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Next: Classification of Columns Up: Symmetries of Columns Previous: Untwisted Symmetry Groups

Twisted Symmetry Groups

We continue to suppose that $\Sigma $ is a symmetry group with $\Sigma_0=\Sigma\cap({\bf SO(2)}\oplus{\bf R})$ .We have $\Sigma\subset H\dot+({\bf SO(2)}\oplus{\bf R})$ where H is given in Table 1 . The canonical projection $\pi:\Gamma\to D_2$ induces a projection $\pi:\Sigma\to H$ .

We say that a symmetry group $\Sigma \subset \Gamma $ is twisted if it is not conjugate to an untwisted symmetry group. Equivalently, there exists an $A\in\pi(\Sigma)$ such that $A\not\in\Sigma$ .

The next lemma states that, without loss of generality, we can always supposethat the element $A=\tau\kappa$ is not responsible for twisting.

Lemma 2.5   Suppose that $\Sigma $ is a symmetry group and that $\tau\kappa\in\pi(\Sigma)$ .Then there is a subgroup of $\Gamma$ that is conjugate to $\Sigma $ and contains $\tau\kappa$ .The conjugacy leaves $\Sigma _0$ unchanged.

Proof: Recall that $\tau\kappa$ acts as -I on ${\bf SO(2)}\oplus{\bf R}$ .By assumption $(\tau\kappa,\theta,t)\in\Sigma$ for some$(\theta,t)\in{\bf SO(2)}\oplus{\bf R}$ .We conjugate by the element $(-\theta/2,-t/2)\in{\bf SO(2)}\oplus{\bf R}$ .Compute that

\begin{displaymath}(1\,,\,(-\theta/2,-t/2))\cdot(\tau\kappa\,,\,(\theta,t))\cdot
(1\,,\,(\theta/2,t/2))=(\tau\kappa\,,\,(0,0)),
\end{displaymath}

as required.

Proposition 2.6   Let $\Sigma $ be a twisted symmetry group. Then either $\Sigma_0={\bf Z}_k$ ,$\Sigma_0={\bf Z}_k\oplus{\bf Z}$ or $\Sigma_0={\bf Z}_k\oplus{\bf N}_{\pi/k}$ .In addition, $\pi(\Sigma)$ is one of the three subgroups ${\bf Z}_2(\tau)$ ,${\bf Z}_2(\kappa)$ and ${\bf D}_2$ .

Remark: The possibility $\Sigma_0={\bf Z}_k\oplus{\bf N}_{\pi/k}$ will be eliminated in the proof of Theorem  2.7.


Proof: It follows from Lemma  2.5 that we can eliminate the subgroups$\Sigma _0$ for which $H={\bf Z}_2(\tau\kappa)$ ,that is we can eliminate ${\bf Z}_k\oplus {\bf L}$ and ${\bf Z}_k\oplus {\bf N}_\omega$ .

Next, suppose that $\Sigma _0$ contains ${\bf SO(2)}$ .As observed in theprevious subsection, $\Sigma $ contains $\tau$ .If $\Sigma $ is larger than ${\bf Z}_2(\tau)\dot+{\bf SO(2)}$ ,then $\pi(\Sigma)={\bf D}_2$ .It follows from Lemma 2.5 that $\tau\kappa\in\Sigma$ and hence $\Sigma={\bf D}_2\dot+\Sigma_0$ .In either case, $\Sigma $ is untwisted. The possibility that $\Sigma _0$ contains ${\bf R}$ can be eliminated similarly. This completes the proof that $\Sigma _0$ is one of the groups ${\bf Z}_k$ ,${\bf Z}_k\oplus{\bf Z}$ or ${\bf Z}_k\oplus{\bf N}_{\pi/k}$ .

Recall that $\pi(\Sigma)$ is a subgroup of ${\bf D}_2$ .If $\pi(\Sigma)={\bf 1}$ ,then $\Sigma=\Sigma_0$ .If $\pi(\Sigma)={\bf Z}_2(\tau\kappa)$ ,then $\Sigma $ is conjugate to ${\bf Z}_2(\tau\kappa)\dot+\Sigma_0$ by Lemma 2.5. Hence, for $\Sigma $ to be twisted, $\pi(\Sigma)$ must be one of the three remaining subgroups of ${\bf D}_2$ .

Theorem 2.7   Up to conjugacy and scaling, there are seven twisted symmetry groups in $\Gamma$ .These are as listed in Table 3 .


  
Table:The 7 twisted symmetry groups $\Sigma \subset \Gamma $ .$\Sigma $ is generated by $\Sigma _0$ together with the generatorsof $\Sigma /\Sigma _0$ .Notation: $\tilde\kappa=(\kappa,(\pi/k,0))$ ,$\tilde\tau=(\tau,(0,1/2))$
\begin{table}\begin{displaymath}
\begin{array}{\vert l\vert l\vert l\vert} \hlin...
...\tilde\tau,\quad \tilde\kappa \\ \hline
\end{array}\end{displaymath}
\end{table}

Proof: By Proposition  2.6, we can assume that $\Sigma_0={\bf Z}_k$ ,${\bf Z}_k\oplus{\bf Z}$ or ${\bf Z}_k\oplus{\bf N}_{\pi/k}$ and that $K=\pi(\Sigma)$ is one of the subgroups ${\bf Z}_2(\tau)$ ,${\bf Z}_2(\kappa)$ or ${\bf D}_2$ .We consider the three possibilities for K in turn.

Suppose that $K={\bf Z}_2(\tau)$ .Then $\sigma=(\tau,(\theta,t))\in\Sigma$ for some $(\theta,t)\in{\bf SO(2)}\oplus{\bf R}$ .Conjugating by $(-\theta/2,0)\in{\bf SO(2)}\oplus{\bf R}$ ,we can set $\theta=0$ .Note that

\begin{displaymath}\sigma^2=(1,(0,2t))\in\Sigma_0.
\end{displaymath}

When $\Sigma_0={\bf Z}_k$ ,it follows that t=0 in which case $\sigma=\tau$ ,and there is no twisting. When $\Sigma_0={\bf Z}_k\oplus{\bf Z}$ ,there is the additional possibility that $2t\in{\bf Z}$ but $t\not\in{\bf Z}$ .Since ${\bf Z}\subset\Sigma$ ,this reduces to the case t=1/2.The argument is more complicated when $\Sigma_0={\bf Z}_k\oplus{\bf N}_{\pi/k}$ .Squaring yields the condition $(0,2t)\in{\bf Z}_k\oplus{\bf N}_{\pi/k}$ .Workingmodulo ${\bf Z}_k\oplus{\bf N}_{\pi/k}$ ,we can choose $\sigma$ so that t=1. But still working modulo ${\bf N}_{\pi/k}$ ,we can replace $\sigma$ by $\sigma=(\tau,(\pi/k,0))$ .Conjugating once again, we have $\sigma=\tau$ and there is no twisting.

The case $K={\bf Z}_2(\kappa)$ is similar. Conjugation reduces to $\sigma=(\kappa,(\theta,0))$ and squaring yields the condition$2\theta\in{\bf Z}_k$ .Twisting occurs when $\theta=\pi/k$ but onlyfor $\Sigma_0={\bf Z}_k$ and $\Sigma_0={\bf Z}_k\oplus{\bf Z}$ . Finally, suppose that $K={\bf D}_2$ .We concentrate attention on the twogenerators

\begin{displaymath}\sigma_1=(\tau,(\theta_1,t_1)) \qquad \sigma_2=(\kappa,(\theta_2,t_2))
\end{displaymath}

of $\Sigma $ modulo $\Sigma _0$ .Since the reflections are orthogonal, we can simultaneously conjugate so that $\theta_1=t_2=0$ .Squaring the generators, we obtain that $\theta_2\in{\bf Z}_{2k}$ and either t1=0, $2t_1\in{\bf Z}$ or $t_1\in{\bf Z}$ dependingon whether $\Sigma_0={\bf Z}_k$ ,$\Sigma_0={\bf Z}_k\oplus{\bf Z}$ or $\Sigma_0={\bf Z}_k\oplus{\bf N}_{\pi/k}$ .The various combinations of generators yieldone untwisted subgroup and one twisted subgroup for $\Sigma_0={\bf Z}_k$ ,and one untwisted subgroup and three twisted subgroups for $\Sigma_0={\bf Z}_k\oplus{\bf Z}$ .Once again, there is no twisting when $\Sigma_0={\bf Z}_k\oplus{\bf N}_{\pi/k}$ .The arguments are similar to the previous cases of K; we replace $\sigma_j$ by untwisted group elements.


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