Root versus coroot translation lattices: A2, B2 and two conventions

The $A_2$ set is finite, nonzero, and spans the sum-zero plane because $e_1-e_2$ and $e_2-e_3$ are independent. Each root reflection swaps two coordinates, so it preserves the set. Every root has squared length $2$, and the dot product of any two listed roots is an integer; hence $2B(\beta ,\alpha )\mathrm{/}B(\alpha ,\alpha )=B(\beta ,\alpha )\in \mathbb{Z}$. No root line contains any listed multiple other than the two signs. Thus ${\mathrm{\Phi }}_{A_2}$ is a reduced crystallographic root system.

The $B_2$ set is finite, nonzero, and spans ${\mathbb{R}}^2$. Reflection in a short root $\pm e_i$ changes the sign of one coordinate; reflection in a long root $\pm e_1\pm e_2$ swaps or swaps-and-negates the two coordinates. These maps preserve the displayed set. Short roots have squared length $1$ and long roots squared length $2$; all dot products of listed roots are integers, so $2B(\beta ,\alpha )\mathrm{/}B(\alpha ,\alpha )$ is integral for either possible denominator. Each root line contains only the two signs. Thus ${\mathrm{\Phi }}_{B_2}$ is a reduced crystallographic root system.

By [F2], the standard walls give translation lattice $Q^{\vee }$. The dual-normal walls are exactly the standard affine walls of the reduced crystallographic root system ${\mathrm{\Phi }}^{\vee }$ from [F9], so [F2] applied to ${\mathrm{\Phi }}^{\vee }$ gives translation lattice $Q^{\vee }({\mathrm{\Phi }}^{\vee })={\sum }_{\alpha \in \mathrm{\Phi }}\mathbb{Z}({\alpha }^{\vee }{)}^{\vee }=Q$ by [F3, F4].

Every $A_2$ root has squared length $2$, so ${\alpha }^{\vee }=\alpha$ and the root and coroot lattices agree.

In $B_2$, the roots $\pm e_i$ have coroots $\pm 2e_i$, while the roots $\pm e_1\pm e_2$ have the same coroots. Since $e_1+e_2=(e_1-e_2)+2e_2$ and $2e_1=2(e_1-e_2)+2e_2$, all these coroots lie in $\mathbb{Z}(e_1-e_2)+\mathbb{Z}(2e_2)$; conversely both displayed generators are coroots. The roots include $\pm e_1,\pm e_2$, and every other root is their integer combination. Therefore $Q=\mathbb{Z}{e}_1+\mathbb{Z}{e}_2$ and $Q^{\vee }=\mathbb{Z}(e_1-e_2)+\mathbb{Z}(2e_2)\subset Q$.

Relative to the basis $(e_1,e_2)$ of $Q$, the two displayed generators of $Q^{\vee }$ are the columns of $M=\left(\begin{array}{cc}1& 0\\ -1& 2\end{array}\right)$, so $\mathrm{\mid }\det M\mathrm{\mid }=2$. By [F5], $Q\mathrm{/}{Q}^{\vee }$ has order $2$, and by [F6] this says $[Q:Q^{\vee }]=2$.

Relative to $(e_1,e_2)$, the basis matrices of $Q$ and $Q^{\vee }$ are $I$ and $M=\left(\begin{array}{cc}1& 0\\ -1& 2\end{array}\right)$. By [F7], their basis-parallelogram areas are $\mathrm{\mid }\det I\mathrm{\mid }=1$ and $\mathrm{\mid }\det M\mathrm{\mid }=2$. For any $x\in E_{B_2}$, write its unique coordinates in the latter basis as $(u_1,u_2)$ and let $m_i=\lfloor u_i\rfloor$ by [F8], so $u_i-m_i\in [0,1)$. Then $x=(m_1(e_1-e_2)+2m_2{e}_2)+((u_1-m_1)(e_1-e_2)+2(u_2-m_2)e_2)$. The first term lies in $Q^{\vee }$ and the second in the half-open parallelogram $P=\{t_1(e_1-e_2)+2t_2{e}_2:0\le t_1,t_2<1\}$. Uniqueness of the integer parts makes this decomposition unique; hence the translates of $P$ by $Q^{\vee }$ partition the plane. Its covolume is twice that of $Q$.

In $B_2$, take the generating root-reflection pair with normals ${\alpha }_1=e_1-e_2$ and ${\alpha }_2=e_2$; their reflections swap coordinates and change one coordinate sign, so they generate the signed permutation reflection group of $B_2$. Their dual roots are ${\beta }_1={\alpha }_1^{\vee }=e_1-e_2$ and ${\beta }_2={\alpha }_2^{\vee }=2e_2$, so ${\mathrm{\Phi }}_{B_2}^{\vee }=\{\pm 2e_i,\pm e_1\pm e_2\}$ is $C_2$, and [F3, F4] give the stated lattice exchange. The normal pairs satisfy $\frac{B({\alpha }_1,{\alpha }_2)}{\mathrm{\parallel }{\alpha }_1\mathrm{\parallel }\mathrm{\parallel }{\alpha }_2\mathrm{\parallel }}=\frac{B({\beta }_1,{\beta }_2)}{\mathrm{\parallel }{\beta }_1\mathrm{\parallel }\mathrm{\parallel }{\beta }_2\mathrm{\parallel }}=-\frac{1}{\sqrt{2}}$. Thus both pairs of reflecting hyperplanes meet at acute angle $\pi \mathrm{/}4$; the product of the two reflections is a rotation through $\pi \mathrm{/}2$ and has order $4$. This proves the diagram statement. All coordinate lists are finite and explicit, and the only interval representatives use the unique integer part; no axiom of choice is used.