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