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?

  • $\begingroup$ $[A,A]\neq0$.${}$ $\endgroup$ Commented May 29, 2019 at 19:46
  • $\begingroup$ $[A,A]\neq 0$ but $[A,AA]= 0$? $\endgroup$ Commented May 29, 2019 at 20:00
  • 1
    $\begingroup$ 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). $\endgroup$
    – user21299
    Commented Feb 10, 2020 at 19:37
  • $\begingroup$ $d_{A}A=F$ is correct. The problem lies in the fact that $A$ is not a tensorial form. Under a gauge transformation, $A$ does not transform like a tensor. It produces an extra Maurere-Cartan form. $\endgroup$
    – Valac
    Commented Jun 16, 2022 at 18:46

4 Answers 4


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.

  • $\begingroup$ 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? $\endgroup$ Commented May 30, 2019 at 16:36
  • $\begingroup$ 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. $\endgroup$ Commented May 30, 2019 at 16:41
  • 2
    $\begingroup$ @annmariecœur The reason for the extra factor $1/2$ is that the connection form $A$ is not a tensorial form. Under a gauge transformation, $F$ transforms like $U^{-1}FU$ while $A$ transforms like $U^{-1}dU+U^{-1}AU$. So the formula $d_{A}=d+[A,\,\,\,]$ does not apply when acting on $A$. This was explained in the book "Differential Geometry-Connections, Curvature, and Characteristic" by Loring Tu. $\endgroup$
    – Valac
    Commented Jun 16, 2022 at 18:51

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.

  • $\begingroup$ Can you explain $d_A F$ in terms of $d_A = d + A \wedge$ and $d_A = d(..) + [A,..]$? $\endgroup$ Commented May 30, 2019 at 16:31
  • $\begingroup$ this is one of the main issues to ask $\endgroup$ Commented May 30, 2019 at 16:32
  • $\begingroup$ @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. $\endgroup$
    – ACuriousMind
    Commented May 30, 2019 at 16:57
  • $\begingroup$ 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 $F$. $\endgroup$ Commented Jan 15, 2021 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 $$ [A, F] = A\wedge F - (-1)^{1*2} F\wedge A =A\wedge F - F\wedge A. $$

  • $\begingroup$ 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. $\endgroup$
    – penovik
    Commented May 29, 2019 at 22:52
  • 1
    $\begingroup$ @AntoninoTravia, thanks! updated the PS portion. $\endgroup$
    – MadMax
    Commented May 30, 2019 at 14:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.