How to show that the structure constant of SU(3) is invariant? So suppose $f_{ijk}$ is the antisymmetric structure constant of SU(3), and $D^8_{ij}(g)$ is the matrices of 8-dimensional adjoint representation of SU(3), then how to show that 
$f_{ijk}$=$D^8_{il}(g)$$D^8_{jm}(g)$$D^8_{kn}(g)$$f_{lmn}$ which shows that the structure constant is indeed invariant.
 A: Well, the relation holds far more generally than SU(3) for the adjoint. 
In fact, you can see it by the respective similarity transformations of the defining (the 3) in the trace defining f,
$$
f_{lek}= \frac{-i}{2} \operatorname{tr} ([\lambda_l,\lambda_j],\lambda_k).
$$
That is, the transformations $\lambda_k\mapsto D^{3~\dagger}(g)\lambda_lD^3(g)$ leave the trace invariant for unitary matrices.
If you just want reassurance strictly within the 8, OK, consider the unitary
$$D^8_{il} =\exp(1\!\!1 +i\theta\cdot T^8)_{il}= \delta_{il}+ \theta^a f_{ail} +... $$ 
You want reassurance, e.g., of invariance to lowest order in θ,
$$
(\delta_{il} +\theta^a f_{ail})(\delta_{jm} +\theta^b f_{bjm})(\delta_{kn} +\theta^c f_{ckn})f_{lmn} = f_{ijk}+  \theta^a (f_{ail}f_{lek}+f_{ajm}f_{imk}+ f_{akn}f_{ion}  ) + O(\theta^2).
$$
But the O(θ) term vanishes by the full antisymmetry of the fs and the Jacobi identity,
$$= f_{ijk} + O(\theta^2),
$$
and, of course, to higher orders, as well, so the expression reduces to just $f_{ijk}$, as inquired.
