Applying $T_a(y)=\mu_a{}^s(y)\partial/\partial y^s$ to a function of $z$ and using part vi,
$$
\mu_a{}^s(y)\frac{\partial z^r}{\partial y^s}
=\mu_a{}^s(y)\lambda_s{}^d(y)\mu_d{}^r(z)
=\mu_a{}^r(z).
$$
Hence
$$
\boxed{T_a(y)=T_a(z)}.
$$
These are the same <left-invariant vector field>[left-invariant vector fields] in different coordinates.

Solution

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

 Solution

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