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Math Help - lie algebra question to show ado's theorem

  1. #1
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    lie algebra question to show ado's theorem

    L is a Lie algebra with basis \{x_1, ..., x_n\} and commutation relations [x_i, x_j] = C^{k}_{ij}x_{k} (*)
    show that the elements X_i = \sum_{j,k}C^{k}_{ji}a^{j}_{k} also satisfy (*) with the a_{ij} the usual basis elements for gl(n) satisfying
    [{a^{i}}_{j}, {a^{k}}_{l}] = \delta^{k}_{j}{a^{i}}_{l} - \delta^{i}_{l}{a^{k}}_{j}

    does this mean i have to show that
    [X_i, X_j] = C^{k}_{ij}X_{k} using the fact that X_i = \sum_{j,k}C^{k}_{ji}a^{j}_{k} (and thus X_j = \sum_{k,l}C^{l}_{kj}a^{k}_{l})? or am i completely on the wrong track?? does [X_i, X_j] = X_{i}X_{j} - X_{j}X_{i}? and how do i use the [{a^{i}}_{j}, {a^{k}}_{l}] = \delta^{k}_{j}{a^{i}}_{l} - \delta^{i}_{l}{a^{k}}_{j} identity if i AM supposed to proceed like that?

    any tips would be most appreciated
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  2. #2
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    Re: lie algebra question to show ado's theorem

    i think the question is actually this: if [x_i,x_j]=\sum_{k=1}^n c_{ij}^k x_k, and

    X_i = \sum_{r,s} c_{ri}^sa_{rs}, \ \ X_j=\sum_{u,v}c_{uj}^v a_{uv},

    then

    [X_i,X_j]=\sum_{k=1}^n c_{ij}^k X_k. \ \ \ \ \ \ \ \ (1)

    for all i,j. to prove this, first recall that the scalars c_{ij}^k are called the structure constants of the Lie algebra L and they satisfy the following relations:

    (i) c_{ij}^k = -c_{ji}^k, for all 1 \leq i,j,k \leq n.

    (ii) \sum_{s=1}^nc_{is}^qc_{jp}^s + c_{js}^qc_{pi}^s + c_{ps}^qc_{ij}^s = 0, for all 1 \leq i,j,p,q \leq n.

    now [a_{rs},a_{uv}]=\delta_{su}a_{rv}-\delta_{rv}a_{us} with (i) and (ii) will give us:

    [X_i,X_j]=\sum_{r,s,u,v} c_{ri}^sc_{uj}^v(\delta_{su}a_{rv}-\delta_{rv}a_{us}) = \sum_{r,s,v}c_{ri}^sc_{sj}^va_{rv}-\sum_{r,s,u}c_{ri}^sc_{uj}^ra_{us}=\sum_{p,q,s} (c_{sj}^qc_{pi}^s - c_{si}^qc_{pj}^s)a_{pq}=\sum_{p,q} \left (\sum_{s=1}^n c_{pi}^sc_{sj}^q-c_{si}^qc_{pj}^s \right)a_{pq}=\sum_{p,q,s}c_{ps}^qc_{ij}^sa_{pq}=
    \sum_{s=1}^n c_{ij}^s \sum_{p,q} c_{ps}^qa_{pq}=\sum_{s=1}^n c_{ij}^sX_s. \ \Box
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