Ley de transformación para la derivada covariante en SU(2)SU(2)SU(2) Yang-Mills

En la página 488 de Peskin y Schroeder, se afirma (énfasis mío):

No es difícil comprobar utilizando (15.27) y (15.21) que, incluso para transformaciones finitas , la derivada covariante tiene la misma ley de transformación que el campo sobre el que actúa.

Estaba tratando de verificar eso.

Esto es lo que he probado:

1. $V\left(x\right)\in SU\left(2\right)^{\mathbb{R}^4}$
2. $\psi\left(x\right)\mapsto V\left(x\right)\psi\left(x\right)$.
3. $D_\mu\left(x\right)\equiv\partial_\mu-igA_{\mu,\,j}\left(x\right)\frac{\sigma^j}{2}$.
4. $igA_{\mu,\,j}\left(x\right)\frac{\sigma^j}{2}\mapsto V\left(x\right)igA_{\mu,\,j}\left(x\right)\frac{\sigma^j}{2}\left[V\left(x\right)^\dagger\right]-V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)^\dagger\right]\right\}$

De este modo:

$$\begin{align} D_\mu\left(x\right)\psi\left(x\right) \mapsto & \left\{\partial_\mu -V\left(x\right)igA_{\mu,\,j}\left(x\right)\frac{\sigma^j}{2}\left[V\left(x\right)^\dagger\right]+V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)^\dagger\right]\right\}\right\}V\left(x\right)\psi\left(x\right) = \\\ &= \left[\partial_\mu V\left(x\right)\right]\psi\left(x\right)+V\left(x\right)\partial_\mu\psi\left(x\right)-V\left(x\right)igA_{\mu,\,j}\left(x\right)\frac{\sigma^j}{2}\psi\left(x\right)+V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right)\psi\left(x\right) = & \\\ &= V\left(x\right)D_\mu\left(x\right)\psi\left(x\right)+\left\{\left[\partial_\mu V\left(x\right)\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right)\right\}\psi\left(x\right) \end{align}$$

Entonces, según tengo entendido,$\boxed{\left[\partial_\mu V\left(x\right)\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right)\stackrel{?}{=}0}$debe ser cero.

Para probar que he usado el hecho de que$VV\dagger=1$:

$$\begin{align} \partial_\mu\left[ V\left(x\right)\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right) &= \\ \partial_\mu\left[ V\left(x\right)1\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right) &= \\ \partial_\mu\left[ V\left(x\right)V\left(x\right)^\dagger V\left(x\right)\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right) &= \\ \left[\partial_\mu V\left(x\right)\right]V\left(x\right)^\dagger V\left(x\right) +V\left(x\right)\left[ \partial_\mu V\left(x\right)^\dagger \right]V\left(x\right) +V\left(x\right)V\left(x\right)^\dagger\left[ \partial_\mu V\left(x\right)\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right) &= \\2\left\{\partial_\mu\left[ V\left(x\right)\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right)\right\}\end{align}$$

¿Es todo esto correcto?