Hi Vincent,<div>The output you show has the following meaning:</div><div><br></div><div>The reindexing of the miller indices of (-k,-h,-l) of the P1 data gives you data in the space group "<span style="font-family:arial,sans-serif;font-size:12.731481552124023px">C 1 2 1 (x-y,x+y,z)</span><span style="font-family:arial,sans-serif;font-size:12.731481552124023px">". This is a special (no lattice translations) version of C2 that can be set to the standard setting by applying the transformation </span><span style="font-family:arial,sans-serif;font-size:12.731481552124023px">(x-y,x+y,z) to the cell. This will subsequently reindex your miller indices as well.</span></div>
<div><span style="font-family:arial,sans-serif;font-size:12.731481552124023px"><br></span></div><div><span style="font-family:arial,sans-serif;font-size:12.731481552124023px">The "C 1 2 1 (</span><font face="arial, sans-serif">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 would incorrectly index your data.</font></div>
<div><span style="font-family:arial,sans-serif;font-size:12.731481552124023px"><br></span></div><div><span style="font-family:arial,sans-serif;font-size:12.731481552124023px">If you go here:</span></div><div><span style="font-family:arial,sans-serif;font-size:12.731481552124023px"><br>
</span></div><div><font face="arial, sans-serif"><a href="http://cci.lbl.gov/cctbx/explore_symmetry.html">http://cci.lbl.gov/cctbx/explore_symmetry.html</a></font><br></div><div><font face="arial, sans-serif"><br></font></div>
<div><font face="arial, sans-serif">and fill out the space group </font><span style="font-family:arial,sans-serif;font-size:12.731481552124023px">C 1 2 1 (x-y,x+y,z) you get this info</span></div><div><span style="font-family:arial,sans-serif;font-size:12.731481552124023px"><br>
</span></div><div><pre style="color:rgb(0,0,0)">List of symmetry operations:
</pre><table border="2" cellpadding="2" style="font-family:Times"><tbody><tr><th>Matrix</th><th>Rotation-part type</th><th>Axis direction</th><th>Screw/glide component</th><th>Origin shift</th></tr><tr><td><tt>x,y,z</tt></td>
<td>1</td><td>-</td><td>-</td><td>-</td></tr><tr><td><tt>-y,-x,-z</tt></td><td>2</td><td>[-1,1,0]</td><td>0,0,0</td><td>0,0,0<br><br></td></tr></tbody></table></div><div><span style="font-family:arial,sans-serif;font-size:12.731481552124023px"><br>
</span></div><div><span style="font-family:arial,sans-serif;font-size:12.731481552124023px">i.e.</span></div><div><span style="font-family:arial,sans-serif;font-size:12.731481552124023px"><br></span></div><div><span style="font-family:arial,sans-serif;font-size:12.731481552124023px">your two-fold axis lies in the xy plane.</span></div>
<div><span style="font-family:arial,sans-serif;font-size:12.731481552124023px"><br></span></div><div><span style="font-family:arial,sans-serif;font-size:12.731481552124023px">I'll contact you off-list with more info.</span></div>
<div><span style="font-family:arial,sans-serif;font-size:12.731481552124023px"><br></span></div><div><span style="font-family:arial,sans-serif;font-size:12.731481552124023px">P</span></div><div><span style="font-family:arial,sans-serif;font-size:12.731481552124023px"><br>
</span></div><div><span style="font-family:arial,sans-serif;font-size:12.731481552124023px"> </span></div><div><br></div><div class="gmail_extra"><br><br><div class="gmail_quote">On 21 November 2012 08:50, vincent Chaptal <span dir="ltr"><<a href="mailto:vincent.chaptal@ibcp.fr" target="_blank">vincent.chaptal@ibcp.fr</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">Dear all,<br>
<br>
I ran "phenix.explore_metric_<u></u>symmetry" with the command:<br>
<br>
phenix.explore_metric_symmetry --unit_cell="97,97,225,78,78,<u></u>68" --space_group=P1<br>
<br>
It output:<br>
...<br>
-------------------------<br>
Transforming point groups<br>
-------------------------<br>
>From P 1 to C 1 2 1 (x-y,x+y,z) using :<br>
* -k,-h,-l<br>
...<br>
<br>
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.<br>
<br>
Naive question: why aren't the two matrices similar?<br>
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?<br>
<br>
Also, in C2 there is a 2-fold axis parallel to b, so reflections (h,k,l) are equivalent to (-h, k, -l).<br>
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?<br>
<br>
thank you<br>
vincent<br>
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</blockquote></div><br><br clear="all"><div><br></div>-- <br>-----------------------------------------------------------------<br>P.H. Zwart<br>Research Scientist<br>Berkeley Center for Structural Biology<br>Lawrence Berkeley National Laboratories<br>
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