Linear independence of cables in the knot concordance group

TL;DR

This paper constructs infinite families of knots with cables that are linearly independent in the knot concordance group, deep within its filtrations, revealing limitations of existing invariants and providing counterexamples to conjectures.

Contribution

It introduces new infinite families of knots with linearly independent cables in the concordance group, surpassing the detection capabilities of known invariants, and offers counterexamples to Kauffman's conjecture.

Findings

Cables form linearly independent sets in the concordance group.

Examples lie arbitrarily deep in the solvable and bipolar filtrations.

Counterexamples to Kauffman's conjecture on slice knots.

Abstract

We produce infinite families of knots $\{K^i\}_{i\geq 1}$ for which the set of cables $\{K^i_{p,1}\}_{i,p\geq 1}$ is linearly independent in the knot concordance group. We arrange that these examples lie arbitrarily deep in the solvable and bipolar filtrations of the knot concordance group, denoted by $\{F_n\}$ and $\{B_n\}$ respectively. As a consequence, this result cannot be reached by any combination of algebraic concordance invariants, Casson-Gordon invariants, and Heegaard-Floer invariants such as tau, epsilon, and Upsilon. We give two applications of this result. First, for any n>=0, there exists an infinite family $\{K^i\}_{i\geq 1}$ such that for each fixed i, $\{K^i_{2^j,1}\}_{j\geq 0}$ is a basis for an infinite rank summand of $F_n$ and $\{K^i_{p,1}\}_{i, p\geq 1}$ is linearly independent in $F_{n}/F_{n.5}$. Second, for any n>=1, we give filtered counterexamples to Kauffman's conjecture on slice knots by constructing smoothly slice knots with genus one Seifert surfaces where one derivative curve has nontrivial Arf invariant and the other is nontrivial in both $F_n/F_{n.5}$ and $B_{n-1}/B_{n+1}$. We also give examples of smoothly slice knots with genus one Seifert surfaces such that one derivative has nontrivial Arf invariant and the other is topologically slice but not smoothly slice.