Shadows

Last week we mentioned the Busemann-Petty problem, which is a question about recovering data from shadows of an object. This sort of inverse problem is very important in mathematics (it pops up in various unexpected guises, for instance in combinatorics) as well as in real life: this is how an MRI works. I want to draw attention to a fairly natural question I found on MO, with a solution which I believe is truly from The Book. 

Various observations from antiquity led to people believing in a (roughly) spherical earth, but the ones pointed out in the MO question beg the question: is it true that the sphere is the only surface all of whose projections are disks?

The corresponding statement in two dimensions is not true. The Releaux triangle is a figure of constant
width, meaning that every projection of it in the plane is a line segment of the same length. (We discussed this in the first blog post!) There are also surfaces of constant width in higher dimensions, meaning that any two parallel bounding set of hyperplanes (touching the boundary) have constant separation. 

Higher dimensional versions of such things exist. A convex body $K\subset\mathbb{R}^n$ all of whose $(n-1)$-dimensional projections have the same $n-1$-content is known as a body of constant brightness, by analogy with bodies of constant width. The theory is very similar to that of bodies of constant width. The surface area measure $dS_K(\mathbf{x})$ takes the place of the support function $h_K$. The brightness in the direction $\mathbf{u}$ is given by $V(K_\mathbf{u}) = \tfrac{1}{2}\int |\langle\mathbf{x},\mathbf{u}\rangle| dS_K(\mathbf{x})$. If we expand $S_K$ in spherical harmonics $\sum_{l,m}s_{l,m}Y_{l,m}$, we get that $V(K_\mathbf{u}) = \sum_{l,m} c_l s_{l,m} Y_{l,m}(\mathbf{u})$, and $c_l=0$ whenever $l$ is odd. Therefore, $K$ has constant brightness if and only if $\frac{S_K(U)+S_K(-U)}{|U|}$ is constant over all Borel sets $U\subset S^{n-1}$ (that is, apart from a constant term, $S_K$ is antisymmetric). From the existence theorem for the Minkowski problem, we can easily construct examples which aren't spheres.

Note also that finitely many circular projections is insufficient, since intersecting finitely many cylinders  would produce a surface having corners and containing some straight line segments.

The fact that you can spin such a surface with all circular projections inside any bounding cylinder is suggestive, but it is also true that you can spin the Reuleaux triangle inside a square, even though it isn't circular.

Now here is the magic proof that the given property characterises the (2-) sphere:

We make some common sense assumptions such as a surface being the boundary of a convex set. In fact, even if one only has access to horizontal projections (i.e. projections to vertical planes), which is roughly what happens in the Earth-Moon case, this is true. One observes that there is a unique north pole N and a south pole S that project to the highest and lowest point respectively in any such projection, and that NS has to be vertical. Let AB be the longest segment with endpoints on the surface. We may assume that its length equals 2 and its midpoint is the origin. Consider projections to the planes that contain AB. Since projections do not increase distances, AB is a diameter of each projection. Hence all projections to this family of planes are unit discs centred at the origin. The intersection of the corresponding cylinders is the unit ball, hence the result. $\blacksquare$

In general, we cannot determine a convex body from the set of shadows (if we don't know the correspondence between shadows and directions of projections).

For example, take a unit ball and cut off three identical tiny caps whose centres form a regular triangle on the sphere and are not on one great circle. Looking at shadows, you cannot tell whether all three or only two caps are removed, because each projection shows you no more than two of them. The same construction works for polyhedra if you start with an icosahedron rather than a ball.

The original MO post asks further intriguing questions apart from the titular one. Thurston's answer also provides a lot of insight, and I am still trying to decipher it. Go check it out!

P.S. I wanted to add a category tag to this but couldn't think of anything so eventually I copied the MO category of mg