This question is closely related to a question of Gowers: http://mathoverflow.net/questions/53471/are-there-any-very-hard-unknots .
I'm thinking about how to create interesting knots from small numbers of local moves on unlinks. The "standard embedded n-component unlink" (let's call it the untangled unlink) is defined as a union of n circles in n disjoint 3-balls; and an unlink is a link which is ambient isotopic to an untangled unlink. In order to get some feel for what is possible and to have a non-trivial example to play with, I'm looking for an unlink which is difficult to untangle. Paraphrasing Greg Kuperberg's formulation:
Can you untangle any unlink with relatively little work, say a polynomial number of geometric moves of some kind? I'm interested in untangling the components one from another, as opposed to to untangling individual components.
I mean this very much in the same sense that Tim Gowers meant his question: Is there a geometric algorithm to untangle components of any unlink, which "makes the unlink simpler" (this is intentionally vague) at each stage? Conversely, is there an unlink which, if it were given to me as a physical object (some tangled loops of rope), I would not be able to disentangle one component from the other without considerable ingenuity?
At the moment, I am most interested in the question for 2-component links and for 3-component links. It's a bit embarassing that I have no intuition at all for what the answer to this question might look like- the "link case" seems completely different from the "knot case".
Edit: Just an aside: experimentally, it's known that fluids made of long closed molecules flow much faster than fluids made of open molecules, although I don't think that the mathematics behind this is understood. But intuitively, it's clear what's going on- closed molecules don't get tangled up in one another as easily as open ones do. So if hard unlinks exist (and in rheology we're talking unlinks with thousands or millions of components), experimentally we can argue that they must at least be rare. Maybe.