# Curvature of a principal bundle and the exterior covariant derivative

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.

• I think you will have your answer once you unravel what you mean, in your formula for $D\mathcal{F}$, by "$\mathrm{d}$" when it is supposed to be operating on elements of $\Omega^r(M;\mathrm{ad}P)$ (and, of course, what you actually mean by $\mathcal{A}$ as "the local form of the connection 1-form $A$"). – Robert Bryant Mar 30 '15 at 8:36
• @RobertBryant: I think they are the same. Loosely speaking: Let $\left\{ e_{i}\right\}$ be a local frame for $ad\, P$. Then $\mathcal{F} = \mathcal{F}^{i}\otimes e_{i}$ and $d_{\nabla}\mathcal{F} = d\mathcal{F}^{i}\otimes e_{i} + \mathcal{F}^{i}\wedge \nabla e_{i} = d\mathcal{F}^{i}\otimes e_{i} + \mathcal{F}^{i}\wedge \omega_{i}^{j} e_{j} = d\mathcal{F} + [\mathcal{A},\mathcal{F}] =\mathcal{D}\mathcal{F}$ where in the last step I have used that the vector bundle is the adjoint bundle of a principal bundle with connection. – Bilateral Mar 30 '15 at 10:51
• What does 'they' refer to in your comment? I was asking you to clarify what you meant by applying $\mathrm{d}$ directly to $\mathcal{F}$ and what you meant by the term $\mathcal{A}$. I claim that neither of these two terms, each taken alone, is a well-defined operator, and, when you do take a local trivialization to define a $\mathcal{D}$ that is independent of local trivialization (as you have done), then you immediately see that $\mathcal{D}$ has the same formula as $\mathrm{d}_\nabla$, so there was no way that they could have failed to be equal, as long as they were both Leibnitz operators. – Robert Bryant Mar 30 '15 at 11:30
• @RobertBryant: When you wrote "I think you will have your answer once you unravel what you mean.." I thought you meant that I had to think about it my self and then I would arrive to the answer :). By "they" I meant $\mathcal{D}$ and $d_{\nabla}$. By $\mathcal{A}$ I mean the local pullback to the base of $A$ by means of a local section. I agree with you: the $d$ in $\mathcal{DF} = d\mathcal{F} + [\mathcal{A},\mathcal{F}]$ is not really well defined, only when you know that it is $d_{\nabla}\mathcal{F}$ it makes sense. However that formula is written like that in many references... – Bilateral Mar 30 '15 at 11:46

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.