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Picture below is from the 19th page of Do Carmo's Riemannian Geometry. In my calculation, $$ \det (d(\pi_2\circ \pi_1^{-1})) =\frac{\partial(y'_1,...,y'_n)}{\partial(y_1,...,y_n)} $$ where , when $i\ne j$
$$ \frac{\partial y_i'}{\partial y_j}= \frac{-2 y_i y_j}{(\sum_{i=1}^n y_i^2)^2} $$ and when $i=j$ $$ \frac{\partial y_i'}{\partial y_i}= \frac{\sum_{i=1}^n y_i^2-2 y_i^2}{(\sum_{i=1}^n y_i^2)^2} $$ In fact, for general $n$, I can't know the sign of $\det (d(\pi_2\circ \pi_1^{-1}))$, but when $n=2$, it is negative. It means the $(R^n, \pi_1^{-1}), (R^n, \pi_2^{-1})$ are not same orientation. Whether I am right ?

enter image description here

Enhao Lan
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  • Maybe read Example 4.5 again? (Since that's what's he's referring to when he says “the simple criterion of the previous example”, and it explains precisely why it works even if the determinant is negative.) – Hans Lundmark Dec 16 '20 at 06:22
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    @HansLundmark I think the "family given" is ${(R^n, \pi_1^{-1}), (R^n, \pi_2^{-1})}$ , but they have different orientation. Therefore, the family given does not determine an orientation. – Enhao Lan Dec 16 '20 at 06:57
  • Oh, that's a good point, you may be right! Probably he didn't even think about checking the sign of the determinant, since the argument in Example 4.5 shows that it can be fixed if needed. – Hans Lundmark Dec 16 '20 at 08:36

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