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Fix the value $w(\infty )=0$ for positive-degree sigma-model lumps. Their holomorphic rational maps can be written $p/q$, where $q$ is monic of degree $k$, $p$ has degree at most $k-1$, and the polynomials have no common zero. The coefficients of $p$ and the nonleading coefficients of $q$ give $2k$ complex parameters; coprimality is an open condition. The topological degree is $k$, obtained by counting generic preimages with multiplicity. Allowing a common factor lowers the degree after cancellation and can describe a degeneration rather than another smooth point of this degree-$k$ parameter space.

For $E=\frac{1}{2}\int (|{\partial }_1\mathbf{n}{|}^2+|{\partial }_2\mathbf{n}{|}^2)$, completing the square gives $E=\frac{1}{2}\int |{\partial }_1\mathbf{n}\pm \mathbf{n}\times {\partial }_2\mathbf{n}{|}^2\pm 4\pi Q$, hence $E\ge 4\pi |Q|$. Saturation yields the Bogomolny equations of the sigma-model lump. With the oriented stereographic projection $\mathbf{n}=(2\mathrm{Re}w,2\mathrm{Im}w,1-|w{|}^2)/(1+|w{|}^2)$, holomorphic rational maps have positive degree and saturate the bound; antiholomorphic maps have negative degree.

For arbitrary maps $\varphi :{\mathbb{R}}^2\to S^2$, integrability of the Dirichlet energy does not imply continuous extension to the one-point compactification. For large $r$, takeand interpolate smoothly to a constant on a bounded disk. Its exterior energy isThe field has no limit because $a(r)$ keeps winding. A continuous extension at the point at infinity instead requires $\varphi (x)\to {\varphi }_{\infty }$ uniformly outside sufficiently large disks. This fixed-limit boundary condition is standard for a sigma-model lump, but is extra information beyond finite energy for arbitrary fields.

Enforce the unit-vector constraint using a Lagrange multiplier $\Lambda (x)$:Integration by parts in a compactly supported variation gives $-\Delta {\varphi }^a+\Lambda {\varphi }^a=0$. Contracting with ${\varphi }^a$ gives $\Lambda ={\varphi }^b\Delta {\varphi }^b$, henceDifferentiating ${\varphi }^a{\varphi }^a=1$ twice also gives $\varphi \cdot \Delta \varphi =-|\nabla \varphi {|}^2$, so an equivalent equation is $\Delta \varphi +|\nabla \varphi {|}^2\varphi =0$. These are the harmonic map equations for the O3 nonlinear sigma model.

For arbitrary maps, the printed finite-energy compactification assertion is too strong. The following counterexample shows that finite sigma-model energy need not give a limit at infinity. Outside a large disk, setInterpolate smoothly to a constant inside the disk. There is no angular dependence, and the exterior energy isNevertheless the field keeps rotating and has no limit as $r\to \infty$. Thus energy integrability alone does not prove a continuous map on the one-point compactification.

The usual sigma-model lump sector adds the boundary condition $\varphi (x)\to {\varphi }_{\infty }\in S^2$ uniformly as $|x|\to \infty$. Defining the value at the added point to be ${\varphi }_{\infty }$ then gives a continuous map $S^2\to S^2$: neighborhoods of the point at infinity are precisely complements of compact subsets of the plane, and uniform convergence supplies continuity there. For the topological degree calculation using differential forms below, take a smooth compactified map, as in the regular lump sector. This boundary/regularity information is additional to finite energy for unrestricted fields.

Write $a_i={\partial }_i\varphi$ and $b_i={ϵ}_{ij}\varphi \times {\partial }_j\varphi$. Since $\varphi \cdot {\partial }_i\varphi =0$ and $|\varphi |=1$,Also the given double-index charge reduces toFor $s=\pm 1$, expanding the nonnegative square givesTherefore the Bogomolny degree bound for the O3 sigma model isThe charge integral is absolutely convergent already for finite-energy fields, because $|\varphi \cdot ({\partial }_1\varphi \times {\partial }_2\varphi )|\le \frac{1}{2}(|{\partial }_1\varphi {|}^2+|{\partial }_2\varphi {|}^2)$. The inequality itself consequently does not need the compactification hypothesis. Equality requires ${\partial }_i\varphi +s{ϵ}_{ij}\varphi \times {\partial }_j\varphi =0$ for $s$ chosen to have the sign of $Q$.

For the topological interpretation, the standard oriented area form on the target sphere isIts pullback is ${\varphi }^{\ast }\omega =\varphi \cdot ({\partial }_1\varphi \times {\partial }_2\varphi )d{x}^1\wedge d{x}^2$. The topological degree is defined by ${\varphi }_{\ast }[S^2]=\mathrm{deg}(\varphi )[S^2]$ in $H_2(S^2;\mathbb{Z})\cong \mathbb{Z}$. Pairing with $\omega$ givesEquivalently, the integer is the signed number of preimages of a regular value. It is invariant under smooth homotopy: if $\Phi :S^2\times [0,1]\to S^2$ is that homotopy, then Stokes theorem and $d\omega =0$ make the difference between the two charge integrals equal to ${\int }_{S^2\times [0,1]}d({\Phi }^{\ast }\omega )=0$. This identifies the charge as the degree charge of an O3 sigma-model lump, rather than merely an arbitrary energy integral.

A nonlinear sigma model describes a field taking values in a Riemannian target rather than an unconstrained linear space. Its static configurations are harmonic maps, critical points of the Dirichlet energy. The standard example for sigma-model lumps is the O3 nonlinear sigma model in two spatial dimensions: a unit vector $\mathbf{n}(x_1,x_2)\in S^2$ hasThe same target is the complex projective line, giving the CP1 nonlinear sigma model description. A lump is a localized finite-energy static solution, whose existence and minimum energy are governed by the degree of its field map.

Impose a fixed angle-independent value ${\mathbf{n}}_{\infty }$ at spatial infinity and require the field to extend smoothly there. The one-point compactification of the plane is a sphere, so the configuration becomes a map $S^2\to S^2$. Its topological degree is the integerThis is the degree charge of an O3 sigma-model lump: the integrand is the pullback of the oriented unit-sphere area form, normalized by its area $4\pi$. Equivalently, the degree counts inverse images of a regular target point with orientation signs. It is unchanged under smooth homotopy with the prescribed boundary value. Consequently fields of different degree belong to distinct topological sectors.

Topology also determines an energy lower bound. The derivatives are tangent to the sphere, so $|\mathbf{n}\times {\partial }_2\mathbf{n}|=|{\partial }_2\mathbf{n}|$. Since
${\partial }_1\mathbf{n}\cdot (\mathbf{n}\times {\partial }_2\mathbf{n})=-\mathbf{n}\cdot ({\partial }_1\mathbf{n}\times {\partial }_2\mathbf{n})$, square completion givesChoosing the sign of $N$ proves the Bogomolny degree bound for the O3 sigma modelFor positive degree, equality holds precisely when ${\partial }_1\mathbf{n}+\mathbf{n}\times {\partial }_2\mathbf{n}=0$; for negative degree the opposite sign applies. These are the Bogomolny equations for the lump. A saturating field minimizes energy in its degree sector and hence solves the static field equation. Explicitly, that equation is $\Delta \mathbf{n}+|\nabla \mathbf{n}{|}^2\mathbf{n}=0$, obtained by varying under ${\mathbf{n}}^2=1$.

The first-order equations become especially simple in stereographic projection. Use $z=x_1+i{x}_2$ andThe round target metric is $4dwd\bar{w}/(1+|w{|}^2{)}^2$. In the one-half energy normalization used here, the stereographic energy of the O3 sigma model and degree areThus $E-4\pi N=8\int |w_{\bar{z}}{|}^2/(1+|w{|}^2{)}^2$. Positive-degree lumps are holomorphic maps, $w_{\bar{z}}=0$, and negative-degree lumps are antiholomorphic maps, $w_z=0$.

A holomorphic sphere map is meromorphic in a stereographic coordinate and hence a rational map $w=p(z)/q(z)$ with coprime polynomials. To see why the rational form suffices, its finitely many coordinate poles can be removed by subtracting their principal parts, leaving a holomorphic function on the compact sphere and therefore a constant. The degree of a rational map of the Riemann sphere isafter canceling common factors. A generic target value $c$ has inverse images solving $p-cq=0$, with the root at infinity counted in homogeneous coordinates when necessary; the number is $N$, and holomorphic local degrees are positive. Poles of $w$ do not make the physical field singular: the reciprocal target coordinate $1/w$ is regular there. Antiholomorphic rational maps have the opposite topological degree and the same minimum energy.

For example, the unit sigma-model lump and size modulus isIts degree is one and radial integration gives $E=4\pi$. The complex center $z_0$ supplies two position parameters, while the magnitude and phase of $\rho$ give size and internal orientation. More generally, fixing $w(\infty )=0$ allows a monic degree-$N$ denominator and a numerator of degree at most $N-1$, with no common roots. Their $2N$ complex coefficients give the based rational-map moduli of sigma-model lumps.

Two-dimensional Derrick scaling leaves this pure-gradient energy unchanged, so degree fixes the minimum energy but not a preferred size. As $\rho \to 0$, the unit-lump energy concentrates and the field limit is singular at its center. A degree cannot change through a smooth boundary-preserving deformation, but concentration can leave that smooth configuration space. Thus topological degree protects the smooth sector and determines the Bogomolny energy, while not by itself stabilizing the lump size. There is no contradiction with the scalar obstruction in Question 1: a curved-target harmonic map obeys a nonlinear constrained field equation, rather than having harmonic unconstrained components.

The kink panel uses the normalization of Question 1. The lump panel has $z_0=0$ and $|\rho |=1$ in the unit sigma-model lump and size modulus; changing its size redistributes the density while keeping its total energy $4\pi$.

For the usual scalar field theory with a nonnegative potential $U$, shifted to zero in the vacuum, write its static energy in $d$ dimensions asApply Derrick scaling ${\varphi }_{\lambda }(x)=\varphi (\lambda x)$. The gradient acquires a factor $\lambda$ and the volume a factor ${\lambda }^{-d}$, givingA smooth finite-energy static solution is a stationary point of this energy, so its Derrick virial identity isFor $d\ge 3$, both terms are nonnegative, forcing $T=P=0$. The field is a constant vacuum: there is no nontrivial finite-energy static soliton in this scalar-gradient-plus-nonnegative-potential model. The assumptions matter: extra derivative terms, gauge fields, time dependence or a potential not bounded below by its vacuum value can change the argument.

For $d=2$, the relation becomes $P=0$ and says nothing further about $T$. A gradient-only energy is scale invariant, so Derrick theorem alone does not rule out nonconstant solutions; sigma-model lumps are examples. For a scalar field with isolated potential minima, $P=0$ constrains the continuous field to one vacuum and hence makes it constant. If the vacuum values form a nontrivial manifold, the scaling relation alone leaves open a nonconstant map into it. In neither case does the two-dimensional scaling argument itself prove existence.

Write $w=u+iv$ and $R=1+u^2+v^2$. Inverting the specified stereographic projection givesThe coordinate takes the value infinity at the south pole, so a general map is described using the extended complex plane and a second chart there. A pole of $w$ is not necessarily a singularity of $\varphi$.

Differentiate the inverse formulas:On squaring and summing, the terms containing $(dR{)}^2$ cancel, leaving $|d\varphi {|}^2=4(d{u}^2+d{v}^2)/R^2$. Thus the stereographic energy of the O3 sigma model isThe same derivatives give $\varphi \cdot ({\partial }_u\varphi \times {\partial }_v\varphi )=4/R^2$, with positive sign: at $u=v=0$ the ordered tangent vectors are $(2,0,0)$ and $(0,2,0)$ at the north pole. The chain rule therefore givesThis is the topological degree for the orientation in the question.

Put $z=x_1+i{x}_2$, $w_z=(w_{x_1}-i{w}_{x_2})/2$ and $w_{\bar{z}}=(w_{x_1}+i{w}_{x_2})/2$. Expanding the squares givesFor $N\ge 0$, subtracting the second expression times $8\pi$ givesfor $N\le 0$, adding $8\pi N$ gives the analogous square with $w_z$. ConsequentlyThe factor is $8\pi$, not $4\pi$, because the printed energy has no prefactor $1/2$. Equality requires a holomorphic map for positive degree, an antiholomorphic map for negative degree, or a constant map for zero energy.

Explicit examples for every integer degree areTo verify both degree and finite energy for $w=z^m$, $m>0$, use polar coordinates: $|w_z{|}^2=m^2{r}^{2m-2}$ and $w_{\bar{z}}=0$. Thenwhere $s=r^{2m}$. Its energy is $8\pi m$. Complex conjugation changes the degree's sign while preserving the energy, and the constant example has degree and energy zero. These sigma-model lumps are smooth sphere-valued maps; the coordinate infinity at spatial infinity is regular in the other target chart.

The degree of a map between oriented manifolds measures how many times the domain covers the target, with signs recording the local local orientation of a manifold. Let $M,N$ be connected, oriented closed manifolds of the same dimension $n>0$. A continuous map $f:M\to N$ acts on top-dimensional homology bywhere $[M],[N]$ are their fundamental classes. Connectedness and the choices of local orientation of a manifold identify $H_n(N;\mathbb{Z})$ with $\mathbb{Z}$. Reversing the orientation of either manifold changes the sign; reversing both does not.

For a smooth map, Sard theorem supplies a regular value $y$. Its inverse image is discrete and, by compactness, finite. At each $x\in f^{-1}(y)$ the differential is an isomorphism; let its sign be $+1$ or $-1$ according to whether it preserves or reverses the chosen local orientations. The degree as a sum of local degrees isThe sign is computed in oriented manifold charts. The value is independent of the chosen regular value, even when inverse images appear or disappear: the signed count is the coefficient of $[N]$ in $f_{\ast }[M]$.

There is a useful local-density expression for the same topological degree. If $\omega$ is a volume form with ${\int }_N\omega =1$, thenThis degree by integration of a pullback volume form follows first by choosing a smooth top-form supported in a small neighborhood of a regular value, where the inverse branches contribute their orientation signs. Any other normalized top-form differs from it by an exact form: integration identifies $H_{\mathrm{dR}}^n(N)$ with $\mathbb{R}$. The integral of its pullback difference vanishes by Stokes theorem. In particular, for any top-form $\eta$, ${\int }_M{f}^{\ast }\eta =\mathrm{deg}(f){\int }_N\eta$.

A homotopy $H:M\times [0,1]\to N$ preserves this integral, since $d\omega =0$ and Stokes theorem givesThus topological degree is a homotopy invariant. It is multiplicative under composition, because the induced maps on homology compose: $\mathrm{deg}(g\circ f)=\mathrm{deg}(g)\mathrm{deg}(f)$. The identity has degree one, a constant map has degree zero for $n>0$, and an orientation-reversing diffeomorphism has degree minus one. An orientation-preserving finite covering map has degree equal to its number of sheets. Nonzero topological degree forces surjectivity, since an omitted point would be a regular value with an empty inverse image.

For the circle, $e^{i\theta }\mapsto e^{ik\theta }$ has degree $k$, positive or negative. This is its winding number, computable as $(2\pi i{)}^{-1}\int f^{-1}df$. The antipodal map of $S^n$ has degree $(-1{)}^{n+1}$: its extension $-I$ on the ambient $(n+1)$-dimensional vector space has that determinant sign and respects the outward-normal convention. A holomorphic map $z\mapsto z^k$, $k\ge 1$, on the Riemann sphere has degree $k$, whereas its complex conjugate has degree $-k$. These examples show how orientation, rather than simply the number of inverse images, determines the integer.

For maps $S^n\to S^n$, topological degree gives the complete homotopy classification ${\pi }_n(S^n)\cong \mathbb{Z}$. The degree does not classify general manifold maps: the identity of the torus and the map induced by the integer matrix $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ both have degree one, but have different induced maps on $H_1(T^2;\mathbb{Z})$ and so are not homotopic. A nonzero-degree map $S^n\to S^n$ cannot extend continuously to $B^{n+1}$, because such an extension would make the boundary map null-homotopic. In the smooth setting, Stokes theorem gives the same obstruction by applying it to the pulled-back normalized volume form.

The hypotheses can be adjusted, but must be stated. For connected oriented noncompact manifolds, a proper map has a degree defined using compactly supported top-forms, and it is invariant under proper homotopies. For manifolds with boundary one uses relative fundamental classes and maps of pairs, or fixes appropriate boundary conditions. Without an integral orientation one can still count inverse images modulo two, obtaining a mod-two degree. The integer integral formula used below assumes the oriented setting.

In classical field theory, these ideas turn continuous fields into quantized topological charges. Suppose a field on ${\mathbb{R}}^d$ approaches one fixed target value at spatial infinity. The one-point compactification makes it a map $\varphi :S^d\to \mathcal{V}$. When the target $\mathcal{V}$ is an oriented closed $d$-manifold, its topological degree labels topological sectors. More generally the sectors are described by homotopy groups; an integer degree is available only when the domain and target have the appropriate dimensions and orientations. Smooth time evolution preserving the boundary condition is a homotopy, so it cannot change the integer. A change requires a singular field, escape from the allowed target, or a change at the boundary.

A normalized closed target $d$-form gives the pullback-volume representation of a topological current. On spacetime, put $\alpha ={\varphi }^{\ast }\omega$. Since $d\alpha ={\varphi }^{\ast }(d\omega )=0$, its dual current is identically conserved, andis independent of time when there is no flux at infinity. This conservation law follows from geometry without using the field equations; it need not arise from a continuous symmetry through Noether theorem.

A concrete example is the O3 nonlinear sigma model in two spatial dimensions. Its unit-vector field $\mathbf{n}$ approaches a constant at infinity, defining $S^2\to S^2$. The normalized area form of the target gives the degree charge of an O3 sigma-model lump:For the energy normalization $E=\frac{1}{2}\int (|{\partial }_1\mathbf{n}{|}^2+|{\partial }_2\mathbf{n}{|}^2)$, the identities $\mathbf{n}\cdot {\partial }_i\mathbf{n}=0$ giveThis is the Bogomolny degree bound for the O3 sigma model. Choosing the sign appropriate to $Q$ makes the square nonnegative; vanishing of the square gives first-order Bogomolny equations and a sigma-model lump saturating the bound. With the oriented stereographic projectionthe maps $w=z^k$ have $Q=k$ and $E=4\pi k$. Their conjugates have $Q=-k$ with the same energy. Holomorphic rational maps have positive degree equal to their degree as rational maps; taking a reciprocal does not reverse the orientation. Antiholomorphic dependence reverses it.

The Skyrme model supplies a three-dimensional example. A field $U:{\mathbb{R}}^3\to \mathrm{SU}(2)$ with $U\to I$ at infinity is a map $S^3\to \mathrm{SU}(2)\cong S^3$. Take $T_i=-i{\tau }_i$ and $U^{-1}dU={\theta }^i{T}_i$, with ${\theta }^1\wedge {\theta }^2\wedge {\theta }^3$ positive. Since $\mathrm{tr}(T_i{T}_j{T}_k)=-2{ϵ}_{ijk}$, the normalized target volume form isThe integral is one on the unit three-sphere. Consequently the Skyrme baryon number as a mapping degree isThis is the topological baryon number in the Skyrme model; the sign has been fixed by the stated orientation and anti-Hermitian generator convention.

A topological charge alone does not guarantee a stable finite-size solution. The degree and energetic stability of a field configuration concern different properties. For a three-dimensional configuration of size $R$, the two-derivative energy scales as $R$, so it can decrease by shrinking while the topological degree remains fixed for every $R>0$. The limit can be singular. The Skyrme term, with four derivatives, scales as $R^{-1}$ and can balance the shrinking tendency. This is the role of Derrick theorem in distinguishing topological obstruction from energetic stability.

For defects, the relevant boundary map can instead be the sphere surrounding a core. A vacuum manifold equal to $S^1$ gives the integer winding number of a vortex; a vacuum manifold $S^2$ gives the degree of a surrounding $S^2$ for a magnetic monopole. This vacuum-boundary degree as a defect charge obstructs extending the normalized vacuum field through the enclosed ball. A nonzero integer therefore forces the field to leave the vacuum manifold somewhere in the core. This construction does not require the field to take one constant value in every direction at infinity.

Degree also appears in four-dimensional gauge theory through a boundary transition function. For an anti-Hermitian SU(2) gauge connection on ${\mathbb{R}}^4$, write $F=dA+A\wedge A$ and assume finite-action boundary behavior $A\to g^{-1}dg$ on the large bounding three-sphere. In the second-Chern conventionthe identity $d\mathrm{tr}(A\wedge dA+\frac{2}{3}{A}^3)=\mathrm{tr}(F\wedge F)$ and the Maurer-Cartan equation giveThis boundary winding representation of Yang-Mills topological charge relates the Second Chern number to the degree of $g:S^3\to \mathrm{SU}(2)$. The Chern-Simons 3-form turns the bulk integral into the boundary winding integral. Conventions which define the instanton number with the opposite trace sign reverse $k$; the integer quantization is unchanged. A Yang-Mills theta term weights a sector by $e^{i\vartheta k}$, giving periodicity $\vartheta \mapsto \vartheta +2\pi$. Thus the same topological degree that counts oriented inverse images also labels field sectors and expresses their quantized charges as integrals of local densities.

There is an exact rational-map description of sigma-model lumps and a restricted variational rational-map description of Skyrmions. The first follows from a Bogomolny equation; the second separates angular and radial dependence in a field that generally does not saturate a Bogomolny bound.

For the exact example, take the two-dimensional O3 nonlinear sigma model with a unit field $\mathbf{n}$, energy $E=\frac{1}{2}\int |\nabla \mathbf{n}{|}^2{d}^{2}x$, and a fixed limit at infinity. Regular fields then have a one-point compactification to maps $S^2\to S^2$. Choose the orientation so that the stereographic fieldhas positive charge when it is holomorphic. Equivalently, this is the CP1 nonlinear sigma model since the target sphere is the complex projective line. Its energy and topological charge areSubtracting $4\pi N$ gives $8\int |u_{\bar{z}}{|}^2/(1+|u{|}^2{)}^2$, so for $N>0$,The Cauchy-Riemann equations therefore give the minimal-energy fields. A holomorphic map from the compactified domain Riemann sphere to the target Riemann sphere is a rational map $u=p(z)/q(z)$, with common polynomial factors cancelled. Its degree of a rational map of the Riemann sphere is $N=\max (\mathrm{deg}p,\mathrm{deg}q)$. Poles of $u$ are coordinate singularities, not singularities of $\mathbf{n}$. For instance, $u=(z-a)/\lambda$ with $\lambda \ne 0$ is an exact unit lump with energy $4\pi$, arbitrary centre $a$, and scale $|\lambda |$. Its density is $4|\lambda {|}^2/(|z-a{|}^2+|\lambda {|}^2{)}^2$, whose plane integral is $4\pi$. For negative charge use antiholomorphic maps. Scale invariance allows arbitrarily small lumps and does not by itself prevent a singular concentration limit in the time-dependent theory.

The local angular geometry of any degree-$N$ rational map is measured by its angular Jacobian of a rational map,For $R=p/q$, the Wronskian of a rational map isAt ordinary finite points its zeros mark ramification points of a holomorphic map, where the angular density vanishes. At a pole use $q/p$ as the target coordinate, and at infinity use $1/z$ as the domain coordinate. A pole of order $m$ is ramified by $m-1$. The Riemann-Hurwitz formula counts total ramification $2N-2$, including infinity, even if the affine Wronskian has smaller degree. For $R=z^N$, $W=N{z}^{N-1}$ accounts for $N-1$ zeros at zero; the reciprocal coordinate shows the other $N-1$ at infinity.

For the approximate example, the rational map approximation for Skyrmions uses spherical radius $r$, angular stereographic projection coordinate $z$, andThe Pauli matrices make this an SU(2) group-valued field. The boundary conditions give a continuous centre and the vacuum value at infinity; the profile must also make the energy finite. Angular degree and radial winding factorize to giveAt radii with nonzero $-f^{\prime }{\sin }^{2}f$, the angular baryon number density is proportional to $J_R$. Thus the Wronskian of a rational map locates zero-density directions and reveals the holes or face directions in shell-like Skyrmion configurations.

In dimensionless Skyrme model units, insert this ansatz into the quadratic and quartic derivative energies. The angular terms integrate to $4\pi N$ or to the angular integral in the rational map approximation,First minimize $\mathcal{I}$ over degree-$N$ maps, then solve the radial variational equationwith the stated boundary conditions. The Cauchy-Schwarz inequality gives $\mathcal{I}\ge N^2$ because $J_R$ has sphere average $N$. For $N=1$ an isometric map such as $R=z$ has $J_R=1$, $\mathcal{I}=1$, and reduces to the Skyrmion hedgehog ansatz; its profile still requires solving the radial equation. For general $N$, angular and radial separation restrict the allowed fields. The minimum within this class is an upper bound on the full sector minimum, not a claim that every exact Skyrmion has this separated form. The original construction is Houghton, Manton and Sutcliffe's arxiv.org/abs/hep-th/9705151.

A rotational symmetry of a rational map must satisfy $R(gz)=hR(z)$, where $g,h$ are the domain and target rotations written as Möbius transformations from SU(2) matrices. Since a target rotation is an isometry, this identity implies $J_R(gz)=J_R(z)$. For $R=z^N$, $R(e^{i\alpha }z)=e^{iN\alpha }R(z)$, giving axial spatial symmetry accompanied by an internal rotation. This explains the axial symmetry of the degree-two toroidal ansatz.

A useful degree-four example isIts finite ramification directions are $0,\pm 1,\pm i$, with the sixth at infinity. They are the six coordinate-axis directions under stereographic projection, so they are the face normals of a cube. The full map has octahedral rotational equivariance: for example $R(iz)=1/R(z)$, and the cyclic-axis generator $g(z)=(iz+1)/(1-iz)$ obeys $R(gz)=e^{2\pi i/3}R(z)$. These generate the cube's rotational group and give a cubic angular density. The associated profile then produces the familiar cubic charge-four approximation. Symmetry of the Wronskian is a useful necessary diagnostic but is not sufficient: replacing this map by $cR$, $c>0$, leaves the same ramification directions, while the identity under $z\mapsto iz$ becomes $cR(iz)=c^2/(cR(z))$, whose target transformation is a sphere rotation only when $c=1$. Equivariance of the entire rational map, rather than symmetry of its critical-point set alone, determines the physical rotational symmetry.

For the O3 nonlinear sigma model with energy $\int |\nabla \mathbf{n}{|}^2/2$, the rational map $w(z)=\rho /(z-z_0)$, $\rho \ne 0$, is a degree-one sigma-model lump. Its energy density is $4|\rho {|}^2/(|z-z_0{|}^2+|\rho {|}^2{)}^2$, whose integral is $4\pi$. The complex center $z_0$ gives its position, while $|\rho |$ gives its size and $\arg \rho$ its internal phase. Varying the size preserves energy. The limit $\rho \to 0$ is singular at the center, so conserved degree under smooth deformations does not prevent concentration.