Is there any Lie algebra structure on the sheaf of sections of adjoint bundle
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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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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
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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
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
asked 15 hours ago
Anonymous
1236
1236
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2 Answers
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up vote
5
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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.
Can you please show how the Jacobi identity follows from functoriality?
– Vít Tuček
14 hours ago
@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
14 hours ago
So monoidal functors are automatically additive?
– Vít Tuček
13 hours ago
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
9 hours ago
add a comment |
up vote
2
down vote
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 $mathfrak{g}otimes mathfrak{g}$.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
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.
Can you please show how the Jacobi identity follows from functoriality?
– Vít Tuček
14 hours ago
@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
14 hours ago
So monoidal functors are automatically additive?
– Vít Tuček
13 hours ago
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
9 hours ago
add a comment |
up vote
5
down vote
accepted
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.
Can you please show how the Jacobi identity follows from functoriality?
– Vít Tuček
14 hours ago
@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
14 hours ago
So monoidal functors are automatically additive?
– Vít Tuček
13 hours ago
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
9 hours ago
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
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.
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.
answered 15 hours ago
Sasha
20k22652
20k22652
Can you please show how the Jacobi identity follows from functoriality?
– Vít Tuček
14 hours ago
@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
14 hours ago
So monoidal functors are automatically additive?
– Vít Tuček
13 hours ago
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
9 hours ago
add a comment |
Can you please show how the Jacobi identity follows from functoriality?
– Vít Tuček
14 hours ago
@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
14 hours ago
So monoidal functors are automatically additive?
– Vít Tuček
13 hours ago
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
9 hours ago
Can you please show how the Jacobi identity follows from functoriality?
– Vít Tuček
14 hours ago
Can you please show how the Jacobi identity follows from functoriality?
– Vít Tuček
14 hours ago
@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
14 hours ago
@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
14 hours ago
So monoidal functors are automatically additive?
– Vít Tuček
13 hours ago
So monoidal functors are automatically additive?
– Vít Tuček
13 hours ago
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
9 hours ago
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
9 hours ago
add a comment |
up vote
2
down vote
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 $mathfrak{g}otimes mathfrak{g}$.
add a comment |
up vote
2
down vote
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 $mathfrak{g}otimes mathfrak{g}$.
add a comment |
up vote
2
down vote
up vote
2
down vote
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 $mathfrak{g}otimes mathfrak{g}$.
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 $mathfrak{g}otimes mathfrak{g}$.
answered 14 hours ago
Vít Tuček
4,96911748
4,96911748
add a comment |
add a comment |
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