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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.

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 Euler sequence on ${\mathbb{P}}^m$ isThe Whitney product formula for the Total Chern class therefore givesin $A^{\ast }({\mathbb{P}}^m)=\mathbb{Z}[H]/(H^{m+1})$.

The Total Chern class of a direct sum of complex vector bundles is the product of their total Chern classes. In each degree, $c_k(E\oplus F)={\sum }_{i+j=k}{c}_i(E)⌣c_j(F)$. The degree-zero class is one, classes beyond the bundle rank vanish, and a trivial bundle has total class one.