order of meromorphic functions on Riemann torus Let $\Lambda$ be a lattice in $\mathbb{C}$ and $f$ be a meromorphic function on $\mathbb{C}$ such that $f(z)=f(z+w)$ for all $z\in\mathbb{C}$ and $w\in\Lambda$. Then I want to know how to prove $\sum_{z\in\mathbb{C}/\Lambda}\text{ord}_z(f)·z=0$, this is an equality in the additive group $\mathbb C/\Lambda$.
Thanks very much!
 A: Your map $f$ defined on $\mathbb{C}$ can be extend to the torus $\mathbb{C}/\Lambda$. In general when you have a meromorphic function $f$ on a Riemann surface, you can define the so called principal divisor of the function
$div(f)=\sum_{p}ord_p(f)\cdot p$
At this point it is necessary to define the Abel map of a complex torus that permit us to consider each divisor as an element of the torus itself, that is an additive group.
In fact the structure of the additive group $\mathbb{C}$ can be induced on the complex torus $\mathbb{C}/\Lambda$ becoming an abelian  group. You have another important abelian group on a Riemann surface $X$, the group of divisors $Div(X)$, the free $\mathbb{Z}$-module generated by the set $X$.
The Abel map is $A: Div(\mathbb{C}/\Lambda)\to \mathbb{C}/\Lambda$ that assigns to each $\sum_{k}a_kp_k$, the element $\sum_{k}a_kp_k+\Lambda$ on the group $\mathbb{C}/\Lambda$.
Your statement is equivalent to prove each meromorphic function $f$ on the torus  satisfies the condition
$A(div(f))=0$
To prove this, we consider the set of zeros $\{p_i\}_{i=1}^n$ and poles $\{q_j\}_{j=1}^m$ of $f$, where we repeat the points according to the order of $f$ in that points. Then we get
$div(f)=\sum_{i=1}^np_i- \sum_{j=1}^mq_j$
However the complex torus is a compact Riemann Surface and it holds the fundamental fact
$deg(div(f))=0$
that permit us to say $n=m$.
This means $div(f)$ can be written as $\sum_{i=1}^n(p_i-q_i)$.
Assume by contradiction that $\sum_{i=1}^np_i\neq  \sum_{i=1}^nq_i$ on the complex torus. Then we can consider two new distinct points $p_0$ and $q_0$ such that
$p_0=(\sum_{i=1}^np_i-  \sum_{i=1}^iq_i)+q_0$
and this time we get $\sum_{i=0}^np_i=\sum_{i=0}^nq_i$. Moreover we can choose $x_i\in \pi^{-1}(p_i)$ and $y_i\in \pi_i^{-1}(q_i)$, where $\pi$ is the quotient map of the complex torus. In. this case we get
$\sum_{i=0}^nx_i-\sum_{i=0}^ny_i=\lambda \in \Lambda$
and so
$\sum_{i=0}^nx_i=\sum_{i=0}^ny_i$
replacing $x_0$ with $x_0-\lambda$.
We are in the situation to built a new meromorphic function on the complex torus:
$R(z):=\frac{\prod_{i=0}^n\theta^{(x_i)}(z)}{\prod_{i=0}^n\theta^{(y_i)}(z)}$
At this point we can observe $R/f$ is a meromorphic function that has only one zero and one pole:
$div(R/f)=div(R)-div(f)=\sum_{i=0}^np_i-\sum_{i=0}^nq_i-(\sum_{i=1}^np_i-\sum_{i=1}^nq_i)=p_0-q_0$
This is a contradiction because $\mathbb{P}^1$ is the only Riemann surface for which there exists a meromorphic function that has only one zero and one pole.
This is not the end of the story because also the converse is true, i.e. if $D$ is a divisor on the torus such that $deg(D)=0$ and $A(D)=0$, the $D$ is a principal divisor, so it is a divisor of a meromorphic function on the torus.
This permit us to say $A: deg^{-1}(0)/Pdiv(\mathbb{C}/\Lambda)\to \mathbb{C}/\Lambda$ is injective and it is called the Abel-Jacobi theorem.
