Computing the volume of a fundamental domain of a lattice

Question

Suppose I have $n$ linearly independent vectors in $\mathbb{R}^m$, say $v_1, .., v_n$. Then $v_1,..., v_n$ form a lattice $\Lambda$ inside a subspace $V$ = $\mathbb{R}v_1 + ... + \mathbb{R}v_n \subseteq \mathbb{R}^m$.

How do we define the volume of a fundamental domain of this lattice $\Lambda$ inside $V$? If $m=n$, then $V = \mathbb{R}^n$ and the volume of a fundamental domain is $|\det V|$, where $V$ is the matrix with $i$-th column as $v_i$. I was wondering if someone could explain me how to this when $n<m$. Thank you very much!

PS I am particularly interested in knowing the formula to compute it, and how this value corresponds to Lebesgue measure of the fundamental domain somehow.

Answer 1

If we rotate $V$ so that it becomes the standard $\mathbb R^n$ (i.e. all $v_i$ have $0$ entries in positions $n+1,\ldots,m$), we can cut off the higher components and compute the determinant of the resulting $n\times n$ matrix. The same can be achieved by adding one by one vectors $v_{n+1},\ldots,v_m$ such that each of these has unit length ans is orthogonal on all preceeding vectors. Then compute the resulting $m\times m$ determinant.

Answer 2

The general formula is $$ vol(V)=\sqrt{\mid \det (<v_i,v_j>)_{i,j}\mid} $$ with the matrix $(<v_i,v_j>)_{i,j}$, for $n\le m$, which is $AA^T$, where $A$ is not a square matrix, if $n<m$. However, $AA^T$ is always square. If $n=m$, then this matrix is $VV^T$, so that $vol(V)=\sqrt{\det(VV^T)}=\sqrt{\det(V)^2}=\mid \det(V)\mid$.