TITLE: Which convex bodies can be captured in a knot?
QUESTION [13 upvotes]: Which convex bodies can be captured in a knot?

This question is based on the discussion in "Is it possible to capture a sphere in a knot?".
We assume that the knot is made from unstretchable, infinitely thin rope.
Comments:

By the construction Anton Geraschenko, the question is equivalent to existence of a graph embedded in the surface of the convex body that locally minimize the total length. Such embedding exists on some convex bodies, for example on an equilateral triangle shown on the diagram (and on anything sufficiently close).



According to the original question a ball cannot be captured (in fact it cannot be captured in a link with 3 components). Moreover a circular disk cannot be captured see my answer (thanks to Wlodek Kuperberg for asking). Likely the same idea works for all convex bodies of revolution. Maybe all convex bodies of general position can be captured.

REPLY [2 votes]: A circular disc cannot be captured.
Likely the same idea can be used to show that any convex body of revolution cannot be captured.
It can be proved using a small variation of the idea as in the original answer.
It is sufficient to show that infinitesimal Möbius tranform $m$ of the disc can shorten the wrapping length while preserving the crossing pattern.
Once it is proved, moving in this direction will eventually allow the disc to escape.
Choose a Möbius tranform $m$ of the unit disc that is close to the identity map.
Denote by $u$ its conformal factor.
Denote by $U(r)$ the average value of $u(x)$ for $|x|=r$.
It is sufficient to show that $U(r)\le 1$ and the inequality is strict for $r<1$.
In this case by rotating the knot and applying the Möbius tranform will shorten the length.
Consider the circle $C_r$ of radius $r$ centered at the origin.
Denote by $r'$ the radius of $m(C_r)$.
Observe that $U(r)=r'/r$.
Evidently $r'\le r$ if $r\le 1$ and  $r'< r$ if $r< 1$, whence the result.