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?


Since $f : S^2 \to S^2$ is an isometry, it maps geodesics to geodesics. In other words, $f$ maps great circles to great circles on $S^2$. Moreover, being an isometry, $f$ preserves the affine separation between points on geodesics, which here is equal to the angular separation between points on great circles.

Now let's extend $f : S^2 \to S^2$ to a map $\hat f : \mathbb R^3 \to \mathbb R^3$, given by $$ \hat f : \ t \mathbf x \mapsto tf(\mathbf x) \ \ \ \ \ \ \ \ \ \ {\rm for \ \ } t \geq0, \ \ \mathbf x \in S^2.$$ Using our observations above, it should be easy to see that $\hat f$ is a linear map. For example:

  • Given two non-antipodal points $t_1 \mathbf x_1$ and $t_2 \mathbf x_2$ in $\mathbb R^3$, there is a unique great circle $C$ on $S^2$ passing through $\mathbf x_1$ and $\mathbf x_2$. The sum $t_1 \mathbf x_1 + t_2 \mathbf x_2$ can be written as $t\mathbf x$ for some $t \geq 0$ and for some $\mathbf x$ on this great circle $C$. The linearity property $\hat f(t_1 \mathbf x_1) + \hat f(t_2 \mathbf x_2) = \hat f(t \mathbf x)$ then follows from the fact that $f$ maps the great circle $C$ to another great circle $f(C)$, and preserves angular separations on these great circles. The case where $t_1 \mathbf x_1$ and $t_2 \mathbf x_2$ are antipodal is even easier...

  • Similarly, the property $\hat f(- t\mathbf x) = - \hat f(t \mathbf x)$ follows from the fact that $f$ maps antipodal points to antipodal points. It then follows that $\hat f( \alpha t \mathbf x) = \alpha \hat f(t \mathbf x)$ for all $\alpha$, positive or negative.

Thus $\hat f$ is a linear map on $\mathbb R^3$. By construction, $\hat f$ also preserves separation from the origin. Hence $\hat f$ is represented by an orthogonal matrix. Restricting $\hat f$ to $S^2$, we conclude that $f$ is also represented by an orthogonal matrix.

[Edit: Yes, the claim that $f$ maps geodesics to geodesics does follow from your local definition of an isometry. A geodesic $t \mapsto \mathbf x (t) \in S^2$ is a path from $\mathbf x(t_0)$ to $\mathbf x(t_1)$ that minimises the length functional $$ L[\mathbf x(t)] = \int_{t_0}^{t_1} dt \sqrt{\left\langle \frac{d \mathbf x(t)}{dt}, \frac{d \mathbf x(t)}{dt} \right\rangle} $$ locally within the space of paths between $\mathbf x(t_0)$ and $\mathbf x(t_1)$. (Think of this as a calculus of variations statement.)

But $f$ is an isometry, so it preserves infinitesimal lengths, and therefore, one would expect that $f$ preserves lengths of paths. Let's check this: \begin{multline} L[f(\mathbf x(t))] = \int_{t_0}^{t_1}dt \sqrt{\left\langle \frac{d f(\mathbf x(t))}{dt}, \frac{d f(\mathbf x(t)}{dt} \right\rangle}\\ = \int_{t_0}^{t_1}dt \sqrt{\left\langle df_{\mathbf x(t)}\left(\frac{d (\mathbf x(t))}{dt}\right), df_{\mathbf x(t)}\left(\frac{d (\mathbf x(t)}{dt}\right) \right\rangle} \\ = \int_{t_0}^{t_1} dt \sqrt{\left\langle \frac{d \mathbf x(t)}{dt},\frac{d \mathbf x(t)}{dt} \right\rangle} = L[\mathbf x (t)]\end{multline} Great! So we can conclude that $t \mapsto \mathbf x(t)$ is a geodesic if and only if its image $t \mapsto f(\mathbf x(t))$ is a geodesic. In the context of this problem, the geodesics are precisely the great circles on $S^2$, so $f$ maps great circles to great circles.

Furthermore, the affine length of a geodesic between its endpoints is given by $L[\mathbf x(t)]$. Since $L[f(\mathbf x (t))] = L[\mathbf x(t)]$ when $f$ is an isometry, we conclude that affine lengths are preserved by isometries. In the context of this question, the affine length between between points on a great circle is equal to their angular separation in radians. ]

  • $\begingroup$ I think the key fact is that "$f$ maps the great circle $C$ to another great circle $f(C)$, and preserves angular separations on these great circles". I avoided using it since I don't know how to prove it. Can you give me an outline of the proof? $\endgroup$ – rmdmc89 May 11 '17 at 3:45
  • 1
    $\begingroup$ @AguirreK I added further explanation of this point. And take a look at Anthony's answer too - I think his answer is very neat! $\endgroup$ – Kenny Wong May 11 '17 at 8:20

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.


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