# Space groups: Herman-Maugin notation, diagrams and Wyckoff positions

I am currently taking an Introduction to Crystallography course. After studying and understanding the punctual symmetry groups (2mm, 4/m 2/m 2/m, 32,...), the flat groups and the diverse elements of symmetry that exist: helical axes, sliding planes, symmetry planes, among others.

I have doubts when building the diagram of a certain spatial group. For example, in an exercise, I am asked to construct the diagram of the spatial group Pmab.

Pmab implies the existence of: an ordinary plane of symmetry perpendicular to the crystallographic axis a, a sliding plane of type a perpendicular to the crystallographic axis b, a sliding plane of type b perpendicular to the crystallographic axis c.

Also, Pmab is an abbreviated notation, because, in reality, there are also two helical binary axes parallel to the axes "a" and "b", and a binary axis parallel to the axis "c".

On the other hand, Pmab belongs to the orthorhombic (rhombic) system and derives from the point symmetry group 2/m 2/m 2/m 2/m.

This is all the information I get from the Herman-Maugin notation. If there is something wrong that I have understood, I hope you will tell me.

Once here I don't know how to start building the diagram, which would look like this: As far as Wyckoff positions are concerned, I am often asked for them, but I am not quite sure what they are.

• How do you know that the two binary axes along "a" and "b" are helical? – Ivan Neretin Jul 3 '19 at 12:53
• I know, because in the orthorhombic system, simplified notation tends to obviate them. But I don't understand why – aprendiendo-a-programar Jul 3 '19 at 13:13
• That would explain how do you know there are axes in the first place. That's right, but that's not what I asked. How do you know they are helical? – Ivan Neretin Jul 3 '19 at 13:45
• I don't know, I said they're helical, because that's how it appears on the diagram. – aprendiendo-a-programar Jul 3 '19 at 14:42
• Well, they are helical all right, but you need to know how to deduce that from "Pmab" alone. – Ivan Neretin Jul 3 '19 at 14:49

Well, knowing crystallography implies being able to deduce the picture above solely from the group's name $$\rm Pmab$$, and that's how it is done.

First you write down the symmetry elements mentioned explicitly in the group's name (that is, $$\rm m,\;a$$, and $$\rm b$$ in our case; in other examples there could have been axes among them, or even worse) and orient those accordingly, which you already know how to do. Then you combine these elements in all possible ways to find out what else is hidden beneath.

Essentially, any symmetry element is but a linear transformation that takes any vector $$\bf\vec x$$ to another vector $$\bf A\vec x+\vec b$$, where $$\bf A$$ is the $$3\times3$$ matrix of our transformation (rotation or reflection or otherwise), and $$\bf\vec b$$ is the shift, that is, a half-translation along the axis for a screw axis, or a half-translation in whatever direction for a glide plane, or $$\bf\vec0$$ in all other cases. The typical matrices are: $$\begin{array}{cc} \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} & \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \\ \end{pmatrix}& \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \\ \end{pmatrix}\\ \text{Reflection \perp X} & \text{Rotation about X}& \text{Center} \end{array}$$

Given two symmetry elements, you simply combine the two transformations by doing some matrix multiplication, which I sincerely recommend, if only to know how it feels. Alternatively, you may rely upon the wisdom of the predecessors, which says:

1. Two perpendicular planes give a twofold axis parallel to both.
2. A plane and a parallel twofold axis give another plane, perpendicular to the plane and parallel to the axis.
3. A plane and a perpendicular twofold axis give a center.
4. Two perpendicular twofold axes give a third axis, perpendicular to both.
5. All of the above remains true if you change
• every "plane" to "plane or a glide plane" and
• every "axis" to "axis or a screw axis".
6. When some of the combined elements have shift in them, the following happens:
• If the shift is parallel to the resulting element, it becomes a part of the said element (that is, the result is going to be a screw axis or a glide plane).
• If the shift is perpendicular to the resulting element, it moves the said element by half the value of the shift (that is, by $$1\over4$$ of a translation).
7. If you end up with a center of symmetry someplace other than at the origin, you relocate your origin there.

Now try to apply the rules 1-7 and see where this gets you.

• Comments are not for extended discussion; this conversation has been moved to chat. – andselisk Jul 4 '19 at 16:07