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For a vector bundle $E$ of rank $r$ on a smooth algebraic variety, algebraic Chern classes have $c_0(E)=1$, vanish for $i>r$, commute with pullback, and obey the Whitney sum formula for Chern classes for every short exact sequence of vector bundles. A line bundle $\mathcal{O}(D)$ has total class $1+[D]$. The projective-bundle splitting construction makes pullback injective and supplies a filtration by line bundles, so these axioms determine all classes as elementary symmetric functions of their first classes. In particular, $c_i(E^{\ast })=(-1{)}^i{c}_i(E)$ and $c_1(L\otimes M)=c_1(L)+c_1(M)$ in the ordinary Chow ring.

For a smooth degree-$d$ projective hypersurface $X\subset {\mathbb{P}}^n$, put $h=c_1({\mathcal{O}}_X(1))$. The cotangent Euler sequence in homogeneous coordinates gives $c({\Omega }_{{\mathbb{P}}^n}^1{|}_X)=(1-h{)}^{n+1}$. The Conormal exact sequence for Kähler differentials and Whitney sum formula for Chern classes therefore yield $c({\Omega }_X^1)=(1-h{)}^{n+1}/(1-dh)$. In codimension $i\le n-1$, its coefficient is $c_i({\Omega }_X^1)={\sum }_{j=0}^i(-1{)}^j(\genfrac{}{}{0ex}{}{n+1}{j})d^{i-j}{h}^i$. Higher classes vanish by rank and dimension; the formal quotient is truncated in the Chow ring.

For the line-projectivization of a rank-$n$ complex bundle, $p^{\ast }E=S\oplus S^{\perp }$. The Whitney sum formula for Chern classes gives $c_t(S^{\perp })=p^{\ast }{c}_t(E)/(1+xt)$ as a formal power series. Comparing coefficients proves the formula. Since $S^{\perp }$ has rank $n-1$, its degree-$n$ Chern class vanishes and yields the monic projective-bundle relation.

// Target: geometry-and-topology.bigb

For the orthogonal complex line flag manifold, take $x,y$ to be the first Chern classes of the duals of its two tautological lines, both in degree two. The Whitney sum formula for Chern classes gives $c_i(L^{\perp })=x^i$. The projective bundle definition of Chern classes gives the second relation, and Leray-Hirsch theorem gives the integral basis $x^i{y}^j$ with $0\le i<n$, $0\le j<n-1$. Polynomial division by the monic relation in $y$ proves that there are no further relations. Multiplying that relation by $y-x$ also gives $y^n=0$.

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.

Use algebraic Chern classes in the Chow ring of a nonsingular algebraic variety $Y$. For a vector bundle $E$ of rank $r$, the classes $c_i(E)\in {\mathrm{CH}}^i(Y)$ are invariant under bundle isomorphism, have $c_0(E)=1$, vanish for $i>r$, and commute with pullback. Their Total Chern class is $c(E)={\sum }_i{c}_i(E)$. The Whitney sum formula for Chern classes applies to every short exact sequence $0\to E^{\prime }\to E\to E^{\prime \prime }\to 0$:Normalization for a line bundle is $c({\mathcal{O}}_Y(D))=1+[D]$, where $[D]$ is the divisor class. In particular, $c_1(L\otimes M)=c_1(L)+c_1(M)$, $c_1(L^{-1})=-c_1(L)$, and the trivial bundle has total class one. These properties determine the classes: pull back to the iterated projective bundle where $E$ has a filtration by line bundles. The projective bundle formula for Chow groups makes this pullback injective, and Whitney multiplication expresses $c_i(E)$ as the elementary symmetric polynomial of the first classes of the line factors. This also gives $c_i(E^{\ast })=(-1{)}^i{c}_i(E)$.

Let $X$ be the smooth degree-$d$ projective hypersurface and set $h=c_1({\mathcal{O}}_X(1))$. The algebraic cotangent bundle has rank $n-1$. Restriction of the cotangent Euler sequence in homogeneous coordinates remains exact, because its terms and quotient are locally free. Thusand Whitney multiplication givesThe defining equation has degree $d$, so the conormal sheaf is ${\mathcal{O}}_X(-d)$. Smoothness makes its differential map injective with locally free quotient, giving the Conormal exact sequence for Kähler differentialsA second application of the Whitney sum formula for Chern classes yieldsThe inverse is the finite geometric series in the graded Chow ring: terms in codimension greater than $n-1$ vanish. Expanding gives every requested class,For example,when the corresponding ranks and codimensions exist. The first expression agrees with the canonical bundle of a smooth projective hypersurface $\det {\Omega }_X^1={\mathcal{O}}_X(d-n-1)$. For a plane curve, $\mathrm{deg}{c}_1=d(d-3)=2g-2$, matching Q4. For a quartic surface in ${\mathbb{P}}^3$, $c_1=0$ and $c_2=6h^2$, whose degree is $24$. These checks also confirm the sign in the cotangent, rather than tangent, formula.

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.

Along the smooth complex curve, there is an exact sequence of complex vector bundlesThe Whitney sum formula for Chern classes gives $c_1(TX){|}_{\Sigma }=c_1(T\Sigma )+c_1(N_{\Sigma /X})$. The complex tangent line has first Chern number $\chi (\Sigma )=2-2g(\Sigma )$. The normal complex line is an oriented real rank-two bundle, whose Euler number is the self-intersection of the curve: a generic normal section has signed zeros equal to the intersections with a small pushed-off copy. Thusand the adjunction formula isEquivalently, since $c_1(K_X)=-c_1(TX)$, it is $2g-2=\Sigma \cdot (\Sigma +K_X)$. The derivation applies to a compact smooth complex curve; its complex orientation fixes all signs.

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.