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Under an infinitesimal gauge transformation $g(x) = 1 - i\alpha{}_i(x)T{}^i$, where $[T{}^a, T{}^b] = if{}^{ab}{}_c T{}^c$, I want to know what happens to the Lagrangian $\mathcal{L} = F{}_{a\mu\nu}F{}_a{}^{\mu\nu}\delta{}^{ab} =: F^2$, with $\text{Tr}(T{}^a T{}^b) = \frac 12 \delta{}^{ab}$. I obtain

$F^2 \rightarrow F^2 + f{}^{ij}{}_a \alpha_i F{}_{j\mu\nu} F{}_k{}^{\mu\nu} \delta{}^{ak}$ .

How is the last term vanishing? Or is the Yang--Mills Lagrangian not gauge-invariant under infinitesimal transformations?

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$$ F^2 \to F^2 + \alpha_i f^{ijk} F_{j\mu\nu} F_k^{\mu\nu} $$ $f^{ijk}$ is completely antisymmetric in all its indices and is being contracted with something that is symmetric in $jk$. Thus, the action is invariant.

We can see that $f^{ijk}$ is totally antisymmetric as follows $$ [T^i , T^j] = f^{ij}{}_k T^k \implies \text{tr} \left( [ T^i , T^j ] T^k \right) = f^{ij}{}_\ell \text{tr} \left( T^\ell T^k \right) = f^{ij}{}_\ell \delta^{\ell k } = f^{ijk} $$ Thus, $$ f^{ikj} = \text{tr} \left( [ T^i , T^k ] T^j \right) = \text{tr} \left( [ T^j , T^i ] T^k \right) = - \text{tr} \left( [ T^i , T^j ] T^k \right) = - f^{ijk} $$ The second equality above is due to cyclity of the trace. Thus, $f^{ijk} = f^{i[jk]}$. But clearly $f^{ijk} = f^{[ij]k}$. Together this implies $f^{ijk} = f^{[ijk]}$.

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  • $\begingroup$ Yes, I forgot that $\delta{}^{ab}$ is the Killing metric on the Lie algebra. Is there a simple proof that $f{}^{ijk} = f{}^{[ijk]}$? $\endgroup$
    – Jens
    May 5, 2016 at 18:27
  • $\begingroup$ @Jens - Answer edited. $\endgroup$
    – Prahar
    May 5, 2016 at 20:42
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    $\begingroup$ Another way to show it: the trace of a comutator always vanishes. $\endgroup$
    – Jens
    May 8, 2016 at 4:21

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