[phenixbb] phenix.explore_metric_symmetry

Peter Zwart PHZwart at lbl.gov
Wed Nov 21 09:29:52 PST 2012

```Hi Vincent,
The output you show has the following meaning:

The reindexing of the miller indices of (-k,-h,-l) of the P1 data gives you
data in the space group "C 1 2 1 (x-y,x+y,z)". This is a special (no
lattice translations) version of C2 that can be set to the standard setting
by applying the transformation (x-y,x+y,z) to the cell. This will
subsequently reindex your miller indices as well.

The "C 1 2 1 (x-y,x+y,z) " is a full space group. The change of basis
matrix stuck at the back of the standard symbol indicates which
transformation is required to get it back to the traditional setting.
INstead of specifying it in x,y,z, you can also make specify the COB on the
unit cell basis vectors. This allows you to do fun things like  P1 (2a,b,c)
to describe the space group P1 with added operator (x+1/2,y,z), i.e. if you

If you go here:

http://cci.lbl.gov/cctbx/explore_symmetry.html

and fill out the space group C 1 2 1 (x-y,x+y,z) you get this info

List of symmetry operations:

MatrixRotation-part typeAxis directionScrew/glide componentOrigin shiftx,y,z
1----y,-x,-z2[-1,1,0]0,0,00,0,0

i.e.

your two-fold axis lies in the xy plane.

P

On 21 November 2012 08:50, vincent Chaptal <vincent.chaptal at ibcp.fr> wrote:

> Dear all,
>
> I ran "phenix.explore_metric_**symmetry" with the command:
>
> phenix.explore_metric_symmetry --unit_cell="97,97,225,78,78,**68"
> --space_group=P1
>
> It output:
> ...
> -------------------------
> Transforming point groups
> -------------------------
> From P 1   to  C 1 2 1 (x-y,x+y,z)  using :
>   *  -k,-h,-l
> ...
>
> I understand that going from P1 to C2, one needs to apply the
> transformation matrix (x-y, x+y,z) on the P1 cell to form the C2 cell, and
> (-k, -h, -l) on the reflections.
>
> Naive question: why aren't the two matrices similar?
> The reciprocal space is the fourier transform of the real space; i was
> thinking that a reorientation matrix in the real space would be kept in the
> reciprocal space. My maths are not that good, and in P1 it is more complex
> than other space groups. Can someone tell me why the matrices are different?
>
> Also, in C2 there is a 2-fold axis parallel to b, so reflections (h,k,l)
> are equivalent to (-h, k, -l).
> In P1, they are not. Applying the above transformation matrix on the
> reflections would give (hP1, kP1, lP1) transforms into (-kP1, -hP1, -lP1),
> and these are equivalent to (kP1, -hP1, lP1)? Is this correct?
>
> thank you
> vincent
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>

--
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P.H. Zwart
Research Scientist
Berkeley Center for Structural Biology
Lawrence Berkeley National Laboratories
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