notation - Gauge-covariance of the Yang-Mills field strength $F_{munu}^a$

Accordingly to Yang-Mills theories, after the introduction of a covariant derivative such that

$$D_\mu = \partial_\mu - igA_\mu, \tag1$$

you can built the kinetic term for the gauge potential $A_\mu$ as

$${\cal L}_A = -\frac{1}{2}tr\{F_{\mu \nu}F^{\mu \nu}\},\qquad F_{\mu \nu} = \frac{i}{g}[D_\mu, D_\nu]. \tag2$$

The action of the covariant derivative transforms under a local group tranformation $\Omega$ in the following way:

$$D_\mu\psi \rightarrow \Omega D_\mu\psi. \tag3$$

And the gauge field as

$$A_\mu \rightarrow \Omega A_\mu\Omega^\dagger + \frac{i}{g}\Omega\partial_\mu \Omega^\dagger . \tag4$$

Introducing Eq. (4) in Eq. (1) you get that under the group tranformations, the covariant derivate transforms as,

$$D_\mu \rightarrow \partial_\mu - ig\Omega A_\mu\Omega^\dagger + \Omega\partial_\mu\Omega^\dagger = \Omega D_\mu\Omega^\dagger + \partial_\mu . \tag5$$

Eq. (5) into definition of strength tensor in Eq. (2) gives

$$F_{\mu\nu} \rightarrow \Omega F_{\mu\nu}\Omega^\dagger + \frac{i}{g}(\partial_\mu(\Omega D_\nu\Omega^\dagger) - \partial_\nu(\Omega D_\mu\Omega^\dagger)) . \tag6$$

Taking $\Omega \simeq 1 - g\theta^iT^i$, with $\{T^i\}$ the set of generators of the group, $\theta^i \in \mathbb{R}$ and $f^{ijk}$ the structure constants:

$$F_{\mu\nu} \rightarrow \Omega F_{\mu\nu}\Omega^\dagger + \frac{i}{g}(\partial_\mu A_\nu^k - \partial_\nu A_\mu^k)T^k + ig^2f^{ajk}T^k[\partial_\mu(\theta^jA_\nu^a) - \partial_\nu(\theta^jA_\mu^a)] . \tag7$$

In Particle Physics books, it is said that

$$F_{\mu\nu} \rightarrow \Omega F_{\mu\nu}\Omega^\dagger \tag{8},$$ but I don't get that in my calculus (Eq. (7)). What am I doing wrong? This result is important because if $F_{\mu\nu}$ doesn't transforms as books say, the kinetic term in Eq. (2) isn't gauge invariant.

Answer

Ok, I think I found the answer. Eq. (5) has to be written as:

$$D_\mu \rightarrow \partial_\mu + \Omega(\partial_\mu \Omega^\dagger) - ig\Omega A_\mu\Omega^\dagger \tag{A}$$

As you can deduce from Eq. (4) that is a consequence of imposing $D_\mu\psi \rightarrow \Omega D_\mu\psi$. Nevertheless, you can re-write Eq. (A) as:

$$D_\mu \rightarrow (\Omega\partial_\mu)\Omega^\dagger - ig\Omega A_\mu\Omega^\dagger = \Omega D_\mu\Omega^\dagger \tag{B}$$

After considering,

$$(\Omega\partial_\mu)\Omega^\dagger = \Omega(\partial_\mu \Omega^\dagger) + \Omega\Omega^\dagger\partial_\mu = \Omega(\partial_\mu \Omega^\dagger) + \partial_\mu \tag{C}$$

So with this, everything makes sense and the strength tensor is perfectly gauge-covariant, keeping in mind Eqs. (B)-(C) when interpreting OP's Eq. (8).