I am confused by the relation between the principal bundle and "usual" approaches to the curvature of a Riemannian manifold $M$ of dim $n$.
In the "usual" approach the Riemann curvature tensor is defined as
$R(X,Y)Z =(\nabla_X\nabla_Y-\nabla_Y\nabla_X) Z $, in components
$R ^a_{\phantom{a}bcd} =e^a R(e_c,e_d)e_b$. It is globally defined on $M$ and can be viewed as an element $\mathcal{R}\in\Omega^2(M,\mathfrak{o}(n))$, $\mathcal{R}=\mathcal{R}^a_b\; e_a\otimes e^b$, with each $\mathcal{R}^a_b$ a 2-form on $M$, $\mathcal{R}^a_b=(1/2)R^a_{bcd} \;e^c\wedge e^d$.
In the principal bundle approach one considers the frame bundle $P$ associated to $M$, an $O(n)$-bundle with base $M$. The curvature $\Omega$ of $P$ is an element of $\Omega^2(P,\mathfrak{o}(n))$, which, being tensorial, can be identified with an element $F_{\Omega}\in \Omega^2(M,\mathrm{ad} P)$, NOT of $\Omega^2(M,\mathfrak{o}(n))$
Provided that what I said above is correct, how do I recover $\mathcal{R}$ from $\Omega$?
I would have expected $F_{\Omega}=\mathcal{R}$, but this cannot be the case since they take values in different spaces.
NOTATION: $\Omega^2(M,\mathfrak{o}(n))$ denotes the space of 2-forms on $M$ taking values in the Lie algebra of $O(n)$; $\mathrm{ad} P$ is the adjoint bundle of $P$.
Best Answer
It's not quite so arcane as you might think. If you take a section $s$ of $P$, this gives you an orthonormal moving frame $e_a$. Pulling back the curvature form $\Omega$ on $P$ by $s$ gives you the $\frak o(n)$-valued curvature $2$-form form $\Omega_s$ on $M$ (which transforms by the Ad-action if you change frame field $s$). And, in turn, $(\Omega_s)^a_b = \frac12 \sum_{c,d}R^a_{bcd}\omega_c\wedge\omega_d$.