Derivation of symbols for symmetry operations from coordinate triplets or matrix pairs

Fischer, W.; Koch, E.

doi:10.1107/97809553602060000523

International
Tables for
Crystallography
Volume A
Space-group symmetry
Edited by Th. Hahn

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International Tables for Crystallography (2006). Vol. A. ch. 11.2, pp. 812-813

Section 11.2.1. Derivation of symbols for symmetry operations from coordinate triplets or matrix pairs

W. Fischer^a and E. Koch^a ^*

^a Institut für Mineralogie, Petrologie und Kristallographie, Philipps-Universität, D-35032 Marburg, Germany
Correspondence e-mail: kochelke@mailer.uni-marburg.de

11.2.1. Derivation of symbols for symmetry operations from coordinate triplets or matrix pairs

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In the space-group tables, all symmetry operations with $[0 \leq w \lt 1]$ are listed explicitly. As a consequence, the number of entries under the heading Symmetry operations equals the multiplicity of the general position. For space groups with centred unit cells, $[w \geq 1]$ may result if the centring translations are applied to the explicitly listed coordinate triplets. In those cases, all w values have been reduced modulo 1 for the derivation of the corresponding symmetry operations (see Section 2.2.9 ). In addition to the tabulated symmetry operations, each space group contains an infinite number of further operations obtained by application of integral lattice translations. In many cases, it is not trivial to obtain the additional symmetry operations (cf. Part 4 ) from the ones listed. Therefore, a general procedure is described below by which symbols for symmetry operations as described in Section 11.1.2 may be derived from coordinate triplets or, more specifically, from the corresponding matrix pairs (W, w). [For a similar treatment of this topic, see Wondratschek & Neubüser (1967).] This procedure may also be applied to cases where space groups are given in descriptions not contained in International Tables. In practice, two cases may be distinguished:

(i) The matrix W is the unit matrix : $[{\bi W} = {\bi I} = \pmatrix{1 &0 &0\cr 0 &1 &0\cr 0 &0 &1\cr}.]$ In this case, the symmetry operation is a translation with translation vector w.

Example

$[\eqalignno{&{\hbox to -8pt{}}x + {\textstyle{1 \over 2}}, y + {\textstyle{1 \over 2}}, z\;\;\Rightarrow \;\;{\bi W} = \pmatrix{1 &0 &0\cr 0 &1 &0\cr 0 &0 &1\cr} = {\bi I},\quad {\bi w} = \pmatrix{{1 \over 2}\cr {1 \over 2}\cr0\cr}\cr &{\hbox to -8pt{}}\!\!\Rightarrow\;\; \hbox{translation with translation vector }{\bf w} = {\textstyle{1 \over 2}{\bf a}} + {\textstyle{1 \over 2}}{\bf b}\cr &\quad(C \hbox{ centring}).}]$

(ii) The matrix W is not the unit matrix: $[{\bi W} \neq {\bi I.}]$ In this case, one calculates the trace, $[\hbox{tr}({\bi W}) = W_{11} + W_{22} + W_{33}]$ , and the determinant, $[\det ({\bi W})]$ , and identifies the type of the rotation part of the symmetry operation from Table 11.2.1.1.

Table 11.2.1.1 | top | pdf |
Identification of the type of the rotation part of the symmetry operation

$[\det({\bi W})]$	$[\hbox{tr}({\bi W})]$
$[\det({\bi W})]$	−3	−2	−1	0	1	2	3
1			2	3	4	6	1
−1	$[\bar{1}]$	$[\bar{6}]$	$[\bar{4}]$	$[\bar{3}]$	m

One has to distinguish three subcases:

(a) W corresponds to a rotoinversion. The inversion point X is obtained by solving the equation $[({\bi W},{\bi w}){\bi x} = {\bi x}]$ . For a rotoinversion other than $[\bar{1}]$ , the location of the axis follows from the equation $[({\bi W},{\bi w})^{2}{\bi x} = ({\bi W}^{2}, {\bi W}{\bi w} + {\bi w}){\bi x} = {\bi x}]$ . The rotation sense may be found either by geometrical inspection of a pair of points related by the symmetry operation or by the algebraic procedure described below.

Example

$[\eqalign{&z,\! -\! y + {\textstyle{1 \over 2}},\! -\! x + {\textstyle{1 \over 2}} \;\;\Rightarrow \;\;{\bi W} = \pmatrix{0 &0 &1\cr 0 &\bar{1} &0\cr \bar{1} &0 &0\cr},\quad {\bi w} = \pmatrix{0\cr {1 \over 2}\cr {1 \over 2}\cr}\cr &\Rightarrow \hbox{tr}({\bi W}) = - 1,\quad \det ({\bi W}) = - 1\cr &\Rightarrow \hbox{fourfold rotoinversion;}\cr &({\bi W}, {\bi w}){\bi x} = {\bi x} \;\;\Rightarrow \;\;\pmatrix{0 &0 &1\cr 0 &\bar{1} &0\cr \bar{1} &0 &0\cr} \pmatrix{x\cr y\cr z\cr} + \pmatrix{0\cr {1 \over 2}\cr {1 \over 2}\cr} = \pmatrix{x\cr y\cr z\cr}\cr &\Rightarrow x = y = z = {\textstyle{1 \over 4}}\cr &\Rightarrow \hbox{inversion point at}\ {\textstyle{1 \over 4} {1 \over 4} {1 \over 4}};\cr &({\bi W}^{2}, {\bi W}{\bi w} + {\bi w}){\bi x} = {\bi x}\cr &\Rightarrow \pmatrix{\bar{1} &0 &0\cr 0 &1 &0\cr 0 &0 &\bar{1}\cr} \pmatrix{x\cr y\cr z\cr} + \pmatrix{{1 \over 2}\cr 0\cr {1 \over 2}\cr} = \pmatrix{x\cr y\cr z\cr}\cr &\Rightarrow x = z = {\textstyle{1 \over 4}},\ y \hbox{ undetermined}\cr &\Rightarrow \hbox{rotoinversion axis at } {\textstyle{1 \over 4}}\; y\; {\textstyle{1 \over 4}};\cr &({\bi W}, {\bi w}) \pmatrix{{1 \over 4}\cr {1 \over 4}\cr z\cr} = \pmatrix{0 &0 &1\cr 0 &\bar{1} &0\cr \bar{1} &0 &0\cr} \pmatrix{{1 \over 4}\cr {1 \over 4}\cr z\cr} + \pmatrix{0\cr {1 \over 2}\cr {1 \over 2}\cr} = \pmatrix{z\cr {1 \over 4}\cr {1 \over 4}\cr}\cr &\Rightarrow \hbox{the rotation sense is negative as verified by}\cr &{\hbox to 14pt{}} \hbox{geometrical inspection.}\cr &\Rightarrow \bar{4}^{-}\ {\textstyle{1 \over 4}}, y, {\textstyle{1 \over 4}};\ {\textstyle{1 \over 4}}, {\textstyle{1 \over 4}}, {\textstyle{1 \over 4}}.}\hfill]$
(b) W corresponds to an n-fold rotation. (W, w) is thus either a rotation or a screw rotation . To distinguish between these alternatives, $[({\bi W}, {\bi w})^{n} = ({\bi I}, {\bi t})]$ has to be calculated. For $[{\bi t} = {\bi o}]$ , $[({\bi W}, {\bi w})]$ describes a pure rotation, the rotation axis of which is found by solving $[({\bi W}, {\bi w}){\bi x} = {\bi x}]$ . For $[{\bi t} \neq {\bi o}]$ , (W, w) describes a screw rotation with screw part $[{\bi w}_{g} = (1/n){\bi t}]$ . The location of the screw axis is found as the set of fixed points for the corresponding pure rotation $[({\bi W}, {\bi w}_{l})]$ with $[{\bi w}_{l} = {\bi w} - {\bi w}_{g}]$ , i.e. by solving $[({\bi W}, {\bi w}_{l}){\bi x} = {\bi x}]$ . The sense of the rotation may be found either by geometrical inspection or by the algebraic procedure described below.

Example

$[{\hbox to -3pt{}}\eqalign{&-z, -x + {\textstyle{1 \over 2}}, y \;\;\Rightarrow \;\;{\bi W} = \pmatrix{0 &0 &\bar{1}\cr \bar{1} &0 &0\cr 0 &1 &0\cr},\quad {\bi w} = \pmatrix{0\cr {1 \over 2}\cr 0\cr}\cr &\Rightarrow \hbox{tr}({\bi W}) = 0,\quad \det ({\bi W}) = 1\cr &\Rightarrow \hbox{threefold rotation or screw rotation};\cr &({\bi W}, {\bi w})^{3} = ({\bi W}^{3}, {\bi W}^{2}{\bi w} + {\bi W}{\bi w} + {\bi w})\cr &\hbox{and}\cr &{\bi W}^{2}{\bi w} + {\bi W}{\bi w} + {\bi w} = {\bi t} = \pmatrix{-{1 \over 2}\cr\hfill {1 \over 2}\cr \hfill {1 \over 2}\cr}\cr &\Rightarrow \hbox{threefold screw rotation with screw part}\cr &\quad\;\; {\bi w}_{g} = {\textstyle{1 \over 3}}\;{\bi t} = \pmatrix{-{1 \over 6}\cr \hfill {1 \over 6}\cr \hfill {1 \over 6}\cr};\cr &({\bi W},{\bi w}_{l}){\bi x} = {\bi x}\cr &\Rightarrow \pmatrix{0 &0 &\bar{1}\cr \bar{1} &0 &0\cr 0 &1 &0\cr} \pmatrix{x\cr y\cr z\cr} + \pmatrix{\hfill {1 \over 6}\cr \hfill {1 \over 3}\cr -{1 \over 6}\cr} = \pmatrix{x\cr y\cr z\cr}\cr &\Rightarrow y = {\textstyle{1 \over 3}} - x;\quad z = {\textstyle{1 \over 6}} - x;\quad x \hbox{ undetermined}\cr &\Rightarrow \hbox{screw axis at } x, {\textstyle{1 \over 3}} - x,{\textstyle{1 \over 6}} - x\cr &\qquad\!\! \hbox{(for the sense of rotation see example below)}.}\hfill]$
(c) W corresponds to a (glide) reflection. The glide character is now found by means of the equation $[({\bi W}, {\bi w})^{2} = ({\bi I}, {\bi W}{\bi w} + {\bi w}) = ({\bi I}, {\bi t})]$ . For $[{\bi t} = {\bi o}]$ , (W, w) describes a pure reflection and the location of the mirror plane follows from $[({\bi W}, {\bi w}){\bi x} = {\bi x}]$ . For $[{\bi t} \neq {\bi o}]$ , (W, w) corresponds to a glide reflection with glide part $[{\bi w}_{g} = {\textstyle{1 \over 2}}\;{\bi t}]$ . The location of the glide plane is the set of fixed points for the corresponding pure reflection $[({\bi W},{\bi w}_{l}) = ({\bi W},{\bi w} - {\textstyle{1 \over 2}}\;{\bi t})]$ and is thus calculated by solving $[({\bi W}, {\bi w}_{l}){\bi x} = {\bi x}]$ .

Example

$[\eqalign{&-\!y + {\textstyle{1 \over 2}}, -\! x, z + {\textstyle{3 \over 4}} \;\;\Rightarrow\;\; {\bi W} = \pmatrix{0 &\bar{1} &0\cr \bar{1} &0 &0\cr 0 &0 &1\cr},\quad {\bi w} = \pmatrix{{1 \over 2}\cr 0\cr {3 \over 4}\cr}\cr &\Rightarrow \hbox{tr}({\bi W}) = 1,\quad\det ({\bi W}) = -1\cr &\Rightarrow \hbox{reflection or glide reflection};\cr &({\bi W}, {\bi w})^{2} = ({\bi W}^{2}, {\bi W}{\bi w} + {\bi w}),\quad{\bi W}{\bi w} + {\bi w} = {\bi t} = \pmatrix{\hfill {1 \over 2}\cr - {1 \over 2}\cr \hfill {3 \over 2}\cr}\cr &\Rightarrow \hbox{glide reflection with glide part}\cr& {\bi w}_{g} = {\textstyle{1 \over 2}}\ {\bi t} = \pmatrix{\hfill {1 \over 4}\cr - {1 \over 4}\cr \hfill {3 \over 4}\cr};\cr &({\bi W}, {\bi w}_{l}){\bi x} = {\bi x}\;\; \Rightarrow\; \;\pmatrix{0 &\bar{1} &0\cr \bar{1} &0 &0\cr 0 &0 &1\cr} \pmatrix{x\cr y\cr z\cr} + \pmatrix{{1 \over 4}\cr {1 \over 4}\cr 0\cr} = \pmatrix{x\cr y\cr z\cr}\cr &\Rightarrow y = - x + {\textstyle{1 \over 4}};\ x, z \hbox{ undetermined}\cr &\Rightarrow \hbox{glide plane } d \hbox{ at } x,\! -\! x + {\textstyle{1 \over 4}}, z\cr &\Rightarrow d({\textstyle{1 \over 4}},\! -\! {\textstyle{1 \over 4}}, {\textstyle{3 \over 4}})\quad x,\! -\! x + {\textstyle{1 \over 4}}, z.}\hfill]$

The sense of a pure or screw rotation or of a rotoinversion may be calculated as follows: One takes two arbitrary points $[P_{0}]$ and $[P_{1}]$ on the rotation axis, $[P_{0}]$ having the lower value for the free parameter of the axis. One takes a point $[P_{2}]$ not lying on the axis and generates $[P_{3}]$ from $[P_{2}]$ by the symmetry operation under consideration. One calculates the determinant d of the matrix $[({\bi v}_{1}, {\bi v}_{2}, {\bi v}_{3})]$ composed of the components of vectors $[{\bi v}_1 = \overrightarrow{P_{0}P_{1}}, {\bi v}_{2} = \overrightarrow{P_{0}P_{2}}]$ and $[{\bi v}_{3} = \overrightarrow{P_{0}P_{3}}]$ . For rotations or screw rotations, the sense is positive for $[d \gt 0]$ and negative for $[d \lt 0]$ . For rotoinversions, the sense is positive for $[{d \lt 0}]$ and negative for $[d \gt 0]$ .

Example

According to the example in (b) above, the triplet $[-z,\! -\!x + {\textstyle{1 \over 2}}, y]$ represents a threefold screw rotation with screw part $[\openup3pt\pmatrix{- {1 \over 6}\cr \hfill {1 \over 6}\cr \hfill {1 \over 6}\cr}]$ and screw axis at $[x,{\textstyle{1 \over 3}} -x,{\textstyle{1 \over 6}} -x]$ . To obtain the sense of the rotation, the points $[0{\textstyle{1 \over 3}}\;{\textstyle{1 \over 6}}]$ and $[{\textstyle{1 \over 6}}\;{\textstyle{1 \over 6}}0]$ are used as $[P_{0}]$ and $[P_{1}]$ on the axis and the points 000 and $[0{\textstyle{1 \over 2}}0]$ as $[P_{2}]$ and $[P_{3}]$ outside the axis. The resulting vectors are $[\eqalign{&{\bi v}_{1} = \pmatrix{\hfill {1 \over 6}\cr - {1 \over 6}\cr - {1 \over 6}\cr},\quad {\bi v}_{2} = \pmatrix{\hfill 0\cr - {1 \over 3}\cr - {1 \over 6}\cr},\quad {\bi v}_{3} = \pmatrix{\hfill 0\cr \hfill {1 \over 6}\cr - {1 \over 6}\cr}\cr &\Rightarrow d = \left|\matrix{\hfill {1 \over 6} &\hfill 0 &\hfill 0\cr - {1 \over 6} &- {1 \over 3} &\hfill {1 \over 6}\cr - {1 \over 6} &- {1 \over 6} &- {1 \over 6}\cr}\right| = {1 \over 72} \;\gt\; 0\cr &\Rightarrow \hbox{the sense of the rotation is positive}\cr &\Rightarrow 3^{+} (- {\textstyle{1 \over 6}},{\textstyle{1 \over 6}},{\textstyle{1 \over 6}})\quad x,{\textstyle{1 \over 3}} - x,{\textstyle{1 \over 6}} - x.}]$

References

Wondratschek, H. & Neubüser, J. (1967). Determination of the symmetry elements of a space group from the `general positions' listed in International Tables for X-ray Crystallography, Vol. I. Acta Cryst. 23, 349–352.Google Scholar

International Tables for Crystallography (2006). Vol. A. ch. 11.2, pp. 812-813