Can the unknotting number of a knot with fixed three-genus be arbitrarily high?

Some background and motivation: the unknotting number $u(K)$ of a knot $K$ (i.e. the minimum number of necessary crossing changes to make the knot trivial) is generally difficult to compute. Kirby's list includes some problems on $u$ (not the above question, though), e.g. it's unknown whether $u$ is additive with respect to connect sum.

Many lower bounds to $u$ can be computed from the Seifert form or Blanchfield form; all of those are subsumed by the algebraic unknotting number, which is equal to or less than $2g(K)$, where $g$ is the three-genus of $K$.

One finds infinitely many examples of knots with $u(K) \geq 2g(K)$, e.g. knots admitting a Seifert matrix $A$ such that $A + A^{\top}$ is the zero-matrix modulo some odd prime.

Brendan Owens found lower bounds for $u$ coming from Heegard-Floer-homology, with which one can show that there are knots $K$ with $g(K) = 1$ and $u(K) = 3$, e.g the pretzel knot $P(3,3,3)$.

But as far as I found, this is as far as one can currently go, and the following questions are open: are there knots with $g(K)=1$, $u(K) > 3$ $?$ Can the difference $u(K) - 2g(K)$ be arbitrarily large?

My wild guess would be that the answers are "yes", realized e.g. by $\lim_{n\to\infty} u(P(3n,3n,3n)) = \infty$; but that we just don't have any idea how to show that.

Are these questions discussed in some book or paper?

  • 1
    $\begingroup$ I take the total silence as a confirmation that nothing is known about this question... $\endgroup$ Dec 6, 2017 at 10:13


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