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Linear independence of cables in the knot concordance group

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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… Click to show full 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.

Keywords: concordance; smoothly slice; slice knots; knot concordance; concordance group

Journal Title: Transactions of the American Mathematical Society
Year Published: 2021

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