Vtyjkyuk

I'm confused about isometries of the flat 2-torus and could't find anything online that cleared my confusion. My problem is the following:

Let $T^2=mathbbR^2/Gamma$ be a 2-torus for $Gamma cong mathbbZ$ a lattice in $mathbbR^2$. I know that any isometry of the euclidean plane which preserves the lattice $Gamma$ induces an isometry of the flat torus $T^2=mathbbR^2/Gamma$ and conversly any isometry of the flat torus lifts to an isometry of the euclidean plane which preserves the lattice $Gamma$.

But my geometric intuition is that any translation in $mathbbR^2$ should descent to an isometry of the flat torus, since translating any "amount" should preserve length of curves on the torus. But when I take for example the standard torus $mathbbR^2/mathbbZ^2$ and a translation $f(x,y)=(x,y+lambda)$ for $lambda notin mathbbZ$, then for any lattice point $(k,l)$ we have $f(k,l)=(k,l+lambda)notin mathbbZ^2$, hence $f$ does not preserve the lattice $Gamma$ and hence does not descent to an isometry of the torus. Why is this?

One has to look carefully at what it means for $f$ to "preserve the lattice". I find that phrase to be a little vague and ambiguous, so I'll avoid it, and instead use a more precise phrase. I suspect that if you the source that you are consulting is written with full and precise definitions (and this is a big "if" when using only on-line sources), then it will have the definition that I write, or something equivalent.

Consider a lattice $Gamma subset mathbb R^2$, meaning an additive subgroup which is discrete and isomorphic to $mathbb Z^2$. We'll say that an isometry $f : mathbb R to mathbb R$ preserves the cosets of $Gamma$ if for every $(k,l) in Gamma$ and every $(x,y) in mathbb R^2$ we have
$$f(x+k,y+l) - f(x,y) in Gamma
$$
Equivalently, $f$ preserves the cosets of $Gamma$ if for every $(k,l) in Gamma$ and every $(x,y) in mathbb R^2$ there exists $(k',l') in Gamma$ such that
$$f(x+k,y+l) = f(x,y) + (k',l')
$$

This condition is necessary for $f$ to induce an isometry of $T^2$.

In fact, even more is true, this condition it is necessary just for $f$ to induce a well-defined function from $T^2$ to itself.

To see why, fix $(x,y) in mathbb R^2$. The point $(x,y)$ is a representative of the coset $(x,y) + Gamma in mathbb R^2 / Gamma = T^2$. In order for $f$ to induce a well-defined function from $mathbb R^2 / Gamma$ to itself, as $(k,l)$ varies over $Gamma$ it must be true that all of the points $f(x+k,y+l)$ are contained in a single coset of $Gamma$. That coset contains the point $f(x+0,y+0)=f(x,y)$ (by taking $(k,l)=(0,0)$), and so it follows that for each $(k,l) in Gamma$ there exists $(k',l') in Gamma$ such that $f(x+k,y+l) = f(x,y) + (k',l')$.

Notice that I have not even talked about isometries, nonetheless I think this addresses your question already: you seem to have overlooked the importance of "preserving cosets".