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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 maps

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First step: any isometry of $S^n$ extends to a map $\mathbb{R}^{n+1} \to \mathbb{R}^{n+1}$. What can you say about this map? –  Qiaochu Yuan Mar 16 '11 at 16:48
    
@QiaochuYuan Could you please give more details about this? Thanks –  dmm Sep 24 '12 at 15:27
    
@dmm: if $f : S^n \to S^n$ is any map whatsoever, it induces a map $f(v) = |v| \left( f(v/|v|) \right)$ from $\mathbb{R}^{n+1}$ to $\mathbb{R}^{n+1}$ (define it to be $0$ at $0$). This is the map you want to show is linear, so what can you say about it? –  Qiaochu Yuan Sep 24 '12 at 18:11
    
@QiaochuYuan: Hmmm I am not sure how to prove that $f$ is linear. My approach was to consider parametrization $\varphi $ around some point $p\in S^n$. Denote by $i:S^n\to R^{n+1}$ inclusion map. Then based on induced metric we have the following $$di_p(u)\cdot di_p(v)=d(i\circ f)_p(u)\cdot d(i\circ f)_p(v),$$ for all $u,v \in R^n$. Based on this equation we should deduce properties of $f$. Any help? –  dmm Sep 24 '12 at 22:42
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2 Answers 2

up vote 6 down vote accepted

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.

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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.

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How are you proposing to prove that statement? –  Mariano Suárez-Alvarez Mar 16 '11 at 17:38
    
Don't worry: you're right. –  t.b. Mar 17 '11 at 12:56
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@Theo, @Anjan: Of course he are right. What he wrote is true. But: The Cartan-Dieudonné theorem states that an element of $SO(n)$ is a product of reflections. It does not say anything about elements of the isometry group of the sphere. To actually do something with it related to the question, you would have to show that an isometry of $S^n$ extends to a linear isometry of $\mathbb R^{n+1}$. But then... you do not need the Cartan-Dieudonné theorem at all because that an isometry of $S^n$ extends to a linear isometry of $\mathbb R^{n+1}$ is precisely what the OP wants! –  Mariano Suárez-Alvarez Mar 17 '11 at 16:39
    
Thanks @Mariano. I would also have done it by extending it to a linear map as Qiaochu commented. –  t.b. Mar 17 '11 at 23:21
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