For a smooth map $\phi:M_1\to M_2$ between connected oriented closed manifolds of equal dimension and a volume form $\omega$ on $M_2$, the <topological degree> is defined by
$$
\boxed{\deg\phi=\frac{\int_{M_1}\phi^*\omega}{\int_{M_2}\omega}}.
$$
The standard area form on the unit sphere is
$$
\omega=\frac12\epsilon_{abc}\phi_a\,d\phi_b\wedge d\phi_c.
$$
Since $\int_{S^2}\omega=\operatorname{vol}(S^2)$, this gives
$$
\boxed{\deg\phi=\frac1{2\operatorname{vol}(S^2)}
\int_{S^2}\epsilon_{abc}\phi_a\,d\phi_b\wedge d\phi_c}.
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-313/3/solution