# Convex bodies with symmetric shadows

Theorem. If all orthogonal projections of a convex body $K \subset \mathbb{R}^n$ onto $2$-dimensional subspaces have a center of symmetry, then $K$ has a center of symmetry.

This is a classic result of Blaschke and Hessenberg (that I just learned thanks to Guillaume's comment.). A short simple proof of it can be found in Bonnesen and Fenchel.

I wonder if it is necessary to know what happens for every orthogonal projection or whether we can get by with less:

Question 1. Let $K \subset \mathbb{C}^{n}$ be a convex body. Assume all orthogonal projections of $K$ onto complex lines have a center of symmetry. Does it follow that $K$ must also have a center of symmetry?

Note. The center of symmetry of the shadows may depend on the subspace containing it.

A similar question is:

Question 2. Let $K \subset \mathbb{C}^{n}$ be a convex body. Assume all orthogonal projections of $K$ onto Lagrangian subspaces have a center of symmetry. Does it follow that $K$ must also have a center of symmetry?

• For the first question, yes. This is a theorem due to Blaschke and Hessenberg, see Theorem 2 in C.A. Rogers, "Sections and projections of convex bodies", purl.pt/2464/1/j-5293-b-vol24-fasc2-art3_PDF/… – Guillaume Aubrun Jan 22 '13 at 14:52
• You're right. Thanks!! The paper you mention seems to use the fact that all projections have the same center, but this is not necessary. I am editing my question to reflect your input. – alvarezpaiva Jan 22 '13 at 15:20

The answers are no to Question 1, and yes to Question 2 (assuming $$n\ge 2$$).
Let $$h$$ be the support function of $$K$$. The projection of $$K$$ to a linear subspace $$L$$ is central symmetric iff the restriction of $$h$$ to $$L$$ is a sum of an even function and a linear function. Here and below "linear" means $$\mathbb R$$-linear.
To construct a counter example to Question 1, begin with the unit ball and its support function $$h(x)=|x|$$. Then, on each complex line, pick a linear function on this line, in such a way that the coefficients of these linear functions vary smoothly from one complex line to another. Let $$f$$ be the union of these linear functions. It is well-defined because the complex lines are disjoint outside 0. The functions above should be chosen so that $$f$$ itself is not linear. Now there is $$\varepsilon>0$$ so small that $$h_1:=h+\varepsilon f$$ is still convex (just because its first and second derivatives are close to the original ones). Then $$h_1$$ is a support function of some convex body, which is a desired counter-example.
In Question 2, we have a function $$f(x)=h(x)-h(-x)$$ which is linear on every Lagrangian subspace. Equivalently, $$f$$ is 1-homogeneous and $$f(x+y)=f(x)+f(y)$$ for all $$\omega$$-orthogonal pairs $$x,y\in\mathbb C^n$$ where $$\omega$$ is the symplectic form. We have to show that $$f$$ is globally linear, i.e., $$f(x+y)=f(x)+f(y)$$ for all $$x,y\in\mathbb C^n$$.
Fix $$x,y\in\mathbb C^n$$ and pick $$u,v$$ from the $$\omega$$-orthogonal complement of $$x,y$$ such that $$\omega(u,v)=\omega(x,y)$$. From linearity on Lagrangian subspaces, $$f(x+y+u+v) = f(x+y)+f(u+v)$$ since $$\omega(x+y,u+v)=0$$, and on the other hand, $$f(x+y+u+v) = f(x+v)+f(y+u) = f(x)+f(y)+f(u)+f(v)$$ since $$\omega(x+v,y+u)=\omega(x,y)+\omega(v,u)=0$$ and $$\omega(x,v)=\omega(y,u)=0$$. Thus $$f(x+y)-f(x)-f(y) = f(u)+f(v)-f(u+v) .$$ The same identity holds for $$-u$$ and $$-v$$ in place of $$u$$ and $$v$$, therefore $$f(x+y)-f(x)-f(y)=0$$, q.e.d.