Obtaining graph knots by twisting unknots (Q1763585)

Let \(K\) be a trivial knot in the 3-sphere \(S^3\) and \(D\) a disk in \(S^3\) meeting \(K\) transversely in the interior, and assume that \(|D\cap K|\geq 2\) over all isotopies of \(K\) in \(S^3-\partial D\). Let \(K_{D,n}\) be a knot obtained from \(K\) by \(n\)-twisting along the disk \(D\). This paper studies when \(K_{D,n}\) is a graph knot, i.e., the complement \(S^3- K_{D,n}\) is a graph manifold. The main theorem of the paper says that if \(K_{D,n}\) is a graph knot, then \(|n|\leq 1\) or the pair \((K, D)\) is an ``exceptional pair'' (Theorem 1.1). Here \((K, D)\) is called an exceptional pair if the link \(K\cup\partial D\) is obtained from the Hopf link by iterated cabling in a specific manner so that \(K\) continues to be a trivial knot. If \((K, D)\) is a special pair, then it is obvious that \(K_{D,n}\) is a graph knot for every integer \(n\). Moreover, if \((K, D)\) is a nonexceptional pair such that \(K_{D,1}\) is a graph knot, then one can obtain, by iterated cabling along \(K\), non-exceptional pairs \((K', D)\) such that \(K_{D',1}\) is a graph knot. So it is natural to ask: if a satellite knot is obtained from a trivial knot by twisting, can every companion knot be obtained in such a manner? A negative answer to this question is also obtained (Proposition 1.3). The proof of the main theorem is based on a detailed analysis of the essential planar surfaces in the exteriors of graph knots (Proposition 3.1) and various results concerning Dehn surgery, including the third author's previous results on twisting of knots.