Given a basis $\mathfrak{B}=\{b_1, b_2, \cdots, b_n\}$ of $\mathbb{R}^n,$ the lattice generated by $\mathfrak{B}$ is the set of all linear combinations with integer coefficients: $$m_1b_1+m_2b_2+\cdots+m_nb_n,\qquad (m_1, m_2, \cdots, m_n)\in\mathbb{Z}^n.$$
Absolute value of the determinant of the vectors in $\mathfrak{B}$ (taking as columns) is called the covolume of the lattice.
It is not difficult to see that the generating basis of a lattice is no unique. However, it says that covolume of a lattice is an invariant.
Can someone give me a simple and rigorous explanation for this fact?
Best Answer
Let $B=(b_1,\dots,b_n)$ and $C=(c_1,\dots,c_n)$ be two bases that generate the same lattice. That means that every $c_i$ can be expressed as linear-combination of the $b_i$'s. Furthermore, the coefficients of this linear-combination must be integers. The same works the otherway around, expressing the $b_i$'s in terms of $c_i$'s.
Now we interpret $B$ and $C$ as real $n\times n$ matrices. By the above argument, there is an invertible matrix $U$, such that such that $UB=C$, and therefore \begin{align}\det{U}\det{B}=\det{C}\end{align}
Now What is $\det U$? The entries of $U$ and $U^{-1}$ are all integers. Therefore both $\det{U}$ and also $\det U^{-1}=\frac{1}{\det{U}}$ must be integers. This can only be true if $\det U=\pm1$. Therefore \begin{align} |\det{B}|=|\det{C}| \end{align}