# Inconsistency between $d_A = d + A \wedge$ and $d_A = d(..) + [A,..]$?

I am confused by something basic stated in this https://physics.stackexchange.com/a/429947/42982 by @ACuriousMind and some fact I knew of. Here $$d_A$$ is covariant derivative.

1. $$d_A A=F$$ --- @ACuriousMind says "The field strength is the covariant derivative of the gauge field."

2. The Bianchi identity is $$d_A F=0.$$

• In the 1st case, we need to define

$$d_A = d + A \wedge \tag{1}$$

So $$d_A A= (d + A \wedge) A= d A + A \wedge A$$

• In the 2nd case, we need to define

$$d_A = d(..) + [A,..] \tag{2}$$

So we get a correct Bianchi identity which easily can be checked to be true $$d_A F= d F+ [A,F]= d (dA+AA)+[A,dA+AA]=0$$

However, eq (1) and (2) look different.

e.g. if we use eq(2) for "The field strength is the covariant derivative of the gauge field.", we get a wrong result

$$d_A A = dA + [A,A] = dA \neq F !!!!$$

e.g. if we use eq(1) for "Bianchi identity", we get the wrong result we get $$d_A F= d F+ A \wedge F\neq 0$$

my puzzle: How to resolve def (1) and (2)?

Could it be that for the $$p$$-form $$d_A \omega = d \omega + \dots,$$ where $$\dots$$ depends on the $$p$$ of the $$p$$-form? How precisely?

• $[A,A]\neq0$.${}$ May 29 '19 at 19:46
• $[A,A]\neq 0$ but $[A,AA]= 0$? May 29 '19 at 20:00
• Hi @annieheart. As I commented in the question you link, $d_A A=F$ is false (except, of course, whenever the lie algebra involved is abelian).
– user21299
Feb 10 '20 at 19:37

There seems to be a lot of confusion regarding the [,] operation. Well, the way I have learned it goes like this. Indeed, the two notations agree since the graded commutator [,] is defined as $$[α,β]=α\wedgeβ-(-1)^{pq}β\wedge α$$ with $$[α,β]=(-1)^{pq+1}[β,α]$$ for $$α\in \Omega^p(M,\mathfrak g)$$ and $$β\in \Omega^q(M,\mathfrak g)$$, where $$\mathfrak g$$ is the Lie algebra of a Lie group $$G$$. Then, in your case $$DA=dA+\tfrac{1}{2}[A,A]=dA+\tfrac{1}{2}(A\wedge A-(-1)^{1\times 1}A\wedge A)=dA+A\wedge A$$ Indeed, the use of $$F=DA$$ tends to be misleading sometimes due to the presence of 1/2. To that I agree, because in general one has in mind that $$DB=dB+[A,B]$$. Moreover, the Bianchi is quite short with this notation since $$DF=dF+[A,F]=\tfrac{1}{2}d[A,A]+[A,dA]+\tfrac{1}{2}[A,[A,A]]$$ Well, $$d[A,A]$$ follows the usual derivation rule, i.e. $$[dA,A]-[A,dA]=-2[A,dA]$$ because $$dA\in\Omega^2(M,\mathfrak g)$$. Then, you can easily prove that $$[A,[A,A]]=0$$ (hint: the graded commutator satisfies a graded Jacobi identity). Taking the aforementioned properties into account, one directly sees that $$DF=0$$.

In an attempt to give some motivation for the introduction of the graded bracket, I think this has to do with a simple fact. Say that $$α,β$$ are just vector valued forms in $$\omega^p(M,V)$$ and $$\Omega^q(M,W)$$ respectively. Then, $$α\wedge β=α^a\wedgeβ^be_a\otimes \tilde e_b$$ where $$e_a$$ is a basis element of $$V$$ and $$\tilde e_a$$ a basis element of $$W$$. You see that the result lies in $$\Omega^{p+q}(M,V\otimes W)$$. Since the operation between Lie algebra elements is the Lie bracket, we can extend this to $$[α,\beta]=\alpha^a\wedge \beta^b[e_a,e_b]$$ where for simplicity consider $$e_a,e_b$$ to be the generators of the algebra $$\mathfrak g$$ with $$α,β$$ as in the beginning (valued in this algebra). Since $$[,]:\mathfrak g\times \mathfrak g\to \mathfrak g$$, the result lies in $$\Omega^{p+q}(M,\mathfrak g)$$. The swapping rule is fairly straightforward since $$[α,\beta]=\alpha^a\wedge \beta^b[e_a,e_b]=(-1)^{pq}\beta^b\wedge\alpha^a[e_a,e_b]=-(-1)^{pq}\beta^b\wedge\alpha^a[e_b,e_a]=(-1)^{pq+1}[\beta,\alpha]$$ Hope I helped a bit.

P.S: $$A\wedge B$$ is not the usual wedge product. If I remember correctly the clear notation is $$A\wedge_{\rho}B$$ where $$(\rho,V)$$ is a representation. Hence, say $$A,B$$ are $$\mathfrak g$$-valued. Then, we consider the adjoint representation, and we can write $$A\wedge_{\mathrm{ad}}B=A^a\wedge B^b\mathrm{ad}(e_a)e_b=A^a\wedge B^b[e_a,e_b]$$ This is why it makes sense to also have such operations between $$\mathfrak g$$-valued and $$\mathfrak p$$-valued forms if $$\mathrm{ad}(\mathfrak g)\mathfrak p=[\mathfrak g,\mathfrak p]\subset \mathfrak g$$ for example.

• Can you clarify one more time why 𝐷𝐵=𝑑𝐵+[𝐴,𝐵] but 𝐷𝐴=𝑑𝐴+(1/2)[𝐴,𝐴], where how we replace $B$ to $A$ to get the later? May 30 '19 at 16:36
• I think that we are putting 1/2 to the graded commutator notation, so that it matches the usual notation when one translates it. At least I never learned another reason for it. This is why I avoid writing $F=DA$. This is because in the usual notation for example, with R-valued forms, e.g. $R^{ab}\neq D\omega^{ab}$ for the curvature form. May 30 '19 at 16:41

The gauge field $$A$$ you mentioned is a Lie algebra-valued 1 form. The covariant derivative of such a form (aka collectively the curvature form) reads $$\nabla A= dA+A\wedge A=dA+\dfrac{1}{2}[A,A].$$ The latter is in some literature referred to as one of the Maurer-Cartan structure equations. If the equality isn’t clear, you may find it helpful to try a wedge product with two 1-forms on $$\mathbb{R}^n$$ first.

When $$A$$’s Lie algebra is abelian such as $$\mathfrak{u}(1,\mathbb{C})\cong\mathbb{R}$$, the commutator vanishes. When it isn’t, such as for the other gauge groups of the standard model, it doesn’t - which leads to extra interactions and lots of ongoing questions.

This really isn't so complicated as the other answers make it seem. The notation $$\mathrm{d}_A = \mathrm{d} + A\ \wedge$$ is supposed to work like this:

For any $$p$$-form $$\omega$$ taking values in a representation $$(V,\rho)$$ of the Lie group $$G$$ for which $$A$$ is algebra($$\mathfrak{g}$$)-valued, we compute $$\mathrm{d}_A \omega= \mathrm{d} \omega + A\wedge \omega$$ by forming the wedge of $$A$$ and $$\omega$$ as forms and letting the components of $$A$$ act on the components of $$\omega$$ through the representation $$\rho$$ (or rather the induced representation $$\mathrm{d}\rho$$ of the algebra if you want to be really pedantic). In coordinates ($$\mathrm{d}x^{i_1...i_p}$$ denotes the suitably normalized wedge of the basic 1-forms $$x^{i_1}$$ through $$x^{i_p}$$):

\begin{align}A\wedge \omega & = A_i \mathrm{d}x^i \wedge \omega_{i_1\dots i_p}\mathrm{d}x^{i_1\dots i_p} \\ & = \left(\rho\left(A_{i_{p+1}}\right)\omega_{i_1\dots i_p} \right)\mathrm{d}x^{i_1\dots i_{p+1}}\end{align}

For $$A\wedge A$$, the representation is the adjoint representation of the Lie algebra on itself through the commutator, and we get $$A \wedge A = [A_i, A_j]\mathrm{d}x^{ij}.$$ Note that since the vector components of $$A$$ are independent as algebra elements, the commutator only vanishes trivially for $$i = j$$.

Now if to obtain the Bianchi identity you write $$\mathrm{d}_A F$$ in components like this, you get a triple commutator that vanishes by virtue of the antisymmetry of the $$\mathrm{d}x^{ijk}$$ and the Jacobi identity.

• Can you explain $d_A F$ in terms of $d_A = d + A \wedge$ and $d_A = d(..) + [A,..]$? May 30 '19 at 16:31
• this is one of the main issues to ask May 30 '19 at 16:32
• @annieheart I know what $A\wedge\dots$ means because I like the notation and use it. You didn't give a definition of $[A,\dots]$ in your post and using a commutator does not seem to make sense for fields in a general representation, so it is not really clear what you expect me to say about a notion you did not define. kospali's answer already gives a definition for $[A,\dots]$ for which all this works out. May 30 '19 at 16:57
• Part of the confusion might be that $A\wedge A$ as you are interpreting it ($A\wedge$ acting on $A$ in the adjoint representation) is different from $A\wedge A$ as it is usually understood, as in the expression $F=dA+A\wedge A=dA+A_\mu A_\nu dx^\mu\wedge dx^\nu$. There is a difference by a factor of two as $[A_\mu,A_\nu]dx^\mu\wedge dx^\nu=2A_\mu A_\nu dx^\mu\wedge dx^\nu$. This makes it so that $d_A A$, as you understand it, is not $F$. Jan 15 at 2:14

The definition of $$d_A$$ varies according to the gauge transformation properties of the object which $$d_A$$ operates on. Contrary to the other answers, here I am highlighting the impact on the definition of $$d_A$$ originated from single vs. double-sidedness of the gauge transformation.

For example, the Dirac spinor transforms as $$\psi \to R\psi,$$ where $$R$$ is the local gauge transformation associated with connection one-form $$A$$. It follows that the covariant derivative has to be defined as $$d_A \psi = (d + A) \psi,$$ so that $$d_A \psi$$ transforms as $$d_A\psi \to R(d_A\psi).$$

On the other hand, the gauge curvature two-form (gauge field strength) $$F = dA + A \wedge A$$ transform as $$F \to RFR^{-1}.$$ In this case, the covariant derivative has to be defined as $$d_A F = dF + [A, F] = dF + A \wedge F - F \wedge A,$$ so that $$d_A F$$ transforms as $$d_AF\to R(d_AF)R^{-1}.$$ Note that there are both $$R$$ and $$R^{-1}$$ in the gauge transformation of $$F$$. The plus sign in $$+A \wedge F$$ stems from the plus sign in gauge transformation $$R^{+1}$$. And the minus sign in $$- F \wedge A$$ stems from the minus sign in gauge transformation $$R^{-1}$$. Whereas there is only $$R$$ in the gauge transformation of Dirac spinor $$\psi$$, thus you have only a positive $$+A\psi$$ in the definition of $$d_A\psi$$.

Of course, if $$F$$ were odd-form, there would be additional sign changes.

After the above preamble, we take a look at how connection one-form $$A$$ transforms $$A \to RAR^{-1} - (dR)R^{-1}.$$

The covariant derivative $$d_AA$$ $$d_AA = dA + A \wedge A = F,$$ transforms as $$d_AA\to R(d_AA)R^{-1}.$$

Oops, now we are in a pretty hairy situation that $$A$$ and $$d_AA$$ transform in different ways!

Back to your main question, the definition of $$d_AA$$ seems like an odd ball, which is just a convenient way to denote $$F$$.

P.S. According to @kospall $$[α,β]=α\wedgeβ-(-1)^{pq}β\wedge α,$$ where $$α$$ and $$β$$ are $$p$$ and $$q$$ forms respectively. Hence $$[A, A] = A\wedge A - (-1)^{1*1} A\wedge A =A\wedge A + A\wedge A = 2 A\wedge A,$$ and $$[A, F] = A\wedge F - (-1)^{1*2} F\wedge A =A\wedge F - F\wedge A.$$

• In your PS, I assume you meant a plus instead of minus? With a minus sign, your ‘commutator’ is vacuously true in all situations and there’s not much of a point to define such a thing. There’s also a physical interpretation as to why non-abelian 1-forms specifically have a non-vanishing commutator. In QED, the gauge bosons are photons and so there is ZERO interaction in QED until you include free/spinor fields precisely because the commutator is zero. In $SU(2)$ and $SU(3)$, since the commutator is nonzero, you have gluon-gluon interaction as well as interaction among $W^{\pm},Z$ gauge bosons. May 29 '19 at 22:52
• @AntoninoTravia, thanks! updated the PS portion. May 30 '19 at 14:00