I am reading Chapter 4 of Ashcroft and Mermin (A&M) in which the basic definitions of a Bravais lattice (BL) — as considered as a purely mathematical entity — are being laid down. One of the definitions of a BL given by A&M goes something like the following:
A Bravais lattice is any subset of $\Bbb{R}^3$ which consists of all integral linear combinations of three non-collinear $a_i \in \Bbb{R}^3$:
$$BL = \{n_1 a_1+n_2 a_2+n_3 a_3 | a_i\in\Bbb{R}^3, n_i \in \Bbb{Z}\}$$
My question is as follows: can the primitive vectors $a_i$ of the Bravais lattice be unique? That is, does there exist a BL for which the set of $\{a_i\}$ generating the BL is unique? Naturally, there are many examples for which this is not the case, but is this not the case for all BLs?
When trying to prove this, I tried to use the usual construction in linear algebra: write an arbitrary element of the BL in the original basis, and transform to a new basis, but I ran into trouble because of the discrete nature of the BL.
Best Answer
As long as you have a 3-by-3 matrix ${M^a}_{b}$ with integer entries, and such that its determinant is unity (so that its inverse is also an integer matrix) you can use this to change the Bravais basis set to $$ {\bf e}'_a = {\bf e}_b{M^b}_{a} $$ and the set $$ n_1 {\bf e}'_1+n_2 {\bf e}'_2+n_3 {\bf e}'_3 $$ is the same set of points as $$ n_1 {\bf e}_1+n_2 {\bf e}_2+n_3 {\bf e}_3. $$ Of course the labels $(n_1,n_2,n_3)$ of the individual points are different.
The set of such transformation matrices is the group ${\rm SL}(3,{\mathbb Z})$, and does not depend on the geometry of the Bravais lattice.