# Infinitesimal gauge invariance of Yang--Mills Lagrangian

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?

$$F^2 \to F^2 + \alpha_i f^{ijk} F_{j\mu\nu} F_k^{\mu\nu}$$ $f^{ijk}$ 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 $f^{ijk} = f^{[ijk]}$.
• Yes, I forgot that $\delta{}^{ab}$ is the Killing metric on the Lie algebra. Is there a simple proof that $f{}^{ijk} = f{}^{[ijk]}$? – Jens May 5 '16 at 18:27