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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

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  • $\begingroup$ @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)? $\endgroup$ – user239970 Nov 8 '19 at 17:56
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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.

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  • $\begingroup$ Thank you for the helpful solution. This is a very neat trick! $\endgroup$ – user239970 Nov 9 '19 at 0:07

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