# $\omega$-triviality of knots?

From the theory of finite type invariants of knots comes the concept of $n$-triviality. A knot is said to be $n$-trivial if there is some projection of the knot and $n$ pairwise-disjoint sets of crossings $S_1,S_2,\ldots,S_n$, such that changing the crossings in every nontrivial subset of $S=\{S_1,\ldots, S_n\}$ turns the knot into the unknot.

An easy example is the trefoil knot with a three-crossing projection. Let $S_1$ and $S_2$ each contain just one crossing from this projection. Than changing $S_1$, $S_2$ or $S_1\cup S_2$ yields the unknot.

This notion of $n$-triviality is quite interesting because Goussarov first proved that a knot is $n$-trivial if and only if all finite type invariants vanish up to degree $n-1$. My question is about generalizing this notion to $\omega$-triviality. For this purpose, it is probably better to regard the crossing changes in $S_i$ as homotopies supported in neighborhoods of arcs connecting the knot to itself (finger moves.) Then one can define a knot to be $\omega$-trivial if there is a pairwise disjoint collection $\{S_1,S_2,....\}$ where each $S_i$ is a set of finger moves on $K$, such that doing any nonempty subcollection $\{S_i: i\in J\}$ for $\emptyset\neq J\subset \mathbb N$, of the finger moves gives you the unknot.

Question: If a knot is $\omega$-trivial, must it be the unknot?

A knot which is $\omega$-trivial would have vanishing finite type invariants of all degrees, so it shouldn't exist, but this question should be a lot easier than the question of whether finite type invariants detect knottedness. Does anybody have any ideas on this question?

• Habiro (in talks, perhaps not in print) has asked this question for links, where C_n-moves take the place of crossing-changes. – Daniel Moskovich Oct 10 '10 at 19:55

• I assume you mean the lower central series of the Torelli group here. I agree that an example such as you describe would be n-trivial for all $n$. But I was hoping that the extra geometric content of $\omega$-triviality would make this an easier question to tackle than the obviously difficult question of whether a knot could be $n$-trivial for all $n$. – Jim Conant Oct 10 '10 at 13:34