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Let $K_0$ and $K_1$ be two knots in $S^3$. We say $K_0$ and $K_1$ are concordant if there exists a smoothly embedded annulus $A \subset S^3 \times [0,1] $ such that $\partial A = -(K_0) \cup K_1$.

Given two non-trivial concordant knots $K_0$ and $K_1$, assume that one of them is hyperbolic, say $K_0$. Is it possible to show that $K_1$ must be hyperbolic?

We may ask the similar question by changing the "hyperbolic" notion with "amphichiral". In other words, does the knot concordance preserve the hyperbolicity or amphichirality of the knot?

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    $\begingroup$ Hyperbolic slice knots. $\endgroup$
    – Anubhav Mukherjee
    Commented Nov 8, 2022 at 6:58

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Neither of these properties are preserved by concordance. As was pointed out in the comments, any hyperbolic (for the first question) or chiral (for the second question) knot which is concordant to the unknot will be a counter-example. For a specific example, the knot 6_1 is slice (concordant to the unknot), but is hyperbolic and is not amphichiral.

In fact, in "Homology cobordisms, link concordances, and hyperbolic 3-manifolds", Myers showed that every knot is concordant to a hyperbolic knot.

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  • $\begingroup$ I thought about these questions, excluding the unknot. Therefore, I updated the question. Rather, I am interested in amphichirality. $\endgroup$
    – Max Schumann
    Commented Nov 8, 2022 at 20:39
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    $\begingroup$ @MaxSchumann I don't see why you'd expect better results by assuming neither is the unknot (after all, you could take both to be slice knots for which neither is the unknot), but in particular notice the last line in the user's answer: every knot is concordant to a hyperbolic knot, so this certainly gives answers to that question. For the second part, you can take an amphicheiral knot and sew in a slice knot which is not amphicheiral. Running in the reverse direction it's an old conjecture of Gordon that any 2-torsion element in the concordance group has an amphicheiral representative. $\endgroup$
    – mme
    Commented Nov 8, 2022 at 21:06

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