----- > [!proposition] Proposition. ([[the crystallographic restriction]]) > Consider a [[discrete subgroup of M2|discrete subgroup]] $\Gamma$ of the [[euclidean isometry|plane isometry group]] $M_{2}$. > > [[analyzing discrete subgroups of M2|We get an]] [[exact sequence]] > > ```tikz > \usepackage{tikz} > \usepackage{amsmath} % Required for \text command in math mode > \usetikzlibrary{matrix, positioning, calc} > > \begin{document} > > \begin{tikzpicture} > \matrix (m) [matrix of math nodes, row sep=1em, column sep=4em] > { > T & M_2 & O_2 \\ > \Gamma \cap T & \Gamma \ \text{discrete} & \overline{\Gamma} \\ > & & \\ > (L,+) & & \\ > }; > \path[-stealth] > (m-1-1) edge (m-1-2) > (m-1-2) edge (m-1-3) > (m-2-1) edge (m-2-2) > (m-2-2) edge (m-2-3); > \node at ($(m-1-1)!0.5!(m-2-1)$) {$\cup$}; > \node at ($(m-1-2)!0.5!(m-2-2)$) {$\cup$}; > \node at ($(m-1-3)!0.5!(m-2-3)$) {$\cup$}; > \node[rotate=270] at ($(m-2-1.south)-(0,1em)$) {$\cong$}; > \end{tikzpicture} > > \end{document} > ``` > and ask, 'what can the [[point group]] $\overline{\Gamma}$ be?' We show that when $L \cong (\mathbb{Z}^{2}, +)$, > 1. Any discrete subgroup of $O_{2}$ must be finite [[finite subgroups of O2 are cyclic or dihedral|hence]] [[cyclic group|cyclic]] or [[dihedral group|dihedral]] > 2. $\overline{\Gamma}$ [[group action|acts on]] $(L,+)$ > 3. Facts $(1)$ and $(2)$ ultimately enforce $\overline{\Gamma} \cong C_{n} \text{ or } \overline{\Gamma} \cong D_{n} \text{ for } n \in \{ 1,2,3,4,6 \}.$ > > [!proof]- Proof. ([[the crystallographic restriction]]) > **1.** Consider the diagram > > $\begin{matrix} > \text{SO}_{2} & \subset & \text{O}_{2} \\ > \cup & & \cup \\ > \overline{\Gamma} \cap \text{SO}_{2} & \subset & \overline{\Gamma} > \end{matrix}$ > Clearly $\overline{\Gamma} / \overline{\Gamma} \cap \text{SO}_{2} \hookrightarrow \text{O}_{2} / \text{SO}_{2} \cong \{ \pm 1 \}$. To show $\overline{\Gamma}$ is finite it is enough to show $\overline{\Gamma} \cap \text{SO}_{2}$ is finite, which it certainly is since it must contain a smallest rotation by [[discrete subgroup of M2|discreteness]] and rotations are mod $2\pi$. Hence as a [[finite subgroups of O2 are cyclic or dihedral|finite subgrop of]] $\text{SO}_2$, $\overline{\Gamma} \cap \text{SO}_{2} \cong C_{n}$, and by finiteness we conclude that in general $\overline{\Gamma} \cong C_{n}$ or $\overline{ \Gamma} \cong D_{n}$. > > So for $\text{O}_2$, $\text{discrete} \iff \text{finite}$. > > **2.** Recall that because $T \trianglelefteq M_{2}$, $M_{2}$ [[group action|acts on]] $T$ by [[conjugate|conjugation]]. And of course $T$ acts trivially on itself by [[conjugate|conjugation]]. Since $\text{O}_{2}=M_{2} / T$, we conclude $\text{O}_{2}$ acts on $T$ by [[conjugate|conjugation]] in the sense that it acts on $(\mathbb{R}^{2}, +) \cong T$ by conjugation ($\rho t_{a}\rho ^{-1}=t_{\rho(a)}$). That is, we lift an element of $\text{O}_{2}$ to an element of $M_{2}$ — it doesn't matter which because any two choices would differ by a [[translation]] and [[translation]]s act trivially on $T$ — and then act on $T$ via the lifted element. > > In the same way, $\overline{\Gamma}$ acts on $(L,+)$ how we would expect: $\rho \in \overline{\Gamma}: a \in L \mapsto \rho(a)$. > > **3.** Pick some nonzero element $a \in L$ of smallest possible length. We know $\overline{\Gamma}$ is [[group isomorphism|isomorphic to]] $C_{n}$ or $D_{n}$, thus it contains the rotation $\rho_{\theta}, \theta=\frac{2\pi}{n}$. By **2** we know $\overline{\Gamma}$ acts on the lattice $L$, so rotating the lattice yields some $b=\rho_{\theta}(a) \in L$. Clearly $\theta$ cannot be too small, then, or else we'll have $\|b-a\| < \|a\|$ where $b-a \in L$, a contradiction to how we defined $a$. The angle at which $\|b-a\|=\|a\|$ and we start avoiding contradiction is $60^{\circ}=\frac{2\pi}{6}$ (equilateral triangle). > ![[CleanShot 2023-11-12 at [email protected]|300]] > Hence $n \leq 6$. But why can't $n=5$? Suppose $n=5$, so that $\theta=\frac{2\pi}{5}$; the [[orbit]] of $a$ under the rotations is visualized below: > ![[CleanShot 2023-11-12 at [email protected]|300]] > > ![[CleanShot 2023-11-12 at [email protected]|300]] > $a \in L$, $b \in L$, $c \in L$. But then $a+c \in L$ with $\|a+c\|<\|a\|$. > ![[CleanShot 2023-11-12 at [email protected]|300]] > > > ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```