Why is the Complete Flag Variety an algebraic variety?
$begingroup$
Let $V$ be a $mathbb C $ - vector space of dimension $n$.
Let's consider the set $Fl(n)$ of all the complete flags $F_{bullet}$ $$F_1 subset F_2 cdots subset F_n$$ where the $F_i$ are subspaces of $V$ with $dim(F_i)=i$ for every $1 leq i leq n$.
Why is this an affine/projective variety? I know that given that we can use the transitive action of $GL(n, mathbb C)$ and deduce that $$Fl(n) simeq GL(n, mathbb C)/B_n$$ where $B_n$ is the subgroup of the upper triangular matrices. But first we need to show that it is an algebraic variety.
Thanks!
linear-algebra algebraic-geometry lie-groups schubert-calculus
edited Dec 25 '18 at 11:46
Matt Samuel
39.2k63770
asked Mar 26 '17 at 20:52
MaffredMaffred
2,670625
$endgroup$
add a comment |
$begingroup$
Let $V$ be a $mathbb C $ - vector space of dimension $n$.
Let's consider the set $Fl(n)$ of all the complete flags $F_{bullet}$ $$F_1 subset F_2 cdots subset F_n$$ where the $F_i$ are subspaces of $V$ with $dim(F_i)=i$ for every $1 leq i leq n$.
Why is this an affine/projective variety? I know that given that we can use the transitive action of $GL(n, mathbb C)$ and deduce that $$Fl(n) simeq GL(n, mathbb C)/B_n$$ where $B_n$ is the subgroup of the upper triangular matrices. But first we need to show that it is an algebraic variety.
Thanks!
linear-algebra algebraic-geometry lie-groups schubert-calculus
edited Dec 25 '18 at 11:46
Matt Samuel
39.2k63770
asked Mar 26 '17 at 20:52
MaffredMaffred
2,670625
$endgroup$
add a comment |
$begingroup$
Let $V$ be a $mathbb C $ - vector space of dimension $n$.
Let's consider the set $Fl(n)$ of all the complete flags $F_{bullet}$ $$F_1 subset F_2 cdots subset F_n$$ where the $F_i$ are subspaces of $V$ with $dim(F_i)=i$ for every $1 leq i leq n$.
Why is this an affine/projective variety? I know that given that we can use the transitive action of $GL(n, mathbb C)$ and deduce that $$Fl(n) simeq GL(n, mathbb C)/B_n$$ where $B_n$ is the subgroup of the upper triangular matrices. But first we need to show that it is an algebraic variety.
Thanks!
linear-algebra algebraic-geometry lie-groups schubert-calculus
edited Dec 25 '18 at 11:46
Matt Samuel
39.2k63770
asked Mar 26 '17 at 20:52
MaffredMaffred
2,670625
$endgroup$
Let $V$ be a $mathbb C $ - vector space of dimension $n$.
Let's consider the set $Fl(n)$ of all the complete flags $F_{bullet}$ $$F_1 subset F_2 cdots subset F_n$$ where the $F_i$ are subspaces of $V$ with $dim(F_i)=i$ for every $1 leq i leq n$.
Why is this an affine/projective variety? I know that given that we can use the transitive action of $GL(n, mathbb C)$ and deduce that $$Fl(n) simeq GL(n, mathbb C)/B_n$$ where $B_n$ is the subgroup of the upper triangular matrices. But first we need to show that it is an algebraic variety.
Thanks!
linear-algebra algebraic-geometry lie-groups schubert-calculus
linear-algebra algebraic-geometry lie-groups schubert-calculus
edited Dec 25 '18 at 11:46
Matt Samuel
39.2k63770
asked Mar 26 '17 at 20:52
MaffredMaffred
2,670625
edited Dec 25 '18 at 11:46
Matt Samuel
39.2k63770
asked Mar 26 '17 at 20:52
MaffredMaffred
2,670625
edited Dec 25 '18 at 11:46
Matt Samuel
39.2k63770
edited Dec 25 '18 at 11:46
Matt Samuel
39.2k63770
edited Dec 25 '18 at 11:46
Matt Samuel
39.2k63770
39.2k63770
asked Mar 26 '17 at 20:52
MaffredMaffred
2,670625
asked Mar 26 '17 at 20:52
MaffredMaffred
2,670625
asked Mar 26 '17 at 20:52
MaffredMaffred
2,670625
2,670625
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
because of that identification we can put a variety (projective) structure on it...comes from projective structure of $textrm{GL}(n,mathbb{C})/B_n$.
there is another identification that this collection of complete flags can be thought of as inside product of $mathbb{G}(1,n) times mathbb{G}(2,n) times cdots times mathbb{G}(n-1,n)$
where $mathbb{G}(r,n)$ Grassmanian variety .
edited Mar 10 at 19:05
davyjones
403313
answered Mar 26 '17 at 21:02
Pinaki SahaPinaki Saha
312
$endgroup$
$begingroup$
I like the second identification! Why is it an algebraic subset of that product?
$endgroup$
– Maffred
Mar 26 '17 at 21:09
2
$begingroup$
Because you can encode the containment of subspaces as polynomial equations in the Plucker coordinates of the subspaces.
$endgroup$
– Mariano Suárez-Álvarez
Mar 26 '17 at 21:45
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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oldest
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active
oldest
votes
$begingroup$
because of that identification we can put a variety (projective) structure on it...comes from projective structure of $textrm{GL}(n,mathbb{C})/B_n$.
there is another identification that this collection of complete flags can be thought of as inside product of $mathbb{G}(1,n) times mathbb{G}(2,n) times cdots times mathbb{G}(n-1,n)$
where $mathbb{G}(r,n)$ Grassmanian variety .
edited Mar 10 at 19:05
davyjones
403313
answered Mar 26 '17 at 21:02
Pinaki SahaPinaki Saha
312
$endgroup$
$begingroup$
I like the second identification! Why is it an algebraic subset of that product?
$endgroup$
– Maffred
Mar 26 '17 at 21:09
2
$begingroup$
Because you can encode the containment of subspaces as polynomial equations in the Plucker coordinates of the subspaces.
$endgroup$
– Mariano Suárez-Álvarez
Mar 26 '17 at 21:45
add a comment |
$begingroup$
because of that identification we can put a variety (projective) structure on it...comes from projective structure of $textrm{GL}(n,mathbb{C})/B_n$.
there is another identification that this collection of complete flags can be thought of as inside product of $mathbb{G}(1,n) times mathbb{G}(2,n) times cdots times mathbb{G}(n-1,n)$
where $mathbb{G}(r,n)$ Grassmanian variety .
edited Mar 10 at 19:05
davyjones
403313
answered Mar 26 '17 at 21:02
Pinaki SahaPinaki Saha
312
$endgroup$
$begingroup$
I like the second identification! Why is it an algebraic subset of that product?
$endgroup$
– Maffred
Mar 26 '17 at 21:09
2
$begingroup$
Because you can encode the containment of subspaces as polynomial equations in the Plucker coordinates of the subspaces.
$endgroup$
– Mariano Suárez-Álvarez
Mar 26 '17 at 21:45
add a comment |
$begingroup$
because of that identification we can put a variety (projective) structure on it...comes from projective structure of $textrm{GL}(n,mathbb{C})/B_n$.
there is another identification that this collection of complete flags can be thought of as inside product of $mathbb{G}(1,n) times mathbb{G}(2,n) times cdots times mathbb{G}(n-1,n)$
where $mathbb{G}(r,n)$ Grassmanian variety .
edited Mar 10 at 19:05
davyjones
403313
answered Mar 26 '17 at 21:02
Pinaki SahaPinaki Saha
312
$endgroup$
because of that identification we can put a variety (projective) structure on it...comes from projective structure of $textrm{GL}(n,mathbb{C})/B_n$.
there is another identification that this collection of complete flags can be thought of as inside product of $mathbb{G}(1,n) times mathbb{G}(2,n) times cdots times mathbb{G}(n-1,n)$
where $mathbb{G}(r,n)$ Grassmanian variety .
edited Mar 10 at 19:05
davyjones
403313
answered Mar 26 '17 at 21:02
Pinaki SahaPinaki Saha
312
edited Mar 10 at 19:05
davyjones
403313
edited Mar 10 at 19:05
davyjones
403313
edited Mar 10 at 19:05
davyjones
403313
403313
answered Mar 26 '17 at 21:02
Pinaki SahaPinaki Saha
312
answered Mar 26 '17 at 21:02
Pinaki SahaPinaki Saha
312
answered Mar 26 '17 at 21:02
Pinaki SahaPinaki Saha
312
312
$begingroup$
I like the second identification! Why is it an algebraic subset of that product?
$endgroup$
– Maffred
Mar 26 '17 at 21:09
2
$begingroup$
Because you can encode the containment of subspaces as polynomial equations in the Plucker coordinates of the subspaces.
$endgroup$
– Mariano Suárez-Álvarez
Mar 26 '17 at 21:45
add a comment |
$begingroup$
I like the second identification! Why is it an algebraic subset of that product?
$endgroup$
– Maffred
Mar 26 '17 at 21:09
2
$begingroup$
Because you can encode the containment of subspaces as polynomial equations in the Plucker coordinates of the subspaces.
$endgroup$
– Mariano Suárez-Álvarez
Mar 26 '17 at 21:45
$begingroup$
I like the second identification! Why is it an algebraic subset of that product?
$endgroup$
– Maffred
Mar 26 '17 at 21:09
$begingroup$
I like the second identification! Why is it an algebraic subset of that product?
$endgroup$
– Maffred
Mar 26 '17 at 21:09
2
2
$begingroup$
Because you can encode the containment of subspaces as polynomial equations in the Plucker coordinates of the subspaces.
$endgroup$
– Mariano Suárez-Álvarez
Mar 26 '17 at 21:45
$begingroup$
Because you can encode the containment of subspaces as polynomial equations in the Plucker coordinates of the subspaces.
$endgroup$
– Mariano Suárez-Álvarez
Mar 26 '17 at 21:45
add a comment |
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