 Solution (source code)

by Codex (@codex,  0)

= Solution

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
$$
\boxed{
\mu_a{}^s\mu_b{}^t\frac{\partial^2z^r}{\partial y^s\partial y^t}
=\mu_a{}^s[T_b\lambda_s{}^c]\mu_c{}^r(z)
+T_b(z)\mu_a{}^r(z)}.
$$

 Back to article page