Understanding Screw Axes in Crystallographic Space Groups

Question

I'm having trouble understanding how a 'screw axis' actually acts as on $\mathbb{E}^3$ as an element of a space group. Everything I talk about here will be in 3 dimensions. nLab defines a space (or crystallographic) group as a subgroup $G$ of the Euclidean group $\mathrm{E}(3)$ for which there exists a lattice $L$ in $\mathbb{E}^3$ such that $G\subseteq\mathrm{Aut}(L)$ and such that $G$ contains all translations by elements of $L$.

Let's consider a very simple example of a space group which contains a screw axis—$P222_1$ (note that I am using Wikipedia as my source for lists of space groups). If I understand Hermann–Mauguin notation correctly (which I probably don't), we choose 3 positive numbers $a,b$ and $c$, and then the group $P222_1$ with respect to this basis is generated by: all translations by elements of the lattice $L=\mathbb{Z}\langle (a,0,0),(0,b,0),(0,0,c) \rangle$; rotation by $\pi$ about the $x$-axis; rotation by $\pi$ about the $y$-axis; and the combination of rotation by $\pi$ about the $z$-axis and translation by $(0,0,c/2)$.

Now, if this is correct, then it is quite clear to me that the orbit of the origin under $P222_1$ is not $L$, but the lattice $L'=\mathbb{Z}\langle (a,0,0),(0,b,0),(0,0,c/2) \rangle$; and $P222_1$ contains all translations by elements of this lattice and preserves this lattice. Thus it is exactly the same as the group $P222$ with respect to the basis $a,b$ and $c/2$, requiring no screw axes to generate it. So $P222_1$ is a completely redundant symbol.

What have I done wrong here? It's driving me crazy.

Answer 1

Hermann–Mauguin notation is sneaky. It tells you the nature of a crystal's symmetry when viewed along certain axes, but not the precise realisations of those symmetries. It does not tell you the positions of rotation and screw axes in the unit cell, for example.

In your case, the symbols $\mathrm P222_1$ do not tell you that the "rotation by $\pi$ about the $x$-axis" should instead be a rotation by $\pi$ about the line $\left(0,0,\frac14\right)+(1,0,0)\lambda$ (which is parallel to the $x$-axis).

With this replacement, the generating set you have described is $\{(x,-y,-z+c/2),(-x,y,-z),(-x,-y,z+c/2)\}$ and correctly describes $\mathrm P222_1$. But now note that the first generator is the composition of the other two modulo the lattice translations. Hence a minimal generating set for $\mathrm P222_1$ is $\{(-x,y,-z),(-x,-y,z+c/2)\}$, showing that there are $4$ fundamental domains in its unit cell.