# Is there a morphism from a locally free sheaf to the dual twisted by determinant bundle?

Let $$X$$ be a smooth projective variety over the field $$\mathbb{C}$$. Let $$E$$ be a locally free sheaf of rank $$r$$ and $$F=E^{\vee}$$, the dual locally free sheaf. If rank $$E$$=2, then we have the isomorphism $$E\simeq F\otimes \text{det }E$$. Do we have an analogous relation in the higher ranks. For example, is there an inclusion $$E\subset F\otimes\text{det}\,E$$?

## 1 Answer

To give a morphism $$E \to E^\vee \otimes \det(E)$$ is the same as to give a bilinear form on $$E$$ with values in $$\det(E)$$. This morphism is injective if the form is generically non-degenerate. It is an isomorphism when the form is everywhere non-degenerate. A necessary condition for this is that $$\det(E^\vee \otimes \det(E)) \cong \det(E^\vee) \otimes \det(E)^r \cong \det(E)^{r-1}$$ is isomorphic to $$\det(E)$$. So, this can be true only if $$\det(E) = O_X$$, or if $$\det(E)$$ is a point of order $$r-2$$ on $$Pic^0(X)$$.

Note also that there may be no nontrivial morphisms. For instance, take $$X = \mathbb{P}^1$$ and $$E = O(-1) \oplus O(-1) \oplus O(-1)$$. Then $$E^\vee \otimes \det(E) \cong O(-2) \oplus O(-2) \oplus O(-2)$$ and $$Hom(E,E^\vee \otimes \det(E)) = 0$$.

• Is the above condition for injectivity or to give a morphism? – user349424 Jan 8 at 12:57
• In other words, is there a morphism which is not injective? – user349424 Jan 8 at 12:59
• There may be no nonzero morphisms, I edited the answer to include an example. – Sasha Jan 8 at 13:18