Proving gauge transformation of non-abelian field strength

Question

For a general Yang-Mills theory, we have the field strength $$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu - ig [A_\mu, A_\nu]$$ I now want to prove that it transforms as $$F_{\mu\nu} \rightarrow U(x) F_{\mu\nu} U^\dagger(x)$$ under the following gauge transformation $$ A_\mu(x) \rightarrow U(x)A_\mu(x) U^\dagger(x) - \frac{i}{g}(\partial_\mu U(x))U^\dagger(x).$$

I have calculated the first term of the field strength under the gauge transformation and I obtain \begin{align} \partial_\mu A'_\nu =& \partial_\mu \bigg[U(x)A_\nu(x) U^\dagger(x) - \frac{i}{g}(\partial_\nu U(x))U^\dagger(x)\bigg] \\ =& \partial_\mu(U(x))A_\nu(x) U^\dagger(x)+ U(x)\partial_\mu(A_\nu(x)) U^\dagger(x) + U(x)A_\nu(x)\partial_\mu (U^\dagger(x)) \nonumber \\ &- \frac{i}{g}\partial_\mu\bigg[(\partial_\nu U(x))U^\dagger(x)\bigg]\\ =& ig \partial_\mu \alpha^a T^a UA_\nu U^\dagger + U (\partial_\mu A_\nu)U^\dagger -ig \partial_\mu \alpha^a UA_\nu U^\dagger T^a + \partial_\mu\partial_\nu \alpha^aT^a \\ =& U (\partial_\mu A_\nu)U^\dagger + ig \partial_\mu \alpha^a\bigg[ T^a UA_\nu U^\dagger - UA_\nu U^\dagger T^a \bigg] + \partial_\mu\partial_\nu \alpha^aT^a \end{align}

I can kind of see that the first term is the one we want and the last term will vanish from the second term in the field strength, but I am very confused about the term with the brackets because we are `multiplying' an element of the Lie algebra with an element of the Lie group and I don't know what to do there. I think I should be using the Jacobi identity somewhere due to this question, but I can't see where to use it.

EDIT: I found that $$\partial_\mu U = \partial_\mu \sum_n \frac{(ig \alpha^a(x)T^a)^n}{n!} = ig \alpha^a(x)T^a \sum_n \frac{(ig \alpha^b(x)T^b)^n}{n!} = ig \alpha^a(x)T^a U,$$ but apparently this is incorrect. I can't really see why though.

Answer 1

Specifically regarding the derivative. By definition: $$ U(x) = \exp(i \alpha_a(x) T^{a}) = \sum_{n=0}^{\infty} \frac{(i \alpha_a(x) T^a)^n}{n!} $$

Taking derivative yields $$ \partial U = \sum_{n=0}^{\infty} \sum_{m=0}^{n} \frac{1}{n!} (i \alpha_a T^a)^m \; ( i \partial \alpha_a \, T^a ) (i \alpha_a T^a)^{n-m-1}. $$

$\alpha_a \, T^a$ and $\partial \alpha_a \, T^a$ don't necessarily commute. You can't pull that term to the front.

Answer 2

You are making heavy weather of something easy. I'll use slightly different conventions but the result is the same.

Start by defining the gauge-covariant derivative $$ \nabla_\mu=\partial_\mu+A_\mu. $$ Now show that the commutator of two of these is equal to $$ [\nabla_\mu,\nabla_\nu]= F_{\mu\nu}. $$ Now use $$ g^{-1} (\partial_\mu+A_\mu) g = (\partial_\mu+A^g_\mu) $$
where $$ A^g_\mu= g^{-1}A_ \mu g+g^{-1}\partial_\mu g, $$ and $$ g^{-1} [\nabla_\mu, \nabla_\nu]g= [g^{-1}\nabla_\mu g, g^{-1}\nabla_\nu g] $$ to see immediately that $F\to g^{-1}Fg$.

With regard to the Maurer-Cartan form $g^{-1} dg$ note that diffferentiating the exponential is non trivial because for a matrix $A$ $$ e^{A+ \delta A}\ne e^A e^{\delta A}= e^A(1+ \delta A+\ldots) $$ because $A$ and $\delta A$ do not commute with each other. Instead we have the Campbell-Baker-Hausdorf formula $$ e^{-A} e^{A+\delta A}= 1+ \int_0^1 e^{-tA} \delta A e^{tA} dt + O[(\delta A)^2] $$ This is why, as @Prahar says, you should just keep the Lie-algebra element $g^{-1}dg$ as one thing.

Answer 3

Here is a derivation using explicit indices for the matrix components of elements of the Lie group and lie algebra. I'm using $g=1$.

For a change of gauge $\varphi:\mathcal M \to G$ the gauge field transforms according to \begin{align} \label{eq:classical_trans_boson} A_\mu{}^a{}_b(x) \to A_{\mu}{}^{\tilde a}{}_{\tilde b}(x):= \varphi^{\tilde a}{}_{a}(x) A_\mu{}^a{}_b(x) \varphi^{-1}{}^{b}{}_{\tilde b}(x) + \varphi^{\tilde a}{}_{c}(x) \partial_\mu \varphi^{-1}{}^{c}{}_{\tilde b}(x)\,. \end{align}

The field strength tensor is given by

\begin{align} F_{\mu\nu}{}^{a}{}_{b} = \partial_{[\mu} A_{\nu]}{}^{a}{}_{b}+ A_{[\mu}{}^{a}{}_{c} A_{\nu]}{}^{c}{}_{b}\,. \end{align}

(with the convention that $F = F_{\mu\nu} dx^\mu \wedge dx^{\nu}$)

Transformation of the field strength

We have \begin{align} F_{\mu\nu}{}^{\tilde a}{}_{\tilde b}(x) = \varphi^{\tilde a}{}_{a}(x) F_{\mu\nu}{}^{a}{}_{ b}(x) \varphi^{-1}{}^{b}{}_{\tilde b}(x)\,. \end{align}

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Proof:

\begin{align} F_{\mu\nu}{}^{\tilde a}{}_{\tilde b}(x) &= \partial_{[\mu} A_{\nu]}{}^{\tilde a}{}_{\tilde b}(x) + A_{[\mu}{}^{\tilde a}{}_{\tilde c}(x) A_{\nu]}{}^{\tilde c}{}_{\tilde b}(x)\,. \end{align} For the first term we have \begin{align} \label{eq:class_first_part} \partial_{[\mu} A_{\nu]}{}^{\tilde a}{}_{\tilde b}(x) &= \partial_{[\mu} ( \varphi^{\tilde a}{}_{a}(x) A_{\nu]}{}^{ a}{}_{ b}(x) \varphi^{-1}{}^{b}{}_{\tilde b}(x)) + \partial_{[\mu} ( \varphi^{\tilde a}{}_{c}(x) [\partial_{\nu]} \varphi^{-1}{}^{c}{}_{\tilde b}(x)]) \nonumber\\ &= (\partial_{[\mu} \varphi^{\tilde a}{}_{a}(x)) A_{\nu]}{}^{ a}{}_{ b}(x) \varphi^{-1}{}^{b}{}_{\tilde b}(x)\nonumber\\ &+ \varphi^{\tilde a}{}_{a}(x) (\partial_{[\mu}A_{\nu]}{}^{ a}{}_{ b}(x)) \varphi^{-1}{}^{b}{}_{\tilde b}(x)\nonumber\\ & + \varphi^{\tilde a}{}_{a}(x) A_{[\nu}{}^{ a}{}_{ b}(x) (\partial_{\mu]} \varphi^{-1}{}^{b}{}_{\tilde b}(x))\nonumber\\ & + \partial_{[\mu} ( \varphi^{\tilde a}{}_{c}(x) [\partial_{\nu]} \varphi^{-1}{}^{c}{}_{\tilde b}(x)])\,. \tag{1} \end{align} For the second term we have \begin{align} \label{eq:class_second_part} A_{[\mu}{}^{\tilde a}{}_{\tilde c}(x) A_{\nu]}{}^{\tilde c}{}_{\tilde b}(x) &= \varphi^{\tilde a}{}_{a}(x) A_{[\mu}{}^{a}{}_{g}(x) \varphi^{-1}{}^{g}{}_{\tilde c}(x) \varphi^{\tilde c}{}_{f}(x) A_{\nu]}{}^{f}{}_{b}(x) \varphi^{-1}{}^{b}{}_{\tilde b}(x)\nonumber\\ &+ \varphi^{\tilde a}{}_{a}(x) A_{[\mu}{}^{a}{}_{g}(x) \varphi^{-1}{}^{g}{}_{\tilde c}(x) \varphi^{\tilde c}{}_{k}(x) [\partial_{\nu]} \varphi^{-1}{}^{k}{}_{\tilde b}(x)]\nonumber\\ &+ \varphi^{\tilde a}{}_{d}(x) [\partial_{[\mu} \varphi^{-1}{}^{d}{}_{\tilde c}(x)] \varphi^{\tilde c}{}_{f}(x) A_{\nu]}{}^{f}{}_{b}(x) \varphi^{-1}{}^{b}{}_{\tilde b}(x)\nonumber\\ &+ \varphi^{\tilde a}{}_{d}(x) [\partial_{[\mu} \varphi^{-1}{}^{d}{}_{\tilde c}(x)] \varphi^{\tilde c}{}_{k}(x)[\partial_{\nu]} \varphi^{-1}{}^{k}{}_{\tilde b}(x)]\nonumber\\ % % &=\varphi^{\tilde a}{}_{a}(x) A_{[\mu}{}^{a}{}_{g}(x) \delta^g{}_f A_{\nu]}{}^{f}{}_{b}(x) \varphi^{-1}{}^{b}{}_{\tilde b}(x)\nonumber\\ &+ \varphi^{\tilde a}{}_{a}(x) A_{[\mu}{}^{a}{}_{g}(x) \delta^g{}_k [\partial_{\nu]} \varphi^{-1}{}^{k}{}_{\tilde b}(x)]\nonumber\\ &-[\partial_{[\mu} \varphi^{\tilde a}{}_{d}(x)] \delta^d{}_f A_{\nu]}{}^{f}{}_{b}(x) \varphi^{-1}{}^{b}{}_{\tilde b}(x)\nonumber\\ &-[\partial_{[\mu} \varphi^{\tilde a}{}_{d}(x)]\delta^d{}_k [\partial_{\nu]} \varphi^{-1}{}^{k}{}_{\tilde b}(x)]\nonumber\\ % % &=\varphi^{\tilde a}{}_{a}(x) A_{[\mu}{}^{a}{}_{f}(x) A_{\nu]}{}^{f}{}_{b}(x) \varphi^{-1}{}^{b}{}_{\tilde b}(x)\nonumber\\ &-[\partial_{[\mu} \varphi^{-1}{}^{k}{}_{\tilde b}(x)] \varphi^{\tilde a}{}_{a}(x) A_{\nu]}{}^{a}{}_{k}(x) \nonumber\\ &-[\partial_{[\mu} \varphi^{\tilde a}{}_{d}(x)] A_{\nu]}{}^{d}{}_{b}(x) \varphi^{-1}{}^{b}{}_{\tilde b}(x)\nonumber\\ &-[\partial_{[\mu} \varphi^{\tilde a}{}_{d}(x)] [\partial_{\nu]} \varphi^{-1}{}^{d}{}_{\tilde b}(x)]\,.\tag{2} \end{align} By comparing (1) and (2), we see that all terms that depend on derivatives of $\varphi$ or $\varphi^{-1}$ cancel each other. Thus, we are left with \begin{align} F_{\mu\nu}{}^{\tilde a}{}_{\tilde b}(x) &= \varphi^{\tilde a}{}_{a}(x) (\partial_{[\mu}A_{\nu]}{}^{ a}{}_{ b}(x)) \varphi^{-1}{}^{b}{}_{\tilde b}(x) + \varphi^{\tilde a}{}_{a}(x) A_{[\mu}{}^{a}{}_{f}(x) A_{\nu]}{}^{f}{}_{b}(x) \varphi^{-1}{}^{b}{}_{\tilde b}(x)\nonumber\\ &= \varphi^{\tilde a}{}_{a}(x) F_{\mu\nu}{}^{ a}{}_{ b}(x) \varphi^{-1}{}^{b}{}_{\tilde b}(x) \,. \end{align}

Remark:

If your gauge group can be represented as a subgroup $G \subset \operatorname{U}(n)$ of the unitary group, then of course $\varphi^{-1} = \varphi^\dagger$. In the case, your formula follows.