I recently attempted a question asking to show the statement $ \mathbb{C} / \Gamma \approx S^{1} \times S^{1} $, $ \Gamma $ is true- this is done easily enough via considering the map $ x+iy \rightarrow (e^{2 \pi ix}, e^{2 \pi iy} ) $ and 1st isomorphism theorem, however Im unsure how, geometrically, we get from one group to the other.
Consider the complex plane with Gaussian integers marked on and the image(s) after multiplying these points by $re^{i \theta} $.
My initial thoughts were:
Just pure rotations leave $ \Gamma $ unchanged with rotations of 90 deg, so I was thinking this might become the property of 1 complete revolution for a fixed 'horizontal' position on the torus (Forming a circular slice of our torus). Then somehow varying $r$ acts as the rotations perpendicular to this.
(Alternatively, I just looked at a single Gaussian integer point and its neighbouring points (a 3x3 square in the plane).I was thinking, due to the infinite nature of the plane (and therefore symmetries), out cosets of $ \Gamma $ are only unique for mapping this point within a circle radius 1/2 around it. I am not sure this is true though)
Any help interpreting how to construct the torus from $ \mathbb{C} / \Gamma$ geometrically would be appreciated.
Consider how addition works in $\mathbb{C}.$ If you add $\frac{1}{2}$ to $\frac{1}{2}+\frac{1}{2}i,$ then you get $1+\frac{1}{2}i.$ I would advise thinking about this in terms of "moving along" the vector starting at $0$ and going to $\frac{1}{2}+\frac{1}{2}i$ and then "moving along" the vector starting at $\frac{1}{2}+\frac{1}{2}i$ and going to $1+\frac{1}{2}i$ (draw a picture!).
Now, what happens in $\mathbb{C}/\Gamma?$ You do the same thing, but you're allowed to "mod out" by integer multiples of $1$ and $i.$ So in some sense we can write $1+\frac{1}{2}i \equiv \frac{1}{2}i \pmod{\Gamma}.$
If you think about this a bit, it becomes clear that no matter where you start and where you end up, at the end you'll always be equivalent to some point in the unit square $[0,1)\times[0,1).$ Ultimately this is what you are proving when you prove that $\mathbb{C}/\Gamma\cong S^{1}\times S^{1};$ it's a restatement of the (perhaps more understandable) fact that $\mathbb{R}^{2}/\mathbb{Z}^{2} \cong (\mathbb{R}/\mathbb{Z})^{2}.$
Now notice that I've written the unit square as $[0,1)\times[0,1)$ (and put aside the fact, for a moment, that I've switched from $\mathbb{C}$ to $\mathbb{R}^{2}$ - I'm sure you can see that it's really no different). Why have I not written $[0,1]\times[0,1]?$ The reason is that $0\equiv 1\pmod{\Gamma}.$ In this sense, we have "glued" the opposite edges of the square together. The bottom edge $[0,1]\times\{0\}$ has been "glued" (by the equivalence relation that comes with quotient-ing) to the top edge $[0,1]\times\{1\}.$ Similarly, the left and right edges have been glued.
Now consider this image, courtesy of the University of Lyons:
and this image from Wikipedia:
and hopefully you should be able to see how this "glued unit square" is, in some sense, the product $S^{1}\times S^{1}$ of two circles.
By the way, those circles arise since $\mathbb{R}/\mathbb{Z}$ is a circle, in the same way that $\mathbb{R}^{2}/\mathbb{Z}^{2}$ is a torus. You glue the ends of the interval $[0,1]$ together.