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?

  • 1
    $\begingroup$ Care to share a code snippet ? $\endgroup$
    – Sektor
    Dec 19, 2013 at 7:27
  • $\begingroup$ I don't have a code yet. I am looking for one, or writing one myself. :) $\endgroup$
    – wonderich
    Dec 19, 2013 at 19:22
  • $\begingroup$ Of course, I shall say the purpose here is definitely NOT to teach me on gauge invariant: which is obvious from the perspective of a finite continuous gauge transformation: $U^\dagger (A-i d) U$ on the C-S term Tr[$A \wedge A +(2/3)A \wedge A \wedge A$]. The purpose here is to teach Mathematica how to simplify nontrivial algebra. This is what I need. :) $\endgroup$
    – wonderich
    Dec 19, 2013 at 22:50


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