# Confusion in known result about moduli space of vector bundle of rank 2 degree 0 vector bundles over smooth curve of genus 2

Theorem: Let $$X$$ be a complete, non-singular algebraic curve of genus $$2$$. Let $$U(2, \Theta)$$ be the space of $$S$$-equivalence classes of semi-stable vector bundles of rank $$2$$ and degree $$\Theta$$. The group $$\Gamma$$ of elements of order $$2$$ in $$J$$ acts on $$PH^0(J^1, L_\Theta^2)$$ in a natural way; let $$A$$ be the associated projective bundle on $$J/\Gamma \cong J$$. Then $$U(2, 0)$$ is canonically isomorphic to $$A$$. In other words, $$U(2, 0)$$ is canonically isomorphic to the space of positive divisors on $$J^1$$ algebraically equivalent to $$2\Theta$$.

Moduli of Vector Bundles on a Compact Riemann Surface

Author(s): M. S. Narasimhan and S. Ramanan

Source: Annals of Mathematics, Second Series, Vol. 89, No. 1 (Jan., 1969), pp. 14-51

Question: Is the projective bundle in theorem comes from some vector bundle on $$J$$. Means is it $$P(E)$$ of some bundle $$E$$ over $$J$$.

EDIT: After I posted my answer, I realized that you were very close to answering your own question---the vector bundle you want is the one whose fiber over $\alpha \in {\rm Jac}(C)$ is $H^{0}(\mathcal{O}(2\Theta) \otimes \alpha)^{\vee}.$ However, this is equivalent via Strange Duality to what I have written below.
In general, if $C$ is a smooth projective curve of genus $g \geq 2$ $U_{C}(r,0)$ is the moduli space of S-equivalence classes of vector bundles of rank $r$ and degree $0$ on $C,$ and we fix a line bundle $L$ on $C$ of degree $g-1,$ the locus $$\Theta_{L} = \{E \in U_{C}(r,0) : h^{0}(E \otimes L) > 0 \}$$ is an ample divisor on $U_{C}(r,0).$ Recall that the fibers of the determinant map ${\rm det} : U_{C}(r,0) \to {\rm Jac}(C)$ are each isomorphic to ${\rm SU}_{C}(r)$, which parametrizes rank-$r$ vector bundles with trivial determinant. For any two line bundles $L,L'$ of degree $g-1$ on $C,$ the restrictions of $\Theta_L$ and $\Theta_{L'}$ to any fiber of ${\rm det}$ are linearly equivalent; the resulting ample line bundle on ${\rm SU}_{C}(r)$, which is denoted by $\mathcal{L},$ is called the $\textit{determinant bundle},$ and it generates ${\rm Pic}(SU_{C}(r)).$
For $k \geq 1$, the $\textit{Verlinde bundle}$ on the Jacobian ${\rm Jac}(C)$ corresponding to the rank-level pair $(r,k)$ (and our choice of $L$, typically a theta-characteristic) is the vector bundle $$\mathbf{E}_{r,k} := {\rm det}_{\ast}\mathcal{O}(k\Theta_{L})$$
For each $\alpha \in {\rm Jac}(C)$ the fiber of $\mathbf{E}_{r,k}$ over $\alpha$ is isomorphic to $$H^{0}({\rm SU}_{C}(r,\alpha),\Theta_{L}|_{{\rm SU}_{C}(r,\alpha)}) \cong H^{0}({\rm SU}_{C}(r),\mathcal{L}).$$ In the case where $g=r=2,$ the paper of Narasimhan-Ramanan that you are reading proves ${\rm SU}_{C}(2) \cong \mathbb{P}^{3}.$ In particular, the determinant bundle $\mathcal{L}$ on ${\rm SU}_{C}(2)$ is exactly the hyperplane bundle $\mathcal{O}(1)$ on $\mathbb{P}^{3}.$ The map ${\rm det}$ then realizes $U_{C}(2,0)$ as a $\mathbb{P}^3$-bundle over ${\rm Jac}(C)$; this is the projectivization of the rank-4 Verlinde bundle $\mathbf{E}_{2,1}.$