Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 49 / 3 / Solution

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.