Orientation of $S^n$ under antipodal map In Riemannian Geometry, doCarmo p.19-20 the author states that the orientation of $S^n$ (using stereographic coordinates) under the antipodal map $A(p)=-p$ is


*

*Reversed for $n$-even,

*Preserved for $n$-odd.


This didn't seem obvious to me at first so I checked it and seem to have come to the opposite conclusion. It would be helpful if someone could check my reasoning and tell me where I'm going wrong:

An orientation for $M$ is a differentiable structure $\{(U_\alpha,x_{\alpha})\}$ satisfying the property that $d({x_{\beta}}^{-1} \circ x_{\alpha})$ has positive determinant for all $\alpha, \beta$ with $x_{\alpha}(U_{\alpha}) \cap x_{\beta}(U_{\beta}) \neq \emptyset.$ If a differentiable structure satisfies this is it said to be orientable.
A map $\phi \colon M_1 \to M_2$ is said to preserve orientation if the union of the old differentiable structure with the new one induced by $\phi$ is orientable.
With these definitions in mind, we look at the stereographic structure on $S^n$ given by $\{(\mathbb{R}^n,{\pi_1}^{-1}),(\mathbb{R}^n,{\pi_2}^{-1})\}$, where $\pi_1(x_1, \ldots , x_{n+1}) = (\frac{x_1}{1-x_{n+1}}, \ldots, \frac{x_n}{1-x_{n+1}})$ and $\pi_2(x_1, \ldots , x_{n+1}) = (\frac{x_1}{1+x_{n+1}}, \ldots, \frac{x_n}{1+x_{n+1}})$.
Now introduce ${\hat{\pi}}_1 = \pi_{1} \circ A$ and ${\hat{\pi}}_2 = \pi_{2} \circ A$ and compute the Jacobian determinant of 


*

*$\pi_1 \circ {{\hat{\pi}}_1}^{-1}$ ... ( $(-\frac{1+x_{n+1}}{1-x_{n+1}} )^n$ )

*$\pi_1 \circ {{\hat{\pi}}_2}^{-1}$ ... ($(-1)^{n}$)

*$\pi_2 \circ {{\hat{\pi}}_1}^{-1}$ ... ($(-1)^{n}$)

*$\pi_2 \circ {{\hat{\pi}}_2}^{-1}$ ... ( $(-\frac{1+x_{n+1}}{1-x_{n+1}} )^n$ )


and conclude that $\phi$ preserves orientation for $n$-even.

The derivative is the linear map $-I$ applied to the tangent space $T_x S^n=x^{\perp}$ (since the map itself is the linear map $-I$ restricted to the sphere). So the determinant of the derivative is $(-1)^n$. Hence orientation reversing just when $n$ is odd. However, $T_x S^n = x^{\perp}$ has opposite orientation from $T_{-x} S^n = (-x)^{\perp}$, even though they are the same subspace of $\mathbb{R}^{n+1}$, since we orient to get $x$ outward unit normal. So actually orientation is reversed just when $n$ is even, preserved just when $n$ is odd.
