A convex polyhedral analog of the pentagram map - MathOverflow

A convex polyhedral analog of the pentagram map

2011-06-21T17:22:38Z

I am wondering if there is a three-dimensional analog of the pentagram map, which maps a convex polygon to another convex polygon. Here's the Wikipedia image:

I am seeking something similar that maps a convex polyhedron to another convex polyhedron (and perhaps works in arbitrary dimensions).

You may know that the pentagram map has been heavily studied, e.g., by Valentin Ovsienko, Richard Schwartz, and Serge Tabachnikov:

"The Pentagram Map: A Discrete Integrable System," Communications in Mathematical Physics Volume 299, Number 2, 409-446.

I would be content with a geometric analog, without duplicating the nice moduli-space properties enjoyed by the pentagram map. In particular, I have a definition in mind, and would like to know if it, or something similar, has been explored in the literature.

Let $P$ be a convex polyhedron in $\mathbb{R}^3$. Let $N(v)$ be the set of vertices of $P$ adjacent to vertex $v$ by an edge (of the 1-skeleton) of $P$. Define $H(v)$ to be the convex hull of $N(v)$, and say a face $f$ of $H(v)$ is an upper face if it is visible from $v$ (in the sense that a ray from $v$ hits $f$ before any other face of $H(v)$). Lastly, define $H^+(v)$ to be the intersection of all the halfspaces including $H(v)$ determined by the upper faces of $H(v)$.

If the points in $N(v)$ are coplanar, $H(v)$ could be considered a doubly covered polyhedron, in which case $H^+(v)$ is just the halfspace (away from $v$) determined by its "top" side.

The map then takes $P$ to $\bigcap_i H^+(v_i)$ for all vertices $v_i$ of $P$.

For example, if $P$ is a cube, $N(v)$ is three points and $H^+(v)$ is the halfspace determined by that triangle, excluding $v$. Thus $P$ gets mapped to an octahedron:

Note that this 3D map is an analog in a sense of the 2D pentagram map, for in the latter $N(v)$ is the pair of vertices adjacent to $v$, and $H^+(v)$ is the halfplane determined by the diagonal connecting them.

Unfortunately the map is only easy to construct "by hand" for regular polyhedra, where it is relatively uninspiring; otherwise I'd include a more interesting example.

Thanks for any pointers to related literature!

Answer by André Henriques for A convex polyhedral analog of the pentagram map

2011-06-21T17:48:04Z

In four dimensions, the convex hull of the points $$ \{(\pm 1,\pm 1,\pm 1,\pm 1)|\text{there is an even number of +'s}\} $$ is congruent to the convex hull of the points $$ \{(\pm 2,0,0,0),(0,\pm 2,0,0),(0,0,\pm 2,0),(0,0,0,\pm 2)\}. $$ Both of them are cross polytopes:

Now, given a 4-polytope that has the same combinatorial type as the 4-dimensional cross-polytope, you can do the following operation:

Take the dual polytope (which has the same combinatorial type as a 4-dimensional hypercube), and then keep only half of its 16 vertices. The convex hull of those 8 vertices has the combinatorial type of 4-dimensional cross-polytope.

Answer by jc for A convex polyhedral analog of the pentagram map

2011-08-25T17:02:57Z

In the preprint entitled "On generalizations of the pentagram map: discretizations of AGD flows" (cited in the now very detailed Wikipedia article rewritten by Richard Schwartz), Gloria Marí Beffa studies an analogue of the pentagram map in higher dimensional projective spaces and related them to integrable PDEs called "Adler-Gel'fand-Dickey flows". This doesn't address your question though because her construction is on polygons in $\mathbb{RP}^m$, not polyhedra. Nonetheless, you may still find the construction geometrically interesting.

The pentagram map and its generalizations are actually defined on a larger set of geometric objects than just polygons. Let $\{x_n\}$ be a map from the integers into projective space such that $x_{n+r}=Mx_n$ for some projective automorphism $M$. These are the positions of the points of an $r$-twisted polygon in $\mathbb{RP}^m$. If $M$ is the identity, then we just have a sequence which iterates over a $r$-gon repeatedly. As is usual, below we will only consider "sufficiently generic" points in the space of twisted polygons.

In the pentagram map, defined for $m=2$, one intersects adjacent diagonals to get a new point, i.e., $T(x_n)$ is the intersection of the segments $\overline{x_{n-1}x_{n+1}}$ and $\overline{x_{n}x_{n+2}}$. Here's fig.1 of Beffa's preprint:

Beffa considers a few different generalizations. The one I'll describe here, which gives the flavor of her constructions is as follows. One intersects the diagonal $\overline{x_{n-1}x_{n+1}}$ with an $m-1$ dimensional linear subspace $\Delta_m$. $\Delta_m$ is fixed by giving a set of distinct integers $k_i\neq\pm1$ for $i=1,\dots,m-1$. $\Delta_m$ is then the $m-1$ dimensional linear subspace specified by the $m$ points: $x_n$ and $x_{n+k_i}$ for $i=1,\dots,m-1$. Here's fig.2 of Beffa's preprint:

In the large $r$ limit, the pentagram map can be interpreted as a discretization of the Boussinesq equation. The Boussinesq equation is apparently one of the simplest of the AGD flows, and Beffa conjectures that a version of her construction with one $k-1$ dimensional subspace and $k-1$ $m-1$ dimensional subspaces is a discretization of a $k$-AGD flow. The generalization above is one case of this conjecture that she has proved.

Unfortunately, I don't know enough about integrable systems of PDEs or AGD flows to really describe the relations between the objects at stake here. Hopefully I didn't get things too wrong...

It seems to me that the key geometric feature of the pentagram map and its analogues are the fact that they commute with projective transformations (at least this is a feature that leads to these remarkable connections with integrable maps). Does your analog also have this property?

I don't know too much about polytopes in projective spaces, but they might be something to ponder. What is a half-space in a projective space, for one?