Prove that the antipodal mapping is an isometry on $S^n$. Help understanding the proof.

Question

Prove that the antipodal mapping $A: S^n \to S^n$ given by $A(p)=-p$ is an isometry.

I know that in order to prove that a map $f$ is an isometry of a smooth manifold $M$ it must hold true that $$\langle v,u\rangle_p = \langle d_f v, d_fu \rangle_{f(p)}$$

Now I found online this document that proves this exercise (Exercise 1.1.). I cannot follow it though. It basically says that in order to prove this we first need to claim that $T_{p}S^n = T_{A(p)}S^n$ and for this it is enough to prove that $T_{p}S^n \subset T_{A(p)}S^n$. Why is this true? In order to show this they claim that $T_{A(p)}S^n \subset \subset T_{A \circ A(p)}S^n = T_pS^n$. What does the $\subset \subset$ mean and why the last relation holds true, i.e. why $T_{A(p)S^n}=T_pS^n$ since these tangent spaces are defined in different points?

Finally, why they later claim that $dA_pv = -v$? I do not understand why this also holds true.

Answer 1

To address the remaining question, the proof of the equation of tangent planes $T_p S^2 = T_{A(p)}S^2$ is obtained by tracing through the definitions and doing some multivariable calculus computations.

First, $S^2$ is the solution set of the equation $$x^2 + y^2 + z^2-1=0 $$ Second, the gradient of the left hand side of this equation at the point $p=(a,b,c)$ is computed to be $$\nabla_p(f) = \langle 2a,2b,2c\rangle $$ and the tangent plane $T_p S^2$ is the perpendicular plane to $\nabla_p(f)$ which is computed to be the set vectors $$T_p S^2 = \{\langle x,y,z \rangle \bigm| ax+by+cz=0 \} $$ Third, the gradient at the point $A(p)=(-a,-b,-c)$ is computed to be $$\nabla_{A(p)}(f)=\langle-2a,-2b,-2c\rangle $$ and $T_{A(p)}S^2$ is the perpendicular plane to $\nabla_{A(p)}(f)$ which is computed to be the exact same set of vectors $$T_{A(p)}S^2 = \{\langle x,y,z \rangle \bigm| ax+by+cz=0 \} $$

Answer 2

As user 7530 suggests, the antipodal map on the sphere is the restriction to the sphere of the $-id$ map on $\mathbb{R}^{n+1}$. Note that $-id$ is linear, and hence, its derivative at a point is also equal to $-id$. More explicitly, for a point $p\in\mathbb{R}^{n+1}$, identify the tangent space $T_p\mathbb{R}^{n+1}$ with $\mathbb{R}^{n+1}$. Then for any $v\in T_p\mathbb{R}^{n+1}$ we have$$d(-id)_p(v)=-v\in T_{-p}\mathbb{R}^{n+1}.$$Consequently, for any $u,v\in T_p\mathbb{R}^{n+1}$ we have$$\langle d(-id)_p(u),d(-id)_p(v)\rangle=\langle-u,-v\rangle=\langle u,v\rangle,$$and so $-id$ is an isometry.

Edit: We show $T_pS^n=T_{-p}S^n$. The argument is essentially the one used in Lee's answer, but spelled out differently. Let $\alpha:(-\epsilon,\epsilon)\to S^n$ satisfy $\alpha(0)=p$. We have$$\langle\alpha(t),\alpha(t)\rangle=const,$$and differentiating at $t=0$, using the Leibniz rule, we obtain$$\langle\dot{\alpha}(0),p\rangle=0.$$Hence,$$T_pS^n\subset p^\bot.$$It now follows from dimension consideration that in fact$$T_pS^n=p^\bot.$$Since $p^\bot=(-p)^\bot$, we are done.

Remarak: As commented bellow, the fact that $T_pS^n=T_{-p}S^n$ may be interesting in general, but is not crucial for this question. Every $M\in O_n(\mathbb{R})$ restricts to an isometry of $S^n$, and the antipodal map is just a particular case.