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

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.

• In differential form notation the term in the bracket would be $D F^a$, $a$ being the index of the internal space. In index notation, this is the first of your suggestions. Commented Aug 30, 2022 at 21:24
• But how do we interprete $(\partial_\mu-igA_\mu^a t^a)F_{\lambda\sigma}^a$ ? $\partial_\mu F_{\lambda\sigma}^a$ is a number but $t^a$ is a matrix? Commented Aug 30, 2022 at 21:27
• P&S certainly define $F_{\lambda\sigma}=t^a F^a_{\lambda\sigma}$ in (15.48). With a group index, you have a spacetime function/gradient operator. Without a group index, you have those functions saturated on generators, so a group representation matrix with such components in its entries. What on earth are you asking? Commented Aug 30, 2022 at 21:59
• No, Peskin and Schroeder never defined $F_{\lambda\sigma}$ to be $t^aF^a_{\lambda\sigma}$. 15.48 never even has $F_{\lambda\sigma}$ in it. Commented Aug 30, 2022 at 23:11
• See Wikipedia — look between the 6th and 7th equations for this convention. How else could the two F’s be related? Commented Aug 31, 2022 at 0:33

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:

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

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 $$(N^2-1)x(N^2-1)$$.

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