Gauge transformation rule for $dA$, where $A$ is the gauge field

Question

Let $G$ be a non-Abelian simple compact gauge group and $\{ t^\alpha\}$ be a normalized set of generators for its Lie algebra $\mathfrak{g}$. Let $C^{\alpha \beta}_\gamma$ be the coupling constant for this set of generators: \begin{equation} [ t^\alpha , t^\beta] = C^{\alpha \beta}_\gamma t^\gamma \text{ where } [,] \text{ is the Lie bracket for } \mathfrak{g} \end{equation}

Now, Let $A:= \bigl( A_{\alpha \mu} t^\alpha \bigr)_{\mu=0,1,2,3}$ be the gauge field for this theory with $\mu$ labeling the spacetime components.

Then, the local action of $G$ on $A$ is given by the formula \begin{equation} A_{\alpha \mu} t^\alpha \to A_{\alpha \mu} \bigl( g t^\alpha g^{-1} \bigr) - i \bigl[\partial_\mu g \bigr]g^{-1} \end{equation} where $g$ is any local $G$-valued mapping.

Now, my question is:

How is the transformation rule for $(dA) = \Bigl( [\partial_\nu A_{\alpha \mu} - \partial_\mu A_{\alpha \nu}] t^\alpha \Bigr)_{\nu,\mu=0,1,2,3}$ given precisely?

My guess is to $\partial_\nu$ on both sides of the above formula for $A$, but I am not sure...

Could anyone please help me with this?

Answer 1

Given $$ A \to g A g^{-1} - i dg g^{-1} $$ where $A$ is the gauge field one-form $A = A_{\alpha \mu} t^\alpha dx^\mu$ and $d$ is exterior derivative, thus $$ dA\\ \to d(g A g^{-1} - i dg g^{-1}) \\ = dg \wedge A g^{-1} + g dA g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1} \\ = g dA g^{-1} + (dg \wedge A g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1}) $$

• For Abelian case, the part in $(\dots)$ drops out, therefore $F=dA$ is covariant/invariant (meaning $F$ transforms as $F \to g F g^{-1} = F$ ) as in Electromagnetism.
• For non-Abelian case, the part in $(\dots)$ is non-zero, therefore you need the extra $A\wedge A$ as in the non-Abelian $F = dA + A\wedge A$ to make $F$ covariant. This is THE major conclusion of Yang and Mills in their 1959 paper.