Relating the Yang-Mills field-strength to the Maxwell tensor in $SU(2)$ gauge theory 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?
 A: 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.
