With $X= X^at_a$, we have the following notation : $D_\mu X = [D_\mu,X] = \partial_\mu X -ig [A_\mu, X]$

The Bianchi identities are written :

$$D_\lambda F_{\mu\nu} + D_\nu F_{\lambda\mu} + D_\mu F_{\nu\lambda} = 0 \tag{1}$$
We may choose $\lambda, \mu, \nu = 0,1,2$, so we have :

$$D_0 F_{12} + D_2 F_{01} + D_1 F_{20} = 0 \tag{2}$$

From the definition of $j$, we have :
$$j^0=F_{12}, j^1=F_{20}, j^2=F_{01}\tag{3}$$

From $(2)$ and $3$, we get :

$$D_\mu j^\mu = D_0j^0+D_1j^1 +D_2j^2 = 0\tag{4}$$

That is :

$$\partial_\mu j^\mu - ig[A_\mu, j^\mu]=0\tag{5}$$

Now, we may look at the $U(2)$ coordinates $(j^\mu)^a$ of $j^\mu$, we get :

$$\partial_\mu (j^\mu)^a +gf^{abc}(A_\mu)_b (j^\mu)_c=0\tag{6}$$
We know, that $f^{0bc}=0$ (because $[t_0,t_b]=0$ for $b=1,2,3$), so we get :

$$\partial_\mu (j^\mu)^0 =0\tag{7}$$

We see, that the current $(j^\mu)^0$ is conserved, and this corresponds to a conserved charge $Q^0 = \int d^2x (j^0)^0(x)$. The conserved $Q^0$ charge comes from the $U(1)$ generator $t_0$, which commutes with the $SU(2)$ generators $t_1,t_2,t_3$

The other currents $(j^\mu)^i$, $i=1,2,3$ are not conserved, because the $SU(2)$ generators $t_1,t_2,t_3$ do not commute with themselves, for instance, we have $\partial_\mu (j^\mu)^1 +g(A_\mu)_2 (j^\mu)_3=0$ (+ cyclic permutations).

This post imported from StackExchange Physics at 2015-09-25 20:27 (UTC), posted by SE-user Trimok