Note that $R_{a,\lambda }\left(a\right)=\lambda a$ and if $v\in a^{\perp }$ (i.e., if $H(v,a)=0\text{),}$ then $R_{a,\lambda }\left(v\right)=v\text{.}$ Now the linearity of $R_{a,\lambda }$ is obvious. Since $R_{a,1}=I,$ (b) would imply that $R_{a,\lambda }$ is invertible and thus (a) follows from (b). To prove (b) we let $v\in V$ and compute $$\begin{array}{ccc}{R}_{a,\lambda }{R}_{a,\mu }\left(v\right)& =& R_{a,\lambda }(v+\left({\mu }^{-1}\right)\frac{H(v,a)}{H(a,a)}a)\\ & =& v+(\lambda -1)\frac{H(v,a)}{H(a,a)}a+(\mu -1)\frac{H(v,a)}{H(a,a)}·\lambda a\\ & =& v+(\lambda \mu -1)\frac{H(v,a)}{H(a,a)}·a\\ & =& R_{a,\lambda \mu }\left(v\right)\text{.}\end{array}$$ For (c) we let $v,w\in V$ and compute $$\begin{array}{ccc}H(R_{a,\lambda }\left(v\right),R_{a,\lambda }\left(w\right))& =& H(v+(\lambda -1)\frac{H(v,a)}{H(a,a)}a,w+(\lambda -1)\frac{H(w,a)}{H(a,a)}a)\\ & =& H(v,w)+(\lambda \bar{\lambda }-1)\frac{H(a,w)H(v,a)}{H(a,a)}\end{array}$$ which implies (c). In (d) we suppose without loss of generality that $\lambda \ne 1\text{.}$ To prove the implication from left to right we evaluate both sides of $R_{a,\lambda }=R_{b,\mu }$ at $a$ to obtain $$\begin{array}{cc}\text{(1)}& \begin{array}{c}\lambda a=a+(\mu -1)\frac{H(a,b)}{H(b,b)}b\text{or}\\ (\lambda -1)a=(\mu -1)\frac{H(a,b)}{H(b,b)}·b\text{.}\end{array}\end{array}$$ Thus $\lambda -1\ne 0$ implies there is a $\gamma \in ℂ$ with $b=\gamma a\text{.}$ Inserting this in (1) we obtain $\lambda -1=\mu -1$ and hence $\lambda =\mu \text{.}$ The implication from right to left is trivial. For (e) we let $v\in V$ and compute $$\begin{array}{ccc}T{R}_{a,\lambda }{T}^{-1}\left(v\right)& =& T(T^{-1}v+(\lambda -1)\frac{H(T^{-1}v,a)}{H(a,a)}a)\\ & =& v+(\lambda -1)\frac{H(T^{-1}v,a)}{H(a,a)}Ta\\ & =& v+(\lambda -1)\frac{H(v,Ta)}{H(Ta,Ta)}Ta\\ & =& R_{Ta,\lambda }\left(v\right)\text{.}\end{array}$$

$\square$

Letting $a=\pi /q_{ij}$ and $b=|\frac{\pi }{2p_i}-\frac{\pi }{2p_j}|$ we see that $a,b\ge 0$ and $a+b\le \pi /2\text{.}$ Hence $\text{cos}\left(a\right)\ge \text{sin}\left(b\right)$ yielding the lemma.

$\square$

It suffices to show that for each $1\le i\le \ell$ we have $S_i^{p_i}=I,$ and that for each pair of distinct integers $1\le i,$ $j\le \ell$ we have $S_i{S}_j{S}_i\cdots =S_j{S}_i{S}_j\cdots$ with $q_{ij}$ factors on each side. The first of these conditions is obviously satisfied from the definition of $S_i\text{;}$ in fact $S_i$ has order $p_i\text{.}$ The second is vacuously satisfied if the graph $\Gamma$ has only one vertex. So, to begin, we assume the graph $\Gamma$ has two vertices. If the vertices are not joined by an edge then using $\{v_1,v_2\}$ as a basis the matrices for $S_1$ and $S_2$ are both diagonal and hence commute. So we can assume $\Gamma$ is $$\begin{array}{c} p1 q p2 \end{array}(q=q_{12})\text{.}$$ The verification of the theorem for this graph occurs in [Cox1962]. We will give a somewhat different argument here. We first compute the eigenvalues of $S_1{S}_2\text{.}$

The characteristic equation of $S_1{S}_2$ is ${\lambda }^2-\text{tr}\left(S_1{S}_2\right)\lambda +\text{Det}\left(S_1{S}_2\right)=0\text{.}$ Now in the basis $\{v_1,v_2\}$ $$S_1=\left(\begin{array}{cc}{\epsilon }_1& \frac{(1-{\epsilon }_1)c}{\text{sin}(\pi /p_1)}\\ 0& 1\end{array}\right),S_2=\left(\begin{array}{cc}1& 0\\ \frac{(1-{\epsilon }_2)c}{\text{sin}(\pi /p_2)}& {\epsilon }_2\end{array}\right),$$ and thus, $$S_1{S}_2=\left(\begin{array}{cc}\frac{{\epsilon }_1+c^2(1-{\epsilon }_1)(1-{\epsilon }_2)}{\text{sin}(\pi /p_1)\text{sin}(\pi /p_2)}& \frac{{\epsilon }_2(1-{\epsilon }_1)c}{\text{sin}(\pi /p_1)}\\ \frac{(1-{\epsilon }_2)c}{\text{sin}(\pi /p_1)}& {\epsilon }_2\end{array}\right)$$ where $c={\{{\text{cos}}^2(\pi /q)-{\text{sin}}^2(\frac{\pi }{2p_1}-\frac{\pi }{2p_2})\}}^{\frac{1}{2}}\text{.}$

Letting ${\theta }_1=e^{{\pi }_i/p_1},$ ${\theta }_2=e^{{\pi }_i/p_2}$ and using $\text{sin}\left(x\right)=\frac{1}{2}(e^{-ix}-e^{ix})$ we see that $\text{tr}\left(S_1{S}_2\right)={\theta }_1{\theta }_2(\frac{{\theta }_1}{{\theta }_2}+\frac{{\theta }_2}{{\theta }_1}-4c^2)\text{.}$ Then putting $\Phi =\frac{\pi }{p_1}-\frac{\pi }{p_2}$ we use $e^{i\Phi }+e^{-i\Phi }=2\text{cos}\hspace{0.17em}\Phi =2-4{\text{sin}}^2(\Phi /2)$ to obtain $$\begin{array}{ccc}\text{tr}\left(S_1{S}_2\right)& =& {\theta }_1{\theta }_2(2-4{\text{sin}}^2(\Phi /2)-4c^2)\\ & =& {\theta }_1{\theta }_2(2-4{\text{cos}}^2(\pi /q))\\ & =& -2{\theta }_1{\theta }_2\text{cos}(2\pi /q)\text{.}\end{array}$$ Hence the characteristic equation of $S_1{S}_2$ is $$\begin{array}{ccc}0& =& {\lambda }^2+2{\theta }_1{\theta }_2\text{cos}(2\pi /q)\lambda +{\theta }_1^2{\theta }_2^2\\ & =& {\lambda }^2+{\theta }_1{\theta }_2(e^{2\pi i/q}+e^{-2\pi i/q})\lambda +{\theta }_1^2{\theta }_2^2\\ & =& (\lambda +{\theta }_1{\theta }_2{e}^{2\pi i/q})(\lambda +{\theta }_1{\theta }_2{e}^{-2\pi i/q})\end{array}$$ yielding the roots $$\begin{array}{ccc}{\lambda }_1& =& e^{\pi i(\frac{1}{p_1}+\frac{1}{p_2}+(1-2/q))}\text{and}\\ {\lambda }_2& =& e^{\pi i(\frac{1}{p_1}+\frac{1}{p_2}-(1-2/q))}\text{.}\end{array}$$ Note that $q\ne 2$ forces ${\lambda }_1\ne {\lambda }_2$ and thus there is an invertible matrix $P$ such that $P^{-1}{S}_1{S}_{2}P=\left[\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right]\text{.}$ If $q$ is even the relation we must check is ${\left(S_1{S}_2\right)}^{q/2}={\left(S_2{S}_1\right)}^{q/2}\text{.}$ But a glance at the expressions for ${\lambda }_1$ and ${\lambda }_2$ reveals that if $q$ is even, then ${\lambda }_1^{q/2}={\lambda }_2^{q/2}\text{.}$ Let $a$ denote this common value. Then $$P^{-1}{\left(S_1{S}_2\right)}^{q/2}P={\left(P^{-1}{S}_1{S}_{2}P\right)}^{q/2}=\left[\begin{array}{cc}{\lambda }_1^{q/2}& 0\\ 0& {\lambda }_2^{q/2}\end{array}\right]=a{I}_2\text{.}$$ Hence ${\left(S_1{S}_2\right)}^{q/2}=a{I}_2$ and thus ${\left(S_2{S}_1\right)}^{q/2}=a{I}_2$ also.

Now we assume $q$ is odd. So here we have $p_1=p_2=$ say $p\text{.}$ Thus ${\epsilon }_1={\epsilon }_2=e^{2\pi i/p}$ and let $\epsilon =e^{2\pi i/p}\text{.}$ Define $T_1=P^{-1}{S}_{1}P,$ $T_2=P^{-1}{S}_{2}P,$ and let $D=P^{-1}{S}_1{S}_{2}P\text{.}$ Finally, put $T_1=\left[\begin{array}{cc}x& u\\ v& y\end{array}\right]\text{.}$

Then $$\begin{array}{ccc}{T}_2=P^{-1}{S}_{2}P& =& P^{-1}{S}_1^{-1}P{P}^{-1}{S}_1{S}_{2}P\\ & =& T_1^{-1}D\\ & =& \left[\begin{array}{cc}y/\epsilon & -u/\epsilon \\ -v/\epsilon & x/\epsilon \end{array}\right]\left[\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right]\\ & =& \left[\begin{array}{cc}y\frac{{\lambda }_1}{\epsilon }& -u\frac{{\lambda }_2}{\epsilon }\\ -v\frac{{\lambda }_1}{\epsilon }& x\frac{{\lambda }_2}{\epsilon }\end{array}\right]\\ & =& \left[\begin{array}{cc}y\gamma & -u\bar{\gamma }\\ -v\gamma & x\bar{\gamma }\end{array}\right]\end{array}$$ where $\gamma =\frac{{\lambda }_1}{\epsilon }=e^{\pi i(1-2/q)}\text{.}$

Now comparing the traces of $S_1$ and $T_1,$ and those of $S_2$ and $T_2$ we obtain the equations $1+\epsilon =x+y$ and $1+\epsilon =\gamma y+\bar{\gamma }x\text{.}$ Thus $x+y=\gamma y+\bar{\gamma }x\text{.}$ Using $\gamma \bar{\gamma }=1$ we obtain $\gamma y-x=\bar{\gamma }(\gamma y-x)\text{.}$ Since $q\ne 2$ we have $\gamma \ne 1$ and thus $\gamma y=x\text{.}$ Since $q$ is odd the relation we must check is ${\left(S_1{S}_2\right)}^{\frac{q-1}{2}}{S}_1=S_2{\left(S_1{S}_2\right)}^{\frac{q-1}{2}}\text{.}$ Clearly it suffices to show $P^{-1}{\left(S_1{S}_2\right)}^{\frac{q-1}{2}}{S}_{1}P=P^{-1}{S}_2{\left(S_1{S}_2\right)}^{\frac{q-1}{2}}P\text{.}$ $$\begin{array}{ccc}{P}^{-1}{\left(S_1{S}_2\right)}^{\frac{q-1}{2}}{S}_{1}P& =& {\left(P^{-1}{S}_1{S}_{2}P\right)}^{\frac{q-1}{2}}{P}^{-1}{S}_{1}P=D^{\frac{q-1}{2}}{T}_1\\ & =& \left[\begin{array}{cc}{\lambda }_1^{\frac{q-1}{2}}x& {\lambda }_1^{\frac{q-1}{2}}u\\ {\lambda }_2^{\frac{q-1}{2}}v& {\lambda }_2^{\frac{q-1}{2}}y\end{array}\right]\\ P^{-1}{S}_2{\left(S_1{S}_2\right)}^{\frac{q-1}{2}}P& =& P^{-1}{S}_{2}P{\left(P^{-1}{S}_1{S}_{2}P\right)}^{\frac{q-1}{2}}=T_2{D}^{\frac{q-1}{2}}\\ & =& \left[\begin{array}{cc}{\lambda }_1^{\frac{q-1}{2}}\gamma y& -{\lambda }_2^{\frac{q-1}{2}}\bar{\gamma }u\\ -{\lambda }_1^{\frac{q-1}{2}}\gamma v& {\lambda }_2^{\frac{q-1}{2}}\bar{\gamma }x\end{array}\right]\text{.}\end{array}$$ Thus we will have the desired equality if and only if $\gamma y=x$ and ${\lambda }_1^{\frac{q-1}{2}}=-{\lambda }_2^{\frac{q-1}{2}}\bar{\gamma }\text{.}$ The first of these conditions was obtained above. For the second we compute ${\left({\lambda }_1{\lambda }_2^{-1}\right)}^{\frac{q-1}{2}}={\left(e^{\frac{-4\pi i}{q}}\right)}^{\frac{q-1}{2}}=e^{\frac{2\pi i}{q}}=-\bar{\gamma }\text{.}$ We now have the theorem for graphs with one or two vertices.

Now consider a graph $\Gamma$ with $\ell \ge 3$ vertices. Let $1\le i,$ $j\le \ell$ be distinct. Put $P=⟨v_i,v_j⟩$ and $Q=P^{\perp }$ $\text{("}\perp \text{"}$ is taken with respect to $H=H\left(\Gamma \right)\text{.}$

Now if $H$ is non degenerate on $P$ we will have $V=P\oplus Q\text{.}$ But by the argument given for the case where the graph had two vertices we see that $S_i$ and $S_j$ satisfy the required relationship when restricted to $P\text{.}$ Now $S_i$ and $S_j$ obviously satisfy the condition on $Q$ as they are both the identity transformation on $Q\text{.}$ Hence $S_i{S}_j{S}_i\cdots =S_j{S}_i{S}_j\cdots$ on the whole space $V\text{.}$

So we are led to assume that $H$ is degenerate on $P\text{.}$ Now $H$ is not identically zero on $P$ as $H(v_i,v_i)=\text{sin}\hspace{0.17em}\pi /p_i\ne 0\text{.}$ In fact we must further have $H(v_i,v_j)\ne 0$ for otherwise $H$ would be non degenerate on $P\text{.}$ Thus we see that $\text{dim}(P\cap Q)=1$ and since $\text{dim}\left(P\right)+\text{dim}\left(Q\right)\ge \ell$ we have $\text{dim}\left(Q\right)\ge \ell -2\text{.}$ If $\text{dim}\left(Q\right)=\ell -1$ we have $V=P+Q$ and there is an $\ell -2$ dimensional subspace $Q\prime$ of $Q$ such that $V=P\oplus Q\prime \text{.}$ Since $S_i$ and $S_j$ are the identity on $Q\prime$ we can argue as in the non degenerate case to obtain the desired result. So we assume $Q$ has dimensional $\ell -2\text{.}$ Hence $V\ne P+Q\text{;}$ therefore there is some basis vector $v_k$ not in $P+Q$ and we have $V=P+⟨v_k⟩+Q\text{.}$ Hence there is an $\ell -3$ dimensional subspace $Q\prime$ of $Q$ such that $V=P\oplus ⟨v_k⟩\oplus Q\prime \text{.}$ Now $S_i$ and $S_j$ are the identity transformation on $Q\prime$ so it suffices to check $S_i{S}_j{S}_i\cdots =S_j{S}_i{S}_j\cdots$ on the subspace $P\oplus ⟨v_k⟩=⟨v_i,v_j,v_k⟩\text{.}$

Recall we are assuming that $H$ is degenerate on $⟨v_i,v_j⟩\text{.}$ Thus $$\text{Det}\hspace{0.17em}\left(\begin{array}{cc}{\alpha }_{ii}& {\alpha }_{ij}\\ {\alpha }_{ij}& {\alpha }_{jj}\end{array}\right)=0\text{.}$$ When expanded the equation is $\text{sin}(\pi /p_i)\text{sin}(\pi /p_j)-{\text{cos}}^2(\pi /q_{ij})+{\text{sin}}^2(\pi /2p_i-\pi /2p_j)=0\text{.}$ Using ${\text{sin}}^2(a+b)-{\text{sin}}^2(a-b)=\text{sin}\left(2a\right)\text{sin}\left(2b\right)$ together with some half angle formulas one obtains the equivalent condition: $$\text{cos}(\frac{\pi }{p_i}+\frac{\pi }{p_j})=\text{cos}(\pi -\frac{2\pi }{q_{ij}})\text{.}$$ Since the arguments on each side are in the interval $[0,\pi ]$ over which cos is one to one we have that $H$ is degenerate on $⟨v_i,v_j⟩$ if and only $\frac{1}{p_i}+\frac{1}{p_j}+\frac{2}{q_{ij}}=1$ [Cox1974, p.110]. The solutions to this equation subject to the restrictions $p_i,p_j\ge 2,$ $q_{ij}\ge 3,$ and $p_i=p_j$ if $q_{ij}$ is odd are given in the table: $$\begin{array}{c}\text{Table 1}\\ \begin{array}{|cccccccc|}\hline q_{ij}& 3& 4& 4& 6& 6& 8& 12\\ p_i& 6& 3& 4& 2& 3& 2& 2\\ p_j& 6& 6& 4& 6& 3& 4& 3\\ \hline\end{array}& (p_i\le p_j)\end{array}$$ Now recall that $v_k$ is not in $P+Q\text{.}$ So in particular $v_k$ is not orthogonal to both $v_i$ and $v_j\text{.}$ Hence the portion of the graph $\Gamma$ involving the $i^{\text{th}},$ $j^{\text{th}},$ and $k^{\text{th}}$ vertices must look like: $$\begin{array}{c} pi pk pj qik qij qjk \end{array}$$ where the dotted lines joining the $i^{\text{th}}$ and $k^{\text{th}},$ and the $j^{\text{th}}$ and $k^{\text{th}}$ vertices indicate that there mayor may not be an edge joining those vertices but that at least one of those two edges actually occurs. Further the numbers $q_{ij},$ $p_i,$ $p_j$ occur in one of the columns of Table 1.

For example, one possibility is $$\begin{array}{c} 3 p 6 q 4 q′ \end{array}$$ $\text{(}{p}_i=3,$ $p_j=6,$ $q_{ij}=4,$ $p_k=p,$ $q_{ik}=q,$ $q_{jk}=q\prime \text{).}$

Let $U=⟨v_i,v_j,v_k⟩\text{.}$ Using the given basis $\{v_i,v_j,v_k\}$ for $U$ the matrix for $S_i$ restricted to $U$ is $$\left[\begin{array}{ccc}\omega & \frac{1-\omega }{\alpha }& x\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$$ while the matrix for $S_j$ restricted to $U$ is $$\left[\begin{array}{ccc}1& 0& 0\\ -\omega \alpha & -{\omega }^2& y\\ 0& 0& 1\end{array}\right]\text{.}$$ Here $\omega =e^{2\pi i/3},$ $\alpha =3^{1/4},$ $x=(\omega -1)\frac{H(v_k,v_i)}{H(v_i,v_i)},$ and $y=(-{\omega }^2-1)\frac{H(v_k,v_j)}{H(v_j,v_j)}\text{.}$ So $$S_i{S}_j{|}_U=\left[\begin{array}{ccc}{\omega }^2& \frac{1-{\omega }^2}{\alpha }& x+\frac{1-\omega }{\alpha }y\\ -\omega \alpha & -{\omega }^2& y\\ 0& 0& 1\end{array}\right],$$ and thus, $${\left(S_i{S}_j{|}_U\right)}^2=\left[\begin{array}{ccc}1& 0& -\omega x+\frac{1-\omega }{\alpha }y\\ 0& 1& -\omega \alpha x+(1-\omega )y\\ 0& 0& 1\end{array}\right]\text{.}$$ Similarly, $$S_j{S}_i{|}_U=\left[\begin{array}{ccc}\omega & \frac{1-\omega }{\alpha }& x\\ -{\omega }^2\alpha & -\omega & -\omega \alpha x+y\\ 0& 0& 1\end{array}\right],$$ and thus, $${\left(S_j{S}_i{|}_U\right)}^2=\left[\begin{array}{ccc}1& 0& -\omega x+\frac{1-\omega }{\alpha }y\\ 0& 1& -\omega \alpha x+(1-\omega )y\\ 0& 0& 1\end{array}\right]\text{.}$$ So we have verified ${\left(S_i{S}_j\right)}^2={\left(S_j{S}_i\right)}^2\text{.}$

A similar computation can be done in the remaining six cases to verify that the desired condition, $S_i{S}_j{S}_i\cdots =S_j{S}_i{S}_j\cdots ,$ is always satisfied.

For $$\begin{array}{c} 6 p 6 q q′ \end{array}$$ one finds that $$S_i{S}_j{S}_i=S_j{S}_i{S}_j=\left(\begin{array}{ccc}0& -1& \omega y+x\\ -1& 0& \omega x+y\\ 0& 0& 1\end{array}\right)\text{.}$$

With $$\begin{array}{c} 4 p 4 q 4 q′ \end{array}$$ one computes that $${\left(S_i{S}_j\right)}^2={\left(S_j{S}_i\right)}^2=\left(\begin{array}{ccc}1& 0& (1-i)(x+y)\\ 0& 1& (1-i)(x+y)\\ 0& 0& 1\end{array}\right)\text{.}$$

In the case of $$\begin{array}{c} p 6 q 6 q′ \end{array}$$ one has that $${\left(S_i{S}_j\right)}^3={\left(S_j{S}_i\right)}^3=\left(\begin{array}{ccc}1& 0& (1-\omega )\sqrt{2}y+({\omega }^2-\omega )x\\ 0& 1& (2-2\omega )y+\sqrt{2}({\omega }^2-\omega )x\\ 0& 0& 1\end{array}\right)\text{.}$$

For $$\begin{array}{c} 3 p 3 q 6 q′ \end{array}$$ one finds that $${\left(S_i{S}_j\right)}^3={\left(S_j{S}_i\right)}^3=\left(\begin{array}{ccc}1& 0& -3\omega (x+y)\\ 0& 1& -3\omega (x+y)\\ 0& 0& 1\end{array}\right)\text{.}$$

If we have $$\begin{array}{c} p 4 q 8 q′ \end{array}$$ we calculate that $${\left(S_i{S}_j\right)}^4={\left(S_j{S}_i\right)}^4=\left(\begin{array}{ccc}1& 0& -4ix+2{\beta }^3(1-i)y\\ 0& 1& -4\beta ix+4(1-i)y\\ 0& 0& 1\end{array}\right)$$ where $\beta =2^{1/4}\text{.}$

Finally, for $$\begin{array}{c} p 3 q 12 q′ \end{array}$$ one finds that $${\left(S_i{S}_j\right)}^6={\left(S_j{S}_i\right)}^6=\left(\begin{array}{ccc}1& 0& 6({\omega }^2-\omega )x-\frac{12\alpha }{\sqrt{2}}\omega y\\ 0& 1& \frac{6\sqrt{2}({\omega }^2-\omega )}{\alpha }x-12\omega y\\ 0& 0& 1\end{array}\right)$$ where $\alpha =3^{1/4}\text{.}$

These verifications complete the proof of the theorem.

$\square$