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For a smooth map $\varphi :M^{\prime }\to M$ and a connection on a vector bundle $E\to M$, the curvature form of a connection on the pullback vector bundle satisfiesas an endomorphism of $E_{\varphi (p)}$. In a frame of a vector bundle written as a row, with section components written as columns, the connection one-form is $\Omega$ and its curvature form of a connection is $d\Omega +\Omega \wedge \Omega$. The pullback connection has matrix ${\varphi }^{\ast }\Omega$; commutation of the pullback of a differential form with the exterior derivative and wedge product of differential forms proves the identity. No pushforward to an ambient vector field is required: $d\varphi (X)$ is a vector field along a map.

The frame bundle of an $n$-dimensional smooth manifold has fiber over $x$ consisting of linear isomorphisms $u:{\mathbb{R}}^n\to T_{x}M$. Right composition by $GL(n,\mathbb{R})$ makes it a principal bundle. A global frame of a vector bundle is a global section of a fiber bundle and trivializes the tangent bundle. A metric restricts the allowed frames to the orthonormal frame bundle.

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.

In a frame of a vector bundle, write a vector bundle covariant derivative as $\nabla (ev)=e(dv+\Gamma v)$. The associated horizontal subspace at $(x,v)$ consists of tangent pairs $(\xi ,-{\Gamma }_x(\xi )v)$. Under a frame change $e^{\prime }=eg$, the connection one-form changes by ${\Gamma }^{\prime }=g^{-1}\Gamma g+g^{-1}dg$, and this horizontal equation changes to the corresponding one in the new frame. Thus the subspaces glue to a unique smooth linear connection on a vector bundle. Its vertical projection gives back the original derivative. The Leibniz rule ensures locality, allowing this construction from an operator initially defined on global sections.

A basis $T_a$ of the Lie algebra of a Lie group gives a global frame of a vector bundle of its tangent bundle by $X_a(g)=(d{L}_g{)}_e{T}_a$. Left translation is a diffeomorphism, so the left-invariant vector fields are independent everywhere. Every Lie group is therefore a parallelizable manifold.

A smooth vector field on a smooth manifold $M$ is a smooth section of a vector bundle $X:M\to TM$ of the tangent bundle projection $\pi$, so $\pi \circ X={\mathrm{id}}_M$. In a coordinate chart it has the form $X={\sum }_i{X}^i{\partial }_i$ with smooth component functions $X^i$.

Suppose a smooth vector bundle isomorphism $\Phi :M\times {\mathbb{R}}^n\to TM$ covering the identity is given. For the standard basis $e_1,\dots ,e_n$ of ${\mathbb{R}}^n$, put $X_i(p)=\Phi (p,e_i)$. These are smooth vector fields, and the linear isomorphism on each fibre makes $X_1(p),\dots ,X_n(p)$ a basis of $T_{p}M$.

Conversely, given such a global frame of a vector bundle, defineIt is smooth, covers the identity and is a linear isomorphism on each fibre. Its inverse is smooth as well: in any coordinate chart, the component columns of the $X_i$ form an invertible smooth matrix $C(p)$, and the fibre coordinates of the inverse are $C(p{)}^{-1}v$. matrix inversion is smooth on the invertible matrices. Thus the tangent bundle is trivial exactly when a global frame exists, the defining property of a parallelizable manifold.

For a Lie group $G$, write $L_h(g)=hg$. A left-invariant vector field is a field satisfyingSuch a field is determined by $v=X(e)\in T_{e}G$, since $X(g)=(d{L}_g{)}_{e}v$. Conversely, this formula defines a left-invariant field, by the chain rule and $L_h{L}_g=L_{hg}$.

It also proves smoothness even if smoothness was not initially assumed. Choose a smooth curve $c(t)$ with $c(0)=e$, $c^{\prime }(0)=v$. ThenMultiplication is a smooth map on a Lie group, so differentiating its coordinate functions in the second variable yields a smooth function of $g$. This gives smooth components for $X$ in every local chart.

Choose a basis $v_1,\dots ,v_m$ of $T_{e}G$, where $m=\dim G$, and left translate it. Each $L_g$ is a diffeomorphism, with inverse $L_{g^{-1}}$, so $(d{L}_g{)}_e$ is a linear isomorphism and the translated fields are a basis at every $g$. The explicit parallelization of a Lie group by left translations isIts inverse sends $w\in T_{g}G$ to $(g,(d{L}_{g^{-1}}{)}_{g}w)$. Identifying $T_{e}G$ with ${\mathbb{R}}^m$ gives the required product bundle. The differential facts used are linearity, the chain rule, $d(\mathrm{id})=\mathrm{id}$ and smooth dependence of the differential of a smooth map on its base point.

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.

The expression $d\varphi (X)$ is in general a vector field along a map, rather than a vector field on $M$. Its intended meaning in the required formula is pointwise evaluation of the vector-bundle-valued differential form $\nabla s$:using the canonical identification $({\varphi }^{\ast }E{)}_p=E_{\varphi (p)}$. This interpretation works for every smooth map, including a constant map or a map with self-intersections.

Choose a frame of a vector bundle over $U$, written as a row $e=(e_1,\dots ,e_r)$. Define its connection one-form $\Omega$ byfor component columns $u$. The pulled-back frame $e^{\varphi }=({\varphi }^{\ast }{e}_1,\dots ,{\varphi }^{\ast }{e}_r)$ is a frame of a vector bundle over ${\varphi }^{-1}U$. For any component column $v$ of smooth functions on that open set, defineHere the matrix entries of ${\varphi }^{\ast }\Omega$ are ordinary pullbacks of a differential form. The displayed operator is linear and satisfies the Leibniz rule, hence is locally a connection on a vector bundle.

To prove these local operators glue, change the original frame of a vector bundle to $\tilde{e}=eh$. Expanding $\nabla (ehu)$ gives the change of frame of a vector-bundle connectionThe pullback of a differential form and the chain rule giveThis is precisely the required transformation law for the pulled-back frame ${\tilde{e}}^{\varphi }=e^{\varphi }(h\circ \varphi )$. Thus the definitions agree and give a global pullback connection.

For $s=eu$, substitute $v=u\circ \varphi$ in the defining formula and use $d(u\circ \varphi )(X)=du(d\varphi (X))$. This proves the required identity. Conversely that identity fixes ${\nabla }_X^{\prime }({\varphi }^{\ast }{e}_a)$; the Leibniz rule then fixes the derivative of every ${\sum }_a{v}^a{\varphi }^{\ast }{e}_a$, proving uniqueness. If the identity is initially imposed only on global sections of a vector bundle, multiply each local frame section by a smooth bump function equal to one near the point in question and supported in $U$, and extend by zero. These global sections agree with the frame locally, so the same uniqueness argument applies. The pullback connection exists and is unique.

The curvature form of a connection is the endomorphism-valued differential two-form defined byIt is alternating in $X,Y$. To see its tensoriality, replacing $X$ by $fX$ produces an extra $-Y(f){\nabla }_{X}s$ from the second term and an extra $+Y(f){\nabla }_{X}s$ from $[fX,Y]=f[X,Y]-Y(f)X$, so these cancel. Alternation gives linearity over smooth functions in $Y$ too. Replacing $s$ by $fs$ produces the additional coefficient $(X(Yf)-Y(Xf)-[X,Y]f)s=0$; all terms involving one derivative of $s$ also cancel. Thus $R^{\nabla }$ belongs to ${\Omega }^2(M;\mathrm{End}E)$.

In the coefficient-column convention of part (b), expanding the definition givesThe first three terms are $d\Omega (X,Y)$, and the last two are $(\Omega \wedge \Omega )(X,Y)$. This proves the Cartan curvature matrix equation $F=d\Omega +\Omega \wedge \Omega$, with matrix multiplication combined with the wedge product of differential forms.

The pullback connection has matrix ${\Omega }^{\prime }={\varphi }^{\ast }\Omega$. Since the pullback of a differential form commutes with the exterior derivative and the wedge product of differential forms, its curvature form of a connection has matrixChanges of frame of a vector bundle conjugate both sides by the same pulled-back transition matrix, so this identity is intrinsic. In fibre notation the answer isBoth sides act on the same fibre $E_{\varphi (p)}$. No claim that $d\varphi$ preserves Lie brackets of vector fields for an arbitrary map is needed.

We use the following existence and uniqueness statement for parallel transport: for a smooth affine connection, a smooth curve $\gamma :[0,1]\to M$, an initial time $t_0$ and a vector $v\in T_{\gamma (t_0)}M$, there is a unique smooth vector field along a map $V$ on the whole curve satisfying $D_{t}V=0$ and $V(t_0)=v$. In a frame of a vector bundle this is the linear ordinary differential equation $\dot{V}=-\Gamma (\dot{\gamma })V$. Its smooth coefficient matrix is bounded on every compact time subinterval in a trivializing neighborhood, and the linear existence theorem gives a solution throughout that subinterval. A finite subdivision of $[0,1]$ into such neighborhoods and uniqueness patch the solutions. No completeness hypothesis on $M$ is needed.

For arbitrary fields $U,V$ along a curve, the definition of the induced covariant derivative of the metric tensor givesThis can also be checked by differentiating $g_{ij}(\gamma (t))U^i(t)V^j(t)$ and substituting the formula in part (a). If $\nabla$ is a metric connection and $V$ is parallel, it follows that $d(g(V,V))/dt=0$. Positivity of the Riemannian metric makes $|V|=\sqrt{g(V,V)}$ constant. Thus parallel transport preserves lengths.

Conversely assume every parallel field along every smooth curve has constant length. Fix $p\in M$ and arbitrary $x,v\in T_{p}M$. Choose a curve through $p$ with velocity $x$ at an interior time $t_0$. Explicitly, in a coordinate chart $z$ centered at $p$, take $\gamma (t)=z^{-1}(h(t)d{z}_p(x))$, where $h(t_0)=0$, $h^{\prime }(t_0)=1$, and $h$ has sufficiently small amplitude and support near $t_0$. A smooth bump function produces such $h$, so the curve is defined on all of $[0,1]$ and remains inside the chart. By the stated parallel transport theorem there is a parallel field with $V(t_0)=v$. Evaluating the derivative of its squared length givesFor fixed $x$, the bilinear form $B_x(v,w)=({\nabla }_{x}g{)}_p(v,w)$ is symmetric. Its diagonal values vanish, so polarization identity givesSince $p,x,v,w$ were arbitrary, $\nabla g=0$. A connection is metric-compatible if and only if all parallel fields have constant length. Polarization also shows that its parallel transport preserves all inner products, not just lengths.

Let $E\to M$ have rank $r$. A vector bundle covariant derivative is a real-linear operatorPairing its one-form slot with a vector field $X$ gives ${\nabla }_{X}s$, with ${\nabla }_{fX}s=f{\nabla }_{X}s$ and ${\nabla }_X(fs)=X(f)s+f{\nabla }_{X}s$.

In the horizontal description, a connection on a vector bundle $A$ is a smooth splitting $T_{a}E=V_{a}E\oplus H_{a}E$, where $V_{a}E=\ker (d\pi {)}_a$, and the horizontal choice is compatible with fibre addition and scalar multiplication. This linearity is essential; an arbitrary nonlinear horizontal splitting is not a vector-bundle connection. Identify a vertical vector at $a\in E_x$ with the element of $E_x$ obtained by differentiating the curve $a+tv$. Let $K_A:T_{a}E\to E_x$ be vertical projection followed by this identification. The induced vector bundle covariant derivative isIn a local frame of a vector bundle $e=(e_1,\dots ,e_r)$, a linear horizontal splitting has the formfor a matrix-valued connection one-form $\Gamma$. Thus for $s=ev$, ${\nabla }^{A}s=e(dv+\Gamma v)$, which directly verifies the Leibniz rule.

Conversely, the Leibniz rule makes $\nabla$ local: if a section vanishes near a point, multiply it by a smooth cutoff function supported in that neighborhood and equal to one near the point to see that its derivative vanishes there too. Local sections can therefore be differentiated by any extension agreeing near the point. Given $\nabla$, define its local connection one-form by $\nabla e_b={\sum }_a{e}_a{\Gamma }_b^a$. The Leibniz rule gives $\nabla (ev)=e(dv+\Gamma v)$. Under $e^{\prime }=eg$, $v^{\prime }=g^{-1}v$, one hasThe tangent fibre coordinate transforms by $w^{\prime }=g^{-1}w-g^{-1}(dg)(\xi )g^{-1}v$, so $w=-\Gamma (\xi )v$ transforms precisely to $w^{\prime }=-{\Gamma }^{\prime }(\xi )v^{\prime }$. Consequently these local horizontal subspaces of a vector bundle connection glue to a smooth linear splitting inducing the given $\nabla$. Its local formula also shows uniqueness. This proves the horizontal connection associated to a covariant derivative construction.

A horizontal lift $\tilde{\gamma }$ of $\gamma$ is a smooth curve in $E$ with $\pi \circ \tilde{\gamma }=\gamma$ and $\dot{\tilde{\gamma }}(t)\in H_{\tilde{\gamma }(t)}E$. In a local frame write $\tilde{\gamma }(t)=e(\gamma (t))v(t)$. Horizontality is exactly the linear ordinary differential equationWe use existence and uniqueness for an ODE with smooth coefficients, and the continuation theorem that a solution extends across a finite interval when it remains bounded inside the coordinate domain. For a linear system $\dot{v}=-C(t)v$ with continuous coefficients bounded by $L$ on a compact time interval, the Gronwall inequality bounds $|v(t)|$ by its initial norm times $e^{L|t-t_0|}$, preventing finite-time fibre escape. Cover a compact subinterval of $I$ by finitely many bundle charts and partition it so each piece lies in one chart. Solve successively, changing fibre frames at the partition points. Coordinate invariance of the horizontal equation and uniqueness make the pieces agree. Exhausting $I$ by compact intervals containing $0$ produces a unique horizontal lift on all of $I$ for every prescribed $a_0$. This is global continuation of linear parallel transport.

The lift equation is linear in its initial value. Hence the endpoint map $P_{\gamma }:E_{\gamma (0)}\to E_{\gamma (1)}$ is linear. Solving the same equation backwards, or transporting along $t\mapsto \gamma (1-t)$, gives its inverse, soThis is parallel transport. For two parallel lifts $s_1(t),s_2(t)$, the assumed metric connection identity extends along the embedded curve and becomesHere $D_t$ is the covariant derivative along a curve. The permitted extension of sections and the tangent vector along the embedded curve justifies applying the bundle identity; equivalently it follows in each local frame. Thereforeso parallel transport is an isometry. This is the principle that parallel transport preserves a fibre metric.

Rescale the ambient coordinates by $q_i=a_i{r}_i$. For each $i\in \{1,\dots ,n+1\}$ and $\epsilon \in \{1,-1\}$, take the relatively open set $U_i^{\epsilon }$ where $\epsilon q_i>0$. Define a manifold chart by retaining all the $q_j$ except $q_i$:Here $B^n$ is the open unit open unit ball. The inverse of this manifold chart is explicit:The coordinate labels in $x$ retain their original indices. All the $a_i$ are nonzero, so this inverse exists; the strictly positive radicand makes it smooth on the whole open open unit ball. Projection and this inverse are continuous, so each manifold chart is a homeomorphism onto an open subset of ${\mathbb{R}}^n$.

These $2(n+1)$ manifold charts cover the ellipsoid, because at every point at least one $q_i$ is nonzero. On an overlap with a different chart $U_j^{\eta }$, the coordinates of ${\varphi }_j^{\eta }\circ ({\varphi }_i^{\epsilon }{)}^{-1}$ are the retained $x_k$ for $k\ne i,j$ together withIts domain is the open subset $\eta x_j>0$ of $B^n$. The displayed functions, and the reverse transition functions obtained by exchanging $i,j$, are smooth. Transitions between identical charts are the identity; opposite-sign charts with the same omitted coordinate do not overlap. The ambient subspace topology is Hausdorff and second countable, as it is inherited from Euclidean space. Thus this coordinate-projection atlas of an ellipsoid is a smooth atlas, proving the ellipsoid is a smooth manifold of dimension $n$. The rescaling $r\mapsto q$ also gives a diffeomorphism with the unit sphere $S^n$.

For the three-sphere, use its identification with SU(2) rather than separate local constructions. With the Pauli matrices ${\tau }_i$, put $T_i=-i{\tau }_i$. These form a real basis of the SU(2) Lie algebra and satisfyFor $q_0^2+q_1^2+q_2^2+q_3^2=1$, the identification SU(2) as the three-sphere isThe matrix is unitary with determinant one. Conversely, every SU(2) matrix has this form, and both the correspondence and its inverse are smooth.

The left translations $L_g(h)=gh$ are diffeomorphisms. Define three left-invariant vector fields byMatrix multiplication makes these vector fields smooth globally. Since $(d{L}_g{)}_e$ is an isomorphism from the SU(2) Lie algebra to $T_g\mathrm{SU}(2)$, the three vectors are linearly independent at every point, and each is nowhere zero. This is the parallelization of a Lie group by left translations.

Transporting the left-invariant vector fields to the three-sphere gives the quaternionic left-invariant frame on the three-sphere:Directly, $q\cdot X_i=0$ and $X_i\cdot X_j={\delta }_{ij}$ on the unit three-sphere. Thus the vectors are tangent and form a global orthonormal frame of a vector bundle. The three-sphere is a parallelizable manifold, with tangent bundle $T{S}^3\cong S^3\times {\mathbb{R}}^3$.

For a vector bundle carrying a smooth nondegenerate symmetric bilinear form of constant metric signature, a pseudo-orthonormal frame is a smooth local frame of a vector bundle whose Gram matrix is the constant diagonal signature matrix $\eta$. It identifies the metric bundle locally with a fixed model ${\mathbb{R}}^{p,q}$. Two such frames differ by a smooth map into the corresponding orthogonal group. Positive-definite signature is included as the ordinary orthonormal case. A global pseudo-orthonormal frame is a section of a fiber bundle of the orthonormal frame bundle and trivializes that principal bundle.

For a smooth vector subbundle $F\subset E$, the quotient has fibers $(E/F{)}_p=E_p/F_p$. Extend a local frame of a vector bundle of $F$ to one of $E$; the classes of the complementary frame vectors trivialize the quotient. Adapted transition functions of a vector bundle are block upper triangular, and their lower-right blocks give the quotient transitions. These are smooth and satisfy the cocycle identity, defining a smooth vector bundle. The projection $E\to E/F$ is a smooth vector bundle morphism.