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?

  • $\begingroup$ $[A,A]\neq0$.${}$ $\endgroup$ May 29 '19 at 19:46
  • $\begingroup$ $[A,A]\neq 0$ but $[A,AA]= 0$? $\endgroup$ May 29 '19 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
    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.

  • $\begingroup$ 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? $\endgroup$ May 30 '19 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. $\endgroup$
    – kospall
    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.

  • $\begingroup$ Can you explain $d_A F$ in terms of $d_A = d + A \wedge$ and $d_A = d(..) + [A,..]$? $\endgroup$ May 30 '19 at 16:31
  • $\begingroup$ this is one of the main issues to ask $\endgroup$ May 30 '19 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
    May 30 '19 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 $F$. $\endgroup$ 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 $$ [A, F] = A\wedge F - (-1)^{1*2} F\wedge A =A\wedge F - F\wedge A. $$

  • $\begingroup$ 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. $\endgroup$ May 29 '19 at 22:52
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    $\begingroup$ @AntoninoTravia, thanks! updated the PS portion. $\endgroup$
    – MadMax
    May 30 '19 at 14:00

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