Exercise 7.5.12

Proof.

(a)

Write $N,S$ the north and south poles, and $A_k$ the point of complex coordinate ${\zeta }_n^k$ in the equatorial plane. The polyhedron $P$ is the set of vertices $$P=\{N,S,A_0,A_1,\cdots ,A_{n-1}\}.$$

The group of symmetry of this polyhedron is

$$G=G_P=\{r\in \mathrm{SO}(3)|r(P)=P\}.$$

$G$ contains the rotation $r=\mathrm{Rot}({\stackrel{\to }{e}}_3,2\pi ∕n)$ of axis $(O,{\stackrel{\to }{e}}_3)$ and angle $2\pi ∕n$, and also the rotation $s=\mathrm{Rot}({\stackrel{\to }{e}}_1,\pi )$. So it contains the set $H=\{e,r,\cdots ,r^{n-1},s,\mathit{rs},\cdots ,r^{n-1}s\}$.

$$G\supset H=\{e,r,\cdots ,r^{n-1},s,\mathit{rs},\cdots ,r^{n-1}s\}.$$

The rotations $r^k=\mathrm{Rot}({\stackrel{\to }{e}}_3,2\mathit{k\pi }∕n),k=0,\cdots ,n-1$ send $A_0$ on $A_k$. So they are distinct and they fix the poles, and the rotations $r^{n-1}\circ s$ are distinct and exchange the two poles, so are distinct of the $r^k$. Consequently $|H|=2n$.

$$\begin{array}{llll}\hfill (r\circ s)(O)& =O=(s\circ r^{-1})(O),& \hfill & \\ \hfill (r\circ s)(P)& =S=(s\circ r^{-1})(P),& \hfill & \\ \hfill (r\circ s)(A_0)& =A_1=(s\circ r^{-1})(A_0),& \hfill & \\ \hfill (r\circ s)(A_1)& =A_0=(s\circ r^{-1})(A_1).& \hfill & \end{array}$$

The 4 points $O,P,A_0,A_1$ are not coplanar, so form an affine frame, so the two rotations $r\circ s,s\circ r^{-1}$ gives the same image to the points of this frame are identical.

As $r$ is of order $n$, as $s$ is of order 2, and as $r\circ s=s\circ r^{-1}$, $H$ is a subgroup of $G$ isomorphic to the diedral group $D_{2n}$.

We show that if $n\ne 4,n\ge 3$, this inclusion is an equality.

Let $\rho \in G,\rho \ne e$, a rotation in $G$. $\rho$ fixes the gravity center $O$ of $P$, thus $\rho (O)=O$.

Write $A_k^{\prime }=\rho (A_k)$. As $\rho$ is an isometry, $A_0^{\prime }{A}_1^{\prime }=A_0{A}_1=|{\zeta }_n-1|$.

Moreover

$$\begin{array}{llll}\hfill A_0^{\prime }{A}_1^{\prime }& =A_0{A}_1=|{\zeta }_n-1|=|e^{2\mathit{i\pi }∕n}-1|& \hfill & \\ \hfill & =\sqrt{{\left(\cos <!--\; FUNCTION\; APPLICATION\; -->\frac{2\pi }{n}-1\right)}^2+{\sin <!--\; FUNCTION\; APPLICATION\; -->}^2\frac{2\pi }{n}}& \hfill & \\ \hfill & =\sqrt{2\left(1-\cos <!--\; FUNCTION\; APPLICATION\; -->\frac{2\pi }{n}\right)}& \hfill & \\ \hfill & =2\sin <!--\; FUNCTION\; APPLICATION\; -->\frac{\pi }{n}& \hfill & \end{array}$$

Note that $P{A}_k=S{A}_k=\sqrt{2}$.

As $n\ge 3$,

$2\sin <!--\; FUNCTION\; APPLICATION\; -->\frac{\pi }{n}=2$ is impossible since $\sin <!--\; FUNCTION\; APPLICATION\; -->\frac{\pi }{n}\le \sin <!--\; FUNCTION\; APPLICATION\; -->\frac{\pi }{3}<1$.

$2\sin <!--\; FUNCTION\; APPLICATION\; -->\frac{\pi }{n}=\sqrt{2}⟺\sin <!--\; FUNCTION\; APPLICATION\; -->\frac{\pi }{n}=\sin <!--\; FUNCTION\; APPLICATION\; -->\frac{\pi }{4}⟺n=4$

Suppose that $n\ne 4$. With a reductio ad absurdum, if $A_0^{\prime }$ was a pole, then $A_0^{\prime }{A}_1^{\prime }=\sqrt{2}$ (if $A_1^{\prime }=A_k$) or $A_0^{\prime }{A}_1^{\prime }=2$ (if $A_1^{\prime }$ is the opposite pole). In both cases, this is impossible, as previously proved.

Consequently $\rho (A_0)=A_k,k=0,1,\cdots ,n-1$. The same argument proves that the image of $A_i$ is in the polygon $\{A_0,,\cdots ,A_{n-1}\}$, so $\rho$ is a permutation of the vertices of this polygon, thus sends $\{P,S\}$ over $\{P,S\}$. Therefore $\rho$ fixes these two poles, or exchanges them.

$\text{∙}$

case 1: if $\rho (P)=P$, then since $\rho (O)=O,\rho (A_0)=A_k$, $\rho$ is the rotation $r^k=\mathrm{Rot}({\stackrel{\to }{e}}_3,2\mathit{k\pi }∕n)$ of axis $\mathit{OP}$ ( $1\le k\le n-1$), or the identity ( $k=0$).

$\text{∙}$

case 2: if $\rho (P)=S$, then $(\rho \circ s)(P)=P$, and by case 1, $\rho \circ s=r^j,j=0,\cdots ,n-1$, that is $\rho =r^j\circ s$.

In both cases, $\rho \in H$, therefore

$$G=H=\{e,r,\cdots ,r^{n-1},s,\mathit{rs},\cdots ,r^{n-1}s\}\simeq D_{2n}.$$

In the case $n=4$, we have proved in Exercise 10 that $G\simeq S_4$.

(b)

The modification of the polyhedron given in the text (or more simply the suppression of the south pole $S$) implies then $\rho (N)=N$, so it remains only the case 1 in the preceding discussion, so $G=⟨r⟩\simeq C_n$.

(c)

$\text{∙}$ The irregular tetrahedron $T=\{(1,0,0),(0,1,0),(0,0,1),(\frac{1}{9},\frac{4}{9},\frac{8}{9})\}$, set of four non coplanar points 4, inscribed in $S_2$ (since ${\left(\frac{1}{9}\right)}^2+{\left(\frac{4}{9}\right)}^2+{\left(\frac{8}{9}\right)}^2=1$) and its symmetry group is $G_T=\{e\}\simeq C_1$.

$\text{∙}$ Complete an isosceles non equilateral triangle in the equatorial plane by the two poles:

$P=\{(1,0,0),(\frac{4}{5},\frac{3}{5},0),(\frac{4}{5},-\frac{3}{5},0),(0,0,1),(0,0,-1)\}$ has the symmetry group $G_P=⟨s⟩\simeq C_2$, where $s=\mathrm{Rot}({\stackrel{\to }{e}}_1,\pi )$.

$\text{∙}$ The rectangle in the equatorial plane is completed by the two poles:

$R=\{(\frac{4}{5},\frac{3}{5},0),(\frac{4}{5},-\frac{3}{5},0),(-\frac{4}{5},-\frac{3}{5},0),(-\frac{4}{5},\frac{3}{5},0),(0,0,1),(0,0,-1)\}$ has the symmetry group $G_R=⟨\sigma ,\tau ,\xi ⟩$, where $\sigma ,\tau ,\xi$ are the rotations of angles $\pi$ and axes ${\stackrel{\to }{e}}_1,{\stackrel{\to }{e}}_2,{\stackrel{\to }{e}}_3$. $G_R\simeq D_4$ is the Klein four-group.

$\text{∙}$ Let $C$ be a rectangular parallelepiped with square basis inscribed in the sphere $S_2$ :

$C=\{(\frac{\sqrt{40}}{9},\frac{\sqrt{40}}{9},\frac{1}{9}),(-\frac{\sqrt{40}}{9},\frac{\sqrt{40}}{9},\frac{1}{9}),(-\frac{\sqrt{40}}{9},-\frac{\sqrt{40}}{9},\frac{1}{9}),(\frac{\sqrt{40}}{9},-\frac{\sqrt{40}}{9},\frac{1}{9}),(\frac{\sqrt{40}}{9},\frac{\sqrt{40}}{9},-\frac{1}{9}),(-\frac{\sqrt{40}}{9},\frac{\sqrt{40}}{9},-\frac{1}{9}),(-\frac{\sqrt{40}}{9},-\frac{\sqrt{40}}{9},-\frac{1}{9}),(\frac{\sqrt{40}}{9},-\frac{\sqrt{40}}{9},-\frac{1}{9})\}.$

The symmetry group of $C$ is $G_C=⟨r,s⟩=\{e,r,r^2,r^3,s,\mathit{rs},r^{2}s,r^{3}s\}$, where $r=\mathrm{Rot}({\stackrel{\to }{e}}_3,\pi ∕2),s=\mathrm{Rot}({\stackrel{\to }{e}}_1,\pi )$, so $G_C\simeq D_8$.