Show the space of flat connections $(\mathbb T^2)^N / {\mathfrak S}_N$ is a $(N-1)$-complex projective space 
*

*Is it true that 
$$
\mathbb{T}^2/(\mathbb{Z}/2)=\mathbb{CP}^1=S^2?
$$
$$
\mathbb{T}^2/(\mathbb{Z}/2)=\mathbb{CP}^1=S^2?
$$


*I am trying to digest the following statements:


$$
M_{\rm flat} =\mathbb E / {\mathfrak S}_N, 
$$
"The ${\mathfrak S}_N$ is the symmetric group usually denoted as $S_N$ (the Weyl group of $\operatorname{SU}(N)$).
$$ 
\mathbb E := \left\{ (\phi_1,\cdots, \phi_N) \in E^N \equiv {(\mathbb  T^2)}^N; \sum_i \phi_i=0 \right\} . 
$$
Here $E  \equiv {\mathbb  T^2}$ as a 2-torus.
"$O(Np)$ is a line bundle on $E$ associated to the divisor $Np$." Can we clarify the meanings of "divisor" in the context of "line bundle" here? 
The set of unordered points $\{\phi_i\}_{1 \leq i \leq N}$ 
on the torus $E$, that is $M_{\rm flat}=\mathbb E/{\mathfrak S}_N$, is given by
${\mathbb P} H^0(E, O(N p)) $, where ${\mathbb P} H^0(E, O(N p))$ is the projective space
associated to the vector space $H^0(E, O(N p))$. The correspondence is given by
mapping the $N$ points $q_i$ of zeros of a function $F \in H^0(E, O(N p))$ to the points on $M_{\rm flat}=\mathbb E/{\mathfrak S}_N$.
Also, the vector space $H^0(E, O(N p))$ has complex dimension $N$ and hence $H^0(E, O(N p)) \cong \mathbb{C}^N$,
so we get 

$$M_{\rm flat} \cong \mathbb{CP}^{N-1}.$$
  This can be shown by well-known techniques in algebraic geometry."

My question is how to show
$$
M_{\rm flat} =\mathbb E / {\mathfrak S}_N \cong \mathbb{CP}^{N-1}, 
$$


*The case of $N=2$ above is particularly simple, since then we have 
$$
\mathbb E:= \left\{ (\phi_1,-\phi_1) \in E^2 \right\} =E\: , \quad M_{\rm flat} =\mathbb{T}^2/\mathbb{Z}_2 \;.
$$
Refs: See page 8 of this paper for the motivation of asking this question
p.s. It is said this book can be useful, "  P. Griffiths and J. Harris,
Principles of algebraic geometry. Wiley-Interscience [John Wiley &
Sons], New York, 1978.  Pure and Applied Mathematics." Which section shall I look at to solve the above?
 A: $\Bbb T^2/(\Bbb Z/2\Bbb Z) = \Bbb P^1$ just means that a torus is a double branched cover of $\Bbb P^1$ in four points, as explained by the authors. More explicitely, a torus is an elliptic curve with affine part given by $\{ (x,y) \in \Bbb C^2 : y^2 = x(x-1)(x- \lambda) \}$. The projection to the $y$ coordinate induces a map $E \to \Bbb P^1$ branched at $0,1, \infty$ and $\lambda$.
Why $M_{flat}$ is the same as the scalar $\phi_i$ : first since the fundamental group is abelian, the $SU(N)$ representation should factors through an abelian subgroup.We can assume for simplicity it's contained in the maximal torus $diag(e^{\phi_1}, \dots, e^{\phi_n})$ with $\sum \phi_i = 0$. So we get two such scalars, but we are also working up to conjugacy since we look at isomorphism classes of flat connections. This implies that you only need to keep one matrix. I got this explanation by a very nice postdoc from my department :)
Now why $\Bbb E/S_n \cong \Bbb P^{N-1} $ ? 
Let $H^0(E, \mathcal O_E(Np))$ be the vector space of meromorphic functions $f : E \to \Bbb P^1$ with at most a pole of order $N$ at $p$. If $f,g$ are in this space with the same zeroes, then they are proportional, indeed $f/g$ has no poles so is constant. By a non-trivial theorem (Abel-Jacobi) the zeroes will verify $q_1 + \dots + q_N = 0$ i.e we exactly get an injection from $\Bbb P^{N-1} \to \Bbb E/S_n$.
In fact here Abel-Jacobi is not used and the authors give elementary arguments instead (I would suggest to read the next pages after the statement that $\Bbb E/S_n \cong \Bbb P^{N-1}$ where lot of details are presented)
Surjectivity is also proved by the authors page 10. For background in algebraic geometry in particular on curves, I would advice to look at Miranda's book on Riemann surfaces, where also line bundles and divisors are explained. Griffith and Harris is really good but pretty challenging.
