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Let $S^n\subset \mathbb{R}^{n+1}$ be the unit sphere centered at $0$. Prove that the isometris of $S^n$ is $O(n+1)$, the orthogonal maps of $\mathbb{R}^{n+1}$.

Certainly all elements of $O(n+1)$ are isometries, but how to prove they are all the isometries? I want a rigorous proof.

tcchhlegend
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  • You’re assuming the isometrics are smooth maps? – Deane May 22 '21 at 15:05
  • @Deane , a bit more. An isometry must be an automorphism which preserves length and angle. Specifically, let $f:M\rightarrow M$ be an isometry, then the differential $df_p: T_p M \rightarrow T_{f(p)} M$ is an orthogonal linear map. – tcchhlegend May 22 '21 at 15:27
  • I meant: you can assume an isometry is just a map that preserves geodesic distances along the sphere and not assume it’s differentiable. That’s the real definition of an isometry. – Deane May 22 '21 at 16:22

2 Answers2

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$S^{n}$ is a set of all unit vectors in $R^{n}$. Every base of $R^{n}$ can be orthonormalized. And $O(n+1)$ is a set of all isometries between orthonormal bases, so it moves unit vectors to unit vectors.

robin3210
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Let $M$ be a complete connected Riemannian manifold. Choose a point $p\in M$ and an orthonormal frame $E=(e_1,\cdots,e_n)\subset T_pM$. Given an isometry $g\in\operatorname{Iso}(M)$, we can act on $E$ with $g$ to obtain a new frame $gE=(dg(e_1),\cdots,dg(e_n))\subset T_{g(p)}M$. Further, if $g_1E=g_2E$ then $g_1=g_2$ (since the exponential map is surjective and commutes with isometries). Thus, the map $g\mapsto gE$ is injective, and $\operatorname{Iso}(M)$ can be realized as a subset of the bundle of orthonormal frames $OFM$.

The largest possible isometry group is one for which the map $g\mapsto gE$ is surjective, i.e. one which acts transitively on orthonormal frames. This is the case for $O(n+1)$ on the sphere.

Kajelad
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