I'm studying topological monopoles in a $SU(2)$ Yang-Mills theory with spontaneous symmetry breaking, through the book "Topological Solitons", by Manton and Sutcliffe. In section 8.2, the authors relate the Yang-Mills field-strength tensor to the Maxwell field tensor. The former is written, in this representation, as: $$F_{\mu\nu}=\partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu} + [A_{\mu},A_{\nu}],$$ where $A_{\mu}=A_{\mu}^{a}T^a$, the $T^{a}\equiv i\sigma^a$ being the generators of the $su(2)$ algebra. The covariant derivative acts on the Higgs Field $\Phi=\Phi^aT^a$ according to $$D_{\mu}\Phi= \partial_{\mu}\Phi + [A_{\mu},\Phi].$$ Now, consider a region of space-time where one may write $\Phi=h\hat{\Phi}$, where $|\Phi|^2\equiv-\frac{1}{2}\rm{Tr}\Phi^2=1$ and $D_{\mu}\hat{\Phi}=0$. The Maxwell field tensor is defined by the relation $f_{\mu\nu}=-\frac{1}{2}\rm{Tr}(F_{\mu\nu}\hat{\Phi})$. So, in order to find it, I need to solve $D_{\mu}\hat{\Phi}=0$ for the gauge potential and substitute the result in the definition of $F_{\mu\nu}$. I should find: $$A_{\mu}=\frac{1}{4}[\partial_{\mu}\hat{\Phi},\Phi] + a_{\mu}\hat{\Phi},$$ where $a_{\mu}$ is a smooth function, and $$F_{\mu\nu}=\left(\frac{1}{8}\rm{Tr}([\partial_{\mu}\hat{\Phi},\partial_{\nu}\hat{\Phi}]\hat{\Phi}) + \partial_{\mu}a_{\nu} - \partial_{\nu}a_{\nu} \right)\hat{\Phi}. $$

I haven't been able to find the solution for $A_\mu$, nor could I find this form for $f_{\mu\nu}$ through substitution of the correct result and algebraic manipulations, even though it should be straightforward. I'd like some with those manipulations, if possible. Also, as a secondary question, I'd be glad if someone could explain what the condition $D_{\mu}\hat{\Phi}$ means, as in why should it be satisfied in regions other than the vacuum?

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    $\begingroup$ Sorry for not posting a full answer, please observe that $\hat{\Phi} B \hat{\Phi} = \mathrm{Tr}( \hat{\Phi} B ) \hat{\Phi}$ for any adjoint quantity $B=\sum_{a=1}^3 B^a t_a$, (such as the gauge potential). As for your second question, please see the remark in Manton and Sutcliffe following equation (8.70), where they explain that this solution is asymptotically valid in the region outside the core of the monopole, where the gauge field Abelianizes. $\endgroup$ Commented May 12, 2019 at 13:51
  • $\begingroup$ Well, about the second part... it's true, but $D_{\mu}\hat{\Phi}=0 $ is a weaker condition. The results must be valid even inside the core, so that the interpretation of the magnetic field as $b_i=-\frac{1}{2}\epsilon_{ijk}f_{jk}$ still holds (in the sense that it remains a valid interpretation).Also, outside of the core we could just use the stronger condition $D_{\mu}\Phi=0$, which must be satisfied far from the origin. $\endgroup$
    – Othin
    Commented May 14, 2019 at 14:43
  • $\begingroup$ I don't know that book, but is this in the case of a BPS monopole? Do you also have the Bogomol'nyi equation, $\vec{D}\Phi = \vec{B}$? $\endgroup$ Commented May 19, 2019 at 16:32
  • $\begingroup$ Well, the solution is not assumed to be BPS here, it's the general case. $\endgroup$
    – Othin
    Commented May 22, 2019 at 0:36

1 Answer 1


First we need some identities, using the conventions for $su(2)$ from the question:

\begin{align} [[A,B],C] &= 2A\ \mathrm{Tr}(BC) - 2B\ \mathrm{Tr}(AC), \\ [[A,B],[C,D]] &= 2A\ \mathrm{Tr}(C[D,B]) - 2B\ \mathrm{Tr}([A,C]D). \end{align}

If we take a derivative of $\mathrm{Tr}(\hat{\Phi}^2) = -2$, we find that $\mathrm{Tr}(\hat{\Phi}\ \partial_\mu\hat{\Phi}) = 0$. Using the first identity we find:

\begin{align} [[\partial_\mu\hat{\Phi}, \partial_\nu\hat{\Phi}], \hat{\Phi}] = 0 \implies [\partial_\mu\hat{\Phi}, \partial_\nu\hat{\Phi}] &\propto \hat{\Phi} \\ \implies [\partial_\mu\hat{\Phi}, \partial_\nu\hat{\Phi}] &= - \frac{1}{2} \mathrm{Tr}([\partial_\mu\hat{\Phi}, \partial_\nu\hat{\Phi}] \hat{\Phi}) \ \hat{\Phi}, \end{align}

where the constant of proportionality is determined by tracing against $\hat{\Phi}$.

Now, take the commutator of $D_\mu \hat{\Phi} = 0$ with $\hat{\Phi}$:

\begin{align} [\partial_\mu \hat{\Phi}, \hat{\Phi}] &= - [[A_\mu, \hat{\Phi}], \hat{\Phi}] \\ &= -2 A_\mu\ \mathrm{Tr}(\hat{\Phi}^2) + 2\hat{\Phi}\ \mathrm{Tr}(A_\mu \hat{\Phi}) \\ &= 4 A_\mu + 2 \hat{\Phi}\ \mathrm{Tr}(A_\mu \hat{\Phi}). \end{align}

If we define $a_\mu = -\frac{1}{2} \mathrm{Tr}(A_\mu \hat{\Phi})$ we get the expression for $A_\mu$ in the question. The derivation of $F_{\mu\nu}$ is very similar, so I won't write it out here unless someone asks.

As for $D_\mu \hat{\Phi} = 0$, note that the $U(1)$ unbroken by the Higgs is $U = \exp(i\alpha \hat{\Phi})$. To leave this unbroken, $F_{\mu\nu}$ must also be proportional to $\hat{\Phi}$. The Bogomol'nyi equation, $\vec{D}\Phi=\vec{B}$, becomes

$$ \vec{\nabla}h\ \hat{\Phi} + h \ \vec{D}\hat{\Phi} = \vec{b} \ \hat{\Phi}. $$

If we trace this against $\hat{\Phi}$, we find that $\vec{\nabla}h = \vec{b}$. Substituting back in gives $\vec{D}\hat{\Phi} = 0$.

Edit: In the non-BPS case we can apply the same tricks to the equation of motion for $F_{\mu\nu}$ to find $h^2 [D_\mu \hat{\Phi}, \hat{\Phi}] \propto f_\mu{}^\nu D_\nu \hat{\Phi}$. If we take the commutator with $\hat{\Phi}$ and trace against $D^\mu \hat{\Phi}$ we find $\lvert D\hat{\Phi} \rvert^2 = 0$. The time component vanishes for a static solution, so the space components do as well.

  • $\begingroup$ Thanks, that helped a lot. I think I'm almost there now but there remain a term proportional to $[\hat{\Phi}a_{\mu},[\partial_{\nu}\hat{\Phi},\hat{\Phi}]]$ and one with μ,ν interchanged in the calculation of Fμν. Not sure why they should vanish $\endgroup$
    – Othin
    Commented May 21, 2019 at 19:05
  • $\begingroup$ If you pull $a_{\mu}$ out of the commutator and use the first identity, you get something proportional to $(\partial_{[\mu} \hat{\Phi}) a_{\nu]}$. This cancels with a term from $\partial_{[\mu}(a_{\nu]} \hat{\Phi})$. $\endgroup$ Commented May 21, 2019 at 20:17
  • $\begingroup$ Thank you, I had forgotten to differentiate the unit vector in the $a_{\nu}\hat{\Phi}$ terms, so I couldn't get the cancelation. It works now. $\endgroup$
    – Othin
    Commented May 21, 2019 at 20:31

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