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.
 A: I understand what you tried to do but it seems to me that you have two problems, which we can fix:


*

*Did you check that the atlas given by $\pi_1$ and $\pi_2$ is an oriented atlas? In other words, are the orientations given by the charts $\pi_1$ and $\pi_2$ compatible? Concretely, does the map $\pi_2 \circ {\pi_1}^{-1}$ have a positive Jacobian determinant?

*You probably got your Jacobian determinants (1) and (4) wrong.


For the first point, I think that $\pi_2 \circ {\pi_1}^{-1} (x) = \frac{x}{\Vert x \Vert ^2}$, where I denote $x = (x_1, \dots, x_n)$ and $\Vert x \Vert^2 = {x_1}^2 + \dots + {x_n}^2$. It does not seem easy to compute the Jacobian determinant of this map, but I think that it is not too hard to see that its sign is negative. Note that geometrically, this map looks like the "reflection" (inversion) through the unit sphere (though that's not exactly what it is, it's not too surprising that it reverses orientation). This means that your charts are not compatible in terms of orientation. You should take that into account when you conclude from your Jacobian determinants (2) and (3).
On the other hand, just looking at your Jacobian determinant (1) or (4) should be enough in order to determine whether $A$ preserves orientation or not. I think you went wrong in your calculations here. First of all note that your maps $\pi_1 \circ {{\hat{\pi}}_1}^{-1}$ and $\pi_2 \circ {{\hat{\pi}}_2}^{-1}$ are maps $\mathbb{R^n} \to \mathbb{R^n}$, so it does not make sense that there is $x_{n+1}$ in your Jacobian determinants. Unless I'm mistaken, my calculation says that $\pi_1 \circ {{\hat{\pi}}_1}^{-1}(x) = -\frac{x}{\Vert x \Vert ^2}$. You can conclude from my previous remark that the sign of the Jacobian determinant of this map is $(-1)^{n+1}$.
I hope that clarifies and answers your question. If you want a more geometric way to find the same result, I suggest you look here.
A: 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.
