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












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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 $Esimeq Fotimes text{det }E$. Do we have an analogous relation in the higher ranks. For example, is there an inclusion $Esubset Fotimestext{det},E$?






algebraic-geometry







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    0












    $begingroup$


    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 $Esimeq Fotimes text{det }E$. Do we have an analogous relation in the higher ranks. For example, is there an inclusion $Esubset Fotimestext{det},E$?






    algebraic-geometry







    share|cite|improve this question









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      0












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      0





      $begingroup$


      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 $Esimeq Fotimes text{det }E$. Do we have an analogous relation in the higher ranks. For example, is there an inclusion $Esubset Fotimestext{det},E$?






      algebraic-geometry







      share|cite|improve this question









      $endgroup$




      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 $Esimeq Fotimes text{det }E$. Do we have an analogous relation in the higher ranks. For example, is there an inclusion $Esubset Fotimestext{det},E$?






      algebraic-geometry




      algebraic-geometry






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      asked Jan 8 at 10:08









      user349424user349424

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          $begingroup$

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






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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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          $begingroup$

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






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          1












          $begingroup$

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






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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














          1












          1








          1





          $begingroup$

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






          share|cite|improve this answer











          $endgroup$



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







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 8 at 13:17

























          answered Jan 8 at 11:16









          SashaSasha

          5,218139




          5,218139












          • $begingroup$
            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


















          • $begingroup$
            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
















          $begingroup$
          Is the above condition for injectivity or to give a morphism?
          $endgroup$
          – user349424
          Jan 8 at 12:57






          $begingroup$
          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




          $begingroup$
          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




          $begingroup$
          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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