Example 2

6
0 1 0 0 0 0
1 0 0 0 0 0
0 0 0 1 0 0
0 0 1 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
6
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
1 0 0 0 0 0
0 1 0 0 0 0

Used Programs

Solution

1. Write the two matrices in a file 'G' and add

     #g2
   

   or

     #2
   

   as top row.
2. Call

      Symbol G
   

   to compute the family symbol of the group in G.
   It turns out to be 3;2-2;1 .
3. Call

      Bravais_catalog
   

   and input the symbol '3;2-2;1'. It says that there are 3 homogeneously decomposable Bravais groups in the family and that there are altogether 6 + 8 + 4 Z-classes of Bravais groups. Don't calculate the Bravais groups.
4. Call

      Order -o G >> G
   

   It adds a bootom line to the file 'G' containing the order of the group G in factorized form to prepare the file as input for QtoZ.

   Alternatives:

   1. Is_finite -o G >> G has the same effect, but works faster.
   2. Order -O G produces a slightly more complete file, containing all the necessary information.
5. Call

      QtoZ -D G
   

   Here the Q-class of the group G is split up into Z-classes. It turns out that there are 18 Z-classes. Because of the -D-option the representative groups are put into the following files

   G.1.1	G.1.2	G.1.3	G.1.4	G.1.5	G.1.6	G.1.7	G.1.8
   G.2.1	G.2.2	G.2.3	G.2.4	G.2.5	G.2.6
   G.3.1	G.3.2	G.3.3	G.3.4

6. Call

      Bravais_type G.1.1 > type.G.1.1
   

   etc. to

      Bravais_type G.3.4 > type.G.3.4
   

   to compute the Bravais group in the Bravais catalog which is Z-equivalent to the Bravais group of the group in G.1.1, etc. to G.3.4:

   type.G.1.1	type.G.1.2	type.G.1.3	type.G.1.4	type.G.1.5	type.G.1.6	type.G.1.7	type.G.1.8
   type.G.2.1	type.G.2.2	type.G.2.3	type.G.2.4	type.G.2.5	type.G.2.6
   type.G.3.1	type.G.3.2	type.G.3.3	type.G.3.4

   It turns out that each Bravais type in the family is represented with multiplicity one.