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Given a convex surface $S$ in $\mathbb{R}^3,$ you can associate to each point the length of the inward pointing normal contained inside $S.$ Call this quantity $s(x).$ Now, given a positive function $f$ on the sphere $S^2,$ are necessary and sufficient conditions known so that $f$ can be realized as $s$ for some convex realization? There is the obvious uniqueness question, as well. The question can be formulated identically in $\mathbb{R}^n,$ but these sort of questions tend to get harder as $n$ gets larger. On the other hand, it is not obvious what the answer is even for $n=2.$

EDIT There seem to be confusion about what I mean by "the length of inward pointing normal": Take the point $x.$ There is a line $l$ through $x$ normal to $S.$ Since $S$ is convex, $l$ intersects $S$ twice. The distance between the two intersection points is $s(x).$

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Could you clarify what you mean by "the length of the inward pointing normal"? –  Rbega Dec 19 '11 at 22:01
    
Definitely not enough information for uniqueness. –  Anton Petrunin Dec 19 '11 at 23:11
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I assume that one can say that one function on $S$ is less information than the distance function (which is a function on $S \times S,$ so your (@Anton's) statement is plausible, but a proof would be nice. –  Igor Rivin Dec 19 '11 at 23:17
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I think non-uniqueness is rather clear even for $n=2$. Let $f(x)=x+s(x)n(x)$ be the "second intersection point" map (here $n(x)$ is the unit inward normal). Look at the special case $s(x)=const$. In this case the map $f(x)$ is an involution by Frenet formulas. For example, take $s\equiv 2$ and look at a unit semicircle. Flatten it a bit away from endpoints (there is a lot of freedom to do that) and extend this curve to be a closed curve using the map $f$. The curve will remain convex so long as the curvature of the flattened semicircle is close to 1 and will have $s\equiv 2$. –  Vitali Kapovitch Dec 20 '11 at 3:55
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Isn't a curve of constant width another counterexample for uniqueness? –  Dirk Dec 20 '11 at 8:12

3 Answers 3

To follow up on Dirk's observation in the comments, here is a smoothed version of a Reuleaux triangle with $s(x)=c$, as illustrated by the dashed normal chords, which each pass through a corner of the equilateral triangle, on which are centered both the red and the green arcs:
       Smooth Reuleaux Triangle
(If curves with tangent discontinuities are permitted, then already a square has $s(x)=c$.) Of course, the circle also has $s(x)=c$.

For higher dimensions, see the MO question, "Are there smooth bodies of constant width?" (the answer is: Yes).

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To get necessary conditions for $f$ to be a $s$-function for smooth convex sets $C$ in the plane, you can write that given a point $M(x)$ on $C$ the point $P(x):=M(x)+f(x)n(x)$ is on $C$ as well (where $n(x)$ denotes the inward pointaing normal). Then taking two derivatives you can express a link between the curvatures at $M(x)$, at $P(x)$ and derivatives of $f$. Convexity insures nonnegativity of curvatures and should yield some differential inequality on $f$ and its derivatives. You can also use Gauss-Bonnet formula for curves... By the way, it reminds me an interesting paper by Nicolka Fusco and Aldo Pratelli entitled "On a conjecture by Auerbach" where they study convex sets with constant width with this kind of approach...

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Dear Igor,

Since the answer to your problem is "no" I'm not going to dwell on that, but I feel there is something interesting behind your question that does not admit such a clear cut answer.

Two known facts:

1. Consider the manifold of oriented lines in three-space and, in particular, the open set $U$ formed by the lines that pass through the interior of your convex body. If you use the symplectic structure on the space of oriented lines in Euclidean space, then this open set is symplectomorphic to the open unit co-disc bundle of your surface. In other words, you can read from $U$ not only the area of your surface (Cauchy-Crofton formula), but also everything related to its geodesic flow.

2. The set of oriented lines normal to your surface is a Lagrangian submanifold that lies inside $U$.

Now consider the John transform (X-ray transform) of the characteristic function of your convex body. Its value at a given line $\ell$ is the length of the segment of $\ell$ that lies in the convex body. In other words, the distance between the entrance point and the exit point.

Your problem can now be written as follows : Restrict the John transform of the characteristic function of a convex body to the Lagrangian submanifold formed by lines normal to its boundary, is this knowledge sufficient to reconstruct the body (i.e., the characteristic function)? This is now an inversion problem in integral geomety.

Generally speaking, one cannot reconstruct the John transform of a function (of three variables) from its restriction to a two-dimensional submanifold on the space of lines, but you can if you take certain three-dimensional submanifolds in the space of lines. Allow me to change the question:

Is there a geometrically-defined three-dimensional submanifold $\Sigma$ of the set of all oriented lines intersecting the interior of a convex body $K$ such that the John transform of $1_K$ can be reconstructed from its restriction to $\Sigma$?

In two dimensions the analogous question has a trivial answer: you need the two-dimensional submanifold of all lines passing through the convex body because of the inversibility of the Radon transform. In higher dimensions, the question looks interesting.

Going back to your original question: one may argue that the characteristic function $1_K$ is a very special function of three variables and that it "feels" like a function of two variables. So the question remains whether one can reconstruct $1_K$ from the knowledge of its John transform on some two-dimensional submanifold (or on some sort of "smallish set") on the space of lines.

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Thanks! I had an integral-geometric interpretation in mind, but certainly nothing as well-thought out as what you suggest. I will ponder... –  Igor Rivin Feb 12 '12 at 2:24

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