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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?2011-07-24
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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.2011-07-24
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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?2011-07-26
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    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.2011-07-26

2 Answers 2

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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?)2011-07-24
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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 $\phi$.2011-07-24
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    @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.2011-07-24
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    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.2011-07-24
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    To be clear, I am referring only to local behavior in my previous comment when I write $\phi(\theta) = \theta^3$.2011-07-24
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    @Zarrax: it can't happen. That's what I tried to explain in my previous comment.2011-07-24
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    +1 Awesome, very helpful. Any thoughts on what can be said without the differentiability assumption? Also (granting differentiability if desired), can bijectivity be deduced from the preservation of orthogonal directions or does it have to be assumed separately?2011-07-25
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    I am not sure (to both questions). But the $C^1$ answer suggests that in the continuous case, if there are counterexamples they may be difficult to write down explicitly (think the different between low and high regularity Nash embedding theorems). I've tried to remove the bijective assumption. As you can see from above, we can weaken it to locally invertible. I suspect it can be removed.2011-07-25
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    Just one more thought: the proof above is differential geometrical. But if you want to do something in $C^0$, large chunks of the proof has to be topological. One possibility is to lift $\phi: \mathbb{S}^2\to\mathbb{S}^2$ to a map $\tilde{\phi}: SO(3)\to SO(3)$ that factors through $\mathbb{S}^2$. In other words, your question can be phrased as "Let $\tilde{\phi}$ be a continuous, bijective map of $SO(3)$ to itself that factors through $\mathbb{S}^2$, is $\tilde{\phi}$ necessarily a group multiplication." Hum, this may actually be a good question for MathOverflow.2011-07-25
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    I don't think I know what you mean "factors through"? I think of "$\phi:A \rightarrow B$ factors through $C$" as meaning that I can write $\phi$ as the composition of $\phi_1:B \rightarrow C$ and $\phi_2:A \rightarrow B$. But I don't think you could mean that here? If a map $SO(3) \rightarrow SO(3)$ factored through $S^2$ in this sense, it would have to be a pretty crazy map because $SO(3)$ is a 3-manifold and $S^2$ is a 2-manifold; and, how would such a map induce a map $S^2 \rightarrow S^2$? Explain?2011-07-26
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    The preservation of orthogonality means that every set of orthonormal basis of $\mathbb{R}^3$ will be carried to another set of orthonormal basis. The total set of (oriented) orthonormal bases can be parametrised by $SO(3)$. So your map on $\mathbb{S}^2$ induces, by passing to the action on the orthonormal bases, a map on $SO(3)$. By saying that a map on $SO(3)$ factors through $\mathbb{S}^2$ I just mean that you can go in the reverse.2011-07-26
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    In regards to maps on three manifolds factoring through maps on two manifolds, naturally what is meant is that the map commutes with a certain natural projection. Throwing aside the present case (which the structure is, indeed, slightly complicated), you may consider the slightly simpler example of the [Hopf fibration](http://en.wikipedia.org/wiki/Hopf_fibration) which induces a natural map from $\pi: \mathbb{S}^3\to\mathbb{S}^2$. We can say that a map $f$ from the 3-sphere to itself factors through $\mathbb{S}^2$ if there exists a map $g$ on the 2-sphere such that $g\circ\pi = \pi\circ f$.2011-07-26
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    Oh, okay got it, thanks!2011-07-26
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Actually, the answer is positive even if you assume only that $\phi$ is a bijection (no need to assume continuity). The point is that such $\phi$ sends great circles to great circles and antipodal pairs to antipodal pairs. Hence, $\phi$ projects to a bijection $\Phi: RP^2\to RP^2$ which preserves collinearity. Von Staudt proved (the fundamental theorem of projective geometry) that all such maps $\Phi$ are projectivizations of linear transformations $R^3\to R^3$. You can find a proof for instance, in Hartshorne's book "Projective Geometry". Now, it is an elementary exercise to prove that a projective transformation which preserves distances equal to $\pi$ on $RP^2$ has to be an element of $PO(3)$. Hence, $\phi$ is orthogonal.