Meaning of the notation $(D_\nu F_{\lambda\sigma})^a$ in Bianchi's identity

Question

I'm studying Peskin and Schroeder chapter 15, on page 500, we have the Bianchi's identity in nonabelian gauge theory,

$$\tag{15.89} \epsilon^{\mu\nu\lambda\sigma}(D_\nu F_{\lambda\sigma})^a=0$$

Here $\tag{15.45} D_\mu=\partial_\mu-igA^a_\mu t^a$

and $\tag{15.49} F_{\mu\nu}^a=\partial_\mu A^a_\nu-\partial_\nu A^a_\mu+g f^{abc}A_\mu^b A_\nu^c$

My question is, what does $(D_\nu F_{\lambda\sigma})^a$ mean? $F_{\lambda\sigma}$ isn't even defined in Peskin and Schroeder. Does it mean $D_\nu F_{\lambda\sigma}^a$ instead? But then it still doesn't make sense because $t^a$ is a matrix acting on a vector space. Does it mean $D_\nu(F_{\lambda\sigma}^a\psi)$ instead? However, if that is the case then $\tag{15.88} (D^\mu F_{\mu\nu})^a=-g j_\nu^a$ doesn't make sense.

Answer 1

In general, in any given representation, the covariant derivative is defined as:

$${(D_\mu)^A}_B={\left(\delta \partial_\mu-igA^a_\mu (\mathcal{T}_a)\right)^A}_B={\delta^A}_B \partial_\mu-igA^a_\mu {(\mathcal{T}_a)^A}_B$$

Where $\mathcal{T}_a$ and $A, B, C...$ are respectively the generators (matrices) and the indices of the given representation.

This is a general concept, in physics there are two important representations:

• Fundamental Representation:

This is the representation where lives the matter fields, as the fermions.

Conventionally we use $t_a$ for the generators and $i,j,k...$ as the indices for this representation.

For example we have that for $SU(3)$ the $t_a$ are the eight Gell-Mann matrices (actually it's $t_a=\frac{\lambda_a}{2}$), and so given a field $\Psi^j$ in this representation (i.e. the quark field), we have:

$${(D_\mu)^i}_j \Psi^j=\left({\delta^i}_j \partial_\mu-igA^a_\mu {(t_a)^i}_j\right)\Psi^j$$

• Adjoint Representation:

This is the representation where lives the radiation fields, as the gauge bosons.

Conventionally we use $T_a$ for the generators and $a,b,c...$ as the indices for this representation. (actually the choice $a,b,c...$ was already made since we had $A^a_\mu (\mathcal{T}_a)$ in the definition of the covariant derivative.)

So we can take the covariant derivative of the gauge boson field $A^a_\mu$ (the gluon for $SU(3)$):

$${(D_\mu)^b}_c A^c_\rho=\left({\delta^b}_c \partial_\mu-igA^a_\mu {(T_a)^b}_c\right)A^c_\rho$$

Or you could do the same with the curvature of the gauge field $F_{\mu\nu}^a=\partial_\mu A^a_\nu-\partial_\nu A^a_\mu+g f^{abc}A_\mu^b A_\nu^c$, or any other field that is in this representation, similarly we have:

$${(D_\mu)^b}_c F^c_{\rho \sigma}=\left({\delta^b}_c \partial_\mu-igA^a_\mu {(T_a)^b}_c\right)F^c_{\rho \sigma}$$

So, you have to understand that the covariant derivative is a different object in different representations, this is because it has "different indices" and different matrices.

In general it is true that ${(T_a)^b}_c=-i{{f_a}^b}_c$ while $t_a$ are the generators of the group through exponentiation.

For example for SU(N) we have that $i,j,k...$ goes from 1 to $N$ while $a,b,c...$ from 1 to $(N^2-1)$ and so are different $t_a$ and $T_a$, the first are a $NxN$ matrices while the latter are $(N^2-1)x(N^2-1)$.

• Final Answer

At the end $(D_\nu F_{\rho\sigma})^b$ is just a cleaner way to express ${(D_\nu)^b}_c F^c_{\rho \sigma}$.

So the Bianchi identity is: $$\epsilon^{\mu\nu\rho\sigma}{(D_\nu)^b}_c F_{\rho\sigma}^c=\epsilon^{\mu\nu\rho\sigma}\left({\delta^b}_c \partial_\nu-igA^a_\nu {(T_a)^b}_c\right)F^c_{\rho \sigma}=\epsilon^{\mu\nu\rho\sigma}\left(\partial_\nu F^b_{\rho \sigma}-gA^a_\nu {{f_a}^b}_c F^c_{\rho \sigma}\right)=0$$