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For a CW complex $X$, let $X^k$ be its $k$-skeleton. The cellular chain complex isThe isomorphism uses excision and $H_k(D^k,S^{k-1})\cong \mathbb{Z}$, after choosing an orientation for each cell. In degree zero, take $X^{-1}=\varnothing$ and the free group on the vertices. The differential is the compositeThe long exact sequences of skeleton pairs imply $d_{k-1}{d}_k=0$. The cellular boundary formula describes its coefficient at a $(k-1)$-cell as the mapping degree of the attaching sphere after all other $(k-1)$-cells are collapsed. In degree one this is the signed difference of the two endpoints. The cellular homology theorem identifies the homology of this complex with singular homology: relative homology of successive skeletons is concentrated in their cell dimension, and the exact skeleton sequences leave precisely these kernels modulo images. This is the construction, rather than merely a count of cells.

The filtration ${\mathbb{CP}}^0\subset {\mathbb{CP}}^1\subset {\mathbb{CP}}^2$ gives one cell in each of dimensions zero, two and four, since ${\mathbb{CP}}^k\setminus {\mathbb{CP}}^{k-1}\cong {\mathbb{C}}^k$. There are no odd-dimensional cells, so every cellular differential vanishes. ThusThe complex projective line has one zero-cell and one two-cell. Its product with itself has one zero-cell, two two-cells and one four-cell. Its cellular chain complex again has zero differentials, givingA homeomorphism induces isomorphisms on homology groups. The second groups have different ranks, and therefore

Now orient the Complex projective plane by its complex coordinates. The inclusion of its two-skeleton ${\mathbb{CP}}^1$ identifies the fundamental class of a projective line with a generator $h$ of $H_2({\mathbb{CP}}^2;\mathbb{Z})\cong \mathbb{Z}$: the cellular generator has no incoming three-cell boundary and no outgoing boundary.

Take the projective lines $L_0=\{z_2=0\}$ and $L_1=\{z_1=0\}$. A path of unitary coordinate transformations takes one to the other, so their oriented homology classes are both $h$. They meet in the single point $[1:0:0]$. In its affine chart, let $w=z_1/z_0$ and $v=z_2/z_0$. Their tangent spaces are the complex $w$-axis and complex $v$-axis, so the intersection is a transverse intersection. The ordered real basesconcatenate to the complex orientation of the ambient four-manifold. Thus the local sign is $+1$, andThis computes the intersection form using distinct representatives of the same class, avoiding an ill-defined attempt to count a line intersecting itself as a set. Bilinearity now givesEquivalently, the map $H_2\to \mathrm{Hom}(H_2,\mathbb{Z})$ sending $a$ to $\lambda (a,-)$ is an isomorphism. This is exactly a unimodular intersection pairing, and proves projective lines generate a unimodular intersection form. Reversing the ambient orientation would change the matrix to $(-1)$ and would still be unimodular.

A symplectic manifold is a smooth manifold $M$ equipped with a differential two-form $\omega$ such thatThe first condition says it is a closed differential form; the second says it is pointwise nondegenerate. A nondegenerate skew-symmetric matrix has even size, so the dimension is $2n$. Equivalently, ${\omega }^n$ is a nowhere-zero top-degree differential form. It supplies an orientation and the volume form ${\omega }^n/n!$. These are the defining conditions on the symplectic form.

There are no symplectic structures on the Möbius strip. A symplectic form on a surface would be a nowhere-vanishing two-form, hence would provide an orientation. The Möbius strip is nonorientable: transport around its core reverses a transverse direction and therefore reverses any local orientation. This is incompatible with such a two-form. The argument applies whether the boundary is included or only the interior is considered.

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.

Write $w=x^1+i{x}^2$, $z=x^3+i{x}^4$. The Euclidean metric is ${\sum }_a(d{x}^a{)}^2$, and the printed four-form is $4d{x}^1\wedge d{x}^2\wedge d{x}^3\wedge d{x}^4$, so it specifies the usual positive orientation. The normalized metric volume form is one quarter of that expression. The self-dual frame in complex Euclidean coordinates isThese are real and have the required complex combinations. For the Hodge star operator with this orientation,and applying $\ast$ again gives the reverse relations. Thus $\ast {\omega }_i={\omega }_i$. They are linearly independent, while the $+1$ eigenspace of $\ast$ on 2-forms has dimension three, proving they span ${\Lambda }_{+}^2$. The TeX aid incorrectly reads the subscript as $1$.

The ASDYM equations require the self-dual projection of the gauge field strength to vanish. Orthogonality to ${\omega }_1,{\omega }_2$ sets its $(2,0)$ and $(0,2)$ parts to zero; orthogonality to ${\omega }_3$ removes the trace of its $(1,1)$ part. Explicitly, if $F={\sum }_{a<b}{f}_{ab}d{x}^a\wedge d{x}^b$, these conditions are $f_{13}-f_{24}=0$, $f_{14}+f_{23}=0$ and $f_{12}+f_{34}=0$. Sinceand the conjugate equation supplies the other complex component for a real curvature form of a connection, the equivalent system is

To obtain the complex potential reduction of anti-self-dual Yang-Mills, set $D_w={\partial }_w+A_w$, $D_z={\partial }_z+A_z$. The first equation is the integrability condition $[D_w,D_z]=0$. Locally it allows an invertible complex matrix $s$ satisfying ${\partial }_{w}s=-A_{w}s$ and ${\partial }_{z}s=-A_{z}s$. The Yang-Mills gauge transformation $A\mapsto s^{-1}As+s^{-1}ds$ consequently gives $A_w=A_z=0$. This is a complex gauge; a real compact gauge group alone generally cannot implement it.

In this gauge the second equation becomesThus the one-form $A_{\bar{w}}dz-A_{\bar{z}}dw$ is closed with respect to the exterior derivative in the $(w,z)$ directions. The local complex version of the Poincare lemma gives a potential $K$ such thatBecause the gauge transformation is complex, $K$ is generally valued in the complexification of a Lie algebra ${\mathfrak{g}}_{\mathbb{C}}$; the printed $\mathfrak{g}$ must be understood in that sense. The elementary reduction is local, and the transformed fields retain a reality condition inherited from the original real connection. For the usual compact matrix gauge groups and smooth fields on all of ${\mathbb{R}}^4={\mathbb{C}}^2$, the gauge and potential can also be chosen globally if no condition at infinity is imposed. The flat partial connection defines a holomorphic principal bundle on the conjugate complex space. That base is a contractible Stein manifold, so the Oka-Grauert principle gives a global trivialization and hence a global complex gauge. After that trivialization, the conjugate of Stein vanishing for the Dolbeault cohomology of functions gives a global primitive $K$, component by component in ${\mathfrak{g}}_{\mathbb{C}}$. Prescribed framing or decay at infinity requires a separate compatibility check and is not automatically preserved by this gauge.

Substitute the potential into the remaining ASDYM equations component:Therefore all three ASDYM equations reduce to the single ASDYM potential equationThe sign follows directly from $A_{\bar{z}}=-K_w$; changing a potential convention would change the displayed commutator sign. Conversely, this equation and the displayed gauge reconstruction make all three curvature conditions vanish, subject to the inherited reality condition when a real gauge field is required.

A tangent vector at $p$ is a derivation at a point: a real-linear map $v:C_p^{\infty }(M)\to \mathbb{R}$ on germs of smooth functions satisfyingAddition and scalar multiplication preserve this identity, defining the tangent space $T_{p}M$ as a vector space. A derivation annihilates constants. In a manifold chart $x=(x^1,\dots ,x^n)$ centered near $p$, the local factorization $f-f(p)={\sum }_i(x^i-x^i(p))f_i$, with $f_i(p)={\partial }_{i}f(p)$, gives $v(f)={\sum }_{i}v(x^i){\partial }_{i}f(p)$. The factorization follows by integrating the first derivatives along a straight segment in a sufficiently small coordinate ball. ThereforeConversely these coordinate derivatives satisfy the derivation identity, so they really give a basis rather than merely a spanning family.

The cotangent space is the dual vector space $T_p^{\ast }M=(T_{p}M{)}^{\ast }$. Its basis $d{x}^i{|}_p$ is characterized by $d{x}^i({\partial }_j)={\delta }_j^i$. If $\alpha =a_{i}d{x}^i=b_{j}d{y}^j$, the chain rule gives the cotangent coordinate transitionThe cotangent bundle is the disjoint union $T^{\ast }M={\coprod }_{p\in M}{T}_p^{\ast }M$, with projection $\pi (p,\alpha )=p$. For each base manifold chart $(U,x)$, defineIts inverse sends $(z,a)$ to $(x^{-1}(z),a_{i}d{x}^i{|}_{x^{-1}(z)})$. On overlaps the transition is $(x,a)\mapsto (y(x),b)$, with the displayed fibre-linear transformation. Its coefficients and inverse are smooth, so these maps give a smooth atlas in dimension $2n$. Give the total space the topology obtained by transporting the product topology through these charts. Different base points are separated by disjoint base neighborhoods; distinct points over the same base point are separated within one bundle chart. A countable base smooth atlas and countable product bases give second countability. Thus the total space is a Hausdorff, second-countable smooth manifold. The same charts are vector bundle trivializations, since $\pi$ is product projection and their fibre changes are invertible linear maps.

The intrinsic canonical one-form on a cotangent bundle isIn the bundle chart it is $\lambda ={\sum }_i{a}_{i}d{x}^i$. This definition is coordinate independent, hence its exterior derivative is the globally defined smooth differential two-formThis sign follows the fibre-first order in this problem; the equally common position-first symplectic form is $-d\lambda$. The volume formnever vanishes. A smooth manifold is orientable exactly when it admits a nowhere-zero top-degree form; its positive ordered bases determine a consistent orientation. Consequently $T^{\ast }M$ is an orientable smooth manifold, even when $M$ is not. This is cotangent bundle orientation. In dimension zero the same conclusion uses the nowhere-zero zero-form $1$.

In a manifold chart, write a smooth differential form as $\eta ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$, with $I$ increasing. Define its exterior derivative byTo check that this definition is global, change coordinates to $y$. The chain rule expresses $d{x}^i$ as ${\sum }_j({\partial }_j{x}^i)d{y}^j$. Differentiating it gives zero because the second partial derivatives of $x^i$ are symmetric and $d{y}^k\wedge d{y}^j$ is antisymmetric. The coordinate definition satisfies the graded Leibniz rule, so differentiating the transformed expression for $\eta$ leaves exactly ${\sum }_{I}d{a}_I\wedge d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. This agrees with the original formula and proves independence of the manifold chart. The same cancellation of mixed derivatives gives $d^2=0$.

The de Rham cohomology is thereforethe vector space of closed differential forms modulo exact differential forms.

Here is a Green-theorem proof of exactness on a disc. Identify the disc with a round disc centred at zero, and let $\eta =Pdx+Qdy$ be a closed differential form. Set $f(x)={\int }_{[0,x]}\eta$. For $x$ and $x+h$ in the disc, Green theorem on the oriented triangle with vertices $0,x,x+h$ gives zero boundary integral, since ${\partial }_{x}Q-{\partial }_{y}P=0$. HenceTaking $h=\epsilon v$ and dividing by $\epsilon$ yields $d{f}_x(v)={\eta }_x(v)$. The radial integral defining $f$ is smooth, so $\eta =df$ andDegenerate triangles follow by continuity. The same proof works on any star-shaped plane domain, including the whole plane.

Integration requires an orientation: take $M$ to be an oriented compact $n$-dimensional smooth manifold without boundary. The Generalized Stokes theorem says ${\int }_{M}d\beta =0$ for every smooth $(n-1)$-form $\beta$. Every $n$-form is closed by dimension, and changing it by an exact differential form leaves its integral unchanged. Thusis a well-defined linear map $H_{\mathrm{dR}}^n(M;\mathbb{R})\to \mathbb{R}$. Compactness alone does not supply the orientation omitted from the printed integration statement.

For an $n$-dimensional Lie group, choose a nonzero $\lambda \in {\Lambda }^n{T}_e^{\ast }G$ and define the invariant volume form on a Lie group byEach $d{L}_{g^{-1}}$ is an isomorphism, so $\omega$ is nowhere zero and defines an orientation. The composition identity $L_{(hg{)}^{-1}}\circ L_h=L_{g^{-1}}$ gives $L_h^{\ast }\omega =\omega$. Moreover, every left-invariant differential form of degree $n$ is determined by its value at $e$, so the space of such forms is one-dimensional.

Under the supplied invariant-representative assumption, all top-degree de Rham cohomology classes are multiples of $[\omega ]$. For compact $G$, orient it so $\omega$ is positive; then ${\int }_G\omega >0$, whereas the Generalized Stokes theorem makes every exact top form have integral zero. Therefore $[\omega ]\ne 0$ andunder that assumption. For actual compact Lie groups, this argument applies when $G$ is connected. The supplied assumption fails for disconnected groups: on $S^1\times \mathbb{Z}/2$, the two components have independent degree-one classes, while globally left-invariant top forms form only a one-dimensional space. In general the top-degree de Rham cohomology of a compact Lie group is ${\mathbb{R}}^c$, where $c$ is its number of connected components.

For $S^2$, use $U=S^2\setminus \{N\}$ and $V=S^2\setminus \{S\}$. Stereographic projection identifies each with the plane. The disc argument gives $\eta {|}_U=d{f}_U$ and $\eta {|}_V=d{f}_V$ for any closed differential one-form $\eta$. The overlap $U\cap V$ is connected, so $d(f_U-f_V)=0$ makes the difference a constant. Adjust one potential by that constant; they glue to a global smooth function $f$ with $df=\eta$. ThusFinally, $S^3$ is a compact connected Lie group: the identificationidentifies it smoothly with the special unitary group $SU(2)$. Matrix multiplication and inversion supply its smooth group operations. The preceding invariant-volume argument therefore gives

An index-$h$ handle is $D^h\times D^{n-h}$, attached along $S^{h-1}\times D^{n-h}$. Its attaching sphere is $S^{h-1}\times \{0\}$ with its attaching framing of an embedded sphere, and its belt sphere in the new boundary is $\{0\}\times S^{n-h-1}$. A handle decomposition is an ordered sequence of such handle attachments.

A handle slide of one $h$-handle over another replaces its framed attaching sphere by a band sum with a parallel framed copy of the second attaching sphere, along a band in the boundary of the previously attached handles. The second handle stays fixed. The band transports the framing of an embedded sphere. The union of the two handles and their attaching collars can be reidentified by a diffeomorphism, so this changes the decomposition, rather than the underlying manifold. A sequence of these moves, allowing the usual isotopies of attaching data, gives handle decompositions related by slides. Reading the decomposition backwards interchanges attaching spheres and belt spheres; a slide of dual handles is a slide of the corresponding original handles.

For the standard handle-calculus assertion, take $1\le k\le n-2$, so both the relevant attaching and belt spheres are connected. We need the local geometric move behind handle-slide isolation of a cancelling pair. Write $p=A_1\cap B_1$. If $q$ is an intersection of some $A_j$, $j>1$, with $B_1$, choose an arc on $B_1$ from $q$ to $p$, and take a narrow parallel band to the unique sheet of $A_1$ at $p$. Slide $A_j$ over an appropriate parallel copy of $A_1$. Choose the copy's orientation so that the new sheet at $p$ has the opposite local smooth intersection number to the sheet at $q$.

Within a thin neighbourhood of the arc, the old sheet at $q$ and the added sheet at $p$ are the two ends of one band. An isotopy pushes that band through the neighbourhood of $B_1$ and removes both ends. Outside this neighbourhood the original $A_j$ is unchanged, apart from the added copy of the parts of $A_1$ away from $B_1$. Since $A_1$ meets $B_1$ only at $p$, this copy contributes no further intersections with $B_1$. Thus one slide followed by an isotopy removes one original geometric intersection with $B_1$. This is a local band move, not an appeal to handle cancellation or to an algebraic cancellation of signs.

The arcs and bands may be chosen one at a time; in the one-dimensional sphere case, start with an intersection adjacent to $p$, so that the arc interior contains no other intersection. Perturb to make all intersections transverse. The relevant spheres are compact, so there are only finitely many intersections. Repeating the move for every $j>1$ leaves $A_1$ and $B_1$ fixed and givesThe bands can avoid the remaining attaching data; any consequent change to intersections with other belt spheres is allowed at this stage.

For the second stage, view the same interface from the dual decomposition. Now $B_i^{\prime }$ are attaching spheres and $A_j^{\prime }$ are belt spheres. Apply the same local move to each intersection of $B_i^{\prime }$, $i>1$, with $A_1^{\prime }$, sliding that dual handle over the one with attaching sphere $B_1^{\prime }$. This changes the original $k$-handles. It keeps $A_1^{\prime }$ and $B_1^{\prime }$ fixed, while removing the unwanted intersections with $A_1^{\prime }$. The parallel copies and bands may be taken disjoint from all $A_j^{\prime }$, $j>1$, because those spheres are already disjoint from $B_1^{\prime }$ and from $A_1^{\prime }$. Hence their intersections with $B_1^{\prime }$ remain empty. We obtainNo handle has been removed.

There is an implicit index-range qualification in this geometric assertion. If arbitrary zero-handles are included, the second conclusion need not hold: two zero-handles joined by one one-handle have an attaching zero-sphere with one endpoint on each zero-handle belt sphere. Slides preserving this handle inventory cannot make its second endpoint avoid both belts. The dual obstruction occurs for top-index handles. Thus the argument uses the customary intermediate-index setting of the handle-slide lemma; the unqualified assertion at the extreme indices would require additional hypotheses or additional moves.

For the numerical formula, fix coherent orientations for the integer smooth intersection numbers, and letKeep the orientation at the surviving point, so its number remains $\delta$. A slide of a $(k+1)$-handle adds or subtracts the first row of the intersection matrix. Removing all geometric intersections in the first column has net row operationEven if opposite-sign intersections require extra slides, their net signed coefficient is the displayed one. In particular,The dual slides then perform the column operationsSince $C_1^{\prime }=(\delta ,0,\dots ,0{)}^T$, these operations leave the lower-right block unchanged. Thus the final answer, including both stages of slides, isThe remaining block is the Schur complement of the unit pivot $\delta$. If the original orientation is chosen so that $\delta =1$, it is simply $m_{lk}-m_{l1}{m}_{1k}$. For unoriented handles use intersection numbers modulo $2$, with the same formula interpreted in ${\mathbb{F}}_2$.

Orient both spheres in the standard way and normalize their area forms to total area $4\pi$. The degree of a map between oriented manifolds can be obtained in two ways. For a regular value $y$, use the degree as a sum of local degrees:Each inverse image is isolated by the inverse function theorem, and compactness makes the set finite. The determinant is computed in consistently oriented local coordinates. A second method is spherical degree by area pullback:where the last formula represents $g$ as a unit vector in ${\mathbb{R}}^3$. The pullback of a differential form already contains the signed Jacobian determinant; no extra $\sin \theta$ is to be inserted in that last coordinate expression.

To relate the methods, replace $\omega$ by a smooth top-degree differential form with the same total integral supported in a small neighbourhood of a regular value. Two such top-degree forms with equal integral differ by an exact differential form on $S^2$, by its top-degree de Rham cohomology. Their pullbacks therefore have the same integral by Stokes theorem. Over the chosen neighbourhood, $g$ splits into local inverse branches; the change of variables formula makes the contribution of each branch its orientation sign times $4\pi$. Their sum is precisely the first formula. Thus the area integral is an integer and agrees with the signed inverse-image count.

For a nonconstant rational map, first use common-factor reduction of a rational map so $p$ and $q$ are coprime. Write $k=\max (\mathrm{deg}p,\mathrm{deg}q)$ for these reduced polynomials. A generic finite target value $w$ has inverse images at the roots of $p-wq$: avoiding exceptional values makes its degree $k$ and its roots simple. The fundamental theorem of algebra supplies $k$ roots. A holomorphic map has positive real Jacobian determinant $|R^{\prime }{|}^2$ at a regular point, so every local sign is $+1$. HenceThe source leaves coprimality implicit. In an unreduced representation the answer is $\max (\mathrm{deg}p,\mathrm{deg}q)-\mathrm{deg}\gcd (p,q)$, including degree zero for a constant reduced map. For example $(z^2-1)/(z-1)$ extends to $z+1$ and has degree one, although the unreduced maximum degree is two. Exceptional inverse images at infinity or multiple roots do not change the degree of a rational map of the Riemann sphere.

For the rational map approximation for Skyrmions, use stereographic projection $z=\tan (\theta /2)e^{i\phi }$ and the unit target vectorCombine this rational map with a radial profile to form a special unitary group field:where $\mathbf{\sigma }$ are the Pauli matrices. The endpoint values make $U(0)=-1$ independent of angle and $U(\infty )=1$. Appropriate radial behaviour gives an admissible finite-energy field configuration. With $L_i=U^{†}{\partial }_{i}U$, choose the topological baryon number in the Skyrme model conventionSeparating the radial and angular factors givesThus the degree of a rational map of the Riemann sphere supplies the Skyrmion charge.

In conventional dimensionless massless Skyrme model units, its static energy reduces towith the angular Jacobian of a rational mapThe Cauchy-Schwarz inequality gives $\mathcal{I}\ge k^2$. These formulas follow from the radial strain $|f^{\prime }|$ and the two equal angular strains $\sin f\sqrt{J_R}/r$: the quadratic energy sums their squares and the quartic Skyrme term sums their pairwise products of squares. Minimize the angular integral in the rational map approximation over degree-$k$ maps, then minimize the remaining radial energy with the stated endpoints. This replaces a three-dimensional field minimization by finitely many map coefficients and an ordinary differential equation for $f$.

The method constructs a charge-$k$ variational approximation, with topology built in and with rotational symmetry of a rational map translated into combined spatial and isospin rotations. It is efficient for identifying shapes and providing initial data for unrestricted numerical relaxation. Its restrictions are equally concrete: it uses one radial profile and a holomorphic angular map independent of radius, so it cannot represent arbitrary radial-angular correlations, separated clusters, or all deformations. Apart from the degree-one Skyrmion hedgehog ansatz, it generally does not solve the full field equation exactly. Massive-pion terms can be included in the radial functional but do not remove these restrictions, and multi-shell or unrestricted fields may be needed for larger charges. Approximate energy minima and a final collective-coordinate quantization are distinct steps.

For the sphere homology calculation use singular homology, homotopy invariance of homology, and the reduced Mayer–Vietoris sequence, not cellular homology. For a point, its chain group for singular homology is $\mathbb{Z}$ in every nonnegative degree, and its boundary operator is multiplication by ${\sum }_{i=0}^q(-1{)}^i$, hence alternately zero and one in positive degrees. Its homology is therefore $\mathbb{Z}$ in degree zero and zero above. Singular chains of a disjoint union split into the summand chain complexes. Start with $S^0$, the disjoint union of two points: $H_0(S^0;\mathbb{Z})={\mathbb{Z}}^2$, all positive groups vanish, and ${\tilde{H}}_0(S^0;\mathbb{Z})=\mathbb{Z}$. For $n\ge 1$, remove the north and south poles to obtain an open cover $U,V$ of $S^n$. Each member is contractible, while $U\cap V$ has a deformation retraction to $S^{n-1}$, including the disconnected intersection when $n=1$. The reduced exact sequence therefore suppliesAlso $S^n$ is connected, so its reduced degree-zero group vanishes. Induction givesThe separate $S^0$ calculation avoids conflating its two degree-zero generators.

Choose an orientation of $S^n$, with $n\ge 1$, and let $[S^n]$ be its fundamental class. The mapping degree is the integer determined by $f_{\ast }[S^n]=\mathrm{deg}(f)[S^n]$. To define the local degree at $x_0$, set $y_0=f(x_0)$ and require $x_0$ to be isolated in $f^{-1}(y_0)$. Choose an open neighborhood $B$ containing no other point of that fiber. Then $f$ defines a map of pairsThe degree-$n$ relative homology groups on both sides are $\mathbb{Z}$, by a manifold chart and excision. The local orientation of a manifold chooses their generators. The induced map multiplies these generators by an integer, denoted ${\mathrm{deg}}_{x_0}f$. Excision and naturality show independence of the choice of $B$. Differentiability is unnecessary; if $f$ is smooth and $D{f}_{x_0}$ is invertible, the inverse function theorem gives ${\mathrm{deg}}_{x_0}f=\mathrm{sign}\det D{f}_{x_0}$ relative to the orientations. Without isolation in the fiber this pointwise definition need not apply.

For the degree as a sum of local degrees, assume $y\in S^n$ has finite fiber $F=\{x_1,\dots ,x_s\}$. Disjoint coordinate neighborhoods and excision identifyThe absolute fundamental class maps to the tuple of local orientation generators. Under the relative map induced by $f$, the $j$th generator goes to ${\mathrm{deg}}_{x_j}f$ times the generator at $y$, and the resulting map from the direct sum adds these contributions. On the other hand, naturality says that first applying the absolute map $f_{\ast }$ and then passing to the local group at $y$ gives $\mathrm{deg}(f)$ times that generator. ThusIf the fiber is empty the sum is zero: the map factors through the contractible punctured sphere and has degree zero. If every point of a fiber is isolated, compactness makes that closed fiber finite, so the theorem applies. In particular, a regular value of a smooth map gives the signed count of its inverse images. For $n=0$, define a mapping degree on the reduced generator $[p_{+}]-[p_{-}]$: the identity has mapping degree one, the swap of the two points minus one, and either constant map zero. With local orientation signs positive at $p_{+}$ and negative at $p_{-}$, the same sum formula holds. The connected-sphere convention above is the standard positive-dimensional one.

The real tautological line bundle has total space $E=\{(\ell ,v):\ell \in {\mathbb{RP}}^n,v\in \ell \}$ and projection $(\ell ,v)\mapsto \ell$. On the open chart $U_i=\{[x]:x_i\ne 0\}$, the vector $s_i([x])=x/x_i$ is a continuous nowhere-zero section of a vector bundle, independent of the representative of $[x]$. The explicit local trivialization iswith inverse $([x],v)\mapsto ([x],v_i)$. Both maps are continuous and linear on each fiber. The charts cover the base, proving the existence of the required local trivializations.

Let $E\to B$ be a rank-$r$ vector bundle oriented over a commutative ring $R$, with $r\ge 1$. An $R$-orientation is a coherent choice of generator of each fiber's top relative cohomology, or equivalently a Thom class $u_E$. The Thom isomorphism theorem identifies $H^{q-r}(B;R)$ with $H^q(D(E),S(E);R)$ by $a\mapsto {\pi }^{\ast }a⌣u_E$. The disk bundle retracts to $B$, and the relative-to-absolute map becomes multiplication by the Euler class $e(E)$. The pair's long exact sequence therefore becomes the Gysin sequence of a sphere bundle:With real coefficients, the bundle must be oriented in the usual sense. For the computation here use $R={\mathbb{F}}_2$, over which every real vector bundle is oriented, including the nonorientable real tautological line bundle when $n\ge 1$.

The unit sphere bundle of ${\gamma }_n$ is $S^n$: a unit vector determines its line. Put $a=e_2({\gamma }_n)=w_1({\gamma }_n)\in H^1({\mathbb{RP}}^n;{\mathbb{F}}_2)$. For $n\ge 1$, the pullback $H^0({\mathbb{RP}}^n;{\mathbb{F}}_2)\to H^0(S^n;{\mathbb{F}}_2)$ is an isomorphism, so the next connecting homomorphism is zero and multiplication by $a$ is injective on $H^0$. Since the sphere's cohomology vanishes strictly between degrees zero and $n$, exactness makes multiplication by $a$ an isomorphism $H^{q-1}\to H^q$ for $1\le q<n$, and injective for $q=n$. For $q=1$ these assertions also use the just-noted vanishing of the connecting homomorphism from $H^0(S^n)$.