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  1. #1
    Member vernal's Avatar
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    Question in lie algebra

    Please Help

    1. Let L be a complex Lie algebra. Show that L is solvable if and only if L' is nilpotent.

    2. Let L be a Lie algebra and let A be a subalgebra of L. The normaliser
    of A, denoted NL(A), is defined by
    NL(A) = {x ∈ L : [x, a] ∈ A for all a ∈ A} .

    (i) Prove that NL(A) is a subalgebra of L and that NL(A) contains
    A. Show moreover that NL(A) is the largest subalgebra of
    L in which A is an ideal.

    (ii) Let L = gl(n,C) and let A be the subalgebra of L consisting of
    all diagonal matrices. Show that NL(A) = A.
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  2. #2
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    Re: Question in lie algebra

    Quote Originally Posted by vernal View Post
    Please Help

    1. Let L be a complex Lie algebra. Show that L is solvable if and only if L' is nilpotent.

    2. Let L be a Lie algebra and let A be a subalgebra of L. The normaliser
    of A, denoted NL(A), is defined by
    NL(A) = {x ∈ L : [x, a] ∈ A for all a ∈ A} .

    (i) Prove that NL(A) is a subalgebra of L and that NL(A) contains
    A. Show moreover that NL(A) is the largest subalgebra of
    L in which A is an ideal.

    (ii) Let L = gl(n,C) and let A be the subalgebra of L consisting of
    all diagonal matrices. Show that NL(A) = A.
    there is nothing interesting about your second question. it is very straightforward. so i'll answer your first question for now and i'll help you with the second later if you tried and still couldn't do it.

    for any lie algebra L let L^{(1)}=L^1=L' = [L,L] and inductively define L^{(n+1)}=[L^{(n)},L^{(n)}] and L^{n+1}=[L^n,L]. an easy induction shows that (L')^n \supseteq L^{(n+1)}. thus if L' is nilpotent, then L is solvable.
    conversely, suppose that L is solvable. then, by Lie's theorem, L has a basis \mathfrak{B} such that [\text{ad}(a)]_{\mathfrak{B}}, the matrix of \text{ad}(a) with respect to \mathfrak{B}, is upper triangular for every a \in L. it is easy to see that if A and B are two upper triangular matrices, then [A,B]=AB-BA is strictly upper triangular and hence nilpotent. thus \text{ad}([a,b])=[\text{ad}(a), \text{ad}(b)] is nilpotent for all a,b \in L. so we have proved that \text{ad}(c) is nilpotent for all c \in L' and hence, by Engel's theorem, L' is nilpotent.
    Last edited by NonCommAlg; December 18th 2011 at 01:27 AM.
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  3. #3
    Member vernal's Avatar
    Joined
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    Re: Question in lie algebra

    Quote Originally Posted by NonCommAlg View Post
    there is nothing interesting about your second question. it is very straightforward. so i'll answer your first question for now and i'll help you with the second later if you tried and still couldn't do it.

    for any lie algebra L let L^{(1)}=L^1=L' = [L,L] and inductively define L^{(n+1)}=[L^{(n)},L^{(n)}] and L^{n+1}=[L^n,L]. an easy induction shows that (L')^n \supseteq L^{(n+1)}. thus if L' is nilpotent, then L is solvable.
    conversely, suppose that L is solvable. then, by Lie's theorem, L has a basis \mathfrak{B} such that [\text{ad}(a)]_{\mathfrak{B}}, the matrix of \text{ad}(a) with respect to \mathfrak{B}, is upper triangular for every a \in L. it is easy to see that if A and B are two upper triangular matrices, then [A,B]=AB-BA is strictly upper triangular and hence nilpotent. thus \text{ad}([a,b])=[\text{ad}(a), \text{ad}(b)] is nilpotent. so we have proved that \text{ad}(c) is nilpotent for all c \in L' and therefore, by Engel's theorem, L' is nilpotent.
    tanks... You explain very well. tank you very much.
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