The $A_2$ root and weight lattices: $P\mathrm{/}Q$ has order three

Put $z=\cos (\pi \mathrm{/}3)>0$ by [F15]. The double-angle and supplementary identities give $2z^2-1=-z$, hence $(2z-1)(z+1)=0$ and $z=1\mathrm{/}2$. Since $m(s,t)=3$, [F4] gives the Gram matrix $\left(\begin{array}{cc}1& -1\mathrm{/}2\\ -1\mathrm{/}2& 1\end{array}\right)$ in $(e_s,e_t)$. With $a_s=e_s$ and $a_t=e_t$, one has $a_s^{\vee }=2a_s$, $a_t^{\vee }=2a_t$, $a_{ss}=a_{tt}=2$, $a_{st}=a_{ts}=-1$, and the displayed Cartan matrix. The form is positive definite because its quadratic form is $(x-y\mathrm{/}2{)}^2+3y^2\mathrm{/}4$. Thus the chosen scaling is crystallographic, and [F3] shows every crystallographic scaling has $c_s=c_t$, so this is unique up to a common positive multiple.

By bilinearity, the reflection formula [F5] defines linear maps and gives $r_{a_s}(a_s)=-a_s$, $r_{a_s}(a_t)=a_s+a_t$, $r_{a_t}(a_t)=-a_t$, $r_{a_t}(a_s)=a_s+a_t$, $r_{a_s}(a_s+a_t)=a_t$, and $r_{a_t}(a_s+a_t)=a_s$; by linearity the set $R=\{\pm a_s,\pm a_t,\pm (a_s+a_t)\}$ is stable under both reflections. By [F6], $\rho (s)=r_{a_s}$ and $\rho (t)=r_{a_t}$, since the reflection formula is unchanged by scaling its normal; by [F7], every $w\in W$ is a word in $s,t$, so ${\mathrm{\Phi }}_c\subseteq R$. Conversely, $a_s,a_t$ belong to ${\mathrm{\Phi }}_c$, while $-a_s=\rho (s)a_s$, $-a_t=\rho (t)a_t$, $a_s+a_t=\rho (s)a_t$, and $-(a_s+a_t)=\rho (t)\rho (s)a_s$. Thus ${\mathrm{\Phi }}_c=R$.

For $B_2$, set the roots ${\beta }_1={\epsilon }_1-{\epsilon }_2$ and ${\beta }_2={\epsilon }_2$ from the coordinate model [F9]. Their coroots are ${\beta }_1^{\vee }={\beta }_1$, ${\beta }_2^{\vee }=2{\epsilon }_2$. The full coordinate root set in [F9] contains ${\epsilon }_1,{\epsilon }_2$, so $Q(B_2)=\mathbb{Z}{\epsilon }_1+\mathbb{Z}{\epsilon }_2$. Its coroot set consists of $\pm ({\epsilon }_1\pm {\epsilon }_2)$ and $\pm 2{\epsilon }_i$; these span exactly $Q^{\vee }(B_2)=\mathbb{Z}{\beta }_1+\mathbb{Z}(2{\epsilon }_2)$, since both displayed generators are coroots and every listed coroot lies in their span. Thus [F12] gives $P(B_2)=\{(x,y):x-y\in \mathbb{Z},2y\in \mathbb{Z}\}=\mathbb{Z}{\epsilon }_1+\mathbb{Z}\frac{{\epsilon }_1+{\epsilon }_2}{2}$. The nontrivial coset is generated by $({\epsilon }_1+{\epsilon }_2)\mathrm{/}2$, of order two, so $[P(B_2):Q(B_2)]=2$.

For $C_2$, set the roots ${\gamma }_1={\epsilon }_1-{\epsilon }_2$ and ${\gamma }_2=2{\epsilon }_2$ from the coordinate model [F10]. Their coroots are ${\gamma }_1^{\vee }={\gamma }_1$ and ${\gamma }_2^{\vee }={\epsilon }_2$. The full root set in [F10] yields $Q(C_2)=\mathbb{Z}({\epsilon }_1-{\epsilon }_2)+\mathbb{Z}(2{\epsilon }_2)$: the two generators are roots and each other root is an integer combination of them. Its coroot set contains ${\epsilon }_1-{\epsilon }_2$ and ${\epsilon }_2$ and is contained in ${\mathbb{Z}}^2$, so $Q^{\vee }(C_2)=\mathbb{Z}({\epsilon }_1-{\epsilon }_2)+\mathbb{Z}{\epsilon }_2={\mathbb{Z}}^2$. Thus [F12] gives $P(C_2)=\{(x,y):x-y\in \mathbb{Z},y\in \mathbb{Z}\}={\mathbb{Z}}^2$. The quotient is generated by $[{\epsilon }_2]$, which has order two because $2{\epsilon }_2\in Q(C_2)$ and ${\epsilon }_2\mathrm{\notin }Q(C_2)$; hence $[P(C_2):Q(C_2)]=2$.

In each coordinate model, the reflection formula [F5] gives the maps $s_1(x,y)=(y,x)$ and $s_2(x,y)=(x,-y)$ for the displayed $B_2$ and $C_2$ root pairs; the second map is unchanged when its normal is $2{\epsilon }_2$ instead of ${\epsilon }_2$. Both maps are involutions. Their product sends $(x,y)\mapsto (-y,x)$; its square is $-I$ and its fourth power is $I$, so its order is four. Thus both standard systems realize the Coxeter diagram $I_2(4)$, giving the stated $B_2=C_2$ diagram coincidence.

Put ${\alpha }_1={\epsilon }_1-{\epsilon }_2$ and ${\alpha }_2={\epsilon }_2-{\epsilon }_3$. The linear map sending $a_s\mapsto {\alpha }_1\mathrm{/}\sqrt{2}$ and $a_t\mapsto {\alpha }_2\mathrm{/}\sqrt{2}$ is an isometry: the images have squared lengths $1,1$ and inner product $-1\mathrm{/}2$, matching their Gram matrix from [F4]. By step 1.2 it sends ${\mathrm{\Phi }}_c$ to $\frac{1}{\sqrt{2}}\{\pm {\alpha }_1,\pm {\alpha }_2,\pm ({\alpha }_1+{\alpha }_2)\}$, the standard $A_2$ coordinate root system of [F8]. For $v=(1,0,-1)$, its positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_1+{\alpha }_2$; thus [F11] makes ${\alpha }_1,{\alpha }_2$ its simple roots.

The two simple roots have squared length $1$, so [F13] gives their simple coroots $2a_s,2a_t$; hence $Q=\mathbb{Z}{a}_s+\mathbb{Z}{a}_t=\mathbb{Z}{\mathrm{\Phi }}_c$ and $Q^{\vee }=\mathbb{Z}(2a_s)+\mathbb{Z}(2a_t)=2Q$. Since every root in the standard $A_2$ coordinate set has squared length $2$, its coroot lattice is $Q_0$ by [F8,F11,F12,F13]. The isometry of step 2.1 sends $Q$ to $Q_0\mathrm{/}\sqrt{2}$ and $Q^{\vee }$ to $\sqrt{2}{Q}_0$, where $Q_0=\mathbb{Z}{\alpha }_1+\mathbb{Z}{\alpha }_2=\{x\in {\mathbb{Z}}^3:x_1+x_2+x_3=0\}$. By [F12], the image of $P$ is the lattice dual to $\sqrt{2}{Q}_0$, namely $P_0\mathrm{/}\sqrt{2}$, where $P_0=\{x\in E:(x,{\alpha }_1),(x,{\alpha }_2)\in \mathbb{Z}\}$.

For $x=(x_1,x_2,x_3)\in E$, the pairings defining $P_0$ are $x_1-x_2$ and $x_2-x_3$. Thus membership is equivalent to having integral coordinate differences. If those differences are integers, write $x_1=x_3+m$ and $x_2=x_3+n$ with $m,n\in \mathbb{Z}$; the sum-zero condition gives $3x_3=-m-n$, so all coordinates lie in $\frac{1}{3}\mathbb{Z}$. Conversely the displayed conditions make both pairings integral. Hence $P_0=\{x\in (\frac{1}{3}\mathbb{Z}{)}^3:{\sum }_i{x}_i=0,x_i-x_j\in \mathbb{Z}\text{ for all }i,j\}$.

The vectors ${\omega }_1=(2{\epsilon }_1-{\epsilon }_2-{\epsilon }_3)\mathrm{/}3$ and ${\omega }_2=({\epsilon }_1+{\epsilon }_2-2{\epsilon }_3)\mathrm{/}3$ satisfy $({\omega }_i,{\alpha }_j^{\vee })={\delta }_{ij}$, since ${\alpha }_j^{\vee }={\alpha }_j$ by [F13]. They are the fundamental weights by [F14] and form a basis of $P_0$: any $x\in P_0$ has integer pairings with the basis ${\alpha }_1^{\vee },{\alpha }_2^{\vee }$ and therefore is the corresponding integer linear combination of ${\omega }_1,{\omega }_2$. In this basis ${\alpha }_1=2{\omega }_1-{\omega }_2$ and ${\alpha }_2=-{\omega }_1+2{\omega }_2$; hence $[{\omega }_2]=2[{\omega }_1]$ and $3[{\omega }_1]=0$ in $P_0\mathrm{/}{Q}_0$. The quotient is nontrivial because ${\omega }_1\mathrm{\notin }{Q}_0$, so it is cyclic of order three. Therefore $P_0\mathrm{/}{Q}_0\cong \mathbb{Z}\mathrm{/}3$, $[P:Q]=3$, and $\det A=2\cdot 2-(-1)(-1)=3$.

The $A_2$ lattices satisfy $Q⊊P$ and $[P:Q]=3=\det A$, whereas both standard $B_2$ and $C_2$ coordinate systems have weight/root index two. No Choice is used: every step is a finite coordinate calculation on the displayed bases and finite root sets.