Does the tangent bundle of the projective space split?












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Does the tangent bundle of the projective space $mathbb P^n$ over an algebraic closed field $k$ split i.e can be written as direct sum of two vector bundle of positive rank?



For cotangent bundle, the computation for the cohomology or hodge number shows it does not split as $h^{1,1}=1$. However, any higher cohomology of the tangent bundle vanishes.






algebraic-geometry







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$endgroup$

















    1












    $begingroup$


    Does the tangent bundle of the projective space $mathbb P^n$ over an algebraic closed field $k$ split i.e can be written as direct sum of two vector bundle of positive rank?



    For cotangent bundle, the computation for the cohomology or hodge number shows it does not split as $h^{1,1}=1$. However, any higher cohomology of the tangent bundle vanishes.






    algebraic-geometry







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Does the tangent bundle of the projective space $mathbb P^n$ over an algebraic closed field $k$ split i.e can be written as direct sum of two vector bundle of positive rank?



      For cotangent bundle, the computation for the cohomology or hodge number shows it does not split as $h^{1,1}=1$. However, any higher cohomology of the tangent bundle vanishes.






      algebraic-geometry







      share|cite|improve this question









      $endgroup$




      Does the tangent bundle of the projective space $mathbb P^n$ over an algebraic closed field $k$ split i.e can be written as direct sum of two vector bundle of positive rank?



      For cotangent bundle, the computation for the cohomology or hodge number shows it does not split as $h^{1,1}=1$. However, any higher cohomology of the tangent bundle vanishes.






      algebraic-geometry




      algebraic-geometry






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 9 '18 at 3:22









      zzyzzy

      2,5641420




      2,5641420






















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          $begingroup$

          Using Borel-Bott-Weil theorem it is easy to check that
          $$
          Hom(T,T) cong k.
          $$
          This proves that $T$ is not split.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you! What about positive characterestic case?
            $endgroup$
            – zzy
            Dec 9 '18 at 15:37










          • $begingroup$
            In positive characteristic you can use Euler sequence to check that $Hom(T,T) = k$.
            $endgroup$
            – Sasha
            Dec 9 '18 at 18:22











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          2












          $begingroup$

          Using Borel-Bott-Weil theorem it is easy to check that
          $$
          Hom(T,T) cong k.
          $$
          This proves that $T$ is not split.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you! What about positive characterestic case?
            $endgroup$
            – zzy
            Dec 9 '18 at 15:37










          • $begingroup$
            In positive characteristic you can use Euler sequence to check that $Hom(T,T) = k$.
            $endgroup$
            – Sasha
            Dec 9 '18 at 18:22
















          2












          $begingroup$

          Using Borel-Bott-Weil theorem it is easy to check that
          $$
          Hom(T,T) cong k.
          $$
          This proves that $T$ is not split.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you! What about positive characterestic case?
            $endgroup$
            – zzy
            Dec 9 '18 at 15:37










          • $begingroup$
            In positive characteristic you can use Euler sequence to check that $Hom(T,T) = k$.
            $endgroup$
            – Sasha
            Dec 9 '18 at 18:22














          2












          2








          2





          $begingroup$

          Using Borel-Bott-Weil theorem it is easy to check that
          $$
          Hom(T,T) cong k.
          $$
          This proves that $T$ is not split.






          share|cite|improve this answer









          $endgroup$



          Using Borel-Bott-Weil theorem it is easy to check that
          $$
          Hom(T,T) cong k.
          $$
          This proves that $T$ is not split.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 8:20









          SashaSasha

          4,728139




          4,728139












          • $begingroup$
            Thank you! What about positive characterestic case?
            $endgroup$
            – zzy
            Dec 9 '18 at 15:37










          • $begingroup$
            In positive characteristic you can use Euler sequence to check that $Hom(T,T) = k$.
            $endgroup$
            – Sasha
            Dec 9 '18 at 18:22


















          • $begingroup$
            Thank you! What about positive characterestic case?
            $endgroup$
            – zzy
            Dec 9 '18 at 15:37










          • $begingroup$
            In positive characteristic you can use Euler sequence to check that $Hom(T,T) = k$.
            $endgroup$
            – Sasha
            Dec 9 '18 at 18:22
















          $begingroup$
          Thank you! What about positive characterestic case?
          $endgroup$
          – zzy
          Dec 9 '18 at 15:37




          $begingroup$
          Thank you! What about positive characterestic case?
          $endgroup$
          – zzy
          Dec 9 '18 at 15:37












          $begingroup$
          In positive characteristic you can use Euler sequence to check that $Hom(T,T) = k$.
          $endgroup$
          – Sasha
          Dec 9 '18 at 18:22




          $begingroup$
          In positive characteristic you can use Euler sequence to check that $Hom(T,T) = k$.
          $endgroup$
          – Sasha
          Dec 9 '18 at 18:22


















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