Yang-Mills field-strength 2-form and exterior gauge-covariant derivative

Question

I think that my problem is not having a formal definition of how the exterior covariant derivative works. What I know is that the exterior covariant derivative $D_A$ is defined as a generalization of the exterior derivative $d$ $$ D_A:=d+\rho(A)\wedge \ , $$ where $A=A_\mu^I\, dx^\mu \, T_I$ is the gauge-potential field with respect to my Lie algebra $\mathcal{G}$ and $\rho(A)$ is some representation of $A$ depending on what will be the object on which this $D_A$ will act. Greek letters $\mu,\nu,\dots$ are indices of space-time, while uppercase latin letters $I,J,K,\dots$ are internal indices; $\{T_I\}$ are the generators of the Lie algebra.

1. $\phi=\phi^I\, T_I \in \Omega^0(\mathcal{G})$ is a Lie algebra-valued 0-form $$ \begin{aligned} D_A \phi & =d\phi +\rho ( A) \phi =\\ & =\left( \partial _{\mu } \phi ^{I} T_{I} +A_{\mu }^{J} \phi ^{I} \rho ( T_{J}) \blacktriangleright T_{I}\right) dx^{\mu } =\\ & =\left( \partial _{\mu } \phi ^{K} +A_{\mu }^{J} \phi ^{I} c_{JI}^{K}\right) dx^{\mu }\, T_{K} \ , \end{aligned} $$ where I used the fact that $\rho ( T_{J}) \blacktriangleright T_{I}$ is the adjoint representation, so $\rho ( T_{J}) \blacktriangleright T_{I}=[T_J,T_I]=c_{JI}^K\, T_K$ and $c_{JK}^I$ are the structure constants of $\mathcal{G}$. I am interested in the gauge-covariant derivative expressed in terms of components, so here I obtained that $$ \mathcal{D}_\mu \phi^K= \partial _{\mu } \phi ^{K} +A_{\mu }^{J} \phi ^{I} c_{JI}^{K} \ , $$ which seems reasonable.

2. $\omega=\omega_\mu^I\,dx^\mu\, T_I \in \Omega^1(\mathcal{G})$ is a Lie algebra-valued 1-form $$ \begin{aligned} D_A \omega & =d\omega +\rho ( A) \land \omega =\\ & =\partial _{\mu } \omega _{\nu }^{I} dx^{\mu } \land dx^{\nu } T_{I} +A_{\mu }^{I} \omega _{\nu }^{J} dx^{\mu } \land dx^{\nu } \rho ( T_{I}) \blacktriangleright T_{J} =\\ & =\tfrac{1}{2}\left(\left( \partial _{\mu } \omega _{\nu }^{K} -\partial _{\nu } \omega _{\mu }^{K}\right) +\left( A_{\mu }^{I} \omega _{\nu }^{J} -A_{\nu }^{I} \omega _{\mu }^{J}\right) c_{IJ}^{K}\right) T_{K} dx^{\mu } \land dx^{\nu } =\\ & =\tfrac{1}{2}\left(\left( \partial _{\mu } \omega _{\nu }^{K} +c_{IJ}^{K} A_{\mu }^{I} \omega _{\nu }^{J}\right) -\left( \partial _{\nu } \omega _{\mu }^{K} +c_{IJ}^{K} A_{\nu }^{I} \omega _{\mu }^{J}\right)\right) T_{K} dx^{\mu } \land dx^{\nu } =\\ & =\tfrac{1}{2}\left(\mathcal{D}_{\mu } \omega _{\nu }^{K} -\mathcal{D}_{\nu } \omega _{\mu }^{K}\right) T_{K} dx^{\mu } \land dx^{\nu } \ , \end{aligned} $$ where of course now I obtained that the exterior gauge-covariant derivative $D_A\omega$ is related to the anti-symmetrization of the gauge-covariant derivative $\mathcal{D}_{\mu } \omega _{\nu }^{K}$ (as the exterior derivative $d$ is related to the derivative $\partial_\mu$ through $d\omega=\tfrac{1}{2}(\partial_\mu \omega_\nu-\partial_\nu\omega_\mu)dx^\mu \wedge dx^\nu$ by definition - here $d$ has been promoted to $D_A$ and $\partial_\mu$ to $\mathcal{D}_\mu$ in order to obtain gauge-covariant derivatives).

3. What if now I want to use the same reasoning with the gauge-potential $A$ itself? It is a Lie algebra-valued 1-form, so I should follow the same steps of what I have done for $\omega$, but I do not obtain the usual definition of the YM field-strength $F$, which is usually written as $$ F_{\mu\nu}^I=\partial_\mu A_\nu^I-\partial_\nu A_\mu^I + A_\mu^J A_\nu^K c_{JK}^I \ , $$ while, following the same steps of point 2. I obtain $$ \begin{aligned} F=D_{A} A & =dA+\rho ( A) \land A=\\ & =\partial _{\mu } A_{\nu }^{I} dx^{\mu } \land dx^{\nu } T_{I} +A_{\mu }^{I} A_{\nu }^{J} dx^{\mu } \land dx^{\nu } \rho ( T_{I}) \blacktriangleright T_{J} =\\ & =\tfrac{1}{2}\left(\left( \partial _{\mu } A_{\nu }^{K} -\partial _{\nu } A_{\mu }^{K}\right) +\left( A_{\mu }^{I} A_{\nu }^{J} -A_{\nu }^{I} A_{\mu }^{J}\right) c_{IJ}^{K}\right) T_{K} \ dx^{\mu } \land dx^{\nu } =\\ & =\tfrac{1}{2}\left(\left( \partial _{\mu } A_{\nu }^{K} -\partial _{\nu } A_{\mu }^{K}\right) +2A_{\mu }^{I} A_{\nu }^{J} c_{IJ}^{K}\right) T_{K} \ dx^{\mu } \land dx^{\nu } \ , \end{aligned} $$ meaning that I have an extra factor 2 in the second term, i.e. $$ F_{\mu\nu}^I=\partial_\mu A_\nu^I-\partial_\nu A_\mu^I + 2A_\mu^J A_\nu^K c_{JK}^I \ . $$

What am I doing wrong?

Answer 1

Let me first of all give a more precise definition of the curvature and exterior covariant derivative. To start with, lets fix the following data:

• A smooth manifold $\mathcal{M}$
• A principal $G$-bundle $P$, where $G$ is a (finite-dimensional) Lie group with Lie algebra $\mathfrak{g}$.

Now, as you have said, a connection $1$-form is a $\mathfrak{g}$-valued $1$-form, i.e. $A\in\Omega^{1}(P,\mathfrak{g})$ satisfying some extra properties. In order to define the corresponding curvature $2$-form $F^{A}\in\Omega^{2}(P,\mathfrak{g})$ there are several (equivalent) ways. One is to define it directly via the structure equation, which is $$F^{A}:=\mathrm{d}A+\frac{1}{2}[A\wedge A]$$ where $\mathrm{d}$ is just the usual exterior covariant derivative of Lie-algebra valued forms, i.e. $$\mathrm{d}A=\mathrm{d}(A^{a}T_{a}):=(\mathrm{d}A^{a})T_{a}$$ where $A^{a}\in\Omega^{1}(P)$ and $\{T_{a}\}_{a}$ is a basis of $\mathfrak{g}$ and where $[\cdot\wedge\cdot]$ is just the wedge product defined via the commutator, i.e.

$$[A\wedge A]:=\sum_{a,b}(A^{a}\wedge A^{b})[T_{a},T_{b}],$$

where $A^{a}\wedge A^{b}$ is just the standard wedge-product of real-valued forms. Starting from this, it is straight-forward to get the coordinate expression: Take a local section ("local gauge") $s\in\Gamma(U,P)$, where $U\subset\mathcal{M}$ open. Then, we define $$A_{s}:=s^{\ast}A\in\Omega^{1}(U,\mathfrak{g})$$ as well as $$F^{A}_{s}:=s^{\ast}F^{A}\in\Omega^{2}(U,\mathfrak{g}),$$ which are now forms defined on $U\subset\mathcal{M}$. A straight-forward computation then yields $$F_{\mu\nu}^{a}=\partial_{\mu}A^{a}_{\nu}-\partial_{\nu}A_{\mu}^{a}+c_{bd}^{a}A_{\mu}^{b}A_{\nu}^{d}$$ where $F_{\mu\nu}^{a}$ are defined via $F_{s}^{A}=F^{a}_{\mu\nu}T_{a}\mathrm{d}x^{\mu}\wedge\mathrm{d}x^{\nu}$ and $A_{\mu}^{a}$ via $A_{s}=A_{\mu}^{a}T_{a}\mathrm{d}x_{\mu}$. The constants $c^{a}_{bc}$ are the structure constants of the Lie algebra, i.e. $[T_{a},T_{b}]=c_{ab}^{d}T_{d}$.

Now, secondly, you can define the curvature via the covariant derivative $D_{A}$. Let $(\rho,V)$ be some (finite-dimensional) representation of $G$. Then, $D_{A}$ is a (family of) map(s) $$D_{A}:\Omega^{k}(P,V)\to\Omega^{k+1}(P,V)$$ defined via $$(D_{A}\omega)_{p}(v_{1},\dots,v_{k}):=(\mathrm{d}\omega)_{p}(\mathrm{pr}(v_{1}),\dots,\mathrm{pr}(v_{k}))$$ for all $p\in P$, $v_{1},\dots,v_{k}\in T_{p}P$ and $\omega\in\Omega^{k}(P,V)$, where $\mathrm{d}$ on the right-hand side is the standard exterior derivative of $V$-valued forms (as explained above) and where $\mathrm{pr}:TP\to H$ is the projection onto the horizontal tangent space $H_{p}:=\mathrm{ker}(A_{p})\subset T_{p}P$ (the "Ehresmann connection corresponding to $A$").

Now, here is the most important point: You stated that $D_{A}$ can be calculated via the formula $$D_{A}\cdot=\mathrm{d}\cdot+\rho(A)\wedge\cdot.$$ That is in general not true! In fact, it is only true for forms living in the subset $$\Omega^{k}_{\mathrm{hor}}(P,V)^{\rho}\subset\Omega^{k}(P,V)$$

This set consists of all forms $\omega\in\Omega^{k}(P,V)$ satisfying the following two extra properties:

1. "$\omega$" is horizontal": $\omega_{p}(v_{1},\dots,v_{k})=0$ whenever at least one of the tangent vectors $v_{i}$ is vertical (i.e. not contained in $H_{p}$).
2. "$\omega$ is of type $\rho$": $(r_{g}^{\ast}\omega)_{p}=\rho(g^{-1})\circ\omega_{p}$ for all $g\in G$, where $r_{g}:P\to G,p\mapsto p\cdot g$ (the action $\cdot:P\times G\to P$ denotes the right-group action contained in the definition of a principal bundle)

In other words, your "definition" of the covariant derivative is actually only valid for this subset. Now, a connection $1$-form is in general not an element of $\Omega^{1}_{\mathrm{hor}}(P,\mathfrak{g})^{\mathrm{Ad}}$! (*)

To sum up, you cannot use your formula for a connection $1$-form. However, you can actually easily show that

$$D_{A}A\stackrel{!}{=}F^{A}=\mathrm{d}A+\frac{1}{2}[A\wedge A],$$

which then also leads to the correct coordinate expression as explained above.

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* _{However, the curvature is an element of $\Omega^{2}_{\mathrm{hor}}(P,\mathfrak{g})^{\mathrm{Ad}}$ and also the difference of two connection $1$-forms is an element of $\Omega^{1}_{\mathrm{hor}}(P,\mathfrak{g})^{\mathrm{Ad}}$. The latter statement implies that the set of connection $1$-forms $\mathcal{C}(P)$ of a principal bundle is an infinite-dimensional affine space with vector space $\Omega^{1}_{\mathrm{hor}}(P,\mathfrak{g})^{\mathrm{Ad}}$.}