Untwisted Symmetry Groups

Suppose that $\Sigma \subset \Gamma$ is a symmetry group. Then $\Sigma_0=\Sigma\cap({\bf SO(2)}\oplus{\bf R})$ is one of the subgroups listed in Table  1. We say that $\Sigma$ is an untwisted subgroup of $\Gamma$ if $\Sigma$ is conjugate to a subgroup of the form$K\dot+\Sigma_0$ where K is contained in the subgroup H given in Table  1. The untwisted symmetry groups are listed in Table  2.

   

Table:The 22 untwisted symmetry groups $\Sigma \subset \Gamma$

It is not the case that every subgroup $K\subset H$ produces a symmetry group. For example, when $\Sigma_0={\bf SO(2)}\oplus{\bf R}$ ,the only symmetry group $\Sigma$ corresponding to $\Sigma _0$ is $\Sigma=\Gamma$ .(This isindependent of the restriction to untwisted symmetry groups.) To verify this point, observe that ${\bf SO(2)}\oplus{\bf R}$ acts transitively on the cylinder ${\cal C}$ .Hence if $\Sigma$ is the symmetry group of a function $f:{\cal C}\to{\bf R}$ ,then f is the constant function.It follows that f is invariant under $\Gamma$ ,and that the symmetrysubgroup $\Sigma=\Gamma$ .

When $\Sigma _0$ contains ${\bf SO(2)}$ ,the function f isconstant on each horizontal cross-section of ${\cal C}$ and hence automaticallyhas the symmetry $\tau$ .In these cases, the only possibilities are $K={\bf Z}_2(\tau)$ and $K={\bf D}_2$ .Similarly, when $\Sigma$ contains ${\bf R}$ then automatically $\kappa\in\Sigma$ and the only possibilities are$K={\bf Z}_2(\kappa)$ and $K={\bf D}_2$ .

In all other cases, there are no restrictions on K other than thecondition $K\subset H$ .