 Solution

The vectors from the B atom to its three nearest N neighbours can be chosen asIndeed ${\mathbf{n}}_1={\mathbf{d}}_N-{\mathbf{d}}_B$, ${\mathbf{n}}_2={\mathbf{n}}_1-{\mathbf{a}}_1$ and ${\mathbf{n}}_3={\mathbf{n}}_1-{\mathbf{a}}_2$. All three have length $a/\sqrt{3}$. Repeating the two-site basis over $\mathbf{R}=m{\mathbf{a}}_1+n{\mathbf{a}}_2$ produces the honeycomb pattern in the first panel of the original figure in solution.

Solving ${\mathbf{a}}_i\cdot {\mathbf{b}}_j=2\pi {\delta }_{ij}$ givesThe first Brillouin zone is the Wigner–Seitz hexagon of this reciprocal lattice. Its vertices are $(0,\pm 4\pi /(3a))$ and $(\pm 2\pi /(\sqrt{3}a),\pm 2\pi /(3a))$, with independent signs in the latter set. The centre is $\Gamma =(0,0)$; one inequivalent corner is $K=(0,4\pi /(3a))$, with $K^{\prime }=-K$. Corners related by a reciprocal lattice translation are equivalent, leaving two inequivalent valleys.