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In Hilbert's book Geometry and the imagination, he said that

sphere is the only surface which can be generated by rotation in more than one way.

It is quite intuitive, but I can't give a rigorous proof.

How to prove it?

PS: Here rotation means rotating a closed curve with respect to the axis of symmetry of it that is in the same plane.

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    The key element that you have added is the statement about the rotated object being a closed curve (and hence the entire surface has to be bounded).2012-10-28

2 Answers 2

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Let $S$ be a nonempty subset of $\mathbb R^3$ that can be generated by rotating a closed curve $C$ around $a$ and is also invariant under rotation around $b$. Then $S$ is compact and connected because $C$ is compact and connected.

If $a$ and $b$ do not intersect, let $c$ be a line intersecting both $a$ and $b$ perpendicuularly. Then the rotation by $\pi$ around $a$ followed by rotation by $\pi$ about $b$ leave $c$ fixed but translate it by twice the distance between $a$ and $b$. Thus this is the same as a screw operation along $c$ and causes $S$ to be unbounded, contradicting compactness.

If $a$ and $b$ intersect (wlog. in the origin), their rotations generate all of $SO(3)$, as joriki says. Therefore a single point $x\in S$ has as orbit a sphere aroound the origin (or consists of $x$ alone if $x=0$). We conclude that $S$ is the union of concentric spheres.

Since $S$ is compact and connected, this leaves only the possibilities $\tag1 S=\{0\} $ $\tag2S=\{x\in\mathbb R^3 \colon |x|=r\}\text{ for some }r>0$ $\tag3S=\{x\in\mathbb R^3 \colon |x|\le r\}\text{ for some }r>0$ $\tag4S=\{x\in\mathbb R^3 \colon r_1\le |x|\le r_2\}\text{ for some }0 The only case that actually leads to a surface is indeed $(2)$. The other cases may be obtained with suitable exotic curves.

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    OK. I see.. In Hilbert's book I think he is just talking about those ordinary book, which can be imagine:)..after all, thank you!2012-10-29
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A surface of revolution looks the same at all points related by rotations around the axis of revolution. If this is true for two axes, it follows that it looks the same everywhere. That implies among other things that the curvature is the same everywhere, which is only true for the sphere.

The two axes of rotation both have to go through the centre of mass. Rotations about two different axes through the same point together generate all of $SO(3)$, so the surface is invariant under all rotations about axes through that point; that's also true only for the sphere.

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    yeah, I got your point. but not so rigorous...2012-10-28