Isometries of the sphere $\mathbb{S}^{n}$

Question

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.

Answer 1

Here's a different approach than the one Qiaochu is suggesting, assuming you have proved that every point in a Riemannian manifold has a normal neighborhood: A neighborhood for which any two points in the neighborhood can be connected by a unique minimizing geodesic.

First, a lemma:

Suppose $M$ is a (connected) Riemannian manifold and $f:M\rightarrow M$ is an isometry. Suppose further that for some $p\in M$, that $f(p) = p$ and $d_p f = Id$. Then $f = Id$.

Proof (sketch): Let $X \subseteq M$ with $X = \{q\in M|$ $f(q) = q$ and $d_q f = Id \}$. Then $p\in X$ by definition so $X$ is nonempty. We will show $X$ is both open and closed.

$X$ is closed because it's defined by equations. $X$ is open becuase if $q\in X$, using a normal neighborhood $U$ and the fact that $f(exp_q(tv)) = exp_{f(q)}(t$ $d_q f$ $v)$ shows all points $r\in U$ satisfy $f(r) = r$. If $d_r f \neq Id$, draw a small triangle connecting $r, q$ and $exp(tv)$ where $d_r f v \neq v$ and $t$ is very small. Appealing to the fact that vertices of the triangle have unique minimizing geodesics between them will give a contradiction, showing $d_r f = Id$.

Thus, $X$ is both open and closed and so, since $M$ is connected, $X = \emptyset $ or $X = M$. But $p\in X$, so $X = M$. But this says that $f(q) = q$ for all $q\in M$ (and that $d_q f = Id$, but we don't really care about that), so $f$ is the identity map.

With this lemma proved, we get a pretty easy corollary: An isometry is determined by where it sends one point and what its differential does at that point.

This in turn gives a potential solution to our problem: If you can show

1) Every linear orthogonal map is an isometry of $S^n$ and

2) Given any $p, q \in S^n$, and any o.n. bases $\{v_1,...,v_n\}$ and $\{w_1,...,w_n\}$ of $T_p S^n$ and $T_q S^n$ respectively, there is a linear orthogonal map taking $p$ to $q$ and $\{v_1,...,v_n\}$ to $\{w_1,...,w_n\}$,

Then, it follows that you've found all of the isometries of $S^n$.

The proof of 1) is straightforward owing to the fact that the linear orthogonal matrices preserve the metric on $\mathbb{R}^{n+1}$ and $S^n$ inherits it's metric from $\mathbb{R}^{n+1}$.

The proof of 2) is slightly more involved, but not so bad: First, prove that you can assume wlog that $p = q = $ north pole and that $\{v_1,...,v_n\} = \{e_2,...,e_{n+1}\}$ using the standard o.n. basis $\{e_1,...,e_{n+1}\}$ of $\mathbb{R}^{n+1}$. From here, if you think about it, given $\{ w_1,...,w_n\}$, you can explicitly write down the linear orthgonal matrix you want.

Answer 2

Any isometry on $\mathbb{S}^n$ is the composition of atmost $n + 1$ reflections. I think it is followed from Cartan–Dieudonné theorem.Please tell me if I am wrong.