Yang-Mills Bianchi identity in tensor notation vs form notation

I've seen the Yang-Mills Bianchi identity written as both

$$0 = dF^a + f^{abc} A^b \wedge F^c$$

and, in tensor notation, as

$$\epsilon^{\mu\nu\lambda\sigma}D_{\nu} F^a_{\lambda\sigma} = 0.$$

Here as usual $$A = A^a_{\mu} T^a dx^{\mu}$$ and $$F = dA + A\wedge A = \frac12 F_{\mu\nu}^a T^a dx^{\mu} \wedge dx^{\nu}$$. With conventions $$D_{\nu} = \partial_{\mu} + A^a_{\mu} T^a$$ and $$[T^a,T^b] = f^{abc}T^c$$, I am unable to connect the two notations. Any help would be appreciated.

Attempt. In order for all Lorentz indices to be contracted we write
$$0 =(\epsilon_{\mu\alpha\beta\gamma} dx^{\alpha} \wedge dx^{\beta} \wedge dx^{\gamma} ) \epsilon^{\mu\nu\lambda\sigma} (\partial_{\nu} + A^c_{\nu} T^c) F^a_{\lambda\sigma}$$ $$= (dx^{\alpha} \wedge dx^{\beta} \wedge dx^{\gamma})\delta^{\mu\lambda\sigma}_{\alpha\beta\gamma} (\partial_{\nu} F_{\lambda\sigma}^a + A^c_{\nu} T^c F^a_{\lambda\sigma})$$ $$=dF^a + (dx^{\alpha} \wedge dx^{\beta} \wedge dx^{\gamma})\delta^{\mu\lambda\sigma}_{\alpha\beta\gamma}A^c_{\nu} T^c F^a_{\lambda\sigma}$$ where $$\epsilon_{\mu\alpha\beta\gamma}\epsilon^{\mu\nu\lambda\sigma}= \delta^{\mu\lambda\sigma}_{\alpha\beta\gamma}$$ is the generalized Kronecker delta. I am not sure how to proceed in order to produce a term of the form $$f^{abc} A^b \wedge F^c$$.

Notice that field strength transforms under adjoint representation. Moreover, recall that the connection piece of covariant derivative acts on an adjoint represented field in a commutator manner:

$$D_\mu \phi = \partial_\mu \phi + [A_\mu, \phi] = \partial_\mu \phi^a T^a + f^{bca}A_\mu^b \phi^c T^a$$.

With this understanding, the equivalence of the tensor notation and differential form notation looks to be immediate.

Here is an argument why covariant derivative in adjoint representation should be in a commutator form. 1. By definition, a field $$\phi$$ transforms in adjoint representation of a Lie algebra in a form of $$$$\delta \phi = [\phi, \lambda],$$$$ where $$lambda$$ are infinitesimal Lie-algebra valued gauge parameters. The covariant derivative of a field in adjoint represention transforms again in adjoint representation, i.e. we will have $$$$\delta D_\mu \phi = \left[ D_\mu\phi, \lambda \right],$$$$ provided $$$$D_\mu \phi := \partial_\mu \phi + [A_\mu, \phi].$$$$ To show this, consider \begin{align} \delta D_\mu\phi &= [\delta A_\mu , \phi ] + D_\mu \delta\phi \\ &= \left[ [A_\mu, \lambda], \phi \right] + [ \partial_\lambda , \phi ] + \partial_\mu [\phi, \lambda] + \left[ A_\mu, [\phi, \lambda]\right] \\ &= [\partial_\mu\phi, \lambda] + \left[ [A_\mu, \phi], \lambda \right] \\ &= [D_\mu\phi, \lambda], \end{align} where we used Jacobi identity as well as the transformation law of gauge field under gauge transformation, $$$$\delta A_\mu = \partial_\mu \lambda + [A_\mu, \lambda].$$$$ The gauge field strength transforms in adjoint representation because the action is then invariant: $$$$\delta\, \text{tr}(F_{\mu\nu}F^{\mu\nu}) = 2 \text{tr}(F^{\mu\nu}[F_{\mu\nu}, \lambda]) = 0.$$$$ Finally, notice that $$\delta \phi = [\phi, \lambda]$$ is equivalent to $$(T_{a})_{bc} = f_{abc}$$, because $$$$\delta\phi = [ \phi^aT_a, \lambda^bT_b] = \phi^a\lambda^b f_{abc} T^c = \lambda^b(f_{bca})\phi^a T^c,$$$$ then notice that this $$$$\delta\phi^c = \lambda^b f_{bca}\phi^a$$$$ is in a form of $$$$\delta\phi^c = \lambda^b (T^{(R)}_b)_{ca} \phi^a,$$$$ and thus the matrix elements of the generators are just structure constant in adjoint representation.

Answer to the question in the comment:

Let us assume $$$$D_\mu \phi = \partial_\mu \phi + A_\mu\phi,$$$$ as you suggested in the comment and original post. After taking the covariant derivative of the adjoint field $$\phi = \phi^aT_a$$, it should still in the adjoint representation and hence $$$$(D_\mu\phi)_{be} = (D_\mu\phi)^a (T^{(\text{adj})}_a)_{be} = (D_\mu\phi)^a f_{abe}.$$$$ Lets suppose the structure constant are totally antisymmetric (and neglect the subtlety for this to be true). Therefore, we see that the result of the covariant derivative should be antisymmetric in the matrix indices $$b,e$$. However, when we expand the covariant derivative, we get some trouble: $$$$(D_\mu\phi)_{be} = \partial_\mu \phi^a f_{abe} + (A_\mu \phi)_{be}.$$$$ While the first term above looks fine, the second term is problematic because $$$$(A_\mu \phi)_{be} = A_\mu^a \phi^c f_{acd}f_{bde} + (\phi A_\mu)_{be},$$$$ where Jacobi identity is used. On the RHS, the first term is OK because it is antisymmetric under exchange of $$b,e$$, while the second term is making trouble.

However, you now see the way out. The simplest solution to get around is to move the second term to the LHS and so we get anti-symmetric $$b,e$$ on the RHS! After we move the second term to the LHS, we get exactly the commutator and the covariant derivative $$$$D_\mu\phi = \partial_\mu \phi + [A_\mu, \phi].$$$$

• But the definition of $D_{\mu}$ is $D_{\mu}\phi = (\partial_{\mu} +A^a_{\mu} T^a)\phi$, is it not? – Dwagg Aug 11 '19 at 20:22
• So why did you write $D_{\mu} \phi = \partial_{\mu} \phi + f^{bca} A_{\mu}^b \phi^c T^a$? – Dwagg Aug 11 '19 at 20:23
• @Dwagg The connection form always appears in the representation the "target" of the covariant derivative transforms under. In case of the curvature form, this is the adjoint representation, which is why the covariant derivative was written in the form given in your second formula. – Bence Racskó Aug 11 '19 at 22:28
• @BenceRacskó Why does the covariant derivative in adjoint rep act as commutator? Adjoint rep means $(T_a)_{bc} = f_{abc}$. The $a,b,...$ indices are raised / lowered by $\delta_{ab}$ and $f_{abc}$ is antisymmetric. Then, with matrix indices, $$A_\mu^a (T_a)_{bd} \phi^c (T_c)_{de} = A_\mu^a \phi^c f_{abd} f_{cde}$$ $$= - A_\mu^a \phi^c f_{abd} (T_d)_{ce}$$ which still is not quite $f_{bca} A^b_\mu \phi^c T^a$ which I should expect, according to the formula given. – Dwagg Aug 13 '19 at 4:04
• @Dwagg We have $D_\mu\phi=\partial_\mu\phi^AT_A+[A_\mu,\phi]=\partial_\mu\phi^AT_A+A^B_\mu\phi^C[T_B,T_C]=(\partial_\mu\phi^A+C^A_{BC}A^B_\mu\phi^C)T_A$, this is the proper index structure. However if there is an Ad-invariant metric on the Lie algebra, and the generators are orthonormal with respect to it, then the structure constants are totally antisymmetric, and by rearranging the indices you can arrive at the same formula as chichi had. – Bence Racskó Aug 13 '19 at 9:01

The another way from $$0 = dF^a + f^{abc} A^b \wedge F^c\tag{1}$$ to
$$\epsilon^{\mu\nu\lambda\sigma}D_{\nu} F^a_{\lambda\sigma} = 0.\tag{2}$$

From (1) we can write $$\begin{eqnarray} 0&=&dF +A^b \wedge F^c [T^b,T^c]\\ &=&dF+A\wedge F\equiv DF=0 \tag 3 \end{eqnarray}$$

or $$DF^a=0\tag 4$$

Then expand (4) in coordinate basis we have $$0= D_\nu F^a_{\lambda\sigma} dx^\nu\wedge dx^\lambda\wedge dx^\sigma\tag 5$$ This implies $$0= D_\nu F^a_{\lambda\sigma} dx^\mu\wedge dx^\nu\wedge dx^\lambda\wedge dx^\sigma\tag 6$$ $$= D_\nu F^a_{\lambda\sigma} \varepsilon^{\mu\nu\lambda\sigma}d^4x\tag 7$$ This implies directly to eqn (2)

Note: eqn (1) and (2) are different forms of the same equation. It means the exterior covariant derivative of curvature is vanish. If you want the form of a covariant derivative you can de-symmetrized it. We then have

$$D_{[\mu}F^a_{\alpha\beta]} =\partial_{[\mu}F^a_{\alpha\beta]}+f^{abc}A^b_{[\mu}F^c_{\alpha\beta]}\tag 8$$

Which means the covariant derivative has the form $$D_{\mu}\phi^a =\partial_{\mu}\phi^a+f^{abc}A^b_{\mu}\phi^c\tag 9$$