Proof of the finite number of Bravais lattices?

Question

I've been taught that there are a finite number of Bravais lattices in 1, 2 and 3 dimensions. I am wondering if there is a proof of this fact. Maybe this is obvious and I am only missing certain key assumptions that are made in setting the problem?

Answer 1

In fact, there are only finitely many Bravais lattices in every dimension $n\ge 1$. The numbers are $1$, $4$, $14$, $64$ in dimensions $1,2,3,4$. The proof follows from a famous theorem of Jordan,

Theorem(Jordan): The group $GL(n,\mathbb{Z})$ has only finitely many conjugacy classes of finite subgroups.

Bravais lattices correspond here to conjugacy classes of certain finite subgroups, the so called Bravais subgroups. The group $GL(n,\mathbb{Z})$ arises here as the automorphism group of the full lattice $\mathbb{Z}^n$ in $\mathbb{R}^n$.
More generally there are only finitely many crystallographic groups in each dimension $n\ge 1$. This was asked by Hilbert in his list of famous problems of $1900$ (problem $17$), and was answered by Bieberbach $10$ years later.
The proof is easier, of course, if we assume $n\le 3$. Here it suffices just to classify all conjugacy classes of finite subgroups of $GL(n,\mathbb{Z})$ for $n=1,2,3$. The lists can be found in papers of Schulz and Plesken on the classification of crystallographic groups.