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Why is the Complete Flag Variety an algebraic variety?

How is this lower bound of the dimension of a homogeneous algebraic variety reached?Bruhat decomposition of flag variety.Generalization of Bézout theorem in higher dimension.Coset variety of an algebraic groupUnion of affine varieities is a projective variety?A basic question on the complete flag varietyA certain generalization of flag varieties$GL_n(F)$ acts on the flag varietyIrreducibility of “almost” flag varietyUnipotent action on flag variety

3

$begingroup$

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

share|cite|improve this question

edited Dec 25 '18 at 11:46

Matt Samuel

38.8k63769

asked Mar 26 '17 at 20:52

MaffredMaffred

2,680625

$endgroup$

add a comment | 

3

$begingroup$

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

share|cite|improve this question

edited Dec 25 '18 at 11:46

Matt Samuel

38.8k63769

asked Mar 26 '17 at 20:52

MaffredMaffred

2,680625

$endgroup$

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

share|cite|improve this question

edited Dec 25 '18 at 11:46

Matt Samuel

38.8k63769

asked Mar 26 '17 at 20:52

MaffredMaffred

2,680625

$endgroup$

share|cite|improve this question

edited Dec 25 '18 at 11:46

Matt Samuel

38.8k63769

asked Mar 26 '17 at 20:52

MaffredMaffred

2,680625

share|cite|improve this question

edited Dec 25 '18 at 11:46

Matt Samuel

38.8k63769

asked Mar 26 '17 at 20:52

MaffredMaffred

2,680625

edited Dec 25 '18 at 11:46

Matt Samuel

38.8k63769

edited Dec 25 '18 at 11:46

Matt Samuel

38.8k63769

asked Mar 26 '17 at 20:52

MaffredMaffred

2,680625

asked Mar 26 '17 at 20:52

MaffredMaffred

2,680625

1 Answer
1

active

oldest

votes

3

$begingroup$

because of that identification we can put a variety (projective) structure on it...comes from projective structure of $textrmGL(n,mathbbC)/B_n$.

there is another identification that this collection of complete flags can be thought of as inside product of $mathbbG(1,n) times mathbbG(2,n) times cdots times mathbbG(n-1,n)$
where $mathbbG(r,n)$ Grassmanian variety .

share|cite|improve this answer

edited Mar 10 at 19:05

davyjones

401313

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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3

$begingroup$

because of that identification we can put a variety (projective) structure on it...comes from projective structure of $textrmGL(n,mathbbC)/B_n$.

there is another identification that this collection of complete flags can be thought of as inside product of $mathbbG(1,n) times mathbbG(2,n) times cdots times mathbbG(n-1,n)$
where $mathbbG(r,n)$ Grassmanian variety .

share|cite|improve this answer

edited Mar 10 at 19:05

davyjones

401313

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 | 

3

$begingroup$

because of that identification we can put a variety (projective) structure on it...comes from projective structure of $textrmGL(n,mathbbC)/B_n$.

there is another identification that this collection of complete flags can be thought of as inside product of $mathbbG(1,n) times mathbbG(2,n) times cdots times mathbbG(n-1,n)$
where $mathbbG(r,n)$ Grassmanian variety .

share|cite|improve this answer

edited Mar 10 at 19:05

davyjones

401313

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 $textrmGL(n,mathbbC)/B_n$.

there is another identification that this collection of complete flags can be thought of as inside product of $mathbbG(1,n) times mathbbG(2,n) times cdots times mathbbG(n-1,n)$
where $mathbbG(r,n)$ Grassmanian variety .

share|cite|improve this answer

edited Mar 10 at 19:05

davyjones

401313

answered Mar 26 '17 at 21:02

Pinaki SahaPinaki Saha

312

$endgroup$

share|cite|improve this answer

edited Mar 10 at 19:05

davyjones

401313

answered Mar 26 '17 at 21:02

Pinaki SahaPinaki Saha

312

edited Mar 10 at 19:05

davyjones

401313

edited Mar 10 at 19:05

davyjones

401313

answered Mar 26 '17 at 21:02

Pinaki SahaPinaki Saha

312

answered Mar 26 '17 at 21:02

Pinaki SahaPinaki Saha

312