Suppose we have the following Lagrangian density:
$$ L=\epsilon^{\mu\nu\rho}\big(\sum_a A^a_{\mu}(x) \partial_\nu A^a_{\rho}(x)-\sum_{a,b,c}\frac{1}{3} f^{bca} A^a_{\mu}(x) A^b_{\nu}(x) A^c_{\rho}(x)\big) $$ under this infinitesimal transformation + $\dots$ $$ A^a_\mu(x) \to {A^a_\mu}'(x)\equiv (A^a_\mu(x) + f^{abc} \alpha^b(x) A^c_\mu(x) + \partial_\mu \alpha^a(x) +\dots) $$ where $x\equiv(t,x_1,x_2)$, $\epsilon^{\mu\nu\rho}$ is anti-symmetric and cyclic, $f^{abc}$ is anti-symmetric, but may not be cyclic in general.
We end up with $L \to L'$ under $A^a_\mu \to {A^a_\mu}'$.
And my question is:
How to obtain a well-simplified $L'$ using Mathematica ?
The key point is: without knowing the detailed structure $f^{abc}$, but only implement $f^{abc}$'s property to simplify the answer $L'$.
ps. I just took a glimpse at this post -how-to-manipulate-gauge-theory-in-mathematica, but I wonder whether there is a simpler way, since I am only doing pure algebra?
