From: lrudolph@panix.com (Lee Rudolph) Newsgroups: sci.math Subject: Re: 1D knot in 4D? Date: 11 Sep 1998 08:22:02 -0400 orourke@grendel.csc.smith.edu (Joseph O'Rourke) writes: >Can anyone point me to a proof that no one-dimensional curve >can be knotted in four-dimensional space? Several knot theory >textbooks discuss knotting 2-spheres in R^4, but I haven't >found a proof at a 1-sphere cannot be knotted in R^4. Thanks! Well, Charles Giffen seems to still be on vacation, so I'll take a crack at this one. First, you ought to decide what notion of "unknotted" you want to use. There are several potentially different ones (which I *think* can all, with more or less effort, be shown to still be the same in these dimensions; see footnote at the very end for a cautionary tale). I'll prove something, and let you deduce in what sense what I prove implies what you want. Let C be a smoothly embedded simple closed curve in R^4. The space of all straight lines in R^4 is a real manifold V of 6 dimensions (the space of all line segments in R^4 is 4+4=8 dimensional, each straight line determines a subspace of line segments of dimension 1+1=2, these subspaces are pairwise disjoint, so dim V = 8-2=6). Let W be the subset of V consisting of (a) all lines which pass through 2 (or more) points of C, (b) all lines which are tangent to C (at at least one point). Then W is the image by a smooth map of a smooth 2-manifold (namely, the symmetric square of C--the space of all unordered pairs of not necessarily distinct points of C: this space, homeomorphic to the 2-sphere, is not a priori smooth along the "diagonal" copy of C, but it can be made smooth in a natural manner; and the assumed smoothness of C as a subset of R^4 is what you need to show that the map from this space into V, onto W, is smooth). For any hyperplane H in R^4 (3-dimensional linear subspace), let U(H) be the subset of V consisting of all lines perpendicular to H. Then U(H) is a 3-dimensional smooth submanifold of V. Since dim(U(H))+dim(W)=3+2<6=dim V, "by transversality" a generic choice of H will make the intersection of U(H) with W empty. By construction, then, orthogonal projection P from R^4 onto H will take C to a smoothly embedded curve C' in H, and in fact you can find a 1-parameter family of diffeomorphisms of R^4 (starting at the identity) which carries C onto C'. So we've reduced the problem of "unknotting" C in R^4 to the case in which C = C' lies in R^3 contained in R^4. And you know how to do that already! (Project C onto a generic 2-plane in R^3 to get a "knot diagram" of the usual sort. Choose half, or fewer, of the crossings which, if changed, will convert the diagram to a diagram of an unknot. Move a small part of the knot out into the 4th dimension to change those crossings. Done.) I think this proof is about as simple as you can get, unfortunately. It has about the same level of difficulty as an alternative proof using the "Whitney trick". Both rely on some sort of really "global" move (at the end). Cautionary footnote: some sort of smoothness, or some restriction on dimensions, *has* to be included in any general theorem about unknottedness. Consider the complex hypersurface X in C^4 defined by the vanishing of x^2+y^3+z^5 (where (x,y,z,t) are the complex coordinates in C^4, and you notice that t isn't used in the equation). It's a fact that X (which has, of course, no natural structure as a smooth manifold) is homeomorphic to R^6. It's also a fact that the t-axis, all points (0,0,0,t) in C^4, is a subset T of X which is obviously homeomorphic to R^2. Yet this R^2 named T is so badly embedded in that R^6 named X that its complement has non-trivial fundamental group (guess which one [hint: 2, 3, 5!]). Now THAT is knotted. Lee Rudolph