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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 $4\pi$.

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.