# Isometries of $\mathbb{S}^n$

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 $\mathbb{R}^{n+1}.$

• – t.b. Apr 10 '12 at 21:41
• An isometry of $S^n$ with respect to what metric? (One cound give $S^n$ the restriction of the metric of $R^{n+1}$, or give it the intrinsic metric...) – Mariano Suárez-Álvarez Apr 10 '12 at 21:58
• Yes, You will isometry with respect metric is induced by $\mathbb{R}^{n+1}$, thanks! – user27456 Apr 10 '12 at 22:17
• Given an isometry $f$ on $\mathbb S^n$, it seems straightforward to write down how $f$ ought to extend to all of $\mathbb R^{n+1}$: scale the input vector to end up on the sphere; apply $f$, scale back again to its original length. However, showing that the extended $f$ is an isometry (or even a diffeomorphism) at $0$ seems to depend on some global knowledge on how $f$ has to act on all of $\mathbb S^n$ at once. So you need some tool that can extend the local condition that $f$ is an isometry to some useful global facts. And if not geodesics, then what? – hmakholm left over Monica Apr 10 '12 at 22:26
• 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." – Jason DeVito Apr 10 '12 at 22:42

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.

• This answer deals with the metric of the sphere which is the restriction of the Euclidean metric of the ambient space. But I am pretty sure the OP wants isometries of the Riemannian metric induced onb the sphere as a subvariety. – Mariano Suárez-Álvarez Apr 11 '12 at 2:18
• Looking at Henning Makholm's comment to a previous answer, maybe the question was as an isometry of Riemann manifolds. That shouldn't be an issue though, because it is just the same thing. The term $\hat x\cdot\hat y$ is just the cosine of the distance from $\hat x$ to $\hat y$ along the sphere, so is preserved by $f$. – George Lowther Apr 11 '12 at 2:18
• @MarianoSuárez-Alvarez: I updated with that in mind. – George Lowther Apr 11 '12 at 2:29
• I suppose you could ask: How do you know that $g$ is smooth? Well, any distance-preserving map is smooth on any Riemann manifold. Or use what you know about isometries of $\mathbb{R}^{n+1}$ (as a metric space) to imply that $g$ is linear, hence smooth. – George Lowther Apr 11 '12 at 2:32
• How can see that your application is linear?? – user27456 Apr 11 '12 at 15:51

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.

• How do you prove that the matrix is orthonormal? (Without more details, the last sentence of your first paragraph says, effectively, «since what you want to prove is true, it is true» :D ) – Mariano Suárez-Álvarez Apr 10 '12 at 21:48
• In $\mathbb{R}^n$ isometries that sent 0 to 0 preserve the inner product, so $(f(e_i),f(e_j))=(e_i,e_j)=\delta_{ij}$. You're right though it seems like there isn't much to prove..? – Nick Alger Apr 10 '12 at 21:51
• But you do not have an isometry of $\mathbb R^{n+1}$, only an isometry $S^n\to S^n$. – Mariano Suárez-Álvarez Apr 10 '12 at 21:55
• Ahh yes i suppose that's right, though I get the impression this issue is not particularly deep and we're just getting dragged down by definitions. If you want to go overkill, just use the singular value decomposition to show that any matrix which takes the sphere to the sphere is orthogonal. – Nick Alger Apr 10 '12 at 22:05
• ....how do you know it's a matrix? I.e. linear? The question concerns isometries with respect to the intrinsic metric: the distance from $a$ to $b$ is the length of the shortest path on the sphere that goes from $a$ to $b$. If you could prove that every isometry with respect to that metric is the restriction to the sphere of some linear mapping on the whole space in which the sphere is embedded, then you'd be done since it's already shown that every linear mapping under which the sphere is invariant is an isometry on the whole space. – Michael Hardy Apr 10 '12 at 22:58

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...