13a - Lecture 13 Invariant Bilinear Forms on V N : V 1 : x...

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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.

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13a - Lecture 13 Invariant Bilinear Forms on V N : V 1 : x...

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