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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.

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  • $\begingroup$ This article explains in the first pages why the current is conserved in the case of d=3 QED (which is U(1) Yang-Mills) but still is quite different... $\endgroup$ – Federico Carta Oct 26 '13 at 17:35
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    $\begingroup$ You're asking us to do all the work without making an effort yourself. For 1), you write that the current is conserved because of the Bianchi identity. But in the next sentence you ask us to prove the conservation of the current. 2) is really textbook material. It's the same as asking whether there's only one "gluon" $A_\mu$ or several. 3.) is also a textbook question. Many of us know the answer but it's a waste of time to write down all these computations if they're done in every QFT textbook, e.g. Peskin-Schroeder (the chapter on non-Abelian gauge theories). $\endgroup$ – Vibert Oct 26 '13 at 19:21
  • $\begingroup$ I have made an effort. I have done part of the computation (as far as I can get) but can't arrive at the correct answer. I can post it if you want to, or do not believe I first tried and then asked. I have also looked at Peskin-Schroeder but there is nothing similar to this in the whole book. They only treat ordinary Yang Mills in 4 dimension, and not in d=3. If it really is textbook material, could you suggest a book in which they treat Yang-Mills in d=3? I believe that there are 4 different currents, but then I don't understand point 4). $\endgroup$ – Federico Carta Oct 26 '13 at 19:41
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    $\begingroup$ Well, the only difference between $SU(3)$ and your case really lies in the structure constants, correct? In turn, they show up in the expression for $F_{\mu \nu}.$ Try to write down (if you haven't already done so) the different $j_\mu^a$ in terms of the structure constants. Also in general, if you have done some work, you should always post it - this helps people to see what's going wrong and where you need some tips. $\endgroup$ – Vibert Oct 26 '13 at 19:46
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    $\begingroup$ @FedericoCarta : Hints : From Wiki and Bianchi identities (you may replace explicitely the indices $\mu,\nu,\rho$, etc.. by $1,2,3$ if it is clearer for you), you have the equation for "conservation" of your dual current. Look at the difference between $f^{oij}$ and $f^{ijk}$, and you will see the difference between "conservation" of $j^\mu_0$ and "conservation" of $j^\mu_i$ $\endgroup$ – Trimok Oct 26 '13 at 19:53
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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).

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