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For a complex vector bundle $E\to X$ of rank $r$, its Chern classes are integral cohomology classesThey are natural under pullback and obey the Whitney sum formula for Chern classes. For a complex line bundle, $c_1$ is the Euler class of its underlying oriented real two-plane bundle; generally $c_r(E)$ is the Euler class of the underlying real bundle.^[ref]

For example, if $h=c_1(\mathcal{O}(1))$ on Complex projective space, the Euler sequence gives $c(T{\mathbb{P}}^n)=(1+h{)}^{n+1}$, truncated above degree $2n$. For an algebraic vector bundle on a smooth variety, algebraic Chern classes instead belong to the Chow groups; the cycle class map recovers the topological classes over the complex numbers.

Complexification replaces each fiber of a real vector bundle by its tensor product with $\mathbb{C}$ over $\mathbb{R}$. The complexification of a real vector bundle of rank one is a complex line bundle. For the tautological bundle over Real projective space, this complex line bundle squares to the trivial vector bundle.

A complex vector bundle is a vector bundle with complex-linear local trivializations and transition maps. A rank-$r$ complex bundle has an underlying oriented real bundle of rank $2r$, using the orientation of complex bases. Rank-one bundles are complex line bundles, and their higher-rank counterparts have integral Chern classes.

The First Chern class turns the tensor product of vector bundles of complex line bundles into addition in integral cohomology. This follows by computing the universal complex line bundle on the product of two Complex projective spaces and restricting to each factor. The proof retains torsion elements, which differential forms alone cannot detect.

Fix a complex line bundle with a Hermitian metric over a closed oriented Riemannian manifold $X$. For a unitary connection $A$, the real curvature form $i{F}_A$ represents a fixed de Rham cohomology class. The Hodge theorem gives its unique harmonic representative $\eta$. Write $\eta -i{F}_A=d\beta$ with $\beta$ a real one-form. Then $A-i\beta$ is a unitary connection with real curvature form equal to $\eta$.

Two such unitary connections differ by a closed imaginary-valued one-form. A gauge transformation removes exactly those differences whose real periods lie in $2\pi \mathbb{Z}$. Their classes under gauge equivalence of principal connections carry a free transitive action ofwhere the integral cohomology is understood through its image in real cohomology. If the first Betti number is zero, there is just one such equivalence class. The compactness assumption supplies the Hodge theorem used here.^[ref]

// Destination: geometry-and-topology.bigb

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.

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.

Start with any unitary connection ${\nabla }_0$. The Hodge decomposition theorem gives a unique real harmonic differential form $\eta$ representing $[i{F}_{{\nabla }_0}]$. Since they have the same de Rham cohomology class, $\eta -i{F}_{{\nabla }_0}=d\beta$ for a smooth real one-form $\beta$. Set $\nabla ={\nabla }_0-i\beta$. The vector-bundle curvature difference formula givesThis proves existence of a harmonic-curvature unitary line connection for any metric and any complex line bundle.

Two such unitary connections have the same vector-bundle curvature because harmonic representatives are unique. Their difference is therefore $i\gamma$ with $d\gamma =0$. When $b_1(X)=0$, write $\gamma =d\phi$ globally. The bundle gauge transformation $u=e^{i\phi }$ changes $\nabla$ to $u^{-1}\nabla u=\nabla +id\phi$. Thus the harmonic-curvature connection is unique up to gauge. On a general base, closed forms modulo the integral-period forms coming from circle-valued gauge transformations leave the torus $H^1(X;\mathbb{R})/(2\pi H^1(X;\mathbb{Z}))$ of possible gauge classes; the hypothesis removes this freedom.

The reverse-oriented Complex projective plane has intersection form $(-1)$, so $b^{+}=0$, while $b_1=0$. The Hodge star splits harmonic two-forms into self-dual and anti-self-dual subspaces, and their dimensions are the positive and negative indices of the intersection form. Therefore every harmonic two-form is an anti-self-dual two-form for every chosen metric.

By part (b), every complex line bundle has a unitary connection with harmonic real vector-bundle curvature, so this connection is an ASD connection. Conversely, an ASD connection on a complex line bundle has $dF=0$ and $d\ast F=-dF=0$, so its vector-bundle curvature is harmonic. The uniqueness in part (b) therefore applies to all ASD connections on that fixed complex line bundle. Each complex line bundle has exactly one gauge orbit of ASD connections.

Let $a,b\in H^2(S^2\times S^2;\mathbb{Z})$ be the two factor classes, normalized by $a^2=b^2=0$ and $⟨ab,[X]⟩=1$. Choose the complex line bundle $L$ with $c_1(L)=a+b$, for example the tensor product of pullbacks of degree-one complex line bundles from the two factors. ThenIf $L$ had an ASD connection, the real closed form $\alpha =iF/(2\pi )$ would represent this class and satisfy $\ast \alpha =-\alpha$. Hencea contradiction. This complex line bundle admits no ASD connection for any metric of the given product orientation. For the reverse product orientation, take $c_1(L)=a-b$ instead, which then has square $+2$. This is the positive-square obstruction to ASD line connections.

The suggested rank-two construction gives the same obstruction. The split bundle $E=L\oplus L^{\ast }$ has trivial determinant andAn ASD connection on $L$ would induce an ASD connection on $E$, but its instanton number would be $\Vert F_E{\Vert }_2^2/(8{\pi }^2)\ge 0$, contradicting $c_2(E)[X]=-2$.

The complex tautological line bundle over Complex projective space isOn the standard chart $U_i=\{[z_0:\cdots :z_n]:z_i\ne 0\}$, the vector $v_i([z])=z/z_i$ is independent of the chosen representative and gives a continuous nonzero frame. The mapis a local trivialization; its inverse takes the $i$th coordinate of the vector in the fibre. Hence $\lambda$ is a locally trivial complex line bundle.

Its unit sphere bundle is $S^{2n+1}$: a unit vector $v\in {\mathbb{C}}^{n+1}$ corresponds to $([v],v)$. The projection is the Hopf fibration. Regard $\lambda$ as an oriented real rank-two vector bundle using its complex orientation, and put $t=e(\lambda )\in H^2({\mathbb{CP}}^n;\mathbb{Z})$. The Gysin sequence of a sphere bundle gives, for $2\le q\le 2n$,as an isomorphism, since the intervening cohomology groups of $S^{2n+1}$ vanish. It also gives $H^1({\mathbb{CP}}^n)=0$. The CW complex structure has one cell in dimensions $0,2,\dots ,2n$ and none above $2n$, so these groups and products giveThe Euler class of $\lambda$ is the negative of the usual hyperplane generator; replacing $t$ by $-t$ gives the same ring presentation. For $n=0$ the formula reads $\mathbb{Z}$.

Now let $\alpha$ generate $H^2(S^2;\mathbb{Z})$ and let $u_i$ be its pullback from the $i$th factor of $(S^2{)}^n$. The Künneth theorem gives $H^2((S^2{)}^n;\mathbb{Z})={⨁}_i\mathbb{Z}{u}_i$. If a map $f$ were invariant under all factor permutations, write $f^{\ast }\alpha ={\sum }_i{a}_i{u}_i$; naturality under a transposition forces every $a_i$ to have the same integer value $a$. Let $\Delta :S^2\to (S^2{)}^n$ be the diagonal. ThenThe second condition is $f\circ \Delta =\mathrm{id}$, so this must equal $\alpha$, giving $na=1$. For $n>1$ this is impossible in $\mathbb{Z}$, and $\boxed{\text{no such continuous map exists}.}$ This is the diagonal-degree obstruction to a symmetric sphere retraction.

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.

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:

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 $t=0$.

Let $E$ have rank $r\ge 1$, let $p:\mathbb{P}(E)\to X$ be its projectivization of a real vector bundle, and let $\lambda \subset p^{\ast }E$ be its tautological bundle. Put $u=w_1(\lambda )\in H^1(\mathbb{P}(E);{\mathbb{F}}_2)$. The Projective bundle definition of Stiefel–Whitney classes starts from the isomorphismIn particular, $p^{\ast }$ is injective. Expand $u^r$ uniquely in this basis. The coefficients define the Stiefel–Whitney classes through the relationSet $w_0(E)=1$ and $w_i(E)=0$ for $i>r$; for a rank-zero vector bundle set $w(E)=1$. Uniqueness also proves naturality of the Stiefel–Whitney classes under a pullback vector bundle.

To prove the Whitney product formula for Stiefel–Whitney classes, apply the splitting principle for real vector bundles simultaneously to $E$ and $E^{\prime }$. Its injective pullback on mod-two cohomology follows by iterating the Projective bundle definition of Stiefel–Whitney classes. It suffices to compute for a split vector bundle $E={⨁}_{i=1}^r{L}_i$. Put $x_i=w_1(L_i)$.

On $\mathbb{P}(E)$, the coordinate projections of the tautological inclusion $\lambda \hookrightarrow p^{\ast }E$ give sections of the real line bundles ${\lambda }^{\ast }\otimes p^{\ast }{L}_i$ with no common zero. The mod-two Euler class of a real line bundle is its first Stiefel–Whitney class, and the First Stiefel–Whitney class of a tensor product of real line bundles gives $u+p^{\ast }{x}_i$. Multiplication of Euler classes under a direct sum, and vanishing of the Euler class of a vector bundle with a nowhere-zero section, therefore giveComparing with the unique monic relation defining the Stiefel–Whitney classes shows that $w_j(E)$ is the $j$th elementary symmetric polynomial in the $x_i$. Thus $w(E)={\prod }_i(1+x_i)$. A direct sum joins the two lists, so injectivity of the splitting pullback gives

For the immersion obstruction, use the notation $g=(f,f)$: the printed $f\times f$ here means the map from one copy of Real projective space into the product. Write $x=w_1({\gamma }_{\mathbb{R}})$ and let $z$ be the mod-two reduction of $c_1({\gamma }_{\mathbb{C}}^{\ast })$. The mod-two cohomology ring of real projective space and the cohomology ring of complex projective space giveNontriviality in degree two forces $f^{\ast }z=x^2$. Necessarily $k\ge 1$, since the degree-two cohomology of ${\mathbb{CP}}^0$ vanishes.

The standard tangent bundle identities for Real projective space and Complex projective space areFor the underlying real vector bundle of a complex line bundle, its total Stiefel–Whitney class is $1+c_1\mathrm{mod}2$. HenceIf $g$ is homotopic to an immersion $j$, its normal bundle $\nu$ has rank $4k-(2k+1)=2k-1$, andThe Whitney product formula for Stiefel–Whitney classes and homotopy invariance of cohomology yieldThe division is legitimate in the truncated cohomology ring, since $1+x$ is a unit. The degree-$2k$ component isIt must vanish because $2k>\mathrm{rank}\nu$, while $x^{2k}\ne 0$. Therefore $k$ must be odd.

For an oriented real rank-$r$ vector bundle $\pi :E\to X$, let $u_E\in H^r(D(E),S(E);\mathbb{Z})$ be its Thom class. If $s:X\to D(E)$ is the zero section, the Euler class of a vector bundle is

Let $u$ generate $H^{2k}(S^{2k};\mathbb{Z})$. The Poincaré-Hopf theorem givesFor every integer $m$, choose a map $f_m:S^{2k}\to S^{2k}$ of degree $m$. Naturality of the Euler class givesThus every even class belongs to ${\mathcal{E}}_{2k}(S^{2k})$.

More generally, every class $a\in H^{2k}(M;\mathbb{Z})$ on a closed $2k$-manifold is $f^{\ast }u$ for some map $f:M\to S^{2k}$, by the realization of top-dimensional cohomology by a sphere map. Pulling back $T{S}^{2k}$ givesso $2H^{2k}(M;\mathbb{Z})\subseteq {\mathcal{E}}_{2k}(M)$. The inclusion can be strict: every complex line bundle on $S^2$ has an underlying oriented real two-plane bundle, and these bundles realize every integral Euler class. Hence

The odd-dimensional analogue fails. The Euler class of an oriented odd-rank vector bundle is two-torsion, whereas $H^{2k+1}(S^{2k+1};\mathbb{Z})\cong \mathbb{Z}$ is torsion-free. Thus every oriented rank-$(2k+1)$ bundle on $S^{2k+1}$ has zero Euler class, and the nonzero subgroup $2\mathbb{Z}$ cannot lie in ${\mathcal{E}}_{2k+1}(S^{2k+1})$.

Finally, if $E$ and $F$ are oriented bundles of ranks $k$ and $l$, orient $E\oplus F$ by the ordered sum. The Whitney product formula for Euler classes statesTherefore cup product restricts to the asserted map ${\mathcal{E}}_k(X)\otimes {\mathcal{E}}_l(X)\to {\mathcal{E}}_{k+l}(X)$.

It need not be injective. For $X=S^2$ and $k=l=2$, one has ${\mathcal{E}}_2(S^2)=\mathbb{Z}$, so the source contains $\mathbb{Z}\otimes \mathbb{Z}\cong \mathbb{Z}$, but $H^4(S^2)=0$ and the map is zero. It need not be surjective either. For $X=S^2$ and $k=l=1$, every oriented real line bundle is trivial, so both degree-one Euler-class sets vanish, while ${\mathcal{E}}_2(S^2)=\mathbb{Z}$.

Each $U_j$ is isomorphic to $\mathbb{C}$ and is therefore contractible. Every complex line bundle over a contractible paracompact space is trivial, so $L{|}_{U_0}$ and $L{|}_{U_1}$ admit the required trivializations.

On a smooth manifold, the sequence of sheaves of abelian groupsis exact. The additive smooth-function sheaf is a fine sheaf, so the connecting map identifies $H^1(X,{\mathcal{C}}_{\mathbb{C}}^{\infty ,\ast })$ with $H^2(X,\mathbb{Z})$. The former classifies smooth complex line bundles and the connecting class is their First Chern class.

A complex line bundle has a canonically oriented underlying real vector bundle of rank two. Its first Stiefel–Whitney class vanishes, and its second Stiefel–Whitney class is the mod-two reduction of its First Chern class.

Suppose symplectic manifolds $(M_1^{2n},{\omega }_1)$ and $(M_2^{2n},{\omega }_2)$ contain compact codimension-two symplectic submanifolds $V_1,V_2$. A symplectomorphism $V_1\to V_2$ identifies their restricted forms. To form their symplectic sum, their symplectic normal bundles, viewed as complex line bundles, must have opposite First Chern classes under this identification. Choosing a fiber-reversing normal-bundle identification then permits removal of tubular neighborhoods and gluing of their circle boundaries.^[ref]

The local smoothing resembles $xy=\epsilon$ in the two normal directions. After adjusting the forms near the gluing locus, the joined manifold carries a symplectic form. The identification is part of the construction, so the result is not determined solely by the two ambient manifolds.^[ref]

In real dimension four, the surfaces must have matching genus and symplectic area, and their self-intersection numbers must sum to zero. For square-zero surfaces the normal bundles are trivial, giving the symplectic fiber sum along a square-zero surface.