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Let $H$ be the tautological bundle on ${\mathbb{CP}}^1\cong S^2$. One sign convention for the Bott element iswhere $[1]$ is the rank-one trivial vector bundle. It has rank zero, and its Chern character has degree-two term $c_1(H)$, a generator up to sign. Choosing the dual line bundle reverses that sign.

The $n$-fold reduced external product in topological K-theory gives a generator ofThis product uses the smash product of different sphere factors. It is distinct from taking exterior powers of a vector bundle over one fixed base. Multiplication by the Bott element realizes Bott periodicity.^[ref]

Complexification replaces each fiber of a real vector bundle by its tensor product with $\mathbb{C}$ over $\mathbb{R}$. The complexification of a real vector bundle of rank one is a complex line bundle. For the tautological bundle over Real projective space, this complex line bundle squares to the trivial vector bundle.

For two vector bundles $E,F$ over the same base space $X$ and the same scalar field, their direct sum of vector bundles is a bundle with fibersIts total space consists of pairs $(e,f)$ lying above the same $x$, with the topology inherited from the fiber product. If $E$ and $F$ have ranks $r$ and $s$, simultaneous local trivializations give fiber $k^{r+s}$, and their transition functions of a vector bundle combine in block-diagonal matrices. Thus the ranks add. This differs from the tensor product of vector bundles, whose ranks multiply.^[ref]

A geometric example comes from the unit sphere $S^2\subset {\mathbb{R}}^3$. At each $x$, tangent vectors form the plane perpendicular to $x$ and normal vectors form the line spanned by $x$. Their direct sum is all of ${\mathbb{R}}^3$, continuously in $x$. Consequently $T{S}^2\oplus {\epsilon }^1\cong {\epsilon }^3$, where ${\epsilon }^j$ denotes the trivial vector bundle of real rank $j$. Explicitly, $(v,t)\mapsto v+tx$ is the fiberwise isomorphism. Adding one trivial direction can therefore trivialize a nontrivial tangent bundle; the Hairy ball theorem prevents $T{S}^2$ itself from being trivial.

For a real vector bundle $E\to B$ with a fiber metric, the disk bundle isIts fiberwise boundary is the sphere bundle $S(E)=\{\Vert v\Vert =1\}$. The radial homotopy $H_t(v)=(1-t)v$ gives a deformation retraction onto the zero section, so $D(E)$ is homotopy equivalent to $B$.

If $B$ is a smooth manifold without boundary and the rank is positive, the total-space boundary is exactly $S(E)$. When $B$ is a manifold with boundary, the base boundary also contributes: the boundary faces are $S(E)$ and $D(E){|}_{\partial B}$, meeting along $S(E){|}_{\partial B}$. For the trivial vector bundle of rank one over $[0,1]$, the disk bundle is the rectangle $[0,1]\times [-1,1]$. Its top and bottom sides form the sphere bundle; its left and right sides come from the base boundary. The intersections are corners.

A fiber bundle consists of a continuous map $\pi :E\to B$ and a model fiber $F$ such that every point $b\in B$ has an open neighborhood $U$ and a local trivializationThus a nearby family of fibers fits together as a product, while changes between local trivializations can twist those products globally. In the smooth category, $E,B,F$ are smooth manifolds and the local trivializations are diffeomorphisms.

The Möbius line bundle has real-line fibers over a circle. Cutting the circle along a small arc trivializes it, but going once around the circle reverses a fiber coordinate. It is a fiber bundle without being the trivial vector bundle. Local triviality requires a suitable neighborhood of each point; it does not assert a product over every open subset of the base.

For a holomorphically trivial vector bundle, choose a global holomorphic bundle frame. Let the columns of $H_{\alpha }$ be the coordinates of that frame in the prescribed frame over $U_{\alpha }$. If fiber coordinates transform as $s_{\beta }=F_{\alpha \beta }{s}_{\alpha }$, then $H_{\beta }=F_{\alpha \beta }{H}_{\alpha }$. Each matrix is holomorphic and invertible, giving the displayed splitting. Conversely such matrices give agreeing global holomorphic sections forming a frame. Topological triviality alone does not supply this conclusion.

// Destination: electromagnetism.bigb

The Möbius line bundle over the circle iswith projection to $t$ modulo $2\pi$. It is a smooth rank-one vector bundle. A section is represented by a function $s$ satisfying $s(t+2\pi )=-s(t)$, so the intermediate value theorem forces every continuous section to vanish somewhere. Thus it has no nowhere-zero section and is not isomorphic to the trivial vector bundle of rank one.

Over the Grassmannian ${\mathrm{Gr}}_k({\mathbb{C}}^N)$, the complex tautological bundle on a Grassmannian isIts fibre at $W$ is the k-plane $W$ itself. A neighbourhood of a fixed plane consists of graphs of linear maps $A:W\to W^{\perp }$; $(A,w)\mapsto (\mathrm{graph}A,w+Aw)$ gives a local trivialization. The inclusions of coordinate spaces give the infinite Grassmannian and its tautological rank-k bundle ${\gamma }_k^{\infty }$.

Use the customary compact Hausdorff base convention for the classification of complex vector bundles by a Grassmannian; equivalently the argument uses finite subordinate partitions of unity. Give a rank-k complex vector bundle $E\to X$ a continuous fiber metric given by Hermitian inner products. Choose a finite trivializing cover with local orthonormal coordinates $t_i:E{|}_{U_i}\to U_i\times {\mathbb{C}}^k$, and a subordinate partition of unity ${\rho }_i$ with support contained in $U_i$. Defineextending each component by zero off its support. This is continuous and satisfies $\Vert J_x(v){\Vert }^2={\sum }_i{\rho }_i(x)\Vert v{\Vert }^2=\Vert v{\Vert }^2$, so it is a fibrewise embedding. The image planes vary continuously: in local frames, their orthogonal projection matrix is $J(J^{\ast }J{)}^{-1}{J}^{\ast }$. Thus they define a map $c_E:X\to {\mathrm{Gr}}_k({\mathbb{C}}^{kr})$, and $v\mapsto (x,J_{x}v)$ identifies $E$ with $c_E^{\ast }{\gamma }_k^{\infty }$. This proves that every bundle is a tautological pullback vector bundle, by the finite-dimensional embedding of a complex vector bundle.

We next prove the two independence assertions that turn this construction into a bijection. First, homotopic maps have isomorphic pullback vector bundles. Pulling back along a homotopy gives a bundle over compact Hausdorff $X\times I$; embed that bundle in a finite trivial vector bundle by the construction just given. Its image has a continuous family of rank-k orthogonal projection matrices $P(x,t)$. Compactness gives a subdivision $0=t_0<\cdots <t_s=1$ withRestriction of $P(x,t_{j+1})$ to the image at $t_j$ is injective: if it killed $v$, then $v=(P(x,t_j)-P(x,t_{j+1}))v$, impossible by the strict norm bound unless $v=0$. Equal ranks make it an isomorphism, with continuous local inverses. Composing these bundle maps identifies the endpoint vector bundles. The fact that close orthogonal projections identify their image bundles therefore makes the pullback assignment well-defined on homotopy classes.

Conversely, suppose two classifying maps give isomorphic vector bundles. The infinite Grassmannian has its standard weak CW topology; the compact-subset property of CW complexes puts the compact images of the two maps in finite stages. Their tautological pullback vector bundles therefore give two finite-dimensional embeddings $J_0:E\to {\mathbb{C}}^{N_0}$ and $J_1:E\to {\mathbb{C}}^{N_1}$ of the same bundle, after using the vector bundle isomorphism. In orthogonal coordinate blocks interpolate byAt least one coefficient is nonzero, so every $J_t$ is injective and its image planes give a continuous Grassmannian homotopy. One endpoint uses the first block and the other the second block. Moving the latter coordinate copy to the standard one is a fixed unitary change of coordinates, connected to the identity by a path in the unitary group. Such a path exists by diagonalizing the unitary matrix and continuously varying its eigenvalue phases. Hence the original two maps are homotopic in the infinite Grassmannian. This also proves independence of all embedding and metric choices. Together the two directions establishFor arbitrary non-Hausdorff compact spaces without the required numerability, that usual bundle-classification statement needs additional hypotheses; the construction above specifies precisely the conventional setting being used.

Write $B={\mathrm{Gr}}_k({\mathbb{C}}^{\infty })$. For the assumed fiber bundle $U_k\to \nu \to B$ with contractible total space, the long exact sequence of homotopy groups of a fibration has segmentFor $q\ge 2$ this gives group isomorphisms. For $q=1$, the relevant pointed-set part and connectedness of $U_k$ give ${\pi }_1(B)=0$, corresponding to the single component ${\pi }_0(U_k)$. Thus the claimed formula includes its endpoint:Finally the determinant gives a fiber bundle $S{U}_2\to U_2\to S^1$. Since ${\pi }_3(S^1)={\pi }_4(S^1)=0$, its exact sequence yields ${\pi }_3(U_2)\cong {\pi }_3(S{U}_2)$. The given identification $S{U}_2\cong S^3$ and the degree calculation ${\pi }_3(S^3)=\mathbb{Z}$ therefore giveThe equality of free and based homotopy classes uses the simple connectedness just proved. HenceThe classification of rank-two complex vector bundles over the four-sphere is a bijection of isomorphism classes, labelled by the integer clutching function homotopy class after choosing a generator. It is not an assertion that Whitney sum gives a group operation on bundles of fixed rank two.

Let $D_{+}^n$ and $D_{-}^n$ be the two closed hemispheres of the sphere. Use $\gamma$ as the clutching function between the trivial bundles on these hemispheres. The clutching construction definesThe projection to $S^n$ is locally a vector bundle. In a collar of the equator, extend the transition function by keeping it constant in the normal coordinate; this gives explicit vector bundle trivializations across the seam. Since the transition matrices lie in the special orthogonal group, the standard fiber orientations and Euclidean inner products agree on overlaps. Thus this is an oriented rank-$k$ vector bundle with a fiber metric.

For $k\ge 2$, choose a base point $u_0$ on the equator. A constant change of the negative hemisphere frame replaces $\gamma$ by $\gamma (u_0{)}^{-1}\gamma$, so we may suppose $\gamma (u_0)=I$. With $e_k$ the last standard unit vector, define the section obstruction for a clutched vector bundleThe crucial point is that a nowhere-zero section can be normalized using the fiber metric. It is then represented by continuous maps ${\sigma }_{\pm }:D_{\pm }^n\to S^{k-1}$ satisfyingon the equator.

If $f_{\gamma }=0$, the extension-null-homotopy criterion for a sphere extends $u\mapsto \gamma (u)e_k$ to $D_{-}^n$. Take this extension as ${\sigma }_{-}$ and the constant map $e_k$ as ${\sigma }_{+}$. They glue to a nowhere-zero section.

Conversely, suppose such a section exists. Restricting a contraction of $D_{+}^n$ shows that ${\sigma }_{+}{|}_{S^{n-1}}$ is homotopic to a constant. The target sphere is path connected, so the constant can be moved to $e_k$. Multiplying this homotopy pointwise by $\gamma (u)$ gives a homotopy from $\gamma {\sigma }_{+}$ to $\gamma e_k$. The first of these maps equals ${\sigma }_{-}{|}_{S^{n-1}}$ and extends over $D_{-}^n$, hence is null-homotopic. So $\gamma e_k$ is null-homotopic too. A freely null-homotopic based sphere map represents zero in its homotopy group; any change of base point sends the zero class to zero. We have proved both directions:For $n=1$, the same argument reads as the extension criterion for a two-point boundary and uses the trivial set ${\pi }_0(S^{k-1})$ when $k\ge 2$. For $k=1$, the special orthogonal group $SO(1)$ is trivial, so the vector bundle is already a trivial vector bundle; the corresponding constant map to $S^0$ has zero obstruction. These cases require no higher homotopy group calculation.

If $k>n$, then $n-1<k-1$ and the permitted vanishing of lower homotopy groups of a sphere makes $f_{\gamma }=0$. Normalize the resulting nowhere-zero section to a unit section $\sigma$, and setLocal orthonormal frames show that these orthogonal complements form a rank-$(k-1)$ vector bundle. The explicit vector bundle isomorphismhas inverse $w\mapsto (w-⟨w,\sigma ⟩\sigma ,⟨w,\sigma ⟩)$. Orient $V^{\prime }$ so that a positive frame followed by $\sigma$ is positive in $V_{\gamma }$. This is splitting a trivial line from a nowhere-zero section and proves the oriented splittingFinally consider the special orthogonal sphere fibrationThe fiber over $e_{s+1}$ consists exactly of $\mathrm{diag}(B,1)$ with $B\in SO(s)$. Near any unit vector, the Gram-Schmidt process continuously completes it to an oriented orthonormal frame, giving local sections of $p$ and local product fiber bundle charts. In particular $p$ has the homotopy lifting property.

Let $\alpha :S^r\to SO(s+1)$ be based and suppose $r<s$. The composite $p\alpha$ is a based null-homotopic map by the given vanishing of ${\pi }_r(S^s)$. Lift a based null-homotopy of $p\alpha$, beginning at $\alpha$, to a homotopy $G:S^r\times [0,1]\to SO(s+1)$. The lift might initially move its base point $u_0$, but $G(u_0,t)$ always fixes $e_{s+1}$. Right multiplication by $G(u_0,t{)}^{-1}$ corrects this: the homotopystill covers the same sphere homotopy, remains based, and starts at $\alpha$. At its endpoint every matrix fixes $e_{s+1}$, so $\tilde{G}(-,1)$ lies in the included $SO(s)$. Thus every homotopy class in the target is represented by one from $SO(s)$:This proves surjectivity of special orthogonal stabilization below the sphere dimension. For $r=0$, this is the corresponding assertion about path components; the positive-rank special orthogonal groups are connected. The lifting argument also explains the zero term ${\pi }_r(S^s)$ in the long exact sequence of homotopy groups of a fibration.

Under the standard smooth manifold hypotheses, the first answer is yes, although the isomorphism requires a choice. A partition of unity combines Euclidean inner products in manifold charts to produce a Riemannian metric $g$. Nonnegative local weights summing to one preserve positivity, and local finiteness preserves smoothness. The musical isomorphism isIt is smooth and fiberwise invertible, with inverse having local matrix $(g_{ij}{)}^{-1}$. It is therefore a vector bundle isomorphism. There is no preferred such map before a metric or equivalent additional structure is chosen.

The second answer is no. On $S^1=\mathbb{R}/2\pi \mathbb{Z}$, compare the trivial vector bundle $S^1\times \mathbb{R}$ with the Möbius line bundleBoth have rank one. A continuous section of a vector bundle of $L$ is represented on $[0,2\pi ]$ by a continuous function $s$ satisfying $s(2\pi )=-s(0)$. If it were nowhere zero, the intermediate value theorem would prevent its endpoint signs from being opposite. Thus $L$ has no nowhere-zero section. The trivial bundle has the section $1$, and a bundle isomorphism would carry that section to a nowhere-zero section of $L$, a contradiction. Hence

For the First Chern class identity, use finite-dimensional classifying maps. A complex line bundle on a compact Hausdorff space embeds in a finite-dimensional trivial vector bundle, so it is a pullback vector bundle of the tautological bundle on some Complex projective space. For the two complex line bundles, the universal tensor product of vector bundles is classified by the mapIts restriction to either factor pulls the tautological bundle back to the tautological bundle on that factor. The degree-two cohomology of the product is the direct sum of the degree-two cohomology of its factors, so these two restrictions determine its First Chern class. Pulling back to $X$ provesThis proves the integral identity, including possible torsion elements in the cohomology of $X$; an argument using only curvature forms would not detect those torsion elements.

The splitting principle for complex vector bundles says that for a vector bundle $E\to X$ there is an iterated projective bundle $p:F(E)\to X$ for which $p^{\ast }$ is injective on integral cohomology, and $p^{\ast }E$ is a direct sum of complex line bundles. The same assertion holds with rational coefficients, and a common such space can split any finite collection of vector bundles. The injectivity follows at each stage from the projective bundle formula for complex vector bundles; a Hermitian metric splits the resulting filtration into a direct sum.

Write $x_i=c_1(L_i)$ for the Chern roots of the pulled-back vector bundle. Define the degree-$2j$ component of its Chern character byThe Fundamental theorem of symmetric polynomials expresses each power-sum symmetric polynomial ${\sum }_i{x}_i^j$ uniquely as an integral polynomial in the elementary symmetric polynomials, which are the Chern classes of $E$. Thus the definition makes sense on $X$ itself, without choosing actual Chern roots there. On a compact Hausdorff space, each vector bundle is classified by a finite-dimensional Grassmannian; its high-degree Chern character components therefore vanish. Hence this defines a class in the direct sum $H^{2\ast }(X;\mathbb{Q})$ even if $X$ has no finite dimension. On more general spaces an infinite Chern character is instead interpreted in the product of the even-degree cohomology groups.

On a common splitting space, the Chern roots of a direct sum are the combined root lists, whereas the First Chern class identity makes the Chern roots of a tensor product of vector bundles the pairwise sums $x_i+y_j$. ConsequentlyThe injectivity in the splitting principle for complex vector bundles descends both identities to $X$. Additivity extends the Chern character to the Grothendieck group by $\mathrm{ch}([E]-[F])=\mathrm{ch}(E)-\mathrm{ch}(F)$. Multiplicativity and $\mathrm{ch}(1)=1$ then show that the Chern character is a unital ring homomorphism.

For Complex K-theory of complex projective space, let $\gamma$ be the tautological bundle, put $t=1-[\gamma ]$, and set $h=-c_1(\gamma )$. The cohomology ring of complex projective space is $\mathbb{Z}[h]/(h^{n+1})$. The CW filtration gives a cofibrationThe K-theory six-term exact sequence and Complex K-theory of a sphere give, inductively,Here $b_n$ is the image of a Bott element from $S^{2n}$, normalized so that $\mathrm{ch}(b_n)=h^n$. This normalization follows from the multiplicative form of Bott periodicity: the Chern character of the degree-two Bott element is an integral generator, and multiplication yields the top-degree generator on $S^{2n}$. In particular, the abelian group $K^0({\mathbb{CP}}^n)$ is a free abelian group of rank $n+1$.

Assume inductively that $1,t,\dots ,t^{n-1}$ form an integral basis on ${\mathbb{CP}}^{n-1}$. Their lifts, together with $b_n$, form an integral basis on ${\mathbb{CP}}^n$. Sincethe Chern characters of this basis are linearly independent over $\mathbb{Q}$: their first nonzero degrees are $0,2,\dots ,2n$. Thus the Chern character is injective on $K^0({\mathbb{CP}}^n)$. By induction $t^n$ restricts to zero on ${\mathbb{CP}}^{n-1}$, so $t^n=a{b}_n$ for an integer $a$. Comparing Chern characters gives $a=1$. Also $\mathrm{ch}(t^{n+1})=0$, so $t^{n+1}=0$. This proves the integral answer, rather than merely its rationalization:The induction begins with ${\mathbb{CP}}^0$ a point and $t=0$.

For the K-theory transfer of a finite covering, let $V\to Y$ be a vector bundle and defineOn an evenly covered neighborhood the $n$ sheets identify this with a direct sum of $n$ local vector bundles. On overlaps the identifications permute the sheets and apply the transition maps of $V$, so these local descriptions glue to a vector bundle on $X$. This operation respects direct sums, hence extends to a homomorphism of Grothendieck groupsIt obeys the projection formula for the K-theory transfer:For actual vector bundles, this follows fiberwise by distributing the tensor product of vector bundles over the direct sum; it then extends to their Grothendieck groups.

Let $P=p_{!}(1_Y)$ be the permutation vector bundle of the covering. It has rank $n$ on every connected component, but need not be a trivial vector bundle; in particular one cannot replace $p_{!}{p}^{\ast }$ by multiplication by $n$ integrally. Instead writeThe componentwise nilpotence result from the preceding solution gives ${\eta }^N=0$ for some $N$. In the localization of a ring $K^0(X)\otimes \mathbb{Z}[1/n]$, the finite geometric seriesis an inverse. The projection formula for the K-theory transfer givesThus $p^{\ast }(a)=0$ implies $a=0$ after inverting $n$. More explicitly, $(n+\eta {)}^{-1}{p}_{!}$ is a left inverse, proving

For $x\in S^2\subset {\mathbb{R}}^3$, let $F(x)=x^T$, viewed as a nonzero $1$ by $3$ matrix. This defines a smooth map $F:S^2\to X_{1,3}$. Its pulled-back fiber isso the evident fiberwise identity gives an isomorphism of vector bundles $F^{\ast }{E}_{1,3}\cong T{S}^2$.

If $E_{1,3}$ were a trivial vector bundle, its pullback would be trivial. A trivial rank-two bundle has a nowhere-zero section, whereas the assumed form of the Hairy ball theorem says that the tangent bundle $T{S}^2$ does not. Hence $E_{1,3}$ is nontrivial.

A real line bundle is a rank-one real vector bundle. Over a paracompact Hausdorff space, or over a smooth manifold in the smooth category, it is orientable exactly when it is a trivial vector bundle.

To prove the nontrivial implication, choose a fiber metric. An orientation of a vector bundle selects one of the two unit vectors in each fiber. These choices agree continuously in oriented local frames, giving a global nowhere-zero section. The map $(x,t)\mapsto ts(x)$ is then a fiber-linear product isomorphism. Conversely, a trivial vector bundle has the positive standard basis as a continuous orientation of a vector bundle.

The Möbius line bundle over a circle is the basic counterexample to triviality: one circuit reverses the fiber orientation, so no global positive unit section of a vector bundle can be chosen.

A unit nowhere-zero section $\sigma$ of a real vector bundle with a fiber metric defines a rank-one trivial vector bundle. Its fiberwise orthogonal complement is a subbundle: locally complete $\sigma$ to an orthonormal frame. The vector bundle isomorphism $(v,t)\mapsto v+t\sigma$ has inverse $w\mapsto (w-⟨w,\sigma ⟩\sigma ,⟨w,\sigma ⟩)$. For an oriented bundle, orient the complement so that its positive frame followed by $\sigma$ is positive.

The Stiefel–Whitney classes of a real rank-$r$ vector bundle $E\to X$ are characteristic cohomology classesThey are natural under pullback vector bundles and obey the Whitney product formula for Stiefel–Whitney classes, $w(E\oplus F)=w(E)w(F)$. For a real line bundle $L$, the normalization is $w(L)=1+w_1(L)$, with $w_1$ the class classifying its orientation double cover. These properties determine the classes through the splitting principle for real vector bundles.

The first Stiefel–Whitney class vanishes exactly when the vector bundle is orientable. The top Stiefel–Whitney class is its Euler class with mod-two coefficients and obstructs a nowhere-zero section. A trivial vector bundle has total class one; the Möbius line bundle has nonzero $w_1$.

Vanishing of every positive-degree Stiefel–Whitney class need not imply triviality. The tangent bundle of the two-dimensional sphere has $w_1=0$ and $w_2=\chi (S^2)\mathrm{mod}2=0$, although its integral Euler class is nonzero. The classes detect useful obstructions while forgetting information lost by reduction modulo two.

Let $p:\mathbb{P}(E)\to M$ be the projectivization of a real vector bundle, using one-dimensional linear subspaces, and let $L$ be its tautological bundle. At a line $\ell \subset E_x$, differentiating the graph of a nearby line identifies the vertical tangent space with $\mathrm{Hom}(\ell ,E_x/\ell )$. Thus there is a canonical short exact sequence of vector bundlesA fiber metric identifies $p^{\ast }E/L$ with the orthogonal complement ${\omega }_E$. A chosen connection on a vector bundle defines horizontal lifts and splits the short exact sequence, givingThe vertical term is intrinsic, while this horizontal direct-sum identification depends on the chosen connection on a vector bundle. For a trivial vector bundle with its product connection, it reduces to the usual tangent bundle splitting of $M\times {\mathbb{RP}}^{r-1}$.

Choose a fiber metric on a real rank-$r$ vector bundle $E\to B$. Its Thom space is the pointed quotient spaceobtained by collapsing the entire sphere bundle to one basepoint. For a compact Hausdorff space $B$, it is also the one-point compactification of the total space of $E$. A single fiber becomes $D^r/S^{r-1}\cong S^r$.

For the trivial vector bundle $B\times {\mathbb{R}}^r$, the quotient is $B_{+}\wedge S^r$, where $B_{+}$ adjoins a disjoint basepoint and $\wedge$ denotes the smash product. In rank zero the convention is $D(E)/\varnothing =B_{+}$; this keeps the basepoint even though the sphere bundle is empty.

For an oriented rank-$r$ vector bundle over a CW complex, its Thom class restricts to the positive generator on every fiber sphere. Multiplication by that class gives the Thom isomorphismThus the Thom space packages the degree shift of cohomology into one pointed space.^[ref]

A rank-$r$ vector bundle over a topological space $B$ is a fiber bundle $\pi :E\to B$ whose fibers are vector spaces over a fixed field $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$. Each point has an open neighborhood $U$ and a vector bundle trivialization ${\pi }^{-1}(U)\cong U\times {\mathbb{F}}^r$ that is a linear isomorphism on each fiber.

On overlaps the vector bundle trivializations change by $(x,v)\mapsto (x,g_{ij}(x)v)$, where the invertible matrices $g_{ij}(x)$ vary continuously and satisfy the Čech cocycle condition. For a smooth vector bundle, the base and total space are smooth manifolds and these functions are smooth. The fiberwise vector addition and scalar multiplication are therefore globally well-defined.

The tangent bundle collects all tangent spaces of a smooth manifold. Its transition functions of a vector bundle are Jacobian matrices, so the chain rule supplies the cocycle identity. The Möbius line bundle supplies another example: every small patch is a product, but no global frame of a vector bundle exists. A global fiber-linear product isomorphism is the extra condition defining a trivial vector bundle.