Rank of a tangent map related to holomorphic line bundles

Question

Let $L,\,\,J$ be two holomorphic line bundles over a compact Riemann surface $X$ of genus $g_X>0$ such that

(1) $d_1:=\dim H^0\big(\operatorname{Hom}(L,J)\big)>0$ and $d_2:=\dim H^0\big(\operatorname{Hom}(L,J^*)\big)>0$;

(2) $\deg J=0$ and $J^2\ncong{\mathcal O}_X$.

The canonical isomorphic map $\varphi_{1}:\operatorname{Hom}(L,J) \rightarrow \operatorname{Hom}(J^{*},L^{-1})$ is defined by $\varphi_1(s_1)(v^{*})=v^{*}\circ s_1$, where $s_1$ is a local section of $\operatorname{Hom}(L,J)$ and $v^{*}$ is a local section of $J^{*}$. Similarly, we could define another canonical isomorphism $$ \varphi_2:\operatorname{Hom}(L,J^{*})\rightarrow \operatorname{Hom}(J,L^{-1}),\ \ \varphi_2(s_{2})(u)=\operatorname{ev}(u)\circ s_2,$$ where $\operatorname{ev}:J\rightarrow (J^{*})^{*}$ is the evaluation isomorphism, i.e. $\operatorname{ev}(u)(v^{*})=v^{*}(u)$.

Now we could define the following map \begin{align*} \rho\colon H^0\big(\operatorname{Hom}(L,\,J)\big)\times H^0\big(\operatorname{Hom}(L,\,J^*)\big)\rightarrow H^{0}\big(\operatorname{Hom}(L,\,L^{-1}\otimes K)\big)\\ \big(i_{0},\hat{i}_{0}\big)\mapsto \varphi_{2}(\hat{i}_{0})\circ \partial_{J}\circ i_{0}-\varphi_{1}(i_{0})\circ \partial_{J^*}\circ \hat{i}_{0} \end{align*} where $\partial_{J}$ and $\partial_{J^*}$ is the $(1,0)$-part of the Hermitian-Einstein connections $D_{J}$ on $J$ and $D_{J^*}$ on $J^*$, respectively. In fact, $\rho$ is a bilinear holomorphic map. In particular, the complex differential $\rho_*$ of $\rho$ at $(i_{1},i_{2})$ along direction $\big(i_{0},\hat{i}_{0}\big)$ has the form of

\begin{equation} \label{diff} \rho_{*(i_{1},i_{2})}\big(i_{0},\hat{i}_{0}\big)=\varphi_{2}(i_{2})\circ\partial_{J}\circ i_{0}+\varphi_{2}(\hat{i}_{0})\circ\partial_{J}\circ i_{1}-\varphi_{1}(i_{1})\circ\partial_{J^*}\circ \hat{i}_{0}-\varphi_{1}(i_{0})\circ\partial_{J^*}\circ i_{2}. \end{equation}

Question. Assume that $0\neq i_{1}\in H^0\big(\operatorname{Hom}(L,\,J)\big)$, $0\neq i_{2}\in H^0\big(\operatorname{Hom}(L,\,J^*)\big)$ and $J^{2}\neq{\mathcal O}_X$. It follows immediately from the above expression of the complex differential that $(i_1,\, -i_2)\in \operatorname{Ker} \rho_{*(i_1,i_2)}$. Does $\operatorname{Ker}\rho_{*(i_1,i_2)}= \mathbb{C}(i_1,-i_2)$ hold true? (This question arises from our study of cone spherical metrics on compact Riemann surface.)

Partial result. $\max(d_1,\, d_2)\leq \operatorname{rank} \rho_{*(i_1,i_2)}\leq d_1+d_2-1$. Proof. We choose a flat trivializing coordinate of $J\oplus J^*$, the constant transition functions of $J\oplus J^*$ on $U_{\alpha\beta}=U_{\alpha}\cap U_{\beta}$ as $ g_{\alpha\beta}=\begin{pmatrix} a_{\alpha\beta} & 0 \\ 0 & a_{\alpha\beta}^{-1} \end{pmatrix}, $ where $a_{\alpha\beta}\in\rm{U}(1)$. Let $(e_\alpha,e^{*}_\alpha)$ be a holomorphic flat frame of $(J\oplus J^*)|_{U_\alpha}$ with respect to $\{U_{\alpha\beta},g_{\alpha\beta}\}$. Choose a holomorphic frame $l_\alpha$ of $L$ on $U_\alpha$ and the transition functions of $L$ are $\ell_{\alpha\beta}$. Denote $i(l_\alpha)=i_{1}(l_\alpha) + i_2(l_\alpha)=s_{1,\alpha}e_\alpha + s_{2,\alpha}e^{*}_\alpha$ and $i_0(l_\alpha)=u_\alpha e_\alpha$ and $\hat{i}_0(l_\alpha ) = v_\alpha e_\alpha^{*}$. Then we have $ i_{t}(l_{\alpha})=(s_{1,\alpha}+t\ u_{\alpha})e_{\alpha}+(s_{2,\alpha}+t\ v_{\alpha})e^{*}_{\alpha}. $, where $i_t:=(i_1+t i_0, i_2+t\hat{i_0})$.

Since $\rho(i_t)(l_{\alpha})$ is a local section of $L^{-1}\otimes K_{X}$ on $U_{\alpha}$, $\rho(i_t)(l_{\alpha})$ can act on $l_{\alpha}$. Then we obtain the local expression of $\rho_{*(i_{1},i_{2})}\big(i_{0},\hat{i}_{0}\big)$: $$ \frac{\rm d}{{\rm d}t}\bigg|_{t=0}\, \rho(i_t)(l_{\alpha})(l_{\alpha}) =\left(\frac{{\rm d}u_{\alpha}}{{\rm d}z_{\alpha}}\cdot s_{2,\alpha}+\frac{{\rm d}s_{1,\alpha}}{{\rm d}z_{\alpha}}\cdot v_{\alpha}-\frac{{\rm d}v_{\alpha}}{{\rm d}z_{\alpha}}\cdot s_{1,\alpha}-\frac{{\rm d} s_{2,\alpha}}{{\rm d}z_{\alpha}}\cdot u_{\alpha}\right){\rm d}z_{\alpha} $$

Now we set ${\hat {i_0}}=0$, so $v_{\alpha}=0$. That is, we consider $i_{t}=\big(i_{1}+t\ i_{0},i_{2}\big), \ t\in \mathbb{C}$. In this case $\frac{\rm d}{{\rm d}t}\bigg|_{t=0}\, \rho(i_t)(l_{\alpha}).(l_{\alpha})={\rm d}u_{\alpha}\cdot s_{2,\alpha}-{\rm d}s_{2,\alpha}\cdot u_{\alpha}.$ Hence if $\frac{\rm d}{{\rm d}t}\bigg|_{t=0}\, \rho(i_t)=0$, we have ${\rm d}u_{\alpha}\cdot s_{2,\alpha}-{\rm d}s_{2,\alpha}\cdot u_{\alpha}=0$ and $u_{\alpha}=C_{\alpha}\cdot s_{2,\alpha}$, where $C_{\alpha}$ is a complex constant on $U_{\alpha}$. Suppose that $C_{\alpha}\neq0$. Since $$ 0=\frac{\rm d}{{\rm d}t}\bigg|_{t=0}\, \rho(i_t)(l_{\beta})(l_{\beta})={\rm d}u_{\beta}\cdot s_{2,\beta}-{\rm d}s_{2,\beta}\cdot u_{\beta}\quad {\rm on}\quad U_\beta, $$ and $u_{\alpha}=\ell^{-1}_{\alpha\beta}a_{\alpha\beta}\cdot u_{\beta}$ on $U_{\alpha}\cap U_{\beta}$, we obtain $u_{\beta}\neq0$ by recalling $i_2\not=0$. Since $u_{\beta}=C_{\beta}\cdot s_{2,\beta}$, $C_{\alpha}=a_{ \alpha\beta}^2\cdot C_{\beta}$ and $\{U_{\alpha},C_{\alpha}\}$ defines a global holomorphic section of $J^2$, we obtain $J^2=\mathcal{O}_{X}$. Contradiction! Hence $C_\alpha=0$, $i_0=0$ and the restriction of $\rho_{*(i_1,i_2)}$ to $H^0\big(\operatorname{Hom}(L,\,J)\big)\times \{0\}$ is non-degenerate. Hence, we have $\operatorname{rank}\rho_{*(i_{1},i_{2})}\geq d_1$.

Similarly, we can show $\operatorname{rank}\ \rho_{*(i_1,i_2)}\geq d_2$.