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A complex rank-$r$ vector bundle has a canonical real orientation, and its Euler class equals its top Chern class. For a complex line this is $e(L_{\mathbb{R}})=c_1(L)$. The splitting principle for complex vector bundles gives an injective cohomology pullback on which the bundle splits into lines. The Whitney product formula for Euler classes and Whitney sum formula for Chern classes then identify both sides with the product of those first Chern classes, proving the general equality.

Apply the splitting principle for complex vector bundles simultaneously to $E$ and $F$. This can be implemented by successive projective bundle pullbacks; the projective bundle theorem makes their composite pullback injective on integral cohomology. After pullback write $E=L_1\oplus L_2$ and $F=M_1\oplus M_2$, with Chern roots $x_1,x_2$ and $y_1,y_2$. LetSince the First Chern class of a tensor product of complex line bundles is the sum of their first Chern classes, the four roots of $E\otimes F$ are $x_i+y_j$, $1\le i,j\le 2$.

Compute their six pairwise products integrally. The two pairs within a fixed $x_i$ row contributeThe four pairs between the two rows contributeTheir sum is the second elementary symmetric polynomial of the tensor roots. Injectivity of the pullback returns the resulting identity to $X$, givingProducts here are cup products in integral cohomology. This second Chern class of a tensor product of rank-two bundles formula uses no division by two and therefore holds without assuming the integral cohomology of $X$ is torsion-free.

For a complex bundle $E$, write $c_t(E)={\sum }_{j\ge 0}{c}_j^{MU}(E)t^j$, with $c_0(E)=1$ and $c_j(E)=0$ for $j>{\mathrm{rank}}_{\mathbb{C}}E$. The Whitney sum formula for Chern classes in complex cobordism isProducts are in the complex-cobordism cohomology ring; these classes have even degrees $2j$. The splitting principle for complex vector bundles explains the formula: on a splitting space the Chern classes are elementary symmetric functions of the first Chern classes of the line summands, and joining the two lists multiplies their total Chern polynomials.

Use the line convention for the projective bundle:Its relative tautological line bundle isIt is a complex rank-one subbundle of $p^{\ast }\eta$. Choose a Hermitian metric and let $E=\eta (1{)}^{\perp }$, of rank $n-1$, so $p^{\ast }\eta =\eta (1)\oplus E$.

Throughout, $x=c_1^{MU}(\eta (1))$, the tautological line's class, as specified before the final equation. Put $a_k=p^{\ast }{c}_k^{MU}(\eta )$ and $b_k=c_k^{MU}(E)$, with $a_0=b_0=1$. Since $c_t(\eta (1))=1+xt$, Whitney multiplication givesComparison of coefficients gives $a_k=b_k+x{b}_{k-1}$, hence recursively $b_k=a_k-x{b}_{k-1}$. Therefore the Chern classes of a tautological-line complement areand the classes for $k\ge n$ vanish by the rank.

At degree $n$ the same recursion has $b_n=0$, givingMultiplying by $(-1{)}^n$ proves the projective bundle relation in complex cobordismThese alternating signs come from the inverse formal power series $(1+xt{)}^{-1}$. They do not require identifying the First Chern class of a dual line with $-x$, which is generally incorrect in complex cobordism.

The final sentence's designation $x=c_1(\eta )$ must mean the earlier $c_1(\eta (1))$. If instead one substitutes $p^{\ast }{c}_1(\eta )$ literally, the claimed relation is false even in ordinary cohomology. For example, take $\eta =L\oplus L$ over ${\mathbb{CP}}^2$, with $a=c_1(L)$ a generator. Its projective bundle is ${\mathbb{CP}}^2\times {\mathbb{CP}}^1$, and $p^{\ast }{a}^2\ne 0$. With the incorrect substitution $x=2p^{\ast }a$, the rank-two polynomial evaluates toThe tautological-line interpretation above is the one consistent with the bundle decomposition and the stated relation.

Write $M{U}^{\ast }={\Omega }_U^{\ast }$ for complex cobordism. A complex orientation of a smooth map is chosen stable complex normal data, not a complex structure on either manifold separately. More concretely, replace the continuous map by a homotopic smooth map $f_0$, choose an embeddingand give its normal bundle ${\nu }_j$ a stable complex structure on a real vector bundle. The auxiliary dimension can be chosen so the normal rank is even; further stabilization makes the normal bundle actually complex. Equivalence means stabilization by trivial complex summands and homotopy of the oriented embedding data. The stable identityidentifies this as orientation of the relative stable normal bundle. For odd relative codimension one uses an odd auxiliary suspension coordinate; the degree formula below is unchanged. This is the geometric factorization model for complex-oriented maps used in Algebraic and Geometric Topology 18, geometric cobordism discussion.

Let $d=\dim Y-\dim X$ and $r={\mathrm{rank}}_{\mathbb{R}}{\nu }_j=d+N$. The chosen normal orientation gives a Thom class $u_{{\nu }_j}$ of degree $r$ and the Thom isomorphism in complex cobordism. For compact manifolds the Pontryagin-Thom collapse of a tubular neighborhood to its Thom space isFor $a\in M{U}^q(X)$, define the Gysin map in complex cobordism by the compositeThus $f_{!}$ raises degree by the real relative codimension $d$. This definition depends on the specified orientation and is invariant under the stated equivalences of its normal data. The plus sign on a space denotes adjoining a disjoint basepoint.

For the embedding $i:L\hookrightarrow M$ with normal rank $2n$, let $c:M_{+}\to \mathrm{Th}(\nu )$ be its tubular collapse. The Thom construction gives $i_{!}(1)=c^{\ast }{u}_{\nu }$. Restricted to $L$, this collapse is exactly the zero section $z:L\to \mathrm{Th}(\nu )$. ThereforeThis is the self-intersection formula in complex cobordism. For an actual complex rank-$n$ normal bundle its top Chern class in complex cobordism is the zero-section pullback of its Thom class, so the right side is $c_n^{MU}(\nu )$. Equivalently, under the splitting principle for complex vector bundles both are the product of the first Chern classes of the line summands. The same argument gives $i^{\ast }{i}_{!}(a)=a{e}_{MU}(\nu )$ for every $a$.

There is a real distinction if the printed word “stable” is interpreted literally. The Euler class of a stably complex real bundle need not equal $c_n^{MU}$ of its stable complex class. For the diagonal embedding $S^4\hookrightarrow S^4\times S^4$, the normal bundle is $T{S}^4$. It has a stable complex structure withso its stable complex rank-two class is trivial and has second Chern class zero. However the ordinary integral image of $i^{\ast }{i}_{!}(1)$ is $e(T{S}^4)$, which evaluates to the Euler characteristic $2$. Thus the always-valid answer is the cobordism Euler class. The requested top-Chern identification is valid for an actual complex normal bundle, or if “top Chern class” here is explicitly being used as the Euler-class terminology for the chosen normal orientation.

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 a complex vector bundle $E$, define the kth Adams operation by the kth Newton polynomial in its exterior powers:After applying the splitting principle for complex vector bundles and writing $E=L_1\oplus \cdots \oplus L_r$, the Newton identities giveThis formula proves additivity. Applied simultaneously to splittings of $E$ and $F$, it also givesThus ${\psi }^k$ extends to a natural ring endomorphism of $K^0(X)$ and, for a line bundle,

Write $X={\mathbb{CP}}_k^{k+2}$ and let $x=[\overline{\gamma }]-1$. Its reduced K-theory is freely generated by $x^k,x^{k+1},x^{k+2}$, with $x^{k+3}=0$, and restriction to the bottom cell sends $x^k$ to a Bott element $u\in {\tilde{K}}^0(S^{2k})$ and the two higher powers to zero. If $r:X\to S^{2k}$ retracts the bottom-cell inclusion, thenfor some integers $a,b$. Naturality and the Adams operation on a Bott class giveSince ${\psi }^2(x)=2x+x^2$, comparison modulo $x^{k+3}$ first in degree $k+1$ and then in degree $k+2$ givesThe first equality makes $k$ even. Then $3k+5$ is odd, so integrality of $b$ forces $8\mid k$; it is also congruent to $2$ modulo $3$, so $3\mid k$. Consequently

The splitting principle for complex vector bundles says that for every complex vector bundle $E\to X$ there is a map $p:F(E)\to X$ such that $p^{\ast }$ is injective on cohomology andsplits into complex line bundles. Write $x_i=c_1(L_i)$ for the formal Chern roots.

Define the Chern character after this injective pullback byEach homogeneous component is a symmetric polynomial in the $x_i$ with rational coefficients, hence a polynomial in the elementary symmetric functions $c_j(E)$. It therefore descends uniquely to $H^{\mathrm{ev}}(X;\mathbb{Q})$ and depends only on $E$. Seton the Grothendieck group $K^0(X)$; additivity under direct sums makes this well defined.

If $E$ has roots $x_i$ and $F$ has roots $y_j$, then $E\otimes F$ has roots $x_i+y_j$. ConsequentlyIt also sends the trivial line to $1$, so it is a unital ring homomorphism.

For $S^{2n}$, a generator of ${\tilde{K}}^0(S^{2n})$ is the $n$-fold exterior product of the degree-two Bott element. The Chern character respects exterior products, and the degree-two Bott element has Chern character equal, up to sign, to the integral generator of ${\tilde{H}}^2(S^2;\mathbb{Z})$. Its $n$-fold product maps to the integral top-dimensional generator. Hence the Chern character on an even-dimensional sphere is integral.

Let the formal Chern roots of $E\to S^{2n}$ be $x_1,\dots ,x_r$ and write $p_n={\sum }_i{x}_i^n$. Sinceall lower Chern classes $c_1(E),\dots ,c_{n-1}(E)$ vanish. The Newton identities then reduce toThe degree-$2n$ term of the Chern character is thereforeIts evaluation on the fundamental class is an integer by integrality of the reduced Chern character. Thusis divisible by $(n-1)!$, proving the Divisibility of the top Chern number on an even-dimensional sphere.

For a rank-$r$ complex vector bundle and $u=c_1({\lambda }^{\ast })$, the cohomology of its projective bundle is free over the cohomology of the base on $1,u,\dots ,u^{r-1}$. Consequently pullback is injective, which yields the splitting principle for complex vector bundles by iteration.

The splitting principle for complex vector bundles makes pullback injective in integral cohomology and splits both bundles into lines with Chern roots $x_1,x_2$ and $y_1,y_2$. The tensor product has roots $x_i+y_j$. Expanding their six pairwise products gives the displayed formula. This integral computation does not divide by two, so it remains valid when the cohomology has two-torsion. In particular, taking $F$ trivial gives $c_2(E\oplus E)=c_1(E{)}^2+2c_2(E)$.