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In 1970 or so, Klee asked if a convex body in $\mathbb R^n$ ($n\ge 3$) whose maximal sections by hyperplanes in all directions have the same volume must be a ball. The counterexample in $\mathbb R^4$ is trivial and can be described as follows:

Let $f:[-1,1]\to\mathbb R$ be continuous, strictly concave and satisfy $f(-1)=f(1)=0$. For every such function, let $Q_f$ be the body of revolution given by $y^2+z^2+t^2\le f(x)^2$. Then $Q_f$ and $Q_g$ have the same maximal sections in every direction if (and, actually, only if) $f$ and $g$ are equimeasurable, i.e., $|\{f>t\}|=|\{g>t\}|$ for all $t>0$. (Of course, there are plenty of concave functions equimeasurable with $\sqrt{1-x^2}$).

With some extra work, one can construct something like this in $\mathbb R^n$ when $n$ is even though I do not know any similarly nice geometric description of such bodies for $n\ge 6$.

What I (and my co-authors) are currently stuck with is the case of odd $n$ (say, the usual space $n=3$). In view of such simple example in $\mathbb R^4$, I suspect that we are just having a mental block. Can anybody help us out?

fedja
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  • what about $n=2$? – Will Jagy Jul 15 '11 at 02:14
  • Planar domains of constant width have been well-known for a long time (normally one thinks of them as having constant projections but all their projections are realized as sections as well). – fedja Jul 15 '11 at 02:42
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    http://en.wikipedia.org/wiki/Curve_of_constant_width –  Jul 15 '11 at 02:43
  • That's right. I knew about curves of constant width, I did not connect that properly with your question. – Will Jagy Jul 15 '11 at 02:46
  • Can this help (Meissner bodies): http://en.wikipedia.org/wiki/Meissner%27s_tetrahedron#Meissner_bodies ? –  Jul 15 '11 at 03:01
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    Not really :(. The bodies of constant width (=constant maximal 1-dimensional sections) are not hard to construct because basically you just have to relate pairs of points. When you raise the section dimension, you have to deal with integral transforms. I suspect that this simple R^4 construction went unnoticed because the customary way is to represent convex bodies by their radial or support functions even for the bodies of revolution and the equimeasurability condition is a total mess in such terms. So, it seems like we need a fresh look with some little twist... – fedja Jul 15 '11 at 03:29

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To those who are still interested: we've finally made it but it's so ugly that a nice alternative approach will be always welcome :) We are still stuck with Bonnesen's question about the possibility to recover a convex body from the volumes of its maximal sections and projections in odd dimensions, so some help will be appreciated. The even-dimensional case can be found here. I feel a bit like a student asking for help with his homework, of course, but why not? We all get stuck now and then :). This should really be a comment but it's too long to fit the number of characters restriction.

fedja
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  • More affirmation of Klee's prowess as a problem-poser! May I ask: Is the essence of your proof in odd dimensions concentrated (in the intellectual sense) in $\mathbb{R}^3$? – Joseph O'Rourke Jan 04 '12 at 02:24
  • It doesn't matter for us whether it is 3,5,7, or any other odd number: there is no simplification or complication when going from one odd dimension $d>1$ to another, if that's what you meant. However, as I said, the odd dimension case is harder than the even one due to the square root popping up everywhere and the non-locality of the Radon transform. – fedja Jan 04 '12 at 12:23