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?
Best Answer
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.