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Unformatted text preview: Lecture 10 We show that there are infinitely many nonisomorphic lie algebras g such that Dim g = 3 and Dim[ g , g ] = 2. We start by fixing an element z ∈ g [ g , g ], and examining the map ad z : [ g , g ] → [ g , g ]. We have already seen that [ g , g ] is abelian, and that ad z is a bijection from [ g , g ] to itself. We split into two cases. Case 1 ad z is not diagonalizable, in which case we find that g is unique up to isomorphism. Case 2 ad z is diagonalizable. In this case choose a basis B = { x,y } for [ g , g ] such that ad z ( x ) = λx and ad z ( y ) = μy for some λ,μ ∈ F . Claim 1. Let g and g be as in Case 2, with corresponding eigenvalues λ,μ and λ ,μ respectively. Then g ≡ g if and only if { λ μ , μ λ } = { λ μ , μ λ } . Proof. First, we show that for a lie algebra g , the set { λ μ , μ λ } is uniquely deter mined by g . Let w ∈ g [ g , g ], and consider ad x . We see that w = ax + by + cz , with c 6 = 0. Therefore, since [ g , g ] is abelian, we have [ ax + by + cz,x ] = [ cz,x ] = cλx and [ ax + by + cz,y ] = [ cz,y ] = cμy. Therefore, x and y are eigenvalues with corresponding eigenvalues cλ and cμ . This shows that the set of ratios of eigenvalues { cλ cμ , cμ cλ } = { λ μ , μ λ } is uniquely determined by g . Now, suppose that φ : g → g is a lie algebra isomorphism. Then since φ (ad z ( v )) = ad φ ( z ) we see that φ ◦ ad z = ad φ ( z ) ◦ φ , which implies that ad φ ( z ) = φ ◦ ad z ◦ φ 1 . Therefore, since ad z and ad φ ( z ) are conjugate, they have the same eigenvalues and hence { λ μ , μ λ } = { λ μ , μ λ } ....
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 Summer '10
 Staff
 Linear Algebra, Algebra, Eigenvalue, eigenvector and eigenspace, Orthogonal matrix, Lie algebra, e+

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