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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.

If $L$ is the complexification of a real vector bundle of the tautological bundle and $\mu =[L]-1$, thenRealizing Real projective space as the circle bundle of the square of the tautological bundle on Complex projective space gives this through the K-theory Gysin sequence of a sphere bundle.

Let $L$ be the complexification of a real vector bundle of the tautological bundle over Real projective space ${\mathbb{RP}}^{2n+1}$. For $k\ge 1$, the K-theory Euler class of $L^{\oplus k}$ is $-2^{k-1}([L]-1)$, andThe K-theory Gysin sequence of a sphere bundle gives an extension by $\mathbb{Z}$, which splits as an exact sequence of abelian groups.

Let $i:{\mathbb{RP}}^d\hookrightarrow {\mathbb{CP}}^d$ be the standard inclusion. If $h=c_1(\mathcal{O}(1))\mathrm{mod}2$ and $a=w_1({\gamma }_d)$, thenThe restricted complex tautological bundle is the complexification of a real vector bundle ${\gamma }_d{\otimes }_{\mathbb{R}}\mathbb{C}$. Its underlying real bundle is ${\gamma }_d\oplus {\gamma }_d$. The Whitney product formula for Stiefel–Whitney classes gives total class $(1+a{)}^2=1+a^2$, while the Stiefel–Whitney class of the underlying real bundle of a complex line identifies its second class with the reduced First Chern class. Dualizing changes the integral sign but has no effect modulo two. Thus the full induced map on cohomology sends $h^j$ to $a^{2j}$.

A real rank-$d$ vector bundle $p:E\to M$ is a continuous family of $d$-dimensional real vector spaces, locally trivial by homeomorphisms $p^{-1}(U)\cong U\times {\mathbb{R}}^d$ which are linear on every fiber. Transition matrices are continuous maps into ${\mathrm{GL}}_d(\mathbb{R})$. The Grassmannian ${\mathrm{Gr}}_k({\mathbb{R}}^n)$ is the space of $k$-dimensional linear subspaces; its topology can be defined by identifying each plane $W$ with its orthogonal projection matrix $P_W$. Local graph charts identify nearby planes with graphs of maps $W_0\to W_0^{\perp }$.

The tautological bundle is $E_{\mathrm{taut}}=\{(W,v):v\in W\}\to {\mathrm{Gr}}_k({\mathbb{R}}^n)$. In graph charts a basis of $W_0$ gives a continuously varying basis of each graph, proving local triviality. The universal quotient bundle has fiber $Q_W={\mathbb{R}}^n/W$. The map $[v]\mapsto (I-P_W)v$ identifies it continuously with the orthogonal-complement bundle $W^{\perp }$, whose graph charts similarly supply local frames. Thus $Q_{\mathrm{taut}}$ is a rank-$(n-k)$ quotient vector bundle.

For the given $E\to M$, at each $m$ choose a basis of $E_m$ and global sections of a vector bundle realizing its basis vectors, using the evaluation-surjectivity assumption. Those $d$ sections remain independent on a neighbourhood of $m$, since the corresponding determinant in a local trivialization is nonzero at $m$ and continuous. Compactness gives a finite cover by such neighbourhoods. Collect all the selected sections as $s_1,\dots ,s_N$; they span every fiber. They define a fiberwise surjective vector bundle morphismIts kernels have constant dimension $N-d$ and form the kernel bundle of a surjective vector bundle morphism. More concretely, in a local frame of $E$ the map has a continuous full-row-rank matrix $A(m)$ and its kernel projection isThis varies continuously and is unchanged by an invertible change of target frame. The Grassmannian as projection matrices therefore makesa continuous map. The fiber quotient map induced by $F_m$ is an isomorphism ${\mathbb{R}}^N/\ker F_m\to E_m$. The same local matrix formula gives continuously varying right inverses, so these fiber isomorphisms give a vector bundle isomorphism. This proves the classification of vector bundles by a universal quotient bundle:The displayed equality in the classification statement is understood as bundle isomorphism, not literal equality of total spaces.

Finally, the universal quotient bundle need not have a nowhere-zero section. Take ${\mathrm{Gr}}_1({\mathbb{R}}^2)={\mathbb{RP}}^1$. Its quotient identifies with $W^{\perp }$, a Möbius line bundle. Write $W_{\theta }=\mathrm{span}(\cos \theta ,\sin \theta )$, with $\theta$ modulo $\pi$. A section of the complement line has the form $a(\theta )(-\sin \theta ,\cos \theta )$, and being single-valued requires $a(\theta +\pi )=-a(\theta )$. If $a$ is nonzero anywhere, the intermediate value theorem forces a zero between that point and its translate by $\pi$; if it is nowhere nonzero the conclusion is immediate. Hence this rank-one quotient bundle has no nowhere-zero section.

For nonzero finite-dimensional $U$ and $V$, the intended assertion is true. Write $a=\dim U$, $b=\dim V$, and replace $W$ by $\mathrm{im}\varphi$, of dimension $r$. Injectivity on the stated slices implies $\varphi (u\otimes v)\ne 0$ whenever $u,v\ne 0$. Consequently the bilinear map defines a continuous map of Complex projective spacesThe rank $r$ is at least one, since one such nonzero tensor has nonzero image.

Let $L_U,L_V,L_W$ be the respective tautological bundles. Fiberwise, $\varphi$ identifies $L_U\otimes L_V$ with $F^{\ast }{L}_W$. For the positive hyperplane classes $x=c_1(L_U^{\ast })$, $y=c_1(L_V^{\ast })$ and $z=c_1(L_W^{\ast })$ (zero when the projective space is a point), the first Chern class of a tensor product of complex line bundles givesThe Künneth theorem, together with part (a), identifies the product cohomology ring withThere are no Tor terms because the factor groups are free. In particular, its monomials $x^i{y}^j$ with $0\le i<a$, $0\le j<b$ form an integral additive basis. In top degree,If $r\le a+b-2$, the target relation $z^r=0$ would imply $(x+y{)}^r=0$ and hence contradict this nonzero top power. ThusWhen $a=b=1$, the already-established inequality $r\ge 1$ gives the same conclusion. The complex bilinear dimension bound is sharp: multiplication of complex polynomials of degrees less than $a$ and $b$ has target dimension $a+b-1$, is injective in either nonzero fixed factor, and its image spans every monomial in that target.

Literal zero-space qualification. The printed assertion does not explicitly exclude zero vector spaces. If $U=0$, $V={\mathbb{C}}^2$ and $W=0$, its slice-injectivity hypothesis is vacuous, while the claimed inequality would be $0\ge 1$. Thus, with zero spaces permitted, this is a counterexample to the assertion exactly as written; the proof above supplies the usual nonzero finite-dimensional interpretation.

Here is an intrinsic definition that also proves well-definedness. For a complex line bundle $L$, define its First Chern class to be the Euler class of its canonically oriented underlying real rank-two bundle: pull its integral Thom class back along the zero section after forgetting relative support. The complex orientation fixes the sign, so this construction makes no arbitrary choice of generator.

For a rank-$r>0$ complex vector bundle $E$, let $\pi :\mathbb{P}(E)\to X$ be its projective bundle of lines, let $S\subset {\pi }^{\ast }E$ be the tautological bundle, and put $h=-c_1(S)=c_1(S^{\ast })$. On every fiber ${\mathbb{CP}}^{r-1}$, $h$ is the positive degree-two generator. The Leray-Hirsch theorem says that if globally defined cohomology classes restrict to a free basis of the cohomology of every fiber, then multiplication by those classes identifies the cohomology of the total space with a free module over the base. Applied here, it givesThe finite trivializing cover in the question is sufficient for this application: the assertion holds on each trivializing open set by the Künneth theorem, since the fiber has finite free cohomology, and the Mayer–Vietoris sequence and the Five lemma glue it over the finite cover. The same local argument constructs the oriented Thom class used for line bundles. It does not require a choice of classifying map.

There are therefore unique classes $c_i(E)\in H^{2i}(X;\mathbb{Z})$ such thatDefine $c_0(E)=1$ and $c_i(E)=0$ for $i>r$. For rank zero the Total Chern class is $1$. This is the projective bundle definition of Chern classes. Existence and uniqueness follow by expressing $h^r$ in the displayed free module basis, with degrees determining each coefficient. The projective bundle and tautological bundle are intrinsic to $E$, and the line Thom class is uniquely fixed by orientation. Hence the resulting Chern classes do not depend on trivializations or other auxiliary choices. For a line bundle the relation is $h+c_1(E)=0$, agreeing with the original normalization. Pulling back this unique relation also proves naturality. This standard construction and the sum theorem are treated in Vector Bundles and K-Theory, Section 3.1.

The requested result is the Whitney sum formula for Chern classes:Here $c(E)=1+c_1(E)+\cdots +c_r(E)$, and the formula holds for complex vector bundles over a common base. In particular, a trivial bundle has total Chern class $1$.

Now take $B={\mathbb{CP}}^{n-1}$ and let $L\subset B\times {\mathbb{C}}^n$ be its tautological bundle. The standard Hermitian inner product gives the rank-$(n-1)$ complex bundle $Q=L^{\perp }$, with $L\oplus Q\cong {\underline{\mathbb{C}}}^n$. The orthogonal complex line flag manifold in the question is precisely $\mathbb{P}(Q)$: over a first line $\ell$, the second line is any line in ${\ell }^{\perp }$. Local orthonormal frames give this identification as a fiber bundle, with fiber ${\mathbb{CP}}^{n-2}$.

Let $x=c_1(L^{\ast })$ on $B$, also writing $x$ for its pullback to $\mathbb{P}(Q)$. By part 3(a), $H^{\ast }(B;\mathbb{Z})=\mathbb{Z}[x]/(x^n)$. The Whitney sum formula for Chern classes givesLet $S_2$ be the second tautological line on $\mathbb{P}(Q)$, and set $y=c_1(S_2^{\ast })$. Thus $x,y$ are exactly the pullbacks of the positive hyperplane classes from the two projective factors. The projective bundle definition of Chern classes for $Q$ givesTogether with $x^n=0$, this gives a surjective graded ring mapThere are no additional relations. Indeed the second relation is monic of degree $n-1$ in $y$, so polynomial division makes its source free over $\mathbb{Z}[x]/(x^n)$ on $1,y,\dots ,y^{n-2}$. The projective bundle formula for complex vector bundles gives exactly the same free basis on the target. The map sends each basis element to its corresponding basis element, and is therefore an isomorphism. ConsequentlyThis is the cohomology ring of the orthogonal complex line flag manifold. Its additive basis is $x^i{y}^j$ with $0\le i<n$, $0\le j<n-1$. As a symmetry check, multiplying the second relation by $y-x$ gives $y^n-x^n=0$, so $y^n=0$ as expected from the second projection. For $n=2$, every line has a unique orthogonal line, and the relations become $x^2=0$, $x+y=0$, giving the ring of ${\mathbb{CP}}^1$. If $n=1$, the space is empty and the second relation is $1=0$, so the printed formula still gives the zero cohomology ring.

Put $F={\mathbb{F}}_2$. Under the usual bundle convention of a paracompact base, choose a fiber metric and let $D(E),S(E)$ be the disk bundle and sphere bundle of the rank-$n$ real vector bundle. A mod-two Thom class is a relative cohomology classwhose restriction to every fiber pair $(D^n,S^{n-1})$ is the nonzero generator. Equivalently it is a class in $H^n(E,E\setminus B;F)$, where $B$ is identified with the zero section. No orientation choices are needed over $F$.

The Thom isomorphism theorem asserts that this class exists and that, for every $r$, the mapis an isomorphism. The disk bundle retracts to the zero section $s:B\to D(E)$, so $H^r(D(E);F)\cong H^r(B;F)$. Define the mod-two Euler class $e_2(E)=s^{\ast }u\in H^n(B;F)$, where the pullback includes the passage from relative to absolute cohomology. Under the Thom identification, the map from relative to absolute cohomology sends $a$ to $a⌣e_2(E)$. Substituting into the pair's long exact sequence in cohomology gives the Gysin sequence of a sphere bundle:Here ${\pi }_{!}$ is the connecting homomorphism followed by ${\Phi }^{-1}$. Thus the sequence and the cup-product map have been derived from the Thom theorem, not just stated. The Thom/Gysin constructions are also discussed in Hatcher's Vector Bundles and K-Theory, §3.2.

The original PDF specifies the tautological bundle on ${\mathbb{RP}}^n$; the occurrences of ${\mathbb{R}}^n$ in this part of the TeX transcription have lost the projective-space symbol. Let $L\to {\mathbb{RP}}^n$ be this real line bundle. Its sphere bundle consists of pairs $(\ell ,v)$ with $v$ a unit vector in the line $\ell$, so $S(L)\cong S^n$, with projection the antipodal double covering. Write $t=e_2(L)$.

The additive mod-two groups, obtainable from the one-cell-in-each-dimension cellular chain complex, areFor $n\ge 1$, both the base and $S^n$ are connected, so ${\pi }^{\ast }:H^0({\mathbb{RP}}^n;F)\to H^0(S^n;F)$ is an isomorphism. Exactness makes the following ${\pi }_{!}$ zero and multiplication by $t$ injective from $H^0$ to $H^1$. For $1\le r<n$, the term $H^r(S^n;F)$ vanishes, so the same Gysin sequence makesinjective. Since these groups are one-dimensional, the maps are isomorphisms. Hence $1,t,\dots ,t^n$ are the respective nonzero generators, while $t^{n+1}=0$ by dimension. The mod-two cohomology ring of real projective space is thereforeFor $n=0$ this is simply $F$, with $t=0$.

For integral coefficients, the additive groups areIndeed, the cellular homology of real projective space has boundary multiplication by $2$ in even positive chain degrees and zero in odd degrees, and the integral cochain differential is its transpose. It remains to determine the products; the additive groups alone do not do so.

For $n\ge 2$, the coefficient sequence $0\to \mathbb{Z}\stackrel{2}{\to }\mathbb{Z}\to F\to 0$ has a Bockstein homomorphism $\beta$. Because $H^1({\mathbb{RP}}^n;\mathbb{Z})=0$ and $H^2({\mathbb{RP}}^n;\mathbb{Z})=\mathbb{Z}/2$, exactness makes $a=\beta (t)$ the nonzero integral degree-two class. Reduction $\rho :H^2({\mathbb{RP}}^n;\mathbb{Z})\to H^2({\mathbb{RP}}^n;F)$ is injective: its kernel is the image of multiplication by $2$, which is zero. ThusMore generally, reduction is injective in each positive even degree, and its compatibility with the cup product gives $\rho (a^j)=t^{2j}$. Consequently $a^j$ is the nonzero generator whenever $2j\le n$. Also $2a=0$, and powers beyond the dimension vanish.

If $n=2m$ is even, these powers and the unit account for all groups, so the integral cohomology ring of real projective space isIf $n=2m+1$ is odd, add an integral top-dimensional orientation class $b$ of degree $2m+1$. Its reduction is the nonzero class $t^{2m+1}$. Every product $ab$ and $b^2$ is zero by dimension, and $b$ has infinite additive order. HenceThese formulas include $m=0$: the even space is a point with ring $\mathbb{Z}$, and the odd space is a circle with ring $\mathbb{Z}[b]/(b^2)$, $|b|=1$. The polynomial presentations are interpreted with the displayed grading and products; no extra positive-degree products are left unspecified.

The tautological bundle $\mathcal{O}(-1)$ over Complex projective space has fibre over $[\ell ]$ equal to the line $\ell \subset {\mathbb{C}}^{r+1}$. Define $\mathcal{O}(1)=\mathcal{O}(-1{)}^{\ast }$ and the twisting sheaf on projective space, viewed analytically as a holomorphic line bundle, by

For ${\mathbb{CP}}^1$, use coordinates $z=Z_1/Z_0$ and $w=Z_0/Z_1=1/z$. The tautological frames $(1,z)$ and $(w,1)$ differ by $z^{-1}$, so frames $e_0,e_1$ for $\mathcal{O}(d)$ satisfy $e_1=z^d{e}_0$. A global holomorphic section is therefore described by entire functions $s_0(z),s_1(w)$ withWrite $s_0(z)={\sum }_{k\ge 0}{a}_k{z}^k$. Holomorphicity of $s_1$ at zero permits only the powers $w^{d-k}$ with $d-k\ge 0$. For $d\ge 0$, this says $s_0$ is a polynomial of degree at most $d$; for $d<0$, every coefficient must vanish. Equivalently, when $d\ge 0$ the sections are homogeneous polynomials of degree $d$ in $Z_0,Z_1$. Thus

Apply Serre duality for compact complex manifolds to $\mathcal{L}\otimes K_M$, where $K_M$ is the canonical bundle:We can show the group on the right vanishes directly from the embedding. Since ${\mathcal{L}}^{-1}=i^{\ast }\mathcal{O}(-1)$, the tautological bundle inclusion gives an inclusion of holomorphic vector bundlesA global holomorphic section of ${\mathcal{L}}^{-1}$ consequently determines $r+1$ global holomorphic functions on $M$. Compactness and the maximum modulus principle make these functions constant on every connected component. On such a component the section is therefore a constant vector $v$ lying in the tautological line $i(x)$ for every $x$.

If $v\ne 0$, all the points $i(x)$ equal $[v]$. This contradicts the fact that $i$ is an embedding of a component of positive dimension $n\ge 1$. Hence $v=0$ on every component, and $H^0(M,{\mathcal{L}}^{-1})=0$. Duality proves

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 $t=0$.

Let $E$ have rank $r\ge 1$, let $p:\mathbb{P}(E)\to X$ be its projectivization of a real vector bundle, and let $\lambda \subset p^{\ast }E$ be its tautological bundle. Put $u=w_1(\lambda )\in H^1(\mathbb{P}(E);{\mathbb{F}}_2)$. The Projective bundle definition of Stiefel–Whitney classes starts from the isomorphismIn particular, $p^{\ast }$ is injective. Expand $u^r$ uniquely in this basis. The coefficients define the Stiefel–Whitney classes through the relationSet $w_0(E)=1$ and $w_i(E)=0$ for $i>r$; for a rank-zero vector bundle set $w(E)=1$. Uniqueness also proves naturality of the Stiefel–Whitney classes under a pullback vector bundle.

To prove the Whitney product formula for Stiefel–Whitney classes, apply the splitting principle for real vector bundles simultaneously to $E$ and $E^{\prime }$. Its injective pullback on mod-two cohomology follows by iterating the Projective bundle definition of Stiefel–Whitney classes. It suffices to compute for a split vector bundle $E={⨁}_{i=1}^r{L}_i$. Put $x_i=w_1(L_i)$.

On $\mathbb{P}(E)$, the coordinate projections of the tautological inclusion $\lambda \hookrightarrow p^{\ast }E$ give sections of the real line bundles ${\lambda }^{\ast }\otimes p^{\ast }{L}_i$ with no common zero. The mod-two Euler class of a real line bundle is its first Stiefel–Whitney class, and the First Stiefel–Whitney class of a tensor product of real line bundles gives $u+p^{\ast }{x}_i$. Multiplication of Euler classes under a direct sum, and vanishing of the Euler class of a vector bundle with a nowhere-zero section, therefore giveComparing with the unique monic relation defining the Stiefel–Whitney classes shows that $w_j(E)$ is the $j$th elementary symmetric polynomial in the $x_i$. Thus $w(E)={\prod }_i(1+x_i)$. A direct sum joins the two lists, so injectivity of the splitting pullback gives

For the immersion obstruction, use the notation $g=(f,f)$: the printed $f\times f$ here means the map from one copy of Real projective space into the product. Write $x=w_1({\gamma }_{\mathbb{R}})$ and let $z$ be the mod-two reduction of $c_1({\gamma }_{\mathbb{C}}^{\ast })$. The mod-two cohomology ring of real projective space and the cohomology ring of complex projective space giveNontriviality in degree two forces $f^{\ast }z=x^2$. Necessarily $k\ge 1$, since the degree-two cohomology of ${\mathbb{CP}}^0$ vanishes.

The standard tangent bundle identities for Real projective space and Complex projective space areFor the underlying real vector bundle of a complex line bundle, its total Stiefel–Whitney class is $1+c_1\mathrm{mod}2$. HenceIf $g$ is homotopic to an immersion $j$, its normal bundle $\nu$ has rank $4k-(2k+1)=2k-1$, andThe Whitney product formula for Stiefel–Whitney classes and homotopy invariance of cohomology yieldThe division is legitimate in the truncated cohomology ring, since $1+x$ is a unit. The degree-$2k$ component isIt must vanish because $2k>\mathrm{rank}\nu$, while $x^{2k}\ne 0$. Therefore $k$ must be odd.

Use the standard complex K-theory Thom class convention. For a rank-$r$ complex vector bundle, its K-theory Euler class is the pullback of the K-theory Thom class along the zero section, namelyHere ${\Lambda }^j$ denotes an exterior power, and a Hermitian metric identifies the dual bundle with the conjugate vector bundle. Choosing ${\lambda }_{-1}(E)$ instead changes the K-theory Euler class by the unit $(-1{)}^r[\det E]$, so its kernel and cokernel computations are unchanged.

For the disk bundle $D(E)$ and sphere bundle $S(E)$, the K-theory Thom class identifies $K^i(D(E),S(E))$ with $K^i(X)$. The exact sequence of this pair becomes the K-theory Gysin sequence of a sphere bundle:The groups are two-periodic by Bott periodicity.

View Real projective space as the circle bundle of ${\gamma }^{\otimes 2}$ over Complex projective space, where $\gamma \to {\mathbb{CP}}^n$ is the tautological bundle. Indeed, $S^{2n+1}/\{\pm 1\}={\mathbb{RP}}^{2n+1}$ maps to $S({\gamma }^{\otimes 2})$ by $[v]\mapsto v\otimes v$; in each fiber squaring identifies exactly $v$ and $-v$.

Use Complex K-theory of complex projective space in the formThe K-theory Euler class of ${\gamma }^{\otimes 2}$ is $1-[\gamma {]}^{-2}$. Multiplying it by the unit $-[\gamma {]}^2$ gives $1-[\gamma {]}^2=2t-t^2$. The K-theory Gysin sequence of a sphere bundle therefore givesUnder the projection $q:{\mathbb{RP}}^{2n+1}\to {\mathbb{CP}}^n$, the pullback vector bundle $q^{\ast }\gamma$ is the complexification of a real vector bundle $L$ of the tautological bundle of Real projective space: the fiber map sends $[v,z]$ to $zv$. Thus $t$ pulls back to $1-[L]$. Put $\mu =[L]-1$. The relation $L^{\otimes 2}\cong 1$ gives ${\mu }^2=-2\mu$, and $t^{n+1}=0$ gives $2^n\mu =0$. HenceFor completeness, these relations give precisely the abelian group $\mathbb{Z}\oplus \mathbb{Z}/2^n$. In the basis $t,\dots ,t^n$, the image of multiplication by $2t-t^2$ is generated by $2t-t^2,2t^2-t^3,\dots ,2t^n$; its relation matrix is triangular with determinant $2^n$, and the relations express every $t^j$ as $2^{j-1}t$. Thus there is exactly one cyclic torsion subgroup of order $2^n$.

If $n\ge 1$ and $a=a_0+a_{1}t+\cdots +a_n{t}^n$ is in the kernel, successive coefficients of $(2t-t^2)a$ give $2a_0=0$, $2a_1-a_0=0$, and so on, up to $2a_{n-1}-a_{n-2}=0$. Hence $a_0=\cdots =a_{n-1}=0$, with $a_n$ arbitrary. For $n=0$ the multiplier is zero on $R=\mathbb{Z}$. In all cases,

Finally let $E=L^{\oplus k}$ with $k\ge 1$, over $B={\mathbb{RP}}^{2n+1}$. Since $L^{\ast }\cong L$, multiplicativity of the K-theory Euler class givesOn $K^0(B)=\mathbb{Z}\oplus (\mathbb{Z}/2^n)\mu$, the image of multiplication by this K-theory Euler class is the subgroup generated by $2^{k-1}\mu$: multiplication by an integer produces every such multiple, while multiplication by $\mu$ only adds multiples of $2^k\mu$. ThereforeMultiplication by $\mu$ is zero on $K^{-1}(B)\cong \mathbb{Z}$, since $2^n\mu =0$ and this abelian group has no nonzero torsion elements. Thus the K-theory Gysin sequence of a sphere bundle gives a short exact sequenceIt splits as an exact sequence of abelian groups because its quotient is a free abelian group. ConsequentlyThe convention $\mathbb{Z}/1=0$ covers $n=0$ and $k=1$. For example, when $k=1$ the sphere bundle is $(S^{2n+1}\times S^1)/\{(v,z)\sim (-v,-z)\}$, a mapping torus of the antipodal map; its Topological K-theory has no nonzero torsion elements, agreeing with the formula. If $k=0$ is allowed, the rank-zero vector bundle has an empty unit sphere bundle, so $K^0(S(E))=0$ instead.