# Polystable vector bundle contains a prescribed line bundle as a line subbundle

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 $$L \hookrightarrow J\oplus J^{-1}$$

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 $$L\hookrightarrow J\oplus J^{-1}$$.
• 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 Nov 11, 2022 at 8:20
• Here I use $L \hookrightarrow J \oplus J^{-1}$ to denote $L$ as a line subbundle of $J \oplus J^{-1}$. Nov 11, 2022 at 8:22
• 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.
• 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 $L^{-1} \otimes J=\mathcal{O}_X(q_1), L^{-1} \otimes J^*=\mathcal{O}_X(q_2)$. Jan 30 at 19:19