Geodesics on the flat torus... I want the set of closed geodesics on the flat torus i.e. $\mathbb{T}= \mathbb{E}^2 / \mathbb{Z}^2$.
That would be


*

*All vertical lines $x=t,\quad t\in [0,1)$

*All horizontal lines $y=t,\quad t\in [0,1)$

*The two diagonal lines $x=y$ and $x=-y$


Is that all, or am I missing something? 
 A: You're still missing a lot.
Besides the $+1$ slope diagonal line $x=y$, there are all the parallel diagonal lines $x=y+a$.
Besides the $-1$ slope diagonal line $x=-y$, there are all the parallel diagonal lines $x=-y+a$.
And besides all of those, for each rational number $r = \frac{m}{n}$ there is the diagonal line $x = \frac{m}{n} y$ as well as all the parallel diagonal lines $x = \frac{m}{n} y + a$.
A: There are infinitely many more diagonal ones because they "wrap around" the torus/closed square.
Take your square and draw the perpendicular bisector through two opposite sides making two smaller rectangles.  Draw a diagonal of one rectangle.  This meets the median at one vertex.  Now "wrap" this diagonal by going to the corresponding point on the opposite side and then drawing the diagonal ofvthe second rectangle that's parallel to the first.  This ends at the opposite corner from where you started but wraps back to the starting point making a closed geodesic.
You can apply this wrapping procedure to other lines as well.  You should find that any wrapped diagonal  line is a closed geodesic iff its slope is a ________ number (you should be able to fill in the blank).
