= Solution
A smooth <fiber bundle> $\pi:E\to M$ with typical fiber $F$ is locally a product: every base point has a neighborhood $U$ and a <local trivialization> $\pi^{-1}(U)\cong U\times F$ commuting with projection to $U$. Its overlap maps act smoothly on $F$. A <principal bundle> with structure <Lie group> $G$ has a free right $G$-action, each fiber is one orbit, and the <local trivializations> are equivariant for right multiplication on $G$. A smooth <section of a fiber bundle> is a map $s:M\to E$ with $\pi\circ s=\operatorname{id}_M$.
For a <principal bundle>, a global <section of a fiber bundle> gives the map
$$
\Phi:M\times G\longrightarrow P,\qquad\Phi(x,g)=s(x)g.
$$
The free transitive action on each fiber makes $\Phi$ bijective. In a local trivialization, write $s(x)=(x,a(x))$; then $\Phi(x,g)=(x,a(x)g)$ and its inverse is $(x,k)\mapsto(x,a(x)^{-1}k)$. Both are smooth and $\Phi$ is equivariant. Conversely, a product <principal bundle> has the section $x\mapsto(x,e)$. This proves \b[a principal bundle is trivial exactly when it has a global section]. The argument depends on the free transitive action: an arbitrary bundle section need not trivialize its bundle.
For an $n$-dimensional <smooth manifold>, the <frame bundle> has fiber over $x$ consisting of ordered bases of $T_xM$, equivalently linear <isomorphisms> $u:\mathbb R^n\to T_xM$. Right composition by $GL(n,\mathbb R)$ makes it a <principal bundle>. If a metric has signature $(p,q)$, its <pseudo-orthonormal frame bundle> consists of the isometries $u:\mathbb R^{p,q}\to T_xM$, with structure group $O(p,q)$. In the positive-definite case this is the usual <orthonormal frame bundle>. Choosing space and time orientations reduces the Lorentzian structure group to its corresponding connected subgroup.
For <Minkowski spacetime>, use a fixed origin and a standard Lorentzian frame. Every pair consisting of a point and a pseudo-orthonormal frame is the image of this reference pair under a unique affine isometry. Thus, for the full groups,
$$
\boxed{\operatorname{Fr}_\eta(\mathbb R^{1,3})\cong ISO(1,3)=\mathbb R^{1,3}\rtimes O(1,3),\qquad\mathbb R^{1,3}\cong ISO(1,3)/O(1,3).}
$$
Here the first expression denotes the total space of the <orthonormal frame bundle>, not merely the Lorentz group. Projection sends an affine isometry to the image of the reference point, and the right action of the stabilizer changes its frame.
Realize <de Sitter spacetime> of radius $\ell$ as $\eta_{AB}X^AX^B=\ell^2$ in $\mathbb R^{1,4}$. The ambient <orthogonal group> is transitive on this hyperboloid. Its point stabilizer acts on the Lorentzian tangent space as $O(1,3)$; choosing a tangent orthonormal frame removes that stabilizer completely. Consequently
$$
\boxed{dS_4\cong O(1,4)/O(1,3),\qquad O(dS_4)\cong O(1,4)\longrightarrow dS_4.}
$$
For oriented, time-oriented frames, replace these groups by $SO_0(1,4)$ and $SO_0(1,3)$, and similarly use the proper orthochronous Poincaré group for <Minkowski spacetime>. The isometries are uniquely determined by their action on a point and its frame, which explains these bundle identifications.
Finally, choose a basis $v_1,\ldots,v_n$ of the <Lie algebra> $T_eG$. The <left-invariant vector fields> $X_i(g)=(dL_g)_ev_i$ form a basis at every $g$, because $L_g$ is a <diffeomorphism>. They give a global <section of a fiber bundle> of the <frame bundle>, so that bundle is trivial. More explicitly,
$$
\boxed{TG\cong G\times\mathfrak g,\qquad(g,v)\longmapsto(dL_g)_ev.}
$$
Thus every <Lie group> is a <parallelizable manifold>; connectedness is not needed for this construction.
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