The infinitesimal change $dy$ can be written
$$
dy^s=\mu_d{}^s(y)\theta^d,
\qquad
\theta^d=\lambda_s{}^d(y)dy^s.
$$
Right multiplication of $g(y)$ by $g(\theta)$ is also right multiplication of $g(z)=g(x)g(y)$ by the same element. Part v therefore gives
$$
dz^r=\mu_d{}^r(z)\lambda_s{}^d(y)dy^s,
$$
so
$$
\boxed{\frac{\partial z^r}{\partial y^s}
=\mu_d{}^r(z)\lambda_s{}^d(y)}.
$$


Solution

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

 Solution

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