Math. Ann. 271,381-400 (1985) 9 SI0ringer-Vertag 1985 Rotation Numbers of Products of Circle Homeomorphisms Mark Jankins I and Walter D. Neumann 2 Department of Mathematics, College of Charleston, Charleston, SC 29424, USA Department of Mathematics, University of Maryland, College Park, MD 20742, USA Introduction Let H = Homeo § (S 1) be the group of orientation preserving homeomorphisms of the circle. Our main question is the following: for given numbers 71 and Y2, what are the possible rotation numbers of a product qlq2 of elements ql, q2 of H with rotation numbers 71, Y2? What if some or all of ql, q2, qlq2 are required to be conjugate to rotations? Our original motivation was the question of which Seifert fibered 3-manifolds admit transverse foliations, which we discuss in Sect. 7. The answers turn out to be much more subtle than we originally expected. We can make the question more precise by working in the universal covering group/7 of H. This is the group of homeomorphisms of F, which lift from an orientation preserving homeomorphism of S 1 =R/Z, that is /7= (Q:R~RIQ monotonically increasing, Q(r+ 1) = Q(r) + 1 for all r ~ R}. (/7 is simply connected since it is a convex subset of IR•.) For y ~ ~-. define sh(~) ~ H by sh(7)(x)=x+~, x~R. The center of/7 is Z = {sh(n)ln e Z} and/7/Z = H. For Q e/7 the (Poincar6) rotation number of Q is defined as Rot(Q)= lim 1-[Q*(x)-x], xeR, 11400 n where Qn means Q ..... Q (n times). As is well known, this limit exist and is independent of x. The rotation number of an element q ~ H is defined as rot(q) =Rot(Q) (modZ) e R/Z where Q is a lift of q, but we will never actually use this. Rotation number is a conjugacy invariant. Clearly Rot(sh(~))= 7. However, not every element of rotation number 7 is conjugate to sh(v). Our question above is the case n = 3 of the: 1 Researchsupported in part by a College of Charleston Research Grant 2 Research partially supported by the NSF and by the Max-Planck-Institut Air Mathematik in Bona