Flat tori as a riemannian product A flat torus is defined as the torus with the metric inherited from its representation as the quotient $\mathbb{R}^2/\Lambda$ where $\Lambda$ is a discrete subgroup of $\mathbb{R}^2$ which is isomorphic to $\mathbb{Z}^2$.
So, I have a couple of questions:
1)Which is the relation between the choice of two different lattices $\Lambda_1$ and $\Lambda_2$ and their associated flat tori $T_{\Lambda_i}:=\mathbb{R}^2/\Lambda_{i}$, with $i=1,2$. In other words, I would like to know if is there a criteria to know when two lattices induce isometric flat tori?
2)In Besse's book Einstein Manifolds, in page 286, when he  speaks about holonomy and the De-Rham Theorem it's said:
A flat torus is not in general a Riemannian product (globally). If we take the lattice $\Lambda$ generated by $u=(0,1)$ and $v=(x,y)$ will never be a product if $x\in (0,\frac{1}{2})$ and $x^2 + y^2 \ge 1$. How can I prove this?
 A: 1.) It's clear that any rotation or reflection of a lattice produces an isometric torus. So we can assume that our lattices have bases of the form $\{(t,0),(x,y)\}$ where $t>0$ and $y>0$. We can of course also change the basis so that $0 < x \le t$. I then claim that the classes of metrics on the torus are in bijection with the choices of $t,x,y$ satisfying these conditions. You can prove this by calculating the distances of points in the lattice closest to the origin. These correspond to the lengths of the shortest closed geodesics on your flat torus, which are clearly isometry invariants.
2.) Topologically, the only way a torus can decompose as a (nontrivial) product is as the product of circles. So a metric torus, if it were a product, would have to be given as $S^1_a \times S^1_b$ where $a,b>0$ denote the lengths of the circle factors. It follows that the two shortest closed geodesics on the torus are just the circle factors intersecting orthogonally. But for a lattice as you've given, the two shortest closed geodesics come from $u$ and $v$ and intersect at angle $\arctan(y/x) \not= \pi/2$.
