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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.

Use integral cohomology and let $k\ge 1$. On the projective bundle $Y=\mathbb{P}(E)$ define the complex tautological line bundlePut $t=e(\mathcal{L})\in H^2(Y;\mathbb{Z})$, using the canonical complex orientation. On each fibre, $t$ is the Euler class of the tautological line over ${\mathbb{CP}}^{k-1}$, so $1,t,\dots ,t^{k-1}$ restrict to an integral basis of its cohomology.

Here is the finite-cover Leray-Hirsch theorem proof in this case. For each open set $U\subseteq X$, defineIf $E$ is trivial on $U$, its projective bundle is $U\times {\mathbb{CP}}^{k-1}$ and $\mathcal{L}$ is pulled back from the tautological line on the second factor. The Künneth theorem makes ${\Phi }_U$ an isomorphism, since the fibre has finite free integral cohomology. The same holds for every open subset of $U$.

Compactness of $X$ provides a finite trivializing cover $U_1,\dots ,U_N$. Induct on its size. If the result holds on $V=U_1\cup \cdots \cup U_{N-1}$, it holds on $U_N$ and on $V\cap U_N$, both lying in a trivializing chart. Form the diagram of Mayer–Vietoris sequences for the base, with the finitely many degree shifts on the left, and the total space on the right. Naturality of pullback and multiplication by the global even-degree classes $t^i$ makes the diagram commute. The Five lemma gives the isomorphism on $V\cup U_N$. ThusThis is the claimed free module statement, with its graded degree shifts made explicit.

The module basis expresses $t^k$ uniquely using the lower powers, with homogeneous coefficients. Define the Chern classes $c_j(E)\in H^{2j}(X;\mathbb{Z})$ by the unique relationThe pullbacks are suppressed when $f$ is regarded as a polynomial over $H^{\ast }(X)$. Evaluation at $t$ gives a surjective map $H^{\ast }(X)[T]\to H^{\ast }(Y)$. Since $f(T)$ is monic, monic polynomial division over a ring writes any polynomial as $q(T)f(T)+r(T)$ with degree of $r$ below $k$. If its evaluation is zero, module independence forces every coefficient of $r$ to vanish. Hence the kernel is exactly the ideal generated by $f$, provingThe even-degree coefficients are central in the graded commutative algebra, so this division and ideal statement also apply when the base has odd-degree cohomology. Uniqueness of the coefficients proves their naturality under pullback, by pulling back the relation and using the same module basis. With the hyperplane convention $u=-t$, the relation has the usual all-positive Chern coefficients; the alternating signs here correspond to the tautological line itself.

Now suppose $E={⨁}_i{L}_i$. The sections of a projective bundle $s_i:X\to Y$ choose the line $(L_i{)}_x\subset E_x$. They satisfy $s_i^{\ast }\mathcal{L}\cong L_i$, so $s_i^{\ast }t=e_i$ and pulling back the relation gives $f(e_i)=0$.

To obtain the full factorization over the possibly torsion-containing base ring, also use the associated open chartsThey contain the images of the sections $s_i$ and cover $Y$. Projection identifies $\mathcal{L}{|}_{V_i}$ with ${\pi }^{\ast }{L}_i{|}_{V_i}$, so $t-{\pi }^{\ast }{e}_i$ restricts to zero on $V_i$. The long exact sequence of the pair lifts this class to $H^2(Y,V_i)$. The relative cup product of the $k$ lifted classes lies inIts absolute image is ${\prod }_i(t-{\pi }^{\ast }{e}_i)$, so that product vanishes. The polynomial ${\prod }_i(T-e_i)$ is monic of degree $k$ and lies in the kernel of evaluation. Subtracting the monic generator $f(T)$ leaves degree below $k$, and module independence again makes the difference zero. ThereforeThe open-cover argument proves the factorization without a non-zero-divisor assumption on the differences $e_i-e_j$.

A Chern number turns local curvature data into a global integer. Let $E\to M$ be a complex vector bundle over a compact oriented manifold without boundary, with a unitary connection represented locally by an anti-Hermitian matrix-valued one-form $A$. Its curvature form of a connection is $F=dA+A\wedge A$. The Chern-Weil theory representative of the total Chern class isThe coefficients are closed differential forms. They represent the images of integral Chern classes in de Rham cohomology. On an oriented $2r$-manifold, a product $c_{i_1}\cdots c_{i_s}$ with $i_1+\cdots +i_s=r$ pairs with the fundamental class to give a Chern number, an integer independent of the unitary connection.

For example,For an $SU(N)$ bundle with $N\ge 2$, $\mathrm{Tr}F=0$. On an oriented four-manifold the Second Chern number is thenusing the fundamental matrix trace and the stated anti-Hermitian convention. The trace-product correction is required for a general $U(N)$ bundle; it cannot simply be omitted. The sign of a physical instanton number also depends on the trace and orientation convention, so these conventions must accompany the formula.

The local origin of closure is the Bianchi identity $D_{A}F=0$ and the cyclic matrix trace: $d\mathrm{Tr}(F\wedge F)=0$. Independence of the connection one-form follows more concretely by varying a family $A_t$. Since ${\dot{F}}_t=D_{A_t}{\dot{A}}_t$, the Chern-Weil connection transgression isIts integral on a closed four-manifold is zero by the Generalized Stokes theorem. This proves connection independence of $q$; the integrality is the global Chern class statement, not merely a consequence of the local formula.

The local primitive of the Second Chern form is the Chern-Simons three-formHere we continue to use $SU(N)$, so $c_2=\mathrm{Tr}(F\wedge F)/(8{\pi }^2)$. The cubic coefficient is forced by the exterior derivative. In the graded cyclic matrix trace, $d\mathrm{Tr}(A^3)=3\mathrm{Tr}(dA\wedge A^2)$ and $\mathrm{Tr}(A^4)=0$, the latter because cycling one degree-one factor past the other three changes its sign. ThereforeMatrix-valued differential forms require both matrix order and the graded signs; treating all factors as commuting scalars would lose this derivation.

The Chern-Simons three-form depends on a local trivialization and is not itself gauge-invariant. For the Yang-Mills gauge transformation convention $A^g=g^{-1}Ag+g^{-1}dg$, set $u=g^{-1}dg$. The gauge change of the Chern-Simons three-form isThe last term is closed by the Maurer-Cartan equation. On a closed three-manifold its integral is an integer with the fundamental $SU(N)$ normalization. Consequently the Chern-Simons integral is naturally defined modulo integers, while its exponential $\exp (2\pi i\ell \int Y)$ is invariant under large Yang-Mills gauge transformations for integer level $\ell$.

This also explains why a nonzero Chern number is compatible with $dY=c_2$: $Y$ need not be a globally defined three-form. On $S^4$, trivialize over two hemispheres and let $A_N=A_S^g$ on their common equator $\Sigma =S^3$, oriented as the boundary of the northern hemisphere. The Generalized Stokes theorem and the gauge-change formula giveFor $SU(2)\cong S^3$, this is the degree of the transition map, with compatible group orientation; for $SU(N)$ it is the corresponding integer in ${\pi }_3(SU(N))$. Thus the Second Chern number measures the obstruction to choosing one trivialization over the whole four-sphere.

A simpler First Chern class example is a line bundle over $S^2$. Write $A=-ia$ and choose local real potentialsThey have common curvature $da=(k/2)\sin \theta d\theta \wedge d\varphi$, so ${\int }_{S^2}{c}_1=(2\pi {)}^{-1}\int da=k$. Their difference $a_N-a_S=kd\varphi$ corresponds to the transition function $e^{-ik\varphi }$, which is single-valued exactly when $k\in \mathbb{Z}$. This illustrates how the global integer arises from patching, rather than from an arbitrary flux normalization.

In physics, these constructions distinguish topological sectors of Yang-Mills instantons and relate four-dimensional characteristic densities to three-dimensional boundary actions. The Chern-Simons three-form itself gives a metric-independent gauge action in three dimensions. On a closed manifold its first variation isso its classical equation is $F=0$. The common thread is that a local expression in the connection one-form records global topology: curvature produces the invariant Chern number, while its local Chern-Simons three-form primitive retains gauge and boundary information.

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.

The quaternionic projective space ${\mathbb{HP}}^n$ is the space of one-dimensional right quaternion subspaces of ${\mathbb{H}}^{n+1}$. Equivalently it is the quotient of the unit sphere $S^{4n+3}$ by simultaneous right multiplication by unit quaternions. Its coordinate filtration has one open cell ${\mathbb{H}}^k\cong {\mathbb{R}}^{4k}$ in each dimension $4k$, for $0\le k\le n$. Hence its cellular cohomology is $\mathbb{Z}$ in those dimensions and zero otherwise.

Let $\mathcal{H}$ be the quaternionic tautological line bundle. Its unit sphere bundle is $S^{4n+3}\to {\mathbb{HP}}^n$, with fibre $S^3$. The Gysin sequence of a sphere bundle shows that multiplication by its Euler class $e\in H^4$ is an isomorphism from $H^{4k}$ to $H^{4k+4}$ for $k<n$. Choose the generator $u=-e$. Its powers generate every nonzero positive degree, giving the cohomology ring of quaternionic projective spaceThis also accounts for $n=0$.

First take $n=2$. Under the coordinate inclusion $i:{\mathbb{CP}}^2\to {\mathbb{HP}}^2$, the pulled-back quaternionic line is the quaternionic extension of the complex tautological line $L$. As a complex rank-two bundle it is $L\oplus \overline{L}$: a transition scalar $c\in \mathbb{C}$ acts on the two complex coordinates of a quaternion by $c$ and $\overline{c}$. Put $x=c_1(L^{\ast })$, the degree-two generator of the cohomology ring of complex projective space. The Whitney sum formula for Chern classes givesHere the Euler class of a complex vector bundle is its top Chern class, using the complex orientation. In particular this degree-four pullback has coefficient one; it is not a multiple of larger absolute value. The compatible tautological bundles on the projective filtrations give the same equality for every $n$, and multiplicativity then determines the whole ring map:It is zero whenever $2k>n$. These facts are the complex inclusion into quaternionic projective space.

For an odd prime $p$, the Steenrod reduced powers are natural stable cohomology operationsThey satisfy $P^0=1$, the Cartan formula, $P^{i}a=0$ when $2i>|a|$, and $P^{i}a=a^p$ when $|a|=2i$. In particular, on ${\mathbb{CP}}^{\infty }$ one has $P^{1}x=x^p$ and $P^{i}x=0$ for $i>1$. The Cartan formula and the binomial theorem givePass to the infinite projective spaces, where $i^{\ast }:H^{\ast }({\mathbb{HP}}^{\infty };{\mathbb{F}}_p)\to H^{\ast }({\mathbb{CP}}^{\infty };{\mathbb{F}}_p)$, $u\mapsto x^2$, is injective. The equality just obtained determines the Steenrod powers on quaternionic projective space; restricting to the finite spaces giveswith coefficients modulo $p$ and powers above $n$ set to zero. In particular $P^{1}u=2u^{(p+1)/2}$, $P^{2}u=u^p$, and $P^{i}u=0$ for $i>2$. The infinite-space argument matters: the finite inclusion cannot detect those degrees for which $x^{2k}=0$.

Finally put $Q={\mathbb{HP}}^{15}/{\mathbb{HP}}^{12}$ and $S={\Sigma }^{48}{\mathbb{HP}}^3$. Both have reduced cohomology $\mathbb{Z}$ in degrees $52,56,60$ and zero otherwise. Choose integral generators $a_{13},a_{14},a_{15}$ for $Q$ whose pullbacks under the quotient map are $u^{13},u^{14},u^{15}$, and suspended integral generators $b_1,b_2,b_3$ for $S$ from $u,u^2,u^3$.

Use $p=5$. Naturality for the quotient and the formula above giveStability under the suspension isomorphism instead givesAny homotopy equivalence $h:Q\to S$ would induce isomorphisms on the rank-one integral groups, so $h^{\ast }{b}_1=\epsilon a_{13}$ and $h^{\ast }{b}_3=\delta a_{15}$ with $\epsilon ,\delta \in \{1,-1\}$. Reducing modulo five and commuting $h^{\ast }$ with $P^1$ would require $\epsilon =2\delta$ in ${\mathbb{F}}_5$. Neither $1$ nor $-1$ equals $2$ or $-2$ modulo five. ThereforeThe essential point is that integral generator signs constrain Steenrod comparisons. Arbitrary changes of basis over ${\mathbb{F}}_5$ could rescale these two nonzero coefficients into agreement; a genuine equivalence must also preserve the integral lattices, where only the two signs are available.

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 compact space $X$, $K^0(X)$ is the Grothendieck group of isomorphism classes of finite-rank complex vector bundles under direct sum. Tensor product descends to this group and makes it a ring, with $[{\mathbb{C}}_X]$ as its unit.

The hypothesis $E_0\oplus {\mathbb{C}}_X\cong E_1\oplus {\mathbb{C}}_X$ says that $E_0$ and $E_1$ define the same stable class. Rank-$d$ bundles are classified by maps to $BU(d)$, and the stabilization $BU(d)\to BU$ is $2d$-connected. Since the finite CW complex $X$ has dimension $k\le 2d$, stabilization is injective on $[X,BU(d)]$. Thus stable cancellation for complex vector bundles gives

Now let $E_0,E_1$ have rank $d$ over Complex projective space ${\mathbb{CP}}^d$ and have the same Chern classes. Their Chern characters agree because each component of $\mathrm{ch}(E)$ is a universal rational polynomial in the Chern classes. The ringis torsion-free, while the Chern character becomes an isomorphism after tensoring with $\mathbb{Q}$; it is therefore injective. Hence $[E_0]=[E_1]$ in K-theory, so after adding trivial bundles they are isomorphic. Since ${\mathbb{CP}}^d$ has real dimension $2d$, cancellation applies once more:

From the tautological exact sequence and the Whitney formula in the Chern class formalism,Thus $c_i(\xi )$ is a sum of terms ${\zeta }^j{p}^{\ast }{c}_{i-j}(E)$ with $0\le j\le i$. The projective-bundle pushforward satisfiesThe projection formula now makes every term vanish when $i<e$. For $i=e$, only the term ${\zeta }^e{p}^{\ast }\alpha$ survives, giving

For a real vector bundle $E$, its Pontryagin classes are the even-index Chern classes of its complexification with the displayed alternating signs. They lie in integral cohomology of degrees $4k$ and obey a total-class product formula under direct sums. If a metric connection on a vector bundle has curvature matrix $R$, its first real representative is $p_1(E)=-\mathrm{tr}(R\wedge R)/(8{\pi }^2)$. The Bianchi identity makes this representative closed, and Chern-Weil connection transgression makes its de Rham cohomology class independent of the connection. This normalization gives degree-four A-hat form $-p_1/24$.

For a rank-$r$ complex vector bundle, let $h=c_1(S^{\ast })$ for the tautological line $S$ on its projective bundle. The Leray-Hirsch theorem gives the free basis $1,h,\dots ,h^{r-1}$ over base cohomology. The uniquely determined coefficients in the displayed monic relation define the integral Chern classes. The line normalization uses the Euler class of the canonically oriented underlying real two-plane bundle, and uniqueness makes this definition intrinsic and natural under pullback. No choice of a trivializing cover or classifying map remains in the definition.