Prove $D_\mu\phi^\dagger=(D_\mu\phi)^\dagger$ [Question closed, the statement was not true] In gauge transformation, $D_\mu$ was defined to be $\partial_\mu-igA_\mu$.
However, I have hard time to see that $D_\mu\phi^\dagger=(D_\mu\phi)^\dagger$ without ambiguity.
(A comparable example in QED might be found here Why define $D_\mu = \partial_\mu -ieA_\mu$ with the electric charge $e$?)
Could you prove $D_\mu\phi^\dagger=(D_\mu\phi)^\dagger$ in a formal setting/mathematically?
 A: The definition of the covariant derivative depends on what it is acting. If the object is the field of an electron, then, as we well know, the electric charge is $-e$ the covariant derivative $\textit{applied}$ to the field (with the signature $(+,-,-,...,-)$) is
$$ D_\mu\phi= \partial_\mu\phi-iqA_\mu\phi =\partial_\mu\phi+ieA_\mu\phi.$$
In general, if the field you are considering has electric charge $Qe$, with $Q$ a real number, then
$$ D_\mu\phi=\partial_\mu\phi+iQeA_\mu\phi.$$
With the real coordinate system and $Q\neq 0$,
$$ (D_\mu\phi)^\dagger=\partial_\mu\phi^\dagger-iQeA_\mu\phi^\dagger 
\neq 
D_\mu\phi^\dagger =\partial_\mu\phi^\dagger+iQeA_\mu\phi^\dagger .$$
The point is: the explicit expression for the covariant derivative with respect to a symmetry depends on the representation of the field it is acting on. The representations of the electromagnetic symmetry group, $U(1)$, are classified by a real number, aka the electric charge. The term with the vector field in the covariant derivative must have the appropriate charge, otherwise it would not transform as a covariant derivative under a gauge transformation.
Now to your question. The field $\phi$ has, say, charge $Qe$, with $Q$ a number. The complex conjugate of the field has opposite charge, namely $-Qe$. Considering what I say above, the covariant derivative for $\phi^\dagger$ must go with $-Qe$ and not with $Qe$.
