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In the International Tables for Crystallography for each crystallographic group an asymmetric unit is supplied (mathematicians call this a fundamental domain of the group). This region is a bounded polyhedron in $\mathbb{R}^3$ that is given a a bunch of inequalities.

What I fail to extract from the IT is the coordinate system used. So my question is

How do I read off the coordinate system used for a group from the International Tables for Crystallogaphy?

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    $\begingroup$ see if this helps: mcl1.ncifcrf.gov/dauter_pubs/284.pdf $\endgroup$ – DavePhD Nov 24 '14 at 15:34
  • $\begingroup$ Isn’t is a purely mathematical question? Shouldn’t it migrate? (There is some technique for moving threads across StackExchange sites.) $\endgroup$ – Incnis Mrsi Nov 24 '14 at 18:50
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    $\begingroup$ @IncnisMrsi: Unfortunately, this is not a purely mathematical question. It is a questions about tables used by crystallographers and solid state scientists, and about particular information contained in it. Mathematicians normally don't use this table at all. $\endgroup$ – eins6180 Nov 24 '14 at 19:17
  • $\begingroup$ @DavePhD: Thanks for the link. I had a closer look at the article but it didn't answer the questions completely. It stated that the coordinates of the fundamental domain are given in terms of the basis vectors of the lattice. But this still leaves my question open, how can I determine the coordinate system used for the given crystallographic group? $\endgroup$ – eins6180 Nov 27 '14 at 17:49
  • $\begingroup$ After a brief look, it seems that they always use a cartesian x,y,z system. Do you have any examples where this is not true? It might also help to include an entry from the tables in the question as an image. I think it's not very clear what exactly is asked in this question. A rephrasing/expanding and an example might help. $\endgroup$ – Heterotic Nov 29 '14 at 14:43
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The ambiguity is resolved by choosing the coordinates first. So:

1)Set up a cartesian system with x,y,z coordinates.

2)Pick the group you would like to study. Let's say $P4_32_12$ from page 1151 of this document: http://mcl1.ncifcrf.gov/dauter_pubs/284.pdf that DavePhD recommended.

3)You can then see that the asymmetric unit is given by $$0\leq x\leq1,\quad 0\leq y\leq1,\quad 0\leq z\leq\frac18$$ in this coordinate system that you have already chosen.

A few more comments:

i)If you would like to move to a different coordinate system you are of course free to do that. You can apply a transformation that changes the coordinates and then you will find the asymmetric unit in the new coordinates. For example, if $$x=\frac{1}{\sqrt2}(x'+y')$$ $$y=\frac{1}{\sqrt2}(-x'+y')$$ $$z=z'$$ then in the new coordinates the asymmetric unit will be $$0\leq \frac{1}{\sqrt2}(x'+y')\leq1,\quad 0\leq \frac{1}{\sqrt2}(-x'+y')\leq1,\quad 0\leq z'\leq\frac18.$$

ii)The other thing that might be confusing is the following senario: You start with a system $x,y,z$ in which the asymmetric unit is described as $0\leq x\leq1,\quad 0\leq y\leq1,\quad 0\leq z\leq\frac18$ and your friend starts with coordinates $x',y',z'$ in which the asymmetric unit is described as $0\leq x'\leq1,\quad 0\leq y'\leq1,\quad 0\leq z'\leq\frac18$. Can you both be correct? Absolutely! Your descriptions are isomorphic and all you have to do to compare any results is just find the transformation that takes you from one coordinate system to the other. In other words, having infinitely many choices is an advantage! Not a problem! Just pick one before doing anything else.

Hope this helps.

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  • $\begingroup$ Thanks for you answer. I am not sure if it is entirely correct, though. You are claiming that I can choose any (cartesian) coordinate system whatsoever. But don't we need to take the Bravais type of the space group into account? And how does the choice relate to non-primitive basis? Can you maybe give a reference for what you've written? $\endgroup$ – eins6180 Dec 2 '14 at 9:25
  • $\begingroup$ I decided to award you the bounty because otherwise it would have expired soon. Even though I am not sure about your answer so far. $\endgroup$ – eins6180 Dec 2 '14 at 10:56
  • $\begingroup$ Thanks. The only reference I used is the one by DavePhD. My answer is m interpretation of what I read there. $\endgroup$ – Heterotic Dec 2 '14 at 11:01

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