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0

$begingroup$

Let $G = mathrmGL_n(k)$ for some field $k$. We have $G = coprod_win WBwB$, where $B leq G$ is a Borel subgroup (group of upper triangular matrices) and $W$ is a set of permutation matrices (or more generally, Weyl group of $G$).

I'm trying to prove this myself, without using any hard theory of algebraic groups. I proved that the double cosets are disjoint in a purely combinatorial way. I proved that $Bw_1 cap w_2B = emptyset$ for any $w_1neq w_2$, by showing that for any $b_1, b_2in B$, there exists $i, j$ where $(i, j)$-th entry of $b_1 w_1$ is nonzero but the same entry $w_2b_2$ is zero. However, I can't show that their union is the whole $G$.
I saw that for general reductive groups, one can show that the union forms a subgroup of $G$ by showing that $bw_1b'w_2b'' in Bw_3B$ for some $w_3$, and consider their dimensions.
However, I want to find some elementary proof without using any algebraic geometry, but only uses linear algebra or some combinatorial way.
Is there any such proof?
I know how to prove it for $n =2$, but this method (just case by case) is hard to be applied for $n geq 3$.

algebraic-groups

share|cite|improve this question

asked Mar 27 at 20:03

Seewoo LeeSeewoo Lee

7,2692930

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, see this MO-question, or in this textbook.
  $endgroup$
  – Dietrich Burde
  Mar 27 at 20:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DietrichBurde Thanks! I needed the proof something like in that textbook.
  $endgroup$
  – Seewoo Lee
  Mar 28 at 0:13
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $G = mathrmGL_n(k)$ for some field $k$. We have $G = coprod_win WBwB$, where $B leq G$ is a Borel subgroup (group of upper triangular matrices) and $W$ is a set of permutation matrices (or more generally, Weyl group of $G$).

I'm trying to prove this myself, without using any hard theory of algebraic groups. I proved that the double cosets are disjoint in a purely combinatorial way. I proved that $Bw_1 cap w_2B = emptyset$ for any $w_1neq w_2$, by showing that for any $b_1, b_2in B$, there exists $i, j$ where $(i, j)$-th entry of $b_1 w_1$ is nonzero but the same entry $w_2b_2$ is zero. However, I can't show that their union is the whole $G$.
I saw that for general reductive groups, one can show that the union forms a subgroup of $G$ by showing that $bw_1b'w_2b'' in Bw_3B$ for some $w_3$, and consider their dimensions.
However, I want to find some elementary proof without using any algebraic geometry, but only uses linear algebra or some combinatorial way.
Is there any such proof?
I know how to prove it for $n =2$, but this method (just case by case) is hard to be applied for $n geq 3$.

algebraic-groups

share|cite|improve this question

asked Mar 27 at 20:03

Seewoo LeeSeewoo Lee

7,2692930

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, see this MO-question, or in this textbook.
  $endgroup$
  – Dietrich Burde
  Mar 27 at 20:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DietrichBurde Thanks! I needed the proof something like in that textbook.
  $endgroup$
  – Seewoo Lee
  Mar 28 at 0:13
  

  
  
  
  

add a comment | 

$begingroup$

Let $G = mathrmGL_n(k)$ for some field $k$. We have $G = coprod_win WBwB$, where $B leq G$ is a Borel subgroup (group of upper triangular matrices) and $W$ is a set of permutation matrices (or more generally, Weyl group of $G$).

I'm trying to prove this myself, without using any hard theory of algebraic groups. I proved that the double cosets are disjoint in a purely combinatorial way. I proved that $Bw_1 cap w_2B = emptyset$ for any $w_1neq w_2$, by showing that for any $b_1, b_2in B$, there exists $i, j$ where $(i, j)$-th entry of $b_1 w_1$ is nonzero but the same entry $w_2b_2$ is zero. However, I can't show that their union is the whole $G$.
I saw that for general reductive groups, one can show that the union forms a subgroup of $G$ by showing that $bw_1b'w_2b'' in Bw_3B$ for some $w_3$, and consider their dimensions.
However, I want to find some elementary proof without using any algebraic geometry, but only uses linear algebra or some combinatorial way.
Is there any such proof?
I know how to prove it for $n =2$, but this method (just case by case) is hard to be applied for $n geq 3$.

algebraic-groups

share|cite|improve this question

asked Mar 27 at 20:03

Seewoo LeeSeewoo Lee

7,2692930

$endgroup$

share|cite|improve this question

asked Mar 27 at 20:03

Seewoo LeeSeewoo Lee

7,2692930

share|cite|improve this question

asked Mar 27 at 20:03

Seewoo LeeSeewoo Lee

7,2692930

asked Mar 27 at 20:03

Seewoo LeeSeewoo Lee

7,2692930

asked Mar 27 at 20:03

Seewoo LeeSeewoo Lee

7,2692930