CARAT Introduction / Examples / Find the dual pairs of Bravais groups in family 1;1;1;1
Author: Oliver Heidbüchel
Description: CARAT Introduction: Example 6 - Find the dual pairs of Bravais groups in family 1;1;1;1
Keywords: CARAT, crystallography, bravaisgroup, bravaisgroups, bravais groups, bravais group, mathematics, algorithms, tables, algorithm, table, examples, example, Bravais_catalog, Datei, Bravais_inclusions, Bravais_type, Symbol, Tr_bravais, dual, dual pair, family

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### Example 6

Find the dual pairs of Bravais groups in family 1;1;1;1

### Used Programs

Bravais_catalog/Datei, Bravais_inclusions, Bravais_type/ Symobl -i, Tr_bravais

### Solution

1. Call

                Bravais_catalog

   and input 1;1;1;1 as family symbol.
   It turns out that there are nine Z-classes of Bravais groups in the family only one of which is homogeneously decomposable. Though we might not need all of these groups explicitly, we write all of them in the file 'b'. Split 'b' up into nine files 'b1', 'b2', 'b3', 'b4', 'b5', 'b6', 'b7', 'b8' and 'b9', so that 'bi' contains 1;1;1;1_1_i .
2. Call

                Bravais_catalog

   and input 4-1 as family symbol. Write all Bravais groups in this family to a file 'B'. There are two groups in file 'B' now. We want the Bravais group of the form represented by the unit matrix, which is the first group. The idea is that any of the subgroups of this group is Z-equivalent (actually equal) to its transposed.
3. Call

                Bravais_inclusions B > Bin
                grep 1\;1\;1\;1 Bin

   We get the file 'Bin' and the output

                Symbol: 1;1;1;1  homogeneously d.: 1 zclass: 2
                Symbol: 1;1;1;1  homogeneously d.: 1 zclass: 1
                Symbol: 1;1;1;1  homogeneously d.: 1 zclass: 5
                Symbol: 1;1;1;1  homogeneously d.: 1 zclass: 9

   We find that the groups 1;1;1;1_1_i with i = 1, 2, 5, or 9 contain the unit matrix in their form space (at least up to Z-equivalence).
   Hence these groups are Z-equivalent to their transposed groups. Note that an odd number of groups still have to be paired. Hence one group must be Z-equivalent to itself, without fixing a form Z-equivalent to the identity form.
4. Call

                Tr_bravais b3 > b3t
                Bravais_type b3t > b3type

   to find that 1;1;1;1_1_3 and 1;1;1;1_1_6 are paired. Similarly one finds that 1;1;1;1_1_4 and 1;1;1;1_1_8 are paired and that 1;1;1;1_1_7 is paired to itself (without fixing I_4).
   Doing this we get the files b3t, b3type, b4t, and b4type.

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last change: 11.09.2000