projection map $p:\Bbb{R}^n\to \Bbb{T}^n$ locally preserves the volume

Question

Let $\Bbb{T}^n:=\Bbb{R}^n/\Bbb{Z}^n$ be the $n-$dimensional torus.

Denote the projection map by $p,$ i.e. $p:\Bbb{R}^n\to \Bbb{T}^n.$

I am reading a document which say

It's intuitive that $p$ preserves locally the volume

I am not sure why it's intuitive?

Answer 1

I'd say that it wasn't intuitive at all.

Instead, it's probably more reasonable to say "The volume of a region $A$ of the torus is defined to be the volume of $Q \cap p^{-1}(A)$, where $Q$ is the unit cube $0 \le x_1, \ldots, x_n \le 1$.

With this definition, it's clear that $p$ locally preserves the volume."

Answer 2

Presumably you want your Haar measure on $\Bbb T^n$ to be normalized, in which case it's quite intuitive because you know that the unit cube has measure $1$. This is in fact part of the definition of the determinant function as the "unique alternating, multi-linear form that takes the value $1$ on the unit cube."

Since our volume form on $\Bbb R^n$ is induced by the determinant it's natural since the space is homogeneous, so that if you have a set completely contained within a fundamental domain--or at least with points all from distinct classes in the fundamental domain--the measure is just given by the measure in $\Bbb R^n$. This is also borne out by Fubini's theorem,

$$\int_{\Bbb R^n}1_S\,dx = \int_{\Bbb Z^n}\int_{\Bbb R^n/\Bbb Z^n}1_{\bar{S}}dh\,dg$$

The measure on $\Bbb Z^n$ being counting measure and the measure on $\Bbb T^n$ being the local Haar measure. Since $S$ only has representatives in one copy of the fundamental domain, the outer integral is trivialized.

Answer 3

I am not sure, but you can observe that $R^n$ is the universal cover, of the torus. This means that the map is locally a homemorphisim. Maybe that what they meant.

Answer 4

It's a matter of definitions and conventions. Both Euclidean space and the torus have (uncountably) many different volume forms/measures. However, as the torus is the quotient of Euclidean space by a discrete action, it may make sense, in this context, to equip Euclidean space with a volume form which is preserved by the action (there are still uncounatbly many of those). Doing so, the torus inherits a natural volume form which makes the projection volume-preserving.

Another way to reach the same result, is to start by choosing your favorite volume form on the torus, and then pull it back to Euclidean space. The pulled back volume form is automatically invariant under the action.

Anyway, "intuitive" is not the right word.

Edit: We now show why a pulled back volume form is automatically invariant under the action. Let $\omega$ be any volume form on the torus, and let $\psi:\mathbb{R}^n\to\mathbb{R}^n$ be a deck transformation. So, by definition, we have$$p\circ\psi=p,$$which yields$$(p\circ\psi)^*\omega=p^*\omega,$$or equivalently,$$\psi^*\left(p^*\omega\right)=p^*\omega.$$