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Combination and reversion of mappings

The combination of 2 symmetry operations follows the procedure of Section 2.2. In analogy to equations (2.2.5) to (2.2.8) one obtains

\begin{displaymath}
\tilde{\mbox{\textit{\textbf{x}}}}=\mbox{\textit{\textbf{U}}...
...xtit{\textbf{x}}}+\mbox{\textit{\textbf{u}}};\rule{3.7cm}{0cm}
\end{displaymath} (4.2.1)

\begin{displaymath}
\tilde{\tilde{\mbox{\textit{\textbf{x}}}}} = \mbox{\textit{...
...{\textbf{x}}}}+\mbox{\textit{\textbf{v}}}; \rule{3.7cm}
{0cm}
\end{displaymath} (4.2.2)

\begin{displaymath}\tilde{\tilde{\mbox{\textit{\textbf{x}}}}}=\mbox{\textit{\tex...
...tit{\textbf{u}}})+\mbox{\textit{\textbf{v}}};\rule{2.2cm}{0cm}
\end{displaymath} (4.2.3)

\begin{displaymath}
\tilde{\tilde{\mbox{\textit{\textbf{x}}}}}=\mbox{\textit{\te...
...tbf{W}}}\mbox{\textit{\textbf{x}}}+\mbox{\textit{\textbf{w}}}.
\end{displaymath} (4.2.4)

These equations may be formulated with matrix-column pairs:

\begin{displaymath}
\tilde{\tilde{\mbox{\textit{\textbf{x}}}}}=(\mbox{\textit{\t...
...W}}},\,\mbox{\textit{\textbf{w}}})\mbox{\textit{\textbf{x}}}.
\end{displaymath} (4.2.5)

Note that in the product (V,v)(U,u) the operation (U,u) is performed first and (V,v) second. Because of writing point coordinates and vector coefficients as columns, in the combination of their mappings the sequence is always from right to left.

By comparing equations (4.2.4) and (4.2.5) one obtains

\begin{displaymath}
(\mbox{\textit{\textbf{W}}},\,\mbox{\textit{\textbf{w}}})=(\...
...,\mbox{\textit{\textbf{V\,u}}} + \mbox{\textit{\textbf{v}}}).
\end{displaymath} (4.2.6)

This law of composition for matrix-column pairs is not easy to keep in mind because of its asymmetry. It would be easy if the resulting matrix part would be the product of the original matrices and the resulting column the sum of the original columns. However, the column u of the operation, which is applied to the point $X$ first, is multiplied with the matrix V of the second operation, before the addition is carried out. In the next section a formalism will be introduced which smoothes out this awkwardness.

The multiplication of matrix-column pairs is associative, because

$\displaystyle (\mbox{\textit{\textbf{W}}},\,\mbox{\textit{\textbf{w}}})((\mbox{...
...textit{\textbf{v}}})(\mbox{\textit{\textbf{U}}},
\,\mbox{\textit{\textbf{u}}}))$ $\textstyle =$ $\displaystyle (\mbox{\textit{\textbf{W}}},\,\mbox{\textit{\textbf{w}}})(\mbox{\...
...ox{\textit{\textbf{V}}}\mbox{\textit{\textbf{u}}}+\mbox{\textit{\textbf{v}}}) =$  
  $\textstyle =$ $\displaystyle (\mbox{\textit{\textbf{W}}}\mbox{\textit{\textbf{V}}}\mbox{\texti...
...tbf{W}}}\mbox{\textit{\textbf{v}}}+
\mbox{\textit{\textbf{w}}}),\rule{3em}{0ex}$ (4.2.7)

and on the other hand,

$\displaystyle ((\mbox{\textit{\textbf{W}}},\,\mbox{\textit{\textbf{w}}})(\mbox{...
...textit{\textbf{v}}}))(\mbox{\textit{\textbf{U}}},\,
\mbox{\textit{\textbf{u}}})$ $\textstyle =$ $\displaystyle (\mbox{\textit{\textbf{W}}}\mbox{\textit{\textbf{V}}},\,\mbox{\te...
...extit{\textbf{w}}})(\mbox{\textit{\textbf{U}}},\,
\mbox{\textit{\textbf{u}}}) =$  
  $\textstyle =$ $\displaystyle (\mbox{\textit{\textbf{W}}}\mbox{\textit{\textbf{V}}}\mbox{\texti...
...tbf{W}}}\mbox{\textit{\textbf{v}}}
+\mbox{\textit{\textbf{w}}}).\rule{3em}{0ex}$ (4.2.8)

By comparison of both expressions one finds

\begin{displaymath}
(\mbox{\textit{\textbf{W}}},\,\mbox{\textit{\textbf{w}}})((\...
...}))
(\mbox{\textit{\textbf{U}}},\,\mbox{\textit{\textbf{u}}}). \end{displaymath} (4.2.9)

Associativity is a very important property. It can be used, e.g., to find the value of a product of matrix-column pairs without any effort. Suppose, that in the above triple product of matrix-column pairs, \( (\mbox{\textit{\textbf{W}}},\,\mbox{\textit{\textbf{w}}}) = (\mbox{\textit{\textbf{V}}},\,\mbox{\textit{\textbf{v}}})^{-1} \) holds and the upper sequence of multiplications is to be calculated. Then, due to the associativity the second equation may be used instead. Because $(\mbox{\textit{\textbf{V}}},\,\mbox{\textit{\textbf{v}}})^{-1}(\mbox{\textit{\t...
...\textit{\textbf{v}}})=(\mbox{\textit{\textbf{I}}},\,\mbox{\textit{\textbf{o}}})$ is the identity mapping, the result `(U,u)' is obtained immediately.

A linear mapping is a mapping which leaves the origin fixed. Its column part is thus the o column. According to equation (4.2.6) any matrix-column pair can be decomposed into a linear mapping (W,o) containing W only and a translation (I,w) with w only:

\begin{displaymath}
(\mbox{\textit{\textbf{W}}},\,\mbox{\textit{\textbf{w}}}) = ...
...}})(\mbox{\textit{\textbf{W}}},\,\mbox{\textit{\textbf{o}}}).
\end{displaymath} (4.2.10)

The linear mapping has to be performed first, the translation after that.

Question: What is the result if the translation (I,w) is performed first, and the linear mapping (W,o) after that, i.e. if the factors are exchanged ?

Before the reversion of a symmetry operation is dealt with, a general remark is appropriate. In general, the formulae of this section are not restricted to crystallographic symmetry operations but are valid also for affine mappings. However, there is one exception. In the inversion of a matrix W the determinant $\det{(\mbox{\textit{\textbf{W}}})}$ appears in the denominator of the coefficients of $\mbox{\textit{\textbf{W}}}^{-1}$, see Subsection 2.6.1. Therefore, the condition $\det{(\mbox{\textit{\textbf{W}}})}\ne 0$ has to be fulfilled. Such mappings are called regular or non-singular. Otherwise, if $\det{(\mbox{\textit{\textbf{W}}})}~=~0$, the mapping is a projection and can not be reverted. For crystallographic symmetry operations, i.e. isometries W, always $\det{(\mbox{\textit{\textbf{W}}})}=\pm 1$ holds. Therefore, an isometry is always reversible, a general affine mapping may not be. Projections are excluded from this manuscript because they do not occur in crystallographic groups.

Now to the calculation of the reverse of a matrix-column pair. It is often necessary to know which matrix C and column c belong to that symmetry operation C which makes the original action W undone, i.e. which maps every image point ${\tilde X}$ onto the original point $X$. The operation C is called the reverse operation of W. The combination of W with C restores the original state and the combined action CW maps $X \rightarrow \tilde{X} \rightarrow X$. It is the identity operation I which maps any point $X$ onto itself. The operation I is described by the matrix-column pair (I,o), where I is the unit matrix and o is the column consisting of zeroes only. This means

\begin{displaymath}\mbox{\textit{\textbf{x}}}=(\mbox{\textit{\textbf{C}}},\,\mbo...
...xtbf{x}}}} + \mbox{\textit{\textbf{c}}} \hspace{1cm} \mbox{or} \end{displaymath}
\begin{displaymath}
\mbox{\textit{\textbf{x}}}=\mbox{\textit{\textbf{I}}}\mbox{\...
...bf{C}}}\mbox{\textit{\textbf{w}}}+\mbox{\textit{\textbf{c}}}.
\end{displaymath} (4.2.11)
Equation (4.2.11) is valid for any coordinate triplet x. Therefore, the coefficients on the right and left side are the same. It follows

\begin{displaymath}\mbox{\textit{\textbf{C}}}\mbox{\textit{\textbf{W}}}=\mbox{\t...
...bf{w}}}+\mbox{\textit{\textbf{c}}}=\mbox{\textit{\textbf{o}}}, \end{displaymath}

\begin{displaymath}
\mathit{i.\,e.}\ \ (\mbox{\textit{\textbf{C}}},\,\mbox{\text...
...,-\mbox{\textit{\textbf{W}}}^{-1}\mbox{\textit{\textbf{w}}}).
\end{displaymath} (4.2.12)
This equation is as unpleasant as is equation (4.2.6). The matrix part is fine but the column part is not just $-\mbox{\textit{\textbf{w}}}$ as one would like to see but $-\mbox{\textit{\textbf{w}}}$ has to be multiplied with $\mbox{\textit{\textbf{W}}}^{-1}$. The next section will present a proposal how to overcome this inconvenience.

It is always good to test the result of a calculation or derivation. One verifies the validity of the equations $(\mbox{\textit{\textbf{W,\,w}}})^{-1}
(\mbox{\textit{\textbf{W,\,w}}})=(\mbox{\...
...textbf{W,\,w}}})^{-1}=(\mbox{\textit{\textbf{I}}},\,\mbox{\textit{\textbf{o}}})$ by applying equations (4.2.6) and (4.2.12). In addition in the following Problem 2A the results of this section may be practised.

Problem 2A. Symmetry described by matrix-column pairs.

For the solution, see p. [*].

In Vol. A of International Tables for Crystallography the crystallographic symmetry operations A, B, ... are referred to a conventional coordinate system and are represented by matrix-column pairs (A,a), (B,b), .... Among others one finds in the space-group tables of IT A indirectly, see Section 4.6:

\( (\mbox{\textit{\textbf{A}}}, \mbox{\textit{\textbf{a}}}) = \left(
\begin{arr...
...array} \right),
\left( \begin{array}{r} 1/2 \\ 1/2 \\ 1/2 \end{array} \right) \) and \( (\mbox{\textit{\textbf{B}}}, \mbox{\textit{\textbf{b}}}) = \left(
\begin{arr...
...0 \end{array} \right),
\left( \begin{array}{r} 0\\ 0\\ 0 \end{array} \right). \)

Combining two symmetry operations or reversion of a symmetry operation corresponds to multiplication or reversion of these matrix-column pairs, such that the resulting matrix-column pair represents the resulting symmetry operation.

The following calculations make use of the formulae 4.2.6 and 4.2.12.
Can one exploit the fact that the matrices A, B, C, and D are orthogonal matrices ?

Questions

(i)
What is the matrix-column pair resulting from
(B, b)(A, a) = (C,c) ?
(ii)
What is the matrix-column pair resulting from
(A, a)(B, b) = (D,d) ?
(iii)
What is (A,a)$^{-1}$ ?
(iv)
What is (B,b)$^{-1}$ ?
(v)
What is (C,c)$^{-1}$ ?
(vi)
What is (D,d)$^{-1}$ ?
(vii)
What is (B,b)$^{-1}$(A, a)$^{-1}$ ?
For another question of this Problem, see p. [*].


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Next: () matrices Up: The description of mappings by ... Previous: Matrix-column pairs

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