Is there any Lie algebra structure on the sheaf of sections of adjoint bundle












3














Let $X$ be an irreducible smooth projective variety over $mathbb{C}$.
Let $G$ be an affine algebraic group over $mathbb{C}$.
Let $p : E_G longrightarrow X$ be a holomorphic principal $G$-bundle on $X$. Let $ad(E_G) = E_G times^G mathfrak{g}$ be the adjoint vector bundle of $E_G$ associated to the adjoint representation $ad : G longrightarrow End(mathfrak{g})$ of $G$ on its Lie algebra $mathfrak{g}$. The fibers of $ad(E_G)$ are $mathbb{C}$-linearly isomorphic to $mathfrak{g}$.
Consider $ad(E_G)$ as a sheaf of $mathcal{O}_X$-modules on $X$.



Question: Is there any $mathcal{O}_X$-bilinear homomorphism
$[,] : ad(E_G)times ad(E_G) to ad(E_G)$ giving a Lie algebra structure on the sheaf $ad(E_G)$?






ag.algebraic-geometry principal-bundles







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    3














    Let $X$ be an irreducible smooth projective variety over $mathbb{C}$.
    Let $G$ be an affine algebraic group over $mathbb{C}$.
    Let $p : E_G longrightarrow X$ be a holomorphic principal $G$-bundle on $X$. Let $ad(E_G) = E_G times^G mathfrak{g}$ be the adjoint vector bundle of $E_G$ associated to the adjoint representation $ad : G longrightarrow End(mathfrak{g})$ of $G$ on its Lie algebra $mathfrak{g}$. The fibers of $ad(E_G)$ are $mathbb{C}$-linearly isomorphic to $mathfrak{g}$.
    Consider $ad(E_G)$ as a sheaf of $mathcal{O}_X$-modules on $X$.



    Question: Is there any $mathcal{O}_X$-bilinear homomorphism
    $[,] : ad(E_G)times ad(E_G) to ad(E_G)$ giving a Lie algebra structure on the sheaf $ad(E_G)$?






    ag.algebraic-geometry principal-bundles







    share|cite|improve this question

























      3












      3








      3







      Let $X$ be an irreducible smooth projective variety over $mathbb{C}$.
      Let $G$ be an affine algebraic group over $mathbb{C}$.
      Let $p : E_G longrightarrow X$ be a holomorphic principal $G$-bundle on $X$. Let $ad(E_G) = E_G times^G mathfrak{g}$ be the adjoint vector bundle of $E_G$ associated to the adjoint representation $ad : G longrightarrow End(mathfrak{g})$ of $G$ on its Lie algebra $mathfrak{g}$. The fibers of $ad(E_G)$ are $mathbb{C}$-linearly isomorphic to $mathfrak{g}$.
      Consider $ad(E_G)$ as a sheaf of $mathcal{O}_X$-modules on $X$.



      Question: Is there any $mathcal{O}_X$-bilinear homomorphism
      $[,] : ad(E_G)times ad(E_G) to ad(E_G)$ giving a Lie algebra structure on the sheaf $ad(E_G)$?






      ag.algebraic-geometry principal-bundles







      share|cite|improve this question













      Let $X$ be an irreducible smooth projective variety over $mathbb{C}$.
      Let $G$ be an affine algebraic group over $mathbb{C}$.
      Let $p : E_G longrightarrow X$ be a holomorphic principal $G$-bundle on $X$. Let $ad(E_G) = E_G times^G mathfrak{g}$ be the adjoint vector bundle of $E_G$ associated to the adjoint representation $ad : G longrightarrow End(mathfrak{g})$ of $G$ on its Lie algebra $mathfrak{g}$. The fibers of $ad(E_G)$ are $mathbb{C}$-linearly isomorphic to $mathfrak{g}$.
      Consider $ad(E_G)$ as a sheaf of $mathcal{O}_X$-modules on $X$.



      Question: Is there any $mathcal{O}_X$-bilinear homomorphism
      $[,] : ad(E_G)times ad(E_G) to ad(E_G)$ giving a Lie algebra structure on the sheaf $ad(E_G)$?






      ag.algebraic-geometry principal-bundles




      ag.algebraic-geometry principal-bundles






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 10 at 15:37









      Anonymous

      1286




      1286






















          2 Answers
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          5














          A principal $G$-bundle gives a monoidal functor from the category of representations of $G$ to the category of vector bundles. In particular, it takes the morphism
          $$
          [-,-] colon mathfrak{g} otimes mathfrak{g} to mathfrak{g}
          $$
          of $G$-representations (for the adjoint action) to a morphism of vector bundles
          $$
          [-,-] colon ad(E_G) otimes ad(E_G) to ad(E_G).
          $$
          By functoriality, it is skew-symmetric and satisfies the Jacobi identity, hence provides the sheaf $ad(E_G)$ with a Lie algebra structure.






          share|cite|improve this answer





















          • Can you please show how the Jacobi identity follows from functoriality?
            – Vít Tuček
            Dec 10 at 16:25










          • @VítTuček: The Jacobian identity says that the sum of three maps $ad(E_G) otimes ad(E_G) otimes ad(E_G) to ad(E_G)$ vanishes. These maps come from three maps $mathfrak{g} otimes mathfrak{g} otimes mathfrak{g} to mathfrak{g}$. The sum of the latter maps is zero, hence so is the sum of the former maps.
            – Sasha
            Dec 10 at 16:41










          • So monoidal functors are automatically additive?
            – Vít Tuček
            Dec 10 at 17:25










          • No, certainly not. You need additivity and the functor also needs to be symmetric monoidal, not just monoidal, to preserve skew-symmetry and the Jacobi identity.
            – Qiaochu Yuan
            Dec 10 at 21:38



















          2














          Yes. It boils down to natural isomorphism $ad(E_G) otimes ad(E_G) simeq E_G times^G (mathfrak{g}otimes mathfrak{g})$ which allows you to compose tensor product of sections with the bracket on $mathfrak{g}otimes mathfrak{g}$.






          share|cite|improve this answer





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5














            A principal $G$-bundle gives a monoidal functor from the category of representations of $G$ to the category of vector bundles. In particular, it takes the morphism
            $$
            [-,-] colon mathfrak{g} otimes mathfrak{g} to mathfrak{g}
            $$
            of $G$-representations (for the adjoint action) to a morphism of vector bundles
            $$
            [-,-] colon ad(E_G) otimes ad(E_G) to ad(E_G).
            $$
            By functoriality, it is skew-symmetric and satisfies the Jacobi identity, hence provides the sheaf $ad(E_G)$ with a Lie algebra structure.






            share|cite|improve this answer





















            • Can you please show how the Jacobi identity follows from functoriality?
              – Vít Tuček
              Dec 10 at 16:25










            • @VítTuček: The Jacobian identity says that the sum of three maps $ad(E_G) otimes ad(E_G) otimes ad(E_G) to ad(E_G)$ vanishes. These maps come from three maps $mathfrak{g} otimes mathfrak{g} otimes mathfrak{g} to mathfrak{g}$. The sum of the latter maps is zero, hence so is the sum of the former maps.
              – Sasha
              Dec 10 at 16:41










            • So monoidal functors are automatically additive?
              – Vít Tuček
              Dec 10 at 17:25










            • No, certainly not. You need additivity and the functor also needs to be symmetric monoidal, not just monoidal, to preserve skew-symmetry and the Jacobi identity.
              – Qiaochu Yuan
              Dec 10 at 21:38
















            5














            A principal $G$-bundle gives a monoidal functor from the category of representations of $G$ to the category of vector bundles. In particular, it takes the morphism
            $$
            [-,-] colon mathfrak{g} otimes mathfrak{g} to mathfrak{g}
            $$
            of $G$-representations (for the adjoint action) to a morphism of vector bundles
            $$
            [-,-] colon ad(E_G) otimes ad(E_G) to ad(E_G).
            $$
            By functoriality, it is skew-symmetric and satisfies the Jacobi identity, hence provides the sheaf $ad(E_G)$ with a Lie algebra structure.






            share|cite|improve this answer





















            • Can you please show how the Jacobi identity follows from functoriality?
              – Vít Tuček
              Dec 10 at 16:25










            • @VítTuček: The Jacobian identity says that the sum of three maps $ad(E_G) otimes ad(E_G) otimes ad(E_G) to ad(E_G)$ vanishes. These maps come from three maps $mathfrak{g} otimes mathfrak{g} otimes mathfrak{g} to mathfrak{g}$. The sum of the latter maps is zero, hence so is the sum of the former maps.
              – Sasha
              Dec 10 at 16:41










            • So monoidal functors are automatically additive?
              – Vít Tuček
              Dec 10 at 17:25










            • No, certainly not. You need additivity and the functor also needs to be symmetric monoidal, not just monoidal, to preserve skew-symmetry and the Jacobi identity.
              – Qiaochu Yuan
              Dec 10 at 21:38














            5












            5








            5






            A principal $G$-bundle gives a monoidal functor from the category of representations of $G$ to the category of vector bundles. In particular, it takes the morphism
            $$
            [-,-] colon mathfrak{g} otimes mathfrak{g} to mathfrak{g}
            $$
            of $G$-representations (for the adjoint action) to a morphism of vector bundles
            $$
            [-,-] colon ad(E_G) otimes ad(E_G) to ad(E_G).
            $$
            By functoriality, it is skew-symmetric and satisfies the Jacobi identity, hence provides the sheaf $ad(E_G)$ with a Lie algebra structure.






            share|cite|improve this answer












            A principal $G$-bundle gives a monoidal functor from the category of representations of $G$ to the category of vector bundles. In particular, it takes the morphism
            $$
            [-,-] colon mathfrak{g} otimes mathfrak{g} to mathfrak{g}
            $$
            of $G$-representations (for the adjoint action) to a morphism of vector bundles
            $$
            [-,-] colon ad(E_G) otimes ad(E_G) to ad(E_G).
            $$
            By functoriality, it is skew-symmetric and satisfies the Jacobi identity, hence provides the sheaf $ad(E_G)$ with a Lie algebra structure.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 10 at 15:51









            Sasha

            20.2k22654




            20.2k22654












            • Can you please show how the Jacobi identity follows from functoriality?
              – Vít Tuček
              Dec 10 at 16:25










            • @VítTuček: The Jacobian identity says that the sum of three maps $ad(E_G) otimes ad(E_G) otimes ad(E_G) to ad(E_G)$ vanishes. These maps come from three maps $mathfrak{g} otimes mathfrak{g} otimes mathfrak{g} to mathfrak{g}$. The sum of the latter maps is zero, hence so is the sum of the former maps.
              – Sasha
              Dec 10 at 16:41










            • So monoidal functors are automatically additive?
              – Vít Tuček
              Dec 10 at 17:25










            • No, certainly not. You need additivity and the functor also needs to be symmetric monoidal, not just monoidal, to preserve skew-symmetry and the Jacobi identity.
              – Qiaochu Yuan
              Dec 10 at 21:38


















            • Can you please show how the Jacobi identity follows from functoriality?
              – Vít Tuček
              Dec 10 at 16:25










            • @VítTuček: The Jacobian identity says that the sum of three maps $ad(E_G) otimes ad(E_G) otimes ad(E_G) to ad(E_G)$ vanishes. These maps come from three maps $mathfrak{g} otimes mathfrak{g} otimes mathfrak{g} to mathfrak{g}$. The sum of the latter maps is zero, hence so is the sum of the former maps.
              – Sasha
              Dec 10 at 16:41










            • So monoidal functors are automatically additive?
              – Vít Tuček
              Dec 10 at 17:25










            • No, certainly not. You need additivity and the functor also needs to be symmetric monoidal, not just monoidal, to preserve skew-symmetry and the Jacobi identity.
              – Qiaochu Yuan
              Dec 10 at 21:38
















            Can you please show how the Jacobi identity follows from functoriality?
            – Vít Tuček
            Dec 10 at 16:25




            Can you please show how the Jacobi identity follows from functoriality?
            – Vít Tuček
            Dec 10 at 16:25












            @VítTuček: The Jacobian identity says that the sum of three maps $ad(E_G) otimes ad(E_G) otimes ad(E_G) to ad(E_G)$ vanishes. These maps come from three maps $mathfrak{g} otimes mathfrak{g} otimes mathfrak{g} to mathfrak{g}$. The sum of the latter maps is zero, hence so is the sum of the former maps.
            – Sasha
            Dec 10 at 16:41




            @VítTuček: The Jacobian identity says that the sum of three maps $ad(E_G) otimes ad(E_G) otimes ad(E_G) to ad(E_G)$ vanishes. These maps come from three maps $mathfrak{g} otimes mathfrak{g} otimes mathfrak{g} to mathfrak{g}$. The sum of the latter maps is zero, hence so is the sum of the former maps.
            – Sasha
            Dec 10 at 16:41












            So monoidal functors are automatically additive?
            – Vít Tuček
            Dec 10 at 17:25




            So monoidal functors are automatically additive?
            – Vít Tuček
            Dec 10 at 17:25












            No, certainly not. You need additivity and the functor also needs to be symmetric monoidal, not just monoidal, to preserve skew-symmetry and the Jacobi identity.
            – Qiaochu Yuan
            Dec 10 at 21:38




            No, certainly not. You need additivity and the functor also needs to be symmetric monoidal, not just monoidal, to preserve skew-symmetry and the Jacobi identity.
            – Qiaochu Yuan
            Dec 10 at 21:38











            2














            Yes. It boils down to natural isomorphism $ad(E_G) otimes ad(E_G) simeq E_G times^G (mathfrak{g}otimes mathfrak{g})$ which allows you to compose tensor product of sections with the bracket on $mathfrak{g}otimes mathfrak{g}$.






            share|cite|improve this answer


























              2














              Yes. It boils down to natural isomorphism $ad(E_G) otimes ad(E_G) simeq E_G times^G (mathfrak{g}otimes mathfrak{g})$ which allows you to compose tensor product of sections with the bracket on $mathfrak{g}otimes mathfrak{g}$.






              share|cite|improve this answer
























                2












                2








                2






                Yes. It boils down to natural isomorphism $ad(E_G) otimes ad(E_G) simeq E_G times^G (mathfrak{g}otimes mathfrak{g})$ which allows you to compose tensor product of sections with the bracket on $mathfrak{g}otimes mathfrak{g}$.






                share|cite|improve this answer












                Yes. It boils down to natural isomorphism $ad(E_G) otimes ad(E_G) simeq E_G times^G (mathfrak{g}otimes mathfrak{g})$ which allows you to compose tensor product of sections with the bracket on $mathfrak{g}otimes mathfrak{g}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 10 at 16:24









                Vít Tuček

                4,99711748




                4,99711748






























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