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

1. ## 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

2. ## 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:

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