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# 13a - Lecture 13 Invariant Bilinear Forms on VN V1 x1 x2 y1...

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Lecture 13 Invariant Bilinear Forms on V N : V 1 : β x 1 x 2 , y 1 y 2 = x 1 y 2 - x 2 y 1 . V 2 : β ( x, y ) = Tr( xy ) x, y sl 2 To see that such β is invariant, compute β ( z · x, y ) and β ( x, z · y ) and add: β ( z · x, y ) = Tr([ z, x ] y ) = Tr( zxy ) - Tr( xzy ) β ( x, z · y ) = Tr( x [ z, y ]) = Tr( xzy ) - Tr( xyz ) β ( z · x, y ) + β ( x, z · y ) = Tr( zxy ) - Tr( xyz ) = 0 . V N : For general N , let v 0 , 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 β B ( V N , V N ) is invariant. Then by invariance under h , 0 = β ( h · v a , v b ) + β ( v a , h · v b ) = β (( N - 2 a ) v a , v b ) + β ( v a , ( N - 2 b ) v b ) = 2( N - ( a + b )) β ( v a , v b ) . Therefore, β ( v z , v b ) = 0 unless a + b = N. Using invariance under e - , we calculate λ n := β ( v N - n , v n ): n = β ( v N - n , v n ) = β ( v N - n , e - · v n - 1 ) = - β ( e - · v N - n , v n - 1 ) = - β (( N - n + 1) v N - n +1 , v n - 1 ) = - ( N - n + 1) λ n - 1 1

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So λ n satisfies the recurrence relation λ n = - N - n + 1 n λ n - 1 . Solving the recurrence relation gives λ n = ( - 1) n N n λ 0 . This formula for λ n implies λ N - n = ( - 1) N - n N N - n λ 0 = ( - 1) N λ n , so that β ( v n , v N - n ) = ( - 1) N β ( v N - n , v n ). From this, we see that β ( · , · ) is symmetric when- ever n is even and antisymmetric whenever n is odd.
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13a - Lecture 13 Invariant Bilinear Forms on VN V1 x1 x2 y1...

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