= Solution
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 $\mathbf n(x_1,x_2)\in S^2$ has
$$
E[\mathbf n]=\frac12\int_{\mathbb R^2}\bigl(|\partial_1\mathbf n|^2+|\partial_2\mathbf n|^2\bigr)\,d^2x.
$$
The 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 integer
$$
N=\frac1{4\pi}\int_{\mathbb R^2}\mathbf n\cdot(\partial_1\mathbf n\times\partial_2\mathbf n)\,d^2x.
$$
This 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 gives
$$
E=\frac12\int|\partial_1\mathbf n\pm\mathbf n\times\partial_2\mathbf n|^2\,d^2x\pm4\pi N.
$$
Choosing the sign of $N$ proves the <Bogomolny degree bound for the O3 sigma model>
$$
\boxed{E\geq4\pi|N|.}
$$
For 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 $z=x_1+ix_2$ and
$$
\mathbf n=\frac{(2\operatorname{Re}w,\,2\operatorname{Im}w,\,1-|w|^2)}{1+|w|^2},\qquad \partial_z=\tfrac12(\partial_1-i\partial_2),\qquad\partial_{\bar z}=\tfrac12(\partial_1+i\partial_2).
$$
The round target metric is $4\,dw\,d\bar w/(1+|w|^2)^2$. In the one-half energy normalization used here, the <stereographic energy of the O3 sigma model> and degree are
$$
E=4\int\frac{|w_z|^2+|w_{\bar z}|^2}{(1+|w|^2)^2}\,d^2x,\qquad
N=\frac1\pi\int\frac{|w_z|^2-|w_{\bar z}|^2}{(1+|w|^2)^2}\,d^2x.
$$
Thus $E-4\pi N=8\int|w_{\bar z}|^2/(1+|w|^2)^2$. 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> is
$$
\boxed{N=\max(\deg p,\deg q)}
$$
after 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> is
$$
w(z)=\frac{\rho}{z-z_0},\qquad \rho\ne0,\qquad
\mathcal E(z)=\frac{4|\rho|^2}{(|z-z_0|^2+|\rho|^2)^2}.
$$
Its 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 $\rho\to0$, 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 \b[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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-56-kink-and-lump.png]
{title=Adjacent-vacuum phi-six kink profiles and the localized energy density of a degree-one sigma-model lump}
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 $4\pi$.
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