Automorphisms of $mathbb{F}_{n}$.
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Let $mathbb{F}_n =mathbb{P}_{mathbb{P}^1}(mathcal{O}(n)oplusmathcal{O})$ with $ngeq1$. I was trying to calculate $Aut(mathbb{F}_{n}) = G$. It is not hard to see that an automorphism maps fibres (of the $mathbb{P}^1$-bundle) to fibres so there is a morphism $G rightarrow Aut(mathbb{P}^1) = PGl_2$. My question is about the kernel, I have read that this semi-direct product of $mathbb{C}^*$ with $H^{0}(mathbb{P}^1, mathcal{O}(n))$. This doesn't look so random since $mathbb{F}_n = mathbb{P}(mathcal{O}(n)oplus mathcal{O})$, I can see how $ mathbb{C}^*$ acts by multiplication on the trivial factor, but how does the semi-direct product act?
algebraic-geometry
asked Nov 20 at 20:38
Nick L
1,15110
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show 1 more comment
up vote
1
down vote
favorite
Let $mathbb{F}_n =mathbb{P}_{mathbb{P}^1}(mathcal{O}(n)oplusmathcal{O})$ with $ngeq1$. I was trying to calculate $Aut(mathbb{F}_{n}) = G$. It is not hard to see that an automorphism maps fibres (of the $mathbb{P}^1$-bundle) to fibres so there is a morphism $G rightarrow Aut(mathbb{P}^1) = PGl_2$. My question is about the kernel, I have read that this semi-direct product of $mathbb{C}^*$ with $H^{0}(mathbb{P}^1, mathcal{O}(n))$. This doesn't look so random since $mathbb{F}_n = mathbb{P}(mathcal{O}(n)oplus mathcal{O})$, I can see how $ mathbb{C}^*$ acts by multiplication on the trivial factor, but how does the semi-direct product act?
algebraic-geometry
asked Nov 20 at 20:38
Nick L
1,15110
I am not sure how standard or well-known the terminology $mathbb{F}_n$ is but maybe just for the convenience of the users, the question should begin with definitions of (new) notations.
– random123
Nov 21 at 5:33
1
Ok, I did this.
– Nick L
Nov 21 at 10:31
When you write $Aut(mathbb{F}_n)$ I understand that you mean the automorphism of the entire scheme?
– random123
Nov 21 at 11:16
1
Just the group of biregular morphisms of this variety.
– Nick L
Nov 21 at 11:25
The kernel looks like the set of all automorphism of the projective bundle over $mathbb{P}^1$. This is same as(? I am not totally sure of this statement) automorphism induced by automorphism of the vector bundle $mathcal{O} oplus mathcal{O}(n)$. This can be easily calculated to be what is required.
– random123
Nov 21 at 15:43
|
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $mathbb{F}_n =mathbb{P}_{mathbb{P}^1}(mathcal{O}(n)oplusmathcal{O})$ with $ngeq1$. I was trying to calculate $Aut(mathbb{F}_{n}) = G$. It is not hard to see that an automorphism maps fibres (of the $mathbb{P}^1$-bundle) to fibres so there is a morphism $G rightarrow Aut(mathbb{P}^1) = PGl_2$. My question is about the kernel, I have read that this semi-direct product of $mathbb{C}^*$ with $H^{0}(mathbb{P}^1, mathcal{O}(n))$. This doesn't look so random since $mathbb{F}_n = mathbb{P}(mathcal{O}(n)oplus mathcal{O})$, I can see how $ mathbb{C}^*$ acts by multiplication on the trivial factor, but how does the semi-direct product act?
algebraic-geometry
asked Nov 20 at 20:38
Nick L
1,15110
Let $mathbb{F}_n =mathbb{P}_{mathbb{P}^1}(mathcal{O}(n)oplusmathcal{O})$ with $ngeq1$. I was trying to calculate $Aut(mathbb{F}_{n}) = G$. It is not hard to see that an automorphism maps fibres (of the $mathbb{P}^1$-bundle) to fibres so there is a morphism $G rightarrow Aut(mathbb{P}^1) = PGl_2$. My question is about the kernel, I have read that this semi-direct product of $mathbb{C}^*$ with $H^{0}(mathbb{P}^1, mathcal{O}(n))$. This doesn't look so random since $mathbb{F}_n = mathbb{P}(mathcal{O}(n)oplus mathcal{O})$, I can see how $ mathbb{C}^*$ acts by multiplication on the trivial factor, but how does the semi-direct product act?
algebraic-geometry
algebraic-geometry
asked Nov 20 at 20:38
Nick L
1,15110
asked Nov 20 at 20:38
Nick L
1,15110
edited Nov 21 at 10:47
asked Nov 20 at 20:38
Nick L
1,15110
asked Nov 20 at 20:38
Nick L
1,15110
asked Nov 20 at 20:38
Nick L
1,15110
1,15110
I am not sure how standard or well-known the terminology $mathbb{F}_n$ is but maybe just for the convenience of the users, the question should begin with definitions of (new) notations.
– random123
Nov 21 at 5:33
1
Ok, I did this.
– Nick L
Nov 21 at 10:31
When you write $Aut(mathbb{F}_n)$ I understand that you mean the automorphism of the entire scheme?
– random123
Nov 21 at 11:16
1
Just the group of biregular morphisms of this variety.
– Nick L
Nov 21 at 11:25
The kernel looks like the set of all automorphism of the projective bundle over $mathbb{P}^1$. This is same as(? I am not totally sure of this statement) automorphism induced by automorphism of the vector bundle $mathcal{O} oplus mathcal{O}(n)$. This can be easily calculated to be what is required.
– random123
Nov 21 at 15:43
|
show 1 more comment
I am not sure how standard or well-known the terminology $mathbb{F}_n$ is but maybe just for the convenience of the users, the question should begin with definitions of (new) notations.
– random123
Nov 21 at 5:33
1
Ok, I did this.
– Nick L
Nov 21 at 10:31
When you write $Aut(mathbb{F}_n)$ I understand that you mean the automorphism of the entire scheme?
– random123
Nov 21 at 11:16
1
Just the group of biregular morphisms of this variety.
– Nick L
Nov 21 at 11:25
The kernel looks like the set of all automorphism of the projective bundle over $mathbb{P}^1$. This is same as(? I am not totally sure of this statement) automorphism induced by automorphism of the vector bundle $mathcal{O} oplus mathcal{O}(n)$. This can be easily calculated to be what is required.
– random123
Nov 21 at 15:43
I am not sure how standard or well-known the terminology $mathbb{F}_n$ is but maybe just for the convenience of the users, the question should begin with definitions of (new) notations.
– random123
Nov 21 at 5:33
I am not sure how standard or well-known the terminology $mathbb{F}_n$ is but maybe just for the convenience of the users, the question should begin with definitions of (new) notations.
– random123
Nov 21 at 5:33
1
1
Ok, I did this.
– Nick L
Nov 21 at 10:31
Ok, I did this.
– Nick L
Nov 21 at 10:31
When you write $Aut(mathbb{F}_n)$ I understand that you mean the automorphism of the entire scheme?
– random123
Nov 21 at 11:16
When you write $Aut(mathbb{F}_n)$ I understand that you mean the automorphism of the entire scheme?
– random123
Nov 21 at 11:16
1
1
Just the group of biregular morphisms of this variety.
– Nick L
Nov 21 at 11:25
Just the group of biregular morphisms of this variety.
– Nick L
Nov 21 at 11:25
The kernel looks like the set of all automorphism of the projective bundle over $mathbb{P}^1$. This is same as(? I am not totally sure of this statement) automorphism induced by automorphism of the vector bundle $mathcal{O} oplus mathcal{O}(n)$. This can be easily calculated to be what is required.
– random123
Nov 21 at 15:43
The kernel looks like the set of all automorphism of the projective bundle over $mathbb{P}^1$. This is same as(? I am not totally sure of this statement) automorphism induced by automorphism of the vector bundle $mathcal{O} oplus mathcal{O}(n)$. This can be easily calculated to be what is required.
– random123
Nov 21 at 15:43
|
show 1 more comment
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I am not sure how standard or well-known the terminology $mathbb{F}_n$ is but maybe just for the convenience of the users, the question should begin with definitions of (new) notations.
– random123
Nov 21 at 5:33
1
Ok, I did this.
– Nick L
Nov 21 at 10:31
When you write $Aut(mathbb{F}_n)$ I understand that you mean the automorphism of the entire scheme?
– random123
Nov 21 at 11:16
1
Just the group of biregular morphisms of this variety.
– Nick L
Nov 21 at 11:25
The kernel looks like the set of all automorphism of the projective bundle over $mathbb{P}^1$. This is same as(? I am not totally sure of this statement) automorphism induced by automorphism of the vector bundle $mathcal{O} oplus mathcal{O}(n)$. This can be easily calculated to be what is required.
– random123
Nov 21 at 15:43