Pull Back Manifolds: Checking Exercise on p67/p68 & p69

[latentcorpse]

In the notes attached in this thread:
https://www.physicsforums.com/showthread.php?t=457123

How do we go about the exercise at the bottom of p67/top of p68?

And secondly, at the top of p69, he giveas the example and invites us to check that $(\phi^* g)_{\mu \nu} = diag ( \sin^2{\theta})$

However I find that
$(\phi^* g)_{\mu \nu}= \left( \frac{ \partial y^\alpha }{ \partial \theta} \right) \left( \frac{\partial y^\beta}{ \partial \phi} \right) \delta_{\alpha \beta} = \left( \frac{ \partial y^\alpha }{ \partial \theta} \right) \left( \frac{\partial y^\alpha}{ \partial \phi} \right) = \begin{pmatrix} \cos{\theta} \cos{\phi} & \cos{\theta} \sin{\phi} & -\sin{\theta} \end{pmatrix} \begin{pmatrix} -\sin{\theta} \sin{\phi} \\ \sin{\theta}\cos{\phi} \\ 0 \end{pmatrix}$
$=-\cos{\theta} \cos{\phi} \sin{\theta} \sin{\phi}+\cos{\theta} \cos{\phi} \sin{\theta} \sin{\phi}=0$

Where have I gone wrong?
Thanks

 

[NateD]

The mistake you made is in the last step. You assumed that $\partial y^\alpha/\partial \phi = (0, 0, 0)$, when in fact it is $(-\sin{\theta}\sin{\phi}, \sin{\theta}\cos{\phi}, 0)$. Thus, when you take the product of the two vectors, you get $\sin^2{\theta}$.