Verifying the Relation in Yang-Mills Theory with a Scalar Field

[Othin]

I'm trying yo verify the relation
\begin{equation}
[D_{\mu},D_{\nu}]\Phi=F_{\mu\nu}\Phi,
\end{equation}
where the scalar field is valued in the lie algebra of a Yang-Mills theory. Here,
\begin{equation}
D_{\mu}=\partial_{\mu} + [A_{\mu},\Phi],
\end{equation}
and
\begin{equation}
F_{\mu,\nu}=\partial_{\mu} A_{\nu} - \partial_{\nu}A_{\mu} + [A_{\mu}, A_{\nu}].
\end{equation}
When the scalar field is an n-tuple, so that the second term in the covariant derivate does not have the brackets, I get the correct result. But here, my calculations give
\begin{equation}
[D_{\mu},D_{\nu}]\Phi=\left(\partial_{\mu} A_{\nu}- \partial_{\nu}A_{\mu}\right)\Phi + \left[[A_{\mu}, \cdot ] , [A_{\nu}, \cdot]\right]\Phi
\end{equation}
The problem is that I don't see how the last term of this equation could become the remaining part of the Field Strength, and I don't see any other possibility! Expanding it, I find
\begin{equation}
\left[[A_{\mu}, \cdot ] , [A_{\nu}, \cdot]\right]\Phi = [A_{\mu}, [A_{\nu}, \Phi]] + [A_{\nu}, [\Phi, A_{\mu}]],
\end{equation}
which, by Jacobi's identity, can be written
\begin{equation}
\left[[A_{\mu}, \cdot ] , [A_{\nu}, \cdot]\right]\Phi =[[A_{\mu},A_{\nu}],\Phi],
\end{equation}
and that's the closest I get. What is wrong with this result? Is it somehow equivalent to the right one? If so, how?

 

[MathematicalPhysicist]

You should check Baez's textbook called: "Gauge Fields, Knots and Gravity" the answer should be there.

 

[Orodruin]

With your notation, $X\Phi = [X,\cdot]\Phi = [X,\Phi]$ and so it should be quite clear that $[[A_\mu,A_\nu],\Phi] = [[A_\mu,A_\nu],\cdot] \Phi = [A_\mu,A_\nu] \Phi$.

If you are still uncomfortable, write out the gauge indices and use the structure functions.

 

[fresh_42]

You will have it easier if you write $[A_\mu,.]=\operatorname{ad}(A_\mu)$. Since $\operatorname{ad}$ is a Lie algebra homomorphism, you have $\operatorname{ad}([A_\mu,A_\nu])=[\operatorname{ad}A_\mu,\operatorname{ad}A_\nu]$ which means your equation (6) is already the same expression as in equation (3).

 

[Othin]

With your notation, $X\Phi = [X,\cdot]\Phi = [X,\Phi]$ and so it should be quite clear that $[[A_\mu,A_\nu],\Phi] = [[A_\mu,A_\nu],\cdot] \Phi = [A_\mu,A_\nu] \Phi$.

If you are still uncomfortable, write out the gauge indices and use the structure functions.

But doesn't $[X,\Phi]=X\Phi - \Phi X?$

 

[Orodruin]

But doesn't $[X,\Phi]=X\Phi - \Phi X?$

You have to separate notation in terms of matrix multiplication and notation in terms of the representation acting on the representation space.

 

[fresh_42]

But doesn't $[X,\Phi]=X\Phi - \Phi X?$

The trouble starts with the sloppy notation of $D_\mu$. It better should have been
$$D_\mu = \partial_\mu + \operatorname{ad}(A_\mu)\, : \,\Phi \longmapsto \partial_\mu \Phi + [A_\mu,\Phi]$$
i.e. given in a way that separates mapping and variable.

 

[Othin]

You will have it easier if you write $[A_\mu,.]=\operatorname{ad}(A_\mu)$. Since $\operatorname{ad}$ is a Lie algebra homomorphism, you have $\operatorname{ad}([A_\mu,A_\nu])=[\operatorname{ad}A_\mu,\operatorname{ad}A_\nu]$ which means your equation (6) is already the same expression as in equation (3). I only would drop a note on why $[A_\mu,\partial_\nu \phi]=0$ which is silently used.

I do see the problem with the notation, I'll try using the one you suggested instead. But still, if equation (6) is the same expression as that in (3), why isn't it $F_{\mu\nu}=\partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu} +\rm{ad} ([A_{\mu},A_{\nu}])$? Isn't $\rm{ad} ([A_{\mu},A_{\nu}])\Phi\neq [A_{\mu},A_{\nu}]\Phi$?

 

[Orodruin]

I do see the problem with the notation, I'll try using the one you suggested instead. But still, if equation (6) is the same expression as that in (3), why isn't it $F_{\mu\nu}=\partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu} +\rm{ad} ([A_{\mu},A_{\nu}])$? Isn't $\rm{ad} ([A_{\mu},A_{\nu}])\Phi\neq [A_{\mu},A_{\nu}]\Phi$?

If you write everything on matrix form (with the matrices being in the Lie algebra of your gauge group), then
$$
\operatorname{ad}([A_\mu,A_\nu]) \rho(\Phi) = \rho([[A_\mu,A_\nu],\Phi]),
$$
where $\rho(\Phi)$ is the representation of your field $\Phi$ as a column vector (which is in the space that $\operatorname{ad}(\ldots)$, which is a $n\times n$ matrix with $n$ being the dimension of the Lie algebra, acts on).

For definiteness, take $SU(2)$ with generators $\sigma_a$ and commutation relations $[\sigma_a,\sigma_b] = 2i\epsilon_{abc}\sigma_c$. You would have
$$
[A_\mu, \Phi] = A_\mu^a \Phi_b [\sigma_a,\sigma_b] = 2i A_\mu^a \epsilon_{abc} \sigma_c \Phi_b,
$$
which by definition would be equal to $\operatorname{ad}(A_\mu)\Phi = (\operatorname{ad}(A_\mu)\Phi)^c \sigma_c$ and therefore
$$
\operatorname{ad}(A_\mu)_{cb} = 2i A_\mu^a \epsilon_{abc}.
$$
Note that $\operatorname{ad}(A_\mu)$ is a 3x3 matrix whereas $A_\mu$ is a 2x2 matrix.

 

[fresh_42]

I do see the problem with the notation, I'll try using the one you suggested instead. But still, if equation (6) is the same expression as that in (3), why isn't it $F_{\mu\nu}=\partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu} +\rm{ad} ([A_{\mu},A_{\nu}])$?

I think it should, but he didn't use the $\operatorname{ad}$ notation at all, so it's missing everywhere. But
\begin{align*}
[D_\mu,D_\nu] &= [\partial_\mu +\operatorname{ad}A_\mu,\partial_\nu +\operatorname{ad}A_\nu] \\
&= \underbrace{[\partial_\mu,\partial_\nu]}_{=0} + [\partial_\mu,\operatorname{ad}A_\nu]-[\partial_\nu,\operatorname{ad}A_\mu]+[\operatorname{ad}A_\mu,\operatorname{ad}A_\nu]\\
&= \partial_\mu \operatorname{ad}A_\nu - \partial_\nu \operatorname{ad}A_\mu - \underbrace{(\operatorname{ad}A_\nu) \partial_\mu}_{=0} + \underbrace{(\operatorname{ad}A_\mu) \partial_\nu}_{=0} + \operatorname{ad}[A_\mu,A_\nu]\\
&=\partial_\mu \operatorname{ad}A_\nu - \partial_\nu \operatorname{ad}A_\mu + \operatorname{ad}[A_\mu,A_\nu]
\end{align*}
and if we simply drop the notation of $\operatorname{ad}$ then we have
$$
[D_\mu,D_\nu] = F_{\mu,\nu}
$$
but I admit that I do not know why $\operatorname{ad}A_\mu (\partial_\nu\Phi) =[ A_\mu,\partial_\nu\Phi ]=0$. I suppose that $\partial_\nu\Phi$ is scalar.

Isn't $\rm{ad} ([A_{\mu},A_{\nu}])\Phi\neq [A_{\mu},A_{\nu}]\Phi$?

 

[Othin]

but I admit that I do not know why $\operatorname{ad}A_\mu (\partial_\nu\Phi) =[ A_\mu,\partial_\nu\Phi ]=0$. I suppose that $\partial_\nu\Phi$ is scalar.

Must it be zero, tough? I thought I'd get rid of it by using the product rule to write $\partial_{\mu}\operatorname{ad}A_{\nu}\Phi=\operatorname{ad}A_\mu (\partial_\nu\Phi) + (\partial_{\mu}\operatorname{ad}A_\nu )\Phi$.

 

[fresh_42]

Must it be zero, tough? I thought I'd get rid of it by using the product rule to write $\partial_{\mu}\operatorname{ad}A_{\nu}\Phi=\operatorname{ad}A_\mu (\partial_\nu\Phi) + (\partial_{\mu}\operatorname{ad}A_\nu )\Phi$.

It would help to know what lives where. As defined in your first post, equation (2), $D_\mu=\partial_\mu +\operatorname{ad}A_{\nu}$ or $D_\mu(\Phi)=\partial_\mu\Phi +[A_{\nu},\Phi]$ which implies, that $\partial_\mu$ (projection) and $\operatorname{ad}A_{\nu}$ (commutator) are operators acting on $\Phi$, I suppose a field. If so, what should the difference be between $\partial_\mu \operatorname{ad}A_{\nu}\Phi$ and $(\partial_\mu \operatorname{ad}A_{\nu})\Phi\,?$ I only used the definition of commutators, wrote $[A_{\nu},\Phi]=\operatorname{ad}A_{\nu}(\Phi)$ and applied the distributive law.

$(\operatorname{ad}A_{\nu})(\partial_\mu \Phi) = [A_{\nu},\partial_\mu(\Phi)]= A_{\nu} \cdot \partial_\mu(\Phi) - \partial_\mu(\Phi) \cdot A_{\nu}$ so what is $\partial_\mu (\Phi)$ and does it commute with $A_{\nu}\,?$

 

[Orodruin]

You need to decide what you mean by $\Phi$. Do you mean the 2x2 matrix that is an element of the Lie algebra, or do you mean the column vector with three components. In your equation (1), it means the representation as a column vector with three components just as $F_{\mu\nu}$ is a 3x3 matrix.

 

[Othin]

It would help to know what lives where. As defined in your first post, equation (2), $D_\mu=\partial_\mu +\operatorname{ad}A_{\nu}$ or $D_\mu(\Phi)=\partial_\mu\Phi +[A_{\nu},\Phi]$ which implies, that $\partial_\mu$ (projection) and $\operatorname{ad}A_{\nu}$ (commutator) are operators acting on $\Phi$, I suppose a field. If so, what should the difference be between $\partial_\mu \operatorname{ad}A_{\nu}\Phi$ and $(\partial_\mu \operatorname{ad}A_{\nu})\Phi\,?$ I only used the definition of commutators, wrote $[A_{\nu},\Phi]=\operatorname{ad}A_{\nu}(\Phi)$ and applied the distributive law.

$(\operatorname{ad}A_{\nu})(\partial_\mu \Phi) = [A_{\nu},\partial_\mu(\Phi)]= A_{\nu} \cdot \partial_\mu(\Phi) - \partial_\mu(\Phi) \cdot A_{\nu}$ so what is $\partial_\mu (\Phi)$ and does it commute with $A_{\nu}\,?$

Well, $\Phi$ is here a scalar field valued in the underlying Lie Algebra. The notation is that of the book Topological Solitons from Niicholas Manton and Paul Sutcliffe (for example, equation (8.40) and (8.41)). I too used the distributive law, but I don't see why $\rm{ad}(A_{\mu})\partial_{\nu}\Phi$, so I tried to expand the first term.

 

[Othin]

You need to decide what you mean by $\Phi$. Do you mean the 2x2 matrix that is an element of the Lie algebra, or do you mean the column vector with three components. In your equation (1), it means the representation as a column vector with three components just as $F_{\mu\nu}$ is a 3x3 matrix.

The expressions were taken from Topological Solitons from Niicholas Manton and Paul Sutcliffe .$\Phi$ is stated to be an element of the Lie Algebra, as is $A_{\mu}$. Wouldn't $F_{\mu\nu}$ be a 2x2 matrix (in the case of SU(2)) since $[A_{\mu},A_{\nu}]$ is?

 

[Orodruin]

No. The field tensor is an operator on the representation. If you write it as a 2x2 matrix you need to take care to properly identify how it acts on the Lie algebra (which is through the adjoint representation so it becomes a commutator).

 

[fresh_42]

Well, $\Phi$ is here a scalar field valued in the underlying Lie Algebra. The notation is that of the book Topological Solitons from Niicholas Manton and Paul Sutcliffe (for example, equation (8.40) and (8.41)). I too used the distributive law, but I don't see why $\rm{ad}(A_{\mu})\partial_{\nu}\Phi$, so I tried to expand the first term.

The difficulty is that equation (2) has $\Phi$ on the right but not on the left. If it were part of the definition of $D_\mu$, then where has it gone to in equation (1) which is simply $[D_\mu,D_\nu]=F_{\mu,\nu}$. Hence $\Phi$ must be the variable, and not part of the function. If it would be part of the function $D_\mu$ then it would occur three times on the LHS of equation (1) and only once on the RHS. All this indicates it's a variable and equation (2) an ill-definition, i.e. should read $D_\mu=\partial_\mu + \operatorname{ad}A_\mu =\partial_\mu + [A_\mu,\,.\,]\,.$

If it is a column vector as suggested by @Orodruin , then $\partial_\mu \Phi$ is a single component which is interpreted as $(\partial_\mu \Phi)\cdot I$ which does commute with any $A_\nu$.

If it is a matrix and in the same domain as $F_{\mu,\nu}$ then I do not see how equation (1) should hold with the definition of $F_{\mu,\nu}$ in equation (3): too many $\Phi$ on just one side of the equation (1).

 

[fresh_42]

I'm not sure whether it will help or confuse you even more, but here are some actual calculations of various situations, i.e. of the different roles as vector or matrix, depending on the context (section 3 and 4, equations (3) and (5)): https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/

 

[Orodruin]

If it is a column vector as suggested by @Orodruin , then ∂μΦ∂μΦ\partial_\mu \Phi is a single component which is interpreted as

No it is not. If you take the derivative of a column vector that depends on spacetime position you get a column vector.

 

[fresh_42]

No it is not. If you take the derivative of a column vector that depends on spacetime position you get a column vector.

Thanks for correction. I thought that $\partial_\mu$ was a basis vector, i.e. the projection on the $\mu$-th component.

But then it isn't clear why $[A_\nu,\partial_\mu \Phi] = 0$ which I think is necessary (see post 10). Or I misinterpreted equation (1) completely and do not understand the role of $\Phi$.

 

[Othin]

The difficulty is that equation (2) has ΦΦ\Phi on the right but not on the left.

Oh, that's right. It's my bad here. It should read $D_{\mu}\Phi=\partial_{\mu}\Phi + [A_{\mu},\Phi]$. That's how the book states it. I used the right one on my calculations though, so it shouldn't be the source of the problem.

 

[fresh_42]

Oh, that's right. It's my bad here. It should read $D_{\mu}\Phi=\partial_{\mu}\Phi + [A_{\mu},\Phi]$. That's hot the book states it.

That was what I read. $D_\mu = \partial_\mu + \operatorname{ad}A_\mu$. In this case equation (1) is equivalent to $(\operatorname{ad}A_\mu) \partial_\nu \stackrel{(*)}{=} \partial_\mu (\operatorname{ad}A_\nu)$. One way to get this is $A_\mu (\partial_\nu \Phi) =(\partial_\nu \Phi) A_\mu$. However, only $(*)$ is needed.

 

[Orodruin]

Thanks for correction. I thought that $\partial_\mu$ was a basis vector, i.e. the projection on the $\mu$-th component.

But then it isn't clear why $[A_\nu,\partial_\mu \Phi] = 0$ which I think is necessary (see post 10). Or I misinterpreted equation (1) completely and do not understand the role of $\Phi$.

Sorry, I cannot answer this without writing a lot of LaTeX and I an on my phone. I might not have time to sit down by the computer until sometime Monday. Remind me if I forget.

 

[Orodruin]

So let's try from the beginning of Yang-Mills theory, which is the classical basis for the standard model once quantised, but let us keep things classical here.

Yang-Mills theory is a classical gauge theory based on some Lie group $\mathcal G$. In many cases, it will be useful to go back to the Abelian case $\mathcal G = U(1)$, which is just classical electromagnetism, to try to figure out what is going on. For simplicity, we will only work with a single scalar field $\Phi$ and the gauge field $A_\mu$.

We start by assuming that $\Phi$ is a section of the fibre bundle with the base space being spacetime and the fiber being a vector space on which we have a natural action by some representation of the gauge group $\mathcal G$ (by "scalar" we are referring to the coordinate transformation properties, not the gauge group transformation properties). This is typically referred to as $\Phi$ "being in the $x$ representation of $\mathcal G$", with $x$ being something that identifies the representation. The naive Lagrangian for just a scalar field that is not in some vector space, but only
$$
\mathcal L_\Phi = |\partial_\mu \Phi|^2 - m^2 |\Phi|^2 + V(\Phi),
$$
where $V(\Phi)$ is some potential. What we are really interested in here is the kinetic term (the one with the space-time derivatives). There is now the typical problem of interpreting what $\partial_\mu \Phi$ would mean as the standard definition of derivatives would require taking the difference between the values of $\Phi$ at different points and taking the limit when the points approach each other. To resolve this, we need a connection field $A_\mu$ that is a linear map from the fiber at one point to itself and the corresponding covariant derivative is $D_\mu = \partial_\mu + \rho(A_\mu)$, where $\rho$ is a representation of the gauge group Lie algebra on the representation space. In essence, the first term (the derivative) represents a change in the components of the fiber whereas the connection field term represents how the bases at nearby points connect to each other. It is quite common not to write out $\rho$, but instead just write $A_\mu$, which I think is what has caused some confusion in this case.

The commutator of the covariant derivatives $F_{\mu\nu} = [D_\mu,D_\nu]$ represents the curvature form and, being a commutator of linear operators on the fiber, is a linear operator on the fiber with the additional property of not depending explicitly on derivatives of the scalar field components.

Let us consider a few special cases. First up, the Abelian case $\mathcal G = U(1)$, i.e., electromagnetism. Since it is Abelian, $U(1)$ only has one-dimensional irreps and any scalar field $\Phi$ just is a section of the fiber bundle with the fiber being the complex numbers $\mathbb C$. Linearly mapping from $\mathbb C$ to $\mathbb C$ means that $A_\mu$ is just a number (for each $x$) and since numbers commute and the derivatives commute
$$
[D_\mu,D_\nu]\Phi = [\partial_\mu + A_\mu,\partial_\nu + A_\nu]\Phi = (\partial_\mu A_\nu - \partial_\nu A_\mu)\Phi
$$
for all $\Phi$ (note that $[\partial_\mu, A_\nu]\Phi = \partial_\mu(A_\nu\Phi) - A_\nu \partial_\mu \Phi = (\partial_\mu A_\nu)\Phi$).

Now to the non-Abelian case. Let us look first at the case where $\Phi$ is in the fundamental representation of some matrix group $\mathcal G$. The covariant derivative is then actually $D_\mu = \partial_\mu + A_\mu$ with $A_\mu$ being in the Lie algebra of $\mathcal G$ (i.e., representing the Lie algebra on itself). The commutator of the covariant derivatives is then (again, the partial derivatives commute)
$$
[D_\mu, D_\nu]\Phi = ([\partial_\mu, A_\nu]+[A_\mu,\partial_\nu]+[A_\mu,A_\nu])\Phi.
$$
The commutators between $\partial_\mu$ and $A_\nu$ follow in the same way as before and this is therefore
$$
[D_\mu,D_\nu] = \partial_\mu A_\nu - \partial_\nu A_\mu + [A_\mu,A_\nu],
$$
which is going to be in the Lie algebra of $\mathcal G$ because each individual term is.

Finally, let us look at the case when $\Phi$ is in the adjoint representation, i.e., it takes values in the Lie algebra of $\mathcal G$. For clarity, let us write $\Phi$ when we refer to the matrix and $\tilde \Phi$ when we refer to its representation as a column vector given some basis of the Lie algebra. The covariant derivative is now
$$
\widetilde{D_\mu\Phi} = D_\mu \tilde\Phi = \partial_\mu \tilde \Phi + \operatorname{ad}(A_\mu) \tilde \Phi
$$
or, equivalently,
$$
D_\mu \Phi = \partial_\mu \Phi + [A_\mu,\Phi].
$$
Using the latter, we find that
$$
D_\mu D_\nu\Phi = \partial_\mu D_\nu \Phi + [A_\mu,D_\nu\Phi]
=\partial_\mu\partial_\nu \Phi + \partial_\mu [A_\nu, \Phi] + [A_\mu,\partial_\nu\Phi] + [A_\mu,[A_\nu,\Phi]].
$$
Noting that, by the product rule for derivatives, $\partial_\mu[A_\nu,\Phi] = [\partial_\mu A_\nu,\Phi] + [A_\nu,\partial_\mu\Phi]$, we therefore end up with
$$
D_\mu D_\nu \Phi =
\partial_\mu\partial_\nu \Phi + [\partial_\mu A_\nu,\Phi] + [A_\nu,\partial_\mu\Phi] + [A_\mu,\partial_\nu\Phi] + [A_\mu,[A_\nu,\Phi]].
$$
The terms $\partial_\mu\partial_\nu \Phi + [A_\nu,\partial_\mu\Phi] + [A_\mu,\partial_\nu\Phi]$ are symmetric in $\mu \leftrightarrow \nu$ and will therefore disappear when we take the commutator between the covariant derivatives. We find that
$$
[D_\mu,D_\nu]\Phi = [\partial_\mu A_\nu,\Phi]+[A_\mu,[A_\nu,\Phi]] - [\partial_\nu A_\mu,\Phi] - [A_\nu,[A_\mu,\Phi]],
$$
which by the Jacobi identity for the commutators can be rewritten
$$
[D_\mu,D_\nu]\Phi = [\partial_\mu A_\nu - \partial_\nu A_\mu + [A_\mu,A_\nu],\Phi] = [F_{\mu\nu},\Phi].
$$
If we were to write this for $\tilde\Phi$ instead we would have
$$
[D_\mu,D_\nu]\tilde\Phi = \operatorname{ad}(\partial_\mu A_\nu - \partial_\nu A_\mu + [A_\mu,A_\nu]) \tilde\Phi = \operatorname{ad}(F_{\mu\nu}) \tilde \Phi,
$$
where $\operatorname{ad}(F_{\mu\nu})$ is a matrix and $\tilde\Phi$ a column vector. Now, physicists usually do not bother to write out the $\operatorname{ad}$ or to make a notational distinction between $\Phi$ and $\tilde\Phi$ and write the last step in both of those equations as $F_{\mu\nu}\Phi$. The reason for this is that in the more general case $F_{\mu\nu}$ is seen as a matrix acting on the column vector that represents the fiber regardless of what the representation is.

In both cases, it represents the action of an element in the Lie algebra on $\Phi$.

Does this clear things up a bit?

Edit: Fixed mistakingly writing $\mathbb G$ instead of keeping consistently to $\mathcal G$.

 

[Orodruin]

I think it should, but he didn't use the $\operatorname{ad}$ notation at all, so it's missing everywhere. But
\begin{align*}
[D_\mu,D_\nu] &= [\partial_\mu +\operatorname{ad}A_\mu,\partial_\nu +\operatorname{ad}A_\nu] \\
&= \underbrace{[\partial_\mu,\partial_\nu]}_{=0} + [\partial_\mu,\operatorname{ad}A_\nu]-[\partial_\nu,\operatorname{ad}A_\mu]+[\operatorname{ad}A_\mu,\operatorname{ad}A_\nu]\\
&= \partial_\mu \operatorname{ad}A_\nu - \partial_\nu \operatorname{ad}A_\mu - \underbrace{(\operatorname{ad}A_\nu) \partial_\mu}_{=0} + \underbrace{(\operatorname{ad}A_\mu) \partial_\nu}_{=0} + \operatorname{ad}[A_\mu,A_\nu]\\
&=\partial_\mu \operatorname{ad}A_\nu - \partial_\nu \operatorname{ad}A_\mu + \operatorname{ad}[A_\mu,A_\nu]
\end{align*}
and if we simply drop the notation of $\operatorname{ad}$ then we have
$$
[D_\mu,D_\nu] = F_{\mu,\nu}
$$
but I admit that I do not know why $\operatorname{ad}A_\mu (\partial_\nu\Phi) =[ A_\mu,\partial_\nu\Phi ]=0$. I suppose that $\partial_\nu\Phi$ is scalar.

So, returning to this, the problem is that you have written $\partial \operatorname{ad}(A)$ (skipping the indices) to mean $\partial (\operatorname{ad}(A)\Phi)$ when acting on $\Phi$, when what is in equation (3) really represents $(\partial \operatorname{ad}(A))\Phi$, with the derivative acting only on ad(A). Thus, it is not the terms you have marked as zero that are zero by themselves, they are canceled by the corresponding terms from $\partial \operatorname{ad}(A)$ when applying the product rule for derivatives.

 

[fresh_42]

Thanks. The line

note that $[\partial_\mu, A_\nu]\Phi = \partial_\mu(A_\nu\Phi) - A_\nu \partial_\mu \Phi = (\partial_\mu A_\nu)\Phi$

contains my misunderstanding, namely, the wrong or missing parentheses in my interpretation.

I already feared you would beat me with an avalanche of indices as you said:

Sorry, I cannot answer this without writing a lot of LaTeX and I an on my phone. I might not have time to sit down by the computer until sometime Monday. Remind me if I forget.

So I'm happy that it was so easy and embarrassed that I missed the point.

 

[Orodruin]

I am happy some part of the post was useful. I spent the better part of the morning on it.

 

[fresh_42]

I am happy some part of the post was useful. I spent the better part of the morning on it.

Well, the serious read will take some more time than to figure out my mistake.
It at least got you an additional like plus I will not complain (I already did in an insight) about

It is quite common not to write out $\rho$, but instead just write $A_\mu$

although I think it would generally help especially if the vectors are simultaneously matrices themselves, or if more than one representation is considered. I often wrote $\dot{A}_\mu$ on my board to save time to mark it as $\rho(A_\mu)$ but I admit that this is even more confusing in a physical context.

 

[Othin]

wow, thanks guys! This is very comprehensive and I think it really solved my problem!

 

[Orodruin]

Well, the serious read will take some more time than to figure out my mistake.
It at least got you an additional like plus I will not complain (I already did in an insight) about

although I think it would generally help especially if the vectors are simultaneously matrices themselves, or if more than one representation is considered. I often wrote $\dot{A}_\mu$ on my board to save time to mark it as $\rho(A_\mu)$ but I admit that this is even more confusing in a physical context.

It is actually very confusing in a physical context as a dot usually is taken to mean some sort of time derivative ... In general, there are so many implied indices in QFT that you would go mad if you had to write them all out...

 

[Orodruin]

Also, just a notational issue: The notation should be $F_{\mu\nu}$, not $F_{\mu,\nu}$. It is quite common to use $,\mu$ as additional subscripts instead of writing out $\partial_\mu$ in front.