What is a general linear subspace?
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When it comes to defining a general plane with rispect to a line in $mathbb R^3$, I can think at this definition as : take any plane not containing the line.
Reading Fulton's "Young Tableau" I can't understand this situation in a proper way: we work in the Grassmanian $Gr(V,d)$ and have two schubert varieties $Omega_{lambda}(F_{bullet})$ and $Omega_{lambda}(tilde F_{bullet})$, where $tilde F_{bullet}$ is the opposite flag to $F_{bullet}$. We take a $textbf{general linear subspace}$ $L subset V$ to define another Schubert variety $Omega(L)$.
What does he exactely mean with $textbf{general}$ linear subspace? Is there any relation with the given flags or?
Thanks in advance!
linear-algebra algebraic-geometry algebraic-topology intersection-theory schubert-calculus
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add a comment |
$begingroup$
When it comes to defining a general plane with rispect to a line in $mathbb R^3$, I can think at this definition as : take any plane not containing the line.
Reading Fulton's "Young Tableau" I can't understand this situation in a proper way: we work in the Grassmanian $Gr(V,d)$ and have two schubert varieties $Omega_{lambda}(F_{bullet})$ and $Omega_{lambda}(tilde F_{bullet})$, where $tilde F_{bullet}$ is the opposite flag to $F_{bullet}$. We take a $textbf{general linear subspace}$ $L subset V$ to define another Schubert variety $Omega(L)$.
What does he exactely mean with $textbf{general}$ linear subspace? Is there any relation with the given flags or?
Thanks in advance!
linear-algebra algebraic-geometry algebraic-topology intersection-theory schubert-calculus
$endgroup$
$begingroup$
I think that here "general" means "any", "arbitrarily chosen"
$endgroup$
– avs
Feb 14 '17 at 19:17
add a comment |
$begingroup$
When it comes to defining a general plane with rispect to a line in $mathbb R^3$, I can think at this definition as : take any plane not containing the line.
Reading Fulton's "Young Tableau" I can't understand this situation in a proper way: we work in the Grassmanian $Gr(V,d)$ and have two schubert varieties $Omega_{lambda}(F_{bullet})$ and $Omega_{lambda}(tilde F_{bullet})$, where $tilde F_{bullet}$ is the opposite flag to $F_{bullet}$. We take a $textbf{general linear subspace}$ $L subset V$ to define another Schubert variety $Omega(L)$.
What does he exactely mean with $textbf{general}$ linear subspace? Is there any relation with the given flags or?
Thanks in advance!
linear-algebra algebraic-geometry algebraic-topology intersection-theory schubert-calculus
$endgroup$
When it comes to defining a general plane with rispect to a line in $mathbb R^3$, I can think at this definition as : take any plane not containing the line.
Reading Fulton's "Young Tableau" I can't understand this situation in a proper way: we work in the Grassmanian $Gr(V,d)$ and have two schubert varieties $Omega_{lambda}(F_{bullet})$ and $Omega_{lambda}(tilde F_{bullet})$, where $tilde F_{bullet}$ is the opposite flag to $F_{bullet}$. We take a $textbf{general linear subspace}$ $L subset V$ to define another Schubert variety $Omega(L)$.
What does he exactely mean with $textbf{general}$ linear subspace? Is there any relation with the given flags or?
Thanks in advance!
linear-algebra algebraic-geometry algebraic-topology intersection-theory schubert-calculus
linear-algebra algebraic-geometry algebraic-topology intersection-theory schubert-calculus
edited Jan 5 at 11:19


Matt Samuel
37.8k63665
37.8k63665
asked Feb 14 '17 at 19:05


MaffredMaffred
2,680625
2,680625
$begingroup$
I think that here "general" means "any", "arbitrarily chosen"
$endgroup$
– avs
Feb 14 '17 at 19:17
add a comment |
$begingroup$
I think that here "general" means "any", "arbitrarily chosen"
$endgroup$
– avs
Feb 14 '17 at 19:17
$begingroup$
I think that here "general" means "any", "arbitrarily chosen"
$endgroup$
– avs
Feb 14 '17 at 19:17
$begingroup$
I think that here "general" means "any", "arbitrarily chosen"
$endgroup$
– avs
Feb 14 '17 at 19:17
add a comment |
1 Answer
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$begingroup$
Usually in algebraic geometry, one says that something is generic if it holds for every choice outside a well defined (but maybe not easy to determine) proper Zariski closed set depending on the problem.
For instance, the quadratic equation $ax^2+bx+c=0$ has two distinct solutions for generic choice of $a,b,c$ because the cases where it does not have two distinct solutions are the ones satisfying $b^2−4ac=0$ that is a Zariski closed subset of the space (say $mathbb{C}^3$) where $(a,b,c)$ lives.
One can think of "general" as "randomly chosen", or "almost every", with respect to some honest probability (technically, any measure that is absolutely continuous with respect to the Lebesgue measure would work).
In this context, I would guess that there is some Zariski open condition that the linear subspace $L$ should satisfy in order for the construction to work. Maybe some transversal intersection property?
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Usually in algebraic geometry, one says that something is generic if it holds for every choice outside a well defined (but maybe not easy to determine) proper Zariski closed set depending on the problem.
For instance, the quadratic equation $ax^2+bx+c=0$ has two distinct solutions for generic choice of $a,b,c$ because the cases where it does not have two distinct solutions are the ones satisfying $b^2−4ac=0$ that is a Zariski closed subset of the space (say $mathbb{C}^3$) where $(a,b,c)$ lives.
One can think of "general" as "randomly chosen", or "almost every", with respect to some honest probability (technically, any measure that is absolutely continuous with respect to the Lebesgue measure would work).
In this context, I would guess that there is some Zariski open condition that the linear subspace $L$ should satisfy in order for the construction to work. Maybe some transversal intersection property?
$endgroup$
add a comment |
$begingroup$
Usually in algebraic geometry, one says that something is generic if it holds for every choice outside a well defined (but maybe not easy to determine) proper Zariski closed set depending on the problem.
For instance, the quadratic equation $ax^2+bx+c=0$ has two distinct solutions for generic choice of $a,b,c$ because the cases where it does not have two distinct solutions are the ones satisfying $b^2−4ac=0$ that is a Zariski closed subset of the space (say $mathbb{C}^3$) where $(a,b,c)$ lives.
One can think of "general" as "randomly chosen", or "almost every", with respect to some honest probability (technically, any measure that is absolutely continuous with respect to the Lebesgue measure would work).
In this context, I would guess that there is some Zariski open condition that the linear subspace $L$ should satisfy in order for the construction to work. Maybe some transversal intersection property?
$endgroup$
add a comment |
$begingroup$
Usually in algebraic geometry, one says that something is generic if it holds for every choice outside a well defined (but maybe not easy to determine) proper Zariski closed set depending on the problem.
For instance, the quadratic equation $ax^2+bx+c=0$ has two distinct solutions for generic choice of $a,b,c$ because the cases where it does not have two distinct solutions are the ones satisfying $b^2−4ac=0$ that is a Zariski closed subset of the space (say $mathbb{C}^3$) where $(a,b,c)$ lives.
One can think of "general" as "randomly chosen", or "almost every", with respect to some honest probability (technically, any measure that is absolutely continuous with respect to the Lebesgue measure would work).
In this context, I would guess that there is some Zariski open condition that the linear subspace $L$ should satisfy in order for the construction to work. Maybe some transversal intersection property?
$endgroup$
Usually in algebraic geometry, one says that something is generic if it holds for every choice outside a well defined (but maybe not easy to determine) proper Zariski closed set depending on the problem.
For instance, the quadratic equation $ax^2+bx+c=0$ has two distinct solutions for generic choice of $a,b,c$ because the cases where it does not have two distinct solutions are the ones satisfying $b^2−4ac=0$ that is a Zariski closed subset of the space (say $mathbb{C}^3$) where $(a,b,c)$ lives.
One can think of "general" as "randomly chosen", or "almost every", with respect to some honest probability (technically, any measure that is absolutely continuous with respect to the Lebesgue measure would work).
In this context, I would guess that there is some Zariski open condition that the linear subspace $L$ should satisfy in order for the construction to work. Maybe some transversal intersection property?
answered Feb 14 '17 at 21:38
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$begingroup$
I think that here "general" means "any", "arbitrarily chosen"
$endgroup$
– avs
Feb 14 '17 at 19:17