I recently learned of an interesting result 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.

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, $M_{X,SU(2)}$ 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!


1 Answer 1


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.)

  • $\begingroup$ Thanks, I understand now. Is there a way of describing the degree -1 line subbundles in terms of the SU(2) representation? $\endgroup$ Aug 8, 2012 at 16:59
  • 1
    $\begingroup$ I think there is no way because that would give an explicit identification of the character variety with the moduli space of (semi)stabile holomorphic structures. This seems to be as difficult as computing the monodromy of a given flat connection (e.g. A fuchsian system) explicitly. $\endgroup$
    – Sebastian
    Aug 8, 2012 at 17:22
  • 1
    $\begingroup$ You may find it very instructive to work out the identification from the NS theorem in the case of genus one and structure group $SU(2)$. $\endgroup$ Aug 8, 2012 at 18:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.