Explanation of Puzzle models

> The paper Triple multiplicities for sl(r + 1) and the spectrum of the exterior algebra of the adjoint representation by Bernstein and Zelevinsky introduce a triangle to count the to the Richardson--Littlewood coefficient (=the number of such triangles)

The three sides are the informatioin from Young diagrams involved (or the coefficient in front of the roots). One important condition is that for each hexagon , we have a +b = x + y,a + c = x + z,b + c = y + z. 
 
> I believed that the indices are just the number of boxes of i inserted in the j-th row (so is a record of Richardson--Littlewood rule). But I am a little confused with the numbering now (the proof in the paper is a direct construction of bijection). 
 
> The main trick is, the condition on hexagon can be translated geometrically, that a,b,c,x,y,z can be the lengths of sides of a hexagon whose angles are all 120° (maybe should be called the fundamental theorem of honeycomb :-D) 
 
> If we put the indices in the triangle to the honeycomb itself, and make the number be the length, we will get the honeycomb model, 
=> 
> The more interesting thing is the puzzle. The Poincar\'e duality
 
> Then let us draw them up with the given length (note that it is not mirror-symmetric) 
=>
and fulfill the rest by puzzles, we get the model of puzzles. 

More examples

 

 

 

 

 

 New picture

 

 

 

 

 

 

 
 
posted @ 2021-01-04 17:33  XiongRui  阅读(77)  评论(0)    收藏  举报