The bounded complex of a polyhedral decomposition

Question

Let $\mathscr{P}$ be a polyhedral decomposition of a real vector space $V$. By that I mean that $\mathscr{P}$ is a finite set of polyhedra in $V$ satisfying the following three properties:

1. The union of the polyhedra in $\mathscr{P}$ is equal to $V$.
2. If $P\in \mathscr{P}$ and $Q$ is a face of $P$, then $Q\in\mathcal{P}$.
3. If $P,Q\in\mathscr{P}$, then $P\cap Q$ is a (possibly empty) face of both $P$ and $Q$.

Question: Is the union of the bounded elements of $\mathscr{P}$ contractible?

If $\mathscr{P}$ consists of the set of faces of an essential hyperplane arrangement, then a positive answer is stated in Exercise 1.7.a of Stanley's lecture notes https://www.cis.upenn.edu/~cis6100/sp06stanley.pdf, though I don't know the reference for a proof. Note that, when $\mathscr{P}$ does not come from a hyperplane arrangement, then the bounded complex need not be pure.

Answer 1

Below I construct a deformation retract of $V$ onto the bounded sub-complex of $\mathscr P$, proving that the latter is contractible.

For a polyhedral complex $\mathscr P$ resp. a polyhedron $P\in\mathscr P$ I write $\mathscr P^b\subseteq\mathscr P$ resp. $P^b\subseteq P$ for its bounded sub-complex. Also, by a retract of a polyhedron $P$ (or polyhedral complex $\mathscr P$) I mean a strong deformation retract $\phi_P: P\times[0,1]\to P^b$ from $P$ onto $P^b\subseteq P$. Such clearly exist if $P$ is bounded (just use the identity) or if $P$ is a half-infinite edge. But below I explain how to define such for all polyhedra, in such a way that they are compatible on their boundaries and so can be glued together to form a retract for all of $\mathscr P$. The latter is therefore homotopy equivalent to $\bigcup_{P\in\mathscr P} P=V$, hence contractible.

Proof.

• Each $P\in\mathscr P$ can be written as $P=P^b+C_P$, where $C_P$ is the recession cone of $P$. Fix a choice of generalized barycentric coordinates (e.g. mean value coordinates, Wachspress coordinates, ...), i.e. for each point $x\in P$ we have a canonical decomposition $x=x_b+x_C$ with $x_b\in P^b$ and $x_C\in C_P$ that only depends on the shape of $P$ and the location of $x$ relative to it (see also below). Then set $\phi_P(x,t):=x_b+(1-t)x_C$.

• For $P,Q\in\mathscr P$ the above definition ensures that $\phi_P(x,t)=\phi_Q(x,t)$ for all $x\in P\cap Q$, so that we can extend the retract to all of $\mathscr P$.

$\square$

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Some notes on "the canonical choice"

As aluded to in the comments, it is sufficient to construct a section $s_P:P\to P^b\times C_P$ for the projection

$$\pi_P:P^b\times C_P\to P,(x_b,x_C)\mapsto x_b+x_C,$$

(i.e., $\pi_P\circ s_P=\mathrm{id}_P$) that matches with prescribed sections $s_\sigma:\sigma\to\sigma^b\times C_\sigma$ on the faces $\sigma\subseteq P$. Here is one idea for how to do this.

The recession cones of polyhedra in $\mathscr P$ form a full fan $\mathscr F:=\{C_P\mid P\in\mathscr P\}$ of $V$. For each cone $C_P\in\mathscr F$ choose a ray $r_P\subseteq C_P$. Now, for a point $x\in P$ let $x_b'$ be the intersection of $\partial P$ with the translated and inverted ray $-r_P+x$ (which exists since the linearity space of $C_P$ is empty). In particular, $x_b'$ is in some face $\sigma\subseteq \partial P$. Then set $x_b:=s_\sigma(x_b')$ and $x_C:= x-x_b$.