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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 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 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.