Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 313 / 3

For a smooth map $\varphi :M_1\to M_2$ between connected oriented closed manifolds of equal dimension and a volume form $\omega$ on $M_2$, the topological degree is defined byThe standard area form on the unit sphere isSince ${\int }_{S^2}\omega =\mathrm{vol}(S^2)$, this gives

For a generic target value $y\in {\mathbb{CP}}^1$, its finite preimages solveWhen $m\ne n$ this has $\max (m,n)$ roots after the missing roots or poles at infinity are counted; when $m=n$, a generic $y$ again gives degree $m$. Holomorphic maps preserve orientation at regular preimages, so every local sign is positive. Hence

For $f(z)=z^m$,The pullback area is thereforeThe same form integrates to $2\pi$ on the target sphere, so

Using $z=x+iy$, the energy isBecause $|\varphi |=1$, both derivatives are tangent to $S^2$ and $|\varphi \times {\partial }_y\varphi |=|{\partial }_y\varphi |$. Completing the square givesThe final integral is $4\pi |\mathrm{deg}\varphi |$ by the topological degree formula. ThusEquality holds exactly when the appropriate first-order Bogomolny equation is satisfied:with the sign chosen to match the degree.