Proving that the kinetic energy of the gluon fields $G_{\mu}^{a}$ is $SU(3)$-invariant? In QCD, we introduce the gluon-fields $G_{\mu}^{a}$ when defining the covariant derivative $D_{\mu} \equiv \partial_{\mu} + ig_{S}G_{\mu}^{a}T^{a}$ to make the free-particle Lagrangian $
\mathcal L_{\text{free}} \equiv \bar{\psi}\left( i\gamma^{\mu}D_{\mu} - m\right)\psi$ $SU\left( 3\right)$-invariant. For the invariance to hold, the gluon fields $G_{\mu}^{a}$ have to transform as following: $$\left( G^{a}_{\mu} \right)' = G^{a}_{\mu} - \partial_{\mu}\alpha^{a}\left( x\right) - g_{S}f^{abc}\alpha^{b}\left(x\right)G_{\mu}^{c} \qquad [1]$$
Now, we also need a kinetic term for the gluon fields $G_{\mu}^{a}$:
$$\mathcal L_{\text{kin}} \equiv -\frac{1}{4}G^{a}_{\mu\nu}G^{a\mu\nu}, \qquad G^{a}_{\mu\nu}\equiv\partial_{\mu}G_{\nu}^{a}-\partial_{\nu}G_{\mu}^{a}-g_{S} f^{abc}G_{\mu}^{b}G_{\nu}^{c} \qquad [2].$$
I would like to show the $SU(3)$-invariance of $\mathcal L_{\text{kin}}$, but I fail along the way. I think my biggest problem is that I wrote down the complete expressions for $G^{a}_{\mu\nu}$ and also $G^{a\mu\nu}$, but I get really long expressions and I'm not sure what to do with these. I have terms $\propto \alpha^{c}G^{d\mu}\alpha^{d}G^{e\nu}$, and the same just with the indices $\mu$ and $\nu$ down, and I'm not sure what to do with them.
Could sb please guide me through the proof?
 A: This is quite a tedious computation, if you want to perform it this way. I would propose an alternative approach (but if you're really interested in doing the calculation in this particular manner, I'll just erase this answer). My line of reasoning is the following: it is easily provable that $\mathcal{L}_\mathrm{kin}$ is invariant under the gauge transformation $(G_{\mu})^\prime= \mathsf{U} G_{\mu} \mathsf{U}^{\dagger}-\mathrm{i}/{g_S}\left(\partial_{\mu} \mathsf{U}\right) \mathsf{U}^{\dagger}$. What you wrote in [1] is actually the corresponding infinitesimal transformation of $G_\mu^a$ from $G_\mu=\sum_a G_\mu^a t^a$ (derivation sketched below, if interested). Thus, it seems reasonable to state that since $\mathcal{L}_\mathrm{kin}$ is invariant under the whole gauge transformation $\mathsf{U}=\mathsf{exp}\{\mathrm{i}g_S\alpha\}$, it should remain invariant under the infinitesimal $\mathsf{U}\approx 1+\mathrm{i}g_S\alpha$, which gives the transformation from [1].
(Gauge invariance under $\mathsf{SU}(3)$)
For a $\mathsf{SU}(3)$ transformation $\mathsf{U}=\mathsf{exp}\{\mathrm{i}g_S\alpha\}$, it is known that the gauge fields, which are elements of the Lie group $\mathsf{SU}(3)$, transform as
$$(G_{\mu})^\prime= \mathsf{U} G_{\mu} \mathsf{U}^{\dagger}-\frac{\mathrm{i}}{g_S}\left(\partial_{\mu} \mathsf{U}\right) \mathsf{U}^{\dagger}\tag{1}\label{gaugetransf}$$
whereas the field strength tensor $G_{\mu\nu}\equiv \partial_\mu G_\nu-\partial_\nu G_\mu+\mathrm{i}g_S[G_\mu,G_\nu]$ gauge transforms as
$$(G_{\mu \nu})^\prime = \mathsf{U} G_{\mu \nu} \mathsf{U}^{\dagger}$$
The kinetic term
$$\mathcal{L}_\mathrm{kin}=-\frac{1}{2}\mathsf{Tr}\{G_{\mu \nu}G^{\mu \nu}\}\tag{2}\label{lkin1}$$
is obviously gauge invariant due to the invariance of the trace under cyclic permutations and unitarity of the gauge transformation. This are all known facts.
(Infinitesimal gauge invariance for $\mathfrak{su}(3)$-valued gauge fields)
Instead of working with the gauge fields as elements of the group $\mathsf{SU}(3)$, it is customary to work with them in a particular representation of the associated algebra $\mathfrak{su}(3)$. As any Lie algebra, $\mathfrak{su}(3)$ is just the tangent vector space around identity at the underlying manifold of $\mathsf{SU}(3)$. If $t_a$ are the generators of the $\mathfrak{su}(3)$ algebra, they form a basis. Thus, any $\mathfrak{su}(3)$-valued gauge field may be expanded as
$$G_\mu=\sum_a G_\mu^a t^a$$
where of course $G_\mu$ will be representation dependent but $G_\mu^a$ won't. Consequently
$$G_{\mu\nu}=\sum_a G_{\mu\nu}^a t^a\quad\mathrm{where}\quad G_{\mu\nu}^a=\partial_{\mu} G_{\nu}^{a}-\partial_{\nu} G_{\mu}^{a}-g_S f^{a b c} G_{\mu}^{b} G_{\nu}^{c}$$
The last result may easily be deduced by using the $\mathfrak{su}(3)$ commutation relations $[t^a,t^b]=\mathrm{i}f^{abc}t^c$. This is precisely your equation [2].
Working infinitesimally and expanding the gauge transformation around identity, namely $\mathsf{U}\approx 1+\mathrm{i}g_S\alpha$ with $\alpha=\sum_a\alpha^a t^a$, the gauge transformation of the fields from \eqref{gaugetransf} becomes
$$(G_\mu)^\prime\approx G_\mu-\partial_\mu\alpha+\mathrm{i}g_S[\alpha,G_\mu]$$
which after using $G_\mu=\sum_a G_\mu^at^a$ and $\alpha=\sum_a \alpha^a t^a$ along with $[t^b,t^c]=-\mathrm{i}f^{abc}t^a$ gives exactly the transformation you wrote in [1]
$$(G^{a}_{\mu})^\prime\approx G^{a}_{\mu} - \partial_{\mu}\alpha^{a}\left( x\right) - g_{S}f^{abc}\alpha^{b}\left(x\right)G_{\mu}^{c}\tag{4}\label{infgaugetransf}$$
Since the kinetic Lagrangian from \eqref{lkin1} is gauge invariant under the $\mathsf{SU}(3)$ gauge transformation \eqref{gaugetransf}, the same Lagrangian expressed as
$$\mathcal{L}_\mathrm{kin}=-\frac{1}{2}G_{\mu\nu}^aG^{a\mu\nu}\underbrace{\mathsf{Tr}\{t^at^a\}}_{\displaystyle 1/2}$$
remains invariant under the infinitesimal variant of the gauge transformation of \eqref{gaugetransf}, as written in \eqref{infgaugetransf}.
