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An equilateral triangle obeys the crystallographic restriction theorem, but it is not a part of 2d crystal structure. What symmetry does it lack? Why can't it be a Bravais lattice?

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It is impossible to make a Bravais lattice of triangular cells (equilateral or not). It is the essence of the Bravais lattice concept that it is based on translational symmetry (in 2D, along two independent directions). In order to build a Bravais lattice with an equilateral triangular cell, one would need to ensure that, by rigidly displacing a triangle along two of its sides, it is possible to have a tessellation (i.e., a covering) of the plane without overlapping and without holes. This is not possible with triangles. One needs parallelograms or other 2D figures that do not need rotations after displacement to tassell the plane.

What could be misleading is that, in 2D, there is the so-called triangular lattice. This is a misname. The elementary cell is a rhombus with $60^{\circ} $ acute angles. The Wigner-Seitz cell is instead a regular hexagon made of six equilateral triangles. However, no cell is a triangle. If we see it as a crystalline structure, it can be thought as a triangular Bravais lattice with a basis made by two equilateral triangles sharing one side. But that it is not a Bravais lattice of triangles.

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It is a possible 2d lattice structure, it just so happens that in the case of equilateral triangles we can observe them to tile together into hexagons. A hexagonal lattice is sometimes called a triangular lattice. https://en.wikipedia.org/wiki/Hexagonal_lattice

We can see clearly how the lattice vectors of a hexagon can form the lattice vector of an equilateral in the following figure

graphic addition of the two lattice vector of a hexagonal lattice

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  • $\begingroup$ Thanks for clarification $\endgroup$ Nov 1, 2022 at 11:23

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