The orbits of semisimple groups in the projective space
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Let $G=Rcdot S$ be a connected reductive algebraic group where $S$ is normal semisimple and $R$ is central.
Suppose that $G$ acts algebraically on the protective variety $mathbb P(V)$. I wanna understand the $S$-orbits in $mathbb P(V)$. They are closed in $G$-orbits, but why they are flag varieties?
algebraic-geometry lie-groups algebraic-groups
asked Aug 4 at 2:46
Ronald
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up vote
4
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favorite
Let $G=Rcdot S$ be a connected reductive algebraic group where $S$ is normal semisimple and $R$ is central.
Suppose that $G$ acts algebraically on the protective variety $mathbb P(V)$. I wanna understand the $S$-orbits in $mathbb P(V)$. They are closed in $G$-orbits, but why they are flag varieties?
algebraic-geometry lie-groups algebraic-groups
asked Aug 4 at 2:46
Ronald
1,6361821
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $G=Rcdot S$ be a connected reductive algebraic group where $S$ is normal semisimple and $R$ is central.
Suppose that $G$ acts algebraically on the protective variety $mathbb P(V)$. I wanna understand the $S$-orbits in $mathbb P(V)$. They are closed in $G$-orbits, but why they are flag varieties?
algebraic-geometry lie-groups algebraic-groups
asked Aug 4 at 2:46
Ronald
1,6361821
Let $G=Rcdot S$ be a connected reductive algebraic group where $S$ is normal semisimple and $R$ is central.
Suppose that $G$ acts algebraically on the protective variety $mathbb P(V)$. I wanna understand the $S$-orbits in $mathbb P(V)$. They are closed in $G$-orbits, but why they are flag varieties?
algebraic-geometry lie-groups algebraic-groups
asked Aug 4 at 2:46
Ronald
1,6361821
edited Aug 16 at 23:46
asked Aug 4 at 2:46
Ronald
1,6361821
asked Aug 4 at 2:46
Ronald
1,6361821
asked Aug 4 at 2:46
Ronald
1,6361821
1,6361821
add a comment |Â
add a comment |Â
1 Answer
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I think I know now the answer of my question:
Since $G$ is algebraic and acts algebraically on $mathbb P(V)$, then $G$ can be represented as an algebraic
subgroup of $PGL(V)$. The commutator $G'$ is -by chevalley's theorem an algebraic subgroup of $PGL(V)$. (i.e. $G'$ is zariski-closed in $PGL(V)$). Since $G$ is reductive, then $G'=S$ and hence $S$-orbits are open in their closures in $mathbb P(V)$. But $S$ is normal in $G$, then all orbits are closed in the compact sets.
Therefore, an $S$-orbit is compact implies that the isotropy group is parabolic. Hence, $S$ orbits are flag manifolds.
answered Aug 19 at 4:33
Ronald
1,6361821
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I think I know now the answer of my question:
Since $G$ is algebraic and acts algebraically on $mathbb P(V)$, then $G$ can be represented as an algebraic
subgroup of $PGL(V)$. The commutator $G'$ is -by chevalley's theorem an algebraic subgroup of $PGL(V)$. (i.e. $G'$ is zariski-closed in $PGL(V)$). Since $G$ is reductive, then $G'=S$ and hence $S$-orbits are open in their closures in $mathbb P(V)$. But $S$ is normal in $G$, then all orbits are closed in the compact sets.
Therefore, an $S$-orbit is compact implies that the isotropy group is parabolic. Hence, $S$ orbits are flag manifolds.
answered Aug 19 at 4:33
Ronald
1,6361821
add a comment |Â
up vote
1
down vote
I think I know now the answer of my question:
Since $G$ is algebraic and acts algebraically on $mathbb P(V)$, then $G$ can be represented as an algebraic
subgroup of $PGL(V)$. The commutator $G'$ is -by chevalley's theorem an algebraic subgroup of $PGL(V)$. (i.e. $G'$ is zariski-closed in $PGL(V)$). Since $G$ is reductive, then $G'=S$ and hence $S$-orbits are open in their closures in $mathbb P(V)$. But $S$ is normal in $G$, then all orbits are closed in the compact sets.
Therefore, an $S$-orbit is compact implies that the isotropy group is parabolic. Hence, $S$ orbits are flag manifolds.
answered Aug 19 at 4:33
Ronald
1,6361821
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I think I know now the answer of my question:
Since $G$ is algebraic and acts algebraically on $mathbb P(V)$, then $G$ can be represented as an algebraic
subgroup of $PGL(V)$. The commutator $G'$ is -by chevalley's theorem an algebraic subgroup of $PGL(V)$. (i.e. $G'$ is zariski-closed in $PGL(V)$). Since $G$ is reductive, then $G'=S$ and hence $S$-orbits are open in their closures in $mathbb P(V)$. But $S$ is normal in $G$, then all orbits are closed in the compact sets.
Therefore, an $S$-orbit is compact implies that the isotropy group is parabolic. Hence, $S$ orbits are flag manifolds.
answered Aug 19 at 4:33
Ronald
1,6361821
I think I know now the answer of my question:
Since $G$ is algebraic and acts algebraically on $mathbb P(V)$, then $G$ can be represented as an algebraic
subgroup of $PGL(V)$. The commutator $G'$ is -by chevalley's theorem an algebraic subgroup of $PGL(V)$. (i.e. $G'$ is zariski-closed in $PGL(V)$). Since $G$ is reductive, then $G'=S$ and hence $S$-orbits are open in their closures in $mathbb P(V)$. But $S$ is normal in $G$, then all orbits are closed in the compact sets.
Therefore, an $S$-orbit is compact implies that the isotropy group is parabolic. Hence, $S$ orbits are flag manifolds.
answered Aug 19 at 4:33
Ronald
1,6361821
answered Aug 19 at 4:33
Ronald
1,6361821
answered Aug 19 at 4:33
Ronald
1,6361821
answered Aug 19 at 4:33
Ronald
1,6361821
1,6361821
add a comment |Â
add a comment |Â
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