Confusion about thinking about a (flat) torus as a quotient

Question

I feel slightly embarrassed to ask this, but I've managed to thoroughly confuse myself about the following.

Consider $\mathbb{R}^2$ together with the lattice $\Lambda=\{(n,m): n,m\in \mathbb{Z}\}$.

Clearly, the torus $\mathbb{T}^2=\mathbb{R}^2/\Lambda$ admits the obvious flat metric that makes it a square torus. Geometrically, this corresponds to taking the fundamental domain $[0,1]\times [0,1]$ with the metric $dx^2+dy^2$ and identifying the appropriate sides.

My confusion is that one could also think of the domain given by the parallelogram $\{(x,y): 0\leq y \leq 1, y\leq x \leq y+1\}$ (i.e. determined by the elements $(1,0)$ and $(1,1)$ that also generate the lattice). If I understand things correctly, on this domain the metric $dx^2+dy^2$ would induce a different metric on the quotient given by identifying the appropriate sides than the metric you would get by taking the appropriate quotient metric (which here would correspond to $dx^2-dxdy-dydx+2dy^2$).

This is causing me some confusion and doubt about how well I actually understand the situation.

I guess for a concrete question: Is there something special about the square as a fundamental domain that makes the quotient metric given by identifying opposite sides agree with the quotient metric of the lattice? Or am I misunderstanding things?

Answer 1

So thinking about this some more I realized that the correct answer is that both the square quotient and the parallelogram quotient are isometric. The part that confused me was that none of the possible isometries lift to an isometry of $\mathbb{R}^2$ and instead look something like

$$ (x,y)\mapsto \left\{ \begin{array}{cc} (x,y) & 0\leq x\leq 1 \\ (x-1, y) & 1\leq x \leq 2. \end{array} \right. $$ which maps the parallelogram to the square and is a smooth isometry on the quotient.

Answer 2

If you choose two different lattices $\Lambda_1$ and $\Lambda_2$, the resulting tori $\Bbb R^2/\Lambda_1$ and $\Bbb R^2/\Lambda_2$ will be diffeomorphic, sure. Both of them inherit flat metrics from $\Bbb R^2$ (because lattices act on $\Bbb R^2$ by translations, and translations are isometries for the standard flat metric in $\Bbb R^2$). Then, with such quotient flat metrics, the two tori are locally isometric for very general reasons, but this is not natural (in a very precise sense). If you can map one fundamental domain for $\Lambda_1$ to a fundamental domain for $\Lambda_2$ via a rigid motion of $\Bbb R^2$, then such rigid motion survives in the quotient as a global isometry between the two tori, and I suppose this is as general as it gets. I.e., taking generators of the lattices, the matrix of base change should be in ${\rm SO}(2,\mathbb{Z})$.

In your example, the two tori are not isometric simply because the two parallelograms playing the role of the fundamental domains do not have the same internal angles, say, so you cannot move one to the other via a rigid motion of the plane.

Compare this with the situation where you think of $\Bbb R^2$ as $\Bbb C$. The tori will be diffeomorphic, but they'll be biholomorphic to each other (as Riemann surfaces) precisely when the matrix of base change is in ${\rm SL}(2,\mathbb{Z})$.