In page 488 of Peskin and Schroeder, it is stated (emphasis mine):
It is not difficult to check using (15.27) and (15.21) that, even for finite transformations, the covariant derivative has the same transformation law as the field on which it acts.
I was trying to indeed verify that.
This is what I have tried:
- $V\left(x\right)\in SU\left(2\right)^{\mathbb{R}^4}$
- $\psi\left(x\right)\mapsto V\left(x\right)\psi\left(x\right)$.
- $D_\mu\left(x\right)\equiv\partial_\mu-igA_{\mu,\,j}\left(x\right)\frac{\sigma^j}{2}$.
- $igA_{\mu,\,j}\left(x\right)\frac{\sigma^j}{2}\mapsto V\left(x\right)igA_{\mu,\,j}\left(x\right)\frac{\sigma^j}{2}\left[V\left(x\right)^\dagger\right]-V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)^\dagger\right]\right\}$
Thus:
\begin{align} D_\mu\left(x\right)\psi\left(x\right) \mapsto & \left\{\partial_\mu -V\left(x\right)igA_{\mu,\,j}\left(x\right)\frac{\sigma^j}{2}\left[V\left(x\right)^\dagger\right]+V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)^\dagger\right]\right\}\right\}V\left(x\right)\psi\left(x\right) = \\\ &= \left[\partial_\mu V\left(x\right)\right]\psi\left(x\right)+V\left(x\right)\partial_\mu\psi\left(x\right)-V\left(x\right)igA_{\mu,\,j}\left(x\right)\frac{\sigma^j}{2}\psi\left(x\right)+V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right)\psi\left(x\right) = & \\\ &= V\left(x\right)D_\mu\left(x\right)\psi\left(x\right)+\left\{\left[\partial_\mu V\left(x\right)\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right)\right\}\psi\left(x\right) \end{align}
So as far as I understand, $\boxed{\left[\partial_\mu V\left(x\right)\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right)\stackrel{?}{=}0}$ should be zero.
To prove that I have used the fact that $VV\dagger=1$:
\begin{align} \partial_\mu\left[ V\left(x\right)\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right) &= \\ \partial_\mu\left[ V\left(x\right)1\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right) &= \\ \partial_\mu\left[ V\left(x\right)V\left(x\right)^\dagger V\left(x\right)\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right) &= \\ \left[\partial_\mu V\left(x\right)\right]V\left(x\right)^\dagger V\left(x\right) +V\left(x\right)\left[ \partial_\mu V\left(x\right)^\dagger \right]V\left(x\right) +V\left(x\right)V\left(x\right)^\dagger\left[ \partial_\mu V\left(x\right)\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right) &= \\2\left\{\partial_\mu\left[ V\left(x\right)\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right)\right\}\end{align}
Is this all correct?