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

Classification of Subgroups of ${\bf SO(2)}\oplus{\bf R}$.

In this section, we classify the closed subgroups of ${\bf SO(2)}\oplus{\bf R}$ up to scaling and conjugacy in $\Gamma$. Also, we compute the normalizers of these subgroups in $\Gamma$.

Lemma 2.1   Suppose that C is a compact subgroup of ${\bf SO(2)}\oplus{\bf R}$. Then $C\subset{\bf SO(2)}\oplus{\bf 1}$.

Proof: If $(\theta,t)\in{\bf SO(2)}\oplus{\bf R}$ and $t\neq0$, then $(\theta,t)$ generates a noncompact subgroup of ${\bf SO(2)}\oplus{\bf R}$ (isomorphic to ${\bf Z}$). It follows that $(\theta,t)\not\in C$.

Proposition 2.2   Suppose that G is a closed connected subgroup of ${\bf SO(2)}\oplus{\bf R}$. Then, up to conjugacy and scaling, G is one of the subgroups

\begin{displaymath}{\bf SO(2)}\oplus{\bf R},\quad {\bf SO(2)}\oplus{\bf 1}, \quad {\bf 1}\oplus{\bf R}, \quad {\bf L}, \quad {\bf 1},
\end{displaymath}

where

\begin{displaymath}{\bf L}=\{(t,t)\in{\bf SO(2)}\oplus{\bf R}:t\in{\bf R}\}.
\end{displaymath}

Proof: If $\dim G = 2$, then connectivity implies that $G={\bf SO(2)}\oplus{\bf R}$. If $\dim G=1$, then connectivity implies that G is group isomorphic to either ${\bf SO(2)}$ or ${\bf R}$. In the first case, it follows from Lemma 2.1 that $G={\bf SO(2)}\oplus{\bf 1}$. In the second case, there is a smooth isomorphism $h:{\bf R}\to G\subset{\bf SO(2)}\oplus{\bf R}$. This isomorphism is given by $h(t)=(\theta_0t,a_0t)$ for some $(\theta_0,a_0)\in{\bf SO(2)}\oplus{\bf R}$ (defined as h(1)). By assumption $a_0\neq0$. If $\theta_0=0$, then $G={\bf 1}\oplus{\bf R}$. If $\theta_0\neq0$, then by axial scaling we can arrange that $a_0=\theta_0$ and $G={\bf L}$.

>From now on, we use the abbreviations ${\bf R}={\bf 1}\oplus{\bf R}$ and ${\bf SO(2)}={\bf SO(2)}\oplus{\bf 1}$. The proper closed subgroups of ${\bf SO(2)}$ are given by ${\bf Z}_k$, $k\ge1$: the subgroup of rotations of the cylinder through angles which are multiples of $2\pi/k$. In addition, we set ${\bf Z}\subset{\bf R}$ to be the subgroup of unit axial translations of the cylinder generated by the element $(0,1)\in{\bf SO(2)}\oplus{\bf R}$. Finally, for any $\omega\in{\bf R}$, we define

\begin{displaymath}{\bf N}_\omega = \{(\omega n,n)\in{\bf SO(2)}\oplus{\bf R}: n\in{\bf Z}\}.
\end{displaymath}

Of course, ${\bf N}_0={\bf Z}$.

Theorem 2.3   Up to axial scaling and conjugacy, the closed subgroups $\Sigma_0\subset{\bf SO(2)}\oplus{\bf R}$ are listed in Table 1.


  
Table: Classification of closed subgroups $\Sigma _0\subset \Gamma $ up to scaling and conjugacy. The normalizers are given by $N(\Sigma _0)=H\oplus ({\bf SO(2)}\oplus {\bf R})$
\begin{table}\begin{displaymath}
\begin{array}{\vert c\vert ll\vert l\vert} \hli...
... {\bf N}_{\pi/k} && {\bf D}_2 \\ \hline
\end{array}\end{displaymath}
\end{table}

Proof: Since $\Sigma _0$ is abelian, we can write $\Sigma_0\cong C\oplus{\bf Z}^p\oplus{\bf R}^q$ where C is compact and $p,q\ge0$. Clearly, $p+q\le1$. By Lemma 2.1, $C={\bf SO(2)}$ or $C={\bf Z}_k$.

Assume that $C={\bf SO(2)}$. Since ${\bf SO(2)}\oplus{\bf R}$ is connected, the only subgroup satisfying $\dim\Sigma_0=2$ is $\Sigma_0={\bf SO(2)}\oplus{\bf R}$. Suppose next that $\dim\Sigma_0=1$. We claim that $\Sigma_0={\bf SO(2)}$ or $\Sigma_0={\bf SO(2)}\oplus{\bf Z}$. Choose the smallest positive $t\in{\bf R}$ such that there is $\theta\in{\bf SO(2)}$ with $(\theta,t)\in\Sigma_0$. Since $(\theta,0)\subset\Sigma_0$, it follows that $\Sigma_0={\bf SO(2)}\oplus t{\bf Z}$, where $t{\bf Z}$ is the subgroup of ${\bf SO(2)}\oplus{\bf R}$ generated by (0,t). By making an axial scaling, we can set t=1 so that $\Sigma_0={\bf SO(2)}\oplus{\bf Z}$.

Now assume that $C={\bf Z}_k$. If $\dim\Sigma_0=1$, then it follows from Proposition 2.2 that $\Sigma_0={\bf Z}_k\oplus{\bf R}$ or $\Sigma_0={\bf Z}_k\oplus {\bf L}$. If $\dim\Sigma_0=0$, then either $\Sigma_0={\bf Z}_k$ or $\Sigma_0\cong{\bf Z}_k\oplus{\bf Z}$. In the latter case, we can choose a generator $(a,b)\in{\bf Z}\subset{\bf SO(2)}\oplus{\bf R}$ with smallest b>0. Making an axial scaling, we can suppose that the generator is of the form $(\omega,1)$ for some $\omega\in{\bf R}$. In other words, $\Sigma_0={\bf Z}_k\oplus
{\bf N}_\omega$. Note that ${\bf Z}_k\oplus {\bf N}_{\omega+2\pi/k}={\bf Z}_k\oplus {\bf N}_\omega$, so we can suppose that $\vert\omega\vert\le\pi/k$. Using formula (2.1) we compute that

\begin{displaymath}\tau\cdot(\omega t,t)\cdot\tau^{-1}=(-\omega t,t),
\end{displaymath}

where $\tau\cdot(\omega t,t)$ is an abbreviation for $(\tau,(0,0))\cdot(1,(\omega t,t))$. Hence up to conjugacy, we may suppose that $\omega\ge0$. The case $\omega=0$ is the distinguished case ${\bf N}_0={\bf Z}$.

Proposition 2.4   The normalizers of the subgroups $\Sigma_0\subset{\bf SO(2)}\oplus{\bf R}$ have the form

\begin{displaymath}N(\Sigma_0)=H\dot +({\bf SO(2)}\oplus{\bf R}),
\end{displaymath}

where the subgroup $H\subset{\bf D}_2$ is as given in Table 1.

Proof: Since ${\bf SO(2)}\oplus{\bf R}$ is abelian, it is clear that ${\bf SO(2)}\oplus{\bf R}\subset N(\Sigma_0)$. Hence $N(\Sigma_0)=H\dot+({\bf SO(2)}\oplus{\bf R})$ for some subgroup $H\subset{\bf D}_2$. We compute that $A\cdot (\theta,t)\cdot A^{-1}$ is the element $A(\theta,t)\in{\bf SO(2)}\oplus{\bf R}$. Hence, H consists of those elements $A\in{\bf D}_2$ that preserve $\Sigma _0$. The element $\tau\kappa$ acts as -I on ${\bf SO(2)}\oplus{\bf R}$ and so is always contained in H. It follows that $H={\bf Z}_2(\tau\kappa)$ or $H={\bf D}_2$. It now suffices to determine whether or not $\tau$ preserves $\Sigma _0$, that is, whether or not $\Sigma _0$ is preserved by the transformation $(\theta,t)\mapsto
(-\theta,t)$.


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