The two realizations of $I_2(4)$: $B_2$ and $C_2$ with their lattices and duality

Put $u=st$. The relators in [F8] give $s^2=t^2=1$ and $u^4=1$; also $sus=u^{-1}$. Replacing $t$ by $su$ and moving each $s$ to the right with $su=u^{-1}s$, every word reduces to $u^k$ or $u^{k}s$, with $0\le k<4$. Every $W$-element is such a word by [F9], so $W$ has at most eight elements and is finite. In coordinates $x{e}_s+y{e}_t$, the form is $x^2-\sqrt{2}xy+y^2=(x-\frac{\sqrt{2}}{2}y{)}^2+\frac{1}{2}{y}^2$, so it is positive definite.

The single edge is a tree with label $4$. Rooting first at $s$ and then at $t$, [F3] gives $c=(1,\sqrt{2})$ and $c^{\mathrm{\prime }}=(\sqrt{2},1)$; both are crystallographic. Their scaled simple roots and coroots are $a_s=e_s$, $a_t=\sqrt{2}{e}_t$, $a_s^{\vee }=2e_s$, $a_t^{\vee }=a_t$ and $a_s^{\mathrm{\prime }}=\sqrt{2}{e}_s$, $a_t^{\mathrm{\prime }}=e_t$, $a_s^{\mathrm{\prime }\vee }=a_s^{\mathrm{\prime }}$, $a_t^{\mathrm{\prime }\vee }=2e_t$. Using [F1] and [F5] gives $A=\left(\begin{array}{cc}2& -1\\ -2& 2\end{array}\right)$ and $A^{\mathrm{\prime }}=\left(\begin{array}{cc}2& -2\\ -1& 2\end{array}\right)=A^T$.

For $c$, [F4] gives $r_s(a_s)=-a_s$, $r_s(a_t)=d:=a_t+2a_s$, $r_t(a_t)=-a_t$, and $r_t(a_s)=p:=a_s+a_t$. By linearity of the reflections, $r_s(p)=p$, $r_t(p)=a_s$, $r_s(d)=a_t$, and $r_t(d)=d$, so $R=\{\pm a_s,\pm a_t,\pm p,\pm d\}$ is stable under $r_s,r_t$. By [F6], these maps are linear and invariant under nonzero rescaling of their normals; since $a_s=c_s{e}_s$ and $a_t=c_t{e}_t$, $r_s=r_{a_s}$ and $r_t=r_{a_t}$. Then [F7] identifies them with $\rho (s),\rho (t)$, and [F9] gives ${\mathrm{\Phi }}_c\subseteq R$. For $w=r_s{r}_t$ one has $w(a_s)=p$, $w(a_t)=-d$, $w(p)=-a_s$, and $w(d)=a_t$, hence $w^2=-I$ on the basis $(a_s,a_t)$. The four positive listed vectors are $a_s,a_t$, $p=\rho (t)a_s$, and $d=\rho (s)a_t$; applying $w^2$ supplies their negatives. Therefore ${\mathrm{\Phi }}_c=R$.

Put ${\beta }_1={\epsilon }_1-{\epsilon }_2$, ${\beta }_2={\epsilon }_2$, ${\gamma }_1={\epsilon }_1-{\epsilon }_2$ and ${\gamma }_2=2{\epsilon }_2$. In $B_2$, choose $v=(2,1)$, which pairs nontrivially with every root in [F11]. Its positive roots are ${\epsilon }_2$, ${\epsilon }_1-{\epsilon }_2$, ${\epsilon }_1$ and ${\epsilon }_1+{\epsilon }_2$; the latter two are ${\beta }_1+{\beta }_2$ and ${\beta }_1+2{\beta }_2$, while ${\beta }_1,{\beta }_2$ are not sums of two listed positive roots. Thus they are simple by [F16]. In $C_2$, the same $v$ pairs nontrivially with every root in [F12] and gives positive roots ${\epsilon }_1-{\epsilon }_2$, $2{\epsilon }_2$, ${\epsilon }_1+{\epsilon }_2$ and $2{\epsilon }_1$; the latter two are ${\gamma }_1+{\gamma }_2$ and $2{\gamma }_1+{\gamma }_2$, while ${\gamma }_1,{\gamma }_2$ are not sums of two listed positive roots. Thus ${\gamma }_1,{\gamma }_2$ are simple by [F16]. The reflections in the first roots swap the coordinates and those in the second roots negate the second coordinate by [F6]; the second reflection is the same for normals ${\epsilon }_2$ and $2{\epsilon }_2$. Their product is a quarter-turn of order four. Therefore both coordinate root systems have Coxeter diagram $I_2(4)$.

In $A_2$, put ${\alpha }_1={\epsilon }_1-{\epsilon }_2$ and ${\alpha }_2={\epsilon }_2-{\epsilon }_3$. The vector $v=(1,0,-1)$ pairs nontrivially with every root in [F10], and its positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_1+{\alpha }_2$, so [F16] makes ${\alpha }_1,{\alpha }_2$ simple. All roots have squared length $2$, so their coroots equal the roots by [F15], and $Q_0=\mathbb{Z}{\alpha }_1+\mathbb{Z}{\alpha }_2=\{x\in {\mathbb{Z}}^3:{\sum }_i{x}_i=0\}$. By [F14] the dual weight lattice is $P_0=\{x\in E:(x,{\alpha }_1),(x,{\alpha }_2)\in \mathbb{Z}\}$. If $m=x_1-x_2$ and $n=x_2-x_3$, then $m,n\in \mathbb{Z}$ and $x=m{\omega }_1+n{\omega }_2$, where ${\omega }_1=(2{\epsilon }_1-{\epsilon }_2-{\epsilon }_3)\mathrm{/}3$ and ${\omega }_2=({\epsilon }_1+{\epsilon }_2-2{\epsilon }_3)\mathrm{/}3$; these vectors are in $P_0$, and every $x\in P_0$ has the same pairings with ${\alpha }_1,{\alpha }_2$ as $m{\omega }_1+n{\omega }_2$, so equality follows because ${\alpha }_1,{\alpha }_2$ span $E$. Thus they form a basis. In this basis ${\alpha }_1=2{\omega }_1-{\omega }_2$ and ${\alpha }_2=-{\omega }_1+2{\omega }_2$, so $P_0\mathrm{/}{Q}_0$ is generated by $[{\omega }_1]$, with $3[{\omega }_1]=0$ and $[{\omega }_1]\mathrm{\ne }0$ because ${\omega }_1\mathrm{\notin }{Q}_0$. Thus $[P(A_2):Q(A_2)]=3$.

For $c^{\mathrm{\prime }}$, [F4] gives $r_s(a_s^{\mathrm{\prime }})=-a_s^{\mathrm{\prime }}$, $r_s(a_t^{\mathrm{\prime }})=p^{\mathrm{\prime }}:=a_s^{\mathrm{\prime }}+a_t^{\mathrm{\prime }}$, $r_t(a_t^{\mathrm{\prime }})=-a_t^{\mathrm{\prime }}$, and $r_t(a_s^{\mathrm{\prime }})=d^{\mathrm{\prime }}:=a_s^{\mathrm{\prime }}+2a_t^{\mathrm{\prime }}$. The further images are $r_s(p^{\mathrm{\prime }})=a_t^{\mathrm{\prime }}$, $r_t(p^{\mathrm{\prime }})=p^{\mathrm{\prime }}$, $r_s(d^{\mathrm{\prime }})=d^{\mathrm{\prime }}$, and $r_t(d^{\mathrm{\prime }})=a_s^{\mathrm{\prime }}$, so $R^{\mathrm{\prime }}=\{\pm a_s^{\mathrm{\prime }},\pm a_t^{\mathrm{\prime }},\pm p^{\mathrm{\prime }},\pm d^{\mathrm{\prime }}\}$ is stable under both generators. The normal-scaling identity from step 1.3 applies to this scaling as well. For $w^{\mathrm{\prime }}=r_s{r}_t$, $w^{\mathrm{\prime }}(a_s^{\mathrm{\prime }})=d^{\mathrm{\prime }}$, $w^{\mathrm{\prime }}(a_t^{\mathrm{\prime }})=-p^{\mathrm{\prime }}$, $w^{\mathrm{\prime }}(p^{\mathrm{\prime }})=a_t^{\mathrm{\prime }}$ and $w^{\mathrm{\prime }}(d^{\mathrm{\prime }})=-a_s^{\mathrm{\prime }}$, hence $w^{\mathrm{\prime }2}=-I$. The positive listed vectors are generator roots or their images: $a_s^{\mathrm{\prime }},a_t^{\mathrm{\prime }}$ are generator roots, $p^{\mathrm{\prime }}=r_s(a_t^{\mathrm{\prime }})$, and $d^{\mathrm{\prime }}=r_t(a_s^{\mathrm{\prime }})$. Applying $w^{\mathrm{\prime }2}$ supplies their negatives. As in 1.3, ${\mathrm{\Phi }}_{c^{\mathrm{\prime }}}=R^{\mathrm{\prime }}$.

In $B_2$, the roots ${\epsilon }_1,{\epsilon }_2$ generate $Q(B_2)={\mathbb{Z}}^2$. The coroots of $\pm {\epsilon }_i$ are $\pm 2{\epsilon }_i$; the mixed roots are their own coroots. These coroots span exactly $Q^{\vee }(B_2)=\mathbb{Z}({\epsilon }_1-{\epsilon }_2)+\mathbb{Z}(2{\epsilon }_2)$: both displayed generators occur, and every other coroot is an integer combination of them. In $C_2$, the roots ${\epsilon }_1-{\epsilon }_2$ and $2{\epsilon }_2$ generate $Q(C_2)$, and the remaining roots lie in that span. Its coroot set contains ${\epsilon }_1-{\epsilon }_2,{\epsilon }_2$ and is contained in ${\mathbb{Z}}^2$, so $Q^{\vee }(C_2)={\mathbb{Z}}^2$. Hence $Q^{\vee }(B_2)=Q(C_2)$ and $Q(B_2)=Q^{\vee }(C_2)$.

By [F14], the dual of $Q^{\vee }(B_2)$ is $P(B_2)=\{(x,y):x-y\in \mathbb{Z},2y\in \mathbb{Z}\}$; writing $y=k\mathrm{/}2$, $x-y=m$ gives $P(B_2)=\mathbb{Z}{\epsilon }_1+\mathbb{Z}({\epsilon }_1+{\epsilon }_2)\mathrm{/}2$. Since $Q(B_2)={\mathbb{Z}}^2$, the quotient is generated by the nontrivial class of $({\epsilon }_1+{\epsilon }_2)\mathrm{/}2$; it is not in $Q(B_2)$ and its double lies in $Q(B_2)$, so it has order two. For $C_2$, $Q^{\vee }(C_2)={\mathbb{Z}}^2$, so $P(C_2)={\mathbb{Z}}^2$; the quotient by $Q(C_2)=\mathbb{Z}({\epsilon }_1-{\epsilon }_2)+\mathbb{Z}(2{\epsilon }_2)$ is generated by $[{\epsilon }_2]$, which is nonzero because ${\epsilon }_2\mathrm{\notin }Q(C_2)$ and has order two because $2{\epsilon }_2\in Q(C_2)$. With the simple systems of step 1.4, [F15] gives coroots ${\beta }_1^{\vee }={\beta }_1$, ${\beta }_2^{\vee }=2{\epsilon }_2$, ${\gamma }_1^{\vee }={\gamma }_1$, ${\gamma }_2^{\vee }={\epsilon }_2$. In the row-coroot convention [F17], $({\beta }_2,{\beta }_1^{\vee })=-1$, $({\beta }_1,{\beta }_2^{\vee })=-2$, $({\gamma }_2,{\gamma }_1^{\vee })=-2$ and $({\gamma }_1,{\gamma }_2^{\vee })=-1$, yielding Cartan matrices $\left(\begin{array}{cc}2& -1\\ -2& 2\end{array}\right)$ and $\left(\begin{array}{cc}2& -2\\ -1& 2\end{array}\right)$, both with determinant $2$.

The map $\psi$ in the Statement is an isometry: the images of $e_s,e_t$ have squared lengths $1,1$ and inner product $-1\mathrm{/}\sqrt{2}=-\sqrt{2}\mathrm{/}2$, which matches [F5]. It sends $a_s,a_t,p,d$ to $({\epsilon }_1-{\epsilon }_2)\mathrm{/}\sqrt{2}$, $2{\epsilon }_2\mathrm{/}\sqrt{2}$, $({\epsilon }_1+{\epsilon }_2)\mathrm{/}\sqrt{2}$, $2{\epsilon }_1\mathrm{/}\sqrt{2}$; it sends $a_s^{\mathrm{\prime }},a_t^{\mathrm{\prime }},p^{\mathrm{\prime }},d^{\mathrm{\prime }}$ to ${\epsilon }_1-{\epsilon }_2$, ${\epsilon }_2$, ${\epsilon }_1$, ${\epsilon }_1+{\epsilon }_2$. By [F11,F12], these are exactly $\frac{1}{\sqrt{2}}\mathrm{\Phi }(C_2)$ and $\mathrm{\Phi }(B_2)$; [F13] states that these coordinate root sets are reduced crystallographic systems. Positive scaling preserves those axioms: finiteness, spanning and reducedness are preserved, reflection normal lines are unchanged, and Cartan integers are unchanged by a common scalar. Thus both ${\mathrm{\Phi }}_c$ and ${\mathrm{\Phi }}_{c^{\mathrm{\prime }}}$ are reduced crystallographic root systems. The coroots of $\pm 2{\epsilon }_i$ in $C_2$ are $\pm {\epsilon }_i$, while mixed roots have squared length $2$ and are their own coroots; hence $\mathrm{\Phi }(C_2{)}^{\vee }=\mathrm{\Phi }(B_2)$. Since $(\lambda \mathrm{\Phi }{)}^{\vee }={\lambda }^{-1}{\mathrm{\Phi }}^{\vee }$ for $\lambda >0$ by [F15], $\psi ({\mathrm{\Phi }}_c^{\vee })=\sqrt{2}\mathrm{\Phi }(B_2)$ and $\psi ({\mathrm{\Phi }}_{c^{\mathrm{\prime }}})=\frac{1}{\sqrt{2}}\psi ({\mathrm{\Phi }}_c^{\vee })$.

All calculations use the fixed two-generator data and explicit finite coordinate root sets. No Choice is used.