Pretzel knots with unknotting number one

Abstract

We provide a partial classification of the 3-strand pretzel knots K = P (p, q, r) with unknotting number one. Following the classification by Kobayashi and Scharlemann-Thompson for all parameters odd, we treat the remaining families with r even. We discover that there are only four possible subfamilies which may satisfy u(K) = 1. These families are determined by the sum p + q and their signature, and we resolve the problem in two of these cases. Ingredients in our proofs include Donaldson's diagonalization theorem (as applied by Greene), Nakanishi's unknotting bounds from the Alexander module, and the correction terms introduced by Ozsváth and Szabó. Based on our results and the fact that the 2-bridge knots with unknotting number one are already classified, we conjecture that the only 3-strand pretzel knots P (p, q, r) with unknotting number one that are not 2-bridge knots are P (3, −3, 2) and its reflection.

Weight Elements of the Knot Groups of Some Three-Strand Pretzel Knots

“…Also, P(−3, 3, ±2) is the only one that has unknotting number one among P(±3, 3, m) with |m| > 1 [2]. (In fact, P(−3, 3, ±2) is expected to be the only three-strand pretzel knot with unknotting number one that is not 2-bridge [2].) Hence, our theorem gives infinitely many new hyperbolic knots that admit infinitely many weight elements.…”

“…Except for P(±2, ∓3, ∓3) and P(±2, ∓3, ∓5), which are torus knots, the knots in Theorem 1.1 are hyperbolic [7]. Also, P(−3, 3, ±2) is the only one that has unknotting number one among P(±3, 3, m) with |m| > 1 [2]. (In fact, P(−3, 3, ±2) is expected to be the only three-strand pretzel knot with unknotting number one that is not 2-bridge [2].)…”

Alternating knots with unknotting number one

“…However, if K is non-alternating, then only one of Σ(K) or Σ(K) is known to bound a sharp manifold [32], so Theorem 4.2 can only obstruct the sign of an unknotting crossing, cf. [5].…”

A biological application for the oriented skein relation

“…sign. We will limit the pretzel knots that we are looking at further by using the result of [6] that P (p, q, r) is a 2-bridge knot if and only if at least one of p, q, or r is equal to ±1. Since all 2-bridge knots are alternating, we won't include these knots in this subsection.…”