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Theorem 5.14.2 of Mathematical Gauge Theory by Hamilton states that the curvature form $F^A_M\in\Omega^2(M,Ad(P))$ satisfies the third form of the Bianchi identity $d_AF^A_M=0$ where $F^A_M$ is the form in $\Omega^2(M,AD(P))$ given by the correspondence between horizontal $k$-forms of type Ad on $P$ with values in $\mathfrak{g}$ and forms on $M$ with values in $Ad(P)$. The proof of this is basically applying this isomorphism to the bianchi identity $dF^A+[A,F^A]=0$. I am interested in how to show that $d_AF^A_M\in\Omega^2(M,AD(P))$ actually corresponds to $dF^A+[A,F^A]$. It has to do with the fact that with respect to a local gauge $s:U\rightarrow P$ we can write $(d_A\omega)_s=d\omega_s+A_s\wedge \omega_s$ and the fact that for $1$-forms the wedge product "is the commutator". Any help is appreciated

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To answer this question, it is useful to look how the exterior covariant derivative $\mathrm{d}_{A}$ acts on form on $P$ and not on $M$. Let us fix the following data.

  1. A smooth manifold $M$.
  2. A principal $G$-bundle $P$ over $M$ with $\mathfrak{g}=\mathrm{Lie}(G)$.
  3. A finite-dimensional representation $(V,\rho)$ of $G$ and let $E:=P\times_{\rho}V$ be the associated vector bundle.

Let $A\in\Omega^{1}(P,\mathfrak{g})$ be a connection $1$-form. Now, there are two common ways to define the object $\mathrm{d}_{A}$, i.e. the exterior covariant derivative induced by $A$. One approach is to first of all define a connection $\nabla^{A}$ on $E$ using a parallel transport. Then, one defines $\mathrm{d}_{A}$ to be the exterior covariant derivative induced by $\nabla^{A}$. This approach is for example followed in Hamilton's book Mathematical Gauge Theory. However, there is another approach, which can be found in many other books (e.g. Baum's Eichfeldtheorie), which is equivalent. In this approach, one first constructs a natural derivative on forms on $P$, which preserves horizontal forms, and then one induces $\mathrm{d}_{A}$ on forms on $M$ using the usual isomorphism. The connection $\nabla^{A}$ is then recovered as the operator $\mathrm{d}_{A}$ acting on zero-forms.

In more details: First of all, we would like to define a natural derivative on $\Omega^{k}(P,V)$. A natural canditate is to take the usual exterior derivative $\mathrm{d}:\Omega^{k}(P,V)\to\Omega^{k+1}(P,V)$. However, in the language of principal bundles, it is more natural to have an object preserving horizontal vector fields. Hence, we consider \begin{align*} \mathrm{D}_{A}:\Omega^{k}(P,V)\to\Omega^{k+1}(P,V) \end{align*} which we define for all $k\in\mathbb{N}_{0}$ and for all $\omega\in \Omega^{k}(P,V)$ by \begin{align*} (\mathrm{D}_{A}\omega)_{p}(v_{1},\dots,v_{k}):=(\mathrm{d}\omega)_{p}(\mathrm{pr}_{H}(v_{1}),\dots,\mathrm{pr}_{H}(v_{k})) \end{align*} for all $p\in P$ and $v_{1},\dots,v_{k}\in T_{p}P$, where $\mathrm{pr}_{H}:TP\to H$ denotes the projection onto the Ehresmann connection $H$ defined by $A$. Now, it turns out that this operator has the following very nice property:

$D_{A}$ maps $\Omega^{k}_{\mathrm{hor}}(P,V)^{\rho}$ to $\Omega^{k+1}_{\mathrm{hor}}(P,V)^{\rho}$. Furthermore, for $\omega\in\Omega^{k}_{\mathrm{hor}}(P,V)^{\rho}$, the following formula holds $$D_{A}\omega=\mathrm{d}\omega+A\wedge_{\rho}\omega,$$ where $\wedge_{\rho}$ is the natural wedge product $\wedge_{\rho}:\Omega^{k}(P,\mathfrak{g})\times\Omega^{l}(P,V)\to\Omega^{k+l}(P,V)$ defined by the pairing $\langle\cdot,\cdot\rangle_{\rho}:\mathfrak{g}\times V\to V,(X,v)\mapsto \rho_{\ast}(X)v$, i.e. $A\wedge_{\rho}\omega:=\sum_{a,b}(A^{a}\wedge\omega^{b})\langle T_{a},e_{b}\rangle_{\rho}$ for basis $(T_{a})_{a}$ of $\mathfrak{g}$ and a basis $(e_{a})_{a}$ of $V$.

Secondly, thanks to the natural isomorphism

$$\Omega^{k}_{\mathrm{hor}}(P,V)^{\rho}\cong\Omega^{k}(M,E)$$

the operator $D_{A}$ induces an operator $\mathrm{d}_{A}:\Omega^{k}(M,E)\to\Omega^{k+1}(M,E)$. This is the same operator as defined using the strategy of parallel transports used in Hamilton's book. Usually, people denote both $D_{A}$ and $\mathrm{d}_{A}$ by the same symbol, but I wanted to avoid confusions. In particular, you see that under the isomosphism above, you get the following correspondence:

$$D_{A}\omega=\mathrm{d}\omega+A\wedge_{\rho}\omega\quad \mapsto \quad\mathrm{d}_{A}\omega_{M}$$

where $\omega_{M}\in\Omega^{k}(M,E)$ is the form obtained from $\omega\in\Omega^{k}_{\mathrm{hor}}(P,V)^{\rho}$ via the isomorphism above. Now, if you take $(\rho,V)$ to be the adjoint representation $(\mathrm{Ad},\mathfrak{g})$ of $G$ (and hence $\mathrm{Ad}_{\ast}=\mathrm{ad}$), you get the formula

$$D_{A}\omega=\mathrm{d}\omega+[A,\omega]$$

which is a global version of the local formula you very mentioning. In particular, you see that $\mathrm{d}_{A}F_{M}^{A}$ really corresponds to $D_{A}F^{A}=\mathrm{d}F^{A}+[A,F^{A}]$ as you claimed.

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  • $\begingroup$ Ok this really really helps, however I am still struggling to see why in the case of the adjoint representation we get the commutator. Following the definition if both $A$ and $\omega$ are written as $A^a\otimes T_a$ and $\omega^b\otimes \omega_b$ we get $ A\wedge_{\rho}\omega:=\sum_{a,b}(A^{a}\wedge\omega^{b}) \rho_*(T_{a})T_{b}=\sum_{a,b}(A^{a}\wedge\omega^{b}) [T_{a},T_{b}]$ is this equal to $[A,\omega]$? $\endgroup$
    – PunkZebra
    Sep 29, 2023 at 10:27
  • $\begingroup$ Moreover in this approach we have actually \emph{defined} $d_A$ as the composition of $D_A$ with the isomorphism right? $\endgroup$
    – PunkZebra
    Sep 29, 2023 at 10:29
  • $\begingroup$ Yes it is. The notation $[\cdot,\cdot]:\Omega^{k}(M,\mathfrak{g})\times\Omega^{l}(M,\mathfrak{g})\to\Omega^{k+l}(M,\mathfrak{g})$ is the natural wedge product induced by the Lie bracket $[\cdot,\cdot]:\mathfrak{g}\times\mathfrak{g}\to\mathfrak{g}$. By definition, $[\alpha,\beta]=\sum_{a,b}(\alpha^{a}\wedge\beta^{b})[T_{a},T_{b}]$ where $\alpha=\sum_{a}\alpha^{a}T_{a}$ for $\alpha^{a}\in\Omega^{k}(M)$ and similar for $\beta$. (I usually prefer the notation $[\cdot\wedge\cdot]$ to stress that it is a mix of wedge product and Lie bracket.) $\endgroup$ Sep 29, 2023 at 10:29
  • $\begingroup$ @PunkZebra yes exactly, it is the composition with the isomorphism, or better, the conjugation, i.e. $\mathrm{d}_{A}=I^{-1}\circ D_{A}\vert_{\Omega^{\bullet}_{\mathrm{hor}}(P,V)^{\rho}}\circ I$ where $I:\Omega^{k}(M,E)\to\Omega^{k}_{\mathrm{hor}}(P,V)^{\rho}$ denotes the isomorphism. $\endgroup$ Sep 29, 2023 at 10:30
  • $\begingroup$ I cannot stress how much you've helped me, thank you (after thoroughly reviewing everything I'll accept your answer) $\endgroup$
    – PunkZebra
    Sep 29, 2023 at 10:35

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