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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 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

On a compact oriented $2r$-manifold without boundary, an integral product of Chern classes of total complex degree $r$ pairs with the fundamental class to give a Chern number. A unitary connection supplies Chern-Weil theory representatives, but the integer is independent of that connection. The Second Chern number on a four-manifold is one important example; the integral of the First Chern class over a closed surface is another.

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 $y^n=0$.

Coordinate inclusion of complex vectors sends a complex line to its quaternionic span. If two such vectors span the same quaternionic line, a nonzero complex coordinate forces their scalar ratio to be complex, so this map is injective. The pulled-back quaternionic tautological line bundle splits as $L\oplus \overline{L}$ with complex first Chern classes $-x,x$. Its top Chern class, and thus Euler class, is $-x^2$. With $u$ the negative Euler generator, $i^{\ast }u=x^2$ and $i^{\ast }(u^k)=x^{2k}$. The map on infinite projective spaces is injective on cohomology; a finite inclusion loses the degrees with $2k>n$.

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.

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 an SU(2) connection one-form $A=A_{\mu }d{x}^{\mu }$ with Skew-Hermitian generators $T_a=-i{\sigma }_a/2$, so $\mathrm{tr}(T_a{T}_b)=-{\delta }_{ab}/2$. Absorb the gauge coupling into $A$, and write its gauge curvature as $F=dA+A\wedge A$. The Yang-Mills gauge transformation convention is $A^h=h^{-1}Ah+h^{-1}dh$, giving $F^h=h^{-1}Fh$.

In four dimensions the Second Chern form and Second Chern number of the associated fundamental rank-two bundle areHere $M$ is a closed oriented four-manifold. This sign follows by expanding the total Chern class $\det (1+iF/(2\pi ))$ and using $\mathrm{tr}F=0$. The first Chern class vanishes for SU(2); the second is the degree-four characteristic class relevant here. With the same curvature convention the commonly positive self-dual instanton number isBoth sign conventions occur; fixing them at the outset avoids identifying opposite-oriented integers.

The form is gauge-invariant by conjugation and cyclicity of the matrix trace. The gauge-theory Bianchi identity $D_{A}F=0$ makes it closed. To see that its integral is independent of the connection, vary $A$:On a closed $M$ the integral of this exact form vanishes. This is Chern-Weil connection transgression, showing that $k_C$ depends on the bundle's topology rather than on a particular gauge field.

Locally the same four-form is an exterior derivative. The Chern-Simons 3-formsatisfiesFor verification, differentiating gives $\mathrm{tr}(dA\wedge dA+2dA\wedge A\wedge A)$; expansion of $F\wedge F$ gives these terms and $\mathrm{tr}{A}^4$. Graded cyclicity moves the first one-form through the other three, changing the sign, so $\mathrm{tr}{A}^4=0$. This local exactness does not make the Second Chern number zero on every closed manifold: on a nontrivial bundle the gauge potentials and ${\omega }_3$ are only patchwise defined.

For the usual finite-action sector on ${\mathbb{R}}^4$, impose pure-gauge behavior $A\to h^{-1}dh$ at infinity. Compactification gives a bundle over $S^4$, with transition map $h:S^3\to SU(2)$. The Maurer-Cartan equation gives $d(h^{-1}dh)=-(h^{-1}dh{)}^2$, henceWith the outward boundary orientation, Stokes theorem therefore givesThese are opposite choices of the winding number, and either becomes the topological degree of $h$ after specifying the orientation of the target SU(2) sphere. Thus integrality also follows from ${\pi }_3(SU(2))=\mathbb{Z}$. Smooth deformations preserving the asymptotic sector cannot change it.

The positive Euclidean Yang-Mills action isDefine $\Vert X{\Vert }^2=-\int \mathrm{tr}(X\wedge \ast X)$. Since the Hodge star has square one on two-forms in Euclidean dimension four,Choosing $s=\mathrm{sgn}Q$ gives the Yang-Mills instanton Bogomolny boundThe self-dual Yang-Mills equations and their anti-self-dual version imply the Yang-Mills equations $D_A\ast F=0$ by the Bianchi identity. Their finite-action solutions are Yang-Mills instantons.

For a concrete self-dual example, the BPST instanton with center $a$ and size $\rho >0$ haswhere ${\eta }_{\mu \nu }^a$ are self-dual 't Hooft symbols. Their contraction is ${\sum }_{a,\mu ,\nu }({\eta }_{\mu \nu }^a{)}^2=12$. Thus $F_{\mu \nu }^a{F}_{\mu \nu }^a=192{\rho }^4/(|x-a{|}^2+{\rho }^2{)}^4$, andThis gives $S=8{\pi }^2/g^2$ and $Q=1$. Reversing duality gives $Q=-1$. The free center, size and framed gauge orientation illustrate the instanton moduli space.

On a three-dimensional slice $\Sigma$, there is no four-form Chern integral intrinsic to the slice. Its transgression is the Chern-Simons number of a gauge field, in the instanton-charge conventionIt is a real number for a general connection, not necessarily an integer, and is not strictly gauge invariant. Set $\theta =h^{-1}dh$ and $\bar{\theta }=dh{h}^{-1}$. Substituting $A^h=h^{-1}Ah+\theta$, using $d\theta =-{\theta }^2$ and graded trace cyclicity, gives the gauge change of the Chern-Simons three-formThe mixed terms collect into the displayed exact form; the last term is also fixed by the pure-gauge case $A=0$. Therefore on a closed $\Sigma$, or with boundary conditions making the exact-form integral vanish,Thus $N_{\mathrm{CS}}\mathrm{mod}\mathbb{Z}$, or $\exp (2\pi i{N}_{\mathrm{CS}})$, is gauge invariant. Gauge maps homotopic to the identity have $w=0$; maps of nonzero winding are large transformations even when they equal the identity at infinity.

For a Yang-Mills vacuum on ${\mathbb{R}}^3$, zero magnetic energy requires $F_{ij}=0$. On simply connected space this is a pure gauge $A=h^{-1}dh$, and $N_{\mathrm{CS}}=w(h)$ is an integer after fixing the asymptotic trivialization. Quotienting only by transformations homotopic to the identity leaves integer-labelled vacuum representatives. Allowing all winding transformations identifies them classically; quantum states can instead transform by a phase, producing the theta vacuum. A three-dimensional pure-gauge representative carries a winding number but is not a localized positive-energy monopole. In fact no static finite-energy lump in three-dimensional pure Yang-Mills theory exists with the usual decay conditions: $A_i^{(s)}(x)=s{A}_i(sx)$ gives $E(s)=sE(1)$, and stationarity forces $E=0$.

Finally, on an oriented spacetime slab $[t_{-},t_{+}]\times \Sigma$, with no side-boundary contribution, the transgression identity givesA Yang-Mills instanton therefore interpolates between vacuum representatives whose Chern-Simons numbers differ by its integer charge. In a four-dimensional quantum theory a Yang-Mills theta term weights such sectors by $e^{i\theta Q}$, making $\theta$ periodic modulo $2\pi$. In an intrinsically three-dimensional theory an action $2\pi \ell N_{\mathrm{CS}}$ has a gauge-invariant phase $e^{iS}$ only for integer level $\ell$ under all winding gauge maps. These statements distinguish the gauge-invariant integer four-dimensional Chern charge from the three-dimensional connection-dependent quantity defined modulo integers.

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.

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.

The orthogonal group acts transitively on unoriented lines in ${\mathbb{R}}^n$. The stabilizer of the line spanned by the first coordinate isThus $G/H={\mathbb{RP}}^{n-1}$; using only $O(n-1)$ would stabilize a vector and give a sphere instead. For the universal contractible free $G$-space $EG$, the associated bundle $EG/H\to EG/G$ gives the Serre fibration $G/H\to BH\to BG$. Its total space identifies up to homotopy with $BO(1)\times BO(n-1)$.

If $n$ is odd, $n-1$ is even and the rational cohomology of ${\mathbb{RP}}^{n-1}$ is just $\mathbb{Q}$ in degree zero. This follows from the integral projective cellular differentials, alternating zero and multiplication by two, with the even-dimensional top class killed rationally. The cohomological Serre spectral sequence therefore has only the rowwith all other rows zero. The local system in degree zero is trivial because the fiber is connected. There can be no differential or extension, and the multiplicative edge map $H^{\ast }(BO(n);\mathbb{Q})\to H^{\ast }(BH;\mathbb{Q})$ is an isomorphism of rings.

Also $BO(1)={\mathbb{RP}}^{\infty }$ has rational cohomology $\mathbb{Q}$ in degree zero only, by the same cellular calculation. The Künneth theorem identifies $H^{\ast }(BH;\mathbb{Q})$ with $H^{\ast }(BO(n-1);\mathbb{Q})$. The inclusion fixing the first coordinate factors through $H$, setting its $O(1)$ coordinate to one; hence the resulting isomorphism is the actual restriction along this inclusion:This includes $n=1$, with $BO(0)$ a point, and proves rational odd-rank stabilization of orthogonal classifying spaces.

For even $n$, the fiber is odd-dimensional real projective space and has a further rational class in degree $n-1$. The spectral sequence has a possible additional row, now with determinant-sign local coefficients: an orientation-reversing orthogonal matrix acts with degree $-1$ on the covering sphere and hence on the odd-dimensional projective space. Therefore the one-row collapse argument is unavailable; this top row must not be treated as automatically untwisted.

There is an actual failure of injectivity in every even rank. Write $n=2m$. The universal class $p_m\in H^{4m}(BO(2m);\mathbb{Q})$ is nonzero: pull the universal bundle back to the sum of the real underlying bundles of $m$ complex lines over $({\mathbb{CP}}^{\infty }{)}^m$. If their first Chern classes are $x_j$, the Pontryagin formula $p_m(E)=(-1{)}^m{c}_{2m}(E\otimes \mathbb{C})$ gives $p_m={\prod }_j{x}_j^2\ne 0$. On $BO(2m-1)$ the restricted bundle is ${\epsilon }^1\oplus {\gamma }_{2m-1}$; its top complex Chern class $c_{2m}$ vanishes by the rank bound and the trivial summand. Thus $p_m$ restricts to zero. The natural restriction is not injective for even $n$, as the extra fiber cohomology already warns.

Use projectivization by quotients, and setThe fibre over $t$ parametrizes one-dimensional quotients of $E_t$, so it is ${\mathbb{P}}^{n-1}$. The standard line bundles areThus $L$ is the fiber class of a rational normal scroll of the ruling and $M$ is the universal quotient line bundle of ${\pi }^{\ast }E$. In additive divisor notation we also write $L,M$ for their first Chern classes. For $n\ge 2$ these generate the Picard group; for $n=1$, $F={\mathbb{P}}^1$ and $M=a_{1}L$. This fixes the sign convention for the rational normal scroll.

Put $s={\sum }_i{a}_i$. The projective bundle formula for Chow groups givesHere the first relation comes from the one-dimensional base, and the second is the projective-bundle relation with $c_1(E)=sL$ and higher Chern classes zero on the base. Integration over a fibre gives $L{M}^{n-1}=1$. Therefore all top-degree mixed intersection numbers areIn particular, for integers $u,v$,For the canonical divisor of a smooth variety, the relative Euler sequencehas determinant ${\mathcal{O}}_F(nM-sL)$. Combining its dual determinant with $K_{{\mathbb{P}}^1}=-2L$ gives

A common twist does not change the abstract projective bundle. Choose an integer $b$ such that $d_i=a_i+b\ge 1$ for every $i$. The complete linear system of divisors $|M+bL|$ hasIts map to ${\mathbb{P}}^N$, where $N=s+nb+n-1$, has coordinate blocksHere $[s_0:t_0]$ is a point of the base and the ${\lambda }_i$ are quotient coordinates in a local trivialization of $E(b)$. These expressions have the correct transition factors and define the global map. Any nonzero block recovers the base point through its degree-$d_i$ rational normal curve. Once that point is known, the blocks recover the projective fibre coordinates. On the open where the first coordinate of a nonzero block is nonzero, the ratio of the next coordinate to the first recovers $t_0/s_0$; the analogous last-coordinate chart recovers $s_0/t_0$. The remaining coordinate ratios then recover the fibre coordinates regularly. Thus the map identifies $F$ with its closed image and is a projective embedding, not merely a pointwise injection.

The hyperplane class on this image is $M+bL$, soConsequently every such natural embedding hasThe tautological choice $b=0$ is an embedding when all $a_i>0$. If the $a_i$ are only nonnegative, $|M|$ is basepoint-free but can contract the projective subbundle arising from zero summands; one must distinguish that scroll image from the smooth abstract scroll.

Geometric quantization. A classical symplectic manifold supplies observables as real functions, their Poisson bracket, and Hamiltonian flows. Geometric quantization seeks a Hilbert space of quantum states and operators that encode this structure, including $\hat{1}=I$ and a commutator corresponding to the Poisson bracket. A direct prescription cannot make every classical product become an operator product without qualification; operator ordering and the choice of physically admissible observables matter. The aim is to build the state space and operators from the geometry rather than from a preferred coordinate chart.

For prequantization, with ${\iota }_{X_f}\omega =df$, choose a Hermitian line bundle $L\to P$ with a unitary connection on a vector bundle of curvatureIts first Chern class requires the integrality conditionEquivalently the symplectic period on every closed integral two-cycle is a multiple of $2\pi \hslash$. Conversely an integral class supplies a line bundle, and a unitary connection can be adjusted by a one-form to have the required curvature. Locally choose $\theta$ with $\omega =-d\theta$; then $\nabla =d-i\theta /\hslash$. Potentials differing by an exact one-form change the local phase of the section. Integrality is precisely what makes these local descriptions glue consistently.

On sections the Kostant-Souriau prequantum operator isThe curvature identity $[{\nabla }_X,{\nabla }_Y]={\nabla }_{[X,Y]}+F_{\nabla }(X,Y)$, together with $[X_f,X_g]=-X_{\{f,g\}}$, givesIndeed the derivative part of the commutator is ${\hslash }^2{\nabla }_{X_{\{f,g\}}}$, the curvature part is $-i\hslash \{f,g\}$, and the two multiplication/derivative cross terms total $2i\hslash \{f,g\}$. These combine to the displayed operator identity. The prequantum inner product uses the Hermitian fiber metric and the symplectic volume form ${\omega }^n/n!$, with appropriate square-integrability and operator-domain conditions.

This is not yet the desired quantum state space: a prequantum section depends on all $2n$ phase-space variables, whereas a wavefunction normally depends on only $n$ configuration variables. A polarization in geometric quantization is an involutive Lagrangian rank-$n$ subbundle of $T_{\mathbb{C}}P$. Polarized sections obey ${\nabla }_{X}s=0$ along that distribution. The symplectic form vanishes on it, so its restricted curvature vanishes and these differential conditions are locally compatible.

For $T^{\ast }Q$, the vertical polarization is spanned locally by ${\partial }_{p_i}$. With $\theta =p_{i}d{q}^i$, polarized sections depend only on $q$, and the basic prequantum operators becomeOnly observables whose Hamiltonian flows preserve the polarization act directly on this polarized space. A generic quadratic kinetic energy does not preserve the vertical polarization, so obtaining its quantum operator needs an additional construction rather than simply restricting every prequantum operator. A compatible complex structure can instead provide a Kähler polarization, yielding holomorphic-type states, as for oscillator variables. The Blattner-Kostant-Sternberg pairing compares suitable state spaces of different polarizations and can implement some transported observables.

Finding a useful global polarization is a genuine problem: it may not exist as a nonsingular real foliation; leaf holonomy can obstruct globally nonzero flat sections; singular leaves and compact leaves impose further conditions. A metaplectic correction uses half-forms to improve the inner product and coordinate transformation law and account for familiar zero-point shifts, when its additional square-root data exist. Thus prequantum integrality, polarization, Hilbert-space completion and observable domains are separate steps, each with geometric constraints.

Applications of Stokes's theorem. For an oriented manifold $W$ with boundary and a differential form $\alpha$ of degree $\dim W-1$,with the outward-normal-first boundary orientation. Locally the identity is integration by parts in a half-space; a partition of unity reduces the general manifold to that local calculation, and contributions along internal patch boundaries cancel. The Generalized Stokes theorem unifies the fundamental theorem of calculus, circulation/flux formulas and the relation between local conservation laws and conserved integrated charges.

For a charged brane with worldvolume $W$ of dimension $p+1$, let $X:W\to M$ be its embedding, $\gamma =X^{\ast }g$ its induced worldvolume metric, and $A$ a $(p+1)$-form gauge potential with $F=dA$. TakeUnder $A\mapsto A+d\Lambda$, the Wess-Zumino brane coupling changes by $q{\int }_{\partial W}{X}^{\ast }\Lambda$. It is invariant for a closed worldvolume, or for suitable boundary conditions; open branes can require compensating boundary terms.

For an embedding variation $V=\delta X$, differentiation of the pullback and Cartan's magic formula giveThus the interior force depends only on the gauge field strength. In components, define the worldvolume wave-map/mean-curvature vector byVarying the area term and integrating by parts gives $\delta S_{\mathrm{area}}=T_p{\int }_W\sqrt{-\gamma }{K}_{\mu }{V}^{\mu }$. Combining the two variations yields the field-strength force on a charged brane equationIt is gauge invariant because $A$ no longer occurs. Its tangential projection vanishes identically by antisymmetry, consistently with worldvolume reparametrization invariance. At $p=0$, the chosen Lorentzian worldline convention has $K=-\ddot{X}$ in proper time and recovers the usual Lorentz-force sign.

For topological conservation, let the spatial manifold compactify to a closed oriented $n$-manifold $\Sigma$, and let a field $U:\Sigma \to N$ take values in a closed oriented $n$-dimensional target with normalized volume form $\Omega$, ${\int }_N\Omega =1$. ThenThis is the degree of a map between oriented manifolds: for a regular value it is the signed sum of inverse images. To relate the integral to that count, use a normalized target form supported near a regular value and change variables on its inverse-image neighborhoods. Any two normalized top forms on a connected closed oriented target differ by an exact form, whose pulled-back integral vanishes by the Generalized Stokes theorem. Hence the value is independent of this localization and equals the integer signed count.

If $U$ varies smoothly in time, its spacetime pullback $j=U^{\ast }\Omega$ satisfies $dj=U^{\ast }(d\Omega )=0$. Apply Stokes theorem to the spacetime slab between two slices: the difference of their charges is minus the flux through the spatial boundary. With no flux at infinity, $Q$ is conserved independently of the field equations. Equivalently a smooth homotopy cannot change the degree. Singularities, altered boundary conditions or nonzero boundary flux are the ways the argument can fail.

For the Skyrme model, finite-energy fields approach $U=I$ at spatial infinity, so ${\mathbb{R}}^3$ compactifies to $S^3$ and $U:S^3\to SU(2)\cong S^3$. With the conventional target orientation chooseThis is a normalized closed target volume form. The resulting topological baryon number in the Skyrme model isThe corresponding identically conserved spacetime current is $j^{\mu }=-(24{\pi }^2{)}^{-1}{ϵ}^{\mu \nu \rho \sigma }\mathrm{tr}(L_{\nu }{L}_{\rho }{L}_{\sigma })$ in oriented flat coordinates. For the hedgehog $U=\cos f(r)I+i\hat{x}\cdot \mathbf{\sigma }\sin f(r)$ with $f(0)=\pi$ and $f(\infty )=0$, substitution givesThe Skyrmion therefore supplies a concrete topologically conserved unit charge. Topological conservation alone does not guarantee a stable finite-size static solution: the positive Skyrme term, with four derivatives, opposes collapse of the two-derivative energy. Under size rescaling, these contributions behave as $E_2\propto \lambda$ and $E_4\propto {\lambda }^{-1}$, allowing a nonzero equilibrium size while preserving the degree.

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.

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.