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Suppose I have a map $\phi: S^2 \rightarrow S^2$ and I know that

a) $\phi$ is continuous and bijective

b) If $a$ and $b$ subtend an angle of $\pi / 2$ at the center of the sphere, then so do $\phi(a)$ and $\phi(b)$.

Does it follow that $\phi$ is an isometry of the sphere?

If yes, can you sketch a proof?

If no, can you furnish an example of a non-isometry $\phi$ satisfying the above?

(Also, in either case, is the requirement that $\phi$ be bijective redundant?)

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Also (meta) what are appropriate tags for this question? –  Ben Blum-Smith Jul 24 '11 at 16:58
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Personally, I'm curious as to whether or not this can be generalized to all automorphisms of hypersurfaces that preserve orthogonality between surface normals. That sounds like it'd make a very nice standalone theorem. –  anon Jul 24 '11 at 19:16
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My immediate thought was that any failure had to be along the lines of the Banach-Tarski paradox, that you could partition the sphere into orbits of the $\pi/2$ rotation operators. But those operators have continuum many components, so a proof of isometry would be that you can fix every point with them. Maybe you could prove that for any two points there are enough paths consisting of $\pi/2$ segments connecting them? –  Ross Millikan Jul 26 '11 at 3:04
    
Taking in mind what @Ross said, I wonder if it is even possible to drop the continuity assumption if we maintain that the map is bijective. –  Willie Wong Jul 26 '11 at 14:13
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1 Answer 1

up vote 9 down vote accepted

In the case that you are willing to assume that the map is once differentiable, the answer is yes.

Sketch of proof:

There is a duality between great circles on $\mathbb{S}^2$ and lines in $\mathbb{R}^3$: you can identify a great circle on $\mathbb{S}^2$ with plane in $\mathbb{R}^3$ that contains the great circle, and then identify with the line that is normal to the plane. In other words, you can write $C_v = \{ w\in \mathbb{S}^2 | w\cdot v = 0 \}$ for any $v\in\mathbb{S}^2$.

The preservation of orthogonality means, therefore, that your map $\phi$ sends great circles to great circles.

Furthermore, we observe that fixing a point $p\in \mathbb{S}^2$, we can identify its tangent directions with the collection of all great circles through it. For $\eta,\omega\in T_pM$, the angle between them can be measured by the angle between their corresponding great circles, which is the same as the angle between their corresponding dual vectors.

So: if $\phi$ is $C^1$, the differential $d\phi$ defines a linear map between the tangent spaces. That $\phi$ preserves orthogonal directions now implies that $d\phi$ preserves orthogonal directions. Hence $d\phi$ must be conformal! (Since it is linear and preserves orthogonality.) So $\phi$ is a conformal map of the sphere. But recall back that $\phi$ preserves all great circles--any conformal automorphism of the sphere that preserves all great circles must be an isometry.

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Question: Are there any issues if $d\phi$ is not injective? Technically $d\phi$ could even be the zero map somewhere (or is there some reason this cannot occur?) –  Zarrax Jul 24 '11 at 19:25
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It cannot occur because $\phi$ is a bijection: any simple arc (and thus any tangent vector) has a unique preimage under $\phi$. –  Alexei Averchenko Jul 24 '11 at 20:42
    
@Zarrax: Part of the proof actually already shows that $d\phi$ cannot be not injective: if $v\neq w$ nonzero such that $d\phi(v) = d\phi(w) \neq 0$, then their corresponding great circles must be overlapping, so $\phi$ is no longer bijective. If $d\phi|_p = 0$, then $d\phi = 0$ along the entire circle "orthogonal to $p$". So $\phi$ maps that entire circle to a single point, again making $\phi$ itself non-injective. –  Willie Wong Jul 24 '11 at 20:49
    
The situation I had in mind was where $\phi(x)$ takes a great circle to another, but you have a situation like (restricting to a circle) $\phi(\theta) = \theta^3$ i.e. $d\phi$ is zero or has rank 1 at some point $p$ but could be nonsingular at nearby points. –  Zarrax Jul 24 '11 at 20:56
    
To be clear, I am referring only to local behavior in my previous comment when I write $\phi(\theta) = \theta^3$. –  Zarrax Jul 24 '11 at 21:04
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