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"Rubber band" domains are defined below.

(i) Is this class of domains already named? If so, what is the name?

(ii) What results about rubber band domains are known/published?

This class of domains seems to be relevant to the "Lion and Man" pursuit problem.

The crucial condition in the following definition is (d).

Definition. A set $D$ will be called a rubber band domain if it satisfies the following conditions.

(i) $D$ is an open bounded connected subset of $\mathbb R^3$. Its boundary is smooth.

(ii) "Every rubber band is contractible to a point." More precisely, if K is a subset of $D$, it is homeomorphic to the unit circle $U$ and has a finite length (i.e., $K$ is a rectifiable Jordan curve) then there exists a continuous function $M(x,t)$ defined on $U \times [0,1]$ such that

(a) for every $t \in [0,1)$, $M(x,t)$ is a homeomorphism between $U$ and its range

(b) the range of $M(x,1$) is a single point in $D$

(c) for any $t \in [0,1]$, the range of $M(x, t)$ is a rectifiable Jordan curve

(d) if $s < t$ and $s,t$ belong to $[0,1]$ then the length of the range of $M(x,s)$ is not smaller than the length of the range of $M(x,t)$.

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    $\begingroup$ Why the "algebraic geometry" tag? $\endgroup$
    – Laurent Moret-Bailly
    Commented Sep 27, 2010 at 20:06
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    $\begingroup$ Any reason you've restricted to subsets of $\mathbb R^3$? It seems the Riemannian manifolds $M$ such that any loop in $M$ is connected to a trivial loop by a path on which the arc-length function is monotonic more or less captures what you're after. No? $\endgroup$
    – AndrewLMarshall
    Commented Sep 27, 2010 at 20:46
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    $\begingroup$ So conditions (a-c) just guarantees that $D$ is simply connected. Condition (d) is similar, but stronger, than the requirement that there are no locally length minimizing geodesics. (The sphere is a rubberband domain, the sphere without the south pole is not.) I wonder if Dick Palais will be able to say more? $\endgroup$
    – Willie Wong
    Commented Oct 12, 2010 at 10:07

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Any convex polyhedron is a rubber-band domain if you drop the requirement that $D$ is open.

To see this, take a loop, and push it around until it becomes a geodesic. Now translate the entire loop in a direction orthogonal to the loop, keeping the length of the loop constant, until it bumps into a vertex. It won't unexpectedly intersect itself while you are doing this. Now if you push a part of the loop past that vertex, the loop decreases in length since the polyhedron is convex. Now push it around again until it becomes a geodesic, and repeat. There are only finitely many possible lengths for a closed geodesic that doesn't intersect itself, so this process eventually finishes.

More generally, I think any convex surface should be a rubber band domain.

Edit: I was thinking about this problem today, and it finally hit me!

If $D$ is the complement of an object (intersected with a large ball containing the surface), then this is exactly the condition that there is no knot that can't be slipped off the object. So by the answers to the question I posed here, we have the results that:

(1) The complement of a ball (intersected with the interior of a larger ball) is a rubber band domain.

(2) The complement of an equilateral triangle (intersected with the interior of a large ball) is not a rubber band domain.

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