One solution is this: Each C^oo curly loop (which we assume is parametrized by arclength) alpha: R/(2pi R) -> R^3 has a well-defined continuously varying ordered right-handed triple of orthonormal vectors at each of its points, hence a closed curve in the rotation group of 3-space, SO(3). (Since, SO(3) is the same space as the space of all such right-handed orthonormal triples. Topologically SO(3) is the real projective 3-space P^3.) The triple is the Frenet frame, (T(s),N(s),B(s)) for the point alpha(s). Any closed curve in SO(3) = P^3 belongs to one of two homotopy classes; loops in the different classes cannot be continuously deformed into one another. (This result is famous in physics; perhaps Veit or Gene can elaborate.) It's easy to check that the circle and the twice-traversed circle give distinct homotopy classes of closed curves in SO(3). And a regular homotopy between any two curly loops (through curly loops) would result in a homotopy between the closed curves in SO(3), implying that the two closed curves are in the same homotopy class, which they aren't. Hence the two curly loops mentioned cannot be regularly homotopic through curly loops. —Dan
On Jul 10, 2015, at 4:07 PM, Dan Asimov <asimov@msri.org> wrote:
Let a C^oo closed curve in R^3 be called a "curly loop" if its curvature is nowhere vanishing.
Two curly loops are for instance A) the unit circle in the plane and B) the unit circle in the plane traversed twice around.
PUZZLE: Can A and B be continuously deformed one into the other — through curly loops — so that at all stages of the deformation the tangent directions vary continuously?