Given a Lie algebra $L$ with bracket given by:
$$
[T_a, T_b] = f_{ab}^c T_c
$$
the map $\phi : L \to \text{End}(V)$, where $V = R^n$ is defined as:
$$
\phi(T_a)_{bc} = f_{ac}^b
$$
where $bc$ denotes the $bc$-element of $\phi(T_a)$. In order to prove that this is a representation of $L$, I use the definition:
$$
\phi([T_i, T_j]) = [\phi(T_i), \phi(T_j)]
$$
Using the first equation for the LHS, I obtain:
$$
\phi([T_i, T_j]) = \phi(f_{ij}^a T_a) = f_{ij}^k \phi(T_k) = f_{ij}^k \left(f_{kc}^b\right)_{bc} = \left( f_{ij}^k f_{kc}^b \right)_{bc}
$$
where $(.)_{bc}$ is a matrix with elements $bc$. Since $f_{ij}^k$ is just a number (if indices are given), i figured that the second step is justified. For the RHS, I obtain:
$$
[\phi(T_i), \phi(T_j)] = [(f_{ic}^b)_{bc}, (f_{jc}^b )_{bc}] = \left( f_{ik}^b f_{jc}^k – f_{jk}^b f_{ic}^k \right)_{bc}
$$
So in order for LHS = RHS, I get the tensor equation:
$$
f_{ij}^k f_{kc}^b = f_{ik}^b f_{jc}^k – f_{jk}^b f_{ic}^k
$$
At this part, I'm completely stuck. Am I at all on the right track? It looks to me that the equality won't hold unless further constrictions are made on $f_{ab}^c$. I tried assuming that $f$ is anti-symmetric, but even then, I couldn't switch indices in such a way that the equality would hold.
Best Answer
Observe that, by definition, $\phi(T)$ is the matrix corresponding to the linear transformation $v\mapsto [T,v]$, with $v\in L$, where $L$ is viewed as a vector space with basis $\{T_i\}_i$. This representation of $L$ has a special name, it is the adjoint representation. Normally one writes $\phi(T)v=\mathrm{ad}(T)(v)$. That the adjoint representation is indeed a representation is equivalent to the Jacobi identity, and this equivalence you can prove without using any indices at all, or even working with a fixed basis.