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 $ad(E_G)$?


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.

  • $\begingroup$ Can you please show how the Jacobi identity follows from functoriality? $\endgroup$ – Vít Tuček Dec 10 '18 at 16:25
  • $\begingroup$ @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. $\endgroup$ – Sasha Dec 10 '18 at 16:41
  • $\begingroup$ So monoidal functors are automatically additive? $\endgroup$ – Vít Tuček Dec 10 '18 at 17:25
  • $\begingroup$ 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. $\endgroup$ – Qiaochu Yuan Dec 10 '18 at 21:38

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 $\mathfrak{g}\otimes \mathfrak{g}$.


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