there is a sequence of several transformations, the overall changes can be calculated by multiplication of the 4 × 4 matrices.

Let [\specialfonts{\bbsf P}_{\rm 1},{\bbsf P}_{\rm 2},\ldots] be the 4 × 4 matrices expressing the basis vector and origin changes of several consecutive cell transformations. [\specialfonts{\bbsf P}_{\rm 1}^{\rm -1}, {\bbsf P}_{\rm 2}^{\rm -1},\ldots] are the corresponding inverse matrices. Let a, b, c be the starting (old) basis vectors and [{\bf a}'], [{\bf b}'], [{\bf c}'] be the (new) basis vectors after the consecutive transformations. Let [\specialfonts{\bbsf x}] and [\specialfonts{\bbsf x}'] be the augmented columns of the atomic coordinates before and after the transformations. Then the following relations hold:[\specialfonts({\bf a}',{\bf b}',{\bf c}',{\bi p}) = ({\bf a},{\bf b},{\bf c},{\bi o})\, {\bbsf P}_{\rm 1}{\bbsf P}_{\rm 2},\ldots \hbox{\quad \rm and \quad} {\bbsf x}' = \ldots {\bbsf P}_{\rm 2}^{\rm -1}{\bbsf P}_{\rm 1}^{\rm -1}{\bbsf x} .][{\bi p}] is the vector of the origin shift, i.e. its components are the coordinates of the origin of the new cell expressed in the coordinate system of the old cell; [{\bi o}] is a zero vector. Note that the inverse matrices have to be multiplied in the reverse order.

Example 1.6.5.1

Take a transformation from a cubic to a rhombohedral unit cell (hexagonal setting) combined with an origin shift of [{\bi p}_1=(\,{\textstyle{1\over 4}},{\textstyle{1\over 4}},{\textstyle{1\over 4}}\,)] (in the cubic coordinate system), followed by transformation to a monoclinic cell with a second origin shift of [{\bi p}_2=(-{\textstyle{1\over 2}},-{\textstyle{1\over 2}},0)] (in the hexagonal coordinate system). From Fig. 1.6.5.1[link] (left) we can deduce[{\bf a}_{\rm hx} = {\bf a}_{\rm cb} - {\bf b}_{\rm cb}, \quad{\bf b}_{\rm hx} = {\bf b}_{\rm cb} - {\bf c}_{\rm cb}, \quad{\bf c}_{\rm hx}= {\bf a}_{\rm cb}+{\bf b}_{\rm cb}+{\bf c}_{\rm cb},]which in matrix notation is[\eqalign{({\bf a}_{\rm hx},\,{\bf b}_{\rm hx},\,{\bf c}_{\rm hx}) & = ({\bf a}_{\rm cb},\,{\bf b}_{\rm cb},\,{\bf c}_{\rm cb}){\bi P}_1\cr & = ({\bf a}_{\rm cb},\,{\bf b}_{\rm cb},\,{\bf c}_{\rm cb})\pmatrix{\hfill 1 & \hfill 0 & \hfill 1 \cr \hfill -1 & \hfill 1 & \hfill 1 \cr \hfill 0 & \hfill -1 & \hfill 1}.}]In the fourth column of the 4 × 4 matrix [\specialfonts {\bbsf P}_{\rm 1}] we include the components of the origin shift, [{\bi p}_1=(\,{\textstyle{1\over 4}},{\textstyle{1\over 4}},{\textstyle{1\over 4}}\,)]: [\specialfonts{\bbsf P}_{\rm 1} = \pmatrix{{\bi P}_1{\hskip -4pt}&{\vrule height 8pt depth 4pt}\hfill&{\hskip -4pt} {\bi p}_1\cr\noalign{\hrule} {\bi o}{\hskip -4pt}&{\vrule height 8pt depth 4pt}\hfill&{\hskip -4pt}1} =\pmatrix{\phantom{-}1 & \phantom{-}0 & 1{\hskip -4pt}&{\vrule height 10pt depth6pt}\hfill&{\hskip -4pt}{\textstyle{1\over 4}}\cr -1 & \phantom{-}1 & 1 {\hskip -4pt}&{\vrule height 10pt depth 6pt}\hfill&{\hskip -4pt}{\textstyle{1\over 4}}\cr \phantom{-}0 & -1 & 1 {\hskip -4pt}&{\vrule height 10pt depth 6pt}\hfill&{\hskip -4pt}{\textstyle{1\over 4}}\cr\noalign{\hrule}\cr \phantom{-}0 & \phantom{-}0 & 0 {\hskip -4pt}&{\vrule height 11pt depth 4pt}\hfill&{\hskip -4pt} 1}.]The inverse matrix [{\bi P}_1^{-1}] can be calculated by matrix inversion, but it is more straightforward to deduce it from Fig. 1.6.5.1[link]; it is the matrix that converts the hexagonal basis vectors back to the cubic basis vectors:[\eqalign{{\bf a}_{\rm cb} &= {\textstyle{2 \over 3}}\,{\bf a}_{\rm hx} + {\textstyle{1 \over 3}}\,{\bf b}_{\rm hx} + {\textstyle{1 \over 3}}\,{\bf c}_{\rm hx},\quad{\bf b}_{\rm cb} = -{\textstyle{1 \over 3}}\,{\bf a}_{\rm hx}+{\textstyle{1 \over 3}}\,{\bf b}_{\rm hx}+{\textstyle{1 \over 3}}\,{\bf c}_{\rm hx},\cr{\bf c}_{\rm cb} &= -{\textstyle{1 \over 3}}\,{\bf a}_{\rm hx}-{\textstyle{2 \over 3}}\,{\bf b}_{\rm hx}+{\textstyle{1 \over 3}}\,{\bf c}_{\rm hx},}][\eqalign{({\bf a}_{\rm cb},{\bf b}_{\rm cb},{\bf c}_{\rm cb}) & = ({\bf a}_{\rm hx},{\bf b}_{\rm hx},{\bf c}_{\rm hx})\,{\bi P}_1^{-1}\cr & = ({\bf a}_{\rm hx},{\bf b}_{\rm hx},{\bf c}_{\rm hx})\pmatrix{ {\textstyle{2 \over 3}} & -{\textstyle{1 \over 3}} &-{\textstyle{1 \over 3}} \cr {\textstyle{1 \over 3}} &\phantom {-}{\textstyle{1 \over 3}} &-{\textstyle{2 \over 3}}\cr {\textstyle{1 \over 3}} & \phantom {-}{\textstyle{1 \over 3}} & \phantom {-}{\textstyle{1 \over 3}} }.}]The column part (fourth column) of the inverse matrix [\specialfonts{\bbsf P}_{\rm 1}^{\rm -1}] is [-{\bi P}_1^{-1}{\bi p}_1] (cf. equation 1.2.2.6): [\let\normalbaselines\relax\openup3pt -{\bi P}_1^{-1}{\bi p}_1 = -\pmatrix{ {\textstyle{2 \over 3}} & -{\textstyle{1 \over 3}} & -{\textstyle{1 \over 3}}\cr {\textstyle{1 \over 3}} & \phantom{-}{\textstyle{1 \over 3}} & -{\textstyle{2 \over 3}}\cr{\textstyle{1 \over 3}} & \phantom{-}{\textstyle{1 \over 3}} &\phantom{-}{\textstyle{1 \over 3}} } \pmatrix{{\textstyle{1 \over 4}}\cr {\textstyle{1 \over 4}} \cr {\textstyle{1 \over 4}}} = \pmatrix{ \phantom{-}0\cr \phantom{-}0\cr -{\textstyle{1 \over 4}}}.]Therefore, [\specialfonts{\bbsf P}_{\rm 1}^{\rm -1}] is[\specialfonts{\bbsf P}_{\rm 1}^{\rm -1}= \pmatrix{{\bi P}_1^{-1}{\hskip -4pt}&{\vrule height 8pt depth 4pt}\hfill&{\hskip -4pt} -{\bi P}_1^{-1}{\bi p}_1\cr\noalign{\hrule} {\bi o}{\hskip -4pt}&{\vrule height 8pt depth 4pt}\hfill&{\hskip -4pt}1} =\pmatrix{{\textstyle{2 \over 3}} & -{\textstyle{1 \over 3}} & -{\textstyle{1 \over 3}}{\hskip -4pt}&{\vrule height 10pt depth6pt}\hfill&{\hskip -4pt}\phantom{-}0\cr {\textstyle{1 \over 3}} & \phantom{-}{\textstyle{1 \over 3}} & -{\textstyle{2 \over 3}} {\hskip -4pt}&{\vrule height 10pt depth 6pt}\hfill&{\hskip -4pt}\phantom{-}0\cr{\textstyle{1 \over 3}} & \phantom{-}{\textstyle{1 \over 3}} & \phantom{-}{\textstyle{1 \over 3}} {\hskip -4pt}&{\vrule height 10pt depth 6pt}\hfill&{\hskip -4pt}-{\textstyle{1\over 4}}\cr\noalign{\hrule}\cr0 & \phantom{-}0 & \phantom{-}0 {\hskip -4pt}&{\vrule height 11pt depth 4pt}\hfill&{\hskip -4pt} \phantom{-}1}.]Similarly, for the second (hexagonal to monoclinic) transformation, we deduce the matrices [{\bi P}_2], [{\bi P}_2^{-1}] and the column part of [\specialfonts{\bbsf P}_{\rm 2}^{\rm -1}], [-{\bi P}_2^{-1}{\bi p}_2], from the right image of Fig. 1.6.5.1[link]:[\let\normalbaselines\relax\openup1pt\eqalign{({\bf a}_{\rm mn}, {\bf b}_{\rm mn}, {\bf c}_{\rm mn}) &= ({\bf a}_{\rm hx}, {\bf b}_{\rm hx}, {\bf c}_{\rm hx})\, {\bi P}_2\cr &= ({\bf a}_{\rm hx}, {\bf b}_{\rm hx}, {\bf c}_{\rm hx})\pmatrix{2 & 0 & {\textstyle{2\over 3}}\cr 1 & 1 & {\textstyle{1\over 3}} \cr 0 & 0 & {\textstyle{1\over 3}} } \cr ({\bf a}_{\rm hx}, {\bf b}_{\rm hx}, {\bf c}_{\rm hx}) &= ({\bf a}_{\rm mn}, {\bf b}_{\rm mn}, {\bf c}_{\rm mn}) {\bi P}_2^{-1}\cr &= ({\bf a}_{\rm mn}, {\bf b}_{\rm mn}, {\bf c}_{\rm mn})\pmatrix{\phantom{-}{\textstyle{1\over 2}} & 0 & -1 \cr -{\textstyle{1\over 2}} & 1 & \phantom{-}0 \cr \phantom{-}0 & 0 & \phantom{-}3} }][\let\normalbaselines\relax\openup1pt -{\bi P}_2^{-1}{\bi p}_2 = -\pmatrix{\phantom{-}{\textstyle{1\over 2}} & 0 & -1 \cr -{\textstyle{1\over 2}} & 1 & \phantom{-}0 \cr \phantom{-}0 & 0 & \phantom{-}3} \pmatrix{-{\textstyle{1\over 2}}\cr -{\textstyle{1\over 2}} \cr \phantom{-}0 } = \pmatrix{ {\textstyle{1\over 4}}\cr {\textstyle{1\over 4}}\cr 0}.]For the basis vectors of the two consecutive transformations we calculate: [\eqalign{\specialfonts{\bbsf P}_{\rm 1}{\bbsf P}_{\rm 2} &= \let\normalbaselines\relax\pmatrix{\phantom{-}1 & \phantom{-}0 &1{\hskip -4pt}&{\vrule height 10pt depth4pt}\hfill&{\hskip -4pt}{\textstyle{1\over 4}}\cr -1 & \phantom{-}1 &1{\hskip -4pt}&{\vrule height 10pt depth 4pt}\hfill&{\hskip -4pt}{\textstyle{1\over 4}}\cr \phantom{-}0 &-1 & 1{\hskip -4pt}&{\vrule height 10pt depth 4pt}\hfill&{\hskip -4pt}{\textstyle{1\over 4}}\cr\noalign{\hrule}\cr \phantom{-}0 & \phantom{-}0 & 0{\hskip -4pt}&{\vrule height 11pt depth 4pt}\hfill&{\hskip -4pt} 1}\let\normalbaselines\relax\pmatrix{2 & 0 &{\textstyle{2\over3}}{\hskip -4pt}&{\vrule height10pt depth4pt}\hfill&{\hskip -4pt}-{\textstyle{1\over2}}\cr 1 & 1 &{\textstyle{1\over3}}{\hskip -4pt}&{\vrule height 10pt depth4pt}\hfill&{\hskip -4pt}-{\textstyle{1\over2}}\cr 0 &0 & {\textstyle{1\over3}}{\hskip -4pt}&{\vrule height 10pt depth 4pt}\hfill&{\hskip -4pt}\phantom{-}0\cr\noalign{\hrule}\cr 0 & 0 & 0{\hskip -4pt}&{\vrule height 11pt depth 4pt}\hfill&{\hskip -4pt}\phantom{-} 1}\cr&=\let\normalbaselines\relax\pmatrix{\phantom{-}2 & \phantom{-}0 &1{\hskip -4pt}&{\vrule height 10pt depth4pt}\hfill&{\hskip -4pt}-{\textstyle {1\over 4}}\cr -1 & \phantom{-}1 &0{\hskip -4pt}&{\vrule height 10pt depth 4pt}\hfill&{\hskip -4pt}\phantom{-}{\textstyle {1\over 4}}\cr -1 &-1 & 0{\hskip -4pt}&{\vrule height 10pt depth 4pt}\hfill&{\hskip -4pt}\phantom{-}{\textstyle {3\over 4}}\cr\noalign{\hrule}\cr \phantom{-}0 & \phantom{-}0 & 0{\hskip -4pt}&{\vrule height 11pt depth 4pt}\hfill&{\hskip -4pt} \phantom{-}1},}][\eqalign{({\bf a}_{\rm mn}, {\bf b}_{\rm mn}, {\bf c}_{\rm mn},{\bi p})&= ({\bf a}_{\rm cb}, {\bf b}_{\rm cb}, {\bf c}_{\rm cb}, {\bi o}) \specialfonts{\bbsf P}_{\rm 1} {\bbsf P}_{\rm 2}\cr &=({\bf a}_{\rm cb}, {\bf b}_{\rm cb}, {\bf c}_{\rm cb}, {\bi o})\let\normalbaselines\relax\pmatrix{\phantom{-}2 & \phantom{-}0 &1{\hskip -4pt}&{\vrule height 10pt depth6pt}\hfill&{\hskip -4pt}-{\textstyle {1\over 4}}\cr -1 & \phantom{-}1 &0{\hskip -4pt}&{\vrule height 10pt depth 6pt}\hfill&{\hskip -4pt}\phantom{-}{\textstyle {1\over 4}}\cr -1 &-1 & 0{\hskip -4pt}&{\vrule height 10pt depth 6pt}\hfill&{\hskip -4pt}\phantom{-}{\textstyle {3\over 4}}\cr\noalign{\hrule}\cr \phantom{-}0 & \phantom{-}0 & 0{\hskip -4pt}&{\vrule height 11pt depth 4pt}\hfill&{\hskip -4pt} \phantom{-}1},}]which corresponds to[{\bf a}_{\rm mn} = 2{\bf a}_{\rm cb} -{\bf b}_{\rm cb}-{\bf c}_{\rm cb},\quad{\bf b}_{\rm mn}={\bf b}_{\rm cb}-{\bf c}_{\rm cb},\quad{\bf c}_{\rm mn}={\bf a}_{\rm cb}]and an origin shift of [{\bi p}=(-{\textstyle {1\over 4}},{\textstyle {1\over 4}},{\textstyle {3\over 4}}\,)] in the cubic coordinate system. The corresponding (cubic to monoclinic) coordinate transformations result from[\specialfonts\eqalign{{\bbsf x}_{\rm mn} &= {\bbsf P}_{\rm 2}^{\rm -1} {\bbsf P}_{\rm 1}^{\rm -1}{\bbsf x}_{\rm cb}\cr \let\normalbaselines\relax\pmatrix{x_{\rm mn}\vphantom{\vrule height 2pt depth11pt}\cr y_{\rm mn}\vphantom{\vrule height 2pt depth11pt}\cr z_{\rm mn}\vphantom{\vrule height 2pt depth9pt}\cr\noalign{\hrule}\cr1}&= \let\normalbaselines\relax\pmatrix{\phantom{-}{\textstyle{1 \over 2}}& 0 &-1{\hskip -4pt}&{\vrule height 10pt depth6pt}\hfill&{\hskip -4pt}{\textstyle {1\over 4}}\cr {\textstyle -{1\over 2}} & 1 &\phantom{-}0{\hskip -4pt}&{\vrule height 10pt depth 6pt}\hfill&{\hskip -4pt}{\textstyle {1\over 4}}\cr \phantom{-}0 &0 & \phantom{-}3{\hskip -4pt}&{\vrule height 10pt depth 6pt}\hfill&{\hskip -4pt}0\cr\noalign{\hrule}\cr \phantom{-}0 & 0 & \phantom{-}0{\hskip -4pt}&{\vrule height 10pt depth 6pt}\hfill&{\hskip -4pt} 1}\let\normalbaselines\relax\pmatrix{{\textstyle{2 \over 3}} & -{\textstyle{1 \over 3}} & -{\textstyle{1 \over 3}}{\hskip -4pt}&{\vrule height 10pt depth6pt}\hfill&{\hskip -4pt}\phantom{-}0\cr {\textstyle{1 \over 3}} & \phantom{-}{\textstyle{1 \over 3}} & -{\textstyle{2 \over 3}} {\hskip -4pt}&{\vrule height 10pt depth 6pt}\hfill&{\hskip -4pt}\phantom{-}0\cr{\textstyle{1 \over 3}} & \phantom{-}{\textstyle{1 \over 3}} & \phantom{-}{\textstyle{1 \over 3}} {\hskip -4pt}&{\vrule height 10pt depth 6pt}\hfill&{\hskip -4pt}-{\textstyle{1\over 4}}\cr\noalign{\hrule}\cr0 & \phantom{-}0 & \phantom{-}0 {\hskip -4pt}&{\vrule height 11pt depth 4pt}\hfill&{\hskip -4pt} \phantom{-}1}\let\normalbaselines\relax\pmatrix{x_{\rm cb}\vphantom{\vrule height 2pt depth11pt}\cr y_{\rm cb}\vphantom{\vrule height 2pt depth11pt}\cr z_{\rm cb}\vphantom{\vrule height 2pt depth9pt}\cr\noalign{\hrule}\cr 1}\cr &= \let\normalbaselines\relax\pmatrix{0 & {\textstyle -{1\over 2}} &{\textstyle -{1\over 2}}{\hskip -4pt}&{\vrule height2pt depth8pt}\hfill&{\hskip -4pt}\phantom{-}{\textstyle {1\over 2}}\cr 0 & \phantom{-}{\textstyle {1\over 2}} &{\textstyle -{1\over 2}}{\hskip -4pt}&{\vrule height 10pt depth 6pt}\hfill&{\hskip -4pt}\phantom{-}{\textstyle {1\over 4}}\cr 1 &\phantom{-}1 & \phantom{-}1{\hskip -4pt}&{\vrule height 10pt depth 6pt}\hfill&{\hskip -4pt}{\textstyle -{3\over 4}}\cr\noalign{\hrule}\cr 0 & \phantom{-}0 & \phantom{-}0{\hskip -4pt}&{\vrule height 11pt depth 4pt}\hfill&{\hskip -4pt} \phantom{-}1}\let\normalbaselines\relax\pmatrix{x_{\rm cb}\phantom{\vrule height 2pt depth11pt}\cr y_{\rm cb}\phantom{\vrule height 2pt depth6pt}\cr z_{\rm cb}\vphantom{\vrule height 12pt depth7pt}\cr\noalign{\hrule}\cr 1},}]which is the same as [\displaylines{x_{\rm mn} = -{\textstyle {1\over 2}} \,y_{\rm cb}-{\textstyle {1\over 2}}\, z_{\rm cb}+{\textstyle {1\over 2}},\quad y_{\rm mn}= {\textstyle {1\over 2}}\, y_{\rm cb}-{\textstyle {1\over 2}}\, z_{\rm cb}+{\textstyle {1\over 4}},\cr z_{\rm mn}= x_{\rm cb}+y_{\rm cb}+z_{\rm cb}-{\textstyle {3\over 4}}.}]

[Figure 1.6.5.1]

Figure 1.6.5.1 | top | pdf |

Relative orientations of a cubic to a rhombohedral cell (hexagonal setting) (left) and of a rhombohedral cell (hexagonal setting) to a monoclinic cell (the monoclinic cell has an acute angle β).








































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