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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?

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    $\begingroup$ 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/… $\endgroup$ – Guillaume Aubrun Jan 22 '13 at 14:52
  • $\begingroup$ 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. $\endgroup$ – alvarezpaiva Jan 22 '13 at 15:20
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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.

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