Geometric Intuition behind the genus of complex torus

Question

This question may be really dumb, but my geometric intuition is really bad.... Fix $w_{1},w_{2}\in\mathbb{C}$ two complex numbers that are linearly independent over $\mathbb{R}$, and consider the Lattice $$L:=\mathbb{Z}w_{1}+\mathbb{Z}w_{2}=\{m_{1}w_{1}+m_{2}w_{2}:m_{1},m_{2}\in\mathbb{Z}\}.$$ The quotient group/space $X:=\mathbb{C}/L$ is a compact Riemann surface, and is called a complex torus.

For $z\in\mathbb{C}$, consider the closed Parallelogram $$P_{z}:=\{z+\lambda_{1}w_{1}+\lambda_{2}w_{2}:\lambda_{i}\in [0,1]\}.$$ We also have the canonical projection $\pi:\mathbb{C}\longrightarrow\mathbb{C}/L$ that sends each element $z\in\mathbb{C}$ to the coset $z+L$.

I have proved that $\pi$ maps $P_{z}$ onto $X$, for any $z\in\mathbb{C}$.

The book "Algebraic Curves and Riemann Surfaces'' by Rick Miranda says the following:

In fact, $X$ has topological genus one, topologically is a simple torus. This is most easily seen by consider $X$ as the image of the parallelogram $P_{0}$ under the map $\pi|_{P_{0}}$, the opposite sides are identified together, and no other identifications are made, giving the familiar construction of the torus.

I know how to prove that $X$ has genus one, because it is fairly easy to construct a homeomorphism from $X$ to $\mathbb{S}^{1}\times\mathbb{S}^{1}$. However, I am not understanding the gluing thing mentioned by Miranda above.

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So, we have that $P_{0}=\{\lambda_{1}w_{1}+\lambda_{2}w_{2}:\lambda_{1,2}\in [0,1]\}$, and thus the quotient map mods out the case when $\lambda_{1},\lambda_{2}=0$ or $1$.

Does this mean that we actually only identify four points on this parallelogram, namely, $0,w_{1},w_{2}$ and $w_{1}+w_{2}$? instead of identifying all the points on opposite sides?

I am sorry if the answer to my equation is easily obvious and I am not seeing it...

Answer 1

Two points $\lambda_1 w_1+\lambda_2 w_2, \lambda'_1 w_1 + \lambda'_2 w_2\in P_0$ have the same image under $\pi$ if and only if one of the following is true:

1. $\lambda_1 = \lambda'_1$ and $\lambda_2,\lambda'_2\in\{0,1\}$.
2. $\lambda_2 = \lambda'_2$ and $\lambda_1,\lambda'_1\in\{0,1\}$.
3. $\lambda_1,\lambda_2,\lambda'_1,\lambda'_2\in\{0,1\}$.
4. $(\lambda_1, \lambda_2) = (\lambda'_1, \lambda'_2)$.

The first identifies the edge from $0$ to $w_1$ with the edge from $w_2$ to $w_1+w_2$, while the second identifies the edge from $0$ to $w_2$ with the edge from $w_1$ to $w_1+w_2$.

This does not only identify the four points $0,w_1,w_2,w_1+w_2$ but also for example $\pi(\tfrac 1 2 w_1) = \pi(\tfrac 1 2 w_1 + w_2)$.

This is exactly the description of the torus as $[0,1]^2/{\sim}$ with opposite edges identified, under the homeomorphism $P_0\cong [0,1]^2$ given by $\lambda_1 w_1 + \lambda_2 w_2 \mapsto (\lambda_1, \lambda_2)$.