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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 $
\begin{table}\begin{displaymath}
\begin{array}{\vert l\vert lll\vert} \hline
\Si...
...f Z}_k\oplus {\bf N}_{\pi/k}) \\ \hline
\end{array}\end{displaymath}
\end{table}

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 $K\subset H$ .


next up previous
Next: Twisted Symmetry Groups Up: Symmetries of Columns Previous: Classification of Subgroups of
Marty Golubitsky
2001-01-29