Curvature of a principal bundle and the exterior covariant derivative

Question

I am sorry if this is too elementary; I had posted it on math.stack but no one answered.

Let $P\to M$ a principal fibre bundle with fibre $G$, and let $A\in \Omega^{1}(P)\otimes\mathfrak{g}$ be a connection on $P$, where $\mathfrak{g}$ is the Lie algebra of $G$. Associated to every connection there is a curvature $F\in\Omega^{2}(P)\otimes\mathfrak{g}$ defined as

$F = DA$

where $D\colon \Omega^{r}(P)\otimes\mathfrak{g}\to \Omega^{r+1}(P)\otimes\mathfrak{g}$ defined as $D\Omega(X_{1},\dots, X_{r+1}) = d\Omega(X^{H}_{1},\dots, X^{H}_{r+1})$. The superscript $H$ denotes the projection to the horizontal distribution given by the connection and $d\colon \colon \Omega^{r}(P)\to \Omega^{r+1}(P)$. Now, $F$ induces (using local sections), a two form

$\mathcal{F}\in\Omega^{2}(M;\mathrm{ad} P)$

taking values in the adjoint bundle of $P$, which is a vector bundle of rank $dim\, \mathfrak{g}$. The two-form $\mathcal{F}$ satisfies

$\mathcal{D}\mathcal{F} = d\mathcal{F} + [\mathcal{A},\mathcal{F}] = 0$

where $\mathcal{A}$ is the local form of the connection one-form $A$. The two-form $\mathcal{F}$ is a particular case of a form taking values in a vector bundle with a connection. For this kind of forms there is a natural derivative $d_{\nabla}$, the exterior covariant derivative. In the case of $\mathcal{F}$ the vector bundle is $\mathrm{ad}\, P$ and the connection $\nabla$ is the induced on the adjoint bundle by $A$. My question is, what is the relation between $\mathcal{D}$ and $d_{\nabla}$? Are they the same?

Thanks.

Answer 1

First, there is a canonical isomorphism $\Phi$ between $\Omega^k(M; Ad P)$ and $\Omega^k_{Ad, h}(P; \mathfrak{g})$, where the subscripts signify that the form is horizontal and of type $Ad$ (i.e. equivariant). This isomorphism is canonical and does not need a local trivialization. To illustrate the idea, consider the case $k=0$. A section $\varphi$ of the adjoint bundle defines a function $\Phi(\varphi)$ on $P$ with values in $\mathfrak{g}$ by $\varphi(m) = [p, \Phi(\varphi) (p)]$ where $\pi(p) = m$. In your notation, $\Phi^{-1}(F) = \mathcal{F}$.

Next, the exterior differential in the adjoint bundle (indeed in every associated bundle) is defined by making the following diagram commutative: $$\begin{matrix} \Omega^k_{Ad, h}(P; \mathfrak{g}) & \overset{D}{\to} & \Omega^{k+1}_{Ad, h}(P; \mathfrak{g}) \\ \downarrow & & \downarrow \\ \Omega^k(M; Ad P) & \overset{d_{\nabla}}{\to} & \Omega^{k+1}(M; Ad P) \end{matrix}$$ (the vertical arrows being the isomorphism $\Phi$)

Thus $D F = 0$ coincides with $d_{\nabla}\mathcal{F} = 0$ under the isomorphism $\Phi$.

Finally, there is also the viewpoint via local trivializations. Locally, we can identify the connection form with a Lie algebra valued 1-form $\tilde{A}$ on $M$ and the curvature is a 2-form $\tilde{F}$ on $M$ with values in the Lie algebra. It is an easy calculation to see that $D$ locally takes the form $\widetilde{D F} = d \tilde{F} + [\tilde{A}, \tilde{F}]$, i.e. the local representant of $DF$ is given by the right-hand side. Note that this expression only makes sense locally, since the differential of $\mathcal{F}$ as a vector-valued 2-form is not well-defined (without a connection). Nonetheless, $d_\nabla$ is always defined and can be applied to $\mathcal{F}$ without any problems.