(Euclidean) open orbit in an irreducible real algebraic set
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Let $tau:GL(n,mathbbR) rightarrow GL(V)$ be a rational representation of the general linear group of degree $n$ on a finite-dimensional real vector space $V$. Let $C$ be an irreducible real algebraic set in $V$ such that the action of $GL(n,mathbbR)$ on $V$ induced by $tau$ leaves the set $C$ invariant.
We endow $C$ with the subspace topology inherited from (Hausdorf) Euclidean topology of $V$. Supose there is a point $vin C$ such that the orbit $GL(n,mathbbR)⋅v$ is an open subset of $C$.
My questions is: is it true that the orbit $GL(n,mathbbR)⋅v$ must be a Zariski-open subset of $C$?
ag.algebraic-geometry algebraic-groups group-actions real-algebraic-geometry
asked Aug 19 at 21:10
AleAlvAlwaysDIEZ
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up vote
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Let $tau:GL(n,mathbbR) rightarrow GL(V)$ be a rational representation of the general linear group of degree $n$ on a finite-dimensional real vector space $V$. Let $C$ be an irreducible real algebraic set in $V$ such that the action of $GL(n,mathbbR)$ on $V$ induced by $tau$ leaves the set $C$ invariant.
We endow $C$ with the subspace topology inherited from (Hausdorf) Euclidean topology of $V$. Supose there is a point $vin C$ such that the orbit $GL(n,mathbbR)⋅v$ is an open subset of $C$.
My questions is: is it true that the orbit $GL(n,mathbbR)⋅v$ must be a Zariski-open subset of $C$?
ag.algebraic-geometry algebraic-groups group-actions real-algebraic-geometry
asked Aug 19 at 21:10
AleAlvAlwaysDIEZ
261
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Let $tau:GL(n,mathbbR) rightarrow GL(V)$ be a rational representation of the general linear group of degree $n$ on a finite-dimensional real vector space $V$. Let $C$ be an irreducible real algebraic set in $V$ such that the action of $GL(n,mathbbR)$ on $V$ induced by $tau$ leaves the set $C$ invariant.
We endow $C$ with the subspace topology inherited from (Hausdorf) Euclidean topology of $V$. Supose there is a point $vin C$ such that the orbit $GL(n,mathbbR)⋅v$ is an open subset of $C$.
My questions is: is it true that the orbit $GL(n,mathbbR)⋅v$ must be a Zariski-open subset of $C$?
ag.algebraic-geometry algebraic-groups group-actions real-algebraic-geometry
asked Aug 19 at 21:10
AleAlvAlwaysDIEZ
261
Let $tau:GL(n,mathbbR) rightarrow GL(V)$ be a rational representation of the general linear group of degree $n$ on a finite-dimensional real vector space $V$. Let $C$ be an irreducible real algebraic set in $V$ such that the action of $GL(n,mathbbR)$ on $V$ induced by $tau$ leaves the set $C$ invariant.
We endow $C$ with the subspace topology inherited from (Hausdorf) Euclidean topology of $V$. Supose there is a point $vin C$ such that the orbit $GL(n,mathbbR)⋅v$ is an open subset of $C$.
My questions is: is it true that the orbit $GL(n,mathbbR)⋅v$ must be a Zariski-open subset of $C$?
ag.algebraic-geometry algebraic-groups group-actions real-algebraic-geometry
ag.algebraic-geometry algebraic-groups group-actions real-algebraic-geometry
asked Aug 19 at 21:10
AleAlvAlwaysDIEZ
261
asked Aug 19 at 21:10
AleAlvAlwaysDIEZ
261
asked Aug 19 at 21:10
AleAlvAlwaysDIEZ
261
asked Aug 19 at 21:10
AleAlvAlwaysDIEZ
261
asked Aug 19 at 21:10
AleAlvAlwaysDIEZ
261
261
add a comment |Â
add a comment |Â
1 Answer
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No, not at all. Take for $V$ the space of quadratic forms on $mathbb R^n$, let $v=sum_ix_i^2$ be the standard form, and $C=V$. Then $GL(n,mathbb R)v$ is the set of positive definite forms. So, it is Hausdorff open but not Zariski open. NB: This example even works for $n=1$.
answered Aug 19 at 21:38
Friedrich Knop
10.5k23054
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
No, not at all. Take for $V$ the space of quadratic forms on $mathbb R^n$, let $v=sum_ix_i^2$ be the standard form, and $C=V$. Then $GL(n,mathbb R)v$ is the set of positive definite forms. So, it is Hausdorff open but not Zariski open. NB: This example even works for $n=1$.
answered Aug 19 at 21:38
Friedrich Knop
10.5k23054
add a comment |Â
up vote
7
down vote
No, not at all. Take for $V$ the space of quadratic forms on $mathbb R^n$, let $v=sum_ix_i^2$ be the standard form, and $C=V$. Then $GL(n,mathbb R)v$ is the set of positive definite forms. So, it is Hausdorff open but not Zariski open. NB: This example even works for $n=1$.
answered Aug 19 at 21:38
Friedrich Knop
10.5k23054
add a comment |Â
up vote
7
down vote
up vote
7
down vote
No, not at all. Take for $V$ the space of quadratic forms on $mathbb R^n$, let $v=sum_ix_i^2$ be the standard form, and $C=V$. Then $GL(n,mathbb R)v$ is the set of positive definite forms. So, it is Hausdorff open but not Zariski open. NB: This example even works for $n=1$.
answered Aug 19 at 21:38
Friedrich Knop
10.5k23054
No, not at all. Take for $V$ the space of quadratic forms on $mathbb R^n$, let $v=sum_ix_i^2$ be the standard form, and $C=V$. Then $GL(n,mathbb R)v$ is the set of positive definite forms. So, it is Hausdorff open but not Zariski open. NB: This example even works for $n=1$.
answered Aug 19 at 21:38
Friedrich Knop
10.5k23054
answered Aug 19 at 21:38
Friedrich Knop
10.5k23054
answered Aug 19 at 21:38
Friedrich Knop
10.5k23054
answered Aug 19 at 21:38
Friedrich Knop
10.5k23054
10.5k23054
add a comment |Â
add a comment |Â
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