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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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This is not true in general. For example, you can easily see that a plane can be generated by rotations about any line perpendicular to it. Do you mean the surface to be bounded? Or the rotations to be about lines through the origin? –  tomasz Oct 28 '12 at 11:48
@tomasz, I edited my post and explained the 'rotation' here, is it clear?.. the object to be rotated is a curve, not a surface, and the curve and the axis must lie in the same plane.. –  hxhxhx88 Oct 28 '12 at 12:25
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). –  tomasz Oct 28 '12 at 18:47
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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 $$\tag1 S=\{0\} $$ $$\tag2S=\{x\in\mathbb R^3 \colon |x|=r\}\text{ for some }r>0$$ $$\tag3S=\{x\in\mathbb R^3 \colon |x|\le r\}\text{ for some }r>0$$ $$\tag4S=\{x\in\mathbb R^3 \colon r_1\le |x|\le r_2\}\text{ for some }0<r_1<r_2.$$ 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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Elegant proof! But I'm not familiar with $SO(3)$, why it is generated by two rotations?.. –  hxhxhx88 Oct 28 '12 at 16:52
It is mentioned without reference e.g. here. –  Hagen von Eitzen Oct 28 '12 at 17:05
OK.. Do you know any reference about the characterization of $SO(3)$?.Particularly, about its closed subgroups... I want to learn it more.. –  hxhxhx88 Oct 28 '12 at 17:41
BTW..Is there any argument that can rule out the condition (3) and (4)? I think maybe we can argue that the Lebesgue measure of a curve and the surface rotated by it is zero? However, maybe the measure of a curve must not be 0, for example, the Peano curve... –  hxhxhx88 Oct 28 '12 at 17:43
That's what I meant with "may be obtained with exotic curves". Yes, you can take a spacefilling curve of an annulus and rotate it to obtain such a thickened sphere. The question is: Do these beasts match the definition of surface as used in Geometry and the imagination? Or is the definition of curve in that context suitably restricted, e.g. to smooth or rectifyable curves? –  Hagen von Eitzen Oct 28 '12 at 17:54
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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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Why are you assuming that the axes intersect at all (and that the center of mass exists)? The surface could be unbounded for all we know. –  tomasz Oct 28 '12 at 11:42
Then again, I think the task does have the implicit assumption that the rotations are about lines through the origin. –  tomasz Oct 28 '12 at 11:47
yeah, I got your point. but not so rigorous... –  hxhxhx88 Oct 28 '12 at 16:56
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