Complexity of equal 0-surgeries (2401.06015v2)

Abstract: We say that two knots are friends if they share the same 0-surgery. Two friends with different sliceness status would provide a counterexample to the 4-dimensional smooth Poincar\'e conjecture. Here we create a census of all friends with small crossing numbers c and tetrahedral complexities t, and compute their smooth 4-genera. In particular, we compute the minimum of c(K)+c(K') and of t(K)+t(K') among all friends K and K'. Along the way, we classify all 0-surgeries of knots of at most 15 crossings. Moreover, we determine for many friends in our census if their traces are equivalent or not. For that, we develop a new obstruction for two traces being homeomorphic coming from symmetry-exceptional slopes of hyperbolic knots. This is enough to also determine the minimum value of c(K)+c(K') among all friends K and K' whose traces are not homeomorphic.