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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 $E\subset F\otimes\text{det}\,E$?

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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 $Hom(E,E^\vee \otimes \det(E)) = 0$.

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

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