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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.

Represent unit-radius de Sitter spacetime by the spacelike unit hyperboloid in five-dimensional Minkowski spacetime. Appending its position vector as the last column of an oriented pseudo-orthonormal frame produces a matrix in $SO(4,1)$. Projection to that last column identifies $SO(4,1)$ with the oriented orthonormal frame bundle, a principal bundle with group $SO(3,1)$. The full bundle allowing both orientations is $O(4,1)$ with group $O(3,1)$. Requiring a time orientation as well selects the corresponding identity components.

A smooth principal bundle with structure Lie group $G$ can be specified by a locally trivial map $\pi :E\to B$ whose fiber coordinate changes have the form $(x,g{)}_i\mapsto (x,h_{ji}(x)g{)}_j$, with the usual transition functions of a principal bundle cocycle on triple overlaps. In each chart define $(x,g)\cdot k=(x,gk)$. On overlaps, $h_{ji}(x)(gk)=(h_{ji}(x)g)k$, so these local right actions agree. They therefore give a global smooth right group action, free and transitive on every fiber. Equivalently, this action together with equivariant local triviality is the definition of a principal bundle.

Use five-dimensional Minkowski spacetime with ambient metric $\eta =\mathrm{diag}(-1,1,1,1,1)$ and realize unit-radius de Sitter spacetime as $\eta (x,x)=1$. Its tangent space is $x^{\perp }$, of Lorentzian signature $(3,1)$. An oriented pseudo-orthonormal frame $(e_0,e_1,e_2,e_3)$ of $x^{\perp }$, together with the last column $x$, is precisely a matrixConversely the last column of such a matrix lies on the hyperboloid, and its other columns give that frame. ThusRight multiplication by $\mathrm{diag}(h,1)$ fixes $x$ and changes the tangent frame by $h$. It is free and transitive on frames over $x$, proving the principal bundle description. Smooth local choices of frames provide its local trivializations. The dimensions are $10=4+6$.

This is the oriented frame bundle of de Sitter spacetime. Allowing orientation reversal gives $O(4,1)$ and structure group $O(3,1)$; imposing a time orientation as well selects $S{O}_0(4,1)$ and $S{O}_0(3,1)$. For radius $a$, take the last column $x/a$.

One definition of a principal bundle $P\to M$ uses local trivializations ${\varphi }_i:{\pi }^{-1}(U_i)\to U_i\times G$ whose overlap maps arewith smooth $G$-valued transition functions of a principal bundle satisfying the cocycle condition. This definition displays the global right group action rather than assuming it: if ${\varphi }_i(p)=(x,g_i)$, define ${\varphi }_i(p\cdot h)=(x,g_{i}h)$. On an overlap,so the local definitions agree. The resulting action is smooth, free and transitive on each fibre, and respects the projection.

If $s:M\to P$ is a global section of a fiber bundle, set $F(x,h)=s(x)\cdot h$. The free and transitive fibre action makes $F:M\times G\to P$ bijective. In a local trivialization, write $s(x)=(x,a_i(x))$; then $F(x,h)=(x,a_i(x)h)$ and its inverse is $(x,g)\mapsto (x,a_i(x{)}^{-1}g)$. Both are smooth and respect the action and projection. Conversely a principal bundle isomorphism $M\times G\to P$ sends $x\mapsto (x,e)$ to a global section. Hence

For a pseudo-Riemannian manifold of metric signature $(p,q)$, its orthonormal frame bundle has fibre at $x$ consisting of linear isometries $u:{\mathbb{R}}^{p,q}\to T_{x}M$. Its structure group is $O(p,q)$ and its right action is $u\cdot h=u\circ h$. Two such frames differ by a unique element of $O(p,q)$, proving the required free and transitive fibre action. A global section is exactly a global pseudo-orthonormal frame. Orientation and time orientation, when specified, restrict the structure group and the allowed frames.

Realize unit-radius four-dimensional de Sitter spacetime asIts tangent space is $X^{\perp }$, with Lorentzian metric signature $(3,1)$. If $e_0,e_1,e_2,e_3$ is a pseudo-orthonormal frame at $X$, append $X$ as the final column. The resulting five-by-five matrix preserves $\eta$, and conversely every such matrix supplies a point $X$ and a frame at that point. Thus the full frame bundle is $O(4,1)$, with projection $g\mapsto g{e}_4$ and fibre group $O(3,1)$.

The printed $SO(4,1)$ statement requires the orientation restriction. Choose the tangent orientation by requiring the appended matrix to have determinant one. Thenis precisely the oriented frame bundle of de Sitter spacetime, with right structure group $SO(3,1)$ embedded as $\mathrm{diag}(h,1)$. If future time orientation is also required, use $S{O}_0(4,1)$ and $S{O}_0(3,1)$. Calling $SO(4,1)$ the full bundle without specifying orientation would exclude valid frames of the opposite orientation.

Finally write the global coordinatesDirect differentiation gives the de Sitter spacetime metric $-d{\tau }^2+{\cosh }^2\tau g_{S^3}$. The assumed global frame of a vector bundle on $S^3$ can be orthonormalized smoothly by Gram-Schmidt orthogonalization for the round positive metric. More explicitly, view $S^3$ as the unit quaternions: $Y_1(q)=qi$, $Y_2(q)=qj$, $Y_3(q)=qk$ form a smooth global orthonormal tangent basis. SetThese have inner products $g(E_0,E_0)=-1$, $g(E_i,E_j)={\delta }_{ij}$ and $g(E_0,E_i)=0$. They are smooth everywhere because $\cosh \tau$ never vanishes. Choose the spatial orientation consistently, reversing one $Y_i$ if necessary, and choose $E_0$ future-directed. This global frame on four-dimensional de Sitter spacetime yields a global section of the full frame bundle and also of either restricted bundle. Yes: the de Sitter frame bundle admits a global section and is trivial.

Let ${\eta }_{ab}=g(e_a,e_b)$ be the constant diagonal metric matrix in a local pseudo-orthonormal frame, and define ${\theta }_{ab}={\eta }_{ac}{\theta }^c{}_b$. The metric compatibility of the Levi-Civita connection givesThus the lowered connection matrix is antisymmetric, including in indefinite signature. Ordinary matrix antisymmetry of ${\theta }^a{}_b$ itself is not the corresponding assertion in indefinite signature.

Let ${\omega }^a$ be the dual coframe, and write a vector field as $Y=e_b{\omega }^b(Y)$. ThenSubtract the same expression with $X,Y$ exchanged and subtract ${\omega }^a([X,Y])$. By the definition of the exterior derivative, this gives the components of the torsion tensor:The Levi-Civita connection is torsion-free, so Cartan's first structure equation is

Use $R(X,Y)={\nabla }_X{\nabla }_Y-{\nabla }_Y{\nabla }_X-{\nabla }_{[X,Y]}$ and define its matrix of curvature two-forms by $R(X,Y)e_b=e_a{R}^a{}_b(X,Y)$. Applying two covariant derivatives to $e_b$ yieldsThe first three terms are $d{\theta }^a{}_b(X,Y)$ and the last two are $({\theta }^a{}_c\wedge {\theta }^c{}_b)(X,Y)$. Therefore Cartan's second structure equation isThe associated Riemann curvature tensor components satisfy $R^a{}_b=\frac{1}{2}{R}^a{}_{bcd}{\omega }^c\wedge {\omega }^d$ with this convention.

For a position-dependent change of pseudo-orthonormal frame, write the frame row as $E^{\prime }=E\Lambda$, with ${\Lambda }^T\eta \Lambda =\eta$. The dual coframe obeys ${\omega }^{\prime }={\Lambda }^{-1}\omega$. Applying the Leibniz rule to $\nabla (E\Lambda )$ givesThe connection one-form has an inhomogeneous term because differentiating the moving basis produces $d\Lambda$. The curvature two-form instead transforms homogeneously. Indeed, curvature is tensorial in the vector to which it is applied, so $R(X,Y)(E\Lambda )=(R(X,Y)E)\Lambda =E^{\prime }{\Lambda }^{-1}R(X,Y)\Lambda$. ConsequentlyEquivalently, substituting the transformed connection into $d{\theta }^{\prime }+{\theta }^{\prime }\wedge {\theta }^{\prime }$ and using $d{\Lambda }^{-1}=-{\Lambda }^{-1}(d\Lambda ){\Lambda }^{-1}$ cancels all derivative terms and gives the same result. This transformation law holds for a general invertible frame change as well; the orthogonality condition ensures both frames remain pseudo-orthonormal.