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Let $X$ be a compact Riemann surface of genus $g \geq 1$, and let L be a line bundle over $X$ with $-g < \deg L \leq -\frac{1}{2}g$. Can we always find a flat line bundle $J \in \operatorname{Pic}^0(X)$, i.e. $\deg J = 0$, such that $L$ could be realized as a line subbundle of $J \oplus J^{-1}$?

I guess there exists some line bundle $L$ with $\deg L \in (-g, -\frac{1}{2}g]$, such that for any $J \in \operatorname{Pic}^0(X)$, we could not find an inclusion $L \hookrightarrow J\oplus J^{-1}$.

I try to prove my guess, but failed. Does anyone have references or suggestions?

Remark: the following table is a collection of results I have obtained

Degree of the prescribed line bundle $L$ Existence of the flat line bundle $J$
$-\frac{1}{2}g < \deg L < 0$ For generic $L$, $\not\exists J \in \operatorname{Pic}^0(X)$, such that $L \hookrightarrow J\oplus J^{-1}$
$-g < \deg L \leq-\frac{1}{2}g$ ?
$-2g < \deg L \leq -g$ For any $L$, $\exists\, J \in \operatorname{Pic}^0(X)$, such that $L \hookrightarrow J\oplus J^{-1}$
$\deg L \leq -2g$ For any $L$, and any $J \in \operatorname{Pic}^0(X)$, there always exists an inclusion $L \hookrightarrow J\oplus J^{-1}$
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No, such a line bundle does not exist. The condition $L\hookrightarrow J\oplus J^{-1}$ is equivalent to $h^0(L^*\otimes J)>0$ and $h^0(L^*\otimes J^{-1})>0$. Given $L$ of degree $-d$, the locus of $J\in \operatorname{Jac}(X) $ with these properties is the intersection in $\operatorname{Jac}(X) $ of the subvarieties $V_d$ and $-V_d$, where $V_d$ is the locus of line bundles $L(E)$ for all effective divisors $E$ of degree $d$. $\ V_d$ and $-V_d$ have dimension $d$, and cohomology class $\ \theta ^{g-d}/(g-d)!$, where $\theta \in H^2(\operatorname{Jac}(X) ,\mathbb{Z})$ is class of the principal polarization (this is the "Poincaré formula"). Since $d\geq \frac{g}{2} $, the class $[V_d]\cdot [-V_{d}]$ is nonzero, hence $V_d\cap (-V_d)\neq \varnothing$. Any $J$ in the intersection satisfies $L\hookrightarrow J\oplus J^{-1}$.

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  • $\begingroup$ Thank you. But I'm a bit confused on the first claim "the condition $L\hookrightarrow J\oplus J^{-1}$ is equivalent to $h^0(L^*\otimes J)> 0$ and $h^0(L^*\otimes J^{-1})>0$". I have an extreme example on $X$ of genus $g=2$: Let $p \in X$ and $L = \mathcal{O}_X(-p)$ be the local free sheaf which has a zero at $p$. Then $\deg L = -1$. If $J = \mathcal{O}_X$ is the trivial line bundle over $X$, then $h^0(L^*\otimes J) > 0$ and $h^0(L^*\otimes J^{-1})>0$, but $L$ is only a subsheaf of $J \oplus J^{-1}$, not a subbundle since the image of fiber $L|_p$ is zero in the bundle $J\oplus J^{-1}$. @abx $\endgroup$
    – swalker
    Commented Nov 11, 2022 at 8:20
  • $\begingroup$ Here I use $L \hookrightarrow J \oplus J^{-1}$ to denote $L$ as a line subbundle of $J \oplus J^{-1}$. $\endgroup$
    – swalker
    Commented Nov 11, 2022 at 8:22
  • $\begingroup$ Ah, sorry, I hadn't realized you want $L$ to be a subbundle. That's much more subtle: you want your divisor $D$ in $V_d\cap (-V_d)$ to be such that the linear systems $\lvert L^*(D) \rvert$ and $\lvert L^*(-D) \rvert$ contain divisors with disjoint support. I would guess that this is always possible, but probably hard to prove. $\endgroup$
    – abx
    Commented Nov 11, 2022 at 16:50
  • $\begingroup$ Yes, it is really hard to prove for me. Thank you for offering your insight to me, but do you know any relevant references? @abx $\endgroup$
    – swalker
    Commented Jan 27, 2023 at 23:24
  • $\begingroup$ There exists a line bundle $L$ of degree $-1$ on $X$ of genus $2$ (i.e. $\deg L=-g/2$) such that $L$ could not be a line subbundle of $J\oplus J^*$. We know that $|K_X|$ defines a projective line $E$ in $Sym^2(X)$ and $j:Sym^2(X)\to Jac(X)$ collapses $E$ to $K_X\in Jac(X)$, but otherwise it is bijective(see p. 203 in Riemann surfaces by Donaldson). Let $p\in X$ such that $K_X\neq\mathcal{O}_X(2p)$ and take $L=\mathcal{O}_X(-p)$. If $L$ is a line subbundle of $J\oplus J^*$, then $\exists q_1\neq q_2\in X$ such that $L^{-1} \otimes J=\mathcal{O}_X(q_1), L^{-1} \otimes J^*=\mathcal{O}_X(q_2)$. $\endgroup$
    – swalker
    Commented Jan 30, 2023 at 19:19

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