Why is the covolume of a lattice is unique?

Question

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?

Answer 1

• 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}

Answer 2

Here is a purely algebraic explanation of this fact. If we have two bases of an lattice, then the transition matrices from the first to the second basis and from the second to the first are integer. Hence, their determinants are $1$ or $-1$.

I don't think a more detailed explanation is needed.