TITLE: Is a knotted trivalent graph determined by its set of unzips?
QUESTION [10 upvotes]: A knotted trivalent graph (KTG) is a framed embedding of a trivalent graph in $\mathbb{R}^3$, modulo ambient isotopy. I believe they were first considered (in the quantum topological context, at least) in a paper by D. Thurston. There is an "unzip" move on edges, turning an "H" pattern into a pair of edges (Thanks to Kea for the hand-drawing of this image).

A knotted theta graph unzips in three ways (one unzip for each edge), giving rise to three possible 2-component framed links, related to one another by handleslide (Kirby 2). The following statement looked obvious at first to me, but now I'm beginning to doubt it's even true, and I have no idea how to prove it or to find a counterexample. It `feels' well-known.

Question: Given two KTG's (say knotted theta graphs for simplicity), all of whose unzips coincide (i.e. given an edge e in one, there exists and edge f in the other, such that unzipping along e and along f give ambient isotopic results), does it follow that the KTG's are themselves ambient isotopic? Or do there exist distinct KTG's all of whose unzips coincide?

REPLY [6 votes]: A framed theta graph (ribbon surface) is really a thrice puncture sphere embedded in $S^3$ (pants). A stronger question is whether one may have two pants whose boundaries give the same 3-component link? In fact, suppose the interiors of the two pants are disjointly embedded in the link complement. The union is a genus 2 surface. Conversely, given a genus 2 surface, one may choose a pants decomposition of it to get such a pair. I claim that if one takes a genus 2 surface bounding an incompressible surface to one side (so the other side is compressible), a generic pants decomp will give inequivalent theta graphs. One may assume the link is hyperbolic without symmetries. If they were equivalent, there would be an isometry taking one pants to the other. This isometry would be finite order, giving a symmetry of the link complement, a contradiction. It remains to find such a surface and pants decomp, but I'm sure any sufficiently complicated asymmetric example will work.