154 N. Rivier and J.-F. Sadoc
Fig. 8.5.Bouligand’s overlap–gap transformation between two Archimedean lat-
tices. Unit cells in grey. (a) A square lattice decorated as a topological square–
triangle lattice 3^2. 4. 3 .4(overlap,z=5).(b) By removing one vertex out of five
from the square lattice, one obtains a topological square–triangle lattice 3^2. 4. 3. 4
(gap,z= 5). The square unit cell has area
√
5 ×√
5 = 5. The vertices removed from
the original lattice are noted aso, and are replaced by a diamond square. (c)The
other Archimedean,z= 5 alternative is 3^3. 42 , but it is much less isotropic. The unit
cell is not square, with an area
√
2 ×(√
2+1/√
2) = 3, and one vertex out of three
has been removed. (d) The only Archimedean alternative is between square lattices
(z= 4). The square unit cell has area
√
2 ×√
2 = 2 (one vertex out of two has been
removed)