Define the shadow of a convex polyhedron $P$ in direction $u$ to be the orthogonal projection of $P$ onto a plane whose normal is $u$. The shadow is a convex $k$-gon. I am wondering to what degree $P$ is determined by knowing the value of $k$ for all the distinct shadows. Let me try to make precise what I mean by "the combinatorics of the shadows."

On a unit sphere $S$, mark all the directions parallel to a face of $P$. Equivalently, intersect $S$ with planes through its center parallel to each face of $P$. Here is the result for a regular tetrahedron:
Each cell $c$ of this arrangement of great circles corresponds to a distinct shadow in the sense that, for all $u \in c$, the shadow in direction $u$ is a $k$-gon, with $k$ fixed. In this example, the shadows for a cell are either triangles or quadrilaterals, 3-gons or 4-gons.

Now form a graph, the shadow graph $G_S$, the dual graph of the arrangement: each node is a cell, with an edge connecting cells that share a positive-length arc. Label each node with the integer $k \ge 3$ if that cell's shadows are $k$-gons. For the tetrahedron example, $G_S$ has 14 nodes, labeled as shown below:
Now imagine you are given the labeled $G_S$ as input. I would like to know to what extent $P$ is determined.

(1) Does this arrangement and/or its dual graph have a name in the literature? I've seen the term Gaussian sphere in the early computer vision literature, but that term is not prevalent. I feel I may be missing a key search phrase.

(2) Does $G_S$ determine $V$, $E$, and $F$, the number of vertices, edges, and faces of $P$, i.e., the $f$-vector?

(3) Given $G_S$, can you find some representative $P$ that realizes those combinatorics?

(4) More generally, what is the class of $P$ that realizes a given $G_S$?

(4a) A very specific version of this question is: Are all the $P$ that yield the $G_S$ illustrated above tetrahedra?

(5) The question generalizes to polytopes in $\mathbb{R}^d$, which is the root of my question.

Perhaps some of these questions are easier under genericity assumptions, e.g., no two faces are parallel. Edit. For example, under the assumption that no three faces are parallel to a line (which implies no two faces are parallel [thanks to Giovanni Viglietta for the correction]), $G_S$ has $m=F^2-F+2$ nodes for $F$ faces, so $F=\frac{1}{2}(1+\sqrt{ 4m -7})$, which partially solves (2) above.

I'd appreciate pointers to relevant literature!

  • $\begingroup$ See related question (at the end of) mathoverflow.net/questions/39127/… To what extent are surfaces determined by their projections? $\endgroup$ – Joel David Hamkins Feb 11 '11 at 13:38
  • $\begingroup$ I believe your sample graph has a mistake: for reasons of symmetry, the lower left and lower right edges should connect to the 4's on the two sides, not to the 3's. $\endgroup$ – Bill Thurston Feb 11 '11 at 14:11
  • $\begingroup$ @Bill: Whoops, you are correct! Fixed now. $\endgroup$ – Joseph O'Rourke Feb 11 '11 at 14:31
  • $\begingroup$ How do you define "shadow graph" if not all faces are triangles? $\endgroup$ – Anton Petrunin Feb 11 '11 at 18:16
  • $\begingroup$ @Anton: For example, for a cube, the sphere gets cut into eight octants, each becomes a node of degree three, and each cell of this arrangement gets labeled "6" because the shadows are hexagons. $\endgroup$ – Joseph O'Rourke Feb 11 '11 at 18:36

This doesn't directly answer your question, I fear, but it's at least tangentially related. Suppose we keep slightly more information than the combinatorics of $G_S$, and label each node as well with the area of the corresponding cell on the sphere. Let $A$ be the sum of the areas corresponding to nodes originally labeled "3" (where the projection is a triangle), and $T$ be the sum of the areas of all nodes. Then it's shown in "Angles as Probabilities" by David V. Feldman and Daniel A. Klain, American Mathematical Monthly, October 2009, that $2\pi A/T$ is the sum of the solid inner angles of the vertices of the original tetrahedron -- that is, you can interpret that angle sum as the probability that an orthogonal projection to a random plane is a triangle.


Here's another tangentially-related answer. There is a nice characterization of "equiprojective polyhedra" due to Masud Hasan and Anna Libiw. (A convex polyhedron is "equiprojective" if, for some $k$, the general shadow is a $k$-gon.)

  • $\begingroup$ Yes, thanks David! For example, a cube is 6-equiprojective: all the nodes would be labeled "6" in $G_S$. They recently proved that any equiprojective polyhedron has at least one pair of parallel faces. They have a new arXiv paper: arxiv.org/abs/1009.2252 $\endgroup$ – Joseph O'Rourke Feb 11 '11 at 18:05

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