$A : \mathbb{R}^n \to \mathbb{R}^n \implies A(\mathbb{Z}^n)=\Gamma$ on theTorus

Question

In the Analysis on Manifolds via the Laplacian page $51$, it is indicated that if $A : \mathbb{R}^n \to \mathbb{R}^n$ be so that $A(\mathbb{Z}^n)=\Gamma$, then $\text{Vol} (\mathbb{T})=\det A$. Could anyone be able to give me a hint how to solve that?

Answer 1

If $\Omega=\{\omega_1,\dots,\omega_n\}$ generate $\Gamma$, then $\operatorname{vol}(\Bbb T)$ will be the area spanned by $\Omega$, which is $\det\Omega$. As each $a_i=A(e_i)$ is $\Omega(v)$ for some $v\in\Bbb Z^n$, it then follows that $\det A$ is an integer multiple of $\det \Omega$. But it can't be negative if $A$ is positive definite, and it can't be zero because $A$ is surjective onto $\Bbb R^n$.

We can reduce the problem to the case when $\Omega$ is the identity matrix, and $\Gamma=\Bbb Z^n$, by composing $A$ with $\Omega^{-1}$. Then we are proving that if $A(\Bbb Z^n)=\Bbb Z^n$, then $\det A=1$. In this form it is even clearer that $\det A$ is an integer, because $A$ is an integer matrix.

To finish the proof, note that $A^{-1}$ satisfies the same conditions as $A$: it is positive definite, and $A^{-1}(\Bbb Z^n)=A^{-1}(A(\Bbb Z^n))=\Bbb Z^n$. Thus $\det A^{-1}=\frac1{\det A}$ is also an integer, so $\det A=\det A^{-1}=1$.

Answer 2

The volume of a parallelepiped is given by the determinant, and the parallelepiped is a fundamental domain for the quotient torus.