# Field-strength tensor after symmetry breaking of $SU(2)$

I am currently trying to understand the paper https://iopscience.iop.org/article/10.1088/0034-4885/41/9/001 and I am stuck on eq. (4.29).

Basically what I understand is that, you have your field strength tensor (for $$SU(2)$$ gauge group) $$G_{\mu\nu}^a = \partial_\mu W_\nu^a - \partial_\nu W_\mu^a -e \epsilon^{abc}W^b_\mu W^c_\nu ~~~~(1)$$ Where $$a$$ is a group index (running from $$1,2,3$$). We consider a monopole solution $$\phi^a$$ satisifying $$(D_\mu \phi)^a =\partial_\mu \phi^a -e \epsilon^{abc}W^b_\mu \phi^c=0 \text{ and }\phi^a \phi^a =a^2$$ Where $$a$$ is a constant (radius of vacuum manifold). Then one of the solution to above equation is $$W^a_\mu = \hat{\phi}^a A_\mu +\frac{1}{e}\epsilon^{abc}\hat{\phi}^b \partial_\mu \hat{\phi}^c~~~~(2)$$ Where $$\hat{\phi}^a \equiv \frac{1}{a}\phi^a$$. Then in the paper, authors insert eq(2) into eq(1) and find that $$G_{\mu\nu}^a = \hat{\phi}^a F_{\mu\nu}~~~~(3)$$ Where $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu + \frac{1}{e}\hat{\phi}^a \left(\epsilon^{abc}\partial_\mu \hat{\phi}^b \partial_\nu \hat{\phi}^c\right)$$ But if I try this I find $$2\partial_{[\mu} W_{\nu]}^a = 2\hat{\phi}^a \partial_{[\mu} A_{\nu]}+2A_{[\nu}\partial_{\mu]}\hat{\phi}^a+\frac{2}{e}\epsilon^{abc}\partial_\mu \hat{\phi}^b\partial_\nu \hat{\phi}^c~~~~(4)$$ and $$e\epsilon^{abc}W^b_\mu W^c_\nu= 2 A_{[\nu}\partial_{\mu]}\hat{\phi}^a -\frac{1}{e}\hat{\phi}^a \hat{\phi}^b \left(\epsilon^{bcd}\partial_\mu \hat{\phi}^c\partial_\nu \hat{\phi}^d\right)~~~~(5)$$ Combining eq(4) and eq(5) almost gives me eq(3) but with extra term $$\frac{2}{e}\epsilon^{abc}\partial_\mu \hat{\phi}^b\partial_\nu \hat{\phi}^c$$. I'm not sure why I can't get eq(3).

Thank you

• @CosmasZachos thank you for the reply. I'm not sure where I can use the identity to simplify eq(4) or eq(5)? Is it the second term of eq(5)? – user239970 Nov 8 '19 at 17:56

The last term of (5), $$\hat{\phi}^a \hat{\phi}^b \epsilon^{bcd}\partial_\mu \hat{\phi}^c\partial_\nu \hat{\phi}^d= \delta^{ae} \epsilon^{bcd} \hat{\phi}^e \hat{\phi}^b \partial_\mu \hat{\phi}^c\partial_\nu \hat{\phi}^d ~.$$
Now recall the identity $$\delta^{e[a} \epsilon^{bcd]} =0,$$ since you cannot antisymmetrize four indices!
Consequently $$\delta^{ea} \epsilon^{bcd} \hat{\phi}^e \hat{\phi}^b \partial_\mu \hat{\phi}^c\partial_\nu \hat{\phi}^d = \delta^{eb} \epsilon^{cda} \hat{\phi}^e \hat{\phi}^b \partial_\mu \hat{\phi}^c\partial_\nu \hat{\phi}^d - \delta^{ec} \epsilon^{dab} \hat{\phi}^e \hat{\phi}^b \partial_\mu \hat{\phi}^c\partial_\nu \hat{\phi}^d + \delta^{ed} \epsilon^{abc} \hat{\phi}^e \hat{\phi}^b \partial_\mu \hat{\phi}^c\partial_\nu \hat{\phi}^d \\ = \epsilon^{cda} \partial_\mu \hat{\phi}^c\partial_\nu \hat{\phi}^d,$$ since the last two terms involve a unit vector dotted onto its gradient.