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For each positive integer n let E(n) denote n-dimensional Euclidean space and let the term "n-dimensional convex body" mean a compact convex subset of E(n) whose interior (with respect to E(n)) is non-empty. It is easy to see that all orthogonal projections (onto a plane) of a 3-dimensional convex body are 2-dimensional convex bodies. Among the responses to mathoverflow.net question No. 39127 about the shape of the earth was a very nice proof of the following statement. "If B is a 3-dimensional convex body all of whose orthogonal projections are closed disks of the same diameter, then B is the 3-dimensional closed ball which has that diameter". Is the following generalization of this statement still true? "If B is a 3-dimensional convex body all of whose orthogonal projections are pairwise mutually congruent, then B is the 3-dimensional closed ball which has the same diameter as its images have". This would be a nice theorem if it were true but I suspect that it might not be.

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2 Answers 2

up vote 8 down vote accepted

I think this is true. A proof would go like this:

First prove that the body must be an ellipsoid.

Without loss of generality you may assume that your body $B$ contains the origin as an interior point. Consider now the euclidean unit sphere $x^2 + y^2 + z^2 = 1$ and on each tangent plane consider the orthogonal projection of $B$ to the tangent space. This gives you a continuous Finsler metric on the two-sphere, but I'm not going to use that. It's just that the proof below also proves that if all the unit tangents discs of a Finsler metric on the two-sphere are affinely equivalent, then all the tangent discs must be ellipsoids (i.e. the metric must be a Randers metric in Finsler lingo).

For each point $x$ in the sphere consider the set $S_x$ of all euclidean isometries between the tangent space at the North pole $n$ and the tangent space at $x$ that send the projection of $B$ onto $T_n S^2$ to the projection of $B$ onto $T_xS^2$. By hypothesis, this set is non empty and if the projections are not ellipses this set is finite. In fact, if the projections are not ellipses, the union of the $S_x$ as $x$ ranges over the sphere is a finite covering of the sphere. This covering must be a finite disjoint union of spheres and therefore it admits a continuous section $s$. In turn, this means that we have a continuous way to choose an isometry from $T_n S^2$ to $T_x S^2$ for every $x \in S^2$ and we can use this to construct a nowhere zero vector field on the sphere, which is impossible. Hence, all projections are ellipses and it is well-known (Blaschke?) that if al projections are ellipses, the body must be an ellipsoid.

Now that you know that the body is an ellipsoid, you can conclude that the only way all projections are congruent is if its three semi-axis are all equal.

Addendum. There is a little gap in the proof above: since the section $s$ assings to every point $x$ an euclidean (affine) transformation between the tangent plane at $n$ and the tangent plane at $x$, it may send a non-zero vector to zero. However, this is easily fixed as follows: redefine the Finsler metric by letting the unit tangent disc at $x$ be the projection of $B$ translated so that its center of mass is at zero. The euclidean transformations between tangent spaces sending one projection to another will now be linear.

Idea behind the proof. The basic hypothesis in the problem boils down to saying that there is a distinguished class of linear transformations between pairs of tangent spaces to the $2$-sphere. Assume that the case where this class of transformations is infinite is completely understood (in this case this happens if and only if the projections are euclidean discs. I took ellipses by mistake---another gap---, because I was thinking about the more general result with affine transformations instead of congruences). Also assume that when the class of transformations is not infinite, they are a fixed finite number that does not depend on the pair of subspaces. In this case, we can construct a covering of the sphere and arrive at a contradiction. This is all very rough, but as I mentioned in the comment, this idea is a baby version of Gromov's solution to almost every case of a conjecture of Banach. I recommend looking at that paper. It's a beautiful application of the topology of vector bundles to convex geometry.

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Many thanks for your answer. I hope that my limited background in metric and convex geometry as well as the books that I have at my disposal will enable me to achieve a sufficient understanding of your proof. Does this result appear in the literature or is it possible that it might be new? –  Garabed Gulbenkian Jul 27 '13 at 18:18
I don't think the result appears as such in the literature, but the proof boils down to a basic principle I learned from what is possible Misha Gromov's first paper: On a geometrical hypothesis of Banach, Izv. Akad. Nauk SSSR 31 (1967), 1105-1114. I'll describe the principle in a modified form for the two-sphere so that the idea of the proof becomes clear. –  alvarezpaiva Jul 27 '13 at 21:01

The answer to this question is easily obtained as a Corollary of the main result in the paper "SO(2)-CONGRUENT PROJECTIONS OF CONVEX BODIES WITH ROTATION ABOUT THE ORIGIN" by BENJAMIN MACKEY (arXiv:1309.7101).

Theorem 1. Let K, L ⊂ R^3 be convex bodies containing the origin as an interior point such that for every ξ ∈ S^2, the projection K|ξ⊥ can be rotated about the origin into L|ξ⊥ . Then either K = L or K can be obtained by reflecting L about the origin.

The proof of the theorem in the paper is not easy, however, we can apply the theorem in the following way: apply a rigid transformation to B and let B´ the transformed body, by the theorem we have that either B´ is a translation of B or a reflection of B about the origin. Since this is true for every rigid motion, then B must be a Euclidean ball.

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