Thanks for that Reinhard.
The order of the -3 operation being 6 was the key to fix all (most) of
my problems. I'll probably implement the augmented matrix methodology
in my code, as it looks like it simplifies keeping track of the
translation vector.
I've also just reserved my library's copy of Giacovazzo for me; I've
been told that should help out.
Just for completeness, and the record: From another email I got, in
general, the order of a rotation N is N. The order of a rotoinversion
-N is N if N is even, otherwise it is of order 2N.
I'm following the generation algorithm outlined by Shumeli (corrected
page numbers in [1]); for Fd-3c:1, the three generators are I3Q999,
P4C993, and P2D939 == -z -x -y; 3/4 3/4 3/4, -y x z; 3/4 0 1/4, and y
x -z; 3/4 1/4 3/4
I'll see if I can figure out how to alter the generated position order
to match the ITA; at this point in time, the fact that I have the
correct positions is enough, but it would be good to fix the order for
completeness
Ta a lot!
Matthew
[1] Acta Cryst (1984), A40, 559-567
On Tue, 8 Nov 2022 at 18:06, Reinhard Neder <reinhard.ne...@fau.de> wrote:
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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