I recently learned of an [interesting result](http://www.ams.org/mathscinet-getitem?mr=242185) of Narasimhan and Ramanan from 1969, which says that moduli space of rank two vector bundles with trivial determinant on a curve $X$ of genus two is naturally isomorphic to $\mathbb PH^0(\operatorname{Pic}^1X,\mathcal L\_\Theta^{\otimes 2})$ (this vector space has dimension four). I'd like to understand this isomorphism better in the context of the [Narasimhan--Seshadri theorem](http://www.ams.org/mathscinet-getitem?mr=0184252).


First, fix a closed **topological** surface $X$ of genus two. Let:
$$M\_{X,SU(2)}=\operatorname{Hom}(\pi\_1(X),SU(2))//SU(2)$$
denote the $SU(2)$ character variety of $X$. For a complex structure $\sigma$ on $X$, let $M\_{X,\sigma,\operatorname{rk}2}$ denote the moduli space of rank two vector bundles on $X$ with trivial determinant (actually, there is a technical stability/equivalence relation which should be included, but I will ignore this). According to the [Narasimhan--Seshadri theorem](http://www.ams.org/mathscinet-getitem?mr=0184252), $M\_{X,SU(2)}$ and $M\_{X,\sigma,\operatorname{rk}2}$ are naturally diffeomorphic (again, there are some qualifications to this which I am ignoring; in particular both spaces are usually singular).


Now I want to recall the isomorphism $M\_{X,\sigma,\operatorname{rk}2}\to\mathbb PH^0(\operatorname{Pic}^1X,\mathcal L\_\Theta^{\otimes 2})$. To any rank two vector bundle $E$, we consider the subvariety $C\_E$ of $\operatorname{Pic}^1X$ consisting of bundles $\xi$ for which there is an exact sequence:
$$0\to\xi\to E\to\xi^{-1}\to 0$$
(that is, $E$ is an extension of $\xi^{-1}$ and $\xi$). Then Narasimhan and Ramanan prove that $C\_E$ is a divisor on $\operatorname{Pic}^1X$ and is linearly equivalent to $2\Theta$. This gives a map $M\_{X,\sigma,\operatorname{rk}2}\to\mathbb PH^0(\operatorname{Pic}^1X,\mathcal L\_\Theta^{\otimes 2})$, which Narasimhan and Ramanan go on to show is an isomorphism. (That was only a rough outline).


OK, now let's reinterpret the map $M\_{X,\sigma,\operatorname{rk}2}\to\mathbb PH^0(\operatorname{Pic}^1X,\mathcal L\_\Theta^{\otimes 2})$ in terms of the Narasimhan-Seshadri theorem. Remember that $\operatorname{Jac}X$ is $\operatorname{Hom}(\pi\_1(X),U(1))$. Thus for a homomorphism $\rho:\pi\_1(X)\to SU(2)$ (corresponding to a vector bundle of rank two), we can define the subvariety $C\_E$ as those $U(1)$ representations $\alpha:\pi\_1(X)\to U(1)$ for which $\rho$ can be conjugated to have the form:
$$\left(\begin{matrix}\alpha&\beta\cr 0&\alpha^{-1}\end{matrix}\right)$$
This is a **subset** of $\operatorname{Hom}(\pi\_1(X),U(1))=\operatorname{Jac}X$. Now according to Narasimhan and Ramanan, it should be a **subvariety** of $\operatorname{Jac}X$ for *any* complex structure on $X$. This seems a bit unlikely to me, because there is a large moduli of complex structures on $X$. Also, somehow I've constructed naturally $C\_E\subseteq\operatorname{Jac}X$, but according to the construction in Narasimhan and Ramanan I should be getting $C\_E\subseteq\operatorname{Pic}^1X$, which is really not the same thing canonically.


I suppose I'm getting confused in applying the Narasimhan-Seshadri theorem. Any assistance is appreciated!



Rank two vector bundles on a curve of genus two
Dear unknown, by NR, there exists always a family of degree -1 line sub bundles for every holomorphic vector bundle of rank two with trivial determinant on a genus 2 surface. As they have degree -1, they do not admit flat connections. This means that when your representation $\rho$ has the upper triangular form you wrote down, the representation $\alpha$ will correspond to a degree $0$ subbundle $L$ and not to a degree -1 subbundle. Of course, this implies that your bundle $E$ is only semi-stable and not stable, or equivalently, $\rho$ is reducible, which is not the generic case. (In that case, $C\_E$ can be explicitly computed as $L\Theta\cup L^\*\Theta$ which clearly depends on the Riemann surface structure.) 


