4
$\begingroup$

Let $S$ be some subset of $\mathbb{R}^3$ and $\text{Conv}(S)$ be its convex hull. Say $S$ has property $\mathcal{P}$ if each point $x \notin S$ lies on some line contained in $\mathbb{R}^3 \setminus S$ such that there is a ruled surface in $\mathbb{R}^3 \setminus S$ giving a homotopy to a line in $\mathbb{R}^3 \setminus \text{Conv}(S)$. Intuitively you can think of this as saying that $S$ can be cut out on a bandsaw. Is there a name for this property or a simpler equivalent form?

My motivation for this is wanting to define the $\mathcal{P}$-hull of a subset $S \subset \mathbb{R}^3$ to be the smallest set containing $S$ that has property $\mathcal{P}$ (which is well-defined since the intersection of sets with property $\mathcal{P}$ has property $\mathcal{P}$). For example if $S$ is the wireframe of a compound of two tetrahedra as below, its convex hull is a cube, but its $\mathcal{P}$-hull is a stellated octahedron.

enter image description here

(I would also be interested in the stronger property where the ruled surface is replaced with a planar region between the lines. Intuitively the shapes with this property would be those that can be cut out on a bandsaw with only straight cuts.)

$\endgroup$
11
  • $\begingroup$ Apollonius (of Perga), of conic section fame, would be proud! $\endgroup$
    – Hudjefa
    Commented Mar 8 at 12:33
  • 1
    $\begingroup$ (+1) Just to note, at risk of mathematical pedantry: (i) A convex set such as a ball has property $\mathcal{P}$, but "requires infinitely many bandsaw cuts"; (ii) a cube "sliced partway through by a helicoid" has property $\mathcal{P}$, but cannot be cut by a physical bandsaw because a real blade is not a line; (iii) certain unbounded regions, such as the "outside" of a right angle between two half-planes can arguably be cut by a bandsaw, but the convex hull is all of space, so such a region does not have property $\mathcal{P}$ as currently stated. $\endgroup$
    – Andrew D. Hwang
    Commented Mar 8 at 15:01
  • 1
    $\begingroup$ Defining a hull will be difficult as, unlike with convex sets, there isn't a unique minimal $\mathcal{P}$-set covering any given set. For example, in $\Bbb{R}^2$, take a simple circle (not a disc, but its boundary). The convex hull, a disc, will have property $\mathcal{P}$ (trivially), as will a hemisphere with that circle as a boundary. If we take a proper subset of either set, we must tear a hole, which will prevent property $\mathcal{P}$. So, both sets could both lay claim to being a $\mathcal{P}$-hull of the circle. $\endgroup$
    – Theo Bendit
    Commented Mar 8 at 15:17
  • 1
    $\begingroup$ @Chris I somehow misread your first reply, twice. I thought you were agreeing that it had property $\mathcal{P}$! Anyway, as I understood it, every line that passes through the convex hull of the hemisphere, but not the hemisphere itself, has a homotopy to a line not passing through the convex hull. There are no such lines, so the property vacuously holds. What am I missing? $\endgroup$
    – Theo Bendit
    Commented Mar 9 at 1:42
  • 1
    $\begingroup$ @TheoBendit Ah I see what you're saying. Property $\mathcal{P}$ requires every point $x \notin S$ to lie on some line in $\mathbb{R}^3 \setminus S$, so there's no way for the property to hold vacuously. In particular, since there are no such lines for points in the interior of the convex hull of the hemisphere, it does not satisfy property $\mathcal{P}$. $\endgroup$
    – Chris
    Commented Mar 9 at 1:45

0

Reset to default

You must log in to answer this question.