Differentiate the identity in part vi with respect to $y^t$, multiply by $\mu_a{}^s(y)\mu_b{}^t(y)$, and use $\lambda\mu=I$. The product rule gives
$$
\boxed{
\mu_a{}^s\mu_b{}^t\frac{\partial^2z^r}{\partial y^s\partial y^t}
=\mu_a{}^s[T_b\lambda_s{}^c]\mu_c{}^r(z)
+T_b(z)\mu_a{}^r(z)}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 302 / 2 / viii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-302/2/viii/solution