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Consider 2-d surfaces in 3-d (at the suggestion of a comment, let's say closed connected 2-dim smooth manifolds, embedded in dimension 3) with finite area. A sphere has the interesting property that if you take any two parallel planes that both intersect the sphere, and measure the area of the part of the sphere between the two planes, then the answer only depends on how far apart the two planes are, regardless of the positions or orientations of the planes. Does this property automatically guarantee that a 2-d surface (defined as above) in 3-d is a sphere? My intuition is that the answer is yes. If so, is there some easy way to see it?

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Would assuming the 2d surface to be a general quadric suffice? – Torsten Hĕrculĕ Cärlemän Jul 22 '13 at 20:54
Doesn't the same property hold for an (infinite) cylinder as well? – svenkatr Jul 22 '13 at 20:55
@svenkatr If the planes are vertical you get infinite area for a vertical cylinder, but if the planes are not vertical you get finite area so I'm inclined to say no. However I meant surfaces with finite area so I'll update my question to reflect that – user2566092 Jul 22 '13 at 21:12
A sphere with a single point removed would also satisfy the property. So perhaps by surface we mean "closed 2-dim smooth manifold, embedded in dimension 3, without a boundary"? – RghtHndSd Jul 22 '13 at 21:36
@rghthndsd: no I meant a closed manifold, which by definition is a compact manifold without boundary. A manifold, to me, can have boundary if it likes (unless it's clear from the context that it shouldn't have one). – tomasz Jul 23 '13 at 0:17
up vote 7 down vote accepted

Let $\alpha>0$ be the number such that the surface area between two planes at distance $d$ apart is $\alpha d$. Also, let $A$ be the total area of the surface. Then every projection of the surface onto a line is an interval of length $A/\alpha$ (this property, being of constant width, was already noted by rghthndsd and Andrew D. Hwang).

We may assume that the center of mass of the surface $S$ in question is the origin. Projecting $S$ onto a line through the origin does not shift the center of mass. Therefore, this projection is an interval of length $A/\alpha$ with midpoint at $c$. It follows that the support function of the convex hull of $S$ is constant, namely $ A/(2\alpha)$. Therefore, the convex hull of the surface is a solid sphere (a ball) $B$, because a convex body is determined by its support function.

Suppose $p\in \partial B\setminus S$. Since $S$ is a compact subset of $B$, and $p$ is an extreme point of $B$, it follows that $p$ is separated from $S$ by a plane. But then $p$ cannot be in the convex hull of $S$, a contradiction. We conclude that $\partial B\subseteq S$. But if a (connected) surface contains a sphere, it is a sphere. $\quad \Box$

Alternative, analytic proof. Let $\nu$ be the surface measure of $S$ and define $$\widehat \nu(\xi)=\int \exp(-i\xi\cdot x)\,d\nu(x),\quad \xi\in\mathbb R^3 \tag1$$ i.e., the Fourier transform of $\nu$. Since the integrand depends only on $\xi\cdot x$, the integral (1) can be written in terms of the projection of $\nu$ to the line with direction $\xi$ (i.e., pushforward of $\nu$ under the projection map). By assumption, this pushforward is the linear Lebesgue measure restricted to the interval $[-A/(2\alpha),A/(2\alpha)]$ and multiplied by $\alpha$.

Consider the sphere with radius $A/(2\alpha)$ centered at the origin. Let $\mu$ be a multiple of its surface measure, normalized by $\mu(\mathbb R^3)=A$. Then the projection of $\mu$ to every line is the same as the corresponding projection of $\nu$. By the above, $\widehat \mu=\widehat \nu$. Since the Fourier transform is injective, it follows that $\mu=\nu$ and thus $S$ is the sphere. (And $\alpha=\sqrt{\pi A}$, by the way.) $\quad\Box$

Remark. We don't need any assumptions on the topology: this is a fact about measures. It could be stated as: a Radon measure in $\mathbb R^3$ for which every linear projection is a fixed multiple of Lebesgue measure on some interval is a multiple of the surface measure of a sphere. (I don't know if fixed can be removed, i.e., if we can allow the coefficient to depend on direction of projection.)

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Nice! I'd overlooked the key observation that projecting $S$ onto a line through the origin doesn't shift the center of mass. – Andrew D. Hwang Jul 24 '13 at 13:39
By the way, the first proof works in all dimensions: it shows that any closed hypersurface with the described property must be a sphere. Since in dimensions $n\ne 3$ the sphere does not have this property, no hypersurface does. – 40 votes Jul 24 '13 at 15:38
Nice, thanks!!! I had a feeling this didn't need to go to the "research" level like on MathOverflow -- it's nice to see such short and simple proofs, even if the required concepts are a bit advanced. – user2566092 Jul 24 '13 at 19:39
@40 votes Also nice to know per your comment, that this is something not only special about the sphere but also the specific dimension. Seems crazy that only one possible dimension and only one surface of codim = 1 in that dimension has the property. – user2566092 Jul 24 '13 at 19:45

The claim is true for a closed surface of rotation; the proof is a nice application of the area formula for a surface of rotation, the fundamental theorem of calculus, and solving ODEs by separation of variables (all first-year calculus tools).

Offhand, the general case looks Difficult.

Edited to add: Here's a sketch for surfaces of rotation. For brevity, let's say $S$ has constant-area zones if the area of the portion of $S$ between two parallel planes that intersect $S$ depends only on the distance between the planes.

Assume $S$ is swept out by revolving the graph $y = f(x)$ of a positive, $C^1$ function about the $x$-axis. The area of the zone $a \leq x \leq b$ is, with $ds$ denoting the arc length element of the profile curve, $$A(a, b) = 2\pi \int_a^b y\, ds = 2\pi \int_a^b f(x) \sqrt{1 + f'(x)^2}\, dx.$$ Consequently, $S$ has constant area zones if and only if there is a real number $C > 0$ such that $A(a, b) = 2\pi C(b - a)$ for all $a$ and $b$ in the domain of $f$.

Fixing $a$ and differentiating with respect to $b$ gives $$f(b)\sqrt{1 + f'(b)^2} = C.$$ In particular, $f(b) \leq C$ for all $b$, since the radical is no smaller than $1$.

If $f(b) = C$ for all $b$ in some interval, the corresponding portion of $S$ is a cylinder. Otherwise we may assume $f(b) < C$ in some interval. Rearranging the preceding equation gives $$\frac{f(b) f'(b)}{\sqrt{C^2 - f(b)^2}} = 1,$$ which is easily integrated to $f(b) = \sqrt{C^2 - b^2}$ (up to translation in $b$), and the corresponding portion of $S$ is a zone of a sphere of radius $C$. (Side comment: Note the non-uniqueness of the solution under the initial condition $f(b_0) = C$!)

Combining all this, if $S$ is a surface of rotation having constant-area zones (with respect to planes orthogonal to the axis of rotation), then $S$ is either an infinite cylinder, a half-infinite cylinder with a hemispherical end cap, a bounded cylinder with two hemispherical end caps, or a sphere.

Among these surfaces, only the sphere has the requested (stronger) property of having constant-area zones with respect to arbitrary parallel planes.

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Cool, I'll +1 this if either you give a proof sketch (I'm well-versed in all the math machinery you mentioned, so I should be able to apply the appropriate tools where needed to fill in details of a brief proof sketch), or if I figure out the proof on my own, or enough other people +1 to indicate it's correct. – user2566092 Jul 23 '13 at 20:43
Looks like one of your coveted upvotes is in order, OP! – The Chaz 2.0 Jul 23 '13 at 22:05
cool thanks, +1 =) – user2566092 Jul 23 '13 at 22:12

Slightly too long for a comment. Here is some progress (I hope!):

Let $S$ be the surface which satisfies the properties in the OP, let $a$ and $b$ be any two points of $S$ which are of maximum distance apart and let this distance be $d$. Take any point $p$ (doesn't have to be in $S$), any line $\ell$ through $p$, and let $T_1$ and $T_2$ be two planes with normal $\ell$ starting off ``at infinity'' and slide them toward each other along $\ell$. Stop sliding them as soon as they hit at least one point in $S$. Then by simple rotation of the planes comparing them to $a$ and $b$, one can show:

(i) They each hit exactly one point;

(ii) The two points they hit are of maximal distance $d$ apart; and

(iii) The planes $T_1$ and $T_2$ are the tangent planes of these points.

It seems unfathomable anything with such a property can not be a sphere, but I haven't been able to show that yet. One can imagine there are points which are not obtainable by this process (think of a sphere with a dent in one side, although this is ruled out by (ii)).

Edit: Fix any point $p$ in $\mathbb{R}^3$ and define a map $\varphi : S^2 \rightarrow \mathbb{R}^3$ as follows. Identifying $p$ with the center of $S^2$, a point $s \in S^2$ determines a line through $p$ along with an orientation ("up" is in the direction of $s$). Sliding a plane normal to this line down "from infinity", stopping when we hit a (unique!) point in $S$ gives the point in $\mathbb{R}^3$, which we define as being $\varphi(s)$.

Let $s \in S^2$. By considering the tangent plane of $S$ at the point $\varphi(s)$, one can easily see that the map $\varphi$ is continuous. Indeed, one can consider a function on $S$ sending a point to its distance from the tangent plane. Considering this function on the closed region of $S$ which is away from an $\varepsilon$ open neighborhood of $\varphi(s)$, it attains a minimum. If this minimum is zero, then the tangent plane hits in two points ($\varphi(s)$ and wherever the minimum occurs) and this contradicts (iii) above. Therefore we may find a $\delta > 0$ such that altering the point $s \in S^2$ "by $\delta$" will leave $\varphi(s)$ inside this $\varepsilon$ neighborhood.

Let $S_0 = \varphi(S^2)$. Then this set has the following properties:

(i) $S_0$ is connected; and

(ii) There exists a constant $d$ such that for every point $a \in S_0$, the max over $|a-b|$ for $b \in S$ is always $d$.

I would like to guess that these two properties force $S_0$ to be a subset of a sphere. It would then follow that $S_0 = S$ and, because $\varphi$ sends a point of $S^2$ to the point with the same tangent plane, that $\varphi$ is the identity map (scaled to the appropriate radius).

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This is a nice idea! Your idea shows that any such surface $S$ must have constant width in the following sense: For every line $\ell$ in $\mathbf{R}^3$, the image of orthogonal projection of $S$ onto $\ell$ has the same length. That's not of itself enough to show $S$ is a sphere, of course, but I'm starting to suspect my offhand impression was too pessimistic. – Andrew D. Hwang Jul 23 '13 at 12:25
Incidental to the original question, but on the subject of solids of constant width:… – Andrew D. Hwang Jul 23 '13 at 14:29

You should see the paper "Kim, Dong-Soo; Kim, Young Ho Some characterizations of spheres and elliptic paraboloids. Linear Algebra Appl. 437 (2012), no. 1, 113–120. "

It seems that this questions is due to Blaschke, indeed, the more intresting case is considering only a fixed distance d between the parallel planes.

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A useful class of compact convex sets with non-empty interior called symmetric convex bodies can give a geometric example of this. A convex body $K \subset \mathbb{R}^n$ is symmetric if $k \in K \Leftrightarrow -k \in K$, meaning that antipodal points can be used as a sort of basis for computing the surface area of the region of $K$ that you describe. The answer by Andrew also works, but convex bodies are not necessarily generated as a surface of rotation, so this is a more general class of compact convex sets you are interested in.

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Can you give an example of a symmetric convex body in $\mathbf{R}^3$ (not a solid ball) whose boundary has the property requested? That would be interesting. :) – Andrew D. Hwang Jul 23 '13 at 12:14
I believe you are saying that if K is a symmetric convex body (after being centered at the origin), and the boundary of K satisfies the "equal area slices" property I stated, then K must be a ball. Right? Then I agree, this class includes additional surfaces besides surfaces of rotation (e.g. a surface of a box). Can you provide a bit more of a proof outline? I'd like to +1 this but I don't understand from your brief overview how a proof would go. – user2566092 Jul 23 '13 at 20:40

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