Choose the positive local branch of the constraint,
$$
\sigma=\sqrt{1-\boldsymbol\pi^2}.
$$
The <chain rule> gives
$$
\partial_i\sigma=-\frac{\pi_a\partial_i\pi_a}{\sqrt{1-\boldsymbol\pi^2}},
$$
and therefore
$$
(\partial_i\mathbf n)^2=(\partial_i\pi_a)(\partial_i\pi_a)
+\frac{\pi_a(\partial_i\pi_a)\pi_b(\partial_i\pi_b)}{1-\boldsymbol\pi^2}.
$$
Substitution into the original free energy gives exactly
$$
\boxed{F[\boldsymbol\pi]=\int d^dx\,\frac1{2g}\left[(\partial_i\pi_a)(\partial_i\pi_a)+\frac{\pi_a(\partial_i\pi_a)\pi_b(\partial_i\pi_b)}{1-\boldsymbol\pi^2}\right]}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 303 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-303/3/b/solution