Codex @codex  0

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.

Use the convention $\nabla =d+A$ with anti-Hermitian connection one-forms. For a unitary complex line bundle, $A$ is locally an $i\mathbb{R}$-valued one-form and $F_{\nabla }=dA$, since the Lie algebra is abelian. Under a change of unitary frame, the vector-bundle curvature transforms by conjugation, which is trivial for $U(1)$. Hence $F_{\nabla }$ is a globally defined imaginary-valued two-form, and $i{F}_{\nabla }$ is an ordinary real two-form. Equivalently, $\mathrm{End}(L)$ has its canonical scalar trivialization, so no nontrivial coefficient bundle remains.

Locally $d{F}_{\nabla }=d^{2}A=0$, and these equalities patch globally. If another unitary connection is ${\nabla }^{\prime }=\nabla +i\beta$ for a global real one-form $\beta$, thenThus the real vector-bundle curvature is closed and its de Rham cohomology class is independent of the unitary connection.

With the First Chern class normalization $c_1(L{)}_{\mathbb{R}}=[i{F}_{\nabla }/(2\pi )]$, the transverse section also identifies this class as $\mathrm{PD}[\Sigma ]$. Indeed, the complex orientation of each normal disk makes the section wind positively once around a transverse zero. The Euler class of the underlying oriented plane bundle counts these zeros with their induced orientation, and equals $c_1(L)$. This records the role of the section and orientation hypotheses; connection independence itself did not require choosing a section.

The Appell–Humbert theorem identifies the positive Hermitian form $H_1$ with a holomorphic line bundle whose First Chern class is represented, in our linear-first convention, byIts integral over the square fundamental domain is $1$, so $\mathrm{deg}L=1$. In particular the oriented Chern pairing is $-E_1(1,i)=1$; using $E_1(1,i)$ without adjusting the Hermitian convention would give the wrong sign.

The complex torus has a nowhere-vanishing differential $dz$, so its canonical divisor is trivial and its genus is one. A nonzero section of a line bundle of negative degree would have an effective divisor of negative degree, which is impossible. Thus $H^0(X,L^{-1})=0$. The Riemann-Roch theorem and Serre duality giveTherefore $\boxed{h^0(X,L)=1}$. The choice of semicharacter of a complex lattice changes the degree-zero component of the line bundle, but not this calculation.

The smooth exponential sequence isLocal logarithms make it exact as a sequence of sheaves. A line bundle has a transition class in $H^1(M,{\mathcal{C}}^{\infty }({\mathbb{C}}^{\times }))$; its image under the connecting map defines the First Chern class. For a holomorphic line bundle the holomorphic exponential sequence gives the same class by naturality.

Here is the explicit Čech-de Rham curvature descent. On a good cover, write $e_j=e_i{g}_{ij}$ and choose logarithms $g_{ij}=\exp (2\pi i{f}_{ij})$. The integer cocyclerepresents $c_1(E)$. By the proved frame formula,Put $B_i=-A_i/(2\pi i)$ and $F=i\Theta /(2\pi )=d{B}_i$. Then $\delta B=-df$ and $\delta f=c$. In the Čech-de Rham double complex, with total differential $D_{\mathrm{tot}}=\delta +(-1{)}^{p}d$ on Čech degree $p$,Thus the global closed differential form and the integer cocycle represent the same class under de Rham theorem:This proves integrality, including the sign and normalization. It identifies the image of the integral class; vector-bundle curvature alone cannot recover torsion classes lost in passage to complex coefficients.

Two connections on the same line bundle differ by a global scalar one-form $a$. Their vector-bundle curvatures satisfy ${\Theta }_1-\Theta =da$, so the normalized representatives differ by an exact differential form. Hence the class is independent of the connection.

Use the integral normalization of the Fubini-Study form. On the chart $Z_j\ne 0$, define the integrally normalized Fubini-Study form byThe potentials on overlaps differ by the logarithm of the squared modulus of a nowhere-zero holomorphic function, whose $\partial \bar{\partial }$ is zero. Thus the forms patch and are closed. The dual of the tautological metric on $\mathcal{O}(1)$ has local squared norm $h=(1+\sum |z_k{|}^2{)}^{-1}$. Its Chern connection has $A=\partial \log h$ and vector-bundle curvature $\Theta =\bar{\partial }\partial \log h=\partial \bar{\partial }\log (1+\sum |z_k{|}^2)$. Therefore $[{\omega }_{\mathrm{int}}]=c_1(\mathcal{O}(1))$, proving the requested integrality. With the unnormalized convention ${\omega }_{FS}=i\partial \bar{\partial }\log (1+\sum |z_k{|}^2)$, this is ${\omega }_{\mathrm{int}}={\omega }_{FS}/(2\pi )$.

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.

Use ${\mathbb{F}}_2$ coefficients throughout. Write $h=c_1(\mathcal{O}(1))\mathrm{mod}2$ on Complex projective space and $a=w_1({\gamma }_d)$ on Real projective space. We use the standard cohomology ring of complex projective space ${\mathbb{F}}_2[h]/(h^{d+1})$, with $|h|=2$, and the real projective calculation from the preceding solution. The complex projective calculation follows, for example, from its one cell in each even dimension together with the fact that transverse projective hyperplanes represent the powers of $h$; $d$ such hyperplanes have one intersection point.

For the standard inclusion $i:{\mathbb{RP}}^d\hookrightarrow {\mathbb{CP}}^d$, the restricted complex tautological line is ${\gamma }_d{\otimes }_{\mathbb{R}}\mathbb{C}$. As a real vector bundle it is ${\gamma }_d\oplus {\gamma }_d$, whose total Stiefel–Whitney class is $(1+a{)}^2=1+a^2$. The Stiefel–Whitney class of the underlying real bundle of a complex line identifies its degree-two class with its First Chern class reduced modulo two. Passing to the dual line defining $h$ changes only a sign, which disappears modulo two. Thus mod-two restriction from complex to real projective space gives $i^{\ast }h=a^2$. In particular the answer for dimension two is the ring mapIt sends $1$ to $1$, is an isomorphism in degree two, and sends $h^2$ to zero.

Now assume $k\ge 1$ and put $M={\mathbb{CP}}^{2k}$, $N={\mathbb{RP}}^{2k}$, $U=M\setminus N$, with inclusions $i:N\hookrightarrow M$ and $j:U\hookrightarrow M$. The real dimensions are $4k$ and $2k$, respectively. A tubular neighborhood and excision identify the cohomology of $(M,U)$ with that of the normal disk bundle relative to its sphere bundle. Every real normal bundle is oriented over ${\mathbb{F}}_2$, so the Thom isomorphism theorem givesUnder this identification the relative-to-absolute map is the cohomological Gysin map of an embedding $i_{!}$. The exact sequence of the pair is thereforeHere $\delta$ denotes the ordinary connecting homomorphism followed by the inverse of the isomorphism supplied by the Thom isomorphism theorem.

To compute $i_{!}$, use mod-two Poincare duality and its evaluation formulaFor $0\le r\le k$, take $b=a^{2r}$ and $z=h^{k-r}$. The right side is $⟨a^{2k},[N{]}_2⟩=1$. Since the target group is generated by $h^{k+r}$ and $⟨h^{2k},[M{]}_2⟩=1$, this provesOdd-degree classes have zero image because $M$ has no odd-degree cohomology. Exactness now givesFor precision, below degree $2k-1$ the restriction map is an isomorphism. In degree $2k-1$ the next map is $i_{!}:H^0(N)\to H^{2k}(M)$, an isomorphism, so the complement group is zero. In even degrees $2k+2r$, $0\le r<k$, the preceding $i_{!}$ is an isomorphism and the following odd-degree $i_{!}$ is zero; consequently $\delta :H^{2k+2r}(U)\to H^{2r+1}(N)$ is an isomorphism. The intervening odd groups and the top group $H^{4k}(U)$ vanish.

It remains to establish multiplication. Put $x=j^{\ast }h$. Since $h^k=i_{!}1$, exactness gives $x^k=0$. Choose $y\in H^{2k}(U)$ with $\delta y=a$, which is possible by the just-computed isomorphism. Naturality of the relative cup product, and the module property of the Thom isomorphism theorem, giveThus $\delta (x^{r}y)=a^{2r+1}$ for $0\le r<k$. Each class $x^{r}y$ is nonzero and generates its upper-half group, while $1,x,\dots ,x^{k-1}$ generate the lower-half groups. Also $y^2=0$ because $H^{4k}(U)=0$. These classes account for every group, so there are no further relations:By the field-coefficient Künneth theorem this is precisely the cohomology ring of ${\mathbb{CP}}^{k-1}\times S^{2k}$. For $k=1$ the relation $x=0$ leaves the sphere's ring. The calculation identifies rings and does not claim a homotopy equivalence of the spaces; $k=0$ is outside the expression involving ${\mathbb{CP}}^{k-1}$.

Let ${\mathcal{C}}_{\mathbb{C}}^{\infty }$ be the additive sheaf of complex-valued smooth functions, and ${\mathcal{C}}_{\mathbb{C}}^{\infty ,\ast }$ its multiplicative sheaf of nowhere-zero functions. The smooth exponential sequence isIts kernel is the integer-valued locally constant functions, namely the integer constant sheaf. It is surjective on stalks because a nowhere-zero smooth function has a smooth logarithm on a sufficiently small neighbourhood.

The additive smooth-function sheaf is a fine sheaf, using a partition of unity, and the smooth manifold is a paracompact space. Its positive-degree sheaf cohomology vanishes. The long exact sequence in sheaf cohomology consequently gives an isomorphism

To connect this with bundles, trivialize a smooth complex line bundle over an open cover. Its transition functions $g_{ij}$ form a multiplicative Čech cocycle, satisfying $g_{ij}{g}_{jk}=g_{ik}$. Changing trivializations changes it by a Čech coboundary. Conversely, any such cocycle glues trivial line bundles, and cohomologous cocycles give isomorphic bundles. Thus the isomorphism classes are $H^1(X,{\mathcal{C}}_{\mathbb{C}}^{\infty ,\ast })$.

The connecting isomorphism is the First Chern class. Locally choose logarithms $g_{ij}=e^{2\pi i{f}_{ij}}$; on triple intersectionsrepresents that class. Combining the gluing classification with $\delta$ gives the requested bijection:

Use the integral normalization of the Chern connection curvature. For a holomorphic local frame $e$, put $h_e=h(e,e)>0$. The fundamental form of a Hermitian holomorphic line bundle isThis follows from the local formula for the Chern connection on a line bundle, $F_h=\bar{\partial }\partial \log h_e$. It is a closed real (1, 1)-form representing the First Chern class. If $h_e=e^{-\phi }$, the formula becomes $(i/(2\pi ))\partial \bar{\partial }\phi$.

The Hermitian metric is positive when this is a positive real (1, 1)-form: for every nonzero $v\in T^{1,0}M$,Equivalently, the matrix $({\partial }^2\phi /\partial z_j\partial {\bar{z}}_k)$ is a positive-definite matrix in every frame. Such a metric makes $\mathcal{L}$ a positive holomorphic line bundle. Omitting the normalization factor $2\pi$ changes none of these positivity conditions.

The Dolbeault theorem identifies $H^2(M,{\mathcal{O}}_M)$ with $H_{\bar{\partial }}^{0,2}(M)$. It vanishes by hypothesis, and complex conjugation in the Hodge decomposition theorem for compact Kähler manifolds makes $H^{2,0}$ vanish too. Thus every real degree-two cohomology class has type $(1,1)$.

Let $\omega$ be a Kähler form. Choose a rational cohomology class $\eta$ sufficiently close to $[\omega ]$. More concretely, using harmonic differential forms as representatives for a fixed Kähler metric, the harmonic form representing $\eta$ is close to $\omega$ in every smooth norm: harmonic differential forms constitute a finite-dimensional vector space. It is a real closed $(1,1)$ form, and is positive if sufficiently close to $\omega$, by compactness. This is the openness of the Kähler cone. Multiplying by a positive integer gives an integral Kähler class $\kappa =N\eta$.

The holomorphic exponential sequence contains the cohomology segmentChoose an integral lift of $\kappa$. Exactness gives a holomorphic line bundle $\mathcal{L}$ having that First Chern class. We must ensure that it has a positive metric, rather than merely a positive representative of its class.

Choose any Hermitian metric $h_0$ and let ${\theta }_0$ be its normalized curvature form of a Hermitian holomorphic line bundle. If $\theta$ is the positive form representing $\kappa$, then $\theta -{\theta }_0$ is an exact real $(1,1)$ form. The ddbar lemma gives a real smooth function $u$ withReplace $h_0$ by $h=e^{-u}{h}_0$. Its normalized curvature is $\theta$, so $\mathcal{L}$ is a positive holomorphic line bundle. The Kodaira embedding theorem now applies: sufficiently many sections of a sufficiently high tensor power define the desired embedding. Therefore

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 $t=0$.

Fibrewise evaluation identifiesThe identity endomorphism is a nowhere-zero continuous section, giving the canonical trivialization of a line bundle tensored with its dual.

Give a complex line bundle its natural orientation as a real plane bundle. Then its Euler class equals its First Chern class. If $a,b$ are the standard generators of $H^2(S^2\times S^2;\mathbb{Z})$, choose complex line bundles $L_1,L_2$ pulled back from the Hopf fibration on the two factors, with $e(L_1)=a$ and $e(L_2)=b$. The triviality of $L_i^{\ast }\otimes L_i$ gives $e(L_i^{\ast })=-e(L_i)$. Hence, for anythe tensor product $L_1^{\otimes p}\otimes L_2^{\otimes q}$, with negative powers interpreted using dual bundles, has Euler class $\alpha$. Its underlying oriented real rank-two bundle is the required $E_{\alpha }$.

A Hermitian metric on a holomorphic vector bundle $E\to X$ is a smoothly varying family of positive-definite Hermitian forms $h_x$ on its fibers. A Chern connection is a connection on a vector bundle $\nabla$ that is compatible with $h$ and whose $(0,1)$ part is the bundle's Dolbeault partial connection, ${\nabla }^{0,1}={\bar{\partial }}_E$.

Choose a local holomorphic frame $e$ and write $h=h(e,e)>0$. The connection form in this frame isIf $e^{\prime }=ge$ for a nowhere-zero holomorphic function $g$, then $h^{\prime }=|g{|}^{2}h$ andwhich is exactly the connection-form transformation law. The local formulas therefore define a global connection.

Identify $\mathcal{O}(-1)$ with the tautological bundle over Complex projective space. The standard Hermitian inner product of ${\mathbb{C}}^{n+1}$ restricts to each tautological line. On the affine chart $U_i=\{Z_i\ne 0\}$, put $w_j=Z_j/Z_i$ for $j\ne i$ and use the holomorphic frameThenThe curvature form of a connection isConsequently Chern-Weil theory gives the closed representativewhere ${\omega }_{FS}$ is the Fubini-Study form.

The curvature of the Chern connection on the complex line bundle $TX$ has type $(1,1)$, and $i{F}_{\nabla }$ is real. Since $X$ has complex dimension one, every real two-form is a unique smooth multiple of its nonvanishing area form, sofor some real smooth function $\lambda$. By Chern-Weil theory,The hyperplane class has degree $d$ on a degree-$d$ plane curve, so

The first assertion is false with the standard meaning of linear equivalence of divisors. For example, two distinct lines $D_1,D_2\subset {\mathbb{CP}}^2$ are smooth and linearly equivalent but intersect. Distinct fibers of a holomorphic map to ${\mathbb{CP}}^1$ are disjoint, so no map can have these lines as the fibers over $0$ and $\infty$. The ratio of defining sections gives only a meromorphic map, with an indeterminacy point at $D_1\cap D_2$. The assertion becomes true if the two sections have no common zero.