 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 $\pi \circ s={\mathrm{id}}_M$.

For a principal bundle, a global section of a fiber bundle gives the mapThe 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 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 $T_{x}M$, equivalently linear isomorphisms $u:{\mathbb{R}}^n\to T_{x}M$. 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 $u:{\mathbb{R}}^{p,q}\to T_{x}M$, 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,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 ${\eta }_{AB}{X}^A{X}^B={\ell }^2$ 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. ConsequentlyFor oriented, time-oriented frames, replace these groups by $S{O}_0(1,4)$ and $S{O}_0(1,3)$, 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 $v_1,\dots ,v_n$ of the Lie algebra $T_{e}G$. The left-invariant vector fields $X_i(g)=(d{L}_g{)}_e{v}_i$ 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,Thus every Lie group is a parallelizable manifold; connectedness is not needed for this construction.