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Determination of the matrix-column pair

In this section it is assumed that not only the kind of symmetry operation is known but also its details, e.g. it is not enough to know that there is a 2-fold rotation, but one should also know the orientation and position of the rotation axis. At first one tries to find for some points $X$ their images $\tilde{X}$ under the symmetry operation. This knowledge is then exploited to determine the matrix-column pair which decribes the symmetry operation.

Examples will illustrate the procedures. In all of them the point coordinates are referred to a Cartesian coordinate system, see Section 1.2. The reader is recommended to make small sketches in order to see visually what happens.

In the system (4.1.1) of equations there are 12 coefficients to be determined, 9 $W_{ik}$ and 3 $w_j$. If the image point $\tilde{X}$ of one point $X$ is known from geometric considerations, one can write down the 3 linear equations of (4.1.1) for this pair of points. Therefore, writing down the equations (4.1.1) for 4 pairs (point $\rightarrow$ image point) is sufficient for the determination of all coefficients, provided the points are independent, i.e. are not lying in a plane. One obtains a system of 12 inhomogeneous linear equations with 12 undetermined parameters $W_{ik}$ and $w_j$. This may be difficult to solve without a computer. However, if one restricts to crystallographic symmetry operations, the solution is easy more often than not because of the special form of the matrix-column pairs.

Procedure 1

In many cases it may be possible to apply the following strategy, which avoids all calculations. It requires knowledge of the image points of the origin $O$ and of the 3 `coordinate points' $A$: 1,0,0; $B$: 0,1,0; and $C$: 0,0,1.

(1)
The origin Let $\tilde{O}$ with coordinates $\tilde{\mbox{\textit{\textbf{o}}}}$ be the image of the origin $O$ with coordinates o, i.e. $x_{\circ}=y_{\circ}=z_{\circ}=0$. Examination of the equations (4.1.1) shows that $\tilde{\mbox{\textit{\textbf{o}}}}$ = w, i.e. the column w can be determined separately from the coefficients of the matrix W without any effort.
(2)
The coordinate points We consider the point $A$. Inserting 1,0,0 in the equations (4.1.1) one obtains $\tilde{x}_i=W_{i1}+w_i$ or $W_{i1}=\tilde{x}_i-w_i$, $i=1, 2, 3.$ The first column of W is separated from the others, and for the solution only the known coefficients $w_i$ have to be subtracted from the coordinates $\tilde{x}_i$ of the image point $\tilde{A}$ of $A$. Analogously one calculates the coefficients $W_{i2}$ from the image of point $B$: 0,1,0 and $W_{i3}$ from the image of point $C$: 0,0,1.
Evidently, the sought after coefficients can be determined without any difficult calculation.

Example 1

What is the pair (W,w) for a glide reflection with the plane through the origin, the normal of the glide plane parallel to c, and with the glide vector g = 1/2,1/2,0 ?

(a)
Image of the origin $O$: The origin is left invariant by the reflection part of the mapping; it is shifted by the glide part to 1/2,1/2,0 which are the coordinates of $\tilde{O}$. Therefore, w = 1/2,1/2,0.
(b)
Images of the coordinate points. Both the points $A$ and $B$ are not affected by the reflection part but $A$ is then shifted to 3/2,1/2,0 and $B$ to 1/2,3/2,0. This results in the equations

3/2 = $W_{11}$ + 1/2, 1/2 = $W_{21}$ + 1/2, 0 = $W_{31}$ + 0 for $A$ and

1/2 = $W_{12}$ + 1/2, 3/2 = $W_{22}$ + 1/2, 0 = $W_{32}$ + 0 for $B$.

One obtains $W_{11}=1,\ W_{21}=W_{31}=W_{12}=0,\ W_{22}=1$, and $W_{32}=0$.

Point $C$: 0,0,1 is reflected to 0,0,$-1$ and then shifted to 1/2,1/2,$-1$.

This means $1/2=W_{13}+1/2,\ 1/2=W_{23}+1/2,\ -1=W_{33}+0$ or $W_{13}=W_{23}=0$, $W_{33}=-1$.

(c)
The matrix-column pair is thus

W = $\left(
\begin{array}{rrr}1&0&0 \\ 0&1&0 \\ 0&0&\bar{1} \end{array} \right)$ and w = $\left(\begin{array}{c} 1/2 \\ 1/2 \\ 0
\end{array} \right)$.

Tests for the correctness of the result are always advisable: Each point $x,\,y,\,0$ is mapped onto the point $x+1/2,\,y+1/2,\,0$, i.e. the plane $x,\,y,\,0$ is invariant as a whole; each point $x,\,x,\,z$ is mapped onto $x+1/2,\,x+1/2,\,-z$, i.e. the plane $x,\,x,\,z$ is also left invariant as a whole. Both results agree with the geometric view.

Example 2 [Draw a diagram !]

What is the pair (W,w) for an anti-clockwise 4-fold rotoinversion $\bar{4}$ if the rotoinversion axis is parallel to c, and 1/2,1/2,1/2 is the inversion point ?

(a)
The anti-clockwise 4-fold rotation maps the origin onto the point 1,0,0; the following inversion in 1/2,1/2,1/2 maps this intermediate point onto the point 0,1,1, such that $w_1=0,\ w_2=1,\ w_3=1$.
(b)
For the other points: $1,\,0,\,0\rightarrow
1,\,1,\,0\rightarrow 0,\,0,\,1; \\ 0,\,1,\,0\rightarrow ...
... 1,\,1,\,1; \hspace{6mm} 0,\,0,\,1\rightarrow
1,\,0,\,1\rightarrow 0,\,1,\,0.$

The equations are $0=W_{11}+0;\ 0=W_{21}+1;\ 1=W_{31}+1; \\ 1=W_{12}+0;\ 1=W_{22}+1;\
1=W_{32}+1;\ \\ 0=W_{13}+0;\ 1=W_{23}+1;\ 0=W_{33}+1.$

(c)
The result is W = $\left( \begin{array}{ccc}
0&1&0 \\ \bar{1}&0&0 \\ 0&0&\bar{1} \end{array} \right)$; w = $\left( \begin{array}{c} 0 \\ 1 \\ 1 \end{array} \right)$.

The resulting matrix-column pair is checked by mapping the fixed point 1/2,1/2,1/2 and the point 1/2,1/2,0. Their images are 1/2,1/2,1/2 and 1/2,1/2,1 in agreement with the geometric meaning of the operation.

Procedure 2

If the images of the origin and/or the coordinate points are not known, other pairs `point-image point' must be used. It is difficult to give general rules but often fixed points are appropriate in such a case. In addition, one may exploit the different transformation behaviour of point coordinates and vector coefficients, see Section 4.4. Vector coefficients `see' only the matrix W and not the column w, and that may facilitate the solution. Nevertheless, the calculations may now become more involved. The next example is not crystallographic in the usual sense, but related to twinning in `spinel' mineral.

Example 3

What is the pair (W,w) for a 2-fold rotation about the space diagonal [111] with the point 1/2,0,0 lying on the rotation axis ?

It is not particularly easy to find the coordinates of the image $\tilde{O}$ of the origin $O$. Therefore, another procedure seems to be more promising. One can use the transformation behaviour of the vector coefficients of the direction [111] and other distinguished directions. The direction [111] is invariant under the 2-fold rotation, and the latter is described by the matrix part only, see Section 4.4. Therefore, the following equations hold

\begin{displaymath}
1=W_{11}+W_{12}+W_{13},\
1=W_{21}+W_{22}+W_{23},\ 1=W_{31}+W_{32}+W_{33}. \end{displaymath} (5.1.1)

On the other hand, the directions [1$\bar{1}$0], [01$\bar{1}$], and [$\bar{1}$01] are perpendicular to [111] and thus are mapped onto their negative directions. This means

\begin{displaymath}
\hspace*{-1em}\left. \begin{array}{*{3}{r@{\hspace{0.5em}}c
...
..._{32}-W_{33},&\ \ -1&=&-W_{31}+W_{33},
\end{array} \right\}.
\end{displaymath} (5.1.2)

From the equations (5.1.2) one concludes

\( \begin{array}{rclrclrcl} W_{12}&=&W_{13},&W_{21}&=&W_{23},&
W_{31}&=&W_{32} \\ W_{11}&=&W_{13}-1,&W_{22}&=&W_{21}-1,&W_{33}&=
&W_{32}-1. \end{array} \)

Together with equations (5.1.1) one obtains

$ W_{11}=W_{22}=W_{33}=-1/3;\ W_{12}=W_{13}=W_{21}=W_{23}=W_{31}=
W_{32}=2/3$.

Thus, W = $\left( \begin{array}{rrr}
-1/3&2/3&2/3\\ 2/3&-1/3&2/3\\ 2/3&2/3&-1/3 \end{array} \right).$

The point 1/2 0 0 is a fixed point, thus
$1/2=-1/3\cdot1/2+2/3\cdot0+2/3\cdot0+w_1$,
$0=2/3\cdot1/2-1/3\cdot0+2/3\cdot0+w_2$, and
$0=2/3\cdot1/2+2/3\cdot0-1/3\cdot0+w_3$.

The coefficients of w are then: $w_1=2/3,\ w_2=-1/3,\ w_3=-1/3.$

There are different tests for the matrix: It is orthogonal, its order is 2 (because it is orthogonal and symmetric), its determinant is $+1$, it leaves the vector $[1\,1\,1]$ invariant, and maps the vectors $[1\,\bar{1}\,0], [0\,1\,\bar{1}]$, and $[\bar{1}\,0\,1]$ onto their negatives (as was used for its construction). The matrix-column pair can be tested with the fixed points, e.g. with $1/2,\,0,\,0$; with $1/2,\,0,\,0+1,\,1,\,1=3/2,\,1,\,1$; or other points on the rotation axis.

Problem 1B. Symmetry of the square.        For the solution, see p. [*].

Problem 1A, p. [*], dealt with the symmetry of the square, see Fig. 3.4.1.

There are 2 more questions concerning this problem.

(v)
Calculate the matrix-column pairs of the symmetry operations of the square.
(vi)
Construct the multiplication table of the group of the square. [The multiplication table of a group $\mbox{$\mathcal{G}$}$ of order $N$ is a table with $N$ rows and $N$ columns. The elements of the group are written on top of the table and on the left side, preferably in the same sequence and starting with the unit element. In the intersection of the $i$th row and the $k$th column the product ${\sf W}_i$${\sf W}_k$ is listed for any pair of indices $1,\,1
\leq i,\,k \leq N,\,N$. The complete table is the multiplication table].

Are there remarkable properties of the multiplication table ?


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