TITLE: Isometries of $\mathbb{S}^2$
QUESTION [6 upvotes]: Definition: let $S,S'\subset\mathbb{R}^3$ be surfaces and a diferentiable function $f:S\to S'$. $f$ is said to be an isometry if for every $p\in S$ we have $<df_p(v), df_p(w)>=<v,w>$ whenever $v, w\in T_pS$

I'm trying to prove that if $f:\mathbb{S}^2\to\mathbb{S}^2$ is an isometry, then $f$ is an orthogonal linear transformation.
What I've done so far was to prove that $df_p$ is an orthogonal linear transformation for every $p\in S$. But I really don't know how to conclude that $f$ it self is an orthogonal transformation.
What is the trick?

REPLY [7 votes]: Here's another approach. The idea is that there so many orthogonal linear transformations that there can't possibly be any other isometries.
It's easy to show that orthogonal linear transformations are isometries of the sphere, and also that they act transitively on isometric 1-jets - that is, if $L : T_p S^2 \to T_q S^2$ is a linear isometry of tangent spaces, then there is an orthogonal linear transformation $f$ such that $df_p = L$. (First perform any rotation bringing $p$ to $q$, and then note that rotations/reflections fixing $q$ have derivatives inducing the entirety of $O(T_q S^2).$) It turns out that this transitivity is all we need to know that we in fact have all the isometries:

If two isometries $f,g$ of a connected manifold satisfy $df_p = dg_p$ for a single point $p$, then $f=g$.

Since an arbitrary isometry must share its derivative at a point with some orthogonal linear transformation, it is thus in fact equal to that map.
You can prove the highlighted result using some basic properties of geodesics: let $\gamma : [0,1] \to M$ be an arbitrary geodesic starting at $p$, and note that both $f \circ \gamma$ and $g \circ \gamma$ are also geodesics, since isometries send geodesics to geodesics. Moreover, our assumption means that these geodesics have the same initial position and velocity, and thus they are equal by uniqueness. Thus we have $f(\gamma(1)) = g(\gamma(1))$. Since there are plenty of geodesicsⁱ, we get $f = g$ everywhere.
ⁱTechnically speaking we either need to assume completeness (all pairs of points are joined by a geodesic) or do a boring open-closed argument here. For $S^2$ we know that great circles are geodesics, and that any two points lie on a great circle, so it's fine.