Let K be a polyhedral oriented knot in S3 and N(K) be a regular neighborhood of K. If S3 ,N(K) can be constructed by attaching a single 2-handle to a genus two handlebody, then there is a homeomorphism of S3 onto itself mapping K onto itself and reversing the orientation of K. We prove the title. A somewhat more careful statement is the following. THEOREM. Let K c S3 be a polyhedral knot and let N(K) be a regular neighborhood of K. IfS3 N(K) can be constructed by attaching a single 2-handle to a genus two handlebody, then K is invertible. PROOF. By a meridian of the knot K we mean a polyhedral disk D in N(K) with aD c aN(K) and N(K) N(D) is homeomorphic with a ball. (N(D) is a regular neighborhood of D in N(K).) We show how to construct a self-homeomorphism of S3 N(K) which maps the boundary of a meridian of the knot onto its inverse. It is easy to see that such a homeomorphism can be extended to an involution of S3 taking the (oriented) knot to its inverse. Let H2 be a genus two solid handlebody, let D, and D2 be meridian disks for H2 and let y be the simple closed curve on H2 to which a 2-handle B is attached to get S3 N(K). Let m be the boundary of a meridian of K. Without loss of generality we may assume that m c aH2 y. Let h: H2 -* H2 be a rotation of H2 through 180' about its axis. (Think of the standard picture of H2. The axis passes through both holes of the handlebody.) Now h induces the symmetry q as defined in [O & S, p. 248]. Thus h can be assumed to map y and m onto themselves while reversing their orientations (see also [B & H, ?5]). The underlying reason for this is that the rotation inverts the Lickorish twists that generate the homeotopy group of the surface. Clearly then h may be extended to a homeomorphism h which maps B onto itself while reversing orientation. This completes our proof. Note. This result shows that knots such as 820 and 10132 from the table of knots in [Rolf] are invertible. This does not appear evident from the presentations given. These knots are not torus knots or 2-bridge knots. The knot 810 does not have a complement with Heegaard genus 2 but it is invertible. That it does not have such a complement follows from the fact that its second elementary ideal is proper [Fox]. Of course, the result above means that the noninvertible pretzel knots of Trotter Received by the editors February 18, 1980 and, in revised form, April 21,1980. 1980 Mathematics Subject Classification. Primary 57M25. ( 1981 American Mathematical Society 0002-9939/81/0000-01 34/$01.50 501 This content downloaded from 157.55.39.35 on Tue, 30 Aug 2016 04:18:33 UTC All use subject to http://about.jstor.org/terms