TITLE: Isometries of the sphere $\mathbb{S}^{n}$
QUESTION [21 upvotes]: Got this as homework and I don't know how to tackle this. Help please!
Prove that the isometries of $\mathbb{S}^{n} \subset \mathbb{R}^{n+1}$, with the induced metric, are restrictions to $\mathbb{S}^{n}$ of the linear orthogonal transformations.

REPLY [9 votes]: Trivially any orthogonal map restricts to an isometry of the n-sphere. Now suppose you have $f: \mathbb{S}^n \rightarrow \mathbb{S}^n$ an isometry. If it's gonna be the restriction of an orthogonal map $F: \mathbb{R}^{n+1} \rightarrow \mathbb{R}^{n+1}$, a very reasonable candidate for $F$ is the map:
\begin{equation}
F(x)=f(\frac{x}{|x|})\cdot |x|, x\neq 0; F(0)=0 
\end{equation}
Let's prove $F$ is an orthogonal map.
The distance between any two distinct points $x$, $y$ in $\mathbb{S}^n$ is just the length of the shortest of the two arcs of great circle joining them (this does not need any connection theory or geodesics, just observe that this great circle is the intersection of $\mathbb{S}^n$ and the plane $\pi$ containing $0_{\mathbb{R^{n+1}}}$, $x$ and $y$, so for any other curve joining $x$ and $y$ in $\mathbb{S}^n$ not contained in $\pi$, its tangent vector will have a nonzero component normal to $\pi$, hence you can keep minimizing its length without moving the endpoints). Since $f$ is an isometry, it preserves the length of curves, so it preserves the distance in $\mathbb{S}^n$. What happens it that the length of this arc of great circle is the (smaller) angle in radians formed by $x$, $y$. Now (make a drawing if needed) take any $p$, $q$ in ${\mathbb{R^{n+1}}}$, since $F$ preserves the norm, the triangles formed by $0_{\mathbb{R^{n+1}}}$, $p$, $q$ and $0_{\mathbb{R^{n+1}}}$, $F(p)$, $F(q)$ are congruent (side-angle-side and this became Plane Geometry!). Hence, $F$ preserves the distance in ${\mathbb{R^{n+1}}}$ and
\begin{equation}
\langle F(p),F(q)\rangle=\frac{1}{2}(|F(p)|^2+|F(q)|^2-|F(p)-F(q)|^2)=  
\frac{1}{2}(|p|^2+|q|^2-|p-q|^2)=\langle p,q\rangle.
\end{equation}
So, $F$ also preserves the inner product in $\mathbb{R^{n+1}}$. Observe now that since $\mathbb{S}^n\cup\{0\}\subset \text{Im }F$, $F$ is onto. Thus, for all $p,q,r\in\mathbb{R}^{n+1}$ and $\alpha, \beta\in\mathbb{R}$
\begin{equation}
\langle F(\alpha p + \beta q)- \alpha F(p)-\beta F(q),F(r)\rangle = \langle \alpha p + \beta q,r\rangle - \alpha \langle p, r\rangle -\beta \langle q,r\rangle=0.
\end{equation}
Then, $F$ is a linear orthogonal map.