# Semistable sheaves on rationally connected manifolds

Let $(X,H)$ be a polarized projective manifold of dimension $n$ defined over $\mathbb C$, and let $\mathcal E$ be a reflexive sheaf on $X$.

If for every subsheaf $\mathcal F \subset \mathcal E$ the following inequality holds $$\frac{c_1(\mathcal F)\cdot H^{n-1}}{\mathrm{rank}(\mathcal F)} \le \frac{c_1(\mathcal E)\cdot H^{n-1}}{\mathrm{rank}(\mathcal E)}$$ then we say that $\mathcal E$ is semistable.

Mehta-Ramanathan proved that if $\mathcal E$ is semi-stable then its restriction to a sufficiently general curve $C$ cut out by elements of $|mH|$, $m \gg 0$, is also semistable.

Question. If we further assume that $X$ is rationally connected can we ensure the existence of a very free morphism $i: \mathbb P^1 \to X$ such that $i^*\mathcal E$ is semistable?

Recall that $i: \mathbb P^1 \to X$ is a very free morphism if and only if $i^* TX$ is ample.

As observed by mdeland in the comments, $i^* \mathcal E$ is semistable if and only if $$i^* \mathcal E \simeq \mathcal O_{\mathbb P^1}(k)^{\oplus \mathrm{rank}(\mathcal E)} .$$ In particular $\deg(i^* \mathcal E)$ is divisible by $\mathrm{rank}(\mathcal E)$.

If we start with an arbitrary very free morphism $i: \mathbb P^1 \to X$ we can compose it with a rational map $\varphi : \mathbb P^1 \to \mathbb P^1$ of degree $m \cdot \mathrm{rank}(\mathcal E), m \in \mathbb N$, to obtain a new morphism $i' = i \circ \varphi$. The original question seems to be related to the problem.

Problem. Describe the splitting type of a general deformation of $i' > : \mathbb P^1 \to X$ for $m \gg 0$.

Any ideas/references on how to tackle this kind of problem ?

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i^*E is semistable on P^1 iff it splits into a direct sum of OO(a) (all of the same degree). In particular, you would have to have that deg(i)*deg(E) was divisible by rank(E). As far as I know, we don't have too many techniques to arrange this "balanced" splitting condition. Even if X is rat conn we don't know how to show the existence of a rat curve so that i^TX is semistable. – mdeland Mar 16 '11 at 13:44
While I can't offer much insight into a proof, I am very interested in this question. I have a recent preprint where I (roughly) show general rank r semistable Steiner bundles on P^n pull back to balanced bundles on general degree r rational curves. I'm afraid the proof is strongly based on the special form of these bundles, however. – Jack Huizenga Mar 16 '11 at 16:16
That is nice, Jack! I think it goes to show that even is X = PP^n and even if you make strong assumptions about the minimal resolution of E, it could/will still be difficult to prove that E pulls back to a semistable bundle on some rational curve. – mdeland Mar 16 '11 at 18:37
Yes, I agree that the question seems difficult. Although, in my work proving that the pullbacks are balanced also gives the first proof that those bundles are semistable--potentially one could infer the balanced splitting result more easily if semistability was already known. – Jack Huizenga Mar 16 '11 at 19:53