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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.