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Crystal operators

0

$begingroup$

Define the operator $s_i$ on tableaux:

Consider letters $i$ and $i + 1$ in row reading word of the tableau.

Successively “bracket” pairs of the form (i + 1, I ).

Left with word of the form $i^r (i + 1)^s$.
Then

$$s_i(i^r(i+1)^s)=i^s(i+1)^r$$

Show that:

(a) $s_i^2(b)=b$,

(b) $s_is_j(b)=s_js_i(b)$ if $|i-j|>1$

(c) $s_is_i+1s_i(b)=s_i+1s_is_i+1(b)$

Part (a) is clear. Beacuse after two operations, powers remain unchanged. Part (b) is also clear because $s_i$ and $s_j$ do not interfere with each other as long as $|i-j|>1$.

Part(c) is known as braid property. It seems to be a little messy to show this property.

combinatorics representation-theory algebraic-combinatorics

share|cite|improve this question

asked Mar 18 at 19:12

S_AlexS_Alex

1929

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What does the second line mean? I am confused by "...in row reading word of the tableau."
  $endgroup$
  – TomGrubb
  Mar 18 at 19:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Starting from the lowest row of tableau, write the numbers inserted in the boxes of the tableau from left to write. Then go to the upper row and so on so forth.
  $endgroup$
  – S_Alex
  Mar 18 at 19:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I find it a bit difficult to understand the presentation here, but this question feels familiar. Have you tried searching for earlier questions about braids?
  $endgroup$
  – Peter Taylor
  Mar 19 at 11:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  can you clarify what you mean by "bracketing"? Do you mean that if you have some subword of the form $(i,i,i+1,i,i+1)$, then $s_i$ acts according to the rule $(i+1,i+1,i+1,i,i)$? (so you don't change the $(i+1,i)$ entry?)
  $endgroup$
  – David Hill
  Mar 19 at 16:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If we have $(i+1,i+1,i,i)$, first we bracket $(i+1,i)$ then remove it from our sequence. Then $(i+1,i)$ remains. Again we bracket $(i+1,i)$ and remove it from the sequence.
  $endgroup$
  – S_Alex
  Mar 19 at 17:22
  

  
  
  
  

 | 
show 4 more comments

0

$begingroup$

Define the operator $s_i$ on tableaux:

Consider letters $i$ and $i + 1$ in row reading word of the tableau.

Successively “bracket” pairs of the form (i + 1, I ).

Left with word of the form $i^r (i + 1)^s$.
Then

$$s_i(i^r(i+1)^s)=i^s(i+1)^r$$

Show that:

(a) $s_i^2(b)=b$,

(b) $s_is_j(b)=s_js_i(b)$ if $|i-j|>1$

(c) $s_is_i+1s_i(b)=s_i+1s_is_i+1(b)$

Part (a) is clear. Beacuse after two operations, powers remain unchanged. Part (b) is also clear because $s_i$ and $s_j$ do not interfere with each other as long as $|i-j|>1$.

Part(c) is known as braid property. It seems to be a little messy to show this property.

combinatorics representation-theory algebraic-combinatorics

share|cite|improve this question

asked Mar 18 at 19:12

S_AlexS_Alex

1929

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What does the second line mean? I am confused by "...in row reading word of the tableau."
  $endgroup$
  – TomGrubb
  Mar 18 at 19:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Starting from the lowest row of tableau, write the numbers inserted in the boxes of the tableau from left to write. Then go to the upper row and so on so forth.
  $endgroup$
  – S_Alex
  Mar 18 at 19:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I find it a bit difficult to understand the presentation here, but this question feels familiar. Have you tried searching for earlier questions about braids?
  $endgroup$
  – Peter Taylor
  Mar 19 at 11:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  can you clarify what you mean by "bracketing"? Do you mean that if you have some subword of the form $(i,i,i+1,i,i+1)$, then $s_i$ acts according to the rule $(i+1,i+1,i+1,i,i)$? (so you don't change the $(i+1,i)$ entry?)
  $endgroup$
  – David Hill
  Mar 19 at 16:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If we have $(i+1,i+1,i,i)$, first we bracket $(i+1,i)$ then remove it from our sequence. Then $(i+1,i)$ remains. Again we bracket $(i+1,i)$ and remove it from the sequence.
  $endgroup$
  – S_Alex
  Mar 19 at 17:22
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

Define the operator $s_i$ on tableaux:

Consider letters $i$ and $i + 1$ in row reading word of the tableau.

Successively “bracket” pairs of the form (i + 1, I ).

Left with word of the form $i^r (i + 1)^s$.
Then

$$s_i(i^r(i+1)^s)=i^s(i+1)^r$$

Show that:

(a) $s_i^2(b)=b$,

(b) $s_is_j(b)=s_js_i(b)$ if $|i-j|>1$

(c) $s_is_i+1s_i(b)=s_i+1s_is_i+1(b)$

Part (a) is clear. Beacuse after two operations, powers remain unchanged. Part (b) is also clear because $s_i$ and $s_j$ do not interfere with each other as long as $|i-j|>1$.

Part(c) is known as braid property. It seems to be a little messy to show this property.

combinatorics representation-theory algebraic-combinatorics

share|cite|improve this question

asked Mar 18 at 19:12

S_AlexS_Alex

1929

$endgroup$

share|cite|improve this question

asked Mar 18 at 19:12

S_AlexS_Alex

1929

share|cite|improve this question

asked Mar 18 at 19:12

S_AlexS_Alex

1929

asked Mar 18 at 19:12

S_AlexS_Alex

1929

asked Mar 18 at 19:12

S_AlexS_Alex

1929