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












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









$endgroup$

















    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









    $endgroup$















      0












      0








      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 8 at 10:08









      user349424user349424

      34317




      34317






















          1 Answer
          1






          active

          oldest

          votes


















          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












          Your Answer








          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065998%2fis-there-a-morphism-from-a-locally-free-sheaf-to-the-dual-twisted-by-determinant%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          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












          $begingroup$

          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

          5,208139




          5,208139












          • $begingroup$
            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


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065998%2fis-there-a-morphism-from-a-locally-free-sheaf-to-the-dual-twisted-by-determinant%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Probability when a professor distributes a quiz and homework assignment to a class of n students.

          Image
          0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
          Read more

          Aardman Animations

          Image
          Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
          Read more

          Are they similar matrix

          Image
          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
          Read more