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Let $H$ be a very ample line bundle on $\mathbb P^2$ given by $\mathcal O_{\mathbb P^2}(1)$. Let $E$ be a rank $r$, slope stable (w.r.to $H$) vector bundle on $\mathbb P^2$. Is it true that there does not exist any nontrivial homomorphism between $E$ and $E \otimes \mathcal O_{\mathbb P^2}(-3)$?

On the contrary, if we assume that $f$ is a nontrivial morphism between them amd $I$ is the image, then rank of $I$ is strictly less than both the bundles (since the morphism is nontrivial) and slope of $I$ is less than or equal to the slope of $E \otimes \mathcal O_{\mathbb P^2}(-3)$ which is strictly less than slope of $E$. But I don't see how to achieve a contradiction from here.

At this point, I have an alternative argument in mind which is as follows: if $f \in Hom (E, E \otimes \mathcal O_{\mathbb P^2}(-3))$, then $\text{det}(f) \in H^0(\mathbb P^2, \mathcal O(-3r))=0$. Can we say that $f \in Hom (E, E \otimes \mathcal O_{\mathbb P^2}(-3)) \subset Hom(E,E) \cong \mathbb C$ as $E$ is simple. If $f$ is nonzero, its an automorphism of $E$ such that $\text{det}(f)=0$. Can we conclude from there that $f$ must be zero?

My main two points is that:

$(i)$ Is it true that $Hom (E, E \otimes \mathcal O_{\mathbb P^2}(-3)) \subset Hom(E,E)$? (Is this obvious to see?)

$(ii)$ If $f$ is nonzero, automorphism of $E$ such that $\text{det}(f)=0$. Can we conclude from there that $f$ must be zero?

Any suggestion is welcome.

Sherlock
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1 Answers1

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Upper question: It is a general fact that for stable bundles $E,F$ we have $Hom(E,F)\neq 0$ implies $\mu(E) < \mu(F)$.
Now if $\mu(E)=d/r$, then your $F :=E\otimes \mathcal{O}(-3)$ has smaller slope by computing the first Chern class $c_1(E\otimes \mathcal{O}(-3))=c_1(det(E\otimes\mathcal{O}(-3)))=c_1(det(E)\otimes \mathcal{O}(-3)^r)=c_1(E)-3r c_1(H)$ and the last term is negative because $H$ was positive. F has smaller slope because same rank, but smaller $d$ so you can conclude that $Hom(E,F)=0$.

(1): I don't know, what should that map be? for fixed $\phi \in Hom(E,E\otimes \mathcal{O}(-3))$ there is a natural map in the other direction, the Lie bracket $[\_,\phi]: Hom(E,E) \rightarrow Hom(E,E\otimes \mathcal{O}(-3))$ which is in general neither injective nor surjective...

(2): Note that an automorphism $f: E \rightarrow E$ induces by functoriality of $\det$ another automorphism $\det(f)$ acting on $\det(E)$ which is certainly non-zero because $f$ is invertible.

Simonsays
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  • I read in a published paper that on a fano threefold for a simple vector vector $E$ on a Fano threefold $(i)$ holds. So you mean on plane, one cannot adapt the same technique? – Sherlock Jun 05 '23 at 15:44
  • That had nothing to do with the base. Usually it’s proven for curves but works more generally, see eg Le Portiers book on vector bundles section stability – Simonsays Jun 05 '23 at 15:59
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    I'm a bit confused about your last comment. In the answer you say that For $(1)$ In general, you think it is false but in your last comment you say that Usually it’s proven for curves but works more generally. Do you mean that $(1)$ stands for plane but need not necessarily for more generalized smooth projective variety? – Sherlock Jun 05 '23 at 16:15
  • sorry misread your comment. I thought you were referring to the first fact of the answer. Concerning (1): Perhaps I was wrong, because using your claim, it should be true as $\mathbf{P}^2$ is a Fano surface. I'll delete my answer for (1) for now and think again, sorry for that mess!! – Simonsays Jun 05 '23 at 16:37