[Jmol-developers] [ jmol-Bugs-2107269 ] CML reader loses crystal lattice information if no symmetry

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Bugs item #2107269, was opened at 2008-09-12 13:13
Message generated for change (Comment added) made by danwilson4
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Category: File Input/Output
Group: None
Status: Open
Resolution: None
Priority: 5
Private: No
Submitted By: dan (danwilson4)
Assigned to: Bob Hanson (hansonr)
Summary: CML reader loses crystal lattice information if no symmetry

Initial Comment:
Jmol (11.6RC14) seems to lose the lattice parameters information if space group information isn't present.

In the attached CML file (UO3.cml), the system has a primitive cell, and hence it's not valid to attach a spaceGroup="..." attribute to the <symmetry> tag. However, in such cases, Jmol doesn't read or display the lattice parameters, assuming the system is molecular.







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>Comment By: dan (danwilson4)
Date: 2008-09-14 11:20

Message:
I guess there are two points here.

First, the supplied CML file for UO3 has it in its primitive setting,
rather
than in its conventional (centred) setting. So it has symmetry, and is
related
to one of the 230 space groups, but it is not one of the 230 space groups,
and
hence doesn't have a H-M symbol to insert into the spaceGroup attribute.

The other, broader, point is that very few of our simulation codes know
anything about H-M/Hall symbols, so more often than not, they simply
don't
have enough information to add the spaceGroup attribute themselves. Jmol
should still display the lattice in these circumstances, treating the
system as P1.

(For the record, CASTEP doesn't output the spaceGroup attribute — I have
a
post-processing utility which reads and interprets the <transform3>s,
then
inserts the spaceGroup attribute into the markup.)

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Comment By: Bob Hanson (hansonr)
Date: 2008-09-13 18:41

Message:
Are you saying that the system is no known space group?

I see in that file:

    <symmetry>
      <transform3>
1.000000000000e0 0.000000000000e0 0.000000000000e0 0.000000000000e0
 0.000000000000e0 1.000000000000e0 0.000000000000e0 0.000000000000e0 
0.000000000000e0 0.000000000000e0 1.000000000000e0 0.000000000000e0 
0.000000000000e0 0.000000000000e0 0.000000000000e0
1.000000000000e0</transform3>
      <transform3>
1.000000000000e0 0.000000000000e0 0.000000000000e0 0.000000000000e0 
0.000000000000e0 -1.000000000000e0 0.000000000000e0 0.000000000000e0 
0.000000000000e0 0.000000000000e0 -1.000000000000e0 0.000000000000e0 
0.000000000000e0 0.000000000000e0 0.000000000000e0
1.000000000000e0</transform3>
      <transform3>
1.000000000000e0 0.000000000000e0 0.000000000000e0 0.000000000000e0
 0.000000000000e0 -1.000000000000e0 0.000000000000e0 0.000000000000e0
 0.000000000000e0 0.000000000000e0 1.000000000000e0 0.000000000000e0 
0.000000000000e0 0.000000000000e0 0.000000000000e0
1.000000000000e0</transform3>
      <transform3>
1.000000000000e0 0.000000000000e0 0.000000000000e0 0.000000000000e0 
0.000000000000e0 1.000000000000e0 0.000000000000e0 0.000000000000e0 
0.000000000000e0 0.000000000000e0 -1.000000000000e0 0.000000000000e0 
0.000000000000e0 0.000000000000e0 0.000000000000e0
1.000000000000e0</transform3>
    </symmetry>

That to me reads:

  x,y,z;x,-y,-z;x,-y,z;x,y,-z;

But when I use those symmetry operators with this data set I get
nonsense.

In the wurtzite example you sent, these transformation matrices are very
odd
they are in fractional coordinates, but the rotational components use
angles defined
in terms of cartesian coordinates. So they are useless to Jmol. 

Since Jmol then cannot read the transform3 entries, it must depend upon a
defined 
unit cell and proper spacegroup name. 

Say again why it would be inappropriate to have a spacegroup defined
here.

Bob




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