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In global coordinates the de Sitter spacetime metric is $-d{\tau }^2+{\cosh }^2\tau g_{S^3}$. A global frame of a vector bundle orthonormal for the round metric $Y_1,Y_2,Y_3$ on $S^3$ gives the displayed global pseudo-orthonormal frame. For example, identify $S^3$ with the unit quaternions and use $qi,qj,qk$. Choosing the spatial orientation and the future direction consistently gives a section of a fiber bundle for the oriented, time-oriented frame bundle. The principal bundle trivialization by a global section then proves that bundle is trivial.

A principal bundle $\pi :P\to B$ with structure group $G$ has a smooth free right $G$-action, each fibre is a single orbit, and it is locally equivariantly isomorphic to $U\times G\to U$. A representation $\rho :G\to GL(V)$ defines the associated vector bundleLocal sections of $P$ identify $E$ with $U\times V$, and changes of section act by $\rho$ on the fibre coordinates.

A principal bundle is trivial if and only if it has a global smooth section. Given $s:B\to P$, the map $(b,g)\mapsto s(b)g$ is an equivariant isomorphism $B\times G\to P$: freeness and transitivity give its unique inverse on each fibre, and local trivializations make that inverse smooth. Conversely a product bundle has section $b\mapsto (b,e)$. This proves principal bundle trivialization by a global section. A contractible paracompact base is a sufficient condition for triviality of such a bundle; local sections alone are insufficient on a general base.

A manifold is parallelizable when its tangent bundle is trivial, equivalently when it possesses a global smooth frame. The circle $S^1$ is an example, with its nowhere-vanishing unit tangent field. The sphere $S^3$ is another: regard it as the unit quaternions. At a unit quaternion $q$, the three tangent vectors $qi,qj,qk$ are independent and smoothly depend on $q$. Quaternion multiplication preserves the norm, so these form an orthonormal frame. This gives the quaternionic left-invariant frame on the three-sphere.

For any Lie group $G$ of dimension $n$, choose a basis $E_1,\dots ,E_n$ of $T_{e}G$. Left translations defineThese fields are smooth; at every $g$, the differential of the diffeomorphism $L_g$ is invertible, so they are a basis of $T_{g}G$. Hence every Lie group manifold is parallelizable, as in parallelization of a Lie group by left translations.

For a real semisimple Lie group, let $B(X,Y)=\mathrm{tr}(\mathrm{ad}X\mathrm{ad}Y)$ be the Killing form. It is nondegenerate and invariant under the adjoint action. Left translating $B$ gives a bi-invariant pseudo-Riemannian metric. For left-invariant fields, the Koszul formula reduces toThus ${\nabla }_{X}Y=[X,Y]/2$. The curvature convention of Question 1 givesTo obtain the Ricci tensor, trace the linear map $X\mapsto R(X,Y)Z$. Since $[[X,Y],Z]=\mathrm{ad}Z\mathrm{ad}YX$,This proves the Killing-form Einstein metric construction for every real semisimple group. If $G$ is compact semisimple, $-B$ is positive definite and its Ricci tensor is $-\frac{1}{4}B=\frac{1}{4}g$, giving a Riemannian Einstein metric. For a noncompact semisimple group the general Killing-form construction is indefinite. Thus the unrestricted assertion is understood in the pseudo-Riemannian sense appropriate here; positive definiteness is not obtained by simply taking $-B$ in the noncompact case.

A fiber bundle with base $B$, fiber $F$ and structure group $G$ acting on $F$ from the left consists of a projection $\pi :E\to B$ and local trivializations ${\varphi }_i:{\pi }^{-1}(U_i)\to U_i\times F$. On overlaps they satisfyThe smooth transition functions are valued in $G$. A principal bundle is the case $F=G$ with the left multiplication action on itself; its transition functions multiply the group coordinate from the left.

In each principal chart define $(x,h)\cdot a=(x,ha)$. This defines a global right action because $g_{ij}(x)(ha)=(g_{ij}(x)h)a$: coordinate changes commute with right multiplication. It is free, and acts transitively on each fiber. Thus the claimed right action follows from the left transition convention rather than being an additional choice.

If $s:B\to P$ is a global section of a fiber bundle, define $\Phi :B\times G\to P$ by $\Phi (x,a)=s(x)a$. Freeness and transitivity give exactly one $a$ for every point in the fiber, so $\Phi$ is bijective and equivariant. In a local trivialization where $s(x)$ has group coordinate $h_i(x)$, the inverse group coordinate is $h_i(x{)}^{-1}h$, proving smoothness of the inverse. Hence $\Phi$ is a principal-bundle trivialization. Conversely, the identity coordinate in a product trivialization gives $s(x)=\Phi (x,e)$. ThereforeThis is the principal bundle trivialization by a global section criterion. It is special to principal bundles; a section alone need not trivialize an arbitrary fiber bundle.

For the quotient examples, assume the Lie subgroups are closed when smooth homogeneous-space statements are intended. The map $G\to G/H_1$ is a principal bundle with structure group $H_1$ and right action $g\cdot h=gh$; $G/H_1$ is a homogeneous space for left multiplication by $G$. Similarly, $G\to H_2\backslash G$ has free transitive left $H_2$ action on its fibers. With the usual principal right-action convention one writes $g\cdot h=h^{-1}g$. The quotient $H_2\backslash G$ is a homogeneous space for right multiplication by $G$. These quotients are groups only when the respective subgroups are normal. Nonclosed immersed Lie subgroups can produce non-Hausdorff quotients, so an arbitrary Lie subgroup does not automatically give a smooth Hausdorff base.

Left $H_2$ and right $H_1$ multiplication commute, so taking their orbit equivalence relations in either order gives the same double coset setThis equality does not assert that the double quotient is a manifold or that either stage is a principal bundle. The $H_2\times H_1$ action $(a,b):g\mapsto ag{b}^{-1}$ can have stabilizers: they correspond to $H_2\cap g{H}_1{g}^{-1}$. If the induced action is free and proper, its quotient is a smooth principal-bundle quotient; without those hypotheses orbit dimensions can change. For example, with $G=SO(3)$ and both subgroups the rotations fixing the third axis, double cosets are parametrized by $e_3\cdot g{e}_3\in [-1,1]$. Interior and endpoint stabilizers differ, so this is not a uniform free principal-bundle construction.

For a principal bundle $P$ and a left $G$ action on $F$, the associated bundle iswith projection $[p,f]\mapsto \pi (p)$. A local principal section gives coordinates $[s_i(x),f]\leftrightarrow (x,f)$, and its changes give precisely the left action of the principal transition functions on $F$. A linear action on a vector space gives an associated vector bundle.

For a metric of signature $(r,s)$ on an $n$-manifold, an orthonormal frame at $x$ is an isometry $u:{\mathbb{R}}^{r,s}\to T_{x}M$. The orthonormal frame bundle $O(M,g)\to M$ has structure group $O(r,s)$, acting on the right by $u\cdot a=u\circ a$. Its associated bundle for the standard representation is the tangent bundle, sinceis well defined: $(ua)(v)=u(av)$. Every tangent vector has exactly one such associated equivalence class, and local frame coordinates show that this is a vector-bundle isomorphism. The full frame bundle works similarly with $GL(n,\mathbb{R})$.

Finally, choose a basis of the Lie algebra of a Lie group $G$. Its left translates are a global frame, yielding the explicit tangent bundle trivializationThis proves parallelization of a Lie group by left translations and triviality of the full frame bundle. Choosing an inner product at the identity and translating it gives a left-invariant Riemannian metric; an orthonormal basis translates to a global orthonormal frame. The global section criterion therefore trivializes the corresponding orthonormal frame bundle as well. For any other positive-definite metric on $G$, smoothly applying the Gram-Schmidt process to this global frame gives the same conclusion. Thus the tangent and frame bundles of every Lie group are trivial; the orthonormal statement is with respect to a chosen metric as specified above.