Hi Matthew,
Your initial description of the symmetry operations and their
multiplication is just fine.
Your error is at the assumption that the triple application of a 3bar
rotation is the unit matrix.
The 3bar operation ( you call it 3Q) is the combination of the 3-fold
rotation with a center of symmetry
into a single symmetry operation. The cycle for this operations is
actually of length six.
To "visualize" this, take a regular cube with the 3-fold rotation axis
along the body diagonal.
For a normal 3-fold rotation you turn the cube around the body diagonal
by 120 degrees and after
3 such operation you are back at the original position.
For the 3bar rotation the first operation moves a side of the cube by
120 degrees and then "inverts" it
to the opposite side. Try it out with a dice, with all even sides up.
The first operation of the "6" then
ends up on the lower side of the dice at at the "5"; then the "5" is
copied into the "4" etc.
The matrix formalism actually becomes easier if you follow the augmented
notation, as described in more recent issues of the IT, or in the
teaching edition.
A symmetry operation in these issues uses letters W and w as P and Q are
used for unit cell transformations
but that is a trivial, irrelevant detail. Anyway, a symmetry operation
(W,w) can be written as a single
4x4 matrix in the form :
( W11 W12 W13 w1 )
( W21 W22 W23 w2 )
( W31 W32 W33 w3 )
( 0 0 0 1 )
This "augmented" matrix allows to use the standard matrix
multiplications rules for matrices of rank two.
So with two symmetry operations W and M in augmented form, the
combination is simply
O = W*M
Cyclic application of the 3bar symmetry operation with translation
vectors : -z -x -y; 3/4 3/4 3/4 will give :
0 0 -1 | 0.75
-1 0 0 | 0.75
0 -1 0 | 0.75
0 1 0 | 0.00
0 0 1 | 0.00
1 0 0 | 0.00
-1 0 0 | 0.75
0 -1 0 | 0.75
0 0 -1 | 0.75
0 0 1 | 0.00
1 0 0 | 0.00
0 1 0 | 0.00
0 -1 0 | 0.75
0 0 -1 | 0.75
-1 0 0 | 0.75
1 0 0 | 0.00
0 1 0 | 0.00
0 0 1 | 0.00
Where the translation components are restricted to the interval [-1,1]
Another few points:
Several space groups Fd-3c included are listed in the ITA with two
origin choices. Origin choice 1 has
the origin at the intersection of the higher symmetry elements like the
rotation axes, while choice
two has the origin at the center of symmetry. This will change several
of the generator matrices.
-z -x -y; 3/4 3/4 3/4 is usually not a generator matrix for Fd-3c (228);
origin choice 1.
If you want to reproduce the atom coordinates in the exact sequence as
in the ITA, you need to use the
generators as listed in the ITA BUT apply the centering operations last!
For F3-3c the standard generators are:
! 2 ( 0, 0, 1/2) 0,1/4,z
! 2 ( 0, 1/2, 0) 1/4,y,0
! 3 x,x,x
! 2 (1/2, 1/2, 0) x,x-1/4,3/8
! -1 3/8,3/8,3/8
! t (0 , 1/2, 1/2)
! t (1/2, 0, 1/2)
for Origin choice 2:
! 2 ( 0, 0, 1/2) 1/8,3/8,z
! 2 ( 0, 1/2, 0) 3/8,y,1/8
! 3 x,x,x
! 2 (1/2, 1/2, 0) x,x-1/4,0
! -1 0,0,0
! t (0 , 1/2, 1/2)
! t (1/2, 0, 1/2)
Note for example the difference in the 5th generator -1 = 1bar.
Another note, the generators in the ITA are listed such that starting
from coordinates x,y,z you apply
the 1st generator to these coordinates. Then you can apply the second
generator to all previously
generated position and so on. You do not need a recursive algorithm to
generate all positions. for
special positions, you can terminate once a previous coordinate is
reproduced.
Feel free to check out the generators and the generation of symmetry
matrices in my DISCUS code
(generate_mod.f90 and spcgr_apply.f90 at
https://github.com/tproffen/DiffuseCode
Hope this helps
Reinhard Neder
Am 08.11.22 um 04:42 schrieb Matthew Rowles:
Hi all
For various reasons, I'm writing some code to generate the general
positions of the space groups. I'm trying to follow the implementation
of Shmueli in their 1984 paper and ITB chapter[1]
My generated positions are fine for all except about 10 settings, and
I think that I might have an issue with how I'm generating the various
matrices which have improper rotations with translation vectors.
I've deduced that a matrix, P, with translation vector, t, multiplied
by Q and u, can be represented as (P,t)*(Q,u) = (PQ, Pu + t), and so
I assume that (P,t)^2 is just (PP, Pt+t).
Following equation 12 from the 1984 paper, I can see that, (for
example) from space group P4(1)32, one of the generating matrices is
(4C, 393), and if I follow the above schema, I can generate (4C, 393),
(4C, 393)^2, (4C, 393)^3, and (4C, 393)^4==(1A,000) (the identity).
{(-y+1/4, x+3/4, z+1/4), (-x+1/2, -y, -z+1/2), (y+1/4, -x+1/4, z+3/4),
(x,y,z)} (see the end for matrices)
I can't get the same cycle when I start with an improper rotation with
a non-zero translation vector.
For example:
I3Q999 from Fd-3c. -> Improper rotation of matrix 3Q (z,x,y) with a
translation vector of (3/4,3/4,3/4)
From Table 4 in the 1984 paper, I know that this is (-z+3/4, -x+3/4,
-y+3/4), so the "improper" rotation only affects the signs of the
matrix, not the translation
But I think I'm missing out something fundamental, as (I3Q999)^3 !=
(P1A000)
eg: {(-z+3/4, -x+3/4, -y+3/4), (y, z, x), (-x+3/4, -y+3/4, -z+3/4)}
I3Q999 = (P,t)
[ 0 0 -1 | 3/4 ]
[ -1 0 0 | 3/4 ]
[ 0 -1 0 | 3/4 ]
(I3Q999)^2 = I3Q999 * I3Q999 = (P,t)*(P,t) = (PP, Pt + t)
[ 0 1 0 | 0 ]
[ 0 0 1 | 0 ]
[ 1 0 0 | 0 ]
(I3Q999)^3 = I3Q999 * I3Q999 * I3Q999 = (P,t)*(PP, Pt + t) = (PPP, PPt
+ Pt + t)
[ -1 0 0 | 3/4 ] [ 1 0 0 | 0 ]
[ 0 -1 0 | 3/4 ] != [ 0 1 0 | 0 ]
[ 0 0 -1 | 3/4 ] [ 0 0 1 | 0 ]
There aren't any examples of improper transformations in the papers, I
don't know enough group theory to get enough out of the rest of the
international tables, and I don't know the notation of Zachariasen [2].
Can anybody help point me in the right direction?
Thanks
Matthew Rowles
[1] Acta Cryst (1984), A40, 567-571 and International Tables, Vol. B,
Chapter 1.4
[2] Theory of X-Ray Diffraction in Crystals (1967), Ch. 2.
P4C393 = (P,t)
[ 0 -1 0 | 1/4 ]
[ 1 0 0 | 3/4 ]
[ 0 0 1 | 1/4 ]
(P4C393)^2 = (P,t)*(P,t) = (PP, Pt + t)
[-1 0 0 | 1/2 ]
[ 0 -1 0 | 0 ]
[ 0 0 1 | 1/2 ]
(P4C393)^3 = (P,t)*(PP, Pt + t) = (PPP, PPt + Pt + t)
[ 0 1 0 | 1/4 ]
[ -1 0 0 | 1/4 ]
[ 0 0 1 | 3/4 ]
(P4C393)^4 = (P,t)*(PPP, PPt + Pt + t) = (PPPP, PPPt + PPt + Pt + t)
[ 1 0 0 | 0 ]
[ 0 1 0 | 0 ]
[ 0 0 1 | 0 ]
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Prof. Dr. Reinhard Neder
Kristallographie und Strukturphysik
Universitaet Erlangen
Staudtststr. 3; 91058 Erlangen
tel. +49-9131-8525191
fax +49-9131-8525182
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