Are any of the Conway knots slice?

Problem: Prove that none of the Conway knots are slice.

Piccirillo is well-known for having shown that the Conway knot is not slice. The Conway knot $C$ is a certain Conway mutant of the Kinoshita-Terasaka knot $KT$. The latter is a symmetric union and hence slice. Many knot invariants have trouble telling a knot apart from its mutant. For example, we know thanks to Kotelskiy--Watson--Zibrowious that Rasmussen's $s$ invariant is invariant under mutation. Therefore \[s(C) = s(KT) = 0,\] providing no information about the sliceness of $C$. The Alexander polynomial is also insensitive to mutation, so \[\Delta_C(t) = \Delta_{KT}(t) = 1,\] which implies that $C$ is topologically slice by Freedman. These explain why the problem Piccirillo solved was tricky.

Perhaps less well known is that $C$ is merely the first member of a bi-infinite family of Conway knots. For integers $r\geq 2$ and $n\geq 1$, Kinoshita and Terasaka studied the knots $KT_{r,n}$ shown in the pictured below (the number in boxes indicate positive half twists):


These are all symmetric unions with Alexander polynomial 1. The Conway knot $C_{r,n}$ is the result of mutating $KT_{r,n}$ in the sphere indicated by the red circle by 180 degree rotation about the axis perpendicular to the plane (Ozsvath--Szabo studied the knot Floer homologies of these knots early on, proving that in fact bi-graded HFK is sensitive to mutation; though $\delta$-graded HFK isn't as shown by Zibrowious). The famous Conway knot is just the simplest of these, $C = C_{2,1}$. The problem is asking to prove that the other Conway knots in this family are also non-slice!

To prove that $C$ is non-slice, Piccirillo found another knot $J$ with the same 0-trace as $C$ and with $s(J) \neq 0$. The latter shows that $J$ isn't slice. On the other hand, the 0-trace completely detects whether a knot is slice, so $C$ isn't either. To find the knot $J$, Piccirillo used the fact that $C$ has unknotting number 1. That isn't going to be true of the other Conway knots, so it might be difficult to find 0-friends for these. Perhaps a different strategy altogether is needed.


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