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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.

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.

For a smooth manifold $M$ of dimension $n$, the tangent space $T_{p}M$ can be defined as the space of derivations at $p$: linear maps $v$ on germs of smooth real functions satisfying $v(fg)=f(p)v(g)+g(p)v(f)$. Equivalently, $v$ is the velocity of a smooth curve through $p$, with curves identified when their coordinate velocities agree. A chart $(x^1,\dots ,x^n)$ supplies the basis ${\partial }_i{|}_p$, so $v=v^i{\partial }_i{|}_p$.

The tangent bundle is the disjoint union $TM={\coprod }_{p\in M}{T}_{p}M$, with projection $\pi (v)=p$. Its smooth structure is defined by the local local trivializationsOn overlapping charts, the change of trivialization isThese are smooth changes of coordinates on a $2n$-dimensional smooth manifold, linear and invertible in each fibre. Thus $TM$ is a rank-$n$ vector bundle, and its projection is a submersion. Fibrewise addition, scalar multiplication and the zero section are smooth. A smooth section of a vector bundle $X:M\to TM$, satisfying $\pi \circ X={\mathrm{id}}_M$, is exactly a vector field; locally $X=X^i{\partial }_i$, with components transforming by the displayed Jacobian matrix. Its action on functions is a derivation, and the commutator of two such derivations is their Lie bracket.

The dual bundle of $TM$ is the cotangent bundle $T^{\ast }M$. Its smooth sections of a vector bundle are differential 1-forms, locally $\alpha ={\alpha }_{i}d{x}^i$. The coefficient transformation is the inverse transpose of that for tangent vectors, ensuring that $\alpha (X)$ is a well-defined smooth function. Tensoring these two bundles gives the tensor bundlesTheir sections of a vector bundle are tensor fields; their transition functions of a vector bundle are the corresponding tensor products of the tangent and cotangent transformations. For example, the bundle $\mathrm{End}(TM)=TM\otimes T^{\ast }M$ contains fields of linear endomorphisms, while $S^2{T}^{\ast }M$ contains symmetric bilinear fields. Taking alternating covariant tensors gives ${\Lambda }^k{T}^{\ast }M$, whose sections are differential forms. The wedge product, exterior derivative and pullback of a differential form make these bundles central to integration, Generalized Stokes theorem and de Rham cohomology.

A nowhere-vanishing section of ${\Lambda }^n{T}^{\ast }M$ chooses an orientation of a smooth manifold; such a section exists exactly when $M$ is orientable. Without an orientation of a smooth manifold, one can still integrate sections of the density bundle $|{\Lambda }^n{T}^{\ast }M|$, whose coordinate changes use the absolute Jacobian determinant. A Riemannian metric is a positive-definite section $g$ of $S^2{T}^{\ast }M$. It identifies $TM$ with $T^{\ast }M$ by the musical isomorphisms $X\mapsto g(X,\cdot )$ and gives the Riemannian volume densityOn an oriented smooth manifold, this density corresponds to a volume form.

Another organizing construction is the frame bundle $FM$. A frame over $p$ is a linear isomorphism $u:{\mathbb{R}}^n\to T_{p}M$, and the right action $u\cdot A=u\circ A$ makes $FM$ a principal bundle with general linear group $GL(n,\mathbb{R})$ as its structure group. Given a group representation $\rho :GL(n,\mathbb{R})\to GL(V)$, the associated bundle $FM{\times }_{\rho }V$ is obtained from pairs $(u,v)$ by the relation $(uA,v)\sim (u,\rho (A)v)$. The standard representation reconstructs $TM$; dual, tensor and exterior representations reconstruct the associated bundles described above. A Riemannian metric reduces the frame bundle to the orthonormal frame bundle, with orthogonal group $O(n)$ as structure group; an orientation of a smooth manifold reduces it further to the special orthogonal group $SO(n)$. A global section of $FM$ is a global frame, so it exists exactly when $M$ is a parallelizable manifold. Local triviality does not imply a global trivialization: the Hairy ball theorem prevents even one nowhere-zero vector field on $S^2$.

A connection on a vector bundle differentiates its sections in tangent directions and defines parallel transport. For $TM$, the Levi-Civita connection is distinguished by compatibility with a Riemannian metric and vanishing torsion tensor. Its coefficients are not tensor components, because a coordinate change introduces second derivatives; the difference of two connections is a section of $T^{\ast }M\otimes \mathrm{End}(TM)$. The Riemann curvature tensor is a genuine tensor measuring the failure of covariant derivatives to commute.

Finally, a smooth map $f:M\to N$ gives the pullback tangent bundle $f^{\ast }TN$. Its sections are vector fields along a map, including the velocity of a curve and Jacobi fields along a geodesic. For an embedded submanifold $i:M\hookrightarrow N$, the derivative identifies $TM$ with a subbundle of $i^{\ast }TN$, and the quotient is the normal bundle. With an ambient Riemannian metric, the normal bundle is identified with the orthogonal complement of $TM$ in $i^{\ast }TN$; the second fundamental form records the normal component of the ambient derivative of tangent fields. These constructions connect the bundle description to both intrinsic and extrinsic differential geometry.

For a connected smooth Riemannian manifold $(M,g)$ without boundary, let $I(M)$ be the group of all global diffeomorphisms $f$ with $f^{\ast }g=g$, endowed with its compact-open topology. Equivalently these maps preserve the intrinsic distance; the Myers-Steenrod theorem supplies smoothness of distance-preserving bijections and makes $I(M)$ a finite-dimensional Lie group acting smoothly. The group is metric-dependent: a topological manifold can carry a very symmetric metric or one with no continuous symmetries. Connectedness matters here. With infinitely many identical disconnected components, permutations of those components can produce a group which is not a finite-dimensional Lie group.

An isometry is determined by its value and derivative at one point. To prove this, suppose $f(p)=p$ and $d{f}_p=I$. Preservation of the Levi-Civita connection giveswhere the exponential map is defined locally. Thus $f$ is the identity in a normal neighborhood of $p$. The set where its value and derivative agree with the identity is both closed and, by the same argument, open. Connectedness makes it all of $M$. No completeness is needed for this determination argument.

The stabilizer $I(M{)}_p$ consequently acts faithfully and orthogonally on $T_{p}M$, so embeds in $O(n)$. More generally an isometry takes a chosen orthonormal frame at $p$ to one at its image point, providing an injective map into the orthonormal frame bundle. It is a smooth embedding, and the full isometry action is proper. The orbit-stabilizer dimension formula therefore givesIsotropy is compact. If $M$ is compact then $I(M)$ itself is compact, by the same frame description or Arzela-Ascoli applied to isometries and their inverses. Properness also means that the quotient by isometries has well-controlled local orbit structure, even though orbit dimensions may vary.

The infinitesimal equation is the Killing equation. Differentiating $f_t^{\ast }g=g$ at zero givesConversely a vector field satisfying this equation has local flows preserving the metric. A global one-parameter subgroup requires the field to be complete. On a geodesically complete manifold every Killing field is complete: its length is constant along each of its own trajectories, so a finite-time trajectory has bounded speed and stays in a closed bounded ball. The Hopf-Rinow theorem makes that ball compact, and the ordinary differential equation can then be continued. The caveat is real: on the Euclidean interval $(0,1)$, the field ${\partial }_x$ is Killing but its translations are not globally defined for all time. This illustrates Killing fields on incomplete manifolds need not generate global isometries.

The Killing transport identities determine a Killing field from $X(p)$ and the skew-adjoint map $(\nabla X{)}_p$. Along a geodesic the field is a Jacobi field, so uniqueness for its second-order differential equation propagates these initial data. They have $n+n(n-1)/2$ components, reproducing the dimension bound infinitesimally. The important distinction on an incomplete manifold is between all local Killing solutions and the Lie algebra of complete global isometry flows.

In dimension three the bound is six, and rotational isotropy gives stronger information. The Lie subalgebras of $\mathfrak{so}(3)$ have dimensions zero, one or three: identifying the bracket with the cross product, a two-plane is not closed under brackets. Suppose $\dim I(M)=5$. Since an orbit has dimension at most three, its stabilizer would have dimension at least two, and hence exactly three. Its connected isotropy acts as all of $SO(3)$ on the tangent space. But the orbit tangent space is invariant under that action and would have dimension $5-3=2$, impossible because the standard rotational representation is irreducible. Therefore dimension five cannot occur.

If the dimension is six, each orbit has dimension three and each stabilizer dimension three. All orbits are open, so connectedness makes the action transitive. At each point the isotropy is transitive on tangent two-planes, forcing sectional curvature to be independent of the plane; transitivity makes its value independent of the point. Thus a maximally symmetric connected three-manifold has constant sectional curvature. Homogeneous Riemannian manifolds are complete, so its simply connected cover is the round sphere, Euclidean space or hyperbolic space after normalization. Conversely these three simply connected metrics attain dimension six:Here $O^{+}(3,1)$ preserves the chosen sheet of the hyperboloid; it includes orientation-reversing hyperbolic isometries. Constant curvature alone does not imply that a quotient retains six-dimensional global symmetry.

Dimension four likewise forces transitivity: stabilizer dimension three would give a one-dimensional invariant orbit tangent space, again impossible for full rotational isotropy. The stabilizer thus has dimension one and the orbits are three-dimensional. Typical four-dimensional isometry groups occur for $S^2\times \mathbb{R}$ and ${\mathbb{H}}^2\times \mathbb{R}$, where the factor groups have dimensions three and one. The usual Nil metric has three translation symmetries and one rotational isotropy symmetry. In centred Heisenberg coordinates it iswhich exhibits the horizontal rotations. Its distinct vertical and horizontal Ricci eigenvalues bound continuous isotropy by that rotation group, so its full isometry dimension is four. Standard metrics on $\widetilde{S{L}_2(\mathbb{R})}$ and nonround Berger spheres also have dimension four. The standard Sol geometry has a three-dimensional translation group and only finite isotropy, giving dimension three. These examples show how a preferred vertical direction reduces rotational symmetry.

The isometry dimensions in dimension three are precisely $0,1,2,3,4,6$, all realizable. A compact hyperbolic three-manifold gives zero; the product of a circle and a closed hyperbolic surface gives one. For dimension two take $T^2\times (0,1)$ withIts scalar curvature is the constant $-14$, but its Ricci eigenvalues are the distinct constants $-3,-6,-5$ in the two horizontal directions and the vertical direction. Hence connected isotropy is trivial and possible isometries preserve these line distributions. A local isometry preserving them has $t^{\prime }=t+c$ and would require constant dilations $e^{-c}$ and $e^{-2c}$ of the two circle coordinates. A global circle diffeomorphism with such a constant dilation has degree $\pm 1$, forcing $c=0$. The identity component is therefore exactly the two circle translations. A flat three-torus gives dimension three; $S^2\times \mathbb{R}$ gives four; a simply connected constant-curvature model gives six. This construction distinguishes local homogeneity from the smaller global isometry group.

For quotients, let a complete simply connected Riemannian manifold $X$ cover $M=X/\Gamma$. Every isometry of $M$ lifts to an isometry of $X$ which normalizes the deck transformation group. Conversely, every normalizing isometry descends. Two lifts differ by a deck transformation, giving the isometry group of a Riemannian quotient formulaThus one should not replace the isometry group of a quotient by that of its model geometry. For a flat torus, for example,The second factor is finite, whereas translations give the three-dimensional identity component.

There is a useful curvature test for compact manifolds. Contracting and differentiating the Killing equation gives ${\nabla }^{\ast }\nabla X=\mathrm{Ric}(X)$, with the nonnegative rough-Laplacian convention. Integration yields the integrated Bochner identity for Killing fieldsIf Ricci curvature is negative definite, no nonzero Killing field exists. The compact isometry group then has dimension zero and is finite. This explains why closed hyperbolic manifolds have finite global symmetry, despite the six-dimensional isometry group of their universal cover. Compactness is essential to this conclusion.

For a compact flat Riemannian manifold, the same identity makes every Killing field parallel. Parallel fields are exactly the vectors fixed by linear holonomy. ThereforeApplied to the six flat types of Question 1, this gives dimensions three for the torus, one for each of the four nontrivial cyclic types, and zero for the Hantzsche-Wendt manifold. These are identity-component dimensions; disconnected finite symmetries can still be present. Local curvature symmetry, global deck-group constraints, and completeness thus play different roles in determining the isometry group.

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.

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.

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.

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.