G2 from I2(6): the scaled realization and its twelve roots

Since $V={\mathbb{R}}^{\{s,t\}}$, evaluating functions at $s$ and $t$ shows that $(e_s,e_t)$ is a basis. Put $z=\cos (\pi \mathrm{/}3)$. Since $0<\pi \mathrm{/}3<\pi \mathrm{/}2$, [F5] gives $z>0$. The supplementary and double-angle identities give $-z=\cos (2\pi \mathrm{/}3)=2z^2-1$, so $(2z-1)(z+1)=0$ and hence $z=1\mathrm{/}2$. Applying the double-angle identity at $\pi \mathrm{/}6$ and using positivity again gives $\cos (\pi \mathrm{/}6)=\sqrt{3}\mathrm{/}2$. Thus $B(x{e}_s+y{e}_t,x{e}_s+y{e}_t)=x^2-\sqrt{3}xy+y^2=(x-\frac{\sqrt{3}}{2}y{)}^2+\frac{1}{4}{y}^2$, which is positive for every nonzero $(x,y)$, so $B$ is positive definite. Set $u=st$; then $u^6=1$, $sus=u^{-1}$ and $t=su$. Moving each $s$ to the right shows every word is $u^{k}s$ or $u^k$, with $0\le k<6$, so $\mathrm{\mid }W\mathrm{\mid }\le 12$.

By [F6] and step 1.1, $c_s=1$ and $c_t=2\cos (\pi \mathrm{/}6)=\sqrt{3}$ give a crystallographic scaling. Hence $a_s=e_s$, $a_t=\sqrt{3}{e}_t$, $B(a_s,a_t)=-3\mathrm{/}2$, $a_{st}=B(a_s,a_t^{\vee })=-1$, and $a_{ts}=B(a_t,a_s^{\vee })=-3$, so $A=\left(\begin{array}{cc}2& -1\\ -3& 2\end{array}\right)$. Choosing $t$ as the tree root gives the other scaling $c_s^{\mathrm{\prime }}=\sqrt{3}$, $c_t^{\mathrm{\prime }}=1$; the same Cartan formula gives $A^{\mathrm{\prime }}=\left(\begin{array}{cc}2& -3\\ -1& 2\end{array}\right)=A^T$.

In coordinates $(x,y)$ relative to $(a_s,a_t)$, the reflection formula [F14] gives $r_{a_s}(x,y)=(-x+3y,y)$ and $r_{a_t}(x,y)=(x,x-y)$; each matrix squares to the identity. These maps preserve $R:=\{\pm (1,0),\pm (0,1),\pm (1,1),\pm (2,1),\pm (3,1),\pm (3,2)\}$: on the six displayed positive pairs their respective images are $((-1,0),(3,1),(2,1),(1,1),(0,1),(3,2))$ and $((1,1),(0,-1),(1,0),(2,1),(3,2),(3,1))$, and the images of their negatives are the negatives of these. Conversely $(1,1)=r_{a_t}(1,0)$, $(2,1)=r_{a_s}(1,1)$, $(3,1)=r_{a_s}(0,1)$ and $(3,2)=r_{a_t}(3,1)$. For every orbit vector $\beta =\rho (w)a_i$, the element $\rho (w{s}_i{w}^{-1})$ sends $\beta$ to $-\beta$, so all twelve pairs lie in the orbit of the simple roots. Substituting positive scalar multiples in the reflection formula gives $r_{a_s}=r_{e_s}=\rho (s)$ and $r_{a_t}=r_{e_t}=\rho (t)$; hence $R={\mathrm{\Phi }}_c$ by [F15]. The squared norm of $(x,y)$ is $x^2-3xy+3y^2$, whose values on $(1,0),(0,1),(1,1),(2,1),(3,1),(3,2)$ are respectively $1,3,1,1,3,3$. The product $U=\rho (st)=r_{a_s}{r}_{a_t}$ has matrix $\left(\begin{array}{cc}2& -3\\ 1& -1\end{array}\right)$, with $U^2=\left(\begin{array}{cc}1& -3\\ 1& -2\end{array}\right)$ and $U^3=-I$; hence $U$ has exact order $6$. The six maps $U^k$ are distinct, as are $U^k\rho (s)$, and the two lists are disjoint since their determinants are $1$ and $-1$. Thus $\mathrm{\mid }\rho (W)\mathrm{\mid }\ge 12$; with step 1.1 this gives $\mathrm{\mid }W\mathrm{\mid }=12$ and $\rho$ is injective.

For every nonisotropic $a$, expansion of [F14] gives $B(r_{a}u,r_{a}v)=B(u,v)-\frac{2B(u,a)B(a,v)}{B(a,a)}-\frac{2B(v,a)B(u,a)}{B(a,a)}+\frac{4B(u,a)B(v,a)}{B(a,a)}=B(u,v)$; applying this to $a_s,a_t$ shows the generators of $\rho (W)$ preserve $B$, hence so does every $\rho (w)$. If $\beta =\rho (w)a_i\in {\mathrm{\Phi }}_c$ by [F15], then $B(\beta ,\beta )=B(a_i,a_i)>0$, and conjugating the reflection formula by the $B$-isometry $\rho (w)$ gives $r_{\beta }=\rho (w)r_{a_i}\rho (w{)}^{-1}$, so $r_{\beta }({\mathrm{\Phi }}_c)={\mathrm{\Phi }}_c$. The set ${\mathrm{\Phi }}_c=R$ is finite, nonzero and spans $V$; its root-coroot pairings are integral by [F7] and its reducedness follows from the six distinct root slopes. Therefore ${\mathrm{\Phi }}_c$ is a reduced crystallographic Euclidean root system.

Let $v=6a_s+\frac{10}{3}{a}_t$; the Gram matrix from [F2], [F4] and step 1.1 gives $B(v,a_s)=B(v,a_t)=1$. The vectors $a_s=e_s$ and $a_t=\sqrt{3}{e}_t$ form a basis. Thus for each listed pair $(x,y)$ with $x,y\ge 0$, $B(v,x{a}_s+y{a}_t)=x+y>0$, and the six listed vectors are exactly the positive roots. Neither $a_s$ nor $a_t$ is a sum of two positive roots, since the only positive root with second coordinate zero is $a_s$ and the only one with first coordinate zero is $a_t$; the other positive roots decompose as $a_s+a_t$, $a_s+(a_s+a_t)$, $a_s+(2a_s+a_t)$ and $a_t+(3a_s+a_t)$. Hence $\{a_s,a_t\}$ is a base. Since $B(a_s,a_t)=-3\mathrm{/}2\mathrm{\ne }0$, these two spanning roots cannot belong to different orthogonal parts in a decomposition, while they already span $V$; [F9] therefore gives irreducibility. By [F11], the root system is of type $G_2$.

Every root reflection is $\rho (w)r_{a_i}\rho (w{)}^{-1}$ as in step 3.1, so [F12] gives $W({\mathrm{\Phi }}_c)\subseteq \rho (W)$; conversely $r_{a_s}=\rho (s)$ and $r_{a_t}=\rho (t)$ generate $\rho (W)$, so $W({\mathrm{\Phi }}_c)=\rho (W)$ and it has order $12$. For the second scaling, $a_s^{\mathrm{\prime }}=\sqrt{3}{e}_s=(\sqrt{3}\mathrm{/}2)a_s^{\vee }$ and $a_t^{\mathrm{\prime }}=e_t=(\sqrt{3}\mathrm{/}2)a_t^{\vee }$; because $\rho (w)$ preserves $B$, $(\rho (w)a_i{)}^{\vee }=\rho (w)a_i^{\vee }$, whence ${\mathrm{\Phi }}_{c^{\mathrm{\prime }}}=(\sqrt{3}\mathrm{/}2){\mathrm{\Phi }}_c^{\vee }$ by [F7]. This construction uses only the two fixed generators and finitely many roots, so no form of the Axiom of Choice is used.