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Let $X$ be a compact Riemann Surface of genus 1. Let $Cl_0(X) := \frac{Div(X)}{PDiv(X)}$, where $PDiv(X)$ is the subgroup of principal divisors on $X$. Let $P \in X$ be a fixed point. We have a bijection

$i_p : X \rightarrow Cl_0(X)$

given by $X \in Q \rightarrow [Q-P]$.

Using this bijection one can pullback the (commutative)group structure of $Cl_0(X)$ to X as follows. Let $\mu$ be the multiplication map $\mu : X \times X \rightarrow X$ be defined by

$\mu(Q_1,Q_2) = i_p^{-1}(i_p(Q_1) + i_p(Q_2))$.

Now my question is how to prove that this group structure makes $X$ into a complex Lie group. Thanks.

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    $\begingroup$ That's a good question. Probably the best way is to defines this map relatively, i.e. defining the group structure on $\mathrm{Hom}(Z,X)$ for any complex manifold $Z$. This is done by doing a relative version of what you have written. The other way is to note that $X$ is, as a complex manifold, equal $\mathbb{C}/\Lambda$ for some $\Lambda\subseteq\mathbb{C}$ a lattice. This follows since the universal cover of genus $1$ surface is $\mathbb{C}$, and has fundamental group $\mathbb{Z}^2$, and so you can conclude. Then, try to show that the operation you have defined is the same thing as the $\endgroup$ – Alex Youcis Aug 21 '15 at 18:01
  • $\begingroup$ group operation on $\mathbb{C}/\Lambda$. $\endgroup$ – Alex Youcis Aug 21 '15 at 18:01
  • $\begingroup$ Thanks for the comment. Is it possible for you to explain your comment in some details in an answer or provide some references? $\endgroup$ – random123 Aug 23 '15 at 7:05

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