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How to prove so elementary (elementary = without using the concept of geodesic) that an isometry of $\mathbb{S}^n$ is a restriction on $\mathbb{S}^n$ of an isometry of $\mathbb{R}^{n+1}$ ?

EDIT:

You will isometry with respect metric is induced by $\mathbb{R}^{n+1}.$

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    You need to impose one more condition on your isometries of $\mathbb{R}^{n+1}$ - the isometry of translation does not restrict to an isometry of the sphere (because it doesn't map the sphere to itself). A natural condition is "restriction on $\mathbb{S}^n$ of an isometry of $\mathbb{R}^{n+1}$ *mapping the origin to itself*."2012-04-10

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This isn't too difficult to show directly. Any isometry $f\colon\mathbb{S}^n\to\mathbb{S}^n$ extends easily to a map $g\colon\mathbb{R}^{n+1}\to\mathbb{R}^{n+1}$, (writing $\hat x\equiv x/\Vert x\Vert\in\mathbb{S}^n$ for $x\in\mathbb{R}^{n+1}\setminus\{0\}$) $ g(x)=\begin{cases} \Vert x\Vert f(\hat x),&\textrm{if }x\not=0,\cr 0,&\textrm{if }x=0. \end{cases} $ It is clear that $f$ is the restriction of $g$ to $\mathbb{S}^n$, so all that needs to be done is to show that $g$ is an isometry. In particular, $ \Vert g(x)-g(y)\Vert=\Vert x-y\Vert\qquad{\rm(1)} $ for $x,y\in\mathbb{R}^{n+1}$. Also, $\Vert g(x)-g(0)\Vert=\Vert x\Vert\Vert f(\hat x)\Vert=\Vert x\Vert=\Vert x-0\Vert$, so (1) only needs to be shown for nonzero $x,y$. However, the distance between $x,y$ can be written purely in terms of $\Vert x\Vert,\Vert y\Vert$ and $\hat x\cdot\hat y$, $ \begin{align} \Vert x-y\Vert^2&=\Vert x\Vert^2+\Vert y\Vert^2-2x\cdot y\cr &=\Vert x\Vert^2+\Vert y\Vert^2-2\Vert x\Vert\Vert y\Vert\hat x\cdot\hat y. \end{align}\qquad(2) $ Now, $g$ preserves the norm of any $x\in\mathbb{R}^{n+1}$. Also, $\hat x\cdot\hat y$ is just the cosine of the distance from $\hat x$ to $\hat y$ along the sphere. So, this is preserved by $f$ and, hence, $\widehat{g(x)}\cdot\widehat{g(y)}=f(\hat x)\cdot f(\hat y)=\hat x\cdot\hat y$. Applying (2) to both $\Vert x-y\Vert$ and $\Vert g(x)-g(y)\Vert$ shows that (1) holds and $g$ is distance preserving.

As the (Riemannian) metric is determined by the distance between points (note: if $\gamma$ is a smooth curve in a manifold $X$, $g(\dot\gamma(t),\dot\gamma(t))=\lim_{h\searrow0}h^{-1}d(\gamma(t),\gamma(t+h))$), then as long as it is known that $g$ is smooth, the argument above implies that $g$ is an isometry in the Riemannian sense. It is known that any distance preserving invertible map between $n$-dimensional manifolds is smooth. If you don't want to use such a result, you can instead use the fact that any distance (and origin) preserving map $\mathbb{R}^{n+1}\to\mathbb{R}^{n+1}$ is linear (e.g., see this question), hence smooth.

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    @Eduardo: I added a link to a question where this is shown.2012-04-11
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Denote the isometry f, and consider the matrix with columns $f(e_1), f(e_2), f(e_3), \ldots$, where ${e_i}$ form an orthnormal basis of $\mathbb{R}^{n+1}$, also conveniently lying on the sphere. This matrix is orthonormal and so defines an isometry of $\mathbb{R}^{n+1}$

The intuition here is that a linear transformation takes spheres to ellipsoids, and is uniquely defined by how it does so.

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    You're right, there is$a$subtlety here and dealing with it amounts to George Lowther's answer.2012-05-24
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A correct presentation would be pretty long. Let us start with the positive orthant, all $x_j \geq 0$ in $ \mathbb S^n.$ The fact that the unit sphere is given by by a sum of squares equalling one amounts to direction cosines. That is, given the standard orthonormal basis $e_j,$ we are saying that $ \sum \cos^2 \theta_j = 1, $ where $\theta_j$ is the angle between our favorite point $x$ on the sphere and $e_j.$ This is not a surprise, as $ \cos \theta_j = x \cdot e_j = x_j, $ the $j$th coordinate, and we are just saying the sum of the squares of the coordinates is $1.$

Next, we need a definition of distance on the metric space $ \mathbb S^n.$ I say $ \mbox{AXIOM} \; \; \; d(x,y) = \arccos (x \cdot y). $

Another way of saying $\sum x_j^2 = 1$ is then $ 1 = \sum (x \cdot e_j)^2 = \sum \cos^2 \theta_j = \sum \cos^2 d(x,e_j).$

Here is the part that would take forever to write out:

AXIOM: If I take a finite set $ \{\epsilon_j\}, \; \; \epsilon_j \geq 0, \; \; \sum \epsilon_j = 1, $ there is exactly one point $x$ in the positive orthant such that each $ (x \cdot e_j)^2 = \epsilon_j. $

Sketch: the isometry maps the positive orthant to something....

5:02 pm Pacific time. I will probably pick this up later...