Conserved topological charge for d=3 Yang-Mills. G=U(2)

Question

Consider a pure Yang-Mills lagrangian density $$\mathcal{L}=-\frac{1}{4}F^{\mu\nu}_aF^a_{\mu\nu}$$ with gauge group $U(2)$.

Take the generators for $U(2)$ to be $t_0$, $t_i \ i=1,...,3$ with commutation relations given by $$[t_0,t_i]=0$$ $$[t_i,t_j]=i\epsilon_{ijk}t_k$$ In particular $t_0$ is the generator of the $\mathfrak{u(1)}$ factor in the expansion $\mathfrak{u(2)}\simeq \mathfrak{u(1)}\times \mathfrak{su(2)}$ and $t_i$ are the generators of the Lie Algebra $\mathfrak{su(2)}$.

Now, in $d=3$ the field strenght Hodge-dual is a current $j^{\mu}:=\frac{1}{2}\epsilon^{\mu\nu\rho}F_{\nu\rho}$ and is conserved in virtue of the Bianchi Identity.

The questions are:

1) What is it meant when they say the current is conserved? Is it covariantly conserved (ie $D_{\mu}j^{\mu}$=0) or simply conserved (i.e $\partial_{\mu}j^{\mu}=0$)

2)Do I have just one vector current, or one for each generator of the gauge group? (i.e 4 in this case)

3) Can you explicitly carry out the computation of the conserved current and charge?

4) I am asked to state if the conserved charge arises because of the factor $U(1)$ of the gauge group (which has an algebra generated by $t_0$), because of the factor $U(1)$ which is the cartan subalgebra of $SU(2)$ (generated by $t_3$), or because both of them. [I really don't understand this question, what would you answer? Thanks.]

The part of the computation I did is the following.

$$F^0_{\mu\nu}=\partial_{\mu}A^0_{\nu}-\partial_{\nu}A^0_{\mu}$$ $$F^i_{\mu\nu}=\partial_{\mu}A^i_{\nu}-\partial_{\nu}A^i_{\mu}+g\epsilon^{ijk}A_{\mu}^jA^k_{\nu}$$

Therefore using Bianchi I have

$$0=D_\mu\epsilon^{\mu\nu\rho}F^0_{\nu\rho}=(\partial_{\mu}-igA_{\mu})\epsilon^{\mu\nu\rho}(\partial_{\nu}A^0_{\rho}-\partial_{\rho}A^0_{\nu})$$

while for the other side

$$0=D_\mu\epsilon^{\mu\nu\rho}F^i_{\nu\rho}=(\partial_{\mu}-igA_{\mu})\epsilon^{\mu\nu\rho}(\partial_{\nu}A^i_{\rho}-\partial_{\rho}A^i_{\nu}+g\epsilon^{ijk}A_{\mu}^jA^k_{\nu})$$

What can I do from here? It seems to me that the currents $$j^{\mu}_0=\epsilon^{\mu\nu\rho}(\partial_{\nu}A^0_{\rho}-\partial_{\rho}A^0_{\nu})$$ and $$j^{\mu}_i=\epsilon^{\mu\nu\rho}(\partial_{\nu}A^i_{\rho}-\partial_{\rho}A^i_{\nu}+g\epsilon^{ijk}A_{\nu}^jA^k_{\rho})$$

are both covariantly conserved...

Thanks a lot for answers and clarifications.

Answer 1

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).