3
$\begingroup$

I am not sure this is a well framed question but I would like to know if anything like "taking the power" of a polytope is known.

Imagine this situation where I want to think of such a thing : say one is given the unit square in the all-positive orthant of $\mathbb{R}^2$. Is there a ``simple" (hopefully piecewise linear!) map which will map this square to the square of side $2$ with its vertices at $(-1,-1),(-1,1),(1,1),(1,-1)$.

Had one started with a quarter of a unit disk then one could have taken the map $z^4 : \mathbb{C} \rightarrow \mathbb{C}$ to get that to the unit disk about the origin. What is the closest analogue to this in the world of polytopes (and linear maps?)?

Improve this question
$\endgroup$
12
  • $\begingroup$ Isn't this just "dilation"? $\endgroup$
    – Sam Hopkins
    Commented Oct 18, 2016 at 17:40
  • $\begingroup$ Could you explicitly state which map you think does this? Also I used the square as an example. I am looking for something more general. $\endgroup$
    – gradstudent
    Commented Oct 18, 2016 at 17:45
  • $\begingroup$ The map that dilates space by a factor of two maps the unit square in the positive orthant to the square with vertices $(0,0),(2,0),(0,2),(2,2)$. This is equivalent to your square via translation, and generally people only study polytopes up to translation (or even, affine equivalence) anyways. Dilation is central to Ehrhart theory, one main topic in the study of polytopes (en.wikipedia.org/wiki/Ehrhart_polynomial). $\endgroup$
    – Sam Hopkins
    Commented Oct 18, 2016 at 17:52
  • $\begingroup$ By the way, the product $P \times Q$ of two polytopes is a well-studied concept: if $P$ is of dimension $p$ and $Q$ is of dimension $q$, then $P\times Q$ is of dimension $p+q$. So there is already a notion of the "power" of a polytope, but this is different than what you want here because it increases the dimension. $\endgroup$
    – Sam Hopkins
    Commented Oct 18, 2016 at 17:53
  • $\begingroup$ I don't want to take products. Think of what the map $z^n$ does as a map from the complex plane to itself - it makes $n$ copies of the initial region and stiches them from end to end. One might take a Riemann surface view of this too. Is there an analogue to this in the world of polytopes? $\endgroup$
    – gradstudent
    Commented Oct 18, 2016 at 17:54

1 Answer 1

Reset to default
2
$\begingroup$

Given any $k$ and any polyheda $\newcommand{\bR}{\mathbb{R}}$ $P, Q\subset\bR^n$ with nonempty interiors one can easily produce a $k:1$ covering $P\to Q$ branched over a codimension $2$-locus.

Here it goes. Fix homeomorphisms $F_P: P\to C^{n-2}\times D^2$, $F_Q: Q\to C^{n-2}\times D^2$, where $C^m$ is the cube $[-1,1]^m\subset\bR^n$ and $D^2$ is the unit disk in the plane.

Consider the canonical $k:1$ cover $S^1\to S^1$. Fix a triangulation $\newcommand{\eT}{\mathscr{T}}$ $\eT$ of $S^1$ and denote by $\tilde{\eT}$ its lift $p_k^{-1}(\eT)$. Then we can view $p_k$ as a $k:1$-simplicial covering. The triangulations $\eT$, $\tilde{\eT}$ of $S^1$ extend to triangulations of $D^2$, viewed as the cone over $S^1$. The map $p_k$ then induces a $k:1$ simplicial map $\beta_k:D^2\to D^2$ branched over the origin. Define

$$ \Psi_k: C^{n-2}\times D^2,\;\;\Psi_k(y,z)=(y,\beta_k(z)). $$

Then

$$ \Psi_k: F^{-1}_Q\circ\Phi_k\circ F_P: P\to Q $$

is a $k:1$ map branched over $F_P^{-1}(C^{n-2}\times 0)$.

How to emulate this in the piecewise affine world?

Think of $\eT$ as the triangulation of the boundary $\newcommand{\pa}{\partial}$ $\pa \Pi_3$ of the regular $3$-gon $\Pi_3$ with vertices on $S^1$ and $\tilde{\eT}$ as the triangulation defined by the vertices of a regular $3k$-gon $\Pi_{3k}$. Then $\beta_k$ can be viewed as a piecewise-affine $k:1$ covering map $\Pi_{3k}\to\Pi_3$, branched over the center of $\Pi_3$. Denote by $\Delta_m$ the $m$-simplex

$$ \{0\leq x_1\leq \cdots \leq x_m\leq 1\}\subset \bR^m. $$

Now fix piecewise-affine homeomorphisms

$$ F_P: P\to\Delta_{n-2}\times\Pi_{3k},\;\; F_Q: Q\to\Delta_{n-2}\times \Pi_{3}. $$

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks! Can you kindly let me know of a paper or book where I can read up more on this? $\endgroup$
    – gradstudent
    Commented Oct 18, 2016 at 20:31
  • $\begingroup$ It's easier to prove a more general result: if $P_1,P_2$ are two $n$-dimensional polytopes then there exists a piecewise affine homeomorphism $F: P_1\to P_2$. Here's the main idea. Triangulate the boundaries of $P_i$. Think of them as (geodesic) triangulations of $S^{n-1}$. Then find a geodesic triangulation that is finer than both of these triangulations. $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 18, 2016 at 21:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .