Smooth irreducible subvarieties in an algebraic group that are stable under power mapsDensity and irreducibilityA characteristic-free proof that the action of a connected algebraic group $G$ on the fundamental group of a $G$-variety is trivialReference needed for projection from a linear subspace (projecting cone)Smooth affine algebraic subgroups as complete intersectionsIs there a non-smooth algebraic group scheme in char $p$, all of whose defining relations have degree less than $p$?“Almost-ideals” in the (simple) Lie algebra of an algebraic group?Why is, for a group scheme of finite type, “smooth” (resp. irreducible) equivalent to “geometrically reduced” (resp. geometrically irreducible)?
Smooth irreducible subvarieties in an algebraic group that are stable under power maps
Density and irreducibilityA characteristic-free proof that the action of a connected algebraic group $G$ on the fundamental group of a $G$-variety is trivialReference needed for projection from a linear subspace (projecting cone)Smooth affine algebraic subgroups as complete intersectionsIs there a non-smooth algebraic group scheme in char $p$, all of whose defining relations have degree less than $p$?“Almost-ideals” in the (simple) Lie algebra of an algebraic group?Why is, for a group scheme of finite type, “smooth” (resp. irreducible) equivalent to “geometrically reduced” (resp. geometrically irreducible)?
$begingroup$
Let $X$ be a smooth irreducible subvariety of an algebraic group over a field, assume $X$ is invariant under $n$-th power map for every integer $n$ ($n=0$ means the identity element is in $X$). Must $X$ be a subgroup?
The motivation for this question is the intuition that if a cone in a linear space is smooth, then it shall be a linear subspace.
ag.algebraic-geometry algebraic-groups
asked 9 hours ago
sawdadasawdada
2,1981 gold badge4 silver badges31 bronze badges
$endgroup$
add a comment
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$begingroup$
Let $X$ be a smooth irreducible subvariety of an algebraic group over a field, assume $X$ is invariant under $n$-th power map for every integer $n$ ($n=0$ means the identity element is in $X$). Must $X$ be a subgroup?
The motivation for this question is the intuition that if a cone in a linear space is smooth, then it shall be a linear subspace.
ag.algebraic-geometry algebraic-groups
asked 9 hours ago
sawdadasawdada
2,1981 gold badge4 silver badges31 bronze badges
$endgroup$
add a comment
|
$begingroup$
Let $X$ be a smooth irreducible subvariety of an algebraic group over a field, assume $X$ is invariant under $n$-th power map for every integer $n$ ($n=0$ means the identity element is in $X$). Must $X$ be a subgroup?
The motivation for this question is the intuition that if a cone in a linear space is smooth, then it shall be a linear subspace.
ag.algebraic-geometry algebraic-groups
asked 9 hours ago
sawdadasawdada
2,1981 gold badge4 silver badges31 bronze badges
$endgroup$
Let $X$ be a smooth irreducible subvariety of an algebraic group over a field, assume $X$ is invariant under $n$-th power map for every integer $n$ ($n=0$ means the identity element is in $X$). Must $X$ be a subgroup?
The motivation for this question is the intuition that if a cone in a linear space is smooth, then it shall be a linear subspace.
ag.algebraic-geometry algebraic-groups
ag.algebraic-geometry algebraic-groups
asked 9 hours ago
sawdadasawdada
2,1981 gold badge4 silver badges31 bronze badges
asked 9 hours ago
sawdadasawdada
2,1981 gold badge4 silver badges31 bronze badges
asked 9 hours ago
sawdadasawdada
2,1981 gold badge4 silver badges31 bronze badges
asked 9 hours ago
sawdadasawdada
2,1981 gold badge4 silver badges31 bronze badges
asked 9 hours ago
sawdadasawdada
2,1981 gold badge4 silver badges31 bronze badges
2,1981 gold badge4 silver badges31 bronze badges
add a comment
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add a comment
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1 Answer
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oldest
votes
$begingroup$
This is false in nonabelian unipotent groups. For these groups, the exponential map is algebraic, and an isomorphism. The image of any linear subspace under this map will be smooth, irreducible, and invariant under the $n$th power map. But it will not be a subgroup unless the subspace is closed under the Lie bracket. For instance, the set of matrices of the form
$$1+ beginpmatrix 0 & a & 0 \ 0 & 0 & b \ 0 & 0 & 0 endpmatrix+ beginpmatrix 0 & a & 0 \ 0 & 0 & b \ 0 & 0 & 0 endpmatrix^2/2 + dots = beginpmatrix 1 & a & ab/2 \ 0 & 1 & b \ 0 & 0 & 1 endpmatrix $$ is smooth, irreducible, and closed under $n$th powers since $$beginpmatrix 1 & a & ab/2 \ 0 & 1 & b \ 0 & 0 & 1 endpmatrix^n =beginpmatrix 1 & na & n^2 ab/2 \ 0 & 1 & nb \ 0 & 0 & 1 endpmatrix ,$$ but isn't a subgroup.
Since most reductive groups contain nonabelian unipotent subgroups, this will not be true for reductive groups either.
answered 8 hours ago
Will SawinWill Sawin
72.8k7 gold badges146 silver badges303 bronze badges
$endgroup$
$begingroup$
Thank you! But I think it's true for torus. Is it true for any commutative groups?
$endgroup$
– sawdada
7 hours ago
2
$begingroup$
@sawdada I don't think it's too hard to prove this in characteristic zero using the exponential map. But probably not in characteristic $p$ for additive type groups as then the $n$th power depends only on $n$ mod $p$ so equations like $x = y^2p-1$ will provide counterexamples.
$endgroup$
– Will Sawin
6 hours ago
$begingroup$
@sawdada There is also a purely algebraic proof for toruses, and mroe generally for semiabelian varieties.
$endgroup$
– Will Sawin
4 hours ago
1
$begingroup$
@sawdada We can mod out the variety by the maximal subgroup that stabilizes $X$, and thus assume $X$ is not stable under translation by any nontrivial element. Our goal is then to show $X$ is the identity. We have an $n$th power map $[n]: X to X$. If the generic point has two preimages, then they are each a generic point of $X$, and differ by an $n$-torsion point, so the translate of a generic point of $X$ by an $n$-torsion point lies in $X$, so $X$ is stable under translation by the $n$-torsion point, contradicting the assumption.
$endgroup$
– Will Sawin
1 hour ago
1
$begingroup$
@sawdada So in fact the map has generic degree $1$ and thus has a generic section, hence has a section by normalcy. So the pullback of any class in $H^1$ of the group to $X$ is trivial. This is impossible unless $X$ is a point. (If the intersection of $X$ with the maximal subtorus has positive dimension, you can do this by induction on the dimension of the torus, and otherwise you can project to the abelian variety and use that the cohomology of the abelian variety is generated by $H^1$.)
$endgroup$
– Will Sawin
1 hour ago
|
show 1 more comment
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$begingroup$
This is false in nonabelian unipotent groups. For these groups, the exponential map is algebraic, and an isomorphism. The image of any linear subspace under this map will be smooth, irreducible, and invariant under the $n$th power map. But it will not be a subgroup unless the subspace is closed under the Lie bracket. For instance, the set of matrices of the form
$$1+ beginpmatrix 0 & a & 0 \ 0 & 0 & b \ 0 & 0 & 0 endpmatrix+ beginpmatrix 0 & a & 0 \ 0 & 0 & b \ 0 & 0 & 0 endpmatrix^2/2 + dots = beginpmatrix 1 & a & ab/2 \ 0 & 1 & b \ 0 & 0 & 1 endpmatrix $$ is smooth, irreducible, and closed under $n$th powers since $$beginpmatrix 1 & a & ab/2 \ 0 & 1 & b \ 0 & 0 & 1 endpmatrix^n =beginpmatrix 1 & na & n^2 ab/2 \ 0 & 1 & nb \ 0 & 0 & 1 endpmatrix ,$$ but isn't a subgroup.
Since most reductive groups contain nonabelian unipotent subgroups, this will not be true for reductive groups either.
answered 8 hours ago
Will SawinWill Sawin
72.8k7 gold badges146 silver badges303 bronze badges
$endgroup$
$begingroup$
Thank you! But I think it's true for torus. Is it true for any commutative groups?
$endgroup$
– sawdada
7 hours ago
2
$begingroup$
@sawdada I don't think it's too hard to prove this in characteristic zero using the exponential map. But probably not in characteristic $p$ for additive type groups as then the $n$th power depends only on $n$ mod $p$ so equations like $x = y^2p-1$ will provide counterexamples.
$endgroup$
– Will Sawin
6 hours ago
$begingroup$
@sawdada There is also a purely algebraic proof for toruses, and mroe generally for semiabelian varieties.
$endgroup$
– Will Sawin
4 hours ago
1
$begingroup$
@sawdada We can mod out the variety by the maximal subgroup that stabilizes $X$, and thus assume $X$ is not stable under translation by any nontrivial element. Our goal is then to show $X$ is the identity. We have an $n$th power map $[n]: X to X$. If the generic point has two preimages, then they are each a generic point of $X$, and differ by an $n$-torsion point, so the translate of a generic point of $X$ by an $n$-torsion point lies in $X$, so $X$ is stable under translation by the $n$-torsion point, contradicting the assumption.
$endgroup$
– Will Sawin
1 hour ago
1
$begingroup$
@sawdada So in fact the map has generic degree $1$ and thus has a generic section, hence has a section by normalcy. So the pullback of any class in $H^1$ of the group to $X$ is trivial. This is impossible unless $X$ is a point. (If the intersection of $X$ with the maximal subtorus has positive dimension, you can do this by induction on the dimension of the torus, and otherwise you can project to the abelian variety and use that the cohomology of the abelian variety is generated by $H^1$.)
$endgroup$
– Will Sawin
1 hour ago
|
show 1 more comment
$begingroup$
This is false in nonabelian unipotent groups. For these groups, the exponential map is algebraic, and an isomorphism. The image of any linear subspace under this map will be smooth, irreducible, and invariant under the $n$th power map. But it will not be a subgroup unless the subspace is closed under the Lie bracket. For instance, the set of matrices of the form
$$1+ beginpmatrix 0 & a & 0 \ 0 & 0 & b \ 0 & 0 & 0 endpmatrix+ beginpmatrix 0 & a & 0 \ 0 & 0 & b \ 0 & 0 & 0 endpmatrix^2/2 + dots = beginpmatrix 1 & a & ab/2 \ 0 & 1 & b \ 0 & 0 & 1 endpmatrix $$ is smooth, irreducible, and closed under $n$th powers since $$beginpmatrix 1 & a & ab/2 \ 0 & 1 & b \ 0 & 0 & 1 endpmatrix^n =beginpmatrix 1 & na & n^2 ab/2 \ 0 & 1 & nb \ 0 & 0 & 1 endpmatrix ,$$ but isn't a subgroup.
Since most reductive groups contain nonabelian unipotent subgroups, this will not be true for reductive groups either.
answered 8 hours ago
Will SawinWill Sawin
72.8k7 gold badges146 silver badges303 bronze badges
$endgroup$
$begingroup$
Thank you! But I think it's true for torus. Is it true for any commutative groups?
$endgroup$
– sawdada
7 hours ago
2
$begingroup$
@sawdada I don't think it's too hard to prove this in characteristic zero using the exponential map. But probably not in characteristic $p$ for additive type groups as then the $n$th power depends only on $n$ mod $p$ so equations like $x = y^2p-1$ will provide counterexamples.
$endgroup$
– Will Sawin
6 hours ago
$begingroup$
@sawdada There is also a purely algebraic proof for toruses, and mroe generally for semiabelian varieties.
$endgroup$
– Will Sawin
4 hours ago
1
$begingroup$
@sawdada We can mod out the variety by the maximal subgroup that stabilizes $X$, and thus assume $X$ is not stable under translation by any nontrivial element. Our goal is then to show $X$ is the identity. We have an $n$th power map $[n]: X to X$. If the generic point has two preimages, then they are each a generic point of $X$, and differ by an $n$-torsion point, so the translate of a generic point of $X$ by an $n$-torsion point lies in $X$, so $X$ is stable under translation by the $n$-torsion point, contradicting the assumption.
$endgroup$
– Will Sawin
1 hour ago
1
$begingroup$
@sawdada So in fact the map has generic degree $1$ and thus has a generic section, hence has a section by normalcy. So the pullback of any class in $H^1$ of the group to $X$ is trivial. This is impossible unless $X$ is a point. (If the intersection of $X$ with the maximal subtorus has positive dimension, you can do this by induction on the dimension of the torus, and otherwise you can project to the abelian variety and use that the cohomology of the abelian variety is generated by $H^1$.)
$endgroup$
– Will Sawin
1 hour ago
|
show 1 more comment
$begingroup$
This is false in nonabelian unipotent groups. For these groups, the exponential map is algebraic, and an isomorphism. The image of any linear subspace under this map will be smooth, irreducible, and invariant under the $n$th power map. But it will not be a subgroup unless the subspace is closed under the Lie bracket. For instance, the set of matrices of the form
$$1+ beginpmatrix 0 & a & 0 \ 0 & 0 & b \ 0 & 0 & 0 endpmatrix+ beginpmatrix 0 & a & 0 \ 0 & 0 & b \ 0 & 0 & 0 endpmatrix^2/2 + dots = beginpmatrix 1 & a & ab/2 \ 0 & 1 & b \ 0 & 0 & 1 endpmatrix $$ is smooth, irreducible, and closed under $n$th powers since $$beginpmatrix 1 & a & ab/2 \ 0 & 1 & b \ 0 & 0 & 1 endpmatrix^n =beginpmatrix 1 & na & n^2 ab/2 \ 0 & 1 & nb \ 0 & 0 & 1 endpmatrix ,$$ but isn't a subgroup.
Since most reductive groups contain nonabelian unipotent subgroups, this will not be true for reductive groups either.
answered 8 hours ago
Will SawinWill Sawin
72.8k7 gold badges146 silver badges303 bronze badges
$endgroup$
This is false in nonabelian unipotent groups. For these groups, the exponential map is algebraic, and an isomorphism. The image of any linear subspace under this map will be smooth, irreducible, and invariant under the $n$th power map. But it will not be a subgroup unless the subspace is closed under the Lie bracket. For instance, the set of matrices of the form
$$1+ beginpmatrix 0 & a & 0 \ 0 & 0 & b \ 0 & 0 & 0 endpmatrix+ beginpmatrix 0 & a & 0 \ 0 & 0 & b \ 0 & 0 & 0 endpmatrix^2/2 + dots = beginpmatrix 1 & a & ab/2 \ 0 & 1 & b \ 0 & 0 & 1 endpmatrix $$ is smooth, irreducible, and closed under $n$th powers since $$beginpmatrix 1 & a & ab/2 \ 0 & 1 & b \ 0 & 0 & 1 endpmatrix^n =beginpmatrix 1 & na & n^2 ab/2 \ 0 & 1 & nb \ 0 & 0 & 1 endpmatrix ,$$ but isn't a subgroup.
Since most reductive groups contain nonabelian unipotent subgroups, this will not be true for reductive groups either.
answered 8 hours ago
Will SawinWill Sawin
72.8k7 gold badges146 silver badges303 bronze badges
answered 8 hours ago
Will SawinWill Sawin
72.8k7 gold badges146 silver badges303 bronze badges
answered 8 hours ago
Will SawinWill Sawin
72.8k7 gold badges146 silver badges303 bronze badges
answered 8 hours ago
Will SawinWill Sawin
72.8k7 gold badges146 silver badges303 bronze badges
72.8k7 gold badges146 silver badges303 bronze badges
$begingroup$
Thank you! But I think it's true for torus. Is it true for any commutative groups?
$endgroup$
– sawdada
7 hours ago
2
$begingroup$
@sawdada I don't think it's too hard to prove this in characteristic zero using the exponential map. But probably not in characteristic $p$ for additive type groups as then the $n$th power depends only on $n$ mod $p$ so equations like $x = y^2p-1$ will provide counterexamples.
$endgroup$
– Will Sawin
6 hours ago
$begingroup$
@sawdada There is also a purely algebraic proof for toruses, and mroe generally for semiabelian varieties.
$endgroup$
– Will Sawin
4 hours ago
1
$begingroup$
@sawdada We can mod out the variety by the maximal subgroup that stabilizes $X$, and thus assume $X$ is not stable under translation by any nontrivial element. Our goal is then to show $X$ is the identity. We have an $n$th power map $[n]: X to X$. If the generic point has two preimages, then they are each a generic point of $X$, and differ by an $n$-torsion point, so the translate of a generic point of $X$ by an $n$-torsion point lies in $X$, so $X$ is stable under translation by the $n$-torsion point, contradicting the assumption.
$endgroup$
– Will Sawin
1 hour ago
1
$begingroup$
@sawdada So in fact the map has generic degree $1$ and thus has a generic section, hence has a section by normalcy. So the pullback of any class in $H^1$ of the group to $X$ is trivial. This is impossible unless $X$ is a point. (If the intersection of $X$ with the maximal subtorus has positive dimension, you can do this by induction on the dimension of the torus, and otherwise you can project to the abelian variety and use that the cohomology of the abelian variety is generated by $H^1$.)
$endgroup$
– Will Sawin
1 hour ago
|
show 1 more comment
$begingroup$
Thank you! But I think it's true for torus. Is it true for any commutative groups?
$endgroup$
– sawdada
7 hours ago
2
$begingroup$
@sawdada I don't think it's too hard to prove this in characteristic zero using the exponential map. But probably not in characteristic $p$ for additive type groups as then the $n$th power depends only on $n$ mod $p$ so equations like $x = y^2p-1$ will provide counterexamples.
$endgroup$
– Will Sawin
6 hours ago
$begingroup$
@sawdada There is also a purely algebraic proof for toruses, and mroe generally for semiabelian varieties.
$endgroup$
– Will Sawin
4 hours ago
1
$begingroup$
@sawdada We can mod out the variety by the maximal subgroup that stabilizes $X$, and thus assume $X$ is not stable under translation by any nontrivial element. Our goal is then to show $X$ is the identity. We have an $n$th power map $[n]: X to X$. If the generic point has two preimages, then they are each a generic point of $X$, and differ by an $n$-torsion point, so the translate of a generic point of $X$ by an $n$-torsion point lies in $X$, so $X$ is stable under translation by the $n$-torsion point, contradicting the assumption.
$endgroup$
– Will Sawin
1 hour ago
1
$begingroup$
@sawdada So in fact the map has generic degree $1$ and thus has a generic section, hence has a section by normalcy. So the pullback of any class in $H^1$ of the group to $X$ is trivial. This is impossible unless $X$ is a point. (If the intersection of $X$ with the maximal subtorus has positive dimension, you can do this by induction on the dimension of the torus, and otherwise you can project to the abelian variety and use that the cohomology of the abelian variety is generated by $H^1$.)
$endgroup$
– Will Sawin
1 hour ago
$begingroup$
Thank you! But I think it's true for torus. Is it true for any commutative groups?
$endgroup$
– sawdada
7 hours ago
$begingroup$
Thank you! But I think it's true for torus. Is it true for any commutative groups?
$endgroup$
– sawdada
7 hours ago
2
2
$begingroup$
@sawdada I don't think it's too hard to prove this in characteristic zero using the exponential map. But probably not in characteristic $p$ for additive type groups as then the $n$th power depends only on $n$ mod $p$ so equations like $x = y^2p-1$ will provide counterexamples.
$endgroup$
– Will Sawin
6 hours ago
$begingroup$
@sawdada I don't think it's too hard to prove this in characteristic zero using the exponential map. But probably not in characteristic $p$ for additive type groups as then the $n$th power depends only on $n$ mod $p$ so equations like $x = y^2p-1$ will provide counterexamples.
$endgroup$
– Will Sawin
6 hours ago
$begingroup$
@sawdada There is also a purely algebraic proof for toruses, and mroe generally for semiabelian varieties.
$endgroup$
– Will Sawin
4 hours ago
$begingroup$
@sawdada There is also a purely algebraic proof for toruses, and mroe generally for semiabelian varieties.
$endgroup$
– Will Sawin
4 hours ago
1
1
$begingroup$
@sawdada We can mod out the variety by the maximal subgroup that stabilizes $X$, and thus assume $X$ is not stable under translation by any nontrivial element. Our goal is then to show $X$ is the identity. We have an $n$th power map $[n]: X to X$. If the generic point has two preimages, then they are each a generic point of $X$, and differ by an $n$-torsion point, so the translate of a generic point of $X$ by an $n$-torsion point lies in $X$, so $X$ is stable under translation by the $n$-torsion point, contradicting the assumption.
$endgroup$
– Will Sawin
1 hour ago
$begingroup$
@sawdada We can mod out the variety by the maximal subgroup that stabilizes $X$, and thus assume $X$ is not stable under translation by any nontrivial element. Our goal is then to show $X$ is the identity. We have an $n$th power map $[n]: X to X$. If the generic point has two preimages, then they are each a generic point of $X$, and differ by an $n$-torsion point, so the translate of a generic point of $X$ by an $n$-torsion point lies in $X$, so $X$ is stable under translation by the $n$-torsion point, contradicting the assumption.
$endgroup$
– Will Sawin
1 hour ago
1
1
$begingroup$
@sawdada So in fact the map has generic degree $1$ and thus has a generic section, hence has a section by normalcy. So the pullback of any class in $H^1$ of the group to $X$ is trivial. This is impossible unless $X$ is a point. (If the intersection of $X$ with the maximal subtorus has positive dimension, you can do this by induction on the dimension of the torus, and otherwise you can project to the abelian variety and use that the cohomology of the abelian variety is generated by $H^1$.)
$endgroup$
– Will Sawin
1 hour ago
$begingroup$
@sawdada So in fact the map has generic degree $1$ and thus has a generic section, hence has a section by normalcy. So the pullback of any class in $H^1$ of the group to $X$ is trivial. This is impossible unless $X$ is a point. (If the intersection of $X$ with the maximal subtorus has positive dimension, you can do this by induction on the dimension of the torus, and otherwise you can project to the abelian variety and use that the cohomology of the abelian variety is generated by $H^1$.)
$endgroup$
– Will Sawin
1 hour ago
|
show 1 more comment
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