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Math Help - Lie algebras

  1. #1
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    Lie algebras

    I need to solve two assignments in Lie algebras. These assignments are not very difficult, but my knowledges in Lie algebras aren't good.
    1. Let be a derivation of the Lie algebra . Show that if commutes with every inner derivation, then ()C(), where C() denotes the centre of .

    2. Let x gl(n,F) have n distinct eigenvalues 1..n in F. Prove that eigenvalues of ad are the n scalars - (1i,jn)

    Your prompt reply will be highly appreciated
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  2. #2
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    Quote Originally Posted by Pivych View Post
    I need to solve two assignments in Lie algebras. These assignments are not very difficult, but my knowledges in Lie algebras aren't good.
    1. Let be a derivation of the Lie algebra . Show that if commutes with every inner derivation, then ()C(), where C() denotes the centre of .

    2. Let x gl(n,F) have n distinct eigenvalues 1..n in F. Prove that eigenvalues of ad are the n scalars - (1i,jn)

    Your prompt reply will be highly appreciated
    let L be our Lie algebra. i will write ad_a for inner derivations. so we are given that \delta ad_a(b) = ad_a \delta(b) for all a,b \in L. that means \delta[a,b]=[a,\delta(b)] for all a,b \in L. call this (1). we also have, by definition of a derivation,  \delta[a,b]=[\delta(a),b] + [a,\delta(b)]. call this (2). now (1) and (2) gives you [\delta(a),b]=0 for all a,b \in L. that means \delta(a) is central for all a \in L and we're done.
    for the second question use the fact that a matrix with distinct eigenvalues is diagonalizable. so suppose that \{v_1, \ldots , v_n \} is a basis for F^n such that the matrix of x in this basis is diagonal. that means x(v_i)=\lambda_i v_i for all i. now for any 1 \leq i,j \leq n define e_{ij} \in gl(n,F) by e_{ij}(v_k)=\delta_{jk}v_i for all k where \delta_{jk} is the Kronecker delta. see that these e_{ij} make a basis for gl(n,F) and

    ad_x e_{ij}(v_k)=(xe_{ij}-e_{ij}x)(v_k)=(\lambda_i - \lambda_k)\delta_{jk}v_i=(\lambda_i - \lambda_j)e_{ij}(v_k).

    hence ad_xe_{ij}=(\lambda_i - \lambda_j)e_{ij} and so the matrix of ad_x relative to the basis \{e_{ij}, \ 1 \leq i,j \leq n \} is diagonal and its diagonal entries are \lambda_i - \lambda_j.
    Last edited by NonCommAlg; May 8th 2011 at 07:14 PM.
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