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

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?

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.

* 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 $$\Omega^{1}_{\mathrm{hor}}(P,\mathfrak{g})^{\mathrm{Ad}}$$.