Let us consider a principal bundle $P$, with a Lie-algebra-valued connection one-form $\omega\in\mathfrak{g}\otimes\Omega^1(P)$ and a Lie-algebra-valued curvature two-form $\Omega\in\mathfrak{g}\otimes\Omega^1(P)$. I will now recapitulate several definitions/assumptions:

- Definition of covariant derivative on the principal bundle: $$\mathrm{D}\phi(X_1,\dots,X_{r+1}):=\mathrm{d}_P\phi(X_1^H,\dots,X_{r+1}^H)\,,\tag{1}\label{def cov der}$$ where $\phi\in\mathfrak{g}\otimes\Omega^r(P)$ and by $X^H$ I mean the horizontal component of $X\in T_uP=V_uP\oplus H_uP$, that is $X=X^V+X^H$. Moreover, $\mathrm{d}_P$ is the exterior derivative on $P$.
- The connection has the following property: $$\omega(X^H)=0\,,\tag{2}\label{onH}$$ for all $X^H\in H_uP$ horizontal.
- Definition of curvature: $\Omega:=\mathrm{D}\omega$.
- The curvature satisfies the Cartan structure equation: $$\Omega=\mathrm{d}_P\omega+\omega \wedge\omega\,.\tag{3}\label{cartan}$$

Now, I aim to prove the Bianchi identity, that is, $\mathrm{D}\Omega=0$, using \eqref{def cov der}, \eqref{onH}, and \eqref{cartan}.

- Applying \eqref{def cov der} to the curvature, we obtain $$ \mathrm{D}\Omega(X,Y,Z)=\mathrm{d}_P \Omega(X^H,Y^H,Z^H)\,.\tag{*} \label{*} $$
- Differentiating \eqref{cartan}, we obtain $$ \mathrm{d}_P\Omega =\mathrm{d}_P\omega\wedge \omega-\omega\wedge \mathrm{d}_P=\Omega\wedge\omega-\omega\wedge\Omega\,,$$ so Eq. \eqref{*} can be continued as $$ \mathrm{D}\Omega(X,Y,Z)=\mathrm{d}_P \Omega(X^H,Y^H,Z^H)=(\Omega\wedge\omega-\omega\wedge\Omega)(X^H,Y^H,Z^H)\,.\tag{**} \label{**} $$
- Now, recalling that \eqref{onH}, it is evident that \eqref{**} is zero.

My question pertains to why the Bianchi identity is often written (e.g., Nakahara's "Geometry, Topology and Physics" or here) as: $$ \mathrm{D}\Omega=\mathrm{d}_P\Omega+\omega\wedge\Omega-\Omega\wedge\omega=0\,,\tag{$\star$}\label{Bid} $$ globally on the principal bundle, or $$ D\mathcal{F}=\mathrm{d}\mathcal{F}+\mathcal{A}\wedge\mathcal{F}-\mathcal{F}\wedge\mathcal{A}=0\,, $$ in the local form on the base manifold, where $\mathcal{F}$ and $\mathcal{A}$ are the local curvature and the local connection, respectively. By "local form" on the base manifold, I mean the pullback of a form on $P$ through a local section, once an open covering on the base manifold is chosen.

I don't understand why the Bianchi identity is written like this if we know that the covariant derivative of the curvature is equal to its exterior derivative evaluated on the horizontal components of the vector fields (as in Eq. \eqref{**}).

I realize that, of course, Eq. \eqref{Bid} is always true: on one side we have $\mathrm{D}\Omega=0$, which I proved in these three steps; on the other side we have $\mathrm{d}_P\Omega+\omega\wedge\Omega-\Omega\wedge\omega=0$, which is *of course* true. It is just the exterior derivative of the Cartan equation, which I can rewrite as $\Omega-\mathrm{d}_P\omega -\omega \wedge\omega=0$. Its exterior derivative is obviously still zero. So I have two identities, $\mathrm{D}\Omega=0$ and $\mathrm{d}_P\Omega+\omega\wedge\Omega-\Omega\wedge\omega=0$. How do we prove Eq. \eqref{Bid}? Do I miss a step in which I should prove that $\mathrm{D}\Omega=\mathrm{d}_P\Omega+\omega\wedge\Omega-\Omega\wedge\omega$? And, if they really are equals, why do we bother ourselves to prove that $\mathrm{D}\Omega=0$, instead of just proving that $\mathrm{D}\Omega=\mathrm{d}_P\Omega+\omega\wedge\Omega-\Omega\wedge\omega$ that we already know is zero (for the Cartan structure equation)? Or are we just equating them because they are two things that vanish separately, so they are equal because both are zero? What would stop me from adding more identically-zero terms?

`\rm`

does not take an argument but applies to the remainder of the enclosed block, so that, e.g.,`\rm{d}_P`

is the same as`\rm d_P`

and produces $\rm d_P$ (note the upright P). You probably meant`\mathrm{d}_P`

, which produces $\mathrm{d}_P$. I edited accordingly. $\endgroup$