From: "Charles H. Giffen" Subject: Re: Who is this legend about? Date: Tue, 30 Nov 1999 14:08:51 -0500 Newsgroups: sci.math To: Douglas Zare Keywords: minimum curvature of a space curve Douglas Zare wrote: > > Penny314 wrote: > > > [...] > > > > The theorem which was homework is the computation of the total curvature of a > > simple closed smooth curve in three space. > > Let me clarify this: The total curvature of a _knotted_ smooth circle is at > least 4pi. I've heard that this was just the start of a theory he developed but > I have never seen an elaboration; I'd appreciate it if anyone could share one. > > In a different direction, several people have proved versions of the following: > Given a smooth knotted circle in R^3, there must be some line which passes > through the knot in 4 places. This can be used to prove the 4pi result, but that > would almost certainly be the wrong perspective. > > Douglas Zare I think that the result finally obtained is that the total curvature of a smooth knot is always greater than 2nPi, where n is the braid index of the knot (ie. the smallest n for which the knot can be represented as a closed braid on n strands), and that this bound is sharp. --Chuck Giffen ============================================================================== From: lrudolph@panix.com (Lee Rudolph) Subject: Re: Who is this legend about? Date: 30 Nov 1999 14:45:32 -0500 Newsgroups: sci.math "Charles H. Giffen" writes: >I think that the result finally obtained is that the total curvature >of a smooth knot is always greater than 2nPi, where n is the >braid index of the knot (ie. the smallest n for which the knot >can be represented as a closed braid on n strands), and that this >bound is sharp. Not the braid index, the bridge number (i.e., the smallest n for which the knot can be represented by a smooth curve in R^3 on which one of the coordinate functions has precisely n local maxima; bridge number is always less than or equal to braid index). This bound *is* sharp (given an n-bridge representation of K, you can pull all the maxima up to one level, all the minima-- which are also n in number--down to another level, make all the 2n strings nearly straight between the two levels, and round off each local extremum into a semicircular arc; the 2n hairpin curves at top and bottom each contribute \pi to the total curvature, and the nearly straight strings contribute a negligible amount which can be made to go to 0 by sufficent stretching. Lee Rudolph