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Unformatted text preview: Lecture 13 Invariant Bilinear Forms on V N : V 1 : x 1 x 2 , y 1 y 2 = x 1 y 2 x 2 y 1 . V 2 : ( x,y ) = Tr( xy ) x,y sl 2 To see that such is invariant, compute ( z x,y ) and ( x,z y ) and add: ( z x,y ) = Tr([ z,x ] y ) = Tr( zxy ) Tr( xzy ) ( x,z y ) = Tr( x [ z,y ]) = Tr( xzy ) Tr( xyz ) ( z x,y ) + ( x,z y ) = Tr( zxy ) Tr( xyz ) = 0 . V N : For general N , let v ,v 1 , ,v N be a standard basis for V N so that h v n = ( N 2 n ) v n , e + v n = ( N n + 1) v n 1 e v n = ( n + 1) v n +1 Suppose B ( V N ,V N ) is invariant. Then by invariance under h , 0 = ( h v a ,v b ) + ( v a ,h v b ) = (( N 2 a ) v a ,v b ) + ( v a , ( N 2 b ) v b ) = 2( N ( a + b )) ( v a ,v b ) . Therefore, ( v z ,v b ) = 0 unless a + b = N. Using invariance under e , we calculate n := ( v N n ,v n ): n n = ( v N n ,v n ) = ( v N n ,e v n 1 ) = ( e v N n ,v n 1 ) = (( N n + 1) v N n +1 ,v n 1 ) = ( N n + 1) n 1 1 So n satisfies the recurrence relation n = N n + 1 n n 1 . Solving the recurrence relation gives n = ( 1) n N n ....
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This note was uploaded on 07/08/2011 for the course MAT 6932 taught by Professor Staff during the Summer '10 term at University of Florida.
 Summer '10
 Staff
 Algebra

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