Pushforwad bundle alond a degree 2 map from $P^1$ to $P^1$












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Suppose $f:P^1rightarrow P^1$ is a degree 2 morphism. Let $L$ be a line bundle on $P^1$ (which is equivalent to a $O(n)$), then what is $f_*L$? As for an open set $U$ we can see the preimage is a disjoint of two pieces of open sets, if $U$ is small enough, so the image is a rank 2 bundle. I guess that it should be $O(n)oplus O(n)$, as we can calculate the dimension of sections.










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    $begingroup$


    Suppose $f:P^1rightarrow P^1$ is a degree 2 morphism. Let $L$ be a line bundle on $P^1$ (which is equivalent to a $O(n)$), then what is $f_*L$? As for an open set $U$ we can see the preimage is a disjoint of two pieces of open sets, if $U$ is small enough, so the image is a rank 2 bundle. I guess that it should be $O(n)oplus O(n)$, as we can calculate the dimension of sections.










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Suppose $f:P^1rightarrow P^1$ is a degree 2 morphism. Let $L$ be a line bundle on $P^1$ (which is equivalent to a $O(n)$), then what is $f_*L$? As for an open set $U$ we can see the preimage is a disjoint of two pieces of open sets, if $U$ is small enough, so the image is a rank 2 bundle. I guess that it should be $O(n)oplus O(n)$, as we can calculate the dimension of sections.










      share|cite|improve this question









      $endgroup$




      Suppose $f:P^1rightarrow P^1$ is a degree 2 morphism. Let $L$ be a line bundle on $P^1$ (which is equivalent to a $O(n)$), then what is $f_*L$? As for an open set $U$ we can see the preimage is a disjoint of two pieces of open sets, if $U$ is small enough, so the image is a rank 2 bundle. I guess that it should be $O(n)oplus O(n)$, as we can calculate the dimension of sections.







      algebraic-geometry projective-space line-bundles






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      asked Dec 31 '18 at 9:48









      Peter LiuPeter Liu

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      307114






















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          $begingroup$

          First,
          $$
          f_*O cong O oplus O(-1).
          $$

          Indeed, this should be a rank 2 vector bundle with $H^0 = Bbbk$ and $H^1 = 0$, by Grothendieck theorem any vector bundle on $mathbb{P}^1$ is a sum of line bundles, and so the only option we have is the one written in the right hand side above. Similarly,
          $$
          f_*O(-1) cong O(-1) oplus O(-1).
          $$

          Finally, from the above and projection formula it follows that
          $$
          f_*O(2n) cong O(n) oplus O(n-1),
          qquad
          f_*O(2n-1) cong O(n-1) oplus O(n-1).
          $$






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            1 Answer
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            active

            oldest

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            1 Answer
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            active

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            1












            $begingroup$

            First,
            $$
            f_*O cong O oplus O(-1).
            $$

            Indeed, this should be a rank 2 vector bundle with $H^0 = Bbbk$ and $H^1 = 0$, by Grothendieck theorem any vector bundle on $mathbb{P}^1$ is a sum of line bundles, and so the only option we have is the one written in the right hand side above. Similarly,
            $$
            f_*O(-1) cong O(-1) oplus O(-1).
            $$

            Finally, from the above and projection formula it follows that
            $$
            f_*O(2n) cong O(n) oplus O(n-1),
            qquad
            f_*O(2n-1) cong O(n-1) oplus O(n-1).
            $$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              First,
              $$
              f_*O cong O oplus O(-1).
              $$

              Indeed, this should be a rank 2 vector bundle with $H^0 = Bbbk$ and $H^1 = 0$, by Grothendieck theorem any vector bundle on $mathbb{P}^1$ is a sum of line bundles, and so the only option we have is the one written in the right hand side above. Similarly,
              $$
              f_*O(-1) cong O(-1) oplus O(-1).
              $$

              Finally, from the above and projection formula it follows that
              $$
              f_*O(2n) cong O(n) oplus O(n-1),
              qquad
              f_*O(2n-1) cong O(n-1) oplus O(n-1).
              $$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                First,
                $$
                f_*O cong O oplus O(-1).
                $$

                Indeed, this should be a rank 2 vector bundle with $H^0 = Bbbk$ and $H^1 = 0$, by Grothendieck theorem any vector bundle on $mathbb{P}^1$ is a sum of line bundles, and so the only option we have is the one written in the right hand side above. Similarly,
                $$
                f_*O(-1) cong O(-1) oplus O(-1).
                $$

                Finally, from the above and projection formula it follows that
                $$
                f_*O(2n) cong O(n) oplus O(n-1),
                qquad
                f_*O(2n-1) cong O(n-1) oplus O(n-1).
                $$






                share|cite|improve this answer









                $endgroup$



                First,
                $$
                f_*O cong O oplus O(-1).
                $$

                Indeed, this should be a rank 2 vector bundle with $H^0 = Bbbk$ and $H^1 = 0$, by Grothendieck theorem any vector bundle on $mathbb{P}^1$ is a sum of line bundles, and so the only option we have is the one written in the right hand side above. Similarly,
                $$
                f_*O(-1) cong O(-1) oplus O(-1).
                $$

                Finally, from the above and projection formula it follows that
                $$
                f_*O(2n) cong O(n) oplus O(n-1),
                qquad
                f_*O(2n-1) cong O(n-1) oplus O(n-1).
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 31 '18 at 11:26









                SashaSasha

                5,218139




                5,218139






























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