Zwiebach quick calculation 2.5 I am working through Zwiebach's a first course in string theory. It's been a while since I did any math (or physics!), and I am stuck on the following problem (quick calculation 2.5 in the book):

Consider the plane $(x,y)$ with the identification
  $$
(x,y)\sim(x + 2\pi R, y + 2\pi R).
$$
  What is the resulting space? Hint: the space is most clearly exhibited using a fundamental  domain for which the line $x+y=0$ is the boundary.

My thoughts
Now I understood all the previous examples he gave for identifications, but I can't figure this one out. Why is the line $x+y=0$ a boundary for the domain?
Points from either side of this boundary can be included in the domain no?
 A: It's not a priori clear why $x+y=0$ should be a boundary. But let's investigate it. Draw the line yourself. You could explicitly parametrise this line, call it $L$, for example as $$p(t) = (t,-t)$$ but it's not necessary.
The equivalence relation tells us to identify the points $p \sim \sigma(p)$ where $$\sigma: (x,y) \mapsto (x + 2\pi R,y+2\pi R).$$ 
Now it's an elementary exercise to see that this means that we identify the line $L$ with a parallel line as follows:
$$ L \sim L' = L + (2\pi R,2\pi R).$$
Now it should be straightforward to see that [one choice of] fundamental domain is the area between two such lines. (Apply $\sigma$ and its inverse several times if you're still confused.)
But such a fundamental domain with two boundary lines identified is just the cylinder, except that it's rotated. Of course you can prove it mathematically, but this is simply a piece of paper with its edges glued together - you can visualise it.
A: The answer should be that of a cylinder. To visualize this, draw a grid on the cartesian plane of squares of size $2\pi R$. Now, choose a pair of lines as the $x$ and $y$ axes. Then, construct the $y=x$ line. Now for any point there, look at the grid and where there is an equivalence of points.
Alternatively, using the hint, take a 45$^\circ$ rotation of axis. Then, the new $x$ axis ($x'$) is represented by $1/\sqrt{2}(\hat{x}-\hat{y})$, and the new $y$ axis ($y'$) is represented by $1/\sqrt{2}(\hat{x}+\hat{y})$, where the hats are the old orthonormal vectors. The new equivalence relations now yield $(x',y')\sim (x',y'+\sqrt{2}2\pi R)$.
