# $SU(2)$ gauge symmetry

Take the Lagrangian with one fermion: $$\mathcal{L} = -\frac{1}{4}F^{\mu\nu}_aF^a_{\mu\nu} + \bar{\psi}(i\gamma^\mu D_\mu - m)\psi$$ where the gauge covariant derivative $D_\mu = \partial_\mu+i\frac{g}{2}t^aW^a_\mu$. The Lagrangian is invariant under a local $SU(2)$ transformation: $$\psi(x) \rightarrow \exp \left[-i\theta^a(x)t^a \right]\psi(x)$$ $$W^a_\mu(x) \rightarrow W^a_\mu(x) +\frac{1}{g}\partial_\mu\theta^a(x) + \epsilon^{abc}\theta^b(x)W^c_\mu(x)$$

Often, we say that $W_\mu^a$ transforms according to the adjoint representation of $SU(2)$ but how can we say that based on the previous equation?

Note that the finite transformation of: $$W^a_\mu \to W^a_\mu + \frac{1}{g} \partial_\mu \theta^a + \epsilon^{abc} \theta^b W^c_\mu$$ is: $$W^a_\mu t^a \to g W_\mu^a t^a g^{-1} + \frac{i}{g} \partial_\mu g \tag{1}$$ where: $$g = \exp(-i \theta^a t^a) \;\;\; \text{and} \;\;\; [t^a,t^b] = i \epsilon^{abc} t^c$$ Thus, the first term on the right-hand side of equation $(1)$ transforms under the adjoint representation of the Lie group. The second term does not transform under the adjoint representation, but it should be easy to verify that the transformed gauge field still takes values in the Lie algebra (hint: looking at infinitesimal transformations is the easiest method to verify this).
• @KoObO I am not really familiar with Dynkin indices to be honest. However, a field that takes values in the Lie algebra, i.e. $\phi \in \mathfrak{g}$, will transform under the adjoint representation as $\phi \to g \phi g^{-1}$ and will take again values in the adjoint representation, i.e. $g \phi g^{-1} \in \mathfrak{g}$. The easiest way to understand this is by look at my answer given here. The whole point of the transformation property of the gauge field is that – Hunter Apr 18 '14 at 9:01