Commutator of covariant derivatives to get the curvature/field strength

Question

For notation and convention, please see Gauge theory formalism and Generalizing the covariant derivate for gauge theory.

The covariant derivative can be used to construct curvatures (called field strengths in the Yang-Mills case).

These are defined as

$$[D_\mu,D_\nu]=-\delta(R_{\mu\nu})$$

where

$$R_{\mu\nu}{}^A= 2 \partial_{[\mu}B_{\nu]}{}^A + B_\nu{}^C B_\mu{}^B f_{BC}{}^A$$

The authors state that this can be derived by using the result from before

"Replace $\epsilon_1$ with $B_\mu$ and replace $\epsilon_2$ with $\epsilon$''

\begin{eqnarray*} [\delta(\epsilon_1),\delta(\epsilon_2)] &=& \delta(\epsilon_1{}^A \epsilon_2{}^B f_{AB}{}^C) \\ \delta(B_\mu)\delta(\epsilon) - \delta(\epsilon)\delta(B_\mu)&=& \delta(B_\mu{}^A \epsilon^B f_{AB}{}^C) \\ \end{eqnarray*}

Question: Explicitly, the authors say that by instead now replacing $\epsilon_1$ with $B_\mu$, and $\epsilon_2$ with $B_\nu$, and using the result of the convariance of $D_\mu$, this expression for the curvature can be derived. I am struggling with this. So far I have,

\begin{eqnarray*} [\delta(\epsilon_1),\delta(\epsilon_2)] &=& \delta(\epsilon_1{}^A \epsilon_2{}^B f_{AB}{}^C) \\ \delta(B_\mu)\delta(B_\nu) - \delta(B_\nu)\delta(B_\mu)&=& \delta(B_\mu{}^A B_\nu{}^B f_{AB}{}^C) \\ \end{eqnarray*}

Then just writing out the commutator of the covariant derivatives explicitly I have,

\begin{eqnarray*} [D_\mu,D_\nu] &=& (\partial_\mu - \delta(B_\mu))(\partial_\nu -\delta(B_\nu)) \\ &=& \partial_\mu\partial_\nu - \partial_\mu\delta(B_\nu) - \delta(B_\mu)\partial_\nu +\delta(B_\mu)\delta(B_\nu) -\partial_\nu\partial_\mu + \partial_\nu\delta(B_\mu) + \delta(B_\nu)\partial_\mu - \delta(B_\nu)\delta(B_\mu) \\ &=& \partial_\mu\partial_\nu - \partial_\mu\delta(B_\nu) - \delta(B_\mu)\partial_\nu -\partial_\nu\partial_\mu + \partial_\nu\delta(B_\mu) + \delta(B_\nu)\partial_\mu + \delta(B_\mu{}^A B_\nu{}^B f_{AB}{}^C) \end{eqnarray*}

Thoughts on how to proceed:

1) use the rule from middle/high school with mixed partial second derivatives, $\frac{\partial}{\partial x}\frac{\partial}{\partial y} f = \frac{\partial}{\partial y}\frac{\partial}{\partial x} f.$

2) use the suggested form from the definition of $R_{\mu\nu}$ given by the authors.

\begin{eqnarray*} \delta(2 \partial_{[\mu}B_{\nu]}) &=& \delta(2 \frac{1}{2!}(\partial_\mu B_\nu - \partial_\nu B_\mu)) \\ &=& \delta(\partial_\mu B_\nu) - \delta(\partial_\nu B_\mu) \\ &=& \delta(\partial_\mu) B_\nu + \partial_\mu \delta(B_\nu) - \delta(\partial_\nu) B_\mu - \partial_\nu \delta(B_\mu) \end{eqnarray*}

Not sure not to rearrange these terms to work towards the derivation.

Answer 1

What book are you using? The notation is horribly obscure. Here is the standard and straightforward calculation: $$ \nabla_\mu = \partial_\mu+A_\mu. $$ where $A_\mu$ takes values in the Lie algebra of the gauge group. In other words $$ A_\mu= A_\mu^a \lambda_a $$ where $$ [\lambda_a,\lambda_b]= {f_{ab}}^c \lambda_c $$ and the $A_\mu^a$ are just number-valued functions of position. Now expanding out and using $\partial_{\mu\nu}= \partial_{\nu\mu}$ we have $$ [\partial_\mu +A_\mu,\partial_\nu+ A_\nu]= \partial_\mu A_\nu -\partial_\nu A_\mu +[A_\mu,A_\nu]\\ = \lambda_c(\partial_\mu A^c_\nu -\partial_\nu A^c_\mu + {f_{ab}}^c A^a_\mu A^b_\nu) $$

Answer 2

I want to add on to Mike's answer, because there's some subtlety here about whether we're talking about a commutator of Lie algebra elements or of operators. We should really think of $A_\mu$ as the operator $\text{ad}_{A_\mu}$ in the adjoint representation. Then for $\alpha$ in the Lie algebra we have

$$\partial_\mu \text{ad}_{A_\nu} \alpha = \partial_\mu [A_\nu, \alpha]$$ $$ = [\partial_\mu A_\nu, \alpha] + [A_\nu, \partial_\mu \alpha]$$ $$= \text{ad}_{\partial_\mu A_\nu} \alpha + \text{ad}_{A_\nu} \partial_\mu \alpha$$

or in other words,

$$\partial_\mu \text{ad}_{A_\nu} = \text{ad}_{\partial_\mu A_\nu} + \text{ad}_{A_\nu} \partial_\mu$$

Now the covariant derivative is

$$D_\mu = \partial_\mu + \text{ad}_{A_\mu}$$

These terms are all operators, so the commutator is defined in terms of composition and subtraction

$$[D_\mu, D_\nu] = [\partial_\mu, \partial_\nu] + \partial_\mu \text{ad}_{A_\nu} - \text{ad}_{A_\nu}\partial_\mu$$ $$+ [\text{ad}_{A_\mu}, \text{ad}_{A_\nu}] - \partial_\nu \text{ad}_{A_\mu} + \text{ad}_{A_\mu}\partial_\nu$$ $$= 0 + \text{ad}_{\partial_\mu A_\nu} - \text{ad}_{\partial_\nu A_\mu} + \text{ad}_{[A_\mu, A_\nu]}$$ $$= \text{ad}_{\partial_\mu A_\nu - \partial_\nu A_\mu + [A_\mu, A_\nu]}$$

Or, abusing notation and writing $A$ instead of $\text{ad}_{A}$,

$$[D_\mu, D_\nu] = \partial_\mu A_\nu - \partial_\nu A_\mu + [A_\mu, A_\nu]$$

To be more precise we could say $[D_\mu, D_\nu] = \text{ad}_{F_{\mu\nu}}$, then the above expression is $F_{\mu\nu}$.